This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
every net has at most one limit. Hint. Assume that x and y have no disjoint neighborhoods. Define A as the set of all pairs (Ux ' Uy ) of the neighborhoods of X and y with the order "(U� ' u; ) -< (u; ' u; ) ¢:::::::> (U; C U� ) and (u; C u; ) " . Then to every such pair of neighborhoods we assign a point in their intersection.
In conclusion of this introduction to topology let us note the following. Of course, Exercise 6 is not the only example showing that a topology is more efficient than a metric. We go back to this many times in the book when studying weak topologies, generalized functions and other questions. Nevertheless, there are some important types of conver gence in analysis that cannot be described even by topology: Exercise 10 (0. G . Smolyanov) . On the set of measurable functions defined on the closed interval [0, 1] , the convergence (of a sequence) almost everywhere cannot be described by any topology.
Hint. Use the following two facts: (i) a sequence converging in measure does not necessarily converge almost everywhere, but it contains a subsequence converging almost everywhere; (ii) if a sequence does not converge to a given point in a topological space, then it contains a subsequence lying outside of some neighborhood of this point.
3 . Categories ; first examples
Terminological remark. When speaking about totalities or families of
objects, we use two terms: "set" or "class" . The fact is that we cannot use the same term for all conceivable totalities. For example, if we allow the notion "set of all sets" to make sense, we immediately obtain a series of well known paradoxes, which overshadowed the last years of Georg Cantor. So in the cases where we are not sure that we can use the term "set" (to speak informally, this concerns "too huge totalities" ) , we use the term "class" . Thus, every set is a class, but not vice versa; say, the class of all linear spaces is not a set. This "naive" approach is absolutely sufficient for our needs. However, in formal set theory everything is much more complicated. The explanation of what are sets and classes, and why this "playing with words" saves us from contradictions, exceeds the scope of this book; see,
3.
21
Categories; first examples
e.g. , [13] . (We only hint that it is expedient to define sets as those classes that are allowed to be elements of other classes.) Definition
1. We say that a category JC is defined if
I. A class Ob(/C) is indicated with elements called
objects of the cate
gory JC (usually they are denoted by letters X, Y, . . . , and we write X E /C, having in mind X E Ob(/C) ) . II. For every ordered pair X, Y E JC a set (now indeed a set!) hK (X, Y) is indicated; elements of this set are called morphisms from X to Y, or morphisms between X and Y. (The condition rp E hK (X, Y) can be written as rp : X --+ Y or X � Y, as if we were talking about mappings of sets; below we will see why this notation is useful) . The object X is called the domain, and Y the range of morphism
rp.
Ill. For every triple X, Y, Z E JC and for every pair rp : X --+ Y,
1/J : Y --+
Z (here the range of the first morphism coincides with the domain of the second morphism) a morphism from X to Z is defined; it is called the composition of morphisms rp and 1/J. (This morphism is denoted by 1/J rp or 1/Jrp ; please, pay attention to the order of the symbols. ) o
Moreover, the following two conditions are supposed to be fulfilled: (i) (associativity of composition) For every X, Y, Z, U E /C, rp : X --+ Y, 1/J : Y --+ Z, and x : Z --+ U we have ( x 1/J) rp == x ( 1/J rp) . (In other words, if a morphism ( x 1/J) rp or, which is equivalent, a morphism x ( 1/J rp) makes sense, then these two morphisms coincide.) (ii) For every X E JC there is a morphism 1 x : X --+ X , called the local identity for X, such that for every Y E JC and rp : X --+ Y (respectively, 1/J : Y --+ X), we have rp 1 x == rp (respectively, 1 x 1/J == 1/J ) . (In other words, if rp 1 x makes sense, then it is rp, and if 1 x 1/J makes sense, then it is 1/J.) o
o
o
o
o
o
o
o
o
o
o
o
Thus, a category consists of three ingredients: objects, morphisms, and the law of composition, and they satisfy two axioms, associativity of com position, and the existence of local identities. Of course, as it happens in algebra, associativity allows us to use long expressions like rp 1 rp 2 · · · 'Pn where the parentheses can be placed arbi trarily. But the difference is that not every such expression makes sense. If a category is chosen, we often write h(X, Y) instead of hK (X, Y) . The class of all morphisms of the category /C, i.e. , the union hK (X, Y) o
o
o
22
0. Foundations: Categories and the Like
over all X, Y E /C, is denoted by hK . Sometimes, if there is no danger of misunderstanding, we write rp E JC instead of rp E hK . A morphism such that its domain and range coincide (like, say, local identity) is called an endomorphism. Proposition 1. For every object there is only one local identity. Proof. If 1� is another candidate for local identity in X, then by property
(ii) , 1x 1� == 1� and at the same time 1x 1� == 1x .
•
A category JC is called a subcategory of a category £, if, first, every object and every morphism in JC are respectively an object and a morphism in £, second, composition of morphisms in JC is the same as in £, and, third, every local identity in JC is a local identity in £. A subcategory is said to be full if for every X, Y E JC we have hK (X, Y) == h.c (X, Y) . Now we invite the reader to the zoo of examples of categories. As a rule, we restrict ourselves to defining objects, morphisms, and the law of composition; in all cases the axioms of category are verified trivially. As usual ( cf. the discussion in Subsection 0. 1 ), the first example pro duces a delusive impression of being "silly" . Example 1. Every given class is turned into a category if we declare its elements to be objects and at the same time local identities, and say that there are no other morphisms. Such a category is called discrete. Another example is already "taken from real life" . Example 2. The category of sets Set . Its objects are arbitrary sets, mor phisms are mappings of sets, the composition of morphisms is the usual composition of mappings, and local identities are identity mappings. Actually in the historically first examples of categories their objects were sets with some additional structure, and morphisms were mappings of the sets compatible, in some sense, with this structure. It was the considera tion of these examples that led to the realization of the following fact: in a substantial mathematical theory the definition of objects should be accom panied with the definition of morphisms, and that "morphisms are more important than objects" . The influence of these ideas on modern mathe matics is difficult to overestimate. In all examples of this type the composition of morphisms is their com position as mappings, and local identities are identity mappings. This will always be assumed in the sequel. Example 3. The category Lin of linear spaces (we recall: over CC). Its objects are linear spaces, and morphisms are operators. This category has an important full subcategory FLin consisting of finite-dimensional spaces.
3.
Categories; first examples
23
Note that Lin is not a subcategory of Set since we can define many different structures of linear space on the same set. (Formally, a linear space is not a set, but a pair consisting of a set and a linear structure on it.) Among purely algebraic categories, we mention, in addition to Lin, the category Ab of abelian groups , the category Gr of (all) groups (of course, Ab is a full subcategory in Gr) , and the category Rin of all (not nec essarily unital) rings . Morphisms in these categories are what is usually called homomorphisms of groups and rings, respectively. Some algebraic categories that are important for functional analysis will be mentioned later (see Sections 5.2 and 6.3 ). Example 4. The category Ord of ordered sets. Its objects are ordered sets and morphisms are the so-called monotone mappings, i.e. , order-preserving (the meaning of this must be clear) mappings. Example 5. The category Met of metric spaces. Its objects are metric spaces, and morphisms are continuous mappings. This category is apparently the most important one in the theory of metric spaces. However, sometimes it is reasonable to consider some other categories, in particular, Metu and Met1 . As in Met , their objects are metric spaces, but morphisms in Metu are uniformly continuous mappings, and in Met1 contractions. Of course, Metu is a subcategory in Met , and Met1 is a subcategory in Metu , but these subcategories are not full. Example 6. The category Top of topological spaces . Its objects are topo logical spaces and morphisms are continuous mappings. This category has an important full subcategory HTop consisting of Hausdorff spaces. The latter in turn has a full subcategory consisting of metrizable topological spaces. (Note that here we say metrizable, but not metric. Indeed, Met is not a subcategory of Top for the same reason that Lin is not a subcategory of Set (explain this reason!) .) As in the theory of metric spaces, in real analysis (and in the related top ics of ergodic theory) there are several reasonable approaches to the question of which mappings of measure spaces should be viewed as compatible with the structure of measure space. Respectively, in these areas we can speak about several substantial categories. We mention here only one of them. Example 7. The category Meas . Its objects are measure spaces. As for morphisms, they are classes of equivalent proper measurable mappings (see Section 0. 1 ) between the corresponding spaces (here we have in mind the general principle of real analysis saying that we should not distinguish be tween equivalent mappings) . The composition of morphisms is defined as the equivalence class of compositions of representatives of the initial classes
24
0. Foundations: Categories and the Like
(check that this definition does not lead to misunderstanding) . The axioms of category are easily verified, and obviously, local identities are precisely the equivalence classes of identity mappings. These examples suggest that different "modern" mathematical sciences study their own categories, or to be more precise, classes of categories. Gen erally speaking, this is indeed the case, although with some part of oversim plification. (In particular, the same, or maybe, even greater consideration is given to the so-called functors, which will be defined in Section 7. ) The number of such examples will essentially increase, mostly at the expense of the categories serving functional analysis (see below Sections 1.4,
2.2, 4. 1 , 5.2-5.3, 6.3) .
Quite a few categories allow us to formulate some mathematical results briefly and elegantly. Here is one such example. Example 8.
The standard simplicial category Ll . Its objects are intervals
�n : == {0, . . . , n} of the set of non-negative integers, and morphisms are
non-decreasing mappings. Thus, Ll can be regarded as a full subcategory in Or d.
On the outstanding role of this category in many areas of mathematics you can read, for example, in [16] .
The following example is of rather general nature. Example 9. Let JC be an arbitrary category. Its
dual category is the cate
gory JC 0 that has the same objects as /C, but the set of morphisms hKo (X, Y) is, by definition, hK (Y, X) , and the composition of morphisms 1/J rp in JC0 (if it exists) is defined as the morphism rp 1/J in JC. o
o
This is an extremely useful example. As we will see many times, it allows us to cut by half the number of definitions, theorems, and other mathematical statements. (As a consequence, we save paper. In this way we can claim that categories have applications in national economy. ) An object X of a category JC is called initial (respectively, final) if for every Y E JC the set h(X, Y) (respectively, h(Y, X) ) consists of exactly one element. ( "There is exactly one arrow from X to Y, respectively, from Y to X." ) Note that X is an initial object in JC X is a final object in JC 0 , and vice versa. (This is the first hint to the practical use of the notion of dual category.) An object 0 is called a zero object if it is both initial and final. Of course, in Lin there is a zero object, namely the zero linear space. On the other hand, in Set there are initial and final objects, but they are different. Final objects here are singletons, and the initial object is the
4.
Isomorphisms
25
empty set. (The fact that there is a unique mapping from empty set to any other set follows from the formal definition of the notion of mapping in rigorous set theory; see, e.g. , [17] .) In Ll there is only one final object, namely �o, but, as it can be easily verified, there are no initial objects. 4. Isomorphisms . T he problem of classification of obj ects and morphisms
Among all morphisms of a given category, some classes are especially inter esting. We start with the "best" ones. Everywhere in what follows, JC is an arbitrary category. Definition 1. Let rp : X --+ Y be a morphism in /C. A morphism 1/J : Y --+ X
in the same category is called inverse to rp if 1/Jrp == 1x and rp1/J == 1y . A morphism in JC is called an isomorphism if it has an inverse morphism. Objects X and Y in JC are called isomorphic if there exists an isomorphism from X to Y (or, equivalently, from Y to X - verify this!) . The inverse morphism to rp is usually denoted by rp - 1 . The following result is almost evident, but as we will soon see, it has important corollaries. Theorem 1.
phic.
Every two initial and every two final objects in JC are isomor
Proof. Suppose X and Y are initial objects in /C. The existence of the
necessary arrow in the definition of an initial object implies the existence of morphisms rp : X --+ Y and 1/J : Y --+ X. Consider 1/Jrp : X --+ X; the uniqueness of the arrow implies that 1/Jrp == 1x . Similarly, rp1/J == 1y . Thus, initial objects are isomorphic. Passing to the dual category (again it is useful!) we immediately see that final objects are isomorphic. II Let us see what the abstract definition of isomorphisms turns out to be in various examples of categories. In discrete categories where the objects "do not want to communicate with each other by means of arrows" , there are no different isomorphic objects. An isomorphism in Set is a bijection (i.e. , a one-to-one correspondence) . An isomorphism in Lin is what one usually call a linear isomorphism (i.e. , an operator which is also a bijection) . As the reader has probably guessed, isomorphisms in Met and in Top have a special name: homeomorphisms. An isomorphism in Metu , i.e. , a uniformly continuous mapping with a uniformly continuous inverse mapping, is called a uniform homeomorphism. Finally, an isomorphism in Met1 is a contrac tion having an inverse contraction; of course, this is just an isometry. An isomorphism in Meas between measure spaces (X, J-L ) and (Y, v) is, as you
0. Foundations: Categories and the Like
26
can easily verify, an equivalence class of mappings f : X --+ Y satisfying the following conditions: ( 1) there exist sets of full measure A C X and B C Y such that f maps A bijectively to B; (2 ) if C C X, then (a) C is measurable in X f (C) is measurable in Y; (b) C has zero measure in X f (C) has zero measure in Y. *
*
*
Informally, the meaning of the notion of "isomorphism" is that isomor phic objects are "actually the same" ; in some sense they represent the same object, but "in different clothes" . All that we can say in categorical terms (i.e. , in the language of arrows) about an object, we can also say about each isomorphic object. In every area of mathematics, when studying a particular category, the following natural question arises: is it possible to classify (or, in other words, to describe) objects of that category up to an isomorphism? As we will see later, in some categories this problem is trivial, in others a solution is given by a fundamental theorem (such as, say, the Riesz-Fischer theorem in Section 2.2 ). And there are some categories where this problem is hopeless because it is impossible to cover all the multitude of the emerging cases. But what do we mean by speaking about the solution (or a big advance) of this "classification problem" ? Example 1. As a simple instructive example, let us consider the category FLin we mentioned before. You know, of course, that two finite-dimensional
spaces are isomorphic they have the same linear dimension. Moreover, every such space is linearly isomorphic to cc n for some n. We can express these facts as follows: the set of all integers can be chosen as a complete system of invariants of isomorphism classes for FLin, and for every such invariant n E N we can take cc n as a model for the corresponding object in
FLin.
In the general case, for a given category JC we sometimes can find some where a class M consisting of sufficiently "intelligible" elements, and to associate to every object JC an element in M in such a way that isomorphic objects correspond to the same element. Such a class M is usually called a system of invariants of isomorphisms in /C. We emphasize that the choice of a class should be viewed as appropriate if the class consists of "things well perceived by our intuition" , like natural numbers in the last example, or some "good" sets; otherwise what are the benefits of the construction?
4.
Isomorphisms
27
(One of the most important systems of invariants in analysis is provided by the spectrum of an operator. We discuss this in Chapter 5.) If we are lucky, a system of invariants we find is complete. This means that two non-isomorphic objects always have distinct invariants. Thus, in this case, objects have the same invariant they are isomorphic. In addition, it is desirable to point out for every invariant a "sufficiently simple" object with this invariant. Such an object is often called a model object, or just a model for this invariant. The system of invariants gives us an idea of how rich our category is with "really different objects" and what is their nature. Thus, it is accepted (and apparently mathematicians silently agree with this) that the problem of classification of objects is solved if at least two things are done. First, you must indicate a system of invariants that is as transparent as possible (and you persuade the mathematical community in the latter) . Second, for every invariant you must indicate its model (again, sufficiently transparent) . It would be good if the construction of the invari ant was simple as well, but this depends on the situation, of course. (In functional analysis the exemplary solution for the problem of classi fication is the relevant result on the category of Hilbert spaces; see Theorem 2.2.2 below.) Example 2. Obviously, two objects (sets) in Set are isomorphic these sets have the same cardinality. Thus, a complete system of invariants in Set
is precisely the class of all cardinalities (or cardinal numbers) . Of course, this declaration is, in fact, a tautology. Recall that cardinality was defined precisely as "the property that is common to equivalent sets" (see [8, p. 25] ) . However, the same is true for finite cardinalities: we merely overcame the corresponding psychological barrier in childhood, and now we forgot that this was not easy. (When saying this we stay, of course, on the "naive" point of view (see [8] ) and do not go deep into the "serious" set theory; cf. , e.g. , [7] . ) As a model of a set of a given cardinality, an interval of the transfinite line with this cardinality is usually suggested; the explanation can be found in [6] .
A little more substantial is the classification of objects in Lin. It is a direct generalization of what we have told about FLin. Exercise 1.
(i) Two linear spaces are isomorphic they have the same linear dimension. (ii) Every cardinality is a linear dimension of some linear space.
28
0. Foundations: Categories and the Like
Hint. Suppose A is a set of cardinality m. Consider the set CC� of complex-valued functions on A taking non-zero values only on finite subsets in A. This is a linear space with respect to pointwise operations that has m
as linear dimension.
It is clear from this exercise that we can again take the class of all cardinalities as the complete system of invariants in Lin. Now, however, the invariant of the object is its linear dimension. If we agree to choose for a given cardinality m a model, say A, in the category of sets, then the model of the space of dimension m can be defined as the space C� .
As to the majority of the other examples described above (with the ob vious exception of Ll) , the problem of classification of objects there seems to be hopeless: "there are too many non-isomorphic objects" . At the same time it can be easily solved in some important subcategories of those cat egories. For example, the full subcategory FAb in Ab consisting of finite abelian groups is one of them. (The classical theorem from algebra provides a key for the classification of objects in FA b. Recall this theorem and draw the corresponding conclusions.) In the category Meas there is an important full subcategory consisting of the so-called Lebesgue spaces introduced by V. A. Rohlin in 1949 [18] . These are measure spaces with countable base satisfying some supplementary (not too complicated ) conditions. We do not want to divert the reader with the exact formulations; instead we only note that these spaces often arise in applications. It turns out (and it was proved by Rohlin ) that every Lebesgue space is isomorphic in Meas to one of the spaces of the form [0, 1] Il X, or X, where the �nterval �s endowed with the standard Lebesgue measure, and X is an at most countable set with "counting measure " (assigning measure 1 to every singleton). (Suggest the corresponding system of invariants, using combinations of finite, count able, and continual cardinality. ) In particular, - and this, of course, is the main result - from the point of view of real analysis, there is a unique Lebesgue space without points of non-zero measure, namely the interval of the real line. The Rohlin theorem and similar results (see, e.g. , [19] , [20] ) show that the structure of a measure space is the coarsest among all the substantial structures on a set (see the discussion in [2 1 , pp. 46-47] ) . *
*
*
Digression. Before going further, we introduce some elements of the cat
egorical language. Let JC be a category. A diagram in JC is an arbi trary family of objects in JC and morphisms between them. If X and Y are objects in a diagram, then a path from X to Y is a finite family 'P I : X � ZI ' 'P 2 : ZI � z2 ' . . . ' 'Pn : Zn - I � y of morphisms in this dia gram. (Of course, there can be many different paths in the diagram from one object to another, or no paths at all.) A diagram is called commutative if for every two objects X and Y and for every two paths
0. Then there exists a bounded linear functional f : E � CC of norm 1 such that f ( x) == II x II . Proof. Take Eo : == span { x} and choose a functional f : Eo � CC : AX �----+ A ll x ll ; A E CC. Obviously, li fo II == 1 , and the Hahn-Banach theorem immedi ately gives a functional f with the required properties. • Corollary 1. Let E be a normed space, and x and y two different vectors in E. Then there is f E E* such that f(x) =/=- f(y) . Theorem 3.
This corollary is often expressed by saying that "there are sufficiently many bounded functionals on a normed space" ( enough to distinguish the vectors ) . Here is a more general fact. Exercise 9. Suppose E is a prenormed space, E 1 is a closed subspace in E, and x E E \ E1 . Then there exists a bounded functional f : E � CC such that f l E1 == 0 and f(x) =/=- 0. Hint. Consider a bounded functional AX + y �----+ A on the subspace {Ax + y : y E E1 , A E CC}. Then apply the Hahn-Banach theorem. To realize why these results are substantial, consider for comparison one quite natural linear space in analysis endowed with the metric that is not generated by any norm. Exercise 1 0* . Let E be a linear space ( of cosets ) of functions on [ 0, 1] measurable in the sense of Lebesgue. Endow it with the following metric:
1 - y(t) j dt d(x, y) : = J{ 1 jx(t) o + l x(t) - y(t) i
( check that convergence with respect to this metric is the convergence in measure ) . Then there is no non-zero linear functional on E continuous with respect to this metric. We indicate another useful corollary of the Hahn-Banach theorem.
Let E be a normed space, and Eo a finite-dimensional subspace of E. Then there exists a closed linear complement of Eo in E.
Proposition 3.
1.
1 08
Normed Spaces and Bounded Operators
Proof. We use induction on the dimension of Eo . If dim Eo
== 1 , and Eo ==
span{x}, then, by Theorem 3, there is f E E* with f (x) =I= 0, and taking into account Proposition 2, we see that Ker(f) is the required complement. Now suppose this proposition is true for all subspaces (of all normed spaces) of dimensions up to n, and dim Eo == n + 1. Take arbitrary x E Eo \ {0} and f E E* with f(x) =I= 0. Put EI :== Ker(f) . Then Eoi :== Eo n EI is a subspace in EI with dim Eo I == n. By the induction assumption, there is a linear complement, say, F, of Eoi in EI , and F is closed in EI . Therefore, taking into account the completeness of EI in E, F is closed in E. In addition F + Eo == F + Eoi + span{x} == EI + span{x} == E, and from F n Eo C EI n Eo and F n Eo C F it follows that F n Eo == 0. The rest is • clear. Remark. Soon we shall prove (see Section 2. 1) that every finite-dimensional
subspace of a normed space is topologically complemented. *
*
*
The Hahn-Banach theorem allows us to give a complete description of bounded functionals on one of the most important normed spaces, C[a, b] . It turns out that every such functional is an integral with respect to some complex measure, and its norm coincides with the variation of this measure. To give the exact formulation and the proof of this result, we need some standard definitions and facts of real analysis (i.e. , of the theory of measure and integral) ; cf. [9] or [106] . Let J-L == VI - v2 + i v3 - i v4 ; VI , . . . , v4 > 0 be a complex measure on [a, b] (see preparation to Example 1.11). Then for every piecewise continuous function x : [a, b] � CC (this class is sufficient) the number
is called the Lebesgue-Stieltjes integral of x with respect to the complex mea sure J-L and is denoted by J: x(t)dJ-L(t) . This number, as can be easily shown, does not depend on the choice of the decomposition of J-L into a linear com bination of usual measures. Further, suppose
0 we have I:� 1 l (r l 2 > c2 for all m. Take n E N and consider the operator In : Mn (l2) � Mn ( l2 ) . Choose a matrix x == (xkz ) in Mn ( l2 ) that has only one non-zero column, namely the first one, with X k 1 :==p k ; k == 1 , . . . , n. Then the matrix y : == In x with entries Ykl == I(xkz ) also has only the first column non-vanishing, and Y k 1 == ( k ; k == 1 , . . . , n.
To find the norms of the matrices x E Mn (l2) and y E Mn (l2 ) , we should, according to the discussed procedure, look at the matrices constructed from the operators a :== Jr ( x) and b :== Jc ( x ) in Mn ( B ( l 2 ) ) . In both matrices only the first column is non-vanishing. Further, the matrix a has on the k1th place the operator taking p k to p 1 , and taking the other unit vectors to zero. In other words, we have the operator corresponding to the elementary matrix with 1 on the k 1 th place and zeroes on all the other places:
0...0
a == (an 1 )
0
0
1 0
0
0...0
0
It is clear after the identification with the corresponding operator acting on l 2 ( nN) that n the matrix a takes (� 1 , . . . , � ) to ( � { p 1 , �� p 1 , . . . , �� p 1 ) . Certainly, this means that l l x l l n , defined as I I a I I n , is equal to 1 . At the same time, the matrix b has on the ( k 1 )th place the operator taking p 1 t o ( k and the other unit vectors to zero; in other words, this operator has the matrix with the sequence ( k as the first column, and zeroes on the other places:
0
120
1.
Normed Spaces and Bounded Operators
Therefore, if we consider the concrete system (p 1 , 0, . . . , 0) , we see that the matrix b, after its identification with the corresponding operator on l2 ( nN) , takes this system to the system ( ( 1 , . . . , ( n ) . But the norm of the latter system in l 2 (nN) is JL:�= 1 ll ( k ll 2 > yric. This means that II y II n , defined as II b II n , is not less than yfic. Thus, we see that the operator In takes the matrix x of norm 1 to the matrix y of norm > yric. Hence, I is not completely bounded. • Remark. Analyzing this proof, we see that to be completely bounded, an operator T : l2 ---+ l2 must satisfy the condition L:� 1 II T ( p k ) 1 1 2 < oo . This means, as we will see in Section 3.4, that it must belong to the class of so-called Schmidt operators. Actually the class of completely bounded operators between 12 and l2 is precisely the class of the Schmidt operators (see, e.g. , [37] ) . Exercise 3 . Let l2in be the quantum space generated by the isometric embedding described in Proposition 2 (where E :== 1 2 ) . Then no two of the spaces l2in , 12 , and 12 are completely topologically isomorphic. Remark. Actually, as Pisier had shown, there is a continuum of quantum spaces with l2 "at the first floor" such that no two of them are completely topologically isomorphic.
Here is another elegant example. This time the question is about an operator that seems to be quite good and acts on the "sample" space B( l 2 ) with standard quantum norm. Proposition 4 (Tomyama, [36, Proposition 2.2. 7] ) . Let t be the transpose operator taking
an operator with matrix ( .A k z ) (in a basis of unit vectors} to the operator with matrix (1-t kl :== Az k ) (in the same basis). Then IIT IIn == n, and, as a corollary, the operator t (being an isometric isomorphism on the "first floor") is not completely bounded. Exercise 4 * . Prove the following estimate for the transpose operator:
liTli n
> n.
Hint. Consider the matrix a E M(B(l2 ) ) which at the klth place has the operator with the matrix having 1 at the lkth place and zeroes at the others. This matrix corresponds to the operator on B( l2 ( nN) ) permuting some vectors of the natural orthonormal basis. Hence, II a II n == 1 . At the same time the matrix t n (a) corresponds to the operator with matrix in the same basis having a "submatrix" of size n x n which consists of elements 1 . Consequently, I I a I I n > n. However (and there would not be a substantial theory without this, of course) , quite a few important classes of operators do consist of completely bounded ones. In the sequel, when speaking about (just) bounded operators between quantum spaces, we will have in mind the operators acting between the corresponding normed spaces (i.e. , "bounded on the first floor" ) . Theorem 2 . Let E be a quantum space and f : E ---+ C a bounded funchonal. Then
f
�s ( "automatically ") completely bounded with respect to the standard quantum norm in C (see Example 1}, and ll f ll c b == ll f ll . Proof. Take the matrix a == ( ak z E Mn (E)) and the element fn (a) E M n . Note that for all � == (� 1 , . . . , �n ) , ry == ( rJ1 , . . . , rJn ) E e n we have L � ,l = 1 f (a k z ) �l'r/k == f (u) , where u :== L� ,l = 1 akl �l 'r/k · In Mn , consider the matrix t having � as the left column, and matrix fJ having rJ as the upper row (and the other entries of both matrices are zeroes) . _ Then, obviously, u : == fja � is the matrix with entry u in the upper left corner and zeroes everywhere else. From the property (i) of the quantum norm it follows that ll u ll n == ! l u ll ,
121
7. Invitation to quantum functional analysis
and from (ii) we have
n
II u II n
p (these norms differ from the norm of lp ) · The same is true for LP [a, b] ; 1 < p, with respect to v
2.
128
Banach Spaces and Their Advantages
the norm inherited from Lq [a, b] ; q < p, and for C00 [a, b] with respect to the norm inherited from an arbitrary cn [a, b] ; n == 0, 1 , 2, . . . . The space C[a, b] with integral norm ll x ll :== J: l x ( t ) l dt is also non-Banach, since (up to identifying a function with its coset) it is a dense proper subspace in L 1 [a, b] . (Give other examples.) Running ahead, we note that there are no other kinds of non-Banach spaces: all of them are dense subspaces in Banach spaces (see Corollary 6. 1 below) . As an exercise, it is useful to ask students to prove directly ( i.e. , without using £1 [0, 1]) that C[O, 1] is not a Banach space with respect to the integral norm, by producing a non convergent fundamental sequence. Somebody will always suggest the sequence Xn (t) : == t n , assuming that since it "converges to a discontinuous function" , it cannot converge in this space. This is a good occasion to recall that different types of convergence should not be mixed up.
Now we show that all the finite-dimensional normed spaces are Banach. This will allow us to establish the equivalence between different norms (and many other things) , as we have promised before. The key step in the proof is the following result.
Every normed space of a finite dimension n is topologically isomorphic to CC!, and moreover, every linear isomorphism between these spaces is a topological isomorphism.
Proposition 4.
n. Everything is clear for n == 1. Suppose the statement is true for some n == k, E is a given (k + I)-dimensional normed space, and I : E cc� + 1 a linear isomorphism. Then for some linear basis e 1 , . . . , e k + 1 in E the operator I acts by the formula x == L:7+11 A.z e z �----+ ( A. 1 , . . . , A. k + 1 ) E CC� + 1 . For m == l , . . . , n put Fm :== span { e 1 , . . . , e m - 1 , em + 1 , · · · , ek + 1 } · By the induction assumption, the subspace Fm in E is topologically isomorphic to CC�. The last space is obviously Banach, hence by Proposition 1, Fm is Banach as well. By Proposition 2, Fm is closed in E. Thus Proposition 1.1.4 (with Fm as F) provides us with a constant Cm > 0 such that for every X == L:7+11 A z e z we have A. m I < Cm ll x II . If we put c :== max{ c1 ' . . . ' ck + 1 }, I then II I ( x) II 1 == L:7+11 I A.z l < C II x II . Therefore, I is a bounded operator. At the same time, I- 1 , as an oper ator from cc� + 1 , is also bounded (see Example 1.3.1). The rest is clear. • Proof. Use induction on --+
2. Every two normed spaces of the same finite dimension are topologically isomorphic. Moreover, every linear isomorphism between them is a topological isomorphism.
Corollary
1.
What lies
on
the surface
129
Suppose E is a finite-dimensional normed space. Then (i) E is complete; (ii) every norm on E majorizes every prenorm; (iii) every linear operator from E into a prenormed space is bounded. Proof. As we know (cf. Example 1.3. 1) , all these facts are true for the spaces C! ; n E N. This means that they are true for every normed space Theorem 1 .
topologically isomorphic to one of these spaces. It remains to apply the • previous proposition. Taking Proposition 2 into account, we immediately obtain Corollary 3.
(i) Every finite-dimensional subspace of a normed space
is closed; (ii) every two norms on a finite-dimensional space are equivalent.
As a curious illustration let us note the following Exercise 1 . If a linear space has a countable linear basis (like coo or the space of polynomials do) , then there is no norm taking it to a Banach space. Hint. From Corollary 3 and from the observation that a closed proper subspace of a normed space is rarefied it follows that if we endow this space with a norm, then we obtain a meager set. After that the Bair theorem works. Remark. It is known that all separable Banach spaces have countable linear dimension. Thus, they all are linearly isomorphic ( Lowig ' s theorem; see , e.g. , [39] ) .
Now we can fulfil our promise given in Section 1.6.
Suppose E is a normed space and Eo is a finite-dimensional subspace in E. Then Eo is topologically complemented. Proof. Suppose E1 is a closed linear complement of Eo in E (it exists by Proposition 1.6.3) , and let pr : E --+ E I E1 be the natural projection. Clearly, the restriction T == pr i Eo : Eo --+ E I E1 is a linear homeomorphism, hence a topological isomorphism due to Corollary 2. It remains to note that the operator P == r- 1 pr is a bounded projection to Eo along E1 , and to • use Proposition 1.5. 10. Proposition 5.
o
Here is another application.
Every bounded finite-dimensional operator T between normed spaces E and F is a sum of several one-dimensional bounded oper ators.
Proposition 6.
2.
130
Banach Spaces and Their Advantages
Proof. First assume that both spaces
E and F are finite-dimensional. Let
e 1 , . . . , e n be a linear basis in E. Denote by Tk the operator taking e k to T ( e k ) and taking other basis vectors to zero. Clearly, T == � Tk and each Tk is a one-dimensional operator, which is bounded by Theorem 1 . In the general case consider the operator T : E / Ker ( T ) � F generated by the operator T ( see Proposition 1.5.3) . Its corestriction to Im ( T) .......
is an operator ( moreover, a linear isomorphism ) between finite-dimensional normed spaces. We already know that it is a direct sum of several bounded one-dimensional operators, say, T1 , . . . , Tn . Hence, the one-dimensional operators Tk : == (in)Tk (pr) : E � F ( where in and pr act in an obvious way ) • are bounded, and their sum is certainly T. .......
.......
.......
Many spaces consisting of operators are Banach as well.
If F is a Banach space, then for every prenormed space E the normed {cf. Proposition 1. 3. 2) space B(E, F ) is again a Banach space. In particular, for every prenormed space E the dual space E* is always a Banach space, and if E is Banach, then B(E) is a Banach space as well.
Proposition 7.
Tn be a fundamental sequence in B(E, F ) . Then for every x E E the sequence Tn (x) is, of course, fundamental in F, and thus has a ( unique ) limit, which we denote by T(x). This defines the mapping T : E � F. From the additivity of all Tn and the continuity of addition in F it obviously follows that T is additive. Similarly, from the homogeneity of Tn and continuity of Proof. Let
the multiplication by scalars in F it follows that T is homogeneous. Thus, T is a linear operator. Since Tn is fundamental and hence bounded in B ( E, F ) , for some C > 0 and every x E BE we have II Tn (x) ll < C. As a consequence, for the same x we have II T(x) ll < C. This means that T is bounded ( i.e. , belongs to B(E, F ) ) . It remains to show that Tn tends to T in the operator norm. Take c > 0 and a natural N such that for m, n > N we have II Tm - Tn ll < � Then for every x E BE and for the same m, n we have II Tm x - Tn x ll < � Taking into account that Tm (x) tends to T(x) as m � oo, we obtain that II Tx - Tn x ll < � < � - If we take the supremum over all x E BE , we see that • li T - Tn ll < � < E . The rest is clear. Corollary 4. A Exercise
reflexive normed space is always a Banach space.
2. If E is a prenormed space with a non-zero prenorm, and
F is a normed space, then the completeness of F is not only sufficient but also necessary for the completeness of B(E, F ) .
1.
What lies
on
131
the surface
Hint. If Yn is a fundamental but not convergent sequence in F and f E E* \ { 0 } , then the one-dimensional operators f 0 Yn form a fundamental
sequence which does not converge in B(E, F) .
Why Banach spaces are better than other spaces? Let us postpone the acquaintance with deep results ( mostly connected with the name of Banach; see Section 4) , and look at what lies on the surface. Here is one of these things: in a Banach space "the series are well-summed" .
( i ) ( "Weierstrass test" ) . In a Banach space, every absolutely convergent series converges. ( ii ) If, in a normed space E, every absolutely convergent series con verges, then E is a Banach space.
Proposition 8.
Proof. ( i ) If �C: 1 X k is an absolutely convergent series, then the sequence
of its partial sums ��= 1 x k ; n == 1 , 2, . . is fundamental. The rest is clear. ( ii ) Let X n be a fundamental sequence in E. Clearly, it has a subsequence x� :== X nk such that ll x � + 1 - x � ll < 2\ . From this and from the hypothesis it follows that the series �r 1 ( x �+ 1 - x �) converges in E. Since the nth partial sum of this series is x� + 1 - x� , we conclude that the sequence x � converges. Thus, the sequence X n E E is fundamental and contains a convergent subsequence; this means that the sequence itself converges. • .
We present one of the numerous consequences of this simple fact.
Let E be a Banach space and F a closed subspace in E . Then the normed quotient space E IF (cf. Proposition 1. 1 . 2) is Banach.
Proposition 9.
Proof. Let � � 1 Xn be an absolutely convergent series in E I F. For every n, choose Xn E X n such that I I X n I I < 2 II Xn II . Clearly, the series �� 1 Xn is
absolutely convergent. Hence, by Proposition 8 ( i ) it converges to a vector x E E. Consequently, the series � � 1 X n converges to the coset of x in • E lF. It remains to use Proposition 8 ( ii ) . Another advantage of Banach spaces concerns extensions of operators. Although the corresponding fact has a simple proof, it is so important for applications that we promote it to the rank of a theorem. Theorem 2 ( "extension-by-continuity principle" ) . Let E be a prenormed space, Eo a dense subspace in E, and F a Banach space. Let To be a bounded operator from Eo to F. Then there exists a unique bounded operator T from E to F extending To {i. e., such that T I Eo == To or, equivalently, the following
2.
132
Banach Spaces and Their Advantages
diagram is commutative: Eo
inl � E
T
F
where in is the natural embedding). Further, l i T II == l i To II , and if To is isometric, then the same is true for T. Finally, if E is a normed space, and To is topologically injective, then T is also topologically injective. x E E and a sequence X n E Eo converging to x . The latter is fundamental. Hence from the estimate II To (x m - X n ) II < l i To II ll x m - X n II it follows that the sequence To (x n ) is fundamental as well. Since F is complete, To (x n ) tends to some vector of this subspace, which we denote by T(x) . If x� is another sequence in Eo tending to x, then obviously To(x n ) - To (x�) tends to zero in F. This means that T(x) does not depend of the choice of X n , and thus the mapping T : E --+ F is well defined. Moreover, if x E E0 , then taking X n : == x for all n, we see that for this x the equality T(x) == To (x) holds. This means that a mapping, T extends To. If we take x, y E E and sequences X n and Yn in Eo tending to x and y respectively, we see that from To (x n + Yn ) == To (x n ) + To ( Yn ) and from the continuity of the sum in F it obviously follows that T(x + y) == T(x) + T(y) . Proof. Take
as
Hence, T is additive. The homogeneity is established in a similar way. Thus T is a linear operator. Finally, for the same x and X n we have II T(x) II == limn �oo II To(x n ) II < limn�oo l i To II ll x n II == l i To II ll x II . This means that T is bounded and l i T II < II To II . Since the opposite inequality is obvious, the norms of the two oper ators coincide. From the same equality we see that the estimate II To ( Y ) II > c ll y ll with the equality II To ( Y ) II == II Y II for all y E Eo implies the same es timate and the same equality for T. This proves the results concerning topologically injective and isometric operators. It remains to note that the uniqueness of a continuous operator extending To follows immediately from the density of Eo in E (together, of course, with • the fact that F is Hausdorff) . As a first, comparatively modest, application we propose the following . exercise. Exercise 3. Prove the Hahn-Banach Theorem
separable E without using Zorn ' s lemma.
1.6. 1 for the case of
2.
Categories of Banach and Hilbert spaces
133
Hint. The separability allows us, using the main lemma in the proof of
the Hahn-Banach theorem countably many times, to extend a given func tional to a dense subspace in E. Then the extension-by-continuity principle works. Let us note a special situation where the latter principle is often applied.
Suppose E and F are Banach spaces with dense subspaces Eo and Fo respectively, and T8 : Eo --+ Fo is a topological isomorphism. Then there exists a unique topological isomorphism T : E --+ F such that T8 is its birestriction. If in addition T8 is an isometric isomorphism, then T is also an isometric isomorphism. Finally, if E and F are Hilbert spaces and T8 is a unitary operator, then T is also a unitary operator. Proof. Suppose S8 : == (T8)- 1 , To : Eo --+ F is a coextension of T8 , and So : Fo --+ E is a co extension of S8. If we extend To and So by continu ity, we obtain bounded operators T : E --+ F and S : F --+ E uniquely defined by their birestrictions. Then the operator ST : E --+ E, being the identity operator on a dense subspace Eo , must be the identity operator; similarly, TS : F --+ F is also the identity operator. Thus, T is a topological Proposition 10.
isomorphism. The rest is clear.
II
In fact, the "roots" of the extension-by-continuity principle lie deeper, in metric spaces and uniformly continuous mappings (in other words, in the category Metu ) ·
M be a metric space, Mo a dense subset in M (with inherited metric) , N a complete metric space, and rpo : Mo --+ N a uniformly Exercise 4. Let
continuous mapping. Then there exists a unique uniformly continuous map ping rp : M --+ N extending 'PO · If, in addition, rpo is isometric, then rp is also isometric. Finally, if M is also complete, the image of rpo is dense in N, and rpo is isometric, then rp is also isometric (i.e. , an isomorphism in Met1 ) . 2. Categories of Banach and Hilb ert spaces . C lassification and t he Riesz-Fischer t heorem
It is time to add four new categories of functional analysis to the ones we already know. They seem to be even more important. (i) The category Ban. The objects are Banach spaces and the mor phisms are (arbitrary) bounded operators. (ii) The category Ban1 . The objects are the same as in Ban, but the class of morphisms is smaller: only contraction operators are declared to be morphisms.
2.
134
Banach Spaces and Their Advantages
(iii) The category Hil. The objects are Hilbert spaces, and the mor phisms are bounded operators, as in Ban. (iv) The category Hil1 . The objects are the same as in Hil, and the morphisms are contraction operators as in Ban1 . All we have said in the context of our previously introduced categories about the composition of morphisms and the verification of axioms of cat egory (see the beginning of Section 1.4) can be automatically carried over to the new categories. The interconnection between all the eight categories can be illustrated by the following scheme: Hil u
c
Ban u
c
Nor u
C
Pre u
where the notation c and U has the same meaning as in the scheme in Section 1.4. The first question about the categories we have just introduced, is of course the following: what are isomorphisms in these categories? Clearly, a morphism in Ban or in Hil is an isomorphism � it is an isomorphism in Nor (or in Pre) . And a morphism in Ban1 or in Hil1 is an isomorphism � it is an isomorphism in Nor1 (or in Pre1 ) . Hence, the isomorphisms
in Ban and in Hil are topological isomorphisms, whereas the isomorphisms in Ban1 and in Hil1 are isometric isomorphisms (we recall that both types were characterized in Proposition 1.4.1). According to the agreement in Section 1 .4, isomorphisms in Hil1 are also called unitary isomorphisms or unitary operators.
Now we proceed to a typical question of classification of objects in a category up to an isomorphism, discussed in Section 0.4. Recall that, infor mally, this is the question of how many "really different" objects are there in the category under consideration. For all categories of functional analysis introduced up to now, the question certainly pertains to the classification of objects up to a topological or an isometric isomorphism. It turns out that the categories of Banach and Hilbert spaces behave quite differently. Oversimplifying matters, we can say that, despite the multitude of seemingly different examples, there are surprisingly few Hilbert spaces. At the same time, there is a huge multitude of Banach spaces (to say nothing about prenormed spaces) . The main result is that all infinite dimensional separable Hilbert spaces are in essence the same space, but presented in different ways.
1907) . Let H and K be infinite-dimensional separable Hilbert spaces with orthonormal {Schauder) bases e� and e� ; n E
Theorem 1 (Riesz-Fischer,
2.
135
Categories of Banach and Hilbert spaces
N (see Theorem 1 . 2.4' ) . Then there exists a unzque unitary isomorphism U : H ----+ K such that U(e� ) == e� for all n. Proof. Let Ho : == span { e� ; n E N } and Ko : == span { e� ; n E N } ; by the definition of a Schauder basis, Ho is dense in H, and Ko in K. Obviously, there exists a linear operator U8 uniquely defined by the rule that for every n it takes e� to e� . Take arbitrary x, y E Ho ; they have the form x == Er 1 Az e� and y == Er 1 J.Lz e �' for some A z , J.Lz E CC and m. Consequently, U8 (x) == Er 1 Az e�' and U8 ( Y ) == Er 1 J.Lz e?. Hence m
m
( U8 (x) , U8 (y)) = :2::: A. k J.Lz ( e % , e n = :2::: A. k J.L k , k, = 1 k= 1 and similarly, ( x, y) == 2:: ; 1 A k fl k · This shows that U8 is a unitary isomor phism between H and K. It remains to apply Proposition 1.9. • Corollary 1 . Every infinite-dimensional separable Hilbert space is unitarily isomorphic to l 2 . l
A half-fictional digression. The Riesz-Fischer theorem lies in the foundation of a vast area mathematics studying and / or using operators in Hilbert spaces. These operators are integral part of the apparatus of quantum mechanics. Long ago, in the mid- 1 920s, the words "quantum mechanics" were not used, but physicists already knew very well that there exist phenomena beyond the classical picture of the world by Newton and Laplace. Finding the laws of this new and strange world seemed to be an incredibly difficult task. But two men took the challenge. Two theories were suggested, seemingly having nothing in common with each other, "matrix mechanics" by Heisenberg and "wave mechanics" by Schrodinger. As usual, heated arguments began on which theory is better.
Now we know who of those outstanding physicists was right: both. The fact is that from the present-day point of view, our physicists were working with operators on different Hilbert spaces: Heisenberg worked with l 2 ( thence matrices; cf. the last part of Section 1 .4 ) , while Schrodinger ( if we restrict ourselves to the simplified one-dimensional case ) in L2 (1R) . But they had no any idea of Hilbert spaces, to say nothing about their unitary isomorphism. That is why they did not guess that their theories were essentially the same, but stated in two different languages. Finally, J. von Neumann 3 crossed the t ' s ( more about him will be said later in the book ) by proving the so-called theorem on the uniqueness of canonical commutation rela tions ( see, e.g. , [63] ) . In the simplified form it shows that one of the isomorphisms between l2 and £ 2 (JR) provided by the Riesz-Fischer theorem implements a unitary equivalence of the operators that are central in both theories ( again, using present-day terminology ) . ( As a matter of fact , this is the very same isomorphism that takes the unit vectors to the Hermite functions in Example 1 . 2.8. ) To speak informally, l 2 can be superimposed on £2 (JR) in such a way that the Heisenberg theory turns into the Schrodinger one. 3 John von Neumann (1 903-19 5 7) , a great mathematician of the 20th century. Many math
ematicians think that he is one of the two greatest mathematicians of his time (the other being A. N. Kolmogorov, who was of the same age) . Von Neumann was born in Hungary, spent part of his young years in Germany, and then moved to the United States. According to some of his biographers and historians of mathematics, he seemed to be one of very few great mathematicians who had easy temper.
2.
136
Banach Spaces and Their Advantages
After this theorem it became clear to von Neumann and other most advanced people, like M . Stone,4 that the "right" theory should not be tied to a concrete Hilbert space; instead, it must be based on an abstract Hilbert space. From matrix mechanics and wave mechanics a unified quantum mechanics was born. 5
The Riesz-Fischer theorem can be generalized to any Hilbert spaces, not necessarily separable. To formulate the result, we recall that an orthonormal Schauder basis is another name for a countable total orthonormal system.
2 (for the proof, see [50 , §2.2] ) . (i) Every Hilbert space has a total orthonormal system. (ii) Every two such systems have the same cardinality (called the Hilbert dimension of this space) . (iii) Two Hilbert spaces are topologically isomorphic � they are uni tarily isomorphic � they have the same Hilbert dimension. More over, if e�; v E A and e�; v E A are total orthonormal systems in H1 and H2 respectively, then there exists a unique unitary isomorphism between H1 and H2 , taking e� to e�; v E A .
Theorem
Certainly, every cardinality m can be a Hilbert dimension. Namely, if we take l 2 ( X ) , where X is a set of cardinality m, we get a Hilbert space with the total orthonormal system 8x ; x E X , where if y == x , otherwise. The theorem we have just formulated, together with the observation we made, gives the complete solution to the problem of classification of objects in Hil and Hil 1 . In both categories (and this feature resembles Set and Lin; cf. Example 0.4.2 and Exercise 0.4. 1) the complete system of invariants is the class of all cardinalities. But now the invariant of a given object is not its cardinality or linear dimension, but the Hilbert dimension. If X is a model in Set of a set of cardinality m, then the space l 2 ( X ) can be regarded as a model of a Hilbert space with the invariant m. In particular (this follows from the Riesz-Fischer theorem) , in the full subcategories of Hil and Hil 1 consisting of infinite-dimensional separable spaces, all objects are isomorphic (they have the same invariant, the countable cardinality) , and we can take l 2 as a unique model here. 4 This is the first time Stone's name appears in this book. 5 I t is quite another matter that the creators of both theories remained unconvinced: "My mechanics is better anyway, and let mathematicians say whatever they want" ( see details in [40 ] ). Here is another unfortunate example of the lack of mutual understanding between mathematicians and "practical" physicists.
2.
Categories of Banach and Hilbert spaces
137
However, when passing from Hilbert spaces to more general Banach spaces not even a shadow remains of the idyllic picture we have just ob served. It turns out that there are too many topologically (and the more so, isometrically) non-isomorphic Banach spaces. Too many to have even a faint hope for a complete classification of objects even in Ban. (In practice, math ematicians were sure of this long ago, even before realizing the categorical meaning of the discussed problem.) Nevertheless, many results have been gathered on the existence of the topological and sometimes even isometric isomorphisms between spaces that look quite different at first sight. From the categorical point of view these results give the classification of objects in some full subcategories of Ban or (with more luck) in Ban1 . We have already discussed the most striking and important for applications discovery of that kind, concerning Hilbert spaces. Here is another result showing that the case of Banach spaces £1 ( · ) differs from the case of Hilbert spaces £ 2 ( · ) .
Let ( X, J-L) be a space with measure that has a countable basis. Then the space L 1 (X, J-L) is isometrically isomorphic to one of the following spaces: CC!; n E N; l 1 ; L1 [0, 1 ] ; the Banach direct sum of L 1 [0, 1 ] and CC! ; n E N; the Banach direct sum of L1 [0, 1 ] and l1 . These spaces (and in particular, £ 1 [0, 1 ] and l 1 ) are not isomorphic (even topologically) to each other. Theorem 3.
A similar assertion holds if we replace the subscript 1 by an arbitrary p; 1 < p < oo, not equal to 2 and instead of the Banach direct sum (in other words, l1-sum) consider the lp-sum. For a proof of Theorem 3 see, e.g. , [41] . Of course, Corollary 1 .2 gives a complete classification of objects in the category of finite-dimensional normed (i.e. , finite-dimensional Banach) spaces and all (i.e. , all bounded) operators. The system of all invariants is again the set of integers (dimensions) , like in the pure algebraic category FLin in Section 0.4. The advanced reader can easily verify that the category of finite-dimensional normed spaces and the category of finite-dimensional linear spaces are equivalent . (But apparently the attempt to construct isomorphism between these categories would be a meaningless task.)
3. 1 , we will indicate some results pertaining to Banach spaces of the form C(O) . (We shall see, in particular, that all the spaces C(OCn ) , where ocn is the n-cube, and n runs over natural numbers, Remark. Later, in Section
are isomorphic to each other topologically but not isometrically.)
In general, the majority of facts on the existence of isomorphisms es tablished by various mathematicians up to now, concern classical Banach spaces. As for arbitrary spaces, a series of results obtained recently show
138
2.
Banach Spaces and Their Advantages
that they can behave in a very odd way. (And this, apparently, buries the hope for a reasonable classification of such spaces.) Here is an impressive example. In contrast with all examples of Banach spaces discussed before, the following theorem holds.
There is a Banach space that is not topologically isomorphic to any of its proper subspaces.
Theorem 4 (Cowers, 1994; see [5 1] ) .
Now let us move from the problem of classification of objects to the problem of the next level of complexity, that of classification of morphisms. Obviously, the similarity of endomorphisms and the versions of this notion discussed for general categories in Section 0.4, are characterized for our new categories in the same terms as for categories from the previous chapter. In particular, the similarity of endomorphisms in Ban and in Hil, i.e. , bounded operators acting on Banach ar1d Hilbert spaces, is nothing but their topo logical equivalence. At the same time, similarity of endomorphisms in Ban with respect to Ban1 , and in Hil with respect to Hil1 (see the same sec tion) is precisely their isometric equivalence (called, as we remember, unitary equivalence in the context of Hilbert spaces) . Finally, the similarity of endo morphisms in Ban 1 or in Hil1 is the very same isometric, or, respectively, unitary equivalence of operators, but now not for all operators but only for contraction operators. The classification problem for bounded operators, both up to topological and (as a 1nore rigorous version) up to an isornetric equivalence, is one of the typical problems of functional analysis. We shall return time and again to this problem for various classes of operators, and to examples of isometrically and topologically equivalent operators. Here is one of the simplest examples. Example 1. The operator of bilateral shift in l 2 (Z) is unitarily equivalent
to the operator of multiplication by z in £ 2 (1r) . This equivalence is imple mented by the unitary operator taking a sequence (0, . . . , 1, 0, . . . ) with 1 at the nth place to zn ; n E Z. The following example will be useful later in the study of compact self adjoined operators in Section 6.2. Example
2. Let H be a Hilbert space. Consider an operator T in H,
such that there exists an orthonormal systent�·-e 1 , e 2 , . . . of finite or count able cardinality m in H that consists of eigenvectors of this operator with eigenvalues AI , A 2 , . . . , and Tx == 0 for every x E Ho : == {e 1 , e 2 , . . . }1_. Then T is unitarily equivalent to the operator R : l24Ho � l24Ho , acting on l2 as a diagonal operator T>.. with A : == (A 1 , A 2 , . . . ) , and sending Ho to zero. This equivalence is established by the unitary operator I : H � l24Ho that takes en to pn E l 2 and is the identity operator on Ho .
2.
139
Categories of Banach and Hilbert spaces
As we have seen in Section 1 .4, the question of equivalence ( first of all, isometric equivalence ) of operators is related to their matrix representation. In the context of Hilbert spaces this relation is especially close.
Operators S : H1 � H1 and T H2 � H2 acting on infinite-dimensional Hilbert spaces are unitarily equivalent � they have the same matrix representations in some orthonormal Schauder bases.
Proposition 1 .
Proof.
:
===> .
This follows immediately from Proposition 1 .4.9. {:::=:: . If { a m n ; n E N} is the matrix of operators S and T in orthonormal bases e; and e; respectively, then the Riesz-Fischer theorem implies that for the operator U : H1 � H2 we have
( T ue ; , e � ) == ( Te; , e � ) == amn == ( Se; , e � ) == ( U se ; , Ue � ) == ( U s e ; , e � ) for all m, n E N. From this we have TUe; == USe; for all n E N, hence, • TU == US. Thus, in the context of Hilbert spaces a matrix determines an operator uniquely up to the unitary equivalence. Now we have the following natural question: is there an effective way to verify whether the table of complex numbers is a matrix of a bounded operator in a Hilbert space? No such way was found up to now ( and, possibly never will ) . Instead we know many necessary and many sufficient conditions. Here is an illustration. Exercise 1 . For a table { amn ; n E N} to be a matrix of a bounded operator on a Hilbert space, ( i ) the condition sup { L: � 1 l amn l 2 ; n E N} + sup { L: � 1 l a mn l 2 ; m E N} < oo is necessary, but not sufficient; ( ii ) the condition L:�, n == l l amn l 2 < oo is sufficient, but not necessary. Let us suggest another, more interesting example of unitary equivalence. It is related to matrices that are now infinite in all direction. Such matrices appear when it is natural to number the elements of the orthonormal basis in a Hilbert space H by all integers, as, for instance, in the case of the trigonometrical basis in £ 2 [ - 1r , 1r ] , or the basis of integer powers of z in £ 2 ('Jr) . If we take such a basis en ; n E Z in H, we can consider the matrix of an operator T in this basis by putting a rnn : = (Ten , ern) ; m , n E Z. An important class of these matrices are the so-called Laurent matrices, i.e. , the matrices with arn +k ,n +k = arnn for all integers m , n, k ( "matrices which are constant on all diagonals parallel to the main diagonal" ) . Exercise 2* . A bounded operator on a Hilbert space can be represented by a Laurent matrix in some basis ¢:::::::> it is unitarily equivalent to the operator of multiplication by a bounded measurable function in £ 2 ('Jr) . Hint. Suppose an operator S in £2 ('Jr) is represented by a Laurent matrix in the basis z n ; n E Z. Let f : = I:� (X) a n o z n . From the Laurent property it follows that S acts as the multiplication by f on every function from the linear span of the basis. For -
140
2.
Banach Spaces and Their Advantages
x E L 2 ('lr) there exists a subsequence X m of these functions tending to x , such that Sxm tends to Sx in norm and at the same time almost everywhere . From this we have Sx = f x . By the way, this implies that f is essentially bounded.
every
Remark. This result started a large part of the theory of operators starts, the analysis of the so-called Toeplitz and Hankel operators. The former are represented by matrices ( amn); m, n E N, where am+k ,n +k = amn ( "the lower right quadrant of a Laurent matrix" ) , and the latter by matrices ( am n); m , n E N with a m - k ,n +k = am n ( "the turned-over upper right quadrant of a Laurent matrix" ) . Informally, "Toeplitz matrices" are constant on the diagonals parallel to the main diagonal, and "Hankel matrices" , on the diagonals perpen dicular to the main diagonal. Both classes of operators admit an important representation in terms of function theory, this time with participation of not only real, but also complex analysis. On these operators and their numerous applications see, e.g. , [42] .
Let us go back to classification of endomorphisms in our categories. Ba sically, the situation here is the same as in the question on the classification of objects we discussed before. For a series of important special subcate gories complete solutions have been found. (The main of these results is, of course, the classification of self-adjoint operators up to unitary equivalence, which will be discussed in the last two sections of Chapter 6.) At the same time, operators acting in Hilbert (the more so, in Banach) spaces apparently again are too numerous to be classified. The reader who has done Exercise 0.4.2 knows that operators acting on finite-dimensional linear spaces admit a complete classification, and this fact relies on the existence of the Jordan form. In turn, the Jordan form is related to the position of invariant subspaces of the operator under con sideration. Naturally, one of the first questions in the study of bounded operators on infinite-dimensional spaces was the following: what are invari ant subspaces of these operators? The reader might have noticed that all operators we listed as examples in the previous chapter had invariant sub spaces, and moreover, rather many invariant subspaces. The more amazing was a comparatively recent result that gave a solution to a forty year old problem.
There exists a Banach space and a bounded operator acting on it which has no proper non-zero invariant closed subspaces. Moreover, one can take l 1 for such a Banach space.
Theorem 5 (P. Enflo and C. Read; see [62] ) .
(It is not known up to now whether it is possible to replace here l 1 by l 2 , that is, by an infinite-dimensional separable Hilbert space.) 3 . T heorem on the orthogonal complement and around it
First we recall Proposition 1 .2.9 on nearest vectors in finite-dimensional subspaces of near-Hilbert spaces. The conclusion of this proposition becomes
3. Theorem on the orthogonal complement
141
invalid for infinite-dimensional subspaces of near-Hilbert spaces, even if they are closed. Exercise 1 . Let Ho be the linear span of all basis vectors starting with the second in l 2 , and H the linear span of Ho and the vector x == (1, 1/2, 1/3, . . . ) . Then Ho is closed in H, but there is no vector in Ho nearest to x. Such disgraceful things are impossible in Hilbert spaces.
Let H be a Hilbert space, Ho an arbitrary closed subspace in H: and x a vector in H. Then there is a unique vector in Ho nearest to x.
Proposition 1 .
Proof. Set d : == inf{d(x, z)
: z E Ho} and take a subsequence Yn E Ho such
that ll x - Yn ll � d as n � oo. Due to the parallelogram inequality, for all m, n E N we have the equality II ( X - Ym ) + ( X - Yn ) II 2 + II ( X - Ym ) - ( X - Yn ) II 2 == 2 II X - Ym II 2 + 2 II X - Yn II 2 · The first of these squares of norms is 4 ll x - Ym tYn 11 2 > 4d2 , and the right hand side of the equality tends to 4d2 as m, n � oo. Hence, the non-negative double sequence I IYm - Yn ll 2 == ll (x - Ym ) - (x - Yn ) ll 2 cannot have positive upper limit, so it tends to zero. We have shown that the sequence Yn is fundamental. But H is complete, so Yn tends to some y E Ho . Since ll x - Yll == limn �oo ll x - Yn ll == d, we see that y is a nearest vector to x in Ho . It remains to show that every nearest vector to x in Ho , say z , coincides with y. The same parallelogram inequality, considered for x - y and x - z, g1ves 0
2 y z + 4 x- 2 + IIY - z ll 2 == 2 ll x - Yll 2 + 2 ll x - z ll 2 == 4d2 . Again the first square of the norm is not smaller than 4d2 , hence the second • is not positive, and therefore must be equal to zero. Thus, y == z.
Proposition 1 becomes invalid if we replace the word "Hilbert" with "Banach" . As we saw in Section 1 .2 (when considering the space CC� , the vector P l , and the subspace span{p 2 } ) , there can be many nearest vectors. But sometimes there are no nearest vectors at all. Exercise
2.
(i) Let f : E � CC be a functional on a Banach space, Eo : == Ker(f) and x E E \ Eo. Then there exists a nearest vector to x in Eo the upper bound in the definition of the norm of f ( cf. Proposition 1 .3. 1 (ii)) is attained.
2.
142
Banach Spaces and Their Advantages
( ii ) Functionals without the indicated property indeed exist, and an example of such a functional is
f : C[ - 1 , 1] � CC : z �----+
1° z(t )dt - l{o z(t)dt. 1
-1
Hint. From the definition of the norm of a functional it follows that
· h l f(x) l · "d es Wit - y ) l . Th"IS COinCI ' I f '' == SUPy E Eo l f(x ll x - yll infyEEo ll x - y ll " Returning back to spaces with inner product, we give the following im portant definition.
H be a near-Hilbert space. The orthogonal complement of a vector x E H is the set xl_ == {y E H : y _L x }. The orthogonal complement of a subset M C H is the set M 1_ == {y E H : y _L x for all X E M} . Proposition 2. If M is a subset in a near-Hilbert space H, then ( i ) M 1_ is a closed subspace in H; ( ii ) M c ( M j_ ) j_ Definition 1 . Let
:
•
Proof. ( i ) follows immediately from the algebraic properties and from the • continuity of the inner product, and ( ii ) is clear.
We write x _L M if x _L y for all y E M.
Let H be a near-Hilbert space, Ho a subspace of H, and x E H. Then x _L Ho � the distance between x and Ho is equal to ll x ll {in other words, 0 is the nearest vector to x in Ho). Proof. ===> This follows from the Pythagorean equality ll x - y ll 2 == ll x ll 2 + IIYII 2 , which is true for all y E Ho . For every y E Ho and A E CC we have ll x - A YI I > ll x ll , hence ( x - Ay , x - A Y ) > ( x , x ) . Thus, ( x , x) - A ( y, x) - A ( x, y) + AA ( y, y ) > ( x, x) . Take A == t (x , y ) . Then for all t > 0 we have - 2t l (x, y ) 1 2 + t 2 1 (x, y ) 1 2 ( y, y ) > 0. 0. The rest is For y :/=- 0 and t < (y2, y ) this is possible only if ( x, y ) • clear. Proposition 3. .
{:::=:: .
:
Now we formulate the main result. Theorem 1 ( on the orthogonal complement ) .
Suppose H is a Hilbert space, and Ho is a closed subspace in H. Then H == Ho EB Hd- , i. e., H can be decomposed into a direct sum of Ho and its orthogonal complement.
3. Theorem on the orthogonal complement
143
E H. Let y be the nearest vector to x in Ho (see Proposition 1) . Put z :== x - y; then the norm of z is equal to the distance from x to Ho. Since y E Ho , this coincides with the distance from z to Ho. By Proposition 3, z _L y. Hence, x can be represented as a sum y+z; y E Ho , z E Hr}- . Then, for all YI E Ho , z1 E Hr}- such that x == YI + z1 , we have y - YI == z1 - z and at the same time y - YI _L z1 - z. Since the vector y - YI is orthogonal to Proof. Take x
itself, it vanishes, hence y == YI and z == z1 . The rest is clear.
II
Here are some geometrical corollaries.
For every {not necessarily closed) subspace Ho in a Hilbert space H, the subspace (Hr}-)1_ coincides with the closure of Ho . In particular, if Ho is closed, then (Hr}-)1_ == Ho .
Proposition 4.
Proof. From Proposition 2 it follows that (Hr}-)1_ is a Hilbert space con taining H0 ( : == the closure of Ho) . Let H1 be the orthogonal complement
of Ho in ( Hr}-) 1_ . Clearly, every x E H1 is orthogonal to Ho and Hr}-; hence, x _L x, and x == 0. Therefore H1 == 0. By Theorem 1 , (Hr}-)1_ == Ho EB H1 . The rest is clear. II
Let { xv ; v E A} be a system of vectors in a Hilbert space H such that for every y E H the condition y _L Xv for all v E A implies y == 0. Then this system is total in H.
Proposition 5.
Proof. Put Ho :== span{xv ; v
E A}; then, by the assumption, Hr}- == {0} . In view of the previous proposition, the closure of Ho is {0} 1_ , which coincides • with the whole H. Proposition 6. Let H be a Hilbert space, Ho a closed subspace in H, and J : Hr}- � HIHo the restriction of the natural projection pr : H � HIHo . Then J is an isometric isomorphism. Proof. From the decomposition into the direct sum (provided by Theo
rem 1) it follows that every element x (a coset) in HI Ho contains exactly one vector x E Hr}- , and every mapping taking x to x is a linear operator inverse to J. Further, ll x ll :== inf{ ll x + Y ll y E Ho} coincides with the distance from x to Ho. Hence, by Proposition 3, ll x ll coincides with ll x ll . • The rest is clear. :
Under the assumptions of the previous proposition, the norm in HI Ho is a Hilbert norm.
Corollary 1 .
We now recall operators called projections in Definition 1.5.2.
2. Under the assumptions of Theorem 1, the projection acting from H to Ho along Hr}- is called an orthogonal projection or, in short, Definition
144
2.
Banach Spaces and Their Advantages
orthoprojection to Ho . It is usually denoted by PH0 , or just P if there is no
danger of confusion.
Orthoprojections are much more than just examples of operators. Their outstanding role in the general theory will become clear in Chapter 6. From the Pythagorean theorem we have
An orthoprojection maps the open (respectively, closed) unit ball in H onto the open {respectively, closed) unit ball in Ho . As a corollary, • the norm of a non-zero orthoprojection is equal to 1 .
Proposition 7.
Note that 1-PHo is also an orthoprojection in H; it has He}- as its range. One of the most important applications of the theorem on the orthogonal complement is that it allows one to give a complete description of all bounded functionals on Hilbert spaces. A mapping between prenormed spaces E and F which is a conjugate lin ear isomorphism (see Section 0. 1 ) and at the same time preserves the norms of vectors, is called a conjugate linear isometric isomorphism. Clearly, a con jugate linear isometric isomorphism is precisely a mapping which, regarded as a mapping between Ei and F, is an isometric isomorphism.
2 (Riesz) . Let H be a Hilbert space. Then every vector e E H defines a bounded functional fe : H � CC by the rule x �----+ (x , e ) , and every bounded functional on H is fe for some uniquely determined e E H. The resulting bijection I : H � H* : e �----+ fe is a conjugate linear isometric iso morphism of normed spaces. {In other words, the bijection I is an isometric isomorphism between Hi and H* .) Theorem
e E H the mapping fe is a linear functional, I fe (e) I < II x II II e II and fe ( x) == II e II 2 . Hence this functional is bounded and II fe ll == II e ll . Thus, the mapping I indicated in the statement of the theorem
Proof. Obviously, for each
is well defined. From the conjugate linearity of the inner product in the second argument it follows that this is a conjugate linear operator. Since I preserves the norms, it must be injective. It remains to show that every f E H* is fe for some e E H. Certainly, it is sufficient to consider the case of f :/=- 0. Put Ho :== Ker( f ) ; this is a closed subspace in H of codimension 1. Take (using Theorem 1) an arbitrary e ' E He}- of norm 1 and put e :== f(e' )e'. Then for every y E Ho we have f ( y ) == fe ( Y ) == 0 and, in addition, f(e' ) == f(e' ) ( e ' , e ' ) == ( e ' , f(e' )e' ) == fe (e' ) . We see that the functionals f and fe coincide on a subspace of codimension • 1 and on one supplementary vector. The rest is clear.
3. Theorem on the orthogonal complement
145
The bijection I in the Riesz theorem is called the canonical bijection between the Hilbert space and its dual. We emphasize that this is not a linear, but a conjugate linear isomorphism. Here are several applications. First, we formulate the following simple but useful result. Proposition 8. If H is a Hilbert space, then its conjugate Banach space H*
is also a Hilbert space with respect to the inner product that is well defined by the equality ( Ix, Iy ) :== (y, x) .
Proof. The fact that this inner product is well defined is immediately veri
fied by using the fact that I is a conjugate linear isomorphism. The assertion that the norm in H* is generated by this inner product follows from the fact that the norm in H is generated by the inner product ( · , · ) , and I is an isometry. II Proposition 9. Every Hilbert space H is reflexive.
rp E H** defines a functional g E H* by the formula g (x) : == rp(fx) (where, as before, we put fx :== I(x)) . By the Riesz theorem, g == fe for some e E H. From this, for each x E H we have rp(fx) == fe (x) == ( e, x) == fx(e) . Again applying the Riesz theorem, we see that rp(f) == f ( e ) for all f E H* , i.e. , rp == ae is the evaluating functional (see Section 1.6) . The rest • is clear. Proof. Every
The following application of the Riesz theorem plays an important role in the questions connected with the future Spectral theorem (see Section
6.6) .
Let S : H x H --+ CC be a conjugate-bilinear functional on a Hilbert space. It is called bounded if sup{ I S(x, y ) I ; x, y E BH} < oo. In other words, S, considered on H x Hi , must be a jointly bounded bilinear functional; see Definition 1.4.6. (As a matter of fact, we could say "separately" instead of "jointly" , but we will see this later, in Theorem 4.5.) The indicated supremum is called the norm of the functional S and is denoted by II S II . The set of bounded conjugate-bilinear functionals will be denoted by CBil(H) . We now take T E B(H) and assign to it the mapping Sr : H x H --+ CC : (x, y ) �----+ ( Tx, y ) . From the properties of the inner product it is clear that Sr E CBil(H) , and II Sr ll == II T II (see Proposition 1.3.3) . The conjugate bilinear functional Sr is said to be associated with the operator T. Theorem 3. The mapping taking T to Sr is a bijection between the sets B(H) and CBil(H) . Proof. Obviously, the indicated mapping is injective. So to prove the the
orem, for each S E CBil(H) we must find T E B(H) such that Sr == S.
146
2.
Banach Spaces and Their Advantages
Take x E H and consider f : H � H : y �----+ S(x, y) . Clearly, f is a linear functional with norm II I II :::; l i S II llx l l . By the Riesz theorem, there exists a unique z E H such that S(x, y) == ( y, z ) . In other words, S(x, y) == ( z, y) . Now let us "release" x, and put T : H � H : x �----+ z. Clearly, this mapping is well defined by the rule ( Tx, y) == S(x, y) ; x, y E H. It follows from the linearity of ( · , · ) and S in the first argument that T is a linear operator. Finally, 1 /Tx ll == sup{ I ( Tx, y) l ; y E BH } == sup{ I S(x, y) l ; • y E BH} < II S II IIxll , hence T is bounded. The rest is clear. Formally, the Riesz theorem deals with abstract Hilbert spaces. How ever, one can deduce from it numerous propositions about concrete Hilbert spaces we encounter in various fields of analysis. As an illustration, we can now fulfil the promise given in Section 1 .6.
Let ( X, J-L) be a measure space. Then every function y E L 2 (X, J-L) defines a bounded functional fy on the very same L 2 ( X, J-L) by the formula x �----+ fx x ( t )y ( t ) dJ-L ( t ) . Conversely, every bounded functional f on L 2 (X, J-L) is fy for some uniquely determined y E L 2 (X, J-L) . This bijection I2 : L 2 (X, J-L) � L 2 (X, J-L)* is an isometric isomorphism of normed spaces. Proof. Obviously, for H == L 2 (X, J-L) the canonical bijection I assigns to every y the functional ]y acting by the formula ]y (x) == fx x (t )y ( t ) dJ-L(t ) Proposition 10.
(we write the hat in the notation of functional to avoid possible confusions) . Hence the functional fy, being just f11 , where y is a complex conjugate function to y, is well defined. Thus, we obtain a mapping I2 , and I2 is a linear operator (unlike I) . Further, every f E L 2 (X, J-L)* is JY for some y E L 2 (X, J-L) , and thus f == f11 . From this we conclude that the mapping I2 is surject ive. Finally, for each y E L 2 ( X, J-L) we have ll fy ll == ll f11 == IIY II == II Y II · • The rest is clear. "'
Warning. The proof of the Riesz theorem works, with obvious changes,
in the case of real Hilbert spaces as well. Moreover, the corresponding canonical bijection is a "true" isomorphism between the space and its dual space. So we can speak about identifying these spaces by means of this bijection. However, for our principal field CC we must be very careful with such statements: here I, although it preserves norms, is not even a linear operator. We note here that from the form of the inner product in H* it follows that the mapping I takes every total orthonormal system in H to a system with the same properties in H* . Hence, the Riesz-Fischer theorem immedi ately implies that in the case of a separable Hilbert space H there are "true" unitary isomorphisms between H and H* (and there are many of them, of course) . Moreover, from the more general Theorem 2.2 it follows that the
3. Theorem on the orthogonal complement
147
same is true for an arbitrary Hilbert space. But all these isomorphisms (be ing linear operators) have nothing in common with the canonical bijection, and they depend on the choice of a total orthonormal system in H. Later (in Section 6. 1) we will see that the Riesz theorem lies in the foun dation of one of the major notions in functional analysis-Hilbert adjoint operator. *
*
*
According to the theorem on the orthogonal complement and Propo sition 7, every closed subspace of a Hilbert space is the image of a pro j ection (i.e. , an idempotent operator) with norm 1 . This means that every closed subspace of a Hilbert space is always topologically complemented (see Definition 1 .5. 1 and Corollary 1.5. 1) . In addition, for a topological direct complement of this subspace we can take its orthogonal complement. But if we go over to general Banach spaces, the picture gets out of con trol: closed subs paces without topological direct complements appear. This phenomenon (that was an unpleasant surprise at the time of its discovery) is usually mentioned in a series of the so-called pathological properties of Banach spaces. Moreover, unlike other pathological properties that will be discussed later (see Theorem 3.3.3) , there is a quite classical series of coun terexamples. Theorem 4.
ments:
The following subspaces have no topological direct comple
(i) co in l 00 (the Phillips theorem) ; (ii) the subspace in C[-1r, 1r] consisting of all functions such that all their
Fourier coefficients with negative indices vanish.
If you are curious of how this can be possible, we give here the proof of the first assertion. Proof of the Phillips theorem. First, for each M C N we put Zoo ( M ) : = {� E Zoo : �n = 0, n rt M} . If f E Z� , then /M : = f l z oo ( M · Note that, if the subsets Mk E N; k = ) 1 , 2, . . . are mutually disjoint, then from the definition of norms in Zoo and Z� it evidently follows that I:� 1 II fMk II < II f II · Now suppose that co is topologically complemented . By Corollary 1 .5. 1 , there exists a bounded projection P : Zoo ---+ Zoo with co as the image. Let f( k ) be a functional on Zoo that assigns to every � the kth term of the sequence P(�) E co . Clearly, it is also bounded. Note that f( k ) ( p z ) = 1 for Z = k, and f( k ) ( p z ) = 0 for l -1 k.
We construct by induction a sequence of natural numbers no < n1 < · · and a sequence of embedded infinite number sets M( o ) =:) M ( l ) =:) · . Put no : = 1 and M(O) : = N . Suppose no , . . . , n k and M( o ) , . . . , M ( k ) have already been constructed. Consider an + arbitrary partition of M ( k ) into disjoint infinite sets MS!: l ) ; m E N, and the corresponding ( n'(J+ l ) ; m E N. Since the series of the norms of the last functionals functionals f( n k ) and fM ·
•
'YYl
•
148
2.
Banach Spaces and Their Advantages
ll f ( n(�+ l ) II
� and no , . . . , n k rt M$;: + 1 ) . We put M( k + 1 ) equal to MS/: + 1 ) , then take an arbitrary N E M( k + 1 ) such that N > n k , and put n k + 1 : = N. Denote by ( ( k ) the sequence with ones at the places n k , n k + 1 , . . . and zeros at the other places . Put ( : = ( ( 1 ) . Obviously, ( = pn1 + · · + pnk + ( ( k + 1 ) , ( ( k + 1 ) E M( k + 1 ) , and l l ( ( k + 1 ) II = 1 for all k. Then, f( n k ) (p n l ) = . . . = f( nk ) ( pnk- 1 ) = 0 and f( n k ) ( pnk ) >= �1 . · From this, by the choice of M( k + 1 ) , for every k we have l f( n k ) ( () I = 1 1 + J ( i) . Obviously, the composition ( in ) S : E --+ E is a projection • with the image Eo. So, Corollary 1.5. 1 works again. Taking the above counterexamples into account, we see that already in the context of Banach spaces not every bounded operator has a bounded extension (to say nothing about a norm-preserving one) . In other words, in the Hahn-Banach theorem, the space of scalars cannot be replaced by an arbitrary Banach space. But, of course, in Hilbert spaces complete harmony reigns: Proposition 12. Let H be a Hilbert space, Ho a subspace in H, and F an arbitrary Banach space. Then every bounded operator rpo : Ho --+ F has an extension rp : H --+ F such that II rpo II == II rp II . Proof. Let HI be a closure of Ho . In view of the extension-by-continuity
principle, rpo has an extension 'P I : HI --+ F such that II 'P I II == II 'P II . It remains to put rp : == 'P I P, where P is an orthoprojection in H onto HI . •
4.
Open mapping and uniform boundedness principles
149
As a matter of fact , examples in Theorem 4 hide the fact that the absence of the indicated pathology is a characteristic property of Hilbert spaces. Theorem 5 ( Lindenstrauss-Tzafriri, [109] ) . The following properties of a
Banach space E are equivalent: ( i ) every closed subspace in E has a topological direct complement; ( ii ) E is topologically isomorphic to a Hilbert space {in other words, the norm in E is equivalent to some Hilbert norm). For discussion and references see [43, p. 255] .
4. Open mapping principle and uniform boundedness principle
Classical functional analysis rests upon three pillars-three fundamental the orems connected with the name of Banach. They are the Hahn-Banach theorem, the Banach theorem on the inverse operator, and the Banach Steinhaus theorem. The reader already knows (we hope ) and respects the first of them. In this section we speak about the second and the third the orems, i.e. , those that had given to Banach spaces their name. Apparently, the Banach theorem on the inverse operator is the deepest among these three results. We start with it. The following proposition historically was the final part of the proof of this Banach theorem, but it has some independent applications.
