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O. Then f is uniformly continuous on [a, b], so that we may find 8 > 0 and a finite subset {Xl, X2, ... ,Xk} of [a, b] such that [a,b] = U~=1 I) and Ilf(x) - f(x))11 < E for all X E I) (j = 1, ... ,k), where we are setting I) = (x) - 8, x) + 8) n [a, b] (j = 1, ... , k). By an easy compactness argument, we may now find real-valued, continuous functions e 1 , . .. ,ek on the interval [a, b] such that: 0 ::; e) ::; 1 on [a, b], such that suppe) ~ I) (j = 1, ... ,k), and such that L:~=1 e) (x) = 1 for all x E [a,b]. (Thus we have constructed a so-called continuous partition of unity subordinate to the open covering {It, h ... , Id of [a, b].) Now define k
fo(x)
=
L e)(x)f(x))
(x
[a, b]).
E
)=1
Then fo E C~[a, b] and, for every x E [a, b], we have
Ilf(x) - fo(x)11 =
lit e) (x)(J(x) -
f(x)))II::; teleX) Ilf(x) - f(x))11
f is a continuous linear mapping from Cda, b] into E such that: (i) if T is any bounded lwear operator from E to some Banach space, then b
T ( l f)
(ii) for every f
E
= lb T
0
f ;
Cda, b], we have
Ili I : ; b
f
lb
IIf(x)11 dx ::;
(b - a) Iflla,b] .
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Introduction to Banach Spaces and Algebras
Proof Let Gi := Gi[a, b] be as in the above lemma. For I = glel + ... + gnen E Gi, we define
1 =?; 1 b
We notice that, for such an 1jJ
I
n (
I
b
)
gk ek·
and every 1jJ E E*, we have
(i bI) = ~ (i b9k) 1jJ( ek) = ib 1jJ I . 0
Since, by the Hahn-Banach theorem, E* separates the points of E, it follows that, for every I E Gi, the element I is well defined in E, and is the unique element of E which satisfies (*). It is clear that the map
J:
I
f-7
ib I , Gi E, -4
is a linear mapping from Gi into E. To see that this mapping is also continuous, let I E Gi, and then, using Corollary 3.4(ii), choose 1jJ E E* with 111jJ11 = 1 and
(i I) Ilib III· b
1jJ
Clearly,
Ili
b
It =
=
ib 1jJ I :S ib III(x)11 dx :S (b - a) III[a,bl
J:
0
Thus the map I f-7 I is continuous on Gi and, more specifically, satisfies Thus this map extends uniquely to a continuous clause (ii) for each I E linear mapping from Gda, b] into E; we also write I f-7 I for this extension. That properties (*), (i), and (ii) hold for all I E GE[a,b] all follow byextension from Gi. The uniqueness of I, subject to (*), is immediate from the fact that E* separates the points of E. 0
Gl
J:
J:
We now use Theorem 3.10 to define some E-valued path integrals. Let 'Y: [a, b] - 4 C be a contour, and suppose that F : b] - 4 E is continuous, where E is a Banach space. Define
1
F(z) dz =
lb
F('Y(t)) 'Y'(t) dt.
This definition has all the properties that you would expect by analogy with the case of complex-valued functions (cf. Section 1.11).
113
Banach spaces
Let U be a non-empty, open subset of C, and let E be a Banach space. A function J : U - t E is an analytic (or holomorphic) E-valued function if
J'(z) = lim J(w) - J(z) w ..... z
W -
Z
exists in E (with convergence in the norm topology) for each z E U; J is weakly analytic (or weakly holomorphic) if A 0 J : U - t C is analytic in the usual sense for each A in E*. It can be shown that each weakly analytic function is already analytic, but this will not be important for us because the only property of an analytic function that we shall use is that it is weakly analytic. We now give vector-valued forms of Cauchy's integral theorem (and formula). Theorem 3.11 Let U be a non-empty, open subset oj C, let E be a complex Banach space, and let J be an analytic E-valued Junction on U. Let '"'( be a
contour in U such that n( '"'(; z) = 0 Jor every z E C \ U. Then: (i)
1
J(z) dz = 0 ;
(ii) Jor every w E U \ b]' we have 1 n('"'(; w)J(w) = -2. 1fl
1
J(z) - dz.
'Y Z -
W
Proof To prove that the equations in (i) and (ii) hold, we apply A E E* to each side of each equation, and use Theorem 3.10 and the standard Cauchy's integral theorem and integral formula, Theorem 1.32, to see that the action of A on each side of both of the equations gives equality. The result follows because E* separates the points of E. D The following result will be used later in our account of spectral theory. Theorem 3.12 (Liouville's theorem for vector-valued functions) Let E be a complex normed space, and let J : C - t E be a Junction which is weakly an-
alytic and bounded. Then J is constant. ,Proof For every A E E*, the function A 0 J is a complex-valued entire function on Co Since J is bounded, each function A 0 J is also bounded, and so is constant by the classical Liouville's theorem, Proposition 1.39. Then, for Zl, Z2 E 0 such that U :2 {y E E : Pk(Y - x)
0 and pEP such that q(x) ::; Mp(x) (x E E). It is an easy exercise to see that the topology of a locally convex space E is metrizable if and only if there is some countable collection of seminorms that is equivalent to the given collection P. Indeed, if (Pn)n21 is a sequence of semi norms that defines the topology of E, then a topologically equivalent metric d may be defined on E by, for example, 00
d(x,y) =
L
n=l
1 2n
Pn(x - y) 1 + Pn(x - y)
(x, y
E X).
It is also simple to see that the sequence (Pn)n21 of seminorms may be taken to be pointwise increasing. A locally convex space whose topology is specified by a complete metric is called a Frechet space. All Banach spaces are Frechet spaces. An important example of a Frechet space (that is not a Banach space) is the space (in fact, algebra) O(U) of all complex-valued analytic functions on a non-empty, open subset U of e (or e n see Section 8.4). The topology on O(U) is the standard one-that of local uniform convergence. Indeed, we may write U = U~=l K n , with each Kn compact and Kn C int Kn+l for each n E fiI, and then define
Banach spaces
115
Pn(f) = sup{IJ(z)1 : z E Kn}
(n E N)
for J E C(U), so that each Pn is a semi norm on C(U). The family {Pn : n E N} defines the topology of local uniform convergence on C(U), and hence on O(U). The next result is analogous to the one where E and Fare normed spaces. Lemma 3.13 Let (E; P) and (F; Q) be locally convex spaces, and let T : E be a linear map. Then the following are equivalent: (a) T is continuous on Ei (b) T is continuous at OEi (c) for each q E Q, there are seminorms PI, ... ,Pn E P on E and M such that q(Tx) :'S Mmax{PI(x), ... ,Pn(x)} for all x E E.
-+
F
> 0 D
Corollary 3.14 (Hahn-Banach theorem for locally convex spaces) Let E be a locally convex space, let F be a subspace, and let g be a continuous linear functional on F. Then there is a continuous linear functional J on E such that f I F = g. In particular, Jor any x -1= 0 in E, there is some continuous linear functional J on E wzth J ( x) -1= o. Proof This is immediate from Corollary 3.3.
D
Corollary 3.15 Let E be a locally convex (in particular, a normed) space. Then a linear Junctional), on E is continuous if and only zJ ker). is closed in E.
Remark: the 'if' part of this result does NOT extend to linear mappings that are not into the scalar field. Proof The scalar field of E is OC. Of course, it is clear that the vector subspace ker). is closed whenever). is continuous. Conversely, let a linear map). : E -+ OC have closed kernel, say Z. We claim that ). is continuous. We may suppose that). -1= 0, since otherwise the result is trivial. Then Z -1= E, and so we may choose Xo E E \ Z. Since Z is assumed to be closed, there is then a continuous seminorm P on E such that, setting V = {x E E : p(x) < I}, we . have (xo + V) n Z = 0. Assume towards a contradiction that )'(V) is an unbounded subset of oc. Then, for any JL E OC, there is some x E V with 1).(x)1 > IJLI. But now
). ().tX) x) = JL
and
P
().tX) x) < 1 ,
so that JLx/),(x) E V, and therefore JL E ).(V). Thus, in this case, )'(V) = II(. But then, in particular, there is some v E V with ).(v) = -).(xo), and so with ).(xo + v) = 0, it follows that Xo + v E (xo + V) n Z, contrary to the choice of V. Hence )'(V) is a bounded subset of II(. By the usual argument, there is a constant M > 0 such that 1).(x)1 :'S Mp(x) (x E E), and the continuity of). is proved. 0
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Introduction to Banach Spaces and Algebras
The following result is proved by a slight modification of the proof of Theorem 2.22. Theorem 3.16 Let E be a metrizable locally c~nvex space. Then there is a unique Frechet space structure on the c~mpletio~E of E such that E is isometrically embedded as a dense subspace of E. Thus E is a Frechet-space completzon ofE. 0
3.4 Paired vector spaces. We wish to discuss, mainly, the weak topology on a Banach space and the weak-* topology on its dual. It is useful to set these in a slightly wider context. Let E and F be linear spaces over IK. A pairing of (E, F) is a specified nondegenerate, bilinear form (x, y) f---> (x, y) on ExF. [Here, 'non-degenerate' means: for each x 1= 0 in E, there is some y E F such that (x, y) 1= 0, and, for each y 1= 0 in F, there is some x E E such that (x, y) 1= o. Unlike the finite-dimensional case, the two parts of this condition are not equivalent.] Examples 3.17 (i) Let E be a IK-linear space, and let F be its algebraic dual space, with (x,!) = f(x) for x E X and f E F. Then (., .) is a pairing of (E,F). (ii) Let E be a locally convex space (in particular a normed space), and set F = E*, the 'continuous dual'; the pairing is as in (i). The proof that this pairing is non-degenerate uses the Hahn-Banach theorem. (iii) Let F be a normed space, and take E = F* to be its dual space. Then (F*, F) is a non-degenerate pairing; this pairing is not, in general, a special case of (ii). However, the pairing (F*, F**) is a special case of (ii). (iv) If (E, F) is any pairing, then there is always the 'reverse pairing', in which the bilinear form remains the same, but the roles of E and F are interchanged. 0 Let us use the above notation (x,1) f---> (x,!) for the pamng between a normed space E and its dual space E*, where x E E and f E E*, and also the notation (j, A) f---> (j, A) for the pairing between E* and E**. Then the canonical embedding J : x -+ X, E -+ E**, described on page 108, takes the form
(j,Jx)
=
(x,!)
(x E E, f E E*).
3.5 Weak and weak-* topologies. It is important to know that, associated with any pairing, there is a weak topology. Let (E, F) be a pairing. Then the mappings
Pf:xf--->l(x,!)I,
E-+lR.,
as f runs through F, form a point-separating family of seminorms on E. The associated locally convex topology on E is called the weak topology associated
117
Banach spaces
with the pairing; it is denoted by a(E, F). It is simple to see that the topology a(E, F) is the weakest topology on E that makes all of the above linear functionals p f continuous. In fact we have the following more precise result.
Theorem 3.18 Let (E, F) be a pairing. A linear functional rp on E is continuous for the weak topology a(E, F) if and only if there is some I E F such that rp(x) = (x,I) (x E E), and so (E;rr(E,F))* = F. Proof Let rp be a a(E, F)-continuous linear functional on E. By Lemma 3.13, there exist h, ... , In in F and M > 0 such that Irp(x)1 ::; M max{l(x, h) I, ... , I(x, In)l} In particular, therefore, if we define
e : E --+ ]Kn
(x E E).
by
e(x) = ((x,h), ... , (x,In))
(x
E
E),
we have ker e
AX,
IK x X
-->
X,
and
(x, y)
f-->
X + Y,
X x X
-->
X,
are both continuous. These spaces are generalizations of locally convex spaces: a topological vector space X is a locally convex space if and only if every neighbourhood of Ox contains a convex neighbourhood of Ox. For an account of these spaces, see [144, Chapter 1]. Let E be a Banach space. We have said that E is reflexive if the canonical mapping J : E --> E** is a surjection. Our intuition is that, if E is not reflexive, then E** is 'much bigger' than E. For example, Co is separable, but (co)** = £= is non-separable. A Banach space E is quasi-reflexive if E** / E is finite-dimensional. In fact there are non-reflexive, quasi-reflexive spaces. For example, let J be the James space of Exercise 2.14. Then J** / J is even one-dimensional. Further, J** is isometrically isomorphic to J-but still J is non-reflexive. For a discussion of J, see [2, Section 3.4] and [47, Example 4.1.45]. For a more sophisticated approach to the integration of Banach-space-valued functions, involving, for example, the Bochner integral, the Pettis integral, and the Bartle integral, see [58, Chapter II]. The fact that a weakly analytic Banach-space-valued function is already analytic is proved in [144, Theorem 3.31]. We have mentioned the strong-operator and weak-operator topology on B(E, F). In fact, there are several other useful topologies that can be defined on B(E, F); see [63, Chapter VI, Section 1], for example. Exercise 3.1 Let E be a separable normed space. Prove the Hahn-Banach theorem for E without using the Zorn's-lemma part of the argument. In fact, one can arrange to use the main part of the proof of Theorem 3.1 count ably many times to extend a given functional to a dense linear subspace of E, and then extend the functional to the whole space by continuity. Exercise 3.2 Let X be the space (£nlR, where 1 ::;: p ::;: 00, so that X is just the plane IR X 1R, with the norm II· lip' Let Y be a one-dimensional subspace of X, so that Y is a line through the origin, and let g be a linear functional on Y. Of course there is a norm-preserving extension of g to a linear functional f on X. Discuss whether or not such a functional f is uniquely defined for various values of p. Exercise 3.3 Let L be the left shift operator on £It', as in Exercise 2.11. Prove that there is a continuous linear functional f on £It' such that f(Lx) = f(x) and liminfx n n~oo
for all x =
(Xn )n2:1 E
::;:
f(x)::;: lim sup Xn n~=
CIt'. (Such a functional is called a Banach limit.)
122
Introduction to Banach Spaces and Algebras Here is a hint. For n E N, define
fn(x)
=
!(Xl n
+ ... + Xn)
(x
E
£llf'),
and set Y = {x E £llf' : limn~oo fn(x) exists}. Set p(x) x E £llf', and apply the Hahn-Banach theorem.
limsuPn~oo
fn(x) for
Exercise 3.4 For f,g E C(ll), define
d(j,g) =
11 . If~tL-:-g(t),I"
dt.
Show that d is a metric on C(ll), and that C(ll) is a topological vector space with respect to the metric topology. Show that the only continuous linear functional on this space is the zero functional. Exercise 3.5 Let E = co. Then E* = £1 and E** = £00 (see Example 2.18). Show, directly, that the linear functional f on £ 1 , defined by
00 f(Xl,X2,X3"")=Lx n
((Xn)E£I)
n=l
is continuous (and so weakly continuous), but that it is not weak-* continuous. By considering the subspace ker f, show that there is no analogue of Theorem 3.20, with 'weak' replaced by 'weak-*'. Exercise 3.6 Let E be a Banach space with dual space E*. Set B = (E*)[!]. Prove that B is metrizable (in the weak-* topology) if and only if E is separable. Exercise 3.7 (i) Let E be a Banach space. Show that the dual space E* is reflexive if and only if E is reflexive. (ii) Show that the closed unit ball of a Banach space E is weakly compact if and only if E is reflexive. (iii) Let E and F be Banach spaces. Show that the closed unit ball of B(E, F) is compact in the weak-operator topology if and only if F is a reflexive space.
Separation theorems We now give some geometric versions of the Hahn-Banach theorem. 3.7 Separation of convex sets. In what follows, E is a locally convex space over JR., except where otherwise stated. (The results may also be applied to a complex space by the usual device of restricting multiplication to real scalars.) Lemma 3.25 Let C be a convex, open neighbourhood of OE in E. For every x E E, define J-lc(X) = inf{a > 0: x E aC}.
Then J-lc is a sub linear functional on E and C = {x
E
E : J-lc(x) < I}.
123
Banach spaces
Proof Let x E E. Note first that, since the map A f-+ AX, lR --+ E, is continuous at A = 0, there is some 8 > 0 such that AX E C whenever IAI < 8. This shows that X E JLC whenever IJLI > 8- 1 . Thus JLc(x) is well defined; 0 ::; JLc(x) < 00. Now let x, y E E, and set r = ILc(X) and s = JLc(Y). Then, for each E > 0, there exist a and (3 such that 0 < a < r + E /2 and 0 < (3 < s + E /2 and such that X E aC and y E (3C. Then X
+ Y E aC + (3C =
+ (3)
(a
Ca: (3) C
+
(a!
(3) C)
0, and hence JLc(x + y) ::; JLc(x) + JLc(Y). For each A > 0, we have AX E aAC if and only if X E aC, and so it follows that JLc(AX) = AJLc(X) (A ~ 0). This shows that JLc is a sublinear functional on E. Let X E C. Since the set C is open, (1 + 8)x E C for some 8 > 0, so that x E (1 + 8)-IC and JLc(x) < 1. Conversely, if JLc(x) < 1, then X E tC for some t < 1, so that x E C since OE E C and C is convex. D Theorem 3.26 (Separation theorem) Let E be a (real) locally convex space, and let A and B be non-empty, convex subsets of E with An B = 0. (i) Suppose that A is open. Then there exist f E E* and ~ E lR such that
f(x)
0 such that, for every y E F, there is x E E with Tx = y and Ilxll :s; K Ilyll (so that T is an open mapping) . . Proof We write BR = B(OE; R) for R > O. Then E = U~=1 Bn, and so F = U~=1 T(Bn). By the Baire category theorem, Corollary 1.22, there is some N E 1'\1 such that int T( B N) =1= 0. An elementary argument then shows that T(BN) includes a neighbourhood of OF, say B(OF; 0), where 0 > O. Then, taking R = N/o, we see that T(E[R]) :2 F[1], and now the result follows from Lemma 3.38. D Note that, in the above theorem, we already knew both that F is complete and that T(E) = F. What is gained is the existence of the constant K, and this implies that T is an open mapping. Let E and F be normed spaces. In general, a continuous linear bijection T is not necessarily a linear homeomorphism (see Exercise 3.16), but the following result shows that this is the case when both E and F are Banach spaces; the continuity of the inverse T- 1 then comes 'for free'.
Corollary 3.41 (Banach's isomorphism theorem) Let E and F be two Banach spaces, and let T E B(E, F) be a bijection. Then the inverse map T- I is continuous, i.e. T-1 E B(F, E). Proof Clearly, T- I : F --* E is linear. Let K be the constant specified in the theorem. For each y E F, there is a unique element x E E with Tx = y. By the theorem, Ilxll :s; K lIyll, and so IIT-I(y)11 :::; K Ilyli. This shows that T- I is bounded, and hence continuous. D
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Introduction to Banach Spaces and Algebras
Corollary 3.42 Let E be a vector space that is a Banach space with respect to two norms, II ·11 and III . III· Suppose that there zs a constant C > 0 such that Illxlll :::; C Ilxll (x E E). Then 11·11 and 111·111 are equivalent. 0
Recall from the definition immediately before Proposition 2.23 that an operator T E [3(E, F) is bounded below if and only if there is a constant c> 0 such that IITxl1 2': cllxll (x E E). Corollary 3.43 Let E and F be Banach spaces, and let T E [3(E, F). Then the following assertions are equzvalent: (a) T is injectzve and has closed range; (b) T is bounded below. Proof We may suppose that E f= {O} since the result is trivial in the case where E = {O}. (a) =}(b) Since imT is closed in F, it is a Banach space, and now T defines a linear homeomorphism of E onto im T. By Corollary 3.41, T has a continuous inverse, say S : im T ---+ E. Of course x = (ST) (x) (x E E), so that
Ilxll :::; IISllllTxl1 f= {O}, necessarily IISII f= 0, and so
(x E E).
Since E (b) holds with (b) =}(a) This was proved in Proposition 2.23(i).
c = IISII- 1 . o
Let E be a Banach space, and let S, T E [3(E). Then 5 and T are said to be similar if there is an invertible operator U E [3(E) such that 5U = UT. Proposition 3.44 Let E be a Banach space, and let 5, T E [3(E) be similar operators. Then 5 is bounded below if and only if T is bounded below. Proof Take U E [3(E) with U invertible such that 5U = UT, and set V = U- 1 , so that V5 = TV. Suppose that T is bounded below, so that there exists c > 0 such that IITyl1 2': c Ilyll (y E E). Let x E E, and set y = U.T. Then
cIlxll :::; c1lUIIIIVxli :::; 1lUIIIITVxli = 1lUIII1V5xli :::; 1lUIIIIVIII15xli , and so 5 is bounded below. The result follows. Let 5 and T be non-empty sets, and let j : 5 graph of f is the set
o ---+
T be a function. Then the
Gr(J) = {(8,j(S)) E 5 x T: s E 5}. In the case where 5 and T are topological spaces and j is a continuous map, it is clear that Gr(J) is a closed subset of 5 x T; in general, the converse is not true. In the case where E and F are vector spaces and T : E ---+ F is a linear map, it is also clear that Gr(T) is a vector subspace of the vector space E x F.
133
Banach spaces
Theorem 3.45 (Closed graph theorem) Let E and F be Banach spaces, and let T : E ----> F be a linear mapping. Then T is contznuous if and only if Gr(T) is a , closed subspace of E x F (in its product topology). Proof The only non-trivial point to be proved is that T is continuous whenever Gr(T) is a closed subspace of Ex F. The space E x F is a Banach space, for example for the norm defined by
II(x, y)11 = Ilxll + Ilyll
(x E E, y E F)
(see Exercise 2.5). Since Gr(T) is closed in E x F, the subspace Gr(T) is itself a Banach space. Let 7fE and 7fF be the coordinate projections of E x F onto E and F, respectively; each of 7fE and 7fF is a continuous linear mapping. However, 7fE I Gr(T) maps Gr(T) bijectively onto E, and so, by Banach's isomorphism theorem, Corollary 3.41, 7fE I Gr(T) : Gr(T) ----> E is a linear homeomorphism. But then T =
7fF
0
(7fE
I Gr(T))-1
is continuous, as required.
D
3.12 Quotient norms. We now discuss quotient norms. Let (E;p) be a seminormed space, let M be a subspace of E, and let 7fM : E ----> ElM be the quotient mapping. For every ~ EEl M, define PM(O = inf{p(x) : x E E, 7fM(X) =
o.
Then it is easy to check that PM is a seminorm on ElM; it is the quotient seminorm. The quotient mapping 7f M is an open mapping and is 'seminorm decreasing' . Lemma 3.46 Let (E;p) be a seminormed space, and let M be a subspace of E. Then PM is a norm if and only if M is closed in E. Proof Suppose that M is a closed subspace of E, and let ~ E ElM with o. Then there is a sequence (Xnk:':l in E with 7fM(X n ) = ~ (n E N) and such that p(xn) ----> o. But then Xl - Xn E M (n EN), whilst Xl - Xn ----> Xl. Since M is closed, it follows that Xl E M, and thus ~ = 7fM(Xt} = o. Thus PM is a norm on ElM. Conversely, suppose that PM is a norm on ElM. Let X E M, so that there is a sequence (Xn)n:::>:l in M with Xn ----> x. But then, writing ~ = 7fM(X), we have PM(~) ::; p(x - Xn) ----> 0 for each n E N, so that PM(~) = 0, and hence ~ = 0, i.e. X E M. Thus the subspace M is closed. D PM(~) =
,
In particular, for a seminormed space (E;p), set Eo = {x E E : p(x) = O}, so that Eo is a subspace which is the closure of {O} in (EiP). Then the quotient seminorm on E I Eo is a norm.
134
Introduction to Banach Spaces and Algebras
Notation: In the case where the subspace M is a closed subspace of a seminormed space E, the norm PM is more usually written in a norm notation, for example, as II . II E/ M or just II . II· It is called the quotient norm on ElM.
Lemma 3.47 Let E and F be normed spaces, and let T E B(E, F). Further, set M = ker T, let 7r M : E ----> ElM be the quotient mapping, and let ElM be given the quotient norm. Then there is a unique linear mapping T M : ElM such that TM 07rM = T. Moreover, TM continuous with IITMII:::; IITII.
is
---->
F
injective, im TM
= im T, and TM zs
Proof The existence and uniqueness of the linear map TM with TM 07rM = T, with T M injective, and with im TM = im T is all elementary algebra (assumed to be well known). We write II·II E/M for the quotient norm. Let ~ E ElM. Then, for every x E E such that 7r M (x) = ~, it follows that IITM(~)II
= IITxl1 :::; IITllllxll·
Hence, taking the infimum over all such x, we see that IITM(~)II Thus TM is continuous with IITM I :::; IITII.
:::;
IITIIII~IIE/M. D
Let E and F be normed spaces, and let T E B(E, F). Then we can regard T (E) as a normed space by 'transferring the norm from E I ker T', since the latter space is linearly isomorphic to T(E). In general, this norm is not equivalent to the relative norm from F. We now show that when we take appropriate quotients of Banach spaces, we 'remain within the category'.
Theorem 3.48 (Completeness of quotients) Let E be a Banach space, and let M be a closed subspace of E. Then (ElM; II·II E/M) is a Banach space. Proof The definition of II·II E / M makes it clear that, for any R > 1 (e.g. R = 2 will suffice), we have 7rM(E[RJ) :2 (EIM)[lJ. Hence, by the open mapping lemma, Lemma 3.38(ii), (ElM; II·IIE/M) is complete. D For a direct proof of the above result, see Exercise 3.13.
Corollary 3.49 Let E and F be Banach spaces, let T E B(E, F), let M = ker T, and suppose that im T is closed in F. Then the induced map T M : ElM ----> F is a linear homeomorphism between ElM and im T. Proof By Lemma 3.47, TM is a continuous, injective linear mapping of ElM onto im T. But ElM is complete by Theorem 3.48, and also im T is complete because it is a closed subspace of F. The result now follows from Banach's isomorphism theorem, Corollary 3.41. D
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Banach spaces
Corollary 3.50 (Universal property of £ I) Every separable Banach space is linearly homeomorphic to a quotient space of £ I . Proof Let F be a separable Banach space, and choose a countable, dense subset {en: n E N} of the closed unit ball F[ll' For x = (Xnk;~l E £ I (real or complex according to whether F is a real or complex space), define CXJ
Tx= Lxnen . n=l Note that the series converges in F because L~=l Ilxnenll :S Ilxlll and F is assumed to be complete. Evidently T E B( £ I , F). Let B be the closed unit ball of £ I. Then the construction makes it clear that {en: n E N} ~ T(B) ~ F[ll' so that T(B) is dense in F[ll' Now the open mapping lemma, Lemma 3.38, shows that, in fact, T( £ I) = F. By Corollary 3.49, T induces a linear homeomorphism between £ I I ker T and F. 0 3.13 The separating spaces and a stability theorem. A null sequence in a normed space E is a sequence (Xn)n>l in E such that lim n ---+ CXJ Xn = O. Let E and F be Banach spaces, and let T E £(E, F). To apply the above closed graph theorem, we must show that Gr(T) is closed in E x F. It is easily seen that, to do this, it suffices to prove the following: for each null sequence (X n )n21 in E such that TX n ----+ y in F, it is necessarily the case that y = O. The point is that we can suppose in advance that the sequence (Tx n )n21 converges to some element of F. Indeed, we can reformulate the closed graph theorem in the terminology of a separating space; this reformulation will be important in Part II. Let E and F be normed spaces, and let T : E ----+ F be a linear map. Then the separating space, 6(T), of T is defined to be the set of all y E F such that , TX n ----+ Y for some null sequence (Xn)n>l in E. Equivalently, y E 6(T) if and only if, for each n E N, there exists x E E with Ilxll < lin and IITx - yll < lin. The following theorem is immediate from the closed graph theorem. Theorem 3.51 Let E and F be Banach spaces, and let T : E ----+ F be a linear map. Then 6 (T) is a closed subspace of F, and T is continuous if and only if
6(T) = {O}.
