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Z2/ 2)
(z E
q.
Take r E IR, and integrate
~ = y"2; Fo(t) Set
r = t,
1
00
00
exp
(
itx 
X2) 2""
dx =
and use the fact that f~oo exp
1
00
00
exp
(_x 2 /2)
(
it(x
dx =
dx. + ir)  (X+ir)2) 2
v'21r.
Exercise 2.38 Let f E £2(IR) be the characteristic function of the interval [1,1]. Calculate and deduce that
1,
1
sin2 t _ 2dt 7r.
00
t
00
Exercise 2.39 Let
f
E Coo(IR), and define
1(z) =
~
y 27r
1
00
f(x) e ZX dx
(z
E
q.
00
1
Show that the integral is well defined for each z E IC, and that is an entire function on the complex plane IC; the function is the Laplace transform of f. The Fourier transform of f may be regarded as the restriction of this function to the imaginary axis.
1
3
Banach spaces
Existence of continuous linear functionals The first section of this chapter is concerned with the question of the existence of continuous linear functionals, principally on normed spaces. We have seen that, for a Hilbert space, the question is easily answered by the Riesz representation theorem, Theorem 2.53. For general normed spaces, we must work a little harder, though it remains true that there is always a good supply of continuous linear functionals. 3.1 HahnBanach theorems. For the purpose of proving theorems, at a later stage, on the separation of convex sets, it is useful to consider something a little more general than a norm. Thus, let X be a real vector space. A sublinear functional on X is a mapping p : X > JR. such that: (i) p(x + y) ::; p(x) + p(y) (x, y EX); (ii) p(o:x) = o:p(x) (x E X, 0: > 0). We note that a sublinear functional is allowed to assume negative real values, and that clause (ii) is only required to hold for 0: > o. Clearly, a seminorm on X is a sublinear functional. Theorem 3.1 (HahnBanach theorem: first form) Let X be a real vector space, and let p be a sublinear functional on X. Let Y be a vector subspace of X, and let g : Y > JR. be a linear functional such that g(y) ::; p(y) (y E Y). Then there is a linear functwnal f : X > JR. such that both flY = g and f(x) ::; p(x) (x EX). Proof Apart from a 'general nonsense' bit (involving Zorn's lemma), we just have to carry out a onestep extension. Thus, let Xo E X with Xo rj:. Y, and let
Z = lin{Y,xo} = {y
+ AXo: y E Y,
A E JR.}.
To extend g from Y to a linear functional f on Z, we just have to specify the value of f(xo) = 0:, say, and we must then take
f(y + AXo) = g(y) + AO:
(y
E
Y, A E JR.).
The catch is that we must also ensure that
f(y
+ AXo)
::; p(y + AXo),
i.e.
g(y)
+ AO: ::; p(y + AXo)
(y
E
Y, A E JR.).
This last requirement is obviously satisfied if A = 0; we then consider the two cases where A > 0 and where A < o. In fact, by homogeneity, it is elementary
107
Banach spaces
to see that it suffices to consider the two cases where A = ±1. In any case, it is clearly necessary that a satisfy:
g(yt) + a ::; P(YI + xo) g(Y2)  a ::; P(Y2  xo)
(Yl
E
Y) ,
(Y2 E Y),
i.e. we need g(Y2)  P(Y2  xo) ::; a ::; P(YI + xo)  g(Yl) for all Yl, Y2 E Y. By the completeness of JR, such a choice of a is possible if and only if
g(Y2)  P(Y2  xo) ::; P(YI
+ xo)
 g(Yl)
(Yl, Y2 E Y)
since we may then choose a to be any real number satisfying
sup{g(y)  p(y  xo)} ::; a ::; inf {p(y yEY
+ xo)
With this motivation, we calculate that, for any Yl, Y2
(P(YI
+ xo)
 g(y)}.
yEY
E
Y, we have
 g(Yl))  (g(Y2)  P(Y2  xo)) = p(Yl
+ xo) + P(Y2 
xo)  g(Yl
2: P(YI + Y2)  g(Yl + Y2) 2:
+ Y2)
o.
Thus a suitable choice of a is possible and the 'onestep' extension of 9 from Y to Z can be done. Now for the Zorn's lemma part. Let n be the set of all pairs (E, fE), where E is a vector subspace of X with E :2 Y and fE is a linear functional on E with IE I Y = 9 and f (x) ::; p( x) (x E E). We partially order n by defining (El,Jt) ::; (E 2, h) to mean that both El <:;;; E2 and 12 lEI = II; it is immediate that we have obtained a nonempty poset. Suppose that C = {(Ea, fa)}aEA is a chain in n. Then we set
E=
U Ea,
aEA
and define f : E + JR by flEa = fa for all a (the function f is well defined because C is a chain). Then (E, f) E n, and (E, f) is clearly an upper bound for the chain C. By Zorn's lemma, n has a maximal element, say (F, gF). Assume towards a contradiction that F =I= X. Then choose any Xo E X \ F, and apply the onestep argument to give an extension of 9 F to lin {F, xo}, preserving the domination condition, so contradicting the claimed maximality of (F,gF). Hence F = X, and so, taking f = gF, we are done. 0 Theorem 3.2 (HahnBanach theorem: second form) Let X be a vector space with scalar field lK, and let p be a seminorm on X. Let Y be a subspace of X, and let 9 be a linear functional on Y such that Ig(Y)1 ::; p(y) (y E Y). Then there is a linear functional f on X such that flY = 9 and If(x)1 ::; p(x) for all x E x.
108
Introduction to Banach Spaces and Algebras
Proof (i) Case lK
= R Since g(y) ::; Ig(y)1 ::; p(y)
(y
E
Y),
Theorem 3.1 gives a linear functional f on X such that flY = 9 and f (x) ::; p( x) for all x E X. But then also  f(x) = f( x) ::; p( x) = p(x), so that in fact If(x)1 ::; p(x) for all x E X. (ii) Case lK =
hl(x) = Reh(x) ,
h2(X) = Imh(x)
*
C
(x E V),
we see that both hI and h2 are reallinear functionals on V. Moreover, we have h(ix) = ih(x) (x E V), and so, by equating real or imaginary parts, we obtain the fact that h2 (x) =  hI (ix) (x E V). Conversely, given any reallinear functional hI : V * JR, then
h(x)
=
hl(x)  ihl(ix)
(x
E
V)
gives the unique (complex) linear functional h on V with hI = Re h. The proof of the complex case of the HahnBanach theorem now follows easily. Indeed, we have a linear functional 9 : Y *
Igl(y)l::; Ig(y)1 ::; p(y)
(y
E
Y),
so that the real case gives a reallinear functional h : X * JR with h I Y = gl and Ih(x)1 ::; p(x) (x E X). Define f(x) = h(x)  ih(ix) (x E X). Then, from the last paragraph, f is linear and flY = g. Moreover, if x E X, we may choose a real number such that e iO f(x) is real and nonnegative. Let Xl = eiOx. Then
e
If(x)1 = If(xdl = f(Xl) = h(Xl) ::; P(Xl) = p(x) ,
o
giving the result
Corollary 3.3 Let X be a vector space wzth scalar field lK, let p be a seminorm on X, and let Xo EX. Then there is a linear functional f on X such that f(xo) = p(xo) and If(x)1 ::; p(x) for all x E X. Proof Let Y = {Axo : A ElK}, and define 9 : Y apply the theorem to obtain the stated result.
*
lK by g(>.x o ) = Ap(XO). Now 0
Recall that we made the following definitions on page 46. Let (E; 11·11) be a normed space, and let x E E. Then x E E** is defined by x(J) = f(x) (J E E*); the linear map J : x ft E * E** ,
x,
is the canonical embedding.
Banach spaces
109
Corollary 3.4 Let (E;
11·11)
be a normed space with E =I {a}.
(i) Let F be a subspace of E, and take 9 that f
IF = 9
and
E
F*. Then there exzsts f E E* such
Ilfll = Ilgll·
(ii) Let x E E. Then there is some f E E* such that in particular, E* separates the points of E.
IIfll =
1 and f(x)
= Ilxll;
(iii) The canonical embedding x f7 X of E into its bidual E** is an isometric linear mapping of E onto a subspace of E**; this subspace is closed zf and only if E zs complete. Proof This is immediate from Corollary 3.3.
o
The functional f of clause (i), above, is said to be a normpreserving extension of g. We now give the promised 'conceptually more interesting' proof of Theorem 2.22. Theor~m
3.5 Let E be a normed space, and let J : E + E be its completion. Then E has a unique Banachspace structure such that J is an isometrzc linear mapping.
Proof By Corollary 3.4(iii), taking J(x) = x (x E E) gives an isometric linear mapping J from E into the Banach space E**. Then J : E + J(E) is a version of the completion of E in which the completion is a Banach space and J is an isometric linear mapping. The uniqueness follows by using Lemma 1.12. 0
Suppose that E is itself a Banach space. Then the above isometric linear map J: E + E** may be a surjection, so that J(E) = E**; in this case, we say that E is reflexive. For example, as in Example 2.18, the space CP is reflexive whenever 1 < p < 00. (We shall see in the notes that E** can be isometrically isomorphic to E without E being reflexive.) Theorem 3.6 Let E be a normed space over IK, let F be a vector subspace of E, and suppose that Xo E E with 8 := d(xo, F) > O. Then there exists f E E* with f(x) = 0 (x E F), with f(xo) = 8, and with Ilfll = 1. Proof Set G = FEB IKxo, and define g(x + AXo) = A8 for x E F and A E IK. If A =I 0, we have IIx + Axoll = IAIII(x/A) + xoll : : : IAI8, and so
Ig(x
+ Axo)1 = IAI8 :::; Ilx + Axoll .
It follows that Ig(y)1 :::; Ilyll (y E G), and so 9 E G* with Ilgll :::; 1. For each x E F, we have 8 = Ig(x  xo)1 :::; IIgllllxo  xii, and so 8 :::; Ilg118, whence Ilgll : : : 1. Thus Ilgll = 1. By Corollary 3.4(i), g has a normpreserving extension, say f E E*. The functional f has the required properties. 0
110
Introduction to Banach Spaces and Algebras
Corollary 3.7 Let E be a normed space such that E* zs separable. Then E is separable. Proof We may suppose that E f= {OJ. Let S be the unit sphere of E*. Then S is separable, and so there is a countable, dense subset, say {fn : n E fir}, of S. For each n E fir, choose Xn E E with Ifn(xn)1 > 1/2, and set X = lin{xn : n E fir}. Assume towards a contradiction that X is not dense in E. By Theorem 3.6, there exists f E S with f(xn) = 0 (n E fir), and so
1
2<
Ifn(xn)1 ~ IUn  f)(xn)1
+ If(xn)1
~
Ilfn  fll .
But this is a contradiction for some n E fir because {fn : n E fir} is dense in S. Thus X = E. It follows that linear combinations of the elements Xn with coefficients in iQl or iQl + iQ constitute a countable, dense subset of E. 0 The converse of the above corollary is not true: the space (C 1; II . II h) is separable, but its dual space (C (X); II . I (X») is not separable. Let E be a normed space, and let F and G be vector subspaces of E. Then we have written E = FEB G if E = F + G and F n G = {OJ. Now suppose that F is a closed subspace of E. Then F is complemented in E if there is a closed subspace G of E such that E = FEB G. Proposition 3.8 Let E be a normed space, and let F be a closed subspace of E. (i) Suppose that F is finitedimensional. Then F is complemented in E. (ii) Suppose that F has finite codimension. Then F is complemented in E. Proof (i) Take {el,"" en} to be a basis for F, so that each x E F has a unique representation x = L~=l aJ(x)eJ , where al, ... , an are linear functionals on F. By Corollary 2.20(ii), each a J is continuous. By Corollary 3.4(i), for j = 1, ... , n, there is a normpreserving extension /3J E E* of a J • We set G = n~=l ker /3)" Then G is closed in E and E = FEB G. (ii) Take {el' ... ,en} to be a basis for E / F, and, for each j = 1, ... ,n, choose x J E E with 7rF(X J ) = eJ. Set G = lin{ Xl, ... ,xn }. By Corollary 2.20(i), G is closed in E, and E = FEB G. 0 3.2 Integration of continuous vectorvalued functions. Let E be a (real or complex) Banach space. For a compact interval [a, bJ of IR, we write Cera, bJ for the Banach space of all continuous functions from [a, bJ to E, normed by the uniform norm I . I[a,bj' which is now defined by
Ifl[a,bj
=
sup{llf(x)11 : a ~ x::; b}
U
E
CEra, b]).
(This generalizes the previous usage of the symbol 1·I[a,bj·) Let C~[a, b] be the subset of Cera, b] consisting of all functions with the special form:
111
Banach spaces
f(t) = gl(t)e1
+ ... + gn(t)e n
where n E N and where ek E E and gk is a vector subspace of Cda, b].
E
(a::; t ::; b),
C[a, b] for k
=
1, ... , n. Clearly, C~[a, b]
Lemma 3.9 The space C~[a, b] is dense in (Cda, b]; l'l[a,b])' Proof Let f E CEra, b], and take E > 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 realvalued, 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 socalled 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 <
)=1
E
)=1
(since, for each j E {l, ... , k}, either e)(x) = 0 or Ilf(x)  f(x))11 < E), and so If  fol[a,b] ::; E. This shows that C~[a, b] is dense in (Cda, b]; l'l[a,b])' 0 Theorem 3.10 Let [a, b] be a compact interval in JR., and let (E; 11·11) be a complex Banach space. Then, for each f E Cda, b], there zs a unique element f of E such that
J:
VJ ( lb f)
= lb VJ
0
(*)
f
J:
for every VJ E E*. Moreover, the mapping f f> 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] .
112
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 HahnBanach 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
f7
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 f7 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 f7 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 Evalued 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 complexvalued functions (cf. Section 1.11).
113
Banach spaces
Let U be a nonempty, open subset of C, and let E be a Banach space. A function J : U  t E is an analytic (or holomorphic) Evalued 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 vectorvalued forms of Cauchy's integral theorem (and formula). Theorem 3.11 Let U be a nonempty, open subset oj C, let E be a complex Banach space, and let J be an analytic Evalued 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 vectorvalued 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 complexvalued 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
114
Introduction to Banach Spaces and Algebras
3.3 A note on locally convex spaces. These spaces are not a principal topic of this book, but they will be needed in a few places. In the first place, it is useful to describe the weak and weak* topologies associated with Banach spaces in a suitable context. (These topologies will be discussed in Section 3.5.) We also need to discuss spaces of analytic functions (on open subsets of e and, later, of en); these spaces (and even algebras) will be, in particular, complete, metrizable locally convex spacesusually known as Frechet spaces (or Frechet algebras). A (Hausdorff) locally convex space is a real or complex vector space E equipped with a given collection P of seminorms on E that together separate the points of E, in the sense that, for every x i= 0 in E, there is some pEP such that p(x) i= o. In terms of P, we define a standard topology on E by saying that a subset U of E is open if and only if, for each x E U, there are finitely many PI,·· . ,Pn in P and E > 0 such that U :2 {y E E : Pk(Y  x) <
E
(k = 1, 2, ... , n)} .
We sometimes write (E; P) for the space E with the topology defined by P. It is an easy exercise to prove that, just as for a topology defined by a single norm, addition and scalar multiplication are continuous operations. Without wishing to formalize the idea, if P and Q are two sets of seminorms on E that define the same topology, then they are considered to be equivalent. In fact, we shall always assume that such a family P has the property that max {Pk : k = 1, ... , n} belongs to P whenever PI, ... ,Pn E P; to impose this extra condition does not change the topology. A seminorm q on E is now continuous if there exist M > 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 complexvalued analytic functions on a nonempty, open subset U of e (or e n see Section 8.4). The topology on O(U) is the standard onethat 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 (HahnBanach 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
116
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 Frechetspace 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, 'nondegenerate' 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 finitedimensional case, the two parts of this condition are not equivalent.] Examples 3.17 (i) Let E be a IKlinear 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 nondegenerate uses the HahnBanach theorem. (iii) Let F be a normed space, and take E = F* to be its dual space. Then (F*, F) is a nondegenerate 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 pointseparating family of seminorms on E. The associated locally convex topology on E is called the weak topology associated
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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 <;;; ker rp. It follows that there is a linear functional, say A, on ocn such that rp = A 0 e. But then, by the general form of a linear functional on ]Kn, there are elements aI, ... ,an of OC such that n
A(Zl' ... , zn)
=
Latzt
(Zl' ... , Zn E OC).
t=l
It follows that rp(x)
= (x, I) (x
E
E), where I = L~=l adt
E
F.
D
Now let us consider two key examples of weak topologies.
Example 3.19 (i) Let E be any vector space over OC, and take E' to be its algebraic dual. Then every linear functional on E is a(E, E')continuous.
(ii) Let E be a normed space, and let E* be its Banach dual space. Then we have the topology rr( E, E*). D Let E be a normed space. Then the weak topology on E is the topology a(E, E*) from the pairing (E, E*), and the weak* topology on E* is the topology a(E*, E) from the pairing (E*, E). We note that, although both the weak and weak* topologies are examples of weak topologies associated with pairings, the weak topology on a normed space E always means a(E, E*). Thus the weak topology on E* is a(E*, E**), whereas the weak* topology on E* is a(E*, E). We have
a(E*, E) ::; a(E*, E**) ::;
II· I ,
where I ·11 denotes the norm topology on E* and '::;' implies that the topology on the left is weaker. In fact, Theorem 3.18 shows that the topologies a(E*, E) and
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x
a(E*, E**) are the same if and only if the canonical isometric mapping x ~ of E into E** is surjective, i.e. if and only if E is a reflexive Banach space. We use the adjective 'weakly' to refer to topological properties of a space with respect to the weak topology. For example, let E be a normed space. Then a subset S of E is weakly compact if it is compact with respect to the dE, E*)topology on E. However, if we say just that'S is closed', we mean that it is closed in the norm topology of E; the set S is weakly bounded if for each f E E*, the subset f(S) of lK is bounded. By Theorem 3.18, a linear functional on E is weakly continuous if and only if it is continuous. Similarly, we use 'weak*' (a little awkwardly) to describe properties with respect to the weak* topology. For example, let E and F be normed spaces, with dual spaces E* and F *, respectively. Then a linear map T : E* + F * is weak*continuous (or, more precisely, weak*toweak*continuous) if it is continuous as a map from (E*;a(E*,E)) to (F*;a(F*,F)). We now give an important theorem concerning the weak topology. It will be extended in Corollary 3.28 to the case where F is a convex subset of E.
Theorem 3.20 Let E be a normed space. Then a subspace F of E is closed if and only if it is weakly closed. Proof Of course, since a := a(E, E*) is weaker than the norm topology, certainly F is closed whenever it is aclosed. Conversely, let F be a closed subspace (with F I E), and let Xo E E \ F. Then, using the HahnBanach theorem, we see that there exists f E E* such that f I F = 0, but f(xo) = 1. Set V = {x E E: If(x)1 < I}. Then Xo + V is a weakly open neighbourhood of Xo that does not meet F. Thus E \ F is weakly open, and so F is weakly closed. 0 The next theorem is even more important and fundamental. Theorem 3.21 (BanachAlaoglu theorem) Let E be a normed space, and let B be the closed unit ball of E*. Then B is weak*compact. Proof For each x E E, let Dx = {A ElK: IAI :s: Ilxll}. Then Dx is a compact disc in
=
n{
{(~x)xEE : 6.x+!lY = A~x + fl~y} : x, y
E
E, A, fl E lK},
which is an intersection of closed subsets of D, and hence is closed in D. Thus 'ljJ(B) is compact in D. Finally, this shows that B itself is compact in the weak* topology. 0
Banach spaces
119
Now let E and F be normed spaces, so that B(E, F) is a normed space; the topology on B(E, F) is given by the operator norm, II· II. Let T E B(E,F). For each f E F*, define T* f
= T*(f) = f
0
T,
(x, T* f)
so that
= (Tx, f)
(x E E).
It is evident that T*f E E* and that T* E B(F*,E*). The operator T* is called the dual operator (or mapping) of T. It is easy to see that T* is weak*continuous, and it will be shown in Lemma 3.62(ii) that each weak*continuous operator in B( F * , E*) is the dual of an operator in B( E, F). In the case where E and F are a Hilbert space Hand T E B(H), the dual operator T* is closely related to the adjoint operator of T, also called T*, which was defined in Section 2.14; technically, while the adjoint operator maps H into H, the dual operator maps the dual of H into the dual of H. Let J be the isometric, conjugatelinear map from H onto the Banach space dual of H that was given in Corollary 2.54. Then the adjoint of T is strictly J1T* J, where T* is the dual operator. Proposition 3.22 Let F be a dual Banach space Then F is a complemented subspace of F**. Proof Suppose that F = E* for a Banach space E, and let L: E + E** = F* be the canonical embedding. Then it is immediately seen that P = L* : F** + E* is a projection with P(F**) = F, and so F** = FEB G, with G = ker P, a closed subspace of F**. Thus F is complemented in F**. 0
3.6
Weak and strongoperator topologies. Px(T) =
IITxl1
For x E E, define
(T E B(E,F)).
It is immediately checked that each Px is a semi norm on B(E, F), and so the family {Px : x E E} defines a locally convex topology on B(E, F). This topology is called the strongoperator topology, and is denoted by 'so'. A net (Ty) in B(E, F) converges to T E B(E, F) in this topology, written
T "(
so >
T
,
if and only if T"(x + Tx in F for each x E E. (We shall require convergence only of sequences, so knowledge of nets is not really needed here.) For x E E and f E F*, define Px,j(T)
= I(Tx, f)l
(T E B(E, F)).
It is again immediately checked that each Px,j is a seminorm on B(E, F), and so the family {Px,j : x E E, f E F*} also defines a locally convex topology on
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13(E, F). This topology is called the weakoperator topology, and is denoted by 'wo'. A net (T"() in 13(E,F) converges to T E 13(E,F) in this topology, written T"(
wo
+
T,
if and only if (T"(x, 1) + (Tx, 1) in C for each x E E and f E F*. Clearly, convergence of a net in (13(E, F); 11·11) implies convergence in the locally convex space (13(E, F); so), and this in turn implies convergence in the locally convex space (13(E, F); wo). Theorem 3.23 Let E and F be normed spaces, and let f be a linear functional
on 13(E, F). Then f is socontinuous if and only zf it is wocontmuous. Proof Clearly, if f is wocontinuous, then it is socontinuous. Suppose that f is socontinuous. Then there is a finite set {Xl, ... , xn} in E and E > 0 with If(T)1 < 1 whenever T E 13(E, F) and IITx J II < E (j = 1, ... , n). In particular, f(T) = 0 whenever TXl = ... = TXT> = O. Consider the space Fn; as in Exercise 2.5, this is a normed space for the norm
II(Yl,"" Yn)11 = sup{IIYJII : j = 1, ... ,n}. Define H : T f+ (TXl,'" TXn)' ~(E, F) + Fn, and set l(y) = f(T) whenever = H (T) E Fn. It is clear that f is well defined and continuous on the vector subspace H(13(E, F)) of Fn, and so, by ~orollary 3.4(i), ~as an extension to a continuous linear functional, also called f, on Fn. Clearly, f must have the form
y
1
n
f(y) =
L
fJ(YJ)
(y = (Yl, ... , Yn) E Fn)
J=l for some h, ... ,fn E F*. Thus f(T) = 'L.;=lfJ(Tx J ) (T E 13(E,F)), and so f is wocontinuous. 0 Proposition 3.24 Let E and F be normed spaces.
/Ln
(i) Suppose that Sn ~S and Tn ~T in 13(E, F) and that An /L in Co Then AnSn + /LnTn~AS + /LT.
+
A and
+
(ii) Suppose that Tn ~T in 13(E, F), U E 13(E), and V E 13(F). Then VTnU~VTU in 13(E,F). Proof (i) This is easily checked; it follows because (13(E, F); wo) is a locally convex space.
(ii) This follows because (V SU X, 1) = (SU X, V* 1) for all X E E, f E F*, and S E 13(E, F). 0
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Banach spaces
Notes For a somewhat different approach to the HahnBanach theorem for locally convex spaces, see [144, Chapter 3]. See also [63, Chapter II], [93, Chapter 1, Section 6], and [102, Chapter 1]. It is interesting that Co is a closed subspace of £= that is not complemented in £ =; this is a theorem of Phillips and of Sobczyk, and it is discussed with related results in [2, Section 2.5]. It follows from Proposition 3.22 that Co is not a dual space. Let X be a vector space over IK, and suppose that X is also a Hausdorff topological space for a topology T. Then (X; T) is a topological vector space if the two bilinear maps (A, x)
f>
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 nonseparable. A Banach space E is quasireflexive if E** / E is finitedimensional. In fact there are nonreflexive, quasireflexive spaces. For example, let J be the James space of Exercise 2.14. Then J** / J is even onedimensional. Further, J** is isometrically isomorphic to Jbut still J is nonreflexive. 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 Banachspacevalued functions, involving, for example, the Bochner integral, the Pettis integral, and the Bartle integral, see [58, Chapter II]. The fact that a weakly analytic Banachspacevalued function is already analytic is proved in [144, Theorem 3.31]. We have mentioned the strongoperator and weakoperator 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 HahnBanach theorem for E without using the Zorn'slemma 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 onedimensional 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 normpreserving 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.)
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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 HahnBanach 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 weakoperator topology if and only if F is a reflexive space.
Separation theorems We now give some geometric versions of the HahnBanach 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 Jlc(X) = inf{a > 0: x E aC}.
Then Jlc is a sub linear functional on E and C = {x
E
E : Jlc(x) < I}.
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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)
<:;;;
(a
+ (3)C,
using the fact that C is convex. Thus JLc (x + y) ::; a + (3 < r + s + E. This holds for all E > 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 nonempty, 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) <
~
::; f(y)
(x E A, y E B).
(ii) Suppose that A and B are open. Then there exist f E E* and ~ E lR such that f(x) < ~ < f(y) (x E A, y E B). (iii) Suppose that A is compact and B is closed. Then there exists f E E* such that sup f(x) < inf f(x). xEA
xEB
'Proof (i) Since A and B are nonempty, we may choose ao E A and bo E B. Set Xo = bo  ao. Then C = A  B + Xo is a convex, open neighbourhood of the origin, so that JLc is sublinear on E by the above lemma. Since An B = 0, we have Xo rf:. c, so that JLc(xo) ~ 1. Define g on the onedimensional subspace lRxo by g(AXo) = A (A E lR). Then g ::; JLc on lRxo. By the HahnBanach theorem (first version), there is a linear functional f on E such that f(xo) = g(xo) = 1 and f(x) ::; JLc(x) for all x E E. In particular, if x E C, SO that JLc(x) < 1, we have f(x) < 1; also, if x E C, then f(x) > 1. Thus If(x)1 < 1 on the open neighbourhood C n (C) of 0, so that f is continuous, i.e. f E E*. If x E A and y E B, then x  y + Xo E C, so that
f(x)  f(y)
+1=
f(x  y
+ xo)
::; JLc(x  y
+ xo) <
1.
Hence f(x) < f(y). Thus f(A) and f(B) are disjoint, convex subsets of lR, i.e. they are disjoint intervals, with f(A) < f(B) in an obvious sense, and (as is
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elementary to see) f(A) is open. We may thus take of the interval f(A) to establish (i).
~
as the righthand end point
(ii) Now f(A) and f(B) are disjoint, open intervals in lR, and so the righthand end point ~ of f(A) has the required property. (iii) Since A is compact, there is a convex, open neighbourhood V of 0 E with (A+ V)nB = 0. Then A+ V is open, so, by (i), there is an element f E E* such that f(A + V) and f(B) are disjoint intervals of lR, with f(A + V) open. But f(A) is a compact subinterval of f(A + V), so the result is immediate. 0 We conclude this section with a corollary to the separation theorem that is valid in real or complex locally convex spaces. Corollary 3.27 Let E be a real or complex locally convex space, let B be an absolutely convex, closed subset of E, and let Xo E E \ B. Then there exzsts f E E* such that If(x)1 ::; 1 for all x E B, but such that f(xo) > 1. Proof We apply Theorem 3.26(iii) to the underlying real space of E, taking A to be {xo} (and we multiply the functional by 1). Then there is a continuous, reallinear functional 9 on E with sup g(x)
< g(xo).
xEB
Since B is balanced, Ig(x)1 E g(B) whenever x E B, so that, multiplying 9 by a suitable positive scalar, we may ensure that g(xo) > 1, while Ig(x)1 ::; 1 for all x E B. So, if E is a real locally convex space, we simply take f = g. In the case where E is a complex locally convex space, define
JI(x)
=
g(x)  ig(ix)
(x
E
E).
Then JI E E* with g(x) = ReJI(x) (as in the proof of Theorem 3.2). Now take a real number () such that e iO g( xo) is real and positive, and then define f = ell} JI, so that f(xo) = IJI(xo)1 ~ g(xo) > 1. For every x E B, we have If(x)1 = IJI(x)l· But, since B is balanced, we can choose, for each x E B, a real number (3 such that e ii3 JI(x) is real and nonnegative. Noting that eii3 x E B, it follows that
If(x)1 = JI(e ii3 x) = g(e ii3 x) ::; 1, as required.
o
Corollary 3.28 A convex subset of a normed space is closed if and only if it is
weakly closed.
0
Corollary 3.29 Let E and F be normed spaces. Then a convex set in B(E, F) is woclosed if and only if it is soclosed. Proof This follows easily from Theorem 3.23.
o
Banach spaces
125
Corollary 3.30 (Goldstine's theorem) Let E be a Banach space, and denote by J: E + E** the canonical embedding. Then J(E[l]) is weak* dense in E[lj. Proof Take B to be the weak* closure of J(E[l]) in E[lj. It is easy to check that B is an absolutely convex, closed subset of (E**;a(E**, E*)). Assume towards a contradiction that B <;:; E[lj, and take A E E[lj \ B. By Corollary 3.27, there exists f E (E**;cr(E**,E*))* with If(x)1 :::; 1 for all x E B, but with f(xo) > 1. By Theorem 3.18, (E**;cr(E**,E*))* = E*, and so f E E*, and Ilfll :::; 1. Thus 1 < If(xo)1 :::; Ilfllllxoll :::; 1, a contradiction. Thus B = E[lj, as required. 0
3.8 Extreme points and the KreinMilman theorem. In this section we shall again suppose that the field of scalars of our vector spaces is R In fact, the results apply immediately to complex spaces also, just by restricting to real scalars. (This is because the definitions of convexity and extreme point, already given, use only real scalars t E IT.) Let C be a convex subset of a real vector space E. Recall that a point x E C is an extreme point of C if and only if y = z = x whenever y, z E C and x = ty + (1  t)z for some t E (0,1). This is a slight rephrasing of the definition on page 34. We continue to write ex C for the set of extreme points of C. We say that a subset L of C is an extreme subset of C if and only if it is nonempty and convex and is such that y, z E L whenever y, z E C and ty + (1  t)z E L for some t E (0,1); sometimes such sets are called faces of C. We observe that, for x E C, the set {x} is an extreme subset of C if and only if x is an extreme point of C. Clearly, C itself is an extreme subset of C, and so the collection of extreme subsets of C is nonempty. Theorem 3.31 (KreinMilman theorem) Let E be a real locally convex space, and let K be a nonempty, compact, convex subset of E. Then K is the closed convex hull of its set of extreme points. In particular, K has at least one extreme point. Proof It is, of course, trivial that conv(exK) ~ K. We must show the reverse inclusion. The proof will be broken into a number of steps.
(i) Every closed extreme subset of K contains a minimal closed extreme subset. This is a simple application of Zorn's lemma. Let J be the family of all closed extreme subsets of K. Then J =I 0 because K E J. The family J is partially ordered by inclusion. Let {La}aEA be a chain in (J; ~). Then the set L := n{ La : a E A} is a nonempty (by the finite intersection property of compact sets), compact, convex subset of K, and hence L is closed. Clearly, L is an extreme subset of K, and hence belongs to J. Thus every chain in J has a lower bound, and so, by Zorn's lemma, bis, J has a minimal element; this is a minimal closed extreme subset of K.
