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0. Then f(B(xo, r)) = f(x0) +rf(B(X)) does not contain 0, so f(B(X)) is not the entire ground field. Hence, as we have seen, f is bounded.
Since 1(f) is a translate of K(f), it is dense if K(f) is dense. (b) This is immediate from (a) and Theorem 1(c).
In fact, B(x0, r) fl K(f) = namely
the
I IJxII/r
bound
0 implies a bound on the norm of f,
f(xo)I/r. for some x E X and so Ilfil
=
0
—
Indeed, otherwise
If(x)I >
r)
and f(y) = 0, contradicting our assumption. Now we turn to one of the cornerstones of elementary functional analysis, the Hahn—Banach theorem which guarantees that functionals can be extended from subspaces without increasing their norms. This means that all the questions posed at the beginning of the chapter have reassuring answers. Although the proof of the general form of the Hahn—Banach theorem uses Zorn's lemma, the essential part of the proof is completely elementary and very useful in itself.
Let YCX be vector spaces and let f€X' and gE 1". Iff(y) = for all y E Y (i.e. flY, the restriction off to Y, is g) then f is an extension of g. We express this by writing g C f. A function on a real vector space X is said to be a convex p: X-+ = functional if it is positive homogeneous, i.e. p(zx) = tp(x) for all t 0 g(y)
Chapter 3: Linear functionals and the Hahn—Banach theorem
48
and x E X, and is a convex function (as used in Chapter 1), i.e. if x,y E X, and 0 t 1 then p(tx+(1—Oy) tp(x)+(1—Op(y). By the positive homogeneity of p, the second condition is equivalent to p(x+y) p(x) +p(y) for all x,y E X, i.e. to the subadditivity of p. As customary for the operations on R = we use the convention =
=
that
for all
s
ER;
= 0; and
=
fort>0.
Note that a norm is a convex functional, as is every linear functional. Furthermore, if X = (V. II' fi) is a normed space then a linear functional f E X* and f E X' is dominated by the convex functional NIIxII S —+ R is said to dominate a function N. As usual, a function liffi
if
i/r: S
R if i/i(s)
—+
q'(s) for all S E S.
Lemma 3. Let p be a convex functional on a real vector space X and let fo be a linear functional on a 1-codimensional subspace Y of X. Suppose that fo is dominated by p, i.e. p(y)
fo(Y)
for all y E Y.
Then fo can be extended to a linear functional f E X' dominated by p: f(x)
p(x)
for all x E X.
Proof. Fix z E X (z 1') so that every x X has a unique representation in the form x = y + tz, where y E Y and t E R. The functional f we are looking for is determined by its value on z, say f(z) = c. To prove (1), we have to show that for some choice of c we have
f(y+tz) in other words
f0(y)+tc
p(y+iz)
for all y E Y and t E R.
For t> 0 inequality t=
—s
(2)
gives an upper bound on c, and for
0, (2) becomes
for all y
Y. For s > 0 we have fo(Y) —Sc
C> for all y E Y. The former holds if
p(y—sz) and so
Chapter 3: Linear functionals and the Hahn —Banach theorem c
49
p(y' +z)—f0(y')
for all y' E Y, and the latter holds iff c
—p(y" — z) + fo(Y")
for all y" E Y. Hence there is an appropriate c 1ff
p(y'+z)—f0(y')
(3)
for all y',y" E Y. But (3) does hold since f0(y')+f0(y") =
p(y'+y")
p(y'+z)+p(y"—z),
0
completing the proof. The following theorem is a slight strengthening of Lemma 3.
Theorem 4. Let Y be a subspace of a real vector space X such that X is the linear span of Y and a sequence z1 , Suppose fo E Y' is dom-
mated by a convex functional p on X. Then fo can be extended to a linear functional f X' dominated by p. If X is a real normed space and Jo E r then to has an extension to a functional f on X such that 11111 = Ilfoll. Proof. Set K, = lin{Y,z1,.. . ,z,,}. By Lemma such that C 12 C fUflCtioflalS to C
inated by p. Define f: X
3 we can define linear and each f,, is dom-
by setting f(x) = f X' extends Jo and it is dominated by p. The second part is immediate from the first. Indeed, fo is dominated where N = Ilfoll. Hence there by the convex functional p(x) = R
is an f E Xt extending to and dominated by p. But then f(x) N = Iltoll, implying NfIxII for all x E X, so that Ilfif
p(x) =
11111
IltoIl.
=
0
restriction on Y in Theorem 4 is, in fact, unnecessary. As we shall see, this is an easy consequence of Zorn's lemma, the standard weapon of an analyst which ensures the existence of maximal objects. For the sake of completeness, we shall state Zorn's lemma, but before The
doing so we have to define the terms needed in the statement. A partial order or simply order on a set P is a binary relation
that (i) a
a for every a
P. (ii) if a
b and b
such
c for a,b,c E P
forsomea,bE Pthena = Briefly,
b.
is a transitive and reflexive binary relation on P. We call the
50
Chapter 3: Linear functionals and the Hahn—Banach theorem
pair (P. a partially ordered set; in keeping with our custom concerning normed spaces and topological spaces, (P, is often abbreviated to P. A subset C of P is a chain or a totally ordered set if for all a, b E C we have a orb a. Anelementm E Pisamaximalelejnentof P if m a implies that a = m; furthermore, we say that b is an upper foralisES. Itcan beshownthatthe bound axiom of choice is equivalent to the following assertion. Zorn's lemma. If every chain in a non-empty partially ordered set P has an upper bound, then P has at least one maximal element. 0
The fact that Theorem 4 holds for any subspace Y of X is the celebrated Hahn—Banach extension theorem.
Theorem 5. Let Y be a subspace of a real vector space X and let fo E Y'. Let p be a convex functional on X. If fo is dominated by p on Y, i.e. fo(y) p(y) for every y E Y, then Jo can be extended to a linear functional f E IC dominated by p.
If X is a real normed space and Jo E r then fo has a normpreserving extension to the whole of X: there is a functional f E X* such that fo C f and 11111 = Ilfoll.
= {f,,: y E 1) of all extensions of fo dominated by p: for each y there is a subspace V,, and a linear functional E such that Y C Y,,, fo C and f., is dominated by p. Clearly is 'less than or equal to' fo if the relation 'C' is a partial order on is a non-empty chain (i.e. - a totally = if,,: y E C fe). If Proof. Consider the set
ordered set) then it has an upper bound, namely f E Y', where Y,, and f(y) = f,,(y) if y E Y, (y E Fe). Therefore, by = Zorn's lemma, there is a maximal extension. But by Lemma 3 every maximal extension is defined on the whole of X. The second part follows as before. 0 With a little work one can show that norm-preserving extensions can be guaranteed in complex normed spaces as well. A complex normed space X can be considered as a real normed space; as such, we denote it by XR. We write for the dual of Xft. It is easily checked that the
mapping r: r
defined by r(f) = Ref (i.e. r(J)(x) = Ref(x) for
r
x E X) is a one-to-one norm-preserving map onto
The inverse of r —' is the map c: defined by c(S) (x) = g(x) — ig(ix). This enables us to deduce the complex form of the Hahn—Banach extension theorem.
Chapter 3: Linear functionals and the Hahn—Banach theorem
51
Theorem 6. Let Y be a subspace of a complex normed space X and let Then fo has a norm-preserving extension to the whole of X: fo such that fo C and 11111 = 111011. there is a functional f E
r
Proof.
f
By Theorem 5, we can extend r(f0) to a functional g on XR The complex functional f = c(g)
satisfying lid = IIr(fo)II =
Er 0
extends fo and satisfies IlfIl = IIfo II.
The Hahn—Banach theorem has many important consequences; we give some of them here. Corollary 7. Let X be a normed space, and let x0 X. Then there is a such that f(x0) = IIxoII. In particular, Iixoii C 1ff functional f C C for all g 5(r). ig(xo)P
Proof. We may assume that x0
Let Y be the 1-dimensional sub-
0.
space lin{x0} and define to E VS by f0(Ax0) =
A IixoII.
Then
and
1
its extension f, guaranteed by the& Hahn—Banach theorem, has the
0
required properties.
Corollary 8. Let X be a normed space, and let x0 C X. If f(x0) =
allfErthenx0=0.
0
for
0
The functional f whose existence is guaranteed by Corollary 7 is said to be a support functional at x0. Note that if x0 C 5(X) and f is a support functional at x0 then the hyperplane 1(f) is a support plane of the convex body B(X) at x0; in other words: x0 C B(X) flI(f) and 1(f) contains no interior point of B(X). The norm on X is said to be smooth if every x0 C S(X) has a unique support functional. T
Corollary 7 implies that the map given by Y) —' is an isometry, as remarked after Theorem 2.4, when we
r,
defined the adjoint.
Theorem 9. If X and Y are normed spaces and T C
r
and itrii
Y) then
= 11711.
Proof. As usual, we may and shall assume that X and Y are non-trivial spaces: X {0} and V {0}. We know that fi r 0, liii. Given there is an x0 S(X) such that if Tx011 11111 — e. Let g C S(VS) be a support functional at Tx0: g(Txo) = iITxoll. Then
(Tg)(x0) = g(Tx0) = so that
lIrgIl
and iirii
IlTxoII
IIi1I—€.
11711
52
Chapter 3: Linear functionals and the Hahn—Banach theorem
Given a vector space V with dual V' and second dual V" = (Vt)', there is a natural embedding V V" defined by v v", where v" is defined by v"(f) = f(v) for f C V. Rather trivially, this embedding is an isomorphism if V is finite dimensional. If X is a normed space with dual r, second dual Xt. and x C X, then we write i for the restriction of x" to X*: I is the linear functional on given by 1(f) = f(x) for In other words, with the bracket notation, fC
(I,f) = (f,x) for all f C X*. Since Ii(f)I = If(x)I IlfIllIxIl, we have I C (not just I E (X*)I), and moreover Dxli. The Hahn—Banach theorem implies that, in fact, we have equality here. Theorem 10. The natural map x I is a norm-preserving isomorphism (embedding) of a normed space X into its second dual X**.
Proof. For x C X (x 0), let f be a support functional at x: lifil = 1 and f(x) = ilxII. Then ii(f)i = If(x)i = iixli and 11(1)1 IIIIiiifiI = so
that iIxli
hID.