Let T E --+ F be a bounded operator from a Banach space to a normed space. Suppose that the set T(B�) {i. e., the image of the open unit ball in E) is dense in an open ball 0 B� (with center at 0 and radius 0 > 0) of the space F. Then T(B�) contains all of OB� . Proof. Take y E OB� and 8 E (0, 1) such that y' :== 8- 1 y lies in OB�. By the assumption, there exists x1 E B� such that IIY' - T(x i ) II < (1 - 8 ) 0. Since the set T((1 - 8 ) B� ) ( i.e. , the image of the ( 1 - 8 ) -dilation of the ball B� ) is obviously dense in 0(1 - 8 ) B� ( i.e. , in the (1 - 8 ) -dilation of the ball OB�) , there exists x 2 ; ll x 2 ll < 1 - 8 such that ll y' - T(x1 ) - T(x 2 ) ll < (1 - 8) 2 0. Since T((1 - 8) 2 B�) is also dense in 0 (1 - 8) 2 B� , there is x 3 ; ll x 3 11 < (1 - 8) 2 such that IIY ' - T (xi ) - T (x 2 ) - T(x 3 ) ll < (1 - 8 ) 3 0. Continuing this process, we obtain a sequence of vectors x n ; n E N ; ll xn ll < (1 - 8 ) n - I such that IIY' - T(x1 ) - · · · - T(x n ) II < (1 - 8) n 0. Since the series 2:: � 1 ll xn ll converges and E is complete, the Weierstrass test ( Proposition 1.8) shows that the series 2:: � 1 X n converges to a vector x ' E E. Since T is continuous, T(x ' ) coincides with 2:: � 1 T(x n ) , in other words, with y' . Proposition 1.
:
2.
150
Banach Spaces and Their Advantages
Thus, y' == T( x' ) , and ll x ' ll < 2:: � 1 ll xn ll < 2:: � 0 (1 - 8) n == 8 - 1 . Put x :== 8 x ' ; then, clearly, x E B� and T( x ) == 8y' == y. The rest is clear. • Taking Proposition 1.4.3 into account, we deduce
If the hypothesis of the previous proposition is fulfilled, then the operator T is topologically surjective. Theorem 1 (Open mapping principle) . Let T : E � F be a surjective bounded operator between Banach spaces. Then T(B� ) contains an open ball O B� for some () > 0. As a corollary, 6 T is open and topologically surjective. Corollary 1 .
U � 1 T ( n B� ) .
But F is a Banach space, and therefore, due to the Bair theorem, there exists m such that T ( m B� ) is not rarefied. Hence, it is dense in some open ball V == y + cB� c F. Put () : == 2� and take z E OB� . Since the vectors y + 2mz and, of course, y lie in V, there exist sequences Yn , y� E T ( m B� ) such that the first tends to y, and the second to y + 2mz. Hence, the sequence Zn y� - Yn , obviously belonging to T( 2m B� ) , tends to 2m z , and thus 2!n zn E T(B� ) tends to z. We have shown that T (B� ) is dense in OB� , and it remains to use • Proposition 1. Proof. Since T is surjective, F ==
: ==
Let T : E � F be a bi jective bounded operator between Banach spaces. Then its inverse linear operator r - 1 is bounded. (In other words, every continuous linear isomor phism between Banach spaces is a topological isomorphism.) Theorem 2 (Banach inverse operator theorem) .
Proof. The previous theorem applied to a bijective T means precisely that
r - 1 is continuous at zero.
The rest is clear.
•
The Banach theorem can be reformulated as follows: every morphism in Ban that is bijective as a mapping of sets is an isomorphism. Certainly, the same is true for Hil (but, of course, not for Ban 1 or Hil1 ) . From the proven theorem, taking Propositions 1 . 1 and 1.2 into account , we have
A bounded operator between Banach spaces zs topologically injective � it is injective and has closed image. Corollary 2.
Exercise 1 .
(i) The open mapping principle follows from the Banach theorem. (ii) Under the assumptions of Proposition 1 , F also is a Banach space. 6 See Proposition
1. 4 . 3 .
4.
151
Open mapping and uniform boundedness principles
Hint. In both cases, Propositions 1 .5.3 and 1.5 .4(i) hold and allow us to
restrict ourselves to the case where T is injective. How important is the condition of completeness in the two Banach the orems? The fact that F should be Banach is already shown in Example 1.4.2. The completeness of E is also necessary, although the corresponding counterexamples look more complicated. Exercise 2. Let ( F, II · II ) be an infinite-dimensional Banach space. Then there is another norm II · II ' such that the identity operator 1 : ( F, II · II ' ) � (F, II II ) is bounded, and even a contraction, but has no bounded inverse operator. Hint. There is a linear basis e v ; v E A in F such that II e v II == 1 and inf{ ll e�-t - e v II ; J-L, v E A} == 0. Hence the following norm will fit: II 2: ; 1 A k e vk ll ' : == 2: ; 1 I A k l · There is another theorem which is also often convenient in applications. In fact, it is equivalent to the Banach theorem. In this theorem we use the notion of the graph of a mapping rp : X � Y between sets. By definition, this is the subset r( rp) : == { (x, rp(x)) : X E X} in the Cartesian product X x Y. Obviously, the graph of a linear operator T : E � F is a subspace in E EB F (or, what is the same, in E x F) . ·
3
(Closed graph theorem) . Let T : E � F be a linear operator between Banach spaces. If the graph r(T) of this operator is closed in E EB F considered with the norm of l 1 -sum, then T is continuous. {In other words, if the convergence of X n to x in E and the convergence of T ( xn ) to y in F imply that y == T(x) , then T is continuous.)
Theorem
Proof. By the assumption, r( T ) is a Banach space with respect to the norm inherited from E EB F. Consider s : r(T) � E : ( x, T ( x )) 1---+ Clearly, this
X.
is a bijective bounded (and even a contraction) operator between Banach spaces. By the Banach theorem, s - 1 is also bounded, i.e. , for some C > 0 and for all x E E we have ll ( x, T( x )) ll :== ll x ll + II T( x ) ll < C ll x ll . From this, II for the same x we have II T( x ) ll < (C - 1 ) ll x ll . Exercise 3 . The Banach theorem follows from the closed graph theo
rem. The Banach theorem, as well as its versions, Theorems 1 and 3, has many applications. Here is, perhaps, the simplest one. It looks quite peculiar, asserting something like "an inequality implies an equality" .
Suppose we have two norms II · II and II · II ' in a linear space E, and both spaces (E, II · II ) and (E, II · II ') are Banach. If the first norm majorizes the second, then they are equivalent.
Proposition 2.
2.
152
Banach Spaces and Their Advantages
Proof. The majorization condition evidently means that the operator 1 : (E, II · II ) -t (E, II · II ' ) is bounded. But then, by the Banach theorem, the inverse operator 1 : ( E, II · II ') -t ( E, II · II ) is bounded as well. The rest is
•
clear.
Now we indicate an important application of the Banach theorem to geometry of Banach spaces. Proposition 3. Let E be a Banach space, E. Then every closed linear complement E2
topological direct complement of E1 .
Proof. In our case the mapping
and E1 a closed subspace of of E1 in E {if it exists!) is a
I : E1 EB1 E2 -t E indicated in Proposition
1 .5.7(iii) , is a bounded bijective operator between Banach spaces. By the • Banach theorem, it is a topological isomorphism. Combining this proposition with Corollary 1 .5. 1 , we obtain
A projection of a Banach space to its closed subspace along another closed subspace is (automatically) bounded.
Corollary 3 .
Exercise 4. In Proposition 3 the word "Banach" cannot be replaced
by the word "normed" . Hint. In l 2 consider the subspaces H1 , the closure of span{p 2n ; n E N} and H2 , the closure of span{ ( p 2n + � p2n - l ) ; n E N} , and put H == H1 + H2 (without closure!) . :
*
*
*
Up to now, we have been discussing the concrete meanings of the general categorical notion of an isomorphism in various categories of functional anal ysis. What can we say about the morphisms of the next level of complexity: about retractions and coretractions? In the category of Hilbert spaces the situation is quite transparent. Proposition 4.
Then
Let S : H -t K be an operator between Hilbert spaces.
(i) S is a coretraction in Hil S is an injective operator with closed
zmage; (ii) S is a coretraction in Hil1 S is an isometric operator.
Proof. (i) , (ii) : ===> . Let T be a bounded left inverse to S (i.e. , a morphism in Hil ) . Without loss of generality we can assume that H =/=- 0, hence T =/=- 0.
Then for each X E H we have TSx = X . Therefore, II S x ll > uh ll x ll · Thus, Corollary 1 .4.2 implies that S is topologically injective, and, in particular
4.
Open mapping and uniform boundedness principles
153
(Corollary 2 ) , its image is closed. If, in addition, S and T are contraction operators (i.e. , morphisms in Hil 1 ) , then the last inequality means simply that II Sx ll == ll x ll , i.e. , S is an isometric operator. (i) , (ii) : � - Put Ko : == Im(S) . By the Banach theorem, the operator S0 : == S I Ko is a topological isomorphism on Ko. Suppose P is an ortho projection onto Ko in K, and P0 : == P I Ko . Then, obviously, the operator T : == (S0 ) - 1 P0 is a left inverse to S, and thus S is a coretraction in Hil. If, in addition, S is an isometric operator, then T is a contraction operator, • and thus S is a coretraction in Hil1 . Following the proof of Proposition 4, it is easy to do Exercise 5. For the same S,
(i) S is a retraction in Hil S is a surjective operator; (ii) S is a retraction in Hil1 S is a coisometric operator.
Hint. Suppose S is surjective, and hence topologically surjective, and
So is the restriction of S to Ker(S) j_ . Propositions 1 .5.4 and 3.6 imply that this is a topological isomorphism, and in addition, an isometry, provided S is a coisometric operator. So the coextension of the operator (So) - l to H is a right inverse operator for S. In our remaining categories the description of (co )retractions i s not so transparent: the geometric nature of these objects is too complicated . Exercise 6. In the category Ban
(i) a coretraction is precisely an injective operator such that its image is a closed subspace having a closed linear complement ; (ii) a retraction is precisely a surjective operator such that its kernel has a closed linear complement. In the case of categories Nor and Pre we can make only a not very impressive declaration that coretractions in these categories are topologically injective operators with topologically complemented images, and retractions are topologically surjective operators with topologically complemented kernels (verify this!) . Certainly, what we have said is true for Ban as well, but in this category (and that is the point) the conditions of Exercise 6 imply "all the rest" automatically. Remark. In the category Ban 1 the description of (co )retractions can be given in geo metric terms, as in Ban, but it is more complicated . We have already seen (in the proof of Proposition 4 (ii) : ::::} ) that to be a coretraction in Ban 1 , a contraction operator T : E ---+ F must be isometric, and its image must have a closed linear complement . The latter, as we now know, means that there exists a bounded projection of F onto this image. But gener ally speaking, this is not sufficient: among such projections there should be a contraction (and as a corollary, it will have norm 1 ) . Similarly, T is a retraction in the same category {=:::::} it is isometric and there is a projection of E onto its kernel which at the same time is a contraction operator.
2.
154
Banach Spaces and Their Advantages
Actually, the question on how to describe (co )retractions in Ban1 is among many questions stimulating the interest to subspaces of Banach spaces that are "so well situ ated" that they are images of contraction operators. In particular, in the same geometric terms (this time the question was about the subspaces in B(H) ) the class of the so-called amenable operator algebras was characterized in the 1980s. These algebras play an im portant role in mathematics and mathematical physics ( cf. the end of Section 6.3) . *
*
*
Now we pass to the third fundamental result connected with the name of Banach. Theorem 4 ( Uniform boundedness principle; Banach-Steinhaus ) .
Let E be a Banach space and F a prenormed space. Suppose Tv E � F; v E A is a family of bounded operators. Assume that for every x E E there exists Cx > 0 such that for all v E A we have II Tv ( x ) II < Cx . Then there exists a constant C > 0 such that II Tv ll < C for all v E A. :
The assumed and the stated properties of the operator family here are called, respectively, the pointwise boundedness and the uniform boundedness ( of this family ) . Proof. Assume the contrary. Then we can choose from our family of op
erators a sequence Tn ; n == 1 , 2, . . . such that II Tn ll > n for all n. For each natural k we put Mk : == {x E E : II Tn ( x ) ll < k for all n } . We shall show that every such set is rarefied. Take an open ball U ( x, r) == x + r E� in E and choose an arbitrary n > � (k + Cx) · In view of the inequality II Tn ll > n and the definition of the operator norm, there exists y' E BE such that II Tn ( y' ) II > n; without loss of generality we can assume that y' E B�. Put z : == x + r y'. Then ll x - z ll < ll r y' ll < r, i.e. , z E U(x, r) , and II Tn ( z ) II > II Tn ( r y' ) II - II Tn ( x ) II > r n - Cx > k. By continuity, the same inequality holds after replacing z by an arbitrary point in a certain neighborhood of z, which we call U. Hence, U(x, r) contains an open set, namely the intersection of U ( x, r ) with U, which does not contain points of Mk . This means that Mk is rarefied. Now consider the set M : == U{Mk : k == 1 , 2, . . . } . Being a union of a countable system of rarefied sets, M is meager. By the Bair theorem, it cannot coincide with the entire E. At the same time for all x E E the estimate II Tn ( x ) II < Cx implies that x E Mk for all k > Cx , a contradiction .
•
Certainly, everything that the Banach-Steinhaus theorem says about operators is also true for functionals. The following special case is useful.
5. Banach adjointness functor and other categorical questions
155
Let E be a normed space, and x v ; v E A a family of vectors such that for all f E E* the set of numbers { f ( x v ) ; v E A} is bounded. Then the initial family is bounded with respect to the norm in E.
Proposition 5.
Proof. Consider the family a ( x v ) of functionals on E* , where a : E � E** is the canonical embedding (see Section 1 .6) . By the assumption, this family
is pointwise bounded and thus, by the completeness of E* , it is uniformly II bounded. It remains to recall that a is an isometric operator. The assumption of the completeness of E cannot be omitted. Exercise 7° . The sequence of functionals fn : ( coo , II · ll oo ) � CC; � �----+ n�n is pointwise bounded but not uniformly bounded. We should emphasize that in the Banach-Steinhaus theorem the com pleteness of only one of the spaces is assumed, namely the domain E of our operators; at the same time, in the Banach inverse operator theorem both spaces E and F should be complete. (As for the Hahn-Banach theorem, no completeness of any space is assumed there.) We shall apply the indicated principle many times in this book. Right now we shall give a corollary concerning bilinear operators. Recall (Defi nition 1 .4.6) that there are two approaches to the question which bilinear operators between (pre )normed spaces must be viewed as bounded. More over, already in the classes of normed spaces these two approaches lead to two different classes: separately bounded bilinear operators form a wider class than jointly bounded ones (Example 1 .4.5) . However, in the class of Banach spaces such examples are impossible.
Let E be a Banach space, and F and G arbitrary prenormed spaces. Then every separately bounded bilinear operator n : E F � G is jointly bounded. Theorem 5.
X
Proof. Take y E F and consider the mapping Ry : E � G (see Definition 1 .4.6) . The separate continuity of n implies that the family {Ry : II Y II < 1 }
is a subset in B(E, G) satisfying the assumptions of the Banach-Steinhaus theorem. By this theorem, sup{ II Ry II : II Y II < 1 } < oo , and this number • obviously coincides with II R II in Definition 1 .4.6. 5 . Banach adj ointness functor and other categorical questions
A series of important functors is defined on the categories of functional analysis. To begin with, on these categories, as on any category, act the co- and contravariant functors of morphisms (see Example 0.7. 1 ) . But a noticeable feature of functors in the categories Ban, Nor, and Pre is that
2.
156
Banach Spaces and Their Advantages
they can (and should) be considered with the values not in Set as in the general case, but in the initial category. We shall restrict ourselves to the case of the first of the indicated categories, which is apparently the most important. We have seen that for each E, F E Ban the set B ( E, F) , i.e. , in the gen eral categorical notation, haan (E, F) , has the structure of a Banach space (Proposition 1 . 7 ) . If in Section 0.7, we, respectively, replace haan (E, F) by B ( E, F) in every notation connected with mar-functors, we can make the following observation. Proposition 1. For every E E Ban and for each morphism rp : F � Ban {i. e., for each bounded operator from F to G) the mappings
B(E, rp) : B(E, F)
B(E, G) : 'ljJ
r---t
rp'l/J
B (rp, E ) : B(G, E ) � B ( F, E ) : 'ljJ are bounded operators.
r---t
'l/Jrp
�
and
G in
Proof. This immediately follows from the estimate for the operator norm • of the composition; see Proposition 1 . 3.4.
Proposition 1 implies that every Banach space E generates two functors, a covariant and a contravariant, from Ban to Ban itself, namely
B( E , ?) : F
r---t
B( E , F) ; rp
B (? , E) : F
r---t
B(F, E ) ; rp
r---t
and r---t
B( E , rp) B ( rp, E ) .
We leave it to the reader to give similar definitions of morphism functors acting from Nor to Nor, and from Pre to Pre. Among all of these functors, the main role in analysis belongs to the contravariant functor B(?, CC) : Ban� Ban corresponding to the case E == CC. Definition 1. This functor is called the
Banach star functor) and is denoted by ( * )
Banach adjointness functor (or : Ban� Ban.
Obviously, for an object E in Ban the object (*) ( E ) is just the dual space E* . Respectively, for a morphism T in Ban it is customary to write T* instead of ( * ) (T) . The morphisms we have built (i.e. , operators) are so important that we shall give their direct definition.
G be an operator between Banach (or, more general, normed) spaces. The operator T* : G * � F* acting by the rule f fT is called the Banach adjoint to T . Definition 2. Let T r---t
:F
�
5. Banach adjointness functor and other categorical questions
157
. Thus, for every x E F we have the equality [T* f] ( x) f(Tx) or, 1n other notation, (T * J, x) == (J, Tx) . (1) If you prefer formulas, you can accept this as an initial definition of the ad joint operator. At the same time the action of this operator is well illustrated by the following commutative diagram:
which is easier to remember than equality ( 1) . Remark. We call this functor "Banach adjointness" instead of just "ad
jointness" because later in the book an important role will belong to the notion of the Hilbert adjoint operator (see Definition 6. 1 . 1 ) , and this is not quite the same. Contrary to ( * ) : = B(?, C) , the (covariant) functor B(C, ?) is of no special interest . Exercise 1 . The functor B(C, ?) is naturally equivalent to the identity functor in Ban.
Here are several properties of the Banach adjoint operator. Exercise 2. Let S : E1 --+ E2 and T : F1 --+ F2 be bounded operators between Banach spaces. Then (i) if E1 == F1 and E2 == F2 , then (S + T)* == S* + T* ; (ii) (AS)* == AS* for all A E CC; (iii) II s * II == II s II ; (iv) if E1 == F2 , then (ST)* == T* S* ; ( v) if 1 is the identity operator on E, then 1 * is the identity operator on E* . Hint. Properties (iv) and (v) are encoded in the word "functor" , and (iii) follows from the Hahn-Banach theorem. Note that (iii) implies that the Banach adjoint to a contraction operator is again a contraction. This allows us to introduce an analogous "Banach adjointness functor" acting on the category Ban1 . Taking the star functor twice we obtain the functor (** ) : Ban-+Ban, which is certainly covariant. This functor assigns to every F E Ban its second dual space F** (see Section 1 .6) and to every operator T : F --+ G its so-called second adjoint operator T** : F** --+ G** . It is closely
158
2.
Banach Spaces and Their Advantages
connected with the identity functor 1 in Ban ( cf. Section 0. 7) . We recall the canonical embedding ap : F � F** defined in Section 1.6 for each normed (in particular, Banach) space F.
For every operator T : F � G there is the following com mutative diagram of Banach spaces:
Proposition 2.
In other words, the family { ap : F E Ban} is a natural transformation between the functors 1 and (** ) acting on Ban {see Definition 0. 7. 6). Proof. Since elements of G** are functionals on G* , our goal is to show
that for all x E F and f E G* we have (aa (Tx) , f) == (T** (apx) , f). The construction of the operators ap and aa together with equality ( 1 ) gives (aa (Tx) , f) == ( f, Tx) == (T * f, x) == (apx, T* f) . Now (apparently, this is a psychologically difficult moment) , we substitute T** for T* , T* for T, apx for f, and f for x in the general formula ( 1) . We • see that (apx, T* f) is just (T** (apx) , f) . The rest is clear. Banach adjoint operator is one of the key notions in functional analysis, and we will often discuss its properties. Now we give several illustrations of what it turns out to be for some concrete operators. Exercise 3. Suppose A E Zoo . Then the Banach adjoint operator r; to the diagonal operator T).. (see Example 1 .3.2 ) acting on l2 (and, respectively, on co, l 1 ) coincides, up to isometric equivalence, with the diagonal operator T).. on l 2 (respectively, l 1 , l 00 ) . To be more specific, we have the following commutative diagrams:
where /2 , Io and /1 are isometric isomorphisms indicated in Exercises 1 .6. 11 .6.3. Exercise 4. The Banach adjoint operator T[* to the operator of the left shift Tz (see Example 1.3.3 ) acting on l 2 (respectively, in co, l 1 ) coincides, up to isometric equivalence, with the operator of the right shift Tr in l2 (respectively, l 1 , l 00 ) , and in this assertion we can interchange Tz and Tr.
5. Banach adjointness functor and other categorical questions
159
(Formulate and prove a more detailed statement, writing the corresponding commutative diagrams.) Exercise 5. Find the Banach adjoint operators for the shift operators on the spaces Lp (X) for X : == IR, 1r and p == 1 , 2 (see Example 1 .3.7 ) . Hint. This is the "shift in the opposite direction" . Certainly, the Banach adjoint operator to a topological or isometric iso morphism I also belongs to the same class. You can either verify this di rectly, or use the fact that I* is the value of the functor (* ) on an isomorphism in Ban or in Ban1 . Remark. The converse is also true: if I* is a topological or isometric iso
morphism, then I have the same properties. This fact will be soon suggested as Exercise 10 in the "advanced" part of the text. In the following two examples, E is a closed subspace of a Banach space F. Example 1. The Banach adjoint operator in* : F*
� E* to in : E � F
assigns to each functional on F its birestriction to E. From the Hahn Banach theorem it evidently follows that this operator is a coisometry.
� F* adjoint to pr : F � FjE, acts by the formula f �----+ g , where g : x �----+ f(x + E) . Clearly, this is a contraction operator, and from the fact that every coset of norm < 1 contains a vector of norm < 1 it follows that ll f ll == sup{ l f ( x ) l : x E B�; E } < sup{ lg(x) l : x E B� } == ll g ll . Example 2. The operator pr * : (F/E) *
Thus, pr* is an isometric operator. Here is a more general observation. Exercise 6.
(i) An operator between Banach spaces is isometric its Banach adjoint operator is coisometric; (ii) an operator between Banach spaces that is Banach adjoint to a coisometric operator, is isometric. Hint. The ===:>-part in (i) and the entire statement (ii) are established by arguments similar to those in Examples 1 and 2. Further, from Proposition 1 it follows that if I** is coisometric, then I has the same property. Taking this into account, we deduce the �-part of (i) from (ii) . The following exercise is a "topological" analogue of the previous "met ric" assertion.
2.
160
Banach Spaces and Their Advantages
Exercise 7.
(i) An operator between Banach spaces is topologically injective � its Banach adjoint operator is surjective (or, equivalently, topolog ically surjective) ; (ii) an operator between Banach spaces that is Banach adjoint to a surjective operator, is topologically injective. Hint. The representation of operators as compositions described in Proposition 1 .5.3 allows us to reduce the case under consideration to (co )iso metrjc operators jn the prevjous exercjse.
Remark. As a matter of fact , in both Exercises 6 and 7 part (ii) is true in
"both directions" , just as part (i) . In other words, an operator is coisomet ric (respectively, surjective) � its Banach adjoint operator is isometric (respectively, topologically injective) . But the proofs of the �-part of this criterion known to author, require powerful tools exceeding the scope of the book. The required arguments are presented in [72, Theorem 4. 15] . (Perhaps you will succeed in finding a simpler proof.) Later we will find out which topological properties distinguish the adjoint operators among all operators acting from F* to E* (see Theorem 4.2.3 ) . Observations we have already made allow us to prove the theorem lying in the foun dation of the so-called "Banach homology" , a set of problems on the boundary between functional analysis and homological algebra. Let K be a category. A sequence in K is a diagram of the form Xn - 1 0 and K : == max{ II 'P k II ; k == 1 , . . . , n } we have n
n
n
k= l
k= l
k= l
The rest is clear.
II
Exercise 14. Every finite family of Hilbert spaces has the product and the coproduct in the category Hil.
Hint. The Hilbert direct sum (see Section 1 ) of the given spaces, con
sidered with the corresponding projections, is their product, and considered with the corresponding embeddings, is their coproduct. From Proposition 4.3 we immediately have
Suppose a Banach space E is decomposed into a direct sum of closed subspaces E1 and E2 . Then, together with the corresponding natural embeddings, E is the coproduct of E1 and E2 in Ban .
Corollary 1 .
Trying to extend the results about finite (co )products to infinite families of objects, we come to a lamentable result. Exercise 15* ( cf. Exercises 0.6. 7 and 0.6.8 about somewhat different situation in Met ) . Any infinite family of non-zero objects in Ban has neither
166
2.
Banach Spaces and Their Advantages
product , nor coproduct in this category. The same is true if we replace Ban by Hil. Hint. If, say, the family En ; n E N has a hypothetical product in Ban, then the desired contradiction follows from the choice of F == CC and op erators 'Pn : CC � En with sufficiently rapidly growing norms. In the case of a hypothetical coproduct we can put F == EBP { Ev : v E A} for each p E [1 , oo] when considering Ban and p == 2 for Hil, and after that take the multiples of the corresponding embeddings, again with rapidly growing norms, as 'Pn : En � F. :
:
But the situation sharply improves if we pass from Ban to Ban1 (i.e. , forbid the operators to have large norms) . Exercise 16. Every family of Banach spaces Ev ; v E A has both the product and the coproduct in the category Ban1 . The product of this family is the Banach direct product ((B00 {Ev : v E A} , II · ll rr ) with projections 1rJ.L : EB 00 { Ev : v E A} � EJ.L : f �----+ f (J-L ), and the coproduct is the Banach direct sum ((B 1 {Ev : v E A} , ll · ll n ) with the embeddings iJ.L : EJ.L � (B 1 {Ev : v E A} taking x E EJ.L to the mapping that sends J-L to x, and the other
elements of the index set A to zero. Hint. The proofs of both results are very similar. We restrict ourselves to the first. Suppose we have F and a (contraction!) 'Pv : F � Ev ; v E A. The pair consisting of the Cartesian product E0 of our spaces and the standard family of projections is the categorical product of our family in Lin (Proposition 0.6 ) . Hence, there exists a unique linear operator 1/; 0 : F � E0 making the corresponding diagram commutative for each v E A. It is easy to see that the corestriction 1/J to the Banach direct product is a contraction operator (i.e. , a morphism in Ban1 ) . The rest of the section is addressed to the reader who wants to know more about the categories under consideration. First , the following exercise contrasts with the previous one. Exercise 17* . In the category Hih (even) a family of two non-zero spaces has neither product , nor coproduct.
Hint. Suppose Hk are our spaces and X k E Hk ; ll x k II = 1 (k = 1, 2) . To obtain a contradiction, take Y := C and 'P k : 1 �-----+ ± x k for products, and 'P k : y �-----+ (y, X k ) for coproducts (see Definitions 0.6 . 1 and 0.6.2 ' ) . It turns out that because of the specific features of "Hilbert" geometry (in particular, the parallelogram identity) the required operator 'lj; cannot be a contraction. We now recall the usual approach to the study of a category formed by sets with some additional structure. Usually, such a category is endowed with the forgetful functor and becomes a concrete category in the sense of Definition 0. 7.2. Certainly, the categories of functional analysis are no exception. Here , in particular, one can speak about concrete categories (K , D) , where K is any of the categories Ban, Ban 1 , Hil, Hih , and D is the forgetful functor to Set . As is easy to see , the concrete categories Ban 1 and Hih are not
167
6. Completion
balanced, but the two remaining categories are: for Ban, if you look at it attentively, you see that this fact is simply an equivalent formulation of the Banach theorem. (Show that the wider concrete category (Nor, D) is already not balanced; cf. Example 1 .4 . 2 . ) In conclusion, consider the question about the categorical bases and freedom. First , we describe free objects in (Ban, D) and (Hil, D ) . Exercise 1 8 . In the indicated two concrete categories the free objects are precisely the finite-dimensional spaces, and a (categorical) basis of a finite-dimensional space is just a linear basis.
Hint. The last statement follows from Theorem 2 . l (iii) . Now let E be an infinite dimensional space , and S a basis in E. Then S cannot be finite , because otherwise , taking F := E j span( S) , we obtain operators pr, 0 : E ---+ F that are different morphisms taking S to 0 E F. And if S contains a sequence Xn ; n E N , then there is no bounded operator on E taking X n to nx n . Exercise 19. In the concrete categories (Ban1 , D) and (Hil 1 , D) no space has a (categorical) basis.
Hint. If x is a point of a hypothetical basis S in E, then every mapping
( ii) . Take x E E. By Theorem 1 .5.3, there exists a functional f : E --+ CC of norm 1 such that f(x) == llx ll · Since f is a contraction
operator to a Banach space, by the assumption, there exists a contraction operator 1/J : E --+ F making the above diagram with CC as F and f as rp commutative. In particular, 1/; i (x) == f(x) , hence ll i (x) ll > 11 1/J i (x) ll == llx ll . Since i is a contraction operator and x is arbitrary, i is an isometric operator. Now let E1 be the closure of the image of i. Put F :== E / E1 and take the zero operator as rp : E --+ F. Then the indicated diagram is commutative if we take for 1/J the natural projection E onto F. But it is commutative, in particular, for 1/J :== 0. By assumption, such 1/J is unique. Hence - the natural projection E onto F is a zero operator, and this means that E == E1 . (ii)� (i) . Put Eo :== Im( i ) and for a given rp : E --+ F consider the operator 1/Jo : Eo --+ F well defined (because i is injective) by the rule 1/Jo ( i (x)) :== rp(x) . Since i preserves norms, 1/Jo is a contraction operator just like rp. Evidently, a contraction operator 1/J : E --+ F makes the above-indicated diagram commutative � 1/J is an extension of 1/Jo . But by assumption, Eo is dense in E. Therefore, the existence and uniqueness of such an operator • follow from the extension-by-continuity principle (Theorem 1 .2) . Definition 1 . A pair (E, i ) that has the equivalent properties indicated in Proposition 1 , is called the completion of the normed space E. Remark. Sometimes, we will briefly say "the completion of the space E" , having in mind just the Banach space E. Let us emphasize, however, that
this is just a liberty of speech: the isometric isomorphism i is an integral part of the definition of completion. If the completion ( E, i ) is given, then, identifying every x E E with i ( x) , we can view E as a dense subspace in E; this is often (although, not always) useful. (Psychologically, there must be nothing new for us here: formerly this is how we had learned to view rational numbers as a special case of reals, and still earlier, integers as a special case of rational numbers.)
169
6. Completion
Warning. In old textbooks the sentence "the completion of E is a
complete space containing E as a dense subset" is used as a precise definition of completion. It follows from what we have said earlier that our point of view is somewhat different. In the universal property of the completion the contraction operators were considered; however, there is an analogue involving arbitrary bounded operators.
Let (E, i) be a completion of the space E. Then for each and each bounded operator rp : E --+ F there is a unique bounded operator 1/J : E --+ F such that the diagram in Proposition 1 {i) is commutative. In addition, 11 1/J II == II 'P II ·
Proposition 2. Banach space F
Proof. This evidently follows from the extension-by-continuity principle
•
(Theorem 1 .2) .
Here are several classical examples of completions. If P[a, b] is the space of polynomials on an interval endowed with the uniform norm, then its completion is the pair (C[a, b] , in : P[a, b] --+ C[a, b] ) . The completion of the space C[a, b] considered with the (non-uniform) norm ll x ll 1 :== J: l x(t) l dt is (L 1 [a, b] , i : C[a, b] --+ L1 [a, b] ) , where i assigns to every continuous function its coset in £ 1 [a, b] . (As you see, it is not always convenient to require that E is a part of E. ) The completion of coo considered with the norm II · li P is (lp , in) for 1 < p < oo, and (co , in) for p == oo . In all these examples we, of course, give just one, apparently the simplest, of possible completions. Note that if E itself is a Banach space, then one of its completions is (E, lE) · At the same time, all completions of such E, obviously, are defined as pairs (F, i E --+ F) where i is an arbitrary isometric isomorphism between E and some Banach space. However, various completions of the same space differ only "in form, but not in substance" . :
Suppose ( E 1 , i 1 ) and ( E2 , i 2 ) are two completions of a normed space E. Then there exists a unique isometric isomorphism I : E1 --+ E2 such that the diagram
Theorem 1 (Uniqueness of completion) .
is commutative.
2.
170
Banach Spaces and Their Advantages
(Thus, E I and E2 are not only isometrically isomorphic, but this iso metric isomorphism can be chosen to agree with the operators i i and i2 ) . Proof. (Cf. proof of Theorem 0.6. 1. ) We introduce the category Ban E (some complication of Ban) . Its objects are all pairs ( F, cp ) consisting of a
Banach space F and a contraction operator cp : E � F. The morphisms be tween objects ( FI , 'PI ) and ( F2 , 'P2 ) of this category are contraction operators 1/J : FI � F2 such that the diagram
is commutative. Composition of morphisms in Ban E is their composition in Ban (i.e. , the usual composition of operators) . The axioms of category are easily verified. From the universal property of completions it evidently follows that the pair (E, i ) is a completion of E it is an initial object in Ban E . Hence, from Theorem 0.4. 1 we see that the pairs (E I , i i ) and (E2 , i 2 ) are isomorphic as objects of Ban E . Certainly, the corresponding morphism I is the required • isometric isomorphism. Now we pass to the question of the existence of completions. Unlike the question of uniqueness, there is no general categorical scheme here, and the specific features of the considered construction come to the foreground. Theorem 2 (On the existence of completion) .
a completion.
The first proof. Consider the linear space
Every normed space E has
£ of all fundamental sequences
x == (X I , x 2 , . . . ) of vectors from E with coordinatewise operations. Endow it with the prenorm II x II o == limn �oo II X n II , where II · II is the norm in E (certainly, such prenorm is well defined) . Put Eo :== {x E £ : ll x ll o == 0 } and consider the normed space E :== £f£o (see Proposition 1 . 1 .3 ) . Further, consider the operator j : E � £ assigning to a vector x the constant sequence x' :== ( x, x, . . . ) and put i :== (pr)j : E � E, where pr : £ � E is the natural projection. Our goal is to show that the pair (E, i ) is a completion of the :
space E. First, since j and pr are, obviously, isometric operators, the same is true for i. Further, take an arbitrary y E E; then y == pr ( x ) for some x == (X I , x 2 , . . . ) E £. Since x is fundamental, it easily follows that the
171
6. Completion
elements ( constant sequences ) x� :== j (x n ) tend to x in £ as n � oo , hence the elements i(xn ) == pr(x�) tend to y in E. Thus, Im (i) is dense in E. It remains to show that E is complete. Let Ym ; m == 1, 2, . . . be a fundamental sequence in E, and Ym == pr ( x m ) ; x m == ( x!, x2 , . . . ) E £. For each m == 1, 2, . . . the sequence x m is fundamental. Therefore, there exists a term of this sequence, which we denote by X m E E, such that ll x m - x � ll < � for all sufficiently large n, and thus II X � - x m ll o < � ( in £ ) . In particular ' we see that the sequences x � and x m in £ either converge or do not converge simultaneously, and if they converge, then have the same limits. Now take the sequence x :== (x1 , x 2 , . . . ) ( formed by the chosen vectors of E) . From the relations
ll xz - X m ll == ll x � - x � ll o < ll x � - x l ll o + ll x l - x m ll o + ll x m - x � ll o for all l, m == 1 , 2, . . . , we can easily see that this sequence is fundamental in E, i.e. , x E £. This evidently implies that the sequence x � ( of elements of £ ) tend to x, and thus, as we noted before, the sequence x m has the same limit. But then the given sequence Ym == pr ( x m ) E E also converges, • namely, to the element pr(x) E E. a : E � E** and denote by E the closure of the image of a in E** . Since a is an isometric operator, Propositions 1 .7 and 1.3 show that the pair (E, a l E ) is the required • completion. The second proof. Consider the canonical embedding
Remark. The second proof perhaps is not so instructive as the first. Of
course, it is much shorter, but this is deceptive: it relies on a very powerful tool, the Hahn-Banach theorem; as we remember, the proof of the Hahn Banach theorem requires serious work. Corollary 1 .
Banach space.
Every non-complete normed space is a dense subspace of some
We now discuss with the advanced reader the category theory interpretation of The orem 2. The fact is that the question on the existence of completions, as well as a vast variety of various questions of algebra and analysis, is a question of representability of some functor (cf. Definition 0.7.8) . Let E be a normed space . Consider the contravariant functor :F : Ban1 ---+ Set assign ing to every Banach space F the unit ball in the space B(E, F) and to every morphism
(iii) . Among all representations of u as a sum of elementary
tensors there is a representation with the least number of summands; let it be ��= I X k Q9 Yk · If the vectors XI , . . . , X n are linearly dependent, then one of them, say, XI , has the form ��= 2 A k x k ; A k E CC. Hence u == ��= 2 X k Q9 (A. k YI + Yk ) · By the choice of the initial representation, this is impossible. Therefore, XI , . . . , Xn are linearly independent, and it remains to apply the same arguments to the vectors YI , . . . , Yn . (iii) ===> ( ii) is clear. (ii) ===> ( i) . Take f : E � CC such that f (x i ) :/=- 0 and j(x 2 ) f (x n ) == 0, and g : F � CC such that g (yi ) :/=- 0. Then (f Q9 g) (u) f (xi ) g (y i ) :/=- 0. The rest is clear. •
The following two facts follow from Proposition 1 . Exercise 2 ° . If x, y E E, then x Q9 y == y Q9 x in E Q9 E x and y are. collinear. Exercise 3 ° . If e � ; J-L E A' and e�; v E A" are linear bases in E and F respectively, then e � Q9 e�; (J-L, v ) E A' x A" is a linear basis in E Q9 F. Now we can suggest another important example of the algebraic tensor product.
Let E and F be normed spaces. Then there exists a linear isomorphism7 Gr8 : E* Q9 F � :F(E, F) uniquely defined by the rule that it takes the elementary tensor f Q9 y to the one-dimensional operator f 0 y.
Proposition 2.
Proof. Consider the bilinear operator g :
x
E* F � :F(E, F) taking a pair
(f, y) to f 0 y. The universal property provides an operator Gr8 uniquely
defined by the indicated rule. From Propositions 1.5.5 and 1.6 it evidently follows that Gr8 is surjective. We show that it is injective. Suppose u E E* Q9 F does not vanish. By Proposition 1 (ii) , it can be represented as ��= I x'k Q9 Yk , where YI , . . . , Yn are linearly independent, and x i :/=- 0. Take XI E E such 7
The notation means that this mapping is a birestriction of the so-called Grothendieck op erator Gr, which will be defined at the end of this section.
177
7. Algebraic and Banach tensor products
that xi (x1) == 1 . If T :== Gr8 ( u ) , then Tx1 == Y1 + ��= 2 x'k (x1 ) Yk =/= 0, and • thus T =!= 0. The rest is clear. Our next goal is to pass from the purely algebraic tensor product to its "Banach" version. As a matter of fact, there exist quite a few different versions of the notion of tensor product for general Banach spaces. However, we shall restrict ourselves to one of them, apparently the most important ( cf. the remark after Exercise 5) . (Another version, the most convenient for working with Hilbert spaces, will be presented in the next section.)
E and F be Banach spaces. A pair (8, 0) , where 8 is a Banach space and 0 : E x F --+ 8 a bilinear contraction operator, is called the projective tensor product of E and F, or, as we shall more frequently say, the Banach tensor product of these spaces if for every Banach space G and for a bilinear contraction operator n : E F --+ G there exists a unique linear contraction operator R 8 --+ G such that the Definition 3 ( cf. Definition 1) . Let
:
X
diagram
ExF
is commutative.
el8 � G R
Let (8 1 , 0 1 ) and (8 2 , 02 ) be two projective tensor products of Banach spaces E and F. Then there exists a unique isometric isomorphism I : 8 1 --+ 8 2 such that the diagram Theorem 3 (Uniqueness; cf. Theorem 1 ) .
is commutative. Proof. Proof of Theorem 1 works with obvious "Banach" modifications. Now we use the category Ban E , F , whose objects are pairs (G, R) consisting
of a Banach space G and a bilinear contraction operator R : E x F --+ G. • We leave the remaining details to the reader.
Now we begin preparations for the existence theorem of the Banach tensor product. Suppose we have (arbitrary) prenorms on linear spaces E and F. For each element u of the algebraic tensor product E Q9 F we put ll u ll p :== inf ��= 1 ll x k II IIYk II , where the infimum is taken over all possible representations u == ��= 1 X k Q9 Yk ; X k E E, Yk E F. Clearly, II · l i P is a prenorm.
2.
178
Banach Spaces and Their Advantages
projective tensor product of the given prenorms or simply the projective prenorm on E Q9 F. Definition 4. This prenorm is called the
We shall denote the prenormed space (E Q9 F, II · li p ) by E @p F. Exercise 4. The open unit ball in E @p F is the convex hull of the set { x Q9 y : x E B� , y E B�} . :
x
Let R E F � G be a jointly bounded bilinear operator to a prenormed space. Then the associated operator Ro : E @p F � G is also bounded, and II Ro I == II R II . Proof. Take u == ��= I X k Q9 Yk ; X k E E, Yk E F; then from Ro ( u ) == ��= I R(x k , Yk ) it follows that li Ra ( u ) II < II R II ��= I ll x k II IIYk II . There fore, by the definition of projective prenorm, II Ro ll < II R II . Further, from R(x, y) == Ro (x Q9 y) and the obvious estimate ll x Q9 Y ll < ll x ll IIYII we have that II R II < II Ro II sup{ II x Q9 y II : x E BE , y E BF } < II Ro II . • Corollary 1. For f E E* and g E F* the norm of f Q9 g as a functional on E @p F is equal to II f II ll g II . Prop osition 4. For all x E E and y E F we have ll x Q9 Yll p == ll x ii iiYII · Proof. Obviously, it is sufficient to consider the case where ll x ll > 0 and IIYII > 0. By Theorem 1 .6.3, there exist functionals f E E* and g E F* such that f(x) == ll x ll , g(y) == IIYII , and ll f ll == ll g ll == 1 . Hence, taking Corollary 1 into account, ll x ll IIYII == (f Q9 g ) (x Q9 y) < ll x Q9 Yll p · The rest is clear. • Proposition 3.
Before proceeding further, let us distinguish a useful application of the Hahn-Banach theorem. It is a special case of the statement presented in Chapter 1 as Exercise 1 .6.9.
Let E be a normed space and XI , . . . , X n E E a linearly independent family of vectors. Then there exists f E E* such that f(x ) =/=- 0, f(x 2 ) == · · · == f(x n ) == 0. Proof. Since XI , . . . , X n are linearly independent, a functional with indi cated property exists on the space Eo == span { XI , . . . , Xn } and is bounded by Theorem 1 . 1 (iii) . It remains to extend it to E using the Hahn-Banach • theorem. Proposition 6. If E and F are normed spaces, then E Q9p F is also a normed space. Proposition 5.
i
u E E @p F is non-zero. By Proposition 1 , u has the form ��= I X k Q9 Yk ; X k E E, Yk E F, where XI , . . . , Xn are linearly independent, and YI =/=- 0. Therefore, by Proposition 4, there exist f E E* and g E F*
Proof. Suppose
179
7. Algebraic and Banach tensor products
such that f(x 1 ) =/= 0, f(x 2 ) == · · · == f(xn ) == 0 and g(y1 ) =/= 0. Hence, taking Corollary 1 into account, n
ll f ll ll 9 ll lluii P > l (f 0 g) ( u ) i = L f(x k ) 9 ( Yk ) k= 1
=
l f(xi ) 9 ( YI ) I > 0. •
The rest is clear.
From now on we shall study the case where E and F are Banach spaces. Then, by the previous proposition, E @p F is a normed space. Take its completion and denote it by (E @ F, i) . Then, as was discussed in the previous section, we can assume, without loss of generality, that E @p F is contained as a dense subspace in E @ F, and i is a natural embedding. Introduce also a bilinear operator J : E x F --+ E @ F assigning to a pair (x E E, y E F ) the elementary tensor x Q9 y . From the definition of the prenorm II · li P it immediately follows that {) is a bilinear contraction operator. A
Theorem 4 (Existence of projective tensor product) . For every Banach space G and for every bounded bilinear operator R : E x F --+ G there is a
unique bounded linear operator R : 8 --+ G such that the diagram ExF
Jl8 � G R
is commutative. Herewith, II R II == II R II . In parti cular, the pair (E @ F, J ) is the Banach tensor product of the spaces E and F. Proof. Suppose Ro : E Q9 F --+ G is the operator associated (in the sense of Definition 1 ) with R; then, by Proposition 3, II Ro ll == II R II . By Proposition 6.3, there exists a unique operator R : E @ F --+ G with the necessary
properties.
II
The norm in E @ F will be also denoted by II · li P (or just II · II ) . We call R the operator associated with the {bounded) bilinear operator R, and the indicated property of the pair (E @ F, J ) , the universal property (of the Banach tensor product) . There should be no confusion with similar purely algebraic notions. Further we shall need the following general fact, which is also of inde pendent interest.
Let E be a normed space, and Eo a dense subspace of E . Then for each c > 0 every element x E E can be represented as the sum of an absolutely convergent series � C: 1 x k , where X k E Eo and � � 1 ll x k ll < ll x ll + c.
Proposition 7.
2.
180
Banach Spaces and Their Advantages
x1
Eo such that ll x - XI II < Similarly, there exists x 2 E Eo such that
Proof. Since Eo is dense in E, there exists
E
c/4. Hence II x i II < ll x ll + c /4. ll x - XI - x 2 ll < c/8. Hence, ll x 2 ll < ll x - XI II + c /8 < 3c /8. Further, there exists X 3 E Eo such that ll x - XI - x 2 - x 3 11 < c /16, hence ll x 3 11 < ll x - XI - x 2 ll + c /16 < 3c /16. Continuing this process, we obtain a sequence x k of vectors in Eo such that n x - L X k < cj2n + l and ll xn ll < 3c/2n +l for all n > 2. k= I From this it follows that, first, the series � C: I X k converges to x. Moreover, 00 00 3 1 L ll x k ll < ll x ll + c ( 4 + L 2 k+ 1 ) = ll x ll + E , k= I k= 2 • and this is what we need. Proposition 8. Every u E E@F is representable as the sum of an absolutely convergent series �C: I X k Q9 Yk ; x k E E, Yk E F, k == 1 , 2, . . . . Moreover, llullp == inf � � I ll x k II IIYk II , where the infimum is taken over all possible representations of u in this form. 0 and, using the fact that E Q9 F is dense in E ® F and Proposition 7, represent u in the form � C: I un , where Un E E ® F and � � I llun llp < llullp + c/2. By the definition of the norm II · li p , every Un has the form � �]_ X ni Q9 Yni , where n L ll x ni II IIYni II < llun l i P + 2 .E2n · i= I Proof. Take
c
>
Hence �� I � �]_ ll xni II IIYni II < llullp + c. Since c was arbitrary, llullp is not smaller than the infimum indicated in the formulation. The converse • inequality is obvious. As an exercise, we suggest the following result, which establishes a close relation between taking the space of operators and taking the projective ten sor product. In fact , this connection is the main advantage of the projective tensor product compared to other versions of this construction. Exercise 5 (Adjoint associativity law) . For all E, F, G E Ban there is an isometric isomorphism between the spaces B(E @ F, G) and B(E, B(F, G)) ,
uniquely defined by the rule that it takes an operator rp to the operator 1/J such that ( 1/J ( x)) ( y ) : == rp ( x Q9 y ) . In particular, up to an isometric isomor phism we have (E @ F) *
== B(E, F * ) .
181
7. Algebraic and Banach tensor products x
Hint. The norm of the bilinear operator from E F to G with which rp is associated, coincides with II � II . Remark. Explicit constructions of other versions of the notion of tensor
product of Banach spaces, which are not considered in this book, could be obtained as a result of the completion of the linear space E Q9 F in norms different from the projective one. There exists a very deep theorem due to Grothendieck, 8 asserting that there are precisely 14, not more, not less, "natural" in some reasonable sense versions of such norms, and thus 14 versions of tensor product (see, e.g. , [48, 11.27] ) . Another use of tensor products of Banach spaces in analysis is that they help to describe the passage from functions of one variable to functions of two variables, that have, to speak informally, "the same nature" . This concerns practically all existing tensor products, in particular, those exceeding the scope of this book. For the projective tensor product we discuss here, the spaces LI ( · ) are especially good. Suppose (XI , J-LI ) and (X2 , J-l2 ) are two measure spaces, and let (XI x X2 , J-LI x J-L 2 ) be their product. We assume that it is known that every element of the Banach space LI (Xk , J-l k ) ; k == 1, 2 can be approximated by linear combinations of the characteristic functions of measurable subsets in Xk, and every element in LI (XI x X2 , f.-LI x J-l2 ) is a linear combination of the characteristic functions of sets of the form M x N, where M (respectively, N) is a measurable set in XI (respectively, in X2 ) . Theorem 5 (Grothendieck) .
Up to an isometric isomorphism,
LI (XI , J-LI ) @ LI (X2 , J-L 2 ) == LI (X I
x
x2 , J-LI
x
Proof. Consider the bilinear operator n : LI (XI , J-LI )
J-L 2 ) . X
LI (X2 , J-l 2 ) � LI (XI X x2 , f.-LI X J-l 2 ) taking a pair (x, y) to the function x(s)y(t) ; s E XI , t E x2 . Clearly, II R(x, y) II == ll x ll IIYII ' and hence II R II == 1. Let R : LI (XI , J-LI ) @ LI (X2 , J-l 2 ) � LI (XI X x2 , f.-LI X J-l 2 ) be an operator with the same norm, associated with R. By Proposition 1 . 10, it is sufficient to estab lish that R isometrically maps a dense subspace in LI (XI , J-LI ) @ LI (X2 , J-L2 ) onto a dense subspace in LI (XI X x2 , f.-LI X J-l 2 ) · Put L : == span{xM Q9 XN }, where M and N are various measurable sub sets in XI and X2 respectively, and X is the characteristic function. From the definition of the prenorm II · li P combined with the above-said about approx imations, it follows that every elementary tensor in LI (XI , J-LI ) @ LI (X2 , J-l 2 ) is approximated by elements from L. Therefore, L is dense in LI (XI , J-LI ) @p L I (X2 , J-L2 ) , hence in the entire LI (XI , J-LI ) @ LI (X2 , J-L2 ) · A. Grothendieck ( born in I 9 2 8) , outstanding French mathematician, who made important 8
discoveries in functional analysis, algebraic geometry, and other areas of mathematics.
2.
182
Banach Spaces and Their Advantages
Further, each u E L is evidently representable as a finite linear com bination of L: z ,m A. z , m XMz Q9 X Nm , where the measurable sets Mz and, re spectively, Nm are disjoint. But then, by the definition of the projective measure, !l ull < L: z m I A.z , m iJL I (Mz ) JL2 (Nm ) , and at the same time II R ( u ) ll == II L: z , m A l, m X Mz (s) X Nm (t) l ! == L: z , m I A.z ,m iJL I (Mz ) JL2 (Nm ) · Hence II R ( u ) ll > II u II , and together with II R II == 1 , th is gives II R ( u ) II == II u II . Thus R isometrically maps L to the linear span of functions of the form XM (s) xN ( t) with measurable M and N, and as was noted before, this span • is dense in Ll (Xl X x2 , JL 1 X JL2 ) · The rest is clear. '
*
*
*
The reader remembers that behind a reasonable construction in math ematics, some functor is usually hidden ( cf. the quotation from Eilenberg MacLane in Section 0. 7) . This is definitely true for the construction of Banach tensor product.
Let S : E1 � E2 and T : F1 � F2 be bounded opera tors between Banach spaces. Then there exists a unique bounded operator S @ T : E1 @ F1 � E2 @ F2 such that (S @ T) (x Q9 y) == S(x) Q9 T(y) for all x E E1 , y E F1 . Moreover, II S @ T II == II S II II T II . Theorem 6.
E1 F1 � E2 @ F2 : (x, y) r--+ S (x) Q9 T(y) . Obviously, this is a bilinear operator with norm II S II II T II · Denote by S @ T the corresponding • associated operator. The rest is clear. Proof. Put R :
x
The constructed operator S @ T is called the projective or Banach tensor product of operators S and T. In the special case of functionals, in other words, when E2 == F2 == CC, the operator S @ T maps E1 @ F1 to CC @ CC, and hence is itself a functional, up to the identification of the latter space with CC (by the isometric isomorphism A.1 Q9 A. 2 r--+ A.1 A. 2 ) ) . It is easy to see that this functional extends by continuity the functional S Q9 T defined on the algebraic tensor product E1 Q9 F1 (see Definition 2) . Using the series representation of elements of Banach tensor products as described in Proposition 8, one can show that the tensor product of two surjective operators is a surjective operator (try to do this) . However, the tensor product of two injective operators is not always injective, even if these operators are isometric. But, it is not easy to give a corresponding example ( cf. [48, I.5.8] ) . Now we can introduce a new class of functors acting on the category Ban. It is the second most important class after the class of operator functors (see Section 5) , and it is closely related with the latter.
183
7. Algebraic and Banach tensor products
Let us take E E Ban. To every F E Ban we assign the Banach space E @ F and to every bounded operator T : F1 � F2 the operator lE @ T : E @ F1 � E @ F2 . Obviously, we obtain a covariant functor from Ban to Ban denoted by E@? and called the functor of Banach tensor product (by E from the left) . Similarly, if F is chosen, a standard covariant functor ? @ F arises (the functor of "Banach tensor product by F from the right" ) . The following material, till the end of this section, is for advanced readers. First, we emphasize again that the Grothendieck theorem reflects special geometric properties of the £ 1 ( ) spaces . In other classes of function spaces the projective tensor product does not have such a transparent description. In particular, the projective tensor product of two Lp ( · ) ; p > 1 spaces is not a space of this class. (The corresponding analogue of the Grothendieck theorem requires other types of tensor products; cf. Exercise 8.4 below. ) We will not discuss this anymore, and invite the reader to take for granted a few words about what the projective tensor product does with the spaces C [a , b] . Let D be the notation for the square [a, b] x [a, b] in the coordinate plane. Consider the contraction operator V : C [a , b] ® C [a , b] ---+ C(D) associated with the bilinear operator V : C [a , b] x C [a , b] ---+ C(D) : (x , y) r--+ x ( s ) y ( t ) . It can be shown (we will not do this, and do not ask you to do this; see, e.g. , [47, Theorem II. 5 .9] ) that this operator is injective, and, thus, allows us to identify the tensor product in question with the image of V . At the same time, although the image contains all smooth functions, and hence is dense in C(D) , this operator is not surjective. The space C [a , b] ® C [a , b] is called the Varopoulos space. (Its isometrically isomorphic copy, the function space Im (V) with norm I I V (u) l l : = ! l u ll , often has the same name.) This space plays a significant role in studying some important (and difficult) questions of the theory of Fourier series and harmonic analysis; see, e.g. , [49] (where these spaces are called tensor algebras) . Now we give one of many examples showing how Banach tensor products work in the theory of operators. Here it is very useful that an important class of operators, the so-called nuclear operators, can be characterized in terms of tensor product. -
Definition 5 . An operator T : E ---+ F is called nuclear if it is representable as the sum of an absolutely convergent in B(E, F) series 2::: � 1 Sk of one-dimensional operators. The infimum inf 2::: � 1 I I Sk I I over all such representations of the operator T i s called the nuclear norm of the (nuclear) operator T, and is denoted by I I T I I N ·
The set of nuclear operators from E t o F is denoted by N (E, F) . I f E = F, we write N (E) instead of N (E, E). The following proposition is easily verified.
r--+
I I T I IN Proposition 9. The set N (E, F) is a subspace in B(E, F)}, and the function T on N ( E, F) is a norm on this space; moreover, this norm �s not less than the operator
norm.
•
Note the resemblance of the definition of nuclear norm and the expression for the norm of elements of Banach tensor products in Proposition 8. We now show that this is not a coincidence. Consider the mapping Q : E* x F ---+ B(E, F) : (/, y) r--+ f 0 y. Obviously, this is a bilinear operator with norm 1 . The bounded operator Gr : E* ® F ---+ B(E, F) associated with Q is called the Grothendieck operator (for the pair ( E, F) ) . We see that this operator is uniquely defined by the equality Gr(f @ y) = f 0 y, and it has norm 1 .
184
2. Banach Spaces and Their Advantages
Theorem 7. The image of the Grothendieck operator is precisely N(E, F) , and the core striction Gr0 of this operator to (N(E, F) , II · liN) is a coisometric operator. Proof. Suppose u E E* ® F can be represented as the sum of an absolutely convergent series L:� 1 fk ® Yk ; fk E E* , Yk E F (see Proposition 8) . Then T : = Gr(u) can be represented as the sum of the series L:� 1 fk 0 Yk absolutely convergent in the operator norm. Therefore, T E N(E, F) . Further, every operator T of nuclear norm < 1 can be represented as the sum of a series L:� 1 sk , where the sk are one-dimensional operators, and L:� 1 I I Sk I I < 1 . By Proposition 1 . 5. 5, every Sk has the form /k 0 Yk for some fk E E* and Yk E F, so that I I Sk l l = l l fk ii i i Yk ll · Hence the series L:� 1 /k ® yk converges in E ® F • to some element u, l l u l l p < 1 , and Gr(u) = T. The rest is clear.
From this theorem and Proposition 1 . 5 .4(ii) we immediately get Proposition 10. We have the following commutative diagram:
E* ® F
prl
1 E* ® F/ Ker(Gr) -------+ N(E, F)
where I is an isometric isomorphism. Taking Proposition 1 . 9 into account, we obtain Corollary 2. The space (N(E, F) , I I · l iN) is a Banach space.
For the vast majority of the known examples of Banach spaces E the operator Gr : E* ® F ---+ B(E, F) is injective for all F E Ban, so that its corestriction to N(E, F) is an isometric isomorphism between E* ® F and (N(E, F) , II · l i N ) . Thus, in this situation nuclear operators can be characterized as elements of the tensor product E* ® F. For the case where both spaces are Hilbert , this fact will be proved later (see Theorem 3.4.5) . But there are "bad" pairs of Banach spaces for which Ker(Gr) :I 0 . Some details will be reported later; see Theorem 3. 3.3. Let us concentrate on the case E = F. Among all bilinear functionals on E* x E the following one attracts attention: it takes a pair (/, x) to the number f (x) and is called the natural duality between E* and E or the painng of E* and E. The bounded functional associated with the natural duality is called the tensor trace functional or just the tensor trace and is denoted by ttr : E* ® E ---+ C. Clearly, it is uniquely defined by the equality ttr(f ® x) = f (x ) ; besides, I I ttr II = 1 . If the operator Gr : E* ®E ---+ B(E) is injective, then, assigning to a nuclear operator T the number ttr( u) , where u E E* ® E is such that Gr( u) = T, we obtain a linear functional on (N(E) , I I · liN) of norm 1 . This functional (which is well defined if Ker(Gr) = 0) is also called the trace or, if we want to be precise, the operator trace, and it is denoted by the symbol tr. The reader can guess that all this must be related to the classical notion of the trace of a matrix, the sum of its diagonal elements. Indeed, this is so: Exercise 6* . Let E be a finite-dimensional space. Then
(i) the Grothendieck operator is a linear isomorphism between E* ® E and £(E) ; (ii) for every T E £(E) its operator trace (well defined by (i) ) coincides with the trace of the matrix of this operator in each linear basis.
185
7. Algebraic and Banach tensor products
Hint. Take a basis e 1 , . . . , e n in E and consider the basis e r , . . . , e � in E* such that is 1 if k = l, and 0 if k ":l l. Look at the trace of the element ® e and the trace of a matrix of an operator T E £ (E) in the indicated basis.
Az k ei k
ek:(e z )
For which Banach spaces E is the trace of nuclear operators acting on E well defined? In other words, when, for a given T E N (E) , is the number ttr( u ) independent of the choice of u E E* ® E such that Gr( u ) = T? Running ahead, we note that this happens precisely when E has an outstandingly important geometric property, the so-called Grothendieck approximation property (see Definition 3.3.2 and Theorem 3.3.3 below) . We emphasize one more time ( cf. what we have said earlier about the injectivity of Gr) that for the vast majority of operators, the operator trace is indeed well defined. For Hilbert spaces we shall prove this in the next section. Finishing the discussion of the Banach tensor product, we again look at it through the "categorical glasses" . We suggest that you do the following two simple exercises. First of them reveals the intrinsic alliance between this construction and the construction of completion ( cf. Exercise 6. 1 ) . Exercise 7. Show that each of the existence theorems of this section (Theorems 2 and 4) is equivalent to representability of some functor. Hint. Consider the functor :F from Lin (respectively, Ban) to Set assigning to each G the set of all bilinear (respectively, bilinear contraction) operators from E x F to G.
The second exercise sheds light on the result of Exercise 5 concerning the coincidence of the Banach spaces mentioned there: actually, not only individual spaces coincide, but the "entire" functors. Exercise 8. For E, F, G E Ban,
(i) the composition of the functors B(?, G) (? ® F) is naturally equivalent to the functor B(?, B(F, G) ) ; (ii) the functor B ( E ® F, ? ) is naturally equivalent to the composition B(E, ? ) B(F, ?) . o
o
Remark. The indicated fact is a manifestation of a deep connection between the functors ? ® F and B(F, ?) , which is usually expressed by the words "the first is adjoint to the second" . About the general notion of adjoint functors see, e.g. , [24] . *
*
*
Finally, a few words to the readers who are interested in quantum spaces discussed in Section 1 . 7. In quantum functional analysis tensor products play even more impor tant role than in classical functional analysis. Since there are two substantial "quantum" versions of the notion of a bounded bilinear operator, there exist two substantial "quan tum" analogues of the notion of Banach (i.e. , projective) tensor product. These are the "operator projective tensor product" of Effros-Ruan and Blecher-Paulsen and "Haagerup tensor product" (see [36] ) . The first is defined using the universal property of completely bounded operators, and the second is defined for multiplicatively bounded bilinear oper ators. Each tensor product is a pair consisting of a Banach (i.e. , complete in all floors) quantum space and a bilinear operator of the corresponding type. The operator tensor product resembles the classical Banach tensor product; in partic ular (and this is very important) it admits a natural analogue of the adjoint associativity law. On the contrary, the Haagerup tensor product "®h" is very unusual; it is sufficient to say that it depends on the order of the tensor factors. For instance, l2 ®h l2 = N (l 2 ), and at the same time l2 ®h l2 = K( l 2 ). (Here N is as always, the symbol for the space of nuclear operators, and K the symbol for the space of compact operators, which will be defined
2.
186
Banach Spaces and Their Advantages
in the next chapter) . But such inconveniences are overweighted by the following advan tage of this tensor product: the "Haagerup tensor product" of two surjective operators is surjective, and the product of two injective operators is injective. No other known tensor product of "classical" Banach spaces possesses this feature of Haagerup tensor products (see details, e.g. , in [36] ) .
8 . Hilbert tensor product
Now we leave general Banach spaces and concentrate on Hilbert spaces. First, note that in this context the projective tensor product, remaining an important working instrument, does not always make us happy: the projective tensor product of Hilbert spaces usually is not a Hilbert space. Exercise 1 . Let H and K be Hilbert spaces, and e 1 and e 2 (respectively, e� and e�) orthogonal vectors of norm 1 in H (respectively, in K) . Then the norm of the vector e 1 Q9 e� + e 2 Q9 e� in H @ K is 2 and, as a corollary, the norm in H @ K is not a Hilbert norm. Hint. The norm of the bilinear functional on H x K taking (x , y) to ( x, e 1) (y, e� ) + (x, e 2 ) (y, e� ) is 1 . Nevertheless, there exists a useful version of the construction of tensor product that preserves the "Hilbert nature" of given spaces. This time, to achieve the goal quickly, we shall immediately give an explicit construction. It is based on the fact that the algebraic tensor product of two near-Hilbert spaces has a natural pre-inner product.
Let H and K be near-Hilbert spaces. Then there exists a unique pre-inner product in the linear space H Q9 K such that (xi Q9 YI , x 2 Q9 Y2 ) == (xi , x 2 ) ( YI , Y2 ) . As a corollary, for the correspon ding prenorm we have ll x Q9 Y ll == ll x ll IIYII . Moreover, if H and K are near-Hilbert spaces, then the same is true for
Proposition 1.
H Q9 K.
Proof. If ( · , · ) is a pre-inner product on HQ9K with the indicated properties,
then for have
u, v E H Q9 K; u == ��= l x� Q9 y� , v == ��= l x% Q9 y� we evidently
(1)
n
(u, v ) = L (x �, x f' ) ( yk , yf' ) . = k ,l l
This means that there can exist at most one such pre-inner product. Let us show that one indeed exists. First take a pair x E H, y E K and denote by gx , y H Q9 K � CC the linear functional associated with the bilinear functional ( x1 E H, YI E K) �----+ :
(xi , x) ( YI , y ) .
8. Hilbert tensor product
187
Now take u E H Q9K and consider the mapping F : H x K � CC; (x , y) �----+ gx , y (u). Then, taking an arbitrary representation of u in the form u == 2:�= 1 x k Q9 Yk , we obtain that F(x, y) == 2: �= 1 (x , x k) ( y, yk) · This clearly implies that F is a bilinear functional. Denote by fu H Q9 K � CC the corresponding associated functional. Finally, for u, v E H Q9 K we put ( u, v ) fu (v) . Supp ose u == 2:: �= 1 x� Q9 y� and v == 2:: �= 1 x % Q9 Y% · Then elementary arguments using the linearity of functionals fu and gx"k ' yk" provide equality (1) . Its special case is the equality given in the statement. Further, it easily follows from (1) that ( u, v ) == ( v , u) , ( u, u ) > 0. Finally, from the obvious equality ( u, v ) == fv(u) it follows that ( u, v ) is linear in the first argument. Thus, it is indeed a pre-inner product. It remains to consider the case of inner products, i.e. , near-Hilbert H and K. Suppose u E H Q9 K; u == 2:: �= 1 X k Q9 Yk is non-zero. Take an orthonormal basis e 1 , . . . , e z in span{x1 , . . . , xn } · Each X k has the form of L:;n 1 A/clez ; A kz E CC. Hence, obviously, u == L:;n 1 e z Q9 zz , where zz == 2::�= 1 A kl Yk , and since u =/=- 0, not all zz vanish. We obtain that ---
:
: ==
Thus,
m
m
i ,j = 1
i= 1
(u, u ) =/=- 0, and H Q9 K is a near-Hilbert space.
•
Taking the continuity of the inner product (Proposition 1.2.4) into ac count, this obviously implies the following result.
If a sequence Xn tends to x in H, and Yn to y in K, then • Xn Q9 Yn tends to x Q9 y in (H Q9 K, ( ·, · ) ) . Proposition 2.
From now till the end of the section H and K are Hilbert spaces. Denote by H Q9 K the Hilbert space which is a completion of the near Hilbert space (H Q9 K, ( · , · ) ) (see Proposition 6.3) . . . . . Definition 1. The pair (H Q9 K, 19) , where 19 : H x K � H Q9 K is a bilinear operator acting by the rule (x, y) �----+ x@y, is called the Hilbert tensor product of Hilbert spaces H and K. Apparently, the "advanced" reader must feel a discomfort because of the discrepancy between the definitions of Banach and Hilbert tensor products: the first was given in terms of a universal property, while the second by an explicit construction. In fact , in an exposition of this circle of problems ( exceeding the very preliminary information we present in this book ) the Hilbert tensor product can ( and should ) be defined in the spirit of Definitions 1 and 3. If you are curious, here is the scheme ( cf. [50] ) . Instead of the
2.
188
Banach Spaces and Their Advantages
class of all bounded bilinear operators, in Definition 3, we should consider the class of Schmidt bilinear operators. This is the name for bilinear operators R : H x K ---+ L, where H, K, L are Hilbert spaces, with the following property: for each total orthonormal system { e JL ; JL E A 1 } in H and { ev; v E A2 } in K and each z E L we have L: { I (R(e JL , ev), z) j 2 : JL E A 1 , v E A2 } < oo . (A remarkable fact is that the indicated number depends only on z and not on the choice of orthonormal systems.) The supremum of the indicated numbers over all z E BL is called the Schmidt norm of our bilinear operator. Now we say that the pair (8, 0) , where e is a Hilbert space and e H X K ---+ e a bilinear Schmidt operator, is the Hilbert tensor product of the Hilbert spaces H and K if for each Hilbert space L and for each bilinear Schmidt operator R : H x K ---+ L with Schmidt norm < 1 , there exists a unique linear operator R with operator norm < 1 (i.e. , a contraction operator) such that the diagram :
HxK
el � e
R
L
is commutative. This definition immediately implies the uniqueness theorem in the spirit of Theorems 1 and 3 (formulate it! ) . After that , as a proof of the existence theorem we must explicitly construct the pair ( H ® K, rO) and establish that it indeed has the universal property j ust described. All these arguments are sufficiently important and instructive, but exceed the scope of this book.
Now we suggest two exercises clarifying the nature of the Hilbert tensor product. Suppose (only for simplicity) that H and K are separable, e�; n E N is an orthonormal Schauder basis in the first space, and en ; n E N in the second space.
e� Q9 en ; m, n E N, arranged as a sequence in an arbitrary order, is an orthonormal basis in H Q9 K. Next, for each n == 1 , 2, . . . we put Hn :== { x Q9 en ; x E H }. Clearly, this is a closed subspace in H Q9 K, and there is an isometric (and unitary) isomorphism between H and Hn taking x to x Q9 en . (We can say that Hn ; n E N are isometrically isomorphic copies of the space H.) By analogy, in the same space HQ9K there are closed subspaces Kn that are isometrically isomorphic copies of the space K . Exercise 2. The system
Exercise 3* . Show that there exists a unique unitary isomorphism .
.
.
H Q9 K � (B{Hn ; n E N} (where EB denotes the Hilbert sum defined in Section 1) taking x Q9 en to the sequence ( z1 , z2 , . . . ) ; Zm E Hm such that Zn == x Q9 en and Zm == 0 for m =/=- n. Construct a similar unitary isomorphism . between H Q9 K and (B{Kn ; n E N}. . Hint. There exists a unique operator R : H Q9K � (B{ Hn ; n E N} taking x Q9 y to the sequence z1 , z2 , . . such that Zn :== ( y, en ) x Q9 en . It defines an isometric isomorphism between the dense subspace in H Q9 K consisting of U:
.
.
8. Hilbert tensor product
189
sums of elementary tensors of the form x Q9 e n ; x E . subspace in (B{Hn ; n E N}.
H, n E N and a dense
Now we give an instructive example. Exercise 4 (cf. Theorem 7.5) . Let
spaces, and (XI isomorphism,
X
x2 , f.-LI J-l 2 ) X
(XI, J-LI) and ( X2 , J-L2 ) be two measure
their product. Then, up to an isometric
Hint. Put L :== span{xM Q9 XN }, where M and N are arbitrary measur
able subsets in XI and X2 respectively, and X is the characteristic function; this is a dense subspace in L 2 (XI, J-LI) Q9 L 2 (X2 , f.-L 2 ) · The operator from L2 (XI , f.-LI) Q9 L 2 (X2 , J-L2 ) to L 2 (X I x X2 , f.-LI x J-L2 ) , well defined by the rule (x, y) �----+ x ( s ) y ( t ) , establishes an isometric isomorphism of L onto a dense subspace in L 2 (XI X x2 , f.-LI X J-l 2 ) · One of the main advantages of the Hilbert tensor product that makes it to resemble the Banach tensor product, is that this construction is "func torial" : it is defined not only for the spaces, but for the operators as well. The following theorem is fundamental here.
Let S : HI � H2 and T : KI � K2 be bounded operators between Hilbert spaces. Then there exists a unique bounded operator S Q9 T : HI Q9 KI � H2 Q9 K2 such that (S Q9 T)(x Q9 y ) == S(x) Q9 T(y) for all x E HI, Y E KI. Moreover, l i S Q9 T i l == II S II II T II . Theorem 1 (cf. Theorem 7.6) . .
.
.
Proof. Since the algebraic tensor product
.
HI Q9 KI , i.e. , the linear span of
elementary tensors in HI Q9 KI, is dense in the latter space, there may exist at most one bounded operator satisfying the desired condition. Furthermore, by the identity ll x @ y ll == ll x ll II Y II for the norm in the Hilbert tensor product, for this hypothetical operator S Q9 T we would have .
II S Q9 T II > sup{ II (S Q9 T) (x Q9 y) ll ; x E BH1 , y E BK1 } == sup{ II S(x) II II T(y) ll ; x E BH1 , y E BK1 } == II S II II T II . Hence, our goal is to show that this operator indeed exists, and l i S Q9 T i l < II S II II T II · In fact, everything can be reduced to the following special case. Lemma. There exists a bounded operator SQ91: HI Q9 KI � H2 Q9 KI such that (SQ91) (x Q9 y) == S(x) y for all x E HI , y E KI . Moreover, 11 8 @ 1 11 < II S II . .
.
.
0
.
.
2.
190
Banach Spaces and Their Advantages
Proof. Consider the bilinear operator n
.
: Hl Kl � H2 Q9 Kl : (x, y) X
1---+
Suppose R : H1 Q9 K1 � H2 Q9 K1 is the associated operator. Take u E H1 x K1 , and a representation u == L:�= l X k Q9 Yk · Without loss of the generality, we can assume that the system Y I , . . . , Yn E K1 is or thonormal. (Otherwise, we choose an orthonormal basis in span{y1 , . . . , Yn }, decompose vectors Yk with respect to this basis, and use the bilinearity of the symbol Q9 ) . Then the system X I Q9 Y I, . . . , X n Q9 Yn is orthogonal in H1 Q9 K1 , and S(x 1 ) Q9 Y I, . . . , S(xn ) Q9 Yn is orthogonal in H2 Q9 K1 . Therefore, using the Pythagorean equality (Proposition 1 .2.7) we have
S(x) Q9 y.
n n n 2 L S(x k ) Yk = L II S(x k ) Yk ll 2 = L II S(x k ) ll 2 =l =l k= l k k n n < II S II 2 L ll x k ll 2 = II S II 2 L ll x k Yk ll 2 = II S II 2 II u ll 2 · k=l k= l Thus, R is a bounded operator from the near-Hilbert space H1 Q9 K to the Hilbert space H2 Q9 K1 , and II R II < II S II . Extending it by continuity to the . . whole H1 Q9 K, we obtain the operator SQ9 1 with required properties. • 69
69
69
Now we complete the proof of. theorem . 1 . Similarly . to the lemma, we . obtain a bounded operator l @ T : H2 Q9 K1 � H2 .Q9 K2 such that Q9 y). == x Q9. T(y) for. all x E H2. , y E K1 , and ll l @ T II < II T II . Put (1Q9T)(x . S Q9 T : == ( 1 Q9T)(SQ9 1 ) : H1 Q9 KI � H2 Q9 K2 . By the multiplicative inequality for the operator norm, this operator is bounded, and II SQ9T II < II S II II T II . • The rest is clear. Here is a finite-dimensional illustration. Let e�, . . . , e� and e 1 , . . . , en be orthonormal bases in H and K, respectively, and S : H � H and T : K � K operators given in these bases by the matrices a == ( akz ) and b == (bij ) . For brevity, denote ers : == e� Q9 e8 ; == 1, . . . , m, s == 1 , . . . , n. r
S@T acting on H@K (or, what is the same, in H Q9 K ) is given in the basis e 11 , e 12 , . . . , e 1n , e 21 , . . . , e 2n , . . . , em l , . . . , emn by the block matrix ( akz b), and in the basis e 11 , e 21 , . . . , em l , e 12 , . . . , em2 , . . . , e 1 n , . . . , emn by the block matrix (bija). Exercise 5 ° . The operator
In conclusion, we suggest that advanced readers define the functor of Hilbert tensor product H®? (similarly to the functor of Banach tensor product) , this time acting on the category Hil.
Chapter 3
From Comp act S p aces to Fredholm O p er ators
1 . Compact spaces and relevant functional spaces
A significant part of this and the next section is, most probably, known to
you. However, our goal is to present this information in the form we will need later. In the history of mathematics there have been many results which were originally viewed by their authors as being of auxiliary nature, and thus were modestly called lemmas. But later developments and results that stood behind all this had been leading ( perhaps, people of another generation ) to the discoveries of fundamental mathematical notions. One of most striking examples of such a development is the story of Borel ' s lemma, which led to the notion of compactness. Let (0, ) be a topological space, and � a subset in 0. A subfamily C is called an open covering of this subset if � C U{ U : U E a} . If ao and a are two open coverings of the same �' and ao C a, then ao is called a subcovering of a; we also say that a contains the subcovering ao . Very often when coverings are discussed, the role of � belongs to the whole space n. T
a
T
Definition 1. A topological space
0 is said to be compact if every open
covering of 0 contains a finite ( i.e. , consisting of a finite number of sets ) subcovering. A subset � of an ( arbitrary ) topological space 0 is said to be a compact subset if it is compact as a topological subspace in 0. 191
3. From Compact Spaces to Fredholm Operators
192
The term "compact" seems to be very appropriate and reflects our ev eryday impression of what is compact. The reader will see this gradually, as more material is accumulated. We can also speak about compact subsets without involving the inherited topology. The following proposition is verified immediately. Proposition 1 . A subset � of a topological space 0 is compact � every
open covering of � contains a finite subcovering.
II
It is not difficult to give an equivalent definition of compactness in terms of closed sets ( since they are defined as complements of the open ones ) and in terms of adherent points. With this goal in mind, we say that a family a of subsets in 0 is centered if every finite subfamily of a has a non-empty intersection. Proposition 2.
alent:
The following properties of a topological space 0 are equiv
( i ) n is compact; ( ii ) every centered family of closed subsets in 0 has a non-empty inter
section; ( iii ) every centered family of (arbitrary) subsets in 0 has a common adherent point. Proof. The implication ( i ) � ( ii ) immediately follows from the standard
set-theoretic identities that connect complements, unions, and intersections (write these identities ) . Then, clearly, ( iii ) implies ( ii ) . The converse impli cation follows from the fact that the family of closures of a centered family is again centered, and a common point of these closures is a common adherent point of the initial sets. II
Evidently, as a special case of ( ii ) , we obtain that the sequence V1 =:) V2 =:) V3 =:) • • • of embedded closed non-empty subsets of a compact topological space always has at least one common point. This property is essentially stronger than the "closed embedded subsets principle" in the context of complete metric spaces. The latter, as we recall, requires that the diameters of these sets tend to zero. The advanced reader who has done Exercise 0.2.7, knows that in a topological space the family of open sets can be described in terms of convergent nets. Naturally, he will ask: how can we determine, using the same terms, whether our space is compact? The answer is given in terms of subnets, a reasonable although not so straightforward generalization of the notion of subsequence .