0
Corollary 3.52 Let E be a vector space that is a Banach space with respect to two norms, 11·11 and 111·111. Suppose that y = 0 whenever Xn ----+ 0 in (E; 11·11) and Xn ----+ Y in (E; 111·111). Then 11·11 and 111·111 are equivalent. Proof By the closed graph theorem, the identity map ~ : (E; 11·11) ----+ (E; 111·111) is continuous, and so there is a constant C > 0 such that Illxlll :s: C Ilxll (x E E). By Corollary 3.42, II ·11 and 111·111 are equivalent. 0
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Proposition 3.53 Let E and F be Banach spaces, let T : E --+ F be a linear map, and take Q : F --+ FIS(T) to be the quotient map. Then S(QT) = {O}, QT is continuous, and T-1(S(T)) is a closed subspace of E. Proof Let y E F be such that Qy E S(QT). Then there is a null sequence (X n )n>l with QTx n --+ Qy as n --+ 00. Thus there exists (Yn)n>l in SeT) such that TX n - Y - Yn --+ 0 in F as n --+ 00. For each n E N, there exists Zn E E with Ilznll < lin and IITzn - Ynll < lin. But now (xn - Zn)n21 is a null sequence in E such that T(xn - zn) --+ Y as n --+ 00, and so Y E SeT) and Qy = O. Hence S(QT) = {O}, and so QT is continuous. We have T-1(S(T)) = kerQT, and so T-1(S(T)) is closed in E. 0 Proposition 3.54 Let E and F be Banach spaces, and let T : E linear map.
--+
F be a
(i) Suppose that E1 zs a Banach space and that R E B( E 1, E). Then we have S(TR) l is a null sequence in E, and T(Rxn) --+ Y as n --+ 00, so that Y E SeT). Hence S(TR) 0 and (Znk::1 in E1 with Ilznll ::; K Ilxnll and RZn = Xn for each n E N. Clearly, (Zn)n21 is a null sequence and (TR)(zn) --+ y, so that Y E S(TR). Hence SeT) l be sequences in B(E) and B(F), respectively. Suppose that TRn - SnT E B(E, F) (n EN). Then the sequence ( Sl ... Sn6(T))
n2 1
is a nest in F which stabilizes.
Proof (i) We may suppose that IIRnl1 = IISnl1 = 1 (n EN). Assume towards a contradiction that SnT R1 ... Rn is discontinuous for infinitely many n E N. By grouping the maps R n , we may suppose that the maps SnTUn : En -+ Fn are discontinuous for each n E N, where we are setting Un = R 1 ··· Rn. For each n E N, inductively choose Yn E En such that llYn I < 2- n and IISmTUnlillYnl1 < 2- n whenever m < n, and so that IIS1TU1Y111 ::::: 2 and n-1 IISnTUnYnl1 ::::: n + 1 +
L
IITUJYJII
(n::::: 2).
(*)
J=l
We now define Xn = 2::;:n UJYJ in E for n E N; each series converges because IIUJYJII ::; 2- J (j EN). Now Xl = U1Y1 + ... + UnYn + Xn+1 (n EN), and so n-1
IISnTUnYnl1 ::;
IITxIil + L
IITUJYJII
+ II S nTx n+111
.
(**)
J=l
But IISnTxn+111 ::; 2::;:n+11ISnTUJIIIIYnll ::; 1, and so, from (*) and (**), it follows that n ::; IITx111 for each n E N, a contradiction. Thus there exists N E N such that SnT R1 ... Rn is continuous for each n:::::N. For n E N, set 6 n = 6(TR1··· R n ), and let Qn : F -+ F/6 n be the quotient map. By Proposition 3.54(i), 6 n +1 s: 6 n (n EN), and so (6(TR1··· R n ))n2 1 is a nest in F. By the above result with Sn = Qn+1, there exists N E N such that Qn+1TRl··· Rn is continuous for each n ::::: N. By Proposition 3.54(ii), 6(TRl··· Rn) s: kerQn+! = 6 n+! (n::::: N). Thus 6 n = 6N (n ::::: N).
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Introduction to Banach Spaces and Algebras
(ii) For n E N, define Un = 8 1 ... 8 n T - TR1 ... R n , so that Un+1 = UnRn+1 - 8 1 " , 8n(TRn+1 - 8 n+1T). By hypothesis, U 1 is continuous and T Rn+1 - 8 n+1T is continuous for each n E N, and so, by an immediate induction, each Un is continuous. Thus 6(8 1 ···8n T)
= 6(TR1'" Rn)
By Proposition 3.54(ii), 8 1 " , 8 n 6(T) 8 1 ···8n 6(T)
(n E N).
= 6(81 ... 8 n T) , and so
= 6(TR1 ... Rn)
(n E N).
The result now follows from (i).
0
3.14 Open mapping theorem for Frechet spaces. In our later study of homomorphisms between algebras of analytic functions, we shall need to know that the open mapping theorem, proved in Section 3.11 for Banach spaces (see Theorem 3.40), is also valid for Frechet spaces. It does, in fact, work for even more general complete, metrizable topological vector spaces, but we shall restrict ourselves to the Frechet case. Although, in retrospect, it may seem that the proof is not dissimilar from the proof for Banach spaces, it will be seen that the details are a good deal more fiddly. Let E and F be Frechet spaces, and let T : E ----> F be a linear map such that T is a bijection. In the case where both T and T- 1 are continuous, we again say that T is a linear homeomorphism and that the spaces E and F are linearly homeomorphic. Let E be a metrizable, locally convex space, with its topology given by an increasing sequence, say (Pn)n21, of seminorms. Then, as before, we may suppose that the associated complete metric d is specified by the formula d(x, y)
=
L
n21
1 Pn(x-y) 2n . 1 + Pn(x _ y)
(x,YEE).
The metric is translation-invariant: that is, d(x + z, y + z) = d(x, y), and, in particular, d(x, y) = d(x - y, 0) for all x, y, z E E. Suppose that the topology of E is defined by a metric d, and let a E E. For r > 0, we again write B(a;r)
= {x
E
E: d(x, a)
< r}
for the open ball of radius r around a. It is clear that B(a; r) = a + E[r] , where now E[r] = {x E E: d(x,O) :::; r}. Most of the work for the open mapping theorem for Frechet spaces is contained in the following lemma.
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Banach spaces
Lemma 3.56 Let (E; d) be a Frechet space, let (F; d) be a metrizable, locally convex space, and let T : E ----+ F be a continuous linear mappzng. Suppose that, for every r > 0, there exists some P == per) > 0 with F[p] ~ T(B(O;r)). Then: (i) B(O; p) ~ T(E[a]) for every a > r; (ii) T is a surjective, open mapping; (iii) F is complete.
Proof (i) Let r > 0, let p = per) be as specified in the hypotheses, and then take a > r. For each x E E, it is clear that B(Tx;p) ~ T(B(x;r)), where
p = per). Let y E B(O; p). We shall show that y = Tx for some x E E[a]. Indeed, let (rn)n~l be a sequence in jR+. with rl = r and a = L~=l rn. Take PI = p and, for each n ::::: 2, take Pn with 0 < Pn s: p(rn) and such that Pn ----+ 0 as n ----+ 00. Evidently, for every x E E and n E N, the set T(B(x; rn)) is dense in B(Tx; Pn). We shall next define, inductively, a sequence (.Tn)n~O in E with Xo = 0 and such that: (a) Xn E B(Xn-l; rn) (n::::: 1), and (b) TX n E B(y; Pn+l) (n::::: 0). For n = 0: clause (b) holds since y E B(O;p), and so 0 = Txo E B(Y;PI)' For n = 1: y E F[Pl] ~ T(B(O; rt)), and so there is some Xl E B(xo; rt) with TXI E B(y; P2). Now take n ::::: 2, and assume that we have chosen Xo, ... ,Xn-l such that (a) and (b) hold. We have B(Txn-l; Pn) ~ T(B(xn-l; rn)), and B(Txn-l; Pn) contains y because TXn-1 E B(y; Pn). Thus there is an element Xn E B(Xn-l; rn) with TX n E B(y; Pn+1)' This continues the inductive definition. We have seen that there is a sequence (Xn)n~l that satisfies (a) and (b). For n, k ::::: 1, we have d(xn, Xn+k) s: rn+l + ... + rn+k, so that (Xn)n>l is a Cauchy sequence in E, and thus limn->oo Xn exists, say x = limn->oo Xn. We have d(x,O) s: L~=l rn = a, i.e. x E E[a]' Also, since T is continuous, TX n ----+ Tx in F. However, d(Txn' y) s: Pn+1 (n EN). Hence TX n ----+ y, and so Tx = y, which implies that y E T(E[a]). This proves (i). (ii) It is immediate that T is surjective. Now let U be a non-empty, open subset of E, let y E T(U), and choose some x E U with Tx = y. Next, choose rS > 0 such that x + E[28] ~ U. Set P = p(rS). It follows from (i) that B(O;p) ~ T(E[28]), so that T(U) :2 T(x
+ E[28]) :2 y + B(O; p).
Thus T(U) is open in F. This proves that T is an open mapping. (iii) Let F be the completi(~n of F (see Theorem 3.16). For every r > 0, there exists some P > 0 such that F[p] ~ T(B(O; r)), where the latter closure is now taken in F. Clause (ii) applied to the map T : E ----+ F shows that im T = F. However, imT ~ F, and so F = F. Thus F is complete. 0
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Introduction to Banach Spaces and Algebras
Let E be a locally convex space whose topology is defined by a collection P of seminorms, and let M be a subspace of E. Then the family {PM: pEP} of quotient seminorms on ElM defines a locally convex topology on ElM; clearly, if P can be taken to be countable, then ElM is metrizable in the latter topology. Corollary 3.57 Let E be a Frechet space, and let M be a closed subspace of E. Then the quotzent space ElM is also complete. Proof Let 7rM : E ---> ElM be the quotient mapping. For every p > 0, the definition of the quotient metric in ElM shows that, for every r > p, we have (EIM)[p] ~ 7rM(B(O;r)). The completeness of ElM now follows from Lemma 3.56(iii). 0 Theorem 3.58 (Open mapping theorem for Frechet spaces) Let E and F be Frechet spaces, and let T be a continuous linear mapping with T(E) = F. Let 7r : E ---> E I ker T be the quotient r::ap, and let T : E I ker T ---> F be the unique linear isomorphism such that T = T 0 7r. Then:
(i) T is an open mapping; (ii)
T is
a linear homeomorphism.
Proof (i) Take R > 0, and set r = R12. Since E = U~=l nE[r] , it follows that F = U~=l nT(E[r]). By the Baire category theorem, Corollary 1.22, there is some N E N such that NT(E[r]) = NT(E[r]) has an interior point. Then T(E[r]) has an interior point, say y. It is also true that -y is an interior point of T(E[r])' so that OF = Y + (-y) is an interior point of T(E[r])
+ T(E[r])
~ T(E[r])
+ T(E[r])
~ T(E[R])·
This shows that T(E[R]) is a dense subset of a neighbourhood of OF. It follows from Lemma 3.56 that T is an open mapping. (ii) This follows at once, by using Corollary 3.57 and then applying Lemma 3.56 to the mapping T. 0
Dual operators In this section, we shall study the relationship between Banach spaces and their duals, especially the notion of the dual of a linear operator.
3.15 Annihilators. We start with a simple lemma about subsets of paired linear spaces over a field ][{ (see Section 3.4). Suppose that E and F are paired linear spaces, with the pairing (x, y) f--+ (x, y). We write a = a(E, F) for the weak topology on E associated with the pairing. The examples that will chiefly occupy us are, for a Banach space E, the pairing (E, E*) (in which case a is the weak topology on E) and the reverse pairing (E*, E) (when a is the weak-* topology on E*); see Section 3.5 for an initial discussion of these topologies.
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Banach spaces
For a subset X
E*
M.L. Thus (ElM) * is isometrically iso-
Proof Let f E (EIM)*. Then 7r'M(J) = f 0 7rM E M.L. Also, if 9 E M.L, then there is a unique f E (E I M)* such that 9 = f 0 7rM, i.e. such that 7r~I(J) = g. Thus im 7r'M = M.L. For f E (EIM)*, set 9 = 7r'M(J). Clearly, Ilgll ::; Ilfllll7rMII ::; Ilfll· But also, for any ~ E ElM and c > 0, we may choose x E E with 7rM(X) = ~ and Ilxll ::; 11~11(1 + c). Then f(~) = g(x), so that Ilf(~)11 ::; Ilgllll~ll(l
+ c).
This holds for every c > 0, so that Ilfll ::; Ilgll, and therefore Ilfll the map 7r'M is an isometry, which completes the proof.
= Ilgll. Thus D
Theorem 3.69 Let E be a Banach space, let F be a normed space, and let T E B(E,F).
(i) Suppose that there is some c > 0 such that IIT* fll 2': cllfll (J
E F*). Then and F is a Banach space. (ii) Suppose that F is a Banach space and that im T = F. Then there is some c> 0 such that IIT* fll 2': cllfll (J E F*).
im T
=F
Proof (i) Let B be the open unit ball of E, and let Yo E F \ T(B). By the separation theorem, Corollary 3.27, there is some f E F* with If(Yo)1 > 1 and such that If(Tx)1 ::; 1 (x E B). We see that I(T* f)(x)1 ::; 1 (x E B), and so IIT* fll ::; 1, and now it follows that Ilfll ::; c- 1 , so that IIYol1 > c. This implies that F[lJ E / ker T to be the quotient map. By Corollary 3.49, there is a unique linear homeomorphism To : E / ker T -> F with T = To 0 7r. But then
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Introduction to Banach Spaces and Algebras
T* = 7r* aTo"; by Corollary 3.63(ii), the map To" is a linear homeomorphism, and, by Proposition 3.68, 7r* is isometric. Thus IIT*fll = IITofll;::: II(To)-lllllfll
(f E F*),
and so IIT*fll ;:::cllfll (fEF*),wherec= II(To")-111 = liTo-III. Corollary 3.70 Let E and F be Banach spaces, and let T following assertions are equivalent:
E
0
B(E, F). Then the
(a) imT = F; (b) T* is injective with norm-closed range; (c) there is some c > a such that IIT*fll ;::: cllfll (f E F*); (d) T* is injective with weak-*-closed range. Proof The equivalence of (b) and (c) is immediate from Corollary 3.43, and the equivalence of (a) and (c) comes from Theorem 3.69. Clearly, (d) implies (b). The proof will now be completed by showing that (a) implies that im T* is weak-*-closed. We know (from Lemma 3.61(i)) that (kerT)~ is the weak-*closure of im T*, so that it will suffice to show that (ker T) ~ s;.; im T* . Indeed, let 7r : E ---4 E / ker T be the quotient map, and let To : E / ker T ---4 F be the unique linear homeomorphism such that T = To a 7r. Then
7r* : (E / ker T)*
---4
E*
is an isometric mapping with im 7r* = (ker T)~, and To" is a linear homeomorphism of F * onto (E / ker T) * with T* = 7r* a To" . Now let g E (kerT)~. Then there is a unique 'Y E (E/kerT)* such that 7r*("() = g, and then there is a unique f E F* with To"(f) = 'Y. It follows that g = 7r*("() = (7r* a To") (f) = T*(f) E imT*, so that (ker T)~ s;.; im T*. The proof is complete. Corollary 3.71 Let E and F be Banach spaces, and let T following assertions are equivalent:
0 E
B( E, F). Then the
(a) imT is closed in F;
(b) im T* is weak-*-closed in E* ; (c) im T* is norm-closed in E* . Proof (a) =}(b) Set Y = imT, a closed subspace of F, and let J : Y ---4 F be the inclusion map. Define TI : E ---4 Y by setting Tix = Tx (x E E), so that TI E B( E, Y) with im TI = Y. By Corollary 3.70, the dual operator Ti is injective, with weak-*-closed range. However, T = J a T 1 , and so T* = Ti a J*. But we know that J* : F * ---4 Y* is surjective, so im T* = im Ti and (b) is proved.
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Banach spaces
(b) :=;. (c) This is trivial. (c):=;.(a) Now set Y = imT, and let J : Y ----+ F be the inclusion map. Define Tl : E ----+ Y by setting T1x = Tx (x E E), so that Tl E 13(E, Y) with im Tl = im T and J 0 Tl = T. Now im Tl = Y, so that Ti : Y* ----+ E* is injective. Also, T* = Ti 0 J* and J* : F * ----+ Y* is surjective, so that im Ti = im T*, which is norm-closed in E*. It follows from Corollary 3.70 that im Tl = Y, which is closed in F. Since im T = im T 1 , the proof is complete. 0
3.17 Short exact sequences. The main results of this section may be neatly summarized in the language of exact sequences. Let X, Y, and Z be vector spaces (over the same field OC), and let S : X ----+ Y and T : Y ----+ Z be linear maps. Then the sequence
X~Y~Z is said to be a complex if im S ~ ker T, and exact at Y if im S = ker T. We now write 0 for the vector space {a}. The only linear maps O---+X or X ---+0 are necessarily the zero map, which is usually not written in a diagram, and so a complex
o---+X~Y is exact at X if and only if S is injective; similarly, a complex
Y~Z---+o is exact at Z if and only if T is surjective. A short sequence of vector spaces and linear maps is a sequence
o ---+
S
T
X ---+ Y ---+ Z ---+ 0;
the sequence is exact if it is exact at X, Y, and Z, i.e. it is such that S is injective, im S = ker T, and T is surjective. In particular, it necessarily follows . that, in this case, we have Z ~ Y / im S as vector spaces. Note also that a short sequence, as above, is a complex if and only if TS = O. We now consider complexes of Banach spaces and continuous linear maps, often written £: O---+X~Y~Z---+O. (The symbol £ is just a way of giving a name to the sequence). We note that by 'exactness' of £, we shall continue to mean the purely algebraic notion just explained. We remark that, if £ is a short exact sequence of Banach spaces and continuous linear maps, then necessarily Sand T have closed ranges, being equal, respectively, to ker T and Z. Also, using the open mapping theorem, we see that the induced linear isomorphism Z ~ Y / im S is also a homeomorphism.
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Introduction to Banach Spaces and Algebras
Given E, there is the naturally induced dual complex E*, given by
E* : 0
f--
X*
?
Y*
Z
Z*
f--
0.
Note that, in the case where imS ~ kerT, so that TS = 0, it follows that S*T* = (T S)* = 0* = 0, so that im T* ~ ker S*. Also, if im T* ~ ker S*, then 0= S*T* = (TS)*, so that also TS = 0 and therefore imS ~ kerT. Propositions 3.67 and 3.68 may be stated in terms of particular short exact sequences. Indeed, let E be a Banach space, and let M be a closed subspace. We now write J : M -+ E for the inclusion map and 7r : E -+ E / M for the quotient map. Then we have the 'canonical' short exact sequence: EM: 0 ---->
M
J ---->
E
7r ---->
E/ M
---->
o.
The dual sequence is:
E~I: 0
f--
M*
J:-
E* ~ (E/M)*
f--
O.
From Propositions 3.67 and 3.68, we know that: (i) J* is surjective; (ii) ker J* = M l.. = im 7r*; (iii) 7r* is injective. Thus EM is exact and, even more strongly, 7r* is isometric and J* is a 'quotient map', in the sense that the norm of M* is precisely the quotient norm on E* / M l... Theorem 3.72 Let E be a complex of Banach spaces and continuous linear
maps, as above. Then E is exact if and only if E* is exact. Proof Suppose that E is exact. Then we shall prove that E* is also exact. (i) Exactness at X*. The map S is injective, so that, by Lemma 3.61(ii), im S* is weak-*-dense in X*. But also S has closed range, and so, by Corollary 3.70, (a) =} (d), imS* is weak-*-closed. Hence imS* = X*. (ii) Exactness at Z*. The map T is surjective, so that T* is injective (with weak-*-closed range) by Corollary 3.70, (a) =} (d). (iii) Exactness at Y*. We know that im T* ~ ker S*. As noted in the proof of exactness at Z*, we know that im T* is weak-*-closed. Thus, since we have (im T*) T = ker T = im S, this result follows because
imT* = (imT*)Tl.. = (imS)l.. = kerS*. For the converse, we suppose that E* is exact, and prove that E is exact. (i) Exactness at X. Since kerS = (imS*)T = (X*)T = {O}, E is exact at X. (ii) Exactness at Z. Since T* is injective and has closed range, we have imT = Z, by Corollary 3.70. (iii) Exactness at Y. We know that im S ~ ker T, and also that imS = (im S)l..T = (ker S*) T = (im T*) T = ker T . But im S* = X*, so that, by Corollary 3.71, im S is closed. Thus im S and the proof is complete.
=
ker T, 0
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Banach spaces
Notes The theorems given above are, with the Hahn-Banach theorem, the great pillars of functional analysis. They are expounded, sometimes in more general form, in all the books on functional analysis; see, for example, [63, Chapter II] and [144, Chapter 2]. We proved the Tietze extension theorem in Theorem 3.39. One could ask if the extension go of fa can be 'chosen linearly'. This is answered in certain cases by a theorem of Borsuk. Let K be an infinite, compact, metric space, and let F be a closed subset of K. Then there is a linear map T : C(F) - t C(K) with (Tf) I F = f (j E C(F» and such that 111'11 = 1 and T(XF) = XK. This is related to Milutin's theorem, which states that C(K) is linearly homeomorphic to C(ll) for each uncountable, compact metric space K. See [2, Section 4.4]) for these results. There are many generalizations of the open mapping and closed graph theorems. First, there are versions for more general topological vector spaces than Fn§chet spaces; see [144]. Secondly, one can replace the vector spaces by various so-called topological groups. Thirdly, it is sometimes not necessary for, say, the graph of a linear map to be closed, but only that it be an analytic topological space; an analytic space is defined to be the continuous image of complete and separable metric space. For this latter generalization, see [47, Appendix A]. The separating space and the stability lemma are discussed in some detail in [47, Section 5.2]; some history is given there. The results have many generalizations. Let E and F be Banach spaces, and let l' E B(E, F). Our dual operator 1'* is often called the adjoint of T; see [144, Chapter 4]. It is denoted by T' in some texts, including [47]. The notions of a complex and of an exact sequence arise in many different contexts in mathematics, and are often formulated as holding in a particular 'category'. (This terminology has no connection with that of 'Baire category', in Section 1.7.) Exact sequences arise in algebraic topology, in group theory, and in pure algebra, for example, as well as in Banach space theory. Classic texts on homological algebra that use this language are [119, 120, 164]; for applications within functional analysis, see [92, 93], for example. These notions are developed substantially within the cohomology theory of Banach algebras [47, Section 2.8]. Complexes and exact sequences also arise in the theory of analytic function of several complex variables; we shall make a brief mention of Dolbeault cohomology groups in Section 8.5. lI!#
Exercise 3.13 Let E be a Banach space, and let M be a closed subspace of E. Give a direct proof that (ElM; II·IIEIM) is a Banach space (cf. Theorem 3.48). Indeed, let (Xn + M)n>l be a Cauchy sequence in ElM. Define inductively a subsequence (xnkh2:l of (X n )n2:l such that IIXnk - xnk+l + < 2- k for each k E 1':1. Then inductively choose a sequence (Ykh2: l in E such that, for each k E 1':1, we have Yk E x nk + M and IIYk - Yk+lll < Tk+l . Using Exercise 2.4, show that (Ykh>l has a limit, say Y E E, and then, using Exercise 1.4, that (Xn + M)n2:1 converges to -Y + M.
Mil
Exercise 3.14 Let (Xn)n2:l be a sequence in s, and take p > 1 with conjugate index q. Suppose that the series L~=l XnYn converges for each sequence (Yn)n2:l in £q. Deduce from the uniform boundedness theorem that (Xn)n2:l belongs to £P. Exercise 3.15 Let E be the Banach space co. For n 2': 1, define
Pn
:
(x)
f-4
(Xl, ... ,Xn,O,O, ... ),
E
-t
E,
so that Pn is 'projection onto the first n coordinates'. Note that (Pn )n2:1 is a sequence in B(E) such that Pnx - t X = lEX (X E E), but that IIIE - Pnll = 1 (n E 1':1). This shows
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Introduction to Banach Spaces and Algebras
that we cannot conclude that theorem, Theorem 3.36.
limn~=
'1' in B(E, F) in the Banach-Steinhaus
'l'n
Exercise 3.16 (i) Let E be the normed space (coo; 11·11=). For x = (Xnk:::1 E E, define Tx = (Xnln)n~l' Check that '1' : E --> E is a continuous linear isomorphism, but that '1' is not a linear homeomorphism. Why is this not a contradiction of Banach's isomorphism theorem, Corollary 3.41?
(ii) Let E
= (C l(IT); I· In) and F = (C(IT); I· In)' Define D:ff--tj',
E-->F.
Show that D has a closed graph, but that D is not continuous. Why is this not a contradiction of the closed graph theorem, Theorem 3.45? (iii) Let (E; 11·11) be an infinite-dimensional Banach space, and take (e", : ex E A) to be basis for E, as in Example 1.5(i). We may suppose that Ileall = 1 (ex E A). For x = L: Aaea E E (the sum is finite and uniquely defines x), set Illxlll = L: IAal. Then (E; 111·111) is a normed space. Show that the identity map t : (E; 11·11) --> (E; 111·111) has a closed graph. Clearly, the projection maps 7r(3 : L:A",ea f--t A(3 are continuous on (E; 111·111) for each (3 E A. Assume towards a contradiction that infinitely many of these maps are continuous on (E; 11·11), say 7r(3n is continuous for each n E N. Set En = ker7r(3" (n EN). Then each En is closed in (E; 11·11). Also, for each x E E, we have 7r(3" (x) = 0 for all but finitely many n E N, and so U::-'=l En = E. By the Baire category theorem, Corollary 1.22, there exists no E N such that int Eno =1= 0. But now Eno = E because Eno is a subspace of E. This is a contradiction. Deduce that t: (E; 11·11) --> (E; 111·111) is not continuous. Why is this not a contradiction of Corollary 3.52? Exercise 3.17 Let (Dn)n~l be the Dirichlet kernel of Section 2.17. Recall from page 92 that IF(Dn)lz = 1 (n E Z) and from Exercise 2.35 that IIDnill ~ logn as n --> 00. Deduce from Banach's isomorphism theorem that A(Z) F. Show that S E B((EIM) x G, F) is a linear isomorphism, and apply Banach's isomorphism theorem.