0
In view of Theorem 10 it is natural to consider X as a subspace of
X the whole of X**, i.e. then X is said to be reflexive. We know that X* * is complete
X= even when X is not, so a reflexive space is necessarily complete. However, a Banach space need not be reflexive. For example, 1, is reflexive for 1
Osuchthat[—€x,exJCA.DefinefunctionspandqonXby Proof. We may assume that a =
0, i.e.,
setting, for XE X,
0: xE tA};
p(x) = inf{t It is
0: XE tB}.
q(x) = sup{t
easily checked that p: X
R is
a convex functional and
is a concave functional. Furthermore, as tA fl tB = 0 for t> 0, we have q(x) p(x). Hence, by Corollary 12, there is a non-zero linear functional f E X' such that q(x) f(x) p(x) for all x E X. To complete the proof, note that if x E A and y B then
q: X
f(x)
Hence we may take c =
p(x)
1
q(y)
f(y).
0
1.
As the last result of this chapter, we shall show that the separation theorem gives a pleasant description of the closed convex hull of a set in a normed space. The convex hull co S of a set S in a vector space X is
the intersection of all convex subsets of X containing S, so it is the unique smallest convex set containing S. Clearly,
coS
t1x1
:
S, t,
0 (i = 1,...,n),
t1
=
1
(n =
=
if X is a normed space then the closed convex hull S of S is the intersection of all closed convex subsets of X containing S, so it is the unique smallest closed convex set containing S. As the closure of a convex set S is the closure of co S. is convex,
The following immediate consequence of the separation theorem is the intersection of all closed half-spaces containing S. shows that It is, of course, trivial that S is contained in this intersection.
Theorem 14. Let S be a non-empty subset of a real normed space X. = {x E X: f(x) su2f(s) for allf E
Then
SE.)
S. Then B(x0, r) fl S = 0 for some 0) separate the convex sets B(x0, r) and 5:
Proof. Suppose that x0
r> 0. Let f E X' (f
f(x) for all x E B(x0, r) and y
bounded above,f€ X* and
c
coS. Since the restriction of f to B(xo, r) is O,f(x0) > c.
0
Chapter 3: Linear functionaLc and the Hahn—Banach theorem
56
Throughout the book, we shall encounter many applications of the Hahn—Banach theorem and its variants. For example, the last result will be used in chapter 8. Exercises
1. Let p and q be conjugate indices, with 1 p < Prove that Show also that [Note that this gives a quick = = proof of the fact that for 1
the space is complete.] p 2. Let c be the subspace of consisting of all convergent sequences. What is the general form of a bounded linear functional on c? 3. Let p and q be conjugate indices, with 1 p < Prove that the dual of 1) is Lq(O. 1). is reflexive. Check also 4. Check that for 1
0 for all so
xE S1. we have m >0. By the definition of f, for any xE V we have llxlI
Mllxflj.
0
This theorem has several easy but important consequences.
Corollary 3. Let X and Y be normed spaces, with X finite-dimensional. Then every linear operator T: X —+ Y is Continuous. In particular, every linear functional on X is continuous. Proof. Note that llxD' = llxll + llTxll is a norm on X; since lll and are equivalent, there is an N such that ilxll' Niixll for all x and so
0 Corollary 4. Any two finite-dimensional spaces of the same dimension are isomorphic. Proof. If dim X = dim Y then there is an invertible operator TC Y). As both T and T1 are bounded, X and Y are isomorphic. 0
Chapter 4: Finite-dimensional normed spaces
Corollary 5. Every finite-dimensional space is complete.
Proof. If a space is complete in one norm then it is complete in every equivalent norm. Since, for example, the space is complete, the
0
assertion follows.
Corollary 6. In a finite-dimensional space a set is compact 1ff it is closed
and bounded. In particular, the closed unit ball and the unit sphere are compact.
Proof. Recall that pact set is compact.
is compact, and that a closed subset of a com-
0
Corollary 7 Every finite-dimensional subspace of a normed space is closed and complete. Proof. The assertion is immediate from Corollary 5.
0
In fact, as proved by Frederic Riesz, the compactness of the unit ball characterises finite-dimensional normed spaces. We shall prove this by making use of the following variant of a lemma also due to Riesz. Theorem 8. Let Y be a proper subspace of a normed space X. S(X) whose 0 there is a point x (a) If Y is closed then for every distance from Y is at least 1 —
d(x,Y) = inf{I)x—yII:y€ Y is finite-dimensional then there is a point x E S(X) whose distance from Y is 1.
Proof. Let z E X\Y and set Z = lin{Y, z}. Define a linear functional f: Z R by f(y + Az) = A for y E Y and A E R. Then f is a bounded linear functional since Kerf = V is a closed subspace of Z. By the Hahn—Banach theorem, f has an extension to a bounded linear functional F E X' with IIFII = 11ff > 0. Note that Y C KerF. (a) Let x E S(X) be such that F(x)
(1—€)IIFII.
Then for y E Y we have IIx—yIj
F(x)
F(x—y) Fl
=
1—c.
Chapter 4: Finite-dimensional normed spaces (b)
63
If Y is finite-dimensional then F attains its supremum on the com-
pact set S(X) so there is an x E S(X) such that Rx) = IIF1I. But then for y E Y we have
>F(x-y)F(x)1 UF1I
—
11111
be finite-dimensional subspaces of a normed space, with all inclusions proper. Then there are unit vectors = for all n 2. x1,x2,... such that x,, E X,, and
Corollary 9. Let X1 C X2 C
1
In particular, an infinite-dimensional normed space contains an infinite of 1-separated unit vectors (i.e. with 1 for sequence n
en).
Proof. To find x,, E
apply Theorem 8(b) to the pair (X, Y) =
0 Theorem 10. A normed space is finite-dimensional if and only if its unit
ball is compact.
Proof. From Corollary 6, all we have to show is that if X is infinitedimensional then its unit ball B(X) is not compact. To see this, simply take an infinite sequence x1 , x2,... E B(X) whose existence is 1 for i j. As this sequence guaranteed by Corollary 9: 11x1 —x1fl has no convergent subsequence, B(X) is not compact. 0
Theorem 10 is often used to prove that a space under consideration is finite-dimensional: the compactness of the unit ball tells us precisely this, without giving any information about the dimension of the space.
The above proof of Theorem 10 is based on the existence of a 1 for all i j. Let sequence of unit vectors such that 11x1 —x,II us show that, in fact, we can do better: we can make sure that the inequalities are strict. All we need is the compactness of the unit ball of a finite-dimensional normed space.
Lemma 11. Let x1,... , x,
linearly independent vectors in a real E S(X) normed space X of dimension n 2. Then there is a vector i < n). such that > 1 for all i (1 — be
Proof. We may assume that dim X = n. Let f S(X) be such that i < n. (In other words, we require
f(x1) = 0 for 1
Chapter 4: Finite-dimensional normed spaces
64
K(f) = lin(x1,..
.
and so we have precisely two choices for f: a functional and its negative.) Furthermore, let g X* be such that g(x1) = (1 i < n). Since S(X) is compact, so is
1
for every i
K={xES(X):f(x)= Let we have
K}. Then for 1
x
—x,) =
1.
Since
the choice of x,, tells us that x,, —x, i < n). IIx,,—x111 > 1 for every 1(1
—x1) =
K, so that
i 1 if i j and lin{x1,. .. ,x,j = for n = 1,2
There is another elegant way of finding 1-separated sequences of unit vectors. This time we shall rely on the finite-dimensional form of the Hahn—Banach theorem. Let us choose vectors x1,x2,... E S(X) and support functionals x,x,... as follows. Pick x1 C S(X1) and let E be a support functional at x1 : = 1. Suppose = n, x1,.. . k
0
(i = 1,... ,n).
1=1
Chapter 4: Finite-dimensional normed spaces
a=
a,, and
the volume of D, we have vol D' = Let E be the ellipsoid
Denoting by so
69
1.
E=
E
Then
BC DflD' CE and
(volE)2 (vol D)2
2
—
2a2
—
2a
—
1 1
a is 1. This contradicts the assumption that D was an ellipsoid of minimal volume containing B. 1
Let us turn to the proof of n 112D C B. Suppose that this is not the case, so that B has a boundary point in the interior of n112D. By taking a support plane of B at such a point and rotating B to make this we may support plane parallel to the plane of the axes x2 , x3,. . . , assume that
BC P =
E
lxii
for some c > For
a>
b
> 0 define an ellipsoid Eab E W':
Ea,b so
that (VOl D)/(vol Ea,b)
+ b2
= ab"'. If x C B then x C DflP and so
+ b2 i=2
x? =
+ b2 a
Hence B C
by
and vol Ea,b
+1,2
+b2.
Thus, to complete the proof, it suffices to show that these inequalities
Chapter 4: Finite-dimensional normed spaces
70 are
satisfied for some choice of 0 < b
R
0. R
be a continuous function such that Show that
=
= 0.
be infinitely differentiable. Suppose for every Let f: R such that = 0 for all k Prove that x E R there is a f is a polynomial. and, for n 1, set F,, = 3. Let K = [0,1], X =
{f E X: 3 t E K such that
+ h)
—
f(t)
n V h with t + h E
Deduce that the set of continuous nowhere-differentiable real-valued functions on [0,11 is dense in 4. Prove that if a vector space is a Banach space with respect to two Prove that F,, is closed and nowhere dense in CR(K).
norms then the topologies induced by the norms are either equivalent or incomparable (i.e. neither is stronger than the other).
5. Let V be a vector space with algebraic basis
e1 ,
e2,... (so
and every v E V is a unique linear combination of the dim V = e,) and let II be a norm on V. Show that is incomplete.
6. LetXbeanormed space andSCX. Showthatif{f(x):xES}is bounded for every linear functional f E X then the set S is K for some K and all x E S. (Using fancy terminology that will become clear in Chapter 8, a weakly bounded set is norm bounded.) 7. Deduce from the result in the previous exercise that if two norms on a vector space V are not equivalent then there is a linear functional f E V' which is continuous in one of the norms and discontinuous in the other. bounded: lixil
8. Let X be a closed subspace of L1(0, 2). Suppose for every f E L1(0, 1) there is an F C X whose restriction to (0, 1) is f. 9.