Definition 2 . Let xv ; v E A and yJ.L ; p, E A ' be two nets of elements of a set X. The second net is called a subnet of the first if there exists a mapping r : A ' ---+ A such that for all p, E A' we have x, ( J.L ) = yJ.L , and for each v E A there is p, E A ' such that v -< r ( p, ) .
1.
193
Compact spaces and relevant functional spaces
Theorem 1 ( [13, Chapter 5 , Theorem 2] ) . A topological space n is compact ¢:::::::> every
net of elements of n has a convergent (in 0) subnet.
Clearly, if two topological spaces are homeomorphic, then they are com pact or not compact simultaneously; as one says, compactness is a topo logical property (i.e. , the property that is invariant under isomorphisms in Top) . This is the main difference of this property from the property of completeness of metric spaces, which is not invariant with respect to home omorphisms ( cf. the beginning of Section 2. 1 ) . The first example of a compact space is, of course, a closed interval of the number line (and this is just the content of Borel ' s lemma) . From a course of calculus you know that an arbitrary set in IRn or ccn with Euclidean metric is compact it is closed and bounded. Proposition 2. 1.4 evidently implies that the same is true for sets in arbitrary finite-dimensional normed spaces (we will return to this in the next section) . Further, a discrete topological space is compact it is finite (look at its covering by singletons) . In compensation, an antidiscrete space is always compact. You can easily verify that the complex plane with the Zariski topology is also compact, and this, of course, distinguishes the Zariski topology from the standard topology in e used in analysis. Actually, e n with the Zariski topology is compact for all n, and moreover a stronger condition is true: every sequence of its embedded closed subsets stabilizes. (This shows how different e n looks to an algebraist and an analyst. ) But we will not digress to this; see, e.g. , [52, 2. 1]
Why do mathematicians like compact spaces? First, the following fact lies on the surface.
If f! is a compact space, then every continuous function rp : f! --+ CC is bounded, and the supremum of its absolute value is attained.
Theorem 2 (recall the Weierstrass theorem!) .
Proof. For every A E cc we put U>._ : == {t E n : l rp(t) - A I < 1 } . Clearly, U>.. ; A E CC is an open covering of our compact space; let U>.. 1 , , U>..n be a
finite subcovering. Then for every t E 0 we have l rp(t) l < ��= l I A k l + 1 . The proof of the statement about the supremum repeats word-for-word the proof of the classical Weierstrass theorem. II •
•
•
Thus, if f! is compact, then the function space C(f!) coincides with Cb (f!) (see Example 1 . 1 .6') , and therefore it is a Banach space with respect to the uniform norm. We will see more than once how important these spaces are. A function rp : f! --+ CC on an (arbitrary, for the present) topological space is called vanishing at infinity if for every c > 0 there is a compact subset � in 0 such that l rp (t) l < c as t tf_ �- From Theorem 2 follows
3. From Compact Spaces to Fredholm Operators
194
Every continuous function vanishing at infinity on a topo logical space is bounded, and the supremum of its absolute value is at • tained.
Proposition 3 .
The set of continuous functions on a topological space 0 that vanish at infinity is denoted by Co ( O ) . It is easy to verify that this is a closed subspace in Cb ( O ) , and therefore a Banach space with respect to the uniform norm. If 0 is compact , then certainly Co ( O ) == C(O) . We have already come across typical spaces of that kind among the examples of normed spaces of Section 1 . 1 : these are co (X) (i.e. , Co (X) for the discrete X, in particular, co == Co (N) ) , and also Co (IR) . Let us continue the discussion of nice properties of compact spaces. Proposition
4. A closed subset of a compact space is a compact space. •
Proof. This follows immediately from Proposition 2(ii) . Proposition 5.
A compact subset of a Hausdorff space is closed.
Proof. Suppose we speak about a subset � in 0. Take a point
in 0 \ � Our aim is to show that X is an interior point for 0 \ �- For each y E � denote by Ui and Vy disjoint neighborhoods of x and y . Then the family {Vy y E � }, being a covering of � ' contains a finite subcovering, say, {Vyl ' . . . ' Vyn }. Hence, the set n�= l Uik is a neighborhood of X that does • not intersect �X
:
Let rp : 0 1 � 0 2 be a continuous mapping of a compact space to an arbitrary topological space. Then Im( rp) is a compact subset in 0 2 . (In short: a continuous image of a compact space is compact.) Proposition 6.
Proof. Let a be an open covering of Im(rp) in 0 2 . Then {rp - 1 (U) : U
a} is an open covering of 0 1 . Take its finite subcovering, say, {V1 , . . . , Vn }, and • consider the family { rp(VI ) , . . . , rp(Vn ) }. The rest is clear. E
Combining these facts we obtain a very substantial result. Theorem 3 (Alexandroff1 ) . Let 01 be a compact space, 0 2 a Hausdorff topological space, and rp : 0 1 � 0 2 a continuous bijective mapping. Then this is a homeomorphism {in other words, rp - 1 is automatically continuous). Proof. Since the homeomorphisms are precisely the continuous open bijec tions (see Section 0.2) , our goal is to show that rp is open, or (what is the 1 P. S. Alexandroff ( 1 896- 1 98 2 ) was a prominent Russian mathematician, one of the creators
of topology. Together with his friend P. S. Uryson (the second P. S. , as they used to joke) he introduced the class of topological spaces that they called "bicompact spaces" . Now we call them "compact spaces" .
1.
195
Compact spaces and relevant functional spaces
same, taking the bijectivity of rp into account) that the image of every closed set � in 0 1 is closed in 0 2 . Successively applying Propositions 4, 6, and 5 , we see first that � is compact, second that rp(�) has the same property, II and finally that the latter set is closed. Remark. Note the intrinsic similarity of the Alexandroff theorem and the Banach theorem on the inverse operator. Both describe the situations (that seem to have nothing in common) where a continuous bijective mapping is automatically a homeomorphism. Certainly, neither the compactness of the first space, nor the Hausdorff property of the second can be omitted. (Construct the corresponding ex amples.) The Alexandroff theorem can be generalized in two directions. Proposition 7 ( cf. Corollaries 2.4.2 and 2.4. 1 ) . Let rp 0 1 � 0 2 be a :
continuous mapping of a compact space to a Hausdorff space. Then ( i) if rp is injective, then it is topologically injective; (ii) if rp is surjective, then it is topologically surjective.
Proof. The first statement follows from Theorem 3 and the Hausdorff prop
erty of 0 2 . To verify the second, denote by the topology on 0 2 and by the quotient topology modulo rp (see Section 0.2 ) . Then, by Proposition 6, (0 2 , ) is compact, and thus the identity mapping 1 : (0 2 , ) � ( 0 2 , ) satisfies the hypothesis of Theorem 3. The rest is clear. II T
T
1
T
1
T
*
*
1
T
*
Now our categorical Zoo is supplemented with a new animal of a valuable breed. In Top we consider the full subcategory whose objects are compact Hausdorff spaces and denote it by CHTop. (Topologists are interested also in the category CTop consisting of all compact spaces, but it is considerably less important for functional analysis.) Certainly, the isomorphisms in CHTop are homeomorphisms, but due to Theorem 3, they can be characterized in this category simply as bijective morphisms. (Thus, the Alexandroff theorem has the same meaning for the category CHTop as the Banach theorem does for the category Ban ) . As for other types of morphisms, one can easily verify ( cf. Propositions 0.5 .5 and 0.5 .6) that the monomorphisms in CHTop are precisely injective morphisms, and all surjective morphisms are epimorphisms. Actually, the epimorphisms are completely characterized as surjective morphisms, and this distinguishes the category CHTop from the category HTop and makes CHTop similar to Top (cf. Section 0.5 ) . But we leave such things for the advanced reader.
3. From Compact Spaces to Fredholm Operators
196
The structure of "good" morphisms distinguished in Section 0 . 5 is ideally simple in the categories Set and Lin. Compared to this, the category CHTop gives a picture of the next level of complexity. The following exercise summarizes our earlier observations. Exercise 1 . In the category CHTop
(i) every bimorphism is an isomorphism; (ii) a morphism
0. A subset N in M is called a Hausdorff c -net (or just an c -net) for Mo if for each element x E Mo there is y E N such that d(x , y ) < c (or, in other words, if Mo C U{U(y, c) ; y E N} ) . Definition 2 . A subset Mo of a metric space M is called totally bounded if for every c > 0 in M there is a finite (i.e. , consisting of finitely many elements) c-net for Mo . In particular, for Mo == M we obtain the definition of a totally bounded metric space. The following proposition eliminates the possibility of different interpretations for the words "totally bounded subset" . Proposition 1. A subset Mo of a metric space M is totally bounded � Mo is totally bounded as a metric subspace in M. Proof. The second property means, of course, that for every c
>
0 the set Mo has a finite c-net consisting of the points of Mo . Hence, � is obvious. To establish ===> , take c > 0 and a finite c/2-net YI , . . . , Yn E M for Mo . Take those k; 1 < k < n for which there exists at least one point Zk E Mo
3. From Compact Spaces to Fredholm Operators
202
such that d(zk , Yk ) < c/2; we choose such a point Zk · Further, for every x E Mo there exists at least one k such that d(x, Yk ) < c/2; so for such k the point Zk is determined, and from the triangle inequality it follows that d(x, zk ) < c. Therefore, the points Zk form an c-net for Mo in the very same • Mo .
CC! and consider an arbitrary bounded subset M. Then for some C > 0 and for all x == (xi, . . . , Xn ) E M we have l x k l < C for all k == 1 , . . . , n. For every m E N we denote by Nm the set of all points in CC! with coordinates of the form c r;;:s ' where r and are integers between -m Example 1 . Take
s
to m; certainly, this set is finite. One can easily prove that for every c > 0 the set Nm for m > V2cCn is an c:-net for Mo . Thus, every bounded subset in
CC! is totally bounded. Proposition 2 ( cf. Proposition 1 . 7) . Let rp : MI � M2 be a uniformly continuous mapping of a totally bounded metric space to an arbitrary metric space. Then Im( rp ) is a totally bounded subset in M2 . (Briefly: uniformly continuous image of a totally bounded metric space is itself totally bounded.)
0, and then take 8 > 0 in the definition of uniform continu ity. Let YI , . . . , Yn be a finite 8-net in MI . Then, obviously, rp(yi), . . . , rp (yn ) • is a finite c-net for Im( rp ) in M2 .
Proof. Take c
>
Two uniformly homeomorphic {i. e. , isomorphic in Metu ; see Section 0.4) metric spaces are either simultaneously totally bounded or not. Proposition 3. The closure of a totally bounded subset of a metric space is totally bounded. Corollary 1.
Proof. Obviously, every c/2-net in M for Mo is an c-net for the closure of
•
Mo . The rest is clear.
We can come to the notion of total boundedness in several ways:
4. The following properties of a subset Mo of a metric space M are equivalent: (i) Mo is totally bounded; (ii) every sequence of points in Mo has a fundamental subsequence; (iii) every subset in Mo consisting of points with pairwise distance ex ceeding some constant () > 0 is finite.
Proposition
x 2 , . . . be our sequence. For each m E N, con struct a subsequence xr ' x 2 ' . . . in the following way. For xi ' � we take the initial sequence. Now suppose for some m we have already constructed the sequence x ! , x2 , . . . . Since for Mo there is a finite 1/m-net in M, Mo Proof. (i)====> (ii) . Let
XI ,
X '
•
•
•
2.
Compact metric spaces and total boundedness
203
is contained in the union of a finite family of open balls of radius 1/m. Then one of these balls contains the points x � for an infinite set of indices of some subsequence in xr ' x2 ' . . . ; we denote n, i.e. ' contains allI elements I the latter by x;_n + , x;n+ , . . . . Note that all pairwise distances between the elements of the constructed subsequence are less than 2/m. Now consider the sequence x� , x� , . . . . It is a subsequence in XI , x 2 , . . . , and, obviously, for k > l we have d(x � , x � ) < 2/l; thus, it is fundamental. (ii)===> (iii) . If in Mo there is an infinite subset satisfying (iii) , then the sequence consisting of different points of this subset has no fundamental subsequences. (iii) ===> (i) . Let us assume that for some c > 0 there is no finite c-net for Mo in M. Take an arbitrary XI E Mo . Since {x i } is not an c-net for Mo , there exists x 2 E Mo with d(x i , x 2 ) > c. Since {x i , x 2 } is not an c-net for Mo , there exists X 3 E Mo with d( XI , X 3 ) , d( x 2 , X 3 ) > c. Continuing, we construct an infinite set of points X n E Mo ; n E N such that d(x m , X n ) > c II for m =/=- n, a contradiction. The following proposition justifies the term "totally bounded" .
5. A totally bounded subset Mo of a metric space M is (au tomatically) bounded. Proof. If YI , . . . , Yn E M is a 1- net for Mo, then pairwise distances between II the points of M0 do not exceed max{d(x k , xz ) ; 1 < k, l < n} + 2. Proposition
The converse statement is, certainly, not true. An infinite discrete metric space is bounded, but it is not totally bounded: every c-net in this space for c < 1, obviously, coincides with the entire space.
A totally bounded metric space is separable. It suffices to look at the union of all finite � -nets; n == 1, 2, . . . . •
Proposition 6. Proof.
Now we arrived at a characterization of compact metric spaces.
The following properties of a metric space M are equivalent: (i) M is compact; (ii) M is complete and totally bounded; (iii) every sequence of points in M has a convergent subsequence; (iv) every infinite subset in M has a limit point in M. Proof. (i) ===> ( iv) . Suppose an infinite subset N in M does not have limit points. In particular, this means that N is closed. Then every point x E N has a neighborhood Ux not containing other points of N. But then the Theorem 1.
204
3. From Compact Spaces to Fredholm Operators
family consisting of M \ N and all Ux ; x E N is an open covering of M without a finite subcovering. ( iv ) ====> ( iii ) . Let xn ; n == 1 , 2, . . . be a sequence in M; denote by N the set of elements of this sequence. If N is finite, then our sequence contains a subsequence with coinciding elements, which certainly converges. If N is infinite, then it has a limit point, and thus ( since we are in a metric space! ) , a strict limit point, say, x. But then there exist indices n 1 < n2 < · · · such that d(x, X nk ) < 21k for all k. The rest is clear. ( iii ) ====> ( ii ) . If M is not complete, then we can find a fundamental se quence of points in M that does not have a limit, and after that we see that every subsequence of this sequence does not have a limit either. If M is not totally bounded, then, by Proposition 4, there exists a sequence of points that does not even has a fundamental ( to say nothing about convergent ) subsequence. So, it remains to prove the most laborious implication. ( ii ) ====> ( i ) . Suppose, on the contrary, that there exists an open covering a of the space M that does not have a finite subcovering. Let us construct by induction a system of embedded closed subsets V 1 =:) V 2 =:) • • • in our space with the following properties: a) for every n E N the family a regarded as a covering of the set vn has no finite subcovering of v n ; b ) the diameter of the set vn does not exceed 2 /n. At the first step of induction we, using the total boundedness of M, take a finite 1-net for M, say, Yf , . . . , y� 1 • For k == 1 , . . . , m 1 denote by Vk1 the closure of the open ball U ( Yk , 1 ) . If for every k the family a, considered as an open covering of Vk1 , has a finite subcovering of this set, then, obviously, the union of these subcoverings over all k gives a finite subcovering of the entire M. Hence there exists at least one k for which this is not the case; we put V 1 : == Vk1 . Suppose we have already constructed the sets V 1 , . . . , vn - 1 with both properties a ) and b ) . Using the total boundedness of M, take a finite 1 / n net, say, y]_ , . . . , y�n . For k == 1 , . . . , mn denote by Vkn the intersection of vn - 1 with the closure of the open unit ball U (yJ: , 1 /n ) . The nsame arguments as at the first step of induction ( but with M replaced by v - 1 and the sets Vk1 by the sets vkn ) show that there exists at least none k such nthat wen cannot choose from a a finite sub covering of the set vk . We put v : == Vk ; obviously, this set also has properties a) and b ) . Now recall that M is not only totally bounded, but also complete. Hence, by the principle of nested closed sets, the intersection of all vn consists of exactly one point, say, x . But x lies in some U E a, as does every other point
2.
Compact metric spaces and total boundedness
205
of the space M. Since U is open, it contains an open ball U(x, c) of some radius c > 0. Therefore, due to the condition on the diameters (property b)) , this ball, and thus U as well, contains all sets vn with sufficiently large n. But then as a covering of such vn , contains a sub covering consisting of only one U. We have come to a contradiction with property a) of the constructed sets. • a,
Taking into account Proposition 3 and the fact that the completeness of a subset in a complete metric space is equivalent to the closedness of this subset, we obtain
A subset of a complete metric space is compact it is totally bounded and closed. The same set is relatively compact it is totally bounded.
Corollary 2.
Certainly, if a metric space is not complete, it always has closed totally bounded sets that are not compact, or, what is the same here, relatively compact: it is sufficient to take the set of elements of a non-convergent fundamental sequence. *
*
*
We recall fundamental properties of finite-dimensional spaces mentioned in Theorem 2. 1 . 1 and in Corollaries 2. 1 .2 and 2. 1 .3. (Their "geometrical meaning" is that on a finite-dimensional space all norms are equivalent, and the "categorical meaning" is that all spaces of the same finite dimension are isomorphic to each other as objects in Ban.) Now we can suggest another approach to the proofs of these results, more precisely, of Proposition 2. 1 .4, from which all others easily follow. Proposition 7 ( == Proposition 2. 1 .4 ) .
Every normed space E of finite di mension n is topologically isomorphic to CC!, and every linear isomorphism between these spaces is a topological isomorphism.
J be an arbitrary linear isomorphism from CC! onto E. Since it is certainly bounded (Example 1 .3. 1) , it is sufficient to find c > 0 such that c ll x ll < II J(x) ll for all x E E (see Proposition 1 .4 . 1 ) . Let S be the unit sphere in CC!. Since J is an injective operator, the function f : S � CC : x �----+ II J ( x) II - l is well defined, and it is continuous by the continuity of the norm. But S, being a closed bounded set in CC! , Proof. Let
is compact (Example 1 and Corollary 2 ) . By Theorem 1 .2 (the Weierstrass theorem) , this implies that f(x) is bounded from above by a constant C. It remains to take c : == 1 /C . II *
*
*
3. From Compact Spaces to Fredholm Operators
206
The rest of the section is devoted to the following important class of problems: which subsets are totally bounded and which are compact in a particular normed space? First , we distinguish the following simple result.
In a normed space, the algebraic sum of two totally bounded sets is totally bounded and every dilation of a totally bounded set is totally • bounded.
Proposition 8.
This fact will be useful later, and now we consider the case of finite dimensional spaces. Actually, everything here is ready for an answer. Proposition 9.
Then
Let M be a subset of a finite-dimensional normed space E.
(i) M is totally bounded it is bounded; (ii) M is compact it is bounded and closed. Proof. By Proposition 7 and Theorem 1 .4. 1 (iv) ,
E is uniformly homeomor
phic to CC! , where n == dim(E) . Hence M is uniformly homeomorphic to a bounded subset in CC!. Since the latter is totally bounded if and only if it is bounded (see Example 1 and Proposition 5) , (i) follows from Corollary 1 . Taking into account the completeness of E (Theorem 2. 1 . 1 (i) ) and Corollary • 2, we deduce (ii) . The situation in the infinite-dimensional spaces is not so simple. First, we gather some "empirical" material. Exercise 1 . The unit balls in the spaces C[a, b] , lp and Lp [a, b] ; 1 < p < oo are not totally bounded. Hint. Take functions (or sequences) of norm 1 with disjoint supports and use Proposition 4(iii) . Now let us find which general fact stands behind these observations. The following proposition is a key for the subsequent theorem.
"lemma on the near-perpendicular" ) . Let E be a normed space, and Eo its closed proper subspace. Then for every c > 0 there exists a vector x E Eo (called an c -perpendicular) such that II x II == 1 and d(x, Eo) > 1 - c.
Proposition 10 (known as
E \ Eo . Since Eo is closed, d(y, Eo) > 0. Hence, taking into account the obvious equality d(A.y , Eo) == I A. I d(y, Eo) for each A. E CC, we can assume without loss of generality that d(y, Eo) == 1. Let z E Eo be such that d(y, z ) < 1 + 8, where 8 > 0 is so small that 1/(1 + 8) > 1 - c. Then for xo : == y - z we have ll xo ll < 1 + 8. At the same time, since the sets {xo E Eo } coincide, E Eo } and {y Proof. Take an arbitrary
y
E
u : u
u : u
2.
207
Compact metric spaces and total boundedness
d ( xo , Eo ) == d ( y, Eo ) == 1 . Therefore, it is easy to see that x required €-perpendicular.
: ==
x f ll x ll is the
•
Certainly, if E is a Hilbert space, it has a vector x such that ll x ll == 1 and at the same time d ( x, Eo ) == 1: this is the "true" perpendicular to Eo ; its existence is stated in the theorem on orthogonal complement. But for general Banach spaces, such a luxury is not guaranteed. Exercise 2.
( i ) There are pairs ( a Banach space E, a closed subspace Eo ) such that for each x E E \ Eo we have ll x ll > d ( x, Eo ) . ( ii ) There are coisometric operators, taking the closed unit ball of the
first space to a proper subset of the second space. Hint. In proving ( i ) Exercise 2.3.2 is useful, and ( ii ) follows from ( i ) . Theorem 2 ( Riesz ) .
The following properties of a normed space E are
equivalent: ( i ) the (closed) unit ball in E is compact; ( ii ) the unit ball in E is totally bounded; ( iii ) E is finite-dimensional.
Proof. The implication ( i ) ====> ( ii ) is clear.
( ii ) ====> ( iii ) . Suppose, on the contrary, that E is infinite-dimensional. Choose an arbitrary c; 0 < c < 1. Denote by x 1 an arbitrary vector in E of norm 1. Since the space E1 : == span ( x1 ) is one-dimensional and closed
by Corollary 2. 1 .3, the lemma on the near-perpendicular guarantees the existence of a vector x 2 of norm 1 such that d ( x 2 , E1 ) > 1 - c. Further, since E2 : == span ( x1 , x 2 ) is two-dimensional, the same lemma gives a vector X 3 of norm 1 such that d ( x 3 , E2 ) > 1 c . Now consider E3 : == span ( x1 , x 2 , x 3 ) , and so on. Following these arguments, we obtain a sequence of vectors xn ; n E N in BE for which the pairwise distances between elements are not smaller than 1 - c. By Proposition 4 ( iii ) , this is impossible. II ( iii ) ====> ( i ) immediately follows from Proposition 9 ( ii ) . -
Exercise
3. The properties ( i ) - ( iii ) of the space E are equivalent to the
local compactness of E.
For a series of concrete Banach spaces, their totally bounded ( and thus, by Corollary 2, relatively compact ) subsets can be described in sufficiently transparent terms. We present one of the most important facts of this kind, concerning the space C[a, b] . The following definition is useful here.
3. From Compact Spaces to Fredholm Operators
208
C[a, b] is called equicontinuous if for every c > 0 there exists 8 > 0 such that l f(t') - f(t") l < c for all f E M and t', t" E [a, b] with I t' - t" l < 8. The classical Cantor theorem implies that every finite subset in C[a, b] Definition 3. A subset M E
is equicontinuous. At the same time, it is easy to verify that the set of functions sin nt; n == 1 , 2, . . . is not equicontinuous.
A subset M in C[a, b] is totally bounded it is bounded and equicontinuous.
Theorem 3 ( Arzela ) .
Take c > 0; our goal is to find a finite c-net for M in C[a, b] . By assumption, there is n E N such that l f(t') - f(t") l < c / 3 whenever l t' - t" l < 1/n and f E M. Put t k :== a + (b -na) k ; k == 0, . . . , n and consider the ( linear contraction ) operator T : C[a, b] � cc�+ l , T(f) : == (f(to), . . . , f(tn ) ) . By Proposition 9 ( i ) , the set T(M) c cc�+ l is totally bounded. Hence, taking Proposition 1 into account, there is a finite subset N in M such that T(N) is an c / 3-net in T(M) . We show that N is an c-net for M. To this end, take an arbitrary function f E M and a point t E [a, b] and choose k such that I t - t k l < ljn. From the definition of the set N it follows that there is g E N such that l g( t k ) - f(t k ) l < c / 3. Hence, l f(t) - g (t) l < l f(t) - f(t k ) l + l f(t k ) - g( t k ) l + l g( t k ) - g (t) l < 3c / 3 == c. This means that N is an c-net for M. ===> . Taking into account Proposition 5, it is sufficient to verify that M is equicontinuous. Take c > 0, and let N be a finite c / 3-net for M in C[a, b] . As we mentioned above, the set N of functions, being finite, is equicontinuous. Therefore, there exists 8 > 0 such that the condition I t' - t" l < 8 for all g E N implies l g(t ' ) - g (t") l < c / 3. Take an arbitrary f E M; then for some g E N we have II f - g II < c /3 because for I t' - t" l < 8 we have the estimate I ! ( t' ) - ! ( t " ) I < I ! ( t ' ) - g ( t' ) I + I g ( t' ) - g ( t " ) I + I ! ( t" ) - g ( t" ) I < 3c I 3 == c . • The rest is clear. Proof.
�-
oo
Exercise 4.
( i ) The unit ball in C 1 [a, b] , regarded as a subset in C[a, b] , is totally
bounded, but not compact. ( ii ) The subset in the unit ball in C[a, b] consisting of functions f such that l f (t' ) - f(t") l < I t' - t" l for all t' , t" E [a, b] , is compact. Next we give a description of totally bounded sets in another important class of Banach spaces.
3. Compact operators: General properties and examples
Exercise 5. Let M be a subset in
bounded
{:::::::>
it is bounded, and
209
lp ; 1 < p < oo. Then M is totally
00 ��00 sup { L l enl p : e = (6 , 6 , . . . ) E M } = 0 (for p < oo) ;
n =m+ l sup l �nl : � == (� 1 , �2 , . . . ) E M } == 0 (for p == oo) mlim � oo sup { n >m +l (i.e. , the norms of the "tails" (�m + l , �m+ 2 , . . . ) of the sequences � from M
uniformly tend to zero) .
3 . Compact op erators : General prop ert ies and examples
The previously considered notions, and first of all the notion of total bound edness, allow us to distinguish a very important class of operators. One can advance rather far in studying these operators, and in the case where such operators connect two Hilbert spaces, we may even say that one can completely perceive their nature. Definition 1. An operator T : E � F between normed spaces is called compact 2 if it takes every bounded set from E to a totally bounded set in F. Certainly, the discussed property of an operator is determined by its behavior on only one bounded set.
An operator T : E � F is compact {:::::::> the image of the II unit ball in E is totally bounded. Remark. As we will see later, under certain conditions on E and F, which Proposition 1.
are fulfilled in many cases, the image of the unit ball is not only totally bounded but compact; see Propositions 4.3 and 4.2. 17. This partially justi fies the term "compact operator" .
Later, after becoming familiar with the so-called weak topologies, we will speak about another approach to the notion of compact operator (see Theorem 4.2.4) . From Proposition 2.9 we immediately obtain the following result. Proposition 2. Every finite-dimensional bounded operator is compact. On the other hand, from Theorem 2.2 (the Riesz theorem) we have
No projection onto an infinite-dimensional subspace (see Sections 1 . 5 and 2. 3) and, in particular, no identity operator on an infinite II dimensional normed space, is compact.
Proposition 3.
2 Half a century ago mathematicians used to say "completely continuous operator" instead of "compact operator" .
3. From Compact Spaces to Fredholm Operators
210
The set of compact operators from E to F is denoted by /C(E, F) , and one usually writes /C ( E ) instead of /C (E, E) . Proposition 4.
JC(E, F) is a closed subspace in B(E, F) .
/C(E, F) . Then (S + T) (BE ) , being a part of S(BE) + T(BE ) , is totally bounded by Proposition 2.8. Hence, S + T E /C(E, F) . It is even easier to verify that A.S E /C(E, F) for every A. E CC. It remains to show that every adherent point of the set /C ( E, F) , say S, belongs to JC ( E, F) . Take c > 0 and T E /C ( E, F) such that l i S - T i l < c/2. Proof. Take S, T
E
Then it is easy to see that every c/2-net for T(BE) is an c-net for S(BE) · The rest is clear. •
If F is a Banach space, then JC ( E, F) is also a Banach space for every normed space E. Corollary 1 .
Propositions 2 and 4 together give a useful way of verifying compactness for a wide class of operators.
If an operator between two normed spaces can be approximated in the operator norm by finite-dimensional operators, then it is compact.
Corollary 2.
The following proposition, in addition to its independent importance, is very useful for the verification of compactness of concrete operators.
Suppose E, F, and G are normed spaces, S E B ( E, F ) , and T E B(F, G) . If at least one of these operators is compact, then TS : E G is also compact. Proposition 5.
�
Proof. If T is compact, then everything is clear. If S is compact, then
TS(BE) is totally bounded by Proposition 2.2 and Theorem 1 .4. l (iv) .
•
This proposition implies, in particular, that a topological isomorphism between infinite-dimensional normed spaces is never compact (otherwise the identity operators on these spaces would be compact as well) . To an advanced reader, we recommend the following Exercise 1 (Schauder) . An operator T compact ¢:::::::> the Banach adjoint operator compact . ====}
: E � F between two Banach spaces T* : F* � E* (see Definition 2. 5.2)
is is
Let Y l ' . . . ' Yn be an c I 4-net in T( BE ) . Consider an operator s : F* � e n ' S(f) == (j(y1 ), . . . , f (Yn )) and take a finite subset N C BF * such that S ( N) is an c/4-net in S(B F * ) . Then T* (N) is an c--net in T* (B F * ) . {:::== follows from ====} and from the uniqueness of the canonical embedding of the space into its second dual space (Proposition 2 . 5 . 2) .
Hint.
.
3. Compact operators: General properties and examples
211
Now it is time to recall various concrete examples of operators we have gathered in Section 1 .3. The new typical question is whether these operators are compact.
The diagonal operator T>.. : lp --+ lp where A E l 00 (see Example 1.3.2) is compact A E co . Proof. � - Let A ( n ) : == (A I , . . . , A n , 0, 0, . . . ) . Then the operator T>.. ( n ) is finite-dimensional, and l i T>.. - T>.. ( n ) II == II A - A (n ) lloo --+ 0 as n --+ oo (Example Proposition 6.
,
1.3.2) . Now Corollary 2 works. ====> . If A t/:. co , then there are n 1 < n 2 < · · · such that I An k I > () for some () > 0 and for all k E N. But for all k , l E N; k =/=- l and p < oo we have II T( pnk ) - T( pn z ) ll == II A n k Pnk - Anl Pn 1 ll = I Ank i P + I .Anl i P > 0 the set M : == {t E [a, b] : l f( t ) l > 0} also has positive measure. Therefore the operator T9 Tf, where g ( t) is 1/ f(t) for t E M and zero otherwise, is a projection onto an infinite-dimensional subspace in Lp [a, b] . Although the following exercise is a special case of the subsequent the orem, it would be nevertheless useful for the reader to look at it first. Exercise 3. The operator of indefinite integration (see Example 1.4.5) , acting on one of the spaces E : == L 2 [0, 1] , C[O, 1] , or L1 [0, 1] , is compact. Hint. The case E == C[O, 1] follows from Exercise 2.4(ii) . If E == £ 2 [0, 1] , then from the Cauchy-Bunyakovskii inequality it follows that the set of functions T (BE) is equicontinuous; so we can use the Arzela theorem and the estimate II · II 2 < II · II 00 • Finally, if E == L 1 [0, 1] , then the set T (BE) (although not equicontinuous) consists of functions with total variation < 1.
3. From Compact Spaces to Fredholm Operators
212
To construct a finite c:-net, we divide the interval [0, 1] into many equal parts and find this net among the functions that are constant on the intervals of the partition. Let us now recall a general class of integral operators on £ 2 [a, b] that contains the operator of indefinite integration (see Example 1 .3.6 and sub sequent comments) . Theorem 1.
Every integral operator on L 2 [a, b] is compact.
Proof. We need the following lemma.
Suppose functions en ( t); n E N form an orthonormal basis in L 2 [a, b] . Then the functions Um , n (s, t) : == em (s) en (t); m, n E N (arbitrar ily numbered) form an orthonormal basis in L 2 (D) (on the latter space, see also Example 1 . 3. 6). Lemma.
Um , n consider a function y E
Proof. Elementary computations using the Fubini theorem show that
is an orthonormal system. To verify its totality, L 2 ( D ) that is orthogonal to all Um , n , or, in other words, satisfies the condi tion Io Um , n (s, t)y(s, t)dsdt == 0 for all m, n E N. By Proposition 2.3.5, it is sufficient to establish that y vanishes as a vector in L 2 (D) , i.e. , y(s, t) == 0 almost everywhere on D. Again by the Fubini theorem, we have I: em (s)xn (s)ds == 0, where Xn ( s ) : == J: en ( t) y ( s, t) dt. This means that for every n the function Xn is or thogonal to all em ; m E N in L 2 [a, b] . By the totality of em , we have X n == 0 in L 2 [a, b] , i.e. , for a set Mn of full measure on [a, b] we have I: en (t)y(s, t)dt == 0 for all s E Mn . But then for s E M == n{Mn ; n E N}, by the totality of en , we have y(s, t) == 0 for almost all t E [a, b] . Since M is also a set of full • measure, this means that y == 0 almost everywhere on D. :
T be an integral operator, and K its kernel. For each n == 1 , 2, . . . consider the integral operator Tn with the kernel Kn (s, t) : == 2:: � l= l A k , l e k (s) ez (t), where A k , l : == ( K, uk , l ) are the corresponding Fourier coefficients. Obviously, every function in the image of this operator is a linear combination of the functions e k ; k == 1 , . . . , n. Hence Tn is a finite-dimensional operator. Further, since T - Tn is an integral operator with the kernel K - Kn , the operator norm of T - Tn is not greater than the norm of K - Kn in L 2 (D) (see Example 1 .3.6) . But from our lemma, it evidently follows that K is the limit of the sequence Kn in £ 2 (D) . Thus, T End of the proof of Theorem 1 . Let '
is approximated by finite-dimensional operators, and therefore (Corollary 2) • is compact. Here is one of the numerous corollaries of this theorem.
3. Compact operators: General properties and examples
Exercise 4. An integral operator on
213
L 2 [a, b] with the kernel K(s, t) is
a projection K(s, t) is degenerate and satisfies the equalities indicated in Exercise 1 .5.5. Hint. The kernel of a one-dimensional integral operator has the form f ( s ) g ( t ) for some J, g E L 2 [a, b] . Exercise 5. The differentiation operators (see Example 1 .3.8) are never compact. Now please do the following exercise of general nature. Exercise 6. Every two weakly topologically equivalent operators are simultaneously either compact or not. *
*
*
We have seen how useful and simple is Corollary 2. But the question whether the converse is true is not simple at all. The analysis of this problem led to the following fundamental notion in the geometric theory of Banach spaces.
2 (Grothendieck, 1955) . A Banach space F has the approxi mation property if for every Banach space E every compact operator from E to F can be approximated in the operator norm by finite-dimensional
Definition
operators.
Here is an important example. Proposition 7. Proof. Let
T:
Every Hilbert space has the approximation property. E
H be a compact operator between a Banach space and a Hilbert space. Take c > 0 and consider an c/2-net Y I, . . . , Yn in H for T(BE) · Put Ho : == span{y1 , . . . , Yn } and consider the orthoprojection P : H --+ H onto Ho . Certainly, it is a finite-dimensional operator, and for every k 1 , . . . , n we have Pyk Yk · Hence, taking into account our choice of the c/2-net, we see that for each x E BE there is k such that II Tx - PTx ll < II Tx - Ykll + II P(yk - Tx) ll < 2 11 Tx - Ykll < c. II The rest is clear. ==
--+
==
Gradually it became clear that the overwhelming majority of "classical" Banach spaces (Lp (X, J-L) , C (O ) , . . . ) have the approximation property. For many separable spaces it was a corollary of the following important result (we recommend that advanced readers prove it; see Exercise 7) .
2. If a Banach space has a Schauder basis, then it has the ap proximation property. Theorem
2 14
3. From Compact Spaces to Fredholm Operators
For some time it looked like the question asked by Grothendieck, namely whether every Banach space has the approximation property (the famous approximation problem) would have a positive answer in the near future, at least in the class of separable spaces. But the optimists were wrong.
3 ( P. Enflo, 1972, [54]) . There exist separable Banach spaces, in particular, closed subspaces in co, that do not have the approximation property. As a corollary (see Theorem 2), there are separable Banach spaces without Schauder bases. Theorem
The question of whether every separable Banach space has a Schauder basis, was posed long ago by Banach. For a long time it was, perhaps, the best known problem in the theory of Banach spaces. Thus, Enflo, by proving this theorem, also gave the negative answer to this old problem. All separable Banach spaces without the approximation property known so far have been skillfully constructed examples. But outside the class of separable spaces, such a "black sheep" has been found among well-known objects. It turned out that B(H) , where H is an infinite-dimensional Hilbert space, does not have the approximation property (A. Szankowski, 1981) . By the way, verify that B(H) is indeed non-separable. Exercise 7* . Prove Theorem 2. Hint. Let e 1 , e 2 , . . . be a Schauder basis in a Banach space (F, II · II ). For n == 1, 2, . . . consider the projections Pn : F � F : I:� 1 A k e k �-----+ I:: �=l A k e k and the new norm II · II ' : y �-----+ sup{ I I Pn (Y) I I ; n E N} in F. If x m ; m == 1 , 2, . . . is a fundamental sequence in (F, II · II ' ), then for every n the sequence Pnx m ; m == 1 , 2 , . . . converges with respect to every norm in Fn :== span{ e1 , . . . , en } to some Xn , and for each l == 1 , 2, . . . we have Pnxn+l == Xn . From the properties of a Schauder basis it follows that the sequence Xn is fundamental in the initial space ( F, II · II ), and thus tends to some x with respect to the norm II · II · At the same time, Pn x == Xn ; n == 1 , 2, . . . . Taking into account that Pn are contraction operators on (E, II · II ' ), we can see that x is the limit of xrn in (F, II · II ' ). Therefore, (F, II · II ' ) is a Banach space , and by Proposition 2 .4.2, all Pn are bounded in ( F, I · II ), and their norms do not exceed some constant C.
Now suppose T : E � F is a compact operator, c > 0, and y1 , . . . , Ym is a finite 2( l� C) -net for T(BE) in F. If an integer N is such that for all n > N and k == 1 , . . . , m we have II Yk - PnYk II < 2 ( l� C) , then for the same n and for each x E B E we have IITx - Pn Tx ll < c . Thus T can be approximated by the finite-dimensional bounded operators Pn T. *
*
*
Advanced readers should also know that the approximation property is a very deep notion, which can appear in various forms that do not resemble each other. The following theorem is by no means simple, and we give it without proof. (The argument needed for its proof can be found, e.g. , in the extensive book [48 , Chap. !.5] . ) Theorem 4 (Grothendieck) . The following properties of a Banach space F are equ�valent:
(i) F has the approximation property;
4.
215
Compact operators between Hilbert spaces
(ii) for every compact subset K C F and for every c > 0 there ex�sts a fin�te dimens�onal bounded operator S : F � F such that I I Y - Sy ll < c for all y E K; (iii) the Grothendieck operator Gr : F* 0 F � B(F) (see Sect�on 2 . 7) �s inJective (and thus �s an �sometr�c �somorphism between F* 0 F and the space of nuclear operators N (F)); ( iv) for every u E F* 0 F the condit�on Gr( u ) == 0 implies tr( u ) == 0 (�. e. , the trace of a nuclear operator act�ng on F �s well defined; cf. Section 2. 7}.
Furthermore, if at least one of the spaces E* or then the operator Gr : E* 0 F � B(E, F) �s inJect�ve.
F
has the approx�mat�on property,
4. Compact op erat ors between Hilb ert spaces
Let H and K be Hilbert spaces. Suppose e is a vector in H, and f is the functional on H defined by e according to the rule described in Theorem 2.3.2 ( Riesz ) . Let y be a vector in K. For the corresponding one-dimensional operator we use, together with the notation f 0 y , also the notation e 0 y. Then the multiplication rule for one-dimensional operators indicated in Proposition 1.5.5 will look as follows:
( e 1 0 Y1 )( e 2 0 Y2 ) == ( 2 e 1 ) ( e 2 0 Y1 ) · y
,
The structure of the compact operators between Hilbert spaces is com pletely described by the following theorem. Theorem 1 ( Schmidt 3 ) . Let H and K be Hilbert spaces and T : H � K a
compact operator. Then there exist ( a ) an orthonormal system e� , e� , . . . in H of finite or countable cardi nality; ( b ) an orthonormal system e1, e�, . . . in K of the same cardinality, and ( c ) a (finite or infinite) sequence s 1 > s 2 > · · · of positive numbers with the index set of the same cardinality, tending to zero if it is infinite, such that T can be represented in the form n where, depending on the case, L: n is either a finite sum, or the sum of a convergent series in B(H, K) . Thus, the operator acts by the formula Tx = L Sn (x, e�) e � , n
(1879-1969) ,
3Erhard Schmidt prominent German mathematician, a student of Hilbert. It was his ( and Frechet's ) work where the language inherited from Euclid's geometry began to be widely used in the study of Hilbert spaces of sequences and functions.
2 16
3. From Compact Spaces to Fredholm Operators
where L:n is either a finite sum, or the sum of a series in K. {In other words, T( e�) == sn e� for all n, and T takes every vector orthogonal to all e� to zero). In addition, li T II == s 1 . Before starting the proof, let us emphasize that in the case where the sum T == L: n sn e� 0 e� has infinitely many summands, we do not claim that this series absolutely converges in the operator norm. The latter condition distinguishes a special class of operators, and we will discuss this later. Proof. Certainly, without loss of generality we can assume that
T =/=- 0.
There exists a vector e E H; ll e ll == 1 such that II Te ll == II T II { "a vector where the operator norm is reached").
Lemma 1.
Proof. By the definition of operator norm, there is a sequence X n E BH such that II TX n II -t II T II as n -t oo . Since the set T ( BH ) is totally bounded
and K is complete, the sequence Tx n has a convergent subsequence ( see Proposition 2.4) . Changing the notation if needed, we can assume that Txn tends to some y E K. 0bviously, II y II == II T II . By the parallelogram identity, for every natural m and n we have
ll x m - X n ll 2 == 2 ll x m ll 2 + 2 ll x n ll 2 - ll x m + X n ll 2 < 4 - ll x m + X n ll 2 · Further, II T(x m + Xn ) ll < II T II II x m + x n ll · For m, n -t oo the left-hand side of this inequality tends to II 2 YII == 2 II T II . Since li T II =/=- 0, it follows that ll x m + xn ll 2 tends to 4, and thus ll xm - x n l l 2 tends to zero. Therefore, the sequence X n is fundamental in H, and hence, tends there to a vector e. The rest is clear. • Lemma 2. Let e E H; II e II == 1 be such that II Te II == II T II . Th en fo r ev e ry x E H the condition x _L e implies Tx _L Te. (In other words, T takes { e } j_ to {Te } 1_ . ) Before starting formal arguments, we note that, according to our geomet ric experience in IR3 , this is automatically true. Let, for example, li T II == 1. If x _L e and Tx has an acute angle with Te, then for small t the length of the vector T( e + tx) == Te + tTx must be greater than the length of e + t x ( Figure 1 ) . But this is impossible since T does not increase the norm of a vector. Proof. Assume that the opposite is true: for some x; x
_L e we have ( Tx, Te )
=/=- 0. Replacing, if necessary, x by AX for some A E CC, we can assume that ( Tx, Te ) > 0 ( "Tx and Te form an acute angle" ) . Then for every t > 0 we
4.
217
Compact operators between Hilbert spaces
e
Te
tx Figure 1
have
II T II 2 l i e + tx ll 2 > l i T( e + tx) 11 2 == ( Te + tTx, Te + tTx ) == 11 Te ll 2 + 2t ( Tx, Te ) + t 2 11 Tx ll 2 . Since II e + tx II 2 == 1 + t 2 II x II 2 ( the Pythagorean theorem) and II Te II == II T II , we have 2t ( Tx, Te ) < t 2 ( II T II 2 II x ll 2 - 11 Tx ll 2 ). For sufficiently small t > 0 •
this is impossible, hence we obtain a contradiction.
End of the proof of Theorem 1. Lemma 1 gives at least one vector
e� E H; l i e� II == 1 such that li Te� II == II T II . We take such a vector and put e� : = 1 }11 Te�, 8 1 : = l i T II · Then we put H1 :== { e� } j_ and T1 :== T I H1 : H1 � K. If T1 == 0, then we stop here. Otherwise, since T1 is compact together with T, Lemma 1 gives at least one e� E H1 ; II e� II == 1 such that II T1 e� II == II Te� II == II T1 II . Choose such a vector and put e� : == wf1 1 Te�, 8 2 : == II T1 II . Note that { e�, e� } is an orthonormal system in H, and by Lemma 2, the same is true for the system {e�, e� } in K. In addition, we have 8 1 > 8 2 . Now we put H2 :== {e�, e� } j_ and T2 :== T I H2 : H2 K. If T2 == 0, we stop here; otherwise, using the compactness of T2 , we take e� E H2 ; II e� II == 1 such that II T2 e; ll = II Te; ll = II T2 II , and set e� : = 1 ,z\ 1 re; , 83 : = II T2 II · Now we see that { e�, e�, e� } and hence ( by Lemma 2) { e�, e�, e� } are orthogonal systems, and 8 1 > 8 2 > 8 3 . After that we go over to the space H3 : == { e�, e�, e� } 1_, the operator T3 :== T I H3, etc. �
Continuing this process, we obviously may face two possibilities: 1 . For some n, after constructing orthonormal systems { e� , e�, . . . , e� } in H and {e�, e�, . . . , e�} in K, and positive numbers 8 1 > 8 2 > · · · > 8n , we see that T vanishes on { e�, e�, . . . , e�}j_. Then, obviously, our operator can be represented as a finite sum T == 2:: �== 1 8 k e� 0 e%.
218
3. From Compact Spaces to Fredholm Operators
2. All Tn ; n E N are non-zero operators. In this case our process leads to countable orthonormal systems { e� , e� , . . . } in H and { e�, e�, . . . } in K, and to a sequence of positive numbers 8 I > 82 > · · · . Put () :== limn� oo 8n . Then by the Pythagorean theorem, for every m, n E N the number d(Te'm, Te�) == 11 8m e� - 8n e� ll equals J8� + 8; > /20. Hence, Proposition 2.4 ( iii ) applied to T ( BE ) guarantees that () == 0. For every n we put Sn :== L:�= I 8 k e � 0 e % . Then the operator T Sn vanishes on span { e� , . . . , e�} and is equal to Tn on ( span { e� , . . . , e�} )1_. Hence, li T - Sn ll == II Tn ll == 8n · Since 8 n -----+ 0 as n -----+ oo , the operator T can • be expanded in the series L:C: I 8 k e � 0 e % . The rest is clear.
L: n 8 n e� 0 e� in the formulation of the Schmidt theorem is called the Schmidt series of a ( compact ) operator T, and the numbers 8n , the 8-numbers of this operator. Definition 1. The sum
Certainly, a compact operator is finite-dimensional its Schmidt se ries contains only a finite number of terms. In the proof of the Schmidt theorem we have seen that, generally speak ing, in the construction of the system e� , e�, . . . ( and thus, of the Schmidt series ) , there is some arbitrariness: for instance, there can be many vectors x satisfying the condition II Tx ll == II T II , and we can choose any one of them as e� . However, as we will now show, this arbitrariness is not that big, and, in particular, the numbers 8 n do not depend on the choice of the system e Ii , e I2 , · · · . Suppose that, for some orthonormal systems e� , e� , . . . and e�, e�, . . . , the statement of the Schmidt theorem is true. Since some neighboring num bers 8 n can coincide, for some natural n1 < n2 < · · · we have 8 I == · · · == 8n1 > 8n1 +I == · · · == 8 n2 > 8n 2 +I == · · · > 8nk - 1 +I == · · == 8n k > · · · · For k == 1 , 2, . . . , consider the spaces H k :== span { e n'k - 1 +I , . . . , e� k } ( we put no == 0 here ) . ·
The spaces H k do not depend on the choice of the sys tems e� , e� , . . . and e�, e�, . . . {hence, they depend only on the operator T). Further, for every k the numbers 8n k +I , . . . , 8n k + 1 coincide with the norm of the operator T( k) :== T I k ) , where H( k ) : == ( span { H I , . . . , H k } )1_ . As a corollary, the numbers 8n are uniquely defined by the operator T.
Proposition 1 .
H(
Proof. If T == 0, then
8 k == 0 for all k, and there is nothing to prove. In the case where T =/=- 0 we use induction on k. Here is the start: Lemma. The space H I consists of all x E H for which II Tx ll l i T II ll x ll , and the numbers 8 I , 8 2 , . . . , 8n1 coincide with II T II .
4.
Compact operators between Hilbert spaces
219
x E H1 the L: n s; l (x , e� ) l 2 . Hence, x E H im
Proof. Knowing how our operator acts, we see that for each
following equality is true: 11 Tx ll 2 == plies II Tx ll == s 1 ll x ll . Further, taking into account the Bessel inequality, 11 Tx ll 2 < L:n s r l ( x, e � ) l 2 < s r ll x ll 2 , and the previous sentence, we have s 1 == · · == Sn 1 == li T II . Finally, if II Tx ll == s 1 ll x ll , then for every n we have s; l ( x , e� ) 1 2 == s i I (x , e� ) 1 2 . Therefore ( x , e� ) == 0 as n > n1 , so that 11 Tx ll 2 == 2::� 1 1 s i l ( x, e� ) 1 2 . Hence II Tx ll == li T II ll x ll implies ll x ll 2 == 2:: � 1 1 I ( x, e� ) 1 2 , II and thus x E H 1 (see Proposition 1 .2.9) . The rest is clear. ·
End of the proof of Proposition 1. Suppose we proved the proposition 1 for 1 , . . . , k . Then, together with the spaces H , . . . , H k , the space H( k ) , and as a corollary, the operator T( k ) also depend only on the operator T. In addition, as you can easily see, the operator T( k ) : H( k ) � K acts by the
formula
Hence, if we consider our lemma with H( k) as H and T( k ) as T, then the space Hk +I will play the role of H 1 . Consequently, applying the lemma to this situation, we obtair1 that H k +I == { x E H( k) : II T( k) x ll == II T( k ) II ll x ll }. This means that Hk +I , as well as the operator T( k) (see above) , depend only on the operator T. The same lemma gives the equalities s nk +I == · · · ==
Snk + 1
==
II T( k) II ·
II
There are other beautiful characterizations of s-numbers, and we now present one of them without proof (see, e.g. , [55] ) .
Under the assumptions of the Schmidt the orem we have the following result: For every n == 1 , 2, . . . the set of numbers li T -
· · Put () : == limn � oo Sn . Then by the Pythagorean theorem, for every m , n E N the number d(Te'm , Te�) == ll sm e� - sn e� ll equals Js� + s; > /20. Hence, Proposition 2.4 ( iii ) applied to T (BE ) guarantees that () == 0. For every n we put Sn :== L:�= I s k e � 0 e % . Then the operator T Sn vanishes on span { e� , . . . , e�} and is equal to Tn on ( span { e� , . . . , e�}) 1_. Hence, li T - Sn II == II Tn II == Sn . Since Sn 0 as n -----+ oo , the operator T can • be expanded in the series L:C: I s k e � 0 e%. The rest is clear. ·
.
-----+
L: n sn e� 0 e� in the formulation of the Schmidt theorem is called the Schmidt series of a ( compact ) operator T, and the numbers s n , the s-numbers of this operator. Definition 1 . The sum
Certainly, a compact operator is finite-dimensional its Schmidt se ries contains only a finite number of terms. In the proof of the Schmidt theorem we have seen that, generally speak ing, in the construction of the system e� , e�, . . . ( and thus, of the Schmidt series ) , there is some arbitrariness: for instance, there can be many vectors x satisfying the condition II Tx ll == II T II , and we can choose any one of them as e� . However, as we will now show, this arbitrariness is not that big, and, in particular, the numbers s n do not depend on the choice of the system e Ii ' e I2 ' . . . . Suppose that, for some orthonormal systems e� , e� , . . . and e7, e�, . . . , the statement of the Schmidt theorem is true. Since some neighboring num bers Sn can coincide, for some natural n1 < n 2 < · · · we have S I == · · · == Bn 1 > Bn 1 +I == · · · == Bn 2 > Sn 2 + I == · · · > Sn k-1 + I == · · · == Sn k > · · · · For k == 1 , 2, . . . , consider the spaces Hk : == span { e� k-1 + I , . . . , e� k } ( we put no == 0 here ) .
The spaces H k do not depend on the choice of the sys tems e� , e� , . . . and e7, e�, . . . {hence, they depend only on the operator T). Further, for every k the numbers Sn k + I , . . . , Snk+1 coincide with the norm of the operator T( k) : == T I k , where H( k ) : == ( span { H I , . . . , H k } )1_ . As a corollary, the numbers Sn are uniquely defined by the operator T.
Proposition 1 .
H( )
Proof. If T == 0, then
s k == 0 for all k, and there is nothing to prove. In the case where T =/=- 0 we use induction on k. Here is the start: Lemma. The space H I consists of all x E H for which II Tx ll and the numbers s i , s 2 , . . . , Sn 1 coincide with li T II .
4.
Compact operators between Hilbert spaces
219
x E H1 the L: n s; l ( x, e� ) l 2 . Hence, x E H im
Proof. Knowing how our operator acts, we see that for each
following equality is true: 11 Tx ll 2 == plies II Tx ll == s 1 ll x ll . Further, taking into account the Bessel inequality, 11 Tx ll 2 < L:n s i l ( x, e� ) l 2 < s i ll x ll 2 , and the previous sentence, we have s1 == · · · == Sn 1 == II T II . Finally, if II Tx ll == s1 ll x ll , then for every n we have s; l ( x , e� ) l 2 == s i l (x , e� ) l 2 . Therefore (x , e� ) == 0 as n > n1 , so that 11 Tx ll 2 == 2::� 1 1 s i l ( x, e� ) l12 . Hence II Tx ll == II T II II x ll implies ll x ll 2 == 2:: � 1 1 l ( x, e� ) l 2 , II and thus x E H (see Proposition 1 .2.9) . The rest is clear.
End of the proof of Proposition 1. Suppose we proved the proposition for 1 , . . . , k. Then, together with the spaces H 1 , . . . , H k , the space H( k ) , and as a corollary, the operator T( k ) also depend only on the operator T. In addition, as you can easily see, the operator T( k) : H( k ) � K acts by the
formula
Hence, if we consider our lemma with H( k) as H and T( k ) as T, then the space Hk +I will play the role of H 1 . Consequently, applying the lemma to this situation, we obtair1 that Hk +I == { x E H( k) : II T( k) x ll == II T( k) II ll x ll }. This means that H k +I , as well as the operator T( k) (see above) , depend only on the operator T. The same lemma gives the equalities s nk +I == · · · ==
Snk + 1
==
II T( k) ll ·
II
There are other beautiful characterizations of s-numbers, and we now present one of them without proof (see, e.g. , [55] ) .
Under the assumptions of the Schmidt the orem we have the following result: For every n == 1 , 2, . . . the set of numbers li T - 4> 11 , where 4> runs through all possible bounded operators with the image of codimension n - 1, has the smallest number, and this smallest number coincides with Sn .
Proposition 2 (Allahverdiev) .
Another characterization of s-numbers will be obtained later, after the introduction of Hilbert adjoint operators (see Proposition 6.2. 13) . The following proposition partly justifies the term "compact operator" .
Suppose T : H � K is a compact operator between Hilbert spaces. Then T(BE ) is compact. Proof. Since T(BE ) is totally bounded in K, and K is dense, we only need to verify that T (BE) is closed in K. Suppose L: n s n e� 0 e� is the Schmidt series of our operator. First we show that the set T(BE ) consists of all vectors of the form L: n Sn A n e� , where L: n 1 An l 2 < 1 . Indeed, since the numbers (x , e� ) in the decomposition of the Proposition 3.
3. From Compact Spaces to Fredholm Operators
220
vector Tx in the Schmidt theorem satisfy the Bessel inequality, in the case of II x II < 1 such a vector always has the indicated form. Conversely, every vector of the indicated form is, of course, Tx for x == L: n Ane�, and the latter obviously belongs to BE . Now let y be an adherent point for the set T (BE ) . Since T (BE ) belongs to the closure of the linear span of the vectors e�; n == 1 , 2, . . . , we have < 1. y = L: n en e� for some en E C. Our goal is to verify that L:n l�_;:t n Let N be the number of vectors e� if they form a finite system, and an arbitrary natural number otherwise. Take c > 0; then there exists z == L: n SnAn e� such2 that I An l2 < 1 and l i z - Yll < c. Obviously, l i z - y ll2 == L:n l �n - SnAn 1 . Consequently, for all n in the latter sum we have l �n < !An i + L SnAn l < c . As a corollary, � . Hence Sn Sn
Since c was chosen arbitrarily, this means that clear.
2:::: ;; 1 l �s'l < 1. The rest is n
•
(Later we will be able to obtain a much more general result of that sort; see Proposition 4.2. 17.) *
*
*
From the general categorical point of view the Schmidt theorem is a striking result about classification of morphisms. We recall the notion of weak unitary equivalence of operators on pre Hilbert spaces discussed in Section 1 .4. The classification theorem we suggest as the next exercise is equivalent to the Schmidt theorem and can be considered as one of its formulations. Exercise 1 .
(i) Let T : H � K be a compact operator between Hilbert spaces. Then there exists an ordered family s == (s1 , s 2 , . . . ) of non-increasing positive numbers of finite or countable cardinality m , and Hilbert spaces Ho and Ko such that T is weakly unitarily equivalent to the operator R : l�ffiHo � l�ffiKo, acting on l2 as a diagonal operator T8 and taking Ho to zero. Moreover, the numbers s1 , s 2 , . . . are s-numbers of the operator T, and thus this family of numbers is uniquely defined by the operator. (ii) Two compact operators T : H1 � K1 and S : H2 � K2 be tween Hilbert spaces are weakly unitarily equivalent their families of
4.
221
Compact operators between Hilbert spaces
s-numbers coincide, the space Ker(T) is unitarily isomorphic to Ker(S) , and the space Im(T)j_ is unitarily isomorphic to Im(S) j_ . Hint. (i) Suppose T == L: n sn e � 0 e�, Ho : == Ker(T) and Ko : == Im(T)j_ . Then there is a unique U1 : H � l2ffiHo taking e� to pn and the identity on Ho. Also there exists U2 : K � l2ffiKo with similar properties. The operators ul and u2 make the required diagram commutative. (ii) ===> . If U1 and U2 are unitary operators establishing the weak unitary equivalence between S and T, then they are the required isomorphisms between the kernels and the orthogonal complements of the images. Further, if T L: n sn e� 0 e� , then U1 e� , U2 e� , and the very same Sn play the analogous role for S. After that Proposition 1 works. (ii) {:::=:: . By (i) , both operators are weakly unitarily equivalent to the same operator. ==
Thus, every compact operator between Hilbert spaces is uniquely defined up to the weak unitary equivalence by the following data: a) a (finite or infinite) sequence s1 > s 2 > · · · of s-numbers; b) the Hilbert dimension of its kernel, and c) the Hilbert dimension of the orthogonal complement of its image (see Theorem 2.2.2) . In other words (cf. general discussion in Section 0.4) , the complete system of invariants of the weak unitary equivalence for this class of operators consists of triples (s, a , /3) , where s is a non-increasing sequence of positive real numbers tending to zero (or finite) , and a and f3 are cardinal numbers. A simplest model (Section 0.4) of a compact operator with the invariant (s, a , /3) is R : l2ffiHo � l2ffiKo. Here m is the greatest natural number for which Sm > 0, or the countable cardinality if there is no such number. Further, Ho and Ko are Hilbert spaces with Hilbert dimensions a and f3 respectively (say, Ho : == l 2 ( X ) and Ko : == l 2 (Y) , where X has the cardinality a , and Y has the cardinality /3) , and R acts by the formula presented in the previous exercise. *
*
*
Certainly, the operator between Hilbert spaces is finite-dimensional its set of s-numbers is finite. Other conditions on its s-numbers distinguish other classes of compact operators. Here is one of the most important cases.
Schmidt operator (or a Hilbert-Schmidt operator) if its s-numbers satisfy the condition L: n s; < oo . The Schmidt norm of such an operator is l i T l i : == JL: n s; . The set of Definition 2. A compact operator T is called a s
Schmidt operators is denoted by S(H, K) , and we write S(H) instead of S(H, H) . The class of operators we have just introduced can also be defined with out appealing to s-numbers. For clarity, we will show this in the case where
3. From Compact Spaces to Fredholm Operators
222
both spaces are separable and infinite-dimensional. Until further notice, we assume that our spaces have these properties. If a given operator is finite-dimensional, so that its "Schmidt series' ' is a '\:"" nN= 1 sn e In O e IIn , we can arb"t1 rar1" 1y augment the vectors eI1 , . . . , eIN fin1"te sum L...to a countable orthonormal system in H by vectors e � + 1 , e � + 2 , . . . , and the vectors e 7 , . . . , e'Jv, to a countable orthonormal system in K by vectors e'Jv+ 1 , e'Jv+ 2 , . . . , and put SN + 1 == S N + 2 == · · · : == 0. Then every compact operator, no matter what its image is, is represented as a "true" series 2:� 1 sn e � 0 e� convergent in the operator norm, which we still call the Schmidt series of this operator. This agreement allows us to avoid tedious repetitions related to the case of finite-dimensional operators in the future. Theorem 2.
The following properties of a bounded operator T : H � K
are equivalent: (i) T is a Schmidt operator; (ii) for some orthonormal basis d� , d� , . . . in H we have L: r 1 II Td� II 2 < . 00 '
(iii) for an arbitrary orthonormal basis d� , d�, . . . in H we have
00 L 11 Td� ll 2 < oo; k= 1 (iv) for the matrix ( a kl ) of the operator T with respect to some orthonor mal bases in H and K we have L: r,z = 1 l a kz l 2 < oo (see Definition 1.4. 5); ( v) for the matrix ( a kl ) of the operator T with respect to arbitrary or thonormal bases in H and K we have L: r,z= 1 l akz l 2 < oo . Finally, if T has these properties, the sums of all these series coincide and they are equal to II T II � {that is, to 2: � 1 s;) . Proof. The implications (iii) ====> ( ii) and (v)====> ( iv) are obvious. Further,
by the Parseval equality (Theorem 1 .2.4(ii) ) , for some orthonormal bases d� , d� , . . . in H and d7 , d� , . . . in K we have II Td� 11 2 == L: r 1 I ( Td� , dt ) 1 2 , and therefore (1)
00
00
00
k= 1
k ,l = 1
k l-- 1 '
From this we have the equivalences (ii){::::::> (iv) and (iii) {::::::> (v) . Now to prove the equivalence of all the five conditions we only need to justify the implications (i) ====> (iii) and (ii)====> (i) . We do this in several steps.
4.
223
Compact operators between Hilbert spaces
Suppose that T is a compact operator with Schmidt series 2:: � 1 sn e � 0 e� . Then for every orthonormal basis d� , d� , . . . in H the num bers 2: � 1 s; and L: C: 1 11T d� ll2 simultaneously either exist or do not exist, and if they exist, they coincide. Proof. For vectors Td� and e� from the corresponding orthonormal systems
Lemma 1.
we have
( Td� , e � ) = ( L sm ( d� , e � ) e � , e � ) = L sm ( d� , e � ) ( e � , e � ) = s n ( d� , e �) . m m Taking into account that 11 Td� ll2 == 2:: � 1 I ( Td� , e�) l 2 , we see that the sums L: C: 1 11 Td� ll2 and L:r,n = 1 s; l ( d�, e �) l 2 are simultaneously finite or not, and if they are finite, they coincide. Let N be an arbitrary natural number. Then, by the Parseval equality, oo N N N
L L s� i ( d� , e�) l 2 = L s � ll e� ll 2 = L s� .
n= 1 n= 1 n= 1 k = 1 This implies that L:C:n= 1 s; l ( d� , e�) l 2 < oo 2:: � 1 s; < oo, and these sums coincide. The rest is clear. • '
For a bounded operator T H � K and an orthonormal basis d� , d� , . . . in H we have l iT II < JL: C: 1 II Td� 11 2 .
Lemma 2.
:
( Certainly, this lemma is interesting only if the latter sum is finite. ) Proof. For every x
E
H we have Tx == L: C: 1 ( x , d� )Td� . Hence, taking into
account the Cauchy-Bunyakovskii inequality and Parseval equality, we have 00
II Tx ll < L i ( x , d� ) I II Td� ll k= 1
The rest is clear.
00
00
00
k= 1
k= 1
k= 1 II
If for some orthonormal basis d� , d� , . . . in H we have the in equality L: C: 1 II Td� II 2 < oo, then T is compact. Proof. For every k, define the finite-dimensional operator Tk by the formula x � 2:: 7 1 ( x , d�) Td� . ( Thus, Tk coincides with T on the first k vectors of our basis and takes the remaining vectors to zero. ) Then L: r 1 II ( T - Tk) d� ll2 == L: r k+ 1 11Td� ll2 . Hence, using the previous lemma for T - Tk as T, we obtain Lemma 3.
that T == limk� oo Tk in the operator norm. Therefore, T is approximated • by finite-dimensional operators and thus ( Corollary 3.2) is compact.
3. From Compact Spaces to Fredholm Operators
224
End of the proof of Theorem 2. It remains to note that the implication
(i)===> (iii) is contained in Lemma 1, and the implication (ii)===> ( i) follows from this lemma together with Lemma 3. Thus, the proof of the equivalence of the conditions (i)-(v) is completed. As for the last statement of the theorem about the equality of indicated sums, it evidently follows from the • same Lemma 1, taking into account formula ( 1 ) . Remark. We see (Theorem 2 (v) ) that the Schmidt operators are completely
described in terms of matrices. In fact, this is the property characterizing this particular class of operators; nobody knows how to describe other classes of operators in terms of matrices.
4. (i) S(H, K) is a subspace in JC(H, K) , and thus a linear space; (ii) II · l i s is a Hilbert norm in S(H, K) . The inner product ( - , ·) gener ating this norm (and unique by the polar identity) is as follows: if we take arbitrary orthonormal bases in H and K, then for Schmidt operators S and T we have
Proposition
00
L a k t b kt , k ,l = l where ( akz) and ( bkz) are the matrices of these operators with respect to these bases ; (iii) (S(H, K) , II · l i s ) is a Hilbert space. ( S, T)
=
Proof. Choose orthonormal bases in H and K, and assign to every com
pact operator from H to K the matrix of this operator with respect to these bases. Since our matrices are functions of two natural arguments, we obtain a mapping from JC (H, K) to the space of all double sequences, and this mapping is evidently an injective linear operator. By Theorem 2, this operator is a bijection (3 : S(H, K) --+ l 2 ( N x N) , and this obviously implies (i) . Further, from the same Theorem 2 it follows that for the standard norm II · l l 2 in l 2 ( N x N) we have l l f3(T) l l 2 l i T l i s · Certainly, this means that II · l i s is a norm in S(H, K) , and (3 is an isometric isomorphism between normed spaces (S(H, K) , II · l i s ) and ( l 2 ( N x N ) , II · l l 2 ) . But the norm in the latter space is generated by the inner product (a , b) :== L: rz -- 1 a kzbkz; this implies (ii) . Finally, from the fact that l2 (N x N) is a Hilbert space and (3 is an isometric isomorphism, we obtain (iii) . • ==
'
Remark. Later, when Hilbert adjoint operators will be introduced, we will
obtain another characteristic property of Schmidt operators and a formula for the inner product in S(H, K) that is not connected with concrete or thonormal bases (see Corollary 6.2.2(i) in the sequel) .
4.
225
Compact operators between Hilbert spaces
It turns out that the Schmidt operators acting on the space L 2 [a, b] are already known, but under a different name.
5. An operator on L 2 [a, b] is a Schmidt operator � it is an integral operator (see Example 1 . 3. 6). The mapping K � TK , where TK is the integral operator on L 2 [a, b] with kernel K, is a unitary isomorphism between Hilbert spaces L 2 ([a, b] [a, b] ) and S(L 2 [a, b] ) .
Proposition
x
2:: � 1 s n e� 0 e� is our operator; now the vec tors e� and e� are square-integrable functions. Put un ( r, t) : == e� (r)e�(t); r, t E [a, b] . From the Fubini theorem it evidently follows that Un is an orthonormal system in L 2 (D), where D : == [a, b] [a, b] . Since the series 2:: � 1 s; converges, the series 2:: � 1 sn Un converges in L2 (D) to a function K == K (r , t) . Consider the integral operator TK with kernel K, and for an arbitrary natural number N consider the integral op erator TKN with kernel KN == 2:: � 1 S n Un · Then for every x E L 2 [a, b] for almost all r E [a, b] we have Proof. ===> . Suppose
T
==
x
N
1a nL= Sn un (r, t)x(t)dt 1 = L s n e �(r) 1 e� (t)x(t)dt = L s n ( x, e� ) e �(r). a n= n=
[TKN (x)] (r) =
b
N
b
1
N
1
This means that TKN (x) == 2:: � 1 s n ( x, e� ) e� , i.e. , TKN == 2:: � 1 s n e� 0 e� . Now we see that by the Schmidt theorem, T == limN-H)() TKN , and at the same time, the estimate II TK - TKN II < I l K - KN II ( see again Example 1.3.6) shows that TK == limN�oo TKN . Thus, T == TK. ¢== . Let TK be an integral operator with kernel K. Choose two ar bitrary orthonormal bases { e� , e� , . . . } and { e�, e�, . . . } in £ 2 [a, b] . Then the matrix of the operator TK with respect to these bases is, obviously, akl :== f0 K(r, t)ukz (r, t)dt dr, where ukz (r, t) : == e % (r)e� (t) . On the other hand, U kl is an orthonormal basis in £2 (D) ( the arguments we used in the proof of Theorem 3. 1 work with obvious modifications ) , so that the numbers akl are the Fourier coefficients of the element K E2 £ 2 (D) with respect to this basis. Therefore, by the Parseval equality, II K II == L: �m= 1 l anml 2 . Apply ing Theorem 2, we see that TK is a Schmidt operator, arid II TK II s == II K II . II The rest is clear. Corollary 1.
( i ) An operator between two separable Hilbert spaces is a
Schmidt operator � it is weakly unitarily equivalent to an integral operator on L 2 [a, b] .
3. From Compact Spaces to Fredholm Operators
226
( ii )
An operator acting on a separable Hilbert space is a Schmidt opera tor � it is unitarily equivalent to an integral operator on L 2 [a, b] .
This corollary can be viewed as an analytical description of the Schmidt operators. We give another interpretation of the Schmidt operators, this time as elements of Hilbert tensor products. For a vector x E H, denote by x* E H * the functional defined by this vector. Recall that for a Hilbert space H the dual space H* is again a Hilbert space with respect to the inner product ( x*, y *) :== (y, x ); x, y E H ( see Proposition 2.3.8) .
There is a unitary isomorphism between the Hilbert spaces H* @ K and S(H, K) taking an elementary tensor x* Q9 y to the one-dimen sional operator x 0 y ; x E H, y E K. Theorem 3.
Proof. The space
H* @ K contains a dense subspace H* Q9 K, i.e. , the
algebraic tensor product of H* and K. The space S (H, K) contains the subspace :F(H, K) , which is again dense because every Schmidt operator is a limit in the Schmidt norm of partial sums of its Schmidt series. Consider the linear isomorphism Gr8 : H * Q9 K � F(H, K) ( see Proposition 2.7.2) . Take u E H * Q9 K; u == 2:: �= 1 x 'k Q9 Yk and choose an orthonormal basis ei, . . . , e :n in span { xi, . . . , x�}. Then u == 2:: � 1 e 'k @ Zk for some z1 , . . . , Zm E K. From the definition of the inner product in H* Q9 K ( see Section 2.8) it follows that the system e 'k Q9 zk ; k == 1 , . . . , m is orthogonal. Therefore, ll u ll 2 == 2::� 1 ll e'k Q9 zk ll 2 == 2:: ; 1 llzkll 2 . At the same time T :== Gr8 (u ) == 2:: � 1 e k 0 Zk · Since Tx == 0 for every x _L span { e1 , . . . , em }, from Theorem 2 it follows that l i T II � == 2:: � 1 II Te k 11 2 == 2:: � 1 II Zk 11 2 . Thus , Gr8 is an isometric, and hence a unitary isomorphism between dense subspaces in H* @ K and S(H, K) . It remains to use the extension-by-continuity principle • ( in the form described in Proposition 2. 1 . 10) . Now, consider another special class of compact operators. In the follow ing definition H and K are arbitrary Hilbert spaces.
nuclear operator ( or a trace class operator ) if for its s-numbers we have L: n Sn < oo . The nuclear norm of such an operator is the number II T II N :== L: n Sn . The set
Definition 3. A compact operator T : H � K is called a
of nuclear operators is denoted by N(H, K) , and we write N(H) instead of N(H, H) . Here the advanced reader may notice that we have already used the same notation and the same name in another context ( cf. Definition 2. 7. 5) , where the s-numbers were not even mentioned. We will soon show that the two definitions of nuclearity agree. However, for now we pretend that we forgot about the definition in Section 2 . 7 and, speaking about nuclearity, we have in mind the definition just given.
4.
227
Compact operators between Hilbert spaces
Again, for clarity we restrict ourselves to the case where our Hilbert spaces are separable and infinite-dimensional.
Let T be a nuclear operator acting on a Hilbert space H, and 2: � 1 sn e� 0 e� its Schmidt series. Suppose e1 , e 2 , . . . is an orthonormal basis in H and ( akz ) is the matrix of the operator T in this basis. Then the series L:r 1 akk {composed from the diagonal elements of our matrix) absolutely converges, and its sum is equal to 2: � 1 Sn ( e�, e� ) . In particular, this sum does not depend on the choice of orthonormal basis.
Proposition 6.
Proof. Consider the series L: �k = 1 s n ( e k , e� ) ( e�, e k ) · It is absolutely conver'
gent since L:r 1 I ( e k , e� ) I I ( e�, e k ) I is the inner product of two vectors in l2 of norm 1 ( modules of the corresponding Fourier coefficients ) and L: n Sn < oo . Hence, the corresponding double series are absolutely convergent, and their sums coincide. By the equality Te k == 2:: � 1 Sn ( e k , e� ) e� we have
f; f� sn (ek , e�) (e� , ek ) f; (f� sn (ek , e�) e� , ek ) f; (Tek , ek) · =
=
At the same time, decomposing e� and e� with respect to the basis e k ; k == 1 , 2, . . . , we see that L:r 1 ( e k , e� ) ( e� , e k) == ( e� , e� ) , and therefore summa tion in another order gives the number 2:: � 1 s n ( e� , e� ) , which does not depend on e k . The rest is clear. • Proposition 6 makes the following notion well defined.
4. The sum of the diagonal elements of the matrix of a nuclear operator T in an arbitrary orthonormal basis of the space H is called the trace of this operator and is denoted by tr (T) . Definition
Again, we will soon show that this definition agrees with the definition of trace in Section 2.7.
Let T be a compact operator and S a bounded operator on H, and suppose ST and TS are nuclear operators. Then tr (ST) == tr (TS ) {the trace of a product does not depend on the order of factors).
Proposition 7.
Proof. Suppose
T == 2:: � 1 sn e� 0 e�. Then for every n we have 00
sn ( Se � , e �) = ( 2 J s k Se�, e� ) e % , e� ) = ( TSe� , e� ) . k= 1 Since ST(x) == 0 for every x _L span {e� : n E N} , we have tr (ST) 2:: � 1 ( STe�, e� ) . On the other hand, Im (TS) C span {e� : n E N} , hence tr (TS) == 2:: � 1 ( TSe�, e� ) . It remains to sum over n the first and the last ( STe � , e�)
=
expressions in the above-written chain of equalities.
II
3. From Compact Spaces to Fredholm Operators
228
The following facts about nuclear operators should be known to all read-
ers.
( i ) N(H, K ) is a subspace ( [55, Chapter III, §§8, 9] ) . in S (H, K ) {hence, a linear space), and II · li N is a no rm on N(H, K ) . Moreover, N(H, K ) is a Banach space with respect to this norm. ( ii ) The composition of two Schmidt operators, as well as the compo sition {in any order) of a nuclear and a bounded operator, is a nuclear operator. ( iii ) The trace is a unique (up to a scalar factor) functional f on N(H) that is continuous in the nuclear norm and satisfies the condition f(ST) == f(TS) for all one-dimensional operators S and T.
Proposition 8
Some of these results will be proved later in "small print" . But there are many things the reader can establish right now as a useful exercise ( and without special effort ) . Here is some
2:: � 1 sn e�
Information to think over. If T == 0 e� E N(H, K ) , then- for every orthonormal systems d� in H and d% in K the double se ries (d� , e� ) (e% , d�) is absolutely convergent. Changing the order of summation, we see that I (Td� , d%) I < II T II N· Therefore, if S, T E N(H, K ) and S + T == s�d� 0 d% , then == ((S + T ) d� , d%) < II S II N + II T II N·
I:� 1 L:r 1 sn
I:� 1
L: r 1 L: r 1
I:� 1 s�
If T E S (H , K ) , S E S ( K, L) , ST == 2:: � 1 s n e� 0 e� , and en is an orthonormal basis in K, then sn == L:r 1 ank bkn , where ank :== ( Se k , e� ) and bkn : == ( Te�, e k ) · Theorem 2(v) implies that the sum of the double series L:�k = 1 ank bkn is the inner product of the double sequences in l2 (N x N) . Suppose T E N(H, K ) , S E B ( K, L) , ST == 2:: � 1 sn e� 0 e�' , and T == I:� 1 t k f£ 0 f�, where L:r 1 t k < oo . Then it is easy to see that I:� 1 s n I: � 1 L:r 1 t k ( e� , !£ ) \Sf�, e�') . Let us change the order of summation. For every k from the Cauchy-Bunyakovskii inequality we obtain the estimate 2: � 1 l ( e�, f£ ) 1 1 ( S j�, e�' ) l < II S II · Hence, 2:� 1 sn < II S II L:r 1 t k , so that ST E N(H, L) and II ST II N < II S II II T II N· Finally, for every x, y E H it is easy to see that tr ( x 0 y ) == ( y, x ) . Let the functional f : N(H) --+ C have the properties from ( iii ) . Then, multiplying some one-dimensional operators in a different order, as described in Proposition 1.5.5, we see that f( xO y ) == f (z O z ) ( y, x ) == f(zO z ) tr ( x O y ) for all x, y, z E H, li z II == 1. Thus, if we take z E H; li z II == 1 and put A : == f(z 0 z) , then f == A ( tr ) on F(H) , which is a dense subspace in (N(H) , II · li N ) , as follows from the form of the Schmidt series. '
==
4.