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Banach spaces
Exercise 3.20 Define 1': ((Xl,X2),Y) f-+ (Xly,X2Y), ]R2 x]R --+ ]R2. Show that l' is a continuous, bilinear surjection, but that there is an open neighbourhood U of (( 1, 1),0) in ]R2 x ]R such that T(U) is not open in ]R2. Thus there is no 'open mapping theorem for bilinear mappings'. Exercise 3.21 Let E be a Banach space, and suppose that F and G are vector subspaces of E such that E = F EEl G algebraically. Then the direct sum is topological if the projections from E onto both F and G are continuous. Show that the graph of the projection Pc of E onto G is
Gr(Pe)
=
{(x,y) E E x E: y E G, x - Y E F}.
Now suppose that both F and G are closed in E. Deduce from the closed graph theorem that the direct sum is topological. Exercise 3.22 Let E be a Banach space, and suppose that F and G are closed vector subspaces of E such that E = F + G algebraically. Prove that there is a constant a > 0 such that each x E E can be written as x = y + z, where y E F, z E G, and
liyll + Ilzll :s: a Ilxll·
Exercise 3.23 Let (E; II ·11) be an infinite-dimensional Banach space. (i) Use the fact that every vector space has a basis to show that there are discontinuous linear functionals on E. (ii) Let f be a discontinuous linear functional on E. Fix a E IC with a #- 1 and Xo E E with f(xo) = 1, and then set
Illxlll", = Ilx -
af(x)xoll
(x E E).
Show that 111·111", is a complete norm on E. (Hint: By Proposition 3.8(i), there is a closed subspace G of E such that E = ICxo EEl G; the projections onto ICxo and G are continuous.) Now take a,(3 E IC\ {I} with a #- (3. Show that 111·111", and 111·111,6 are not equivalent. Exercise 3.24 In the text, we deduced Banach's isomorphism theorem for Banach spaces, Corollary 3.41, from the open mapping theorem, Theorem 3.40, and we deduced the closed graph theorem, Theorem 3.45, from Banach's isomorphism theorem. In fact these three theorems are essentially equivalent. Indeed, first deduce Banach's isomorphism theorem from the closed graph theorem by considering the graph of the inverse mapping. Then deduce the open mapping theorem by using Lemma 3.47. Deduce an isomorphism theorem and a closed graph theorem for Frechet spaces from the open mapping theorem, Theorem 3.58. Exercise 3.25 Let E and F be Banach spaces, and take l' E B(E, F). (i) Prove that the map 1'** E B(E**, F**) is surjective if and only if Tis surjective, and that 1'** is injective with closed range if and only if l' is injective with closed range.
(ii) Take F = Co and E = F* = £ 1, so that E* = F** = £00. As in Proposition 3.22, there is a projection P : E** --+ E such that E** = E EEl G, where G = ker P ab=a·b,
AxA-+A,
known as multiplication or product, which satisfies the following axioms for all a,b,c E A and every,\ E lK: (i) a(bc) = (ab)c; (ii) '\(ab) = ('\a)b = a('\b); (iii) a(b + c) = ab + ac and (a + b)c = ac + be. Thus, an (associative) algebra is an algebraic structure that is both a ring and a vector space, where the addition of the ring is the same as the vector addition and multiplication by scalars relates to the ring multiplication by the axiom (ii) just given. The structure (A, . ) is a semigroup; it is the multiplicative semigroup of A.
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An algebra A is commutative if its ring multiplication is commutative, so that ab = ba (a, bE A); an algebra A has an identity element, say 1 or lA, provided that this element satisfies 1a = a1 = a for every a in A. It is evident that an identity element is unique whenever it exists. An algebra with an identity is a unital algebra. For example, let E be a vector space, and let £(E) be the space of linear endomorphisms of E, as on page 43. For S, T E £(E), the composition of Sand T is ST, defined by (ST)(x) = S(Tx) (.1: E E). The identity operator on E is denoted by I or Ie. Then £(E) is an algebra with respect to this product, and Ie is an identity of the algebra. Let A be an algebra. An ideal (or, more precisely, a two-sided ideal) I of A is a subset of A such that: (i) I is a vector subspace of A;
(ii) both AI and so the product in the algebra £ 1 (S) extends the product in S. A semigroup algebra £ 1 (S) is commutative if and only if S is abelian. The particularly important case is when S is a group, say G; now we say that (£l(G); 11.11 1 ; *) is the group algebraof G. For example, let G = (Z; +). Then
£1(Z)
=
{f =
(an)nEz :
Ilflll = n'f;= lanl < oo} ,
with the above convolution product specified by 8m * 8n = 8m +n (m, n E Z). Next, suppose that w : S ----+ jR+. is a weight on a semigroup S, so that
w(st) ::; w(s)w(t)
(8, t
E
S).
Then
£l(S,W)
= {f
E CS
:
Ilfllw:= i.::{lf(s)lw(s): 8
It is again easy to verify that (£ 1 (S, w); weighted semigroup algebra on S.
II . IIw ; *)
E S}
< oo} .
is a Banach algebra; it is a D
Example 4.5 Let n E N. Then, as in Section 2.3, we denote by Mn the set of all n x n matrices over C, identified with £(cn). Then Mn is an algebra with ,an identity. It is a Banach algebra with respect to any of the (equivalent) norms described in Part I. An element of Mn will often be written as T = (a,]). Note that such a matrix is invertible if and only if det T -I=- 0; the function det is continuous on M n , and the set of invertible matrices is a dense, open subset of Mn. It is an easy exercise to check that Mn is a simple algebra. D The next example is the 'infinite-dimensional generalization' of the previous example.
Example 4.6 For each complex Banach space E, the algebra B(E) of all bounded linear operators on E, equipped with the operator norm and composition as
Introduction to Banach Spaces and Algebras
160
product, is another important example of a unital Banach algebra. This algebra is a subalgebra of £(E). It is, of course, a non-commutative algebra (provided that dim E > 1). A central role will be played by certain closed subalgebras of B(H), where H is a Hilbert space. Let Xo E E and fo E E*. Then, as in Example 2.17, the rank-one operator
Xo ® fo : x
f---+
fo(x)xo,
E
--t
E,
belongs to B(E). The space of continuous finite-rank operators on E is F(E). For each T E B(E), Xo E E, and fo E E*, we have
To (xo®fo) =Txo®fo
and
(xo®fo)
0
T=xo®T*(fo),
(*)
and so we see that F(E) is an ideal in B(E). We claim that F(E) is the minimum (with respect to inclusion) non-zero ideal in B(E). Indeed, let J be a non-zero ideal in B(E), and take S E J with S =I- O. Then there exist non-zero elements Xo and Yo in E with Sxo = Yo. Choose fo E E* with fo(Yo) = 1, take T = Xl ® h, where Xl E E and h E E*, and set RI = Xl ® fo and R2 = Xo ® h· Then R I , R2 E B(E), and
R ISR2x = h(x)RIYO = h(X)XI = Tx
(x E E),
so that T = R I SR 2 E J. It follows that F(E) l be a sequence in G(A), and suppose that an ----+ a E A as n ----+ 00. Then a Eo G(A) if and only if SUPnEN lIa;;-lll < 00. (ii) Suppose that a E 8G(A). Then there is a sequence (bn)n>l in A such that Ilbnll = 1 (n EN), but such that bna ----+ 0 and ab n ----+ 0 as n ----+ 00. In partzcular, a has neither left nor right inverse. Proof (i) Suppose that a E G(A). Then, by Corollary 4.12, a;;-l ----+ a-I as n ----+ 00, so that certainly sUPnEN Ila;;-lll < 00. Conversely, suppose that (a;;-l)n>l is bounded, say Ila;;-lll :::; K (n EN). Choose N E N such that Klla - aN11 < 1. Then
a=aN+(a-aN)=aN(l+aj/(a-aN)) .
Since Ilai\/(a - aN)11 :::; Klla - aN11 < 1, a is invertible by Lemma 4.10. [Remark: if a ~ G(A), then it follows that no subsequence of the above sequence (1Ia;;-lllk:::l can be bounded, so that even Ila;;-lll ----+ 00.] (ii) Let a E 8G(A). Then, since G(A) is an open subset of A, it follows that a ~ G(A), but that there is a sequence (an)n>l in G(A) such that an ----+ a. By (i), we may suppose that Ila;;-lll----+ 00 as n ----+ 00. Define bn = a;;-l/lla;;-lll (n EN). Then IIbnll = 1 (n E N) and IIbnall = Ila;;-l(a - an)
+ llllla;;-lll-l
so that bna ----+ O. Similarly, ab n ----+ O.
::::: Iia - anll
+ Ila;;-lll-l
----+ 0
as
n ----+
00,
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Banach algebras
To see that a can have neither left nor right inverse, just notice that if, say, ba = 1, then we have 1 = Ilbnll = Ilbabnli :::; Ilbllllabnll -+ 0 as n -+ 00, a contradiction. [Remark: this latter argument in fact proves the stronger result that a can have neither left nor right inverse in any algebra B in which A is included as a closed subalgebra.] 0
Theorem 4.14 Let A be a unital Banach algebra, and let J be a proper (left) ideal of A. Then its closure] is also a proper (left) ideal. Further, every maximal ( left) ideal of A is closed. Proof Certainly] is a (left) ideal. Since J i- A, we have J n G(A) = 0. But G(A) is open by Corollary 4.11, and non-empty (since lA E G(A)). Then also ] n G(A) = 0, so that] i- A. Let M be a maximal (left) ideal in A. Then M is also a (left) ideal in A, and clearly M ~ M i- A. By the maximality of M, we have M = M, i.e. Mis closed. 0 The fact that maximal ideals in Banach algebras with identities are closed does not extend to Frechet algebras; for this, see Example 4.53, below. Recall that, if E is a Banach space, then an operator T E B(E) is said to be invertible if and only if there is some S E B(E) such that ST = TS = Ie, where IE is the identity operator on E (see Section 2.5). It is clear that this is precisely the same as T being an invertible element of the Banach algebra B(E). The following theorem summarizes some special cases of Lemma 4.10 and of Corollaries 4.11 and 4.12. We shall write G(E) for the group of invertible, bounded linear operators on E.
Theorem 4.15 Let E be a Banach space. (i) Let T E B(E) with IIIe - Til < 1. Then T E G(E). (ii) The set G(E) is an open subset of B(E). (iii) The map T f-+ T- 1 is a homeomorphism of G(E) onto itself.
0
4.5 The spectrum. We note that, in this section, it is important that all algebras are over the complex field, C. Let A be any complex algebra with an identity, and let a E A. Then the spectrum of a, denoted by SPA (a) (or usually just Sp a) is defined as: SPA (a) =
{A
E
C: Al - art G(A)}.
We remark that, in the trivial case where A = {O}, the zero algebra, then I = 0 and the unique element of A is invertible, so that Sp A(O) = 0. In the case where A is a complex algebra without an identity and a E A, we define Sp A a = Sp A+ a . We have the obvious remark that, in this case, 0 E Sp a.
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Introduction to Banach Spaces and Algebras
Let A be an algebra. Then an idempotent in A is an element p such that p2 = p. Of course 0 is an idempotent, and the identity of a unital algebra is an idempotent; these are the trivial idempotents. But there may be other, non-trivial, idempotents. Suppose that A has an identity 1 and that p is an idempotent in A. Then 1 - p is also an idempotent. Here is a simple remark. Proposition 4.16 Let A be an algebra.
(i) Let p be an idempotent in A. Then Spp ')l-p. Then (>'l-p)b = b(>'l-p) = (>. - >.2)1. Thus, if >. -=I- 0,1, necessarily>. Sp p.
rt
(ii) It is sufficient to note that 1 + ba is invertible if and only if 1 + ab is invertible. For this, suppose that c(l + ab) = (1 + ab)c = 1. Then (1 - bca)(l + ba) = 1 + ba - bca - bcaba = 1 + ba - bca - b(l - c)a = 1,
o
and also (1 + ba)(l - bca) = 1.
An element a in an algebra A is nilpotent if there exists n E N such that an
= o. For example, the matrix T =
(~ ~)
is such that T2 = 0, and so T is nilpotent in the algebra M 2. It is clear that Sp a . E C with>' -=I- o. Then the inverse of >'1 - a is
1(
>.
a an-I) 1+-+···+-- . >. >.n-I
An element a in an algebra A is said to be quasi-nilpotent if either Sp a = 0 or Sp a = {O}. We shall write Q(A) for the set of quasi-nilpotent elements of A; in general, Q(A) is not a vector subspace of A. Every nilpotent element is quasi-nilpotent; we shall soon see some quasi-nilpotent, non-nilpotent elements of Banach algebras. Let A and B be two complex algebras with identities, and let 8 : A ----> B be a unital homomorphism. Then clearly 8(G(A)) SPA:a. D
Theorem 4.19 (Gel'fand-Mazur theorem) Let A be a normed division algebra over C Then A ~ C Proof Let a E A. By Corollary 4.18, Spa =I- 0, and so there exists an element, say J-L, in Spa. We have J-Ll - a tJ- G(A). But then, since A is a division algebra, J-Ll - a = o. The injective mapping A f---+ Al is thus an isomorphism of C onto A (and this map is clearly an isometry if A is unital). D
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Introduction to Banach Spaces and Algebras
Example 4.20 Let A be an algebra of complex-valued functions on some set X (with the algebraic operations defined pointwise on X). Suppose that A can be given some Banach-algebra norm, I . II. Then every function in A is bounded. Indeed, IJ(x)1 :'S IIJII for all J E A and x E X; this is immediate from Theorem 4.17 since it is clear that J(x) ESp J (x EX). 0 Note that, for a normed algebra A, it may be that the spectrum of an element of A is neither bounded nor closed in C. For example, let A = O(q be the algebra of all entire functions on C, so that A is a normed algebra for the uniform norm I·I~. Then
G(A) = {J E A : Z(J) = 0}
and
Sp J
= J(q
(J E A) ,
as is immediately checked. Now set exp(z) = e Z (z E q, so that exp E A. Then SPA(exp) = exp(q = C \ {O}. Proposition 4.21 Let A be a Banach algebra, let a E A, and let U be an open neighbourhood oj Sp a in C. Then there exists 6 > 0 such that Sp b c U whenever lib - all < 6. Proof We may suppose that A is unital. Assume to the contrary that the result is false. Then there is a sequence (bn)n~l in A with limn->oo bn = a and such that there exists An E Sp bn \ U for each n E N. Since (b n )n>l and hence (An)n>l - is bounded, so is (A n )n>l, - has a convergent subsequence. Thus we may suppose that An -> IL, say. Clearly, IL 1- U because U is open. But then AnI - bn -> ILl - a E G(A) as n -> 00. Since G(A) is open, AnI - bn E G(A) for sufficiently large n E N, contrary to the fact that An E Spb n for each n E N. 0 Hence the result holds. Lemma 4.22 (Spectral mapping property for polynomials) Let A be a complex algebra wzth an identity, let a E A, and let p be a complex polynomial. Then
Spp(a)
=
{p(A) : A E Spa}.
Proof We may suppose that degp = n 2': 1 because the result is trivial for a constant polynomial. For every A E C, simple algebra shows that p(A)I- p(a) = (AI- a)b for some bE A such that ab = ba. It is immediate that p(A) E Spp(a) whenever A E Spa. Conversely, let IL E C, and suppose that AI, ... , An are the roots (not necessarily distinct) of p( A) - IL in C. Thus
p(A) - IL
=
C(A - Ad ... (A - An)
(A
E
q,
where C is a non-zero constant. It follows that
p( a) - ILl = C( a - All) ... (a - An 1) EA. But then, in the case where ~ E Sp p( a), it follows that Ak E Sp a for at least one k = 1, ... , n, and hence ~ = p(Ak) E p(Sp a). 0
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Banach algebras
Let A be a Banach algebra, and let a PA(a) (or just pea)) of a to be
A. We define the spectral radius
E
PA(a) = sup{I.\1 :.\
E
Spa}.
From Theorem 4.17, it is immediate that pea) ::; Iiali. For example, pea) = 0 if and only if a is a quasi-nilpotent element of A. We now give the famous Beurling-Gel'fand spectral radius formula for an element in a Banach algebra.
Theorem 4.23 (Spectral radius formula) Let A be a Banach algebra, and let a E A. Then PA(a) = lim Ilanll l / n = inf Ilanll l / n . n--->oo nEN
[Remark. It is elementary to prove the existence of the limit limn--->oo Ila n III/n and to show that it is equal to infnEN Ila n III/n just by using the obvious fact that Ilaffi+nli ::; Ilaffilillanil (m,n EN). However, the existence of this limit emerges anyway in the proof to follow.] Proof We mula would Let .\ E that pea) ::; Now let
may suppose that A is unital (it being clear that the proposed foryield the same value for any two equivalent norms on A). Sp a. Then, as an easy application of Lemma 4.22, .\n E Sp an, so Ilanll l / n (n EN). Thus pea) ::; infnEN Ilanll l / n .
g(z) = (1 - za)-l
(Izl
< 1Ip(a))
(where we define 9 on all of p(a), so that liz ~ Spa, and hence 1 - za is invertible in A. Thus the element g(z) is always well defined. By Corollary 4.12, 9 is continuous, and so, for each R > p(a), we have
MI/R(g)
:=
sup{llg(z)11 : Izl = 11R}
oo Ilanll 1 / n ::; R. It follows that limsuPn--->oo Ilanll 1 / n ::; pea).
Introduction to Banach Spaces and Algebras
170 We have therefore shown that
p(a) ::::: inf Ilanll l / n ::::: liminf Ilanll l / n ::::: lim sup Ilanll l / n ::::: p(a). n~oo
nEN
n---+CX)
Thus limn->oo Ilanll l / n exists and equals p(a).
o
For example, limn->oo Ilanll l / n = 0 for each a E Q(A). Also,
p(a 2) = p(a)2
(a E A).
Corollary 4.24 Let A be a Banach algebra, and let a, bE A with ab = ba. Then
p(ab) ::::: p(a)p(b). Proof Since ab = ba, we have (ab)n = anb n (n E N), so that lI(ab)nlll/n::::: Ilanlll/nllbnlll/n
(n E N).
o
The result is now immediate from Theorem 4.23.
Another proof of this last corollary will be given below, and there is a further proof in Corollary 4.48(ii), using the theory of characters on commutative Banach algebras. By using Lemma 4.8, we see that there is another description of the spectral radius which will be useful in a few places. Corollary 4.25 Let (A; II ·11) be a Banach algebra, and let E be the set of all algebra-norms on A that are equivalent to II . II. Let a E A. Then
PA(a) = inf{lllalll :
111·111 E
E}.
Proof Certainly, by Theorem 4.17, PA(a) ::::: inf{lllalll : 111·111 E E}. Now take M > PA(a). We have PA(M-1a) < 1, and so it follows from Theorem 4.23 that S := {M-na n : n E N} is bounded, and this set is certainly closed under multiplication. By Lemma 4.8, there is some III . III E E such that Illslll ::::: 1 for every s E S. In particular, Illalll ::::: M. The result follows. 0 Corollary 4.26 Let A be a Banach algebra, and let a, b E A with ab
p(ab) ::::: p(a)p(b) and p(a
+ b)
::::: p(a)
= ba. Then
+ p(b).
Proof Fix E > 0, and set u = aj(p(a) +E) and v = aj(p(b) +E). Since uv = vu, the set S = {uJ,vk,uJv k : j,k E N} is a bounded semigroup in (A,·). By Lemma 4.8, there is an algebra-norm III . Ilion A that is equivalent to II . II such that Illalll < p(a) + E and Illblll < p(b) + E. But now
p(a + b) ::::: Ilia + bill::::: Illalll + Illblll < p(a) + p(b) + 2E. This holds true for each E > 0, and so p(a + b) ::::: p(a) + p(b). Similarly, we have p(ab) ::::: p(a)p(b).
o
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Banach algebras
4.6 The spectrum of an operator and invariant subspaces. a vector space, and let T E £(E). For A E e, we define
E(A) = {x
E
Let E be
E: Tx = AX}.
As in elementary linear algebra, A is an eigenvalue of T if E(A) -=I- {O}; in this case, E(A) is the corresponding eigenspace and each x E E(A) with x -=I- is a corresponding eigenvector of T. The eigenvalues of T form the point spectrum, Spp(T), of T. Let E be a Banach space, and let T E B(E). An invariant subspace for T is a closed vector subspace F of E such that Tx E F (x E F), i.e. T(F) 1 in H with Ilxnll = 1 (n E N) and with TX n - AX n ---* 0 as n ---* 00. But then
A = n----+(X) lim (Txn' xn)
lim (xn' Txn) =
=
n---+(X)
>:,
so that A is real. Thus SpT ~ JR, and so SpT ~ [-IITII, IITlll. To show that at least one of ± IITII belongs to Sp T is equivalent to showing that IITI12 E Sp (T2), i.e. we must show that IITI12 is an approximate eigenvalue of T2. For this, take (Xnk::>1 in SH such that IITxnl1 ---* IITII as n ---* 00. Then II(T2 -IITI12)XnI1 2 = IIT2Xnl12 :::: IITI14
+ IITI14 -
+ IITI14 -
211TI1211 Txnl1 2
211TI1211Txnl12
---*
0
as
n
---* 00,
as required. (ii) Since U is unitary, IIUxl1 = Ilxll for all x E H, and so 11U11 = 1 and Sp U ~ {z : Izl :::: I}. But also U- 1 is unitary, and so Sp U- 1 ~ {z : Izl :::: I}. Clearly, SpU- 1 = {Z-1 : z E SpU}, so in fact SpU ~ {z: Izl = I}. Now assume towards a contradiction that Sp U = 0, and consider the operator
T = i(U + IH)(U - I H )-1 (noting that U - IH is invertible since 1
rf- Sp U by hypothesis). Then
T* = -i(U* + IH)(U* - I H )-1 = -i(U* = -i(IH + U)(IH - U)-1 = T, so that T is self-adjoint, and clearly U elementary from (i) that Sp U = giving the result.
{
+
X - .i : X -1
x
(T
E
+ IH)U((U* + iIH )(T
- I H )U)-1
- iIH )-1. Then it is
Sp T } =I- 0 ,
o
175
Banach algebras
4.8 The spectrum of a compact operator. Let E be a non-zero Banach space, and consider an operator T E K(E), so that T is a compact operator. Compact operators were introduced in Example 2.17. Our aim is to describe Sp T. As in Section 4.6, E(A)is the eigenspace corresponding to an eigenvalue A and Spp(T) is the point spectrum of T. Lemma 4.32 Let E be a Banach space, let T E K(E). Take A E SpT with Ai- o. Then:
(i) E(A) is finite-dimensional; (ii) im(T - AI) is closed in E; (iii) either there exists x E E with x f E E* with f i- 0 such that T* f = Af·
i- 0
such that Tx
= AX, or there exists
Proof (i) As above, E(A) is an invariant subspace for T. Set S = T I E(A). Then S E K(E(A)). For X E E(A), we have S(X/A) = x, and so S is a surjection. By Proposition 3.64, E(A) is finite-dimensional.
(ii) By Proposition 3.8(i), E(A) is complemented in E, and so there is a closed subspace G of E such that E = E(A) EEl G. Define
S :x
f----+
AX - Tx ,
G
---*
E,
so that S E B(G,E), S is injective, and imS = im(T - AI). We shall show that S is bounded below. Assume to the contrary that S is not bounded below. Then there exists a sequence (Xn)n>l in Se with SX n ---* O. Since T is compact, we may suppose that (TXn)n>1 is convergent in E, say TX n ---* Xo as n ---* 00. It follows that AX n ---* Xo, so that Xo E G and, further, SXo = limn->(X) ASX n = O. Since S is injective, Xo = O. But Ilxnll = 1 (n EN), and so Ilxoll = IAI > 0, a contradiction. By Proposition 2.23(i), im(T - AI) is closed in E. (iii) Assume that T - AI is an injection and that im(T - AI) is dense in E. By (ii), T - AI is a surjection, and hence a bijection. By Banach's isomorphism theorem, Corollary 3.41, T -AI E G(E), a contradiction ofthe fact that A E Sp T. Suppose that T - AI is not an injection. Then there exists X E E with x i- 0 such that Tx = AX. Suppose that im(T - AI) is not dense in E. Then there exists f E E* with f i- 0 such that f 0 (T - AI)(X) = 0 for all X E E, and this implies that T* f = Af. 0 Lemma 4.33 Let E be a Banach space, and let T E K(E). Suppose that (An)n~l is a sequence of distinct complex numbers and that (Xn)n~l is a sequence of nonzero vectors in E such that TX n = AnXn (n EN). Then An ---* 0 as n ---* 00. Proof We first note that {xn : n E N} is linearly independent. For assume inductively that {Xl, ... ,xn } is linearly independent, and then assume that Xn+l
= aixi + ... + anxn
Introduction to Banach Spaces and Algebras
176 for some al, ... ,an E C. Then
0= (An+1I - T)(Xn+l) = al(A n+1 - AI)XI
+ ... + an(An+1
- An)Xn,
and so al = ... = an = 0 because An+1-AJ -I- 0 for j = 1, ... , n, a contradiction. Thus {Xl, ... , Xn+1} is linearly independent, and the induction continues. Assume to the contrary that An f+ O. Then, by passing to subsequences, we may suppose that IAnl 2: E for some E > O. For n E N, set
Ln = lin{xI"'" x n }, a finite-dimensional, and hence closed, vector subspace of E. Since {xn : n E N} is linearly independent, Ln m, the vector Zm,n := (Yn - A:;;ITYn) + X;;,1TYm belongs to L n -
l .