Show that there is a constant c such that our function F can always be chosen to satisfy IIF1I clifli. Let 1 p,q and let A = (a11)' be a scalar matrix. Suppose for every x = (x1)° the series is convergent for
Chapter 5: The Baire category theorem
82
Show that the Ely, where y, = map A: I,, —p 'q' defined by x y, isa bounded linear map. [0,1] (n = 1,2,...) be uniformly bounded continu10. Let ous functions such that
every 1, andy =
j
dx
c
0, (n = 1,2,...) and
for some c > 0. Suppose
c,,q,,,(x) n=1
for
everyxE [0,1]. Prove that n1
11.
cn
and y = (y1)° set
For two sequences of scalars x = (x,y) =
x•y,. i=1
I 0. For every x E K there is a natural number > f0(x) is a non-negative continuous function, As that x has an open neighbourhood (i, such that if y C U then
0. for which {x E L: If(x)l If f and g are functions on the same space, define Af. f+g and fg by pointwise operations. With these operations, C(L) is a commutative functions; similarly.
algebra. This definition,
algebra has a unit, the identically
I
function.
By
C C0(L) C C(L) and CC(L) and C0(L) are subalgebras
of C(L). We know that the vector space C(L) is a Banach space with the uniform (supremum) norm: Ilfil
= sup{lf(x)I: x E L}.
Theorem 6. Let L be a locally compact Hausdorff space. The function
is a norm on C(L) and the space C(L) is complete in this norm; is a dense subspace of C0(L) is a closed subspace of C(L) and C0(L). Furthermore, forf,g E C(L) one has (1)
Ilfil and
I. the identically I function, has norm 1.
is dense in C0(L). Proof. The only assertion needing proof is that e}. Then K0 is = (x E L: If(x)I Given f E C11(L) and >0, set be an open neighbourhood of x with compact. For x E K0 let for Then K0 C compact closure (ix and so K0 C some x1 ,. E K(,. Setting K1 = compact set such that K0 C tnt K1. .
.
we find that K1 is a
By Urysohn's lemma (Theorem 2) there is a continuous function and 0 on K1\lnt K1. Extend h to a funch: K1 [0, 1] that is I on tion g on L by setting it to be 0 outside K1: h(x)
ifxEK1,
0
Then g E
and so gf E
Clearly IIgf—fII
An algebra A which is also a normed space and whose norm satisfies (1) is called a normed algebra. If the algebra has an identity e and the norm of e is I then it is called a unital normed algebra. If the norm is
complete then the algebra is a Banach algebra. We saw in Chapter 2 the set of bounded linear operators on a normed space X, is that is a unital a unital normed algebra and if X is complete then
Chapter 6: Continuous functions on compact spaces
93
Banach algebra. The trivial part of Theorem 6 shows that C(L) is a commutative unital Banach algebra, C0(L) is a commutative Banach is a commutative normed algebra. algebra and are defined analogously; they The spaces CR(L), and consist of the appropriate continuous real-valued functions. These spaces are not only real normed algebras but they are also lattices under the natural lattice operations fvg and fAg. Given real-valued functions f and g on a set S, define new functions fvg and fAg on S by setting, for x E S,
(fvg)(x) = max{f(x),g(x)}
and
We call fvg the join off and g and fAg
(fAg)(x) = min{f(x),g(x)}. the
meet off and g. The join
and the meet are the lattice operations.
Of course, the join of two functions is just their maximum and the and meet is just their minimum. It is clear that CR(L), are all closed under the lattice operations. For a set A of bounded functions on a set S. the uniform closure A of the space of A is the closure of A in the uniform topology on bounded functions on S with the uniform norm (Example 2.1 (ii)). Thus a function f on S belongs to A if for every 0 there exists a g E A such that If(x) <e for all x E S. In this chapter we shall study the uniformly closed subalgebras of C(L), C0(L), CR(L) and so we are particularly interested in uniform approximations by elements from various subsets of these algebras. As the next result shows, the lattice operations are very useful when we look for such approximations. Theorem 7. Let K be a compact Hausdorif space and let A be a subset of which is closed under the lattice operations. Then A, the uniform closure of A, is precisely the set of those continuous (real-valued) functions on K that can be arbitrarily approximated at every pair of points by functions from A.
Remark. Note that A is not assumed to be a subalgebra, not even a subspace, it is merely a set closed under the lattice operations, i.e. it is a sublattice of CR(K).
Proof. It is obvious that the uniform closure A is at most as large as claimed.
Suppose then that f can be approximated by elements of A at every pair of points. We have to show that f E A.
Chapter 6: Continuous functions on compact spaces
94
Let e > 0. The existence of approximations of f means precisely that E A such that for x, yE K there exists a function
f(x)—E Let
<E•
0 there is a 8 > 0 such that If(x) — f(y) <e whenever d(x, y)
0
and r =
4s(1
—k). Define f:
X by
f(x) = x+€(x). Then there is an open neighbourhood U of 0 such that the restriction of
f to U is a homeomorphism of U onto D,(X). Proof. Define F: D,(X) X
X by F(x,y) = x—e(y).
F satisfies the conditions of Theorem 3. Indeed, (3) holds by assumption; furthermore, if x E thefl Then
IIF(x,0)D and
= 11xe(0)II < r+r = s(1—k),
so (4) also holds. Hence, by Theorem 3, there is a unique function
Chapter 7: The contraction-mapping theorem
g: D,.(X)
105
D3(X) such that g(x) = F(x,g(x)) = x—e(g(x)),
i.e. x = g(x) + €(g(x)) = f(g(x))
for all x E Dr(X). Set U = g(D,.(X)). To complete the proof, we have to check that U is an open neighbourhood of 0 and f: U —+ D,(X) is a homeomorphism. Clearly,f is a one-to-one map of U onto D,(X). Since g is unique, the continuity off implies that U =7 '(D,(X)) is open. As g is also continuous,fis indeed a homeomorphism. Finally,f(0) = e (0) E D,.(X) and so
0
OEMU.
From Theorem 4 it is a small step to the inverse-function theorem for Banach spaces.
In defining differentiable maps, it is convenient to use the 'little oh' notation. Let X and Y be Banach spaces. Given a function a: D —+ Y, where D C X, we write
a(h) =
o(h)
if for every e > 0 there is a S > 0 such that Ila(h)II
eflhfl
h E D and lihil 0.
An inner-product space is a vector space V together with a positive definite hermitian form on V. This positive-definite hermitian form is said to be the inner product or scalar product on V and we shall write it
(.,.) (and, x,y, z
E V and
occasionally, as (,
for
all scalars A
and
= A(x,z)+p.(y,z); (ii) (y,x) = (x,y); (iii) (x,x) 0, with equality if x =
)).
Thus (,) is such that for all
we have
(i)
0.
More often than not, it will not matter whether our inner-product space is a complex or a real inner-product
space.
As the complex case
tends to look a little more complicated, usually we shall work with
132
Chapter 9: Euclidean spaces and Hilbert spaces
complex spaces. By Theorem 1, if (•,•) is an inner product on a vector space V then lxii = (x,x)'12 is a norm on V. A normed space is said to be a Euclidean space or a pre-Hilbert space if its norm can be derived from an inner product. A complete Euclidean space is called a Hubert space. Clearly every subspace of a Euclidean space is a Euclidean space and every closed subspace of a Hilbert space is a Hubert space. The Cauchy—Schwarz inequality states that l(x,y)l lIxIlIlyll; in particular, the inner product is jointly continuous in the induced norm. The inner product defining a norm can easily be recovered from the norm. Indeed, we have the following polarization identities: 4(x,y)
Ox +y112— Ox —y02+ ilIx+ iyll2—illx— iyll2
(3)
if the space is complex, and 2(x,y) =
Ilx+y112—11x112—11y112 =
(4)
the real case. Therefore in a Euclidean space we may, and often shall, use the inner product defining the norm. For this reason, we use the terms 'Euclidean space' and 'inner-product space' interchangeably, and we may talk of orthogonal vectors in a Euclidean space. The complex polarization identity (3) has the following simple extension. If T is a linear operator (not necessarily bounded) on a complex in
Euclidean space then 4(Tx,y) = (T(x+y),x+y)—(T(x—y),x—y)
+i(T(x+iy),x+iy)—i(T(x—iy),x—iy).
(3')
This implies the following result.
Theorem 2. Let E be a complex Euclidean space and let T E such that (Tx,x) = 0 for all x E E. Then T = 0.
Proof. By (3') we have (Tx,y) =
0
for all x,y E E. In particular,
OTxll2=(Tx,Tx)=OforallxeE,soT=0. It
be
0
is worth pointing Out that Theorem 2 cannot be extended to real
Euclidean spaces (see Exercise 2).
In a Euclidean space, the theorem of Pythagoras holds and so does the parallelogram law.
Theorem 3. Let E be a Euclidean space. If x1,...,x,, are pairwise orthogonal vectors then
Chapter 9: Euclidean spaces and Hubert spaces
133
2
xl
=
E then
Furthermore, if x,y
IIx+y112+
= 211x112+211y112.
Proof. Both (5), the Pythagorean theorem, and (6), the parallelogram law, are immediate upon expanding the sides as sums of products. To spell it out, ,
/,,
2
=
\i=1
a
\
a
I
1=1
x.) = =
(x1 , x,)
(x1 ,x1) +
i#j (x1,x,)
11x1112
= and
I!x+y112+ Ix—y112 = (x+y,x+y)—(x—y,x—y) =
211x112+211y112.
In fact, the parallelogram law characterizes Euclidean spaces (see
Exercise 3). This shows, in particular, that a normed space is Euclidean iff all its two-dimensional subspaces are Euclidean. Furthermore, a complex normed space is Euclidean iff it is Euclidean when considered as a real normed space. The last assertion is easily justified without the above characterization of Euclidean spaces. Indeed, if E is an Euclidean space then Re(x.y) is
a real inner product on the underlying real vector space of E and it defines the same norm on E. Conversely, if V is a complex vector space is an inner product on and (.,•) is a real inner product on V (i.e. Vft) such that the induced norm satisfies IIAxII = A IIIxH for all x E V and A E C then
(x,y) = (x,y)—i(ix,y) = (x,y)+i(x,iy)
is a complex inner product on V defining the original norm
and
satisfying Re(x,y) = (x,y). Examples 4. (i) Clearly, (x,y) = is an inner product on 12" and so 12" is an n-dimensional Hubert space.