Compact operators between Hilbert spaces
229
Now we state a fundamental theorem describing the action of functionals on the introduced spaces of compact operators. Theorem 4 (Schatten-von Neumann, [68, Chapter II, §1] ) . Let H and K
be (arbitrary} Hilbert spaces. Then (i) Every nuclear operator T : K --+ H defines a bounded functional fr on the space JC(H, K) (with the operator norm) by the rule fr (S) : == tr (ST ) , and every bounded functional on JC(H, K) has the form fr for a unique T E N(K, H) . The resulting bijection IK : T �----+ fr is an isometric isomorphism of the space (N(K, H) , II · l i N ) onto JC(H, K)*. (ii) Every bounded operator T : K --+ H defines a bounded functional fr on the space N(H, K) {with the nuclear norm} by the rule fr (S) : == tr (ST) , and every bounded functional on (N(H, K) , II · l i N ) has the form fr for a unique T E B(K, H) . The resulting bijection IN : T �----+ fr is an isometric isomorphism of the space B(K, H) {with the operator norm) onto (N(H, K) , II . l l N )*. (iii) Every Schmidt operator T : K --+ H defines a bounded functional fr on the space S(H, K) {with the Schmidt norm} by the rule fr(S) : == tr (ST ) , and every bounded functional on (S(H, K) , II · l i s) has the form fr for a unique T E S(K, H) . The resulting bijection Is : T �----+ fr is an isometric isomorphism of the space (S(K, H) , II · l i s) onto (S(H, K) , II · l i s)*. Certainly, the fact that (S(K, H) , II · l i s) and (S(H, K) , II · l i s )*) are iso
metrically isomorphic, immediately follows from the Riesz-Fischer theorem: both spaces are separable and Hilbert. It is essential that the operators in S(K, H) define functionals on S(H, K) precisely by the formula using the trace. As to the second of these results, later it will turn out to be a simple corollary of the interpretation of the space of nuclear operators in terms of Banach tensor products . We discuss this at the end of this section (see Exercise 2) .
Note the inner similarity between these three assertions and the de scription of actions of functionals on the most important function spaces (Exercises 1.6. 1-1.6.3) . The space JC(H, K) behaves like c0 , N(H, K) like l 1 , and S(H, K) like l 2 . Actually it is advisable to treat each of these spaces of operators as an operator version of the corresponding space of sequences; then many features in their behavior become predictable. By the way, we have already observed another example of the similarity: the composition of two Schmidt operators is a nuclear operator, and this is an analogue of the fact that the coordinatewise product of two sequences in l 2 belongs to l 1 . Remark. If we restrict ourselves to diagonal operators on l 2 (Example 1.3.2) , then, informally speaking, the Schatten-von Neumann theorem turns into a unification of Exercises 1.6. 1-1.6.3. Indeed, for an arbitrary (i.e. , just
3. From Compact Spaces to Fredholm Operators
230
bounded) diagonal operator T>.. , A can be an arbitrary sequence in l 00 • At the same time, a diagonal operator is a Schmidt operator {=:::} A E l 2 , and it is a nuclear operator {=:::} A E l 1 . Moreover, tr(T>.. T11) , whenever this number is defined, coincides with 2:: � 1 An f.1n , i.e. , the same sum that was used for the description of functionals on the spaces of sequences. The spaces lp for other p E [1 , oo ) also have an operator version: they are the so-called Schatten-von Neumann classes of order p. Namely, by def inition, a compact operator T : H � K belongs to this class if its s-numbers satisfy the condition 2:: � 1 s� < oo . Knowing the spaces lp and their rela tionships, one can predict many things about their operator versions: for instance, the composition ST, where S and T belong to the Schatten-von Neumann classes of order p and q , respectively, is a nuclear operator if 1 Ip + 1 I q == 1. In addition, operators of order q act "as functionals gen erated by the trace" on the class of order p, etc. For details about these classes (that go beyond the scope of this book) ; see, e.g. , [55] or [56] . Advanced readers should also know about an interpretation of nuclear operators be tween Hilbert spaces in terms of tensor products. This is an analogue of Theorem 3 proved earlier. Besides , such readers should know some corollaries of this fact. In particular, we will show that in the context of Hilbert spaces the definitions related to nuclearity are equivalent to "general Banach" definitions in Section 2 . 7. Warning. However, before this has been done, we give the nuclear operators and nuclear norms the meaning indicated in this section ( see Definition 3) . For simplicity we assume, as before, that the Hilbert spaces H and K are separable and infinite-dimensional ( so that we can always speak about infinite Schmidt series ) . In what follows , for each x E H we denote by x* the corresponding functional on H acting by the formula x* (y) = (y, x) ( see Section 2.3) . Theorem 5. The set N(H, K) of nuclear operators ts a subspace tn K(H, K) , and the
nuclear norm I I · I IN ts a norm there. Moreover, there ts an tsometrtc tsomorphism between the space (H* 0 K, l l · l l p ) (where I I · I I P ts the proJecttve norm defined tn Section 2. 7) and the space (N(H, K) , I I · I I N ) , taking an elementary tensor x* @ y to a one- dimensional operator x O y ; x E H, y E K . Proof. Let Gr : H* 0 K ---+ B(H, K) be the Grothendieck operator defined in Section 2 . 7. It acts on the elementary tensors in the way we indicated. As we will see, our desired ison1etric isomorphism will be the corestriction of this operator to its image. Lemma 1 . Suppose u E H* 0 K and S
:=
e� , e� , . . . tn H and e � , e � , . . . tn K we have
Gr (u ) . Then for every orthonormal system
L I ( S e� , e � ) l
n= 1
.
Suppose Ker(S) == 0, but Im(S) # E. Put En : == Im( Sn ) ; n == 1 , 2, . . . . Then, obviously, for all n we have En + I == S (En ) C En , and every En is an invariant subspace of the operator S, and consequently, of T as well. Take an arbitrary x E E \ E1 . Since S is injective, Sx does not belong to the set of vectors of the form Sy; y E E1 . This means that E2 # E1 . The same argument applied to a vector in E1 \ E2 shows that E3 # E2 . Similarly, E4 # E3 , and eventually, En +I is a proper subspace in En for every n. Let Sn , Tn : En � En be the birestrictions of operators S and T to En . Evidently, Sn == lEn - Tn , and Tn is compact together with T. By Theorem 2, E1 is closed, hence is a Banach space. Therefore, S1 satisfies the assumptions of Theorem 2, and this guarantees that E2 is closed. Continuing these arguments, we establish that all En are closed. Now, using the lemma about near-perpendicular, we take in each En a vector Xn such that l l x n ll == 1 and d(x n , En +I ) > 1/2. Then, obviously, Tx n - Tx n + l == X n - z , where z == Sx n + X n +l - Sx n +I E En +l · Since d(Tx n , Txn +I ) > 1/2 and, by Proposition 2.4(iii) , the set {Tx n ; n E N} is not totally bounded, and we obtain a contradiction with the fact that T is compact. n � - Now assume that Im(S) == E, but Ker(S) # 0. Put E :== Ker( Sn ) ; n == 1, 2, . . . . Then for all n we have En+ I == {x E E : Sx E En } and En C En +I . Taking x E E 1 \ 0 and using the surjectivity of S, we see that x == Sy for some y , which necessarily lies in E2 \ E 1 . Similarly, y == Sz for z E E3 \ E2 , and so on. Continuing these arguments, we see that En is a proper subspace in En + I for every n. Since S is continuous, the inverse image of every closed set is closed. Successively applying these arguments, we establish that all En are closed. Using the lemma about near-perpendicular, in every En ; n > 2 we take a vector Xn such that l l xn l l == 1 and d(xn , En - I ) > 1/2. Then the same Proof.
4 An alternative in mathematics, well in real life, means an obligatory choice between two incompatible possibilities ( for example, "To be , or not to be?" ) . as
as
3. From Compact Spaces to Fredholm Operators
238
arguments as at the end of the �-part, give d(Txn , Tx n - 1 ) > 1/2. Thus, • we obtain a contradiction with the compactness of T. Theorem 4. The index of the operator S {which, as we already know, is
Fredholm} is zero. Proof. Let E 8 be an arbitrary linear complement of Im(S) in E. It is finite-dimensional together with Ker(S) , and thus closed. Further, let Es and P be the same subspace and projection as in the proof of Theorem 2, and Q : == lE - P. Obviously, Q is a projection onto Ker(S) along Es , which is bounded together with P.
For every operator R : Ker(S) � E the operator S : E � E : x �----+ SPx + RQx has the form lE - T , where T is a compact operator.
Lemma.
A
A
A
A
Proof. Put T : == lE - S. This operator acts by the formula x �----+ x A
- RQx. Thus, taking into account that x == Px + Qx, we have T == TP + Q - RQ. By Proposition 3.5, the operator TP is compact together with T, whereas Q and, as can be easily seen, RQ are bounded finite-dimensional • operators; thus, they are compact as well. The rest is clear. SPx
End of the proof of Theorem 4. Combining the lemma with the Fred
holm alternative, we see that for every R : Ker(S) � E 8 the corresponding operator S is injective {=:::} it is surjective. But as one can easily verify, Ker(S) == Ker(R) , and Im(S) == Im(S) EB Im(R) . Hence, if Ind(S) > 0, or, in other words, dim Ker(S) > dim E 8 , then, we can take a surjective but not in jective R, and we obtain that Ker(S) # 0, and at the same time Im(S) == E. On the other hand, if lnd(S) < 0, or in other words, dim Ker(S) < dim E8 , then, taking an injective but not surjective R, we obtain that Ker(S) == 0, and at the same time Im(S) # Im(S) EB E 8 == E. In both cases we obtain a • contradiction. So the only possibility is that Ind(S) == 0. A
The following exercise generalizes the above theorem. Exercise 4. Suppose an operator S : E � F between Banach spaces has the form I - T, where I is a topological isomorphism, and T is compact. Then S is a Fredholm operator, and lnd(S) == 0. Hint. The operator S :== I - 1 S is a compact permutation of the identity operator. It has the same kernel as S, and I is an isomorphism between the images of these operators. One of the most intriguing problems of geometry of Banach spaces , open as of now, 5 is as follows: are there Banach spaces E "so pathological" that B(E) = span ( l , K(E) )? By the theorems proved above, every operator acting on such a hypothetical space is either compact or Fredholm of index 0. 5 May 2 4 , 2 00 1 .
5. Fredholm operators and the index
23 9
None of the concrete Banach spaces mentioned in this book has this property. To make sure of this, it is sufficient to find a Fredholm operator on a given space with a non-zero index, or an operator which is neither compact, nor Fredholm. In the role of the latter one we can take a projection for which both the kernel and the image are infinite-dimensional. In our concrete spaces we have plenty of such projections. The difficulty in this problem is that , as we already know, there are Banach spaces so complicated that they have no projections with that property. In other words (see Corollary 2.4.3) , they do not have closed infinite-dimensional subspaces with an infinite dimensional closed linear complement ( ! ) . About this group of problems, see , e.g. , [59] .
Now let us pay homage to those questions of analysis that eventually led to Theorems 2-4. The equation of the form (1)
x(s)
- 1b K(s, t)x(t)dt = y(s) ,
where K E L 2 (D) and y E L2 [a, b] are given functions and x E L 2 [a, b] is a known function, is called ( since the 19th century ) an integral equation of the second kind. 6 ( Such equations are considered in function spaces other than £ 2 [a, b] , but we restrict ourselves to the latter spaces. ) If y == 0, then the equation is called homogeneous ; certainly, it has the form (2)
x(s)
- 1b K(s, t)x(t)dt = 0.
From the algebraic point of view the considered equation is a special case of an operator equation. This is the name for an equation of the form Sx == y, where S : E � F is an operator between two linear spaces, y a given vector in F, and x an unknown vector in E. For y == 0 such an equation is called a homogeneous operator equation. Certainly, in the case we consider here we have E == F == L2 [a, b] , and S == 1 -T, where T is an integral operator on L2 [a, b] with kernel K(s, t) . When one considers operator equations in general, and the integral equa tions of the second kind in particular, two typical questions arise: 1 ) For which y in the right-hand side does our equation have a solution, and if it does, what is the description of the set of all solutions? 2) What is the set of solutions of a homogeneous equation; in particular, does a homogeneous equation have a non-zero solution? Obviously, both questions have adequate translations into the geomet rical language of the theory of operators: 1 ) What is the image of S, and what is the full inverse image of every vector y ? 6 Integral equations of the first kind are equations of the form are more complicated, and we will not discuss them.
J: K( s , t) x (t)dt == y ( s ) . They
240
3. From Compact Spaces to Fredholm Operators
2) What is the kernel of S, and, in particular, is S an injective operator? These two questions are closely related to each other: obviously, if y E Im(S) and y == Sx for some x E E, then the full inverse image of y (i.e. , the set of all solutions of the corresponding operator equation with y as the right-hand side) is {x + z : z E Ker(S) } . The following theorem in the theory of integral equations, which we deliberately formulate in an old-fashion style, is actually a special case of Theorems 2-4. Theorem 5 (sometimes called the triple Fredholm theorem) .
Let {1} be an
integral equation of the second kind. Then (i) There is a finite family X I ( s) , . . . , Xm ( s) of linearly independent so lutions of the homogeneous equation {2} such that every solution of (2} is a linear combination of solutions X I ( s) , . . . , Xm ( s) . (ii) There is a finite family ZI ( s) , . . . , Zn ( s) of linearly independent func tions in L 2 [a, b] such that the right-hand sides y(s) for which the equation (1) has at least one solution are precisely those functions for which J: y(t)zk (t)dt == 0 for k == 1 , . . . , n . (iii) m == n {i. e., the number of solutions in {i} coincides with the num ber of linearly independent functions in {ii}}.
Thus , integral equations of the second kind, being an evident object of infinite dimensional analysis, behave surprisingly similarly to finite systems of linear equations with the same number of variables. Major consequences for the entire mathematics arose from the fact that Hilbert was among people surprised by the Fredholm paper ( cf. , e.g. , [38, p. 22 1] ) .
H :== L 2 [a, b] , consider the operator S :== 1 -T, is the integral operator with kernel K(s, t) . Clearly, part (i) is
Proof. In the Hilbert space
where T equivalent to the statement that the kernel is finite-dimensional; therefore, this is a direct corollary of Theorems 2 and 3. 1. Clearly, m == dim Ker(S) . Let us now "modify" part (ii) . We do this in several steps. The first equivalent formulation is obvious: - There is a linearly independent system of vectors ZI , . . . , Zn in H, such that Im(S) == (span{zi , . . . , zn } )l_ . Since the subspace in the right-hand side of the latter equality is closed, by Proposition 2.3.4 we can express this as follows: - The image of the operator S is closed, and its orthogonal comple ment is finite-dimensional. Finally, establishing the equality dim Im(S) j_ == dim H/ Im(S) Propo sition 2.3.6 shows that if we take into account Proposition 1, the latter
5. Fredholm operators and the index
241
statement is equivalent to the fact that codimH Im(S) is finite. Therefore, the indicated codimension is precisely n. Hence part (ii) also follows from Theorems 2 and 3. 1 , and (iii) follows from these theorems together with Theorem 4. • Remark. In fact, Fredholm made another important contribution, antici
pating a valuable fact we present in Exercise 8. He connected the solution of integral equation (1) with the solution of the so-called adjoint integral equation, where the kernel K(s, t) is replaced by K* (s, t) : == K ( t , s) . Later we will speak about this in the general context of Hilbert adjoint operators; see Proposition 6. 1. 10. We now recall the category Banj/C introduced at the very beginning of this section: we will keep the promise we gave there.
6 (S. M. Nikol ' skii) . Let S : E --+ F be an operator between Banach spaces. Then S is a Fredholm operator {=:::} there exists a bounded operator R : F --+ E such that RS == lE - T1 , where T1 E /C(E) , and SR == lp - T2 , where T2 E /C ( F ) . {In other words, S is Fredholm {=:::} its coset S + JC(E, F ) is an isomorphism in Banj/C). Moreover, the operator R can be chosen in such a way that T1 and T2 are finite-dimensional.
Theorem
�-
Suppose Es is a closed linear complement of Ker(S) in E (cf. proof of Theorem 2) , P1 is a projection onto Es along Ker(S) , Q 1 : == 1E - P1 , Fs is a linear complement of Im(S) in F, P2 is a projection onto Im(S) along Fs, and Q 2 : == lp - P2 . All these projections (cf. the same proof) are bounded. Denote by sg : Es --+ Im(S) the corresponding corestriction of the operator S. It is a bijective bounded operator, and since Im(S) is closed, by the Banach theorem sg is a topological isomorphism. Put R : F --+ E : x � (Sg)- 1 P2x. Clearly, RS == P1 , and SR == P2 ; thus RS == lE - Q1 and SR == lp - Q 2 . Since Q 1 and Q 2 are finite-dimensional operators, we obtain the implication � together with the last statement of the theorem. ¢== . According to Theorem 2, RS and SR are Fredholm operators. Consequently, Ker( RS) (containing Ker( S) ) is finite-dimensional. At the same time Im(SR) (contained in Im(S)) has finite codimension in F. The • rest is clear. Proof.
Exercise 5 * . Let Ban/ F be the category defined similarly to Ban/ K, but taking the quotient modulo finite-dimensional operators instead of compact ones. Then for an operator S between Banach spaces E and F, its coset S + K ( E , F) is an isomorphism in Ban/K (i.e. , S is a Fredholm operator) ¢:::::::> the coset S + F ( E , F) is an isomorphism in BanfF.
3.
242
From Compact Spaces to Fredholm Operators
(We see that although there are many more compact operators than finite-dimensional ones, the property of "being an isomorphism up to compact perturbations" is equivalent to the property of "being an isomorphism up to the finite-dimensional perturbations" . )
The Nikol ' skii theorem allows us to expose one more (in addition to Theorem 1) advantage of index compared to dim Ker(S) and codimp lm(S) . Proposition 3 (Stability of the index under compact perturbations) . If S
is a Fredholm operator and T a compact operator between Banach spaces E and F, then S + T is also a Fredholm operator and Ind(S + T) == Ind(S) .
Proof. According to the ::::=:> - part of the previous theorem, there are opera tors R, T1 , and T2 with indicated properties. Then R(S T) == lE - T1 RT and (S T) R == lp - T2 TR. From this, taking into account the ¢=
+
+
+
+
part of the previous theorem, it evidently follows that S + T is a Fredholm operator. Further, from the very same theorem, where R is used as the initial operator, it follows that R is Fredholm as well. Combining Theorem 1 with the fact that the operators RS and R( S + T) satisfy the hypotheses • of Theorem 4, we see that lnd(S) == - lnd(R) == lnd(S + T) . Thus, the subset in B(E, F) consisting of Fredholm operators contains together with each S the whole coset S + IC(E, F) , and the index is constant on this coset. In Section 5.3, where topological properties of the operator composition will be discussed, we will come across another stability property of Fredholm operators, this time with respect to perturbations that are small in the operator norm. There we will find some further facts on the structure of the set of Fredholm operators.
6. A Fredholm operator has index 0
it is represented as a sum of a topological isomorphism and a compact (and even a finite dimensional) operator. Hint. The required finite-dimensional operator T isomorphically maps Ker(S) onto a linear complement of Im(S) . Exercise
{=:::}
We suggest that the reader who has learned about exact sequences (see Section 2.5) , will translate the definition of a Fredholm operator into the language of "Banach homo logical algebra" . Exercise 7° . An operator S an exact sequence
(3)
·
·
·
:
E ---+ F is Fredholm {::=:::::} in the category Ban there is
s f-- O f-- C f-- F f-E f-- K f-- 0 f--
with finite-dimensional C and K
.
Here is the concrete use of this point of view.
·
·
·
5. Fredholm operators and the index
243
Exercise 8 . If S : E ---+ F is a Fredholm operator, then S* : F* ---+ E* has the same property, and dim Ker( S* ) == codimp im( S) , codimE* Im(S* ) == dim Ker(S) . As a corollary, Ind(S* ) Ind(S) . Hint. This is a simple corollary of Theorem 2 . 5 . 1 . ==
-
Exercise 9* . The converse is also true: if S* is a Fredholm operator, then the same is true for S.
Hint. The difficulty is in establishing that Im( S) is closed : if it is done, we can apply Theorem 2 .5 . 1 to the exact sequence of the form (3) , where K :== Ker(S) and C :== F/ Im( S) . Since S* * is a Fredholm operator (see above) and in view of the properties of the canonical embedding a , it is sufficient to show that S* * (E) == S* * ( E1 ) is closed , where E1 is a closed linear complement of E n K in E. Take X n E E1 such that S** (xn ) tends to y E F* * . Suppose Eo is a closed linear complement of K in E* * , P is a projection onto Eo along K , and Q :== 1 - P. If Xn is bounded, everything is fine: we can assume that Qxn tends to some z in the (finite dimensional! ) space K. At the same time, since S** Pxn ---+ y; n ---+ oo, and Im( S* * ) is closed, the Banach theorem shows that Pxn converges in Eo to some z. Therefore X n converges in E 1 , and y == S** (limn-+ex) Xn ) · If, on the contrary, the norms of xn are infinitely increasing, and x� :== Xn / l l xn l l , then S* * x� (i.e., S** Px� ) tends to zero, and the same is true for Px� . Taking into account that dim K < oo, we can assume that Qx� tends to some u E K, and hence x� tends to the same u . We have a contradiction with the choice of E1 . (Perhaps, you will be able to find a simpler proof. )
Chapter 4
Polynormed S p aces , Weak Topologies , and Generalized Functions
1 . Polynormed spaces
Up to now it was sufficient to consider only one norm ( or a prenorm ) on a linear space. We used this structure to describe a wide class of various types of convergence in analysis, like uniform convergence, mean conver gence, mean-square convergence, etc. But there are many other natural important types of convergence that cannot be described in this way. Here are some examples.
C 00 [a, b] of infinitely smooth ( == infi nitely differentiable ) functions on an interval [a, b] . The so-called classical convergence in C 00 [a, b] is defined as follows: a sequence X n tends to X
Example 1. Consider the linear space
"classically" if for every k == 0, 1 , . . . the sequence of kth derivatives x �k ) converges to x ( k ) uniformly on [a, b] . ( As usual, we define the Oth derivative of x as x itself: x ( O) == x). Note that, informally, from this convergence one can derive several types of convergence in the theory of generalized functions ( we shall see this soon ) and in differential geometry. Example 2. Let 0 (]IJ) 0 ) be the linear space of holomorphic functions on the
open unit disk ]IJ)0 in the complex plane. Consider the so-called Weierstrass convergence in 0 (]IJ)0 ) : a sequence Wn tends to w in the sense of Weierstrass if for any closed subset K c ]IJ)0 the sequence of the restrictions wn i K tends
245
246
4.
Polynormed Spaces and Generalized Functions
to w i K uniformly on K. This is the standard convergence used in complex analysis.
c00 (of all sequences) consider the coordi natewise (or simple) convergence: � ( n) tends to � if for every k the sequence
Example 3. In the linear space
n �k ) tends to �k ·
Suppose that the class of convergent sequences in a linear space E is given. Then we say that a prenorm II · II on E generates this convergence if for any sequence X n in E the following is true: Xn converges to x in E in the announced sense {=:::} X n converges to x in the prenormed space ( E, II · II ) . Exercise 1 .
(i) There is no prenorm in C00 [a, b] generating the classical conver gence. (ii) There is no prenorm in 0 (]IJ)0 ) generating the Weierstrass conver gence. (iii) There is no prenorm in c00 generating the coordinatewise conver gence. Hint. In all these situations the limit is unique. So it is sufficient to prove that the convergence cannot be generated by a norm (instead of a prenorm) . (i) If Xn tends to zero classically, then the same is true for x�. So for our hypothetical norm, ll xnll � 0 implies ll x� ll � 0. But look at the sequence
Xn ( t ) :== n leenntt II "
(ii) Similar arguments work in complex analysis: if Wn converges in the sense of Weierstrass, then the same is true for w� . (iii) Take Pn / II Pn ll · However, these three types of convergence, as well as many others that are not generated by one (pre )norm, can be adequately described using a structure of the "next level of complexity" -polynormed spaces. What this actually means is as follows. Definition 1 . A linear space E equipped with a family of prenorms
ll · ll v ;
v
E A, or, to be more precise, a pair (E, ll · l v ; v E A) consisting of a linear space E and a family of prenorms ll · ll v ; v E A on it, is called a polynormed space. 1
(Of course, it would be more accurate to say "polyprenormed" instead of "polynormed" , but let us have a pity on our language.) 1 This notion is related to the notion of locally convex space that can be found in many
textbooks. This connection will be considered later, in one of the remarks.
1.
Polynormed spaces
247
If ( E, ll · ll v ; v E A) is a polynormed space, then any prenorm on E of the form max { II · ll v1 , , II · ll vn }, where v1 , . . . , Vn is a finite set of indices from A, will be called an accompanying prenorm ( for the family II · ll v ; v E A) . Each space ( E, II · II ) , where II · II is an accompanying prenorm, is called an accompanying prenormed space ( for the polynormed space ( E, ll · ll v ; v E A) ) . Of course, ( semi ) normed spaces are special cases of polynormed spaces (they correspond to the family consisting of only one prenorm ) . If the family of prenorms is countable, i.e. , A == N, we say that the space is countably .
.
•
normed.
If E is equipped with a family of prenorms, then the restrictions of these prenorms to a subspace F in E endows F with the structure of polynormed space called a polynormed subspace in E. Now, following reader ' s expectations, we collect a list of examples. Example 4. Take an arbitrary linear space E and equip it with all existing
prenorms on E ( i.e. , all the functions satisfying conditions ( i ) and ( ii ) of Definition 1.1.1). Such polynormed spaces are called the strongest. ( As we shall see soon, their role in functional analysis resembles the role of discrete spaces in topology. )
C 00 [a, b] in Example 1 becomes polynormed when equipped with the family of norms II · l i n , where ll x lln :== max { l x ( k) ( t) l ; k == 0, . . . , n; a < t < b }. ( Thus, the nth norm here is precisely the norm of coo [a, b] as a subspace in en [a, b] .) We should note that this example is Example 5. The space
extremely important for the future theory of generalized functions-it plays the role of an "embryo" of standard spaces of test functions. Example
6. The space O(U) of holomorphic functions on a domain U in the
complex plane becomes polynormed if equipped with the family of prenorms II · I lK ; K E A, where A is the family of all closed bounded subsets in U, and ll w ii K :== max { l w ( z ) l ; z E K} . ( By the way, which of these prenorms are norms? ) These spaces and their multidimensional analogues are polynormed spaces considered in complex analysis. Example 7. The space c00 becomes polynormed if equipped with the family of prenorms II · l i n, where II � l i n : == l �n l ·
8. Consider the linear space B ( E, F) , where E and F are pre normed spaces. In Proposition 1.3.2 we had endowed it with the standard Example
prenorm, but right now let us forget about it and make this space poly normed in the following way. Take A : == E and define a family of prenorms ll · ll x; x E E on B ( E, F) by putting II T II x to be equal to II T (x ) ll , the prenorm of T ( x ) in F . This family of prenorms in B ( E, F) is called the strong-operator family of prenorms and is denoted by so.
4.
248
Polynormed Spaces and Generalized Functions
Remark. The special case of this polynormed space when F == C defines a family of prenorms on the dual space E* . It plays an important role in the theory of weak topologies and generalized functions; this will be discussed in details in subsequent sections of this chapter.
Example 9. The same space B(E, F) can be made polynormed in a different
way. Namely, take A == E x F* and equip B(E, F) with the family of prenorms II · ll x , f ; x E E, f E F* by putting II T II x , J equal to l f(T(x)) l . This family of prenorms in B( E, F) is called the weak-operator family of prenorms and is denoted by wo. The case where E == F == H is a Hilbert space plays a special role here. In this case the canonical bijection between H and H* allows us to consider the index set A :== H x H and the family of prenorms II T II x , y == I ( Tx, y ) l . :
:
Remark. Both families of prenorms in B(H) , so and w o , are indispensable in the theory of operator algebras. In particular, they participate in the definition of one of the most important classes of operator algebras, the so-called von Neumann algebras ( see Definition 6.3.5 below ) . In addition, these families will be useful in Sections 6.6-6.8 in questions related to the spectral theorem.
Other important examples of polynormed algebras will be considered later in this chapter. As we know, every (pre)normed space is automatically (pre)metric, and thus a topological space. As for polynormed spaces, we shall describe a natural way to turn them into topological spaces. (As we shall see later, these topological spaces are not necessarily metrizable.) This procedure resembles the topologization of metric spaces by indicating open balls. Let (E, I · ll v ; v E A) be a polynormed space. Take r > 0, x E E and choose a finite set of indices v1 , . . . , Vn . With every such system we associate the set Ux , vl ,··•,Vn , r == { y E E II Y - x ll vk < r; k == 1, . . . ' n }, i.e. , an open ball of radius r centered at x in the accompanying prenormed space (E, max{ II · ll v1 , , II · ll vn }. Every such set is called a standard open ball in E. Note that Ux , v1 , ... , vn , r == n�= l Ux , vk , r · Further, if M is a subset in E, we call a point x E M an interior point of this set if it is contained in M together with some standard open ball Ux , v1 , .. . , vn , r· Finally, a set U in E is said to be open if every point of U is interior. As you have already guessed, this terminology is justified by the following :
:
•
•
•
The class of open subsets in E defines a topology. This topology contains all the open sets of every accompanying space for E.
Proposition 1 .
Proof. From the evident inclusion
Ux , vl ,·· · ,Vn ,r n UX,J.Ll ,· · ·,J.Lm ,S
�
Ux , vl ,···,Vn ,J.Ll , ·· ·,J.Lm ,t
1.
249
Polynormed spaces
(where all the indices belong to A and t : == min{r, s } ) we see that the intersection of every two (and thus, of every finite number) of open subsets is open. This establishes condition (iii) from Definition 0.2. 1. The rest is clear. • Remark. It is easy to see that the described topology can be given also by
each of the following equivalent definitions:
1) It is the only topology for which the family of standard open sets Ux,v,r is a subbase (see Proposition 0.2. 1). 2) It is the weakest topology for which all the prenorms II · l l v , v E A
are continuous. 3) It .is the weakest topology containing all open sets of all accompany1ng spaces.
The topology defined in this way is called the topology generated by the family of prenorms l l · l l v ; v E A. In particular, the topology in B(E, F ) (and thus in B(H) ) generated by the strong or weak family of operator prenorms, is called, respectively, the strong- or weak-operator topology and is denoted by the same symbol so or wo. Let us formulate the following obvious result.
In a polynormed space E the following holds: the addition E x E � E ( x , y) �----+ x + y is a continuous map of topological spaces; the multiplication by scalars C x E � E : (A, x ) �----+ AX is a contin uous map of topological spaces; if U is an open set in E, then for every M C E the algebraic sum U + M is open in E; in particular, for any x E E the shift U + x is open in E; if U is an open set in E, then for any A E C the dilation AU is open in E. •
Proposition 2.
(i) (ii) (iii) (iv)
:
(In fact, the last two properties follow from the first two, but we shall not need it.) From this already we can see that not every topology on a linear space is generated by a family of prenorms. So it would be timely to ask, how can these topologies be characterized? We have already considered a similar question for prenormed spaces, and the information obtained there will be useful now.
250
4.
Polynormed Spaces and Generalized Functions
E is generated by a family of prenorms � there exists a family of subsets in E satisfying the Exercise 2. A topology defined on a linear space
following conditions: ( i ) every set U in this family is convex, balanced, and contains a non zero vector from each one-dimensional subspace in E; ( ii ) the corresponding system of sets Ux,t :== {x + ty : y E U} (where x E E, t > 0, and U runs through our family ) is a basis of our topology. Hint. Consider Minkowski ' s functionals of these sets ( see Exercise 1.1.6) .
Remark. A linear space equipped with a topology satisfying these two conditions is called locally convex. This term is standard in the literature, but we shall not use it . The difference between the notions of polynormed and locally convex space is of the same sort as the difference between metric and metrizable spaces . Locally convex spaces are special cases of topological vector spaces , the most general class of linear spaces with reasonable topologies. This is the name for a space with a topology such that the algebraic operations are jointly continuous, i.e. , satisfy the conditions (i) and (ii) of Proposition 2. The overwhelming majority of topological vector spaces in analysis are locally convex (i.e., "polynormed" ) . Nevertheless, there are some exceptions. For example, the convergence in measure in the space of measurable functions (see Exercise 1 .6. 10) is the convergence with respect to some metric , and the corresponding topology provides an example of a topological vector space that is not locally convex.
After we defined topology in polynormed spaces, we can discuss a series of standard questions about this topology. We shall see that every topo logical property we consider has an adequate and sufficiently transparent description in terms of prenorms. Let us start with the notion of conver gence.
A sequence X n tends to x in a polynormed space (E, II · ll v ; v E A) � for any v E A we have ll x - X n ll v � 0 as n � oo {in other words, X n tends to x in every accompanying polynormed space).
Proposition 3.
Proof.
===> .
E A and c x - x n v � 0 as n �
For any
v
> 0 we have X n
E Ux,v,c
for sufficiently
large n, i.e. , ll oo . ll Take a neighborhood Ux of x . It contains a standard ball {::::::= . Ux,v1 , . . . ,vn ,r · For every k == 1, . . . , n, ll x - X n ll vk � 0 as n � oo . Hence, for sufficiently large n we have II x - X n II vk < r for every k == 1 , . . . , n. This • means that X n lies in Ux,v1 , . . . ,vn ,r and hence in Ux . In particular, we see that the convergence of a sequence in the poly normed space c oo [a, b] is precisely the classical convergence. The conver gence in O ( U) is the uniform convergence on any compact set ( this type of convergence is called the Weierstrass convergence, by analogy with the case of U == JI)) 0 ) . The convergence in c00 is the coordinatewise convergence.
1.
Polynormed spaces
25 1
Thus, we failed when trying to use a single prenorm ( see Exercise 1 ) , but succeeded with a family of prenorms. Recall that the pointwise convergence in C[a, b] cannot be defined by a metric, to say nothing about a norm ( Exercise 0.2. 1 ) . But this type of convergence can be adequately described in terms of a polynormed space. Indeed, if E is an arbitrary space of functions on a set X, then the pointwise convergence in E is obviously the convergence in the polynormed space ( E, II · II t ; t E X) , where II x II t : == I x ( t) l In the polynormed spaces ( B ( E, F) , so) and ( B(E, F) , wo) ( Examples 8 and 9) the statement that Tn tends to T means precisely that for every x E E ( respectively, for every x E E, f E F * ) T(x n ) tends to T(x) ( respectively, f(Tx n ) tends to f(Tx) ) . In the first case we again obtain the pointwise ( or, maybe we should say here "vectorwise" ) convergence. In addition, note that the convergence of Tn to T in (B(H) , wo) , where H is a Hilbert space with a chosen orthonormal basis, implies the convergence of every matrix entry of Tn to the corresponding matrix entry of T. As for the strongest spaces ( Example 4) , the convergence in these spaces can be described in purely algebraic terms:
E is a strongest polynormed space, X n is a se quence in E, and x is a vector. Then X n tends to x the following two Exercise 3* . Suppose
�
conditions are fulfilled:
( i ) all vectors X n and x belong to a finite-dimensional subspace F of E; ( ii ) for some ( and hence, for each ) basis e k ; k == 1 , . . . , m in F in the n corresponding expansions X n == 2:: � 1 A i ) e k , x == 2:: � 1 A k e k we n have >. i ) --+ >. k ( n --+ oo ) for every k.
Hint. If the vectors Yn : == x X n are linearly independent, then we can extend this system to a linear basis in E ( Exercise 0. 1 . 2) . After that we can construct a prenorm ( and even a norm ) in E such that IIYn ll 1 for all n . -
The reader knowing what a convergent net is, may notice that in Proposition 3 se quences (i .e. , nets with the domain N) can be replaced with arbitrary nets. Our proof remains true up to the obvious changes. (Nevertheless, give an accurate proof. ) But the result of Exercise 3 essentially uses the fact that we deal with sequences , and it cannot be extended to arbitrary nets. (Give a counterexample.)
A polynormed space (E, II · ll v; v E A) is Hausdorff as a topological space for every x E E, x # 0 there exists v E A such that ll x ll v > 0 . {In other words, we can distinguish elements of E by prenorms ll x ll v · )
Proposition 4.
�
252
4.
Polynormed Spaces and Generalized Functions
Let y and z be two different vectors in E. Take x : == y z and choose the index v indicated in the statement. Then clearly the neighborhoods Uy , v,r and Uz , v, r of y and z are disjoint for r < ll x l v /2 . :::::=::> . Take x E E \ { 0} . We use the fact that 0 has a neighborhood that does not contain x. Then x is not contained in some standard ball, say, Uo ,v1 , ... , vn , r · But this means precisely that for some k; 1 < k < n we have Proof.
{:::::::= .
ll x ll vk > r.
•
Now we see that all the spaces in our examples are Hausdorff; the only possible exceptions are (B(E, F) , so) and (B(E, F) , wo) : they are Hausdorff {=:::} F is normed. Which operators are compatible with the structure of a polynormed space? The following theorem answers this question. You are already pre pared to perceive it since Theorem 1.4. 1 gives the answer in the special case where spaces are prenormed (see also Proposition 1.3. 1) .
The following properties of an operator T between polynormed spaces (E, II . II IL ; 11 E r) and ( F, II . ll v ; E A), are equivalent: (i) for each v E A there exists an accompanying prenorm I · II ' on E and a constant C > 0 such that for every x E E we have II T(x) ll v < C ll x ll ' {in other words, T is bounded as an operator between prenormed spaces (E, II · II ') and (F, II · ll v)) ; (ii) for each accompanying prenorm II · II in F there exists an accompa nying prenorm 1 · 11' on E and a constant C > 0 such that for every x E E we have II T(x) II < C ll x ll ' {in other words, T is bounded as an operator acting between prenormed spaces (E, II · I ') and (F, I · II )); (iii) T is continuous at zero {with respect to the topologies generated by the given families of prenorms); (iv) T is continuous (with respect to the same topologies).
Theorem 1 .
lJ
II · II == max { ll · l l vk ; k == 1 , . . . , m }. Then for every k there exists an accompanying prenorm I · I � in E and Ck > 0 such that for every x E E we have II T(x) ll vk < Ck ll x ll � · Clearly, the assertion is true for the accompanying prenorm II · II ' : == max{ ll · II � ; k == 1, . . . , n } and C == max { Ck ; k == 1, . . . , n } . Proof. (i)::::=::> (ii) . Put
:
(ii) ::::=::> ( iv) . Take x E E. We must show that T is continuous in x. Let U be a neighborhood of the point T(x) in F. It contains a standard open ball U� o) == {y E F : II Y - Tx ll < r} for some accompanying prenorm I · I in F and some r > 0. By hypothesis, there exists an accompanying prenorm II · II ' in E such that T is bounded as an operator between (E, II · I ') and (F, II · I ). Consequently (Theorem 1.4. 1) , T is continuous. But then there
1.
253
Polynormed spaces
exists a neighborhood V of x in the space (E, I · II ') such that T(V) lies in U� o ) and hence in U. It remains to note that V is an open set in the polynormed space (E, II . II IL ; M E r) . The implication (iv)::::=::> (iii) is obvious. (iii) ======> ( i) . Take v E A and a neighborhood Uo ,v, I of zero in F. By hypothesis, there exists a neighborhood W of zero in E such that T(W) C Uo ,v, l · We know that w contains a standard open ball wj 0) = {X E E : l l x l l ' < 8} , where II · II ' is some accompanying prenorm in E. Thus T, regarded as an operator between prenormed spaces (E, 1 1 · 1 1 ') and (F, ll · l l v ) , maps some dilation of the open unit ball to a bounded set. Obviously, such an operator must be bounded. • Now we can answer the question concerning the comparison of different topologies generated on the same linear space by two different systems of prenorms. We have already discussed this when studying prenormed spaces, and the information we have obtained there suggests the answer.
1.4.2) . A family of prenorms II . I IL ; M E r in a linear space E majorizes a prenorm 11 · 11 in E if there is a finite set of indices Definition 2 ( cf. Definition
. . . , f-ln such that the accompanying prenorm max{ I · 1 1 111 , , II · l l �tn } majorizes II · I · Furthermore, we say that a family of prenorms majorizes another family of prenorms if the first family majorizes every prenorm in the second family. Finally, we say that two families of prenorms are equivalent if each of them majorizes the other. Here is an instructive example. /L l ,
•
Exercise 4. In the space
•
•
B (E, F ) , where E and F are prenormed
spaces, the family consisting of one operator prenorm majorizes the strong operator family, and the strong-operator family of prenorms majorizes the weak-operator family. If E and F are infinite-dimensional normed spaces, then these three families of prenorms are not pairwise equivalent. From Definition 2 we see that if we add to a given family of prenorms the maxima of all its finite subfamilies, then the new family of prenorms is equivalent to the initial one. Note that any finite family of prenorms II · Il k ; k == 1 , . . . , n is equivalent to one prenorm, namely to their maximum.
Suppose we have two families of prenorms on a linear space. Then the topology generated by the first family is not weaker than the topology generated by the second family the first family majorizes the second. As a corollary, two families of prenorms generate the same topology � they are equivalent.
Proposition 5.
�
1.4.5 can be transferred with obvious • changes from prenormed to the polynormed spaces. Proof. The proof of Proposition
4.
254
Polynormed Spaces and Generalized Functions
Now we can easily answer the question of when the topology of a poly normed space can be generated by just one prenorm (in this case the space is called prenormable) . Moreover, with some additional effort , we can an swer the question of when the topology can be defined by a premetric that in general is not generated by one prenorm. (Of course, the prefix "pre-" can be omitted if the initial family of prenorms satisfies the hypothesis of Proposition 4.)
6. A polynormed space is prenormable norms is equivalent to a finite subfamily.
Proposition
�
its family of pre
Proof. As we have said before, every finite family of prenorms is equivalent
to one prenorm (namely, their maximum) . This immediately implies the sufficiency. The necessity follows from the previous proposition and from the following obvious observation: if a family consisting of one prenorm is equivalent to some family of prenorms, then the first family is equivalent to • a finite subfamily of the second family.
1. A countably normed space (E, II · l in; n E N) such that II · lln+ l majorizes II lin , but these prenorms are not equivalent, cannot be prenormable. Of course, C 00 [a, b] is an obvious example of a countably normed space with this property, and the same is true for C00 after replacing II li n by max { II · II I , · · · II · II n} · Exercise 5. The polynormed space O ( U ) is not prenormable. Hint. The family of prenorms in O(U) is equivalent to the family { II · II Kn ; n == 1 , 2, . . . }, where K1 C K2 C · · · is a sequence of compact sets in U such that U == U� 1 Kn . Corollary
·
·
'
The prenormability criterion can be reformulated in the following terms. We call a subset in a polynormed space bounded if it is bounded in every accompanying prenormed space. (In our exposition we shall not use this notion frequently, but it comes to the forefront if we go sufficiently deep into the theory of topological vector spaces; see [60] or [61] .) Exercise
6. A polynormed space is prenormable
bounded open set.
�
it contains a
Now let us discuss the case of finite-dimensional spaces. We have seen that every two norms in such a space are equivalent (Corollary 2. 1.3(ii) ) . The following statement is a generalization of this fact to the polynormed spaces.
1.
255
Polynormed spaces
Exercise 7. Every two families of prenorms generating Hausdorff topo
logies on a finite-dimensional space are equivalent. In particular, every such family of prenorms is equivalent to some (and thus, any) norm. Hint. The main step is to show that each "Hausdorff" family of prenorms {x E I I · l l v ; v E A majorizes any norm I I · I I in E. Consider the sphere S E I I x I I 1 } with topology generated by the norm, and for any x E S, take the index v ( x ) E A such that l l x l l v ( x ) > 0. Since E is finite-dimensional, every norm always majorizes every prenorm. Hence there exists an open covering { Ux : x E S } of our sphere such that I I Y I I v ( x ) > 0 for all y E Ux . But the sphere is compact (we again use that E is finite-dimensional) . If we take a finite subcovering { Ux 1 , , Uxn } , we obtain that the prenorm max{ ll · l l v ( xk ) ; k 1 , . . . , n} is a norm. :�
�
:
•
•
•
�
Now we suggest that you work with the condition of premetrizability. Exercise 8* . A polynormed space is premetrizable
�
the family of
prenorms is equivalent to a countable subfamily. Hint. ¢::== . If (E, I I · l i n ; n E N) is a countably polynormed space, then its topology can be defined by the premetric d ( x , y ) : = 2::.::: � 1 2 n c � lln ) . (Notice that although this premetric is invariant under translations, it is not generated by any prenorm. This follows, in particular, from the fact that diameter of E is not greater than 1 . ) :::::=::> . If the topology is generated by a premetric, then the neighborhood of zero { x : d(O, x ) < � } contains a standard open ball centered at zero; this means that to every n we can associate a finite subfamily of our family of prenorms. The union of all these subfamilies over all n E N is the countable subfamily we are looking for.
:�;��
Here is an illustration. Exercise 9 ° . Polynormed spaces C 00 [a, b] , O( U ) , and Coo are metriz able. At the same time, infinite-dimensional strongest polynormed spaces are not premetrizable. Hint. Regarding O(U) , see the hint to Exercise 5. Now suppose E is a strongest space, en ; n E N is a linearly independent system of vectors in E, and II l i n ; n E N are some prenorms. Then we can extend en to a basis of E and construct a prenorm (and even a norm) II · I I such that for any n we have l l en ll > n max{ llen ii i , · · · , l l en lln } · ·
Now we would like to say a few words about a certain class of polynormed spaces playing an important role in the general theory and applications. Definition 3. A polynormed space is called a Frechet space if its topology can be defined by a complete metric that is invariant under shifts (i.e. , a metric satisfying the condition d ( x , y) == d (x - y, 0) for all x, y) .
256
4.
Polynormed Spaces and Generalized Functions
( Verify that c oo [a, b] and all other examples of metrizable polynormed spaces men tioned above are Frechet spaces. ) Frechet spaces possess many properties typical for Banach spaces . For example, some fundamental results like the Banach theorem on the inverse operator or the Banach Steinhaus theorem are valid for Frechet spaces . In general, it is a good idea to view Frechet spaces as the next reasonable generalization of Banach spaces . For details, see, e.g., [63] , [60] , and [61] . In fact, the requirement that the metric is invariant with respect to the shifts in Definition 3 can be omitted. This follows from a very deep theorem by V. Klee [103] . *
*
*
We now return to the operators discussed in Theorem 1 . Remark on terminology. In the context of polynormed spaces, the op
erators that satisfy the hypotheses of Theorem 1 will be called continu ous. As for the term "bounded" , we reserve it for operators that preserve bounded sets. Actually, outside the special case of prenormed spaces, not every bounded operator is continuous ( see, for example, [60, 11.8]) . Here are several examples of continuous operators. Some of them are presented as exercises.
differentiation operator D on the polynormed space (c oo [a, b] ; II · l i n ); n == 1 , 2, . . . that assigns to every function its deriva tive. From the obvious inequality II D(x) lln < ll x lln + l ; n == 0, 1 , . . . it follows that this is a continuous operator on c oo [a, b] . On the other hand, it is use Example 10. Consider the
ful to note that this operator is not continuous on any of the accompanying spaces of c oo [a, b] . Exercise 10. The similarly defined differentiation operator in
O(U) is
also continuous. it follows that Hint. From the integral formula w '( z ) = 2!-i J"Y if K and L are closed sets in U and a contour 'Y goes around K and lies in L, then for some C > 0 we have ll w ' IIK < C ll w ii L·
(����2d(
Exercise 1 1 . The operator of multiplication by an infinitely smooth
function in C 00 [a , b] and the operator of multiplication by a holomorphic function in O(U) are continuous.
(B(E, F) , so) , choose x E E, and consider B ( E, F) --+ F T �----+ T(x). From the equality
Example 1 1 . Take the space
the evaluation operator Tx II Tx(T) II == II T I x it follows that this operator is continuous. Note that the continuity is preserved after endowing B(E, F) with the operator prenorm. At the same time the same operator from (B( E, F) , wo) to F is, generally speaking, not continuous ( explain why ) . :
:
1.
257
Polynormed spaces
Here is another observation which is important in the spectral theory (see Sections 6.5-6. 7) and in the theory of operator algebras. Consider again B(E, F) and take operators S E B(E) and R E B(F) . Since the composition of continuous operators is continuous, we obtain the maps Ms : B(E , F ) � B(E , F) : T r--+ TS and R M : B(E , F) � B(E , F) : T r--+ RT. Clearly, they are linear operators; they are called operators of composition (with S from the left and with R from the right) . Proposition 1.3.4 immediately shows that both these operators are continuous with respect to the operator prenorm in B(E , F) . Moreover, we have Proposition 7.
(i) Operators Ms and R M are continuous as opera-
tors in (B(E , F) , so) ; (ii) the same is true if we replace (B(E , F) , so) with (B( E , F) , wo) .
II · llx ; x E E is a prenorm belonging to the strong-operator family and T E B(E , F) . Then II Ms (T) II x == II TS(x) ll == II T II sx and IIR M(T) IIx == II RT(x) ll < II R II II T IIx · If II · llx , f ; x E E, f E F* is a prenorm from the weak-operator family, then for the same T we have II Ms (T) llx , f == l f(TSx) l == II T II sx ,J and IIR M(T) II x ,J == l f (RTx) l == I ( R * f) (Tx) l == II T IIx,R* f· Thus, in both cases we have the estimates needed in Theorem 1. (Note that
Proof. Suppose
instead of a finite family of prenorms now one prenorm fits; in both cases the required inequality turns out to be equality with the constant C == 1.) •
We can characterize the strongest polynormed spaces and also the poly normed spaces with zero prenorm (it is natural to call them the weakest polynormed spaces) in terms of operators as follows. Exercise 12 ° ( cf. Exercise 0.2.5 about discrete and antidiscrete topo logical spaces) . (i) A polynormed space E is the strongest (up to the equivalent family of prenorms) � for any polynormed space F every linear operator from E to F is continuous. (ii) A polynormed space E is the weakest � for any polynormed space F every linear operator from F to E is continuous. Other instructive examples of continuous operators between polynormed spaces will appear later in connection with weak topologies and generalized functions. *
*
*
Now we have two new categories of functional analysis: (i) The category Pol. Its objects are polynormed spaces and mor phisms are continuous operators.
4.
258
Polynormed Spaces and Generalized Functions
(ii) The category HPol. It is a full subcategory in Pol with Hausdorff polynormed spaces as objects. To avoid possible misunderstanding, we emphasize that a linear space with two different families of prenorms gives rise to two different objects in Pol, even if the two families are equivalent (and consequently, generate the same topology on E) . Obviously, the category Pre described before is a full subcategory in Pol, and Nor is a full subcategory in HPol (and thus, in Pol as well) . Exercise 13 ° . There is a full subcategory in HPol ( and thus, in Pol) which is isomorphic ( see Subsection 0. 7) to Lin: this is the category of strongest polynormed spaces . There is another subcategory in Pol with the same property: this is the category of weakest polynormed spaces.
Thus, "one may consider linear algebra
as
part of the theory of polynormed spaces ."
Note in addition that in HPol there is another important subcategory, Fr, consist ing of Frechet spaces ( see Definition 3) . Its behavior resembles the behavior of the full subcategory Ban. But the category Fr is beyond the scope of this book.
Clearly, just as in Pre, the isomorphisms in these categories are con tinuous operators having continuous inverse operators. So we preserve for them the same name: topological isomorphisms. (Categories that generalize the category Pre1 , and consequently the notion of isometric isomorphism, have not appeared in mathematics so far, and it is not likely that they will be needed in the future.) The set of morphisms between objects E and F of our new categories, i.e. , the set of continuous operators between polynormed spaces, will (again as in Pre) be denoted by B(E, F) . The set of continuous functionals on E again is denoted by E* . Proposition 8 (cf. Proposition
£( E, F) , and thus a linear space.
1.3.2) . The set B(E, F) is a subspace in
S, T E B(E, F) and choose an accompanying prenorm II · II in F. By Theorem 1 , there exist accompanying prenorms II · II ' and II · II " in E such that S (respectively, T) is bounded as an operator from (E, II II ' ) (respectively, (E, II II ") ) to (F, II · II ). Hence, both operators are bounded as operators from (E, max{ II II ' , II II "}) to (F, II · II ) , and the same is true for their sum. Again by Theorem 1 we have S + T E B(E, F) . We leave to the reader the verification of the fact that the elements in B(E, F) can be multiplied by scalars. • Proof. Take
·
·
·
·
The linear space E* will be called dual to E (by analogy with the case when E is prenormed) .
1.
Polynormed spaces
259
As an optional material (even for the advanced reader) , we describe some additional properties of the categories Pol and HPol. First of all, Proposition 8 shows that we can define morphism functors on these categories with range in Lin. The reader can easily restore the details by analogy with the morphism functors B(E, ?) , B(?, E) : Ban ---+ Ban discussed in Section 2.5. Here you may feel dissatisfied . Why is the set B( E, F) endowed with the structure of linear space only? Can we make it an object of the same category Pol like we did when considering morphisms in Ban? It turns out that there is no "canonical" way to make B(E, F) a polynormed space. Instead, there are many ways to do that , and each has its advantages and disadvantages. We describe just the simplest way (used in the theory of generalized functions) ; it is inspired by Example 8. Let E and (F, II . llv; v E A) be polynormed spaces. Take r : == E X A and for any pair (x E E, v E A) consider the function II · llx,v : B(E, F) ---+ IR:+ , where II T II x,v : == II T(x) llv · Obviously, this is a prenorm. This family of prenorms and the corresponding topology on B(E, F) are said (following the same example) to be strong-operator. Now you can easily check the following. For every polynormed space E, the same constructions as those used when defining the functors with range in Lin, but now with strong-operator families of prenorms in B(E, F) , give covariant and contravariant functors from Pol to Pol (and from HPol to HPol) . We should, however, keep in mind that these functors are not extensions of the functors defined for Ban in Subsection 2 . 5 . This follows from the fact that even if E and F are infinite-dimensional Banach spaces , these functors give non-normed spaces (explain why! ) . Many properties of Pre and Nor are preserved when passing to more general cate gories Pol and HPol. In particular, the notion of topological direct sum is extended from prenormed spaces to polynormed spaces (Definition 1 . 5 . 1 ) , as well as related questions of describing (co )retractions (see Exercise 1 .5 .3) . The same is true for the characterization of continuous projections in terms of a decomposition of the space into a topological direct sum (Proposition 1 .5 . 1 0 and Corollary 1 .5 . 1 ) . Finally, the results on the characterization of mono- and epimorphisms (and their " extreme" generalizations) are carried on literally from Pre to Pol (Proposition 1 . 5 . 1 1 ) , and from Nor to HPol (Exercise 1 .5 .6) . You can easily restore the details. Here is something more interesting. In some aspects the categories Pol and HPol behave much better than, for example, Ban (and they rather resemble Ban 1 ) . As we know, only finite families of objects in Ban have the (co )product (Exercise 2 . 5 . 1 5 ) . It turns out that in our new categories any family of objects has the product and the coproduct . Let E�.� ; v E A be a family of polynormed spaces , and II · II JL ; J.l E A v the family of prenorms on E�.� . Take the linear space X {E�.� ; v E A} (i.e., the product of our family in Lin ) and endow it with the family of prenorms 111 · 111� ; J.l E A�.� , v E A put ting 111!111 � :==
ll f ( v ) II JL ·
Exercise 14. The space X { Ev ; v E A} endowed with the family of prenorms 111 · 111� ; J.L E A�.� , v E A and with the projections 1rv (see Subsection 0.6 ) is a product of the family Ev ; v E A in Pol. If, in addition, all our spaces are Hausdorff, then X { Ev ; v E A} is a product in HPol.
Now consider the linear space E == ffi{Ev ; v E A} (i.e., the coproduct of our family in Lin ) . For every v E A let us agree to identify Ev with the subspace in E consisting of f such that f (J.L) == 0 for J.l =1- v (see Section 0.6 ) . Further, consider in E the family � of all prenorms II · II such that the restriction of II · II to each Ev is majorized by the family of prenorms II · II JL ; J.l E Av of the space Ev .
4.
260
Polynormed Spaces and Generalized Functions
Exercise 15* . The space ffi{Ev ; v E A} endowed with the family of prenorms � and with the inj ections iv (see Subsection 0.6) is a coproduct of the family E�.� ; v E A in Pol. Moreover, if all our spaces are Hausdorff, then it is a coproduct in HPol as well. Example 12. Consider a countable family of copies of C. The product of this family is the polynormed space c(X) with the projections 1r : c(X) ---+ C : � �-----+ � , and its coproduct is Coo with the strongest family of prenorms and the injections : C ---+ coo : A �-----+ (verify both statements! ) .
n
in
n
.Apn
Concluding the discussion of the categories Pol and HPol, we note that they have sufficiently many free objects for the existence of the freedom functor. We can regard these categories as concrete categories (see Subsection 0. 7) if for any E E Pol we take its underlying set as DE. Exercise 16* . Suppose K is Pol or HPol, and E
E
Ob(K) . Then
(i) E is a free object in K ¢:::::::> the family of prenorms of E majorizes any prenorm (in other words , E is the strongest polynormed space up to a topological iso morphism) ; (ii) there exists a freedom functor :F : Set---+ K that associates with every set S the linear space :F(S) (see Example 0. 7.4) equipped with the strongest family of prenorms.
2. Weak topologies
In this section we shall concentrate on one special way to define a family of prenorms, namely using appropriate families of functionals. The topologies arising here play an important role in studying general polynormed spaces, and also allow us to better understand objects of classical functional analysis, namely normed spaces. We formulate, for further references, a special case of Theorem 1.1 deal ing with functionals. Until further notice, (E, II . 11 1-L ; 11 E r) is a polynormed space.
The following properties of a functional f : (E, 11 · 11 1-L ; 11 E f) � C are equivalent: (i) there is a finite set of indices /1 1 , . . , f.1n and a constant C > 0 such that for each x E E we have l f(x) l < C max { ll x ii J.L 1 , • • • , ll x ii J.Ln } {in other words, f is bounded as a functional on the accompanying pren o rmed space ( E, max { II · II J.L 1 , · · · , II · II J.Ln } )}; ( ii) f is continuous at zero; (iii) f is continuous. • Theorem 1 .
.
We emphasize that for the continuity of a functional on a polynormed space it is necessary and sufficient that this functional is continuous with respect to at least one accompanying prenorm, and not with respect to all of them. (A frequent mistake on exams!)
2.
261
Weak topologies
How many continuous functionals are there on polynormed spaces? The Hahn-Banach theorem gives a complete answer to this question. Theorem 2 ( cf. Corollary 1 .6. 1 ) .
A space (E, II · II IL; /1 E r) is Hausdorff {::::::} for each vector x E E; x # 0 there exists f E E* such that f ( x) # 0, or, equivalently, for every vectors x , y E E; x # y there is f E E* such that f(x) # f(y) . ===> .
Proposition 1 .4 gives 11 with ll x ll 11 > 0, and Theorem 1 .6.3 gives a non-vanishing functional on x bounded with respect to the prenorm II · 11 11• Then Theorem 1 works. ¢::=: . If f(x) # 0 for f E E* , then Theorem 1 gives at least one 11 c • ll x ll 11 =/= 0. Then Proposition 1 .4 works.
Proof.
For our future study of weak topologies we need several standard facts from linear algebra, which we present here with complete proofs.
Let T E --+ F be an operator between linear spaces such that for some finite family of functionals !I , . . . , fn on E the condition fi (x) · · · fn (x) 0 implies Tf 0 {in other words, n �== I Ker(fk ) C Ker (T) ) . Then the image of T is finite-dimensional. Proof. In e n consider the subspace of the rows (!I ( X ) ' . . . ' fn ( X ) ) ; X E E. Consider vectors X I , . . . , X m such that the rows (/I (x k ) , . . . , fn (x k ) ); k 1 , . . . , m form a basis of this subspace. Then for every x E E there are AI , . . . , A m E C such that fz (x) L "!: I A k fz (x k ) for all l 1 , . . . , n. Hence, x L "!: I A k X k belongs to the kernel of all fz , and therefore, the kernel of • T. Thus Tx E span (Tx i , . . . , Tx m ) for each x E E. The rest is clear. Corollary 1 . If !I , . . . , fn is a finite family of functionals on a linear space E, then the quotient space E I n �== I Ker(fk ) is finite-dimensional. :
Proposition 1 . �
�
�
�
�
�
�
-
As an application, try to obtain one of the many proofs of the following "intuitively clear" fact.
E be an infinite-dimensional Hausdorff polynormed space. Then the dual space E* is also infinite-dimensional. Hint. For every !I , . . . , fn E E* there exists f E E* that does not vanish on n�== I Ker(fk ) · Proposition 2. Let E be a linear space, /I , . . . , fn functionals on E, and f another functional on E. Suppose that for every x E E the condition fi (x) · · · fn (x) 0 implies f(x) 0 {i. e., n�== I Ker(fk ) C Ker(f)). Then f E span ( fI , . . . , fn ) · Exercise 1 . Let
�
�
�
�
262
4.
Polynormed Spaces and Generalized Functions
� e n : X 1---+ ( !I ( X ) ' . . . ' fn ( X ) ) and the corresponding injective operator T : E I Ker(T) � e n : X + Ker(T) 1---+ Tx . (We came across such a construction in Section 1 .5.) From our assumption, Ker(T) C Ker(f) . Hence f generates a functional f : E I Ker(T) � e : x + Ker(T) �----+ f (x) . By the injectivity of T, there exists a functional g such Proof. Consider the operator T : E �
�
that the diagram
E I Ker(T)
T
-
en
� e /.
is commutative. It remains to put A k for all x E E we have -
f(x) == f (x
: ==
g(p k ) ;
k == 1 , . . . , n and to see that
+ Ker(T) ) == gT( x + Ker(T) ) �
n
n
k= I
k= I
= g C�= !k (x) p k ) = L A k fk (x) . • Suppose again that E is a linear space, and E0 a space of linear func tionals on E (i.e. , a subspace in the algebraic dual space E � to E) . We say that E0 is sufficient if for every non-zero vector x E E there exists f E E0 such that f (x) # 0. For instance, Theorem 2 means in these terms that a polynormed space E is Hausdorff {=:::} continuous functionals on E form a sufficient space.
Suppose E0 is sufficient, XI , . . . , Xn is a linearly indepen dent system in E, and A I , . . . , An are arbitrary complex numbers. Then there exists f E E0 such that f(x k ) == A k ; k == 1 , . . . , n .
Proposition 3.
Proof. Consider the operator s : E0
�
en :
X
1---+
(f (x i ) , . . . ' f ( x n ) ) .
We have to show that S(E0) == e n . Suppose this is not the case. Then there exists a non-zero functional g on e n vanishing on S(E0 ) . For x : == E�= I g(p k ) x k E E and for every f E Eo we have f(x) == E�= I g(p k ) f (x k ) == g( L�= I f(x k ) P k ) . Since E�= I f(x k ) P k is (f (xi ) , . . . , f ( x n ) ) E S(E0 ) , the choice of g guarantees that f(x) == 0. At the same time from g # 0 we have that not all numbers g(p k ) vanish, and thus, x =/= 0. The functional f E E0 was chosen arbitrarily, hence we came to a contradiction with the fact that • E0 is sufficient. In the following definition, E is (still just) a linear space, and A is a set of linear functionals on E (i.e. , a subset in E � ) .
2.
263
Weak topologies
Definition 1 . A family of prenorms
{ II · l i t; f E A} on E, where ll x llt
l f(x) l , is said to be a A-weak family of prenorms ,
:�
and the corresponding
topology generated on E is called the A -weak topology.
We immediately note the following Proposition 4. The indicated family of prenorms is equivalent to the family { II · l i t; f runs over span(A) in E� } .
A k fk , then for x E E we have l g (x) l < C max { l fk ( x ) l ; 1 , . . . , n}, where C n max{ I A kl ; k 1 , . . . , n } . The rest is clear. •
Proof. If g
k
�
�
E�== l
:�
�
Thus, every time we speak about a A-weak topology, we can assume that A is a subspace in E � . We had already encountered two spaces with a family of prenorms of indicated type: these are c00 in Example 1 . 7 and (B(E, F) , wo) in Example 1.9. (Describe A in both cases.) Here are two principal classes of examples. Definition 2. Let (E, II · l v; v E A ') be a polynormed space. A weak (just weak, without mentioning any set) family of prenorms in E is the family { II · l i t; f E E* } , where ll x ll t l f(x) l . This family, as well as the cor responding generated topology in E, which also is called weak, is shortly denoted by w . Thus, the introduced family is a A-weak family of prenorms in E if we take the subspace E * in E � as A. :�
From this definition one can easily deduce
For every polynormed space E the weak topology in E is not stronger (i. e., no finer} than the initial one.
Proposition 5.
Proof. Look at the condition of continuity of a functional in Theorem 1 . It
precisely means that the prenorm I · l i t ; f E E* is majorized by the initial • family of prenorms. Definition 3. Again let ( E, II · l v ; v E A') be a polynormed space. The weak* (read "weak-star" ) family of prenorms in E* (note that this time in the dual, not in the original space!) is the family { II · ll x; x E E} , where l f ll x l f(x) l . This family, as well as the topology in E* generated by it (also called weak*), is denoted by w* . :�
This is a special case of Definition 1. Namely, like in the case of normed spaces (see Section 1 .6) , every vector x E E defines a functional ax on E* by the formula f �----+ f ( x), called evaluating functional. Various evaluating functionals form a subspace in ( E* ) � . If we denote it by A, then clearly the introduced family is precisely the A-weak family of prenorms in E* .
4.
264
Polynormed Spaces and Generalized Functions
When speaking about A-weak and, in particular, weak or weak* conver gence, we have in mind the convergence in the corresponding topology. Again, let E and A be the same as in Definition 1 . Note the following property of A-weak topologies, sharply distinguishing them from the normed ones.
6. Let E have infinite dimension. Then, for each A, every neighborhood of zero in the A-weak-topology of E contains a nontrivial and even an infinite-dimensional subspace.
Proposition
Proof. Such a neighborhood contains a standard open ball centered at zero:
for some functionals f1 , . . . , fk E A and r > 0 it has the form { x E E : I fk ( x) I < r; k == 1 , . . . n, n } . Therefore, it always contains the kernel of the operator T : E --+ e : X 1---+ (f1 (x ) , . . . fn (x) ) acting from an infinite • dimensional space to a finite-dimensional one. The rest is clear. '
Proposition 1 .4 immediately implies the following result.
space E endowed with a A-weak topology is Hausdorff • {=:::} the space span ( A ) is sufficient.
Proposition 7. A
This fact, combined again with Proposition 1 .4 and the tautology that a functional is non-zero if it is non-zero on some vector, yields Corollary 2.
( i ) For an arbitrary polynormed space E, the space (E* , w* ) is Haus
dorff. ( ii ) If E is a polynormed space, then the space (E, w ) is Hausdorff {=:::} E (with the initial topology) is Hausdorff. Exercise 2. A linear space
E is premetrizable in the A-weak topology
{=:::} the linear dimension of the space span ( A ) is at most countable. Hint. Use Exercise 1 .8.
Remark. From Exercise 2 it follows in particular that the weak topology
on an infinite-dimensional normed space and the weak* topology on the dual space to an infinite-dimensional Banach space are not metrizable ( cf. Exercise 2. 1 . 1 ) . Functionals that are continuous in the A-weak topology are called A weakly continuous. It is not difficult to distinguish them from all linear functionals:
functional f E E � is A-weakly continuous {=:::} it is a linear combination of functionals in A.
Proposition 8. A
2.
265
Weak topologies
A is a subspace in E� (see Proposition 3) . {:::::: . Using the tautology I f ( x) I < II x II f , we can apply Theorem 1 . ===> . By the same theorem, there are /I , . . . , fn E A and C > 0 such that lf(x) l < C max{ I !I (x) l , . . . , l fn (x) l } . But then the functionals /I , . . . , fn • and f satisfy the hypothesis of Proposition 2. The rest is clear. Corollary 3. Let E be a polynormed space. Then (i) functionals on E continuous in the weak topology are precisely those continuous in the initial topology; (ii) functionals on E* continuous in the weak* topology, are precisely the evaluating functionals.
Proof. We can assume that
In different concrete situations that we shall encounter later in the book it will be important to know whether a particular subspace E0 in E* is dense with respect to the weak* topology. Here is the way to determine this.
If E0 is sufficient, then it is dense in E* with respect to the weak* topology.
Proposition 9.
Proof. Take g E E* and an arbitrary neighborhood of g in the weak* topol
ogy. This neighborhood contains a standard open ball U9,xl , . . . , xn , r , and thus contains all h E E* where h (x k ) == g (x k ) ; k == 1 , . . . , n. Our goal is to find among those h a vector from E0 • If the system X k consists of zeroes, everything is clear. Otherwise, it contains a maximal linearly independent subsystem; without loss of gener ality we can assume that this is X I , . . . , x m ; m < n. The sufficiency con dition together with Proposition 3 provides f E E0 such that f ( x k ) == g ( x k ) ; k == 1 , . . . , m. Further, for every l == 1 , . . . , n the vector xz has the form E � I J1k X k ; /1 k E C. Hence, f (x z ) == E � I l1 k f ( x k ) == E � I J1kg (x k ) == • g (xz ) . The rest is clear. Exercise 3. If E is Hausdorff, then the converse is true.
Let T : E --+ F be a continuous operator between polynormed spaces. Then it is easy to see that a linear operator T* : F* --+ E* acting by the formula f �----+ JT is well defined. In other words, T* is defined by the formula [T* f] ( x) == f (Tx) or, which is equivalent, by the commutative diagram T F E ----�
� �
T
( cf. Definition 2.5.2) .
266
4. Polynormed Spaces and Generalized Functions
The operator T* is continuous with respect to the weak* topologies in F* and E* .
Proposition 10.
Proof. Take a prenorm
II
· ll x ; x E E in (E* , w* ) and look at the prenorm
II · ll rx in (F* , w*) . From the construction of T* it immediately follows that • l i T* f ll x == ll f ll rx · Replacing "==" by " 1 . We note that in V there is an increasing chain of subspaces VN : == { cp E V : cp ( t) == 0 for l t l > N; N == 1, 2, . . . } . Of course, V == U{VN : N ==
1 , 2, . . . } .
We want to make V a polynormed space. A prenorm II · II in V is called admissible if for every N E N there are n E Z + and C > 0 such that for every cp E VN we have ll cp ll < C ll cp ll � ) . In other words, a prenorm is admissible if on every space VN it is majorized by some standard prenorm (depending on N) .4 We make V a polynormed space equipped with the family of all admissible prenorms. This family will often be denoted by d. Remark. For every N the space VN is polynormed with respect to the standard family of prenorms II · II � ) , n E Z + . It is easy to see ( cf. Theorem 1.1 and Proposition 1.5) that the family d is the "strongest" (i.e. , biggest) among all possible families of prenorms on V for which all the embeddings VN � V are continuous. Certainly, every standard prenorm in V is admissible. Here are more interesting examples. Example 1 . Let a(t) ; t E 1R be an arbitrary continuous everywhere positive function. For cp E V we put l cp ll a : == max{ a ( t) l cp ( t) l ; t E JR } . Obviously, II · ll a is an admissible prenorm (and even a norm) in V, and in the above estimate for every N we can take n == 0. 4 The same condition can be described, up to equivalence, by a frightening formula [29, p. 99] , but, fortunately, we will not need it.
3. Spaces of test functions and generalized functions
Example 2 . For
279
1
cp E V let us put ll cp ll :== 2::: � l cp ( k ) (k) l . Obviously, II · II
is also an admissible prenorm. However, this time in the above estimate the order n of the derivative must increase unboundedly as N grows. Note a few obvious properties of the family of admissible prenorms:
( i ) If II · I l k ; k == 1, . . . , m are admissible prenorms, then II · II : == max { ll · I l k ; k == 1 , . . . , m } is also admissible. ( ii ) If a prenorm is majorized by an admissible one, then it is admissible as well. ( iii ) If II · II is admissible, then the same is true for the prenorm 111 · 111 defined by the formula lll cp lll :== ll cp' ll · •
Proposition 1.
Now we pass to the next space. We say that an infinitely smooth function cp(t) ; t E 1R is rapidly decreasing if for every p, q E Z + the function tPcp ( q) (t) is bounded, or, in other words, if cp( t) and all its derivatives decrease faster than every negative power of every polynomial as l t l � oo . The most famous among rapidly decreasing functions is, perhaps, the Gauss function 2 t e-2 , so popular in probability theory. ( In these lectures it will also play an important role later in the discussion of the Fourier transform. ) Thus, we introduce
The space S, also called the Schwartz space. It consists of all rapidly
decreasing infinitely smooth functions. We make it polynormed by intro ducing the family of prenorms ( actually, norms, as can be easily verified ) II . ll p, q ; p , q E z + , where ll cp ll p , q :== max{ l tPcp ( q) (t) l : t E JR } . This family will be often denoted by s. Finally, here is
The space £. It consists of all infinitely smooth functions ( no matter how they grow ) . The family of prenorms we define on this space is extremely
simple: it consists of all the standard prenorms, and only them. We shall denote this family by e.
Being polynormed, all three spaces V, S, and £ automatically become topological spaces. Let us discuss which sequences converge there. We shall say that a sequence of infinitely smooth functions cpm classically converges to cp on a subset M C 1R if cp};:) uniformly converges to cp ( n ) on M for all n == 0, 1 , 2, . . . ( cf. similar usage of this term in Example 1 . 1) . Then it follows from Proposition 1.3 that the convergence of a sequence cpm to cp in £ is precisely the classical convergence on each interval in JR. FUrther, from the very same proposition and from the choice of prenorms in S it easily follows that cpm tends to cp in S {=:::} for every p == 0, 1, 2, . . . , the sequence
4.
280
Polynormed Spaces and Generalized Functions
tP cpm ( t ) classically converges to tP cp ( t ) on the real line. As for V, the answer in this case will probably surprise you.
sequence cpm tends to cp in (V, d) {=:::} the following condi tions are fulfilled: ( i ) all cpm vanish outside some interval {in other words, belong to some V N} ; ( ii ) on this interval (or, equivalently, on the entire real line} cpm clas sically converges to cp.
Theorem 1. A
=====>-
( i ) . Suppose that, on the contrary, for every natural N there are cpm N and tN ; l tN I > N such that cpm N (tN) # 0. Passing, if necessary, to a sufficiently rarefied subsequence, we can assume that cpN(tN) # 0, and the sequence tN; N == 1 , 2, . . . either unboundedly increases or unboundedly decreases. Obviously, there is a continuous function a(t) ; t E 1R such that a(tN) == l cp N ( tN) l - 1 ; N == 1 , 2, . . . and a( t) > 0 for all t. Since cp is finitary, the function a(t) l cp(t) - cpN(t) l is equal to 1 at the point tN for every N except the first few. Hence for the same N and for the norm ll · ll a from Example 1, ll cp - cpN IIa > 1. Since ll · ll a belongs to the family d, Proposition 1 .3 forbids the convergence of the sequence cpm to cp, a contradiction. =====>- ( ii ) . Since all standard prenorms belong to d, cpm classically con verges to cp on every interval. Taking ( i ) into account, this provides the classical convergence of cpm to cp on the entire real line. {:::::::= . Our goal is to show that cpm tends to cp with respect to every admissible prenorm II · II · Take N indicated in ( i ) . By the definition of admissible category, for some n E z + and C > 0 we have ll cpm - cp ll < C ll cpm - cp ll � ) . But by ( ii ) , the numbers in the right-hand side of this • inequality tend to zero. The rest is clear. Proof.
From Proposition 1 .4 it evidently follows that all the three spaces V, S, and £ are Hausdorff. From Proposition 1 .6 it is also easy to see that they are not normed. To advanced readers, we suggest the following Exercise 1 . £ and metrizable.
S are metrizable, and even Frechet spaces, whereas
Hint. Use Exercise 1 .8. For
V,
V
is not
you can apply Theorem 1 instead.
Now we shall discuss the mutual positions of all three spaces of test functions and compare their topologies. Clearly, V c S c £. Later we shall need the following technical res11lt. Proposition 2 ( "on a hat" ) . For every interval [a, b] C 1R and for each c > 0 there exists a function ra , b ,c ( t ) in v {the so-called "hat") such that
3. Spaces of test functions and generalized functions
0 < ra, b ,c (t)
0 such that F( ct) == 0 as l t l > � and C fiR F(ct)dt == 1. Now it is easy to verify that the function b+c/2 l ra,b,c (t) : == c a-c/2 r(c(s - t))ds
{
is what we need. Proposition 3. dense in (S, s) .
• The space V {hence also S) is dense in ( £, ) , and V is e
£ (respectively, cp E S) a sequence 1/Jm E V tending to cp with respect to every prenorm from e (re spectively, from s) . Let us show that we can take 1/Jm : == cpFm, where rm(t) : == F- 1 , 1 , 1 (!) (see Proposition 2). We see that on every interval [ - N, N] the function cp - 1/Jm vanishes for m > N, and therefore for the same m and for all n == 0, 1, . . . we have ll cp - 1/Jm ll �) == 0. Thus, we established the required convergence in the space £. If cp E S, then, clearly, l l cp - 1/Jm ll p,q == max{ l tP (cp - 1/Jm)(q ) (t) l : l t l > m } . From the Leibniz formula it follows that (cp - 1/Jm)(q) == cp(q) - cp(q ) rm + :Ek= l Ck
0 such that Proof. It is sufficient to find for a given
cp
E
-1 q ll cp - '1/Jm ll p,q < max { 2 l tP
0 such that l f ( 'P ) I < CI I 'P I �N) for all 'P E £. :
:
:::=::::> Combining Theorem 2. 1 and Proposition 1, we see that there exists an admissible prenorm II · I in V such that l f ( 'P ) I < II 'P II for all 'P E V. The rest is clear. (i) , ¢::=: This condit ion means that the prenorm II · l i t ; II 'P II t l f ( 'P ) I Proof. (i) ,
.