It follows that
IIT(A:;;IYn) - T(A;;/Ym) II = llYn - zm,nll > 1/2. Hence no subsequence of (T(A;;IYm))m>1 converges. This is a contradiction of the compactness of T because IIA;;IYm l/E (m EN). Thus An ----+ 0 as n ----+ 00. D
f::::
Theorem 4.34 Let E be a Banach space, and let T E K(E). Then SpT is either a finite set or an infinite, countable set with 0 as a lzmzt point. We have o E Sp T whenever E is infinite-dimensional. Each non-zero A E Sp T is an isolated point of Sp T and an ezgenvalue, and the corresponding eigenspace E(A) is finite-dimensional. Proof In the case where E is infinite-dimensional, T ~ G(E), for otherwise E K( E), and so 0 E Sp T. Since Sp T is a non-empty, compact subset of C, the conclusion will follow from the fact that every non-zero point of Sp T is isolated in Sp T. Take A E Sp T with A -I- O. Assume to the contrary that A is not isolated. Then there is a sequence (An)n2':1 of distinct points of Sp T with An ----+ A as n ----+ 00. By Lemma 4.33, only a finite number of the maps AnI - T fail to be injective. By Lemma 4.32(iii), only a finite number of the maps AnI* - T* are injective. However, by Schauder's theorem, Theorem 3.65, T* is compact, and so, by Lemma 4.33 applied to T*, we obtain a contradiction. Thus every non-zero point of Sp T is isolated. We have A E SpT. By Theorem 4.28, A E SPapT, and so there exists (Xn)n2':1 in SE such that (T - AI)X n ----+ 0 as n ----+ 00. Since T is compact, we may suppose that limn--->oo TX n = Y E E. But now limn--->oo AXn = Y with Ilyll = IAI > 0, so that y -I- O. Clearly, limn--->oo ATxn = Ty, and so Ty = )..y. Thus).. E SpT is an eigenvalue. By Lemma 4.32, E()") is finite-dimensional. D
Ie
a
177
Banach algebras
We shall now prove the following striking result of Lomonosov on hyperinvariant subspaces for compact operators. Theorem 4.35 (Lomonosov's theorem) Let E be an infinite-dimensional Banach space, and let T E K(E) with T '" o. Then T has a proper hyper-invariant subspace. Proof We first remark that we can suppose that T is quasi-nilpotent. For otherwise there exists A E Sp T with A '" 0; by Theorem 4.34, A is an eigenvalue, and the eigenspace E(A) is finite-dimensional, and hence a proper hyper-invariant su bspace for T. Let 21 = {T} C = {S E 8(E) : ST = T S}, so that 21 is a unital subalgebra of 8(E). Assume towards a contradiction that there is no proper, closed subspace of E which is invariant for each S E 21. For x E E, set 21x = {Sx : S E 21}. The assumption implies that, for each x '" 0, we have 21x = E, for otherwise 21x '" E would be a proper invariant subspace of E for each S E 21. For y E E, we write B(y) for the open ball B(y; 1). We choose a vector y E E such that the compact, convex set C := T(B(y)) does not contain the zero vector OE, say r := inf{llzll : z E C} > o. For each x E C, choose Ax E 21 with Axx E B(y), and then take an open neighbourhood Ux of x such that Ax(Ux ) ~ B(y). Since C is compact, there are finitely many operators AI, ... ,An E 21 such that n
C ~
U A;I(B(y)). )=1
We have Ty E C, and so there exists jl E {1, ... ,n} with A)ITy E B(y). But then T A)l Ty = A)l T2y E C, and so there exists j2 E {I, ... , n} with A)2A)IT2y E B(y). Continuing in this way, we obtain, for each mEN, vectors Ym
Set k
= A)m ... A)l Tmy
= max{IIA 1 11, ... , IIAnll}. 1
E
B(y) .
Then r ~ IIYml1 ~ k m IITmllllyl1 (m EN), and
= lim rl/m ~ lim k IIT m ll l / m = kp(T) = 0, rn----+CX)
m~CX)
the required contradiction.
D
4.9 Spectrum relative to a subalgebra. about spectra relative to closed subalgebras. For a Banach algebra A and a E A, let
There are a few useful results
RA(a) = C \ SPAa.
Recall that RA(a) is an open subset of the complex plane C and that RA(a) :2 {A E C :
IAI > PA(a)} :2
The set R A (a) is called the resolvent set of a.
{A
E
C : 1>'1 > lIall}·
178
Introduction to Banach Spaces and Algebras
Theorem 4.36 Let A be a unital Banach algebra, let B be a closed subalgebra of A with 1A E B, and let a E B. Then SPAa t;;; SPBa and RB(a) is a relatively open-and-closed subset of RA (a). Proof Clearly, RB(a) = {A E RA(a) : (AlA - a)-l E B} t;;; RA(a), so that Sp A a t;;; Sp Ba. Also, RB (a) is certainly open relative to RA (a) since it is even an open subset of Co By Corollary 4.12, the map A f---> (AlA - a)-l is a continuous mapping from R A (a) into A. Since B is closed in A, it follows that R B (a) is closed relative to RA(a). 0
We now need an elementary topological remark. Let K be a non-empty, compact subset of C. Then C \ K is open, and each of its components is also open. The set of these components is either finite or countably infinite; precisely one of these components is unbounded. Corollary 4.37 Let A be a unital Banach algebra, let B be a closed subalgebra of A with 1A E B, and let a E B.
(i) Let U be a component of RA(a). Then either U is also a component of RB(a) or Un RB(a)
=
0.
(ii) The open sets RA(a) and RB(a) have the same unbounded component with respect to C. (iii) 8SPBa t;;; 8SPAa. Proof (i) This is immediate from Theorem 4.36.
(ii) Let U be the unbounded component of RA(a). Then Un RB(a) ;;2 {A E C :
IAI > Ilall}·
By (i), U must be a component of RB(a), and it is then clearly the unbounded component. (iii) Let A E 8SPBa = SPBa n RB(a). Assume towards a contradiction that RA(a). Then). E RB(a) because RB(a) is closed relative to RA(a). But this contradicts the fact that A E Sp Ba, and therefore)' E SPA a. Since also A E RB(a) t;;; RA(a), we have A E 8SPAa. 0 ). E
Remark: By Corollary 4.37(i),(ii), the compact set SPBa is the union of SPA a with certain bounded components of RA(a). In particular, if RA(a) is connected, then Sp Ba = SPA a. We state a special case of this explicitly. Corollary 4.38 Let A be a unital Banach algebra, let B be a closed subalgebra of A with 1A E B, and let a E B. Suppose that SPA a C JR. Then SPBa = SPAa. 0
Moreover, if B is taken to be as small as possible, then all the 'holes' in SPA a are filled in. This is the content of the next corollary.
179
Banach algebras
Let K be a non-empty, compact subset of C. Then the polynomial hull of K is defined to be the set
K=
{z
E
C : Ip(z)1 :::; IplK for all polynomials p};
here, I· IK is the uniform norm on K. We say that K is polynomially convex if K = K. It is clear that, for any compact K m, is an isometry. Also, a + J E G(m), and hence m is the required extension. The converse is obvious. Exercise 4.8 Again, let Mn be the algebra of all n x n matrices over IC. The matrix units ofM n are the matrices E'J for i,j = 1, ... ,n; here, E'J is the matrix with 1 in the (l, J )th-position and 0 elsewhere. Show that the centre 3 (Mn) of Mn consists of the multiples of the identity matrix and that lin{E11 , ... , Enn} is a maximal commutative subalgebra of Mn. Exercise 4.9 Let H be the Hilbert space g 2, and consider the left and right shift operators, Land R, of Exercise 2.11 as elements of B(H). Show that RL = I H , but that LR i= I H. Calculate the spectra, sets of eigenvalues, and approximate point spectra of Land R. Exercise 4.10 Fix p 2: 1, and set E = gPo For n E N, let en be the characteristic function of {n}, as in Section 2.3. Let (Wn)n~l be a sequence in (0,1], and define T E B(E) by requiring that Ten = wnen+l (n EN). Then T is a weighted right shift operator on E. It is clear that IITII S 1 and that n{1,n(E) = {O} : n EN}. Show that Sp T contains 0 and that T has no eigenvalues. For ( E 'f, define U( E B(E) by requiring that U(e n = (ne n (n EN). Show that U( E G(E) and that (TU( = U(T, so that (T is similar to T. Deduce that SpT and SPapT are invariant with respect to rotation about the origin, and that Sp T is a disc (possibly degenerate) centred at the origin. Express the spectral radius of T in terms of the numbers W n . Operators of the above type can be used as examples of operators which do or do not belong to many different classes; see Exercise 5.10 and [114, Section 1.6].
185
Banach algebras
Commutative Banach algebras 4.11 Characters and ideals. Throughout this section, A will usually be a non-zero, commutative Banach algebra, but initially A will denote a more general algebra. Let A be a unital algebra. We shall write MA for the set of all maximal ideals of A. In the case where A is commutative, the quotient of A by a maximal ideal is a field. Let A be a unital algebra, and let B be a simple algebra. Then ker e is a maximal ideal in A whenever A ----> B is an epimorphism; in particular, ker 'P is a maximal ideal in A for each character 'P on A.
e:
Theorem 4.46 Let A be a commutative, unital Banach algebra. Then cI> A and the mapping 'P f---4 ker'P is a bijection from cI> A onto MA.
-I- 0
Proof Certainly ker 'P is a maximal ideal in A for each 'P E cI> A. Suppose that 'P,1/) E cI> A with ker'P = ker 1/J. Then, for each a E A, we have a - 'P(a)1 E ker'P = ker1/J, and so 1/J(a - 'P(a)l) = 0, i.e. 1/J(a) = 'P(a). Thus 'P = 1/J, and so the specified map is an injection. It remains to show that every maximal ideal is the kernel of a character on A. But, for each M E MA, the quotient AIM is both a field and a Banach algebra in its quotient norm. By the Gel'fand-Mazur theorem, Theorem 4.19, AIM ~ C, so that the quotient homomorphism Q : A ----> AIM 'is' a character, with ker Q = M. Thus the specified map is a surjection. Since MA -I- 0, we have cI>A -I- 0. D Let A be any Banach space, taken with the zero product, so that, in this product, ab = 0 (a, bE A). Then A is a commutative Banach algebra without an identity. Clearly, cI> A = 0; every vector subspace of co dimension 1 is a maximal ideal in A. Corollary 4.47 Let A be a commutative, unital Banach algebra, and let a
(i) a E G(A) if and only if'P(a) (ii) Spa (iii) p(a) (iii) a
E
= =
-I- 0 for
E
A.
each 'P E cI>A.
{'P(a) : 'P E cI>A}.
sup{I'P(a)1 : 'P
E cI>A}.
Q(A) if and only if'P(a)
= 0 for each 'P
E cI>A.
Proof (i) By elementary algebra, a E G(A) if and only if Aa = A (since A is commutative). But Aa is an ideal in A, and so it is proper if and only if it is included in some maximal ideal. Thus a E G(A) if and only if a does not belong to any maximal ideal of A. Hence, by the theorem, a E G(A) if and only if 'P(a) -I- 0 for each 'P E cI>A. D (ii), (iii), and (iv) These are now very simple deductions. We now give another proof of a more general version of Corollary 4.26.
186
Introduction to Banach Spaces and Algebras
Corollary 4.48 Let A be a Banach algebra, and let a, b E A with ab
(i) Sp (a + b) ~ Spa + Spb and p(a + b) ::; p(a) (ii) Sp (ab) ~ Spa· Spb and p(ab) ::; p(a)p(b).
= ba. Then:
+ p(b);
Proof We may suppose that A is unital. Let C = {a, b}CC. Then, by Corollary 4.42, Spca = SPA a, etc. Hence, we may apply Corollary 4.47(ii) to the commutative, unital Banach algebra C to see the results. 0
We next consider characters on commutative Banach algebras; recall from Theorem 4.43 that each character on a Banach algebra is continuous. We begin with some examples. Example 4.49 Let A = C(K), with K a non-empty, compact Hausdorff space, so that, as before, A is a commutative, unital Banach algebra. For every x E K there is the character cx, defined by
cxU) = J(x) U E C(K)) , i.e.
Cx
is 'evaluation at x', with corresponding maximal ideal Mx = {J E C(K) : J(x) = O}.
We shall now see that these are the only characters (equivalently, maximal ideals) of A. In fact, let M be a maximal ideal of C(K), and assume towards a contradiction that, for every x E K, there is a function gx E M such that gx(x) i- o. By continuity, gx i- 0 on an open neighbourhood, say U(x), of x. Since K is compact, there are finitely many points, say X1, ... ,X n , such that K = U~l U(x,). But then n
g:= L ,=1
n
Igx,1 2
=
Lgx,gx, EM ,=1
and g(x) > 0 for all x E K, which immediately implies that g E G(C(K)). Thus M is not proper, contrary to hypothesis. Hence there exists x E K such that M ~ NIx. Since Jl1 is a maximal ideal, necessarily M = Mx. 0 Example 4.50 Let A = A(~), the disc algebra. Once again, for each z E ~, there is the corresponding evaluation character cz, defined as before. Again, we shall see that these are all the characters on A, but the proof is quite different. Let cp be a character on A. Then, setting z = cp(Z), where Z is the coordinate functional, we have Izl ::; IZIb. = 1, so that z E ~. Then it is clear that, for each polynomial p, we have cp(p) = p(z). But, as remarked above, it is a consequence of Fejer's theorem that the set of polynomials is dense in A(~) (this shows that A is polynomially generated by Z), and cp is continuous, so that cpU) = J(z) U E A), and hence cp = Cz. 0
Banach algebras
187
Example 4.51 Let K be a non-empty, compact subset of A is the relativization to 1> A of the weak-* topology on A *. This topology is sometimes called the weak-* topology on 1> A, and written 0"(1) A, A). It is easily seen to be the weakest topology on 1> A that makes each of the maps cp I----t cp(a), 1> A ----> C, for a E A, continuous.
189
Banach algebras
Let A be a Banach algebra. Then we shall call the topological space ( A of A *. From the definition of the Gel'fand topology, it is clear that each E C (1> A)'
a
Theorem 4.59 (Gel'fand representation theorem) Let A be a unital Banach algebra. Then the Gel 'fand transform
9:af---+a,
A----+C(1)A),
is a continuous, unital homomorphism. Suppose that A is commutative, and take a E A. Then a E G(A) if and only if a(ip) -10 for all ip E 1>A, and SPAa = SpC(1)A)a = {a(ip) : ip E 1>A} . Proof This is essentially just a translation of some of our earlier results.
D
Let A be a Banach algebra without an identity, so that 1> A is a locally compact space. Suppose that 1> A -I 0. Then it is easily checked that E C o( A), and that the map 9: a f---+ a, (A; 11·11) ----+ (CO(A); l'I1>A)' is a continuous homomorphism with 11911 ::; 1.
a
192
Introduction to Banach Spaces and Algebras
Let A be a Banach algebra with A -I=- 0. Then the homomorphism Q is the Gel'fand representation of A. The range of Q is sometimes denoted by 11, so that 11 = {a: a E A} is a natural Banach function algebra on A. Of course, 11 is isomorphic to AI ker Q, and we are giving 11 the quotient norm. In general, Q is neither injective nor surjective. In the case where A is commutative, it is easy to describe the kernel of Q. Recall that an element a of A is quasi-nilpotent if and only if Spa = {O}, and that we write Q(A) for the set of quasi-nilpotent elements of A. Corollary 4.60 Let A be a commutative, unital Banach algebra. Then kerQ = {a E A: zp(a) = 0 (zp E An = n{M: M E MA} = Q(A).
o
Example 4.61 In Example 2.18, we defined a Banach sequence space on each of Nand Z. A Banach sequence algebra on Z is a Banach algebra A such that coo(Z)
f
* g on
Ll(JR.) by
(*)
We explain the above formula. The product f * g is certainly defined on all of JR. by (*) when f, g E Coo (JR.); it is easy to check that f * g then belongs to Coo(JR.) and that we obtain a bilinear map
T: (f,g)
f->
f
* g,
Coo(JR.) x Coo(JR.)
----+
Coo(JR.) c Ll(JR.).
It is also easily checked by writing a repeated integral in the different order that IIT(f,g)lll ~ Ilflllllglll (f,g E Coo (JR.)) , and so IITII ~ 1. As in Exercise 2.16, we extend this bilinear map to a bilinear map
T:
Ll(JR.) x Ll(JR.)
----+
Ll(JR.)
with the same norm. We then set f * g = T(f,g) (f,g E Ll(JR.)). The formula (*) can also be applied directly to functions f,g E Ll(JR.) to define f * g 'almost everywhere' as an element of Ll(JR.); this requires the theory of Lebesgue integration. In the case where f and g are Riemann-integrable and at least one of f and g is bounded, (f * g)(x) is indeed defined by the integral in (*) for all x E R It is then easily checked that (Ll(JR.); 11.11 1 ; *) is a commutative Banach algebra. 0 Example 4.66 In a similar way, we define the convolution product on Ll (11') by the formula
(f
* g)(B) =
1 2n
-
1271" f(B -
~)g(~) d~
(B
E
[0,2nD
0
for f, g E Ll (11'), where we again write f( B) for f( e iO ). Again, (Ll (1!'); I . 111 ; * ) is a commutative Banach algebra; the inequality Ilf * gill ~ Ilflll Ilglll follows explicitly from Proposition 2.37(ii). Note that various formulae that arose in Sections 2.9, 2.16, and 2.17 are actually examples of this convolution product. For example: in Lemma 2.42(ii), we have fr = f * Prj in Lemma 2.81, we have SN(f) = f * DN; in Lemma 2.83, we have I7N(f) = f * K N . (In each case, we are using the notation of the reference.) 0
196
Introduction to Banach Spaces and Algebras
Example 4.67 Now consider the Banach space Ll(lR+); we regard this space as a closed subspace of Ll(lR) by extending each I E Ll(lR+) to be equal to 0 on the negative half-line lR-· = (-00,0). In this case, the product of two elements I and 9 of Ll(lR+) is given by the formula
(J
* g)(x) =
l
x
I(x - t)g(t) dt
(x
clearly Ll(lR+) is a closed subalgebra of Ll(lR). A weight function on lR+ is a function w : lR+
w(s
+ t)
:::; w(s)w(t)
E
--+
(*)
lR+);
lR+· such that w(O) = 1 and
(s, t E lR+).
(cf. Example 4.52). For example, the functions t f---+ e-a-t for a > 0 and t for Ct ::::: 1 are continuous weight functions on lR+. Let w be a continuous weight function on lR+, and let
Ll(lR+,W) = {I:
1IIIIw =
1
00
II(t)lw(t)dt
0, we have I(t) = 0 (0:::; t :::; 0). Then it is simple to verify that f*k, the kth convolution power of I, vanishes on [0, kO], so that IN = 0 for N > 1/0, i.e. I is a nilpotent element of A. But this implies that I E J(A). In fact, it is clear that the set of such nilpotent elements is dense in V, and so, since J(A) is closed, J(A) "2 V. We now easily deduce that J(A) = V, so that V is a radical Banach algebra, and that V is the unique maximal ideal of A, with the corresponding D unique character cp being given by cp('x1 + 1) =,X (,X E C, I E V).
197
Banach algebras
On page 86, we defined the Fourier transform
F: f
f---+
(!(n))nEZ,
Ll(11')
----+
CO(Z);
we have noted that F is a continuous linear operator with IIFII :::; 1. Further, F is an injection, but not a surjection; the range of the mapping F is A(Z), a Banach space with respect to the norm transferred from L 1 (11'). As we noted on page 89, the trigonometric polynomials are mapped onto coo(Z), and so coo(Z) is a dense subspace of A(Z). Theorem 4.69 The Fourier transform F: (Ll(1I'); *)
----+ (co(Z); .) is a monomorphism, and A(Z) is a Banach sequence algebra. The characters on Ll(1I') have the form f f---+ !(n) for a unique nEZ, the character space of Ll(1I') is homeomorphic to Z, and (Ll(1I'); *) is semisimple.
Proof Let
f, 9
E C(1I') and n E Z. Then
----1127r (J * g)(e)e- .mlJ de f * g(n) = 27r
0
=
2~fo27r (2~fo27r f(e-1/;)9(1/;)d1/;)e- inlJ de
=
(2~ fo27r f(e _1/;)e- in (IJ-1/;) de) (2~ fo27r g(1/;)e- in 1/; d1/;)
= f(n) . g(n), and so F(J * g) = F(J) . F(g). This latter equation also holds for f, 9 E Ll(1I') by continuity. Thus A(Z) is a subalgebra of (Co(JR); 1·1 1R ; .) is a homomorphism with IIFII = 1. Proof First, suppose that f E Coo(JR), say suppf ~ [-R,R] for some R > o. There is a constant K > 0 such that le z - 11 :::; K Izl whenever Izl :::; R. Let Yl, Y2 E JR with IYl - Y21 :::; 1. Then
I[(Yl) - [(Y2) I: :;
1:
If(t)lle- i (Yl-Y2)t
-
11 dt
:::; K IYl - Y211: Itllf(t)1 dt,
and so lis uniformly continuous on R Clearly,
1~1R
Cb(JR), a space defined in Example 2.4(iii). It follows from Proposition 2.21 that we have a continuous linear mapping F: (Ll(JR); 11.11 1 ) ----> (Cb(JR); 1·1 1R ) with IIFII :::; 1. Now let fELl (JR) have the form f = X[a,bJ. Then f(y) =
l
b
E
• . e- 1yt
1 iby _ e- iay ) dt = _(e-
(y
Y
a
Thus l[(y)1 :::; 2/lyl
:::; Ilflll' and so 1
---->
0 as IYI
----> 00,
so that 1
E
E
JR \ {O}) . Co(JR)· It follows that
1 E Co(JR) whenever f is a simple function in U (JR). Since the simple functions are dense in (Ll(JR); 11.11 1 ), the range of F is contained in Co(JR). Let f = X[O,lJ· Then Ilflll = 1 and 11(0)1 = 1, so IIFII ::::: 1. Hence IIFII = 1. The proof that F(f * g) = F(f) . F(g) (f,g E Ll(JR)) is very similar to the corresponding proof in Theorem 4.69. Thus F is a homomorphism. 0 The range of the Fourier transform is called A(JR), so that A(JR) ~ Co (JR). We shall see soon that F is actually an injection. We leave it as an easy exercise to check the following properties of the Fourier transform. For each f E Coo(JR) and a E JR, we define Saf(t) = f(t - a) and fa(t) = af(at) for t E JR, and then we extend the maps f 1-+ Saf and f 1-+ fa to be elements of B(Ll(JR)) by continuity.
199
Banach algebras
Proposition 4.71 Let f E Ll(JR) and a E lR. Then:
f( -y)
(i) f(y) =
(y
E
JR);
(ii) Saf E Ll (JR) and Saf(y) = e- iay f(y) (y E JR) ; (iii) the function a
f--->
Saf, JR
---+
(Ll(JR); 11.11 1), is continuous;
(iv) fa E Ll(JR) with Ilfaill = Ilflll and ia(y) = 1(y/a) (y E JR);
(v) if 9
:=
Zf
E
Ll(JR), then
1 is differentiable on JR,
and
(1)' = -ig.
0
Let f E £l(JR), and take a E lR. For h > 0, take Xh to be X[a.a+h]/h, so that each Xh E Ll(JR) with Ilxhlll = 1. We claim that
Saf= lim f*Xh h--->O+
in (Ll(JR);II·11 1)·
First, suppose that f = X[c,d], where c < d. Then a trivial calculation shows that the graph of X[c,d] * Xh is the trapezium which takes the value 1 on the interval [c+a + h, d + aJ, which is linear on the two intervals [c+ a, c+ a + h] and [d + a, d + a + h], and which is 0 elsewhere, and so the area given by the difference Sa(X[c,d]) - (X[c,d] * Xh) is at most 2h. Thus the claim holds for this function. The claim now follows from Proposition 2.16, taking TnU) = f * Xhn (n E N) and TU) = Saf for any sequence (hn)n>l in JR+. with limn--->oo h n = 0 in that result. A vector subspace E of Ll(JR) is translation-invariant if Saf E E whenever fEE and a E JR. The following result is immediate from our claim. Proposition 4.72 A closed ideal in Ll (JR) is translation-invarwnt.
Let
f
E
0
Coo (JR), and define 00
F(e)=2Jr
L
f(e+2k7r)
(eE[-Jr,Jr]).
(*)
k=-oo
Then it is clear that the sum is finite, that F E C[-Jr, JrJ, and that 11F111 ::::: Ilfll l · For each nEZ, we have 1
F(n) = 2Jr
j7r-7r F(e)e- mo de = k~OO j7r-7r f(e + 2k7r)e- inO de = f(n), 00
1
and so F = I Z. By continuity, this formula also holds for f E Ll(JR). It follows that A(JR) 0 and define fa as above; the corresponding function in Ll(11') as given in (*) is Fa, so that 00
2:
Fa(e) = 27ra
+ 2ka7r)
f(ae
(e
E
[-7r,7r]).
k=-oo
Then Fa(n) = f(nja) (n E Z). Suppose that suppf C [-27rR, 27rRJ, where R > 0, and that a > R. Then clearly Fa(e) = 27raf(ae) (e E [-7r,7r]), and so IWaill = Ilfll l · Now let f E Ll(IR), and take c > o. Then there exist 9 E C oo (lR) and hE Ll(lR) with f = 9 + hand Ilhlll < c, so that IIHalll::::: Ilhail l
=
Ilhlll < c.
For a sufficiently large, we have IWaill ~ IIGall l - c
Ilglll - c ~ Ilflll - 2c,
=
and so lim a -+ oo IWaill = Ilfll l · Theorem 4.73 The Fourier transform:F: £1(IR) ----. Co(lR) is an injection.
1
Proof .-!ake f E Ll(lR) with = o. For each a > 0, we have Fa(n) and so Fa = o. By Theorem 4.69, Fa = 0, and so
Ilflll whence
f
=
= 0
(n E Z),
= a-+oo lim IWaill = 0, o
0 in L l (IR).
Thus we can identify Ll (IR) with the subalgebra A(IR) of C o(lR) (with the 'transfered norm' on A(IR)). We wish to show that the character space of this algebra is (homeomorphic to) IR; we require a lemma. Lemma 4.74 Let 'ljJ: IR ----. C be a continuous function such that 1jJ(0) = 1 and
1jJ(s + t) = 1jJ(s)1jJ(t) Then there exists
Z E
C such that 1jJ(t)
(s, t
E
= e zt (t
Proof Since 1jJ is continuous, there exists J
>
IR). E
(*)
IR).
0 with
0:
t E IR, we have
o:1jJ(t) =
:=
f:
1jJ
i- o.
For each
10(i 1jJ(s)1jJ(t)ds= 10(Ii 1jJ(s+t)ds= It+1i 1jJ(u)du; t
the right-hand side is a differentiable function of t, and so 1jJ is differentiable. Set Z = 1jJ' (0). By differentiating both sides of (*) with respect to s and setting s = 0, we see that 1jJ'(t) = z1jJ(t) (t E IR). Thus 1jJ(t) = e zt (t E IR). 0
201
Banach algebras
Theorem 4.75 The character space of L1(JR) can be homeomorphically identified with JR in such a way that each character on L1(JR) has the form f f--+ J(y) for some unique y E R
Proof Let be the character space of L1 (JR). Since A(JR) is an algebra of functions on JR, each functional 0, set
V(n, r) = {cpy E : sup le- iyt Itl::on
11 < r}
.
We claim that these sets V (n, r) form a base of neighbourhoods of CPo in the Gel'fand topology of . Clearly, each set V(n, r) is an open neighbourhood of CPo.
202
Introduction to Banach Spaces and Algebras
Now let N = {!py E : l!py(f)) -!Po(f))1 < c (j = 1, ... ,m)} be a basic open neighbourhood of !Po in , where il, ... , 1m E Ll(JR) and c > o. We may suppose that il, ... , 1m E Coo(JR). Take n E Nand r > 0 such that supp IJ 1
and
f(x)::::: 1
(x E K).
By replacing f by (J + a . 1)/(1 + a) for suitable a > 0, we may suppose that f E E+, and so IflK ::::: 1. But now 1'\0(J)1 : : : 1, a contradiction. Thus
conv(K)
=
K E.