(ii) Also, (x, ') =
is an inner product on 12 and so 12 is a
separable infinite-dimensional Hilbert space. (iii) Let E be the vector space of all eventually zero sequences of complex numbers (i.e. x = (x,)° belongs to E if x = 0 whenever i is
Chapter 9: Euclidean spaces and Hubert spaces
134
sufficiently large), with inner product (x,y) = dense subspace of 12; it is an incomplete Euclidean space. (iv) It is immediate that
(f,g) is
= Ja
Then E is a
f(t)g(t) dt
an inner product on the vector space C([a, b]); the norm defined is
the 12-norm \1/2
/ 11f112
and
=
(j
If(t)12d1
not the uniform norm. This is an incomplete Euclidean space (see
Exercise 14).
(v) Also,
(f,g) is
= Ja
the inner product on L2(O, 1) =
E
1]: f is measurable and
101 If(i) 12
dt
=OforallxE K)
and the annihilator of a set L C X* in X (or the preannihilator of L) is = {x
X: (x,f) =
0
for ailfE L}.
= K0 and Show that if K and L are subspaces then = where K° is the polar of K and °L is the prepolar of L. Show also
that = (linK)a = (linK)0
=
and
a,, =
a(ljfl L)
=
a(lin L)
= °(lin L).
2. Give examples showing that for a Banach space X and a subspace
L C r, the sets °L and L° need not be equal under the natural inclusion X C Y). Show that X and Y be normed spaces and T E is the closure of Im T. °(Ker 4. A subspace U of a normed space V is said to be an invariant sub-
space of an operator S E if SU C U, i.e. Su E U for all u U. Let X be a normed space and T E Show that a closed subspace Y of X is an invariant subspace of T 1ff Y° is an invariant subspace of r. 5. Let X and Y be Banach spaces and T E Y). Prove that Im T is closed iff Im T* is closed.
6. Let X be a Banach space. Show that for T E T"/n! converges in norm to an element of exp T. Show also that (exp T) * = exp r and if mutes with T then
the series denoted by
S
(exp S)(exp T) = (exp T)(expS)exp(S+ T).
In the exercises below, H denotes a Hubert space.
com-
Chapter ii: Adjoint operators
7. Let T be a bounded linear operator on a Hubert space H. Show that T has an eigenvector iff r has 1-codimensional closed invariant subspace.
8. Let H be a Hilbert space and P E
a projection: P2 =
P.
Show that the following are equivalent: (a) P is an orthogonal projection; (b) P is hermitian; (c) P is normal;
(d) (Px,x) = IIPxII2 for all x E H. 9. Let U E be a unitary operator. Show that Im(U—I) = Im(U*_!) (I) Ker(U—I) = (ii) For n
and
deduce
that
1 set
= Show that
PMX for every x E H, where M = Ker(U—I).
(One expresses this by saying that 5,, tends to
in the strong operator topology.) 10. Show that if T E is hermitian then exp iT is unitary. 11. The aim of this exercise is to prove the Fuglede—Putnam theorem. Suppose that R,S, T E with R and T normal and RS = ST. (i) Show that
(exp R) S = S(exp T). show that
(ii) By considering exp(R* — R) S exp( T—
IRexpR*)Sexp(_r)II (iii) For! E
IlSil.
and A E C set F(A)
= f(exp(AR*)Sexp(_Ar)).
Show that F(A) is an analytic function and IF(A)I
IlfilliSli for every A C. Apply Liouville's theorem to deduce that F(A) = F(0) = f(S) for every A and hence that
exp(AR*)S = Sexp(Ar).
(iv) Deduce that R*S =
Sr.
166
Chapter 11: AdjoinS operators
Notes
The notion of an adjoint operator was first introduced by S. Banach, Sur les fonctionelles Iinéaires Ii, Studia Math., 1 (1929), 223—39. Our treatment of adjoint operators is standard. The Gelfand—Nalmark theorem was proved in I. M. Gelfand and M. A. Nalmark, On the embedding of normed rings into the ring of operators in Hubert space, Mat. Sbornik N.S., 12(1943), 197—213; for a thorough treatment of the subject see S. Sakai, and Springer- Verlag, New York-Heidel-
berg-Berlin, 1971. For the Fuglede-Putnam theorem in Exercise 11, see chapter 41 in P. R. Halmos, introduction to Hilbert space and the theory of spectral multiplicity, Chelsea, New York, 1951.
12. THE ALGEBRA OF BOUNDED UNEAR OPERATORS
In this chapter we shall consider complex Banach spaces and complex unual Banach algebras, as we shall study the spectra of various elements. Recall that a complex unital Banach algebra is a complex algebra A with an identity e, which is also a Banach space, in which the algebra structure and the norm are connected by lieU = 1 and llabII x* in A such that Ilaillibil for all a,b E A. If there is an involution x = x*+ye, = Ax*, (xy)* = y*x* and llx*xfl = = x, 11x112 then
As we noted earlier, if X is a complex
A is a
is a complex unital Banach algebra, and if H is Banach space then a complex Hilbert space then is a
An element a of a Banach algebra A is
invertible
(in A)
if
ab = ba = e for some b E A; the (unique) element b is the inverse of a, and is denoted by a1. The spectrum of a E A is oA(a) = SpA(a) =
{A
E C: Ae — a is not invertible in A},
and the resolvent set of a is 8A(a)
C\UA(a). A point of ÔA(a) is said to
be a regular point. The function R: ö(a) A given by R(A) = (Ae — a)1 is the resolvent of a. The element ROt) is the resolvent of a at A or, with a slight abuse of terminology, the resolvent of a. The prime example of a Banach algebra we are interested in is the algebra of bounded linear operators on a complex Banach space X; so our algebra elements are operators. In view of this, if T then we define the spectrum and resolvent set of T without any reference to =
{A
E C: Al— T is not invertible}
where I is the identity operator on X, and
p(T) = C\o(T). 167
168
Chapter 12: The algebra of bounded linear operators
If A is a complex unital Banach algebra then A can be considered to the algebra of all bounded linear operators be a subalgebra of acting on the Banach space A, with the element a corresponding to the operator La of left multiplication by a (so that a La, where A is invertible then so La(X) = ax for every x E A). In particular if a with inverse La'. Conversely, if S E is the inverse La E of La, so that X
= (LaS)X = a(Sx)
for every x E A, then with b = Se we have ab = 1 and so a(ba — e) = (ab)a—a = 0. Hence ba—e E KerLa and so ba = e. Thus b is the inverse of a. Also, Ae — a is invertible 1ff A!— La is invertible. Hence crA(a) =
Although the spectrum of an operator T E how T fits into the algebraic structure of
depends only on is of considerable interest to see how the action of T on X affects invertibility. In particular, we may distinguish the points of cr(T) according to the reasons why it
A!— T is not invertible.
What are the obstruction to the invertibility of an operator S E
By the inverse-mapping theorem, S is invertible 1ff KerS = (0) and ImS = X. Thus if S is not invertible then either KerS (0) or ImS X (or both, of course). Of these, the former is, perhaps, the more basic obstruction. Accordingly, let us define the point spectrum of T E =
The elements of
{A
E C: Ker(AI— T)
as
(0)}.
are the eigenvalues of T; for an eigenvalue
the non-zero vectors in Ker(A!— T) are called eigenvectors with eigenvalue A. Furthermore, Ker(AJ— T) is the eigenspace of Tat A. AE
Clearly
C o-(T).
then the two conditions If X is finite-dimensional and S E KerS = (0) and ImS = X coincide. Hence a finite-dimensional space.
However, if X is infinite-dimensional then we may have KerS = (0) and Im S X, so the point spectrum need not be the entire spectrum. More precisely, by Theorem 11.10 , A E o(T) if either Im(A1— T) is not dense in X or A!— T is not bounded below: there is no 0 such that II(AI— T)xII €IIxIl for every x E X. In the former case A is said to belong to the compression spectrum T), and in the latter case, A is
Chapter 12: The algebra of bounded linear operators
169
said to belong to the approximate point spectrum of T, denoted by Uap(T). In other words, =
E C: there is a sequence
{A
C S(X) such that (Al—
Sometimes
Clearly o(T) C
O}.
is called an approximate eigenvector with eigenvalue A. T) and
0(T) =
0ap(T)U(Tcom(T).
the points of the spectrum are classified further: the residual spectrum is 7r(T) (Tcom(T)\C7p(T) and the continuous spectrum is Sometimes
= (7(T)\(Ucom(T)U(Tp(fl). Thus
o(T) = 0p(T)U0c(T)UUr(T), with the sets on the right being pairwise disjoint.
It is immediate from the definition that the approximate point spec-
is a closed set; the point spectrum
trum closed.
need not be
is a non-empty closed subset of the disc We shall show that The latter assertion is an immediate consequence
{A E C: IA I
of the following simple but important result.
Theorem 1. Let TE
r(T)}, relation (3) tells us that the Laurent series
L
n=0
convergent for IA > r(T). Hence, recalling the formula for the radius of convergence, we find that r(T) 0 is
It is easily seen that the spectral radius is an upper semicontinuous function of the operator in the norm topology; in fact,
r(S+7')
r(S)+ lfl
for all S, T E
(see Exercise 8). It is worth recalling that all the results above are true for the spectra of elements of Banach algebras, not only of elements of In the simplest of all Banach algebras, C, every non-zero element is invertible. In fact, C is the only Banach algebra which is a division algebra.
Theorem 10 (Gelfand—Mazur theorem) Let A be a complex unital Banach algebra in which every non-zero element is invertible. Then
A=C.
Chapter 12: The algebra of bounded linear operators
175
Proof. Given a E A, let A E 0(a). Then A —a (= Al—a) is not inverti-
0
bleandsoA—a=0,i.e.a=A.
is invertible 1ff know from Theorem 11.11 that T E is invertible. Hence A!— T is invertible if Al— T is. 1'. E Therefore, recalling that for a Hilbert space H, the dual H is identified A!— T is with H by an anti-isomorphism, we get that for T E invertible if (A!— T)* = A!— r is invertible. Finally, recalling from
We
Theorem 11.8(b) that for a normal operator T we have IIT"II = 1, we have the following result. n
11111"
for
Theorem 11.
(a) For a Banach space X and an operator T E u(T') = 0(T). (b) For a Hilbert space H and an operator T E
we have
we have
a(r) = conju(T) = {A: A C (7(T)}. (c) If T is a normal operator on a Hilbert space then r(T) = II Tfl.