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Introduction to Banach Spaces and Algebras
(ii) Every minimal closed extreme subset of K is a singleton. Let L be a closed extreme subset of K such that L contains at least two distinct points, say x and y. Since E* separates the points of E, there is some element J E E* such that J(x) =I J(y); without loss of generality, we may suppose that J(x) < J(y). Since L is compact, J attains its supremum on L; define Mf
= {z
E
L : J(z)
= maxI}. L
Then Mf is clearly a nonempty, closed, convex subset of L; also x rf M f , so that M f is a proper subset of L. We claim that Mf is also an extreme subset of K. For suppose that y, z E K and t E (0,1) are such that ty + (1  t)z EMf. Then, since M f <;;: Land L is an extreme subset, we have y, z E L. But now, by the definition of M f , we have: maxJ L
= J(ty + (1  t)z) = tJ(y) + (1  t)J(z)
~ maxJ. L
So there must be equality throughout, i.e. J(y) = J(z) = maxL J, and hence y, z EMf. Thus M f is an extreme subset of K that is a proper subset of L, and hence L is not minimal. It follows that every minimal closed extreme subset of L is a singleton. (iii) It follows immediately from (i) and (ii) that every closed extreme subset of K contains an extreme point of K. (iv) Completion of the proof. Define C = conv(exK). Then C is a closed (in fact, compact), convex subset of K. Assume towards a contradiction that C =I K, and choose any Xo E K \ C. Then, by Corollary 3.27 of the separation theorem, there is some J E E* such that sup J(x) < J(xo).
c
But then, just as in the proof of (ii), the set Mj = {x E K : J(x) = SUPK I} is a closed extreme subset of K for which Mj n C = 0. It follows from (iii) that Mf contains an extreme point, contrary to the fact that exK C C. Thus we conclude that C = K. 0 There is a partial converse to the KreinMilman theorem. Theorem 3.32 Let K be a compact, convex subset oj a locally convex space E, and let X be a closed subset oj K such that K = conv(X). Then X ::2 exK. Remark. The reason that this result is only a partial converse to the KreinMilman theorem is the requirement that the subset X be closed. For example, the closed unit disc ~ in E = ]R.2, with boundary 'II', has 'II' = ex~. But then, if we take X = ~ \ 'II', it is still true that II = conv(X), but it is not true that 'II' <;;: X. We first isolate certain technicalities in a lemma; the proof is left as a (fairly simple) exercise.
Banach spaces
127
Lemma 3.33 (i) Let K l , ... , Kn be compact, convex sets in a locally convex space. Then the convex hull of U~=l K, is also compact. (ii) Let Xo be an extreme point of a convex set C, and set Xo = 2:::1 t,x" where each t, 2'" 0, where 2::~=1 t, = 1, and where each x, EC. Then Xo = x, for 0 some i E {I, ... , n}. Proof of Theorem 3.32 Let Xo be any extreme point of K, take p to be a continuous seminorm on E, and define V = {x E E : p( x) :::; I}. Then V is a closed, absolutely convex neighbourhood of 0E. By the compactness of X, there is a finite subset {Xl, ... ,xn } of X such that X <;;: U~=l (x, + V) n K. Let Y = conv (U~=l (x, + V) n K). Since each set (x, + V) n K is compact and convex, it follows from Lemma 3.33(i) that Y is compact, and therefore closed. Hence K = conv(X) <;;: Y. By Lemma 3.33(ii), it follows that Xo E (x, + V) n K for some i E {I, ... , n}. Thus, for every continuous seminorm p on E, we have Xo E X + V, where V = {x E E : p(x) :::; I}, i.e. (xo + V) n X I 0 (since V = V). But such sets V form a basic family of neighbourhoods of OE (from the definition of the topology of a locally convex space). Hence Xo E X = X, since X was required to be closed. We have shown that ex K <;;: X. 0 Notes The functional /Lc of Lemma 3.25 is called the Minkowski functional of C. The KreinMilman theorem, and various extensions, are discussed in many books, including [63, Chapter V], [102, Chapter 1], and [144, Chapter 3]. A Banach space has the KreinMilman property if every closed and bounded (not necessarily compact) convex subset is the closed convex hull of its extreme points. This is closely related to a certain RadonNikodjm property of Banach spaces. Indeed, a dual Banach space has the KreinMilman property if and only if it has the RadonNikodym property. This is further related to the important BishopPhelps theorem that asserts the following. Let K be a closed, bounded, convex subset of a Banach space E. Then the collection of functionals in E* which attain their maximum value on K is dense in E*. For these results, see [58, Chapter VII]. Exercise 3.8 Let K be a compact, convex subset of a normed space E. Show that ex K is a Goset in E. Exercise 3.9 Let K be the convex hull in ]R3 of the set that consists of the points (1,0,1) and (1,0, 1) and of the circle {(Xl,X2,0) : xi + x~ = I}. Show that K is compact, but that ex K is not compact. Exercise 3.10 Show that the sequence (1,1,1, ... ) is an extreme point of the closed unit ball of the Banach space c of convergent sequences. Show that the closed unit ball ofthe sequence space (co; II· lie,,') has no extreme points. Deduce that Co is not the dual of any Banach space. Exercise 3.11 Let K be a compact Hausdorff space. Describe the set of elements of the unit sphere of (C(K); I ·IK) that are extreme points of the closed unit ball C(K)[l]. Exercise 3.12 Let K be a compact subset of ]Rn, where n 2: 1. Show that each point of K is a convex combination of at most n + 1 extreme points of K.
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Category theorems 3.9 The uniform bounded ness theorem. We now turn to results for which completeness of the space is normally essential. The starting point is the Baire category theorem, which we proved as Theorem 1.21; here we give some applications. The first, the famous uniform boundedness theorem, comes from the beginnings of Banachspace theory. Theorem 3.34 (Uniform boundedness theorem) Let E be a Banach space, let {Ea}aEA be a family of normed spaces, and let Ta : E ~ Ea be a bounded linear mapping for each a E A. Suppose that sUPaEA IITaxl1 < 00 for every x E E. Then sUPaEA IITall < 00. Proof For each n EN, let Fn
=
{x E E: sup II TaX II :::; n} aEA
=
n
{x E E: II TaX II :::; n}.
aEA
By hypothesis, E = U:=l Fn· Since each Ta is continuous, every Fn is a closed set (being an intersection of closed sets). By Corollary 1.22, there is an integer N with int FN i 0. Thus there are Xo E FN and <5 > 0 such that {x : IIx  xoll :::; <5} <;::; FN, and so, for all x E E with IIx  xoll :::; <5 and all a E A, we have IITaxll :::; N. Now take x E E with IIxll :::; <5. By writing x = (x + xo)  Xo, we deduce that IITaxll :::; IITa(x
+ xo)1I + II Taxo II
:::; 2N
(a E A).
It follows that IITall :::; 2Nj<5 (a E A). This gives the result.
o
Corollary 3.35 Let E be a normed space. Then a subset of E is bounded if and only if it is weakly bounded. Proof Let X be a weakly bounded subset of E. Using the canonical isometric embedding J : E ~ E** described above, we may regard X as a pointwisebounded subset of the bidual space E**. Thus this corollary is immediate from the uniform bounded ness theorem. 0 Theorem 3.36 (BanachSteinhaus theorem) Let E and F be Banach spaces, and let (Tn)n21 be a sequence in B(E,F). Suppose that (Tnx)n21 converges in F for each x E E, and set Tx
Then T
E B(E, F)
=
lim Tnx
(x E E).
n+CXJ
and IITII :::; liminfn+CXJ IITnll.
Proof It is immediately checked that T : E ~ F is a linear mapping. For each x E E, we have sUPnENIITnxll < 00, and so, by the uniform boundedness theorem, Theorem 3.34, sUPnENIITnll < 00, say M = sUPnENIITnll. But now IITxll =
lim IITnxl1 ::; M Ilxll
n+CXJ
(x E E) ,
and so T is bounded. Thus T E B(E, F). Clearly, IITII ::; liminfn ..... CXJ IITnll.
0
Banach
space~
129
A bilinear map was defined in Exercise 2.16. Theorem 3.37 Let E be a Banach space, and let F and G be normed spaces. Then each separately continuous, bilinear map from E x F into G is continuous. Proof Let T : E x F + G be a separately continuous, bilinear map. Take y E F, and define
E+G,
Ty:xftT(x,y),
so that Ty E B(E,G). For each x E E, the family {Ty : y E F[l]} satisfies the hypotheses in the above uniform bounded ness theorem because the map y ft T(x, y), F + G, is bounded, and so, by that theorem, there exists M > such that IITyl1 ::; M for each y E F[l]. It follows that
°
IIT(x,
y)11 ::; M Ilxllllyll
(x E E, y E F),
and so, by Exercise 2.16, T is a continuous bilinear map.
D
3.10 The open mapping lemma. Our next aim is a certain 'open mapping theorem'. As a preliminary, we prove an 'open mapping lemma'. This is a, not completely standard, name for that part of the proof of the open mapping theorem that does not use the Baire category theorem; it is a useful result in its own right. We say that a subset Y of a metric space (Z; d) is kdense in a subset X of Z if, for every x E X, there is some y E Y with d(x, y) ::; k. Of course, if Y ~ Z is such that Y ~ X, then Y is kdense in X for each k > 0. The following is our 'open mapping lemma'. Lemma 3.38 (Open mapping lemma) Let E be a Banach space, let F be a normed space, and let T E B(E, F). Suppose that R > and k E (0,1) are such that T(E[R]) is kdense in F[l]. Then:
°
(i) T is surjective and, moreover, for every y E F, there is some x E E with
Ilxll ::; and Tx
(1 ~ k) Ilyll
= y (so that T is an open mapping),
(ii) F is complete. Proof (i) Let y E F. The result is trivial if y = 0, and hence we may suppose that Ilyll = 1, and so y E F[l]. Choose Xo E E[R] such that IITxo  yll ::; k. Then IlkI(TxO  y)11 ::; 1, so that there is an element Xl E E[R] such that IITxI  kl(y  Txo)11 ::; k, i.e.
Ily 
(Txo
+ kTxI)1I
::; k 2 .
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Introduction to Banach Spaces and Algebras
We continue to follow this idea to obtain a sequence
Ily  ~krTxrll ~ kn+ l
(xn)n~O
in E[RJ with
(n EN).
By the completeness of E, we see that the series 2::~0 FX r converges in E, say = 2::~0 k r x r · But now Tx = y. Also,
x
Ilxli
~ Rfk
C~ k)
=
r
r=O
liyli,
and so (i) follows easily. (ii) Let F be the completion of F. Then, regarding T : E + F ~ F as a mapping into F (so that, easily, T(E[RJ) is idense in F[lJ for each £ E (k,I)), we deduce from (i) that T(E) = F. This shows that F = F, and so F is already complete. D We give an immediate application. In the following theorem, we write
If Ix = sup{lf(x)1 : x
E
X}
and
Igly = sup{lg(y)1 : y
E
Y}
for f E Cb(X) and g E Cb(y). Theorem 3.39 (Tietze extension theorem) Let X be a normal topological space,
and let Y be a nonempty, closed subspace of X. (i) Let go : Y + [1, 1] be a continuous function. Then there is a continuous function fo : X + [1,1] such that fo I Y = go· (ii) Let go E Cb(y). Then there exists fo E Cb(X) such that fo I Y = go and Ifol x = Igoly· Proof (i) Set E
=
C~(X) and F
=
C~(Y), and define T : E + F by setting
Tf
=
flY
so that T E B(E, F). Now let g E F[lJ' so that 1
~
g(y)
A = {y E Y : 1
~
~ g(y) ~  ~},
(J E E) , 1 (y E Y). Let
B = {y E Y :
~ ~ g(y) ~ 1 }
.
Then A and B are disjoint, closed subsets of X, and so, by Urysohn's lemma, Theorem 1.26, there is a continuous function k : X + II with k I A = 0 and k I B = 1. Set h = (2k  1)/3. Then h E E[1/3J is such that h I A = 1/3 and hi B = 1/3. We see that ITh  gly :S 2/3, and so T(E[1/3]) is 2/3dense in F[l]·
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Banach spaces
By Lemma 3.38, it follows that, for each 9 E F, there is a function fEE such that flY = 9 and 1/3 If Ix :s; (1 _ 2/3)
Igly = Igly .
In particular, for each go E c~(Y) with go(Y) ~ [1,1]' there is fo E C~(X) with fo I Y = go and Ifol x :s; 1, so that fo(X) ~ [1,1]. (ii) Now take go E Cb(y). The case where go = 0 is trivial, so, by scaling, we may suppose that Igo Iy = 1. Clearly, there exist two continuous functions 91,g2 : Y * [1,1] with go = g1 + ig2. By (i), there exist continuous functions /1, h : X * [1, 1] with h I Y = g1 and h I Y = g2; set h = It + ih· For z E C, define 1j;(z) = z when Izl :s; 1 and 1j;(z) = z/Izl when Izl ~ 1, so that tP : C * ~ is a continuous map; fo = 1j; 0 h has the required properties. D 3.11 Open mapping and closed graph theorems. In this section, we shall present two central theorems about linear mappings between Banach spaces.
Theorem 3.40 (Open mapping theorem) Let E and F be Banach spaces, and let T E B(E, F) with T(E) = F. Then there exists K > 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. T1 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 IITI(y)11 :::; K Ilyli. This shows that T I is bounded, and hence continuous. D
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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 nonempty 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.
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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 nontrivial 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.
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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 T1(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 T1(S(T)) = kerQT, and so T1(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) <:;;; SeT); further, S(TR) = SeT) whenever R is a surjection. (ii) Suppose that F1 is a Banach space and that S E B(F, H). Then
S(S(T))
= S(ST).
In particular, ST is continuous if and only if S(S(T))
= {O}.
Proof (i) Let Y E S(TR). Then there is a null sequence (X n )n21 in E1 with (TR)(xn) + Y as n + 00. But now (RX n )n>l is a null sequence in E, and T(Rxn) + Y as n + 00, so that Y E SeT). Hence S(TR) <:;;; SeT). Suppose that R is a surjection, and take Y E SeT). Then there is a null sequence (X n )n21 in E with TX n + Y as n + 00. By the open mapping theorem, Theorem 3.40, there are K > 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) <:;;; S(TR), and so SeT) = S(TR). (ii) Let Y E SeT). Then there is a null sequence (X n )n21 in E with TX n + y. But then STx n + Sy, and so Sy E S(ST). Hence S(S(T)) <:;;; S(ST). Since S(ST) is closed, S(S(T)) <:;;; S(ST). Let Q : F + F IS(T) be the quotient map. By Proposition 3.53, QT is continuous. Let Q1 : F1 + FdS(S(T)) be the quotient map. Clearly, kerQ <:;;; ker(Q1 S ), and so it follows from Lemma 3.47 that there is a continuous linear operator S : FIS(T) + FdS(S(T)) such that SQ = Q1S, Since S(QT) is continuous, so is Q1ST, and hence S(Q1ST) = {O}. But, by the first part of the proof, Q1S(ST) c S(Q1ST), and so S(ST) <:;;; kerQ1 = S(S(T)). Hence S(S(T)) = S(ST). 0 We now come to the important stability theorem. A sequence (Sn)n2:1 of subsets of a fixed set S is a nest if Sn+1 ~ Sn (n EN); a nest (Sn)n>l stabilizes if there exists N E N such that Sn = SN (n::::: N). 
Banach spaces
137
Theorem 3.55 (Stability theorem) Let E and F be Banach spaces, let (En)n>o be a sequence of Banach spaces such that Eo = E, and let Rn E B(En ,En  1) for each n EN. Take T E £(E, F).
(i) Let (Fn)n>l be a sequence of Banach spaces, and let Sn E B(F, Fn) for each n E N. Suppose that SmT R1 ... Rn is contmuous whenever m < n m N. Then there exists N EN such that SnTRl ... Rn is continuous for each n ::::: N. Further, the sequence (6(TR1··· R n ))n2 1 is a nest in F which stabilizes. (ii) Let (Rn)n>l and (Sn)n>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 n1 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 n1
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(xy) 2n . 1 + Pn(x _ y)
(x,YEE).
The metric is translationinvariant: 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(Xnl; 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, ... ,Xnl such that (a) and (b) hold. We have B(Txnl; Pn) ~ T(B(xnl; rn)), and B(Txnl; Pn) contains y because TXn1 E B(y; Pn). Thus there is an element Xn E B(Xnl; 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 nonempty, 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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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, we define its annihilator Xl.. to be xl.. = {y
E
F: (x, y) = 0 (x
E
X)};
E
E: (x, y)
= 0 (y
E
Y)}.
similarly, for Y <;;: F, we set yT
= {x
Evidently yT is a rr(E, F)closed linear subspace of E and Xl.. is a dF, E)closed linear subspace of F. In the following lemma, all topological notions in E refer to the topology rr = dE, F); we recall that (E; rr) is a locally convex space.
Lemma 3.59 Let E and F be paired vector spaces. Then: (i) for subsets Xl and X 2 of E, Xl <;;: X 2 =}xt ::2 xi:; for subsets YI and Y2 of F, YI <;;: Y2 =}ylT ::2 Y2T; (ii) for each closed subspace L of E, L = Ll.. T; (iii) for subsets X of E, X l..T = lin X; (iv) for subsets X of E, Xl.. = Xl..Tl.. and Xl.. T = Xl..Tl..T; (v) for subsets X of E, Xl.. = {O} if and only if lin X = E, and Xl.. = F if and only if X = {O}.
Proof Clause (i) is trivial, and (iv) is a simple exercise. (ii) Clearly, for every X <;;: E, we have X <;;: Xl.. T. Let L be a closed subspace of E. Then, for each x E E \ L, by the HahnBanach theorem for locally convex spaces, Corollary 3.14, and using Theorem 3.18, there is an element y E F with yELl.. and (x, y) i O. So x 1 Ll.. T. This shows that Ll.. T <;;: L, so that L = Ll.. T. (iii) Let L be a closed subspace of E with X <;;: L. Then Xl.. T <;;: Ll.. T = L, and so Xl.. T is the closed linear span of X. (v) Suppose that Xl.. = {O}. Then Xl..T = {O}T = E, so that linX = E by (iii). The converse is clear, since it is obvious that, for every subset X of E, we have Xl.. = (lin X)l... Now let Xl.. = F. Then X <;;: Xl.. T = FT = {O}, so X = {O}. The converse is ~M.
0
The following corollary is proved by applying the above lemma to the pairings (E, E*) and (E*, E). Recall that, for a subset of a normed space E, 'closed' means 'normclosed'; for a subset of E*, 'weak*closed' means 'closed in the weak* topology'. We recall, from Theorem 3.20, that a linear subspace of a Banach space E is weakly closed if and only if it is normclosed.
Corollary 3.60 Let E be a Banach space, and take X <;;: E and Y <;;: E* . (i) Xl..T is the closed linear span of X in E; yTl.. is the weak*closed linear span ofY in F. (ii) Xl.. = {O} if and only iflinX = E; Xl.. = E* if and only if X = {O}; yT = {O} if and only if the weak*closed linear span of Y is F; yT = E if and only if Y = {O}. 0
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Introduction to Banach Spaces and Algebras
Let E and F be Banach spaces, and let T E B(E, F). Recall that the dual operator T* E B(F *, E*) was defined in Section 3.5 by T* f = f 0 T (f E F*). Lemma 3.61 Let E and F be Banach spaces, and let T E B(E, F). Then: (i) kerT = (imT*)T and kerT* = (imT)L; (kerT)L is the weak*closure of imT* in E* and (kerT*)T is the closure ofimT in F;
(ii) T is injectzve if and only if im T* is weak*dense in E*; (iii) T* is injective if and only if im T is dense zn F. Proof (i) Let x E E. Then x E (im T*) T if and only if, for every f E F *, we have (f 0 T)(x) = O. But, by the HahnBanach theorem for the space F, this holds if and only if Tx = 0, i.e. if and only if x E ker T. Similarly, for every f E F*, f E (imT)L if and only if f(Tx) = 0 (x E E), i.e. T* (f) E EL = {O}, that is f E ker T*. The statements about (ker T)L and (ker T*) T are now immediate from Corollary 3.60.
(ii) The map T is injective if and only if kerT = {O}, and so, by (i), if and only if (imT*)T = {O}. By Corollary 3.60(ii), this is equivalent to imT* being weak*dense in E* (noting that imT* is already a linear subspace of E*). (iii) The map T* is injective if and only if ker T* = {O}, and so, by (i), if and only if (im T)L = {O}. By Corollary 3.60(ii), this is equivalent to im T being 0 dense in the Banach space F. x
Let E be a Banach space. Recall that the canonical isometric embedding X, E > E**, was defined by setting x(f) = f(x) (f E E*).
f+
Lemma 3.62 Let E and F be Banach spaces. Then:
(i) the mapping T f+ T* is an isometric linear mapping from B(E, F) into B( F * , E*) (so that its image is a closed subspace of B( F * , E*)); (ii) for each S E B(F *, E*), there is a (necessarily unique) T E B(E, F) such that S = T* if and only if S is weak*continuous; (iii) if G is another Banach space and R E B(F, G), then (RoT)* = T* 0 R*; (iv) (Ie)* = IE, the identity mapping on E*. Proof (i) It is very simple to see that the mapping T f+ T* is linear; we must show that IITII = IIT*II. Thus, let T E B(E,F). For each f E F*, we have
IIT* fll = Ilf
0
Til
:s; Ilf1111T11
using Proposition 2.14(ii), so that IIT* II :s; IITII· Now take E > 0, and choose x E E with Ilxll = 1 and IITxl1 > IITII  E. By Corollary 3.4(ii) , there is some f E F* with IIfll = 1 and f(Tx) = IITxll, Le. (T* f)(x) = IITxll. Hence IIT*II ~ IIT*fll ~ I(T*f)(x)1 This holds for every
E
> 0, so that IIT* II
~
=
IITxl1 > IITIIE.
IITII.
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Banach spaces
(ii) Let T E B(E, F). We shall show that the operator T* E B(F *, E*) is weak*continuous. Thus, let x E E, and set Ux = {g E E* : Ig(x)1 < I}. We note that finite intersections of sets such as Ux (as x runs through E) form a neighbourhood base at 0 E* for the weak* topology on E*. However, for each f E F *, we have T* f E Ux provided that If(Tx)1 < 1. It follows that T* is weak*continuous. Conversely, let S E B(F *, E*), and suppose that S is weak*continuous. We must show that there is some T E B(E, F) such that S = T*. Let x E E. Then the mapping x : E* + K is weak*continuous on E*. But then x 0 S is a weak*continuous linear functional on F *. Hence, for each x E E, there is a unique element, say Tx, in F such that (x 0 S)(I) = f(Tx), i.e. such that (Sf)(x) = f(Tx) (I E F*). It follows that T is linear and that IITxll :::;
IISIIIIxll
(x E E) .
Thus T E B(E, F). Clearly, T* = S. Finally, the map T is uniquely defined by the fact that T* = S. (iii) and (iv) These are simple exercises.
0
Corollary 3.63 Let E and F be Banach spaces.
(i) Let T T** (x)
E
B(E, F), so that T** E B(E**, F**). Then for every x E E,
= f(;) (or, informally stated, T** I E = T).
(ii) A map T in B(E, F) is a linear homeomorphism if and only if T* is a linear homeomorphism in B( F * , E*), in which case (T*) 1 = (T 1 ) * . Proof (i) Let x E E. Then, for each g E F *, we have
T**(x)(g) = (x 0 T*)(g) = x(g
so that T**(x)
=
Tx,
0
T) = g(Tx) = CTx)(g) ,
as required.
(ii) Suppose that T is a linear homeomorphism. Then, writing S = T 1 , we see that S E B(F, E), that ST = Ie, and that TS = IF. By Lemma 3.62, it follows that T*S* = (Ie)* = I E * and S*T* = (IF)* = IF*. Thus T* is a linear homeomorphism, and (T*)l = (T 1 )*. Conversely, suppose that T* : F * + E* is a linear homeomorphism. It follows at once from Lemma 3.61 that T is injective and that im T is dense in F. Because of the open mapping theorem, it suffices to prove that im T is closed, since then imT = F, and hence T is a linear homeomorphism. Indeed, since T* is a linear homeomorphism, it follows from the first paragraph of this part of the proof that T** : E** + F** is a linear homeomorphism. There is thus a constant c > 0 such that IIT**(A)II ;::: cliAIl for every A E E**. But then, by (i), for every x E E, we have IITxll ;::: cllxll. It follows from Corollary 3.43, (b) =? (a), that T is injective (which we already knew) and has closed , range, as required. 0
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Introduction to Banach Spaces and Algebras
Proposition 3.64 Let E and F be a Banach spaces, and let T Suppose that im T is closed in F. Then im T zs finitedimensional.
E
K(E, F).
Proof Take B to be the closed unit ball of the Banach space im T, and let (Yn)n?l be a sequence in B. By the open mapping theorem, Theorem 3.40, there exists K > such that, for each n E N, there exists Xn E E with TX n = Yn and Ilxnll ::; K. Since T is compact, (Yn)n?l has a convergent subsequence. By Theorem 1.19, (b) =} (a), B is compact. By Theorem 2.31, imT is finiteD dimensional.
°
Theorem 3.65 (Schauder's theorem) Let E and F be a Banach spaces, and let T E B( E, F). Then T is compact if and only if the dual operator T* E B (F * , E*) is compact. Proof Let B = E[l) be the closed unit ball in E. Suppose that T is compact, so that X := T(B) is a compact space, and let (fn)n?l be a sequence in F[;); set gn = fn I X (n E N) and :F:= {gn : n EN}, so that :F is a bounded set in (C(X); I· Ix ). Since Ifn(Y)  fn(x)1 ::;
Ily  xii
(n E N, x, Y E F),
the family :F is equicontinuous. It follows from the ArzelaAscoli theorem, Theorem 1.20, that:F is compact in C(X), and so, by Theorem 1.19, (a)=}(b), there is a subsequence, say (gnJ )J?l, of (gn)n?l that converges uniformly on X. Thus (fn JoT) J?l converges uniformly on B, and so (T* (fn,)) J?l converges in E*. This shows that T* is compact. Conversely, suppose that T* is compact. Then T** E B(E**, F**) is compact. Thus T**((E**)[I)) is totally bounded in F**. By Corollary 3.63(i), T(B) is totally bounded in F, whence T(B) is compact by Theorem 1.19, (c)=}(a). Thus T is compact. D Corollary 3.66 Let H be a Hilbert space, and let T E B(H). Then the adjoint T* E B(H) ofT is compact if and only ifT is compact. D
3.16
Duals of subspaces and quotients.
Proposition 3.67 (Dual of a subspace) Let E be a Banach space, let M be a closed subspace of E, let J : M + E denote the inclusion map, and finally let 7r M : E* + E* / M.l denote the quotient mapping. Then: (i) J* : E* + M* is the restriction to M, i.e. J*(f) = f I M (f E E*); (ii) ker J* = M.l and im J* = M*; (iii) there is a unique linear homeomorphism ¢M : E* /M.l + M* such that J* = ¢ M 0 7r M, and then ¢ M is an isometry, so that M* is isometrically isomorphic to E* / M.l .
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Banach spaces
J for all f E E*. (ii) That ker J* = M.L is immediate from the definition. It is also immediate from the HahnBanach theorem that every 9 E M* has a normpreserving extension to a functional f E E*, so that im J* = M*. (iii) The existence of the linear homeomorphism 'l/JM follows from (ii), Lemma 3.47, and Corollary 3.49. The fact that 'l/JM is isometric follows because, by the HahnBanach theorem, Ilgll = inf{llfll : f E E*, f I M = g} (g E M*). D Proof (i) This is immediate because J* (J) =
f
0
Proposition 3.68 (Dual of a quotient) Let E be a Banach space, let M be a closed subspace of E, and let 7rM : E > ElM denote the quotient map. Then 7r'M: (EIM)*
is a lmear isometry, with im 7r~1 morphic to M.L.
>
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 <;;; T(c 1 B). But then the open mapping lemma, Lemma 3.38, shows that F is complete and that im T = F. (ii) Take 7r : E > 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 = liToIII. 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 normclosed 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 normclosed 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 normclosed 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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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 HahnBanach 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 socalled 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)
f4
(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 BanachSteinhaus
'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:fftj',
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 infinitedimensional 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 ft 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) <;; co(Z). Exercise 3.18 Let E and F be the Banach spaces (C(IT); I· In) and (Cl(IT); 11.11 1 ), respectively, in the notation of Exercise 2.10, so that Ilflll = If In + II'ln (f E C l(IT». For fEE, define
T(f)(t)
=
it
f
(t E IT).
Show that '1' is an injective, bounded linear map from E into F with range the linear subspace M = {g E F : g(O) = O} of F. Is the map '1'1 : M > E continuous? Exercise 3.19 Let E and F be Banach spaces, and letT E B(E, F) be such that the range T(E) of T has finite codimension in F. Prove that T(E) is closed in F. Indeed, let M = kerT, and define 'I'M E B(EIM, F), as in Lemma 3.47. There is a (necessarily closed) finitedimensional linear subspace G of F such that F = 'I'(E) EVG. Define S: (x + M, y) ft 'lM(X) + y, (ElM) x G > 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 infinitedimensional 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 <;;; E**. Let T: E + F be the identity map, so that T is injective. Show that kerT** = G, so that 1'** is not injective.
4
Banach algebras
Elementary theory In many of the examples of Banach spaces so far considered, there is a natural notion of 'multiplication' of the elements of the space, in addition to the vectorspace operations. Most obviously this applies in the space A = C(K), where K is a compact, Hausdorff space (see Example 2.4(i)). Here functions f and 9 in C(K) may be multiplied pointwise on K (just as the linear operations were defined pointwise). Another very important special case is the space A = B(E) of bounded linear operators on a Banach space E, with multiplication defined as composition (as in Example 2.15(i)). In both cases, we see immediately that the norm is submultiplicative with respect to the product, in the sense that
Ilabll :::; Ilallllbll
(a,b E A).
(*)
This inequality is analogous to that giving the subadditivity of a norm. The study of spaces with this additional operation of multiplication satisfying (*) turns out to be very important. The combination of vectorspace operations with multiplication comes in a structure known as an (associative) algebra. Just as we have assumed some elementary knowledge of vector spaces, so we shall assume some modest knowledge of algebras or, at least, of rings. The actual knowledge assumed will be very small. We shall briefly recall the basics. 4.1 Algebraic background. Let A be a vector space over a scalar field lK. Then A is an algebra if it also has an operation
mA:(a,b)f'>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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Introduction to Banach Spaces and Algebras
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 twosided ideal) I of A is a subset of A such that: (i) I is a vector subspace of A;
(ii) both AI <;:: I and I A <;:: I.
Here, for a subset S of an algebra A, we write as = lin{ax : XES}, etc. More generally, for nonempty subsets Sand T of A, we set
ST = lin {ab : a
E
S, bET} ;
further, we set S2 = SS, and then we define the sets sn for n E PI by induction via Sk+l = SkS (k E PI). The term 'ideal' on its own will always mean a 'twosided ideal'. Further, a left (respectively, right) ideal of A is a subspace I such that AI <;:: I (respectively, I A <;:: 1). It is immediately checked that each space In is a (left) ideal in A whenever I is a (left) ideal. An algebra A is simple if A2 =1= {O} and if the only ideals in A are {O} and A. For example,
a
E
A \ L. Then there exists b E A with
1
+ ba E
L.
Proof Set J = {ba + L : b E A} = Aa + L. Then J is a left ideal in A with J ;2 L, and so J = A because L is maximal. Thus there exist b E A and x E L with 1 = ba+x. Then 1 +ba = x E L. 0
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Banach algebras
Let A and B be algebras. A map e : A + B is a homomorphism if linear map such that e(ab) = ea· eb (a,b E A).
e is
a
(Occasionally such maps are called 'algebra homomorphisms'.) In this case, kere is immediately seen to be an ideal of A. A homomorphism e : A + B is an embedding or a monomorphism if it is injective, and then we often regard A as a subalgebra of B; a bijective homomorphism is an isomorphism, and an isomorphism from A to itself is an automorphism. Two algebras A and Bare isomorphic, written A ~ B, if there is an isomorphism from A onto B. In the case where A and B have identities 1A and 1B, respectively, e is a unital homomorphism whenever e(1A) = lB. Suppose that I is an ideal of A. Then the quotient space AI I has a natural algebra structure well defined by the formula (a+I)· (b+I)=ab+I
(a,bEA);
in this case, the quotient mapping 1f : A + AI I is a surjective homomorphism with ker 1f = I. Suppose that e : A + B is a homomorphism of algebras with ker e = I. Then there is a unique embedding AI I + B such that e = 1f.
e:
eo
4.2 Definitions and basic examples. Let A be an associative algebra over a scalar field lK. By an algebranorm on A is meant a mapping a I> I all of A into lR+ such that:
(i) (A; 11·11) is a normed space over K; (ii) Ilabll : : : Ilallllbll (a, bE A). Clause (ii) says that II· I is submultiplicative. Similarly, a semi norm p on an algebra A is submultiplicative if p(ab) ::::: p(a)p(b) (a, bE A). A normed algebra is a pair (A; 11·11), where A is a nonzero algebra and 11·11 is a given algebranorm on A. A Banach algebra is a normed algebra that is complete in its norm (i.e. is a Banach space). Recall from Section 3.3 that a Frechet space is a locally convex space whose topology is given by a complete metric. The topology of such a space can be defined by a sequence (Pn)n~l of seminorms, and we may suppose that the sequence is pointwise increasing. Let A be an algebra. Then A is a Fhkhet algebra if there is a sequence (Pn)n>l of submultiplicative seminorms on A such that (A; (Pn)n~l) is a Frechet space. More generally, A is an LMC algebra (or locally multiplicativelyconvex algebra) if there is a family P of submultiplicative seminorms on A such that (A; P) is a locally convex space; such an LMC algebra is complete if, for each net (a,,) in A such that (p( a" )) is a Cauchy net for each PEP, there exists a E A such that p(a"  a) + 0 for each pEP. By a unital normed (or Banach) algebra, is meant a normed (respectively, Banach) algebra with an identity element 1 such that: (iii) 11111 = 1.