We now return to the general case. The map f f--+ f is an isometry, and so we may suppose that K = rcA).
I f(A),
A ---- C(f(A»,
206
Introduction to Banach Spaces and Algebras
Let Ao
E
K A, and take a basic neighbourhood U of 0 in A *, say U = {A
E
A* : IA(f,)1 < 1 (i = 1, ... , n)},
where II, ... , fn E A. Set V = {A E C(K)* : IA(f,)1 < 1 (i = 1, ... , n)}. There is an extension flo E C(K)* of AO with Ilfloll = flo(1) = 1. By the above result, there is III E cony K with fll - flo E V. Set Al = fll I A. Then Al E cony K and Al - Ao E U. This holds for each such U, and so Ao E conv(K). By Theorem 4.80, ro(A) = r(A), and so conv(ro(A)) = KA. Now take Xo E ro(A). Then there exists f E A with f(xo) = 1 and If(x)1 < 1 for each x E K with x of Xo. Set Ko = {A E KA : A(f) = I}. Then Ko is a nonempty, closed extreme subset of KA, and so, as in the Krein-Milman theorem, Theorem 3.31, Ko contains an extreme point of K A . Since ex KA 00
as n
--> 00,
but that IhnllR
*
XI-I,I]'
(n E N).
= 1 (n EN). Deduce from this that
A(JR) =I- Co(JR). Exercise 4.22 Let E be a closed subspace of the convolution algebra (LI(JR); *). Prove that E is translation-invariant if and only if E is an ideal.
211
Banach algebras
Exercise 4.23 Set IT = {z = x + iy E C : x > o}. For fELl (JR+) (see Example 4.67), we now define the Laplace transform £(f) by
£(f)(z) =
1
00
f(t)e- zt dt
(z E IT).
Show that £(f) E Co(IT) and that £(f) is analytic on IT. Show that
£ : (LI(JR+); *)
---->
(Co (IT); .)
is a monomorphism, and that the character space of LI(JR+) can be identified with IT in such a way that each character on L1(JR+) has the form f f---> £(f)(z) for some ZEIT. Exercise 4.24 Let w be a continuous weight function on JR+. Show that the limit p = limt~oo W(t)l/t exists. Let LI(JR+,W) be the commutative Banach algebra defined in Example 4.67. Show that LI(JR+,W) is semisimple whenever p > 0 and radical whenever p = o. Exercise 4.25 Let A be a natural uniform algebra on a metrizable, compact space K, and let Xo E K. Show that Xo is a peak point for A if and only if there exist a,j3 with o < a < 13 < 1 such that, for each neighbourhood U of xo, there exists f E A with IflK ::; 1, with f(xo) > 13, and with If(y)1 < a (y E K \ U). Exercise 4.26 Let A be a natural uniform algebra on a compact set K, and let Show that 8Sp(f) ~ f(r(A)).
f
E
A.
Exercise 4.27 Let K and L be the closed discs in C, each ofradius 1, and with centres at (0,1) and (0, -1), respectively, so that K n L = {(O,O)} and P(K U L) is natural on K U L. Show that (0,0) is a peak point for P(K U L), but that there is no polynomial that peaks at (0,0). Exercise 4.28 Let K = {(z, t) E C x JR : Izl ::; 1, 0::; t ::; I}, and let A be the set of functions in C(K) such that the function z f---> f(z,O) is analytic on int~. Show that A is a natural uniform algebra on K with r(A) = K. Identify the peak points for A, and note that they form a proper subset of K.
Runge's theorem and the holomorphic functional calculus 4.14 Runge's theorem. Let K be a non-empty, compact subset of Co We first recall the notations P(K), R(K), and A(K) from Example 4.3. These are uniform algebras on K, and R(K) and A(K) are natural. For a E l in OK converges to 0 if there is an open neighbourhood U of K such that each fn is in O(U) and limn---+<Xl fn = 0 on a compact neighbourhood of K in U. Theorem 4.83 (Runge's theorem) Let K be a non-empty, compact subset ofe, and let A be a subset of e \ K that meets every bounded component of e \ K. Let f E OK. Then there is a sequence (rn)n~l of rational functions each having all its poles in A such that r n -+ f uniformly on K. Proof Let the function f be analytic on the open neighbourhood U of K. By Proposition 1.31, there is a contour, say "I, in U \ K that surrounds K in U. Let E be the closed subspace of R(K) generated by {un: a E e \ K}. Since the function a f---> Un, e \ K -+ E, is continuous, the integral
2~i j
f(a)u n da
defines an element, say g, in E. But, for each Z E K, the map h f---> h(z), E -+ e, is a continuous linear functional on E, so that we see from Theorem 3.10 that
.j
1 g(z) = -2 7rl
f(a)un(z) da = f(z)
(z E K),
'Y
now using the Cauchy integral formula. This proves that f I K = gEE. Set B = R(K)(S), the closed subalgebra of R(K) generated by S, where S = {Z, u" : A E A}. Then, by Lemma 4.82, Un E B for every a E e \ K. It follows that E r;;;; B. o The conclusion of the theorem now follows easily.
213
Banach algebras
Corollary 4.84 Let K be a non-empty, compact subset of re, and let A be a subset of re \ K that meets every bounded component of re \ K. Then the set of all rational functions havmg all their poles in A is dense in R( K). Proof We apply the theorem with having all its poles in re \ K. Corollary
4.85
f
taken to be an arbitrary rational function
D
Let K be a non-empty, compact subset of rc. Then:
(i) R(K) = O(K) ; (ii) R(K)
= P(K) if and only if re \ K is connected.
D
Let U be a non-empty, open subset of rc. We write R(U) for the subalgebra of O(U) consisting of all rational functions having their poles in re \ U. Corollary 4.86 (Runge's theorem for open sets) Let U be a non-empty, open subset of rc. Then R(U) is dense m O(U) in the topology of local uniform convergence. Proof Let f E O(U). We first claim that, for each non-empty, compact subset K of U and each c > 0, there is some r E R(U) with If - rlK < c. Indeed, choose n E N such that, for all z E K, we have both Izl ::; nand dist(z, re \ U) 2 lin, and then define
L = {z
Ere:
Izl ::; n, d(z,re \ U) 2 lin}.
Clearly, L is a compact subspace of re with U ~ L :2 K. Suppose that V is a bounded component of re \ L. Then av ~ L, and so Izl ::; n (z E V). But this implies that V must meet re \ U, since otherwise, for every z E V, we would have d(z, re \ U) 2 deL, re \ U) 2 lin, so that z E L, a contradiction. Thus, every bounded component of re \ L meets re \ U. We may then apply Theorem 4.83 to L with a set A ~ re \ U to see that f may be uniformly approximated on L, and so also on K, by functions in R(U). This gives the claim. As on page 114, we write U = U:=l K n , with each Kn a compact subset of U and Kn C int K n+ 1 (n EN). By the claim, for each n E N, there is a function rn E R(U) with If - rnlKn < lin. Clearly, rn ----+ f in the topology of local uniform convergence on O(U). D
4.15
The holomorphic functional calculus. Recall that, for each algebra A with an identity and for each a E A, we may define the element p( a) in A for each polynomial p; the map
e :p
1----*
p( a) ,
q Xl
----+
A,
is a unital homomorphism with e(x) = a. The map e is a 'polynomial calculus' for the element a. We wish to extend this idea to define f (a) for elements a in a
214
Introduction to Banach Spaces and Algebras
Banach algebra A and functions J that are in a unital algebra B that is strictly larger than qXj in such a way that the map 8 : J f---4 J(a), B ~ A, is a unital homomorphism, still with 8(X) = a. We start this process of defining more general functional calculus maps with a modest first step involving rational functions. Let A be a Banach algebra with an identity, let a E A, and take U to be an open neighbourhood of Spa in .1 - a)-l d>.. "Y
r(a), R(U)
--+
A, is continuous.
Proof By elementary algebra, there is an analytic A-valued function h on U with r(>')1 - r(a) = (>.1 - a)h(>') (>. E U). By Cauchy's theorem, we have J"Y h(>') d>' = 0, so that
~1 2m
r(>.)(>.1 - a)-l d>' = "Y
r(a)~ 1(>'1 27rI
a)-l d>' = r(a)
"Y
by Lemma 4.87. The function>. f---> Ra(>') = (>.1 - a)-l is continuous, and hence bounded, on the compact set b]' and so it is clear from the formula for r(a) that Ilr(a)11 :::; m Irlh]' where m is a constant not depending on r. This implies that the homomorphism r f---> r(a), R(U) --+ A, is continuous. D We can now establish what is called the holomorphic functional calculus in one variable for Banach algebras.
Theorem 4.89 (Holomorphic functional calculus) Let A be a unital Banach algebra, let a E A, let U be an open neighbourhood of Sp a in C. Then there is a unique continuous, unital homomorphism 8 a : O(U) --+ A such that 8 a (Z) = a. Set C = {a}ee. Then: (i) for every contour"( that surrounds Spa in U and every f E O(U), we have 1 . 1 f(>.)(>.1 - a)-l d>.; 8 a (f) = -2 7rI "Y (ii) 8 a (r) = r(a) (r E R(U)); (iii) im 8 a 0, there exists no EN with Ila n - all < min{c, 112M}. For each n ~ no, we have
II(AI - an) - (AI - a)11 :::;
~ II(AI -
a)-I
r
l
(A
E
[,,(D,
and so, by equation (*) in the proof of Corollary 4.12,
II(Al- an)-I - (AI - a)-III:::; 2M 2 c. It follows that (AI - an)-I ----+ (AI - a)-I uniformly on ["(], and so the result D follows from the formulae involving integrals for f(a n ) and f(a).
Proposition 4.94 Let A and B be unital Banach algebras, and suppose that T : A ----+ B is a continuous, unital homomorphism. Then, for any a E A and f analytic on some neighbourhood of SPA a, we have T(J(a))
Proof Since SPA a Then
~
= f(Ta) .
Sp BTa, f is also analytic on a neighbourhood of Sp BTa.
e:f
f---4
T(J(a)) ,
O(U)
----+
B,
is a continuous, unital homomorphism such that e(Z) so that T(J(a)) = 8 T (a)(f) = f(Ta).
= Ta. Hence
e
=
8 T (a), D
The next theorem discusses the behaviour of the functional calculus with respect to compositions.
Theorem 4.95 Let A be a unital Banach algebra, let a E A, and let f E OSPAa· Then (g 0 f)(a) = g(f(a)) (g E OSPAf(a)). Proof Set b = f(a). By Theorem 4.89(iv), Spb = f(Spa), and so, for any 9 analytic on a neighbourhood V of Sp b, it follows that 9 0 f is well defined and analytic on some neighbourhood of Sp a. Define e : O(V) ----+ A by 8(g) = 8 a (g 0 f) (g E O(V)). Then clearly e is a continuous, unital homomorphism; also, Z 0 f = f, and so 8(Z) = 8 a (f) = b. It follows from the uniqueness assertion in Theorem 4.89 that 8 = 8b, and so (g
as required.
0
f)(a)
= (}(g) = 8b(g) = g(b)
(g E O(V)) , D
219
Banach algebras
Recall that an element p in an algebra A is an idempotent if p2 = p. Theorem 4.96 Let A be a unital Banach algebra, and let a E A be such that Sp a is the dzsjoint union of two non-empty, closed subsets K and L. Then there is a non-trivzal idempotent pEA such that ap = pa and such that Sp(pa)=KU{O}
and
Sp(a-pa)=LU{O}.
Proof There are open neighbourhoods U of K and V of L, respectively, with Un V = 0. Define f on U U V to be the characteristic function of U, so that f is an idempotent in O(U U V). Set p = GaU). Then p is an idempotent in A; by Theorem 4.S9(iii), ap = pa. By Theorem 4.S9(iv), Spp = f(Spa) = {O, I}, and so p is a non-trivial idempotent. Also, by Theorem 4.S9(iv), Sp (pa) = K U {O} andSp(a-pa)=LU{O}. D Corollary 4.97 Let E be a Banach space, and let T E B(E). Suppose that SpT is the disjoint union of two non-empty, closed subsets K and L. Then there are non-zero projections P, Q E B(E) such that P + Q = IE, PQ = QP = 0, T P = PT, and such that P(E) and Q(E) are proper invariant subspaces of E such that Sp 13(p(E))(T I P(E)) = K and Sp 13(Q(E))(T I Q(E)) = L. Proof Take P E B(E) as specified in Theorem 4.96, and set Q = IE - P, so that Q2 = Q; all is clear, save perhaps that Sp 13(P(E)) (T I P(E)) = K. Set Ko = Sp 13(P(E)) (T I P(E)); we shall show that Ko = K. Let A E K, and assume that A rf. Ko. Then there exists A E B(P(E)) with (AlE - T)Ax
= A(AlE - T)x (x
E
P(E)).
Let g(fJ) = 0 for fJ in a neighbourhood of K and g(fJ) = (A - fJ)-l for fJ in a neighbourhood of L, so that 9 E OSpT. Then g(T)(AlE - T)
Define B = AP
E
(B
=
(AlE - T)g(T)
= Q.
B(E). Then
+ g(T))(AIe -
T) = (AlE - T)(B
+ g(T))
= Ie,
and so A rf. Sp T, a contradiction. Thus K ~ Ko. Conversely, take A E C \ K. Define h(fJ) = 0 for fJ in a neighbourhood of L and h(fJ) = (A - fJ)-l for fJ in a neighbourhood of K not containing A. Then h(T)(AlE - T) = (AlE - T)h(T) = P. Set R = h(T) I P(E) and S = T I P(E). Then R(Alp(E) - S) = (Alp(E) - S)R = Ip(E) , and so A rf. Ko. Thus Ko
~
K.
o
220
Introduction to Banach Spaces and Algebras
Let E be a Banach space, and let T E K(E). We know from Theorem 4.34 that each non-zero ,X E SpT is an eigenvalue and that the eigenspace E('x) is finite-dimensional. By Proposition 3.8(i), there is a projection P).. : E ----) E('x); of course, SPS(E()"))P).. I E('x) = {A}. Let A be a Banach algebra with an identity, and let a E A. We have previously defined exp a, sin a, and cos a in A. Now regard exp, sin, and cos as entire functions on C. Then these functions belong to Osp a, and clearly exp a
= e a (exp) ,
etc. Note that exp(ia) = cosa + isina, for example. We shall sometimes write exp A = {exp a : a E A} .
Proposition 4.98 Let A be a unital Banach algebra. Then:
(i) for a, bE A with ab = ba, we have exp(a + b) = (expa)(expb); (ii) for every a E A, we have expa E G(A), with inverse exp(-a).
Proof (i) The identity of A is lA. For every n n
Xn
=
lA
+
L
k=l
k
Yn
= lA+
L
N, set n
bk k!'
n
~!'
E
Zn
lA
k=l
n Ilbll k Bn=L~'
= ~~, ~ k! k=O
+L
(a ;,b)k
k=l
and set An
=
en
=
t
(Hall + Ilbll)k k!
k=O
k=O
Then, by elementary algebra, since ab = ba, we have n
XnYn - Zn
= L
aJkaJbk,
J,k=l
where aJk :::: 0 for all j, kEN. Therefore, n
IIXnYn - Znll :::; L aJkllaWllbll k J,k=l
But limn~oo(AnBn - en) limn---;oo IIXnYn - Znll = O. The result follows.
= AnBn - en·
ellallellbll - e(llall+llblll
0, and so we see that
(ii) Take b = -a in (i), it being clear that expO = lA.
0
221
Banach algebras
We shall now show that some elements in a Banach algebra have square roots and logarithms. Recall that we write lR - = {x E lR : x ::; O}; also, we shall denote by II the open right-hand half-plane, so that
II = {z = x
+ iy E C : x > O} ,
as in Exercise 4.23. In the next two results, we set D
= C \ lR-.
Theorem 4.99 Let A be a unital Banach algebra, and let a E A be such that Sp a cD. Then there is a unique element b E A such that both b2 = a and Spb c II. Further, bE {a}C n {a}CC. Proof For ZED, we may write Z = re iO with r > 0 and
J(reiIJ)
-Jr
< () < Jr. Set
= rl/2eiO/2
for such zED. Then it is well known that J is the unique analytic function on D such that J(z)2 = Z (z E D) and J(I) = 1; note that J(z) E II for all zED. Set b = J(a). By Theorem 4.95, b2 = a, and this implies that b EA. Also, Sp b = J(Sp a), so that Sp b c II. Again from Theorem 4.89(iii), b E {a}C n {a }CC. Now let C E A satisfy c2 = a. Then c commutes with a = c2 , so that also c commutes with b, and hence
(b - c) (b + c) = b2
-
c2 = 0 .
By Corollary 4.48(i), Sp (b + c) ~ Sp b + Sp c. In addition, suppose that c also satisfies Spc c II. Then Sp(b+c) c II; in particular, 0 (j. Sp(b+c), and so b+c is invertible in A. Hence c = b, and so b is uniquely specified by the prescribed conditions. 0 Under the conditions of Theorem 4.99, we shall occasionally refer to b as the principal square root of a. Theorem 4.100 Let A be a unital Banach algebra, and let a Sp a cD. Then there exists b E A with exp b = a. Proof Let log be the unique analytic function
exp(f(z)) = z
(z E D)
J on D
and
E
A be such that
such that
J(I) = 0,
o
and set b = log a E A. By Theorem 4.95, a = exp b. In the above case, we set b = loga. Further, we set
ac'
=
exp((b)
(( E
q.
Then (aC, : (E C) is a semigroup in (A, .), and the map (f--> ac', C entire A-valued function.
--+
A, is an
222
Introduction to Banach Spaces and Algebras
Now suppose that a E A with p(l - a) uniquely specified by the formula
n). Proof For kEN, take Tk to be the representation of A on Ek defined by (Tka)(O = a . ~ (a E A, ~ E Ek). Fix n EN, and set B = Et n ... n E;" a Banach space. For k > n, set Uk
=
{b E B : det(Tkb) -1= O};
since det is a continuous function on the space £(Ek ), the set Uk is open in B. Since n ... n E;, Cl Et, there exists bk E Uk. For each bE B \ Uk, we have b = limm~(X)(b + bk/m), and b + bk/m E Uk for sufficiently large mEN, and so bE Uk. Thus Uk is dense in B. By the Baire category theorem, Theorem 1.21, there exists an E B with an E Uk (k > n). Clearly, an . Ek = Ek (k > n), as required. 0
Er
Representation theory
239
Notes For an introductory account of the theory of modules, see [47, Section 1.4] and many books on algebra, including [97]. For an introduction to the theory of Banach left A-modules and Banach A-bimodules, see [47, Section 2.6], where a somewhat different terminology is used; for an extensive account, couched in the language of representations, see [126]. Many examples of Banach A-bimodules are given in [47]; the concept is ubiquitous in more advanced work on Banach algebras. There are many results about derivations on particular Banach algebras in [47]. Here are two examples; for another example, see Section 5.7. (i) Let Ql be a Banach operator algebra on a Banach space E, and let D : Ql -+ B(E) be a derivation. Then there exists T E B(E) such that D(A) = AT - TA (A E Ql) [47, Theorem 2.5.14].
(ii) Let G be a locally compact group. Then the following question was discussed and only partially resolved in [47, Section 5.6]: Does every continuous derivation from Ll(G) to itself have the form D : f f-+ f * f-t - f-t * f for some measure f-t on G? Subsequently, this question has been resolved positively by Losert in [118]. The question whether or not all continuous derivations from a Banach algebra A into a specific Banach A-bimodule are inner is the first step in the enormous cohomology theory of Banach algebras; see [47, Section 2.8] for a preliminary account. A Banach algebra A is amenable if all continuous derivations from A into each dual Banach A-bimodule are inner. This concept was introduced by Johnson [100], where it was proved that a group algebra L 1 (G) is an amenable Banach algebra if and only if G is an amenable locally compact group. Intrinsic characterizations of amenable Banach algebras are given in [47, Theorem 2.9.65]. For example, the Banach algebras C(K) are all amenable. A recent advance is a theorem of Runde [147]: for each p:O:: 1, the Banach algebra B(fP) is not amenable. The Singer~Wermer theorem, Theorem 5.7, was first proved in 1955 in [150]. In this paper, it was conjectured that it was not necessary to assume that the derivation be continuous. This conjecture was finally proved by Thomas in 1988 in [159]: for each derivation on a commutative Banach algebra A, we have D(A) ~ J(A). See [47, Theorem 5.2.48] for a proof. A 'non-commutative Singer~Wermer theorem' would establish that, for each derivation on a Banach algebra A, we have D(P) ~ P for each primitive ideal P; this is true for continuous derivations [148]. This question has defied solution for 55 years; the strongest partial result is in [160]. Let A be an algebra with an identity. What we have called a primitive ideal should, more precisely, be called a left primitive ideal in A because it is defined using left Amodules or, equivalently, maximal left ideals. Similarly, there are right primitive ideals in A, defined using right A-modules or maximal right ideals. In pure algebra, there are left primitive ideals that are not right primitive, but it seems to be a long-standing open question, going back to at least [31, page 125], whether or not the classes of left primitive and right primitive ideals coincide in every unital Banach algebra. Let A be an algebra, not necessarily with an identity. A proper left ideal I in A is modular if there exists an element U E A such that a - au E I for each a E A; in this case, u is a right modular identity of I. A maximal modular left ideal in A is a maximal element in the family of proper, modular left ideals in A. Since each left ideal containing a modular left ideal is also a modular left ideal, a maximal modular left ideal in A is a maximal left ideal. It follows from Zorn's lemma that each proper, modular left ideal is contained in a maximal modular left ideal. We may define the Jacobson radical, J(A), of A without reference to A+: it is equal to the intersection of the maximal modular left ideals of A. This gives the same ideal as that defined in the text, but this version may be a little 'cleaner' and more general. Different radicals from the Jacobson radical are sometimes considered: see [47, 126]. For example, let A be an algebra with an identity. Then the strong radical of A is the
240
Introduction to Banach Spaces and Algebras
intersection of the maximal ideals of A and the prime radical of A is the intersection of the prime ideals of A. (A proper ideal P of an algebra A is prime if, for ideals I and J in A, either I l' . a.
f->
a .T
Exercise 5.2 Let A be an algebra, let E be an A-bimodule, and let D : A ----> E be a derivation. (i) Suppose that pEA is an idempotent. Prove that p . Dp . p = 0, and that Dp = 0 if, further, p . Dp = Dp . p. (ii) Suppose that a E A with a . Da = Da . a. Prove that D(a n ) = nan~l . Da for each n E N. Exercise 5.3 (i) Let A be a natural Banach function algebra on a non-empty, compact Hausdorff space K, and let x E K. Show that a linear functional on A is a point derivation at Ex if and only if d(l) = 0 and dUg) = 0 whenever 1,g E Mx = kerE x . (ii) Let C 1 (II) be the Banach function algebra of continuously differentiable functions introduced in Exercise 4.12(i), so that A is a natural Banach function algebra on II. Show that every continuous point derivation at EO has the form 1 f-> 001'(0) for some a E C. Also show that the space of discontinuous point derivation at EO is infinite-dimensional. (iii) Prove that all point derivations on the disc algebra A(~) are continuous. Exercise 5.4 Let A be a unital Banach algebra, and let E be an irreducible A-module. Extend Jacobson's density theorem by showing that, for linearly independent sequences (6, ... , ~n) and ("11, ... , "In) of elements in E, there is some a E A such that (expa) .
~k
=
"Ik
(k
= 1, ... ,n).
Representation theory
241
Automatic continuity There are remarkable results in which algebraic properties related to Banach algebras, Banach modules, and linear maps between them actually force the relevant map to be continuous. This is called the 'automatic continuity' of the linear map. A basic example of this phenomenon was given in Section 4.10: every character on a Banach algebra is automatically continuous. A particular aspect of automatic continuity theory concentrates on the 'uniqueness-of-norm' property for Banach algebras, and we shall now discuss thisi in the present section, we shall consider commutative Banach algebras, and then, in some later sections, we shall turn to general Banach algebras. The first such result is very easYi it is almost immediate from the fact that characters on Banach algebras are continuous. Theorem 5.25 Let A and B be Banach algebras, and suppose that B zs commutative and semisimple. Then every homomorphism from A into B is automatically continuous. Proof Let () : A ~ B be a homomorphism. We use the language of the separating space 6(()), which was introduced on Section 3.13 i by Theorem 3.51, we must show that 6( ()) = {O}. Thus, suppose that (an)n>l is a null sequence in A with ()(a n ) ~ bin B. Then, for every character tp on B, tp 0 () is either a character on A or is identically zeroi in either case, it is continuous. Thus tp(()(a n )) = (tp 0 ())(a n ) ~ 0 as n ~ 00. But also tp is continuous on B, so that tp(()(a n )) ~ tp(b) as n ~ 00. It follows that tp(b) = 0 for every tp E 1 is a nest which stabilizes. Similarly, (6(D) . an··· at}n>l is a nest which stabilizes. D
Proposition 5.32 Let A be a Banach algebra with identity such that A zs a separating ideal in itself. Then AI J(A) is finite-dimensional. Proof Let E be an irreducible left A-module. By Corollary 5.3, E is a Banach left A-module for a suitable norm. Assume towards a contradiction that the space E is infinite-dimensional, say {~n : n E N} is a linearly independent set of distinct elements in E. By Proposition 5.23, there is a sequence (an)n>l in A with an+l ... al . ~n = 0 and an··· al . ~n =I- 0 for each n E N. Define In = Aan ··· al (n EN). Since Aan · .. al . ~n = E, we have In . ~n = E. Since Aan+l ... al . ~n = {O} and E is a Banach left A-module, I n+l . ~n = {O}. Thus I n+1 £.; In (n EN). This contradicts the fact that the nest (In)n>l stabilizes. Thus each primitive ideal in A has finite co dimension in A. Assume towards a contradiction that J(A) is not a finite intersection of primitive ideals of A. Then it is clear that there is a sequence (Pn)n>l of primitive ideals in A such that PI n· . ·nPn ~ Pk for each k > n in N, say Pk ~ Et (k EN), where (Ek)k>l is a sequence of finite-dimensional, irreducible Banach left Amodules. By -Proposition 5.24, for each n E N, there exists an E Ef n ... n E;, with an . Ek = Ek (k > n). Again, set In = Aan ··· al (n EN). For n E N, we have In+! 0, and set U = {A E C: 1,\1 < p(a) +c}, so that U is open and Sp a C U. By Proposition 4.21, there exists 8 > 0 such that Sp be U whenever lib - all < 8. But then p(b) < p(a) + c whenever lib - all < 8. Second proof. Let a E A, and take c > o. By Corollary 4.25, there is a norm III . III which is equivalent to II . II and such that (A; III . III) is a Banch algebra and Illalll < p(a) + c. But then, by the continuity of 111·111 at a, there exists 8 > 0 such that p(b) ~ Illblll < p(a) + c whenever lib - all < 8. Third proof. For each n E N, set fn(a) = IlanIl I / n . Then certainly each function fn : A -+ 1R+ is continuous on A. From Theorem 4.23, p(a) = infn fn(a). Then just this fact (Le. that p is the pointwise infimum of a collection of continuous
249
Representation theory
functions) implies the upper semi-continuity of p. For let a E A and E > O. Then there is some N EN such that fN(a) < p(a) + E/2. By the continuity of fN at a, there is then a 5 > 0 such that, for every b E A with lib - all < 5, we have p(b) ::; fN(b) < fN(a) + E/2 < p(a) + E. D As you will have guessed from the discussion of semi-continuity, the spectral radius function is not in general continuous. Of course, if A is a commutative Banach algebra, then PAis even uniformly continuous, since then
IPA(a) - PA(b)1 ::; PA(a - b) ::;
Iia - bll
(a,b E A).