:�
.
is admissible. Now Theorem 2. 1 works. (ii) , :::=::::> This follows from Theorem 2. 1 and the estimate max{ ll · ll ��k ) ; max{n k; k 1 , . . . , m } and N k 1, . . . , m } < II · I �N) ' where n max{Nk ; k � 1 , . . . , m } . • ( ii) , ¢::=: This is clear. .
:�
�
:
�
.
One says that a generalized function f has finite order if in the estimates of Proposition 6(i) the "order of derivative" n does not depend on N. The minimum of all such n is called the order of the generalized function f. Now let us discuss what guises a generalized function might have. Call a measurable function on the real line locally integrable if it is in tegrable on every interval with respect to standard Lebesgue measure. The set of locally integrable functions (more precisely, of their cosets modulo coincidence almost everywhere) is denoted by Lioc . Clearly, Lioc is a linear space with respect to the pointwise operations (more precisely, pointwise on the sets of full measure in 1R ) . Let f be a locally integrable function. Consider the mapping f V � C 'P r--+ fiR f(t)'P(t)dt. Clearly, j is a functional. Moreover, for 'P E V N N N ( ) we have the estimate l f('P) I < C II 'P II o , where C J_ N l f(t) l dt. Thus, by Proposition 6, f is a generalized function. Definition 2. A generalized function is called regular if it has the form f for some locally integrable function f. Otherwise, it is called singular. A
:
:
A
:�
A
A
A
Clearly, the functional f does not change if we replace f by an equivalent function. Hence, we have the well-defined mapping i Lioc � V* f r--+ f, which, certainly, is a linear operator. (From this moment, following A
:
:
284
4. Polynormed Spaces and Generalized Functions
the tradition of real analysis, we use expressions like "a locally integrable function f from Lioc ." As always, if you remember what stands behind them, this does not result in a confusion.) Thus, a generalized function is regular {=:::} it lies in Im( i ) . Certainly, regular generalized functions have order 0. To move forward, we need the following fact from real analysis.
Let f be an integrable function on a closed interval [a, b] such that I: f ( t ) cp ( t ) dt 0 for every cp E V vanishing outside [a, b] . Then f 0 almost everywhere on [a, b] .
Proposition 7.
�
�
Proof. Without loss of generality we can assume that our function takes
real values. Consider an arbitrary interval [c, d] C [a, b] and its characteristic function Xc, d · Take the hats Fn (t) rc+ I /n , d - I /n ,I /n ( t). Then the sequence Fn al most everywhere tends to Xc, d on [a, b], and everywhere we have l f ( t ) Fn ( t ) l < l f( t ) l . By the Lebesgue theorem, :�
d
jc f(t)dt = 1b f (t ) c,d (t ) dt = nl�m 1b f (t ) n (t ) dt = 0. x
a
00
a
T
In particular, I: f(t)dt 0 for all x E [a, b]. Differentiating this identity • with respect to x we see that f ( x ) 0 almost everywhere on [a, b]. �
�
Here is the first application.
The operator i is injective; in other words, two locally inte grable functions generate the same generalized function {=:::} they are equiv alent.
Proposition 8.
Proof. Let a locally integrable function
A
f be such that f
0. From the construction of f and Proposition 7 it immediately follows that f vanishes almost everywhere on each interval of the line. Hence, it vanishes almost • everywhere on the whole line. A
�
Thus, i is a bijection between Lioc and the subspace in V* consisting of regular generalized functions. Identifying these two spaces by means of this bijection (a usual "psychological device" for a mathematician) , we can as sume that Lioc is a part of V* . Since Lioc contains, to speak informally, "the majority" of functions we encounter in analysis, this identification essentially justifies the term "generalized function" . In what follows, speaking about a "generalized function f ( t) ; f E Lioc " (say, a polynomial, or an exponent) , we will have in mind the regular generalized function i(f) corresponding to f.
3. Spaces of test functions and generalized functions
285
Here is the first, and historically most famous, example of a singular generalized function.
8 : V � C : cp �----+ cp(O) is called the Dirac 8-function, or just the 8-function. Theorem 2. The Dirac 8-function zs a singular generalized function of order 0. Definition 3. The functional
Proof. The fact that this generalized function has order zero immediately
N
follows from the estimate l 8(cp) l < ll cp ii � ) for all cp E V and N. Suppose it is regular and corresponds to some f E Lioc . Then for every interval [a, b] not containing 0, and for every cp E V vanishing outside [a, b] we have J! f(t)cp(t )dt == fiR f(t ) cp(t)dt == 8(cp) == 0. By Proposition 7, f == 0 almost everywhere on [a, b]. But the line is the union of a countable family of such intervals with the singleton {0}. Hence, f == 0 almost everywhere on JR, and therefore 8 == f is a zero functional. But, by definition, 8 ( cp) is non-zero for cp(O) # 0, a contradiction. • A
Remark. Many physicists are convinced that the 8-function is indeed a genuine function 8(t) such that 8(t) == 0 for t # 0 and 8(0) is "so large" that fiR 8(t)dt == 1 and, "as a corollary" , fiR 8(t)cp(t)dt == cp(O) for each cp E V. But
what is admissible for physicists, is forbidden for mathematicians. Again:
The 8-function is not a function, despite its name!
The same formula cp �----+ cp(O) defines a functional in S* and in £* . We denote such functionals by 8 and call them 8-functions-this will not lead to a confusion. The following class of generalized functions contains both regular gen eralized functions and a 8-function. Let BORN be the family of Borel subsets of the interval [ - N, N], BOR00 :== U{BORN : N == 1 , 2, . . . } , and J-L : BORoo � C a set function such that for every N the restriction /-LN of J-L to BORN is a complex measure on the corresponding interval. Exercise 4. The mapping f : V � C : cp �----+ fiR cp(t)dJ-L(t) , where the integral is understood as J: cp(t)dJ-L N ( t ) for sufficiently large N, is a well defined generalized function. It is regular {=:::} for each N the measure /-LN is absolutely continuous with respect to Lebesgue measure. Exercise 5. The generalized function f has order zero {=:::} it is gener ated by some measure J-L from the previous exercise. Hint for ===> . From the Riesz Theorem 1 .6.4 it follows that the restric tion of f to VN acts as an integral with respect to some complex measure /-LN ·
4.
286
Polynormed Spaces and Generalized Functions
In fact, generalized functions may appear in very different constructions. Here is another example. Below, P.V. denotes the integral in the sense of principal value. Exercise 6. The mapping cp E V �----+ P.V. fiR
0, O(t) 0, t < 0. �
For every cp and sufficiently large
() ' ( cp)
�
- l 0( t)cp' (t)dt
=
{
N we have -
fooo cp' (t)dt
=
- cp( t) � �
�
cp(O) .
Thus, the generalized derivative of the Heaviside function is the Dirac 8function. Exercise 1 * . Suppose we are given a function on the real line that has
bounded variation on every interval. Find its generalized derivative. Answer: It must be among the "measures" in Exercise 3.4. Remark. For a specific example, take the Cantor scale. 5
5The Cantor scale is a continuous non-decreasing function on the closed interval [0 , 1 ] , closely related to the Cantor set ( see footnote on p. 44) . This function, being continuous, is uniquely determined by its values on the open intervals that were deleted in the process of constructing the Cantor set. It is constant on each of these intervals, and equals 1 / 2 on the interval of rank 1 ( cf. footnote on p. 44) , equals respectively 1 /4 and 3/4 on the intervals of rank 2 , 1 /8, 3/8, 5/8, and 7/8 on the intervals of rank 3, and so on. The "classical" derivative of the Cantor scale is
4.
29 1
The structure of generalized functions
2.
Find the generalized derivative of the function ln l t l . Answer: This is the "principal value integral" from Exercise 3.5. As you know, sometimes it happens that an ordinary function is dif ferentiable almost everywhere, but cannot be recovered from its derivative. ( The same Cantor scale is the classical example. ) The theory of generalized function saves us from this unpleasant situation. Exercise 3. Let f be a generalized function. Then f' == 0 {=:::} f is constant ( the precise meaning of such a statement was discussed above ) . As a corollary, two generalized functions that have the same generalized derivative coincide up to a constant summand. Hint. The functional f vanishes on all cp'; cp E V, i.e. , on all cp such that fiR cp(t)dt == 0. Hence, f is the constant f(cpo) for every cpo such that Exercise
fiR cp0 (t)dt == 1 .
If f is a generalized function, then its generalized primitive is a general ized function g such that g' == f. Here is another advantage of generalized functions as compared with ordinary functions. Exercise 4. Every generalized function has a generalized primitive, and two such generalized primitives are equal up to a constant summand. Hint. Take cpo E V such that fiR cpo(t)dt == 1 . As a primitive for f we can take g cp �----+ -! (�) , where �(t) is the unique primitive for cp(t) ( fiR cp(t)dt)cpo(t) lying in V. Remark. This fact initiates the theory of differential equations for gener alized functions, which serves as a powerful apparatus of modern analysis and physics. You can read about it, say, in the classical monographs [72], [73], [74], or in the excellent textbook [70]. :
-
*
*
*
We have been discussing the differentiation in V* . But the main con struction can be almost literally transferred to tempered generalized func tions. Exercise 5 ° . Theorem 1 remains true if we replace V with S and V * with S* . The operator D acting on S* turns out to be a birestriction of the same operator acting on the ( larger ) space V*. Another classical operation in analysis, multiplication by a sufficiently "good" function, can also be transferred from ordinary functions to gener alized ones. -
the function that is equal to zero almost everywhere. In our "generalized" sense it is the Cantor measure. You should admit that such a "derivative" provides much more information about the initial function.
4.
292 Exercise 6. Let
1/;(t) ; t
Polynormed Spaces and Generalized Functions
E
1R be an infinitely smooth function. Then
the operator M'l/J : V � V (see Exercise 3.3) admits a unique weakly* continuous biextension to an operator M'l/J : V* � V* (called the operator of multiplication by 1/J in V* ) . If, in addition, 1/J and all its derivatives are tempered, then the statement remains true if we replace V with S. Hint. This time M'l/J is precisely the weak* adjoint operator for M'l/J. We recall the topological characterization of compactly supported gener alized functions (Proposition 3.9) . They have another, perhaps more visual, description. We define a support of a generalized function f to be an arbitrary closed set M c 1R such that f ( cp) == 0 whenever cp == 0 in some neighborhood of M. In this case it is sometimes said that f is concentrated on M, or that f vanishes outside M. (Thus, the statement "f vanishes at a point" has no meaning, but "f vanishes on an open set" has reasonable interpretation.) Remark. Among all supports of a generalized function there is the minimal
one, namely, the intersection of all supports. (Try to prove this! ) People often speak about the support have in mind this minimal support.
2. A functional f E V*
belongs to £* {=:::} it has a compact support. {In other words-and this is not a joke!-a compactly supported generalized function is precisely a generalized function with compact sup port.) Proposition
:::::=::> .
From Propositions 3.9 and 3.6 (ii) it follows that for some N and n we have l f(cp) l < ll cp ll }t' ) . Hence, if cp ( t ) == 0 for t tf_ [ - N, N] , then f ( cp ) == 0. {::::::= . By the assumption, there is N such that f == 0 outside [ - N , N] . Hence, if we put � : == F- N, N, l (see Proposition 3.2) , then for each cp E V we have f(cp - cp � ) == 0, and thus f ( cp ) == f ( cp � ) . But cp � E V N + l · There fore, by Proposition 3.6(i) , there are n and C > 0 such that l f(cp � ) l < C ll cp � ll }t'+ l ) . From the Leibniz formula for higher derivatives it easily fol lows that ll cp � ll }t' + l ) < C' ll cp ll }t' + l ) for some C' > 0. Thus, l f(cp) l < CC' II cp l r+ l ) , and therefore the functional f is continuous with respect to the topology inherited from £. It remains to apply again Proposition 3.9. • Proof.
Theorem 3 below describes the structure of generalized functions with compact support. It shows that in some sense they "do not go far away from the regular ones" . But first we have to prove an important preliminary result (Theorem 2) .
4.
293
The structure of generalized functions
Suppose f E V* and N E N. Then there exist n E Z + and a square-integrable function h(t) on [-N, N] such that f(cp ) h(t)cp ( n) (t)dt
Theorem 2.
=
for every cp E VN .
1:
By Proposition 3.6(i) , there exists n - 1 (it is now more convenient to speak about n - 1 rather than n) such that our functional, considered on VN , is bounded with respect to the norm II · 11�)1 . Take the corresponding n and for arbitrary cp, 'ljJ E VN put ( cp, 'l/J ) :� fiR cp( n) (t)'ljJ( n ) (t)dt. Lemma. ( , ·) is an inner product in VN . The norm II · ll 2,n defined by the N equality II 'P II 2 ,n JI N l cp(n) (t) l 2 dt (i. e., the norm of the corresponding near-Hilbert space} majorizes the standard norm II · 11�)1 in VN . Proof. Clearly, (· , · ) is a pre-inner product, hence II · ll 2 ,n is a prenorm in VN. For k � 0, 1 , . . . , n - 1 we put ll cp ll� :� max{ l cp ( k ) (t) I : t E [-N, N] } . Since II · 11�)1 � max{ ll · II� ; k � 0, . . . , n - 1 } , it is sufficient to show that ll · ll 2 ,n majorizes 11 · 11 � _ 1 , and for every k � 0, 1 , . . . , n - 2, 11 · 11 �+ 1 majorizes II · II � · Take cp E VN. For every k we have cp ( k) (-N) � 0, so that cp( k ) (t) � J� N cp ( k+ 1 ) (s)ds; t E [-N, N] . For k � n - 1 this equality together with the Cauchy-Bunyakovskii inequality for square-integrable functions implies Proof.
-
:=
I 'P c n - l ) (t) l
0 such that II Tx - Ax ll > c ll x ll for all x E E {in other words, T - Al is topologically injective). Then A E ar (T) . Proof. This directly follows from the fact that the image of a topologically • injective operator between Banach spaces is closed. From this we immediately have Proposition 3. Suppose A E ac(T) . Then there exists a sequence X n E E; ll xn ll 1 such that Tx n - AX n --+ 0 as n --+ oo . • The points of the spectrum can be classified in other ways, and the most important of these ways is apparently as follows. A point A is called an essentially singular point of an operator T if T - Al is a non-bijective oper ator that is not even a Fredholm operator. The subset in a(T) consisting of essentially singular points of the operator T, is called the essential spectrum of T and is denoted by ae (T) . Thus, a(T) is divided into two parts, ae (T) and its complement, according to the "remoteness of T - Al from a bijective operator." Note that Proposition 3.5. 1 gives the inclusion ac(T) C ae (T). ==
5. At the Gates of Spectral Theory
300
It is easy to show (do this) that the essential spectrum is also an invariant of the topological equivalence. Exercise 1 . The spectrum of an operator T : E --+ E coincides with the spectrum of the adjoint operator T* : E* --+ E* , i.e. , a(T*) == a(T) . In addition, (i) if A E ar (T) , then A E ap (T*); (ii) if A E ap(T) , then either A E ap(T*) or A E ar (T* ); (iii) if A E ac(T) , then either A E ac(T* ) or A E ar (T*) ; moreover, the latter inclusion is not possible if E is reflexive. Hint. The operation (* ) , being a functor after all, preserves topological isomorphisms. Therefore, a(T) => a(T*). Put S : == T - A 1 . (i) If x does not belong to the closure of Im(S) , then S* f == 0 for each f such that f i m ( ) == 0 and f(x) # 0. (ii) If Sx == 0 for x # 0, and f(x) # 0, then f does not belong to the closure of im(S*). (iii) If Im(S) is dense in E, then S* is injective. Hence, if it is not surjective, then S** (as well as S*) is a topological isomorphism. Then Proposition 2.5.2 works. Combining all this we see that a(T) == a(T*). 1 i
S
The reader who has done Exercises 3.5.8 and 3.5.9 will conclude from them that the essential spectra of T and T* also coincide: ae (T* ) = ae (T) .
Certainly, for T == A 1 ; A E C we have a(T) == ap(T) == { A }, and the same is true for ae (T) , provided E is infinite-dimensional. Somewhat more interesting is the following Proposition 4. Let Eo and E1 be non-zero closed subspaces in E, and P a projection onto Eo along E1 . Then a ( T) == aP ( T) == { 0, 1 } , 0 ¢: ae ( T) {:=:::> dim E1 < oo and {1 } E ae (T) {:=:::> dim Eo < oo . Proof. If we look at the action of P on vectors of our subspaces, we shall conclude that 0, 1 E ap(T) . If A # 0, 1, then we put Q : == 1 - P, and after that ( 1 - A) - 1 P - A - 1 Q will be the inverse operator for P - A 1 . The rest is clear. • The next result is not so simple. Theorem 1 . Let T : E --+ E be a compact operator acting in an infinite dimensional space. Then a(T) consists of 0 and at most countable set of eigenvalues for which 0 is the only possible limit point {if this set is infinite). 1 However, taking into account the previous remark in small print, advanced readers may note that the coincidence of u(T) and u(T * ) is a special case of Exercise 2 .5.10.
1.
Spectra of operators and their classification. Examples
301
Moreover, for each non-zero eigenvalue the subspace of the corresponding eigenvectors is finite-dimensional. Finally, ae (T) == {0} . From the fact that T is not a Fredholm operator (Proposition 3.5.2) it follows that 0 E ae (T) , and hence 0 E a(T) . Now take A E a(T) \ {0}. Then T - A1, as well as 1 - A - 1 T == -A - 1 (T - A1) , does not have the inverse operator. Thus, one of the two possibilities is true: either Ker(1- A - 1 T) I= 0, or Im(1 - A - 1 T) I= E. But A - 1 T is compact; hence, it follows from the Fredholm alternative that the second possibility implies the first. From this we see that the common kernel of the operators T - A1 and 1 - A - 1 T cannot vanish, which means that A is an eigenvalue of T. FUrther, the space of the corresponding eigenvectors, clearly, coincides with this kernel. Therefore, by Theorem 3.5.2 this kernel is finite-dimensional. By the same theorem, A t/: ae(T) . It remains to prove that a(T) is at most countable, and 0 is its only possible limit point. Assume that at least one of these statements is not true. Then there exists a sequence An of points in the spectrum that tends to some () I= 0; without loss of generality we can assume that I A n l > 0/2 for all n, and all An are distinct. We already know that all the A n are eigenvalues. Suppose Tx n == An Xn for X n I= 0; n == 1 , 2, . . . ; then, as we remember from linear algebra, the system Xn is linearly independent. Put En : == span{x 1 , . . . , x n }; clearly, it is an invariant subspace of the operator T. Using the lemma on the near-perpendicular, we choose, for every n > 1 , a vector Yn E En such that ll Yn II == 1 and d(yn , En - 1 ) > 1/2. Since codimEn En - 1 == 1 , we have that Yn == f.1n Xn + Zn for some f.1n E C \ {0} and Zn E En - 1 · Therefore, for every k == 1, . . . , n- 1 we have Tyn - TYk == f.1n An Xn + Tzn - TYk == A n Yn - Un , where Un : == An Zn - Tzn + Tyk E En - 1 · Hence, II TYn - TYk ll == I A n i iiYn - An 1 un ll > () /4. Thus, T maps the bounded set { yn ; n E N} to a set which is not totally • bounded. This contradicts the compactness of T. Proof.
*
*
*
We now look at our standard examples of operators and find their spec tra. At this point we restrict ourselves to the questions lying on the surface. Some things that we could have proved right now, but without much ele gance, we will obtain afterwards as immediate applications of some general results. Exercise 2. Let T11; 11 == (11 1 , /1 2 , . . . ) be a diagonal operator on lp; 1 < p < oo . Then a(T) is the closure of the set {Mn }, and ap(T) is this set itself. FUrthermore, a(T) \ ap(T) is equal to ac(T) for p < oo , and to ar (T) for p == oo . Finally, (for all p) ae (T) is the set of limit points of the sequence 11 ·
5. At the Gates of Spectral Theory
302
Hint. Put S :== T - Al. If A # f.1n for every n, then the subspace coo lies in Im( S) (and is dense in lp for p < oo ) . If A is a limit point of the sequence f.1 , then II S( pn ) ll can be as small as we wish, and in the case of l 00 we have d(�, Im(S)) > 1 for � :== (1, 1, 1 , . . . ) . The rest is clear, taking Exercises 1.4.3 and 3.5.2 into account. Exercise 3. Let (X, 11 ) be a measure space, and Tt the operator of multiplication by f E L00 (X, 11 ) in Lp (X, 11 ) ; 1 < p < oo . Then u(TJ ) is the set f(X)ess of essential values of the function f (i.e. , such numbers A that for every c > 0 the measure of the set {t E X : l f(t ) - A I < c } is positive) . Further, in the notation ZA :== {t E X : f(t) == A } we have up (TJ ) == {A : 11 ( ZA ) > 0 } , and u (TJ ) \ up (TJ ) coincides with uc (TJ ) for p < oo and with ur (TJ ) for p == oo . Finally, if (X, 11 ) is an interval [a, b] with usual Lebesgue measure, then ue(TJ ) coincides with all u(TJ ) . In particular (and, apparently, this is the most instructive case) , for the operator of multiplication by the independent variable in L 2 [a, b] we have u(TJ ) == uc (Tj ) == ue(TJ ) == [a, b] . Hint. If 11(ZA ) > 0, then for every g E Lp (X, 11 ) \ {0 } vanishing out side ZA , we have Tt g == Ag . If 11 ( ZA ) == 0, then S :== Tt - Al is injective. However, if in addition A E f(X)ess , then, taking the sets Yn :== {t E X : l f (t ) - A I < 1 / n } and their characteristic functions Xn , we see that S(xn/ll xnll ) --+ 0 for n --+ oo . Moreover, if p < oo , then the subspace U� {g E Lp (X, 11 ) : g l yn == 0 } lies in Im(S) and is dense in Lp (X, J1 ) . If, on the contrary, p == oo , then d(h, Im(S)) > 1 for h(t) 1. The rest is clear, taking Exercises 1.4.3 and 3.5.2 into account. Exercise 4. The point spectrum of the operator of left shift in lp ; 1 < p < oo coincides with JI)) 0 for p < oo , and with JI)) for p == oo . The residual spectrum of the operator of right shift in lp ; 1 < p < oo coincides with JI)) 0 for 1 < p < oo , and with JI)) for p == 1. If p == oo , the residual spectrum of the right shift contains JI)) 0 . Finally, the point spectrum of the right shift is empty for all p. Hint. If � == (1, A, A 2 , . . . ) lies in lp, then Tz� == A�. If the same � defines a functional f� : 1J �----+ 2:: � �n 1Jn on lp, then every vector in Im(Tr - Al) belongs to the kernel of this functional. The rest follows from Exercise 1. (Certainly, all that we said in Exercises 2 and 4 about lp ; 1 < p < oo is true for co as well; check this!) Remark. These exercises already show that each possibility indicated in conditions (ii) and (iii) of Exercise 1 can indeed occur (give the corresponding examples) . 1
1
The following general observation often helps in studying spectra.
1.
303
Spectra of operators and their classification. Examples
(a generalization of Proposition 1). Suppose for S : E --+ E and T : F --+ F the operators S - A1 and T - 111 ; A , 11 E C are weakly topologically equivalent. Then A E a(S) {=:::} 11 E a(T) , and the same is true for ap , ac , ar , or ae in place of a. Hint. Look at the diagram in the definition of weak topological equiva lence. Exercise 5
Here is a typical application. Let T be any of the shift operators acting on Zp , Zp (Z) , or (for a # 0) in Lp (lR) ; here we assume that 1 < p < oo . Then for each A E 1r the operators T - 1 and T - A 1 are weakly topologically equivalent. As a corollary, each of the sets a(T) , ap (T) , ac (T) , ar ( T) , and ae (T) either contains the entire 1r or is disjoint from 'lr. Hint. In the case of Tz : Zp --+ Zp the desired equivalence is implemented by the pair of diagonal operators defined by the sequences (A, A 2 , . . . ) and (A 2 , A 3 , ) Similar pairs work for Tr and Tb. In the case of Ta : Lp (JR) --+ Lp(JR) the desired pair consists of the operators of multiplication by the exponentials ei ct with parameters c depending on A and a. Exercise 6.
•
•
•
•
(Which alternative actually occurs in each concrete shift, will be clarified in Section 3. However, we already know something from Exercise 4, and will learn more in the next exercises.) 7* . The residual spectrum of the right shift operator on Zoo also contains the entire JI)) . Hint. For every 1} E Im(Tr - 1 ) we have limn � oo � l:�= I 1Jk == 0. As a corollary, the distance from (1, 1, 1, . . . ) E Z oo to Im(Tr - 1) is 1. Exercise
If p == oo , then the point spectrum of the shift operators on Zp (Z) and (for a # 0) in Lp (lR) contains 'lr. If p == 1 , then the residual spectrum of the same operators contains 1r. of the functional Hint. If p == 1, then Im( Tb - 1) is contained in the kernel 00 � r--+ 2:: � _ 00 �n , and Im(Ta - 1 ) in the kernel of x r--+ J 00 x(t)dt. The rest is clear. Exercise 8 .
The point spectrum of the shift operator by a Lp ( 'lr ) contains all numbers an ; n E Z. Exercise 9 ° .
E
1r in
Later (see Exercises 3.3-3.6 and 3.9) , we will give further information about the spectra of concrete operators. However, from Theorem 1 we al ready know a great deal about the integral operators, since they are compact.
304
5. At the Gates of Spectral Theory
2. Somet hing from algebra: Algebras
To understand better the nature and the role of spectra, it is useful to find out which of their properties are related to analysis, and which have purely algebraic character. (We have to "separate the structures" , as I. M. Gelfand2 used to say in his famous seminar.) Let us first look at spectra from a very general viewpoint of abstract algebra. Definition 1. Let A be a linear space. A bilinear operator m A x A � A is called a multiplication if it satisfies the associativity identity ( ab )c == a(bc), where the notation ab is used instead of m( a , b). A linear space endowed with a multiplication (formally speaking, a pair (A, m ) consisting of a linear space and a multiplication on it) is called a complex associative algebra, or, since we do not consider others, just an algebra. :
Associativity implies that the product a 1 a 2 an does not depend on the way we arrange parentheses in it. The simplest example of algebra is, of course, C, and the main example in this book is the space B(E) of all bounded operators in a Banach space E, equipped with the operator composition as multiplication. But the algebraists have their own favorite toys. The most important of them (and the one of much use for us) is the algebra of polynomials C[t] in a formal variable t (the "C" means that we speak about polynomials with complex coefficients) . The set Mn of matrices of size n x n with usual matrix multiplication is also an algebra, and it is very important both in algebra and analysis. Another instructive example is the algebra e x of all complex valued functions on an (arbitrary) set X with pointwise multiplication. We say that two elements a, b of a given algebra commute if ab == ba. The algebra is called commutative if all its elements commute. (Show that B(E) is commutative {:=:::> dim E == 1.) Definition 2. An element 1A of an algebra A (also denoted by 1 if there is no danger of confusion) is called a unity of this algebra if a1 == 1a == a for every a E A. If an algebra has a unity, it is said to be unital. · · ·
All the above-mentioned examples are, of course, unital; in particular, a unity in B(E) is the identity operator. But there are algebras without unity; the algebra of matrices of the form ( g B ) ; A E C is one of them. Definition 3. Let A be a unital algebra and a E A. An element a[ 1 (respectively, a; 1 ) in A is called a left {right} inverse to a if a[ 1 a == 1
2 1. M. Gelfand (born in 1913 ) , one of the most prominent mathematicians of our time. His
enormous contribution to the science includes the creation of the theory of Banach algebras.
2.
305
Something from algebra: Algebras
(respectively, aa; 1 == 1 ) . An element a - 1 is called (just) an inverse to a if a - 1 a == aa - 1 == 1 . An element that has an inverse is called invertible. Proposition 1 (resembling Proposition 0.5. 1). If an element a E A has a left inverse a[ 1 and a right inverse a; 1 , then a is invertible and a - 1 == a[ 1 == a; 1 . As a corollary, an element of an algebra can have at most one inverse element.
It is sufficient to put parentheses in the expression az- 1 aa; 1 in two • different ways. Proof.
However, sometimes an element may have many left (or right) inverse elements, but no "real" inverse element. Please, give an example, using, say, shifts in l 2 . Proposition 2. Suppose a 1 , . . . , a n E A. Then (i ) if all of them are invertible, then the same is true for a 1 a 2 an ; (ii) if all of them commute, then the converse statement is also true: the invertibility of a 1 a 2 an implies the invertibility of all factors. · · ·
· · ·
(i) Clearly, an 1 an� l a1 1 is the inverse one to a 1 a2 an (ii) We put b :== (a 1 a2 an ) - 1 . Then, by commutativity, we have (ba 1 ak - l ak+ l an ) ak == 1 and ak ( a 1 ak - l ak+ l an b) 1 for all k == 1, . . . , n. It remains to apply Proposition 1. • · · ·
· · ·
· · ·
Proof.
· · ·
· · ·
· · ·
.
· · ·
Here is the general notion of spectrum we promised before. Definition 4. Let A be a unital algebra. A complex number A is called a regular point of an element a E A if the element a A1 is invertible; otherwise, it is called a singular point of a. The set of all singular points of an element a is called the spectrum of this element and is denoted by aA ( a ) or just a (a) . -
We see that the spectrum of a bounded operator on a Banach space E defined in the previous section is precisely the spectrum of it as an element of the algebra B(E) . Remark. One can also define the spectrum of an element of an arbitrary, not necessarily unital, algebra; this can be done using the procedure of the so-called unitization (adjoining a unity) ; see, e.g., [25] or [79] . But we will not need this. What subsets on the complex plane can be the spectra of elements of algebras?
306
5.
At the Gates of Spectral Theory
The spectrum of an element of the algebra e x , i.e. , the spec trum of a given function on X, is, clearly, the set of values of this function. Thus, if X has the continuum cardinality, then every non-empty subset in C is the spectrum of some element of this algebra. Example 2. In the algebra C ( t ) of rational functions of a formal variable t every element has empty spectrum unless it is a multiple of the identity. The same is certainly true for the algebra of meromorphic functions on the plane (recall complex analysis) , and moreover, for every algebra where each non-zero element has the inverse. Example
1.
These examples already show that every subset on the complex plane, including the empty one, can be the spectrum of an element of an algebra. In the next section we will see that this "algebraic permissiveness" disappears in functional analysis. Exercise 1. If a and b are elements of a unital algebra A and a is invertible, then a(ab) == a(ba). Hint: ba - A l == a - 1 ( ab - Al)a. *
*
*
As we have already said many times, after introducing a new structure, this time the structure of algebra, we have to introduce the class of mappings consistent with this structure. Definition 5. A linear operator rp : A --+ B between two algebras is called a homomorphism if rp(ab) == rp(a)rp(b) for all a, b E A. If in addition both algebras are unital, then our homomorphism is called unital whenever rp(lA) == lB. We now formulate the following obvious result. Proposition 3 . If rp : A --+ B is a unital homomorphism and a invertible, then rp( a ) E B is also invertible.
E
A is •
Algebras and their homomorphisms form a category denoted by Alg (with obvious composition law and local identities) . Clearly, isomorphisms in Alg (called isomorphisms of algebras) can be characterized as bijective homomorphisms. The classical example is the isomorphism between B(E) with dim E == n and Mn assigning to every operator its matrix in a given basis. The structure of the category Alg is much more complicated than that of Lin or even Gr. If you are curious, you can do the following Exercise 2* . Monomorphisms in Alg are injective homomorphisms . On the other hand, not every epimorphism is surjective; at the same time the extreme epimorphisms
2.
307
Something from algebra: Algebras
are precisely surjective ones. There are fewer extreme monomorphisms than arbitrary monomorphisms. Every family of objects in Alg have a product and a coproduct . Hint. The embedding in: C[t] � C(t) is an epimorphism, and at the same time a (non-extreme) monomorphism. Products as sets are Cartesian products, while coproducts are the so-called free products, resembling free coproducts of groups.
We recall the standard fact that in "categorically designed" areas of mathematics a valuable information on the structure of an object can be obtained by studying morphisms of this object into some simplest, or at least "well-understood" objects of the category. In the category Lin such a role belongs, of course, to functionals, and in Ban-due to the Hahn Banach theorem-to bounded functionals. As for Alg, the following classes of morphisms are very useful. Definition 6. A homomorphism from an algebra A to C is called a character of A. A homomorphism from A to B(E) is called a representation of A in the Banach space E. Since the algebra B(E) for a one-dimensional E coincides (up to an isomorphism of algebras) with C, it is natural to regard characters as a special case of representations. Remark. Characters are particularly effective in the study of commutative algebras. But if an algebra if "far from being commutative" , then, generally speaking, its characters are of little use. (For example, you can verify that the algebra Mn with n > 1 has no non-zero characters at all.) The fun damental role in the study of algebras passes to representations; for details see, e.g. , [78] or, in the context of functional analysis, [25] . Under the action of homomorphisms spectra change in the following way. Proposition 4. Let rp A --+ B be a unital homomorphism between unital algebras. Then for every a E A we have aB(rp(a)) C aA(a). :
If A E C is a regular point for a, then, by Proposition 3, rp ( a )- A lB == rp( a - AlA) is invertible, and thus, A is a regular point for rp( a). The rest is • clear.
Proof.
Certainly, the spectrum of an element may be preserved under the action of a homomorphism, or it may become smaller (give examples) . *
*
*
We were talking about homomorphisms from arbitrary algebras to some standard algebras. Now let us consider an important class of homomor phisms from a standard algebra to arbitrary ones.
5. At the Gates of Spectral Theory
308
Let p(t) == co + c 1 t + · · · + cn tn be a polynomial (i.e. , an element of the algebra C [t] ) , A a unital algebra, and a E A. Then the element co l + c1 a + · · · + cn an E A is called the value of the polynomial p on a or just a polynomial p of a and is denoted by p(a). The mapping rp : C[t] --+ A : p �----+ p (a) (which is clearly a unital homomorphism) is called the polynomial calculus of a in A. Definition 7.
(The subscript in the notation rp indicates that later we shall come across some other "calculi" . ) The polynomial calculus interacts well with spectra. For a complex valued function f defined on a set M C C, we denote the set { / (A) : A E M } briefly by f ( M ) . In particular, speaking about the set p(M); p E C [t] we regard p as the corresponding function of a complex variable. Theorem 1 (Spectral mapping law for polynomial calculus). Let A be a unital algebra, a E A and p E C[t] . Then a (p (a) ) == p ( ( a) ) . a
Take 11 E C. By the fundamental theorem of algebra, a polynomial p(t) - 11 can be decomposed into a product c(t - AI) · · · ( t - An ) , where AI , . . . , An are roots of this polynomial, n its degree, and c # 0. Since the polynomial calculus is a homomorphism, the element p( a ) - 111 can be decomposed into the product c ( a - Ai l) · · · (a - A n l) . Since all a - A k l; k == 1 , . . . , n commute, Proposition 2 shows that invertibility of p( a ) - 111 is equivalent to invertibility of all a - A k l . Therefore, 11 E o-(p( a )) {:=:::> A k E a ( a) for at least one k {:=:::> A E (a) for at least one root A of the polynomial • p (t) - 11 · The last condition, certainly, means that 11 E p( o-(a)). Proof.
a
Here are some of the simplest examples showing how this theorem works. An element a of an algebra is called idempotent if a 2 == a, and nilpotent if an == 0 for some n. Proposition 5. Let A be a unital algebra, and a E A. Then (i) if a is idempotent and different from 0 or 1, then a ( a) == {0, 1 }; (ii) if a is nilpotent, then a ( a ) == {0} . Proof. (i) Since p( a) == 0 for p(t) :== t 2 - t, for such p we have o-(p(a)) == {0} . Hence, by Theorem 1, A E o-(a) implies A 2 - A == 0, and consequently, o-(a) cannot contain numbers other than 0 or 1 . Moreover, the equality a(a - 1)== 0 and the fact that both factors are non-zero evidently imply that they are both non-invertible. The rest is clear. (ii) Now p( a) == 0 for another polynomial, namely, p(t) :== tn . Hence, by • the same theorem, A E o-(a) implies An == 0.
2.
309
Something from algebra: Algebras
In particular, we obtained a new and almost instantaneous proof of Proposition 1.4 concerning the structure of the spectrum of a projection. Up to now we were considering polynomials of elements of algebras. Now, to speak informally, we shall take the function 1/t of these elements. Proposition 6. Let a be an invertible element of a unital algebra A. Then a(a - 1 ) == { A - 1 : A E a(a)}. {In other words, a ( a - 1 ) == f( a(a)), where j (A) == 1 / A.) Proof. Clearly, a(a - 1 ) , as well as a ( a), does not contain 0. Further, for A # 0 we have a - I - A 1 == - A1a - I ( a - A - l l). Therefore, taking into account Proposition 2, we see that a - - A l is invertible {=:::} a - A- 1 1 is invertible. The rest is clear. • :
One can combine Theorem 1 and Proposition 6 as follows .
=
Exercise 3. Let r (t) p(t) jq(t); p, q E C [t] be a rational function of the formal variable t (we assume that the polynomials p(t) and q(t) are relatively prime) , and a an element of a unital algebra A such that the spectrum of a does not contain the poles of the function r (t) (i .e . , the roots of the polynomial q(t) ) . Then for the element r(a) : = p(a) /q(a) (which exists by Theorem 1 ) 3 we have
a ( r (a))
= r (a(a) ) ,
where r in the right-hand side is viewed as a function of a complex variable. Remark. Let a be an element of a unital algebra A. Then it is easy to verify that a subset C a (t) in C (t) consisting of all rational functions with all poles outside a(a) is an algebra with respect to the same operations as in C(t) . Assigning to every r E C a (t) an element r( a) E A considered in the previous exercise , we obtain a homomorphism lr : C a ( t) ---+ A extending (as a map) the polynomial calculus. It is called the rational calculus of the element a. The meaning of Exercise 3 is that this more general calculus also satisfies the spectral mapping law.
The rational calculus is probably the most general "functional calculus" in pure alge bra. In the next section we pass from algebra to analysis , and this will allow us to speak about a wider class of functions of elements of (this time not "pure" ) algebras.
We introduce several general notions. Definition 8. Let A be an algebra. A subspace B in A is called a subalgebra of A if a, b E B implies ab E B. A subspace I in A is called a two-sided ideal or just an ideal in A if a E I implies ab, ba E I for every b E A. Certainly, every subalgebra B in A is an algebra with respect to the multiplication inherited in the obvious way from A. FUrther, every ideal is a subalgebra. Finally, if the algebra is unital, then from the definition of the ideal it follows that 1 E I {=:::} I == A. Hence, if A is not one-dimensional, then span{l } is a simplest example of a subalgebra which is not an ideal. 3 The element r ( a ) can be called "the value of the rational function r on the element
a '' .
5 . At the Gates of Spectral Theory
310
In every commutative algebra A the subset { aao; a E A} , where ao is a chosen element in A, is an ideal. (Using the Euclidean algorithm, show that there are no other ideals in e [t] .) If X is an arbitrary set, and Y a subset in X, then the set { f E e x : f l y == 0 } is an ideal in e x . (Note that in the case when X is infinite, there are other ideals; find some of them.) In the classical functional analysis the following example is one of the most important. Example 3. The set K(E) consisting, as we remember, of compact oper ators on a Banach space E, is an ideal in the algebra B(E) of all bounded operators on E. This is just the "scientific form" of Proposition 3.3.5. Let us also pay attention to the set K+ (E) :== { A l + T : T E K (E) } . It is certainly a subalgebra in B(E) ( "the subalgebra of compact perturbations of the scalar operators" ) . As we already discussed in Section 3.5, for all concrete Banach spaces mentioned in this book, K+ (E) is strictly smaller than B(E) , but whether it is always true is a well-known open problem.
But we digressed from our main theme here. Let us go back to the pure algebra and introduce one more notion we shall need later. Let A be an algebra, and I an ideal in A; consider the quotient space A/ I. Take a, b E A and choose arbitrary elements a' , b' in the cosets a + I and b + I. Then ab - a' b' == (a - a')b + a'(b - b' ) belongs to I. This means that the coset of the product ab modulo I does not change if we replace a and b with some other representatives of their respective cosets. Thus, by assigning to every pair (a + I, b + I ) the coset ab + I, we define a mapping from A/ I A/ I to A/ I. Clearly, it has all the properties of a multiplication in A/ I. Definition 9. The space A/ I endowed with the indicated multiplication is called the quotient algebra of the algebra A by the ideal I. x
Obviously, the natural projection pr: A A/ I is a homomorphism of algebras. FUrthermore, if A is unital, then A/ I is also unital: lA + I is its identity. The example of a quotient algebra most important for this book is B(E)/K(E) , where E is a Banach space. Such an algebra is called a Calkin algebra and is denoted by C(E) . You can guess that studying this algebra is equivalent to studying operators up to compact perturbations. In particular, the Nikol' skii theorem for the case E == F can be formulated as follows. Proposition 7. An operator S E B(E) is Fredholm {=:::} its coset S + K(E) is invertible in the algebra C(E) . • -t
This in turn implies
3. Banach algebras and spectra
311
The essential spectrum of an operator T E B(E) is precisely the spectrum of the coset T + JC(E) as an element of C ( E ) .
Corollary 1 .
3 . Banach algebras and sp ectra of t heir element s .
Fredholm op erators revisited
In the previous section, we had a gasp in the somewhat rarefied atmosphere of abstract algebra. Now let us see what happens on the "lower floor" , namely consider an operator algebra B(E) not from the pure algebraic point of view, but, as von Neumann used to say, from the point of view of "abstract analysis" . The role of the foundation now goes to the following structure. Definition 1 . A Banach space A with multiplication is called a Banach algebra if this multiplication is related to the norm by the following inequal ity: l l ab l l < l l a l l ll b l l · This inequality is called the multiplicative inequality ( for the norm ) . In particular, we see that the bilinear operator of multiplication is bound ed and its norm is not greater than 1 . Hence, as a special case of Proposition 1.4. 1 1 , we obtain Proposition 1 . In a Banach algebra, if an tends to a, and bn tends to b, • then an bn tends to ab. Remark. Exercise 1 .4.7 shows that the bilinear operator of multiplication is continuous with respect to the Tychonoff topology in A A. But in the future, when speaking about continuity of multiplication, we shall have in mind Proposition 1 . x
Actually the same structure of Banach algebra arises if we require an apparently much weaker relation between the multiplication and the norm. Exercise 1 . Let A be a Banach space endowed with a multiplication which is separately continuous as a bilinear operator. Then one can endow A with a norm II · II ' equivalent to the initial norm and such that (A, II · II ' ) is a Banach algebra. Moreover, if A is unital, then this norm can be chosen in such a way that II l ii ' == 1. Hint. Use Theorem 2.4.5. There are three main sources of Banach algebras in analysis: the theory of operators, the theory of functions, and harmonic analysis. ( You already know something about the first two areas, and if you heard about Fourier series, then you know something about the most standard part of the third area as well. )
312
5. A t the Gates of Spectral Theory
The main Banach algebras in the theory of operators are, as you can guess, B( E ) and K( E) ; they are most studied and important for applications in the case of Hilbert spaces. We recall that both algebras are endowed with the operator norm and composition of operators as multipli cation. Example 2 ( the main Banach function algebra) . This is the Banach space C(O) , where 0 is a compact space ( see Section 3. 1) . This algebra is endowed with the pointwise multiplication. A slightly more general example is the Banach algebra C0 (0) , where 0 is a locally compact space ( again, cf. Section 3. 1). Example 1.
Remark. We should treat algebras in these examples ( first of all Co (O ) and B(H) , where H is a Hilbert space ) with special respect . They are the ranges of two canonical homo morphisms, the "Gelfand transform" and the "universal representation" , both playing the fundamental role in the general theory of Banach algebras ( cf. [25 , p. 1 79] ) . We shall say something about these constructions at the end of this chapter and in Chapter 6.3.
The Banach space l00 and, more generally, L00(X, 11) are also Banach algebras with respect to the pointwise ( in the case of l00 it is better to say coordinatewise ) multiplication. Speaking about L00 ( X, 11), we have in mind coordinatewise multiplication of representatives of the corresponding cosets; certainly, the coset of the product does not depend on the choice of such representatives. ( Actually, these algebras are special cases of algebras from Example 2. The meaning of this strange declaration will be clarified later in Section 6.3.) Example 4. The space cn [ a , b] ( cf. Example 1.1.7) is a Banach algebra with respect to the norm ll x ll ' :== L:�= O �� max{ x ( k ) (t); a < t < b} ( check that it is equivalent to the norm indicated in Example 1 . 1.7) and the pointwise multiplication. Example 3.
In the last chapter of the book we shall discuss typical Banach algebras of harmonic analysis, namely, Banach spaces £ 1 ( G) , where G :== Z, JR, or 'lr, endowed with the so-called convolution multiplication. Here we describe the simplest of them. Example 5 ( one of the guises of the Wiener algebra) . The Banach space l 1 (Z) is a Banach algebra with respect to the multiplication (� 1J ) n :== I: � _ 00 �k 1Jn - k · Here the product of � and 1J is denoted by � 1J and is called the convolution. Verify that the convolution indeed belongs to l 1 (Z) . In Chapter 7 we shall explain to advanced readers another appearance of the Wiener algebra, this time as a certain algebra of continuous functions. *
*
3. Banach algebras and spectra
313 *
*
*
Like Banach spaces, Banach algebras are the objects of two natural cat egories. In the first of them the morphisms are all bounded (as operators) homomorphisms, and in the second, only contraction homomorphisms are allowed. We shall not go deep into the study of these categories. It is nec essary to note only that isomorphisms between objects of the first category, which are called topological isomorphisms of Banach algebras , are, clearly, bounded homomorphisms with bounded inverse; by the Banach theorem, these isomorphisms can be characterized just as bijective bounded homomor phisms. Isomorphisms in the second category, called isometric isomorphisms of Banach algebras are bijective isometric (as operators) homomorphisms. We now present the first substantial result of the theory of Banach alge bras, which will give you an idea of one distinctive feature of this theory. It turns out that in some situations the continuity (i.e. , boundedness) of opera tors between Banach algebras can be deduced from pure algebraic properties. Here is the oldest example of such a "theorem on automatic continuity" . Proposition 2. All characters of Banach algebras are contraction opera tors. Suppose, on the contrary, that we have a character x : A � C such that x(a) == 1 for some a with ll a ll < 1. Then the series 2:: � an converges to some b E A, due to the estimate l an l < ll a ll n and the completeness of A. From the continuity of the multiplication it follows that a + ab == a + limn � oo a l:�=l ak == b. Hence, 1 + x( b) == x(a + ab ) == x(b) , which, of • course, is impossible. Proof.
1
The entire book [79] is devoted to the phenomenon of automatic conti nuity, and there you can find much deeper results. In Section 6.3, we shall offer to our advanced readers another striking result of that kind. Among subalgebras and ideals of Banach algebras, the closed ones are naturally of most interest. The ideal K(E) in B(E) is one of them (see Proposition 3.3.4) . Proposition 3. Let I be a closed ideal of a Banach algebra A . Then the quotient algebra A/ I endowed with the quotient norm is a Banach algebra. Proof.
Then
Our goal is to verify the multiplicative inequality. Take ii , b E A/ I. -
-
-
lliib l == inf{ ll c ll : c E iib} < inf { -ll ab ll : a E -ii , b E b} < inf{ ll a ll ll b ll : a E ii , b E b} == lliill ll b ll · •
5. At the Gates of Spectral Theory
314
In particular, we see that the Calkin algebra (introduced at the end of the previous section) is a Banach algebra with respect to the quotient norm. Let us look at the behavior of invertible objects of Banach algebras. If A is a unital algebra, then the set of its invertible elements is denoted by Inv(A) . By Proposition 2.2, this is a group with respect to the multi plication inherited from A. We shall denote by inv : Inv(A) � Inv(A) the mapping a �----+ a- 1 .
4.
Let A be a unital Banach algebra. Then (i) for each a E A; ll a ll < 1 the element 1 - a is invertible, and the inverse element can be represented as a sum of the series ( 1-a ) - 1 == 1 + 2::� 1 an {"the K. Neumann series "); (ii) the mapping inv is continuous at 1.
Proposition
(i) Consider the formal series 1 + l:C: 1 ak . Since ll ak ll < ll a ll k , the "Weierstrass test" guarantees the convergence of this series to some b E A. Let bn be partial sumsn Iof this series. Then from the obvious equality (1-a)bn == bn (l-a) == 1-a + and from the continuity of the multiplication it follows that (1 - a ) b == b (l a ) == 1. (ii) From the form of the K. Neumann series it follows that for II a ll < 1 we have Proof.
-
Hence, for inv(l ) .
b
:
-
1 - a we have limb�l inv(b)
== lima�o (l
- a) - 1
1
•
Let A be a unital Banach algebra. Then (i) the group Inv(A) is an open subset of A; (ii) the mapping inv is continuous everywhere on Inv(A) .
Proposition 5.
(i) Take a E Inv(A) . Then for every b E A such that ll b ll < l a- 1 11 - 1 , by the multiplicative inequality we have ll a- 1 b ll < 1. Taking into account Proposition 4(i) , we see that a + b == a(l + a- 1 b) E Inv(A) . Thus the whole U(a, r) lies in Inv(A) for r == ll a- 1 11 - 1 . (ii) Take c > 0. By Proposition 4(ii) , there exists 8 > 0 such that for every c E A with l c ll < 8 we have 1 - c E Inv(A) and 11 ( 1 - c) - 1 - 1 11 < c/ ll a- 1 11 · If b is such that ll b ll < 8 /ll a- 1 11 , and, as a corollary, l a- 1 b l < 8, Proof.
3. Banach algebras and spectra
315
then
ll ( a - b) - 1 - a - 1 11 == ll (a (1 - a - 1 b )) - 1 - a - 1 11 == 11 (1 - a - 1 b) - 1 a - 1 - a - 1 11 == 11 (( 1 - a - 1 b) - 1 - 1)a - 1 ll < 11 (1 - a - 1 b) - 1 - 1 ll ll a - 1 ll < l a - 1 l ( c:/ ll a - 1 ll ) == c.
The rest is clear. • Let us see how these facts affect the behavior of spectra. Recall that each subset in C can be the spectrum of an element of an abstract algebra. In the context of Banach algebras the situation is quite different. Theorem 1. Let A be a unital Banach algebra, and a E A . Then a(a) is a compact {i. e. , bounded and closed) set in the disk { z E C : l z l < l a ll }. Proof. We must verify that the set of regular points of an element a con tains all A with I A I > II a II , and that it is open. The first condition follows from the fact that for the indicated A the element 1 - A - I a is invertible by Proposition 4 ( i ) . Hence, the element a - A1 == -A(1 - A- 1 a) is invertible as well. The second condition follows from the fact that the invertibility of an element a - A1 implies the invertibility of elements sufficiently close to a - A1 ( Proposition 5 ( i )) . In particular, all elements a - 111 for 11 sufficiently • close to A are invertible. Example 6. The spectrum of an element f of the algebra C( O ) is, obviously, f(O) , i.e. , the set of values of the function f. In particular, in the algebra C[a, b]; a, b E 1R the spectrum of the function t �----+ t ( independent variable ) is the interval [a, b]. This special case plays an important role in the exposition of fragments of spectral theory in Chapter 6. Proposition 6. Let a be an invertible element of a unital Banach algebra A such that II a ll < 1 and ll a- 1 1 < 1 . Then the spectrum of a is contained in the unit circle. Proof. By the previous theorem, a ( a), a ( a- 1 ) c JI)) . At the same time, by • Proposition 2.6, A E a(a - 1 ) {=:::} A- 1 E a(a). The rest is clear. Corollary 1. Let T be an operator in a Banach space. If it is an isometric isomorphism {in particular, a unitary operator in a Hilbert space), then
a(T) C 'lr.
However, conclusions that are even more important can be derived from the fact that the spectrum of an operator cannot be empty. To prove this fact, we use complex analysis. Unless stated otherwise, everywhere in the sequel, A is a unital Banach algebra, a is an element of A, and Reg ( a ) is the set of regular points of a
5. At the Gates of Spectral Theory
316
(i.e. , C\a ( a ) ) . Due to the compactness of a ( a) , Reg (a) is an open set in C and it contains a neighborhood of the point at infinity. For A E Reg( a) we put R (A) :== (a - A1) - 1 . It evidently follows from Proposition 5(ii) that the mapping R : Reg(a) --+ A : A �----+ R(A ) ( the so-called resolvent function of the element a ) is continuous. Proposition 7 (Hilbert ' s identity) . For every A, 11 E Reg ( a) we have R(A ) - R( M ) == (A - M )R( M )R(A). Multiplying both elements from the left by a - 111 and from the • right by a - A1, we obtain (A - 11 ) 1 in both cases. Proposition 8. For every bounded functional f A --+ C the function W J : Reg(a) --+ C : A �----+ f ( R ( A) ) is holomorphic. Proof.
Take Ao E Reg (a) ; then from the Hilbert identity it follows that for all A E Reg ( a) we have WJ (A = �: (Ao) R( Ao )R(A).
Proof.
l
=
The mappings R and f are continuous, hence the right-hand side has a limit as A --+ Ao , namely, f( R (Ao)) 2 . We have verified the Riemann definition of • a holomorphic function. Theorem 2. The spectrum of every element a of a unital Banach algebra is a non-empty set in C. On the contrary, suppose that a ( a) == 0 , in other words, Reg( a ) == C. Then for every f E A* , w f (A) is an entire analytic function. The continuity of inv at 1 implies that 1 ( 1 - A - 1 a) - 1 ) == 0 . -A lim R A == lim ( ( ) .A-+ oo .A-+ oo Proof.
In particular, lim .A -+ oo W J (A) == 0. Thus, W J is bounded, and by the Liouville theorem, it is constant. Therefore, W J (A) 0. Thus, for each A E C and f E A* we have f( R (A)) == 0. Since f E A* is arbitrary, Theorem 1 .6.3 shows that R (A) 0. This contradicts the fact • that invertible elements in any algebra are different from zero.
Putting A == B(E) in Theorems 1 and 2 and, with Corollary 2. 1 taken into account, also A == C ( E ) , we immediately obtain Corollary 2. Both the spectrum and the essential spectrum of a bounded operator in a Banach space are non-empty sets.
3. Banach algebras and spectra
317
Note that there are no restrictions on the spectrum of an element of a Banach algebra, other than the requirements of compactness and non emptiness. Exercise 2. Every non-empty compact set K c C is a spectrum of some element of a unital Banach algebra. Hint. The spectrum of a function w ( z ) == z as an element of C(K) is K. The same is true for the spectrum of an appropriate diagonal operator on lp (see Exercise 1. 2) . We have come to the theorem that underlies the majority of results of the whole theory of Banach algebras (see, e.g. , [25 , 79 , 80, 101] ) . Theorem 3 (Gelfand-Mazur) . Let A be a unital Banach algebra, and at the same time a division ring {in other words, every non-zero element of A is invertible). Then A coincides, up to an isomorphism of algebras, with the field C . By the previous theorem, for every a E A there is at least one A E C such that a - A1 is not invertible. Since A is a division ring, this means that • a == A1. Hence, A == {A1 ; A E C} . The rest is clear.
Proof.
Now we draw our attention away from facts of general character, and instead use them to advance much further in our study of spectra for concrete operators. Exercise 3 ( cf. Exercises 1 .4 and 1. 7) . The spectrum of the operator of left or right shift on lp ; 1 < p < oo , and on co, is JI)) . In more details, (i) in the case of lp ; 1 < p < oo and co we have ap (Tz ) == ar (Tr ) == JI)) 0 , ac (Tz ) == ac (Tr ) == 'lr, and ar (Tz ) == ar (Tr) == 0 ; (ii) in the case of l1 we have ap(Tz ) == JI))0 , ac (Tz ) == 'lr, ar (Tr) == JI)) , and ar (Tz ) == ap (Tr ) == ac (Tr ) == 0 ; (iii) in the case of Zoo we have ap (Tz ) == ar (Tr ) == JI)) , and ac (Tz ) == ac (Tr ) == ar (Tz ) == ap (Tr ) == 0 . Hint. The result about the entire spectrum follows from Exercise 1 .4 and Theorem 1. For the continuous spectrum, the points of 1r are neither the eigenvalues of the operator under consideration, nor the eigenvalues of the adjoint operator. Hence, Exercise 1 . 1 (i) works. Exercise 4 (cf. Exercise 1 .8) . For all lp ; 1 < p < oo the spectrum of the operators Tb lp (Z) � lp (Z) and Ta Lp (lR) � Lp (IR) ; a # 0 is 'lr. In more details, if 1 < p < oo , then the entire spectrum is continuous, if p == 1 , then it coincides with the residual spectrum, and if p == oo , then it coincides with the point spectrum. :
:
318
5. At the Gates of Spectral Theory
Hint. The location of the spectrum follows from its non-emptiness , Corollary 1, and Exercise 1.6. The same Exercise 1.6 allows us to restrict our consideration to the point 1 E 1r. We immediately see that 1 is an eigen value of our operator only if p == oo. Together with Exercises 1 . 1 (i) and 1.8, this gives the classification of the spectrum in the case of 1 < p < oo. Exercise 5 The essential spectrum of the operators Tz , Tr : lp � lp , Tb : lp (Z) � lp (Z) , and Ta : Lp ( lR ) � Lp( lR) ; a # 0 (everywhere 1 < p < oo ) is 'lr. Hint. As a clarifying example consider the operator Tz . Take A E [))0 and in the Calkin algebra C ( lp ) consider the equality ( (Tz - A1) + JC(lp )) (Tr + JC(lp )) == (1 + JC(lp )) - A(Tr + JC (lp ) ) . By Proposition 4(i) , the right-hand side is invertible. Hence, Proposition 2.7 and the Fredholm property for Tr show that Tz - A1 is a Fredholm operator. Hence, ae ( Tz ) C 'lr, and it remains to use the non-emptiness of the essential spectrum and Exercise 1.6. Exercise 6 ( cf. Exercise 1. 9) . The spectrum of the shift operator by an element a E 1r in Lp ( 'lr ) is as follows: (i) if a == e2 1ri 7: , where m and n kare relatively prime natural numbers, then a(Ta) == ap (Ta) == { e 21r�· n ; k == 0, 1 , . . . n - 1 }; (ii) in the other cases a(Ta) == 'lr. Hint. In the first case, due to the equality T'/: == 1, Theorem 2 . 1 works, and in the second, Theorem 1 works. Remark. Actually, in the "Hilbert" case p == 2 every shift operator figuring in the three previous exercises except Tz and Tr , is unitarily equivalent to some operator of multiplication by a function (which at first sight bears no resemblance to it) . If we knew this by now, we could immediately obtain the corresponding facts about spectra as simple special cases of Exercise 1.3. These unitary equivalences are established by the Fourier operators which we did not discuss yet. We shall consider them in the last chapter (see Proposition 7.4.5) . We suggest another curious application of the theorem on the properties of the spec trum. Exercise 7 ( "on the complexity of quantum mechanics" ) . No Banach algebra has a pair of elements a, b such that for some A E C; A =f. 0 we have ab b a == Al.
-
Hint. Replacing, i f necessary, a with a + ( II a l l + 1 ) 1 , we can assume that a is invertible. Taking into account Exercise 2 . 1 , we have a ( a b ) == a ( a b ) + A. But if a set in C does not change under the shift by A =f. 0, then it is either empty or unbounded. How is quantum mechanics related to all this? As a special case of the obtained result , we see that in a Hilbert space there is no patr of bounded operators S and T
3. Banach algebras and spectra
31 9
sattsfytng the "canontcal commutatton relatton" ST - TS == ilil , where n ts the Planck constant. However, precisely such a pair must enter basic known mathematical models of quantum mechanics (giving , in particular, the mathematical picture of the Heisenberg indeterminacy principle) . That is why we have to look for such pairs among unbounded operators, which are much more difficult to work with than bounded ones ( cf. the epigraph to Chapter VIII in [63] ) .
Now we introduce an important numeric characteristic of the spectrum. Definition 2. Let a be an element of a unital Banach algebra A. The spectral radius of a is the number r ( a) :== max { I A I : A E ( a)}. We see that the spectral radius is defined in pure algebraic terms inde pendently of the norm. Nevertheless, it has an explicit description in terms of the norm. Theorem 4 ( Spectral radius formula ) . For every a in the previous definition a
we have
r ( a ) == nlim -+oo �Proof. Put U :== { A : I A I > r ( a )} and V :== { A : I A I > ll a ll }. By Theorem 1 we have V U. Take an arbitrary f E A* . By Proposition 8, the function W J is holomorphic in U and, as a corollary, in V. But by Proposition 4 ( i ) , for every A V we have R (A) == - A - 1 (1 - A - 1 a ) - 1 == - L A - ( k+ l )ak . k=O Hence W J can be expanded into the Laurent series - L:r 0 A - ( k + 1 ) f(a k ) in C
E
00
But then, by well-known properties of holomorphic functions, the same Laurent series represents our function in U as well. Thus, in particular, for every A E U we have limn-+ oo A- ( n + 1 ) f(an ) == 0. Now take A E U and "release" f. Taking into account Proposition 2.4.5, from the last equality we see that for some C > 0 and all n we have II A- (n+ 1 ) an ll < C, so that � < I A I �- Passing to the upper limit, we see that limn-+ oo � < I A I . Since A is an arbitrary number satisfying the inequality A > r(a), we obtain (1) limn-+ oo � < r ( a). Now take A E (a). Combining Theorem 2. 1 ( for p (t) == tn ) with The orem 1, we see that l A i n < ll an ll · Consequently, I A I < � for every n. Passing to the lower limit, we see that, by the definition of spectral radius, (2) limn-+ oo � > r (a). It remains to combine inequalities ( 1 ) and ( 2 ) . • V.
a
5. At the Gates of Spectral Theory
320
The spectral radius formula can be proved without using holo morphic functions; see [80 , pp. 22-23] . Remark.
The formula we discuss provokes interest to the following class of ele ments in Banach algebras. The behavior of these elements resembles the behavior of nilpotent elements in abstract algebras. Definition 3. We say that an element a of a Banach algebra is topologically nilpotent or quasinilpotent if limn � oo � == 0. Certainly, this condition is equivalent to the fact that the sequence ll an ll tends to zero faster than e - an for every a > 0. From the spectral radius formula we have Corollary 3. Quasinilpotent elements in a Banach algebra are precisely those for which the spectrum consists of zero only. Find the spectrum of a topologically nilpotent element without using the spectral radius formula. Hint. If A # 0, then the series 2:: � A - ( n+ I )a n converges to the inverse of Al - a. Now we can find the spectrum of an old friend. Exercise 9. The operator T of indefinite integration (in £ 1 [0, 1] , £ 2 [0, 1] , and C[O, 1] ; see Example 1.3.5) is a topologically nilpotent operator, and, as a corollary, its spectrum consists of a single point, namely, zero. The same is true for the Volterra operator in L 2 [a, b] (see Example 1.3.6) with essentially bounded kernel K ( s, t) . As a corollary, for such K the integral equation K(s, T)x(T)dT - Ax(s) == y(s) Exercise 8.
0
t 1
has a unique solution in L2 [a, b] for each A E C \ {0} and for every right hand side y E L 2 [a, b] . (This type of integral equations is called the Volterra equations ; cf. the Fredholm integral equations in Section 3.5.) Hint. Let T be the operator of indefinite integration. Then for every in the unit ball of the corresponding function space, rn + I x is a continuous function satisfying the estimate 1 Tn+ 1 x(t) l < t n jn! ; t E [0, 1] . x
This result is true for the Volterra operators with an arbitrary square integrable kernel. But this is much more difficult to prove (see, e.g. , [ 8 1 , exercise 14 7] ) . Remark.
*
*
*
3. Banach algebras and spectra
321
Recall that one can "take polynomials" of every element of an abstract algebra. Our possibilities become wider if we go over from abstract alge bras to Banach algebras: in this situation we can "take some holomorphic functions" of elements. In this book taking entire functions will be sufficient. Consider the set O(U) of holomorphic functions on a domain U of a complex plane, and write 0 instead of O(C) . We know already that O(U) and, in particular, 0 is a polynormed, and as a corollary, a topological space (see Section 4. 1 ) . But this set, certainly, is also an algebra with respect to the pointwise operations. Clearly, these two structures agree. Proposition 9. If a sequence Vn tends to v and Wn to w in O(U) (we mean the Weierstrass convergence; cf. Section 4. 1}, then Vn Wn tends to vw. • Take a unital Banach algebra A and an element a E A. Suppose w : C � C is an entire holomorphic function, and �C: ck z k is its Taylor series. Then the numeric series �C: l ck l ll a ll k converges. This, due to the estimate II an II < ll a ll n andk the Weierstrass test, guarantees the convergence of the series �C: ck a in the Banach space A. We denote the sum of this series by w(a) . Definition 4. The element w(a) E A is called the value of the entire func tion w at the point a. The mapping re : 0 � A : w �----+ w(a) is called the entire holomorphic calculus, or just the entire calculus of a. 0
0
0
In particular, we can speak about the element exp (a) (denoted also by e a ) or, say, about the element sin(a) . They are defined as !e (w) , where in the first case exp(z) , and in the second one, sin(z) is taken as w E 0. Note that the algebra of (formal) polynomials is, up to an isomorphism of algebras, a subalgebra of 0 consisting of "polynomial functions" . Proposition 10. The entire calculus (of every a} is a continuous unital homomorphism of algebras, extending the polynomial calculus. First, it is clear, that 'Ye is a linear operator. Now suppose K c C is the disk of radius r > ll a ll centered at zero. By the classical Cauchy inequality (see, e.g. , [12, p. 1 1 1] ) , for every w E 0; w(z) == �� Cn Zn we have l cn l < ll�llK . Hence 00 00 ll w(a) ll < L cn an < L l cn l ll a ll n < C ll w ii K , n ==O n ==O where C = (� � ���r ) . By Theorem 4. 1 . 1 , le is a continuous operator. Suppose now that w 1 , w 2 E 0 are represented by the Taylor series �C: ck z k and �� dk z k . Consider the partial sums Pn : == ��==O ck z k Proof.
0
:
0
0
0
5. At the Gates of Spectral Theory
322
and Qn :== ��= O dk z k ; n E N. We know that the polynomial calculus is a homomorphism, hence (Pn Qn )(a) == Pn ( a)qn (a). The sequence Pn tends in 0 ( i.e. , in the sense of Weierstrass ) to W I , Qn tends to w2 , and consequently, Pn Qn tends to W I W2 ( by Proposition 9) . Thus, due to the continuity of the mapping w �----+ w (a), we have limn --+ oo Pn ( a ) == W I (a), limn --+ oo Qn (a) == w2 ( a), and limn --+ oo (Pn Qn )(a) == (wiw 2 )( a ). But the multiplication in A is also con tinuous ( Proposition 1 ) ; hence, li mn --+ oo Pn ( a ) qn (a ) == W I ( a )w 2 (a). The rest • is clear. Later ( Section 6.2 ) a special case of this proposition where W I ( z ) exp ( z ) and w 2 ( z ) :== exp ( -z ) will make our life much easier. Corollary 4. For every a the element exp ( a) is invertible with the inverse exp ( -a ) . :
The entire calculus is as good in dealing with spectra as the polynomial calculus.
11. For every w E 0 we have w(a ( a )) C a(w(a)). Proof. Take A E a ( a ) and put wo( z ) :== w( z ) - w(A) E 0 . Since wo(A) == 0, we have wo(z) == (z - A)wi ( z) for some W I E 0 . Since our calculus is a Proposition
homomorphism, we have
w( a ) - w(A)1 == !e(wo) == !e( z - A ) !e(wi) == (a - A1)wi (a). But a - A1 is not invertible, and by the commutativity of the algebra 0 , a - A1 and wi (a) commute in A. Hence, by Proposition 2.2 ( ii ) , w(a) - w(A)1 •
is not invertible. The rest is clear.
Actually, the converse inclusion also holds. We will not prove it, but you should know this fact. Theorem 5 ( spectral mapping law for an entire calculus, [25 , Chapter II, Theorem 2.23]) . For every w E 0 we have w(a(a)) == a(w(a)) . Remark. Thus, we know what are the exponential and the sine of an element in a Banach algebra, in particular, of a bounded operator. But can we speak, for example, about the logarithm of such an element? This is not clear since the function log z can be defined only in domains that are smaller than C. It turns out that the situation is as follows Gelfand, 1 939) : for a holomorphic function w in a domain U C C, one can give a reasonable definition of the element w (a) {=:::::} U contains a(a) . Moreover, the corresponding mapping w �-----+ w (a) from 0 ( U) to A the so-called holomorphic calculus of a in U ) is a homomorphism, for which in addition the corresponding version of the spectral mapping law is true. If U is an open disk, or, more generally, an annulus, then the element w (a) can be easily constructed by analogy with an entire function of a: we should put our element instead of the complex variable into the Taylor or Laurent expansion of a given function. For more complicated domains the element w (a) can be constructed using the vector-valued contour integration technique. For details see, e.g. , [25] .
(
(
(
)
3. Banach algebras and spectra
323
The holomorphic calculus of (one) element of a Banach algebra was defined and its study was almost completely finished by the early 1 940s . But the question of the "right" definition of holomorphic functions of several commuting elements of a Banach algebra and, as a major special case, of several commuting operators in a Banach spaces remained open. This problem was solved in 1970 by J . Taylor using methods of homological algebra, functional analysis, and multidimensional complex analysis . These topics and the main result of this theory-the Taylor theorem on a holomorphic multioperator calculus-are well presented in the book by J . Eschmeier and M . Putinar [77] . Let us now make another digression, this time of general character. Second time in this book the needs of applications compel us to consider the multiplication in polynormed algebras which are not Banach algebras. First , having in mind the future spectral theorem we considered the multiplication in the algebra B( E) endowed with the weak-operator and also the strong-operator family of prenorms (see Examples 4 . 1 .8-4. 1 .9) . Now the algebras O(U) appear. All these algebras belong to the class of so-called polynormed algebras, i.e. , polynormed spaces endowed with multiplication that is continuous in some reasonable sense. In the case of A == (B(E) , s o ) and A == (B(E) , w o ) the multiplication is separately continuous, i.e. , continuous with respect to each argument (Proposition 4. 1 . 7) . At the same time in the case of A == O(U) the multiplication, as you can easily verify, has a stronger property of so-called joint continuity, i.e. , continuity with respect to the Tychonoff topology in A x A. It is not difficult to see that a series of other examples in Chapter 4, first of all c oo [a, b] and the spaces of test functions, also are polynormed algebras with respect to the pointwise multiplication, and some of them with respect to the convolution as well. (But this is important for the applications exceeding the scope of our book.) As a matter of fact, the theory of polynormed (and more general topological) algebras is a part of functional analysis that is closely related to the theory of Banach algebras but at the same time has its own distinctive features . About this circle of problems you can see, e.g. , [82] , [83] , [25] . One of the most important applications, namely to the multi parameter spectral theory, is presented in [77] .
Now we return to the material for all readers and give an overview of main results of the most traditional part of the theory of Banach algebras, the theory of commutative Banach algebras. The core of this theory is that these algebras are, roughly, algebras of functions. A commutative Banach algebra is called semisimple if it has no topo logically nilpotent elements except zero. ( This is a special case of the fun damental notion of semisimple algebra; you can read about this notion in, say, [25] . ) Theorem 6 ( Gelfand, [25 , Chapter IV, Theorem 2. 14] ) . Let A be a semisim
ple commutative Banach algebra. Then there exists an injective contraction homomorphism of this algebra to the algebra Co(O), where 0 is a locally compact (and in the case of unital A, compact) topological space.
The Gelfand theorem shows that every semisimple commutative Banach algebra coincides, up to an isomorphism in Alg, with an algebra of contin uous functions-the image of the mentioned homomorphism. Let us explain how this homomorphism acts. First , where does the mentioned locally compact space comes from? The answer follows . As a set , it consists of all non-zero
324
5. At the Gates of Spectral Theory
characters of our algebra (i.e., we recall, non-zero functionals on A "respecting" the mul tiplication) . By Proposition 2 , it can be identified with a subset in A* , and this allows us to endow it with the topology inherited from the weak* topology in the latter space. The introduced topological space is called the Gelfand spectrum of our commutative Banach algebra; we shall denote it by n. (A) or just by 0. Exercise 1 0. The Gelfand spectrum is a locally compact, and i f A is unital, a compact Hausdorff space .
Hint. If we add the zero character 0 to 0, we obtain a closed, with respect to the weak* topology, subset in B A* . (Moreover, in the unital case, 0 is its isolated point . ) Then the Banach-Alaoglou theorem works. Now we define one of the most important functors in functional analysis. Consider the category UCBA, whose objects are unital commutative Banach algebras and morphisms are unital continuous homomorphisms . Suppose r.p : A ---+ B is a morphism in UCBA. Then it is easy to see that the adjoint operator r.p* : B * ---+ A* maps O. (B) to O. (A) . Denote by O. (r.p) : O. (B) ---+ O. (A) the corresponding birestriction; from Proposition 4.2. 1 0 it follows that this is a continuous mapping. Obviously (taking into account Exercise 10) , the correspondence A �-----+ n. (A) ; r.p �-----+ n. ( r.p) is a covariant functor from the category UCBA to the category CHTop. It is called the Gelfand functor. Remark. To simplify the exposition we have introduced only a "part" of the Gelfand functor. In fact, it is defined on the category of all commutative Banach algebras and all their continuous homomorphisms, and takes values in the category of locally compact spaces (its morphisms are discussed at the end of Section 3 . 1 ) . You can easily restore all the missed details.
Going back to the promised construction of homomorphism, consider a commutative Banach algebra A and its Gelfand spectrum 0. Let us assign to every a E A the function a : 0 ---+ C acting by the rule a( x ) : = x (a ) ; X E 0. Now we can give the detailed formulation of the Gelfand theorem. Theorem 6' . Let A be a commutative Banach algebra, and 0 its Gelfand spectrum. Then
(i) for every a E A the function a(t); t E 0 is continuous and vanishes at infinity; (ii) the mapping FA : A ---+ Co ( O ) : a �-----+ a is a contraction homomorphism; its kernel is the set of all topologically nilpotent elements of algebra A; (iii) for distinct s , t E 0 there exists a E A such that a ( s ) =/= a(t) {the image of FA separates the points of 0). Exercise 1 1 . Prove part ( i) of Theorem 6 ' .
Hint. The functional f �-----+ f ( a ) is continuous on A* with respect to the weak* topol ogy. Its restriction to the compact set in BA* consisting of all characters (including 0) vanishes at the point 0. Definition 5 . The homomorphism FA introduced in this theorem is called the Gelfand transform of the commutative Banach algebra A. Example 7 ( "let well alone" ) . Let A be the algebra C0 (0) , where 0 is a locally compact Hausdorff space. Then, using the Alexandroff theorem , it is easy to prove (try!) that the Gelfand spectrum of this algebra coincides up to a homeomorphism with 0, and FA : A ---+ Co ( O ) is just the identity homomorphism.
3. Banach algebras and spectra
325
The Gelfand transforms of different algebras are compatible. In the following exercise we shall again restrict ourselves to the unital case. Recall the functor C CHTop---+ B an 1 considered in Example 3. 1 . 1 . Obviously, the same construction gives a functor from CHTop to UCBA, which we again denote by C. :
Exercise 12. For every unital continuous homomorphism r.p : A ---+ B between unital commutative Banach algebras the following diagram is commutative:
In other words , the family { FA : A E UCBA} is a natural transformation between the identity functor in UCBA and the composition of functors C o n . : UCBA---+ U CBA. *
*
*
To conclude this section, we return to the Fredholm operators, armed with the knowledge of topological properties of the group of invertible op erators. We must fulfil our promise given in Section 3.5: to continue the discussion of the stability of the index. Denote by f/J(E, F) the set of Fred holm operators between Banach spaces E and F and write f/J(E) instead of f/J(E, E) . Theorem 7 (Stability of index under small perturbations; cf. Proposi tion 3.5.3) . The set f/J(E, F) is open in B(E, F ) , and the function lnd : f/J(E, F) � Z : S �----+ Ind(S) is continuous. Take S E f/J( E, F) . Our first goal is to show that if the norm of T E B(E, F) is small, then the operator S + T is also a Fredholm operator. By the Nikol' skii theorem the coset S + JC(E, F) is an isomorphism in the category Banj/C ; let R + IC(F, E) be the inverse operator. Then, you can easily see, for every T E B(E, F) we have R(S + T) + JC(E) == U + JC(E) and (S + T)R + JC(F) == V + JC(F) , where U :== 1 + RT and V :== 1 + TR. Now we assume that II T il < II R II - 1 , so that II RT II , li TR ll < 1 . Then, by Proposition 4(i) for B(E) and B ( F ) as A, the operators U and V are invertible. As a corollary, the morphism ( S + T) + JC(E, F) of the category Banj/C has the left inverse u- 1 R + JC( F, E) and the right inverse RV- 1 + JC ( F, E) . Hence (by Proposition 0.5 . 1 ) , it is an isomorphism. By the same Nikol' skii theorem, S + T E f/J(E, F) . Now, look at the indices. We know that RS E 1+ /C (E) and R(S + T) E U + JC( E) , where U is invertible. Combining this with Theorem 3. 5 . 1 and Proposition 3.5 .3 we obtain the identities Ind(R) + Ind(S ) == Ind(1) == 0 and Ind(R) + Ind(S + T ) == Ind ( U ) == 0. Thus, Ind(S ) == Ind (S + T) , i.e. , lnd is • a locally constant function. The rest is clear. Proof.
as
5. At the Gates of Spectral Theory
326
However, the dimensions of the kernel and cokernel of a Fred holm operator, taken separately, are not stable under small perturbations. Consider, for example, an arbitrary Fredholm (i.e. , having finite-dimensional kernel) projection P. Among all its "arbitrarily small" perturbations there are operators with zero kernel, namely, operators of the form P + Al, which are invertible for sufficiently small A E C. Remark.