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It is easy to show (see Lemma 4.8, below) that any normed algebra with an
identity may be given an equivalent algebranorm that makes it unital. From now on, all 'normed algebras' will be over the complex field, C, and will not be equal to the zero algebra {O}, unless we say otherwise. We now give some basic examples; more examples will be given later. Example 4.2 Let A be (C(K); I·I K ), the Banach space of all continuous, complexvalued functions on a nonempty, compact Hausdorff space K. Then A is a commutative algebra when the product of functions f, g E C(K) is defined by
(fg)(x) = f(x)g(x)
(x
E
K).
Clearly, IfglK :::; IflK IglK (f,g E C(K)). The constant function 1 is the identity of A. More generally, let K be a nonempty, locally compact Hausdorff space, and let A = (Co(K); I·I K ) be the Banach space of all complexvalued, continuous functions on K that vanish at infinity, as in Example 2.4(ii), again with the pointwise product. Then A is a commutative Banach algebra; it is unital if and 0 only if K is compact. Example 4.3 A uniform algebra on a nonempty, compact Hausdorff space K is a subalgebra A of C(K) that is closed in the normtopology, that contains the constant functions, and that separates the points of K. The latter phrase means that, for each x,y E K with x I y, there exists f E A such that f(x) I f(y). In particular, set ~ = {z E
{J
E C(~):
f I int~ is analytic}.
We remarked that A(~) is closed in (C(~); I'I~), and clearly A(~) is a subalgebra of C(~), and so A(~) is a uniform algebra on ~. More generally, for any compact subset K of
A(K) = {J E C(K) : flint K is analytic}. Again, it is immediate that all the above are uniform algebras. Theorem 2.86 implies that P(~) = A(~). It ",ill follow from Runge's theorem (see Corollary 4.85) that P(K) = R(K) if and only if
Banach algebras
159
Example 4.4 Let S be a semigroup, so that S is a nonempty set together with a map (s, t) I> st, S x S + S, such that (rs)t = r(st) (r, s, t E S). We have defined the Banach space (£I(S); 11.11 1 ) on page 38. Let f,g E £1(S). Then we set (f * g)(t) = l.:)f(r)g(s) : r, s E S, rs = t} (t E S), where we take (f * g)(t) = 0 when there are no elements r, s E S with rs = t. It is easy to verify that (£ 1(S); I . 111 ; *) is a Banach algebra; it is called the semigroup algebraof S, and * is the convolution product. Again, let 8s be the characteristic function of {s} for 8 E S; we shall often identify s with 8" and so regard S as a subset of £ 1 (S). We see that 8s * 8t = 8s i> 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 'infinitedimensional 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 noncommutative 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 rankone operator
Xo ® fo : x
f+
fo(x)xo,
E
t
E,
belongs to B(E). The space of continuous finiterank 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) nonzero ideal in B(E). Indeed, let J be a nonzero ideal in B(E), and take S E J with S =I O. Then there exist nonzero 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) <;;;; J, giving the claim. In Example 2.17, we also introduced the closed subspace K(E) of compact operators on E. It is clear from remarks in that example that K(E) is a closed ideal in B(E). The identity operator IE on E is not compact unless E is finitedimensional, and so K(E) is a proper ideal in B(E) whenever E is infinitedimensional. The closure A(E) of F(E) in (B(E); 11·11) is the collection of approximable operators; this collection is the minimum nonzero, closed ideal in B(E). For many Banach spaces E, including all those that arise in this book, we have A(E) = K(E); for these spaces K(E) is the minimum nonzero, closed ideal of
B(E). A Banach operator algebra on E is a subalgebra Q( of B(E) such that Q( contains F( E) and (Q(; III . III) is a Banach algebra for some norm III· III. For example, A(E) and K(E) are Banach operator algebras on E when given the operator norm. We note that the embedding of a Banach operator algebra (Q(; III . III) in (B(E); II· II) is always continuous; this is a nice application of the closed graph theorem, Theorem 3.45. Indeed, first take X E E and f E E*. Then, by (*), (x ® /)2 = f(x)(x ® /)' and so If(x)1 ::; Illx ® fill. Now take T E Q(. Then, by (*) again,
If(Tx)1 ::;
IllTx ® fill::; IllTllllllx ® fill·
(**)
Let Tn t 0 in (Q(; 111·111) with Tn t T in (B(E); 11·11). Then, by (**), f(Tx) = 0 for each x E E and f E E*, and so T = O. Thus, by the closed graph theorem, the embedding is continuous. 0
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Banach algebras
Example 4.7 Let U be a nonempty, open subset of C, and let O(U) be the space of all complexvalued, analytic functions on U, with the topology of local uniform convergence, as described in Section 3.3. Then it is easy to see that O(U) is a Fn§chet algebra with respect to this topology; all the specified seminorms Pn are clearly submultiplicative. 0
4.3 Elementary constructions. algebra is a Banach algebra.
(i) Every closed sub algebra of a Banach
(ii) Let A be an algebra over a scalar field JK such that A does not have an identity. Then the unitization of A is the unital algebra A+
:=
A EEl JK . 1,
with the obvious product that makes the symbol 1 into the identity. Thus
(a
+ AI) (b + /1,1)
=
ab + JLa + Ab + AJL 1
(a, bE A, A, JL E JK) .
We take A+ = A in the case where A does have an identity. In the case where A is a nonunital normed algebra, A+ is normed as an el  sum, i.e. Iia + Alii = Iiall + IAI (a E A, A E JK).
It is a simple exercise to show that (A+; 11·11) is then a normed algebra, that A is embedded in A+ as a closed ideal, and that A+ is complete if and only if A is complete. (iii) Let J be a closed ideal in a Banach algebra A. Then the quotient algebra AI J is a Banach algebra with respect to its standard quotient norm; the algebra AI J is unital whenever A is unital. (iv) Let A be a unital Banach algebra, and, for each a E A, define La E B(A) by La(b)=ab (bEA). Then it is easily checked that the mapping a f+ La is an isometric, unital isomorphism of A onto a closed subalgebra of B(A). (v) We may use a slight extension of the argument of (iv) to show that a normed algebra A has a completion which is a Banach algebra. By the use of (ii), we may restrict attention to the case where A is a unital normed algebra. Let X = A be the completion of A as a Banach space; this completion exists by Theorem 2.22. For a E A, define La E B(A) (as in (iv)), by La(b) = ab (b E A). Then La extends uniquely to a boun~ed linear operator La on X, and it is very easy to show that the mapping a f+ La is an isometric isomorphism of A onto a Subalgebra of the Banach algebra B(X). The closure of that subalgebra in B(X) is then a Banachalgebra completion of A. (By Theorem 2.22, it is the unique 0 completion of A as a Banach space.)
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There is a result on renorming a normed algebra with an equivalent norm that will be used in a few places. Lemma 4.8 Let (A; 11·11) be a normed algebra, and let S be a subset of A that is bounded and is closed under the product in A. Then there is a submultiplicative norm III . Ilion A such that:
(i) 111·111 zs equivalent to 11·11 ; (ii) Illslll S; 1 for every s E S; (iii) in the case where A has an identity element 1, then 1111111 = 1. Proof By the 'elementary construction' in (ii), above, we may suppose that A has an identity 1 and that 1 E S (for otherwise replace S by S U {1}). Let k = sup{llsll : s E S}; necessarily k ~ 1 because 1 E S. For a E A, define lal = sup{llsall : s E S}. Clearly, 1·1 is a seminorm on A. But also Iiall = 111ali S; lal S; kllall (a E A),
so that I . I is actually a norm on A that is equivalent to II . II, where we note that 111 = sup{llsll : s E S} = k. Now define Illalll = sup{labl : bE A, Ibl = I}. Then III ·111 is a submultiplicative norm on A with 1111111 = 1 because Illalll is just the 'operator norm' of left multiplication by a on (A; 1·1). Taking b = k 1 1, we see that, for every a E A, we have Illalll ~ k 1Ial. Also, for every a, bE A with Ibl = 1, we have labl S; kllabll S; kilalillbil S; klal , so that Illalll S; k lal, and hence 111·111 rv 1·1 rv 11·11· Finally, let a E S. Then, for every s E Sand b E A with Ibl = 1, we have sa E S, and so Iisabil S; Ibl = 1, which implies that also labl S; 1. Thus Illalll S; 1, which completes the proof. 0 4.4 The group of units. Let A be an algebra with an identity 1. Then a E A is left invertible if there exists b E A with ba = 1, and right invertible if there exists b E A with ab = 1. Further, a E A is invertible if it is both left and right invertible. In this case, it is easy to see that there exists a unique element b E A with ab = ba = 1; the element b is called the inverse of a, and it is denoted by a 1. In this case, we shall also say that a is a unit of A, and we shall write G = G(A) for the set of these units. Clearly, 1 E G(A) and G(A) is a group which is a subsemigroup of (A, . ); for a,b E G(A), we have (ab)1 = b 1a 1. For example, G(Mn) = {T E Mn : det T 1= O} for each n E N and, for each compact Hausdorff space K and f E C(K), we have f E G(C(K)) if and only if Z(f) = 0.
163
Banach algebras
Lemma 4.9 Let A be an algebra with an identity 1, and let I be a left zdeal in A. Let a E I with 1+a E G(A). Then 1+a E G(I +C· 1). Proof There exists b E A with (1 + b)(1 + a) = (1 + a)(1 + b) b = a  ba E I, and 1 + b is the inverse of 1 + a in I + C . 1.
= 1. But now 0
We now consider the group of units in a Banach algebra. The first lemma is very elementary, but is absolutely fundamental; it says that elements 'close to the identity' in a unital Banach algebra are also invertible. Lemma 4.10 Let (A; 11·11) be a unital Banach algebra, and let a E A be such that 111  all < 1. Then a E G(A) and 00
al=I+~)Ia)k. k=l
(*)
Proof Since 11(1  a)kll ::::; 111  all k for all kEN and since 111  all < 1, the series in (*) is absolutely convergent, and so convergent, by the completeness of A. (See Exercise 2.4.) Setting n
Sn = 1 +
2::(1 a)k
00
and
s = lim Sn = 1 + n+oo
k=l we have
lisa  111 Iisna
=
limn+oo Iisna 
111 = 11(1 
111.
a)n+lll ::::;
"'(1  a)k , ~ k=l
However,
111 
all n+ 1 ~ 0
as
n ~
00,
so that sa = 1. Also, for each n E N, we have Sna = aS n and so as = sa = 1. Hence a E G(A) and aI = s. 0 A completely equivalent way of formulating the above is the following: for each b E A with IIbll < 1, we have 1  b E G(A) with 00
(1  b)l = 1 +
2:: bk . k=l
Corollary 4.11 Let A be a unital Banach algebra. Then the set G(A) zs open in A. More precisely, if a E G(A) and if lib  all < liliaIII, then b E G(A). Proof Let a E G(A). Then
b=a(ab)=a(1a 1 (ab))
(bEA).
Now suppose that lib  all < liliaIII. Then Ilal(a  b)11 ::::; IlaIlllia  bll < 1, and so it follows from Lemma 4.10 that 1  aI(a  b) E G(A). We see that b has been written as a product of two units, so that also bE G(A). 0
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Introduction to Banach Spaces and Algebras
Corollary 4.12 Let A be a unital Banach algebra. Then the mapping
a
f+
aI,
G(A) + G(A) ,
is a homeomorphism of G(A) onto itself. Proof Set 8(a) = aI (a E G(A)). Then certainly 8 is a bijective mapping of G(A) onto itself. Also, since 8 1 = 8, it suffices to prove that 8 is continuous. Let a E G(A). For b E A with lib  all < 1/21Ia 111, it follows from Corollary 4.11 that bE G(A). Also,
libIII :::; Ilb 1  aIII
+ lIa111 =
Ilb1(a  b)a111
+ Ila111
1
:::; 211blll
+ Ila111,
whence libIII:::; 211a 111 and Ilb 1  aIII:::; Ilb11l11a  blllla111 :::; 211a 111 2 11a  bll,
(*)
o
and so the result follows. Corollary 4.13 Let A be a unital Banach algebra.
(i) Let (an)n>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 + aI 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+(aaN)=aN(l+aj/(aaN)) .
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;;llll
so that bna + O. Similarly, ab n + O.
::::: Iia  anll
+ Ila;;llll
+ 0
as
n +
00,
165
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 nonempty (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, nontrivial, 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 <;;;; {O, I}. (ii) Let a, bE A. Then Sp (ab) <;;;; {O} U Sp (ba). Proof We may suppose that A has an identity, 1. (i) Let>. E C, and set b = (1 >')lp. Then (>'lp)b = b(>'lp) = (>.  >.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 <;;;; {O} for each nilpotent element of an algebra A. Indeed, suppose that an = 0, and take>. E C with>' =I o. Then the inverse of >'1  a is
1(
>.
a anI) 1++···+ . >. >.nI
An element a in an algebra A is said to be quasinilpotent if either Sp a = 0 or Sp a = {O}. We shall write Q(A) for the set of quasinilpotent elements of A; in general, Q(A) is not a vector subspace of A. Every nilpotent element is quasinilpotent; we shall soon see some quasinilpotent, nonnilpotent 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)) <;;;; G(B), and so SPBBa <;;;; SPAa
(a E A).
The following theorem is the fundamental theorem of Banach algebra theory. Recall our convention that a normed algebra is necessarily nonzero.
Banach algebras
167
ThL_ Jm 4.17 Let A be a Banach algebra, and let a E A. Then Spa is a nonempty, compact subset of the dzsc {A E
R(A)  R(JL) = (A  JL)R(A) R(JL)
(A, JL E
(*)
so that R is an analytic Avalued function on C \ Spa, with R'(A) = R(A)2. If IAI> Iiall, then A tJ Spa and IIR(A)II
= IAI I II(1  AIa)lll
>
0
as
IAI
> 00
since 1  Ala > 1 as IAI > 00 and inversion is continuous. Assume towards a contradiction that Spa = 0. Then R is an Avalued entire function with R(A) > 0 as IAI > 00. By Liouville's theorem for vectorvalued functions, Theorem 3.12, we see that R(A) = 0 (A E q, and in particular aI = 0, which is a contradiction. Hence Spa =I 0. D Thus, for a quasinilpotent element a in a Banach algebra A, we have Spa
= {O}.
Corollary 4.18 Let A be a normed algebra, and let a E A. Then Sp a =I 0. Proof This is an easy consequence of the theorem because we clearly have SPAa;;::>SPA:a. D
Theorem 4.19 (Gel'fandMazur 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 JL, in Spa. We have JLl  a tJ G(A). But then, since A is a division algebra, JLl  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 complexvalued functions on some set X (with the algebraic operations defined pointwise on X). Suppose that A can be given some Banachalgebra 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 nonzero 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 quasinilpotent element of A. We now give the famous BeurlingGel'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
MI/R(g)
:=
sup{llg(z)11 : Izl = 11R} <
00.
For Izl < 1/11all (so that Izl < II p(a)), Lemma 4.10 gives 00
g(z) =
L
zna n .
n=O Now, for any A E A* with IIAII = I, set G = A 0 g. Then G is an analytic function of z for Izl < 1Ip(a) (and for all z E
IA(an)1 =
.1
1 12 71"1
Gn~~ dzl ::; R n MI/R(g).
Izl=l/ R z
Hence Ilanll ::; RnM1/R(g) (n EN), so that limsuPn>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 algebranorms 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(M1a) < 1, and so it follows from Theorem 4.23 that S := {Mna 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 algebranorm 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) <;;;; F; F is a hyperinvariant subspace for T if F is an invariant subspace for each S E B(E) for which ST = TS. A subspace F is proper if F =I {O} and F =I E. We shall be interested in the question whether or not every such operator T has a proper invariant subspace. This is a natural first step in an attempt to 'break T into smaller pieces' for the sake of understanding its properties. This question will be discussed further in Section 4.8. For example, let A be an eigenvalue of T. Then E(A) is a nonzero, closed subspace, and E(A) is a hyperinvariant subspace for T; each proper, closed subspace of E(A) is an invariant subspace for T. In particular, suppose that E is a finitedimensional (complex) space. Then each T E .c(E) has an eigenvalue, and hence, in the case where dim E 2: 2, T has a proper invariant subspace. This is the start of a process that allows us to choose a basis of E such that the matrix of T with respect to this basis is uppertriangular. Let E be a complex Banach space, and consider an operator T E B(E). Then the spectrum of T is the subset Sp T of e defined by:
°
Sp T = {A E
e : AI 
T is not invertible} .
en
for some Clearly, SppT <;;;; Sp T. If E were finitedimensional (so that E ~ n EN), then Sp T would be precisely SppT. In the general case, it is clear that the spectrum of T as an operator on E is the same as its spectrum as an element of the Banach algebra B(E), so that Theorems 4.17 and 4.23 apply immediately to the case where the element a in those theorems is a bounded linear operator on a complex Banach space. By Banach's isomorphism theorem, Corollary 3.41, SpT = Sp c(E)T. However, SppT could be empty, or a nonempty, proper subset of Sp T. For example, an operator T E B(E) is quasinilpotent if and only if SpT = {O}. Let P E B(E) be an idempotent, so that p 2 = P. Set
F = {x
E
E: Px = x},
the range of P. Then F is a closed linear subspace of E, and P is said to be a projection onto F. Recall that SpP <;;;; {O, I} for an idempotent P. Suppose further that T E B(E) and that PT = T P. For each x E F, we see that
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Introduction to Banach Spaces and Algebras
Tx = T Px = PTx E F, and so F is an invariant subspace for T and a hyperinvariant subspace for P; F is proper if P i 0 and PiIE' Let (E; 11·11) be a normed space, and take T E 8(E). A complex number A is an approximate eigenvalue of T if there is a sequence (Xnk::':l in E with Ilxnll = 1 (n E N) and such that (T  AI)Xn + 0 as n + 00. Of course, an eigenvalue of T is also an approximate eigenvalue of T. The set of approximate eigenvalues of T is the approximate point spectrum of T, denoted by SPap(T)
or
SPapT.
It is clear that we always have SppT S;;; SPapT S;;; Sp T .
Proposition 4.27 Let E be a Banach space, and let S, T E 8(E) be similar operators. Then Sp S = Sp T and SPapS = SPapT. Proof This follows easily from the definition (on page 132) of similarity for D
~m~~.
Theorem 4.28 Let E be a nonzero Banach space, and let T E 8(E). For each A E C, the following assertions are equivalent: (a) Art SPapT; (b) T  AI is bounded below;
(c) T  AI is injective and has closed range. Moreover, SPapT is a closed subset of Sp T containing the frontier, asp T, and so SPapT is not empty.
Proof The equivalence of (a) and (b) is immediate from the definition, and the equivalence of (b) and (c) is Corollary 3.43. To see that SPapT is closed, we take (A n )n2:1 to be a sequence in SPapT such that An + A E C. For each n E N, there is a vector Xn E E with Ilxnll = 1 such that II(T  AnI)xnll < lin. Clearly, 1
II(TAI)xnll:::::IAAnl++O n
as
n+oo,
and so A E SPap(T). To see that aSpT S;;; SPap(T), take A E aSpT, and set S = AI  T. Then clearly S E aC(E), and so, by Corollary 4.13(ii), there exists a sequence (Sn)n2:1 in B(E) such that IISnl1 = 1 (n E N) and SSn + 0 as n + 00. For n E N, choose Yn E E with IIYnl1 ::::: 2 and IISnYnl1 = 1, and set Xn = SnYn, so that Ilxnll = 1. Then II(T  AI)xnll = IISSnYnl1 ::::: 211SSnii + 0 as n + 00. This shows that A E SPapT.
D
173
Banach algebras
4.7 The spectrum of operators on Hilbert spaces. There are a few interesting results, with elementary proofs, about the spectra of special types of operator on a Hilbert space. Recall that an operator T E B(H) is normal if T*T = TT* and unitary if T*T = TT* = I H . Theorem 4.29 Let H be a Hilbert space, and let T be a normal operator in B(H). Then p(T) = IITII. Proof From Theorem 4.17, we already know that p(T) ::::: IITII for each operator T E B(H). Suppose first that T is selfadjoint. By the C* condition, given in Proposition 2 k k 2.57, we have liT II = IITI12, and then an easy induction gives IIT2 II = IITI12 for each kEN. Thus IITII = IIT2k liT"
>
p(T)
as k
>
(Xl
by the spectral radius formula, Theorem 4.23. It follows that IITII = p(T) for each selfadjoint operator T. Recall that, for every T E B(H), T* is the adjoint of T; we have IIT*II = IITII and hence p(T*) = p(T). Since T*T is selfadjoint and IITI12 = IIT*TII by the C*condition, we have IITI12 = p(T*T). For T normal, p(T*T) ::::: p(T*)p(T) by Corollary 4.24. Thus IITI12 = p(T*T) ::::: p(T)2 ::::: IITI12 ,
o
so that p(T) = IITII.
Thus the only quasinilpotent and normal operator in B(H) is O. We shall shortly, in Theorem 4.31, see a simple, direct proof that Sp T I 0 and that p(T) = IITII when T is a selfadjoint or unitary operator. Let T be a bounded operator on a nonzero Banach space. Then, by Theorem 4.28, SPapT is a closed subset of Sp T. For normal operators T on a Hilbert space, there is a reverse inclusion. Theorem 4.30 Let H be a Hilbert space, and let T E B(H) be a normal operator. Then every point of Sp T is an approximate eigenvalue. Proof For each A E C, the operator T  AI is normal, with (T  AI)* = T*  XI. Hence, by Proposition 2.58(ii), we have II(T  AI)xll = II(T*  XI)xll
(x
E
H),
so that T  AI is bounded below if and only if T*  XI is bounded below. It then follows from Theorem 2.60 that T  AI is invertible if and only if T  AI is bounded below. Hence A E Sp T if and only if T  AI is not bounded below, i.e. if and only if A is an approximate eigenvalue. D
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Introduction to Banach Spaces and Algebras
Theorem 4.31 Let H be a Hilbert space. (i) Suppose that T E B(H) is selfadjoint. Then SpT ~ [IITII, IITlll and at least one of±IITII belongs to SpT (m particular, SpT =I 0). (ii) Suppose that U moreover, Sp U =I 0.
E
B(H) is unitary. Then SpU
~
= {z
]'
E
C: Izl
= I};
Proof (i) We have T = T*. Take A E Sp T. By Theorem 4.30, there is a sequence (Xnk:>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 = {Z1 : 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 selfadjoint, 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 nonzero 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 finitedimensional; (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 finitedimensional.
(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+1AJ 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 finitedimensional, and hence closed, vector subspace of E. Since {xn : n E N} is linearly independent, Ln <;; Ln+1 (n EN). By Riesz's lemma, Lemma 2.30, for each n 2: 2, there exists Yn E Ln with IIYnl1
= 1 and
llYn  xii> 1/2
(x
E
Lnd·
Since Yn E lin{ Xl,' .. , x n }, it follows that (AnI  T)(Yn) E L n l , and so, for each n > 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 infinitedimensional. Each nonzero A E Sp T is an isolated point of Sp T and an ezgenvalue, and the corresponding eigenspace E(A) is finitedimensional. Proof In the case where E is infinitedimensional, T ~ G(E), for otherwise E K( E), and so 0 E Sp T. Since Sp T is a nonempty, compact subset of C, the conclusion will follow from the fact that every nonzero 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 nonzero 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 finitedimensional. 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 infinitedimensional Banach space, and let T E K(E) with T '" o. Then T has a proper hyperinvariant subspace. Proof We first remark that we can suppose that T is quasinilpotent. 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 finitedimensional, and hence a proper hyperinvariant 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}·
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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 openandclosed 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 nonempty, 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 nonempty, 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 <:;;; C, the set K is also compact and K <:;;; K. Then it is easy to see that K is the smallest polynomially convex, compact set containing K. ~ As a temporary notation, we write K for the union of K with all the bounded components of C \ K. It is then clear, using the maximum modulus principle, that K <:;;; K <:;;; K. We shall see very soon that K = K. Let A be a unital algebra, and let aI, ... ,an E A. Then algA {I, al,"" an} is the smallest subalgebra of A containing al,"" an and 1. Clearly, in the case where a,aJ = aJa, whenever i, j E {l, ... ,n}, the subalgebra algA {I, al,· .. ,an} is commutative and consists of the elements p(al, ... , an), where p is a complex polynomial in n variables. In particular, we have defined algA {I, a} for a E A, and this is always a commutative subalgebra of A. Now suppose that A is a unital Banach algebra. We define
A(al,"" an)
=
algA {I, al,"" an},
the closure of algA {I, al,"" an} in A, so that A(al,"" an) is the smallest closed subalgebra of A containing al, . .. ,an and 1. The Banach algebra A is polynomially generated by al,"" an if A(al,"" an) = A. In particular, we have defined A( a) for a E A; A is polynomially generated by a if A( a) = A. More generally, let 5 be a subset of a unital algebra A. Then we shall later write algA (5) for the smallest subalgebra of A containing 5 and 1. In the case where A is a unital Banach algebra, the closure of algA (5) is A( 5); it is the subalgebra of A polynomially generated by 5. Proposition 4.39 Let A be a unital Banach algebra, and let a E A. Then Sp A(a)a =
sp:a .
Proof Suppose that fL tj. SPA(a)' Then (fL1  a)l E A(a). It follows from the definition of A(a) that (fL1  a)l may be approximated by polynomials in a. In particular, there is some complex polynomial, say p, such that
IIp(a) (fLI  a) Define a polynomial q by
1
111 < 2 .
180
Introduction to Banach Spaces and Algebras q(.\)
Then Iq(fl)1 =
1 11
=
p(.\) . (fl  .\)  1
(.\ E
q.
= 1. By Lemma 4.22, q(.\) E SpAq(a) (.\ E SPAa), and so
Iq(.\)1 ::; Ilq(a)11 < 1/2. This shows that fl rf

sr:;a, so that
SPA a <;;; SPA(ala.
By the remark following Corollary 4.37, SPA(ala <;;; SPAa <;;; follows that (*) SPA(ala = SPA a = SPAa,
sr:;a. It thus

o
giving the result.
Let K be a nonempty, compact subset of C To complete the proof that R is just the union of K with all the bounded components of C \ K, we simply take A to be the commutative Banach algebra C(K), and define Z E A to be the coordinate functional (restricted to K), so that SPA Z = K, and now the result is immediate from (*). We obtain the following evident corollary. Corollary 4.40 Let K be a nonempty, compact subset of C Then K is polynomially convex if and only if C \ K is connected. 0
For some purposes the study of the spectrum of an element of a Banach algebra may be reduced to the case of a commutative algebra. There are various ways of doing this; one rather attractive method introduces the idea of a bicommutant. Let A be an algebra with an identity lA, and let X be a subset of A. Then the commutant of X (relative to A) is defined to be
Xc = {b
E
A : ba = ab for all a
E
X} .
The bicommutant of X is XCc = (XC)c. A subset X <;;; A is called commutative if and only if ab = ba for all a, b EX; clearly X is commutative if and only if X <;;; Xc. The centre of A is defined to be N, and it is denoted by 3(A); it is a commutative subalgebra of A. Clearly, Xc is closed when A is a Banach algebra. A subalgebra B of a unital algebra A is inverseclosed if 1A E B and if aI E B whenever an element a E B is invertible in A. In this case, SPBb = SPAb for each bE B. The following lemma is a very easy exercise. Lemma 4.41 Let A be an algebra with an identity, and let X and Y be subsets of A. Then: (i) Xc is an mverseclosed subalgebra of A containing lA, and Xc is closed in the case where A zs a normed algebra; (ii) if X <;;; Y, then Xc :2 YC; (iii) X <;;; XCC and (XCC)C = (XC)CC = Xc;
(iv) if X is commutative, then so is XCC.
o
181
Banach algebras
We say that a subalgebra C of a unital algebra A is a bicommutant subalgebra if C = CCC. By the last lemma, it is clear that C is bicommutant if and only if C = Xc for some subset X of A. By Lemma 4.41(i), such a subalgebra is always inverseclosed, and so it is apparent that Spca = SPA a for every a E C. Hence we see the following corollary. Corollary 4.42 Let A be an algebra with an identity, and let X be a commutative subset of A. Set C = XCC. Then C zs a commutative subalgebra of A containing lA such that Spca = SPAa for every a E C. In the case where A is a Banach algebra, C is closed in A. 0
4.10 The continuity of characters. We close this section with a quite fundamental remark about characters on a Banach algebra. The theory of characters is particularly important for commutative algebras, and this theory will be given in the next section. However, the basic result applies to all Banach algebras. Let A be an algebra. Then a character on A is a nonzero homomorphism cp : A + C; we write A for the set of all characters on A. Suppose that A has an identity 1. Then cp( 1) = 1 for every cp E A, and so a character is a unital homomorphism.
Theorem 4.43 Let A be a Banach algebra, and let cp E A. Then cp is continuous and Ilcpll :s; 1. Suppose that A is unital. Then Ilcpll = 1. Proof Let a E A with Iiall :s; 1, and assume towards a contradiction that Icp(a)1 > 1. Set b = cp(a)la. Then cp(b) = 1, while Ilbll = Icp(a)I 1 Iiall < 1. By Lemma 4.10, lA  b E G(A+), and so there exists c E A such that
(IA  b)(IA  c)
=
lA,
and hence bc = b + c. Applying the homomorphism cp then gives cp( c) = 1 + cp( c), a contradiction. Thus Icp(a)1 :s; 1 whenever Iiall :s; 1, so that cp is continuous, with licpll :s; 1. Suppose that A is unital. Then Ilcpll = 1 because cp(l) = 1 = 11111. 0 Let A be a Banach algebra with an identity, and let cp E A. Then ker cp is a maximal ideal in A. But, by Theorem 4.14, maximal ideals of A are closed, and so it again follows that cp is continuous. Example 4.44 Let A(~) be the disc algebra, and let Ao
= {f
E A(~)
: f(O)
= O}.
Let cp be the character on Ao defined by cp(f) = f(I/2) (f E Ao). Then it is elementary, from Schwarz's lemma, Proposition 1.38, that Ilcpll = 1/2 < 1. 0
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Introduction to Banach Spaces and Algebras
Example 4.45 It is not the case that characters on a normed algebra are automatically continuous. For example, let A = qXl be the algebra of all polynomials in one variable. Then A is a normed algebra with respect to the uniform norm I . I~. The evaluation characters E z : p f+ p( z) on A are continuous if and only if
z
E~.