An example showing that the spectral radius is not a continuous function on 8(£2) will be given in Exercise 5.lD.
5.9 Analytic properties. In this section we shall look at the following situation: A will be a Banach algebra, U will be an open subset of C, and f : U ---+ A will be an analytic A-valued function. We shall see that the function Zf---4PA(J(Z)) ,
U---+lR+,
has some very analytic-like properties, of which important use will be made. (In fact this function is 'subharmonic', and most of the properties that we shall prove in this section are true for subharmonic functions in general; but we emphasize that we still do not generally have continuity of the function-merely upper semi-continuity. ) We start with a trivial, but crucial, observation. Let A be a Banach algebra, and take a E A. As in the third proof of Corollary 5.36, above, let us write fn(a) = Ilanl1 1 / n for n E N. Now consider just the subsequence of N in which n runs through the powers of 2. Specifically, define gn = h n (n EN). Then of course gn(a) ---+ p(a) as n ---+ 00. However, in addition, (gn(a)k::':l is a decreasing sequence. This is just because 11·11 is a submultiplicative function, and so
gn+1(a) =
lIa2n+lIITcn+l) ::; lIa 2n II T (n+l) Ila 2" II Tcn +
1
)
= gn(a)
(n
E
N). '
We now need an elementary lemma which is a variant on Dini's theorem. Lemma 5.37 (Dini's lemma) Let K be a non-empty, compact Hausdorff space, and let (fn)n?l be a decreasing sequence in CIR(K)+. Set
f(x) Then
sUPxEK
fn(x)
---+ sUPxEK
=
lim fn(x)
n->oo
f(x) as n
(x
---+ 00.
E
K) .
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Introduction to Banach Spaces and Algebras
Proof The sequence
(SUPK
fn)n?l is decreasing and
supfn2':supf2':O K
So L := limn--><Xl(suPK fn) exists, and L 2': En
(nEN).
K
=
SUPK
f. For n E N, define
{x E K : f n (x) 2': L} .
Since each f n is continuous, with sup K f n 2': L, we see that each En is compact and non-empty. But also the sequence (En)n?l is decreasing, and so, by the finite intersection property, nnEN En -I- 0. Take Xo EnnEN En. Then fn(xo) 2': L for all n E N. Hence f(xo) = limn--><Xl fn(xo) 2': L, and so SUPK f 2': L. This shows that SUPK f = L = limn--> <Xl (SUPK fn). 0 As a matter of interest, notice that the more familiar Dini's theorem is an immediate consequence of the above lemma.
Corollary 5.38 (Dini's theorem) Let K be a non-empty, compact Hausdorff space, and let (gn)n?l and 9 be continuous, real-valued functions on K such that gn(x) ---> g(x) monotonically for each X E K. Then gn ---> 9 uniformly on K. Proof We apply Dini's lemma to (fn)n?l, where fn
Ign -
=
gl
(n EN).
0
The first application of the above to Banach algebras gives a form of a maximum principle. Recall that oK denotes the frontier of a compact plane set K.
Theorem 5.39 Let K be a non-empty, compact subset of C, and take U to be an open neighbourhood of K. (i) Let E be a complex Banach space, and let f : U ---> E be an analytic function. Then Ilf(z)ll::::: sup Ilf(w)11 (z E K). wEaK
(ii) (Weak spectral maximum principle) Let A be a Banach algebra, and let f : U ---> A be an analytic function. Then
p(J(z))::::: sup p(J(w))
(z E K).
wEaK
Proof (i) Let z E K, and choose X E E* with II xii = 1 and X(J(z)) = Ilf(z)ll· Now X 0 f : U ---> C is an analytic function, so that, by the classical maximum modulus principle,
Ilf(z)11 = l(xof)(z)l::::: (ii) Apply (i) to the function we have
sup wEaK
l(xof)(w)l::::: sup IIf(wll· wEaK
f 2n , and take the 2n th root: for each n E N,
IIf(z)2 nII Tn
:::::
sup
Ilf(w)2n 11 2 -
n •
wEaK
Then, by Dini's lemma, Lemma 5.37, we deduce the required conclusion by letting n -+ 00. 0
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Representation theory
As an immediate application we give the following characterization of the radical of a Banach algebra. Recall that J(A) and Q(A) denote the Jacobson radical and the set of quasi-nilpotents, respectively, of an algebra A.
Theorem 5.40 (Zemanek's characterization of the radical) Let A be a Banach algebra, and let a E A. Then the following are equivalent:
(a) a E J(A); (b) Sp (a + x) = Spx for every x E A; (c) p(a + x) = p(x) for every x E A; (d) p(a + x) = 0 for every x E Q(A). Proof We may suppose that A is unital. (a) =?(b). Let a E J(A), let x E A, and take A tf- Spx. Then x - A1 is invertible in A, and a + x - Al = (x - AI) (1 + (x - A1)- l a), which is invertible, using Theorem 5.9(ii). Thus A tf- Sp (a + x). The reverse inclusion also follows because x = (-a) + (a + x). The implications (b) =?(c) =?(d) are trivial. The main task is to show that (d) =?(a). Thus, let a E A be such that p(a + x) = 0 for every x E Q(A); in particular p(a) = 0 (taking x = 0), i.e. a E Q(A). Now let b E A be arbitrary, and define
F(z) = exp( -zb) a exp(zb)
(z E
q.
Then F is an A-valued entire function, F(O) = a, and F'(O) = ab - ba = [a, b]. Let
G(z) = {
F(Z)-a z
(z ~ 0),
[a, b]
(z
= 0).
Then G is also an A-valued entire function. Since Sp (F(z)) = Spa = {O} for all z E C, hypothesis (d) shows that p(G(z)) = 0 for all z E C \ {O}. By the weak spectral maximum principle, Theorem 5.39(ii), it follows that
p(ab - ba) = p(G(O)) = 0, i.e. that [a, b] E Q(A). We deduce from Corollary 5.18 that a E J(A). This completes the proof.
D
It should be noted that the use of the weak spectral maximum principle in the last argument seems to be quite essential; the semi-continuity of the spectral radius yields only the fact that p( G(O)) 2: o. For the further development of our analytic methods, we need a form of the three-circles theorem. For 0 < Rl < R 2 , define the closed annulus
Ll(Rl' R 2 ) := {z
E
C : Rl :s; Izl :s; R 2 }
.
Introduction to Banach Spaces and Algebras
252
Theorem 5.41 (Spectral three-circles theorem) Let A be a Banach algebra, let 0< Rl < R 2, let U be an open nezghbourhood of 6.(Rl' R 2 ), and let f : U ----7 A be an analytic function. For j = 1,2, set M J = sUPlwl=R, p(J(w)). Then p(J(z)) :::; MiMi- t where t E (0,1)
is
the unique number wzth
(Rl
< Izl < R 2),
Izl =
RiR~-t.
Proof Take z E C with Rl < Izl < R 2, and write r = 14 For p E Z and q E N, we apply Theorem 5.39(ii) to the function w f---+ IwIPp(J(w))q = p(w Pf(w)q) on a neighbourhood of 6.(Rl' R 2 ), and then take the qth root, so obtaining r P/ qp(J(z)) :::; max{ Rf/q M I , R~/q M 2 } .
(*)
Let 0: E lR be such that Rf Ml = R';f M 2 , and then apply (*) to a sequence of pairs (Pn, qn) E Z x N with Pn/qn ----70:. We deduce that rQ;p(J(z)) :::; RrMl = R~M2.
(**)
But log r = t log Rl + (1 - t) log R2 and 0: log Rl + log Ml = 0: log R2 + log M 2, and so we easily deduce from (**) that logp(J(z)):::; tlogMl + (l-t)logM2' which is equivalent to the stated result. 0
Corollary 5.42 Let A be a Banach algebra, take R > 1, and let f be an analytzc A-valued function on a neighbourhood of the annulus 6.(1/ R, R). Then (p(J(I)))2:::;
sup p(J(w))· sup p(J(w)). Iwl=l/R Iwl=R
Proof This is the case of the theorem in which Rl Izl2 = RIR2 = 1.
1/ R, R2
R, and 0
The first application of the theorem gives a Liouville-type theorem; we write the proof to show that, for this result, Corollary 5.42 is sufficient.
Corollary 5.43 Let A be a Banach algebra, and let f be an A-valued entire function. Suppose that p 0 f is bounded on C. Then p 0 f is constant. Proof Let M = sUPzEC p(J(z)) and m = inCEc p(J(z)) 2: 0; we may clearly suppose that M > o. Assume to the contrary that p 0 f is not constant. Then m < M, and so we may choose E > 0 such that m + E < M. Take Zo E C with p(J(zo)) < m + E. By considering f(z + zo) if necessary, we may suppose that Zo = o. By the upper semi-continuity of p 0 f at 0, there exists 8 > such that p(J(z)) < m + E whenever Izl :::; 8.
°
253
Representation theory
Let 0 =I- a E 1 for which lal/R < min(lal, 8) ::; lal < R/lal. Then, by applying Theorem 5.41 (or just Corollary 5.42) to the annulus centred at 0 and with radii jail Rand R/lal, we have
(p(f(a))2::;
sup p(f(w))· sup p(f(w))::; (m + E)M. Iwl=lall R Iwl=Rllal
Thus (m+E)M is an upper bound for (p 0 J)2 on 0 be such that {z E 0, and choose Zl E
B be a o
Notes For more continuity and analytic properties of the spectrum, see [20, Chapter III, Section 4] and the earlier work [18]; this leads to the theory of analytic multifunctions, discussed in [20, Chapter VII]. For a general discussion of harmonic functions, see [136]. The earliest result in this area seems to be Vesentini's theorem, given in [47, Theorem 2.3.32]' which states that PA 0 f : U -> lR is a subharmonic function whenever A is a Banach algebra, U is a non-empty, open set in C, and f : U -> A is an analytic function. The idea of proving Johnson's uniqueness-of-norm theorem by the approach of the present section is contained in Aupetit's important paper [19]; see also [20, Chapter V, Section 5]. Ransford's beautiful simplification of the proof is in [135]. Here is a striking result of Aupetit [18] that is proved by the methods of this section; see [47, Theorem 2.6.28]. Let A be a semisimple Banach algebra, and suppose that there
258
Introduction to Banach Spaces and Algebras
is a real-linear subspace H of A such that A = H + iH and Sp a is finite for each a E H. Then A is finite-dimensional. Theorem 5.40 is due to Zemanek [169]; for an elementary proof, see [170]. The proof of Theorem 5.41 adapts one of the standard proofs of the Hadamard three-circles theorem; Corollary 5.42 is given in [135], and Theorem 5.45 is contained in [19]. It appears that the more special three-circles lemma of Ransford, Corollary 5.42, leads to the inequality PB(1'a? S PA(a)PB(1'a + b), which is slightly weaker than that of Theorem 5.45, but it should be remarked that this latter inequality is still sufficient for the deduction of Corollary 5.46. We shall sketch in Exercise 5.10 an example that shows that the spectral radius function PA is not necessarily continuous on a Banach algebra A. A more complicated example of Muller [123] exhibits a Banach algebra A with Q(A) = {O} such that the spectral radius function is discontinuous at each point of a line in A. In [137]' Ransford proved that there exist S, l' E A = B(£2) such that the map A f--4 PA(T - AS) is discontinuous at almost every point of the open unit disc; the argument avoids the combinatorial intricacies of Muller's example and uses some elementary potential theory. An example of Dixon [60], given as [47, Example 2.3.15], exhibits a semisimple Banach algebra A such that Q(A) B is linear with PB(Ta) S PA(a) (a E A). Then it is not the case that l' is automatically continuous: a counter example to this possibility is given in [47, p. 601]. It is also easy to see that, in that example, ker l' is not closed. But it appears that the question of whether or not it is always the case that 6(1') 0 as n -> 00; (ii) the spectral radius PA is continuous at ao E A whenever Sp ao is totally disconnected. Exercise 5.10 We consider a weighted right shift operator l' on the Hilbert space H = £2, as defined in Exercise 4.10. The operator l' is defined by a sequence (w n ) in (0,1]' and then TOn = W nOn +l (n E N). To specify (W n )n21' note that each n E N has the form 2k(2£ + 1) for some uniquely specified numbers k,£ E Z+; in this case, set Wn = e- k . Then a calculation shows that, for each mEN, we have (WIW2'"
W2 m _l)1/(2"'-1)
> ( [ ( exp
C/+l ))
2
Representation theory
259
Set a = 2:;:1 j /2J+1. Then it follows from the above formula that the spectral radius p('1') satisfies p(T) :::: e- 2 0" > 0, and so the operator T is not quasi-nilpotent. By Exercise 4.10, SpT is a disc with centre 0 and with strictly positive radius. Next, for each k E fiI, define an operator 1", E B(H) by setting 1",e n = 0 when n = 2k(2L' + 1) for some L' E z:+ and 1"e n = 0 otherwise. Then Tk is nilpotent. However, liT - Tkll = e- k (k E fiI), and so 1" --> '1' in B(H). This shows that Q(B(H)) is not closed and that the spectral radius function p is not continuous on the algebra B(H) at the element T. Exercise 5.11 Show that the two formulations of the 'notorious open problem' in Section 5.11 are equivalent.
6
Algebras with an involution
Banach algebras with an involution 6.1
Let A be an algebra over
General Banach *-algebras.
rc.
Then an
involution on A is a mapping
* :a
f---+
a*,
A
-->
A,
such that:
= a for all a E A; (ii) (>.a + p,b)* = "xa* + "jib* for all a, bE A and >., p, (iii) (ab)* = b* a* for all a, b E A. (i) a**
E C;
An algebra with an involution is a *-algebra. Note that, immediately from (ii), 0* = 0 and, from (iii), that, if A has an identity 1, then 1* = 1. It then follows that, if A has no identity, then the involution on A is uniquely extendible to an involution on A+. So, we may suppose without loss of generality that A has an identity. Let A be a *-algebra. A subset S of A is *-c1osed or a *-subset if a* E S whenever a E S. A *-closed subalgebra or ideal in A is a *-subalgebra or *-ideal, respectively. Note that a *-closed left ideal in A is an ideal. A homomorphism () : A --> B between *-algebras is a *-homomorphism if (()a)* = ()(a*) (a E A). Suppose that A is a Banach algebra. Then an involution * on A is isometric if Ila*11 = II all (a E A). The algebra (A; *) is a Banach *-algebra if * is an isometric involution on A. Example 6.1 The following are all examples of Banach *-algebras, as is easily checked. (i) Set A = Co(K), the algebra of all complex-valued, continuous functions which vanish at infinity on a non-empty, locally compact Hausdorff space K, and define
j*(x) = f(x)
(f
E
A, x
E
K),
so that j* = 7 in a previous notation. (ii) Set A = A(,6.), the disc algebra, as in Example 4.3, and define
j*(z) = f(z) (iii) Set A
(f E A, z E ,6.).
= (Ll(JR.); 11.11 1 ), as in Example 4.65, and define j*(t)
= f( -t) (f
E
A, t
E
JR.).
Algebras with an involution
261
(In some notations, this 1* is denoted by /.) Note that F(f*) = F(f) for each and so the Fourier transform F is a *-homomorphism from Ll(lR) into Co(IR).
f E £1 (1R),
The above examples (i), (ii), and (iii) are all commutative algebras. The most important non-commutative example is the following. (iv) Set A = B(H), the algebra of all bounded linear operators on a Hilbert space H, with the involution taken as the Hilbert-space adjoint; see Corollary 2.56. Then A is a Banach *-algebra. A *-representation of a *-algebra A on a Hilbert space H is a *-homomorphism () : A ---. B(H). Our main aim in this section is to construct *-representations of arbitrary Banach *-algebras. (v) Any closed sub algebra of B(H) that is invariant under taking the adjoint is a Banach *-algebra. In particular, let T E K(H), the operator algebra of all compact linear operators on H. By Corollary 3.66, the adjoint T* is compact, and so K(H) is a Banach *-algebra. (vi) As a special case of the above, let A = (a'J) E MIn, the algebra of all n x n-matrices over C. Then the adjoint A* of A is (aJ ,). (The matrix (aJ ,) is sometimes called the conjugate transpose matrix of (a'J)') (vii) Let A be a Banach *-algebra. Suppose that J is a closed *-ideal of A, and set (a + J)* = a* + J (a E A). Then (AjJ; *) is a Banach *-algebra. In particular, for a Hilbert space H, the so-called Calkin algebra B(H)jK(H) is a Banach *-algebra. D On page 76, we defined 'self-adjoint' (or 'hermitian'), 'normal', and 'unitary' for operators on a Hilbert space. We now give more abstract versions of these definitions. Let A be a *-algebra, and let a E A. Then a is self-adjoint or hermitian if a = a*, and normal if a*a = aa*; in the case where A has an identity 1, a is unitary if a*a = aa* = 1. The set of self-adjoint elements of A is a real-linear subspace of A; it is denoted by Asa. In the case where A is commutative, Asa is a real subalgebra of A. The set of unitary elements is denoted by U(A). In the case where A is a Banach *-algebra, Asa and U(A) are closed in A. Proposition 6.2 Let A be a unital *-algebra, and let a E A. Then a is invertible if and only if both a*a and aa* are invertible; if a is normal, then a is invertible if and only if a*a is invertible.
Proof Suppose that a E G(A). Then a* E G(A), and hence a*a, aa* E G(A). Conversely suppose that a*a, aa* E G(A), say b = (a*a)-l. Then ba*a = 1, and a has a left inverse. Similarly, a has a right inverse, and so a E G(A). The remark about a normal element a is immediate, since then a*a = aa*. D Let A be a *-algebra. Then each a E A can be written uniquely as a = b + ic, with b, c E A sa , by taking b = (a + a*)j2 and c = (a - a*)j2i. It follows that A = Asa EB iA sa . Note that a = b + ic is normal if and only if bc = cb. It is also a
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Introduction to Banach Spaces and Algebras
simple, purely algebraic, fact that (in the case where A is unital), a is invertible if and only if a* is invertible, in which case (a*)-l = (a- 1 )*. Hence Sp(a*)
= {:-\: A E
Spa} .
Thus, if a E A sa , then Sp a is symmetrical about the real axis. (N.B.: it does not follow that Sp a " ~ ' - 5 -1,
f(8)8
f 1 (G)
-+
II . III ; *)
in
f 1 (G).
sEC
Verify that * is an involution on f1(G) and that (f1(G); algebra.
11·111;
*; *) is a Banach *-
269
Algebras with an involution
Exercise 6.5 Let §2 be the free semigroup on two generators, u and v, so that elements of §2 are 'words' in u and v. Show that there is a unique involution U on (l1(§2) such that 8t = 8v and 8t = 8u and such that ((l1(§2); 11.11 1 ; 0 is a Banach *-algebra. For each word w E §2, let nu (w) and nv (w) be the number of times that the letters u and v, respectively, occur in the word w. Show that, for each (1, (2 E ~, the continuous linear functional 'P defined by requiring that 'P(8 w ) = (~,,(w)Gv(w) for each word w is a character on (11 (§2), and that each character arises in this way. Deduce that 8uv is self-adjoint, but that Sp 8uv :;;> ~.
C* -algebras 6.4 Elementary theory of C* -algebras. nach algebra with an involution such that
Ila*all =
IIal1 2
A C *-algebra is a (non-zero) Ba-
(a E
A).
The above equality is called the C *-condition. Let A be a C *-algebra. For each a E A, we have IIal1 2 =
Ila*all : : :
Ila*llllall,
and so Iiall ::::: Ila* II. Since a** = a, we deduce that Ila* II = Iiali. Thus A is a Banach *-algebra. Suppose that A has an identity 1. Then 111112 = 111*111 = 11111, so that 11111 = 1, i.e. a C* -algebra with an identity is necessarily a unital Banach algebra. Further, for each unitary u, we have IIul1 2 = Ilu*ull = 11111 = 1, and so Ilull = 1. Let A be a C *-algebra. Then a *-subalgebra of A which is closed in the norm topology of A is called a C *-subalgebra of A. Examples 6.16 (i) Let K be a locally compact Hausdorff space, and consider CoCK), as in Example 6.1(i). It is clear that IfllK = Ifl~ (f E CoCK)), and so CoCK) is a commutative C*-algebra. (ii) Let 8 be a non-empty set, and let X be a non-empty topological space. Then (£00(8); I· Is) and (Cb(X); I· Ix) are commutative C*-algebras: the involution is * : f 1-+ ] . (iii) Let H be a Hilbert space, and consider l3(H); the involution is the map T*, where T* is the Hilbert-space adjoint of T E l3(H), as in Example 6.1(iv). That the C*-condition holds for l3(H) was verified in Proposition 2.57.
T
1-+
(iv) Each C *-subalgebra of a C *-algebra is itself a C *-algebra. In particular, each closed *-subalgebra of l3( H) for a Hilbert space H is a C *-algebra. This includes the example lC(H). D We shall see in the following section that every C *-algebra is (isometrically *isomorphic to) a C *-subalgebra of l3( H) for some Hilbert space H. In the present section, we shall prove the easier theorem that every commutative C *-algebra has the form CoCK) for some non-empty, locally compact Hausdorff space K.
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Introduction to Banach Spaces and Algebras
Lemma 6.17 Let a be a normal element of a C*-algebm A. Then p(a)
= Iiali.
Proof First, suppose that a E A sa , so that a = a*. Then IIa 2 11 = IIal1 2 by the C*-condition; a simple induction then gives Ila 2 " II = Ila11 2 " (n EN), and so
Iiall = Ila 2 " liT"
--->
p(a)
as
n
---> 00
by the spectral radius formula, Theorem 4.23. Thus p(a) = Iiali. For each a E A, a*a is self-adjoint. Suppose that a is normal. Then, by Corollary 4.48(ii), p(a*a) p(a*)p(a), and so
s:
IIal1 2 = Ila*all = p(a*a)
s: p(a*)p(a) = p(a)2 s: lIa11 2 .
Thus p(a) = Iiall, as required.
o
Corollary 6.18 Every C*-algebm is semisimple. Proof Let A be a C *-algebra, and take a E J(A). By Lemma 6.3, a* E J(A), and so also b = (a + a*)/2 and c = (a - a*)/2i belong to J(A). But then band c are quasi-nilpotent as well as self-adjoint, and so Ilbll = p(b) = 0 and IJeII = o. Thus b = c = 0, and so a = o. Hence J(A) = {O}. 0 Corollary 6.19 Let A and B be C*-algebms, and let e : A ---> B be a *-homoII all (a E A). morphism. Then e is continuous and, moreover, Ileall
s:
Proof Let a E A. Then a*a, and hence (ea)*(ea), are normal, and so, by Lemma 6.17, we have
Ileal1 2 = II(ea)*(ea)11 = PB((ea)*(ea)) = PB(e(a*a))
s: PA(a*a) = Ila*all = Ila11 2 , o
giving the result.
It follows that a C *-algebra has a unique complete norm in a strong sense: if A is a C*-algebra with respect to 11·11 and 111·111, then II all = Illalll (a E A).
We can now establish the famous Gel'falld~Naimark theorem for commutative C *-algebras; we recall that the Gel'fand representation theorem for commutative Banach algebras was given in Theorem 4.59. We shall require the following preliminary results, in each of which A is a unital C *-algebra with identity 1. Lemma 6.20 Let f E A* with Ilfll = f(l). Then f(a) E IR whenever a E Asa· Proof Without loss of generality, we may suppose that Ilfll
= f(l) =
1.
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Algebras with an involution
Take a E A sa , and set f(a) = 0: + i,6, with 0:,,6 E R Now define
u(t)
=
tl - ia
(t E JR).
Let t E R Then we have Ilu(t)11 2 = Ilu(t)u(t)*11 = IIt 21 + a 2 11 ::; t 2 + Ila11 2 . But also
t 2 + 0: 2 +,62 + 2,6t = It - io: +,612 = If(u(t))1 2
::;
Ilu(t)112 ,
so that 0: 2 + ,62 + 2,6t ::; Ila11 2. This holds for all t E JR, and so we must have ,6 = O. Thus f(a) E JR. D
Corollary 6.21 (i) 'P(a) E JR (a E A sa , 'P E
IAnl),
n
T-LAJP;"JII=lfnISpT~O as n~oo, J=l where we are using the isometry condition of Theorem 6.26(i). This shows that ~{AP;" : A E SpT} converges to T in (8(H); 11·11). For kEN and x E H, set (Ck ® Ck)(X) = (x, Ck)Ck. Then we have shown that CXl
T = L Ji,kCk ® Ck k=l
in
8(H).
Now consider the formula
f(T)
=
L{f(A)P;" : A E SpT}
(*)
for f E C(SpT). The above argument shows that the sum on the right-hand side of (*) does converge to f(T) in (8(H); 11·11) in the case where, additionally, f(O) = o. Further, in the case where f is the identity function 1 on Sp T, we have
f(T)x = x = L{P;,,(x): A E SpT}
(x
E
H)
by Theorem 2.63, (a) =} (c); since each f E C(SpT) has the form f where g E C(SpT) and g(O) = 0, we see that
f(T)x
=
L {f (A) P;., (x) : A E SpT}
(x
E
=
f(O)l +g,
H)
for each f E C(SpT). Thus (*) is a correct formula when we take convergence in the strong-operator topology. We shall return to this in Section 7.3. The following result complements Corollary 6.19. Proposition 6.29 Let A and B be unital C * -algebras, let () : A ~ B be a unital *-homomorphism, and take a to be normal in A. Then: (i) f(()a) = ()(f(a)) (f E C(SpAa)); (ii) in the case where () is injective, Sp B()a = SPA a, the map () is isometric, and ()(A) is a C*-subalgebra of B.
Algebras with an involution
275
Proof (i) Certainly SPBOa oo lim (Tnx, y)
(x, y E H),
i.e. Tn ~T. Since (Tx, x) = limn->oo(Tnx, x) :::: 0 (x E H), T is positive. For every x E H, the sequence ((Tn x,X))n2:1 is increasing in JR., so that Tn ~ T, and T is an upper bound for the set {Tn: n EN}. If S is also such an upper bound, then clearly (Tnx, x) ~ (Sx, x) (x E H), and so, passing to the limit, (Tx,x) ~ (Sx, x) (x E H), i.e. T ~ S. Thus T = sUPn> 1 Tn· Let n E N. Since 0 ~ Tn ~ T, we have liT - Tnll ::; IITII .;; k, and so II(T - Tn)x11 2 ::; k II(T - Tn)1/2XI12 = k((T - Tn)x,x) for each x E H. Thus Tn ~T.