Now we say a few words about further properties of the topological space ( x, Ty ) 0 for all y E H {:=:::> ( T* x, y ) 0 for all y E H {:=:::> T* x 0 {:=:::> x E Ker(T* ) . Taking into account Propositions 2.3.4 and 3(iv) , we see that this implies Im(T* ) - Im(T * ) _t_t Ker(T** ) _t Ker(T) _t . • The rest is clear. Proposition 5 . If T acts on H, and Ho is an invariant subspace for T, then Ht is an invariant subspace for T* . Proof. Since y E Ho implies Ty E Ho, from ( x, y ) 0 for each y E Ho we • have ( T*x, y ) ( x, Ty ) 0 for the same y. The rest is clear. Let us show that a number of possible properties of a given operator are preserved under the passage to the adjoint operator. Proposition 6. IfT is a one-dimensional operator of the form xOy {cf. the beginning of Section 3.4), then T* is also one-dimensional, and T* y 0 x. Proof. This follows from the elementary verification of the adjointness re • lations. Proposition 7. If T is finite-dimensional, then T* is finite-dimensional as well, and dim(Im(T) ) dim(Im(T*)) . Proof. Since the image Im(T) is closed, Theorem 3 shows that Im(T*) T* (Ker(T*) EB Im(T) ) Im(T* I Im (T) ) . Since Ker(T*) n Im(T) {0 } , we see • that T* I Im( T) must be injective. The rest is clear. Exercise 4. Deduce this fact from Proposition 6. Hint. Take the representation T L:: �= l X k 0 Yk with linearly indepen dent systems {x 1 , . . . , X n } and {y1 , . . . , Yn } · Proposition 8. If T is compact, then T* is compact as well. If in addition T l: n sn e� 0 e� is the decomposition of the operator as in the Schmidt theorem, then T* l: n sn e� 0 e� . ==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
==
1.
333
Hilbert adjointness: First information
The compactness of T* does not require using the Schmidt theo rem. Namely, by Proposition 3.3.7, T is approximated in the operator norm by finite-dimensional operators. In view of Proposition 3(i) , (v) , T* is ap proximated by the adjoints to finite-dimensional operators, which, by the previous proposition, are finite-dimensional well. Now we recall the Schmidt theorem and consider the decomposition of T indicated there. From Proposition 6, taking into account algebraic and topological properties of the Hilbert star, we immediatelyc obtain the required decomposition for T* . Of course, the latter also shows that T* is • approximated by finite-dimensional operators. Proposition 9. If S is a Fredholm operator, then S* is a Fredholm operator as well. Proof. According to the �-part of the Nikol'skii Theorem 3.5.6, there are operators R, T1 , T2 with the properties indicated there. But by Propositions 3 and 8, the operators R* , r;, Ti satisfy the hypothesis of the {::::= - part of • the same theorem. The rest is clear. Now let us go back to the Fredholm integral equations of the second kind (see Section 3.5) . Together with the initial equation Proof.
as
x (s )
(1)
- 1b K ( s , t)x (t)dt = y ( s ) ,
we shall consider the so-called adjoint equation x (s)
(1*)
- 1b K* (s , t ) x (t ) dt = y (s) . as
We have already noted that equation ( 1 ) is the same the operator equation Sx == y in the space L 2 [a, b] , where S ==1-T, and T is the integral operator with kernel K (s , t) . Also, it is clear, taking Proposition 4 into account, that equation ( 1 *) is the same the operator equation S*x == y. Consider the corresponding homogeneous equations as
- 1b K(s, t) x (t)dt = 0, x (s) - 1 b K * (s , t) x (t)dt = 0.
(2)
x (s)
(2* )
Preserving the style of Theorem 3.5.5, let us formulate the promised addi tion. Proposition 10 (Fredholm). Let x1 , . . . , X m and z 1 , . . . , Zn be families of linearly independent functions, occurring in Theorem 3. 5. 5. 1 Then 1 We know that m
=
n, but now it is not important.
334
6. Hilbert Adjoint Operators and the Spectral Theorem
(i) the functions z1 , . . . , Zn are solutions of the homogeneous equation (2*) such that every solution of this equation is a linear combination of the indicated solutions; (ii) the right-hand sides y ( ) for which equation ( 1 *) has at least one solution are precisely those functions for which J: y (t)x k (t) dt == 0 for k == 1 , . . . , m. s
By Theorem 3.5.5 (ii) , Im(S) == span{z 1 , . . . , zn } ..l. Taking the or thogonal complement and using Theorem 2, we have that Ker(S*) == span{z 1 , . . . , zn } , and this is equivalent to (i) . Furthermore, Theorem 3.5.5(i) gives Ker(S) == span{x 1 , . . . , xm } , and Proposition 9 implies that Im(S*) is closed. Hence, by Theorem 2, Im(S*) == • Ker(S)..l == {x 1 , . . . , xm } ..l , and this is equivalent to (ii) . Proof.
Here is another application of Theorem 3. Exercise 5 (cf. Exercises 2.5.6 and 2.5.7) . (i) The adjoint to an isometric operator between Hilbert spaces is coisometric, and vice versa. (ii) The adjoint operator to a topologically injective operator is topo logically surjective, and vice versa. Hint. If V : H1 � H2 is coisometric, then the adjointness relations show that V* takes y to the unique vector x _l Ker(V) for which V x == y. Now we show that some classes of operators defined in terms of Hilbert geometry, can be characterized in purely algebraic terms using the operation of Hilbert star. In a compressed form, what we do looks like the following small algebra-geometry dictionary: unitary operator U * == u - 1 P* == p == p2 orthogonal projection orthogonal reflection J* == J == J - 1 isometric operator V*V == 1 coisometric operator VV* == 1 partially isometric operator WW*W == W Let us explain what all this means.
11.
An operator U : H1 � H2 is unitary {=:::} it is invertible, and its inverse operator coincides with its adjoint operator.
Proposition
1.
Hilbert adjointness: First information
335
===> Since U is invertible and preserves the inner product, for all x E H1 , y E H2 we have ( x, U - 1 y ) == (U x, uu - 1 y ) == ( U x, y ) == ( x, U * y ) . {::::== For the same x, y, (U x, Uy ) == ( x, U * U y ) == ( x, u - 1 U y ) == ( x, y ) . • Proposition 12. An operator P : H � H is an orthogonal projection {=:::} it is idempotent and coincides with its adjoint operator. Proof. ===> Since P is a projection, we have P2 == P. By assumption, Ker(P) _l Im(P) . So for all x, y E H, taking into account that Py - y E Ker(P) , we have ( Px, y ) == ( Px, y + (Py - y) ) == ( Px, Py ) , and similarly ( Py ) == ( Px, Py ) . This implies that P* == P. {::::== Since P2 == P, P is a projection. Taking into account that P* == P, Theorem 3 gives that Ker(P) _l Im(P) . • Warning. From this moment on, everywhere in this chapter "projec tion" means an orthogonal projection: we shall not need other projections. In the next several exercises we ask you to describe the geometric action of an operator, starting from its algebraic properties. Exercise 6. How does J : H � H act if we know that it coincides with its inverse and with its adjoint operator? Answer. There is an orthogonal decomposition H == H+ ffiH_ such that J == 1 on H+ and J == - 1 on H_ . (An operator acting in this way is called an orthogonal reflection.) Hint. Look at the obvious identity x == � (x + Jx) + � (x - Jx) . Exercise 7. How does V : H1 � H2 act if we know that V* V == 1 ? And what if we know that VV* == 1 ? Answer. The first equality describes isometric operators, and the second, coisometric ones. The following class of operators contains all operators from our dictio nary except orthogonal reflections. Exercise 8. How does W : H1 � H2 act if we know that WW*W == W? How does its adjoint operator act? Answer. There are closed subspaces K1 C H1 and K2 C H2 such that W isometrically (i.e. , unitarily) maps K1 onto K2 and takes K/- to zero. At the same time W* maps K2 onto K1 acting as the inverse to W, and takes K;j- to zero. Proof.
.
.
.
x,
.
336
6. Hilbert Adjoint Operators and the Spectral Theorem
Hint. The ugly looking identity is equivalent simply to the fact that W* W is an orthogonal projection, say P. Hence, for K1 : == Im(P) and x , y E K1 we have (Wx , Wy) == (x , y) . The operators we have just talked about are called partially isometric. They play an important role in the theory of operators and operator algebras (classification of factors [4 6], operator K-theory [84], and a series of other questions; cf. Exercises 2.7 and 4.9 below) . Finally, let us recall spectra. Those readers who have done Exercise 5 . 1 . 1 (which is not very simple actually) know how the spectra behave under the operation of Banach star. It is much easier to understand how the spectra and various special subsets of spectra defined in Section 5 . 1 react to the Hilbert star. Exercise 9. Let T : H --+ H be an operator on a Hilbert space. Then a (T* ) == {A : A E a(T) } , and (i) if A E ar (T) , then A E ap (T* ) ; (ii) if A E ap (T) , then A E ap (T* ) or A E ar (T* ) ; (iii) ac (T* ) == {A : A E ac (T) } ; (iv) ae (T* ) == {A : A E ae (T) } . Remark. Both situations in (ii) can actually happen. Give the correspond ing examples. 2. Selfadj oint op erators and their sp ectra. Hilb ert-Schmidt theorem
Let us concentrate on operators acting in a chosen Hilbert space H. Their adjoint operators also act in H (i.e. , belong to the same B(H) ) . Naturally, we have an interesting class of operators, which appears in various problems of mathematics and physics. Definition 1. An operator T E B(H) is called selfadjoint (or Hermitian) if it coincides with its Hilbert adjoint operator (i.e. , T* == T) . The adjointness formulas show that a selfadjoint operator is exactly an operator satisfying the identity (1) (Tx , y) == (x , Ty) ; x , y E H. Proposition 1 .2 allows us to characterize selfadjoint operators as those T for which Sr == (Sr ) * . We come close to another useful characterization, this time in terms of quadratic forms. For an operator T (so far arbitrary) , its quadratic form is defined as the function Qr : H --+ C : x �----+ Sr ( x , x) (or,
2.
337
Selfadjoint operators and their spectra
what is the same, x �----+ (Tx , x) ) . Note that Proposition 1 .2. 1 immediately implies • Proposition 1. T == 0 {:=:::> Q r == 0 (i. e., (Tx, x ) == 0 for all x E H). Remark. Here it is important that we speak about complex linear spaces. For a comparison, look at the operator of rotation through 90 degrees in 1R2 . Proposition 2. An operator T is selfadjoint {:=:::> its quadratic form takes real values {i. e. , ( Tx, x) E lR for all x E H). By the previous proposition, T == T* {:=:::> time, for each S E B (H) we have Qs * == Q s . Proof.
Qr
== QT * . At the same
•
In addition to the selfadjoint operators, the more general, so-called nor mal operators also deserve to be mentioned. These are operators that com mute with their adjoint operators, i.e. , operators T such that T*T == TT* . Obviously, selfadjoint operators and unitary operators are normal. Remark. Actually, much of what we will say about selfadjoint operators it true for normal operators as well. But in this book we pay much less attention to them just because selfadjoint operators are geometrically more visual and appear more often in applications. The set of selfadjoint operators on H is denoted by B(H)s a · Note that, according to Proposition 1 .3, this is a real (but not complex! ) Banach space. Obviously, for every T E B(H) there exists a unique pair Tre, Tim E B(H)s a such that (2) T == Tre + i Tim ; These operators are Tre :== � (T + T*) and Ti m :== J-i (T - T*) . Two other selfadjoint operators that we can construct for every T are T*T and TT* . Remark. A well-justified and fruitful way of interpreting operators on Hilbert spaces is to view them as a profound "non-commutative" gener alization of complex numbers. From this point of view the passage to the adjoint operator corresponds to the passage to the complex conjugate num ber, selfadjoint operators play the role of real numbers, and equality (2) generalizes the algebraic form of complex numbers (and becomes precisely the latter when H :== C ). You may ask what is an analogue of the polar de composition (i.e. , trigonometric form) of complex numbers? As we will see later, there are two possible candidates for the role of polar decomposition, and they differ because of non-commutativity of operator multiplication. But we are not ready to discuss this yet (see Exercises 7 and 4.9 below) . Here we only mention that in the algebra B(H) the role of positive numbers
338
6. Hilbert Adjoint Operators and the Spectral Theorem
belongs to operators of the form T*T, whereas the role of unimodular com plex numbers belongs in some cases to unitary operators, and in other cases to more general partially isometric operators from Exercise 1 . 8. Here are several standard examples. From Exercises 1 .3 you can see that - a diagonal operator T>.. is always normal, and it is selfadjoint {=:::} the sequence A consists of real numbers; - (a more general fact) the operator Tt of multiplication by a function f is always normal, and it is selfadjoint {=:::} f takes real values almost everywhere; - the operators of left and right shifts in l 2 are not selfadjoint, and even not normal: TzTr == 1, and at the same time TrTz is a projec tion with the one-dimensional kernel span(p 1 ) . Proposition 1 .4 immediately implies Corollary 1. An integral operator on L 2 [a, b] is selfadjoint {=:::} its ker nel K(s, t) satisfies the identity K(s, t ) == K(t, s) almost everywhere in the square D . (Such kernels are called symmetric.) We formulate several simple properties of selfadjoint operators . Proposition 3. Suppose T is selfadjoint. Then (i) the eigenvalues of T are real numbers; (ii) the eigenvectors of T corresponding to different eigenvalues are or thogonal; (iii) the kernel and the image are connected by the relations Im(T)..l == Ker( T) and Ker(T)..l == Im(T) - {as always, the bar - means clo sure}; in other words, H == Ker(T) EB Im(T) - ; (iv) if Ho is an invariant subspace under T, then Ht is also an invari ant subspace under T; (v) a subspace Ho in H is invariant under T {=:::} T commutes with the orthogonal projection P : H � H onto Ho ; (v i) II T2 II == II T II 2 ; (vii) r(T) == II T il ; (viii) if S is another selfadjoint operator commuting with T, then ST is also selfadjoint; in particular, every power of T is a selfadjoint operator. Proof. (i) If Tx == Ax; x # 0, then A ( x, x) == ( Tx, x) == (x , Tx) == A (x , x ) . The rest is clear.
2.
Selfadjoint operators and their spectra
339
(ii) If Tx == AX and Ty == J1Y ; x , y # 0, then, taking (i) into account, we have A (x, y) == (x , Ty) == 11 \ x , y) . Hence, A # 11 implies (x , y) == 0. (iii) , (iv) , (vi) , (viii) immediately follow from Theorem 1.3, Proposition 1.5, Theorem 1.2, and Proposition 1.3(iii) . (v, ===> ) Since H == Ho tiJ Hd- , it is sufficient to show that PTx == T Px for x E Ho and for x E Hf . Clearly, in the first case both vectors we compare coincide with Tx , and in the second case, in view of (iv) , they vanish. (v, {::::::= ) If x E Ho , then PTx == T Px == Tx , hence Tx E Ho . • (vii) follows from (vi) and the spectral radius formula.
1
Parts (ii) , (iii) , (vi) , and (vii) in the previous proposition are valid for normal operators. Hint. If T is normal, then II T*x ll == II Tx ll for all x E H; thus, we have (iii) . Therefore, Ker(T*) == Ker(T) and Ker(T - Al) == Ker(T* - Al) , and this is (ii) . Finally, the C*-identity gives II T2 II 2 == II (T 2 )*T2 II == II T*T II 2 == II T II 4 · As for the classical Fredholm theorems on integral equations, they look completely transparent in this context. Proposition 4 (Fredholm) . Suppose Exercise * .
(3)
x (s)
- 1b K (s , t)x (t) dt = y (s)
is an integral equation of the second kind with symmetric kernel and (4)
x (s)
- 1b K ( s , t)x (t) dt = 0
is the corresponding homogeneous equation. Then there exists a linearly independent system X I , . . . , X m of solutions of equation (4) satisfying the following conditions: (i) every solution of equation (4) is a linear combination of X I , . . . , xm ; (ii) the right sides y(s) for which equation {3} has at least one solution are precisely the functions satisfying the equation J: y(t)xk (t) dt == 0 • for all k == 1, . . . , m. Let us now speak about spectra. Those who had done Exercise 1.9, know that the spectrum of a selfadjoint operator is symmetric with respect to the real axis. But actually a much stronger proposition is true: it lies on the real axis. The implications of this fact are so serious that we shall give two proofs for it: "algebraic" (or abstract) and "geometric" (or vector) . Let us start the necessary preparations.
340
6. Hilbert Adjoint Operators and the Spectral Theorem
Suppose T is selfadjoint. Then the operator U :== exp( iT) (see Definition 5. 3.4} is unitary. Proof. By the definition of the operator exponential, U == 1 + E � 1 �! (iT) n . Hence, from the Hilbert star properties we have oo n oo 1 i n U * = 1 + L -1 T = 1 + L ( - iT) n == exp( - iT) . n= l n. n= l n. Hence, by Corollary 5.3.4, U* and U are mutually inverse operators, and • our result follows from Proposition 1 . 1 1 . Proposition 6. Suppose T : H � H is selfadjoint and topologically injec tive. Then it is a topological isomorphism. Proposition 5 .
-,
By assumption, Im(T) is closed, hence, by Proposition 3(iii) , it co • incides with H. The rest is clear. Proposition 7. For each S E B(H) and t > 0 the operator R :== S* S + t1 is invertible. Proof.
By what we had already said, it is sufficient to prove that R is topologically injective. For every x E H we have II Rx ll ll x ll > I (Rx , x ) I == I ( S * Sx , x ) + t (x , x ) I == ( Sx, Sx ) + t (x , x ) > t 11 x ll 2 . • Therefore, II Rx ll > t ll x ll , and after that Corollary 1.4.2 works. Theorem 1. The spectrum of a selfadjoint operator T : H � H belongs to the interval [ - II T II , II T II J and contains at least one of its ends. Proof.
Combining Corollary 5.3. 1 with Proposition 5, we have a(exp(iT) ) C 'lr. Hence, if A E a(T) , then, applying Proposition 5.3.11 to w ( z ) :== exp(i z ) , we see that exp(iA) E 'lr. This obviously implies that • A E JR, and it only remains to use Proposition 3(vii) . Algebraic proof.
Suppose A E C ; A == s + it is such that t # 0. Put S :== T - s1. Then, by Proposition 7, the operator S* S + t1 == S2 + t1 == (S + it1) (S - it1) is invertible, and thus, both factors, commuting with each other, are also invertible (see Proposition 5.2.2) . Hence, it tf. a(S) , and this is equivalent to the fact that A tf. a(T) . Thus, a(T) C JR. Similarly to how it was done in the algebraic proof, we can now complete the proof using Proposition 3(vii) . But there are more elementary tools in our "vector" arguments. Take a sequence X n E H; ll x n ll == 1 such that Geometric proof.
2.
Selfadjoint operators and their spectra
341
II Tx n ll � II T II ; n � oo . Then for every n we have II (T2 - II T II 2 1)xn ll 2 == ( (T2 - II T II 2 1)x n , (T2 - II T II 2 1)x n) == II T 2 xn ll 2 - 2 ( T2 xn , II T II 2 xn) + II T II 4 < 2 IITII 4 - 2 II T II 2 (Txn , Txn) == 2 II T II 4 - 2 II T II 2 11 Txn ll 2 , hence (T 2 - I I T II 2 1)xn � 0 as n � oo . Of course, this means that the oper ator T2 - I I T II 2 1, or, what is the same, the operator (T + II T II l) (T - II T II l) , is not invertible. Therefore, either II T II or - I I T II belongs to a(T) , and it • only remains to recall that r(T) < II T II (Theorem 5.3. 1).
What we have just said is the maximal information one can obtain about the spectra of a general selfadjoint operator. Exercise 2.
(i) Every compact subset of lR is the spectrum of a selfadjoint operator. (ii) Every compact subset of 1r is the spectrum of a unitary operator. (iii) Every compact subset in C is the spectrum of a normal operator. In all these cases the operator can be chosen in such a way that its spectrum coincides with its essential spectrum. Hint. Take the diagonal operator T>.. l 2 � l 2 , where A is the sequence such that {An} - coincides with the given set. Theorem 1 and Proposition 3(iii) immediately imply the following result. • Proposition 8. Selfadjoint operators have no residual spectra. Exercise 3. The same is true for the normal operators. Now we pass to the preparation of a theorem allowing us to completely understand the nature of operators that are compact and at the same time selfadjoint. Theorem 2 (Hilbert-Schmidt). Let T H � H be a compact selfadjoint operator on a Hilbert space. Then there exist a) an orthonormal system e 1 , e 2 , . . . in H of finite or infinite cardi nality, and b) a {finite or infinite) sequence AI , A 2 , . . . of non-zero real numbers with the index set of the same cardinality, tending to zero if it is infinite, such that our operator acts by the following rule: Tx = L >.n ( x, en) en (5) :
:
342
6. Hilbert Adjoint Operators and the Spectral Theorem
where E n means either a finite sum, or a sum of a series in H. {In other words, the en are the eigenvectors for T with eigenvalues An , and T takes every vector orthogonal to all en to zero}. Proof. By the Schmidt Theorem 3.4. 1 , T can be represented as the sum T ==
E n sne� 0 e� . From the selfadjointness of T and Proposition 1 .8 it follows that the same operator can be represented as the sum T == E n sne� 0 e� . . e + .· - eI + eII and e ·. - eI - eII , we see th at Te ± - ± Sne ± . P Utt Ing n n n - n n n - n n Since the s-numbers of T do not increase, for some natural numbers n i < n 2 < · · · we have S I == · · · == Sn 1 > Sn 1 +I == · · · == Sn 2 > Sn2 + I == · · · > s n k- l +I == · · · == s n k > · · · . For k == 1 , 2, . . . , consider the spaces Lt :== span{e�k- l + I , . . . , e�k } (here we put no : == 0) . Obviously, for each e E Lt we have Te == ±s n k e. Hence, by Proposition 3(ii) , all these spaces are mutually orthogonal in H. Consider those subspaces Lt that do not vanish and choose orthonor mal bases in these subspaces. If we collect together all vectors from all these bases and number them arbitrarily, we obtain an orthonormal system, say { e i , e 2 , . . . } . Evidently, every en is an eigenvector for T, and the correspond ing eigenvalue, which we denote by An , coincides with the number ±s k for some k. Moreover, every An appears at most finitely many times. Further, the linear spans of the systems { e I , e 2 , . . . } and { e� , e; , . . . , e1, e� , . . . } obviously coincide. Hence the sequences AI , A 2 , . . . and S I , s 2 , . . . are either both finite, or both infinite. If they are infinite, then from the fact that s n tends to zero it follows that A n tends to zero as well. Finally, take an arbitrary x E H. It can be represented as -
x=
L ( x, en) en n
+
xo ,
where x 0 is orthogonal to all en and, thus, to all e� . Hence, by the Schmidt theorem, Tx 0 == 0, and equality (5) evidently follows from the properties of • T as a continuous operator. The Hilbert-Schmidt theorem admits the following equivalent formula tion; compare it with the result of Exercise 3.4. 1.
Let T : H � H be a compact selfadjoint operator in a Hilbert space. Then there exists a finite or infinite sequence A == (AI , A 2 , . . . ) of real numbers, tending to zero if it is infinite, and a Hilbert space Ho such that T is unitarily equivalent to the operator R l2 EB Ho � l2 EB Ho {where m < oo is the number of the terms of the sequence A) that acts on l2 as the diagonal operator T>.. and takes Ho to zero. Proposition 9 .
:
2.
Selfadjoint operators and their spectra
343
Proof. Take the sequence that appears in the Hilbert-Schmidt theorem as A, and put Ho : == Ker(T) . Then, by the Hilbert-Schmidt theorem, T belongs
to the class of operators in Example 2.2.2. The rest is clear. Exercise Proposition 9.
4.
•
Show that the Hilbert-Schmidt theorem follows from
Recall that for an operator acting in a linear space the multiplicity of an eigenvalue is the dimension of the corresponding subspace of eigenvectors.
Let T be a selfadjoint operator and A n the sequence in the Hilbert-Schmidt theorem. Then (i) the non-zero eigenvalues A of T are elements of the sequence {An}, and the multiplicity of every eigenvalue A coincides with the number of times A occurs in the sequence {An}; ( ii) II T II == max { I A n I ; n == 1 , 2, . . . } .
Proposition 10.
Proof. Evidently, the norm, as well as the family of the non-zero eigenvalues
and the multiplicity of every eigenvalue do not change when we pass to a unitarily equivalent operator. Hence, these characteristics of T are the same as those of the operator R in Proposition 9. Thus, these characteristics coincide with those of the diagonal operator T>.. in this proposition. But every eigenvalue of T>., , say J1 , is obviously one of the numbers An , n == 1 , 2, . . . , and the space of the respective eigenvectors is span { p k : A k == 11} . In addition, l i T>.. II == m ax { I An l ; n == 1 , 2, . . . } (cf. Example 1.3.2) . The rest is clear. • Remark. The fact that the norm of a compact selfadjoint operator is the
greatest absolute value of the eigenvalues follows also from the equality II T II == r (T) and from the fact that the spectrum of a compact operator contains only eigenvalues (and possibly zero) ; see Theorem 5 . 1 . 1 . Finally, the statement in the Schmidt theorem that the operator in ques tion can be decomposed into the sum of one-dimensional operators can be reformulated in the "selfadjoint" case as follows.
Suppose that T H � H, en , and A n are as in the Hilbert-Schmidt theorem. Then T can be represented as T == E n Anen 0 en , where En means either a finite sum, or the sum of the series converging in the operator norm.
Proposition 1 1 .
:
Proof. Equality (5) in the Hilbert-Schmidt theorem can be rewritten as
Tx == E �== l An [en 0 en ] (x) . If the number of terms is finite, everything is clear. If it is infinite and Sk is the partial sum of the corresponding series, then, clearly, T - sk acts by the formula (T - Sk )x == E � k+l An (x , en) en. Now Proposition 10 applied to T - Sk gives l i T - Sk l l == max{ I An l ;
344
6. Hilbert Adjoint Operators and the Spectral Theorem
n == k + 1 , k + 2, . . . } . Hence, l i T - Sk ll tends to zero as k with Ak ·
--+ oo
together
•
While the Schmidt theorem can be regarded as a result on the classi fication up to the weak unitary equivalence, the Hilbert-Schmidt theorem is a result on the classification up to the unitary equivalence (i.e. , up to much more rigid identification) . At the same time it is remarkable that for the class of operators considered, the two kinds of "non-weak" equivalence, unitary and topological, coincide.
:H
--+
H and S : K
--+
K be compact selfadjoint operators. Then they are unitarily equivalent {=:::} they are topologically equivalent {=:::} the sets of their non-zero eigenvalues, taken with multiplic Exercise 5 . Let T
ities, coincide, and Ker(T) is unitarily isomorphic to Ker(S) . Hint. For arbitrary operators their topological equivalence implies the coincidence of the Hilbert dimensions of the spaces of eigenvectors corre sponding to every chosen eigenvalue. (In fact, topological and unitary equivalence coincide on the entire class of selfadjoint operators, but this is more difficult to prove; see Exercise 4. 10 below.) We see that every selfadjoint operator in a Hilbert space is uniquely defined up to unitary equivalence by the following data: a) the (non-ordered) family A of the non-zero eigenvalues, where every eigenvalue occurs as many times as its multiplicity is, and b) the Hilbert dimension of its kernel (see Theorem 2.2.2) . Thus, a complete system of invariants of the unitary equivalence for this class of operators consists of pairs (A, a ) , where A is an at most countable family of non-zero real numbers with possible finite repetitions and with zero as the only possible limit point, and a is a cardinality. A model of the compact operator with the invariant (A, a ) is the operator R : l2 EB K --+ l2 EB K (see Proposition 9) , where A is the given family, arbitrarily ordered, and K is a Hilbert space of Hilbert dimension a (say, l 2 ( X ) , where X is a set of cardinality a ) . All that we have said is true if we replace the unitary equivalence by topological equivalence. *
*
*
For a separable space the Hilbert-Schmidt theorem and the equivalent Proposition 9 can be reformulated more transparently as follows.
Let T : H --+ H be a compact selfadjoint operator in separable Hilbert space. Then
Proposition 12.
a
2.
345
Selfadjoint operators and their spectra
(i) there is an orthonormal basis in H consisting of eigenvectors of T. The corresponding sequence of eigenvalues tends to zero and
consists of real numbers; (ii) T is unitarily equivalent to a diagonal operator T l2 � l2, where m is the Hilbert dimension of the space H, and the sequence A tends to zero and consists of real numbers. :
Proof. Take an orthonormal basis in the formulation of the "general" Hil
bert-Schmidt theorem. Using that Ker(T) is separable, consider an or thonormal basis in this space. If we arbitrarily renumber the union of these two bases, we certainly obtain a required basis in H. The rest is clear. • In particular, it is easy to see that in the separable case the syste1n of invariants of the unitary (as well as topological) equivalence for compact selfadjoint operators also takes a more visual form. Namely, for invariants we can take an arbitrary sequence of real numbers tending to zero, considered up to permutations of its elements. The simplest model of an operator with the sequence A as its invariant is certainly the diagonal operator T>.. : l 2 � l 2 . Now we can fulfill the old promise we gave in Section 1.3. Exercise
6* . The norm of the operator of indefinite integration on
L 2 [0, 1] is ; . Hint. 2 Let us represent our T as US, where S : x �----+ J01 - s x(t)dt, and U is the unitary operator x(t) �----+ y(t) :== x(1 - t) . Hence, II T II == II S II , and since S
is selfadjoint, Proposition 10 reduces the problem to the search of those non zero A E 1R for which the integral equation Sx == AX has a non-zero solution. But these are the same A for which the equation Ax '(t) + x(1 - t) == 0 has non-zero solutions satisfying the condition x(1) == 0. They certainly are the solutions of the equation A 2 x "(t ) + x( t ) == 0 with x( 1) == x' (O) == 0. Knowing the general solution of the latter ordinary differential equation, we see that the non-zero solutions satisfying these boundary conditions exist only for A == ( � + k1r ) - 1 ; k E Z, and for every k they are multiples of cos( ( � + k1r)t) . Thus, II T II < ; , and it remains to verify that Sx == ; x for x(t) :== cos ;t .
In concluding this section, let us go back to the general theory. The Hilbert-Schmidt theorem allows for another important characterization of the s-numbers of compact operators, which also indicates that these numbers do not depend on the choice of the orthonormal systems involved in the Schmidt theorem (see Section 3.4) . 2 This observation simplifies the well-known proof from [81] . It was suggested in a class on functional analysis by Natasha Grinberg, a third-year student at that time.
346
6. Hilbert Adjoint Operators and the Spectral Theorem
13.
Let T : H � K be a compact operator between Hilbert spaces with s-numbers S I > s 2 > · · · . Then the sequence s i , s� , . . . is pre cisely the sequence of non-zero eigenvalues, with multiplicities and in de creasing order, of each of the operators T*T : H � H and TT* K � K.
Proposition
:
Proof. Take T == E n sne� 0 e� . By Proposition 1 .8, T* == E n sne� 0 e� . Hence ' T*Te'n == sn2 e'n and TT* e"n == sn2 e"n for each n · In addition T*Tx == 0 for x _l { e� , e; , . . . } and TT* y == 0 for y _l { e� , e� , . . . } . It remains to apply '
• Proposition 10 to the two operators under consideration. Corollary 2. Let T be a compact operator between two Hilbert spaces, and AI , A 2 , . . . the sequence of eigenvalues of the operator T*T (or, equivalently, TT*) taken with multiplicities. Then (i) T is a Schmidt operator {=:::} E n An < oo ; (iii) T is a nuclear operator {=:::} E n A < oo . Now that we have the Hilbert adjoint operators at our disposal, we can define an inner product in the space S(H, K) of Schmidt operators (see Proposition 3.4.4 (ii) ) using a formula that is not related to orthonormal bases. Namely, for each S, T E S(H, K) we have (S, T) == tr(T* S) . (Thy to prove this formula using Proposition 3.4.8(ii) . )
Now we can show what the promised polar decompositions of the opera tors look like, though at this point we only do it for compact operators. We call a compact operator on a Hilbert space positive if it is selfadjoint and all its eigenvalues are non-negative. (This is a special case of Definition 4.2, which will appear later) . Exercise 7° (polar decomposition of a compact operator) . Suppose T : H � K is a compact operator between Hilbert spaces. Then there exists a partially isometric operator W : H � K and positive operators S : H � H, S' : K � K such that T == WS == S'W. Moreover, if T == E n sne� 0 e� , then our equalities are true for S :== E n sne� 0 e� , S' == E n sne� 0 e� , and W that maps e� to e� and sends all vectors orthogonal to all e� to zero. Under some natural conditions the indicated polar decompositions are . unique. Exercise 8. For the same T, W, S, S', (i) If T == WI SI , where WI is partially isometric, SI is positive, and (Ker(WI))_i == lm(SI ) - , then WI == W and SI == S. (ii) If T == s� WI ' where WI is partially isometric, s� is positive, and (Ker(S� ))_i == Im(WI ) , then WI == W and S� == S'. Hint for (i) . Obviously, WtT == WtWI SI == SI , hence, Sf == T*T == S2 . For the same reason, SI is compact. Since SI is, in addition, positive ,
3. Involutive, C* -, and von Neumann algebras
34 7
Proposition 10 shows the following. A vector e is an eigenvector of S1 with eigenvalue 11 > 0 {=:::} the same e is an eigenvector of sr with eigenvalue /1 2 . Together with the equality T*T == S 2 , this gives S1 == S. A polar decomposition of arbitrary operators will be given later; see Exercise 4. 9. 3. An overview : lnvolutive algebras ,
C*-algebras , and
von Neumann algebras
The set B(H) , already endowed with a rich structure of Banach algebra, has another operation: sending an operator to its Hilbert adjoint. This operation resembles taking the complex conjugate of a number in C. Now it is time to look at what we have obtained, from a more general position. Perhaps, some of the readers do not like that there are many theorems without proofs in this textbook. Then this section will be especially irritat ing for them. Let us try to justify ourselves. The theorems we will speak about are among the most substantial achievements of modern mathematics. At the same time they have simple and transparent statement. We believe that these facts should certainly be mentioned in a text book on functional analysis, and the reader should know about them. As for the proofs, they usually are not simple at all, and together with the necessary preparation would occupy too much space. Therefore, in our opinion, these proofs go beyond the scope of our lectures. You can find them in many other books, e.g. , [86], [25], [72], [50], and [87]. To understand better the forthcoming general definitions, take some examples of algebras (both pure and Banach; see Sections 5.2 and 5.3) from our list: C, C [t] , Mn , B (E) , C 1 [a , b] , l 1 ( Z) , l L ( X, 11) , and put Co ( O ) and B(H) in the most prominent place. (Here 0 is a locally compact topological space, ( X, 11) a measure space, H a Hilbert space, and E an arbitrary Banach space.) We will look at each of these algebras while introducing new notions. First, let us give a purely algebraic oo ,
oo
1. Suppose A is an algebra. A mapping (*) : A --+ A is called an involution in A if (writing a* instead of (*)(a)) for all a, b E A, A E C the Definition
following equalities hold: (i) ( a + b)* == a* + b*; ( ii) (A a)* == A a* ; (iii) ( ab) * == b *a* ; (iv) a** == a.
348
6. Hilbert Adjoint Operators and the Spectral Theorem
An algebra endowed with an involution is called an involutive algebra or, in short, a * -algebra. The element a* is said to be the adjoint to a. All the algebras we have exhibited, save one important exception, have natural involutions. Of course, in the simplest algebra C we have A* :== A. Similarly, in C 1 [a, b] , L00 ( X, 11 ) , and C0 ( 0) , we can put x*(t) :== x( t) , and in l 00 we take �� :== �n · In the algebra of polynomials C [t] the involution is given by p( t) == co + · · · + Cn t n �----+ p * ( t ) :== co + · · · + Cn tn . (This is also the passage to the complex conjugate function if we view polynomials as functions on JR; if we consider them on C, there is a slightly more complicated formula p*(z) :== p(z) .) In the algebra of matrices Mn , for a given a == (akz ) we put a 'kz :== az k· In the Wiener algebra l 1 (Z) we put a� :== a - n · Finally-and this is the main case-the involution in B(H) , as you could have already guessed, is the passage to the adjoint operator. As for the algebra B ( E ) for a non-Hilbert space E, it has no natural involution; sometimes this also happens. Everywhere below in this section A is an involutive algebra. Properties of an involution evidently imply Proposition 1. If A is unital, then 1 * == 1, and for an invertible a E A we • have ( a - 1 )* == ( a*) - 1 . An element a E A is called selfadjoint if a* == a, and normal if a*a == aa*. If A is unital, then an element u in A is called unitary if it is invertible and
u - 1 == u*.
Among all homomorphisms between two *-algebras A and B , those com patible with the involution in a proper way deserve special attention. Definition 2. A homomorphism cp : A � B is called involutive or, shortly, a * -homomorphism if cp(a*) == cp(a)* for all a E A. As a very important special case, a *-homomorphism from A to B(H) is called an involutive representation or a * -representation of the algebra A in the Hilbert space H ( cf. general Definition 5.2.6) . Bijective *-homomorphisms of *-algebras are called involutive isomor phisms or * -isomorphisms . Certainly, they are isomorphisms in the category of *-algebras and *-homomorphisms (give a definition of this category) . Example 1. Clearly, a polynomial calculus of an element a of a unital *-algebra is a *-homomorphism {=:::} a is selfadjoint. In particular, assigning to every p E C [t] the same polynomial viewed as a function on [a, b] , we obviously obtain an injective (as a mapping) *-homomorphism from C[t] to C[a, b] . If we replace [a, b] in this example by an interval [z1 , z2 ] not lying on the real line, we obtain a non-involutive homomorphism from C [t] to
C[z1 , z2 ] .
3. Involutive,
C* - , and von Neumann algebras
Example 2. The mapping
349
cp
: l00 � B(l 2 ) assigning to every sequence A the diagonal operator T>.. is obviously a *-representation of the *-algebra l00 in the Hilbert space l 2 . As a generalization of this example, the mapping cp : L00 ( X, 11) � B(L 2 (X, 11) ) assigning to each essentially bounded measurable function f the operator Tt of multiplication by f is a *-representation of the *-algebra L00 (X, 11) in the Hilbert space L2 (X, 11) . , en be an orthonormal basis in a finite-dimensional Example 3. Let e 1 , Hilbert space H. It is easy to see that assigning to every operator acting on H its matrix in this basis, we obtain a *-isomorphism between the *-algebras B(H) and Mn . .
.
.
Certainly the following proposition is true.
A * -homomorphism maps selfadjoint, normal, and {if we speak about unital homomorphisms} unitary elements to elements of the • same type. A subset M in A is called selfadjoint, or, briefly, a * -subset if a E M implies a* E M. A similar meaning is given to such terms as * -subalgebra, *-ideal, etc. Here is an example. Proposition 2.
4.
Proposition 1 .8 shows that the ideal K(H) in B(H) , where H is a Hilbert space, is a *-ideal. At the same time a subalgebra K(H) + of compact perturbations of scalar operators is a *-subalgebra, but not a *-ideal, if H is infinite-dimensional.
Example
Exercise
1. The sets S(H) and N(H) of Schmidt operators and nuclear
operators, respectively (see Section 3.4) , are also *-ideals in B(H) . The following result is evident. Proposition 3. Suppose I is a * -ideal in A. Then the quotient algebra A/ I is a * -algebra with respect to the involution well defined by the equality
(a + I)* : == a* + I.
•
In particular, the Calkin algebra C (H) : == B(H) /K(H) for a Hilbert space H (see Section 5.2) also has a natural structure of involutive algebra. Let us now proceed from abstract algebra to functional analysis. Definition 3. A Banach involutive algebra (i.e. , an algebra endowed with both a complete norm and an involution) A is called a Banach star-algebra if II a* II == II a ll for all a E A . The following result immediately follows from the definition. Proposition 4. If A is a Banach star-algebra, then the mapping (*) : A � A is continuous, and, in particular, an � a; n � oo implies a� � a*; n � . ()().
350
6. Hilbert Adjoint Operators and the Spectral Theorem
In fact, the converse proposition is true, up to replacing the norm by an equivalent one. Exercise 2. Let A be a Banach algebra with a continuous involution. Then it can be endowed with a norm II · II ' equivalent to the initial one and such that ( A, II · II ') is a Banach star-algebra. Hint. Put ll all ' : == max { ll all , ll a* ll } . Certainly, all involutive algebras in our examples, except Mn and C [t] , are Banach star-algebras with respect to the norms introduced earlier and the involutions introduced in this section. However, Mn can also be easily turned into a Banach star-algebra, and moreover there are many ways to do this. For instance, we can identify this algebra with B(H) (see Example 3) . Or, say, we can define the norm by the equality ll a ll : = ,/'£ �, l = l i akz l 2 • (As for the algebra of polynomials, it cannot be turned even into a Banach space; the readers who have done Exercise 2. 1 . 1 know this. ) Similarly to general Banach algebras (see Section 5.3) , the Banach star algebras are objects of two categories with either bounded, or contraction involutive homomorphisms as morphisms. Isomorphisms of these categories are respectively topological and isometric * -isomorphisms . The meaning of both terms is obvious, as well as, say, of the term contraction or isometric * -homomorphism. (Note that the mappings in Example 2 are isometric *-homomorphisms.) We hope that the following non-trivial fact from the family of "automatic continuity results" will be interesting to our advanced readers ( cf. comments to Proposition 5 .3.2) . Proposition 5 (see [25 , Chapter 4, § 5 . 25] ) . Every * -representation of a Banach star
algebra tn a Hilbert space ts a bounded, and moreover, a contraction * -homomorphism. Exercise 3. Prove that this is true in the case of unital algebras and unital repre sentations.
Hint. If r.p : A ---+ B( H) is our representation, then for a selfadjoint a E A the estimate ll r.p(a) ll < ll a ll can be obtained by consecutive applications of Proposition 2 . 3 (vii) (where T == r.p(a) ) , Proposition 5 .2.4, and Theorem 5.3. 1 . In the general case we must apply the obtained estimate to the element a* a and use Theorem 1 .2 (where T == r.p(a)).
Now we pass to the "best algebras in functional analysis" . The following notion is, apparently, the most important in the entire theory of algebras with involution.
4.
A Banach star-algebra A is called an abstract C*- algebra or, if there is no danger of confusion, just a C* -algebra 3 if for every a E A we Definition
3 Perhaps this term does not sound natural, but it became standard long time ago. Appar ently, it was I . Segal (1 94 7 ) who first introduced it for the operator algebras. The letter "C" seems to be used for emphasizing the role of the algebras in question as non-commutative generalizations
3. Involutive,
C* -, and von Neumann algebras
35 1
have
ll a * a ll == ll a ii 2 This equality is called the (abstract) C* -identity.
4°.
Every involutive Banach algebra where the C*-identity holds is (automatically) a Banach star-algebra, and thus a C*-algebra. We give two main examples, one commutative and one non-commutative. Exercise
Example
5. The algebra Co (O) , where 0 is a locally compact space, and in
particular the algebra C(O) , where 0 is a compact space, are C*-algebras.
6.
The algebra B(H) , where H is a Hilbert space, as well as each norm-closed selfadjoint subalgebra of this algebra, are C* -algebras. This immediately follows from the operator C*-identity (Theorem 1 .2) .
Example
The algebras indicated in the last example are called concrete or operator C* -algebras. Certainly, they include K (H) and, as is easy to verify, the algebra of the operators of multiplication by functions in L2 (X, J1 ) . (The special cases of the latter are the algebra of diagonal operators in l 2 , and the algebra of compact diagonal operators in l 2 .) Certainly, many facts concerning operators which follow from Theorem 1 .2 are true for the elements of an arbitrary C* -algebra. In particular, for selfadjoint elements in C* -algebras we have ll a2 1 1 == l l a ll 2 . (Actually, the related part of Exercises 2. 1 establishes the same for normal elements of these algebras.) And here are counterexamples . Exercise 5. The Banach star-algebra C 1 [a, b] is not a C* -algebra; more over, it is not topologically *-isomorphic to any C*-algebra. The same is true for S(H) and N(H) (see Exercise 1 ) . Hint. Find a sequence of selfadjoint elements an : ll an l l == 1 such that a; � O; n � oo . Remark. The aforesaid is also true for the Wiener algebra l 1 (Z) , but this
is more difficult to establish.
Two results of great importance for the whole mathematics claim that essentially there are no other examples of C* -algebras besides those given in Examples 5 and 6. The first of these results describes all commutative algebras of this class, and the second, arbitrary C* -algebras. Here are the exact statements. of C ( 0) (see Theorem 1 and the discussion of it) , and the star indicates the outstanding role of the involution.
352
6. Hilbert Adjoint Operators and the Spectral Theorem
1 (First Gelfand-Naimark theorem; for the proof, see [25, Chap ter IV, Theorem 7. 13] ) . 4 Every commutative C* -algebra A is isometrically * -isomorphic to an algebra Co ( 0 ) , where 0 is a locally compact topological space; if in addition A is unital, then 0 is a compact space. Theorem
The authors of the theorem give also the concrete form of this isometric *-isomorphism: this is the Gelfand transform r : A � Co (OA ) , where OA is the Gelfand spectrum of our Banach algebra. Remark. The readers following the examples (and they should!) must have
been surprised by the statement of this theorem. What about the algebras L00 (X, J1) and, in particular, l00? Obviously, they are C*-algebras of func tions, but at first sight they do not belong to the class Co (O) . However, they are C0 ( 0 ) algebras in a latent form, up to an isometric *-isomorphism. This "invisible" compact space 0 is an important characteristic of the algebras considered. In particular, up to the indicated equivalence, loo == C(,BN) , where ,BN (the Gelfand spectrum of the algebra Zoo ) is just the Stone-Cech compactification of the discrete space N ( cf. discussion in Section 3. 1 ) . A
Theorem
2 (Second Gelfand-Naimark theorem, 1943; for the proof, see [25,
Chapter IV, Theorem 7.57] ) . Every abstract C* -algebra A is isometrically
* -isomorphic to some concrete {i. e. , operator) C* -algebra. In other words, every C* -algebra has an isometric * -representation in some Hilbert space.
(What is this Hilbert space and how the required representation arises, we explain closer to the end of this section.) An immediate useful consequence of these theorems is that they infor mally show that all we can do with continuous functions, can also be done with the elements of commutative C* -algebras, and all we can do with opera tors on Hilbert spaces, can be done with the elements of general C* -algebras. For instance, in every algebra of the form C0 ( 0 ) the equation x n y has the solution for each non-negative right-hand side and for each natural n. Hence, the same is true for every commutative C* -algebra consisting, say, of operators. Exercise 6* . The spectrum of a unitary element of a unital C* -algebra lies in 1r, and the spectrum of a selfadjoint element is in JR. Hint. In the first of the two proofs of Theorem 2. 1 there is nothing related "specifically to operators" . ==
4 Mark Aronovich Naimark ( 1909-19 78) , a prominent Russian mathematician. He obtained a series of first-class results in functional analysis.
3. Involutive,
C* - , and von Neumann algebras
353
Now we make a few general comments . After more than half a century since the appearance of these theorems, their place in the entire building of mathematics is per haps more important than the creators themselves anticipated. Now these theorems are an integral part of a new important area called non-commutative geometry. One of the "philosophical aspects" of this area is as follows . Speaking informally, the first Gelfand Naimark theorem says that the structure of a commutative C * -algebra is adequately de scribed in purely topological terms of its spectrum. (One can give a precise meaning to this statement using the categorical language. We shall do this later in the part for ad vanced readers. ) Because of this (and now we are even more vague) one can and should view an arbitrary C * -algebra as the algebra of functions on a "non-commutative locally compact space" , or even as a "non-commutative locally compact space" itself. Of course, these are strange words . However, this idea indeed turned out to be extremely fruitful. If you know topology, you can predict properties of non-commutative C * -algebras, and to build useful (in particular, for quantum physics) "non-commutative" versions of classical objects of topology, for instance, "non-commutative sphere" or "non commutative torus" ( cf. [88] ) . At the same time, a higher level of understanding of a series of topological (i.e. , "commutative" ) notions and results is sometimes achieved after dealing with non-commutative C * -algebras. (Moreover, some purely topological facts can obtain more transparent proofs. For instance, the well-known Bott periodicity theorem is proved by I. Cuntz using compact operators; see, e.g. , [84, Theorem 1 1 .2. 1] .) That is why the theory of C * -algebras is also called non-commutative topology. *
*
*
We now turn from the general discussion back to mathematics. Among the operator C* -algebras there is a special class, which was studied much earlier than general C*-algebras (both abstract and concrete) . It was John von Neumann who introduced this class in 1930. One of his main motiva tions was the hope (partially justified later, although not completely) that these algebras, being the "right place for observables in quantum mechanical systems," may put quantum mechanics on a solid mathematical foundation. The algebras we begin to discuss can be defined both in topological and in algebraic terms, and the result will be the same. Recall that in B(H) , where H is a Hilbert space, in addition to the operator norm, there are several structures of a polynormed space. We had already encountered the weak-operator, strong-operator (and, in the case of advanced readers, also ultraweak) families of prenorms and the topologies with the same names, which are generated by these families (see Section 4. 1 and the end of Section 4.2) . For every subset M C B (H) its commutant is the set M ' : == {b E B (H) : ba == ab for all a E M} , and the bicommutant is the set M 1' : == ( M ' ) ' (i.e. , commutant of the commutant) . Evidently, we always have M C M 1' . (von Neumann bicommutant theorem; see [25, Chapter III, Theorem 2.34] ) . Let A be a selfadjoint subalgebra in B(H) containing 1 . Theorem
3
354
6. Hilbert Adjoint Operators and the Spectral Theorem
Then the following properties are equivalent: ( i ) A is closed in the weak-operator topology; ( ii ) A is closed in the strong-operator topology; ( iii ) A '' == A. Exercise
7.
Show that ( iii ) implies ( i ) and ( ii ) .
There are at least four other useful topologies that could have participated in this formulation instead of the two we have just mentioned . The ultraweak topology is one of them (see the end of Section 4.2) ; however, it was not known at the time when the first version of the von Neumann theorem appeared. The proof of the "full" bicommutant theorem, where all the six topologies take part , can be found, e.g . , in [68] . We emphasize that the selfadjointness condition of the considered subalgebra cannot be omitted.
Definition 5. An operator algebra satisfying the assumptions of Theorem 3
is called a von Neumann algebra.
In addition to the algebra B(H) , among the operator algebras we have already met, the algebra of ( all ) diagonal operators on l2 belongs to the class of von Neumann algebras. More generally, the algebra of operators of multiplication by functions in L 2 (X, 11) is also a von Neumann algebra. ( Try to prove this at least for the first algebra. ) Since the topologies participating in Theorem 3 are weaker ( i.e. , coarser ) than the norm topology, every von Neumann algebra is automatically closed in the latter topology, and thus is an operator C* -algebra. It is easy to see that the von Neumann algebras form a much smaller class than the general operator C*-algebras. For instance, the algebra K (H) , and the algebra of compact diagonal operators on l2 are not von Neumann algebras. Exercise
8.
( i ) Both the weak- and the strong-operator closures of the algebra F(H) ( hence of K(H) ) coincide with B(H) . ( ii ) Both the weak- and the strong-operator closures of the algebra of finite-dimensional ( hence of compact ) diagonal operators on l2 coincide with the algebra of all diagonal operators.
Is it possible to characterize the von Neumann algebras in abstract terms, similarly to the way the second Gelfand-Naimark theorem describes the operator C* -algebras? This outstanding problem has remained open for many years. Here is an impressive solution in terms of geometry of Banach spaces. Let E be a Banach space. Another Banach space E* is called predual to E if E coincides with ( E* ) * , up to an isometric isomorphism. For instance,
3. Involutive,
C* -, and von Neumann algebras
355
l 00 has a predual, namely l 1 . At the same time, it can be shown that neither c0 nor, say, K(H) have predual spaces ( cf. Exercise 4. 2.8) . Theorem 4 ( S. Sakai, 1957; see [68, Chapter III, Theorem 3.5]). ( i ) Every von Neumann algebra viewed as a Banach space has a pred ual space, which is determined uniquely up to an isometric isomor phism. ( ii ) If a C* -algebra viewed as a Banach space has a predual algebra, then it is isometrically * -isomorphic to a von Neumann algebra. Sakai explicitly indicated the predual space of a von Neumann algebra. It is an (automatically closed) subspace in A * consisting of functionals that are continuous in ultraweak topology. At the end of Section 4. 2 we have already verified this for the case of
A == B(H) .
The Sakai theorem plays the same role in the theory of von Neumann algebras as the second Gelfand-Naimark theorem does in the theory of op erator C*-algebras. The role of the first Gelfand-Naimark theorem in this analogy belongs to the following fact, which had been found much earlier: Theorem 5 ( von Neumann; for the proof, see [86, Theorem 4.4.4] , [50, Theorem 9.4. 1] ) . Every commutative von Neumann algebra is isometrically * -isomorphic to the algebra L 00 (X, J1 ) , where (X, 11) is a measure space.
Moreover {von Neumann-Halmos), if our algebra acts on a separable Hilbert space, then it is isometrically * -isomorphic either to l� , where n is a finite or countable cardinality, or to L 00 [0, 1] , or to the l 00 -sum of these two algebras.
Remark. This theorem shows, in particular, to what extent the class of
commutative von Neumann algebras is more narrow than the class of com mutative operator C* -algebras ( at least if we consider algebras acting on a separable space ) . We should have in mind that, as it can be easily shown, the second class cannot be more narrow than the class of metrizable compact spaces. The first Gelfand-Naimark theorem suggests to treat C * -algebras as if they were "non-commutative topological spaces" . Continuing this analogy, Theorem 5 suggests to treat von Neumann algebras as if they were "non-commutative measure spaces" . We have already discussed the theory of C * -algebras as "non-commutative topology" . Similar reasons allow us to call the theory of von Neumann algebras "non-commutative measure theory" (or "non-commutative probability theory" ) . Again, this sentence sounds meaning less. However, it reveals a strategy of actions. If we know measure theory (i.e., probability theory) , we can predict the results of the theory of von Neumann algebras: the "proba bilistic" intuition works here. (For instance, the classical Radon-Nikodym theorem from measure theory has a beautiful and important non-commutative version. ) In the opposite direction, the knowledge of the theory of von Neumann algebras helps to better interpret measure theory as a limiting, or "classical" , case. Physicists also obtain something useful: a new view of things and new methods. Here is an example. From this point of view, it
356
6.
Hilbert Adjoint Operators and the Spectral Theorem
is fruitful to interpret quantum mechanical events as orthogonal projections in von Neu mann algebras, the reason being that in the commutative case these projections can be naturally identified with measurable subsets in X , i.e. , with "classical" events. Both "non-commutative areas" , their close mutual relations, and many other related things (in particular, the so-called "non-commutative differential geometry" ) are presented in A. Connes' book Non-commutative geometry [21] . Now we explain to advanced readers the following facts. Behind Theorem 1 , which was formulated as a result about the identification of certain mathematical objects, a stronger result is hidden. It deals with the identification "as a whole" of two important categories . Consider the category of C* -algebras denoted by C *alg. It is clear what are the ob jects of this category; as for the morphisms, we define them as arbitrary *-homomorphisms of algebras of this class. As a matter of fact , they only seem to be arbitrary: all these mappings have extremely good structure. r.p
Theorem 6 (see [25, Chapter IV, Theorem 7 . 83] ) . Suppose
* -homomorphism. Then there ts a commutattve diagram A
0 for every a E A. Such a functional defines a pre-inner product on A by the formula ( a, b ) : == f (b* a) . Put Jf :== {a E A : ( a, a ) == 0} . The procedure we had already learned (see Proposition 1 .2.2) turns the pre-Hilbert space A into a near-Hilbert space if! :== A/ I, and after the completion (see Proposition 2.6.3) , a Hilbert space H i . We would like to define a representation of the algebra A in H 1 , i.e., a homomorphism from A to B( H f ) . However, in the general case we can only define a homomorphism from A to .C(iif ) by the formula a �-----+ T1 : b + Jf �-----+ ab + Jf . The obstacle is that the operators T1 are not necessarily bounded. But (and this is already a non-trivial result) one can show the following: if A is a C* -algebra, then all T1 are automatically bounded and even contraction operators. Therefore, we can extend every T1 by continuity to an operator T1 and obtain a mapping Tf : A ---+ B( H f ) : a �-----+ Tf . It is not difficult to verify that this mapping is a *-representation of the algebra A in H f . It is called the GNS-representation associated with the positive functional f. The GNS-construction, i.e. , the construction of the GNS-representation, works for some other classes of algebras; for instance, all that we have said is true if A is an ar bitrary unital Banach star-algebra. However, since we want to end up with an isometric representation, we must restrict ourselves to the C* -algebras. Thus, we assume that A is a C* -algebra and consider the family of GNS-representa tions of A associated with all possible positive functionals. In practice, it often turns out that we can find an isometric one among these representations. Here are two examples. Example 7. Suppose A : == C[O, 1] and f : a �-----+ J; a(t) dt. Then obviously H i == L2 [0, 1] , and T1 is the operator of multiplication by a function a(t) in L2 [0, 1] . We see that the GNS-representation Tf : C[O, 1] ---+ B(L2 [0, 1]) is isometric. Exercise 9. Suppose A : == B ( H ) , x E H is different from zero, and f : B( H ) ---+ C : i is unitary isomorphic to H, and Tf is an isometric a �-----+ ( a ( x ) , x ) . Then the space H isomorphism of B( H ) onto B( H f ) . Remark. An isometric GNS-representation of a C * -algebra always exists if this algebra is separable. In this case among the functionals on A there is a strictly posttive one, i.e. , a functional f such that f (a* a) > 0 for all a =1- 0. From this we can easily deduce that the representation Tf is injective, hence isometric if we take Theorem 6 for granted. But to construct such a functional one must use rather complicated techniques.
In the general case it can happen that every GNS-representation of our algebra has a non-zero kernel, and thus looses some information about the structure of the algebra. (Show how it happens , say, in the case of A : == eo (X) , where X is a non-countable set . ) So we proceed as follows.
358
6.
Hilbert Adjoint Operators and the Spectral Theorem
Consider the set of all positive functionals such that 11 ! 1 1 == 1 . (Such functionals are called states ; 5 if A is unital, then they are characterized by the condition f(e) == 1 ) . Put H : == E9{H 1 : f is a state} , i.e. , take the Hilbert sum of (an enormous number! ) of Hilbert spaces corresponding to all possible states. This is the Hilbert space promised in Theorem 2. As for the promised representation, it is the mapping T A ---+ B(H) assigning to every a E A the operator Ta taking the "line" ( . . . , x f , . . . ) ; x f E H f to ( . . . , Tf (xJ ) , . . . ) . The verification that this "sum of GNS-representations" is also a * representation is relatively easy. On the contrary, the fact that this representation is isometric requires non-trivial arguments and uses some special properties of C* -algebras. :
Remark. Again, the creation turned out to be more perfect than the creators them selves expected. The approach that Gelfand and N aimark used to identify an abstract C* -algebra with a concrete one (an operator C* -algebra) , had, as the authors initially thought , a shortcoming. The Hilbert space in which the operators act turned out to be very large; it is always non-separable (since there are too many terms in the Hilbert sum) , and the image of the representation of T is only a "small part" of all B(H). This caused a discomfort the authors felt , and provoked a desire to construct more "efficient" represen tation. 6 It took more than twenty years to discover that this , at first sight "inefficient" , representation of C* -algebras as operator algebras has great possibilities. It turned out that every possible representation of a given C* -algebra as an operator algebra can be obtained as a part of this special representation introduced by Gelfand and N aimark. Due to this , the latter is now called the universal representatton. Moreover, this representation allows one to connect the topics that initially had little in common: the theory of von Neu mann algebras (non-commutative measure theory) and the theory of general C* -algebras (non-commutative topology) . To do this we take the operator algebra constructed in the Gelfand-Naimark representation and pass to its bicommutant in B(H) . We obtain the so-called enveloptng von Neumann algebra of the initial (abstract) C* -algebra, which is the desired link between the two theories. Note the remarkable fact that this algebra, as a Banach space, is isometrically isomorphic to the second dual space A* * . About all this, see, e.g. , [68] , [90] , [87] .
*
*
*
Our digression has already taken much time. However, it is tempting to tell you a dramatic story about an old problem of "non-commutative measure theory" . Here is another diversion. After introducing "his" algebras, von Neumann attempted to classify them up to a *-isomorphism. Realizing independent importance of such a work, he believed that this result should be very useful for quantum mechanics. (Explanation of such connections exceeds the scope of this book. We only note that at that time (the early 1930s) von Neumann set a verily ambitious goal. He wanted to put quantum mechanics on a solid mathematical foundation to make it as harmonious as classical mechanics, and ideally, to turn it into a mathematical theory with its own system of axioms, definitions, and theorems: something like quantum-mechanical Euclid ' s "Elements" . ) At the same time he felt that the required mathematical techniques would allow him to "kill several more (purely mathematical) birds" , which we will not describe here (we only mention that this is about group representations and infinite-dimensional analogues of the Wedderburn theorems) .
5In some mathematical models of quantum mechanics they represent ( physical) states of quantum mechanical systems; see, e.g. , [45, pp. 1 3- 1 5] . 6 It was my teacher Mark Aronovich Naimark who told me about this.
3. Involutive,
C* , and von Neumann algebras -
359
At the beginning, everything looked promising. First , Theorem 5 formulated above, gave a full description of commutative von Neumann algebras. (As we said before, this theorem was the origin of the non-commutative measure theory. ) As for the general von Neumann algebras, von Neumann realized that their study can be reduced to consideration of "elementary cells" , or "bricks" , from which every von Neumann algebra can be built with the help of a natural procedure-the so-called direct integral of operator algebras (there is no sense to speak about it in details now) . Here is the definition of these cells. Definition {a E A : ab
6. ==
A von Neumann algebra A is called a factor if its center, i.e. , the set ba for all b E A} , consists of scalar multiples of the identity operator.
Note that the factors participating in the von Neumann decomposition of commutative von Neumann algebras (i.e. , according to Theorem 5, algebras of the form Loo (X, J.l ) ) are simply C , and there are as many of them as the cardinality of X is. But what are these factors? In the early 1 930s only one type of factors was known: B(H) for a Hilbert space H. (You can verify that this is indeed a factor. ) It was natural to conjecture that there are no other factors at all. But in 1 935 one of the major discoveries of the 20th century happened: von Neumann and his colleague P. Murray found a factor that is completely different from B(H) . (One of the realizations of this factor is described in [92] . ) Continuing their investigations, Murray and von Neumann showed that factors are divided into three types depending on the behavior of projections in them. More precisely, to every space that is the image of such a projection, one can assign a non-negative real number or oo, the so-called relative dimension. (Here we assume for simplicity that H is separable.) If a factor is such that the relative dimension of the images of projections can take natural values or, possibly, oo, then it belongs to type I; these are the algebras B( H) and nothing more. The next level of complexity is represented by "factors of type II" , for which these numbers take values everywhere in the interval [0, 1] (the above mentioned Murray-von Neumann factor is one of those) or in the extended positive real axis IR+ U { oo } . Finally, the third and the last logically possible case is that the relative dimensions of the projection spaces take the values 0 (for the zero subspace) and oo (for all other subs paces) . These are "factors of type III" . They have the most complicated structure. (The fact that these factors really exist became evident only five years after the discovery of factors of type II. ) Well, there are more factors than was initially expected, but do they admit a full description? It turned out that "the more you work, the more you have to do. " Soon it was realized that already among the factors of type II there are many that are mutually non-*-isomorphic, and still no one could figure out whether there are non-isomorphic factors of type III. The first example of such kind appeared only in 1957 (L. Pukanszky) , and it was extremely complicated technically. By that time von Neumann was terminally ill, and it looks like he did not find out about this example. The following important step was made ten years later, when R. Powers constructed a continuous family of mutually non *-isomorphic actors of type Ill indexed by a parameter A E (0, 1 ) . But it was the French mathematician Allain Connes who made a revolutionary dis covery in 1973. By that time it was realized that the problem of classification of all factors is hopeless: there are too many of them. And the first Connes ' achievement was that he distinguished a substantial class within which it was reasonable to formulate the classification problem. As it befits a fundamental notion, the class of factors and, more generally, the class of von Neumann algebras distinguished by Connes, can be character ized by many different and at first sight dissimilar ways. These algebras are often called amenable (the term inherited from harmonic analysis and homological theory of Banach
360
6.
Hilbert Adjoint Operators and the Spectral Theorem
algebras) . Depending on the choice of the way leading to this notion, these algebras are also called hyperfinite , injective, semidiscrete, etc. The shortest way to give the definition is , following Connes, to use the notions from the geometry of Banach spaces: a von Neu mann algebra A C B(H) is called amenable if there is a projection of norm 1 in B(H) that has A as its image (in other words, the natural embedding of A into B(H) is a coretraction in Ban 1 ; cf. the discussion in Section 2.4) . Amenable factors turned out to be the class where a complete classification was achieved . Everything was clear with the factors of type I (they are, of course, all amenable) : they are completely determined by their dimension as a linear space. Connes proved that there are only two non-isomorphic amenable factors of type II: the one discovered by Murray and von Neumann, and another one . As for the amenable factors of type III, Connes divided them into subtypes III.x , where A runs through the interval [0, 1] . He showed that for each A E (0, 1 ) there is one factor, namely, that found by Powers. There are many non-isomorphic factors of type III 0 , but all of them are described in sufficiently transparent terms of the so-called "Krieger factors" . The only difficulty Connes faced were factors of type III 1 (by the way, not long ago they came into fashion in quantum field theory) . Connes proved that the "Araki-Woods factor" known by that time is an amenable factor of this type, but he could not say whether there are other amenable factors of the same type. This last gap in the Connes theory was filled in 1982 by Danish mathematician Uffe Haagerup. He proved that there are no other factors, and thus brought the problem of classification of amenable factors to conclusion.
4 . Continuous functional calculus and posit ive operators
Theorem 2. 1 on the location of spectra is the first serious step in studying selfadjoint operators. Now we can go much further. Until further notice, T is a selfadjoint operator on a Hilbert space H and a :== a(T) is its spectrum; as we recall, it is a non-empty compact set in JR. We have already defined polynomials in arbitrary elements of an algebra. Moreover, if the algebra is a Banach algebra, we can take entire holomorphic functions of its elements. Our next goal is to show that we can take arbitrary continuous functions of a selfadjoint operator provided these functions are defined on intervals (and, more generally, arbitrary subsets) in 1R containing the spectrum. As always, for each p E C [t] we denote by P i a the restriction of p (as a function of real variable) to a. Let us distinguish the following preparatory
1. For every p E C [t] we have ll p(T) II == II P i a lloo · Proof. We first assume that p is a selfadjoint element in C[t] . Then from Proposition 3.2, taking into account Example 3. 1 , it follows that the operator p(T) has the same property. Combining Theorems 2. 1 and 5.2. 1 , we see that ll p(T) II == max{ I A I : A E a(p(T) )} == max{ j p(A) j : A E a } == II P i a lloo · Now we pass to the general case. Since the polynomial q : == p*p is selfadjoint, from what we have just said it follows that ll q(T) II == ll q la lloo · Propo sition
4.
Continuous functional calculus and positive operators
But by the C*-identity and Example 3. 1, we have llq (T) II == ll p* ( T )p( T) II == ll p ( T) *p( T) II == At the same time, it is obvious that
361
ll p (T) II 2 -
llqlalloo == max { l p*p(A) I ; A E a} == max { l p (A)p (A) I ; A E a} == I P i a ll� •
The rest is clear.
In the following definition, [a, b] is an interval in JR, and t denotes the restriction to [a, b] of the "independent variable" t �----+ t.
1. Continuous functional calculus or just continuous calculus of T on [a, b] is a unital homomorphism rc : C[a, b] � B(H) such that 'Yc ( t ) == T. Theorem 1 (on continuous functional calculus) . If [a, b] contains a, then the continuous calculus rc of T on [a, b] exists and is unique. Furthermore, (i) 'Yc is an involutive homomorphism of * -algebras; (ii) l i re ( ! ) II == II !Ia II for every f E C[a, b] and, as a corollary, rc is a contraction operator; (iii) Ker( !c ) == {f E C[a, b] : / Ia == 0}; (iv) if a bounded operator on H commutes with T, then it commutes with the operator rc (f) for every f E C[a, b] . On the other hand, if [a, b] does not contain then there is no continuous calculus of T on [a, b] . Definition
a,
Proof. Suppose
a C [a, b] . Denote by P the subset in C[a, b] consisting of
the restrictions of polynomials. Certainly, it is a *-subalgebra. Denote the restriction of p E C[t] to [a, b] again by p; this will not lead to a confusion. Since the continuous calculus 'Yc of T on [a, b] is a unital homomorphism, we have rc (P ) == p(T) for all p E C[t] . From the continuity of 'Yc and the Weierstrass approximation theorem it follows that there is at most one such calculus. Now, consider the mapping 'Yo : P � B(H) : p �----+ p(T) (coinciding with the polynomial calculus if we identify P with C [t] ) . Certainly, it is a unital *-algebra *-homomorphism taking t to T. Proposition 1 guarantees that this is a contraction operator with respect to the norm in P inherited from C[a, b] . Since B(H) is a Banach space, the continuity extension principle gives a unique contraction operator 'Yc extending 'Yo to C[a, b] . Let us look at its properties. Suppose f, g E C[a, b] , a sequence Pn E P tends to f, and a sequence Qn E P tends to g . Taking into account the continuity of 'Yc and the continuity
6.
362
Hilbert Adjoint Operators and the Spectral Theorem
of the multiplication in C[a, b] , we see that rc (Pn Qn ) tends to !c (fg). At the same time, since the polynomial calculus is an homomorphism and the multiplication in B(H) is continuous, we have that rc (Pn Qn ) == [pnqn ] (T) == Pn (T) qn (T) == !c (Pn ) !c (Qn ) tends to !c (f) !c (g). This proves that rc is a homomorphism (and, thus, it is a continuous calculus of T on [a, b] ) . Again by the continuity of rc and the "star property" of C[a, b] we see that rc (P�) tends to rc (f*), and at the same time, since the polyno mial calculus on T is involutive (Example 3.1) and B(H) is a star-algebra, !c (P�) == p� (T) == Pn(T) * == !c (Pn )* tends to !c (f)*. Hence, !c is an involu tive homomorphism. Finally, again from Proposition 1 , we have limoo II Pnla lloo == ll f lalloo · lim l rc (Pn ) ll == n� limoo II Pn (T) II == n� ll rc ( / ) 11 == n�oo This proves ( ii) and, as a corollary, (iii) . If some S E B(H) commutes with T, then, obviously, Sp(T) == p(T)S for every polynomial p. Hence, if f E C[a, b] and Pn E P tends to f in C[a, b] , then Src (f) == S(limn� oo Pn (T)) == limn� oo (Spn (T)) == limn� oo (Pn (T)S) ==
!c (f)S.
It remains to prove the last statement. If rc exists, then, by Proposition 5.2.4, a(T) C a ( t ) , and the latter set is [a, b] (Example 5.3.6) . The rest is • clear. Continuous calculi on different intervals are compatible.
Suppose C [c, d] C [a, b] , rc and !� are continuous calculi of T on the intervals [a, b] and [c, d] respectively, and C[a, b] � C[c, d] is the restriction f �----+ / I [c , d] . Then rc == !�
Proposition 2.
a
T.
Proof. The mapping
T :
!� obviously, satisfies the definition of continuous T,
calculus of T on [a, b] . Hence, by the uniqueness of the latter, it coincides • with rc · From this moment let us agree that if f is defined on an arbitrary subset in 1R containing the interval [a, b] ::J a, then we denote by f(T) the operator rc (f l [a , b] ) · In view of Proposition 2, this operator does not depend on the choice of the interval. Exercise 1. Let T be a compact selfadjoint operator. If f(O) == 0, then f(T) is compact as well. If, in addition, H is infinite-dimensional, then the converse is true. Speaking about the continuous calculus of operators, we restricted our selves to functions defined on intervals; this simplifies the arguments and is completely sufficient for further applications. But we could have taken
4.
Continuous functional calculus and positive operators
363
an arbitrary compact set containing a, including a itself, as the domain of our functions. In the following exercise � is such a compact set, t is the restriction of the independent variable to � ' and by continuous calculus of T on � we mean a bounded unital homomorphism rf : C(�) � B(H) such that rf (t) == T. Exercise 2. The assertion of Theorem 1 remains true after replacing [a, b] by � - As a corollary, ,g (i.e. , rf for � :== a) is an isometric operator. Hint. If [a, b] is an interval containing � ' then every f E C(�) is a restriction of some f E C[a, b] . Taking into account the known form of Ker(rc) , this allows us to define the mapping f �----+ rc(f) . It is the required calculus. Remark. The result of this exercise in the case of � : == a can be regarded as a special (very special) case of the first Gelfand-Naimark theorem. In this situation the considered C*-algebra is {p (T) : p E C [t] } , i.e. , the minimal operator-norm-closed algebra containing T. Let us find out what the general notion of continuous calculus turns out to be for some concrete classes of operators. In the next examples and exercises, M is a subset in 1R containing an interval where the spectrum of the given operator lies, and f is a continuous function on M. Example 0. Suppose T : == A1 (A E C) . Then a == { A} , and f(T) == j (A) 1 for every f. Example 1. Let T : == P be a projection in H different from 0 and 1. As we remember, in this case a == { 0, 1}. Then f ( T) == g ( T) for each g E C[O, 1] coinciding with f at the points 0 and 1. If we take the linear function /(0) (1 - t) + f(1)t as g , we see that f(P) == f(O)Q + f(1)P, where Q :== 1 - P. Example 2. Suppose H :== L 2 ( [a, b] , 11) , where 11 is a Borel measure on [a, b] and T an operator of multiplication by t (independent variable) . Then the mapping 'Yc : C[a, b] � B(L 2 ([a, b] , 11)) taking every f to the operator Tt of multiplication by f is a continuous calculus of T on [a, b] . Thus, in the considered case, f(T) coincides with Tt · Exercise 3. Suppose T :== T>.. ; A == (AI , A 2 , . . . ) is a selfadjoint diagonal operator on l 2 . Then M contains all An , and f(T) is a diagonal operator T11 with l1n : == /(An ) · Exercise 4 (Generalization of Exercise 3) . Suppose (X, 11) is a mea sure space and T : == T'P is a selfadjoint operator of multiplication by cp E L 00 (X, 11) in L 2 (X, 11) . Then a contains cp(t) for almost all t E X, and f(T) is the operator of multiplication by the function t �----+ f ( cp ( t )). -
364
6. Exercise
Hilbert Adjoint Operators and the Spectral Theorem
5. Suppose T is a selfadjoint compact operator on H, and the
vectors e n and numbers An are the same as in the Hilbert-Schmidt theorem. Then M contains all A n , and, in the case of infinite-dimensional H, contains 0 as well; f(T) takes e n to j(An )e n and is equal to f(O)x on every x such that x _l e n ; n == 1 , 2, . . . . Let us now compare the continuous calculus we have just constructed with the entire holomorphic calculus from Section 5.3. As we remember, entire holomorphic functions can be constructed for arbitrary elements of Banach algebras. Now the initial requirements are more rigid: we consider the algebra B(H) and a selfadjoint operator in it. But in compensation we have many more functions of an element: for instance, we can speak about the "module of T" , or the "cube root of T" , etc. Remark. In fact, what is essential in the construction of continuous calculus
is that we consider a C*-algebra A and a normal (not necessarily selfadjoint) element a in it. In this situation, for a continuous function f on a compact set in C containing a(a) , one can define the element f(a) (see, e.g. , [33] ) . But this is the limit of possible generalizations.
The fact that we use similar symbols w(T) for w E 0 (see Section 5.3) and f(T) for f E C[a, b] in this section will not cause a confusion. Exercise 6 (Compatibility of functional calculi) . Suppose re is an entire calculus of T as an element of B(H) , and [a, b] is an interval containing a. Then the following diagram: 0
1
C[a, b]
�B(H)y
where j : w �----+ w I [a, b] , is commutative. Hint. The set of polynomials is dense in 0 , and the set of their restric tions to [a, b] is dense in C[a, b] . Note an attractive feature of the continuous calculus that brings it closer to the calculi we considered before. Theorem 2 (spectral mapping law for continuous calculus) .
For each inter
val [a, b] ; a C [a, b] C 1R and for each f E C[a, b] we have a(f(T)) == f(a(T) ) .
Proof. First, let us show that a(f(T) ) C f(a(T) ) . Suppose
A ¢: f (a(T) ) ,
i.e. , f( t ) # A for t E a(T) . Consider a function (f( t ) A) - 1 on a(T) . As any other continuous function on a(T) , it can be extended to a continuous func tion, say g , on [a, b] (this evidently follows from a representation of 1R \ a(T) -
4.
Continuous functional calculus and positive operators
365
as a union of disjoint intervals) . Then the function g (t) (f(t) - A) - 1 van ishes at a (T) , and thus (Theorem 1) belongs to Ker( !c ). Hence, g(T) (f (T) A l) == 1, and this means that A ¢: a ( f (T)) . Let us prove the reverse inclusion. Suppose f ( s) ¢: a ( f (T)) for s E a (T) . In other words, the operator f (T) - f (s )l has an inverse in B(H) , say, S. For every n E N consider the function 9n E C[a, b] that is equal to 1 at s, 0 for I t - s l > � , and is linear on the intervals [s - �, s] and [s, s + �]. Since rc is a homomorphism and ll 9nlloo == ll9nl a (T) lloo == 9n (s) == 1, Theorem 1 shows that
l gn (T) II == l gn (T)(f(T) - f(s)l)S II < ll gn (T) (f(T) - f(s)l) II I S II · Since rc is a contraction operator, we see that ll gn (f - f(s)) lloo > 11111 • 0 for a But, by the choice of gn , we have ll 9n (f - f(s)) lloo • contradiction. 1 ==
�
n � oo ,
Here is another useful observation: the identification of operators by the topological (or, as a special case, unitary) equivalence implies the respective identification of the values of a continuous function on these operators.
3.
Suppose I : H � K is a topological Hilbert space isomor phism implementing a topological equivalence between selfadjoint operators T : H � H and S : K � K. Then for every f E C[a, b], where [a, b] is an interval containing the spectrum of these operators {by Proposition 5. 1. 1 they have the same spectrum), the same I implements a topological equivalence between the operators f ( T) and f ( S) . Proof. Consider the mapping �� : C[a, b] � B(K) : f r--+ I f (T) I - 1 . Evi Proposition
dently, it has the properties of a continuous calculus of S on [a, b]. Conse quently, by the uniqueness of the latter, I f (T) I - 1 is nothing but f(S). •
One of numerous applications of continuous calculus is that it allows us to characterize from different points of view a very important class of operators. Whereas selfadjoint operators behave like real numbers, these operators behave like non-negative numbers. Definition 2. An element
a
of an (arbitrary) unital involutive algebra is called positive (notation: a > 0) if it is selfadjoint and a ( a ) C JR + . In particular, a bounded operator on a Hilbert space H is called positive if it is positive as an element of the involutive algebra B(H) . Certainly, every (orthogonal) projection is positive. The following propo sition gives a huge set of examples of positive elements.