0
In Theorem 4.43, we proved the almost trivial fact that all characters on a Banach algebra are continuous. Now suppose that A is a commutative Frechet algebra. It is a remarkable fact that it is not known whether or not every character on A is necessarily continuous. An algebra (A; T) is a topological algebra if (A; T) is a Hausdorff topological vector space and the multiplication rnA : A x A ~ A is continuous. We say that topological algebra A is functionally continuous if every character on A is continuous, and so we are asking whether or not every commutative Frechet algebra is functionally continuous. Notes The foundations of the theory of Banach algebras were laid down by Gel'fand in the harsh conditions of Moscow in 1941. Subsequent expositions include [31, 47, 82, 92, 140] and Part III of [144]. The standard texts on uniform algebras are [79] and [152]; for recent results on semigroup algebras, see [52]. A more general version of the renorming result, Lemma 4.8, is given in [47, Proposition 2.1.9]. Indeed, let A be an algebra, let 11·11 be a norm such that (A; 11·11) is a Banach space and multiplication on A is separately continuous, and let S be a bounded semigroup in (A, .). Then there is a norm 111·111 on A such that 111·111 is equivalent to 11·11 and (A; 111·111) is a Banach algebra such that Illslll ::; 1 (s E S). Let E be a Banach space. We have defined Banach operator algebras in Example 4.6, and given some examples. Several further examples are given in [47, Section 2.5]. A recent advance in our subject is the construction by Argyos and Haydon [16] of an exotic, infinitedimensional Banach space E for which K(E) EEl CIe = B(E); the existence of such an example answers a question that was at least 40 years old. Let E be a Banach space, and let T E B(E). Then we have defined SpT and SPapT. Several other subsets of Sp l' have been identified and discussed; for example, SPsu 1', the surjectivity spectrum, is the set of A E C such that l'  AlE is not a surjection, and SPcomT, the compression spectrum, is the set of A E C such that (1'  AlE)(E) is not dense in E. For a clear and detailed account of these spectra, see [114]. The study of the spectrum of a compact operator is very classical; see [63, Theorem VII.4.5] or [144, Theorem 4.25], for example. This study has many applications in the theory of differential and integral equations. We have given the Gel'fandMazur theorem in Theorem 4.19 for normed division algebras. It also holds for locally convex (F)algebras, and, in particular, for Frechet algebras; see [47, Theorem 2.2.42]. The derivation of the spectral radius formula uses a simplification from [162] of earlier proofs; our proof avoids the use of the uniform boundedness theorem, which is often used at this point. There is a somewhat different proof of the fundamental theorem of Banach algebras in [31, Section 5, Theorem 8]; the first proof of the general case is in [81], following an earlier result for a special case by Beurling [29]. The study of the spectrum of a compact operator is a classical theory, and much more is known than we have revealed; see [63, 88, 144], for example. Lomonosov's theorem is from [116]; our proof is due to Hilden; see [71] for a more general result. Let S be a commutative subset of a unital algebra A. Then it follows from Zorn's lemma that S is contained in a maximal (with respect to inclusion) commutative subset,
183
ach algebras
say M, of A. Clearly, M = Me and JIv[ is a unital subalgebra of A; for each a E M, we have Sp Ma = Sp A a. Thus in results like Corollary 4.42 we could work with M, rather than C. For a more expansive treatment of this result and other algebraic background to Banach algebra theory, see [47]. The seminal work on Frechet algebras, and more general LMC algebras, is [122]; an early text is [168]. For remarks on these algebras, see [47, Section 2.2] and [93, Chapter 4], where they are termed 'polynormed algebras'. Even more generally, The question whether or not every commutative Frechet algebra is functionally continuous was specifically discussed in [122], and so it is often called Michael's problem; it seems that this question was even discussed by Mazur and his students around 1938, so it is very old. For a strong partial result of Arens, see [47, Corollary 4.10.11]; for a sufficient condition, see the remarkable result of Dixon and Esterle [61], given as [47, Corollary 4.10.16], which introduces a deep question in the theory of entire functions of several complex variables. For a further partial result, see [55]. There are various papers in the literature which claim, explicitly or implicitly, a positive solution to Michael's problem, but none seems to have convinced the community. All the results in this section are quite standard, indeed canonical. Exercise 4.1 Let (A; 11·11) be a normed algebra, and let S be a vector subspace or an ideal in A. Show that S is a vector subspace or an ideal in A, respectively. Exercise 4.2 Let A be a unital Banach algebra, and set 2l = Mn(A), the algebra of all n x n matrices over A. Show that 2l is a unital Banach algebra with respect to the obvious product and the norm given by II(a'l)11 = max{lladll
+ ... + Ilamil
: i = 1, ... ,n}.
Exercise 4.3 Show that the two matrices
(~ ~)
and
(
~ ~)
are both nilpotent in M 2 , but that neither their sum nor their product is quasinilpotent in M 2 • Exercise 4.4 Let A be a unital Banach algebra. Show that 1 is an extreme point of the closed unit ball A[1I. Exercise 4.5 Let (A; II·ID be a normed algebra with unit sphere S, and let a E A. Then a is a topological divisor of 0 if inf{llabll
+ Ilball
: b E S}
=
o.
Prove that every element in the frontier of G(A) is a topological divisor of O. Exercise 4.6 Let (A; II·I!) be a normed algebra. A Banach algebra (B; 111·111) is a Banach extension of (A; II . I!) if there is an isometric embedding of (A; I! . I!) into (B; III . III).
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Introduction to Banach Spaces and Algebras
Let A and B be Banach algebras, let I be a closed ideal in B, and let 0 : A be a continuous homomorphism. Set m= A EB I, with the norm II(a,x)11 = Iiall and define a product on
+ Ilxll
(a
E A,
x
E
>
B
1),
m by setting
(al,a2)(b 1 ,b2) = (ala2, b1 b2 + O(al)b2 + b1 0(a2))
(al,a2 E A, b1 ,b2 E 1).
Show that (m; 11·11) is a Banach extension of A. Exercise 4.1 Prove the following theorem of Arens. Let A be a unital, commutative Banach algebra, and let a E A. Then there is a Banach algebra extension mof A such that a E G(2l) if and only if a is not a topological divisor of O. Indeed, suppose that a is not a topological divisor of 0, and set
B = {b =
~ aJXJ : aJ E A
(j
E Z+), Ilbll
=
~ Ila II < J
00 }
Then (B; II '11) is a unital Banach algebra which is a Banach extension of A. Next, set J = (IA  aX)B, a closed ideal in B, and set m = BIJ. For e E A, we have Ilc  (IA  aX)pll 2: lIell for each polynomial p = 2:;=0 aJXJ E B, and so the natural embedding c f+ e + J, A > 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 )thposition 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].
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Banach algebras
Commutative Banach algebras 4.11 Characters and ideals. Throughout this section, A will usually be a nonzero, 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 f4 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'fandMazur 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.
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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 nonempty, 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 nonempty, compact subset of
~
= q[X]] be the algebra of all formal sums of the form 00
LanX n , n=O
where
000,001,'"
E
xm . xn = x m+n for all m, n 00
a
+b
L(an + f3n)X n , n=O
a· b
=
~ (~akf3nk) Xn;
we note in particular that the inner sum in the formula for the product is a finite sum, despite the fact that elements of ~ are infinite sums, and so the product is well defined. Let m E Z+. We write 7rm : 2::=oa n X n It am; these are the coefficient functionals. Further, the maps m
00
Pm: LanX n n=O
It
L lanl, n=O
~
+
1R+,
are submultiplicative semi norms on ~. The algebra ~ has a unique maximal ideal consisting of the elements of the form 2::=oa n X n with 000 = 0, and its unique character is the map 7r0. We may identify the algebra qX] of elements 2::=oa n X n E ~ such that an = 0 eventually with the algebra of all polynomials in one indeterminate. It can be shown that the algebra ~ is not a Banach algebra with respect to any algebranorm. However, it is clear that ~ is a Frechet algebra with respect to the sequence (Pm : m E Z+) of seminorms. These seminorms define the topology Tc of coordinatewise convergence on ~. A Banach algebra of power series is a Banach algebra (A; 11·11) such that A is a unital subalgebra of ~ with qX] <;;; A and such that all of the maps 7rm : (A; I . II) +
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Introduction to Banach Spaces and Algebras
j!l(Z+,W)
=
{a ~anxn =
E
J: Iiall w:= ~Ianlw(n) < oo} .
It is easily checked that (j!l(Z+,W); 11·1Iw) is a Banach algebra of power series. It is a weighted semigroup algebra on Z+ in the sense of Example 4.4. Clearly, j! l(Z+, w) is polynomially generated by X, and Ilxnllw = Wn (n EN), so that p(X) = lim w(n)l/n. n>CXJ
The above examples of weights show that we can have either p(X) > 0 or p(X) = 0; in the latter case, X is a quasinilpotent element, and indeed each element 2::::=0 anxn of j! 1(Z+, w) with ao = 0 is quasinilpotent. In particular, we obtain commutative Banach algebras A for which Q(A) "I {O}. 0
Example 4.53 Let U be a nonempty, open subset of C, and let O(U) be the Frechet algebra of analytic functions on U, as in Example 4.7. We first claim that 1>O(U) = U, in the sense that every character on O(U) is given by evaluation at a point of U. Indeed, let cp E 1>O(U), and set z = cp(Z). Then z E U, for otherwise there exists 9 E O(U) with (zl  Z)g = I, giving an immediate contradiction. For each f E O(U), there exists 9 E O(U) with f = f(z) + (Z  zl)g in O(U), and so cp(f) = f(z; this gives the claim. It follows that the Frechet algebra O(U) is functionally continuous. Let I be the subset of O(C) consisting of all functions f with the following property: there exists n E N such that f(k 2 ) = 0 for each k:::::: n. It is clear that I is an ideal in O(C). For n E N, consider the canonical product CXJ
fn(z) =
IT (1  :2)
(z
EC),
k=n
where n E N. It is standard (for example, see [143, Chapter 15]) that the infinite product converges and that Z(fn) = {k 2 : k : : : n}, so that fn E I. Further, for each compact subset K of C, the sequence (fn)n>l converges uniformly on K to the constant function 1. Thus 1 E I, and henc;:: I = O(C) and I is dense in O(C). Further, I is contained in (many) maximal ideals of O(C). This shows that there may be dense maximal ideals in a commutative Frechet algebra. 0 Now let A be any Banach algebra. By Theorem 4.43, we have 1> A ~ A *, the Banachspace dual of A. The Gel'fand topology on 1> 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 It 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 (
{ip E where al, ... , an
E
lip(a))1 < 1 (j = 1, ... ,nn,
A.
Theorem 4.54 Let A be a commutative, unital Banach algebra. Then
{ip E B : ip(1) = 1,
ip(ab)  ip(a)ip(b) = 0 (a, bEAn,
and so
o
Let A be a Banach algebra without an identity, and let A+ be the unitization of A. Then it is easy to see that we can identify
190
Introduction to Banach Spaces and Algebras
Proof Write 1](x) = Cx (x E K), so that 1] : K > A. Since A has an identity, the space A is compact; since A separates the points of K, the map 1] is injective. For each y E K, the character 1](Y) = Cy E A has a neighbourhood base in the Gel'fand topology consisting of sets of the form
u= where
{'P E A : 1'P(fk)  h(y)1 < 1 (k = 1, ... , n)},
h, ... , fn EA. Then 1]I(U)
= {x
E K : Ih(x)  fk(y)1
<
1 (k
= 1, ... , n)},
which is an open subset of K because h, ... , fn are continuous. Thus 1] is continuous, and so 1](K) is a compact, and therefore closed, subset of A. By Lemma 1.14, 1] is a homeomorphism onto its range. 0 Let A be a Banach function algebra on a nonempty, compact Hausdorff space K. Then we say that A is natural in the case where the map of the above lemma is a surjection onto A. Thus, we can say that, for a natural Banach function algebra A on K, each maximal ideal has the simple form
Mx
= {f E A : f(x) = O}
for some x E K. For example, the algebras C(K), A(Do), and R(K), are all natural Banach function algebras on the compact spaces on which they are defined. We now give a more general version of this observation. Theorem 4.56 Let A be a Banach function algebra on a nonempty, compact Hausdorff space K. Then A is natural on K if and only if, for each finite subset {h,···, fn} of A for which n~=1 Z(f,) = 0, there exist gl,···, gn E A such that L~=1 f,g, = 1. Proof Recall from page 23 that Z(f) = {x E K : f(x) = O} for f E A. Suppose that A is natural on K, and take functions h, ... , f n E A such that n~=1 Z(f,) = 0. Set I = L~1 f,A, so that I is an ideal in A. By Theorem 4.46, I is not contained in any maximal ideal of A, and so I = A. The existence of the required functions gl, ... ,gn E A follows. Conversely, suppose that A is not natural on K, and choose 'P E A \ K. For each x E K, there exists fx E A with 'P(fx) = 0 and fx(x) = 1. Since K is compact, there exist h, ... ,Jn E ker'P with n~=1 Z(f,) = 0. Clearly, for each gl, ... ,gn E A, we have 'P(L~l f,g,) = 0, and so L~=1 f,g, =I 1. 0
Recall that a Banach function algebra A on a nonempty, compact Hausdorff space is selfadjoint if and only if 7 E A whenever f E A. Corollary 4.57 Let A be a selfadjoint Banach function algebra on a compact Hausdorff space K. Then A is natural if and only if 1/ f E A whenever f E A with Z(f) = 0.
191
Banach algebras
Proof Suppose that A is natural. Then certainly 1/ f E A whenever f E A with Z(f) = 0. Conversely, suppose that 1/ f E A whenever f E A. Take iI, ... , f n E A such that n~=l Z(f.) = 0, and set h = L~=l f.f.; we see that h E A because A is selfadjoint. Clearly, Z(h) = 0, and so l/h E A. But now L~l f.g. = 1, where g. = f./h (i = 1, ... , n), and so A is natural by Theorem 4.56. D Corollary 4.58 Let A be a Banach function algebra on a compact Hausdorff space K. Suppose that the uniform closure A is natural on K. Then A is natural on K if and only if 1/ f E A whenever f E A with Z(f) = 0. Proof We shall show that A is natural if 1/ f E A whenever f E A with Z(f) = 0. Take iI, .. ·, fn E A such that n~=l Z(f.) = 0. Then there exist gl,"" gn E A with L~=l f.g. = 1. For i = 1, ... , n, choose h. E A such that Ig.  h.I K < l/n(lf.I K + 1). Then
It
f•h • ~
11K';
t
If.IK Ig. ~ h.IK <
1,
and so Z(L~=l f.h.) = 0. Thus there exists h E A with h L~l f.h. = 1. The result follows from Theorem 4.56. D Let A be a Banach algebra such that 1> A function 1> A + C by
a:
a(ip) = ip(a)
I 0.
For each a
E
A, we define the
(ip E 1>A)'
a
Thus the mapping a f+ is simply the canonical embedding of A into A ** followed by restriction to the subspace 1> 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
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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 quasinilpotent if and only if Spa = {O}, and that we write Q(A) for the set of quasinilpotent 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) <;;; Ace z; similarly we have Banach sequence algebras on N. Of course, each such Banach algebra is commutative. For example, fix p E [1, (0). Then the Banach space P with coordinatewise multiplication of sequences, is easily seen to be a Banach sequence algebra on N, with coo as a dense subalgebra. Let A be Banach sequence algebra on N or Z. Then each of the projections 7r m : x f+ Xm is obviously a character on A, and hence continuous. Now let zp E A. Since eme n = 0 for m I= n, clearly zp( en) I= 0 for at most one value of n. In the case where Coo or coo(Z), respectively, is dense in A, there exists a unique n in N or Z with zp(e n ) I= 0 (because zp I= 0), and then zp(e n ) = 1 and zp = 7rn' Thus A can be identified with N or Z, respectively. For zp E A, with zp identified with n, we see that {zp} = N E A : I1/;( en)  zp( en) I < 1}, and so zp is isolated in A. This shows that the Gel'fand topology on A is discrete, and so A is homeomorphic to N or Z, respectively. The Gel'fand transform is the identity map. 0
e
Example 4.62 This example describes the Wiener algebra, W Indeed,
W
= W('Jr) =
{f
=
n'%:;oo anZn E C('Jr) : Ilflll
:=
= W('Jr).
nf;oo lanl <
oo} ,
where, as before, Z : Z f+ Z is the coordinate functional, now on 'Jr. The algebraic operations are defined pointwise on 'Jr, so that W('Jr) is easily checked to be a subalgebra of C('Jr). But it is not a closed subalgebra of (C('Jr); 1·11['): indeed, it is immediate from the StoneWeierstrass theorem that W('Jr) is uniformly dense in C('Jr). Another common way of describing W is to regard it as the algebra of absolutely convergent Fourier series.
Banach algebras
193
Clearly, Z E G(W) and IIZI11 = IIZ 1111 = 1. Let cp be a character on W, and set cp(Z) = (. Then cp(Zl) = (l, so that both 1(1 ~ 1 and 1(11 ~ 1, i.e. ( E 11'. Since finite sums of the form L:=N an zn are dense in W (this shows that W is polynomially generated by Z and Zl), we can now easily show (as in Example 4.50), that we have
cp(f) = I(()
(f
E W).
Thus every character on W is given by evaluation at a point of 11'. We now note the important fact that the Banach algebra W is also the group algebra on (2:; +) (see page 159) 'in disguise'. We recall that, as a Banach space, £ 1(2:) consists of all functions I : 2: + C such that 111111 := L {11(n) I : n E 2:} < 00, and that (£ 1(2:); * ) is a commutative Banach algebra for a product satisfying the rule that 8m * 8n = 8m +n for all m, n E 2:. Then the map 00
(): (an)nEZ
f+
L
anZ n ,
(£1(2:); *)
+
(W; .),
n=oo
is an isometric isomorphism, as is very easily checked. Thus, to establish properties of either (£1(2:); *) or (W; .), it is sensible to focus on the presentation of the algebra that is more convenient for the problem at hand. For example, we see immediately that the character space of (£ 1 (2:); * ) is (homeomorphic to) the circle 11'. We see that the above map () is exactly the Gel'fand transform of (£ 1 (2:); * ), and that () is a prototype of a Fourier transform map from a group algebra (£l(G); *) (or (L1(G); *)) into an algebra (Co(r); .); here G is a (locally compact) abelian group and r is its 'dual group'. For more on this, see Section 4.12 and the notes to that section. 0 From the simple argument in the last example, we can immediately deduce the following classical result of Wiener. Theorem 4.63 (Wiener's theorem) Let I E C(1I') be such that I has an absolutely convergent Fourier series. Suppose that I( () I= 0 (( E 11'). Then the function 1/1 also has an absolutely convergent Fourier senes. 0 Historically, the above proof of Wiener's theorem played a considerable part in arousing interest in Banachalgebra methods in analysis. We note that a direct calculation of the Fourier coefficients of 1/1 seems to be intractable. Let A be a commutative Banach algebra. Then the ideal defined by J(A)
= n{M: M
E
MA+}
is the (Jacobson) radical of A. (For a definition in a more general context, see Section 5.3.) Clearly, J(A) = Q(A) is a closed ideal in A. The algebra A is
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Introduction to Banach Spaces and Algebras
semisimple if J(A) = {O}. Further, A is radical if J(A) = {O}; in this case, A cannot have an identity. Evidently a commutative, unital Banach algebra is semisimple if and only if its Gel'fand representation is injective. For example, each of the examples C(K), any uniform algebra (so, in particular, the disc algebra A( ~) and the algebra R( K)), and also the Wiener algebra W(1I') is semisimple. Indeed, every Banach function algebra A is semisimple. In each case, the Gel'fand representation is simply the inclusion map into the appropriate algebra C( A). Each of A(~), R(K) (when int K i= 0) and W(1I') provide examples in which the representation is not surjective. In the case of the Weiner algebra W(1I'), the image of the Gel'fand representation is not even closed in C( A). A Banach space with the zero product is a radical Banach algebra. An example of a commutative Banach algebra which is not semisimple was given within Example 4.52, and a further example will be given in Example 4.68. Proposition 4.64 Let A be a unital Banach algebra which is polynomially generated by an element a E A. Then the map cp
f+
cp(a) ,
A + SPAa,
is a homeomorphism. Proof Let the specified map be denoted by 7]. It is clear from Corollary 4.47(ii) that 7] is a continuous surjection of A onto SPA a. Suppose that cp, 'IjJ E A with 7]( cp) = 7]( 'IjJ). Then cp(p( a)) = 'IjJ(p( a)) for each polynomial p, and so cp and 1/) agree on a dense sub algebra of A. Since both cp and 'IjJ are continuous, they agree on A, and so cp = 'IjJ. Thus the map 7] is also an injection, and hence, by Lemma 1.14, 7] is a homeomorphism. D 4.12 The Banach algebras L1(1I') and L1(1R). In this section we shall discuss two basic commutative Banach algebras which are, as we mentioned, prototypes of the 'group algebras of a locally compact group'. These algebras have many elegant properties and many subtleties; they also provide the theoretical underpinning of many applications, for example in the solution of differential equations. We gave our definition of the Banach spaces L1(1I') and L1(1R) (always taken with the norm II· Ill) in Section 2.6 as the completions of C(1I') and Coo (IR), respectively, and explained how to define the integrals
I(f)
=
~
r 27r io
2rr
f(e) de
(f
E
L1(1I'))
and
I(f)
=
1<Xl f(t) dt <Xl
(f
E
L1(1R)) ,
so that I is a continuous linear functional with 11111 = 1 on L1(1I') and L1(1R), respectively; for this, we extended the usual (Riemann) integrals on C(1I') and Coo (IR). Note that, as suggested earlier, we have introduced a factor of 1/27r
Banach algebras
195
into the definition of the integral on [0, 2n], which is identified with 11'. We also noted in Exercise 2.7 that the simple functions can be regarded as dense vector subspaces in Ll(lI') and Ll(JR.). Example 4.65 We define the convolution product (f,g) the formula
(f
* g)(x) =
1:
f(x  t)g(t) dt
(x
E
JR.).
f>
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 Riemannintegrable 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
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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 halfline 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+ eat 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 <
f+
et~
oo} .
(Here we are considering 'Lebesgue measurable functions' I, so a little Lebesgue theory has crept into our text, but we could work through completions of Coo(lR), as before.) For I, 9 E Ll (lR+, w), define 1* 9 as in (*). Then it is easy to check that (Ll(lR+,W); 11·ll w ; *) is a commutative Banach algebra. D
Example 4.68 To give an example of another commutative Banach algebra A in which J(A) =I {O}, we let V be the Banach space (Ll(lI); 11.11 1), with multiplication defined as (J, g) f+ I * g, where now I * 9 is the 'choppedoff' convolution product of I and 9 specified by
(J
* g)(x) =
l
x
I(x  t)g(t) dt
(0:::; x :::; 1).
It is easy to verify that V is a commutative Banach algebra without an identity; it is the Volterra algebra. Indeed, V is an obvious quotient of the Banach algebra (Ll(lR+); 11·111; *) by a closed ideal. We set A = V+, the unitization of V. Suppose now that I E V and that, for some 0 > 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(e1/;)9(1/;)d1/;)e inlJ de
=
(2~ fo27r f(e _1/;)e in (IJ1/;) 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
F: f
f+
f,
L2(1R)
+
L2(1R),
and we established some of its properties in Theorem 2.94. In particular, we showed that, for a continuous function f E Ll(lR) n L2(1R), we have
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Introduction to Banach Spaces and Algebras
[(t)
= _1_ ,J27r
/00 00 f(x)e
ixt
dx.
We now define F(f)(y) = [(y) = 1 : f(t)e iyt dt
(f
Ll(JR)).
E
(Thus we no longer require that f be continuous; the integral is defined as in Section 2.6. Also, we have changed our notation slightly and have removed the arbitrary constant 1/,J27r; the latter was a natural normalizing factor in the case where f E L2(JR), but is now not so aesthetic in the context of Banach algebras.) Theorem 4.70 (RiemannLebesgue lemma) Let f E Ll(JR). Then 1 E Co(JR), and F: (Ll(JR); 11.11 1 ; *) > (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 (YlY2)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 translationinvariant 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 translationinvarwnt.
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
j7r7r F(e)e mo de = k~OO j7r7r 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) <;;; Co (IR). For take (an)nEz E co(Z) \ A(Z) (see Exercise 3.17). Then there exists 9 E Co(lR) such that g(n) = an (n E Z). Assume that there exists f E Ll(JR) with = g. Then there exists F E Ll('JI') with F(n) = !(n) = an (n EN), a contradiction.
1
200
Introduction to Banach Spaces and Algebras
Now take a > 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 righthand 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
. Then there exists fo E L1 (JR) such that
1j;(t) = cp(Sdo)j
(t E JR).
Let s, t E R Since Ss+d * f = Ssf * Sd for each f E L1(JR), as is easily checked, we have 'I/)(s + t)cp(fO)2 = 1j;(s)cp(fo)1j;(t)cp(fo), and so 1j;(s + t) = 1j;(s)1j;(t). Certainly 1j;(0) = 1. By Proposition 4.71 (iii) and the fact that
(t
E
L
JR) ,
and so a is differentiable on JR and 1j;(t) = a'(t) (t E JR). Let t E R Since IIX[t,t+hl/hll l = 1 (h E JR+ e ), it follows that 11j;(t) I :::; 1. Since 1j;(t)1j;(t) = 1, we have 11j;(t)1 = 1, and so Rez = O. Thus 1j;(t) = e iyt (t E JR) for some unique yE R Take a, b E JR with a < b, and set X = X[a,bl' Then
cp(X) = a(b)  a(a) = lb a'(t) dt = lb 1j;(t) dt
= Ib e iyt dt = a
Joo x(t)e iyt dt = x(y) . 00
Since the linear span of such functions X[a,bl is dense in (L1(JR); 11.11 1 ), we see that
onto R We still have to verify that the Gel'fand topology of coincides with the usual topology on R For this, it is sufficient to show that the mapping
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.
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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 <:;;; [n, n] and r III) III < c for j = 1, ... , m. Then it is immediately checked that V(n, r) <:;;; N, giving the claim. But now, for each n E Nand r > 0, we see that !Py E V(n, r) if and only if IYI < (2/n)sinl(r/2), and so the two topologies do coincide. 0 We see that the Gel'fand and Fourier transforms of Ll (JR) again coincide when we have made the identifications of the above theorem. 4.13 Boundaries and peak sets. Let A be a natural Banach function algebra on a compact space K. A subset E of K is a peak set for A if there exists I E A such that
I(x)
=
1
(x
E
E)
and
II(x)1 < 1 (x
E
K \ E);
the function I peaks on E. A point Xo E K such that {xo} is a peak set is a peak point, and the set of peak points is denoted by ro(A). A subset E of K is a boundary for A if, for each I E A, there exists x E E with II(x)1 = III K ; a closed boundary is a closed subspace of K that is a boundary. Clearly, each boundary contains ro(A). For I E A, define the maximum set of I by
M(f) = {x
E
K: II(x)1 = III K
}·
We shall show that, in the general case, the intersection rcA) of all the closed boundaries for A is itself a boundary for A (and so rCA) 1= 0). Let x E K. Clearly, x E rCA) if and only if, for each neighbourhood U of x, there exists I E A with M(f) <:;;; U, and so rCA) = rCA), where 11 is the uniform closure of A in C(K). We first give some examples. Examples 4.76 (i) Let (K; d) be a nonempty, compact, metric space, and let A = C(K). For Xo E K, the function Y f7 1/(1 + d(xo, y)) belongs to A and peaks at xo, and so ro(A) = rCA) = K. (ii) Let A = A(~) be the disc algebra, as in Example 4.3. The maximum modulus theorem, Proposition 1.35, shows that '][' is a closed boundary for A. However, take Zo E ']['. Then the function (Z + zol)/2zo E A peaks at zo, and so every point of'][' is a peak point. Thus clearly ro(A) = rcA) = ']['. (iii) Let K be a nonempty, compact subset of C, and let A = R(K), as in Example 4.51. By the maximum modulus theorem, 8K is a closed boundary for A. Take Zo E 8K, and let U be a neighbourhood of Zo in
Banach algebras
203
Lemma 4.77 Let A be a natural Banach function algebra on a nonempty, compact Hausdorff space K, and take 11, ... , fn EA. Set
v = V(J"
.. ,fn) = {x
E
K : IfJ(x)1 < 1 (j = 1, ... , n)}.
Then either V n E =I 0 for each boundary E, or else F \ V is a closed boundary for A whenever F is a closed boundary for A. Proof Suppose that there is a closed boundary F such that the closed set F\ V is not a boundary. We must show that V n E =I 0 for each boundary E. There exists f E A with If IF = IflK = 1, but with IflF\v < 1. By replacing f by a suitably high power of itself, we may suppose that
If fJ IF\V < 1
(j = 1, ... , n) .
For each j = 1, ... , n, we have If fJ (x) I < 1 (x E V), and hence If fJ IF < 1. Thus IffJIK < 1 because F is a boundary. Take Xo E M(f), so that If(xo)1 = 1. Then IfJ(xo)1 < 1 (j = 1, ... , n), and so Xo E V. Hence M(f) <:;; V, and so VnE =I 0, as required. 0 Theorem 4.78 Let A be a natural Banach function algebra on a compact space K. Then r(A) is a closed boundary for A. Proof Certainly f(A) is a closed subset of K (at this stage, possibly empty). Assume towards a contradiction that r(A) is not a boundary. Then there exists f E A with IflK = 1 and M(f) n r(A) = 0. Let y E M(f). Then there is a basic open neighbourhood, say of the form V = V(J, ,... ,fn)' of y and a closed boundary E for A with V n E = 0. By Lemma 4.77, F \ V is a closed boundary whenever F is a closed boundary. However, these sets V together form an open cover of the compact space M(f); let {VI, ... , Vn } be a finite subcover of this open cover. Certainly K is a closed boundary; hence K \ VI is a closed boundary, and, repeating the argument n  1 times, K \ (VI U· .. U Vn ) is a closed boundary. Since M(f) <:;; VI U ... U Vn , this is a contradiction. Thus r( A) is a boundary. 0
The set r(A) is called the Shilov boundary of A. We now wish to prove that ro(A) is a boundary, and hence dense in r(A), whenever K is metrizable and A is a uniform algebra. Lemma 4.79 Let A be a natural uniform algebra on a compact space K. Suppose that f,g E A are such that: (i) IflK = 1; (ii) IgIM(J) = 1; (iii) L =I 0, where
L = {x Then there exists h
E
E
K : g(x)
A such that M (h)
= f(x) = I}.
= L.
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Introduction to Banach Spaces and Algebras
Proof We may suppose that f(x) = 1 (x E M(f)), for otherwise we replace f by (1 + f)/2 E A without changing anything else. Set m = IgIK' so that m 2 1. If m = 1, set h = (1 + g)f /2, and then h E A and M(h) = L, as required Now suppose that m > 1. For n E N, set
Kn = {x E K: 1 + Tn(m  1) :::; Ig(x)1 :::; 1 + Tn+l(m  I)}. Each Kn is closed (possibly empty) in K and, by hypothesis (ii), If(x)1 any x E Kn. Further,
U{Kn : n E N}
=
< 1 for
{x E K: Ig(x)1 > I}.
For each n E N, there exists Pn E N with IfP" (x)1
< 1/2 for any x
E Kn
Define
=
h = g + 4(m  1) L T n f P"
;
n=l
we note that the series converges in A because IfP" IK We now consider two cases for a point x E K. First, suppose that Ig(x)1 :::; 1. Then
Ih(x)l:::; Ig(x)1
+ 4(m 
= 1 (n
EN).
=
1) LTn If(xW" :::; 1 + 4(m  1). n=l
This inequality is strict unless Ig(x)1 = If(x)1 = 1. But f(x) = 1 when If(x)1 = 1, and so the inequality is strict unless also g(x) = 1. Thus, for such a point x, we have Ih(x)1 = 1 + 4(m  1) if and only if x E L. Secondly, suppose that Ig(x)1 > 1. Then x E Kr for some r E N, and so Ig(x)1 :::; 1 + 2 r + 1 (m  1) and If(xW' < 1/2, which gives
Ih(x)1 :::; Ig(x)1
+ 4(m 
1) (Tr If(xW'
+ LTn) nfr
=
Ig(x)1
+ 4(m 
< 1 + Tr+l(m 
+ 1  Tr) 1) + 4(m  1)(T r  1 + 1  Tr)
1) (Tr If(xW'
=
1 + 4(m  1).
We conclude that IhlK = 1 + 4(m  1) and M(h) = L, as required.
0
Theorem 4.80 Let A be a uniform algebra on a compact, metrizable space. Then the set ro(A) of peak points is a boundary for A, and ro(A) = r(A). Proof We suppose that A is a uniform algebra on K, a compact, metrizable space. Let fo E A. We must show that M(fo) contains a peak point. Take J to be the family of all maximum sets contained in M (fo) (so that J I 0), and order J by inclusion, so obtaining a partially ordered set. Let It
Banach algebras
205
be a chain in J, and set M = n{N : N E It}. Certainly M =I 0 because K is compact. Since K is metrizable, M is in fact the intersection of countably many sets in It, say M = n{M(h n ) : n EN}, where Ihnl K = 1 (n EN). Choose Yo E M. By multiplying each h n by a suitable constant of modulus 1, we may suppose that hn(yo) = 1. Set h = L~=lhn/2n, so that h E A and h(yo) = 1 = IhIK. Since Ihn(y)1 < 1 for y E K \ M(h n ), we see that M(h) <;;; M, and so M(h) is a lower bound for It in (J; <;;;). By Zorn's lemma, bis, J has a minimum element, say M(J), where f E A. We may suppose that IflK = 1. Assume towards a contradiction that M(J) contains more than one point. Then there exists 9 E A that is not constant on M(J); again we may suppose that IgIM(f) = 1. By multiplying f and 9 by constants of modulus 1, we may suppose that f(x) = g(x) = 1 for some x E M(J). Thus L <;; M(J), where L is as in Lemma 4.79. However, by Lemma 4.79, L E J, a contradiction of the minimality of M(J). Hence M(J) is a singleton, say M(J) = {xo}. Clearly, Xo E M(Jo) n fo(A), and so f 0 (A) is a boundary for A. Certainly fo(A) <;;; f(A). But fo(A) is a closed boundary for A, and so fo(A) ;;2 f(A). Thus fo(A) = f(A). D Let A be a unital Banach algebra. Then we set
KA = {'\ E A*:
PII = '\(1) =
I},
so that KA is a closed, convex subset of (A*)[lJ' and hence KA is compact in (A*;O"(A*,A)). By Theorem 4.43, A <;;; KA. In particular, let A be a uniform algebra on a compact, Hausdorff space K. Then we regard K as a closed subset of KA, In the next theorem, closures are taken in the weak* topology.