-->
0
as
n
--> 00
0
Algebras with an involution
283
Notes There is huge number of texts on C·-algebras. For an introductory account, see [125], [47, Section 3.2], and [56, Chapter 1], which continues with many interesting examples. For a major account, which nevertheless begins at a rather elementary level, see the four volumes of Kadison and Ringrose, beginning with [102, 103]. For more advanced texts, see [154] and later volumes, and also [126, 127]. As one of the many texts relating operator algebras to a branch of physics, see the volumes of Bratteli and Robinson, beginning with [34]. An important analogue of C· -algebras in the context of topological algebras is the class of algebras called 'b· -algebras' in [5, 59], 'pro-C·algebras' in [17], and 'locally C·-algebras' in [78]. We regret that we have not discussed the important topic of bounded approximate identities in C·-algebras. From such a discussion, it follows that AI I is a C·-algebra whenever I is a closed ideal in a C·-algebra A [47, Theorem 3.2.21]. Given this, it follows from Proposition 6.29 that the range of a *-homomorphism between C· -algebras is always a C· -subalgebra. See also [102, Theorem 4.1.9]. A C· -algebra A is a von Neumann algebra if, as a Banach space, A is isometrically isomorphic to a dual of another Banach space. For example, the C· -algebra £ is a von Neumann algebra; for each Hilbert space H, the C·-algebra B(H) is a von Neumann algebra. It is a theorem of Sakai that each von Neumann algebra can be represented as a unital C· -subalgebra A of B( H) for some Hilbert space H such that A CC = A (see [103, 10.5.87] and [154, III.3.5]). For a substantial theory of von Neumann algebras, see [103] and [127, Section 9.3]. Considerations of the numerical range of an element of a unital Banach algebra lead to a striking geometric characterization of C· -algebras in the Vidav-Palmer theorem; see [31, Section 38] and [127, Theorem 9.5.9]. For more on the strong-operator and weak-operator topologies on B(H) for a Hilbert space H, see [102, Section 5.1]; in fact, there are several other interesting topologies on B(H). A unital C· -subalgebra of B(H) is a von Neumann algebra if and only if it is closed in the weak-operator topology on B(H). Let A be a C· -algebra. Then every derivation from A into a Banach A-bimodule is automatically continuous [47, Corollary 5.3.7]; this is a theorem of Ringrose [141]. In Section 5.11, we discussed the 'dense range problem', which asks if each homomorphism 8 : A ---> B, where A and B are Banach algebras with 8(A) = Band B semisimple, is necessarily continuous. This problem is open even when A is a C·algebra, and even when both A and B are C· -algebras; various partial solutions are given in [3] and [70]. It is also not known if every epimorphism from a C· -algebra onto a Banach algebra is automatically continuous. There is a vast theory on when a C· -algebra is amenable, and on the cohomological properties of C· -algebras and of von Neumann algebras; see [149], for example. (Xl
Exercise 6.6 A topological space X is completely regular if, for each closed subset F of X and each x E X \ F, there exists f E Cb(X) such that f I F = 0 and f(x) = 1. Let X be a completely regular space, and let (3X denote the character space of the commutative C·-algebra Cb(X). Show that X is homeomorphic to a dense subset of (3X, and that every function in Cb(X) extends to a continuous function on {3X. The space {3X is the Stone-Cech compactincation of X. Show that {3N is a compact, extremely disconnected space (in the sense that every open subset of (3N has an open closure) and that I{3NI = 2'. For more on {3X and especially (3N, see [83, 95]. Exercise 6.7 Let T be a compact, normal operator on a Hilbert space H. We have shown that Tx = L),xpA(x) : ,x E SpT}
Introduction to Banach Spaces and Algebras
284
in H for each x E H. However, the series is not necessarily absolutely convergent. Indeed, take H = £2 and Tx = (anx n ), where an -> 0 in IR+-. Then Tis compact and self-adjoint, the eigenvalues of l' are just the numbers an, and Pn is the projection onto the nth coordinate. So the above sum becomes 1'x = L:~= I anxnen for x E £ 2. Show that there exist (x n ) E £2 and (an) E Co such that absolute convergence of the series fails. Exercise 6.8 Let T be a compact operator on an infinite-dimensional Hilbert space H. Then 1'*1' is compact and self-adjoint, and so there is an orthonormal sequence (ukh~l in K := ker(T*1')1- such that, for each kEN, we have T*Tuk = fi,kUk for some fi,k E 0 (k EN), and then set Vk = VJik and Vk = Vi:ITuk for kEN. Show that the following hold: (i) (vkh~l is an orthonormal basis for K; (ii) x = L:~=l (x, Uk)Uk (x E K) ; (iii) Tx = L:~=l Vk(X, Uk)Vk (x E H); (iv) T*x = L:~=l Vk(X, Vk)Uk (x E H). Exercise 6.9 Prove the following Russo-Dye theorem. Let A be a unital C *-algebra. Then AlII = conv U(A). Indeed, first use the polar decomposition of Corollary 6.40 to show that, for each a E A with Iiall < 1 and each U E U(A), the element (a + u)/2 belongs to convU(A). Thus (a + x)/2 E conv U(A) for each x E conv U(A). Now take a E A with Iiall < 1, and define al = 1 and an+l = (a + an )/2 (n EN). Then (an)n~l is a sequence in convU(A) and limn~oo an = a, and so a E conv U(A). Exercise 6.10 Prove the following slight generalization of Fuglede's theorem. Let A be a C * -algebra. Suppose that a E A, that band c are normal elements of A, and that ab = ca. Then ab* = c* a. Exercise 6.11 Let A be a unital C* -algebra. The extreme points of SeA) are the pure states of A. Show that SeA) is the weak-*-closed convex hull of the set of pure states. Show that a non-zero linear functional on a commutative C * -algebra is a pure state if and only if it is a character. Exercise 6.12 Let A be a unital Banach algebra, and let a E A have numerical range W(a) and numerical radius w(a). Prove the following: (i) sup{Re.>.:.>. E W(a)} = limt~o+(111 + tall- l)/t; (ii) sup{Re.>. : .>. E W(a)} = sUPt>o(log Ilexp(ta)ll)/t = limt~o+(log Ilexp(ta)II)/t; (iii) if w(a) ::; 1, then Ilexp(za)1I ::; e (z E 11'); (iv) we have
a
=
1 -2' 7rl
1 11
dz exp(za)-2; Z
(v) II all /e::; w(a) ::; lIall. Exercise 6.13 Consider the left and right shift operators Land R on the Hilbert space H = £2, as in Exercise 2.11. Note that L* = R. Show that L n -> 0 in (8(H);so), but that R n ft 0 in (8(H); so). Deduce that the map T f--+ T* on 8(H) is not strongoperator continuous, and hence that the strong-operator and weak-operator topologies do not coincide on 8(H).
7
The Borel functional calculus
Our aim in this chapter is to generalize the continuous functional calculus of the previous chapter to give a 'Borel functional calculus' and a 'spectral theorem'. To do this we require a preliminary discussion of an integral; this discussion is given in the first section, and then our main theory will be developed in the second section of this chapter. Our development of the theory eschews any knowledge of measure theory.
The Daniell integral In the theory of the Daniell integral, we show how to 'integrate' rather a lot of functions without previously introducing any measure theory, or the notion of a Lebesgue integral. 7.1 Baire functions and monotone classes. Let X be a non-empty set. As in Example 1.4, the space (JR x ; :::;) is a poset with the ordering: f :::; 9 if and only if f(x) :::; g(x) (x E X). For E C;;;; JRx, we write E+ for the set of functions fEE with f(X) C;;;; JR+. We shall consider a sequence (ft) == (ft)t>1 in JRx and an individual function f in JR x . We shall write ft l' f (or ft l' f as i ----+ (0) to mean that: (i) (ft) is an increasing sequence in (JR x ; :::;) ; (ii) ft(x) ----+ f(x) as i ----+ 00 for every x E X. The notation ft 1 f is defined analogously. We say that, in either case, f is the pointwise monotone limit of the sequence (ft). Suppose that ft l' f and gt l' 9 and that >., j1 E JR+. Then it is easy to see that >'ft + J19t l' >.f + j1g. Now let M be a subset of JR x . Then M is a monotone class on X if it is closed under pointwise-monotone limits: i.e. if (ft) is contained in M and if either ft l' f or ft 1 f, then also f E M. It is clear that JRx itself is a monotone class, and also that the intersection of any collection of monotone classes on X is a monotone class. Hence, for each subset E of JRx, there is a unique smallest monotone class M on X with M :2 E. This smallest M is called the monotone class generated by E. Let K be a non-empty, locally compact Hausdorff space, and set L = CO,IR (K), the space of continuous, real-valued functions that vanish at infinity, so that L C;;;; JRK. Then the set of (real-valued) Baire functions on K is defined to be the monotone class generated by L. A real-valued Baire function on K is simply a member of this set; a complex-valued function on K is a Baire function if and only if its real and imaginary parts are both real-valued Baire functions.
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Introduction to Banach Spaces and Algebras
In all interesting cases (e.g. K = II), it is elementary to see that Cffi. (K) is not itself a monotone class, so that the set of real-valued Baire functions is strictly larger than Cffi.(K). For example, the characteristic function of any non-empty, open subset of II is the pointwise limit of an increasing sequence of continuous functions, and so a real-valued Baire function. Also, be warned that there is generally no useful description of an arbitrary Baire function on K; this fact has a profound effect on methods of proof. Because our interest will be entirely in bounded functions, we have a slightly modified notion of 'monotone class'. Let X be a non-empty set. A subset M of IR x is a bounded-monotone class if it is closed under bounded monotone limits, in the sense that, if (f,) is an increasing sequence in M with f, ::; c (i E N) for some c > and if f, f, then f E M, and similarly for decreasing sequences. For each subset E of IR x , there is a unique smallest bounded-monotone class containing E; it is the bounded-monotone class generated by E. For the remainder of this section, we take K to be a non-empty, locally compact Hausdorff space, and set L = Co,ffi.(K). We write Bffi.(K) and B(K)for the sets of bounded, real-valued Baire functions on K and of bounded Baire functions on K, respectively. Clearly, Bffi.(K) and B(K) are real- and complexvector spaces, respectively.
°
r
Lemma 7.1 (i) The set Bffi.(K) is the bounded-monotone class generated by L.
(ii) The set Bffi.(K)+ is the bounded-monotone class generated by L+. Proof (i) From the definition of the Baire functions, it is clear that Bffi.(K) is a bounded-monotone class on K that contains L. Now suppose that M is any such bounded-monotone class, and define E to be the set of all those real-valued functions f on K such that, for every c > 0, we have (f /\c) V (-c) E M. It is evident that E is a monotone class that includes L, and therefore it includes the set of all real-valued Baire functions on K. But then, if f E Bffi.(K), we have fEE. However, for sufficiently large c > 0, necessarily f = (f /\ c) V (-c) E M.
(ii) This is similar, and is left as a simple exercise.
o
Theorem 7.2 Let K be a non-empty, locally compact Hausdorff space. Then B(K) is a C * -subalgebra of (ROO(K); I·IK)' Also, Bffi.(K) is a sublattice of Rli{'(K). Proof Let f E Bffi. (K), and then take M (f) to be the set of all those functions 9 E Bffi.(K) such that each of f + g, f V g, and f /\ 9 is in Bffi.(K). It is clear that M(f) is a bounded-monotone class and that M(f) ::2 L for each f E L. Thus, by Lemma 7.1(i), M(f) is the whole of Bffi.(K). However, 9 E M(f) if and only if f E M(g), so that M(g) ::2 L for every 9 E Bffi.(K). Thus, for every f,g E Bffi.(K), we see that f + g, f /\g, and fV g are all in Bffi.(K). Also, )..f E Bffi.(K) for each f E Bffi.(K) and)" E IR. In particular, Bffi.(K) is a sublattice of £li{'(K).
287
The Borel functional calculus
A similar argument (first considering non-negative functions, and then using Lemma 7.1(ii)) shows that BIlt(K) is closed under the pointwise product. Now suppose that (fn)n?1 is a sequence in BIlt(K) with fn --+ f uniformly. By replacing each fn by fn -If - fnlK . 1, we may suppose that fn ~ f (n EN). By replacing each fn by II V··· V fn, we may suppose that fn+! :::: fn (n EN). Now fn i f, and so f E BIlt(K). Thus BIlt(K) is closed in (£IR'(K); I·I K )· Since B(K) = BIlt(K) EEl iBIlt(K), the extension to complex-valued functions, by combining real and imaginary parts, is now immediate. D
Corollary 7.3 Let K be a non-empty, locally compact Hausdorff space. Then
SPB(K)f
=
f(K)
(f
E
B(K)).
Proof Let f E B(K). For each x E K, the map ex : f ~ f(x) is a character on B(K), and so f(K) .f) = >.1(f) (>. :2: 0, fEU).
(iv) 1(f) ~ 1(g) whenever f,g E U with f ~ g.
Proof (i) and (iv) follow immediately from Lemma 7.6, and (iii) is a simple exercise. For (ii), we may take ft = f for all i E N. D E £;o(K), and suppose that there exists (ft) in U with ft i f· Then fEU and 1(ft) ----+ 1(f).
Lemma 7.8 Let f
Proof For each i E N, choose (g;)J?l in L with!!? i ft as j ----+ 00, and then define h j = gi v··· V g~ (j EN). Then (hJ)J?l is an increasing sequence in L with ~ h J ~ fJ for i = 1, ... ,j.
g;
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Introduction to Banach Spaces and Algebras
First, let j ----> 00 to deduce that f, :::; limJ--->CXl hJ :::; f, and then let i ----> 00 to see that f :::; limJ--->CXl h J :::; f· Thus hJ i f, so that fEU. Analogously, from the inequality I(gn :::; I(h J ) :::; J(fJ) (i = 1, ... ,j), we see that lim HCXl J(f,) :::; J(f) :::; limJ--->CXl J(fJ) , so that limJ--->CXl J(f,) = J(f). 0 Now let -U = {f : - fEU}. For f E -U, we define l(f) = -J( - 1). (In fact, -U is then precisely the set of f E CiR such that f, 1 f for some decreasing sequence in L, and the definition of 1 is equivalent to setting l(f) = lim,--->CXl I(f,) for any such sequence.) Suppose that f E (-U) n U. Then l(f) = J(f) because f + (-1) = 0 and J is additive on U. (This applies in particular to all f E L.) The set -U, with the mapping L has properties exactly analogous to the properties of U and J given in Corollary 7.7. Thus we can conclude that L O. For each i
o :s:
f, - f,-l
E
N, choose h,
:s:
h,
and
E
-
U with -
I(h,) < IU, - f,-d
c
+ 2'
.
Let n E N. Then fn :s: 2::~=1 h, := H n , say, and Hn E U+. Further, Hn /\ c E U+ and Hn /\ c;::::: fn, and (Hn /\ C)n21 is an increasing sequence, bounded above by c. Set H = limn--->oo Hn /\ c, so that H ;::::: f and Hn /\ c i H. By Lemma 7.8, we have H E U+, and n
I(H) = lim I(Hn /\ c) n----+CX)
:s: lim sup LI(ht):S: lim IUn) + c < 00. n----+oo n----+oo ,=1
Now choose N so that I(H) < IUm) + 2c (m;::::: N). Since fN E L'fR(K), there exists 9 E -U with g:S: fN and with [(g) > IUN) - c. We now see that 9 :s: fN :s: fm :s: f :s: H (m;::::: N), and I(H) < [(g) + 3c. Thus f E L'fR(K), so that L'fR(K) is a bounded-monotone class, and, further, IU) = limn--->oo IUn) , so that the space L'fR(K) has the monotone-convergence property. 0 Corollary 7.14 We have B'R,(K)
1R,
is a non-negative, Borel measure on Sp T. We wish to show that, for every .1: E H, we have
(Tx, x) =
r
Z d/Lx,
JSpT
where Z here denotes the coordinate function ). f---+ ). on Sp T. This is often written symbolically (or literally, if the appropriate integral has been defined), as
T =
r
Z dP =
JSpT
r ). dP-x .
(** )
JSpT
The proof of (**) is easier than you might expect. The secret is to be more ambitious, and to prove that, for every 9 E B(SpT), we have ((3T(g)X, x)
=
r
iSpT
9 d/Lx
(x
E
H)
(***) .
301
The Borel functional calculus
For this, consider, first, the very special case in which 9 = XE, where E is a Borel subset of SpT. Then, by definition, f3r(g) = f3T(XE) = peE), and so
(f3T(g)X, x) = (P(E)x, x) = fLx(E) =
r
IE dfLx .
JSpT
By the linearity of f3T, it then follows that (***) holds whenever 9 is any simple Borel function on Sp T. But every 9 E B(Sp T) is the uniform limit of simple Borel functions, and so the general result follows. The special case (**) follows by taking 9 = Z, since f3T(Z) = T. Notes For a related approach to the Borel functional calculus, see [128, Section 4.5J; for a different approach, see [102, Section 5.2J. There are many accounts of the spectral theorem for normal operators; see, for example, [64, Section X.2J, [88], [102, Section 5.2], [144, Chapter 12J. Our account of this calculus has been functional-analytic, and has avoided measure theory; most other accounts depend on the theory of regular Borel measures. There has been a substantial industry of obtaining results analogous to the spectral theorem for operators on a Hilbert space that are not necessarily normal and for operators on Banach spaces that are not Hilbert spaces; there are many problems to be overcome. A very important thread is the study of spectral operators; see [65, 114], where many applications of spectral theory can be found. Despite vast effort by many outstanding mathematicians, it is still an open question whether a general operator l' E B(H) has a non-trivial, closed invariant subspace. (Here, H is an infinite-dimensional, separable Hilbert space.) This is the famous invariant subspace problem for Hilbert spaces. Counter examples are known in the analogous case where l' is an operator in B(E) for a Banach space E that is not a Hilbert case: these examples, which are very sophisticated, are due to Enflo [68J and Read [138, 139J. There is a huge number of partial results on the above invariant subspace problem. For accounts, see [39, 71, 131], for example. Here is one striking theorem, from [36J. Let T E B(H) satisfy 111'11 ::; 1, and suppose that 'f 0, there is a finite set {PI"'" Pn } of pairwise orthogonal projections in B(H) with '£;=1 PJ = IH and >\1, ... , An E C such that
IIT-'£;=IAJPJII ::; E. Indeed, for j = 1, ... ,n, take PJ = XJ(T), where XJ is the characteristic function of a suitable 'half-open' square in C. (iii) Show that the linear span of the orthogonal projections is II . II-dense in B(H). Exercise 7.8 Let H be a Hilbert space, and take T E B(H) with 0 ::; T ::; 1. Find a sequence (Pn )n2';1 of pairwise commuting projections such that T = '£~=1 2-n P n .
s
seJqa~le 4:l eue pue salqe!JeA xaldwo:l leJaAas
III lJed
8
Introduction to several complex variables
In Section 4.15, we developed the holomorphic functional calculus for a single element of a unital Banach algebra. In the final part of this work, we shall develop the important analogous result: a 'several-variable holomorphic functional calculus' for finitely many elements of a unital (commutative) Banach algebra. The single-variable functional calculus used a number of results from the theory of analytic functions of one complex variable; these results are covered in undergraduate courses, and were assumed to be known. Naturally, the several-variable functional calculus will rely on results from the theory of analytic functions of several complex variables. These results are considerably harder, and are rarely covered in undergraduate courses. Thus we shall first give an account of this theory that is sufficient for our purposes-and indeed goes a little beyond what is strictly necessary.
Differentiable functions in the plane Before looking at functions of several complex variables, we shall need to look at some properties of smooth functions in the plane. Again, we shall do a little more than is strictly needed for the applications to follow because the results seem to be of considerable interest in themselves. 8.1 Theorems of Green and Cauchy. Let U be a non-empty, open subset of the complex plane C. Then the space of functions j : U ----+ C such that both the partial derivatives oj lox and oj loy exist on U is denoted by P D l(U). It is clear that P D 1 (U) is an algebra of functions on U. For j E P D 1 (U), we define the following extremely convenient notation:
oj == oj == ~ (OJ _ i OJ) , oz 2 ox oy
8j== oj Oz
==~(Oj +i Oj ), 2
ox
oy
where z = x + iy E U. Clearly, a and 8 act as derivations from P D 1 (U) into CU. We also introduce the space C 1 (U) of functions in P D 1 (U) for which oj I ox and oj loy are both continuous on U. A function j : U ----+ C is real-differentiable at a E U if and only if there is a linear function La : ]R2 ----+ ]R2 and a function c : U ----+ C such that
j(z) = j(a)
+ La(z - a) + c(z)lz - al (z
E U),
(*)
306
Introduction to Banach Spaces and Algebras
where E( z) ----> E( a) = 0 as Z ----> a in U. The space of real-differentiable functions on U is denoted by D l(U), so that D l(U) is an algebra of functions on U. Clearly, Di(U) ~ PDi(U), and
La(z - a) = (z - a)af(a)
+ (z -
a) 8 f(a)
(z E U)
for fED 1 (U). It is a standard exercise that C 1 (U) ~ D 1 (U) ~ C(U). However, there are real-differentiable functions whose partial derivatives are not continuous. The Cauchy-Riemann criterion for analyticity becomes: fED 1 (U) is analytic (i.e. complex-differentiable) on U if and only if 8 f = 0 on U. Moreover, in the case where f is analytic on U, we have of = 1', the ordinary complex derivative of f, on U. We now come to a form of Green's theorem. We shall give a proof under conditions that make clear its link with Cauchy's theorem, Theorem 1.32(i). The area (or two-dimensional Lebesgue measure) of a compact plane set K is denoted by rn(K). Theorem 8.1 (Cauchy-Green theorem for a rectangle) Let R be a closed rectangle in 0. Since 8 J is continuous at a, we see that there exists no E N such that 18 J(z) - 8 J(a)1 < ry (z E R(n») for each n ;::: no; also, we may suppose that Cn < ry (n;::: no). We then deduce that
Iw(R(n»)1 ::; 2ry bC 4n
+ 2ry (b + c)(b2 + c2 )1/2 4n
Cry
4n
(n;:::no),
°: ;
say, where C is independent of nand ry. Thus k/4n ::; Cry/4 n (n;::: no), so that k ::; Cry. This holds for each ry > 0, and so k = 0, as required. D
°: ;
We now apply Theorem 8.1 to obtain a version of Cauchy's integral formula. First, recall that the function z f---' 1/ z is locally integrable on C with respect to two-dimensional Lebesgue measure, in the sense that
JJK-lzlr dxdy
< ... < ip
~
n}
to denote both a basis of AP (C n ) (as an (;) -dimensional vector space), and also for the corresponding basis of £p(U) as a free CCXl(U)-module of degree (;). Also, we write n
£(U)
= L EB£P(U) p=o
for the space (module) of all smooth mappings w : U ----+ A(C n ). The space £(U) becomes a complex algebra under the pointwise product of forms:
(w /\ ".,)(z) = w(z) /\ ".,(z)
(w,,,., E £(U), z E U)
The product /\ on £(U) is the exterior product. N ow recall the definition of the exterior derivative namely
8j
=
t :: k=1
dZ k
(! E £0(U))
8
£ 0 (U)
£ 1 (U),
----+
.
k
We next extend the definition of 8 to the whole of £(U). First, for each p 2:: 1, -
-p+1
-P
-P
we define 8 : [, (U) ----+ [, (U). A generic element of [, (U) is, uniquely, a sum of monomial forms, and these monomials have the form
w = g dZ'1 /\ dZ'2 1\ ... /\ dz,l' ' where 1
~
i1
< ... < ip -
8w
-
~
= (8g)
nand g E CCXl(U). For such a form, we define
_
_
_
-p+1
1\ dZ'1 /\ dZ'2 /\ ... /\ dz,p E [,
(U).
Then the map 8 is uniquely extended, by additivity, to a complex-linear mapping ----+ £P+l(U). This having being done for each p 2:: 1, the mapping 8 may then be again uniquely extended by additivity to a linear endomorphism of£(U). It is called the exterior derivative. We have, in particular, the sequence of spaces and linear mappings:
£p(U)
-
[,: {
0
-------+
O(U)
-0
-------+[,
a
-1
a
a
-P
(U) -------+ [, (U) -------+ .•• -------+ [, (U)
~ lP+l(u) ~ ... ~ ~(U)
-------+
o.
322
Introduction to Banach Spaces and Algebras
Example 8.17 We take n
-2 (U) dimE
=
(2) 2
= 1.
= 2,
-
-1
and then calculate 8 : E (U)
+ hdz2
For w = gdz 1
E
-1 E (U),
---->
-2
E (U), so that
where g, hE C=(U), we
have
-8w = (8 - 2 g) dz 2 !\ az1
+ (8- 1 h) dZ 1 !\ az2 = (8h Ozl
-
89 ) dz 1 !\ az2.
Oz2
It thus appears that the definition of 8 w has been so arranged that and only if 82 g = 8 1 h. In particular,
8 2 j = 0 (f
E
8w =
0 if
= 8 2 j /8Z 2 0z 1 .
D
EO(U) = C=(U));
note that this is just a re-writing of the equation 8 2 j / Ozl Oz2
Returning to the general case, for exactly the same reason we have the following lemma. -
-
Lemma 8.18 For the mapping 8: E(U)
---->
-
-2
E(U), we have 8
=
O.
D
In particular, therefore, the sequence E is a complex (so that, as in Section 3.17, the image of each mapping in E is included in the kernel of the next one). We now make the definitions, first, that a p-form w (for p ? 0) on U is 8-closed if and only ifaw = 0, and, secondly, that the p-form w (for p? 1) is 8 - exact if and only if there is some (p - I)-form 7] such that w = 7]. Let ZP(U) and BP(U)
a
be the spaces of all a-closed p-forms on U (for p ? 0) and of all 8-exact pforms (for p ? 1), respectively; for convenience, we also set BO(U) = {O}, so that BO(U) is the image of the mapping 0 ----> O(U) in the complex E. We have remarked that ZO(U) = O(U). The complex E is called the Dolbeault complex of U; the group (in fact, complex vector space)
HP(U)
=
ZP(U)/ BP(U)
(p? 0),
is called the p th Dolbeault cohomology group of U. For us, this will just be a name: we shall not be developing any formal cohomology theory. Our interest will be to prove just that, for suitably shaped open subsets U of we have HP(U) = {O} (or, equivalently, that BP(U) = ZP(U)) for each p ? 1. This says that the inhomogeneous Cauchy-Riemann equations and their higher-degree analogues are solvable on such open sets U. Towards the above programme, we shall need a few comments on the mapping properties of forms; we shall restrict our attention to analytic mappings.
en,
323
Introduction to several complex variables
Theorem 8.19 Let U and V be non-empty, open subsets of en and em, respectively, and let fJ : U ---4 V be an analytic mapping. Then there is a unique mapping fJ * : £ (V) ---4 £ (U) such that: -0
(i) for f E £ (V) == COO(V), we have fJ *(f)
=
f
0
fJ;
(ii) fJ * is a complex-linear mapping; (iii) fJ *(w 1\ "7) = fJ *w 1\ fJ *"7 (w, "7 E £(V)) ; (iv) fJ*(8w)
= 8(fJ*w) (w
E £(V)).