0
Let us introduce another bounded subset of the complex plane associated with a linear operator. Given a Banach space X and an operator TC
the (spatial) nwnerical range of T is
V(T) = {(Tx,f): XE X, f C X, lIxIl = 0111 = f(x) = 1}. With the notation used before Lemma 8.10,
V(T) = {f(Tx): (x,f)
fI(X)}.
Thus to get a point of the numerical range, we take a point x of the unit sphere S(X), a support functional f at x, i.e. a point of the unit sphere
S(Xt) taking value I at x, and evaluate f at Tx. It is clear that the numerical range depends on the shape of the unit ball, not only on the algebra If T is an operator on a Hubert space then V(T) is just the set of values taken by the hermitian form (Tx, x) on the unit sphere: V(T) = {(Tx,x): lixil = 1}. Nevertheless, the numerical range can be easier to handle than the spectrum and is often more informative. It is clear that, just as the spectrum, V(T) is contained in the closed disc of centre 0 and radius 11711. Even more, the closure of V(T) is sandwiched between 0(T) and this disc. But before we show this, we prove that can be only a little bigger than V(T). Theorem 12. For we have TC
a complex Banach space X and an operator
Chapter 12: The algebra of bounded linear operators
176
C V(T),
V(T) C
where V(T) is the closure of V(T).
Proof. The first inclusion is easily seen since if x E S(X), f E = (x, T'f) = and (x,f) = 1 then i E S(X**), (f,i) = I and where, as earlier, i denotes x considered as an element of To see the second inclusion, let C so that there are f C and C S(X**) such that (f,p) = 1 and = Let 0 < e
0.) Deduce that if A C aeo V(T) and (Al— T)2z = 0 then (Al—T)z = 0. is said to be hermitian if V(T) C R. (Note 21. An operator T E space; for a Hubert space this definition cointhat X is a Banach
cides with the usual (earlier) definition of a hermitian operator on a Hubert space.) Thus T is hermitian if both iT and —iT are dissipative. Prove that T is hermitian if iiexp(iT)xH
= lxii
and x C X, where
for all r C
S'
expS
=
for S C
22. Let H be
a
Hilbert space and let S C Show that
V(S) C
(Sx,y) and
be such that
(Sx,x)"2(Sy,y)'12
deduce that IISxIi2
for all x,y
C
(Sx,x)iiSii
H. Deduce that if 0
V(S)
then 0 C a(S).
Chapter 12: The algebra of bounded linear operators
183
23. Show that if S is a hermitian operator on a Hubert space then V(S) = coo(S).
(Note that the assertions in the last two exercises are easily deduced from Theorem 11 (c) as well.)
24. Prove that the result in the previous exercise holds for a normal operator S on a Hilbert space. 25. Let M be a proper ideal of a unital Banach algebra B. Show that the closure of M is also a proper ideal. Deduce that every maximal ideal of B is closed.
26. Let M be a maximal ideal of a complex unital Banach algebra B. Show that B/M is also a complex unital Banach algebra.
The aim of the next five exercises
is
to prove the commutative
Gelfand—Nalmark theorem. In these exercises A is a commutative complex unital Banach algebra.
27. Show that if M is a maximal ideal of A then AIM is isometrically isomorphic to C. 28. Let h: A —' C be a non-zero homomorphism. Show that fihil = 1. 29. Let At be the set of maximal ideals of A. By Exercise 27, At may be identified with the set of non-zero homomorphisms h : A —÷ C. By Exercise 28, this is a subset of B(A*). Give At the relative weak* topology (i.e. the topology induced by the weak* topology on A*). Endowed with this topology, we call At the maximal ideal space of A. Prove that At is a compact Hausdorff space. 30. For x E A the Gelfand transform of x is the function 1: At C defined by i(h) = h(x), where h E At is considered as a homomorphism h : A C. Show that the spectrum o(x) is the range of 1. 31. Show that if A is a commutative unital C-algebra with maximal ideal space At then the Gelfand transformation x '—b maps A isometrically and isomorphically onto C(At), the commutative algebra of continuous functions on the compact Hausdorff space
i
At.
32. Let x=
be the algebra of all (doubly infinite) complex sequences such that =
x,j
with convolution product xy = z, where
Chapter 12: The algebra of bounded linear operators
184
Zn = is a commutative Banach algebra. Show also that the maximal ideal space of 11(Z) can be identified with the unit cirC is defined by cle T = {z E C: Izi = 1}, where z:
Show that
z(x) = 33.
Deduce from the result in the previous exercises that if f(t) =
0
x,j where
=
Let H be a Hilbert space and let T E e > 0 there is an invertible operator S IISTS'II ments such that But then, in particular, E if n m, and so has no convergent subse— TYmI! > quence, contradicting the compactness of T. We claim that, in fact, Y0 = Suppose that this is not so. Then Then, there is an m such that Let U E there exists a pOint v such that as Su SYm, I'm + i u—v E KerS, contradicting Su = Sv. But then S(u—v) = 0 and so 0 our assumption. Consequently = I'0, i.e. SX = A', as claimed. The
=
proof is essentially complete. The bounded map S: X —
1—1 map of the Banach space X onto itself and so, by the
mapping theorem, S is
invertible.
X
is a
inverse-
0
Chapter 13: Compact operators on Banach spaces
193
Let us restate the information contained in Theorems 4 and 7 about the spectrum of a compact operator as a single result.
Theorem 8. Let T be a compact operator on an infinite-dimensional Banach space. Then (T) = {O,A1,A2,...}, where the sequence Ai ,A2,... (of non-zero complex numbers) is either finite or tends to 0; furthermore, every A. is an elgenvalue of T, with finite-dimensional
0
eigenspace.
With some more work, we can gel more detailed information about the structure of compact operators. If T is compact and S = 1— T then Im S is a finite-codimensional closed subspace of X; even more, for a suitable n 1, X is the direct sum of KerS" and ImS". We prove this, and a little more, in the following theorem.
Theorem 9. Let X be a Banach space, T E and S = I— T. Set Nk = KerSk and Mk = ImS" (k = 0,1,...), where = Then is an increasing nested sequence of finite-dimensional subspaces and (Mk
is
a decreasing nested sequence of finite-codimensional subspaces.
There is a smallest n 0 such that N,, = Nm for all m n. Furthermore, M,, = Mm for all m n, Xis the direct sum of M,, and N,,, and M,, is an automorphism of M,,. (1— T)", we see that = I— where Tk is compact. Hence, by Lemma 5, Mk is a closed subspace of X. Clearly N0 C N1 C ... and M0 J J ...; furthermore, we know that each
Proof. By expanding 5" =
Nk is finite-dimensional.
As in the proof of Theorem 7, Lemma 6 implies that there is a smallest n such that N,, = N,, + and there is also a smallest m such that a and Mm = Mm' for all Mm = Mm+i. Then N,, = N,,. for all a' Let us turn to the main assertions of the theorem. We prove first that N,, fl M,,. As y E M,,, we have y = S"x for N,, fl M,, = {0}. Let y some x E X. But as y N,,, S"y = 0 and so = 0. Hence x E N2,, = N,,, implying S"x = 0. Thus y = S"x = 0, showing that
N,,flM,, =
{0}.
We claim that for p =
max{n, m} we have X = N,, Indeed, given x E X, we have But = and so there is a such that = Sex. Hence x—y N,, and so vector y Could we have p > a? x = y + (x — y) shows that X = N,, + Clearly not, since then M,, would strictly contain and so we would
Chapter 13: Compact operators on Banach spaces
194
have
#
Thus p =
{0}.
n
and X is the direct sum of
and
is finite-dimensional (see Exercise 4.20). Finally, = = and as
=
=
N1
C
= {0}.
Hence, by the inverse-mapping theorem, the restriction of S to
is
invertible.
0
Putting Theorems 8 and 9 together, we arrive at the crowning achievement of this chapter: a rather precise description of the action of a compact operator. Theorem 10. Let X be an infinite-dimensional Banach space and let T be a compact linear operator on X. Then o(T) = {0,A1,A2,...}, where the sequence A1,A2,... is either finite or tends to 0. For every A = A there is an integer kA I and closed subspaces NA = N(A; T) and = M(A; T) invariant under T, such that NA is finite-dimensional and MA X = The restriction of Al— T to MA is an automorphism of MA, NA
and for
=
A
= Ker(AI—
A=
A
Ker(Al—
we have NA C MM.
Proof. Only the last claim needs justification. The operator T maps MA into itself and NA into itself. Furthermore, T)INA is an autoof the finite-dimensional space NA since if we had (id— T)x = 0 for some x E NA (x 0) then we would have 0 for every n I, contradicting = (AJ—T)'1x = 0. Consequently, for every n 1, and so NA C MM. = NA morphism
0 There
is no doubt that Theorem 10 is a very beautiful theorem. At
first sight it is not only beautiful but very impressive as well: it seems to come close to giving us a very fine decomposition of the space into a direct sum of generalized eigenspaces. Unfortunately, this is rather a mirage: the theorem cannot even guarantee that our compact operator has a non-trivial closed invariant subspace, let alone give a direct-sum decomposition. In fact, non-trivial closed invariant subspaces do exist, as we shall prove in Chapter 16. However, to prove the existence of invariant subspaces we shall need some results to be proved in Chapter 15.
Before we turn to that, in the next chapter we shall show that a
Chapter 13: Compact operators on Banach spaces
195
best possible decomposition can be guaranteed if we deal with a compact normal operator on a Hubert space. Exercises 1.
A=
li,.: lxi
{x = (x,)°
a compact subset of (1 Prove 2. Let K be a closed and bounded subset p 0 there is an n such that K is compact if and only if for every lx1V
IfW(()1
1=0
Show
that Xk =
1), II
Ilk) is a Banach space and the formal
identity map i: Xk —p Xk_I (f J) is a compact operator. Y), where X is infinite-dimensional. Show. that the 4. Let T E closure of TS(X) = {Tx: x X, lixil = 1} in Y contains 0. (HINT: 1 C S(X) such that for Consider a sequence n m.) be an orthonormal basis of a Hilbert space H and let 5. Let
Y), where Y is
T
6. Let X be every
a normed space.
Show that
a normed space such that for every finite set
0,
0.