Theorem 4.81 Let A be be a uniform algebra on a compact space K. Then conv(fo(A)) = KA and fo(A) <;;; exKA <;;; f(A). Proof Clearly, conv(fo(A)) <;;; KA; we shall first prove the converse in the special case where A = C(K). Set E = ClR(K), a reallinear space. Take'\ E C(K)* with 11,\11 = '\(1) = 1. We borrow a result from Lemma 6.20 that shows that '\(J) E lR for all fEE, and hence that ,\ I E is a reallinear functional; clearly,\ lEE KE = {JL E e* : IIJLII = JL(1) = I}. Assume towards a contradiction that there exists'\o E KE with'\o conv(K). By Theorem 3.26(iii), there exists A E (E*; O"(E*, E))* = E such that A('\o) > 1 and SUPxEK A(cx) ::::: 1, and so there exists fEE with
rt
'\0(J) > 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»,
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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 KreinMilman theorem, Theorem 3.31, Ko contains an extreme point of K A . Since ex KA <;:; K, there exists x E K with x E exKA n Ko. Thus f(x) = 1, and so x = Xo. Hence Xo E exKA. We have shown that ro(A) <;:; exKA. D Notes A more general version of Theorem 4.46 states that, for each algebra A, the map 'P f+ ker'P is a bijection from CPA onto the set of maximal modular ideals of co dimension 1 in A [47, Proposition 1.3.37]. A maximal ideal in even a commutative, unital algebra does not necessarily have co dimension 1. For example, there are many examples of fields_ven ordered fields [47, Definition 1.3.61]of large dimension, and {OJ is a maximal ideal in these fields. These 'large' fields do have an important role in Banach algebra theory; see [47, 51]. Indeed, there are 'very large' algebras analogous to algebras of formal power series [51]. We have defined a 'Banach algebra of power series' (A; 11·11) within Example 4.52; the definition requires that the maps 7l"m : (A; 11·11) > C are all continuous. In fact, it is a recent result [55] that this timehonoured phrase is redundant. For a discussion of Banach algebras of power series, including a determination of their family of closed ideals in certain cases, see [47, Section 4.6]. We have defined the product in the Banach algebras Ll(1r) and Ll(JR) without using the theory of Lebesgue integration. Take G to be 1r or JR. The formulae given also define f * 9 directly for f,g E Ll(G). For this, one must show using Fubini's theorem that f * 9 E Ll(G) (and hence that IU * g)(x)1 < (Xl almost everywhere) for each f,g E Ll(G). Care must be taken over some measuretheoretic points. See [143, Section 7.13], for example. For f ELI (JR +) with f =1= 0, define aU)
= sup{ 8 ~ 0 : f I [0,8] = 0 almost everywhere} = inf supp f .
In Example 4.68, we essentially used the obvious fact that aU * g) ~ aU) + a(g) for f,g E L 1 (JR+) with f,g =1= O. In fact, it is a deep theorem of Titchmarsh (see [47, Theorem 4.7.22], for example) that aU * g) = aU) + a(g) for f,g E L1(IR+) with f,g =1= O. This implies that (Ll(IR+,W); *) is an integral domain for each continuous weight function w on 1R+. Let G be a locally compact group, so that G is a group with a locally compact topology and the group operations are continuous with respect to the topology. For a E G and f on G, define (Saf)(S) = f(a1s) (s E G). Then there is a socalled left Haar measure m on G such that m =1= 0 and fG Saf dm = fG f dm for all a E G and f E Ll(G) = Ll(G, dm). For example, the Haar measure on (IR; +) is just the
Banach algebras
207
Lebesgue measure. The convolution product of formula
(f
* g)(s) =
fa
I, gEL 1 (G)
l(t)g(C l s) dm(t)
is now defined by the
(s E G).
In this case, (L 1(G); II . 111 ; *) is a Banach algebra; it is commutative if and only if G is abelian. It is shown in [47, Corollary 3.3.35] that Ll(G) is always semisimple. Next, suppose that G is abelian, so that G is an LeA group. A character on G is a group morphism from G to 1I'. The set of all continuous characters on G forms a group with respect to the pointwise product; this is called the dual group of G, and it is often denoted by r = G. The dual group r is itself an LCA group for a suitable topology, and the famous Pontryagin duality theorem states that the dual group of r can be identified with G. For example, it is implicit in our work that Z = 1I', that 'if = Z, and that i. is a 'second copy' of R The Fourier transform of 1 E L1(G) is the function on r defined by
1(r) =
fa 1
(s)')'( s) dm(s)
(r
E
r, 1 E
L1(G)).
1
The (generalized) RiemannLebesgue lemma states that E Co(r), and then the Fourier transform is the map :F : 1 f> Ll(G) > Co(r). As in each of the specific examples that we have considered, this map is a monomorphism and the character space of Ll(G) can be identified with r in the sense that each character on Ll(G) has the form 1 f> i( 'Y) for some (unique) 'Y E r. The identification of the character space of Ll(G) with r is a homeomorphism, and, with this identification, the Gel'fand transform coincides with the Fourier transform. The details of the above are given in various texts, including [47, 84, 94, 106, 145]. Our approach to the specific examples is rather close to that in [82]. In general, it is not true that E Ll(r) when 1 E L1(G). However, if E L 1 (r), there is an 'inversion formula' that recovers 1 from j Let K be a nonempty, compact subset of the plane. We have remarked that it is very easy to see that the uniform algebra R(K) is natural on K. It is true, but harder to prove, that the uniform algebra A(K) is also natural on K; this is a theorem of Arens, proved in [47, Theorem 4.3.14] and [79, Chapter II, Theorem 1.9]. There is a substantial study of when any two of the algebras R(K), A(K), and C(K) are equal in [79, Chapter VIII]; these questions encompass a considerable amount of classical approximation theory. Mergelyan's theorem [79, Chapter II, Theorem 9.1] asserts that P(K) = A(K) whenever e \ K is connected, and the HartogsRosenthal theorem [79, Chapter II, Corollary 8.4] asserts that R(K) = C(K) whenever K has plane measure zero. Indeed, it is a theorem of Bishop [79, Chapter II, Theorem 11.4] that R(K) = C(K) if and only if the set of points that are not peak points has plane measure zero. An example for which R(K) i A(K) is given in Exercise 4.16, below; criteria involving the 'capacity' of sets for R(K) = C(K) and for R(K) = A(K), and various exotic examples, are given in [79, Chapter VIII] and [152]. Conditions for a point in 8K to be a peak point for R(K) or A(K) are given also in [79, Chapter VIII]. There is one obvious example of a natural uniform algebra on the closed unit interval I, namely C(I). It is a conjecture that goes back to Gel'fand that this is the only natural uniform algebra on I. Let A be such an algebra. We do not even know if every point of 1 must be a peak point for A; however, we shall see in Corollary 9.12 that necessarily r(A) = I. Further, we do not know whether or not necessarily A = C(I) if we assume that ro(A) = I. This question is related to that of polynomial approximation on curves in en; see [152, Section 30J. Here is a related tantalizing open question of considerable
1,
1
1
208
Introduction to Banach Spaces and Algebras
antiquity. Let A be a natural uniform algebra on the closed unit disc ~. Does the Shilov boundary r (A) necessarily meet (or contain) the topological boundary '][' ? The proof in Theorem 4.80 is essentially due to Bishop. In the nonmetrizable case, one can replace 'peak point' by 'strong boundary point'; the set of these points is dense in r(A). Let A be be a natural uniform algebra on a compact, Hausdorff space K. Then we showed in Theorem 4.81 that ro(A) <;;; ex KA. In fact, ex KA is equal to the set of strong boundary points for A (and hence to ro(A) in the case where K is metrizable). This set ex KA is called the Choquet boundary for A; it is characterized in several different ways in [47, Theorem 4.35]. A famous theorem of Choquet is relevant here. It states the following. Let K be a metrizable, compact, convex subset of a locally convex space E, and let Xo E K. Then there is a probability measure IL on K such that f(xo) = K fdlL (f E E*) and IL is supported on exK. For a natural Banach function algebra A on a compact, metrizable space K, the set ro(A) is not necessarily a boundary for A [43], but it is still true that ro(A) = r(A) [47, Corollary 4.3.7]. The socalled peakpoint conjecture was outstanding for a long period. Let K be a compact, metrizable space. This conjecture stated that the only natural uniform algebra A for which ro(A) = K is A = C(K). In other words, for each proper, natural uniform algebra on K, there exists x E K that is not a peak point for A. This conjecture was shown to be false by Cole in his thesis [41]; see also [152]. An explicit counter example was given by Basener [26] a little later; it is the uniform algebra R(K) for a certain compact subset K of the unit sphere in C 2 • See [152, Example 19.8]; for related results of Feinstein, see [76]. We have given various standard results for commutative Banach algebras. Some, but not all, extend to Fn§chet algebras and other topological algebras. For example, there is an appropriate Gel'fand representation theorem for complete, commutative LMC algebras. However, there are examples that show that certain statements do not carry through; see [47, Section 4.10].
J
Exercise 4.11 (i) Let K be a nonempty, compact Hausdorff space, and let F be a closed subset of K. Set
J(F) = {J E C(K) : f I F = O} . Show that J(F) is a closed ideal in C(K), and that every closed ideal in C(K) has this form. (ii) Let K and L be nonempty, compact Hausdorff spaces. Prove that the Banach algebras C( K) and C( L) are isomorphic as algebras if and only if the topological spaces K and L are homeomorphic. Exercise 4.12 (i) Take kEN, and let A = (Ck(H); II· Ilk) be the Banach space introduced in Exercise 2.10. (See also Exercise 2.19.) Show that A is a natural Banach function algebra on H with respect to the pointwise multiplication of functions. (ii) Let A = CCXl(H) = n{Ck(H) : kEN} be the space of infinitely differentiable functions on H. Show that A is a Fn§chet algebra with respect to the sequence (II '1I kh2':l of seminorms on A and pointwise multiplication of functions. Show that A is functionally continuous, and that all characters on A are given by evaluation at a point of H.
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Banach algebras
Exercise 4.13 Let (K; d) be a compact metric space, and take ex with 0 < ex S 1. Then let A = Lip"K be the Banach space of Lipschitz functions on K; this space was introduced in Exercise 2.13. Show that A is a natural, selfadjoint Banach function algebra on K with respect to the pointwise multiplication of functions. Exercise 4.14 (i) Let bv be the space of sequences x = (Xn)n2:1 in co(N) such that 00
Ilxll bv = Ilxli oo
+L
IXn+1  xnl <
00.
n=l
Show that (bv; 11·ll bv ) is a Banach sequence algebra on N. (ii) We defined the James space (J;N) in Exercise 2.14. Let x,y E J. Show that xy E J with N(xy) S 2N(x)N(y), and hence conclude that J is a Banach sequence algebra on N for a norm equivalent to N. Exercise 4.15 Let U be a nonempty, open subset of the complex plane IC, and let (HOO(U); 1·l u ) be the Banach space that was introduced in Exercise 2.12. Prove that this space is a Banach algebra with respect to the pointwise multiplication of functions. This Banach algebra is studied extensively in the monograph of Garnett [80]. Let Ez : J f+ J(z) be the evaluation character on HOO(U) for z E U, and let
+ ... + IJn(z)1
: z E U} > 0,
there exist gl, ... ,gn E H oo (U) such that 2:: 1J,g, = 1. The fact that this condition is indeed satisfied in the case where U is the open unit disc in IC is the famous corona theorem of Carleson [37]; for an exposition of this theorem, based on an approach of Wolff, see [80, Chapter VIII, Section 2], and, for a simpler proof, see [28]. Exercise 4.16 A Swiss cheese is a nonempty, compact subset of IC obtained from ~, the closed unit disc, by deleting a sequence (Dn)n2:1 of open discs, with each Dn contained in int~; we also require that Dm n Dn = 0 whenever m, n E N with m =1= n and that 2:~=1 Tn < 1, where Tn is the radius of Dn. We then set
K
=
~
\
U{ Dn : n E N} ,
so that K is a compact subset of ~. Give the details to show that, starting with a countable, dense subset S of int~, an inductive construction gives a sequence (Dn)n2:1 satisfying the above conditions, with the centre of each Dn in S, and so that K has empty interior. It follows that A(K) = C(K) in this case. However, we claim that R(K) =1= C(K). To see this, consider the continuous linear functional
A:J
f+
1
8~
J(z) dz 
f= 1
J(z) dz
n=l 8Dn
on C(K). Verify that A is indeed an element of C(K)* and that (j, A) = 0 for each rational function J with poles off K, and hence that A I R(K) = O. However, show that
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Introduction to Banach Spaces and Algebras
(z, A) This shows that
Z
=
27ri (1  ~ rn) =I o.
~ R(K), and so R(K)
=I C(K).
Exercise 4.17 (i) We have discussed the algebra ~ = C[[X]] of formal power series in one variable in Example 4.52. This algebra is an integral domain, and so we know from standard courses in algebra that it has a quotient field. Verify that this quotient field can be identified with the field q(X)) of Laurent series in one variable; the latter is the algebra of formal series of the form L~=no anX n , where no E Z and each an E
(ii) Let IC{ X} be the collection of absolutely convergent power series, so that IC{ X} consists of the formal power series L~=o anXn E ~ such that L~=o Ian Ien < 00 for some 10 > O. Verify that IC{ X} is a subalgebra of ~ and that its quotient field, the field qX) of meromorphic functions, is a subfield of q (X)). Is there a norm II . II such that (IC{X}; 11·11) (respectively, (qX); 11·11)) is a normed algebra? Exercise 4.18 Let w be a weight function on Z+ , and let A = eI (Z+ ,w) be the Banach algebra of power series, as defined in Example 4.52. Set P = PA(X) = limn>oo w(n)l/n. Show that SPA X = {Z E IC : Izi ::; p} = iPA, say SPA X = K. Show that A <;;; P(K) = A(K) and that A is semisimple whenever p> O. Exercise 4.19 For t E (0,1), set f(t) = 1/0 and get) = 1/yff"=t, and, for t:O:: 1 and t ::; 0, set f(t) = get) = O. Show that f and g are Riemannintegrable on JR, but that the integral for (f * g)(1) does not converge. Exercise 4.20 Let A be a Banach algebra. A (sequential) approximate identity in A is a sequence (a n )n2:1 in A such that aa n > a and ana > a as n > 00 for each a E A. Show that each 'approximate identity' as defined in Exercise 2.36 is an approximate identity in the Banach algebra (L I ('f); II . III ; * ). Exercise 4.21 Let n E N. Compute explicitly the function hn := XIn,n] Show that h n = fn (n EN), where
fn(t) = sin tsin(nt) 4t 2
Show that
Ilfnlll
> 00
as n
> 00,
but that IhnllR
*
XII,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 translationinvariant 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 nonempty, 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
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Introduction to Banach Spaces and Algebras
Proof (i) Set B = R(K)(Z, U/3), a closed subalgebra of A = R(K), and take U to be the component of e \ K to which a and f3 belong. Since K = SpAZ, the set U is a component of e \ SPAZ. But u/3 E B, so that f3 ~ SpBZ, and then, by Corollary 4.37(i), Un Sp BZ = 0. Thus a ~ SpBZ, i.e. Un E B. (ii) By Corollary 4.37(ii), the unbounded component of e \ K does not meet SpcZ for any closed subalgebra C of A that contains Z. In particular, this applies with C = P(K), and therefore Un E P(K) for every a in this unbounded component. 0
Let K be a nonempty, compact subset of C. Then we denote by OK the algebra consisting of all equivalence classes of functions analytic on some neighbourhood of K. [To be precise, for f, 9 E OK, we set f rv 9 to define the equivalence relation if there is an open neighbourhood W of K with f I W = 9 I W E O(W); further, if f E O(U) and 9 E O(V), where U and V are open neighbourhoods of K, we define f + 9 and fg pointwise on Un V; the algebraic operations are compatible with the equivalence relation, and so the family of equivalence classes is an algebra.] The elements of OK are called germs of analytic functions. There is a way of defining a topology on the algebra OK by regarding it as the inductive limit of the Fn§chet algebras O(U) as U runs through the family of open neighbourhoods of K. For us, it is sufficient to say that a sequence (fn)n>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 nonempty, 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.
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Banach algebras
Corollary 4.84 Let K be a nonempty, 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 nonempty, 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 nonempty, 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 nonempty, 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 nonempty, 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 f4 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
r
f4
r(a) ,
R(U)
~
A,
is a homomorphism. It will be shown in the next lemma that this mapping is continuous, and then Corollary 4.86 will give a unique extension to a continuous, unital homomorphism 8 a : O(U) ~ A such that 8 a (Z) = a. Notice that, for every r E R(U), it is clear that r(a) E C := {a}CC, the bicommutant of {a} in A; since C is closed, it will then also follow that 1m 8 a ~ C. Now let a E A. Then we may define 00
exp a == ea =
L n=O
an n! '
where a O = 1. Note that the convergence of this series is an immediate consequence of the completeness of A, with the inequality Ilanll : : ; Iiali n (n EN). In a similar way, we may define sin a and cos a in A. More generally, we may define an element J(a) E A for each entire function J: if J has the Taylor expansion J(z) = L~=o An Zn for z E C, then J(a) = L~=o Anan. It is easily checked that the map
8: J
f4
J(a) ,
is a unital homomorphism with 8( Z)
O(C)
~
A,
= a.
Lemma 4.87 Let A be a Banach algebra with an identity, and let a E A. Take "I to be a contour in C \ Sp a such that n( "I; z) = 1 Jor all z E Sp a. Then
2~i I(A1 y
a)l dA
=
1•
Proof Let I be the lefthand side of the above equation, noting that the existence of this integral is ensured by Theorem 3.10 and Corollary 4.12. Take
215
Banach algebras
R > Iiall, and let fR be the circle given by fR(t) = Re it (0 :::; t :::; 27r). Then n(fR;z) = 1 for all Z E Spa, and thus Theorem 3.11(i) shows that
I=~ 27rI
r (>.Ia)ld>..
irR
But, on the circle [rR], we have (>.1  a)l = 2:::=0 >.(n+l)a n , with uniform convergence. Termwise integration then immediately gives I = 1, the identity of A. D
Lemma 4.88 Let A be a Banach algebra with an identity, and let a E A. Let U be an open neighbourhood of Sp a, and let "( be a contour that surrounds Sp a zn U. Then, for every r E R(U), we have r(a) = Further, the homomorphism r
~1 2m
f>
r(>') (>.1  a)l d>.. "Y
r(a), R(U)
+
A, is continuous.
Proof By elementary algebra, there is an analytic Avalued 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 <:;;; {ay n {a}ee = {ay n C; (iv)
E
<1>c);
216
Introduction to Banach Spaces and Algebras
Proof It is clear from our preliminary remarks and Lemma 4.88 that there is a continuous, unital homomorphism 8 a : O(U) 4 A such that 8 a (Z) = a, and hence such that (ii) holds. Now suppose that () is any such map. Then, by elementary algebra, it follows from the fact that ()(Z) = a that we must have ()(r) = r(a) = 8 a (r) for every r E R(U). By Corollary 4.86, R(U) is dense in O(U), and so () = 8 a . Thus 8 a is uniquely specified by the stated conditions. It also follows that the formula of (i) holds for every f E O(U). Since Z f = f Z in O(U), we have a8 a (J) = 8 a (J)a, and so im8 a ~ {aVo It is further clear that r(a) E C for every r E R(U), and so, since C is closed in A, we have (iii). It remains to prove (iv). Let f E O(U) and
(2~i
1
2~i
f(A)(A1  a)l dA) =
1
f(A)(A 
and so
=
8;.' (J I V)
(J
E
E
A. Suppose that U
O(U)) .
Proof The mapping f 1+ 8:; (J I V), O(U)  4 A, is evidently a continuous homomorphism mapping Z to a. The result follows from the uniqueness statement in Theorem 4.89. 0
217
Banach algebras
This last lemma justifies us in writing 'Sa' rather than 8~: if we simply have a function J analytic on an unspecified neighbourhood of Spa, then Sa(f) is well defined by setting it equal to 8~ (f) for any open neighbourhood U of Sp a in C on which J is defined. Thus we have a unital homomorphism
8 a : Osp a
+
A
such that 8 a (Z) = a, and this homomorphism is continuous when OSpa has the topology described on page 212. Further, 8 a is the unique continuous homomorphism with the specified properties. The convention of writing J(a) := 8 a (f) is likewise unambiguous. In preparation for the next proposition, we remark that, if A is a commutative, unital Banach algebra and at, a2 E A satisfy at = a2, then it follows that SPA (at) = SPA(a2). Let U be an open set in C with U ::::l SpA(at}. Then there are the two functional calculus homomorphisms 8 al : J f+ J(at) and 8 a2 : J f+ J(a2) from O(U) into A. Proposition 4.92 Let A be a commutative, unital Banach algebra, and take elements at, a2 E A with at = a2. Let U be an open set in C with U ::::l SPA (at). Suppose that J E O(U) is such that
J(at) = J(a2) ,
while
f'(z)
# 0 (z
E
SpA(at}).
Then at = a2. Proof Write K = SPA(at) = SPA(a2), and let "( be a contour in U \ K that surrounds K, but does not surround any point of C \ U. Then we have
0= J(at)  J(a2)
=
=
1 ~1 ~
J(z)((zl at}t  (zl  a2)t) dz
2m
'Y
2m
'Y
J(z)(zl  at)t(zl  a2)t(at  a2) dz
=u·(at a2), where
u=
~ 2m
1
J(z)(zl  at}t(zl  a2)t dz.
'Y
Let t.p E
.j J(z)(z  a)2
1 t.p(u) = 2 7rl
'Y
dz
= a2, say a = t.p(at} = t.p(a2). = 1'(a) # 0
since a E K. It follows from Theorem 4.59 that u is invertible in A, so that
at  a2 =
o.
D
218
Introduction to Banach Spaces and Algebras
Proposition 4.93 Let A be a unital Banach algebra, and let a E A. Suppose that f E OSpa and that an + a in A. Then f(a n ) + f(a). Proof We may suppose that f E O(U), where U is an open neighbourhood of Sp a in
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
f4
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 nonempty, closed subsets K and L. Then there is a nontrivzal idempotent pEA such that ap = pa and such that Sp(pa)=KU{O}
and
Sp(apa)=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 nontrivial idempotent. Also, by Theorem 4.S9(iv), Sp (pa) = K U {O} andSp(apa)=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 nonempty, closed subsets K and L. Then there are nonzero 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 nonzero ,X E SpT is an eigenvalue and that the eigenspace E('x) is finitedimensional. 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 righthand halfplane, 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 Avalued 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
<
1. Then Spa
c
D, and loga is
10ga=_I=(1a)n n=l
n
Thus we obtain the following result, which will be referred to later. Proposition 4.101 Let A be a unital Banach algebra. Then expA contains the neighbourhood {a E A : 111  all < I} of 1. 0 The following is an attractive curiosity. Theorem 4.102 (Kahane and Zelazko) Let A be a commutative, unital Banach algebra, and let f be a linear functional on A with f (1) = 1 and such that f (exp a) =f. 0 (a E A). Then f E A· Proof We first claim that f is continuous and that Ilfll = 1. For assume that there exists a E A with Iiall < 1 and f(a) = 1. By Proposition 4.101, there exists bE A with exp b = 1 a. By hypothesis, f(la) = f(exp b) =f. 0, a contradiction. This gives the claim. For a E A, set G (z) = f (exp za) (z E q. Then G is an entire function, say
=f
G(z)
(~ z:~n) = ~ znf(,a n )
(z E
q.
N ow we see that IG(z)1
s::
I= Izl n=O
n
Ii,a n.
r = exp(lzlllall)
(z
E
q.
(*)
By hypothesis, G has no zeros, and so, by Proposition 1.37, G = exp F for some entire function F. Since G(O) = 1, we may suppose that F(O) = O. By (*), ReF(z) s:: Iziliall (z E q. By Theorem 1.40, F = aZ for some a E C, and so
= exp(az) =
n
anz L ,n. 00
G(z)
(z E
q.
n=O
Thus f(a n ) = an = f(a)n (n E N) by comparing coefficients in the two expansions of G(z); in particular, f(a 2 ) = f(a)2. This implies that f E A. 0
223
Banach algebras
4.16 The principal component of the group of units. This section follows on from Section 4.4; it comes at the present point because it makes some use of the functional calculus. Let A be a Banach algebra with an identity, and let G = G(A) be the group of units of A, topologized as a subset of A. We know that G is an open subset of A and that the mapping a f+ a 1, G  t G, is a homeomorphism. Of course, we also know that the multiplication map (a, b) f+ ab, G x G  t G, is continuous because it is the restriction of the multiplication map A x A  t A. It then follows that, for any given a E G, the maps b f+ ab and b f+ ba are homeomorphisms of G onto itself. (These statements may be summarized by saying that G is a topological group; but we shall not assume any knowledge of the theory of such structures. ) We define the principal component Go = Go(A) of G to be the component of G that contains the identity element 1 of A. Lemma 4.103 Let A be a unital Banach algebra. Then expA <;;; Go. Proof Let a E A. Consider the mapping f : t s,t E 1I, we have
Ilf(s)  f(t)11
=
f+
exp(ta), 1I
t
G. For each
Ilexp(ta) (exp (s  t)a  1)11 :::; ellall (elstiliall  1) ,
and so f is continuous by the continuity of the real exponential function. Thus f defines a continuous path in G from 1 = f (0) to exp a = f (1). This shows that expa EGo. D Lemma 4.104 Let A be a unital Banach algebra. For each a E G(A) and bE A such that lib  all < Iialll l , we have alb E exp A. Proof As in Corollary 4.11, b = a(1  a l (a  b))). Since Ilal(a  b) II < 1, we D have alb = 1  alea  b) E expA by Proposition 4.101. Theorem 4.105 Let A be a unital Banach algebra. Then: (i) Go is an openandclosed, normal subgroup of G(A), and the components of G are precisely the cosets of Go in G(A); (ii) Go consists of all finite products of exponentials, so that
Go = {exp(adexp(a2)·· ·exp(ak): al, ... ,ak E A, kEN};
(iii) in the case where A is commutative, Go = exp A. Proof (i) Set G = G(A). Recall that the continuous image of a connected topological space is connected. Define fL: (a, b)
f+
ab l
,
Go x Go
t
G.
Then fL is continuous, so that fL(G o x Go) is a connected subset of G. Also, fL(G o x Go) contains 1 = fL(l, 1). Thus fL(G o x Go) ~ Go, and this shows that
Introduction to Banach Spaces and Algebras
224
Go is a subgroup of G. Similarly, for each bEG, the mapping D:b : a f+ b1ab is a continuous map, and D:b (1 A) = 1A, so that again D:b ( Go) ~ Go. Thus Go is a normal subgroup of G. That Go is closed in G is a general result of pointset topology; it holds simply because Go is a component of G. This completes the proof of (i), except for showing that Go is open and giving the description of the components of G. We shall first prove clause (ii). (ii) By Lemma 4.103, exp A ~ Go. Let E be the set of all finite products of exponentials. Evidently, E is a subgroup of G, and so, by (i), E ~ Go. We shall next show that E is open in G, from which the rest of the proof will quickly follow. Thus let a E E, say
a
= exp(al)exp(a2)···exp(ak),
where aI, ... , ak E A. But, by Lemma 4.104, if b is sufficiently close to a, then we have b = aexp(c) for some c E A. Hence also bEE, so that E is open. But then every (left) coset of E is also open because each mapping a f+ ba, G + G, is a homeomorphism of G. It follows that G \ E is also open because it is the union of all the cosets of E other than E itself. Hence E is closed in G. We have shown that E is a nonempty, openandclosed subset of the connected set Go, so in fact E = Go. It then follows that each coset of Go in G is also connected and openandclosed in G. Thus the cosets of Go are precisely the topological components of G. This proves (ii), and completes the proof of (i). (iii) This is immediate from (ii) and Proposition 4.98(i).
D
Notes For alternative treatments of the holomorphic functional calculus, see [47, Section 2.4]. See also [31,32, 79, 126, 143]. We have touched on its applications of the functional calculus to the study of Sp T when T E B(E). This is the beginning of the spectral theory of linear operators; see [63, 114] for much more. The theorem of Kahane and Zelazko, Theorem 4.102, is based on [104]. For an elementary proof, see [142]. Let A be a Banach algebra with an identity. Then the quotient group G(A)/Go is called the index group of A; the quotient map from G(A) onto G(A)/Go is an epimorphism of groups. This index is discussed in [62]. For 'automatic continuity' and uniqueness results for the holomorphic functional calculus, see [44] and [158]. Exercise 4.29 We used Runge's theorem, Corollary 4.86, in the proof of Theorem 4.89. Here is an alternative approach, which avoids an appeal to Runge's theorem. Let A be a Banach algebra with an identity, let a E A, and take U to be an open neighbourhood of Sp a in C. For j E O(U), define 8 a (f) as in clause (i) of Theorem 4.89. Then all the specified properties of the map 8 : O(U) > A follow as before, save for the fact that 8 a (fg) = 8 a (f)8 a (g) whenever j,g E O(U). For this, first choose a contour 1'1 that surrounds Spa in U to specify 8 a (f). Let V be the open set bounded by bd, and choose a contour 1'2 that surrounds Sp a in V to specify 8 a (g). Then
225
Banach algebras
Use the facts that Ra()..)  Ra(P,) = ()..  P,)Ra()..)Ra(P,) and that
!
f()..)(p,  )..)1 d)"
~2
to deduce that indeed 8 a(fg)
= 0
and
= 8 a(f)8 a(g)
g()..)
=!
g(p,)(p,l  a)I dp,
~1
when f,g E O(U).
Exercise 4.30 Let A be a unital Banach algebra, so that expA <;::: G(A). (i) Let a E A. Prove that expa = limn~=(1 + a/n)n. (iii) Let A = MIn, the algebra of n x n matrices. Show that exp A = G(A). (ii) Let A = G(T). Show that expA =I G(A). Indeed, G(A) has count ably many components, each containing exactly one of the functions Zk, where k E Z. Exercise 4.31 This exercise requires some knowledge of algebraic topology. Let K be a nonempty, compact Hausdorff space. Prove that the index group of the Banach algebra G(K) is naturally isomorphic to HI (K; Z), which is the first Cech cohomology group of K with integer coefficients. Exercise 4.32 Let A be a commutative Banach algebra. An analytic semigroup in A on TI is an analytic Avalued function z f> a Z , TI > A, with a Z+ w = aZa w (z, wE TI). (i) Let (a Z : z E TI) be an analytic semi group in A. Show that a ZA = aA (z E TI). (ii) For z E TI and t E JR+, define
where r is the Gamma function. Show that /z E £I(JR+), and calculate the Laplace transform (see Exercise 2.39) of F for z E TI. Deduce that (F : z E TI) is an analytic semigroup in (£I(JR+); *). (This semi group is called the fractional integral semigroup.) (iii) For z E TI and t E JR, define 1 GZ(t) = 271'1/2 Z I/2 exp
(_t
2
~
)
Show that G Z E £I(JR), and calculate the Laplace transform of G Z for z E TI. Deduce that (G z : z E TI) is an analytic semigroup in (£I(JR); *). (This semigroup is called the Gaussian semigroup.)
5
Representation theory
Representations and modules 5.1 Modules. Let A be a complex algebra, and let E be a complex vector space. A representation of A on E is a homomorphism T : a f+ Ta from A into the algebra £(E) of all linear endomorphisms of E. These representations are of great importance in pure algebra, especially when E is finitedimensional, so that a representation can be regarded as a homomorphism into an algebra of matrices. A related theory is very important in Banachalgebra theory; however, usually we replace the finitedimensional space E by an infinitedimensional Banach space and consider continuous homomorphisms from our Banach algebra into 13(E). Given a representation T of an algebra A on E, we can give E the structure of a (left) Amodule by defining
a .
~ =
(Ta)(O
(a
E
A,
~ E
E).