Proof Set fJ = (fJI, ... , fJm), and write (WI, ... , w m) for the coordinates in For a generic p-form won V, say
w
=L
em.
g[ dill'1 1\ dill'2 1\ ... 1\ dw,p ,
[
where g[ E coo(V), I
= (i l , ... ,ip), and 1 :::::
fJ *w = L(g[
0
fJ) 8ll'1
il
< ... < ip::::: m, define
1\ ... 1\ 8ll,p
,
[
and then extend fJ * by additivity to the whole of leV). The proof that fJ * has the required properties is a matter of simple verification. For example, (iv) is essentially just the 'chain rule' for differentiation: first, we see from (i) that
_
fJ*(af) = fJ*
(m L
af
Ow dw)
)=1)
) =
Lm(af Ow
0
)_
_
fJ all) = a(f OfJ)
)=1)
for f E COO(V), and then we extend this to all wE leV).
o
Remarks. (i) This 'dual mapping' fJ * also preserves the degree of forms, i.e. fJ * (£P (V)) ~
£P (U)
for each p ~ O. This is because
8f
is a I-form for each
f E coo(U). (ii) An important special case of the above theorem arises when U is an open subset of V and ~ : U ---4 V is the inclusion map. Then ~ * : £ (V) ---4 "£ (U) is just restriction of forms-i.e. restriction of each coefficient function. We then normally write w I U instead of ~ * (w). Before proceeding to show that HP(U) = {O} (p ~ 1) for certain open sets U in en, we give a simple consequence of such a result; it will be used later.
Theorem 8.20 Let U be a non-empty, open subset of en, and let V and W be open subsets of U with U = V U Wand V n W -1= 0. Take p 2: 0, and suppose that HP+1(U) = {O}. Then, for every w E ZP(V n W), there exist a E ZP(V) and (3 E ZP(W) such that w = a - (3 on V n W.
324
Introduction to Banach Spaces and Algebras
Proof It follows from Theorem 8.5 that there exists a function 0 E C=(U) with O(U) c IT and with supp 0 c V and supp(l - 0) ~ W. (Thus {O, 1 - O} is a smooth partition of unity on U, subordinate to the open covering {V, W}.) -p -p Now define 0:1 E [; (V) and (31 E [; (W) by:
O:I(Z) = {~l and (31(Z) =
-
(z E V n W), (z E V \ W);
O(z))w(z)
{~O(z)w(z) (Z
(z
E E
V n W), W \ V).
Then we do have w(z) = O:I(Z) - (31(Z) for Z E V n W, but, of course, the forms 0:1 and {3J will not generally be 8 -closed; we must modify them to achieve this. On V n W, we have 80:1 - 8 {31 = 8w = o. Thus there is a well-defined -MJ - form 17 E [; (U) specified by setting 17 1 V = a0:1 and 171 W = a{31. We then -2 -2 have a17 = 0 on U because a17 is locally equal to either a 0:1 or a (31. But, by -p hypothesis, 1{P+l(U) = {O}, and so there is some "y E [; (U) such that 17 = aT We then define 0:
= 0:1 -
("y
1
V)
on V,
{3 = {31 - ("y W) 1
on W.
On the intersection V n W, we still have 0: - {3 = 0:1 - (31 = w, whilst on V, for example, we have 80: = 80:1 - 8"Y = 80:1 -17 = o. Thus 0: E ZP(V) and, similarly, (3 This completes the proof.
E
ZP(W).
o
We record what is, for us, the most important special case of the above theorem, that in which p = O. Corollary 8.21 Let U be a non-empty, open subset of <en, and let V and W be open subsets of U wzth U = V U Wand V n Wi' 0. Suppose that 1{ I(U) = {O}. Then, for every f E O(V n W), there exist 9 E O(V) and h E O(W) wzth f=g-honVnW. 0 Notes For general introductions to the theory of analytic functions of several complex variables, see [85, 86, 96, 111, 133]' for example. It is a central fact that there are many key differences between the several-variable theory and the one-variable theory; for an attractive discussion of 10 key differences, see [111, Section 0.3]. For example, a non-empty, open set U in en is a domain of holomorphy if there do not exist non-empty, open sets V and W, with W connected, such that V IIIK; the set U is holomorphically convex if there is a sequence (Km)m~1 of compact, holomorphically convex subsets such that U = U Km. Then U is holomorphically convex if and only if it is a domain of holomorphy. Here is another example of a key difference between e and en for n 2:: 2. The Riemann mapping theorem says that every proper, simply connected, open subset of e is biholomorphic to the open unit disc. However, there is no analogous result in en for n 2:: 2: for example, by a classical result of Poincare, there is no biholomorphic mapping from the unit polydisc onto the unit polyball [111, Appendix to Section 1.4]. Let U be a non-empty, open set in en. Then the Frechet algebra O(U) is functionally continuous, and the character space A}'
Of course, for n = 1, it follows from Corollary 4.47(ii) that we simply recover the spectrum SPAal of al. In fact, from Theorem 4.46, we also have SPAa =
{(>'1'
00.
,An)
E
en :
t
A(Ak 1 - ak)
of.
A} ,
k=l
which is an extension of the original definition of the spectrum. Note that, for a commutative, unital algebra A, a = (aI, ... ,an) E An, and p E qX] = qXI"'" Xn], we can define p(a) =P(al, ... ,an ); the map 8 a : p f---+ p(a), qX] --+ A, is a unital homomorphism such that 8 a (XJ ) = aJ (j = 1,oo.,n). Again, we wish to extend 8 a to have a larger domain. The next lemma summarizes some immediate consequences of the above definition of SPA a; it is an easy exercise.
Introduction to several complex variables
331
Lemma 8.28 Let A be a commutatzve, unital Banach algebra, let n E N, and let a E An. Then: (i) SPAa zs a non-empty, compact subset ofre n ;
(ii) Jrm,n(SPA(al, ... ,an )) =SPA(al, ... ,a m ) (m= 1, ... ,n); (iii) SPA(al, ... ,an ) ~ rr~=lSpA(ak); (iv) SpAP(a) = p(SpAa) (p E qX]).
o
8.9 Polynomial hulls. We now extend to the space ren a definition already made for the case where n ~ 1 on page 179. Let K be a compact subset of ren. Then the polynomial hull K of K is defined to be:
R = {z E ren : Ip(z)1
::; IplK for all polynomials p E qX]} .
The set K is defined to be polynomially convex if and only if K = K. Since the coordinate projections Zm (for m = 1, ... , n) are themselves polynomials, it is clear that R is always compact. Also, K ~ Rand R is polynomially convex. Suppose that Pl, ... , Pr E qX], and set
K = {z E
ren : Ip] (z) I ::; 1
(j = 1, ... , r)} .
If K is compact, then clearly K is polynomially convex. In particular, a closed polydisc is polynomially convex. By Corollary 4.40, a compact subset K of re is polynomially convex if and only if re \ K is connected, and so polynomially convex sets are specified topologically. For K c re n , we may still deduce from the maximum modulus principle, Corollary 8.13, that ren \ K is connected whenever K is polynomially convex. However, the converse assertion may fail when n 2: 2. For example, the compact set K = {(z, 0) E re 2 : Izl = I} is such that re 2 \ K is connected, but R = {(z,O) E re 2 : Izl ::; I} :2 K. This set K is homeomorphic to the set L = {(z, z) E re 2 : Izl = I}, but L is polynomially convex. Recall that a Banach algebra A is polynomially generated by al,"" an E A if A( al, ... , an) = A, in the notation of page 179. The reason why an approach to function theory via polynomial convexity is appropriate for applications to Banach algebras is contained in the following simple result. Theorem 8.29 Let A be a commutative, unital Banach algebra, and let a E An.
(i) S71ppose that A is polynomially generated by al, ... , an in A. Then Sp Aa is polynomially convex, and the mapping
a: r.p >--> (r.p(al)""
,r.p(an )),
A
is a homeomorphism. (ii) In the general case, SPA(al, ... ,an)(a) =
s;:a.
--+
SPAa,
Introduction to Banach Spaces and Algebras
332
Proof (i) Certainly a : q>A ----> SPAa is a continuous surjection. As in Proposition 4.64, is an injection, and hence a homeomorphism. Let A = (AI, ... , An) E en \ Sp A(a). Then there are bl , ... , bn E A with L:~=l bk(Ak 1 - ak) = 1. Since A(al,"" an) = A, there are PI,··· ,Pn E qX] such that
a
111-
~Pk(al, ... ,an)(Ak1-ak)11 < l.
Set q(Zl, ... , zn) = 1- L:~=l Pk(Z)(Ak - Zk) EqX]. Then q(Al, ... , An) = 1. By Lemma 8.28(iv), Iq(z)1 ::::; Ilq(a)11 < 1 (z E SPAa), and so A ~ This shows that SPA a is polynomially convex.
sp:a.
(ii) Trivially, Sp Aa t;;; Sp A(al, ... ,a n ) (a), and so, by (i),
sp:a.
sp:a t;;; Sp A(al, . ,an) (a).
Conversely, take A = (AI, ... , An) E en \ Then there exists P E qX] with Ip(A)1 = 1, but PA(p(a)) = sup{lp(z)1 : Z E SPA a} < 1. But the spectral radius is unchanged in passing to a closed sub algebra (by Theorem 4.23), so that also A ~ SPA(al, ... ,an)(a). 0 Example 8.30 In Example 4.3, we described some standard uniform algebras on a non-empty, compact subset K of C. Now suppose that K is a non-empty, compact subset of en. Then there are entirely analogous examples on K. Indeed, P(K), R(K), and O(K) are the uniform closures in C(K) of the restrictions to K of the algebras of all polynomials on en, of all rational functions p/q, where P and q are polynomials and 0 tI. q(K), and of the functions which are analytic on some neighbourhood of K, respectively, and the algebra A(K) is defined to be A(K) = {f E C(K) : flint K is analytic}.
Clearly, we again have P(K) t;;; R(K) t;;; O(K) t;;; A(K) t;;; C(K). It is easy to see that the character space of P(K) can be identified with the polynomial hull of K, and so 'up to homeomorphic identification' we have q>P(K)
=
SPP(K)(Zl,"" Zn)
= R.
To determine the character spaces of R(K), O(K), and A(K) is more challeng~g.
0
We shall now leave Banach algebras for a while and turn to the study of polynomially convex sets. 8.10 Polynomial polyhedra. Again, throughout this section, n E Ii will be fixed. So far, we have defined polynomial convexity only for compact subsets of en. We now say that an open subset U t;;; en is polynomially convex if and only if, for every compact subset K of U, then also R c u.
333
Introduction to several complex variables
A polynomial polyhedron in
P
{z
=
E ~ :
en
is an open subset P of
IPr (z) I < 1
(r
=
en of the form
1, ... , k)} ,
where ~ is some open polydisc and PI, ... ,Pk are polynomials. The closure of a polynomial polyhedron is compact. Evidently, an open polydisc is an example of a polynomial polyhedron. Lemma 8.31 (i) Every polynomial polyhedron is polynomially convex. (ii) Let U be an open neighbourhood of a compact, polynomially convex set K. Then there is a polynomial polyhedron P such that K cPS;:; U. (iii) Let U be a non-empty, open, polynomially convex subset of Then there exist compact, polynomially convex sets Km for mEN with U = U:=I Km and Km C int K m+ 1 (m EN).
en.
o
Proof This is an easy exercise.
We aim to prove that H PCP) = {O} for p 2': 1 and every polynomial polyhedron Peen. A key step towards this is contained in the following famous lemma. We now write D = {z E e : Izl < I} for the open unit disc in C. Lemma 8.32 (Oka's lemma) Let U be a non-empty, open subset of that p 2': 0 and that HP+1(U x D) = {O}. Take
(z, w, 'P2(Z), ... , 'Pk(Z)) ,
Since U'P2," ,'Pk x D = (U X inductive hypothesis gives
Dk. D)'P2, ... ,'Pk and since U x Dk = (U x D) X D k-
JLo(ZP(U x Dk)) for each p ;:::
o.
=
U'P2"",'Pk x D
ZP(U'P2"",'Pk
X
D)
----*
U
X
l ,
the
Introduction to several complex variables
335
As in the proof of Corollary 8.33(ii), HP(U'P2"",'Pk x D) = {O} (p ~ 1). But now Corollary 8.33(i) shows that HP(U'Pl, .. ,'Pk x D) = {O}. Define J.LI: z
>---*
(Z,'PI(Z)) ,
U'Pl"",'Pk
----+
U'P2, .. ,'Pk X D.
Then
J.L{(ZP(U'P2, .. ,'Pk
X
D))
= ZP(U'Pl, ... ,'Pk)
for P ~ O. Now J.L = J.Lo 0 J.LI, and so J.L* = J.Li 0 J.La. This implies the result by showing that J.L*(ZP(U x Dk)) = ZP(U'Pl"",'Pk) for every P ~ O. D Corollary 8.35 Let II be an open polydisc in
en,
let PI, ... ,Pk be polynomials,
and let P be the polynomial polyhedron P = llPl, .. ,Pk := {z Ell: IPr(z)1 < 1 (r = 1, ... , k)}.
Then HP(P) such that
= {O} (p
~ 1). Further, Jor J E O(P), there exists FE O(ll
J(z) = F(Z,PI(Z), ... ,Pk(Z))
(z
E
X
Dk)
P).
Proof Since II x Dk is an open polydisc in e nH , it follows from Theorem 8.23 that HP(ll x Dk) = {O} (p ~ 1). But now the result follows immediately from Corollary 8.34. D
en, and let V and W be open subsets of P with P = V U Wand V n W 1= 0. Then, Jor every J E O(V n W), there exist g E O(V) and h E O(W) with J = g - h on V n W. Corollary 8.36 Let P be a non-empty polynomial polyhedron in
Proof By Corollary 8.35, HI (P) = {O}, and so this is immediate from Corollary 8.21. D
For the following results, we recall that O(U) is a Frechet algebra for the topology of local uniform convergence; topological notions always refer to this topology. Corollary 8.37 Let Q be a polynomial polyhedron m
en,
let PI, ... ,Pk be poly-
nomials, and set P = QPl"",Pk := {z Then, Jor each J
E
Q : IPr(z)1 < 1 (r O(Q
X
Dk) such that
J(Z) = F(Z,PI(Z), ... ,Pk(Z)) = (F
0
J.L)(z)
E
O(P), there exists F
= 1, ... , k)}.
E
(z
E
P).
The map J.L* : F >---* F 0 J.L, O(Q X Dk) ----+ O(P), is a continuous, surjective homomorphism of Frechet spaces, and so there is a topological and algebraic isomorphism O(P) ~ O(Q x Dk)/ ker J.L* .
336
Introduction to Banach Spaces and Algebras
Proof For the existence of F, we note just that Q x Dk is a polynomial polyhedron in n + k , and then apply Corollaries 8.35 and 8.34. It is clear that /1* is a continuous, surjective homomorphism; the conclusion concerning the isomorphism then follows from the open mapping theorem for Frechet spaces, Theorem 3.58. 0
e
In Theorem 8.39, below, we shall identify the space ker /1* of the above corollary. Corollary 8.38 (Oka-Weil theorem) Let U be an open, polynomially convex subset oJ en. Then the set oJ polynomial Junctions on U is dense in 0 (U). Proof First, consider the case of a polynomial polyhedron, say, P = .6. P1 , ... ,Pk' where .6. is an open polydisc in en and PI, ... ,Pk are polynomials. By Corollary 8.35, for each J E O(P), there exists a function F E 0(.6. X Dk) such that J(z) = F(Z,PI(Z), ... ,Pk(Z)) (z E P). By Theorem 8.11, F has a power-series expansion on the polydisc .6. x Dk, say, 21 2,,)1 )k' F( Zl,···,Zn,WI,···,Wk ) -- ~ L....,a21 .. 2,,)1 ... )k ZI "'Zn WI '''W k '
with local uniform convergence on .6. x Dk. Hence
J(Z) =
L a21 ...
)k
Zl1 ... Z~" PI (z)11 ... pk(z)1k
(z
E
P),
with local uniform convergence on P. The partial sums of this series then provide the approximating polynomials to the function J. Now take U to be an arbitrary open, polynomially convex subset of en, and let K be a compact subset of U, so that R c U. By Lemma 8.31(ii), there is a polynomial polyhedron P, with K A, such that, in an obvious notation, B(ZJ)=aJ
(j=I, ... ,n),
B(VVr)
= br
(r
= 1, ... ,k).
For F E O(~ x Dk), set (fJ* F)(z)
= F(Z,Pl(Z), ... ,Pk(Z))
(z E U),
as in Corollary 8.37. By Corollary 8.37, fJ* : O(~ x Dk) --'> O(P) is a continuous, surjective homomorphism, and 0 (P) ~ 0 (~ X Dk) / ker fJ *. There will thus be a (unique) continuous homomorphism, say 'ljJ : O(P) --'> A such that the diagram O(~
X
Dk)
M'i~ O(P)
'ljJ
• A
is commutative if and only if ker fJ* O(P) be the restriction map, and let O(P) --> A be the homomorphism given by case (ii). Then let e a = e;: 0 R; the map e a has the required properties. 0
e;: :
9.2 General version. Our aim now is to remove the condition of polynomial convexity imposed on the neighbourhood U of Sp a in the last lemma; unfortunately, this process involves some arbitrary additional choices, and so the simple uniqueness statement of Lemma 9.1 is lost. The extension uses an ingenious idea of Arens and Calderon, which we give as a separate lemma. Lemma 9.2 (Arens-Calderon lemma). Let A be a commutative, umtal Banach algebra, let a = (a1, ... , an) E An, and let U be an open neighbourhood of Sp A a in en. Then there is a finite subset, say {a n+1, ... , am}, of A such that SPA a ~ SPA(a\, .,am)(a) cU. Further, there is a polynomial polyhedron P in em with SPA(a1, ... ,am) C P and 7rn ,m(P) ~ U.
with SPA a C ~. Take fl = (fl1, ... ,fln) E ~ \ U. Then fl tf- SPAa, and so there are elements b1 , .•• , bn E A such that
Proof There is an open polydisc
~
n
L br (flr1 -
ar )
=
l.
r=1
Set F(,L):= A(a1, ... ,an ,b 1, ... ,bn ). Then fl tf- SPF(/1)(a). Since this latter set is closed, there is an open neighbourhood V(fl) of IL with V(,L) n SPF(/1) (a) = 0. The set ~ \ U is compact, and so finitely many of the neighbourhoods V(fl) cover ~ \ U; for each such V(fl), there is a corresponding set {b 1 , ... , bn } of elements of A. Let {a n+1, ... , am} be the union of all these finite sets, and set F = A(a1, ... ,an ,an+1, ... ,am ). Then SpF(a) n (~ \ U) = 0. Certainly SPF(a) C ~ (for example, by using Theorem 8.29(ii)), and so it follows that SPF(a) C U. The final clause follows from Theorem 8.29(ii) and Lemma 8.31(ii). 0
Notation. Take 1 ~ n ~ m. For an open subset U of en, there is a continuous homomorphism 7r;:,m : O(U) --> O(U x e m - n ) defined by setting 7r;:,m(f)(Z1"", zm) = f(z1"", zn)
(J
E
O(U)).
We say that the mapping 7r;;',n is defined by 'ignoring coordinates' Zn+l,"" Zm· We can now give our statement of the holomorphic functional calculus in several variables for a fixed element of An.
343
Holomorphic functional calculus
Theorem 9.3 (Holomorphic functional calculus) Let A be a commutative, unital Banach algebra, let a = (al, ... , an) E An, and let U be an open neighbourhood of SPA a in C n . Then there is a continuous, unital homomorphism 8~ == 8 a : O(U) -+ A such that; (i)8 a (ZJ)=aJ (j=l, ... ,n)i (ii) cp(8 a (f)) = f(cp(at), ... , cp(a n )) (f E O(U), cp E
A with z = a('P)' Thus J(G(z)) = J(g('P)) = al('P) = Zl; also J'(G(z)) =1= O. By another use of the inverse function theorem, Proposition 1.36, each z E ~ has an open neighbourhood, say N(z), on which is defined an analytic function, say G z , which is unique subject to the conditions that: (a) Gz(z) = G(z);
(b) (f 0 Gz)(w) = WI (w E N(z)). By shrinking N(z), if necessary, we may then also suppose that: (c) Gz(w) = G(w) (w E N(z) n~). For simplicity, take N(z) to be an open ball N(z) = B(z; 30(z)). Define
N = U{B(z; o(z)) : z E ~}, so that N is an open neighbourhood of ~. If B(Zl; o(zt)) n B(Z2; O(Z2)) =1= 0, then, without loss of generality, we may suppose that o(zt) ::; O(Z2)' But then B(Zl;O(Zt)) ~ B(Z2;30(Z2))' so that G Z2 = G Z1 on the ball B(Zl;O(Zl)), and thus, in particular, on the set B(Zl;O(Zt)) nB(Z2;o(Z2))' We have shown that there is a well-defined function G E O(N) such that G I B(z; o(z)) = G z I B(z; o(z)) (z E ~). Then certainly G I ~ = G, and J(G(w)) = WI (W EN). This proves Step 2. Let 8 : O(N) -+ A be a continuous functional calculus homomorphism such that 8(ZJ) = a J (j = 1, ... ,m), so that, in particular, 8(Zt) = a1 = a. Define b = 8(G) EA. By the composition theorem, Theorem 9.4, we have f(b) = f(8(0)) = (f
Clearly,
b=
0
O)(a) = a1 = a.
g. Thus we have proved the existence of the required element b.
Introduction to Banach Spaces and Algebras
348
It remains to prove the uniqueness statement involving b. However, we know that the function J is analytic on a neighbourhood of g( A) = Sp b and that 1'(z) f. 0 (z E Spb)), and so this is immediate from Proposition 4.92. 0
Examples 9.8 Let A be a commutative, unital Banach algebra. (i) Let a E G(A), and suppose that (for some p ::::: 2) there exists 9 E C( A) with gP = Ii on A. Then there is a unique element b E A such that bP = a and b = g. [Take J(z) = zP in Theorem 9.7: clearly Jog = Ii and, since a is invertible, 1'(z) f. 0 (z E Spa).] (ii) Let a E G(A), and suppose that Ii has a continuous logarithm on A, in the sense that Ii E exp C( A)' Then there is a unique element b E A such that exp b = a and b = g. [This is just like (i), but with J(z) = eZ .] (iii) We remark that by taking a = 0 and 9 = XK (where K is an open-andclosed subset of A) and J(z) = z2 - z for z E C, we obtain an alternative proof of Shilov's idempotent theorem. 0
9.5 The Arens-Royden theorem. Another viewpoint on Shilov's idempotent theorem is to see it as the start of a programme to describe the topology of the character space A explicitly in terms of algebraic features of A: it sets up a bijection between the set of idempotent elements of A and the set of openand-closed subsets of A. The Arens-Royden theorem takes this a stage further: it says that Hl(A;Z) ~ G(A)/expA. Here, as usual, G(A) is the group of units of A and expA = {e a : a E A}. The notation H 1 ( A; Z) means the first integral cohomology group of A: if you know what that is, fine, but, if not, do not worry. Set C = C( A) (so that c = A)' Then what will be proved (in a slightly more explicit form) is that
G(A)/ exp A
~
G(C)/ exp C.
The fact that G(C)/ exp C, which is manifestly a topological invariant of A, can also be identified with HI ( A; Z) is then a matter of pure topology, and must be outside the scope of this book. Let A be a commutative, unital Banach algebra, with Gel'fand transform g. Then it is clear that 9(G(A)) G(C)/ exp C, induced by the Gel 'fand representation oj A, is an isomorphism.
349
Holomorphic functional calculus
Proof The proof falls naturally into two parts. First, we note that 9 is mjective: i.e. if a E G(A) and if E expC, then a E expA. But this is an immediate consequence of Example 9.8(ii), above. It thus remains to prove that 9 is surjective: i.e. given f E G(C), there is some a E G(A) with f /a E expC. Thus, let f E G(C), so that f is a continuous function on the compact Hausdorff space A with Z(f) = 0. Set
a
c
= inf{lf( 0, and, for each g E C with Ig - fl"), 171, 175, 220 E ~ F, 227 E*,45 E[r],35 (E;P),114 lO(u), 318 ll(U),318 lP(U),320 l(U),321
370 ex, 186, 189 en, 47
f * g, 195 f, i f, 285 f(k),66 34 F tB G, 34, 110, 151 (F), 48 F, 89,100 F(E,F),47 F(E), 47, 160 J' = IC[[X]], 187, 210 J'n = IC[[X1, ... ,Xn]], 315 A, 181, 189
F+ G,
Index of symbols !inS, 33 linX,35 lip",K,58 £1 (JR.), 194, 260 £1(JR.+,w), 196,211,247 £1 ('Jr), 194 £1(JR.+),196 £P[a, b], £P('Jr), £P(JR.), 53 Lip",K, 58, 209 AP, A, 319 £(E,F),42 £(E),42 £(f), 211
MA,185 J1. *w, 323
G(A),162 Go(A),223 G(E), 165 Gr(f), 132 Q, 191 [,],30 r(A), 202 I'o(A), 202 H H
MI m ,n,43 MIn, 43, 159 norms
·11,34 .11 1 , 37 ·11=,38
·ll w ' 188 ·ll p ' 35, 40
(K;Z),225
. 11 p,q '
1 (A;Z),348
'l s ,36
I
H=(U), 58, 209 1{,98 1{P(U),322 imT,42 inner product (-, . ), 70 int~, 7 Ie, 42 J(f), 289 1(f),290 ll,7
,l(A), 193, 232 ,l*(A),264 kerT,42 R, 179, 331 KA, 205,279 KN,93 K(E, F), 46, 56 K(E), 47, 160, 176 {! l(s), 38 {!1(Z+,w),188 {!2, 72 {!=,38 {! (S), {!R"(S), 36 e=(Z),38 {!P 39 40
=
ee.:
50
n(,; z), 30 N,7
O(U), 31, 114, 161, 188, 246, 314 w 1\ 'f/, 321 OK, 212 Ord,15 pairing (-, .), 116 P(K), R(K), O(K), A(K), 158, 332 PD leU), 305, 313 PA, 220, 294 P(S),7 7r,';"n (f), 342 1Q,1Q+,7 Q(A), 166, 233
R(K),187 R(U),213 RaP. ), 167 RA(a), 177 JR. JR.+ JR.- JR.+. 7 JR. ,9:285' , PA(a),169
x
s, 37, 47 suppf,309 S(A),263 SE,35
ISBN 978-0-19-920654-4
1111111 III
9 780199 206544