A C X and
X has a decomposition
X=
as a direct sum of two closed subspaces, such that M is
finite-
dimensional,
d(a,M)
for every a
0 and every x E X, where PN tion
onto N. Show that
is the canonical projec-
is the norm closure of
i.e.
Chapter 13: Compact operators on Banach spaces
196
every compact operator on X is the operator-norm limit of finite rank operators. 7. Let X be a Banach space with a Schauder basis )'. Show that is the closure of be a sequence of non-zero complex numbers tending to 8. Let 0. Show that cr(T) = {0,A1,A2,. .} for some complex operator T .
on some compact Banach space X. Show that if all A, are real then X can be chosen to be a real Banach space.
9. Let X and Y be Banach spaces, let T E JE
Y), and let
Y) be invertible. Show that Im(J— T) is closed in Y and
has finite codimension.
10. Let X be a complex Banach space, and let T E be such that is compact for some n 1. Show that o(T) = {0, A1, A2,. . where the sequence A1 , A2,... is finite or tends to 0, and every A is an elgenvalue of T. What is the relationship between the sub.
spaces N(A; T) and
Ta)?
11. Let X1 , X2,... be Banach spaces and let X = having norm direct sum of these spaces, with x = Ilixill
be the
=
T
Let
T
E E
for every n.
and let 7, —' T in the 12. Let X be a Banach space, (1, C is relatively comoperator-norm topology. Show that is the unit ball of X. Show also that if pact, where B = B(X) A then A C o(T). 13. Let E be the space C[0, 1], endowed with the Euclidean norm
= 11f112
1
(j
If(x)12 dx)
i.e. T let T be as in Example 1 (iii). Show that T C maps the unit ball of the Euclidean space E into a relatively compact set. Show that T C 14. Let H be a Hilbert space and T C 0 whenever x,, converges weakly to 0, i.e. whenever (x,, , x) —' 0 for every x C H. 15. Construct a compact operator T on 1,, (1 p co) such that cr(T) = {0} and 0 is not an eigenvalue of T. and
Chapter 13: Compact operators on Banach spaces
197
16. Let c3 , c2,... be non-negative reals and
{x E 12: x =
C
for every k}.
xkI
Show that if C is a compact subset of '2 then Ck sequences (Ck
is
0.
For what
C compact?
17k. Let K be a compact subset of a normed space. Show that K is contained in the closed absolutely convex hull of a sequence tending to 0: there is a sequence x,, 0 such that K C C, where
C=
: n = 1,2,...}
A1X1
:
1A11
n=
= Notes
Compact operators were first introduced and applied by Hubert, Grundzuge einer ailgemeinen Theorie der linearen Integraigleichungen,
Leipzig, 1912, and F. Riesz, Les systémes d'équations a une infinite d'inconnus, Paris, 1913, and Uber lineare Funktionalgleichungen, Acta Math., 41(1918), 71—98. In presenting the Riesz theory of compact operators we relied on Ch. xi of J. Dieudonné, Foundations of Modern Analysis, Academic Press, New York and London, 1960, xiv + 361 pp. Theorem 3 is due to J. Schauder, Uber lineare vollstetige funktional Operationen, Studia Math., 2 (1930), 185—96. The first solution of the approximation problem was published by P.
Enflo, A counterexample to the approximation problem in Banach spaces, Acta Math., 30 (1973), 309—17; a simplified version of the solution is in A. M. Davie, The approximation problem for Banach spaces, Bull. London Math. Soc., 5 (1973), 261—6.
14. COMPACT NORMAL OPERATORS
In the previous chapter we saw that for every compact operator T on a Banach space X, the space can almost be written as a direct sum of generalized eigenspaces of T. If we assume that X is not merely a Banach space, but a Hubert space, and T is not only compact but compact and normal, then such a decomposition is indeed possible — in fact, there is a decomposition with even better properties. Such a decomposition will be provided by the spectral theorem for compact normal operators: a complete and very simple description of compact normal operators. Thus with the study of a compact normal operator on a Hilbert space we arrive in the promised land: everything fits, everything works out beautifully, there are no blemishes. This is the best of all possible worlds.
We shall give two proofs of the spectral theorem, claiming the existence of the desired decomposition. In the first proof we shall make use of some substantial results from previous chapters, including one of the important results concerning the spectrum of a compact operator. The second proof is self-contained: we shall replace the results of the earlier chapters by easier direct arguments concerning Hilbert spaces and normal operators. To start with, we collect a number of basic facts concerning normal operators in the following lemma. Most of these facts have already been proved, but for the sake of convenience we prove them again.
be a normal operator. Then the following
Lemma 1. Let T E assertions hold. (a) =
for every x E H.
(b) KerT= Kerr. (c)
198
= 11711" for every n
1.
Chapter 14: Compact normal operators (d) r(T) =
199
11711.
then Ker(AI— T) I Ker(j&I— T).
(e) If A
(f) For every A E C, both Ker(AI— T) and (Ker(A1— T))1 are invari-
ant under both T and r. (g) If H is the orthogonal direct sum of the closed subspaces H0 and H1 invariant under T then with T0 = TIH0 and T1 = nH1 we have max{ll Toll,
11111 =
T, is a normal operator on
and
Proof. (a)
As rr =
II T111}
we have
(rTx,x)
=
(TTx,x) =
(b) By part (a), we have Tx = 0
if rx = 0.
llTxIl2
= (Tx,Tx) =
(c) If S E
= rh-i1 (i = 1,2).
with
(rx,rx)
= llrxll2.
is hermitian then
llSxll2
= (Sx,Sx) = (SaSX,x) = (S2x,x)
IlS2llllxll2.
From this it follows that 115211 = 11S112, and by induction = 11S112'". This implies that IISII = IISII" for every n
on m we get
1. As rr
is hermitian, 11Th2 =
= II(Tr)hh =
=
(d) By (c) and the spectral-radius formula (Theorem 12.9),
r(T) =
urn
II
= urn
11Th
=
for a complex Banach (In fact, (c) and (d) are equivalent: if S E for every n 1.) space X then r(S) = IISII if IlShI =
(e) If Tx = Ax and Ty = T) = Ker(jiI— yE A(x,y) = (Tx,y) =
then ry = jiy because by (b) we have Therefore
(x,ry) = (x,1y) =
then (x,y) = 0. (f) As Al— T commutes with T and r, Ker(AI— T) is invariant under
and so if A
both Tand r. Also, let (x,y) = 0 for all yE Ker(AI—T). Then, since Ker(Al—T) is Invariant under T, for y E Ker(Ai— T) we have (Tx,y)
= (x,ry) = 0.
Hence Tx E (Ker(A1— T))'. Similarly,
200
Chapter 14: Compact normal operators
(Px,y) = (x, Ty)
=0
for every y E Ker(Al— T) and so Px = (Ker(AI—
(h) Letx = h0+h1, with li E H, (i = 0,1). Then llx112 = 11h0112+11h1112,
Tx = Th0+Th1 = T01z0+Th1
and IITxll2 = lIToholI2+ llT1h1ll2 + lIT1 11211h1
max{II
112
II T1112}(11h0112+ 1lh1112)
= max{11T0112, 11T1112}11xlI2.
Thus
max{II T111, II T2!I}. The reverse inequality is obvious.
Finally, as H0 and H1 are invariant under T, it follows that are invariant under P. H0 =
H1 =
0
It is worth emphasizing that Lemma I is a collection of elementary and simple facts, except for part (d), which is based on the spectralradius formula. Let us see then the first incarnation of the spectral theorem, claiming the existence of a spectral decomposition for a compact normal operator. Theorem 2. Let T E be a compact normal operator. For an eigenvalue A of T, let HA = Ker(T— Al) be the eigenspace of T belonging to A, and denote by PA the orthogonal projection onto HA. The operator T has countably many non-zero eigenvalues, say A1 , A2 Furthermore, dim HAk for every k, the projections PA are orthogo1, and nal, i.e. "Ak"A, = 0 if k (1) k
where the series is convergent in the norm of
Proof. By Theorem 13.8 and Lemma 1(e), we have to prove only (1). Given >0, choose n 1 such that lAkI <e fork> n. Set
H1
S, = SIH1 (i = 1,2), we have T0
and
= S0 and S1 = 0.
and Therefore, by
Chapter 14: Compact normal operators
201
Lemma 1(g), lIT—SO = max{JIT0—Soll, llT1—S111} = IITill.
But T1 is a compact normal operator and so, by Theorem 13.8, llT1ll is
precisely the maximum modulus of an eigenvalue of T1. As every
eigenvalue of is an eigenvalue of T, by our choice of n we have 0 Hence (1) does hold. IIT1II Let us state two other versions of the spectral theorem.
Theorem 3. Let T be a compact hermitian operator on an infinitedimensional Hilbert space H. Then one can find a closed subspace of H, a (finite or countably infinite) orthonormal basis of and a sequence of complex numbers v,, 0, such that if x = where E then Tx =
Proof. Let A1,A2,... and HA1,HA2,... be as in Theorem 2. Take a (necessarily finite) orthonormal basis in each and let be the union of these bases. Let H0 be the closed linear span of the orthonorand set v,, = Ak if x,, E HAk. mal sequence 0 Corollary 4. Let T be a compact normal operator on a Hilbert space H. Then H has an orthonormal basis consisting of eigenvalues of T. 0 In fact, compact normal operators are characterized by Theorem 2 (or Theorem 3). Let {x7: y E f) be an orthonormal basis of a Hubert space H, and let T be such that Tx,, = Then T is compact 1ff (2)
for every E > 0 (see Exercise 2).
Our proof of Theorem 2 was based on two substantial results: Theorem 13.8 concerning compact operators on Banach spaces, and the spectral-radius formula. We shall show now how one can prove Theorem 2 without relying on these results. It is a little more convenient to prove Theorem 2 for compact hermuian operators; it is then a simple matter to extend it to normal operators.
Recall that the numerical range V(T) of a Hilbert space operator
TE
is
Chapter 14: Compact normal operators
202
{(Tx,x): x E S(H)} and the numerical radius v(T) is
v(T) = sup{IAI : A E V(T)}.