We may also speak about the algebra A acting on the space E (via the homomorphism T). We are not supposing any knowledge of module theory, but it is often convenient to use some of the language of this subject. Indeed, a vector space E is an Amodule for a product . if the map (a,~)
f+
a . ~,
A x E
>
E,
(*)
is bilinear and ab . ~ = a . (b . 0 (a, bE A, ~ E E). (Strictly, we should say 'left Amodule'.) For example, a complex vector space is a ((>module in the obvious way, and so the notion of an Amodule generalizes that of a vector space. Let E be an Amodule, and let a E A. Then
a . E = {a . ~ : ~
E
E}
and
E.L = {a
E
A : a . E = {O}}.
A representation T : a f+ Ta, A > £(E), is faithful if the map T is injective; equivalently the Amodule E is faithful if E.L = {O}. Suppose that (A; I '11) is a normed algebra, (E; 1·1) is a normed space, that T : A > £(E) is a representation, and that each Ta is a bounded linear operator on E, so that Ta E 13(E). Then we say that the homomorphism T is a normed representation of A (or that E is a normed Amodule). Thus, in this case, we require that, for each a E A, there is a constant Ka > 0 such that
la . ~I
:::; Kal~1
(~E
E).
Representation theory
227
This normed representation or the Amodule is said to be continuous if the mapping T : (A; 11·11) > (8(E); 11·11) is continuous, where 11·11 also denotes the operator norm on (E; 1·1), or, equivalently (using Exercise 2.16), if the bilinear map in (*) is continuous, and so now we require that there be a constant K > 0 such that la . ~I ::; K lIalll~1 (a E A, ~ E E). It is easy to see that, in this case, we may replace the norm on E by an equivalent
norm to ensure that K
= 1,
la . ~I
::;
and hence that lIalll~1
(a
E
A, ~ E E);
(**)
we shall suppose that this has been done. In the case where (E; 1·1) is a Banach space and (**) holds, we say that E is a Banach left Amodule. The great advantage of the module viewpoint is that we have the obvious (algebraic) notions of module homomorphism, isomorphism, submodu1e and quotient module. Let A be an algebra, and let E be an Amodule Then F <:;; E is a submodu1e of E if F is a vector subspace of E and a . ~ E F whenever a E A and ~ E F; in this case, the quotient vector space ElF is an Amodule for the operation a . (~+ F) = a . ~ + F (a E A, ~ E E). Let E and F be Amodules. Then a map T E £(E, F) is a module homomorphism if T( a . ~) = a . T~ (a E A, ~ E E) . Two Amodules E and F are isomorphic if there is a bijective module homomorphism T : E > F; we write E c:::: F in this case. Suppose that A is a normed algebra, that E is a normed Amodule, and that F is a closed submodule of E. Then, in the quotient norm, ElF is also a normed Amodule. The closed submodules of ElF correspond bijectively to those closed submodules of E that contain F. Suppose that E is a continuous normed Amodule. Then the normed Amodules F and ElF are also continuous. For example, let A be an algebra, and let I be a left ideal in A. Then I and AI I are left Amodules. In the case where A is a Banach algebra and I is a closed left ideal, I and AI I are Banach left Amodules. Let A be an algebra, and let E be an Amodule. For ~ E E, set A .
~
= {a .
~
:a
E
A},
so that A . ~ is a submodule of E. A vector ~o E E such that E = A . ~o is a cyclic vector for E; the module E is cyclic if there is a cyclic vector, and E (or the corresponding representation) is irreducible (or simple) if it is nonzero and has no submodules other than {O} and E itself. Evidently, a nonzero module E is irreducible if and only if every nonzero vector in E is cyclic. Thus, in the case where A has an identity 1, we have 1 . ~ = ~ (~ E E) for an irreducible module E.
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Introduction to Banach Spaces and Algebras
Example 5.1 Let E be a complex Banach space, and let Q( ~ 8(E) be a Banach operator algebra on E, as in Example 4.6. In particular, consider Q( = 8(E) itself. Then E is, of course, an Q(module in a natural way, just taking T . x = Tx
(T E Q(, x E E) .
We shall show that E is an irreducible Q(module. Thus take x, y E E with x =I o. We must show that there is some T E Q( with Tx = y. But this follows at once from the HahnBanach theorem. Indeed, there is an element J E E* with J(x) = 1, and we may define T by setting Tz = J(z)y (z E E), so that T = y ® J E Q(. Evidently Tx = y, as required. 0 Let A be a Banach algebra with an identity, 1. Then the leftregular representation of A is the homomorphism
T :a
f+
La,
A
t
8(A) ,
where, as before, La(b) = ab (b E A) for each a E A. Clearly, this representation is continuous, and it has 1 as a cyclic vector. In treating A as a (left) Amodule, we always intend this representation. For the leftregular representation, a submodule is just a left ideal of A. Let L be such a left ideal. Then AlLis also cyclic, with cyclic vector the coset 1 + L. It follows from Proposition 4.1 that the module AlLis irreducible if and only if L is a maximal left ideal. Since every maximal left ideal in a Banach algebra is closed, the module AlLis then a Banach left Amodule.
Proposition 5.2 Let A be an algebra with an identity, and let E be an irreducible Amodule. Then E ~ AIL for some maximal left ideal L of A. Proof Let 0 =I ~o E E, so that A . ~o = E. Set L = {a E A: a . ~o = O}. Then a f+ a . ~o, A t E, is a surjective module homomorphism with kernel L. So L is a left ideal and E ~ AIL. But L is maximal because the module E, and therefore AIL, is irreducible. 0 Corollary 5.3 Let A be a Banach algebra with an identity, let E be an zrreducible Amodule, and let ~o E E with ~o =I o. For every ~ E E, define: I~I = inf{llall : a E A, a . ~o =
O·
Then 1·1 zs a complete norm on E such that 10, . ~I ::::; Ilalll~1 (a E A, ~ E E). Proof The ideal L of the above proof is closed in A, and so AlLis a Banach algebra with respect to the quotient norm. The given norm on E is just this quotient norm transferred to E by the isomorphism of Proposition 5.2. 0 The above corollary shows that every irreducible Amodule E over a unital Banach algebra A is a Banach left Amodule for a suitable norm on E.
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Representation theory
5.2 Bimodules and derivations. Let A be a complex algebra, and let E be a complex vector space. Then E is an Abimodule with respect to bilinear maps (a,~) f+ a . ~ and (a,~) f+ ~ • a from A x E to E whenever the following hold for all a, b E A and ~ E E:
(i) a . (b . ~) = ab . ~; (ii) (~ . a) . b = ~ . ab; (iii) a . (~ . b) = (a . ~) . b. For example, the algebra A itself, and each ideal in A, are Abimodules with respect to the product map in A. Suppose that A is a subalgebra of an algebra B. Then B is an Abimodule with respect to the product in B. Now suppose that (A; 11·11) is a Banach algebra and that (E; 1·1) is a Banach space such that E is an Abimodule with respect to the above maps. Then E is a Banach Abimodule if the two bilinear maps are continuous. In this case, by changing to an equivalent norm on E, we may and will suppose that
la . ~I
::;
Ilalll~l,
I~· al ::; Ilalll~1
(a E A, ~ E E).
For example, a Banach algebra A itself, and each closed ideal I in A, are Banach Abimodules with respect to the product map in A, and the quotient space AI I is a Banach Abimodule for the products
a . (b
+ I)
=
ab + I,
(b
+ I)
. a = ba + I
(a, bE A).
Examples 5.4 Let A be a Banach algebra. (i) Let E be a Banach left Amodule. We define an action of A on the right by ~ . a = 0 (a E A, ~ E E). Then it is clear that E is now a Banach Abimodule. (ii) Let E be a Banach Abimodule. Then the dual space E* of E is a Banach Abimodule for the operations defined by (a, J) f+ a . f and (a, J) f+ f . a, where
(x, f . a) = (a . x, 1) ,
(x, a . 1) = (x . a, 1)
(a
E
A, x
E
E,
f
E
E*) .
In this case, E* is the dual bimodule of E. (iii) Suppose that A is commutative, and take E to be a Banach left Amodule. We define an action of A on the right by ~ . a = a . ~ (a E A, ~ E E). Then it is again clear that E is a Banach Abimodule. In this case, we say that E is a Banach Amod ule. (iv) Let cp E
C such that d(ab)
a . z = z . a = cp(a)z
(a
E
as is very easily verified; this bimodule is called
A, z
E
<eip.
o
Let A be an algebra, and let E be an Abimodule. Then a derivation from A to E is a linear map D E £(A, E) such that
D(ab) = a . Db + Da . b (a, bE A). For example, let ~ E E, and set 0e(a) = a . ~ ~ . a (a E A). Then oe : A > E is easily checked to be a derivation; these derivations are called inner derivations.
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Introduction to Banach Spaces and Algebras
In particular, consider the case where E = A (and the module operations are the product in A). Then a derivation D : A > A is called a derivation on A. Examples 5.5 (i) Consider the Abimodule C
= 'P(a)d(b) + 'P(b)d(a)
(a, bE A);
such maps are called point derivations at 'P. For example, let A = A(~), 'P = EO, and d(f) = 1'(0) (f E A). Then d is a continuous point derivations at EO. (ii) Let A = C l(n) and E = C(n). Then E is a Banach Amodule with respect to the usual pointwise product of functions, and the map D : f f+ 1', A > E, is a continuous derivation. (iii) Let U be a nonempty, open subset of C. Then D : f f+ I' is a continuous derivation on the Frechet algebra O(U). 0 The following version of Leibnitz's rule is proved by an easy induction; by convention, DO is the identity operator on A. Proposition 5.6 Let A be an algebra, and let D be a derivation on A. Then, for each n EN, we have Dn(ab) =
:t (~)
(Dka)(Dnkb)
(a, bE A).
k=O
o
Theorem 5.7 (Singer and Wermer) Let A be a Banach algebra, and let D be a continuous derivation on A. Then exp D is a continuous automorphism on A. In the case where A is commutative, D(A) ~ J(A). Proof As before, exp D E B(A) is defined by Dn
L (!n. 00
(exp D)(a) =
(a E A).
n=O
Now take a, bE A. Then we calculate that (expD)(ab)
=
~ D~~b) = ~ ~! ~ (~)(Dka)(Dnkb)
~~ 1 k 1 nk = L L k! D a . ..D b= n=Ok=O
(f ~~a) (f ~~b) ,=0
}=o
= (exp D) (a) (exp D)(b) , and hence exp D is a homomorphism on A. The inverse of exp D is exp(  D), and so exp D is a continuous automorphism on A.
Representation theory
231
Let z E
Irp(exp(zD))(a)1 ::;
Iiall
(a
E
A).
Let a E A. Then the map j : z f+ rp(exp(zD)(a)), C ~ C, is bounded, and it is clearly entire. By Liouville's theorem, Proposition 1.39, j is constant. In particular, the coefficient of z in the powerseries expansion of j, namely rp(Da), is zero. Thus rp(Da) = 0 (rp E A)' Now suppose that A is commutative. By Corollary 4.60 (applied to A+), D(A) ~ Q(A) = J(A). D The following corollary will be strengthened in Corollary 5.34.
Corollary 5.8 Let A be a commutative, semisimple Banach algebra. Then there are no nonzero, continuous derivations on A. D
5.3 The Jacobson radical. Let A be an algebra, and let L be a left ideal of A. The standard representation of A on AlLis defined by (Ta)(b+L)=ab+L so that T : A by
~
(a,bEA),
£( AI L) is a homomorphism. We define the quotient L : A of A L :A
=
{a
E
A : aA
~
L} .
Thus L : A is the kernel of the above standard representation T of A on AIL, so that L : A is an ideal in A. Suppose that the algebra A has an identity. Then L : A ~ L, and indeed L : A is the maximal ideal that is contained in L. In this case, an ideal P in A is said to be primitive if P = L : A for some maximal left ideal L of A. It is evident that P = E.L is primitive if and only if it is the kernel of an irreducible representation of A on an Amodule E. Also, every maximal ideal M of A is a primitive ideal because M ~ L for some maximal left ideal L, and then M ~ L : A, so that M = L : A by the maximality of M. The algebra A is said to be primitive if {O} is a primitive ideal. In the case where A is a commutative algebra with an identity, the primitive ideals of A are simply the maximal ideals; each character rp E A defines an irreducible module structure on C by the formula
a . z = rp(a)z
(a
E
A, z
E
q.
Let A be a Banach algebra with an identity, and let L be a closed left ideal in A. Then L : A is a closed ideal in A. Thus each primitive ideal in A is closed.
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Introduction to Banach Spaces and Algebras
Let A be an algebra with an identity. Then we define the Jacobson radical, J(A), of A to be the intersection of all the primitive ideals of A or, equivalently, as the intersection of all the kernels of the irreducible representations of A. For an algebra A without an identity, we define J(A) to be J(A+). It is clear that, if A is a commutative Banach algebra, then the present definition of J(A) reduces to the previous one, given on page 193. An algebra A is said to be semisimple if J(A) = {O} and radical if J(A) = A; necessarily a radical algebra does not have an identity. It is immediate that J(A) is always an ideal in an algebra A. We remark that we use the term 'Jacobson radical' because, in pure algebra theory, there are several different radicals that can be defined; some authors just say 'the radical of A' for J(A). Nevertheless J(A) is the only radical that we shall discuss seriously. There is a variety of different characterizations of the Jacobson radical, some intrinsic, and we give some of these in the next theorem. A key point is that the above discussion of representations has been carried out in the context of left Amodules and maximal left ideals. (There is an entirely similar theory involving right Amodules and maximal right ideals.) Thus, at first sight, it appears that the definition of J(A) depends on the choice of 'left' or 'right' in the definition; however we shall see that this is not in fact the case. Let A be an algebra. Then the opposite algebra to A is the algebra B which is the same as A as a vector space, but such that the product in B is taken in the opposite order to that of A. Clearly, (maximal) left ideals of B correspond to (maximal) right ideals of A. The opposite algebra to A is denoted by AOP. Recall from Section 4.4 that G(A) denotes the group of units of an algebra A with an identity, so that G(A) = G(AOP), and that Aa = {ba : bE A}. Theorem 5.9 Let A be an algebra with an identity, and let J Jacobson radical of A. Then:
= J(A) be the
(i) J is the intersection of the maximal left ideals of A; (ii) J = {a
E
A : 1 + Aa
~
G(A)};
(iii) J = {a E A : 1 + aA ~ G(A)}; (iv) J is the intersection of the maximal right ideals of A. Proof (i) Let L be a maximal left ideal. Then J ~ P = L : A ~ L, so that J is contained in the intersection of the maximal left ideals. Conversely, suppose that ao ~ J. Then there is an irreducible representation T: A > £(E) such that Tao of o. Thus there exists ~o E E with ao . ~o of o. Set
L = {a
E
A :a .
~o =
O} .
Then L is a maximal left ideal of A and a ~ L. This proves (i). (ii) Let a E J and b E A. Assume that 1 + ba is not left invertible in A. Then there is a maximal left ideal L of A with 1 + ba E L. However, ba E L (by (i»,
Representation theory
233
and so 1 E L, contrary to the fact that L =1= A. Thus 1 + ba is left invertible in
A. Now choose C E A so that (1 + c)(1 + ba) = 1, so that c = ba  cba E J. By the argument of the previous paragraph, 1 + c is left invertible in A. But we know that 1 + c is right invertible, and so 1 + c is invertible, and hence so is 1 + ba. Thus 1 + Aa c:: G(A). Conversely, suppose that a E A has the property that 1 + ba is invertible (or even just left invertible) for every b E A. Let L be a maximal left ideal of A, so that L + Aa is a left ideal in A containing L. Assume that a if L. Then L + Aa =1= L, and so L + Aa = A by the maximality of L. Thus there is some b E A with 1 + ba E L, which contradicts the left invertibility of 1 + ba. Hence a is in every maximal left ideal, so that a E J by (i). This proves (ii). (iii) This follows from (ii) since, by Proposition 4.16(ii), 1 + ba is invertible if and only if 1 + ab is invertible. (iv) From (ii) and (iii), we see that J(AOP) = J, and so (iv) follows from the equivalence of (i) and (ii). D Corollary 5.10 Let A be an algebra. Then J(A) is an ideal in A and AIJ(A) is semisimple. Proof We may suppose that A has an identity. Set J = J(A), an ideal in A. Let 7r : A + AI J be the quotient map. For each irreducible Amodule E, J is included in the kernel of the representation of A on E, so that E becomes an irreducible AIJmodule in a natural way (by setting 7r(a) . ~ = a . O. Suppose that ~ E J (AI J). Then ~ = 7r( x) for some x E J, and so ~ = O. This shows that J(AI J) = {O}, and so AI J is semisimple. D
Here is an alternative proof that the quotient AI J is semisimple. Take x E J(AI J) and y E AjJ, say x = 7r(a) and y = 7r(b), where a, b E A. Since 1 + xy is left invertible in AI J, there exists some e E A such that the element u := e(1 + ab)  1 E J. But then e(1 + ab) = 1 + u is left invertible in A, so that 1 + ab is left invertible in A. Hence a E J and x = 7r( a) = O. Thus J(AI J) = {O}, and so AI J is semisimple. Let A be an algebra. Recall that we are writing Q(A) for the set of quasinilpotent elements of A. A subset S of A is said to be quasinilpotent if S c:: Q(A). Corollary 5.11 Let A be an algebra. Then J(A) is a quasinilpotent ideal that is the union of all the quasinilpotent left ideals in A. Proof We may suppose that A has an identity. Take a E J(A) and A E
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Introduction to Banach Spaces and Algebras
Corollary 5.12 Let I be an zdeal in an algebra A. Then J(I)
= In J(A).
Proof We may suppose that A has an identity. Take a E J(I) and b E A. Then ba E I and 1 + Iba <;;; 1 + Ia <;;; G(I+) by Theorem 5.9(ii). By the same result, ba <;;; J(I). Thus J(I) is a left ideal in A. Clearly, J(I) is quasinilpotent in A, and so J(I) <;;; J(A) by Corollary 5.11. Now suppose that a E I n J(A) and b E I. Then 1 + ba E G(A), and so 1 + ba E G(I + C . 1) by Lemma 4.9, and so, by Theorem 5.9(ii) again, a E J(I). Hence In J(A) <;;; J(I). 0
We now see that the situation for Banach algebras is particularly satisfactory. Theorem 5.13 Let A be a Banach algebra. Then J(A) is a closed ideal in A, and AI J(A) is a semisimple Banach algebra. Proof We may suppose that A has an identity. We know that J(A) is an ideal in A. By definition, J(A) is the intersection of the primitive ideals of A. However, each primitive ideal of A is closed, and so J(A) is closed. By Corollary 5.10, AI J(A) is semisimple. 0 Example 5.14 Let E be a Banach space, and let 2l be a Banach operator algebra on E, as in Example 4.6. We already know from Example 5.1 that 2l is an irreducible B(E)module, and so 2l is semisimple. As in Proposition 5.2, Lx := {T E 2l : Tx = O}
is a maximal left ideal for each x E E, and
nxEE
Lx = {O}.
o
We have shown in Corollary 4.60 that, for a commutative Banach algebra A, we have J(A) = Q(A). The above example, even in the simplest case where A = M 2 , shows that, for a noncommutative Banach algebra A, it may happen that J(A) S;; Q(A). Indeed, here A is semisimple, but the matrix
(~ ~) is a nonzero nilpotent element of A. This example also shows that a closed, unital, commutative sub algebra of a semisimple algebra need not be semisimple. Indeed, matrices in M2 of the form
(~ ~),
where
A, JL
E
C,
form a closed, unital, commutative subalgebra A of M2 such that J(A) is the onedimensional space of matrices of the form (
~ ~),
where
JL E Co
The next lemma will enable us to give an attractive new proof of the most important result of the next section, the Jacobson density theorem. We continue to suppose that all our Banach algebras are over the complex field.
235
Representation theory
Theorem 5.15 (Harris and Kadison) Let A be a Banach algebra with an zdentity, let L be a maximal left ideal of A, and let a E A. Suppose that La L. Then there exists A E C such that a  Al E L.
s:
Remark: The converse to this lemma is, of course, trivial. It is also clear that, for given L and a, the complex number A is uniquely determined and belongs to Spa. Proof By Theorem 4.14, the maximal left ideal L is closed in A, and so AIL is a Banach Amodule; set E = AIL. We first show that, if ba E L for some b E A, then either a E L or bEL. For suppose that ba E L, but that b 1 L. By Proposition 4.1, there exists e E A such that 1 + eb E L. Hence a + eba E La L. But eba E L, and so a E L. Since La L, we can act on AIL by right multiplication by a, i.e. define Ra E 8(E) by Ra(b+L) = ba+L (b E A). We claim that, if a 1 L, then Ra is an invertible operator. Indeed, Ra is injective, for if Ra (b + L) = L, then ba E L, so that bEL by our second paragraph. Also, Ra is surjective because there exists e E A with ea  1 E L, and so, given b E A, we have Ra(be + L) = b + L. We have shown that Ra is bijective; by Banach's isomorphism theorem, Corollary 3.41, Ra is invertible in 8(E). For each A E C, we have L(a  AI) Land R a)..l = Ra  AI. Certainly Sp Ra =I 0 by Theorem 4.17; take A E Sp Ra. Then R a)..l is not invertible, and so a  Al E L. 0
s:
s:
s:
Corollary 5.16 Let A be a Banach algebra with an identity, let L be a maximal left ideal of A, and let a E Q(A) be such that La L. Then a E L.
s:
Proof By Theorem 5.15, there exists A E C with a  Al E L. But, as noted above, A E Sp a. Since a E Q(A), necessarily A = 0, so that a E L. 0 The first application of Theorem 5.15 will be to prove a result of Zemanek that is an important step in proving his characterization of the radical (see Theorem 5.40). First, we need a simple algebraic lemma. Throughout, for an algebra A, we write [a, bJ = ab  ba (a, bE A) . Lemma 5.17 Let A be a unital algebra, let L be a maximal left ideal of A, and let a E A. Suppose that [b, aJ E Q(A) for every bEL. Then La L.
s:
Proof Assume towards a contradiction that La CZ L. Then there exists c E L with ea (j. L. Again by Propositon 4.1, there exists bE A with 1 + bea E L, and hence also 1 + [be, aJ = 1 + bea  abc E L. Since be E L, the hypothesis implies that [be, aJ E Q(A). But then 1 + [be, aJ is invertible, which contradicts the fact that L I A. Thus La <;;:; L. 0
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Introduction to Banach Spaces and Algebras
Corollary 5.18 Let A be a Banach algebra, and let a E Q(A) be such that [b,aj E Q(A) for every b EA. Then a E J(A). Proof We may suppose that A has an identity. By Lemma 5.17 and Corollary 5.16, a belongs to every maximal left ideal of A, and so a E J(A). 0 5.4 Irreducible representations. We now return to consideration of irreducible representations or modules. It is useful to make some further definitions. Let A be an algebra, and let E be a left Amodule. Take n E N. Then E is ntransitive if, for every linearly independent sequence (6, ... '~n) of elements in E and every sequence (1]1, ... ,1]n) in E, there is some a E A such that a . ~k = 1]k (k = 1, ... , n). Evidently E is irreducible if and only if it is Itransitive. Theorem 5.19 (Jacobson's density theorem) Let A be a Banach algebra with an identity, and let E be an irreducible Amodule. Then E is ntransitive for every n E N. It is convenient to prove first two lemmas. Lemma 5.20 Let A be an algebra with an identity, and let E be an irreducible Amodule. For each n :::: 2, the property of ntransitivity is equivalent to its special case (*)n, defined as follows: (*)n: for every linearly independent sequence (6, ... , ~n) of elements in E, there exists a E A such that a . ~k = 0 (k = 1, ... , n  1) and a . ~n =1= o.
Proof Let (6, ... , ~n) be a linearly independent sequence in of elements E, and let (1]1, ... ,1]n) be a sequence of elements in E. We are supposing that (*)n holds. By renumbering the elements, we see that there is, for each k = 1, ... ,n, an element ak E A with ak . ~J = 0 for all j =1= k and with ak . ~k =1= o. We then use the case n = 1 to find, for each k = 1, ... , n, an element Ck E A with Ck . (ak . ~k) = 1]k, and finally define a = L~=l Ckak E A. Clearly, a . ~k = 1]k (k = 1, ... , n), and so E is ntransitive. 0 The next lemma, proving 2transitivity, is where use will be made of Theorem 5.15, which applies to Banach algebras.
Lemma 5.21 Let A be a Banach algebra with an identity, and let E be an irreducible Amodule. Then E is 2transitive. Proof Let (6,6) be a linearly independent sequence in E. Set L = {a E A : a . 6 = O}, so that L is a maximal left ideal in A. Now choose bE A with b· 6 = 6. Then, for>. E C, we have (b>.I) ·6 = 6>'6 =1= 0, and hence b>'1 ~ L. By Theorem 5.15, it follows that Lb g L, i.e. there is some a E L with ab ~ L. Thus a . 6 = 0, but a . 6 = ab . 6 =1= o. This proves (*)2. By Lemma 5.20, E is 2transitive. 0
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Representation theory
Proof of Theorem 5.19 It remains to prove ntransitivity for every n ~ 3. Take n ~ 3, and make the inductive hypothesis that E is (n  l}transitive; we shall prove that (*}n holds. Thus, let (6, ... ,~n) be a linearly independent sequence of elements in E We set I={aEA:a'~k=O
(k=1, ... ,n2)},
B = 1+
= 0 (k = 1, ... , n  2) and a·
~
= 1],
i.e. a E I ~ B and a . ~ = 1]. But now a . I = a . 7r(O = 7r(1]), and so B acts irreducibly on ElF. Clearly, (7r(~nl),7r(~n)) is linearly independent in ElF and so, by Lemma 5.21, there is some b E B with b . 7r(~nd = 0 and b . 7r(~n) = 7r(~n); i.e. b . ~nl E F and b . ~n  ~n E F. By the inductive hypothesis, there is some c E I with c . ~n = ~n' Since IF = {O}, we have cb . ~k = 0 (k = 1, ... , n  2). Since b . ~nl E F, we have cb . ~nl = O. Also, cb . ~n = C • ~n = ~n I= O. Thus (*)n holds (with a = cb). This continues the induction. By induction, E is ntransitive for each n E N, and this completes the proof of Theorem 5.19. 0 The following is a classical theorem of Wedderburn. Theorem 5.22 (Wedderburn's theorem) Let A be a finitedimensional, semisimple algebra (over
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Introduction to Banach Spaces and Algebras
with PI = J2 + P2' Since h <::; PI, necessarily P2 = PI 12 E PI, and hence P2 E PI n P 2 . Continuing, we eventually obtain Pk1 = jk + Pk with jk E Jk and Pk E H n ... n P k = {a}. Our claim follows. It follows that A ~ MInI EEl· .. EEl MInk' The identity of A is e1 + ... + ek. 0
Proposition 5.23 Let A be a Banach algebra with an identity, and let E be an irreducible Amodule. Suppose that {~n : n E N} is a linearly independent set of distinct elements in E. Then there is a sequence (a n )n21 in A such that an'" a1 . ~m = 0 for m, n E N wzth m < n and an'" a1 . ~n 1= 0 for n E N. Proof Let {17n : n E N} be a linearly independent set of distinct elements in E. By Lemma 5.21, there exists b1 E A such that b1 . 171 = 0 and b1 . 172 1= o. It follows from Theorem 5.19 by induction that there is a sequence (b n )n>l in A such that, for each n ~ 2, we have Ilbnll ~ 2 n , bn . 171 = ... = bn . 17n = 0, and bn . 17n+ 1
Set b =
rt lin {b
t .
17] : i = 1, ... , n  1, j = 1, ... , n
L::1 bt ; the series converges in A.
+ I} .
( *)
Then n1
b . 171
= 0 and b· 17n = ~ bt
.
17n
(n ~ 2).
t=l
Clearly, b· 172 = b1 . 172 1= o. Assume that b· 17m+1 E lin{b· 171,··" b· 17m} for some m ~ 2. This contradicts the case in which n = m of (*), and so {b . 17n : n ~ 2} is a linearly independent set of distinct elements in E. Now choose a1 = 1 so that a1 . ~ = ~ (~E E). Next, choose a2 E A such that a2al . 6 = 0 and {a2al . ~n : n ~ 2} is a linearly independent set of distinct elements in E. Choose a3 E A such that a3a2a1 . 6 = 0 and {a3a2al . ~n : n ~ 3} is a linearly independent set of distinct elements in E. Continue in this way to obtain the required sequence (a n )n21 in A. 0
Proposition 5.24 Let A be a Banach algebra with an identity, and let (E k )k2 1 be a sequence of fimtedimensional, irreducible Banach left Amodules. Suppose that Et n· .. n E;, Cl Et for each k > n in N. Then, for each n E N, there exists an E Et n ... nE;, such that an' Ek = Ek (k > 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 Amodules and Banach Abimodules, 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 Abimodules 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 * ft  ft * f for some measure ft 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 Abimodule 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 Abimodule 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 'noncommutative 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 Amodules or maximal right ideals. In pure algebra, there are left primitive ideals that are not right primitive, but it seems to be a longstanding 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
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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 <:;: P or J <:;: P whenever I J <:;: P; an ideal P is prime if and only if either a E P or b E P whenever aAb <:;: P.) Theorem 5.15 is due to Harris and Kadison [90]. Some results, including Theorem 5.15 and Jacobson's density theorem are true for a wider class of algebras than Banach algebras, and are so formulated in texts on algebra; the key fact that we have used is the fundamental theorem of Banach algebras that the spectrum of an element in such an algebra is always nonempty. Jacobson's density theorem, Theorem 5.19, and the Wedderburn theorem, Theorem 5.22, are classical algebraic results. The latter is a rather special case of the more general Wedderburn~Artin theorem, given in many texts, including [97, Chapter IX, Section 3] and [126, Section 8.1]. The related Wedderburn principal theorem states that a finitedimensional algebra A (over q can be written as A = B EB J(A), where B is a unital subalgebra of A with B ~ A/J(A); see [47, Theorem 1.5.18]; possible generalizations to Banach algebras are discussed in [47]. Exercise 5.1 Let A be a Banach algebra, and let E and F be Banach left Amodules. For a E A and l' E B(E, F), define
(a . T)(x) = a . Tx,
(1'. a) (x) = T(a . x)
(x E E).
Verify that B(E, F) is a Banach Abimodule with respect to the maps (a, 1') and (a, 1') f> l' . a.
f>
a .T
Exercise 5.2 Let A be an algebra, let E be an Abimodule, 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 nonempty, 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 infinitedimensional. (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 Amodule. 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 'uniquenessofnorm' 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
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Introduction to Banach Spaces and Algebras
5.5 Johnson's uniquenessofnorm theorem. In this section, we shall prove a dramatic theorem of B. E. Johnson showing the uniqueness of the norm topology on every semisimple Banach algebra, extending the result for commutative Banach algebras that we have just proved; we shall give Johnson's now classical proof. In Section 5.10, we shall give an alternative proof of this theorem. The following is the main result of Johnson. Theorem 5.27 Let A be a Banach algebra with an identity, and let E be an irreducible normed Amodule. Then E is a continuous Amodule. Proof Let P be the kernel of the representation of A on E, so that P is a primitive ideal in A, and hence P is closed. Then AlP is a Banach algebra, and we may consider E as a faithful, irreducible AlPmodule. Clearly, the continuity of the corresponding representation of AlP would imply the same for A. Thus, without loss of generality, we may suppose that E is a faithful Amodule; i.e. we suppose that A ~ B(E) (of course, we do not hypothesize any relation between the norm of A and the operator norm on B(E)). Next, if dimE < 00, then dimB(E) < 00, and the result is trivial. Thus we suppose that E is infinitedimensional. We are considering the normed space (E; 1·1); let (it; I ·1) be the completion of E. By Proposition 2.21, there is an isometric mapping j : U ~ j(U),
B(E)
+
B(E).
For each ~ E E, define R~ : S ~ S~, B(E) + E, so that R~ is a continuous linear operator. We again set (Ta)(~) = a· ~ (a E A), and define e~ : a ~ a·~, A + E, so that e~ E .c(A, E) and R~ 0 T = e~, where T = joT. Set F = {~E E: e~ E B(A,E)}. Then F is clearly a subspace of E, and also, if ~ E F and b E A, then the map is the composition of the continuous maps a ~ ab on A and e~, so that b . ~ E F. Thus F is a submodule of E. Since E is irreducible, either F = E or F = {O}. First, assume towards a contradiction that F = {O}, and take ~ E E with ~ =I o. Then 6(e~) =I {O}. By Proposition 3.54(ii), 6(T) . ~ =I {O}. There is a linearly independent set {~n : n E N} of distinct elements in E. By Proposition 5.23, there is a sequence (a n )n2:1 in A such that an··· al . ~m = 0 for m, n E N with m < n and an··· al . ~n =I 0 for n E N. This implies that
eb. ~
6(T) . (Tan)··· (Tan)~n =I {O},
6(T). (Tan+l)··· (Tad~n = {O},
(*)
for each n E N. We now apply the stability theorem, Theorem 3.55(ii), taking each En of that theorem to be A, taking Rn : b ~ ban, so that Rn E B(A), taking F to be B(E), with Sn : U ~ _U 0 Tan, so that Sn E B(F), and taking T of that theorem to be the present T; we verify that TRn = SnT, as required.