4.
Unital homomorphisms of unital * -algebras take positive elements to positive elements.
Proposition
366
6. Hilbert Adjoint Operators and the Spectral Theorem
• 5.2.4. a(T) and f > 0, then the
Proof. This immediately follows from Propositions 3.2 and
1. If f E C[a, b] , where [a, b] operator f ( T ) is positive.
Corollary
=:)
In particular, the operator described in the following definition is posi tive. Suppose T > 0 and the function f ( t ) : == Jt is defined on some interval in JR+ containing a(T) . Then the operator f ( T ) is called the arithmetic square root of T and is denoted by VT.
Definition
3.
(As we recall, this operator does not depend on the choice of the inter val.) From the definition it evidently follows that ( Vf') 2 == T (which justifies the name "square root" ) .
7.
VT is a unique positive operator R such that R2 == T. Hint. Suppose Pn ( t ) tends to Jt in C[O, II T II J . If S > 0 is such that S2 == T and Qn ( t ) : == Pn (t 2 ) , then qn (S) == Pn (T) tends to S, and at the same time to VT. Exercise
Theorem
alent:
3.
The following properties of an operator T E B(H) are equiv
T is positive; T == S2 for some positive S; T == S2 for some selfadjoint S; T == S* S for some S E B (H, K) , where K is (generally speaking) another Hilbert space; ( v) the quadratic form of the operator T takes only non-negative values {i. e., (Tx, x ) > 0 for all x E H).
( i) (ii) (iii) (iv)
Proof. (i)� (ii) . It is sufficient to take
S : == Vf'.
(ii) � (iii)� (iv) . Clear. (iv)� (v) . (Tx, x ) == ( S* Sx, x) == ( Sx , Sx ) . (v) � (i) . By Proposition 2.2, T is selfadjoint, hence a(T) C JR. There fore, our goal is to show that for t > 0 the operator T + tl is invertible. For every x E H we have
II (T + t l)x ll ll x ll > I ((T + tl)x, x ) l == (Tx, x ) + t (x , x ) > t 11 x ll 2 . Hence, T + tl is topologically injective, and it remains to use Proposition • 2.6.
4.
367
Continuous functional calculus and positive operators
Whereas selfadjoint operators form a real subspace in B (H) , positive operators form the so-called cone. T, - T >
5.
If S, T 0, then T == 0.
Proposition
>0
and t
> 0,
then S + T
>0
and tS
> 0.
If
Proof. The first assertion immediately follows from Theorem 3(v) . Taking
into account Proposition 2. 1 , the second assertion is true as well.
•
Let us give some applications of square roots. Proposition
6. If T > 0, then II T II
==
sup{ ( Tx, x ) ; x
Proof. This follows from the equalities IITII
( v'Tx, v'Tx)
==
( v'Tv'Tx, x )
8.
==
( Tx, x) ; x
E H.
==
E BH } ·
ll v'T II 2 and ll v'Tx ll 2
•
For every selfadjoint operator T we have max( a(T) ) sup{ (Tx, x) } and min(a (T) ) == inf{ ( Tx , x) } . As a corollary, we have the formula II T II == sup{ I ( Tx, x) l ; x E BH } · Hint. For sufficiently large t > 0 we have max a(T) + t == l i T + t l ll == sup{ ( Tx, x) ; x E BH} + t. Positive operators allow us to fulfil our promise given before. Namely, we can now describe the polar decomposition for an arbitrary bounded operator ( cf. Exercise 2. 7) . Exercise 9* (polar decomposition) . Suppose T : H � K is a bounded operator between Hilbert spaces. Then there exist a partially isometric operator W : H � K and positive operators S : H � H and R : K � K such that T == W S == RW . The pair of operators W, S is uniquely determined by the condition Ker( W ) ..l == Im(S) - , and the pair W, R, by the condition Im( W ) ..l == Ker(S) . Hint. Put S : == VT*T. Then II Sx ll == IITx ll for all x E H. Hence, there exists a partially isometric operator W , uniquely defined by the rule that it takes Sx to Tx and vanishes on Im(S) ..l . The role of R belongs to VT'f*. The polar decomposition helps to establish the important fact that for selfadjoint operators, the two kinds of identification, "soft" topological equivalence and "rigid" unitary (i.e. , isometric) equivalence, coincide ( cf. Exercise 2. 5) . Exercise 10* . Suppose selfadjoint operators T1 : H � H and T2 : K � K are topologically equivalent. Then they are unitarily equivalent. Hint (we follow A. M. Stepin) . Suppose I : H � K implements this topological equivalence, and I == W S is its polar decomposition. Then W must be unitary, and S must be a positive topological isomorphism. Exercise
368
6. Hilbert Adjoint Operators and the Spectral Theorem
Since IT1 == T2 I, the selfadjointness of the initial operators implies that T1 commutes with J* I. Thus (Theorem 1 ) , it commutes with S == Vf*i as well. Hence, WT1 == WST1 S - 1 == T2 IS - 1 == T2 W. Exercise 11 * . Every selfadjoint operator is a linear combination of two unitary operators. As a corollary, every bounded operator is a linear combination of four unitary operators. Hint. If II T II < 1 , then, imitating the representation of a number in the interval [ - 1 , 1] as a sum of two numbers in 1r, consider the operators u :== T±iV1 - T2 . *
*
*
The notion of positive operator allows us to introduce another important structure in B(H) , an order.
4.
Suppose S, T E B(H) . We say that S is less than T and write S < T if T - S > 0 (in other words, ( Sx, x) < (Tx, x ) for all x E H) . Definition
Let us list some properties of the relation " 0 so small that
!
c ma { lgn ( t ) l : a < t < b } < min { l f ( t ) l : A + < t < b } . Then l cgn l < I f I on [a, b], and Proposition 4.8(ii) shows that llcgn (T)x ll • < l l f(T)x l l == 0. Thus 9n (T)x == 0 for every n, and g(T)x == 0. x
We have obtained a very important geometric object. As we will see later, specifying this object is equivalent to specifying the operator itself. Definition 1. The family H>.. ; A E 1R is called the family of subspaces asso ciated with the (selfadjoint) operator T (or the associated family of T) .
The family H>.. has the following properties: for some a, b E 1R we have H>.. == 0 for A < a and H>.. == H for A > b {boundedness); if A < J1 , then H>.. C H11 {monotonicity); for every A we have H>.. == n { H11 11 > A } {continuity from the right); all H>.. are invariant with respect to g(T) for each g E C(JR) and, in particular, with respect to T.
Proposition 2.
(i) (ii) (iii) (iv)
:
5. Spectral theorem as an integral
Proof. (i) If
A < min{a(T) }, then there exists f
373 E
C( JR ) vanishing for
t < A and non-vanishing on a(T) . Then by Theorem 4.2 (on the mapping of spectra) , f(T) is invertible, and thus, H>.. C Ker(f(T) ) == 0. On the other hand, if A > max{ a(T) } , then for every f E C(JR) such that f(t) == 0 for t < A, we have f (T) == 0 (Theorem 4. 1 (iii) ) , and hence H>.. == H.
(ii) The assertion immediately follows from Definition 1 . (iii) Take an arbitrary f vanishing for t < A, and for every J-L > A put f11 ( t ) :== f(t - (J-L - A)) . Then f11 ( t ) tends to f(t) as J-L � A + 0 uniformly on each interval. Hence, f11 (T) tends to f(T) in B(H) as J-L � A + 0. Therefore, if x E H11 (and thus f11 (T) x == 0 for all J-L > A) , then f (T) x == 0. The reverse inclusion follows from (ii) . (iv) Our goal is to show that x E H>.. implies f (T) ( g (T) x ) == 0 for every f E C( JR) , where f ( t) == 0 for t < A. But f (T) ( g (T) x ) == ( f g ) (T) x , and • fg ( t) also vanishes for t < A. The rest is clear. Example 1. Suppose P is a projection onto a subspace K in H. From the structure of the operator f (P) ; f E C(JR) (see Example 4. 1 ) , we see that H>.. == 0 for A < 0, H>.. == K i_ for 0 < A < 1 , and finally, H>.. == H for A > 1 . Example 2 . Suppose T is an operator of multiplication by the independent variable considered in Example 4.2. Then it is clear from that example that H>.. == 0 for A < a, H>.. == {g E L2 [a, b] : g (t) == 0 for almost all t > A} for a < A < b, and finally, H>.. == H for A > b.
1.
Give the explicit form of the family H>.. for the operators in Exercises 4. 3-4.5. Suppose H>.. and K>.. ; A E 1R are families of subspaces in Hilbert spaces H and K respectively. We call them topologically (respectively, unitarily) equivalent if for some topological (respectively, unitary) isomorphism U we have K>.. == U(H>.. ) for all A. We say that U implements the equivalence of the corresponding type. Proposition 3. Suppose T : H � H and S : K � K are selfadjoint operators, H>.. and K>.. being their associated families. Suppose that some U : H � K implements the topological (or, as a special case, unitary) equiv alence of these operators. Then U implements the (same type) equivalence between the associated families of the operators T and S. Exercise
Proof. By Proposition 4.3,
C (JR) . The rest is clear.
f (T) x == 0 {:=:::> f ( S ) (Ux ) == 0 for every f
E
•
Thus, the classes of unitarily equivalent families of subspaces associated with selfadjoint operators are invariants of unitary equivalence (and the reader who has done Exercise 4. 10 knows that this is the same as to say
374
6. Hilbert Adjoint Operators and the Spectral Theorem
"topological equivalence" ) . But we do not know yet whether this system of invariants is full (i.e. , whether unitary non-equivalent operators generate unitary non-equivalent families) . Also we do not know which families of subspaces are associated with selfadjoint operators and which are not. All this will follow from the spectral theorem. However, right now we can already see how important the information about T contained in the family H>.. is. Here are two illustrations. (Later you will be able to obtain both results as simple corollaries of the spectral theorem, but it is instructive to prove them using the tools we already have.) We call A E 1R a point of increase of the family H>.. if every neighborhood of A contains 11 such that H11 # H>.. , and we call it a discontinuity point of this family if (U{H11 : 11 < A}) - is a proper subspace in H>., . In the sequel, we use the following notation. Let H be a Hilbert space, and H1 C H2 C H closed subspaces. We define H2 8 H1 :== Hf n H2 . (In other words, H2 8 H1 is the orthogonal complement of H1 in the space H2 .) Exercise 2.
(i) A number A belongs to a(T) {=:::} it is a point of increase for the family H>., ; (ii) A number A is an eigenvalue for T {=:::} it is a discontinuity point for the family H>.. .
Hint. (i) {::::: . If A ¢: a(T) , then (A - A + ) n a(T) == 0 for some > 0, and all the functions f vanishing for t < A - and equal to 1 for t > A c,
c
c
c
c
give the same operator f(T) . (i) ===> . Suppose (for definiteness) that A == 0 and H11 == Ho for 1 11 1 < 0. Then by Proposition 1, Ker(T+) == Ker(T + 01) + == Ker(T - 01) + · If x E Ker(T+ ) , then (T + 01)x == - (T + 01) - x , ( Tx, x) < -O ( x , x) and II Tx ll > O llx ll · But if x _l Ker(T+ ) , then x E Im(T - 01) - C Ker(T - 01)_ , (T - 01)x == (T - 01) +x, and again II Tx ll > O llx ll · Thus, 0 ¢: a(T) . (ii) {::::: . Suppose x E H>.. 8 H11 for 11 < A. Then (T - A1)+x == 0 and x E [Ker(T - 111) + ]1_ == [Im(T - 111) + ] - C Ker(T - 111) _ . Hence (T - A1)_ x == lim11 � >.. - o (T - 111) _x == 0, and Tx == AX. (ii) ===> . Suppose Tx == AX. Then (T - A1) _x == 0, and since (t - 11) < (t - A) - for 11 < A, we have (T - 111) _x == 0 for the same 11 · Thus (A - 11)x == (T - 111) +x, and x _l Ker(T - 111) + · Now it is convenient to pass from the language of subspaces to the lan guage of projections onto these spaces. Suppose P>.. is a projection onto H>.. . The following notion plays the role of a visiting card for a selfadjoint operator.
5. Spectral theorem as an integral
Definition 2. The family
for the operator T.
375
PA; A E 1R is called the resolution of the identity
1. The resolution of the identity for an operator T has the fol lowing properties: (i) there exist a , b E 1R such that PA == 0 for A < a and PA == 1 for A > b (boundedness); (iii) if A < J1 , then PA < P11 {monotonicity); (iii) for every x E H we have PAx == lim11� A + O P11x (strong-operator continuity from the right); (iv) all PA commute with g(T) for each g E C(JR) and, in particular, with T itself.
Theorem
Proof. (i) , (ii) follow from Proposition 2 together with Proposition
4.9.
(iv) If g is real-valued, then (vi) follows from Proposition 2(vi) together with Proposition 2.3(v) . The general case can be reduced to the real-valued case using the decomposition of g into the real and imaginary parts. (iii) Clearly, it is sufficient to establish that the condition limn� oo /1n == A; f.1n > f.1n+1 > A implies that for every x E H we have limn � oo Pn x == PAx, where Pn :== P11n . Put Hn :== H11n , Ho :== H, Kn :== Hn 1 8 Hn : n == 1 , 2, . . . and denote by Q n the orthogonal projection onto Kn . Then, clearly, for every n we have the decomposition into a Hilbert sum: H == K1 EBK2 EB · · · EBKn EBHn . -
Using the notation Yn :== 2:::: �=1 Q k x, we have x == Yn + Pn x. Further more, the inequality 2:::: �=1 11 Q k x ll 2 < ll x ll 2 , which is guaranteed by the Py thagorean theorem, implies that the series EC: 1 1 Q n x l 2 converges. There fore, the sequence Yn is fundamental. Let y be the limit of this sequence. Then the sequence Pn x tends to x - y. Since Q k x _l Hn for k < n, we have Yn _l Hn . Hence Yn _l HA and y _l HA. FUrthermore, x - Yn == Pn x E Hn , and hence X - Ym E Hn for all m > n. Thus, X - y E n � 1 Hn . And here is the main point: by Proposition 2(iii) , the last space is HA. Therefore, � oo Pn x == x - y == PA(x - y) == PAx - PAy == PAx. nlim
•
1 (Weak-operator continuity of the resolution of the identity from the right) . For every y E H the function A � ( PAx, y ) is continuous from the right. Corollary
x,
*
*
*
376
6. Hilbert Adjoint Operators and the Spectral Theorem
Now we pass directly to the notion of operator-valued Riemann-Stiltjes integral. It is a special case of the notion of vector-valued Riemann-Stiltjes integral, and the latter is worth mentioning by itself. Let us forget about T for a short while. Consider a Banach space E and an interval [a, b] . Let cp be a function on [a, b] with values in E, and f a complex-valued function on the same interval (both cp and f are arbitrary at the moment) . FUrthermore, as in the traditional definition of the Riemann integral, consider various partitions A : a == Ao , AI , . . . , An == b of the interval and put d( A ) : == max{ A k - A k - I ; k == 1 , . . . , n} . Finally, to each partition A we associate the vector n L.A (f, cp) : = L f(A k ) (cp(A k ) - cp(A k - I)) k= I in E, called the Riemann-Stiltjes integral sum corresponding to the partition
A.
Definition
3. Suppose there exists limd( A) �o � A (f, cp) (i.e. , x E E such that 0 there is 8 0 with ll x - � A (f, cp ) ll < c as d(A) < 8) . Then
for every c > > this limit is called the Riemann-Stiltjes integral of the function f ( t) with respect to the vector-valued function cp(t) and is denoted by J: f(t)dcp(t) . Under what conditions on f and cp does the introduced vector-valued integral exist? This question is rather delicate, and we shall not discuss it now. Instead we immediately pass to a special case that is particularly interesting for us. The integrals of our functions will take values in the Banach space B(H) , so we call them operator-valued. Denote the set of all projections in H by P(H) . We consider a mapping JP> : 1R --+ P(H) and agree to write PA instead of JP>(A) . Definition 4. The mapping JP> is called an abstract resolution of the identity (or just a resolution of the identity if there is no danger of confusion) if it has properties (i)-(iii) in Theorem 1. Every interval [a, b] with a and b satisfying condition (i) of this theorem is called an admissible interval (for JP>) . Certainly, Theorem 1 means that the resolution of the identity with respect to a selfadjoint operator is an abstract resolution of the identity. (Soon we will see whether the converse is true. ) For each A, J.-l E JR, A < J.-l , we put Pt, : == P11 - PA . By Proposition 4.9(iv) , this is a projection. Using the same proposition, it is easy to verify that for AI < A 2 < A 3 < A4 we have
(1) (2)
5. Spectral theorem as an integral
377
Suppose JP> is an ( abstract ) resolution of the identity and [a, b] is an admissible interval for this resolution. A function f : [a, b] � C is called a step-function if for some partition A : a == Ao , AI , . . . , An == b this function takes a constant value on every half-open interval (A k , A k + I] ; k == 0, . . . , n - 1. Denote the set of all step-functions on [a, b] by St [a, b] ; clearly, this is a normed ( but not complete ) algebra with respect to pointwise operations and the uniform norm. Note the evident fact that every continuous function on [a, b] is a limit of a uniformly converging sequence of step-functions. Take a step-function h on [a, b] and choose a partition A satis fying the above condition. Denote by I (h) the operator 2::: �= 1 h(A. k )P;: - 1 • Clearly, this definition does not depend on the choice of A, and I : St[a, b] � B(H) is a linear operator. Propo sition
4.
II I( h ) ll
. More over, ( i ) II I: f(A) cliP (A) II < ll f lloo ; ( ii ) J: f(A)diP(A) commutes with all P>.., and hence with all Pt_ ; ).. < J-l ; ( iii ) I: f(A)diP(A) does not depend on the choice of the admissible in terval [a, b] ; ( iv ) the mapping r : C [a, b] � B(H) : f r--+ I: f(A)diP(A) is a continuous calculus on [a, b] for the operator S : == I: AdiP(A) .
Theorem 2.
Proof. By virtue of Proposition 4 and equalities
(1) and (2) , the mapping
B(H) : h r--+ I (h) is a contraction operator and at the same time a unital homomorphism of algebras. Denote by A the closure of St[a, b] in the space of all bounded functions l oo ( [a, b] ) . By Theorem 2. 1.2, I can be extended to a contraction operator from A to B(H) . Denote this exten sion by the same symbol I. Note that A is a subalgebra in l oo ( [a, b] ) , and
I : St[a, b]
�
378
6. Hilbert Adjoint Operators and the Spectral Theorem
the constructed extension I is again a homomorphism of algebras. Indeed, take / , g E A. If hn and h� are sequences of step-functions uniformly tend ing to f and g, then, certainly, the sequence hn h� uniformly tends to fg . Hence, f g E A, and this means that A is a subalgebra. FUrthermore, the sequence I(hn h� ) == I(hn ) I(h� ) tends to I ( fg ) and at the same time (since the multiplication in B( H) is continuous) to I ( f ) I ( g ) as well. Therefore, I : A � B(H) is a unital homomorphism, and its restriction to C[a, b] is a continuous functional calculus of the operator T : == I(t) (see Definition 4. 1 ) . Now we show that the Riemann-Stiltjes integral of every f E C[a, b] exists and coincides with I(f) . Let us choose c > 0 and take 8 > 0 such that I f ( t' ) - f ( t") I < c whenever I t' - t" I < 8. Let A : a == Ao, . . . , An == b be a partition of the interval [a, b] such that d(A) < 8. Take the step-function h with value j(A k ) on each half-closed interval (A k - I , A k ], k == 1 , . . . , n . Then clearly I(h) == �A(f, JP>). On the other hand, ll h - f lloo < c . Hence,
II �A(f, JP>) - I( J ) II == II I (h - / ) I < ll h - Jlloo < c. Since the partition A, d(A) < 8, was arbitrary, this means that the integral J: j ( A)diP(A) exists and is equal to I (f) . Taking into account what we have
said before, this immediately implies (iv) and (i) . Furthermore, it is clear that for each A E 1R and for every partition A the operators �A (f, JP>) and PA commute. This evidently implies (ii) . To establish (iii) , it remains to consider the case where one admissible interval, say, [c , d] , is contained in another, [a, b]. Consider a partition A : a == Ao, A 1 , . . . , A n == b of the larger interval such that the points c and d are among the A k · Then the points of the partition A, starting from c and ending on d, provide a partition, say, A' , of the interval [ c, d] . Comparing the corresponding integral sums, we see that � A differs from �A' by terms of the form j(A k )(O - 0) (at the beginning) and j(A k )(l - 1) (at the end) . Hence, both sums give the same operator. It remains to take a sequence of partitions Am with this property, such that their diameters tend to zero, • and pass to the limit as m � oo .
In what follows, we say that the operator J: j(A)diP(A) is associated with the resolution of the identity JP>. To emphasize that the operator does not depend on the choice of admissible interval, we often use the notation fiR j ( A)diP(A). Here is a useful
5.
Proposition f(t) == 0 if A
.. k ) = 0. Hence all
the terms in � A ( / , JP>)P; vanish. It remains to take a sequence of partitions Am with the indicated property, such that their diameters tend to zero, and • pass to the limit as m � oo . Remark. Perhaps you have noticed that property (iii) of a resolution of the
identity has not been used so far. In particular, we did not need it to prove the existence of the integral in Theorem 2. But now it comes into action.
An (abstract) resolution of the identity JP> is the resolution of the identity with respect to the operator S :== fiR A.diP(A.) (associated to JP>) .
Proposition 6.
Proof. Let H>.. be a family of subspaces associated to the operator S, and
Q>.. the resolution of the identity with respect to S. Our goal is to verify that P>.. == Q>.. for all A. E JR. Lemma.
For every A. , 11 E JR; A. < 11 we have P>.. < Q>.. < P11 •
Proof. Again we can assume that the integral is taken over an admissible
interval [a , b] containing A. and 11 · Suppose f E C(JR) is such that f(t) == 0 for t < A. . Then Proposition 5 considered for a half-closed interval ( a , A.] gives f(S)Pf == 0. Thus, due to the equality P>.. == Pf, we have Im(P>.. ) C Ker(f(S) ) . From this, by the definitions of the spaces H>.. and operators Q>.. , we have Im(P>.. ) C H>.. == Im(Q>.. ) . In view of Proposition 4.9, this gives the first inequality. Now take f E C(JR) such that 0 < f < 1, f(t) == 0 for t < A. and f(t) == 1 for t > 11 · Then certainly H>.. C Ker(f(S) ) . Furthermore, Proposi tion 5 applied to the function 1 - f and the half-closed interval ( 11, b] gives (1 - P11) ( 1 - f(S)) == 0. Hence, f(S)x == 0 implies (1 - P11)x == 0. Thus, Ker(f(S) ) C Ker(1 - P11) == Im(P11) , and Proposition 4.9 gives the second inequality. II
End of the proof of Proposition 6. By the lemma, for A. and 11 oc
curring there and for every x E H we have ( Q>.. x , x) < ( P11x, x) . Now it is time to recall that the family JP>, being a resolution of the identity, is strong-operator continuous from the right. Hence, in particular, ( P>.. x , x) == lim11� >.. + o( P11x, x ) , and thus ( P>.. x , x) > ( Q>.. x , x) for all x. In other words, P>.. > Q>.. . Since the reverse inequality also follows from the lemma, it re mains to apply Proposition 4.7(iii) . II To consider the spectral theorem, we need one last preparation. Again, let T be a selfadjoint operator, H>.. the associated family of subspaces, and JP> the resolution of the identity (now not abstract) associated to T.
380
6. Hilbert Adjoint Operators and the Spectral Theorem
7. Suppose the functions f, 9 E C(JR) coincide on the half closed interval ( A , J1] . Then f (T) P; == 9(T)P; .
Proposition
== 0. Let us represent f in the form !I - /2 ; /I , /2 E C(JR) , where !I (t) == 0 for t < J1 , and !2 (t) == 0 for t > A. Then H11 C Ker(JI (T) ) , hence !I (T)P11 == 0. Further, for every h E C(JR) such that h( t ) == 0 t < A, we have h/2 == 0. Consequently, h(T)f 2 (T) == 0 and Im(f 2 (T)) C Ker(h(T) ) . Since h is arbitrary, this means that Im(J2 (T) ) C HA . Therefore, Im(l-PA ) == Hf C Ker(J2 (T) ) , and /2 (T) (l - PA ) == 0. Hence, f(T)P; == (JI (T) - f2 (T) ) (P11 ( l - PA ) ) == 0 Proof. Clearly, it is sufficient to consider the case 9 as
(taking into account that the projections commute) . The rest is clear.
Suppose !I , !2 E C(JR) take real values, and !I half-closed interval (A, 11] . Then !I (T)P; < !2 (T) P; .
Proposition 8.
.. for S. By the spectral theorem, at least one of the projections P; , where A < J.l, is not compact. Since SP; > AP; (this follows from Proposition 8) , the birestriction of S to Im(P; ) is an invertible operator. Multiplying S by the operator that is "inverse to S on lm ( P; )" and vanishes on Im( P; ) j_ , we see that P; E I. Since dim lm ( P; ) == oo, there is an isometric operator U and a coisometric operator V such that V P; u == 1 . Hence, 1 E I and I == B(H) .
6 . Borel calculus and the spectral theorem as an op erator-valued Leb esgue integral
The ultimate goal of this section is to suggest another, though similar to the one we already have, analytic formulation of the spectral theorem, this time as an operator version of the Lebesgue (not Riemann-Stiltjes) integral. The corresponding integration will be done with respect to the so-called spectral measure, which is geometrically less intuitive than the resolution of the identity from Theorem 5.3. Moreover, the proof itself does not look so natural as the arguments that led us to Theorem 5.3. Instead, it is more like a couple of tricks where a magician takes first a complex measure, and then a conjugate-bilinear functional (rather than a notorious rabbit) from his hat. Of course, this is the author's subjective impression, and the reader may disagree with it. As a compensation, the obtained result turns out to be more power ful than Theorem 5 .3. Together with an integral representation of a given operator we obtain a similar representation for an entire class of functions
383
6. Spectral theorem as a Lebesgue integral
of an operator argument, which is much wider that the class of continuous functions. Let us choose an interval [a, b] of the real line. Recall that a complex valued function on this interval (we do not need more general domains) is called a Borel function if the inverse image of every open set in C is a Borel set. We denote the set of bounded Borel functions on [a, b] by B[a, b] . Note that the mathematicians like these functions because they are integrable with respect to an arbitrary (we mean usual, i.e. , non-negative) Borel measure, and thus, with respect to an arbitrary complex measure. Clearly, B [a, b] taken with pointwise algebraic operations, passage to the complex conjugate function as an involution, and the uniform norm is a Banach involutive algebra and, moreover, a C*-algebra. It is clear also that B[a, b] contains C[a, b] as a norm-closed *-subalgebra. Remark. By the first Gelfand-Naimark theorem, B [a, b] can be viewed as the algebra C(O) on some ( rather complicated ) compact space 0. But we do not need this. ( The way to the spectral theorem using this type of arguments is presented , e.g. , in [50] . )
What is really important is that the algebra B [a, b] has one more struc ture, this time the structure of a polynormed space. Namely, for each com plex measure 11 on [a, b] , consider the prenorm 11 · 11 11 on B [a, b] defined by the equality II f ll 11 :== I I: f( t ) d11 ( t) I · The family of prenorms { 11 · 11 11 11 E M [a, b] } (respectively, the topology generated by this family) is called the weak measure family (respectively, topology) and is denoted by wm. Proposition 1. The subspace C[a, b] is dense in the polynormed space (B [a, b] , wm) {as opposed to the normed space (B [a, b] , II · ll oo )). :
Proof. Consider the mapping
i : (B[a, b] , wm)
�
(M [a, b] * , w*) assigning to every f the functional 'P! 11 r--+ I: f(t)dl1(t) . Clearly, this is an injective op erator. It allows us to identify B [a, b] and C[a, b] with subspaces in M [a, b] *. By the Riesz theorem, the latter space can be identified with the second adjoint space C[a, b] ** . Under this identification the restriction of the map ping i to C[a, b] turns into the canonical embedding of C[a, b] into C[a, b] ** . Thus, we obtain a sequence of embeddings C[a, b] C B [a, b] C C[a, b] ** . As a special case of Proposition 4.2. 14, C[a, b] is dense in (C[a, b] ** , w* ) . Thus, C[a, b] is dense in B [a, b] with respect to the topology inherited from ( C[a, b] ** , w* ) == (M [a, b] * , w*) . It remains to note that this inherited topol ogy on B[a, b] coincides with the weak-measure topology. • :
Now let us choose a selfadjoint operator T on the Hilbert space H. Recall that in addition to the normed topology in B(H) , there are several other topologies defined by various systems of prenorms. Here we shall need the weak-operator topology "wo" (see Section 4. 1) .
384
6. Hilbert Adjoint Operators and the Spectral Theorem
In the following definition and theorem we denote (as we did in Section 4) the spectrum of an operator T by a , and the restriction of the independent variable by t . Definition
1. A Borel functional calculus or just Borel calculus of T on
[a, b] is a unital homomorphism rb : B [a, b] � B(H) that is continuous as an operator between polynormed spaces (B [a, b] , wm ) and (B(H) , w o ) and satisfies the condition rb ( t ) T. =
For further references, we distinguish the two tricks mentioned at the beginning of this section. Recall that for a C [a, b] there exists a continuous calculus of T on [a, b] , and thus, for each f E C[a, b] we have the function of operator argument f ( T) .
Suppose [a, b] contains Then for every x, y exists a unique complex measure f.1x , y such that a.
Proposition 2.
E
H there
b
l f(t ) d�-tx,y (t) = ( f (T) x , y)
(1)
for each f E C[a, b] . Moreover,
IIMx , y II :==
v
a r ( J1x , y )
- part in (ii) . Now our goal is to construct the required calculus under the assumption that C [a, b]. We show that this is the mapping rb B[a, b] -t B ( H ); f r--+ f(T), where f(T) is the operator defined by equality (2) . First of all, let us note that the last assertion in Proposition 3 means that the mapping rb extends the continuous calculus of T to [a, b] (cf. (iii) ) . In particular, it takes the identity in B[a, b] to the identity in B ( H), and t to T. FUrthermore, since the left-hand side of (2) linearly depends on the op erator f(T) applied to x , and the right-hand side linearly depends on J, it a
:
6. Hilbert Adjoint Operators and the Spectral Theorem
386
rb
follows that f(T) linearly depends on f. In other words, is a linear oper ator. From the estimate for l i St II (or, what is the same, ll f(T) II ) established in Proposition 3, it follows that is a contraction operator between the corresponding normed spaces ( cf. ( v) ) . Furthermore, take a prenorm II · ll x , y in the weak-operator system of prenorms in B(H), and consider the prenorm I · II J.Lx , y in the weak-measure system of prenorms in B[a, b]. For each f E B[a, b] we have
rb
l f(T) I x , y = l (f(T)x, y) l =
rb
1b f(t)dp,x,y (t)
=
11/ I Jlx, y ·
By Theorem 4. 1 . 1 , is a continuous operator between the corresponding polynormed spaces. We now prove that is a homomorphism. This requires more work. Take / , g E B[a, b] and x, y E H, and for every Borel set Y C [a, b] put vg , x , y (Y) :== Jy g(t)df.1x , y (t). Clearly, vg , x , y is a complex measure on our interval; similarly, the complex measure VJ, x , y appears. The functions f and g are uniformly approximated by linear combina tions of characteristic functions of Borel sets, and moreover fg == gf; hence
rb
1b f(t)dv9 ,x,y (t) = 1 b f(t)g(t)dP,x,y (t) = 1b g(t)dvf,x,y (t).
fg for f in equality (2) , we have (3) 1 b f(t)dv9,x,y ( t) = ( (fg)(T)x , y) = 1b g(t)dvf,x,y (t). On the other hand, using the relation ( f(T)g(T)x , y ) == (g(T)x , f(T) * y ) and
Substituting
substituting appropriate functions and vectors in equality (2) , we see that (4)
1b f(t)dP,g(T) x,y (t) = ( f(T)g(T)x, y) = 1b g(t)d!Lx,J (T) • y (t).
As we know, a continuous calculus is a homomorphism. Hence, for J, g E C[a, b] , the middle terms of equalities (3) and ( 4) coincide. Therefore, taking a look at the left-hand sides of these equalities, we see that the measures vg ,x , y and /1g ( T)x , y define the same functional on C[a, b]. Consequently, these measures coincide. This, in turn, implies that the integrals in the left-hand sides of our equalities coincide for each f E B[a, b]. If we look at the right-hand sides of these equalities, we see that the integrals there coincide for all f E B[a, b] and g E C[a, b]. Thus, the measures VJ, x , y and f.1x , f ( T * y define the same ) functional on C[a, b] , this time for each f E B[a, b]. Thus, they coincide. Therefore, the integrals in the right-hand sides of (3) and ( 4) are now equal for all / , g E B[a, b]. Consequently, for these same f and g the inner
6. Spectral theorem as
a
387
Lebesgue integral
products in the middle terms of these equalities coincide. Since the vectors x and y were arbitrary, (fg)(T) == f( T )g( T ) , so rb is a homomorphism. Taking into account the established properties of !b , we see that this is a Borel calculus that at the same time satisfies conditions (iii) and (v) . Furthermore, let us substitute a complex conjugate function to f for f in (2) . Then (5)
(f(T)x, y) =
b
b
1 f(t)df-lx ,y (t) = 1 f(t) dJ-lx,y (t),
where 11x , y denotes the measure that as a set function is complex conjugate to 11x , y · At the same time, transposing x and y in (2) , we see that (6)
(f(T) * x, y ) = (x, f(T)y ) =
b
(f(T)y , x) = 1 f(t)df-ly ,x (t).
But we already know that the continuous calculus is involutive. Hence, for f E C[a, b], the left-hand sides of (5) and (6) coincide. Passing to the right hand sides, we see that the measures f.1x , y and /1y , x give the same functional on C[a, b] "by the Riesz recipe" . Consequently, f.1x ,y == /1y , x · But then the right-hand sides of the two equalities coincide for all f E B[a, b]. Thus, for the same f, the left-hand sides also coincide. Since x, y E H are arbitrary, from the form of the involution in B[a, b] it follows that f * ( T ) == f( T ) == f( T) * . This establishes (iv) . It remains to verify the assertion about the kernel of rb · Let us choose an arbitrary interval [c, d] inside U :== [a, b] \ a, and arbitrary x, y E H. For every n E N consider the function on 1R which is equal to 1 on [c, d], vanishes outside ( c - �, d + �), and is linear on [c - �, c] and [d, d + �]. Suppose hn is the restriction of this function to [a, b]. Clearly, hn E C[a, b], and hnla == 0 for sufficiently large n . By Theorem 4. 1 (iii) and equality (1) , for the same n we have I! hn (t)df.1x , y == 0. But, as is easy to see, the sequence hn pointwise tends to X[c,d] (here and below, x is the characteristic function) . By Lebesgue ' s convergence theorem,
1-lx , y [c, d] = 1
b
X[c ,d] (t)df-lx , y = J!__,�
b
1 hn (t)df-lx,y = 0.
Thus, the complex measure f.1x , y vanishes on every interval inside U. Hence, being countably additive, it vanishes on U itself, and I! xu ( t )df.1x , y == f.1x ,y (U) == 0 for every x, y E H. Together with equality (2) , this shows that xu (T) == 0. Hence, if f E B[a, b] is such that ! I a == 0, then, since we have already shown that rb is a homomorphism, we have f ( T ) == (!xu ) (T) == f(T)xu (T) == 0. The rest is clear. •
Remark. The proof of the property of rb to be a homomorphism could have been much shorter if we knew that the multiplication in the algebras (C[a, b] , wm and (B(H) , w o is
)
)
388
6. Hilbert Adjoint Operators and the Spectral Theorem
jointly continuous ( cf. the beginning of the proof of Theorem 5.2) . But , unfortunately, this is not the case (check this!)
Borel calculi on different intervals are compatible:
4. Suppose C [ d] C [a, b], and T : B [a, b] B [c, d] is the restriction map f r--+ f l [c, d] . Then for every f E B[a, b] we have f( T) == ( T f)( T) . Proof. Take an arbitrary 11 E M[c, d], and for each Borel set Y in [a, b] put v ( Y) :== 11 (Y [c, d]). Then, clearly, v E M[a, b], and for each f E B[a, b] ) --+ we have l i T f ll 11 == ll f ll v · This means that the operator T : (B[a, b], (B[c, d], ) is continuous. Now let r' be a Borel calculus of T on [c, d]. Then the operator r :== ) --+ (B(H) , o) is continuous, as a composition of two r ' T : ( B[a, b], continuous operators. Clearly, it also has other properties of a Borel calculus of T on [a, b]. By the uniqueness of the latter, we have f( T ) :== r (f) • !1 T(f) ==: ( T f) (T) . From now on, for an arbitrary Borel function f defined on an arbitrary subset containing the interval [a, b] on the real line, we can denote the operator f l [a , b] (T) just by f(T). We see that this operator does not depend on the choice of the interval [a, b]. a
Proposition
c,
--+
n
wm
wm
wm
w
�
a
Remark. Speaking about the Borel calculus of an operator, we had re
stricted ourselves to functions defined on intervals. But instead of intervals we could take an arbitrary compact set � containing a , in particular, a itself. The reader can easily give the corresponding analogue of Definition 1 and verify the corresponding version of Theorem 1 ; as is easy to guess, the operator f(T) depends only on the restriction of the function f to a . However, for our goal intervals will suffice. If we consider convergent sequences in ( B[a, b], wm ) , then it is possible to say more about the behavior of the corresponding "functions of T" than what immediately follows from Definition 1. Exercise 1 * .
(i) (ii)
fn tends to f with respect to the weak-measure topology in B[a, b] {=:::} fn converges pointwise and is uniformly bounded; in this case the sequence of operators fn (T) converges not only in
the weak- but also in the strong-operator topology of the space B(H) .
6. Spectral theorem as a Lebesgue integral
389
Hint.
(i) To prove part � , you must consider the corresponding se quence of functionals on M[a, b] and apply the Banach-Steinhaus theorem. Part ¢::=:= follows from Lebesgue ' s theorem. (ii) Here you must verify that ll f(T)x - fn (T)x ll 2 == J: I f - fnl 2 df.1x , x (t) and again apply Lebesgue ' s theorem. Remark. The theorem concerning Borel calculus admits a substantial gen
eralization from selfadjoint to normal operators. One can take Borel func tions of normal operators, this time defined on compact subsets in C; these subsets must contain the spectrum of a normal operator, and therefore, they do not necessarily belong to JR. For details, see, e.g. , [33]. Let us give some instructive examples and look at the typical operators considered in the previous sections. Later we will be able to easily and quickly find Borel functions of these operators using the strong tools pro vided by the future version of the spectral theorem ( cf. Exercise 6 below) . But it is useful to do this right now, even if we have to follow possibly a slower way. We especially distinguish the following example, which will b e important later. Example 1 (cf. Example 4.2) . Suppose T is the operator of multiplication by t in H :== Then for every E the operator
L2 ( [a, b], f.1).
f B[a, b] f(T) coincides with the operator Tt of multiplication by f. To see this, it is sufficient to show that the mapping f �----+ f(T) has the properties of a Borel calculus. All we have to verify is the continuity with respect to proper systems of prenorms. Clearly, for every x, y E H we have (f(T)x, y ) = 1b f(t)x(t)y(t)dp, (t) = 1 b f(t)dv (t), where the complex measure v is defined for Borel sets Y by the equality v ( Y) :== Jy x(t)y(t)df.1. This gives ll f ll v == l f(T) II 11 and proves the required continuity.
f(T) in Example 4. 1 and Ex ercises 4.3-4.5, does not change if we replace a continuous function f by a Exercise 2 . The rule defining operators
Borel function.
Remark. From those examples one can see in particular that the Borel
calculus differs from the continuous one in the following aspect: we do not assert here that its kernel consists precisely of the functions vanishing on a . Take, say, the operator T of multiplication by the independent variable in L2 [a, b]. As we know, for this T, a == [a, b] (Exercise 5. 1 .3) . At the same time, it is easy to verify that f(T) == 0 for every f E B[a, b] vanishing almost everywhere on [a, b] with respect to the standard Lebesgue measure.
390
6. Hilbert Adjoint Operators and the Spectral Theorem
We recall Proposition 4.3. It admits the following generalization.
The assertion of Proposition 4.3 remains true if we replace C[a, b] with B[a, b] {i.e., the continuous functions with the Borel ones). Proof. The mapping r£ : B[a, b] � B(K) : f r--+ f (T) I 1 has the proper
Proposition 5.
I
-
ties of a Borel calculus. (Continuity with respect to the appropriate systems II of prenorms follows from Proposition 4. 1. 7.) The rest is clear. *
*
*
We now look at the new possibilities opened after we enlarged the class of functions of T. A very important fact is that now we have a sufficiently wide family of projections; it is much larger than the resolution of the identity from Section 5. In the framework of the theory of operator-valued Lebesgue integral to be introduced in the sequel, this family plays approximately the same role as the family of characteristic functions of measurable subsets does in the classical theory of Lebesgue integral. Let us start with the axiomatization of the properties of this family of projections, which allow us to work with it. Recall that by BOR we denote the class of all Borel sets on the real line, and by P(H) the class of all (orthogonal) projections in H.
P. : BOR � P(H) is called an abstract spectral measure, or just a spectral measure, if (i) P.(0 ) == 0 and P.([a, b]) == 1 for some interval [a, b]; (ii) if Y == Y1 U Y2 , then P.(Y ) == P. ( Y1 ) + P.(Y2 ) (finite additivity) ; (iii) if Y == Y1 Y2 , then P.(Y ) == P. ( Y1 )P.(Y2 ); (iv) if Y == IJ� 1 Yn , then for each x, y E H we have ( P. ( Y)x, y ) == E� 1 ( P. ( Yn ) x, y ) (weak-operator countable additivity) . (Certainly, (iv) means that for arbitrary x, y E H the set function Y r--+ ( P. ( Y)x, y ) is a complex measure.)
Definition 2. A mapping
n
Exercise 3. Prove that properties (ii) and (iii) , as well as the for mula == 0, follow from the countable additivity of the measures r--+ y ) indicated in (iv) .
Y P. ( 0 ) ( P. ( Y )x, Hint. Let us explain why (iii) follows from (iv) . If for orthogonal pro jections Q 1 and Q 2 we have II Q 1 + Q 2 l == 1 , then Q 1 Q 2 == Q 2 Q 1 == 0. Hence ( P.(Y ) + P.(Y1 \ Y ))(P.(Y) + P.(Y2 \ Y)) == P.(Y) 2 . Proposition 6 (strong-operator countable additivity of spectral measure) . L� == EH == IJ�
If Y
1 Yn , then for each x
we have
1 P. ( Yn )x P.(Y)x.
391
6. Spectral theorem as a Lebesgue integral
Proof. Elementary calculations using the properties of a spectral measure
sho'v that
n
n
2
P. (Y)x - L P. (Yk )x = ( P. (Y) x, x) - L ( P. (Yk )x, x) . k= l k =l Hence the desired strong-operator countable additivity follows from the weak-operator one. • We now recall our initial operator T. Take an arbitrary interval [a, b] containing a, consider the Borel calculus of T on this interval, and for each Y E BOR put Pr (Y) : == x� (T) , where X� is the characteristic function of the set Y n [a, b] . By Proposition 4, Pr (Y) does not depend on the choice of the interval [a, b] . Proposition
measure.
7.
The mapping Pr : BOR � P(H) : Y � Pr (Y) is a spectral
Proof. Only property (iv) requires our attention; the remaining properties
immediately follow from algebraic properties of the Borel calculus. Suppose Y == ll � 1 Yn . Take [a, b] ::J a. Replacing the given sets by their intersections with [a, b] if needed, we can assume without loss of generality that they all belong to this interval. Take x, y E H and consider the ( countably additive! ) complex measure 11x ,y constructed in Proposition 2. By the definition of Borel functions of T (see equality (2)) we have ( P. (Y)x, y ) == f.1x , y (Y) and the same equality • with Y replaced by Yn ; n == 1 , 2, . . . . The rest is clear.
3. The constructed mapping Pr : BOR � P(H) is called the spectral measure associated with the operator T or just the spectral measure of the operator T.
Definition
The reader may guess that the spectral measure of an operator T is compatible with the resolution of the identity JP> A � PA of this operator discussed in the previous section. This is true: the first operator-valued function is, in fact , an extension of the second. :
Proposition 8.
For every A E 1R we have PA == Pr(-oo , A] .
Proof. Suppose [a, b]
::J
a(T) ; then Pr (-oo, A]
== Pr [a, A] ==
1 - Pr (A, b] .
Hence our goal is to show that HA == Im(PA) coincides with Ker(Pr (A, b] ) . In other words we have to verify that for every x E H the following equivalence holds: f ( T)x == 0 for all f E C[a, b] such that f(t) == 0 as t < A {:=:::> X ( A ,b] ( T) x == 0.
392
6. Hilbert Adjoint Operators and the Spectral Theorem
For the indicated f we have f == fX ( A , b] · Taking into account that the Borel calculus is a homomorphism, we have f(T)x == f(T) X ( A , b] (T)x == 0. � - For each n == 1 , 2, . . . consider the function fn E C[a, b] that van ishes for t < A, equals 1 for t > A + � , and is linear on [A, A + �] . From the classical Lebesgue theorem it evidently follows that fn tends to X ( A , b] with respect to the weak-measure topology in B [a, b]. But then, by the defini tion of Borel calculus, the sequence ( fn ( T ) x , x ) tends to ( X ( A , b] (T)x, x ) (Pr(A, b]x, x ) == II Pr (A, b]x ll 2 . The rest is clear. • {::::::= .
The reader who has done Exercise 2, will easily find the spectral measures of the following "classical" operators. Exercise 4.
(i) If T is a projection, then Pr (Y ) == 0 for 0, 1 ¢: Y; Pr (Y) == T for 0 ¢: Y, 1 E Y; Pr (Y) == 1 T for 0 E Y, 1 ¢: Y; and finally Pr (Y) == 1 for 0, 1 E Y. (ii) The spectral measure of a selfadjoint diagonal operator T :== TA in l 2 assigns to every Y the diagonal operator To; 0 == ( OI , 02 , . . . ) , where O n == 1 for A n E Y and On == 0 for An t/: Y. (iii) The spectral measure of the selfadjoint multiplication operator T :== Tcp in L2 (X, 11 ) assigns to every Y the operator of multiplication Tx, where x is the characteristic function of the set 'P-I (Y) . (iv) The spectral measure of a selfadjoint compact operator from Ex ercise 4.5 assigns to every Y the orthogonal projection onto HI :== (span{en : An E Y } ) - if 0 ¢: Y, and the orthogonal projection onto HI EB {e i , e 2 , . . . } ..l if 0 E Y. -
We especially distinguish the following Example 2. The spectral measure of the operator of multiplication by
the independent variable in L 2 ( [a, b], 11 ) assigns to every Y the operator of multiplication by XY. This immediately follows from Example 1. How does the introduced characteristic behave if we replace the given op erator by a unitarily equivalent one? From the definition of spectral measure in terms of Borel calculus and from Proposition 5 we immediately obtain
Let T : H H and S : K --+ K be selfadjoint operators with spectral measures Pr and Ps respectively, and let U : H --+ K establish topological (or, as a special case, unitary) equivalence of these operators. Then for every Borel set Y C 1R the same U establishes the corresponding equivalence between operators Pr (Y) and Ps (Y) . In other words, if for a
Corollary
1.
--+
6. Spectral theorem as a Lebesgue integral
unitary operator U the diagram H u
1
K
T
-�
s
commutes, then for every Y the diagram H
1K
U
also commutes.
Pr ( Y )
Ps ( Y )
393
H
1
u
K
H
1K
U
Now we turn to the description of the procedure that, as we will soon see, recovers an operator from its spectral measure. Let P. : BOR � P(H) be a spectral measure ( arbitrary, for time being ) , and [a, b] an interval in condition ( i ) of the definition of a spectral measure. As usual, a Borel function on [a, b] is called a simple function if it takes only a finite number of values, or, in other words, it is a linear combination of characteristic functions of Borel sets. The set of simple functions will be denoted by S [a, b] . Clearly, it is a *-subalgebra in B[a, b] dense with respect to the uniform norm and also with respect to the weak-measure topology. For every f E S[a, b] ; f == E�= l XYk we put Io(f) :== E �= l P. ( Yk ) · Then for each x, y E H the number ( Io(f)x, y ) is equal to the integral of f with respect to the complex measure Y �----+ ( P.(Y)x, y ) . Therefore, it does not depend on the representation of f in this form. Consequently, the same is true for the operator Io ( f ) . Thus, we have a well-defined mapping Io : S[a, b] � B(H) : f �----+ Io(f), which is certainly a linear operator. More over, from the properties of the spectral measure it evidently follows that Io is an involutive homomorphism of *-algebras, and in addition it is unital because P.([a, b] ) == 1. Finally, if we represent f as a linear combination of characteristic functions of disjoint sets, we see that for every x E H the vectors P. ( Yk )x; k == 1, . . . , n are mutually orthogonal. Hence,
A k i 2 II P. ( Yk )x ll 2 < max 1 A k l 2 l x ll 2 = ll f ll� ll x ll 2 · 1 ll lo(f)x ll 2 = L k= l This means that ll lo( / ) 11 < ll f lloo ; in other words, Io is a contraction oper n
ator. Suppose I: B[a, b] � B(H) is the contraction operator extending by continuity the operator Io ( see Theorem 2. 1.2) .
394
6. Hilbert Adjoint Operators and the Spectral Theorem
f E B[a, b] the operator I(f) is called the {Lebesgue} integral of the function f with respect to the spectral measure P. and it is (also) denoted by I: f(t)dP.(t). Thus, I: f(t)dP.(t) is the limit of a sequence of "Lebesgue integral sums" n (m) m L ,\� ) P. ( Y� m) )
Definition 4. For
k= l
for every sequence of simple functions tending to f.
fm
=
:L� (�) ,\�m) Xy:k 11 1 (Y) == 0 ¢=:=:> 11 2 (Y) == 0 ¢=:=:> (A2 EB v) (Y) == 0 ¢=:=:> A 2 (Y) == 0 ¢=:=:> A 2 (X) == 0. • Proposition 3. Suppose the measures J.ln ; n == 1 , 2, . . . are mutually orthogonal. Then there exists a partttton [a, b] == il� 1 Zn such that Zn == supp(J.ln ) · Proof. Construct by induction a certain sequence of sets Xn . By assumption, for each m =/= n there exists Z� == supp(J.lrn ) such that C Z� == supp(J.ln ) . Put X1 : == n� 2 Zf . Clearly, X1 == supp(J.l 1 ) and CX 1 == supp (J.lk ) for all k E N; k =/= 1 . If X 1 , . . . , Xn are already constructed in such a way that for all k == 1 , . . . , n we have Xk == supp (J.lk ) and CXk == supp(J.ll ) for l E N; l =/= k, then we put Xn+ 1 : == CX 1 n · · · n CXn n n� n+ 2 Z� + 1 · Evidently, the previous assertion about Xk and CXk will be true for all k == 1 , . . . , n + 1 . Clearly, the constructed supports Xn ; n == 1 , 2 , . . . are mutually disjoint , and for XCX) : == C(il� 1 Xn ) we have J.ln (ZCX) ) == 0 for all n. It remains to put Z1 : == X1 u XCX) and • Zn : == Xn for n > 1 . Proposition 4. Suppose each of the sequences J.ln and Vn conststs of mutually orthogonal measures. Suppose that for some X and n we have J.ln (X) == 0 but Vn (X) =/= 0 . Then there eX'/,st Y C X and m =/= n such that J.li (Y) == 0 for all i =/= m, vn (Y) > 0, and vt (Y) == 0 for
i =/= n.
Proof. Using Proposition 3, take a partition of the interval into the supports Zk == supp (vk ) ; k == 1 , 2, . . . and put xk : == X n zk . Clearly, Vk (Xn ) > 0 ¢=:=:> k == n. Using again Proposition 3, take a partition of the interval into the supports Zi == supp(J.ll ) ; l == 1 , 2, . . . and put Yi : == Xn n Zi . Then Xn == il � 1 Yi , so that there exists l such that vn (Yz ) > 0. At the same time, of course, J.lrn (Yz ) == 0 for m =/= l. It remains to put y : == Yz . •
Let us add to all this several simple relations between vectors measures J.lx , and some spectral measure P. : Y �-----+ P(Y ) .
x
E H, their personal
8. Proof of the final form of the spectral theorem
407
Proposition 5. If X n Y == 0 , then P(X)x l_ P(Y)y for every x, y E Proof. ( P(X)x, P(Y)y ) Proposition 6. If Z
==
==
H. •
( x, P(X n Y)y ) .
supp ( ll x ) , then x
==
P(Z)x .
Proof. By Proposition 7. 1 applied to Y : == CZ we have P(CZ)x clear. Proposition 7. If /lx l_ /ly , then
Hx
l_
0. The rest is
•
Hy .
Proof. Take Z == supp ( llx ) such that C Z == supp ( /ly ) . Then, by Proposition 6, for every X and Y we have P(X)x == P(X n Z)x and P(Y)y == P(Y n CZ)y. Hence , by Proposition 5, P(X)x l_ P(Y)y. It remains to recall that the indicated vectors form total sets in Hx • and Hy , respectively. Proposition 8 . If y E
Hx ,
then llx
>>
/ly .
Proof. Suppose ll x ( Y ) == 0, or, in other words ( see Proposition 7. 1 ) , P(Y)x == 0. Then P(Y)z == 0 for every z of the form P(X)x; X E BORb · But y, being an element of Hx , can be approximated by linear combinations of vectors of the indicated form. Hence , • P(Y)y == 0 or, what is the same, /ly ( Y ) == 0.
and for some measures /l n ; n == 1 , 2, . . . � e have /lx rv Then there extst vectors Xn E Hx such that /lxn rv /l n and Hx == ffi n Hn , where
Proposition 9. Suppose x E
ffi n /l n · Hn : == Hxn
H
·
Proof. Take the supports Zn of measures /ln from Proposition 3 and put Xn : == P(Zn )x. By Proposition 7. 1 , for each Y and X :== Zn n Y we have llxn (Y) == llx (X) . Hence, /lxn (Y) == 0 {::::::> /lx (X) == 0 {::::::> L k /lk (X) == 0 {::::::> lln (X) == 0 {::::::> /ln (Y) == 0. This means that llxn rv /ln . This, in turn, implies that for m =f. n we have llxrn l_ llxn . Thus ( Proposition 7) , Hm l_ Hn .
Furthermore, every vector from Hx is approximated by linear combinations of vec tors of the form P(Y)x. By the strong-operator countable additivity of the spectral measure ( Proposition 6.6) , these vectors can be approximated by vectors of the form I:: �=l P(Zk )P(Y)x == I:: �=l P(Y)x k , i.e. , by vectors from the algebraic sum of spaces Hn . _ This certainly implies that Hx and ffi n Hn coincide. • *
*
*
Proof of Theorem 7.4. Now, armed with all these facts , we start proving Theorem 7.4.
We say that a vector x E
/lx
>>
/l y ·
H is
large (with respect to T) if for every y E
Lemma 1 . Large vectors exist, and, moreover, for every e E
that e E
Hx .
H
we have
H there exists a large x such
Proof. Consider the set M of all possible families of non-zero personal measures of vectors from H, orthogonal to each other and to the measure /le . It is ordered by inclusion. The standard trick with Zorn's lemma provides us with a maximal element of M; denote it by /ly 13 ; {3 E A. By Proposition 7, all Yf3 are orthogonal to each other and to the vector e. Hence, due to the separability of H, there is at most countable set of them; we denote these vectors by Yn and their personal measures by /ln . Multiplying these vectors by appropriate factors, we can assume that L n 11 Yn ll 2 < oo.
408
6. Hilbert Adjoint Operators and the Spectral Theorem
By the latter inequality, there is a well-defined vector x :== e + L n Yn · If Z == supp (J.le ) and Zn == supp (J.l n ) are the supports described in Proposition 3, then, taking into account Proposition 6. 1 and Proposition 6, we have e == P( Z)e == P( Z)e + L n P( Z)yn == P( Z ) x . Consequently, e E Hx . It remains to show that x is large. Assume the contrary: there exist z' E H and Y such that J.lx (Y) == 0, but J.l z ' (Y) > 0. Put z :== P(Y) z' . Then J.lz (Y) == J.lz' (Y) > 0 and J.lz ( CY) == 0, i .e. , J.lz l_ J.lx . But J.ln , J.le > V00 , and by the same Proposition 1 , there exists a measure J.loo such that V1 rv Voo EB J.loo · The measures J.l n ; n < oo have already been constructed. Now we have to show that this is what we need (recall the formulation) . First we establish that for all n E N we have Vn rv ( E9 � n J.lk ) EB J.loo , by induction on n. For n == 1 everything is clear from the definition of measure J.loo . Suppose the assertion is true for some n, i.e. , Vn rv J.l n EB (ffi� n +l J.lk ) EB J.loo · By the choice of J.ln , we have Vn rv J.ln EB Vn+ l · So the induction step is guaranteed by Proposition 2. Now we use Proposition 9. For every n E N it gives vectors Xnn , Xnn+ l ,(X). . . , Xnoo E Hn such that in the notation Hn k : == Hx nk , we have a decomposition Hn == ffi k = n Hnk EB Hnoo , and J.lk rv J.l x nk for all n < oo , n < k < oo. We can write the following table consisting of these subspaces: •
H11
H22 H12
H33 H2 3 H1 3
•
H3 oo H2 oo H1 oo
Here the Hilbert sum of the kth row from below is Hn : == Hxn , and so the Hilbert sum of all spaces in this table coincides with our initial H. Furthermore, the nth column of this table consists of precisely n spaces , and in particular, the last column consists of the "countable" number of spaces . (Of course, nothing prevents some of the columns from consisting of zeros. ) Now, instead of taking Hilbertn sums of spaces in the rows, let oous do this by columns. · . · Namely, for each n, put Kn :== ffi k = l Hkn · Then certainly H == ffi n = l Kn EBKoo . Recall that Hk n , being generated by a vector Xk n , is invariant with respect to T. Consequently, the corresponding birestriction of the operator T is an operator unitarily equivalent to the operator MJ.L , where J.l is the personal measure of the vector Xk n (see Corollary 7. 1 ) . As a corollary, by Proposition 7.5, this birestriction is unitarily equivalent to the operator MJ.Ln . Hence, taking into account Proposition 7.8, the birestriction of the operator T to Kn ; n == 1 , 2, . . . , oo is unitarily equivalent to the operator n x M J.Ln , and "the whole" T is unitarily equivalent to the Hilbert sum of the latter operators over all n < oo. The ordered spectral picture of the operator T is constructed. It remains to show that the spectral types of the measures participating in the ordered spectral picture of the operator T are uniquely determined. Here the following lemma, of independent interest , is helpful. Lemma 4. Suppose T == ffi n Tn , where Tn is the birestriction of T to the invariant subspace Hn in H, and suppos e that for each n, P.n : BOR ---+ P(Hn ) : Y � Pn (Y) is the spectral measure of the operator Tn . Then P. : BOR ---+ P(H ) : Y � P(Y) , where P(Y) :== ffi n Pn (Y) , is the spectral measure of the operator T .
410
6. Hilbert Adjoint Operators and the Spectral Theorem
Proof. It is easy to verify that P. has the properties of an abstract spectral measure. Consider the operator I : B [a, b] ---+ B(H) participating in the definition of the operator valued integral with respect to this "measure" (see Section 6) , and for every n, take the analogous operator In : B [a, b] ---+ B(Hn ) corresponding to the given spectral measures of the operators Tn . Suppose fm E B [a, b] is a sequence of simple functions uniformly tending to t (i .e. , to the independent variable) . Then, by the spectral theorem, In (fm ) tends to Tn in B(Hn ) for each n. But from the form of P. it is clear that I(fm ) == ffi n in (fm ) , and it is easy to see that I(fm ) tends to $ n Tn , i.e. , to T. This means that T == I: t dP (t) . • Hence, by the same theorem, P. is the spectral measure of our operator.
Let us apply Lemma 4 to an operator of the form M J-£• It is easy to represent the space where it acts as L�· :== ffi n = 1 nL� n EBoo L�oo , where nL� n ; n < oo denotes the Hilbert sum of n copies of the space L� n . Thus , the elements of the space nL�n are the rows f == (f1 , . . . , fn ) ; /k E L� n , and the inner product is defined by the equality ( /, g ) : == I: L � 1 ft (t)gi (t) dJ.ln (t) . Since we know the spectral measures of the operators L� n (see Example 6.2) , the following lemma follows from the previous one. •
•
(X)
•
Lemma 5. The spectral measure of the operator M J-£• ts a mapptng X
�
P (X) , where the latter projectton is untquely defined by the conditton that tt maps ( f1 , . . . , fn ) E nL� n • to ( x x f1 ' . . . ' x x fn ) . *
*
*
We continue the proof of Theorem 7.4. We have two operators MJ-£• and M�.�. . If J.ln rv Vn for all n < oo, then by Proposition 7 . 5 , MJJ. n is unitarily equivalent to Mvn for the same n. Taking into account Proposition 7.8, we see that n x MJ-L n is unitarily equivalent to n X M vn , and finally, MJ-£. and M �.�. are unitary equivalent . Let us prove the converse. Suppose that , on the contrary, some unitary tsomorphism U establishes a unttary equtvalence between MJ-£• and M �.�. , but for some n the measures J.ln and Vn are not equtv alent. Our goal is to deduce from this some evidently absurd corollary. We can obviously assume that J.ln (X) == 0 and vn (X) =1- 0 for some n < oo and some Borel set X . Then we are under the assumptions of Proposition 4, with the only difference that now the set of indices of our measures contains, in addition to natural numbers , also oo. Choose Y and m < oo ; m =1- n provided by this proposition. Denote the spectral measures of the operators MJ-£• and M �.�. described in Lemma 5 by P( ) and Q( ) respectively. By Corollary 6. 1 , for every Z E BORb the diagram ·
·
is commutative. In particular, it shows that for Z C Y the operator U establishes an isometric isomorphism between the subspaces Im (P(Z)) C L�· and Im(Q (Z)) C L�· . But, by Lemma 5 and the choice of Z, the first of them is a subspace in mL�rn consisting of the rows f == (!1 , . . . , fm ) , where all functions vanish J.l m -almost everywhere outside of Z. Denote this subspace by mL� (Z) . In turn, Im(Q (Z) ) is a subspace in nL�n also consisting of the rows vanishing Vn -almost everywhere outside Z (but now there are n functions in these rows) . Denote it by nL'2 ( Z) . Thus, for every Z C Y we have that mL� ( Z) == 0 ¢:::::::> nL'2 ( Z) == 0. Evidently, this is equivalent to the fact that J.lm (Z) == 0 ¢:::::::> vn (Z) == 0. In particular, since vn (Y) > 0 (s ee Proposition 4) , both these spaces nL'2 (Y) and mL� (Y) are different from zero.
8.
41 1
Proof of the final form of the spectral theorem
Recall that m =/= n. Exchanging, if necessary, the measures J.lm and Vn (and consider ing u- 1 instead of U) , we can assume that m > n. For f == ( f1 . . . , fm ) E mL� (Y) and Z C Y denote the row (xz /1 , . . . , xz fm ) by xz f , and use similar notation for the rows from nL2 (Y) . Note that the commutativity of the above diagram for the indicated Z means just that U (xz f) == xz U f . Now (taking into account that the case of m == oo is possible) we choose an arbitrary k E N such that n < k < m, and for every i == 1 , . . . , k take in mL� (Y) the row f i : == (0, . . . , 0, 1, 0, . . . ) with 1 at the ith place. These k rows evidently form an orthogonal system. Hence, putting g i == (gi , . . . , g� ) :== U ( ft ) , we obtain an orthogonal system in nL2 (Y) . ,
Lemma 6. The function Vn , for any i == 1 , . . . , k.
2:7 1 l gi (t) l 2 ; t E Y
does not vanish on a set of full measure for
Proof. If, on the contrary, for some i this function vanishes on Z C Y with vn (Z) > 0, then l l xzg t l l 2 == fz 2: 7 1 l 9z l 2 dvn == 0. But xzg t == xz U (ft ) == U (xz ft ) ; therefore • xz f t == 0 , so that J.lm (Z) == 0 . But then Vn (Z) == 0 , a contradiction. Lemma 7. For each i, j
==
a set of full measure for Vn .
1 , . . . , k; i =/= j the functton
2:7 1 gJ (t)gf (t) ; t E Y
vanishes on
Proof. For every Z C Y the rows xz ft ; i == 1 , . . . , k are orthogonal in mL� (Y) . Hence , the same is true for the rows xzgi == U (xz ft ) in nL2 (Y) . This precisely means that fz 2:7 1 gJ (t)gf (t)dvn (t) == 0 . It remains to use the fact that Z was chosen arbitrarily. •
Now we complete the proof of Theorem 7.4. Combining Lemmas 6 and 7, we see that for almost all t with respect to the measure Vn , the vectors (gi ( t) , . . . , g� ( t)) ; i == 1 , . . . , k of the arithmetic space e n are different from zero and mutually orthogonal. Taking at least one such t, we obtain a linearly independent system in the n-dimensional spaces consisting of k > n vectors. The desired contradiction is obtained. • *
*
*
To understand better the essence of the proof, do the following exercise. Exercise 1 . Under the assumption that T is compact , describe the spaces Hn , Hn k , Kn , and the measures Vn , J.ln arising in the process of constructing the spectral picture. Find the dimension of the space Hn k ·
Recall that the theorem we proved gives a classification up to an isomorphism of objects of one of the major categories in functional analysis. The objects of this category are selfadjoint operators acting in Hilbert spaces, and morphisms between the objects T : H ---+ H and S : K ---+ K are contraction operators R making the diagram
commutative. (If you have forgotten the definitions, see again Sections 0.4 and 1 .4 . )
Chapter 7
Fourier Tr ansform
1 . Classical Fourier transform
In a mathematical analysis course you have been told a bit about the rep resentation of 21r-periodic functions on the real line by Fourier series. 1 Now we address a reader who knows the foundations of both real and complex analysis. Suppose we have a 21r-periodic complex-valued function, say cp( t ) ; t E JR. Suppose it is integrable ( now we can say: in the sense of Lebesgue ) on the interval [ -1r, 1r] . For each n E Z we put Cn : == 2� f1T"1r cp (t) e -int dt and call this number the nth Fourier coefficient of the function cp ( t ) . ( Previously, you probably considered functions taking real values, and instead of exponentials e -int in this formula you had cos nt and sin nt; n == 1 , 2, . . . , and a constant. But here and in what follows, only the "version for grown-ups" will be treated. ) Thus, to every cp E £ 1 [ -1r, 1r] ( as always, speaking about a function we mean its coset ) we assign a family of complex numbers en ; n E Z depending on an integer index, that is, a two-sided sequence. These numbers are called the Fourier coefficients of the function cp ( t) . There is also a reverse process: we take a two-sided sequence en ; n E Z and assign to it a formal functional series 2:: cn eint . ( Note that such a series always converges uniformly and absolutely if 2:: � l cnl < oo. ) A traditional question is how to recover the 00
00
_ 00
1 Joseph Fourier (1 768-1830) , outstanding French mathematician. His life was full of adven tures. He took part in Napoleon's disastrous Egyptian expedition, and even was with him during the famous battle of the Pyramids. ( Gaspard Monge, another outstanding mathematician, was there as well. ) As you may remember, at the climax of the battle Napoleon gave his famous order: "Donkeys and scientists in the middle!" It seems that at those times, the authorities treated representatives of fundamental sciences with more consideration than now.
413
414
7. Fourier Transform
function 'P(t) from this series if the numbers Cn are the Fourier coefficients of this function. In this chapter, however, instead of Fourier series we consider a number of related notions that historically originated from them. Another discovery of Fourier, lying in the foundation of all this, is the so-called Fourier integral. In his classical work, Analytical theory of heat ( 1822 ) , Fourier came to it, roughly speaking, using the following arguments. We were talking about 21r-periodical functions. But is there a reasonable analogue of the Fourier coefficients and Fourier series for other functions on the line? Suppose first that we have a 21rl-periodic function 'P for some l == 1, 2, . . . . Then, passing from 'P to the 21r-periodic function cp(t) : == 'P( lt ) , we see that as an analogue of Fourier series for 'f?l , it is natural to take the 1r 1 · Tn t dt are numbers 1 · Tn t , where c : == 27r � � e e formal series 2:: � _ 00 yc f'P(t) l 7 7 1r indexed by fractions 7 ; n E Z. But what are we going to do with an arbitrary 'P(t), which, generally speaking, may have no period? Imitating ( with some oversimplification ) arguments of the past when people did not care about mathematical rigor, let us do something like a formal passage to the limit as l � oo . Then for a " ( normalized ) Fourier coefficient of the function 'P " depending on00a continuous real parameter s, we take the formal expression �(s) : == 2� f 00 'P(t) e- i st dt. This is a formal Fourier integral, which, certainly, is a genuine function of s if 'P( t) is integrable. As for the analogue of Fourier series, the factor t allows us to recognize in it something like an integral sum, and the formal "passage to the limit" leads to an 00 expression of the form f 00 �(s) e i st dt. The question of whether the latter integral represents the initial function 'P(t) or, to be more careful, how this integral is related to the initial function, will be considered later. First we study ( now with proper mathematical rigor ) the very process of passing from a given function to its Fourier integral. Definition 1. The classical Fourier transform ( on the real line ) or the Fourier integral of a function 'P(t); t E 1R is the function
ist dt; e cp(t) F(cp) ( s ) : = vb { 2 7r jiR
sER
( The factor vk is chosen for convenience; this guarantees that some future results obtain a simpler and nicer form. ) Remark. We have already mentioned heuristic arguments that had led to the Fourier integral as a "continuous analogue of the Fourier coefficients" . But let us emphasize that its role turned out to be much more important than one could expect when it was discovered. Now it is one of the most powerful tools in analysis and differential equations, and-in its sufficiently general sense-in topological algebra and representation theory.
1.
Classical Fourier transform
415
There are several versions of the Fourier transform. They are conceptu ally similar, but formally different. In this book we encounter two of them: the Hilbert Fourier transform and the Fourier transform of tempered gener alized functions. The word "classical" in Definition 1 means that this version of Fourier transform is given by an explicit formula. At the same time, as we shall see later, none of the two versions can be defined by a formula. The tools of classical analysis are not sufficient, and functional analysis must be used. Thrning back to the the theory of Fourier series, we can say that it gives us two other constructions that are similar to but simpler than the Fourier transform on the real line. We continue mentioning them from time to time to set off some properties of our basic notion. Definition 2. Let cp(t) be a function on the unit circle 1r of the complex plane, and suppose it is integrable with respect to the standard (i.e. , gener ated by the arclength) Lebesgue measure. The classical Fourier transform of this function on 1r is the function of an integer argument,
n dt ; n E Z. cp(t) F(cp)(n) : = � c f 2 7r }1r Recall that defining a function cp (t) on the circle 1r is equivalent to defin ing a 21r-periodic function cp( t') on the real line 1R by the rule cp( t' ) :== cp ( ei t' ) .
Evidently, this correspondence takes the classical Fourier transform of a function on the circle to the sequence of Fourier coefficients of the function cp( t') (up to a factor) . Definition 3. Suppose cp( n) E l 1 (Z) (to provide uniformity with the previ ous definitions one can say that cp is an integrable function on Z with respect to the counting measure • ; cf. Section 1 . 1 ) . The classical Fourier transform of this function is the function on 1r defined by the formula
n d (n) ; t E F(cp)(t) : = � cp(n) C f 21r lz en c n , where Cn : = 'lj;(n)). (i.e. , F(cp)(t) is just vk I:� •
1I'
- oo
Again, passing to a 21r-periodic function, we see that this new "Fourier transform" imitates the Fourier series in the situation where this series con verges. The only difference is that (again to provide the uniformity with the two previous definition) we write e - ins instead of e in s . Remark. There is a vague feeling that these three definitions are related, right? In fact, there exists a general notion of Fourier transform, where the role of JR, 'lr, or Z as the domains of the initial functions is played by an arbitrary group in a sufficiently large class. We will discuss this later; it will be presented as an optional material in the last section of this book.
416
7. Fourier Transform
Now we turn back to the initial Definition 1 , and give two important examples. Example 1. Let X a , b be the characteristic function of the interval [a, b] . b a i i s s As you can easily see, F(xa ' b )(s) = �( ee) . Note that this 27r�S function is everywhere continuous and vanishes at infinity ( i.e. , tends to zero as I s I � oo ) . t2
2. Suppose cp(t) :== e - 2 . Take s and consider the holomorphic function e - 2 e- �s z; E C. According to the integral Cauchy theorem, for Example
z2
.
z
every r E 1R \ {0} and N > 0 the integral over the rectangular path -
-N
... -
N
vanishes. As you can easily see, the2 integral of this function over the upper (t +ir) . (t . ) N horizontal interval is - f- N e - 2 e- �s + �r dt, and the integrals over the vertical intervals ( where r is constant ) tend to zero as N � oo . Thus, . (t . ) ) (t+ir) 2 r2 t 2 "t ( 1 1 + + + 2 r r T r � � � F(cp)(s) == v'27f IR e - e - s dt == v'27f e s IR e -2- s dt 82 82 t2 for every r. Putting r : == - s, we have F(cp) (s) = vk/ - 2 fiR. e - 2 dt = e- 2 . Thus, the Fourier transform of the Gauss function ( or, what is the same, of the first Hermite function; see Example 1 .2.8) coincides with the original function.
1
1
What can we say about properties of a function that is the Fourier transform of another function? Recall the Banach space C0 (JR) of continuous functions on the real line vanishing at infinity ( cf. Example 1 . 1 .6"') .
1. For every cp E £1 (JR) the function F ( cp) belongs to Co ( JR) , and the 1mapping F : £1 (JR) � Co (JR) : cp �----+ F ( cp) is bounded operator with norm v'21r .
P roposition
a
s we have the estimate dt cp(t) = vk ll cp ll l· l f I F(cp)(s) i < vk i 2 7r 2 7r jiR for every non-negative cp E L1 (1R) we have I F(xa , b )(O) I
Proof. For every
FUrthermore, == vk ll cp ii i · Taking into account the linearity of the integral, this means that
1.
Classical Fourier transform
417
the correspondence cp �----+ F( cp) is an operator from £ 1 (JR) to the space l 00 (JR) of all bounded functions on R The norm of this operator is vh . Further, Example 1 shows that this operator maps the characteristic functions of intervals and their linear combinations to Co (lR) . But the latter functions approximate every cp E L 1 (1R) ; hence, F(cp) is approximated in the norm of l 00 (1R) , i.e. , in the uniform norm, by functions from Co (lR) . Hence, F(cp) E Co (lR) . •
Co (lR) is called the classical Fourier transform or the classical Fourier operator.
Definition 4. The operator F : £1 (JR) --+
To specify the domain, one sometimes says "Fourier transform (or oper ator) on the real line" and uses the notation FJR . Thus, depending on the context, the term "Fourier transform" may refer to either a concrete function or an operator; however, this will not lead to confusion. Exercise
1.
(i) There is a well-defined operator F1r : £ 1 ('lr) --+ co (Z) : cp �----+ F( cp) with norm vh , the so-called classical Fourier transform on the
circle.
(ii) There is a well-defined operator Fz : l 1 (Z) --+ C('lr ) : cp �----+ F( cp) with norm vh , the so-called classical Fourier transform on the group Z. The following properties of the Fourier operator on the real line play an important role in classical questions of the theory of Fourier transform. Everywhere in the sequel, unless explicitly stated otherwise, the symbol F denotes this operator.
1. (i) (Uniqueness theorem) The operator F is injective {in other words, F(cp i ) == F(cp 2 ) implies cp 1 == cp 2 almost everywhere). (ii) The operator F is not surjective {in other words, there are functions in Co (lR) which are not Fourier transforms of integrable functions}.
Theorem
(At the same time, the image of F is dense in C0 ( 1R ) . This follows, for instance, from the fact that Im( F) contains all rapidly decreasing infinitely smooth functions-this will be proved in the next section.) The proof of this theorem uses delicate arguments of classical and real analysis, and you can find it, e.g. , in [8] or [93] . We will prove part (i) (uniqueness) in Section 3 using the tools of the theory of generalized func tions.
418
7. Fourier Transform
Remark. The two other classical Fourier operators-on 1r and on Z-are
also injective and not surjective, although they have dense image. (Check the latter assertion.) A fundamental property of the Fourier transform, which is a source of a great number of its applications, is that it "exchanges" differentiation and multiplication by the independent variable. The precise meaning of this is described in the following two propositions.
2. Let cp1 E L1 (JR) be a one-time-smooth function {i.e . , a function of the class C }, and let its derivative cp' belong to L1 (JR) . Then F(cp')(s) == isF( cp)( s) . Proof. Since cp is integrable on JR, there exist a n , bn E JR; n == 1 , 2, . . . such that limn � oo an == - oo , limn� oo bn == + oo , and limn � oo cp( an ) limn�oo cp(bn ) == 0. Hence, integration by parts gives bn i ts dt i '(t)e �F(cp' )(s) == 1m cp'(t) e - its dt == nlim cp � oo an b n n lim i cp(t) de - its == nlim � oo cp(t) e - its 1 �n - n�oo n a bn == lim is i cp(t) e - i ts dt == �i s F(cp)( s ). n� oo an Proposition
m lN..
•
Using Proposition 1 , we immediately obtain E L1 (JR) is an n-times-smooth function {i.e., a function If cp n ) and its n derivatives are also integrable, then F(cp (n))(s) == ofn the class e i sn F(cp)(s) . As a corollary, F(cp) (s) == o( l s l - n ) as l s i � oo . Moreover, if cp E £1 (JR) is an infinitely smooth function with integrable derivatives, then F ( cp ) ( s) decreases faster than every negative power of I s 1 . Proposition 3. Suppose cp(t) E L1 (JR) and at the same time tcp(t) E L1 (JR) . Then F ( cp) ( s ) is a one-time-smooth function, and F ( cp )' ( s ) == - iF ( t cp( t )) ( s). Proof. Put a(r) F(tcp(t))(r) ; r E JR. The function of two variables j3(t, r) :== tcp(t) e- irt is integrable in the strip JR x [0, s] for every s E JR.
Corollary
1.
: ==
Hence by the Fubini theorem we have
los a(r)dr = vk los ( l tcp(t)e-irt dt) dr = vk l tcp(t) (los e -irt dr) dt = � { cp (t)( e - ir s - e - ir O )dt == iF( cp)( s) - iF( cp)(O ). 27r jiR
1.