If T E p.4(H) is hermitian, i.e. r =
T,
then its numerical range is real
since
(Tx,x) = (x,rx) = (x,Tx) = (Tx,x)
every x H and so (Tx,x) is real. In fact, T E is hermitian if its numerical range is real. Also, the spectrum of a hermitian operator is real. We shall not make use of any of the results proved about numerical ranges; the next lemma is proved from first principluses. for
Lemma 5. Let T be a hermitian operator. Then 1111 = v(T). Proof. Set = v(T), so that (Tx,x)I have to show that ill v. Given x S(H), let y E S(H) be (Tx,y) = (x, Ty) = IJTxII and so
for every x
,'
H.
We
(I
IITxIIy.
Then
v, as claimed.
0
6. Let U be a compact hermitian operator on H. Then
U has
ITxH =
(Tx,y) =
=
Hence IITxII Theorem an
such that Tx =
I)y112} = v.
i' for every x E S(H) and
so 11Th
eigenvalue of absolute value
Proof. Set
a = inf (Ux,x) lxii
=1
and
b=
sup
(Ux,x)
11111 = I
that = [a,b]. By Lemma 5, flUfl = max{—a,b}. Replacing U by —U, if necessary, we may assume that hUh = b > 0. We have to show that b is an eigenvalue of U. By the definition of b, there is a sequence C S(H) such that —' b. Since U is a compact operator, by replacing by a so
subsequence,
Then
we may suppose that is convergent, say —' b and = I. As
b because
Chapter 14: Compact normal operators
203
=
=
and —, b,
we have —*0.
Therefore =
Then, on the one hand, Yo = bx0 and, on the Consequently we have Ux0 = bx0. Since 0 (in fact, lixoll = 1), b is indeed an eigenvalue of U.
Put x0 = y0/b.
other hand, IIxofl
1
Ux,, —p Ux0.
Let us now see how Theorem 6 may be used to deduce Theorem 2 for compact hermitian operators. For the sake of variety, we restate Theorem 2 in the following form. Theorem 7. Let H be a Hubert space and let U
be a compact hermitian operator. Then there is a (possibly finite) sequence (Ak) of real numbers and a sequence (Bk) of linear subspaces of H such that (a) Ak .—' 0;
(b) dimHk (c)
(d) if x =
Xk+X, where Xk
Hk and
i E H,' for every k, then
Ux = k
Proof. Let A,., (y E I') be the non-zero elgenvalues of U and let II,, be the eigenspace belonging to A7: H,, = Ker(U—A,,I). We know that H,..1H8 if y 8. and, for every Let us show first that dim H,, 0, there are only finitely many A,, with IA,. I e. Suppose not. Then, by taking an orthonormal basis in each H,, with IA., I e, we find that there is an infinite orthonormal sequence such that Ux,, = where does not contain- a convergent subsequence, €. But then contradicting the compactness of U. I
I
Chapter 14: Compact normal operators
204
This implies that the non-zero eigenvalues may be arranged in a sequence (Ak) such that with = Ker(U—Akl) the conditions (a)—(c) are satisfied. Then, as each H,, is invariant under U, so is the closed linear span M of all the H,, and, consequently, so is M1. Denote by U the restriction of U to M Then U E is also a compact hermitian operator.
As a non-zero eigenvalue of U is also a non-zero eigenvalue of U, it follows from the definition of M and from Theorem 6, that U =
0.
If the sequence (A,,) of non-zero eigenvalues is finite then we are done. Otherwise, let x
=
k1
where Xk E 11k and i E M1.
Xk +
Put
Xk + I
and
= k=I =
Then
and
= k=1
AkXk.
x. As
—.
AkXk
= k=1
H,
0
the continuity of U implies that Ux = y, proving (d).
Before we recover from Theorem 7 the full force of Theorem 2, let us
show that compact hermitian operators are rather like real numbers. An operator T E is said to be positive if it is hermitian and i.e. (Tx,x) 0 for every x E H. Note that if T is any V(T) C (bounded linear) operator on a Hilbert space then rr and are positive (hermitian) operators:
(rTx,x) =
(Tx, Tx)
IITxII2
and
(Trx,x)
= I$Tx112.
Theorem 8. A compact positive operator U on a Hubert space has a unique positive square root V. Every hermitian square root of U is compact.
Proof. Let A 1,A2,... be the non-zero eigenvalues of U, let Hk be the eigenspace belonging to A,, and let M be the closed linear span of the Then = KerU and > 0 for every k. Define V E by Vx = if x E Hk and Vx = 0 if x E M1. Then V is a positive square root of U.
Chapter 14: Compact normal operators
206
,..., be commuting compact nonnal operators on a Hilbert space H. Then H has an orthonormal basis consisting of comTheorem 9. Let T1
mon eigenvectors of all the T1.
C C and k = 1,. , n, the eigenspace Ker(pJ — Tk) is invariant under all the 7. Hence H is the orthogonal direct sum of Proof. For every
. .
the subspaces = ('1
All these spaces are finite-dimensional, with the possible exception of the union of ,o. Taking an orthonormal basis of each .
0
these bases will do.
As our final theorem concerning abstract operators in this chapter, let us note that our results, say Theorem 2 or Theorem 3, give a complete characterization of compact normal operators up to unitary equivalence. Two operators T, T C are said to be unitarily equivalent if for some unitary operator U we have T' = U'TU = U*TU, i.e. if they have the same matrix representation with respect to some orthonormal bases. Let X be the collection of functions n: C\{0} {0, 1,2,. . } whose 1} has no accumulation point (i.e. in C there C\{0}: n(A) support {A is no accumulation point other than 0). In particular, the support is finite or countably infinite. The following result is easily read out of Theorem 2 (see Exercise 14). .
Theorem 10. Let H be an infinite-dimensional complex Hilbert space be the collection of compact normal operators on H. For and let TC
and A C C\{0} set
nr(A) = dim Ker(A1— T).
Then the correspondence T
a surjection furthermore, T and T' are unitarily equivalent if ni.. = nr. defines
—÷
0
We close this chapter by showing how the spectral theorems we have just proved enable us to solve a Fredholm integral equation. Let 1 = [a, b] for some a 0} arranged in a decreasing order, with y1 , the corresponding eigenvectors. Putting it another way: let > 0 be the sequence of nonP2 negative eigenvalues repeated according to their multiplicities. Show that = max{(Ux,x): lixil = 1, x 1y1 for i = 1,...,n—
1}.
Show also that = rninmax{(Ux,x): xE H,,...1, lxii = 1},
where the minimum is over all
(n —
1)-codimensional subspaces
H,,1. Finally, show that = maxmin{(Ux,x): x E H,,, lixil = 1},
where the maximum is over all n-dimensional subspaces F,,. 8. Let U be a positive hermitian operator. Show that llUxlI4
(Ux,x)(U2x, Ux)
for every vector x. Deduce from this that hUll = v(U). 9. Let U E be a positive hermitian operator with Ker U = {0}.
Show that there is a sequence of hermitian operators (U,,)° C such that U,, Ux x and UU,,x — x for every x E H. Can one have U,, U
I as well?
10. Let U E be hermitian. Prove that Im U is a closed subspace of H if U has finite rank.
Chapter 14: Compact nor,nal operators
211
Prove that T is 11. Let T E (a) normal 1ff H has an orthonormal basis consisting of eigenvectors of T; (b) hermitian 1ff it is normal and all its eigenvalues are real; (c) positive iff it is normal and all its eigenvalues are nonnegative reals.
be hermitian. Prove that there are unique positive 12. Let U E such that operators U+, U_ E U=
-
and
U.... =
U....
U.k. = 0.
13. Prove the Fredhoim alternative for hermitian operators: Let U be a compact hermitian operator on a Hilbert space H and consider the following two equations: Ux—x = 0
(2)
Ux—x=x0
(3)
and
where x0 E H. Then either (a) the only solution of (2) is x =
0,
and then (3) has a unique
solution,
or
(b) there are non-zero solutions of (2), and then (3) has a solution 1ff x0 is orthogonal to every solution of (2); furthermore, if (3) has a solution then it has infinitely many solutions: if x is a solution of (3) then x' is also a solution if x — x' is a solution of (2). 14. Give a detailed proof of Theorem 10. In particular, check that the f( is a surjection. map
be normal and, as in Theorem 10, for A E C set I for every A E C n7.(A) = dim Ker(AI— T). Prove that nT(A)
15. Let T E
(including A = 0) if there is a cyclic vector for T, i.e. a vector x0 C H such that lin{x0, Tx0, T2x0.. . } is dense in H. .
Notes
There are many good accounts of applications of the spectral theorem for
compact hermitian opertors to differential and integral equations. We followed i. Dieudonné, Foundations of Modern Analysts, Academic Press, New York and London, 1960, xiv + 361 pp. Here are some of the
212
Chapter 14: Compact normal operators
other good books to consult for the Sturm—Liouville problem, Green's functions, the use of the Fredholm alternative, etc: D. H. Griffel, Applied Functional Analysis, Ellis Horwood, Chichester, 1985, 390 pp., I. J. Maddox, Elements of Functional Analysis, 2nd edn., Cambridge University Press, 1988, xii + 242 pp., and N. Young, An Introduction to Hubert Space, Cambridge University Press, 1988, vi + 239 pp.
15. FiXED-POINT THEOREMS
In Chapter 7 we proved the doyen of fixed-point theorems, the contraction-mapping theorem. In this chapter we shall prove some considerably more complicated results: Brouwer's fixed-point theorem and some of its consequences. It is customary to deduce Brouwer's theorem from some standard results in algebraic topology, but we shall present a self-contained combinatorial proof. Before we can get down to work, we have to plough through some definitions.
A flat (or an affine subspace) of a vector space V is a set of the form F = x+ W, where W is subspace of V. If W is k-dimensional then we call F a k-flat. As the intersection of a set of flats is either empty or a flat, for every set S C V there is a minimal flat F containing 5, called the flat spanned by S. Clearly
F
A,x1:
x. E S,
A=
1, n
=
=
be points in a vector space. We say that these points Let x0 , x1 ,. . are in general position if the minimal flat containing them is kdimensional, i.e. if the vectors x1 — x0, x3 — x0,. , X,, — x0 span a kdimensional subspace. Equivalently, they are in general position if = 0 whenever = 0 and = 0 or, = = = in other words, if the points are distinct and {x1 —x0,x2—x0,. . ,x1, —x0} is a linearly independent set of vectors. For 0 k n, let x0,x1,. . . ,Xk be k+ 1 points in R" in general position. The k-simplex o = (x0,x1,. ,Xk) with vertices x0,x1,. . ,Xk is the . .