Representation theory
243
Thus, by Theorem 3.55(ii), (6('T) . (Tan)··· (Tan))n>l is a nest which stabilizes, a contradiction of (*). Secondly, suppose that F = E, so that 0E E 8(A, E) for each ~ E E. For each a E A and ~ E E[l], we have IOE(a)1 = I(Ta)(~)1 ::::: IITall. By the uniform boundedness theorem, Theorem 3.34, there is M > 0 such that IIOd : : : M (~ E E[l]), and so IITal1 : : : M Iiall (a E A). Thus T is continuous, the required conclusion. The proof is complete. D Corollary 5.28 Let A and B be Banach algebras, and suppose that B is semisimple. Let 0: A + B be a homomorphism wzth O(A) = B. Then 0 is continuous. Proof We use the closed graph theorem. Indeed, take (an)n>l to be a null sequence in A with O(a n ) + b in B. Assume towards a contradiction that b "I o. Then there is some irreducible representation T of B on a vector space E with Tb "I o. By Corollary 5.3, E can be given a complete norm I . I such that T becomes a continuous, normed representation. But now ToO is a normed representation of A on E. Since O(A) = B, it follows that ToO is irreducible, and then, by Theorem 5.27, ToO is continuous. This implies that
Tb = lim T(O(a n )) = lim (T n+oo
n+oo
0
O)(a n ) = 0,
which is a contradiction. Thus b = 0, and so 0 is continuous by the closed graph theorem, Theorem 3.45. D The following corollary is Johnson's famous uniquenessoEnorm theorem. Corollary 5.29 Let (A; unique complete norm.
11·11)
be a semisimple Banach algebra. Then A has a
Proof This follows from Corollary 5.28 in the same way as Corollary 5.26 followed from Theorem 5.25 in the commutative case. D
5.6 Eidelheit's theorem. Let E be a Banach space, and let Q( <;;; 8(E) be a Banach operator algebra on E, as in Example 4.6. Then Q( is a semisimple Banach algebra, and so Q( has a unique complete norm. We now give a delightful proof of Eidelheit from as long ago as 1940 that shows a somewhat stronger result in this special case. In the result, the operator norm on 8(E) is denoted by 11·11. Theorem 5.30 (Eidelheit's theorem) Let E be a Banach space, and let subalgebra of 8(E) that contains F(E).
Q(
be a
(i) Suppose that III . III is an algebranorm on Q(. Then there is a constant C such that IITII : : : C IIITIII for each T in Q(. (ii) Any two complete algebranorms on Q( are equivalent.
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Introduction to Banach Spaces and Algebras
Proof (i) Assume towards a contradiction that there is no such constant C. Then there exists a sequence (Snk::':l in ~ such that IllSnll1 = 1 (n E N) and IISnl1 + 00 as n + 00. By the uniform bounded ness theorem, Theorem 3.34, there exists Xo E E such that the sequence (1ISnxoll)n>l is unbounded. By Corollary 3.35, there exists fo E E* such that the sequence (lfo(Snxo)l)n>l is unbounded; set Zn = fo(Snxo) (n EN). Now define T = Xo ® fo, so that T E ~ and T 1= o. Let n E N. Then (TSnT)(x)
= fo(x)(TSn)(xO) = znfo(x)xo = znTx (x
E
E),
and so znT = TSnT, whence IZnllllTll1 :::; IIITII12. Since IIITIII 1= 0, it follows that IZnl :::; IIITIII (n EN), a contradiction of the fact that the sequence (Zn)n21 is unbounded. (ii) Suppose that ~ is a Banach algebra with respect to III . 1111 and III . 1112' Take (Tnk:':l in ~ such that Tn + 0 in (~; 111·1111) and Tn + T in (~; 111.111 2), By (i), Tn + 0 in (S(E); 11·11) and Tn + Tin (S(E); 11·11). Thus T = 0, and so, D by Corollary 3.52,111'111 1 and 111·1112 are equivalent. 5.7 The continuity of derivations. Our aim in this section is to prove a theorem of Johnson and Sinclair on the automatic continuity of derivations on a Banach algebra. Recall that the notion of a nest which stabilizes was introduced on page 136. Let A be a Banach algebra, and let E be a Banach Abimodule. Then E is a separating module for A if, for each sequence (a n )n21 in A, each of ( a1 ... a n . E) n21
and
(E· a n ... a1) n21
is a nest which stabilizes. In the case where E is a closed ideal in A, we speak of a separating ideal in A. Proposition 5.31 Let A be a Banach algebra. (i) Let B be a Banach algebra, and let e : B + A be an epimorphism. Then 6(e) is a separating ideal in A. (ii) Let E be a Banach Abimodule, and let D : A + E be a derivation. Then 6(D) is a separating module for A. Proof Let (a n )n21 be a sequence in A. (i) Since e is an epimorphism, 6(e) is a closed ideal in A and there exists a sequence (b n )n21 in B such that e(b n ) = an (n EN). For n E N, define Rnb
= bbn (b
E
B)
and
Sna
= aa n (a
E
A).
Then (Rn)n21 and (Sn)n21 are sequences in S(B) and SeA), respectively, and eRn  Sne = 0 (n EN). It follows from Theorem 3.55(ii) that (al ... a n 6(e))n21 is a nest which stabilizes.
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Representation theory
Similarly, (6(B)a n ··· at}n21 is a nest which stabilizes. (ii) Clearly, 6(D) is a closed subbimodule of E. For n
Rna = ana
(a E A)
and
Snx = an . x
E
N, define
(x E E),
so that (R n )n21 and (Sn)n21 are sequences in B(A) and B(E), respectively. Also,
(DRn  SnD)(a)
=
D(ana)  an . Da
=
Dan· a
(a
E
A),
and so DRn  SnD E B(A, E) (n EN). It follows from Theorem 3.55(ii) that (al ... an . 6(D))n>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 finitedimensional. Proof Let E be an irreducible left Amodule. By Corollary 5.3, E is a Banach left Amodule for a suitable norm. Assume towards a contradiction that the space E is infinitedimensional, 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 Amodule, 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 finitedimensional, 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+! <;;; Pn+!, and so In+l . En+l = {O}. However, an··· al . En+l = En+l' and so In . En+l = En+l · Thus In+l £.; In (n EN), a contradiction of the fact that (In)n>l stabilizes. Thus J(A) is an intersection of finitely many primitive ideals, each of finite co dimension in A, and so AI J(A) is a finitedimensional algebra. D Theorem 5.33 (Johnson and Sinclair) Let A be a semisimple Banach algebra, and let D be a derivation on A. Then D is automatically continuous. Proof Set I = 6(D). By Proposition 5.31(ii), I is a closed ideal in A such that I is a separating ideal. By Corollary 5.12, I is semisimple, and I is a separating module in I. By Proposition 5.32, I is a finitedimensional algebra.
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Introduction to Banach Spaces and Algebras
Assume towards a contradiction that I =1= {a}. By Theorem 5.22, I contains a nonzero idempotent, say p. Let (a n )n21 be a null sequence in A such that Dan + pas n + 00. Then pan + 0 in I and
D(pa n ) = p(Da n ) + (Dp)a n
+
p2 = P as
n
+ 00.
However, D I I is continuous because I is finitedimensional, and so p = 0, a contradiction. Thus 6(D) = {O}, and so D is continuous. 0 Corollary 5.34 Let A be a commutative, semisimple Banach algebra. Then there are no nonzero derivatwns on A.
Proof This is immediate from Corollary 5.8 and Theorem 5.33
o
Corollary 5.35 The algebras coo(lI) and O(U), where U is a nonempty, open subset of C, are not Banach algebras with respect to any norm. 0 Notes Johnson's proof [98] of his uniquenessofnorm theorem, Corollary 5.29, answering a longstanding open question, was the seed from which much automatic continuity theory grew; see [47]. A second proof of the theorem will be given in Corollary 5.47, below. The theorem of Eidelheit, given as Theorem 5.30, is from [67]. It is natural to consider whether Johnson's uniquenessofnorm theorem extends to Banach algebras with finitedimensional radicals. An example, given as Exercise 5.7 below, will show that this is not always the case. The general case was investigated by Dales and Loy in [49], and further results and examples are given by Pham in [129]. Despite a substantial amount of work on automatic continuity and uniquenessofnorm questions, some obvious questions remain unanswered. For example, let A be a commutative Banach algebra which, as an algebra, is an integral domain, in the sense that ab oJ 0 whenever a and b are nonzero elements of A. Then it is not known whether or not A necessarily has a unique complete norm. Theorem 5.33 of Johnson and Sinclair is from [101]. For more general forms of Corollary 5.29 and Theorem 5.33, see [47, Theorem 5.2.28]. Other results, and some open questions, on the continuity of derivations are given in [23, 54]. It is an unproved conjecture that all derivations from a group algebra L 1 (G) into a Banach L 1 (G)bimodule are automatically continuous. This is proved or stated for some groups G in [47, Section 5.6]; there seems to have been no further progress on this question. It is also an unproved conjecture that all derivations from an amenable Banach algebra A into a Banach Abimodule are automatically continuous. There is a discontinuous derivation from the disc algebra A(~) into a Banach A(~)bimodule, although all point derivations on the disc algebra are continuous (see Exercise 5.3). There is a generalization of this result in [22] and [47, Theorem 5.6.79]. A different technique of proving automatic continuity results flows from the main boundedness theorem of Bade and Curtis [21]. This leads to a theorem of Johnson [99]: for a Banach space E such that E is homeomorphic to its square (see Exercise 2.9), every homomorphism from 8(E) into a Banach algebra is automatically continuous. It also leads to a theorem of Bade and Curtis [21] that shows that a homomorphism from a commutative Banach algebra of the form C(K), where K is a compact Hausdorff space, into a Banach algebra is necessarily continuous on a dense subalgebra of C(K), and which gives a very precise description of such a homomorphism; the 'discontinuity' of e is 'concentrated on a finite singularity set'. It is a theorem of Esterle [74] that every
247
Representation theory
epimorphism from C(K) is automatically continuous. See [47, Section 5.4] for details and further results. The theorem of Bade and Curtis led to the conjecture that every homomorphism from C(K) into a Banach algebra is automatically continuous. However, it was proved independently by Dales [46] and Esterle [73] that, with CH, there is a discontinuous homomorphism from C(K) into certain Banach algebras whenever K is an infinite, compact Hausdorff space. It follows that, for each such K, there is an algebranorm on C(K) that is not equivalent to the usual uniform norm; this answers a question of Kaplansky. The construction of such a homomorphism grew out of an earlier theorem of [10] that showed that the algebra J = iC[[X]] of formal power series in one variable can be embedded into certain Banach algebras; the latter construction used the notion of an element 'of finite closed descent' in a Banach algebra. For more on elements of finite closed descent, see [1113]. Full details of these constructions and a characterization of the commutative Banach algebras that arise as the codomain of a discontinuous homomorphism from a maximal ideal of C(K) into a radical Banach algebra, are given in [47, Section 5.7]. Note that the abovementioned construction of a discontinuous homomorphism requires the assumption of the continuum hypothesis, cn. It is a remarkable theorem of Woodin that this settheoretic assumption is not redundant: there are models of set theory ZFC (in which CH does not hold) such that, in this model, all homomorphisms from C(K) into a Banach algebra are continuous. An account of this result is given in [51].
Exercise 5.5 Let w be a continuous weight function on JR+, and let L1(JR+,W) be the commutative Banach algebra defined in Example 4.67 and discussed in Exercise 4.24. Show that every derivation on L 1 (JR+ ,w) and every epimorphism from a Banach algebra onto L1 (JR+, w) is continuous, and hence that L1(JR+, w) has a unique complete norm. Indeed, let I be a nonzero, closed ideal in L1(JR+,W). Use Titchmarsh's theorem (given in the notes on page 206), to show that I cannot be a separating ideal in L1(JR+,W). Exercise 5.6 Let A be a commutative Banach algebra, and let E be a Banach Amodule. Set 2l = A EEl E, with the product specified by (a,x) . (b,y) = (ab, a . y
+b
. x)
(a,b E A, x,y E E)
and the norm specified by II(a,x)11 = lIall + Ilxll (a E A,x E E). (i) Show that (2l; II '11) is a commutative Banach algebra with J(2l) (ii) Let D : A > E be a derivation, and set
111(a,x)111 = lIall + IIDa  xii Show that (2l;
III . III)
=
J(A) EEl E.
(a E A,x E E).
is a Banach algebra.
(iii) Show that the norms continuous.
11·11
and
111·111
are equivalent on 2l if and only if D is
(iv) Use Exercise 5.3 to construct a commutative algebra with a onedimensional radical which is a Banach algebra with respect to two nonequivalent norms.
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Introduction to Banach Spaces and Algebras
Exercise 5.1 Let A be the sequence algebra (£2,11·112)' so that A is a commutative Banach algebra. Set 2l = A EEl IC as a vector space. For a, bE A and z, W E IC, define
(a,z) . (b,w) = (ab, 0)
and
II(a,z)11 = IIal12
+ Izl
.
Verify that 2l is a Banach algebra with respect to this product and norm, and that J(2l) is the onedimensional space {a} EEl IC. Now let f be a linear functional on £ 2 such that f I £ 1 is the linear functional 00
f : a = (an) and set
f+
L an , n=l
111(a, z)111 = max{ll a I1 2 , If(a)  zl} ((a,z) E 2l). Verify that (2l, 111·111) is also a Banach algebra and that the norm 111·111 is not equivalent to 11·11. A Banach algebra A is strongly decomposable if there is a closed subalgebra B such that A = B EEl J(A). Show that (2l, 111·111) is not strongly decomposable. This exercise is based on an example from 1951 of Feldman [77].
Variation of the spectral radius Let A be a Banach algebra. We have defined the spectrum, SPA a, and spectral radius, PA(a), of an element a of A in Section 4.5. In particular, we proved the fundamental theorem, Theorem 4.17, that SPA a is a nonempty, compact subset of C. We shall now study how this spectrum varies as the element a varies in the Banach algebra, and give some implications of this variation. 5.8 Continuity properties. The first point to notice is that the spectral radius function is upper semicontinuous on A. We shall give three simple proofs of this; we recall that upper semicontinuity was defined on page 18.
Proposition 5.36 Let A be a Banach algebra. Then PAis upper semicontinuous on A. Proof First proof. Let a E A, take c > 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 semicontinuity 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 semicontinuity, 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 Avalued function. We shall see that the function Zf4PA(J(Z)) ,
U+lR+,
has some very analyticlike 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 functionmerely upper semicontinuity. ) 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 nonempty, 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 nonempty. 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 nonempty, compact Hausdorff space, and let (gn)n?l and 9 be continuous, realvalued 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 nonempty, 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
251
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 quasinilpotents, 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 Avalued 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 Avalued 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 semicontinuity 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 threecircles 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 threecircles 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 + (lt)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 Avalued 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 Liouvilletype 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 Avalued 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 semicontinuity 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
(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
p(f(z)) ::; p(f(zo)) Then p
0
(z
E
D).
f is constant on D.
Proof We first show that p 0 f is constant on a neighbourhood of zoo Let R > 0 be such that {z E
m = inf{p(f(z)) : Iz  zol < R/2}, take E > 0, and choose Zl E
0
f is
E = {z ED: p(f(z)) = p(f(zo))} = {z ED: p(f(z)) 2: p(f(zo))}. By the upper semicontinuity of p, the set E is relatively closed in D. By the argument of the last paragraph, E is open. Since Zo E E, we have E =I 0, and so, by the connectedness of D, it follows that E = D. The theorem is proved. 0
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Introduction to Banach Spaces and Algebras
5.10 Aupetit's lemma. We recall a definition from Chapter 3. Let E and F be Banach spaces, and let T : E  F be a linear map. As on page 135, the separating space SeT) of T is the set of all y E F such that TX n  y for some null sequence (Xn)n~l in E. It was noted in Theorem 3.51 that SeT) is a closed subspace of F and that T is continuous if and only if SeT) = {a}. Now suppose that A and B are unital Banach algebras and that e: A  B is a unital homomorphism with e(A) = B. Then it is easy to see that See) is a closed ideal of B. We remark also that, in the above case, certainly Sp B (ea) ~ Sp A(a), so that PB(ea) :::; PA(a), for each a E A. We give the following lemma for any linear mapping T : A  B for which PB(Ta) :::; PA(a) (a E A); by our remark, this includes the case in which T is a homomorphism. Theorem 5.45 (Aupetit's lemma) Let A and B be Banach algebras, and let T : A  B be a linear mapping such that PB(Ta) :::; PA(a) (a E A). Then, for every a E A and every b E SeT), we have
PB(Ta) :::; PB(Ta
+ b) .
Proof (Ransford) Let f :
PB(f(l)) :::; sup (PB(f(W)))t. sup (PB(f(W))/t, Iwl=r
(*)
Iwl=R
where t = log Rj(log R  logr). Let a E A and b E SeT), say (an)n~l is a null sequence in A with Tan  b in B. Take E > O. By Corollary 4.25, we may, by changing to a norm on B equivalent to the original norm (which does not affect the conclusion), suppose that IITa + bll < PB(Ta + b) + E. For each n EN, apply inequality (*) to the function In, where
In(z) = T(a
+ an)  zTa n (z
E
certainly each In :
PB(fn(1)) :::; sup (PB (T(a
+ an)  zTa n )) t
. sup (PA (a
Izl=r
+ an  zan)) It
Izl=R
:::; (1IT(a
+ an)11 + rIITanll)t(lla + anll + Rllanll)lt.
Since In(1) = Ta (n EN), we may let n 
PB(Ta) :::; (11Ta Now, for each fixed r > 0, let R t = log Rj(logR logr)  1, so that
00
to deduce that
+ bll + rllbll)tllaillt. 00
and note that in this case we have
PB(Ta) ::; IITa + bll + rllbll Finally, let r  O. Then PB(Ta) ::; IITa every E > 0, so that the result follows.
(r > 0) .
+ bll < PB(Ta + b) + E.
This holds for 0
255
Representation theory
Corollary 5.46 (Generalized Johnson's theorem) Let A and B be Banach alge
bras, and let T : A + B be a linear s1trjection such that PB(Ta) :s; PA(a) (a E A). Then 6(T) s;:; J(B). In particular, T is continuous whenever B is semisimple. Proof Let b E 6(T), and let c E Q(B). Since T(A) = B, there exists a E A with Ta = (b+c). By Theorem 5.45, PB(c+b) :s; PB(C) = O. By Theorem 5.40, (d)=}(a), bE J(B). In the case where B is semisimple, it follows that 6(T) = {O}, and so T is continuous. D Let A, B, and T : A + B satisfy the conditions of the above theorem. Take bE 6(T). Then, for every sequence (bnk:~l in T(A) such that PB(b n  b) + 0, it follows that PB(bn) + O. But this does not imply that PB(b) = O. Corollary 5.47 Let A and B be Banach algebras with B semisimple, and let be an epimorphism. Then 0 is automatically continuous. D
o: A + B
5.11 A notorious open problem. We would like to offer a few thoughts about a notorious unsolved problem in this area, often called the dense range problem. The problem is as follows:
Let A and B be Banach algebras, with B semisimple, and let 0 : A + B be a homomorphism with dense range, so that O(A) = B. Is it true that 0 is automatically continuous? Remarks (i) By Theorem 5.25, the problem has a positive solution in the case where B is commutative. (ii) An equivalent formulation is to drop the requirements in the above statement that B be semisimple and that 0 have dense range, and ask the following question.
Let A and B be Banach algebras, and let 0 : A it always true that 6(0) s;:; Q(B)?
+
B be a homomorphism. Is
It follows from Exercise 5.9, below, that, for each b E 6(0), the set Spb is connected, and, of course, 0 E Sp b. But this does not exclude the possibility that PB(b) > 0 because the function PB might not be continuous at b. The proof in Theorem 5.45 shows that 6(0) n O(A) s;:; Q(B). (iii) We may divide the problem into two parts: (a) show that kerO is closed; (b) with the additional hypothesis that ker 0 is closed, prove that 0 is continuous. Each of these two subproblems appears to be open.
Elements analytically attached to subspaces. Let E be a complex Banach space, and let F be a vector subspace (not necessarily closed) of E. Then an element x of E is analytically attached to F if there is some open neighbourhood N of
o in
C and an analytic function f : N
+
E such that: (i) f(O)
=
x; and (ii)
Introduction to Banach Spaces and Algebras
256
J(z) E F (z E N\ {O}). We shall denote by a(F) the set of all elements of E that are analytically attached to F. The following comments are almost immediate: (i) F
<;;;;
a(F)
<;;;;
F;
(ii) a(F) is a vector subspace of E, and, if A is a Banach algebra with F a subalgebra of A, then a(F) is a subalgebra of A; (iii) without loss of generality, we can always take N = D = {z E CC : Izl < I}, simply by replacing J(z) by J(l5z) for a sufficiently small positive 15. For example, let E be any infinitedimensional, complex Banach space, and let F be a vector subspace that is the union of a strictly increasing sequence (En)n?l of closed subspaces; we remark that, by the Baire category theorem, Theorem 1.21, the vector subspace F is not closed. (For a specific example, take E = £2 and F = coo.) Now suppose that x E a(F). Then there is an analytic function J : D + E with J(O) = x and with J(z) E F for all zED \ {O}. But then there is some N E N such that J (z) E EN for uncountably many zED \ {O}. It is then elementary that J(z) E EN (z E D \ {O}). But EN is closed, and so also x = J(O) E EN <;;;; F. Hence a(F) = F. Let A be a complex, unital Banach algebra, and let a be element of A such that lIall ::; 1 and pea) =I 0 for each monic polynomial p. Define
J(z) = a(l  za)l
(z
E
D).
It is an elementary exercise to show that {fez) : zED} is a linearly independent subset of A. Define F = lin{f(z) : 0 < Izl < I}. Then it is easy to see that a = J(O) E a(F) \ F. Thus a(F) =I F. Lemma 5.48 Let B be a unital Banach algebra, let F be a vector subspace of B, let a E a(F), and take c > O. Then
pea) ::; sup{p(b) : bE F, lib  all < c}. Proof We may suppose that a rt. F and that there is an analytic function J : D + B with J(O) = a and J(z) E F for 0 < Izl < 1. Choose 0 < 15 < 1 such that IIJ(z)  J(O) II < c whenever Izl < 15. By the weak maximum principle, Theorem 5.39(ii), we have
pea)
=
p(f(O)) ::; sup{p(f(z)) : Izl
=
15} ::; sup{p(b) : bE F, lib  all < c},
o
giving the result. Corollary 5.49 Let A and B be Banach algebras, and let homomorphism. Then 6(e) n a(e(A)) <;;;; Q(B).
e:A
+
B be a
Proof Let b E 6(e) n a(8(A)), and take c > O. By Theorem 5.45, PB(C) < c for every c E e(A) with IIc  bll < c. Hence
PB(b) :S SUp{PB(C) : c E 8(A), IIc  bll < c} ::; c. Since this holds for every c > 0, we have PB(b) = 0, and so b E Q(B).
0
257
Representation theory
There is much that I do not know about the mapping F f+ a(F) when F is a vector subspace of a Banach space or a subalgebra of a Banach algebra. In order to make use of Corollary 5.49 in approaching the dense range problem, it would be necessary to have a useful sufficient condition for an element to belong to a(F). Two specific questions are as follows (with F a subspace of a Banach space E):
Question 1. Is it ever the case that a( a( F))
=I=
a( F)?
a(F) = F? We remark, finally, that the definition of a(F) can be iterated transfinitely (though, obviously, this is not of interest unless the answer to Question 1 is 'Yes'). Let us say that a subspace F of a Banach space E is analytically closed if a(F) = F. Clearly, every closed subspace is analytically closed, while examples given above show that, for a nonclosed subspace, both situations are possible. Question 1 is equivalent to asking whether a(F) is always analytically closed. It is clear that every intersection of analytically closed subspaces is itself analytically closed, and so there is a unique smallest analytically closed subspace of E that includes F: we shall denote that subspace by F"', and shall call it the analytic closure of F. It is clear that, for every subspace F of E, we have Question 2. Can it happen that F
F
~
=I=
a(F)
~
F'"
~
F.
We may also reach F'" 'from below' by transfinite recursion. We define aJ.L(F) for every ordinal JL as follows. (i) Let a1 (F) = a(F); (ii) for every ordinal JL, define aJ.L+1(F) = a(aJ.L(F)); (iii) define aJ.L(F) = U{av(F) : v < JL} for a limit ordinal JL. There is then a least ordinal, say v := v(F), such that av(F) is analytically closed. It is evident that av(F) = F"', as previously defined. If a(F) is not always analytically closed, then we may ask:
Question 3. Does there exist a nonclosed subspace F for which F'" = F? The following is an easy extension of Corollary 5.49. Corollary 5.50 Let A and B be Banach algebras, and let 8 homomorphism. Then 6(8)) n 8(A)'" ~ Q(B).
A
>
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 nonempty, open set in C, and f : U > A is an analytic function. The idea of proving Johnson's uniquenessofnorm 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
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Introduction to Banach Spaces and Algebras
is a reallinear subspace H of A such that A = H + iH and Sp a is finite for each a E H. Then A is finitedimensional. 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 threecircles theorem; Corollary 5.42 is given in [135], and Theorem 5.45 is contained in [19]. It appears that the more special threecircles 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 f4 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) <;; Q(A) = A, so that PA is not continuous on A; this semisimple Banach algebra contains as a dense, continuously embedded subalgebra a radical Banach algebra. The equivalent formulation of the dense range problem in the first remark, above is due to Albrecht and Dales in [3]; see also [47, p. 601]. Suppose that we assume (with B semisimple and T(A) = B) just that T : 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') <;;; Q(B) is open in this more general case. Exercise 5.8 Show that the following conditions on a Banach algebra A are equivalent: (a) J(A) = Q(A); (b) Q(A) + Q(A) <;;; Q(A); (c) Q(A) . Q(A) <;;; Q(A). Exercise 5.9 Let A be a unital Banach algebra, and let a E A. Suppose that K is a nonempty, openandclosed subset of Sp a. Prove that, for each open neighbourhood U of K in C, there exists 0 > 0 such that Sp (a + b) n U =1= 0 whenever b E A with Ilbll < O. Deduce that: (i) Sp a is a connected, compact set containing 0 whenever there is a sequence (a n )n2 1 in A such that an > a and p(an ) > 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 quasinilpotent. 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 complexvalued, continuous functions which vanish at infinity on a nonempty, 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 noncommutative 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 Hilbertspace 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 nmatrices 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 socalled Calkin algebra B(H)jK(H) is a Banach *algebra. D On page 76, we defined 'selfadjoint' (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 selfadjoint 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 selfadjoint elements of A is a reallinear 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 <;;; JR., as is shown by considering the element Z in the disc algebra, with the involution as in Example 6.1(ii); here Z is selfadjoint, but SpZ = fl.) In the case where A is a Banach *algebra, p(a*) = p(a) (a E A). Lemma 6.3 Let A be a *algebra. Then its radzcal J(A) is a *ideal, and so AI J(A) has a natural involution. Proof We may suppose that A is unital. Let a E J(A), and take b E A. Then (1 + ba*)* = 1 + ab*, which is invertible by Theorem 5.9(iii). By the above remarks, 1 + ba* is also invertible, so that a* E J(A) by Theorem 5.9(ii). Since a** = a, the converse holds. 0 Theorem 6.4 (Johnson) Let A be a semisimple Banach algebra with an involution. Then the involution is continuous on A, and A is a Banach *algebra for an equivalent norm. Proof Let 11·11 be the norm on A. The theorem is immediate from Corollary 5.29 because the formula
Illalll := max{llall, Ila*11}
(a E A)
defines a second Banach algebranorm on A, and this must then be equivalent to 11·11. Clearly, (A; 111·111) is a Banach *algebra. 0 Let A be a Banach algebra, and let a E A be such that Sp a n JR. = 0. Recall that we showed in Theorem 4.99 that there is a unique element bE A such that both b2 = a and Sp b c II. The element was called the 'principal square root of a'. There is a simple further point to be made in the case of an algebra with an involution; note that we do not suppose that the involution is continuous in the following result. Again, we set D =
= a* = a, and also
Sp b* = {X : A E Sp b} <;;; II . Hence, by the uniqueness of the principal square root, b* = b.
o
Algebras with an involution
263
Proposition 6.6 Let A be a Banach algebra with an involution, and let a be a normal element of A. Then there is a closed, commutative *subalgebra C of A, containing a and such that Spca = SPA a. Proof We may suppose that A is unital. Take C = {a, a*}Cc. Since {a, a*} is a commutative subset of A, it follows from Corollary 4.42 that C is a closed, commutative subalgebra of A such that Spca = SPA a. Clearly, a E C and C is *closed in A. 0 6.2
Positive linear functionals.
A+
Let A be a *algebra. Then we define
= {ta;aJ: a1, ... ,an
E
A, n E
J=l
N} .
Clearly, (A + , +) is a semigroup and aA + <;;; A + for each a we have 4ab
E
jR+. For a, b E A,
= (b+a*)* (b +a*)  (b  a*)* (b  a*) + i(b+ia*)* (b+ ia*)  i(b ia*)* (b  ia*) ,
and so A2 = linA+. A linear functional f on a *algebra A is positive if and only if f(a*a) :::: 0
(a E A).
In this case, f(A+) <;;; jR+. A positive linear functional f on a unital *algebra A is a state if f(l) = 1; the set of states on A is the state space of A, and it is denoted by S(A). For example, every character on C(K) for a compact Hausdorff space K is a state, and the map T f7 (Tx, Xl is a positive linear functional on 8(H) for each x E H; the latter map is a state if and only if Ilxll = l. Note that, for each b E A and each positive linear functional f on a *algebra A, the functional a f7 f(b*ab) is also a positive linear functional on A. Proposition 6.7 Let A be a *algebra, and let f be a positive linear functional on A. Then: (i) f(a*) = f(a) (a E A2);
(ii) f(a*b) = f(b*a) (a, bE A); (iii) (CauchySchwarz inequality) If( a* b) 12 ::; f( a* a )f(b* b) (a, b E A) ; (iv) in the case where A is unital, If(a)1 2 ::; f(l)f(a*a) (a E A). Proof (i) Let a E A2, say we have a = 'L;=1 AJa;aJ , where A1,'" An E C and al,··.,an E A. Then a* = 'L;=lXJa;aJ, and we know that f(a;a J ) :::: 0 for j = 1, ... , n. The result follows. (ii) This is a special case of (i).
264
Introduction to Banach Spaces and Algebras
(iii) Take a,b E A. Then f((Aa + p,b)*(Aa + p,b)) :::: 0 for each A,p, E Co In particular, take p, = f(a*b) and A = t E lR. Then
t 2f(a*a)
+ 2t If(a*b)1 2 + If(a*b)1 2 f(b*b)
:::: 0,
and this implies the CauchySchwarz inequality. (iv) This is a special case of (iii).
o
Proposition 6.8 Let A be a unital Banach *algebra, and let f be a positive
linear functional on A. Then: (i) If(a*ba)1 ::; f(a*a)p(b) (a E A, bE Asa); (ii) If(a*ba)1 ::; f(a*a)p(b*b)1/2 (a, bE A); (iii) If(a)1 ::; f(l)p(a*a)1/2 ::; f(l) Iiall (a E A); (iv) If(a*a)1 ::; f(l)p(a*a) ::; f(l) IIal1 2 (a E A); (v) f is continuous, wzth Ilfll = f(I), and f(A+) C
jR+.