419
Classical Fourier transform
By the continuity of a ( r) (Proposition 1 ) , the indefinite integral in the be ginning of this chain of equalities is a one-time-smooth function. It remains to take the derivative. •
If tk cp(t) E £1 (JR) for all k n== 20, 1 , . . . , n {for instance, if cp is measurable, bounded and n( t ) == O( l t l - n - ) nas t � oo ) , then F(cp)(s) isn n-times-smooth and F( cp) ( ) ( s) == (- i ) F( t cp( t)) ( s) . If, in addition, t cp(t) E L 1 (1R) for all n E N {for instance, if cp is measurable, bounded, and decreases faster than every negative power oft}, then F ( cp) ( s ) is an infinitely smooth function. Corollary 2.
cp
As for the Fourier transforms on 1r and Z, we can speak only about the smoothness of the initial function, or only about the rate of decrease depending on where it is defined. In this case the smoothness of the given function implies rapid decrease of its Fourier coefficients, or vice versa. Exercise 2. Formulate and prove an analogue of Proposition 2 for the
Fourier transform on 1r and an analogue of Proposition 3 for the Fourier transform on Z. Let us again look at Corollary 2. Recall that if a function decreases more rapidly than negative powers of polynomials, this does not mean that it cannot decrease still more rapidly: for instance, exponentially, or in many other ways. What happens with Fourier transforms in such cases? It is natural to expect that in some reasonable sense they must be "smoother than infinitely smooth functions" . A partial answer to this question is given by the following exercise.
0 we have e b l t l cp(t) E L1 (1R) . Then 1/J : == F(cp), as a function on the real line in the complex plane, can be extended to a holomorphic function in the strip { z E C : - b < Im( z) < b}, namely, '1/J ( z) : = vk JJR cp ( t ) e - i zt dt. Hint. The function a (z) :== fiR tcp(t)e- izt dt is defined and continuous on this strip. Its integral over every interval [z 1 , z2 ] lying in this strip is J21fi(1/;(z2 ) - 1/; (zi)). So its integral over every triangular contour inside this strip vanishes. Hence, a ( z ) and its primitive J21f1f; (z) are holomorphic functions. Exercise 3. Suppose
cp(t) is such that for some b
>
This fact has numerous applications.
cp
Exercise 4. Suppose is as in the previous exercise, and in addition == 0 for all == 0, 1 , 2, . . . . Then == 0 almost everywhere.
n cp fiR tn cp(t)dt Hint. Proposition 3 and Exercise 3 together show that F( cp)
is a re striction to 1R of a function holomorphic in a strip, and that this function
420
7. Fourier Transform
vanishes at the origin together with all its derivatives. Thus, we can apply Theorem 1 ( i ) ( on uniqueness ) . 1.2.
F( cp) == 0, and
Now we recall our old promise about Hermite functions given in Section
The system of Hermite functions Pn (t) e -2 ( see Example 1 .2.8 ) is an orthonormal basis in the Hilbert space L 2 (1R) . Exercise
t2
5.
t2 n Hint. It is sufficient to show that 1/J t e - 2 ; n == 0, 1 , 2, . . . implies2 t 1/J == 0 in £2 (JR) . This follows from the fact that the function cp ( t) : == 1/J ( t) e -2 j_
satisfies the assumptions of the previous exercise.
Finally, consider the extreme case, when the initial function is finitary.
6°. Suppose cp( t ) vanishes outside the interval [ - a, a]. Then its Fourier transform extends to an entire function 1/J(z) satisfying for some Exercise
C > 0 the following estimate:
1 1/J(z) 1 < Ce a l Im ( z) l .
Note that such a growth ( not faster than an exponential on the imaginary axis and on its shifts ) is the strongest of the possible requirements on an entire function bounded on the real line. The fact is that there are no non-constant functions satisfying the latter condition and growing more slowly than e a l Im( z) l for each a > 0 ( see, e.g . , [94] ) .
In addition to the operations cp �----+ cp' and cp �----+ tcp there are two other operations that change places with each other after the Fourier transform, this time without any additional assumptions on the initial function. For a E 1R denote by Ta the operator of shift by a acting in £ 1 (JR) or Co ( lR ) ( cf. Example 1 .3. 7) , and by Ea the operator of multiplication by e- iat ; t E JR, in one of these spaces. Proposition 4.
The following diagrams are commutative:
Proof. The commutativity of the first diagram follows from the equalities
a a + i i i s l
0 put 12a ' l t l < a, 'ljJ ( t) == 0, l t l > a. Then, clearly, ( 'P 'ljJ ) ( t ) == 21a f/+aa 'P( T )dT. Thus, for every t E 1R this convolution is the mean value of the function 'P on the interval [t - a, t + a]. Exercise 2 (Convolution smoothes) . Take 'P E L 1 (1R) and a finitary one-time-smooth function 'lj;. Then the convolution ('P * 'l/J )(t) (that evidently exists for all t) is also a one-time-smooth function, and we have ( 'P 'ljJ ) ' == 'P 'l/J' everywhere on 1R. Hint. First make sure that 'P 'ljJ is continuous. Then using the FUbini theorem, show that the indefinite integral of the function 'P 'l/J ' coincides, up to a constant summand, with 'P 'l/J . *
{
*
*
*
*
*
*
Now we describe the most important situation where the existence of the convolution is guaranteed.
1. Suppose 'f?, 'ljJ E £ 1 (JR) . Then the convolution 'P 'ljJ exists, belongs to L 1 ( 1R ) , and satisfies the estimate II 'P 'l/JII < II 'P II II'l/JII {wh ere II · II is the norm in L 1 ( JR)}.
Theorem
*
*
fiR I 'P( T) I ( fiR l'l/J (t -T) l dt)dT == II 'P II II'l/JII · From real analysis we know that this implies that the function 'f?(T)'lj;(t - T) of two variables is Lebesgue integrable. Hence, the FUbini theorem guarantees the existence of the iterated integral fiR ( fiR I 'P(T) 'ljJ (t -T) I dT)dt, which is also equal to II 'P II II'l/JII · As a corollary, we have the estimate fiR I fiR 'f?(T)'ljJ(t-T)dT i dt < II 'P II II 'l/J I · The Proof. Evidently,
rest is clear.
•
426
7. Fourier Transform
There are other conditions, frequently encountered in practice, for the existence of convolutions. The following exercise describes some of them. Exercise 3.
( i ) Suppose cp E £ 1 (JR) and 'ljJ E L00 ( 1R). Then cp * 'ljJ exists and is defined everywhere; in addition, it is uniformly continuous. The same is true for cp, 'ljJ E £ 2 (JR) . ( ii ) Suppose cp E £ 1 (JR) and 'ljJ E L 2 (1R). Then cp * 'ljJ exists and belongs to £ 2 (JR) . Moreover, for the norms in the corresponding spaces we have the estimate II cp * 'l/J ll2 < I cp ii i ii 'l/J II2 · Hint for ( ii ) . From the assumption and Theorem 1 it follows that the function ( of T) J l cp(t - T) l l'l/J (T) I belongs to L2 (1R). Now we can apply the Cauchy-Bunyakovskii inequality to this function and to the function J l cp(t - T) l . Then the required estimate follows from the estimate in The orem 1 .
We now return to our principal case where both functions lie in L 1 (1R).
The space L 1 ( JR ) is a commutative Banach algebra with respect to the multip lication * : £ 1 (JR) x £ 1 (JR) � L 1 (JR) : ( cp, 'ljJ ) r--+ cp * 'ljJ .
Theorem 2.
Proof. The linearity of the integral immediately implies that
* is a bilinear
operator. Furthermore, suppose that cp 1 , cp 2 , cp 3 E L 1 (1R). Using the Fubini theorem and ( in the fourth equality ) the substitution T - p == T1, we obtain, for almost all t E JR, the equalities
[(cp l * Cf? 2 ) * tp3 ](t) = l (cp l * Cf?2 )( 7 )cp3 (t - T)dT
l ( l Cf?I (p) cp2 (T - p)dp) cp3 (t - T)dT = l Cf? I ( P ) ( l Cf? 2 (T - p )cp 3 (t - T)dT ) dp = l tp 1 ( p ) ( l cp 2 (T ' )cp 3 (t - p - T 1 )dT' ) dp = l Cf? I( P )(cp 2 * cp3 )(t - p )dp =
==
[cp l * ( cp 2 * cp3 )] ( t) .
This shows that the convolution is associative, hence a multiplication. The estimate indicated in Theorem 1 means precisely that multiplication is re lated to the norm by the multiplicative inequality. Thus, we have verified the properties of a Banach algebra. Proposition • l ( i ) guarantees its commutativity.
2.
427
Convolution. Fourier transform as a homomorphism
Exercise 4. The algebra £1 (JR) is a Banach star-algebra with respect
to the involution taking
cp to �( t ) : == cp ( - t ) .
What is notable in this new Banach algebra? First (contrary to the case of l 1 (Z) ) it has no identity, but we suggest that you prove this later. Here we would like to show that £1 (JR) has something that can be viewed as a substitute for that identity, namely, a "collective identity" .
8n
Proposition 2 . Let be a sequence of non-negative integrable functions on 1R such that == 0 outside [ - � , �] and fiR ( T ) dT == 1 . Then for every � E L1 (JR) the sequence * � converges to � in L1 (JR) .
8n
8n
8n
[a, b] . Then 8n * � ( t) == J: 8n ( t - T) dT. Hence this function takes values in [0, 1] , and for sufficiently large n is equal to 1 inside the interval [a + � , b - �] and to 0 outside the interval [a - � , b + �] . Therefore, Proof. Assume first that � is the characteristic function of an interval
Thus, 8n * � tends to � when � is the characteristic function of an interval. By the linearity of convolution with respect to each variable, this is also true when � is an arbitrary step-function. Now let � E £1 (JR) be arbitrary. Take c > 0, and using the fact that the step-functions are dense in £1 (JR) , take a step-function cp such that II � - cp l < � - Let N E N be such that for n > N we have l 8n * cp - cp ll < � Then for the same n we have
ll 8n * � - � II == ll 8n * ( � - cp) + 8n * cp - cp + cp - � II ll 8nll ll cp - � II + ll 8n * cp - cp ll + ll cp - � II ll 8n * cp - cp ll + 2 ll cp - � �� 3 + 2 3 = E.
0.
If an algebra has a bounded approximate identity, then we should be happy. In this case many ( although, not all, of course ) questions about our algebra can be treated as if the algebra were unital. Here is an illustration. Theorem 5 ( Cohen ' s 2 factorization theorem, [80, Chapter I, § 1 1 , Corollary 1 1] ) . If a
Banach algebra has a bounded approxtmate identity, then every element of this algebra can be represented as a product of two other elements.
2 The same Paul Cohen who later became famous for clarifying the situation with the continu urn-hypothesis.
2.
Convolution. Fourier transform as a homomorphism
431
Combining Cohen ' s theorem and Proposition 2, we obtain that every integrable func tion on the real line can be represented as a convolution of two other integrable functtons. Taking into account Exercise 7(iii) , the same is true for integrable functions on the circle. Certainly, the case of algebras like £ 1 (JR) or £ 1 (1r) is most favorable because for these algebras a bounded approximate identity can be found here among usual sequences (not nets of general form) . In such situations one can speak about a countable bounded approxtmate identity. The reader can verify that the algebra Co (O) (see Example 5 .3.2) , which is not unital, has a bounded approximate identity, and each of the algebras C0 (1R) , eo (Z) , and K(H) (for a separable H) has a countable bounded approximate identity. Every C * -algebra has a bounded approximate identity (but it is not easy to prove; see, e.g . , [25] ) . At the same time (and these are sufficiently simple facts) the algebras h and l2 with pointwise multiplication and their "non-commutative analogues" , algebras N (H) and S(H) , have approximate identities but do not have bounded approximate identities. Finally, the subalgebra in C 1 [a, b] (Example 5.3.4) of functions vanishing at a given point of the interval does not have approximate identity, even non-bounded. 3 ° . The Fourier transform as a Gelfand transform. Recall that for every commu tative Banach algebra A, there exists a canonical homomorphism of A to the algebra of continuous functions vanishing at infinity on a locally compact topological space , namely, the Gelfand spectrum of the given algebra. It is the so-called Gelfand transform ( Defini tion 5.3.5) . Certainly, this is true, in particular, for the algebra £ 1 (JR) with convolution multiplication. It turns out that the classical Fourier transform is , up to necessary iden tifications, a concrete realization of the Gelfand transform for £ 1 (JR) . As a motivation, let us first consider a simpler algebra h (Z) . What is its Gelfand spectrum? Obviously, every s E 1r defines a non-zero character (i.e. , an element of the Gelfand spectrum) of our algebra by the formula Xs (e ) == 2::: :' en s n , and distinctn points on the circle correspond to distinct characters. Furthermore, every unit vector p ; n E Z is the nth "convolution power" (positive, negative, or, for 1 == p0 , zero) of the unit vector p 1 . As a corollary, linear combinations of powers of the unit vector p 1 are dense in h (Z) . From this we have that every non-zero character X : h (Z) ---+ C of the algebra considered has the form of Xs for s == (X (p 1 )) - l . (The fact that s E 1r evidently follows from the fact that X does not increase the norm.) Thus, in the notation 0 :== O . ( h (Z) ) , there is a bijective mapping w : 1r ---+ 0 : s �-----+ Xs which, as is easy to see, is continuous . Consequently, by the Alexandroff theorem, w is a homeomorphism, and the Gelfand spectrum of the algebra h (Z) , as a topological space , is the unit circle. Now, consider the mapping C(w) : C(O) ---+ C ( 1r ) that assigns to every function x : 0 ---+ C the function C ( w ) x : s �-----+ x ( w ( s )) . (We have already encountered it in a different form in Example 3. 1 . 1 .) It is easy to verify that this is an isometric isomorphism of Banach algebras. Identifying these two algebras via this isomorphism, we see that the Gelfand transform of the algebra h (Z) is just the Fourier transform multiplied by � To be more formal, there is a commutative diagram - (X)
h (Z)
where
r
/ � C(O) C 1r
:== Fz 1 (Z ) and F :== Fz; .
C (w )
( )
-
432
7. Fourier Transform
We now pass to our principal object , the algebra £ 1 (IR) . We immediately formulate the main result , which must be known to advanced readers. Then we present several exercises giving an idea of how this theorem can be proved . From this moment on, 0 denotes the Gelfand spectrum of the algebra £ 1 (IR) , and is the Gelfand transform of this algebra.
r
Theorem 6 (see [25, Chapter IV, Theorem 3.10] ) . There is a homeomorphism w : lR ---+ 0
assigning to every s E lR the character Xs defined by the formula x. (cp)
1 cp(t)e -ist dt
=
(in other words, the value of Xs at a function r.p is the Fourier transform multiplied by V2:ff of r.p at the point s). In addition, the diagram L 1 (JR)
/ � Co (O) Co C (w)
( IR )
is commutative; here C(w) is a (well-defined} isometric isomorphism of Banach algebras that assigns to a function x E Co (O) the function C(w)x : s � x(w(s)) . Thus , in the identification of the two algebras by means of the isometric isomorphism C(w) , the Gelfand transform of the algebra £ 1 (JR) becomes the Fourier transform. Here is the outline of the proof.
* Exercise 12 . For every X E 0 there exists a unique continuous function X : lR ---+ C such that for each r.p E £ 1 (JR) we have X (cp)
=
1 cp(t)X(t) dt.
This function is defined by the rule t � lim h _. +O X ( k x t , t + h ) . Hint. Take r.p with X (r.p) =/= 0. The equality lim h _.+ o (r.p k xt , t+ h ) = Tt (r.p) (cf. the proof of Theorem 2) shows that for every a < t the derivative of the function a ( r ) : = X (x a , r ) ; r > a at the point t is equal to X(Tt r.p) j X(r.p) (thus, it does not depend on the choice of a or r.p) . It is this derivative that we denote by X ( t) . The continuity of the mapping lR ---+ L 1 (1R) :- t � Tt (r.p) implies that the function X ( t ) is continuous. t Furthermore, since a ( t) = fa X ( T ) dr , we have X ( r.p ) = JR r.p( t )X ( t )dt in the case where r.p is the characteristic function of an interval. But these functions form a total set in £ 1 (JR) , and X is continuous. *
X (t + r )
Exercise 13. The function X takes values in 1f and satisfies the functional equation
(5)
- = X(t)X (r) .
Hint. From Proposition l (ii) and the properties of characters it follows that
X (Tt + r 'P )X (r.p) It remains to divide by X ( r.p ) 2 .
=
X ( Tt r.p)X (Tr r.p) .
The following fact has various applications in analysis, and perhaps you have already come across it . Exercise 1 4 . Every continuous function X : lR ---+ 1r satisfying (5) is of the the form X(t) = for a unique s E JR.
e -ist
3. Fourier transform of generalized functions
433
Hint. Clearly, X (O) == 1 . Hence, for a sufficiently large n E N, the number X ( 2� ) is e-i 8 � for a unique s E lR such that -1r /2 < -s/2 n < 1r /2. Therefore, X (t) == e-i 8 t for all rational, and by the continuity of X, for all real t. The only point in the proof of Theorem 6 that is not covered by these three exercises is that the resulting continuous bijection lR ---+ 0 : X � X8 is a homeomorphism. The continuity of the inverse bijection 0 ---+ lR can be proved directly: we take X8 E 0 and note that if IXr (X a , b ) - X8 (X a , b ) l is sufficiently small for a suitable interval [a, b] , then lr - s l is also sufficiently small. However, we note instead that the initial bijection can be extended to a continuous bijection from the Alexandroff compactification JR + to the space 0 appended by the zero character. (The continuity of this extended bijection follows from the fact that X8 ( cp ) or, what is the same, v'2/rrF ( cp ) (s) tends to zero as l s i ---+ oo. ) Hence, the Alexandroff theorem guarantees that the extended bijection is a homeomorphism, and so the same is true for its restriction to JR. As you may guess , a parallel result is valid for the Fourier transform on the circle: it is also a specialization of the Gelfand transform, but now for the algebra £ 1 ('lf) . The characters of this algebra are indexed not by arbitrary real numbers , as in the case of L 1 (JR) , but by integers. Namely, each n E Z defines the character Xn by the formula
Xn : L1 (1l' ) --+ C : cp
1-+
i cp(t) C n dt ( = .J2;Fr ( cp) (n) )
( "the nth character is the nth Fourier coefficient" ) . The Gelfand spectrum of the algebra L 1 (1r) is a discrete countable topological space, which can be identified with z . Respec tively, the algebra C0 (0. (L 1 (1R) ) ) can be identified with eo (Z) , and in this identification the classical Fourier transform on the circle becomes the Gelfand transform multiplied by v'2/rr . (Draw the corresponding commutative diagram. ) The proof of these results follows the course of the proof of Theorem 6. In particular, obvious analogues of the facts expressed in Exercises 12 and 1 3 are true. The main difference between the cases of the line and of the circle appears when we pass to the explicit description of the functions X. The role of Exercise 14 now belongs to the following
-
Exercise 15. Every continuous function X : 1r ---+ 1r satisfying the functional equation
-
-
-
X(tr) == X(t)X (r) , t- n for a uniquely defined n E Z. Hint. The function r � X( e i r ) on the line satisfies the functional equation (5) . Hence, for some s E lR and for all T E lR we have X( e i r ) == e-i 8 r . In particular, e-i 8 ( r +2 1r ) == e-i 8 r , has the form X( t )
==
and this means that s is an integer.
3. Fourier transform of test functions and of generalized functions
Results on the classical Fourier transform take the most complete and ele gant form if for initial functions we take rapidly decreasing infinitely smooth functions introduced in Section 4.3. As we recall, they form one of the most important spaces of test functions, the Schwartz space S.
1. Suppose cp E S and �( s) : == F( c.p) ( s) . Then � is an in finitely smooth function, and for every p , q E Z + the function sP �( q) ( s) is Proposition
434
7. Fourier Transform
a linear combination of functions of the form F(t k cp ( l ) (t)) ; k == 0, 1, . . . , p, l == 0, 1, . . . , q with coefficients independent of cp. As a corollary, 1/J E S .
1 1/;(s) is infinitely smooth, and for every q E Z+ we have 1/J ( q) (s) == (- 1 )qF(tqcp(t))(s). From this we have, now by Corollary 1.1, that the function sP 1f; ( q) (s) is, up to a constant factor, the Fourier transform • of the function ( tq cp ( t)) (p). The rest is clear. Hence, we can consider the birestriction of the operator F in Definition 1.1 to the pair ( S, S) . The resulting linear operator on S will again be denoted by F ; this will not lead to a confusion. Proof. By Corollary .2,
In the space S ( contrary to £1 (JR) , the initial domain of the Fourier operator ) the operators D cp �----+ cp' and M : cp(t) �----+ tcp(t) act. Hence, the relation between the differentiation and multiplication by an independent variable mentioned in Propositions 1 .2 and 1.3, can be expressed in the following way. :
Proposition 2.
The following diagrams are commutative:
We now recall that S is a polynormed space with respect to the family of prenorms s introduced in Section 4.3. Proposition 3.
The operator F is continuous as an operator in the poly
( S, s) . Take cp E S, 1/;(s)
normed space
Z+ . Proposition 1 evi dently implies that for some cl > 0 we have the estimate II F(cp) ll p ,q < cl max { II F(tk cp ( l ) (t)) lloo : k == 0, 1, . . . , p , l == 0, 1, . . . ' q} , where II · lloo is the norm in Co ( lR ) . But Proposition 1.1 implies that for every pair k, l we have Proof.
: ==
F(cp)(s), and
p,
q E
II F(tk
.. : l 2 --+ l 2 , where A is the sequence ( i ) n l ; n == 1 , 2, . . . . Hint. The unitary equivalence is established by the operator taking the nth Hermite function to the n + 1th unit vector, n E Z + . In fact, the Hilbert Fourier operator is in many cases more convenient than the classical Fourier operator from Section 1 , and it allows us to obtain more complete results. Here is an instructive example. As we said before, it is apparently hopeless to describe the image of the classical Fourier transform in some reasonable terms. Similarly, nobody knows how to describe the Fourier transforms of finitary integrable functions. At the same time, the following deep fact is true. -
Theorem
JR; a
3
( Paley-Wiener; see
[73,
-
Chapter VII, §6.2] ) . For every
aE
0 the following properties of a function 1/;(t) ; t E 1R are equivalent: ( i ) 1/; ( t) is the Hilbert Fourier transform of a square-integrable function on the real line that vanishes almost everywhere outside the interval
>
[ - a, a] ;
( ii ) 1/; ( t) is square-integrable and can be extended to an entire function 1/J ( z ) ; z E c satisfying the estimate 1 1/J(z) I < c ea l Im ( z ) l ' where c > 0
is a constant (depending on 1/J).
Remark. The Hilbert Fourier transform plays an outstanding role in mathematical meth ods of quantum mechanics, where it implements a unitary equivalence between some most important operators. We have to note, however, that these operators are often not the usual bounded operators, but somewhat more complicated mappings , so-called unbounded operators, which may be defined not on the entire L 2 (1R) , but only on some dense sub spaces of it. Nevertheless, the notion of unitary equivalence makes sense for the unbounded operators as well. The following two operators can be distinguished among those as the most important ones: "the operator of coordinate" and "the operator of impulse" . Their domains contain S, and the first operator acts by the rule cp(t) � tcp(t) , whereas the second by the rule cp � icp' . The fact that the Fourier transform implements a unitary equivalence between these operators allows us, in particular, to obtain, as a mathematical theorem, one of the main principles of quantum mechanics , the Heisenberg indeterminacy principle (see, e.g. , [9 3, chapter 7, §6] ) . A great deal of information about the role of Fourier transform in mathematical physics, and in particular, in the axiomatic quantum field theory, is contained in [ 64] .
5 . A little ab out harmonic analysis on groups What is harmonic analysis? With some oversimplification we can say that the objects of this science are function spaces where the operation of shift is defined; first of all, these are spaces of functions on groups with topology. As for the methods of studying these spaces, they are more or less related to the Fourier transform.
5. A little about harmonic analysis on groups
449
The classical part of harmonic analysis has a long and rich history; it studies "under strong magnification" spaces of functions on the three traditional sets with group structure: the circle, the real line, and the integers (and also on their Cartesian powers) . The so-called abstract harmonic analysis (or Fourier analysis on groups) deals with function spaces on groups of much more general nature. In the previous sections of this chapter only the classical harmonic analysis, or to be more specific , its intersection with functional analysis , was discussed. The main attention was paid to Fourier transforms on IR, 1r, and Z. In this section we try to describe, briefly and informally, some preliminary notions and results of abstract harmonic analysis . (A far-from-complete list of references contains the two-volume book [31] , [32] and recently published [96] ; see also [97] , [98] , [99] .) First of all we discuss general notions hidden behind these three "concrete" Fourier transforms. In the previous sections, although not calling things by their right names, we widely used the fact that the three traditional domains of Fourier transform, IR, 1r, and Z, are locally compact abelian groups. We now explain what it is . Definition 1 . Let G be a group and at the same time a Hausdorff topological space. We say that G is a topological group if both group operations, the multiplication m : G x G ---+ G : ( s , t) � s t and the passage to the inverse element inv : G ---+ G : t � t - 1 are continuous mappings. A topological group is called abelian (or commutative) , locally compact , etc. , depending on the properties of the underlying group or topological space.
The topological groups form a category with continuous group homomorphisms as morphisms. In this category there is an important full subcategory consisting of locally compact groups. In its turn, it contains a full subcategory of locally compact abelian groups. We denote the latter category by LCAb. It also contains two important full subcategories. The first consists of discrete groups and, thus, is nothing but the category of abelian groups Ab we already came across in Chapter 0; the second , denoted by CAb, consists of compact abelian groups. Clearly, isomorphisms in all these categories are continuous homomorphisms having continuous inverse homomorphisms. Just as in the category Ban, or, say, UCBA, they are called topological isomorphisms. The following class of morphisms of topological groups deserves special name. Definition 2. Let G be a topological group. A group character of this group is a morphism (i.e. , continuous homomorphisms) of G to the group 1r .
(We say just "character" if there is no danger of mixing it up with algebraic characters, i.e. , the characters of algebras defined in Section 5.2.) set of characters of a topological group G is traditionally denoted by G. Clearly, The this is also an abelian group with respect to the pointwise multiplication. The identity element of this group is the "identity character" ; we call it the - trivial character and denote by 1 . (This will not - lead to a confusion. ) But the set G also has a natural topology. Namely, consider G to be a subset in the linear space C(G) of all continuous functions on G. The latter is endowed with the family of prenorms ll x ii K : = max{ l x (t)-l : t E K} , where K runs over the family of all compact subsets in G. The topology in G inherited from the polynormed space ( C (G) , { I I x I I K } ) is often called the Pontryagin topology ; clearly, the convergence in this topology is the uniform convergence on compact sets. Theorem 1 (see [99 , Chapter II, § 1 , Corollary 2] ) . If G is a locally compact group, then
G is also a locally- compact group with respect to the Pontryagin topology. Moreover, if G is compact, then G is discrete, and if G is discrete, then G is compact. Exercise 1. Prove the part of t his theorem concerning compact and discrete groups.
450
7. Fourier Transform
Hint. If t E 1r \ { 1 } belongs to the right half-plane, then some power of t jumps out of it . Hence, if G is compact and -for X E G we have II X - l i l a < J2, then X _ 1 . If, on the contrary, G is discrete , then G is homeomorphic to a closed subset in the topological product of a family of copies of 1r indexed by elements of G. If the group G is not abelian or not locally compact , then G can be very small, or even consist of only one (trivial) character. Certainly, simple groups have only trivial character; the same is true, say, for the group SU(2, C) of unitary matrices of the second order with determinant 1 . On the other hand, the additive group of the space from Exercise 1 .6.10 also has no (continuous) non-trivial characters. However, for the groups that are abelian and at the same time locally compact, in other words , for the category LCAb, the notion of character, as we will soon see, plays a central role. This role is similar to the one played by bounded functionals in the theory of normed spaces , or to the role of algebraic characters in the theory of semisimple commuta tive Banach algebras. (There are sufficiently many characters, or bounded functionals , or algebraic characters; see below. ) In the context of LCAb the group G has a special name: the "dual group in the sense of Pontryagin" (or just , the "dual group ") of the group G. We now look at what our general definition of character becomes for our three groups. The following result, aside from its almost evident first part, is just a reformulation of Exercises 2 . 1 4 and 2 . 1 5 . Exercise 2. Up to topological isomorphism of topological groups, we have Z = 'lr, lR = IR, and 1r = Z. More precisely, the following mappings are topological isomorphisms : (i) assigning to every X E Z the unique s E 1r such that X(n) = s - n ; n E Z; ( ii) assigning to every X E lR the unique s E lR such that X ( t) = e i s t ; t E JR ; (iii) assigning to every X E j the unique n E Z such that X(t) = t- n ; t E 1r . -
Thus , in all three cases the classical Fourier transform can (up to the factor written as a general formula (1)
F(cp) (X)
:=
l cp(t)X (t)dt.
k ) be
It takes every integrable function on G to a function on G vanishing at the infinity. The resulting Fourier operator acquires a general form F = Fa : £ 1 (G) ---+ Co (G) . Once this formula is written, we would like to extend it to other topological groups. However, we note that in the above concrete situations, in addition to the algebraic and topological structures of Z, 1r, and IR, an essential role is played by another structure, namely a measure. We mean the measures with respect to which we integrate: the stan dard Lebesgue measure on lR and 1r, and the counting measure on Z. Hence , if formula ( 1 ) is to make sense for a group G, we must integrate the functions on G with respect to a measure having the same good properties as the measures in the above examples. At the same time we can ask: why do these concrete measures enter classical definitions? Why are they better than other measures? It turns out that everything will be settled in a perfect manner. The only condi tion is that the group under consideration should be locally compact . To formulate the corresponding result we need some preparation. Let G be a topological space (arbitrary, so far) . When speaking about measures on G, we mean countably additive functions of Borel sets taking their values in the extended pos itive real line [0, oo] . (Thus, measures can have infinite values, like the standard Lebesgue measure on the real line and the counting measure on Z.) Furthermore, it is desirable that the measure be compatible with the topology on G.
5. A little about harmonic analysis on groups
451
Definition 3. A measure J.l on G is called regular if
( i) for every measurable set M of finite measure and for every c > 0, there are an open set U � M and a compact set K C M such that J.l ( U \ K) < c ; ( ii) the measure of every compact set is finite ( reflecting the fact that "such a set is small" ) .
In addition, let G be a group. Then we can speak about shifts of subsets: for M C G and a E G the left shift on a of the set M is the set a M : = { at : t E M} . It is easy to guess that a self-respecting measure must interact with shifts as follows . Definition 4 . A measure J.l on a topological group G is called left-tnvariant if for each measurable set M and each a E G the set a M is also measurable and J.l ( a M) = J.L ( M) .
It is easy to show that the standard Lebesgue measure on the real line and on the circle and the counting measure on the integers are the only, up to a factor, ( left)invariant regular measures . ( This is why they enter the corresponding Fourier transforms . ) The following theorem lies in the foundation of the entire abstract harmonic analysis . It has a rather long history and is related to the names of Haar, von Neumann, and Andre Weil. Theorem 2 ( see [31 , Chapter IV, § 1 5] ) . Let G be a locally compact group. Then there exists a left-invariant regular measure on G, and this measure is unique to a constant
factor. If a topological group satisfies the so-called completeness condition ( in the case of a metrizable group this notion coincides with the usual completeness) , then the converse result is also true: if a group has a left-invariant regular measure, then it is locally compact. The left-invariant measure in this theorem is called the Haar measure. Thus, this measure is implicitly contained in the algebraic-topological structure of the locally compact group: every group "knows its measure" . Remark. If a group G is non-abelian, then the left- and the right-invariant measures on G in general are different . The existence of the latter ( it must be clear what it is) immediately follows from Theorem 2: we have only to replace the given multiplication on G by the "opposite" one ( by definition , we put a o b to be equal to the initial b a ) . However, for a wide class of groups a left-invariant measure is also right-invariant. Such groups are called unimodular. In addition to abelian groups, this class contains all discrete groups and all compact groups. However, say, the group of all matrices of the form ( g � ) with real a =/= 0 and b ( or, what is the same, the group of affine transformations of the real line) is not unimodular.
We fix a Haar measure on every locally compact group; in what follows, speaking about the Haar measure we mean this particular measure. ( The fact that this procedure requires the axiom of choice, will not stop us. ) We denote this measure by m and use the short notation L 1 ( G) for the space £ 1 ( G, m) . Taking Theorem 2 for granted, we give the following general definition . ......
Definition 5. Let G be a locally compact group, G the group of characters of G with Pontryagin topology, and r.p an integrable ( with respect to the Haar measure) function on G. The Fourier transform of this function is the function on G defined by the formula ......
F ( rp ) (X) : =
l rp (t)X (t) dm(t) ; X E G.
452
7. Fourier Transform
The function F ( cp) is continuous on G (you can easily prove this) and vanishes at infinity (this is more difficult) . Thus , we obtain a mapping F : L1 (G) ---+ Co (G) , which, as you can easily see, is a bounded operator of norm 1 ; this is the so-called Fourier operator of the group G . Up to now we were discussing arbitrary locally compact groups. But the Fourier transform becomes an efficient tool only under the assumption that the group is abelian. Only under this assumption the following non-trivial theorem is true. Theorem 3 (Uniqueness theorem; cf. Theorem 1 . 1 , [99 , Chapter II, § 1 , Proposition 3] ) .
Suppose that a locally compact group G is abelian. Then its Fourier operator is injective. In other words, if two Haar-integrable functions on G have the same Fourier transform, then they coincide almost everywhere.
From this, one can - easily deduce the following important fact about the relationship of the groups G and G. Theorem 4 (Sufficiency of the set of characters; see [99, Chapter II, § 1 , Proposition 3] ) . For different s , t E G there exists a character X E G such that X ( s ) =1- X ( t) . -
,.,.......,
�
-
Exercise 3. Deduce this theorem from Theorem 3.
Hint. Take a E G such that a =1- 1 and X( a) = 1 for all X E G. Let U be a neighbor hood of the identity such that a U n U = 0, and x the characteristic function of this neigh borhood. Since the Haar measure is invariant, fa x(t)X (t) dm(t) = fa x ( a - 1 t ) X ( t ) dm ( t) for every X E G. �
Theorem 4 sounds very similar to Corollary 1 . 6 . 1 on the sufficiency of the set of bounded functionals on a normed space. Corollary 1 .6 . 1 is equivalent to the fact that the canonical embedding a : E ---+ E* * of a normed space into its second dual space (see Section 1 . 6) is injective. But in the context of locally compact abelian groups there exists something similar to the canonical embedding. Namely, take G E Ob(LCAb) and consider the dual -- group G. It is also abelian and locally compact , hence we can consider its dual group G, called the second dual group of the group G. Now to each element t E G we can assign the character O: t on G defined by the formula O: t (-X) : = X(t) . Clearly, this character is continuous. So, we obtain a mapping 0: : G ---+ G. Evidently, it is a homomorphism of groups, and by Theorem 4, this homomorphism is injective. However, there is an essential difference between the homomorphism 0: in the theory of topological groups and the operator a in the theory of normed spaces . As we recall, in general, a is not an isomorphism in Nor or even in Ban. In other words, Banach spaces are not necessarily reflexive. But the objects of the category LCAb are always reflexive: Theorem 5 (Pontryagin's duality principle 3 ) . For every locally compact abelian group G -
the homomorphism
0:
-
: G ---+ G is a topological tsomorphism.
As in the case of another result of great value for the whole mathematics , the first Gelfand-Naimark theorem (see Section 6.3) , behind Pontryagin's principle lies identifica tion of some important apparently different categories. We recall the contravariant asterisk functor (* ) acting on the category Ban. There is a similar functor acting on the category LCAb. It is called the duality functor ( ) . To 3 L. S. Pontryagin (1908- 1988) , outstanding Russian mathematician, the author of fundamen
tal results in many areas of mathematics, first of all in the theory of topological groups, algebraic topology, differential equations, and calculus of variations.
5. A little about harmonic analysis on groups
453
...... every group G it assigns the dual group G, and to every morphism f : G1 ---+ G2 the "dual morphism" J : G2 ---+ G1 defined by the formula (]X) (t) : = X(f(t)) ; X E G 2 , t E G1 . By Exercise 1 , this functor takes compact groups to discrete ones, and vice versa. Hence, we can speak about the corestrictions of this functor to the corresponding categories . (i) The category LCAb is dually equivalent to itself (in other words, Theorem 6. it is equivalent to its dual category; cf. Section 0. 7) . (ii) The category Ab is dually equivalent to the category CAb. (iii) These dual equivalences are given by the duality functor (...... ) , or by the corre
sponding corestriction of this functor.
Note that the role of the "inverse" functor ( cf. Definition 0. 7. 7) belongs to the same functor of duality or to its corestriction defined on CAb and taking values in Ab. The ........) : = (...... ) (...... ) belongs to the role of the natural equivalence between the functors 1 and (.... canonical isomorphism a for all G E LCAb (or G E Ab ) . There are two quite different ways to prove Pontryagin's principle (and in both cases there are many interesting things to be found on the roadside) . Pontryagin's original proof was based on deep analysis of the structure of locally compact abelian groups . To speak informally, this proof reduces general groups in this class to the "elementary bricks" , which are IR, 1r, Z, and Z/Z n (see, e.g. , [100, §24] ) . Another proof appeared later and belongs to D . A. Raikov (see, e.g. , [101 , §31] ) . It is based on the notion of Hilbert Fourier transform on an arbitrary group. Is uses a theorem generalizing Plancherel's Theorem 4. 1 , as well as on our recent observations concerning the Hilbert Fourier transforms on 1r and Z. Here is the general Plancherel theorem, also of significant independent interest . ·
Theorem 7. Let G be a locally compact abelian group. There exists a bounded operator F. : £ 2 (G) ---+ £ 2 (G) untquely determined by the fact that on L1 (G) n £2 (G) it coincides
with the ( "classical") Fourier transform from Definition to a positive factor.
5.
This operator is unitary, up
If a group is not abelian, then the Fourier transform is not so useful anymore. We had already discussed the reason: the characters of a general group do not in general distinguish between elements. Instead of characters we must consider rather far-away generalizations, the so-called irreducible unitary representations (of the given group) . These are homo morphisms of the group G to the group of unitary operators acting in a Hilbert space H. They are supposed to be continuous with respect to the strong-operator topology (inherited from B(H) ) , and the word "irreducible" means that the image of the group G consists of operators having no common closed invariant subs paces different from { 0} and H. (If the group is abelian, then it turns out that all Hilbert spaces of the irreducible unitary representations are one-dimensional, and we obtain just usual characters. ) Of course, to work with representations i s much more difficult than t o work with characters. However, they also allow us to solve a series of important problems . The case where the group is compact is especially favorable: in this situation the Hilbert spaces of all the irreducible unitary representations are finite-dimensional. For compact groups the so-called Tannaka-Krein duality principle is established. Its role resembles the role of Pontryagin's principle for abelian groups (see [ 3 1] or [98] ) . This principle allows us to recover the compact group from its irreducible unitary representations. As for the general theory of locally compact groups, one of the main results here is the so-called Gelfand-Raikov theorem [8 7 , Corollary 13.6.6] generalizing the theorem on sufficiency of characters for the abelian case. It says that for every element t =f. 1 of a locally compact group there exists an irreducible unitary representation of this group,
454
7. Fourier Transform
taking t to the operator different from 1 . ( This is also "sufficiency" , but this time of the indicated family of representations. ) A fundamental role in harmonic analysis on groups belongs to the general notion of convolution. Recall that for the three classical groups G = IR, 1r, and Z the space £ 1 (G) becomes a Banach algebra with respect to the convolution multiplication. Let G be an arbitrary locally compact group. Definition 6. The convolution of functions
cp 'ljJ ( t) *
:=
r.p,
'ljJ E £ 1 (G) is the function
L cp ( ) 'ljJ ( - l t) dr. r
r
Similarly to the classical cases, the convolution considered in Section 2 exists for almost all t E G and belongs to L 1 (G) . Further, the Banach space £ 1 (G) is a Banach algebra with convolution as a multiplication. In other words, the operation is associative and satisfies the multiplicative inequality. ( Of course, these facts use the left-invariance of the Haar measure. ) Furthermore, the algebra £ 1 (G) is commutative {::::::> the group G is commutative ( i .e . , abelian ) . Besides, our algebra always has a bounded approximate identity ( see Definition 2.2) , and the "true" identity exists {::::::> G is discrete. In conclusion, let us go back to the Fourier transform on groups. Similarly to our three "classical" groups, after multiplying by some positive number it takes the convolution to pointwise multiplication. In other words, it is a homomorphism between the algebras L 1 (G) ( with convolution multiplication ) and Co (G) . This fact comes to the foreground in the situation where the group G and, consequently, the algebra £ 1 (G) are commutative. Then Theorem 2.6 acquires the following general form. *
Theorem 8 ( see [100, § 3 1 , Theorem 1] or [101 , Corollary 3.7] ) . Let G be a locally compact abelian group. Then the Gelfand spectrum 0 of the algebra L 1 (G) coincides, up
to a homeomorphism, with the underlying topological space of the Pontryagin dual group G . In detail, the mapptng w : G ---+ 0 assigning to each group character X the algebraic character X by the formula
X (cp) : =
L cp(t)X (t)dt
is a homeomorphism. Furthermore, under the corresponding identification of the spaces 0 and G the Gelfand transform of the algebra L 1 (G) turns out to be the Fourter transform on G. -
( We suggest that you draw the corresponding commutative diagram. ) Having stated this beautiful and important theorem , we wish you all the best.
B ibliogr aphy
[1] G . Pisier, A polynomially bounded operator on Hilbert space which ts not stmilar to a contraction. J . Amer. Math. Soc. 10 ( 1 997) , 35 1-369. [2] J . von Neumann, The mathematictan. In: The works of the mind, R. B. Heywood (ed . ) , Chicago Univ. Press, 1 947, pp. 180-1 96 . [3] V . Kodiyalam and V . S. Sunder, Topological quantum field theories from subfactors. Chapman & Hall/CRC Boca Raton, FL, 200 1 . [4] V. F. R . Jones, Subfactors and knots. Amer. Math. Soc. , Providence, RI, 1991 . [5] F. A. Medvedev, Early history of the axtom of chotce. Nauka, Moscow, 1 982. (Russian) [6] P. S. Alexandroff. Introduction to the general theory of sets and functions. Gostekhizdat, Moscow, 1948. (Russian) [7] K. Kuratowski and A. Mostowski , Set theory. North-Holland, Amsterdam, 1967. [8] A. N. Kolmogorov and S. V. Fomin, Elements of the theory of functions and func tional analysts. Nauka, Moscow, 1972; English transl. of 1st ed. , Graylock Press, Rochester, NY, 195 7. [9] M. I. Dyachenko and P. L. Ulyanov, Measure and tntegral. Faktorial, Moscow, 1998. (Russian) [10] V. M . Fedorov, Theory of functtons and functtonal analysis. I. Theory of functions. Mechanics and Mathematics Department , Mosk. Gos. Univ. , Moscow, 2000. (Russian) [1 1]
, Theory of functions and functional analysis. II. Functional analysis. Mechanics and Mathematics Department, Mosk. Gos. Univ. , Moscow, 2000. (Russian)
[12] B. V. Shabat, Introduction to complex analysis. I. Nauka, Moscow, 1 985 . (Russian) [13] J . L. Kelley, General topology. Van Nostrand, Toronto-New York-London, 1 955 1957. [14] R. Engelking, General topology. Panstwowe Wydawnictwo Naukowe, Warsaw, 1985 . [15] S. Lang, Algebra. Addison-Wesley, Reading, MA, 1965. [16] S. MacLane, Categories for the working mathematician. Springer-Verlag, Berlin Heidelberg-New York, 1 971 . [17] R. Goldblatt , Topot. The categorial analysts of logtc. North-Holland, Amsterdam, 1 979.
455
456
Bibliography
[18] V. A. Rohlin, On the fundamental ideas of measure theory. Mat . Sb. 25 ( 67} (1949) , 1 07-1 50; see also Selected works. Moskov. Tsentr Nepr. Mat . Obraz. , Moscow, 1999. (Russian ) [19] J . von Neumann, Uber Funktionen von Funktionaloperatoren. Ann. of Math. 32 ( 1 931 ) , 191-226. [20] I. E. Segal, Equivalences of measure spaces. Amer. J. Math. 73 ( 1 951 ), 275-313. [21] A. Connes, Noncommutative geometry. Academic Press, Orlando, FL, 1 990. [22] Sze-Tsen Hu, Homotopy theory. Academic Press, Orlando, FL, 1 959. [23] S. Eilenberg and S. MacLane, General theory of natural equivalences. Trans. Amer. Math. Soc. 58 ( 1945 ) , 231-294. [24] I. Bucur and A. Deleanu, Introduction to the theory of categories and functors. Wiley, New York, 1 968. [25] A. Ya. Helemskii, Banach and polynormed algebras. General theory, representations, homologies. Nauka, Moscow, 1989; English transl. , Banach and locally convex alge bras. Clarendon Press, Oxford, 1 993. [26] S. I. Gelfand and Yu. I. Manin, Methods of homologtcal algebra. Nauka, Moscow, 1 988; English transl. , Springer-Verlag, Berlin-Heidelberg-New York, 2003. [27] G. E. Bredon, Sheaf theory. McGraw-Hill, New York, 1 967. [28] M. Kashiwara and P. Schapira, Sheaves on manifolds. Springer-Verlag, Berlin Heidelberg-New York, 1 994 . [29] A. A. Kirillov and A. D . Gvishiani , Theorems and problems of functional analysis. Nauka, Moscow, 1 979; English transl. , Springer-Verlag, Berlin-Heidelberg-New York, 1 982. [30] K. Hoffman, Banach spaces of analytic functions. Prentice-Hall, Englewood Cliffs, NJ, 1962. [31] E. Hewitt and K. A. Ross, Abstract harmonic analysis. Vol. 1 , Springer-Verlag, Berlin Heidelberg-New York, 1963. [32] E. Hewitt and K. A. Ross, Abstract harmonic analysis , Vol. 2, Springer-Verlag, Berlin-Heidelberg-New York, 1 970. [33] N. Dunford and J . T. Schwartz, Linear operators. Vol. 1 . Interscience, New York, 1 958. [34] G . Pisier, Similarity problems and completely bounded maps. Springer-Verlag, Berlin Heidelberg-New York, 1 995. [35] V. Paulsen, Completely bounded maps and dilations. Longman, Wiley, New York, 1 986. [36] E. G. Effros and Z .-J . Ruan, Operator spaces. Clarendon Press, Oxford, 2000. [37] G . Pisier, Introduction to the theory of operator spaces. Cambridge Univ. Press, Cambridge, 2002. [38] N. Bourbaki, Topologie generale. Ch. iX, X, Hermann, Paris , 1 948, 1 949. [39] M . M. Day, Narmed linear spaces. Springer-Verlag, Berlin , 1 958. [40] I. E. Segal, Noncommutative geometry, by Alain Cannes (review). Bull. Amer. Math. Soc. 33 ( 1 966) , 459-465 . [41] P. Wojtaszczyk, Banach spaces for analysts. Cambridge Univ. Press, Cambridge, 1991 . [42] A . Bottcher and B . Silberman, Analysis of Toeplitz operators. Akademie-Verlag, Berlin, 1 989 .
Bibliography
457
[43] A. Pelczynski and Cz. Bessaga, Some aspects of the present theory of Banach spaces. in: S. Banach, CEvres . Vol. II, PWN, Warszawa, 1979. [44] S. MacLane, Homology. Springer-Verlag, Berlin-Heidelberg, 1963. [45] 0. Bratteli and D. W. Robinson, Operator algebras and quantum statistical mechanics. Vol. 1 . , Springer-Verlag, Berlin-Heidelberg-New York, 1 979. , Operator algebras and quantum statistical mechanics. Vol. II, Springer[46] Verlag, Berlin-Heidelberg-New York, 1 981 . [47] A. Ya. Helemskii, The homology in Banach and topological algebras. Mosk. Gos . Univ. , Moscow, 1986; English transl. , Kluwer, Dordrecht, 1989. [48] A. Defant and K. Floret , Tensor norms and operator ideals. North-Holland, Amster dam, 1 993. [49] J .-P. Kahane, Series de Fourier absolument convergentes, Springer-Verlag, Berlin Heidelberg-New York, 1970. [50] R. V. Kadison and J . R. Ringrose, Fundamentals of the theory of operator algebras. I, II. Academic Press, London, 1 983, 1 986. [51] W. T. Gowers, A solution to Banach 's hyperplane problem. Bull. London Math. Soc. 26 (1994) , 523-530. [52] R. Hartshorne, Algebraic geometry. Springer-Verlag, Berlin-Heidelberg-New York, 1977. [53] Z. Semadeni , Banach spaces of continuous functions. PWN, Warszawa, 1971 . [54] P. Enflo, A counterexample to the approximation property in Banach spaces. Acta Math. 130 ( 1 973) , 309-317. [55] I. C. Gohberg and M. G. Krein, Introduction to the theory of nonselfadjoint linear operators. Nauka, Moscow, 1 965 ; English transl. , Amer. Math. Soc. , Providence, RI, 1969. [56] A. Pietsch, Operator ideals. VEB Verlag, Berlin, 1978. [57] A. S . Mishchenko, Vector bundles and their applications. Nauka, Moscow, 1 984 . ( Rus sian ) [58] I. Fredholm, Sur une classe d 'equations fonctionnelles. Acta Math. 27 ( 1 903) , 365390. [59] B . Bollobas, The work of William Thimothy Cowers. Proc. Intern. Congress . Math. ( Berlin, 1998) , vol. 1, Documenta Math . , DMV, Berlin, 1 998. [60] H. H. Shaeffer, Topologtcal vector spaces. Macmillan, New York-London, 1 966. [61] A. P. Robertson and W. Robertson, Topological vector spaces. Cambridge Univ. Press , Cambridge , 1964. [62] C. J . Reid, A solution to the invariant subspace problem ion the space h . Bull. London Math. Soc. 17(1985) , 305-317. [63] M. Reed and B . Simon, Methods of modern mathematical physics. Vol. 1, 2nd ed. , Academic Press , New York, 1 980. [64] M. Reed and B. Simon, Methods of modern mathemattcal physics. Vol. 2, Academic Press, New York-London, 1975 . [65] R. E. Edwards, Functional analysis. Holt, Rinehart and Winston, New York-Toronto London, 1965 . [66] T . Gamelin, Uniform algebras. Prentice-Hall, Englewood Cliffs, NJ, 1 969. [67] L. A. Lyusternik and V. I. Sobolev, Elements of functional analysis. Nauka, Moscow, 1 965; English transl. , Frederick Ungar, New York, 1 961 .
458
Bibliography
[68] M . Takesaki, Theory of operator algebras. Vol. 1 , Springer-Verlag, Berlin-Heidelberg, New York, 1979. [69] I. M. Gelfand and G . E. Shilov, Spaces of test and generalized functions. Fizmatgiz, Moscow, 1958; English transl. , Academic Press, New York-London, 1968 . [70] M . A. Shubin, Lectures on equations of mathematical physics. Mosk. Tsentr Nepr. Mat . Obraz. , Moscow, 2001 . (Russian) [71] R. A. Adams, Sobolev spaces. Academic Press, New York, 1 975 . [72] W. Rudin, Functional analysis. McGraw-Hill, New York, 1 973. [73] G. E. Shilov. Mathematical analysis. Second special course. Nauka, Moscow, 1965 . (Russian) [74] S. Mizohata, Theory of equations with partial derivatives. Cambridge Univ. Press, Cambridge, 1 973. [75] A. S. Mishchenko and A. T. Fomenko, Course of differential geometry and topology. Mosk. Gos. Univ. , Moscow, 1980. (Russian) [76] F. H. Vasilescu, Analyttc functional calculus and spectral decompositions. Reidel, Dor drecht , 1 982. [77] J. Eschmeier and M. Putinar, Spectral decompositions and analyttc sheaves. Clarendon Press, Oxford, 1996. [78] R. S . Pierce, Assoctative algebras. Springer-Verlag, New York-Berlin, 1982. [79] H. G. Dales , Banach algebras and automatic continuity. Clarendon Press, Oxford, 2000. [80] F. F. Bonsall and J . Duncan , Complete normed algebras. Springer-Verlag, Berlin Heidelberg-New York, 1973. [8 1] P. R. Halmos, A Hilbert space problem book. Van Nostrand, Toronto-New York London, 1967. [82] Topological algebras. Selected Topics. North-Holland, Amsterdam, 1986. [83] E. Beckenstein, L. Narici, and C. Suffel, Topological algebras. North-Holland, Amsterdam, 1 977. [84] N. E. Wegge-Olsen, K-theory and C* -algebras. Oxford Univ. Press, Oxford, 1993. [85] M. F. Atiyah, K-theory. Benj amin, New York-Amsterdam, 1967. [86] G. J . Murphy, C* -algebras and operator theory. Academic Press, Boston, 1990. [87] J . Dixmier, Les C* -algebres et leurs representations. Gauthier-Villars, Paris, 1969. [88] E. G. Effros, Why the circle is connected: An introduction to quantized topology. Math. Intelligencer 1 1 ( 1989) , no. 1 , 27-34. [89] C* -Algebras: 1 943- 1993. A Fifty Year Celebration, R. S . Doran (ed. ) . Contemporary Math. , vol. 167, Amer. Math. Soc. , Providence, RI, 1 994 . [90] G . K. Pedersen, C* -algebras and their automorphism groups. Academic Press, Lon don, 1 979. [91] G . K. Pedersen, Analysis now. Springer-Verlag, Berlin, 1 989. [92] A. Ya. Helemskii, An elementary realization of a nondiscrete factor. Russian J . Math. Phys. 7 (2000) , no. 2, 187-1 94. [93] G . E. Shilov, Mathematical analysts. Special course. Fizmatgiz, Moscow, 1960. (Rus sian) [94] A. I. Markushevich, The theory of analytic functions. Gostekhizdat, Moscow, 1 950; English transl. , AMS/Chelsea, Providence, RI , 1977.
Bibliography
459
[95] V. Runde, Lectures on amenability. Springer-Verlag, Berlin, 2002. [96] H. Reiter and J. D. Stegeman, Classical harmonic analysis and locally compact groups. Clarendon Press, Oxford , 2000. [97] D . P. Zhelobenko, Compact Lie groups and their representations. Nauka, Moscow, 1 970; English transl. , Amer. Math. Soc. , Providence, RI, 1 973. [98] A. A. Kirillov, Elements of representation theory. Nauka, Moscow, 1 972; English transl. , Springer-Verlag, Berlin-Heidelberg-New York, 1 976. [99] N. Bourbaki, Theortes spectrales. Hermann, Paris, 1 967. [100] L. S. Pontryagin, Conttnuous groups. Nauka, Moscow, 1 974; English transl. , Prince ton Univ. Press, Princeton, NJ, 1 939, 1 958. [101] M. A. Naimark, Normed rings. Nauka, Moscow, 1 968; English transl. , Wolters Noordhoff, Groningen, 1 970. [102] V. A. Zorich, Mathematical analysts, Vol. 2 . Mosk. Tsentr Nepr. Mat. Obraz . , Moscow, 1 998; English transl. , Springer-Verlag, Berlin, 2004. [103] V. L. Klee, Invartant metrics in groups (solution of a problem of Banach}. Proc. Amer. Math. Soc. 3 ( 1 952) , 484-487. [104] N. Th. Varopoulos, Tensor products and harmonic analysis. Acta Math. 1 1 9 ( 1 967) , 51-52. [105] R. Palais, Seminar on the Attyah-Singer tndex theorem. Princeton Univ. Press, Princeton, N J, 1 965. [106] W. Rudin, Real and complex analysis, 2nd ed. McGraw-Hill, New York-Dusseldorf Johannesburg, 1 974. [107] A. Fraenkel and Y. Bar-Hillel, Foundations of set theory. North-Holland, Amster dam, 1 958. [108] Functional analysis, 2nd ed. S. G. Krein (ed. ) , Nauka, Moscow, 1 972 . (Russian) [109] J. Lindenstrauss and L. Tzafriri, On the complemented subspaces problem. Israel J . Math. 9 ( 1 971 ) , 263-269.
Index
*-algebra, 348 *-homomorphism, 348 isometric, 350 *-isomorphism, 348 contraction, 350 isometric, 350 topological, 350 *-representation, 348 *-subalgebra, 349 *-subset, 349 C * -algebra, 350 operator, 351 C * -identity, 330, 35 1 8-function, 285 c--net, 201 €-perpendicular, 206 .X-dilation, 3 lp-sum, 63 of operators, 95 s-numbers, 218 x-shift, 3 absolute value of an operator, 370 adherent point, 10 adjoint associativity law, 1 80 adjointness formula, 329 admissible interval, 376 Alexandroff theorem, 194 algebra, 304 Banach, 311 semisimple, 323 star-, 349 Calkin, 310 involutive, 348 quotient, 310
unital, 304 von Neumann, 354 amenable, 360 enveloping, 358 Wiener, 423 algebraic sum, 3 approximate identity, 430 bounded, 430 countable, 43 1 approximation problem, 214 approximation property, 213 balanced category, 50 balanced set, 3 ball, 248 open, 9 Banach inverse operator theorem, Banach-Alaoglu theorem, 270 Banach-Steinhaus theorem, 1 54 Banach-Stone theorem, 198 base of a topological space, 1 1 basis of an object, 50 Bessel inequality, 74 bicommutant, 353 biextension, 3 bilinear operator completely bounded, 122 contraction, 92 jointly bounded, 92 multiplicatively bounded, 122 Schmidt, 188 separately bounded, 92 bimorphism, 33 birestriction, 3
1 50
461
462 bound,
Index
6
calculus Borel, 384 continuous, 361 on a compact set, 363 entire holomorphic, 32 1 polynomial, 308 rational, 309 canonical isometric operator, 1 10 Cartesian product, 40 category, 21 category of abelian groups, 23 groups, 23 linear spaces, 22 measure spaces, 23 metric spaces, 23 ordered sets, 23 sets, 22 topological spaces, 23 Cauchy-Bunyakovskii inequality, 68 character, 307 group, 449 closed graph theorem, 151 closed set , 1 0 closure, 1 1 codimension, 7 coextension, 3 Cohen's factorization theorem, 430 coisometry, 78 commutant, 353 commutative diagram, 28 compact, 209 compactification, 197 Alexandroff, 200 completion, 168 complex Borel measure, 61 complex-conjugate space, 4 conjugate linear isomorphism, 4 operator, 4 continuous mapping, 1 6 contraction, 9 convergence classical, 245 , 279 coordinatewise, 246 generated by a prenorm, 246 weak, 264 weak* , 264 Weierstrass, 245 , 250 convex set, 3 convolution, 312 of a function and a generalized function, 429
of functions, 424 on a group, 454
of generalized functions, 430 coproduct of a family of objects, 38 corestriction, 3 coretraction, 32 cosimplicial K-object, 51 covering, 191 crust, 278 dense subset , 12 derivative generalized, 289 Sobolev, 289 diagram, 28 diameter, 1 5 Dini condition, 421 direct complement topological, 97 direct product Banach, 127 direct sum Banach, 127 of a family of linear spaces, 41 of subspaces, 97 discrete category, 22 disjoint union, 2 distance between an element and a subset, from a point to a subset, 9 distribution, 282 divisible abelian group, 35 domain of a morphism, 2 1 dual category, 24 dual equivalence of categories, 53 dual space linearly, 4 element adjoint, 348 idempotent, 308 inverse, 305 invertible, 305 nilpotent, 308 normal, 348 positive, 365 selfadjoint, 348 unitary, 348 elementary tensors, 1 75 endomorphism, 22 Enfio theorem, 214 Enfio-Read theorem, 1 40 epimorphism, 33 equation adjoint, 333 Volterra, 320
73
463
Index
tempered, 282 with compact support, 282 Heaviside, 290 locally integrable, 283 rapidly decreasing, 279 resolvent, 316 simple, 393 square-integrable, 67 symmetric, 338 vanishing at infinity, 63 , 193 functional, 4 conjugate bilinear, 66 evaluating, 263 positive, 35 7 regular, 286 functor Banach adjointness (Banach star) , contravariant, 45 covariant, 45 faithful, 46 full, 46 Gelfand, 324 Hilbert adjointness, 330 of continuous functions, 198 weak* star, 275 fundamental sequence, 125
equicontinuous set of functions, 208 equivalent categories, 53 families of prenorms, 253 families of subspaces, 373 measure spaces, 9 prenorms, 87 essentially bounded function, 59 extension, 3 extreme epimorphism, 37 extreme monomorphism, 36 factor, 359 family of prenorms A-weak, 263 compatible, 274 majorizing another family, 253 strong-operator, 24 7, 259 weak, 263 weak* , 263 weak-measure, 383 weak-operator, 248 of spectral types, 404 of subsets centered, 192 of subspaces associated with an operator, 372 final object, 24 forgetful functor, 48 Fourier coefficients, 73, 413 integral, 414 operator, 417 series, 73 transform, 414 Hilbert, 442 , 445 inverse, 421 of a tempered generalized function, on Z, 415 on a group, 45 1 on the circle, 415 Fredholm alternative, 237 theorem, 240 free object, 50 freedom functor, 49, 51 full category, 22 function Borel, 383 finitary, 278 Gauss, 279 generalized, 282 concentrated on, 292 of finite order, 283 regular, 283 singular, 283
439
Gelfand functor, 324 spectrum, 324 theorem, 323 transform, 324 Gelfand-Mazur theorem, 317 Gelfand-Naimark theorem first, 352 second, 352 GNS-representation, 357 Gowers theorem, 138 greatest element, 8 Grothendieck operator, 183 theorem on £ 1 (X, J-l ) , 181 on approximation property, group dual, 450 topological, 449 abelian, 449 locally compact, 449
1 56
214
Hahn-Banach theorem, 1 03 Hellinger's theorem (final form of Hilbert's spectral theorem) , 404 Hermite function, 73 polynomial, 73
464
Hilbert spectral theorem geometric form, 402 in terms of operator-valued Lebesgue integral, 394 in terms of operator-valued Riemann-Stiltjes integral, 380 Hilbert-Schmidt theorem, 341 homeomorphism , 1 7, 25 uniform , 25 homomorphism, 306 involutive, 348 unital , 306 hyperplane, 105 separating subsets, 1 1 3 hyperplane of support , 105 ideal , 309 identity Hilbert 's, 316 image of the mapping, 2 index, 233 inequality multiplicative, 3 1 1 initial object , 24 inner product , 66 integral Lebesgue with respect to a spectral measure, 394 Lebesgue--Stieltjes, 108 integral equation, 239 homogeneous, 239 interior point , 10 invariant for an isomorphism, 26 invariant subspace, 4 inverse morphism , 25 involution, 347 isometry, 9, 78 isomorphic objects, 25 isomorphism, 25 involutive, 348 isometric, 83 complete, 1 18 of (pre) normed spaces, 84 of Banach algebras, 313 of algebras , 306 topological , 83, 258 complete, 1 18 of Banach algebras , 313 unitary of (pre) normed spaces, 84 kernel, 3 kernel (of an integral operator) , 81 Krein-Milman theorem, 273 K uratowski theorem, 197
Index
least element , 8 left inverse morphism, 32 limit of a net , 19 of a sequence, 18 limit point, 10 strict , 10 Lindenstrauss-Tzafriri theorem, 149 linear basis, 7 linear complement , 3 linear dimension, 7 linear isomorphism , 25 linear span, 3 linearly independent system of vectors, 7 mapping bijective, 2 continuous, 9 continuous at a point, 9 injective, 2 isometric, 9 open, 1 7 surjective, 2 topologically injective, 1 7 topologically surjective, 1 7 uniformly continuous, 9 matrix of an operator, 90 maximal element , 6 mean convergence, 60 mean-square convergence, 60 measurable mapping, 8 proper , 8 measurable set , 8 with a countable basis, 8 measure absolutely continuous , 399 Haar, 45 1 left-invariant , 451 personal of a vector , 397 regular, 45 1 spectral , 390 of an operator, 39 1 unimodular, 45 1 measure space, 8 measures equivalent , 399 mutually singular, 404 orthogonal, 404 metric discrete, 13 inherited, 9 metric subspace, 9 Milyutin theorem , 199 Minkowski functional, 64 model, 27 monomorphism , 33
465
Index
mor-functor covariant , 47 multiplication, 304 multiplicative inequality for operator prenorms, 77 multiplicative property of index, 235 multiplicity, 343 natural equivalence, 52 of contravariant functors, 52 natural transformation of covariant functors, 52 neighborhood, 10 net , 19 Nikol'skii theorem , 241 norm, 55 admissible, 164 Hilbert , 70 nuclear, 183, 226 of the nth floor, 1 1 5 operator, 77 quantum, 1 1 5 generated by, 1 1 7 standard, 1 1 6 Schmidt, 188, 221 normed space, 55 open mapping principle, 150 open set , 10, 24 8 operator, 3 n-dimensional, 4 adjoint weak* , 266 associated with a bilinear operator, 1 73 associated with the resolution of the identity, 378 bounded, 76 completely bounded, 1 18 completely contractive, 1 18 completely isometric, 1 18 continuous , 256 contraction, 78 cyclic, 398 diagonal, 79 differentiation, 82 , 2 56 finite-dimensional, 4 Fredholm, 233 generated by, 94 Hermitian, 336 Hilbert-Schmidt , 22 1 homogeneous, 405 idempotent , 98 identity, 78 integral, 81 normal, 337 nuclear, 183, 226 of bilateral shift, 79
of composition, 257 of indefinite integration, 80 of translation by , 82 one-dimensional , 4 positive, 346 Schmidt, 221 selfadjoint , 336 shift , 79 unitary, 84 Volterra, 82 operator equation, 239 homogeneous, 239 operators equivalent isometrically, 89 topologically, 89 unitarily, 90 weakly isometrically, 88 weakly topologically, 88 weakly unitarily, 90 partially isometric , 336 order, 5 discrete, 6 in the algebra B(H) , 368 linear, 6 of a generalized function, 283 ordered set , 6 directed, 1 9 linearly, 6 orthogonal vectors, 71 orthogonalization process , 73 orthonormal system of vectors, 71 .
.
.
Paley-Wiener theorem, 448 parallelogram identity, 70 part of an operator negative, 370 positive, 370 path, 17, 28 Plancherel theorem , 445 point at infinity, 200 discontinuity, 374 extreme, 273 interior, 248 linearly interior, 1 1 3 of increase, 37 4 regular , 297, 305 singular, 297, 305 essentially, 299 polar identity, 66 for a prenorm , 70 polynomial of an element of an algebra, 308 Pontryagin's duality principle , 452 pre-inner product, 66
466 premetric , 14 prenorm, 55 accompanying, 24 7 admissible, 278 generated by a (pre-)inner product , 70 Hilbert , 70 majorizing, 87 of a bilinear operator, 92 operator , 76 projective, 1 78 standard, 278 prenormed space, 55 presheaf, 52 primitive generalized, 291 principle extension-by-continuity, 131 of nested closed subsets, 126 product of a family of objects, 38 proj ection, 98 proper subset , 3 Pythagorean equality, 71 quadratic form, 66 , 336 quantization, 1 1 6 maximal , 1 1 7 minimal, 1 1 7 quotient (pre) norm, 57 quotient space topological, 1 2 Rademacher functions, 72 radius spectral, 319 range of a morphism , 21 rarefied set , 126 reflection orthogonal, 335 representation, 307 involutive, 348 universal, 358 representing object , 54 resolution of the identity, 375 restriction, 3 retraction, 32 Riemann-Stiltjes integral , 376 integral sum, 376 Riesz theorem, 1 09, 1 44 on compactness of unit ball, 207 Riesz-Fischer theorem, 1 34 right inverse morphism, 32 Sakai theorem, 355 Schauder basis, 75
Index
Schmidt series , 218 Schmidt theorem , 2 1 5 sequence 8-shaped, 287, 427 in a category, 160 dual , 160 exact , 160 short , 161 square-integrable, 67 two-sided, 413 rapidly decreasing, 437 set fat , 126 meager, 126 shift of a set , 45 1 similar endomorphisms, 29 simplicial K-object , 51 singleton , 14 space v, 278 £ , 279 s , 279 accompanying, 24 7 Banach, 1 26 countably normed, 24 7 dual, 77 Frechet , 255 Hilbert , 126 locally convex, 250 metric , 9 totally bounded, 201 normed quantized, 1 1 5 underlying, 1 16 polynormed, 246 normal , 275 strongest , 247 weakest , 2 57 pre-Hilbert , 67 predual, 354 premetric , 1 5 bounded, 1 5 prenormable, 254 quantized maximal, 1 1 7 minimal, 1 1 7 quantum, 1 1 5 reflexive, 1 1 1 Schwartz, 279 second dual , 1 10 topological vector, 250 Varopoulos, 183 spectral mapping law for an entire calculus, 322 for continuous calculus, 364 spectral picture, 402
467
Index
ordered, 404 spectral radius formula, 319 spectral type, 399 spectral types independent , 404 spectrum, 297 continuous, 298 essential, 299 Gelfand, 324 of an operator , 305 point , 298 residual, 299 square root (of an operator) , 366 standard simplicial category, 24 state, 358 step-function, 377 strictly positive functional, 357 subalgebra, 309 subbase of a topological space, 1 1 subcategory, 22 subcovering, 191 subnet , 192 subset bounded, 254 compact , 191 in an ordered set bounded, 6 relatively compact , 199 selfadjoint , 349 totally bounded, 201 subspace cyclic, 397 in the algebraic dual space sufficient, 262 polynormed, 24 7 quantum, 1 16 topological, 1 1 topologically complemented, 97 sum Hilbert of Hilbert spaces, 400 of operators , 401 of linear spaces, 3 support , 292 of a Borel measure, 404 tensor product algebraic , 1 73 Banach, 177 functor of, 183 of operators, 182 Hilbert , 187 functor of, 190 projective, 1 77 of prenorms, 178 tensor trace, 184
theorem on Borel functional calculus, 385 on continuous functional calculus , 361 on existence of algebraic tensor product , 1 75 of completion, 1 70 of projective tensor product , 1 79 on injectivity of Fourier transform , 452 on the inversion of Fourier transform, 436 on the stability of index, 325 on the sufficiency of the set of characters, 452 on uniqueness of algebraic tensor product , 1 74 of completion, 169 of product in a category, 39 of projective tensor product , 1 77 topological space, 10 antidiscrete, 12 compact , 191 discrete, 12 Hausdorff, 14 locally compact , 199 metrizable, 13 path-connected, 1 7 separable, 12 topology, 10 A-weak, 263 antidiscrete, 1 2 discrete, 12 Dixmier, 27 5 generated by a (pre )norm, 56 generated by a family of prenorms, 249 inherited, 1 1 Pontryagin, 449 Tychonoff, 42 ultraweak, 275 weak, 263 weak* , 263 Zariski , 16 total subset (system) , 74 trace, 184, 227 trace class operator, 226 trigonometric system , 72 Tychonoff product , 43 theorem, 196 topology, 42 uniform norm , 58 unit ball, 64 unit sphere , 64 unit vector, 4 unity of an algebra, 304
468 universal property of Banach tensor product , 1 79 of the algebraic tensor product , 1 73 variation, 61 vector cyclic , 398 large, 407 von Neumann theorem on bicommutant , 353 on commutative von Neumann algebras , 355
Index
von Neumann-Jordan theorem, 71 Walsh system, 72 weakly similar morphisms, 31 Weierstrass convergence, 245 , 250 test , 131 Zariski topology, 16 zero of a category, 24 operator, 78
Titles in This Series 233 231 230 229 228
Lectures and exercises on functional analysis, 2006 K iyosi Ito , Essentials of stochastic processes, 2006 Akira K ono and Dai Tamaki , Generalized cohomology, 2006 Yu . N. L in'kov , Lectures in mathematical statistics , 2005 D. Zhelobenko , Principal structures and methods of representation theory, 2006
227
Takahiro K awai and Yoshits ugu Take i ,
A . Ya. Helemski i ,
Algebraic analysis of singular perturbation
theory, 2005 226 225
V. M. Manuilov and E. V . Troitsky,
Hilbert C* -modules, 2005
Moduli of Riemann surfaces, real algebraic curves, and their
S. M. Natanzo n ,
superanaloges, 2004 224 223 222 22 1 220 219 218
Stochastic analysis , 2004 Masatoshi Noumi , Painleve equations through symmetry, 2004 G . G . Magaril-Il'yaev and V. M . Tikhomirov , Convex analysis: Theory and applications, 2003 Katsuei Kenmotsu, Surfaces with constant mean curvature, 2003 I. M . Gelfand , S . G . G i ndikin , and M . I . G raev , Selected topics in integral geometry, 2003 S . V . Kerov , Asymptotic representation theory of the symmetric group and its applications to analysis, 2003 Kenji U eno , Algebraic geometry 3: Further study of schemes, 2003 Ichiro Shigekawa,
217 216 215
Tatsuo K i mura,
214
L.
213 212
V. N. Sachkov and V . E. Tarakanov ,
D-modules and microlocal calculus, 2003 Quasipower series and quasianalytic classes of functions , 2002 Introduction to prehomogeneous vector spaces , 2003
Masaki Kashiwara, G . V . Badalyan ,
S.
Grinblat ,
Algebras of sets and combinatorics, 2002
A. V. Mel'n ikov , S . N. Volkov , and
Combinatorics of nonnegative matrices , 2002 M . L . Nechaev, Mathematics of financial
obligations , 2002 211 210 209
Takeo O hsawa,
204
201 200
Cohomological analysis of partial differential equations and Secondary Calculus, 2001 Te S un Han and Ki ngo K obayashi , Mathematics of information and coding, 2002 V. P. Mas lov and G . A . O mel 'yanov , Geometric asymptotics for nonlinear PDE. I, 2001 Shigeyuki M orita, Geometry of differential forms , 2001 V. V. P rasolov and V. M. T ikhomirov , Geometry, 2001
199
Shigeyuki Morita,
198 197 196
Simplicial and operad methods in algebraic topology, 2001 Kenj i Ueno , Algebraic geometry 2 : Sheaves and cohomology, 2001 Yu. N . Lin'kov , Asymptotic statistical methods for stochastic processes, 2001
195 194
Minoru Wakimot o ,
Analysis of several complex variables, 2002 Toshitake Kohno , Conformal field theory and topology, 2002 Yasumasa Nishi ura, Far-from-equilibrium dynamics , 2002 208 Yukio Matsumot o , An introduction to Morse theory, 2002 207 Ken' ichi Ohshika, Discrete groups, 2002 206 Yuj i Shimizu and Kenj i Ueno , Advances in moduli theory, 2002 205 S e iki Nishikawa, Variational problems in geometry, 2001
203 202
A. M . Vinogradov ,
Geometry of characteristic classes, 2001
V. A . S mirnov ,
Valery B.
Infinite-dimensional Lie algebras, 2001 Nevzorov , Records: Mathematical theory, 2001
TITLES IN THIS SERIES 193 1 92
Toshio Nishino, Yu . P. Solovyov
Function theory in several complex variables, 2001 and E. V. Troitsky , C* -algebras and elliptic operators in differential
topology, 2001 191 1 90 189 1 88 187 186 185 184 183 182 181 180 1 79 1 78 1 77 1 76 1 75 1 74 1 73 172 171 1 70 1 69 1 68 167 1 66 165 164 163
Methods of information geometry, 2000 A lexander N . Starkov , Dynamical systems on homogeneous spaces, 2000 M itsuru lkawa , Hyperbolic partial differential equations and wave phenomena, 2000 V. V. B uldygin and Yu . V. Kozache nko , Metric characterization of random variables and random processes, 2000 A . V . Fursikov , Optimal control of distributed systems. Theory and applications, 2000 Kaz uya Kat o , Nobushige K urokawa, and Takeshi Sait o , Number theory 1 : Fermat' s dream, 2000 Kenj i Ueno , Algebraic Geometry 1 : From algebraic varieties to schemes, 1999 A. V. Mel'nikov , Financial markets, 1999 Haj ime Sat o , Algebraic topology: an intuitive approach, 1999 I. S . K rasil'shchik and A . M . Vinogradov , Editors , Symmetries and conservation laws for differential equations of mathematical physics, 1999 Ya . G . Berkovich and E . M. Z hmud ' , Characters of finite groups. Part 2, 1999 A. A . M i lyut i n and N . P. Osmolovskii, Calculus of variations and optimal control, 1998 V. E . Voskresenskil, Algebraic groups and their birational invariants , 1998 M itsuo M orimot o , Analytic functionals on the sphere, 1998 Satoru lgar i , Real analysis-with an introduction to wavelet theory, 1998 L . M . Lerman and Ya. L . Umanskiy , Four-dimensional integrable Hamiltonian systems with simple singular points (topological aspects) , 1998 S . K . God unov , Modern aspects of linear algebra, 1998 Ya- Z he C hen and Lan- C heng Wu, Second order elliptic equations and elliptic systems, 1998 Yu . A . Davydov , M . A. L ifshit s , and N. V. S morod ina, Local properties of distributions of stochastic functionals, 1 998 Ya . G . Berkov ich and E . M . Z hmud ' , Characters of finite groups. Part 1 , 1998 E . M . Land is , Second order equations of elliptic and parabolic type, 1998 Viktor P rasolov and Yuri So lovyev, Elliptic functions and elliptic integrals, 1997 S . K . God unov , Ordinary differential equations with constant coefficient , 1997 J unj iro Noguch i , Introduction to complex analysis , 1998 Masaya Yamagut i , M asayoshi Hata, and J un K igami , Mathematics of fractals, 1997 K e nj i Ueno , An introduction to algebraic geometry, 1997 V . V . lshkhanov , B. B. L ur'e, and D. K. Faddeev , The embedding problem in Galois theory, 1997 E . I . Gordon, Nonstandard methods in commutative harmonic analysis, 1997 S hun- ichi Amari and Hiroshi Nagaoka,
A . Ya . Dorogovtsev, D . S . S i lvest rov , A. V. S korokho d , and M . I. Yadrenko ,
Probability theory: Collection of problems , 1997 162
M. V . Boldin, G . I . S i monova, and Yu . N. Tyurin,
Sign-based methods in linear
statistical models , 1997 161 M ichae l B lank, Discreteness and continuity in problems of chaotic dynamics, 1 997 160 V. G . O smolovskil', Linear and nonlinear perturbations of the operator div, 1997
For a complete list of titles in this series, visit the AMS Bookstore at www.ams.org/bookstore/ .