.
. .
.
following subset of R": 213
Chapter 15: Fixed-point theorems
214
k
k
p.,
=
1,
p.,> 0 for all i
1=0
The skeleton of a is the set {x0 , x1 ,. .. , x,j and the dimension of a is k. Usually we write 0k for a simplex of dimension k and call it a ksimplex. A 0-simplex is called a vertex. A simplex a1 is a face of a simplex a2 if the skeleton of crj is a subset of the skeleton of a2. Note that the closure of the simplex a = (x0, x1,... , in R" is
5=
,Xk] k
k
=
p.
1=0
p., =
x1:
0
1, p.,
i=0
C {0,1,...,n}},
=
i.e. the closure of a is precisely the union of all faces of a, including itself. Also, 5 is precisely x1,. . , X,,, }, the convex hull of the vertices, and a is the interior of this convex hull in the k-flat spanned by .
the vertices. A finite set K of disjoint simplices in
is called a simplicial complex
if every face of every simplex of K is also a simplex of K. We also call K a simplicial decomposition of the set 1K I = U {u: E K}, the body of K. If K is a simplicial complex and a, r E K then the closed simplices 5 and are either disjoint or meet in a closed face of both. We are ready to prove the combinatorial basis of Brouwer's theorem.
Lemma 1. (Sperner's lemma) Let K be a simplicial decomposition of a closed n-simplex 5 = [x0, x1,. .. , Let S be the set of vertices of K and let y: S — {0, 1,... ,n} be an (n+ 1)-colouring of S such that the colours of the vertices contained in a face [x¼, x.R,. . , x•] of a- belong to Call an n-simplex a" multicoloured if the vertices of o" {i0, , i,.}. are coloured with distinct colours. Then the number of multicoloured n-simplices of K is odd. .
Proof. Let us apply induction on n. For n =
0
the assertion is trivial;
so assume that n 1 and the result holds for n — 1. Call an (n — 1)-face of K marked if its vertices are coloured with
0, 1,... , n —1, with each colour appearing once. For an n-simplex a-" E K, denote by m(o-") the number of marked (n — 1)-faces of a-". Note that a multicoloured n-simplex has precisely one marked (n — 1)-
face, and an n-simplex, which is not multicoloured, has either no
Chapter 15: Fixed-point theorems
215
marked face or two marked faces. Therefore the theorem claims that
m(K) =
(1) (1"EK
is
odd.
Now let us look at the sum in (1) in another way. What is the contriE K to m(K)? Jf is not marked, bution of an (n— 1)-simplex the contribution is 0. In particular, if a-" is in a closed (n — 1)-face of = [x0,x1,... then the contribution of is 0. a other than E K is in and is marked then the contribution of tf o" is a face of exactly one n-simplex of K. Furthermore, if u"—1 is in a, i.e. in the interior of the original n-simplex, then
contributes 1 to m(u") if a"1 is a face of a": as there are two such n-simplices cr", the total contribution of to m(K) is 2. Hence, modulo 2, m(K) is congruent to the number of marked (n— 1)-simplices in ö0. By the induction hypothesis, this number is odd. Therefore so is m(K), completing the proof. 0 Given points XO,Xi,.. , of R" in general position, for every point x of the k-dimensional affine plane through x0,x1,. .. there are unique reals , A2,. , A, such that . .
Ac
x=x0+
A.(x1—x0). i=1
such that x Hence, there are unique reals p.o, i,... , = and = 1. These p., (i = 0,1,... ,k) are called the barycentric coordinates of x with respect to (xo,x1 ,... , Xk). Also, if p., = 1
then
p.,x,
E
Furthermore, the closed half-space of
contain-
ing 1k and bounded by the (k—1)-flat spanned by X0,X1,...,Xk_1 is characterized by 0. The barycentric coordinates can be used to define a very useful simplicial decomposition. Given a simplicial complex K, the barycentric subdivision sd K of K is the simplicial decomposition of 1K I obtained as follows. For a simplex a = (x0,x1,. . ,Xk) K set .
k
c,.
thus
plices
=
1
Lxi;
is the barycentre of a-. The complex sd K consists of all simsuch that a proper face of a-i +1 ce,,. . , .
(i=0,l,...,k—1).
Chapter 15: Fixed-point theorems
216
To define the r-times iterated barycentnc subdivision of K, set 1. Thussd1K= sdK. sd°K = Kandsd'K= The mesh of K, written mesh K, is the maximal diameter of a simplex of K. Equivalently, it is the maximal length of a 1-simplex of K. Note that if = (x0,x1,. . ,x,) (i = 0,1,... ,k) are faces of a k-simplex = (x0, x1,. . , and r = , then the diameter of = r is less than k/(k + 1) times the diameter of if. Therefore, if K is any simplicial complex then for every 0 there is an r such that mesh sdrK < €. Let Y be a subset of a topological space X, and let a = {A,,: y E 1'} .
. .
.
be a collection of subsets of X. We call a a covering of Y if Y C UEJ. A,,. Furthermore, a is a closed covering if each A,,
is
closed, and it is an open covering if each A,, is open. In what follows, the underlying topological space X is always Sperner's lemma has the following important consequence.
be a closed covering of a closed n.
Corollary 2. Let {A0,A1,.. . simplex a = [x0, x1,. .
.
, x,,
}
such that each closed face [x¼, x11,.
A.. Then
a is contained in
A.
a
. .
,
x.] of
A is
Proof. As we may replace A by
compact. The compactness of the sets A0, A1 ,. , implies that it suffices to show that for every e > 0 there are points a E A. (i = . .
0,1,...,n) such that Ia,—a11 < e if i
j.
Let K be a triangulation of & such that every simplex of K has diameter less than €; as we have seen, for K we may take an iterated barycentnc subdivision of a. Given a vertex x of K contained in a face . ,x1) of u, we know that x E U.,0 As,. Set y(x) = min{i1 : x E Aj. The colouring y of the vertex set of K satisfies the conditions of Lemma 1 and so K has a multicoloured n-simplex Let then
0.
.
y(a1) = i
But then a1 E A, as required.
(i = 0,1,...,n).
0
From here it is a short step to one of the most fundamental fixedpoint theorems, namely Brouwer's fixed-point theorem. A closed n-cell is a topological space homeomorphic to a closed n-simplex. Theorem 3. (Brouwer's fixed-point theorem) Every continuous mapping of a closed n-cell into itself has a fixed point.
Chapter 15: Fixed-point theorems
217
Proof. We may assume that our n-cell is exactly a closed simplex 0" —p 0" is a continuous map, 0" = [xo,xt,... , x,, 1. Suppose that sending a point =
=
1=0
to
=
(IL;
=
i).
1=0
For each i, let
A= . ,A,,} is a closed covering of 0". If a point belongs to a closed face [xc, x•1,. .. ,x.] of 0" then = 0 x= = 1. Since for i {i0, p4 = 1, there is } and so and so x E Consequently, an index j such that p.4'
Then {A0,A1 ,.
. .
,x,] C U_0 A,,
showing that the conditions of Corollary 2 are satisfied. Thus there is a pointx in all the A; such an xis a fixed point of ç. 0
The following lemma enables us to apply Theorem 3 to a rather pleasant class of spaces, namely the compact convex subsets of finitedimensional spaces, i.e. the bounded closed convex subsets of finitedimensional spaces.
Lemma 4. Let K be a non-empty compact convex subset of a finitedimensional normed space. Then K is an n-cell for some n. Proof. We may assume that K contains at least two points (and hence it contains a segment) since otherwise there is nothing to prove. We may also suppose that K is in a real normed space and hence that K is a compact convex subset of = (IR", II•II) for some n. Further-
more, by replacing R" by the flat spanned by K and translating it, if necessary, we may assume that 0 E mt K. Finally, let 5 be an n-simplex containing 0 in its interior, and define a 5 as follows: for x E R" define homeomorphism K
n(x) = ,zK(X) = inf{t:t> 0, x C tK}, and
Chapter 15: Fixed-point theorems
218
m(x) = mo(x) = inf{t: and
t> 0, x E t&},
for x E K set
ifx=0,
0
1
n(x) j—x m(x)
0
.
Corollary 5. Let K be a non-empty compact convex subset of a finite-
dimensional normed space. Then every continuous map f: K—' K has a fixed point. Proof. This is immediate from Theorem 3 and Lemma 4.
0
Our next aim is to prove an extension of Corollary 5 implying, in particular, that the corollary is true without the restriction that the normed space is finite-dimensional. This is based on the possibility of approximating a compact convex subset of a normed space by compact convex subsets of finite-dimensional subspaces. Unfortunately, the simple lemma we require needs a fair amount of preparation.
Let S = {x1 ,.. • , x,, } be a finite subset of a normed space X. For e > 0 let N(S, e) be the union of the open balls of radius centred at
xI,... k
N(S,€) = U D(x1,€). 1=1
For x E N(S,e) define A(x) = max{0,€—IIx—x1Ij} (i = 1,... ,k) and set A(x) A.(x). !f x E N(S,€) then x belongs to at least one open = ball D(x1,€), and for that index i we have A•(x) > 0. Hence A(x) > 0 for every x N(S, e). Define the Schauder projection : N(S, e) —' co{x1,. . . by
=L
A(x)
Here k
co{x1,. ..,Xk} =
k
Ax1: A
0,
A1 = 1
i=1
is the convex hull of the points x1 , -
, xp: the intersection of all convex sets containing all the points x — 1,... , xk. This convex hull is, in fact, compact, since it is the continuous image of a closed (k — 1)-simplex in is the standard basis of R" Indeed, if = say, then the . -
II -
.
Chapter 15: Fixed-point theorems ek] is a bounded closed subset of Furthermore, ç: co{x1 xk}, given by
closed simplex 5 =
it is compact.
219
[e1
k
k
k
0 and
where A,
A,x1,
A-e,
and so
A- =
1.
1=1
is
a continuous map.
is a continuous map from
Lemma 6. The Schauder projection N(S,E) to co{x1,. .. ,x,,} and IS,E(x)—xII < E
for all x E N(S,e).
Proof. Only (2) needs any justification. If x E N(S, e) then k
A,(x)
=
A,(x)
=
i=I But if A.(x) >0 then 11x1—xll