Proof (i) Let a E A and b E Asa; we may suppose that p(b) < 1, and so Sp (1 ± b) ~ TI. By Proposition 6.5, there exist y, Z E Asa with (1  y)2 = 1  b and (1  Z)2 = 1 + b. Set u = a  ya and v = a  za. Then u*u = a*a  a*ba and v*v = a*a + a*ba. Thus f(a*a) ± f(a*ba) :::: 0, and so If(a*ba)1 ::; f(a*a). (ii) Let a, b E A. By the CauchySchwarz inequality,
If(a*ba)12 ::; f(a*a)f(a*b*ba), and so the result follows from (i). (iii) The first inequality is a special case of (ii); for the second, note that
p(a*a) ::; Ila*all ::; Ila11 2 . (iv) This follows from (iii) because p(a*aa*a) = p(a*a)2 (a E A). (v) Let a E A. Then If(a)1 ::; f(l) Iiall by (iii), whence Ilfll ::; f(l). Since Ilfll :::: f(l), we have Ilfll = f(l). Since f(A+) ~ jR+ and f is continuous, it follows that f(A+) C jR+. 0 Corollary 6.9 Let A be a unital Banach *algebra. Then
S(A) = {f
E
A* : f(A+) ~
Let A be a *algebra, and let define
jR+,
Ilfll =
f(l) = I}.
o
f be a positive linear functional on A. Then we
L j = {a E A: f(a*a) = O}. It is clear from the CauchySchwarz inequality that L j is a left ideal in A and that f(ba) = f(a*b) = 0 (a E L j , bE A). Suppose that A is unital. Then we define the *radical of A by J*(A) = n{L j
:
f
E
S(A)}.
The *algebra A is *semisimple if and only if J*(A) = {o}.
265
Algebras with an involution
Proposition 6.10 Let A be a unital *algebra. Then: (i) J*(A) = n{ker f : f E S(A)}; (ii) J*(A) is a *ideal in A; (iii) Aj J* (A) is *semisimple. Proof (i) Take a E J*(A). For f E S(A), we have a E ker f by Proposition 6.7(iv). Take a E n{ker f : f E S(A)}. For f E S(A) and ,\ E C, we have f(('\l
+ a)*a('\l + a))
=
0,
and so '\f(a*a) +).f(a 2 ) = 0. By taking'\ = 1 and'\ = i, we see that f(a*a) = 0, and so a E Lt. Thus a E J*(A). (ii) Certainly J*(A) is a left ideal in A. Let a E J*(A), and take f E S(A). Then the linear functional g : b 1+ f(aba*) is also positive, and so f(aa*aa*) = 0, whence a* E Lg s;;: J*(A). Thus J*(A) is *closed, and hence a right ideal.
o
(iii) This follows easily.
Proposition 6.11 Let A be a unital Banach *algebra. Then J*(A) is a closed ideal in A, AjJ*(A) is a *semisimple Banach algebra, and J(A) s;;: J*(A). Proof It follows from Proposition 6.8(iv) that ker f is closed for each f E S(A), and so J*(A) is closed. Thus AjJ*(A) is a *semisimple Banach algebra. Let a E J(A). Then a*a E J(A) s;;: Q(A), and so f(a*a) = (f E S(A)) by Proposition 6.8(iv). Thus a E J*(A). 0
°
Corollary 6.12 Let A be a unital Banach *algebra such that A is *semisimple. Then A is semisimple. 0 Theorem 6.13 (Kelley and Vaught) Let A be a unital Banach *algebra. Then J*(A)
=
{a E A: a*a E A+ } .
Proof Take a E A with a*a E A+, and let f E S(A). Then f(a*a) s;;: lR nlR+ by Proposition 6.8(v). Thus f(a*a) = 0, and a E Lt. It follows that a E J*(A). Conversely, suppose that a E J* (A). For a E Asa , let fJ,(a)
= inf{lla + bll : b E
A+}
= d( a, A+).
Then fJ, is a sublinear functional on the real vector space Asa , and so, by the HahnBanach theorem, Theorem 3.1, there is a reallinear functional f on Asa with f(a*a) = fJ,(a*a) and f(b) :::::: fJ,(b) (b E Asa). Extend f to be a complexlinear functional, also called f, on A. For b E A, we have  f(b*b) :::::: fJ,( b*b) = 0, and so f(b*b) ;::: 0. This shows that f is a positive linear functional, and so f(a*a) = because a E J*(A). Thus fJ,(a*a) = 0, and so a*a E A+. 0
°
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Introduction to Banach Spaces and Algebras
6.3 The GNS construction. We now give the socalled GNSconstruction for Banach *algebras; the letters G, N, S honour Gel'fand, Naimark, and Segal, who developed the theory. Let A be a unital Banach *algebra, and let J E S(A), so that L f is a closed left ideal in A. We denote by 7rf : A + AIL f the quotient mapping, and consider the equation (7r f (a), 7rf ( b)) f = J (b* a) (a, b E A) . (* ) Suppose that aI, a2 E A with al  a2 ELf. Then J(b*aI) But also, if bl , b2 E A with bl  b2 ELf, then J(bra)
=
J(a*b l )
=
J(a*b2)
=
JW2a)
=
J(b*a2) (b E A).
(a E A).
Thus the formula (*) gives a welldefined value to (7r f (a), 7rf (b)) f' It is then clear that ( " . ) f is an inner product on AIL f; the corresponding norm on AIL f is denoted by 11·lI f , so that l17rf(a)lI~ = J(a*a) (a E A), and the Hilbert space which is the completion of (AILf; 1I·lI f ) is called Hf; see Section 2.12. For a E A, define a mapping Tfa: AILf + AILf by (Tfa)(7rf(b)) = 7rf(ab)
(b E A).
That is, we make AILf into a left Amodule in the natural way, as in Section 5.1 In this notation, we have the following 'single representation' result, corresponding to a fixed J E S(A). Lemma 6.14 (i) For each a E A, the map Tfa is a bounded lmear operator on (AI L f ; 1I·lI f ), and so extends uniquely to a bounded linear operator (also denoted by Tfa) on H f .
(ii) The mapping Tf : a and IITfall :::::
lIall
f4
Tfa, A + B(Hf), is a unital *homomorphism
(a E A).
(iii) kerTf = Lf : A = {a E A: J(b*a*ab) = 0 (b E A)} ~ L f . (iv) J*(A)
= n{kerTf
: J E S(A)}.
Proof (i) For a, bE A, we have II(Tfa)(7rf(b))IIJ = J(b*a*ab). For b E A, the function g : a f4 J(b*ab) is a positive linear functional on A, and so, using Proposition 6.8(iv), we have J(b*a*ab)
=
g(a*a) ::::: g(1)lIaIl 2 = J(b*b)lIaIl 2 = l17rf(b)IIJ
lIall 2 .
Hence II(Tfa)(7rf(b))lIf = g(a*a)I/2::::: l17rf(b)lIf lIall, and so Tfa E B(AILf)· (ii) Clearly, the map Tf is a unital homomorphism with IITfall ::::: lIall (a E A), and so IITf II = 1. Now take a, b, c E A. Then ((Tfa)(7rf(b)), 7rf(c)) f
and so (Tfa)*
= Tf(a*).
=
J(c*ab)
=
J((a*c)*b)
=
(7rf(b), (Tf(a*))(7rf(c))) f'
Thus T f is a *homomorphism.
267
Algebras with an involution
(iii) This is immediate from the definition of T f . (iv) Let a E kerTf . Then a ELf, and so n{kerTf : f E S(A)} ~ J*(A). Conversely, let a E J*(A), and take f E SeA). Then f(b*a*ab) = 0 (b E A), and so a E kerTf . Thus J*(A) ~ n{kerTf : f E S(A)}. 0 The final step in the proof of our main representation theorem is to fit together the homomorphisms Tf for f E SeA). For this we need to be able to form a 'Hilbertspace directsum' of a collection of Hilbert spaces. The basic idea is quite simple, and we shall be brief over some details. Let (H).,; 11·II).,hEA be a collection of Hilbert spaces. Then the Hilbertian sum, H = L).,EA H)." of these spaces is the space H consisting of all those families
{ (x).,) E
IT H)., : L)., Ilx).,ll~ < oo} .
).,EA
It is an exercise to prove that, with respect to the obvious coordinatewise operations and with inner product given by (x, y)
=
L
(x)." Y).,).,
(x
=
(X).,».,EA, Y
= (Y).,».,EA)'
).,EA H becomes a Hilbert space; by Proposition 2.45, l(x)",y)")"1 ::::: Ilx)"lllly)"11 for oX E A, and so the series used to define the inner product is absolutely summable by Holder's inequality, Theorem 2.8(i). Let A be a unital Banach *algebra, and suppose that T)., is a unital *representation of A on H)., with liT)., II = 1 for each oX E A. For a E A, define
(Ta)((x).,».,EA)
=
(T).,a(x)"»)"EA
((X).,».,EA E H).
Then Ta E B(H) with IIT).,all ::::: IITal1 ::::: Iiall (oX E A), and the map T is a unital *representation of A on H with IITII = 1. It is called the direct sum of the *representations T)." and is denoted by T = E8{T)., : oX E A}. Now let Hf and Tf be as above for f E SeA), and set H = LfES(A) Hf· Then T = E8{Tf : f E S(A)} is a unital *representation of A on H with IITII = 1. Let a E A. It is clear that Ta = 0 if and only if Tfa = 0 (f E SeA»~, and so kerT
=
J*(A)
=
n{Lf : f E S(A)}.
Hence T is an injection if and only if A is *semisimple. Let A be a unital *algebra. Then a *representation T of A on a Hilbert space H is faithful if it is injective, and universal if T is unital and each state on A has the form a ~ (Ta(x), x) for some x E H with Ilxll = 1. Let T be a unital *representation of a unital Banach *algebra on a Hilbert space H, and take x E H with Ilxll = 1. Then f : a ~ (Ta(x), x) f is a state on A and IITa(x)11 2 = f(a*a) ::::: lIa11 2 , and so IITII = 1. Suppose that T is faithful and that a E J*(A). Then Ta = 0, and so a = O. Thus A is *semisimple. By combining the above remarks we establish the following theorem.
268
Introduction to Banach Spaces and Algebras
Theorem 6.15 Let A be a unital Banach *algebra. Then there is a Hilbert space H and a universal *representation T of A on H with IITII = 1. Further, there is a faithful universal *representation if and only if A is *semisimple. 0 Notes A magisterial account of *algebras and Banach *algebras is the twovolume work of Palmer [126, 127J. For a brief introduction, see [47, Sections 1.10, 3.1J and [140, Chapter 4J. There is a variation in the terminology regarding Banach *algebras: the point is that a Banach algebra with an involution is not necessarily a Banach *algebra. For example, for Palmer [127], the term 'Banach *algebra' does not imply that the involution is continuous. There is an example in [47, Theorem 5.6.83], from [45], of a Banach algebra A with a discontinuous involution * and an element a E A such that (expa)* =1= exp(a*). For a study of topological algebras with an involution, see [78J. An involution * on a unital algebra A is symmetric if 1 + a*a E G(A) for all a E A and hermitian if Sp a <;;; lR for each a E Asa. A symmetric involution is always hermitian. In the case where A is a Banach algebra, the converse is true, and this holds if and only if p(a)2 ::; p(a*a) (a E A); see [47, Corollary 3.1.11J. There seems to be no standard notation for what we have called J*(A), the *radical. It is called the prereducing ideal in [127, Definition 9.7.14J. It seems to be unknown whether or not J(A) <;;; J*(A) for every *algebra A. The theorem of Kelly and Vaught is from [108J. For more on Hilbertian sums, see [102, Section 2.6J. The GNS construction is extensively discussed in the literature. Exercise 6.1 Let A be an infinitedimensional Banach space, and define ab = 0 for a, b E A. Then A is a commutative Banach algebra. Let {Xn : n E N} U {y, : 'Y E r} be a basis of A consisting of elements of norm 1. Define
y; = y,
('Y E
r) ,
* X2n
= nX2n1 ,
X;n1
= X2n/n
(n
E N),
and extend * by conjugatelinearity to A. Verify that * is a discontinuous involution on A. Exercise 6.2 Let A be a Banach *algebra. Show that (expa)* = exp(a*) (a E A). Exercise 6.3 Let (H; ( . , . )) be a Hilbert space, and suppose that S, T E .c(H) are such that (Tx, y) = (x, Sy) (x,y E H). Use the closed graph theorem to prove that both Sand T are continuous. Thus it is sensible to define T* for T E .c(H) only if T E B(H). Exercise 6.4 Let G be a group. We defined the group algebra (f 1 (G); Example 4.4. Now define
* : '~ " f(8)8 5 sEC
f> " ~ '  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 selfadjoint, 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 (nonzero) 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 nonempty set, and let X be a nonempty 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 Hilbertspace 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 nonempty, 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 selfadjoint. 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 quasinilpotent as well as selfadjoint, 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.
271
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
D
Proposition 6.23 Let A be a unital C * algebra, and let B be a C * subalgebra of A with lA E B. Then SPBb = SPAb for every b E B. Proof Of course we always have Sp Bb ~ SPA b for each b E B. For the reverse inclusion, it suffices to show that, if b E B is not invertible in B, then it is not invertible in A. Thus let bE B, and suppose that b rf. G(B). By Proposition 6.2, at least one of b*b and bb* is not invertible in B, say b*b rf. G(B). But b*b is selfadjoint, and so, by Corollary 6.22(i), SpAb*b <;;; R By Corollary 4.38, SpBb*b = SPAb*b, and so b*b rf. G(A). Thus b rf. G(A). D Let H be a Hilbert space, and let T E B(H); we write SpT to denote the spectrum of T regarded as a bounded linear operator on H. Clearly, SpT is identical with the spectrum of T when T is regarded as an element of the Banach algebra B(H). By Theorem 6.23, SpT = SPAT for any C*subalgebra A of B(H) that contains T.
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Introduction to Banach Spaces and Algebras
Theorem 6.24 (Commutative Gel'fandNaimark) Let A be a commutative, unital C * algebra. Then the Gel 'fand representation of A is an isometric *isomorphism of A onto C(1)A). Proof Using Corollary 6.21(ii), we already have the formula
;?(cp) = cp(a*) = cp(a) = a(cp)
(a
E
A, cp
E
1>A),
and so the Gel'fand transform g: a f+ a, A + C(1)A), is a *homomorphism. But also each a E A is normal, and so lalA = p(a) = Iiall by Lemma 6.17; this implies that 9 is isometric. The image A = {a : a E A} is a closed subalgebra of C( 1> A) that contains the constants, separates the points of 1> A, and is closed under conjugation. By the StoneWeierstrass theorem, Corollary 2.33, A = C(1)A). Thus 9 is an isometry. 0 In the case where A is a commutative, nonunital C *algebra, we see that 1> A i= 0 and that 9 : A + C o ( 1> A) is an isometric *isomorphism.
Example 6.25 Let S be a nonempty set, and write (3S for the character space of the commutative C *algebra £OO(S). Then £OO(S) is isometrically *isomorphic to C((3S) for a certain compact Hausdorff space (3S; (3S is the StoneCech compactincation of S. See also Exercise 6.6. 0 6.5 The continuous functional calculus. We described the holomorphic functional calculus for an element a in a Banach algebra A in Section 4.15; this was a map 8 a from OSPAa into A such that 8 a (Z) = a. We now show that, for a normal element a of a C* algebra A, continuous functions, rather than just analytic functions, 'operate on' a. We write C*(a) for A(a, a*), the closed C *subalgebra of A polynomially generated by the elements a and a*. Theorem 6.26 (Continuous functional calculus) Let A be a unital C *algebra, and let a be a normal element of A. Then there is a unique unztal *homomorphism 8 a : C(Sp a) + A such that 8 a (Z) = a. Moreover:
(i) 8 a is isometric; (ii) im 8 a
=
C*(a);
(iii) a8 a (g) = 8 a (g)a and Sp(8 a (g)) = g(Spa) for every g E C(Spa). Proof Set C = C* (a), a commutative, unital C *subalgebra of A. By Proposition 6.23, we have Sp a = Spca. Note first that, by the StoneWeierstrass theorem, Corollary 2.35(ii), polynomials in Z and Z* are dense in C(Spa), so that each (necessarily continuous) *homomorphism B: C(Spa) + A with B(Z) = a has imB ~ C and is uniquely defined. The mapping
c + Spa, is a homeomorphism, and so, identifying 1>c with Sp a, we see that the Gel'fand representation of C is an isometric
273
Algebras with an involution
*isomorphism from C onto C(Spa). The mapping 8 a that we seek is then just the inverse of this Gel'fand representation. The required properties of 8 a are immediate from Theorem 6.24. 0
Remarks: (i) The mapping 8 a is called the continuous functional calculus map for the normal element a of A. Frequently, 8 a (g) is written as g(a) whenever gEC(Spa). (ii) The continuous functional calculus extends the holomorphic functional calculus 8 a in the following sense: if f E OSPA a , then f I Spa E C(Spa) and 8 a (f) agrees with its previous definition. This fact is an easy consequence of the uniqueness property of the holomorphic functional calculus. (iii) An important application of the continuous functional calculus is obtained by taking a = T, a bounded normal operator on a Hilbert space H. Then
Ilg(T)11
=
Igl spT
(g
E
C(SpT)).
Corollary 6.27 Let A be a unital C * algebra, and let a be normal in A. Let f E C(Spa). Then f(a) is normal and (gof)(a) = g(f(a)) (g E C(Sp(f(a)))). Proof This follows from Theorem 6.26 in the same way as Theorem 4.95 followed from Theorem 4.89. 0 Corollary 6.28 Let H be a Hilbert space, and let T E 13(H) be a normal operator. Suppose that A is an isolated point ofSpT. Then A is an eigenvalue ofT, and there is an orthogonal projection P : H > E(A). Proof By Corollary 4.97, there is a nonzero projection P E 13(H) such that TP = PT and SP13(p(H))(T I P(H)) = {A}. Since P is the image of a realvalued function in C(Sp T), it follows that P = P*, and hence that PT* = T* P and TP is normal. Thus (T  >.IH)P is normal in A, and p((T  >.IH)P) = O. By Lemma 6.17, TP = AP, and so Tx = AX (x E P(H)). Thus A is an eigenvalue ofT.
0
Let H be a Hilbert space, and let T E 13(H) be both compact and normal. By Theorem 4.34, each nonzero A E Sp T is an isolated point of Sp T and an eigenvalue, and the corresponding eigenspace E(A) is finitedimensional. By Corollary 6.28, there is an orthogonal projection P).. = P; : H > E(A). Again, set T).. = T I E(A), so that T).. = >.IE()..) and Sp 13(E()..))T).. = {A}. It is clear that T).. is a normal operator (or matrix) on E(A), and it is standard linear algebra that there is an orthonormal basis of E(A) (use the GramSchmidt procedure, for example) with respect to which T).. is a diagonal matrix. The union of all these orthonormal bases, together with an orthonormal basis for ker T, is an orthonormal basis for H. Thus we may regard T as the direct sum of the zero operator on a closed subspace of H and a diagonal operator Ef){.\P).. : A E SpT}. Let us be a little more precise about the formula T = Ef){.\P).. : A E Sp T}, given above. Consider the case where Sp T is infinite. Let us enumerate Sp T as
274
Introduction to Banach Spaces and Algebras
(A n )n>l, where the An are distinct and IAn+11 :::; IAnl (n EN), and write En for E(An). For each n E N, choose an orthonormal basis {C mn _ 1 +1, .... 'c mn } of En (where mo = 1), so that (Ck k::: 1 is an orthonormal basis of a closed linear subspace of H, and set Ji,k = An for k = m n1 + 1, ... , m n . Then we have CXl
Tx = LJi,k(X, Ck)Ck k=l
(x
E
H).
Now take n E N, and set fn(z) = Z (izi :::; IAnl) and fn(z) = 0 so that fn E C(SpT). Then T  ~~=1 AJP;,,) = 8 T (fn), and so
(Izl > IAnl),
n
TLAJP;"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 righthand 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 strongoperator 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 <:;; SPAa. Set C = C(SpAa). Then 0 0 Sa and Sea I C are two *homomorphisms from C into B which agree on Z, and so, by the remark on uniqueness in Theorem 6.26, 0 0 Sa = Sea I C. It follows that f(Oa) = O(f(a)).
(ii) Assume towards a contradiction that there exists A E SPA a \ Sp BOa. Take f E C(SPAa) such that f(A) = 1 and f I SPBOa = O. Then f(a) =I 0, but O(f(a)) = 0, a contradiction of the fact that 0 is injective. Thus SPBOa = SPAa. For each b E A, the element b*b is normal, and so it follows from the above that PB((Ob)*(Ob)) = PA(b*b). Now it follows as in Corollary 6.19 that 0 is an isometry. It is then immediate that O(A) is a C*subalgebra of B. D An element a of a C* algebra A is positive, written a :::: 0, if a E Asa and pa C JR+. The collection of positive elements is denoted by Apos. We write a :::: b for a, b E Asa if a  b :::: o. Theorem 6.30 Let A be a C * algebra, and let a E Asa. Then a :::: 0 if and only if a = b2 = b*b for some b E Asa. In the case where a :::: 0, we may choose b so that also b:::: 0, in which case b zs uniquely specified. Proof Suppose that a = b2 for some b E Asa. Then Sp a = {A2 : ). E Sp b} C JR+ because Sp b c JR, and so a :::: o. Conversely, suppose that a :::: O. The nonnegative square root function
Zl/2 : t
f+
tl/ 2
is well defined and continuous on JR+, and hence Zl/2 E C(Spa). By the continuous functional calculus, the element b := Zl/2(a) E A(a) satisfies b E Asa and b2 = a. Note also that Spb = Zl/2(Spa) C JR+, so that b:::: o. Now let c E Asa be such that c2 = a. Then c commutes with a, and so also with b because b E A(a). Set C = A(b,c). Then C is a commutative C*sub algebra of A containing band c (and therefore a). By Theorem 6.23, we have Spcx = Sp AX (x E C). We now apply the commutative Gel'fandNaimark theorem to the algebra C to see that and b are both square roots of a in C(c). If both b,c E Apos, so that b,CE C(o)+, then b = C, and then b = c. This establishes the uniqueness of b. D
c
Corollary 6.31 Let A and B be unital C * algebras, and let 0 : A ; B be a unital *isomorphism. Then O(Asa) = Bsa and 0 I Asa : Asa ; Bsa is an orderpreserving, isometric, reallinear isomorphism. In the case where A is commutative, 0 I Asa : Asa ; Bsa is a realalgebra isomorphism. D
Let H be a Hilbert space, and take T E B(H). Recall from page 77 that T is a positive operator if and only if (Tx, x) 2: 0 (x E H). It is important that this usage does not conflict with the notion of positivity just introduced. That this is indeed the case is established in the next result.
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Introduction to Banach Spaces and Algebras
Proposition 6.32 Let H be a Hilbert space. An operator T E B(H) is a positive operator if and only if it zs positive as an element of the C * algebra B(H). Proof Suppose that T is positive as an element of B(H). Then T E B(H)sa and there is 8 E B(H)sa with 8 2 = T. Then, for every x E H, we have
(Tx, x)
= (8 2 x, x) = (8x,8x) = II 8xll 2
~ 0,
so that T is a positive operator. Conversely, suppose that T is a positive operator on H. Then we have
(Tx, x)
= (x, Tx) = (T*x, x) (x
E H) ,
and so, by Corollary 2.56, T = T*. Since T is normal, for each A E Sp T, there is a sequence (Xn)n~l of unit vectors in H for which (Txn, x n ) + A (see Theorem 4.30). It follows that A ~ 0, and T is therefore positive as an element of the C* algebra B(H). 0 Let T E B(H). Then 8
= T*T
is positive because
(8x, x) = (T*Tx, x) = (Tx, Tx)
~
0
(x E H) .
In fact the same result holds for an element of any C* algebra, but that is decidedly less obvious; it will be proved in Theorem 6.38. Proposition 6.33 Let A be a unital C*algebra, and let a Then there exists b E Asa such that eb = a. Proof We have Sp a
c
~+ •.
~
0 with a E G(A).
Set
f(t)
= logt (t > 0),
so that f E C]R(Spa) and b:= f(a) E Asa. Set g ~+., and so eb = a by Corollary 6.27.
= exp on R Then go f = Z on 0
Proposition 6.34 Each element of a unital C * algebra is a linear combination of four unitary elements. Proof Let A be a C * algebra. It is sufficient to show that each a E (Asa)[lJ is a linear combination of two unitary elements. For such an a, we have Sp a ~ [1, 1]. Define f(t) = t + (t E [1,1]),
iJ1=t2
so that f E C[l, l], Z = (f + 7)/2, and f7 = 7f = 1. Set u a = (u + u*)/2 and uu* = u*u = lA, giving the result.
= f(a). Then 0
The following result is included mainly for the elegance of its proof; also, it will be used in Theorem 7.19.
277
Algebras with an involution
Theorem 6.35 (Fuglede's theorem) Let a be a normal element of a C* algebra A, and let b E A satisfy ba = abo Then also ba* = a*b. Proof We may suppose that A is unital. We have anb = ban for each n EN), and it follows that, for every z E
fez) = exp(za*  za)bexp( (za*  za)) = u(z) bU(Z)l
(z E
q,
so that Ilf(z)11 ::; Ilbll (z E q, and hence f is bounded. But f is an Avalued entire function, and so is constant by Liouville's theorem, Theorem 3.12. In particular, fez) = f(O) = b (z E q, and so exp(za*)b = bexp(za*). Expanding the exponentials and equating the coefficients of z in the two sides of the equation gives a*b = ba*. D 6.6 Representation of general C*algebras. Our main objective in this section is the Gel'fandNaimark theorem, Theorem 6.47, which gives the canonical form of an arbitrary C* algebra; en route to this, we shall show in Theorem 6.38 the key fact that a*a is always positive in a C *algebra. We shall utilize the class of positive linear functionals on a C *algebras A and the state space SeA). We require two further lemmas towards Theorem 6.38.
Lemma 6.36 Let A be a unital C *algebra. (i) Let a E Asa. Then a ::::: 0 if and only if Iltl all::; t for each (respectively, some) t::::: Iiali. (ii) Suppose that a, bE Apos. Then a + b E Apos. Proof (i) We shall use Lemma 6.17. Suppose that a ::::: 0 and t ::::: Iiali. Then Sp a <:: [0, II all], and so we have Sp(tl a) <:: [t ilall ,t] C JR+. Thus Iltl all = p(tl a)::; t. Conversely, take t E JR+ with Iltl all::; t. Then Sp (tl a) <:: [t, t], so that Sp(a  tl) <:: [t,t] and Spa <:: [0,2t]. In particular, a::::: O. (ii) Certainly a + bE Asa. Set s = Iiall and t = Ilbll, so that Ilsl all::; sand Iitl  bll ::; t by (i). But then s + t ::::: Iia + bll and lI(s + t)1  (a
so that a + b ::::: 0 by (i).
+ b)11
::; Iisl  all
+ Iitl 
bll ::; s
+ t, D
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Introduction to Banach Spaces and Algebras
Lemma 6.37 Let A be a unital C*algebra, and take a E A with a*a Then a = o. Proof Set a = b + ie, with b, c E
Asa.
~
o.
By Proposition 4.16(ii),
Sp (aa*) ~ Sp (a*a) U {O} ~ lR.+ , and so aa* ~
o.
But a*a
+ aa* = 2(b 2 + c2), a* a
=
so that
2(b 2 + c2)  aa* ~ 0
by the last lemma. Since also a*a
~
0, we have Sp (a*a)
=
{O}, and hence
IIal1 2 = Ila*all = p(a*a) = o. Thus a
=
o.
o
Theorem 6.38 Let A be a C*algebra. Then a*a
~
0 for every a EA.
Proof We may suppose that A is unital. Set b = a*a and B = A(b), so that B is a commutative C*subalgebra of A. Using the commutative Gel'fandNaimark theorem, Theorem 6.24, for B, we may write b = b+  b, where b+, b ~ 0 in Band b+b = o. Let c = ab. Then
c*c
= ba*ab = b(b+  b)b = _(b)3.
Since b ~ 0 and Sp((b)3) = {t 3 : t E Spb}, we see that c*c By Lemma 6.37, c = 0, so that (b)3 = 0, and then, since b E b = O. Hence a*a = b+ ~ o. Corollary 6.39 Let A be a and A+ = Apos.
C
* algebra. Then the subset
Apos
=
(b)3 ~ o. we have
Asa,
0
is closed in A,
Proof We may suppose that A is unital. By Lemma 6.36(i), we have Apos
= {a
E Asa :
IIlIall . 1  all::::; lIall},
and so it is clear that Apos is closed. By Theorem 6.30, Apos ~ A+. By Theorem 6.38 and Lemma 6.36(ii), we also have A+ ~ Apos. Hence A+ = Apos. 0 Corollary 6.40 (Polar decomposition of an invertible element) Let A be a unital = bu with
C * algebra, and let a E G (A). Then there is a unique decomposition a b 2: 0 and u E U(A). Moreover, b = (aa*)1/2.
279
Algebras with an involution
Proof Since a E G(A), we have aa* E G(A) (by Proposition 6.2) and aa* ~ 0 (by Theorem 6.38). By Theorem 6.30, there exists b ~ 0 with b2 = aa*j clearly bE G(A). Let u = b1a. Then u E G(A) and uu* = b1aa*b 1 = b 1b2 b 1 = 1. Thus 1 u = u* and u E U(A). If also a = cv with c ~ 0 and v E U(A), then aa* = cvv*c = C2 j by the uniqueness statement in Theorem 4.99, c = b, and then v = u. 0 We summarize some results obtained so far. Theorem 6.41 Let A be a C * algebra, and let a E A. Then the following are equivalent:
(a) a E A+ ; (b) a ~ 0, so that a
Apos; (c) a = b2 for some b E Asa ; (d) a = b*b for some bE A; E
(e) a E Asa and Iitl  all ::; t for each t E lR with t ~ Iiali.
o
Corollary 6.42 Each C * algebra is *semisimple. Proof Let A be a C*algebra, and take a E J*(A). Since, by Corollary 6.39, A+ is closed, it follows from Theorem 6.13 that a*a E A+. By Theorem 6.41, 0 (a)=}(b), we have a*a ~ O. By Lemma 6.37, a = O. Thus J*(A) = {O}. Lemma 6.43 Let A be a unital C * algebra, and take f E A *. Then f is positive if and only if Ilfll = f(I). Proof Suppose that f is positive. Then Ilfll = f(l) by Proposition 6.8(v). For the converse, suppose that f E A* with Ilfll = f(I). The result is trivial when f = 0, and so we may suppose that IIfll = f(l) = 1. By Proposition 6.20, f(a) E lR whenever a E Asa. Next, we shall show that f(a) ~ 0 whenever a ~ O. We may suppose that pea) = Iiall = 1, so that Spa <;;; 1I. Then Sp (1  a) <;;; 1I and, by Lemma 6.17, 111  all = p(1  a) ::; 1. Thus 11  f(a)1 = If(1  a)1 ::; 1, so that f(a) ~ O. Now, for each a E A, we have a*a ~ 0 by Theorem 6.38, so that f(a*a) ~ 0 and f is a positive linear functional. 0 Let A be a unital Banach algebra. Recall from page 205 that
KA = {J E A* ; Ilfll = f(l) = I}. Corollary 6.44 Let A be a unital C * algebra. Then SeA)
= {J E A* ; Ilfll = f(l) = I} = K A
.
Proof This is immediate from the definitions and Lemma 6.43.
o
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It is simple to see that SeA) is a convex, weak*compact subset of A*. We now define the numerical range, WA(a) = W(a), of an element a of a unital Banach algebra A: it is the subset
WA(a) = W(a) = {I(a) : f
E
KA}.
The numerical radius of a is WA(a) = w(a) = sup{I).1 : ). E W(a)}, so that w(a) ::; Iiali. Proposition 6.45 Let A be a unital Banach algebra, and let a, b E A. Then:
(i) W(a) is a nonempty, compact, convex subset ofe; (ii) W(a1 + f3a) = a + f3W(a) (a, f3 E
q
and W(a + b)
~
W(a) + Web);
(iii) Spa ~ W(a); (iv) WB(a) = WA(a) for each Banach subalgebra B of A with 1,a E B; (v) pea) ::; w(a) ::; Ilall· Proof Clause (i) follows because W(a) is the image of KA under the continuous linear mapping f f+ f(a) from (A*;O"(A*,A)) into e, and (ii) is clear.
(iii) Let). E Spa. For (, TJ E C, we have (A + TJ E Sp ((a + TJ1), and so I(A+TJI ::; II(a+TJ111· Thus the map f: (a+TJ1 f+ ( ) . + TJ, lin{l,a} + e, is a linear functional on lin{l,a} with f(l) = Ilfll = 1. By the Hahn~Banach theorem, f has a normpreserving extension to an element, also called f, of KA· Thus). = f(a) E W(a). (iv) The restriction map f
f+
fiB maps KA onto K B .
o
(v) This follows immediately. Proposition 6.46 Let A be a unital C * algebra, and let a E A. Then:
(i) Ilall/2 ::; w(a) ::; Iiall ; (ii) a E Asa if and only if W(a) c JR; (iii) a 2 0 if and only ifW(a) c JR+; (iv) there exists f E SeA) with f(a*a)
= Ila11 2.
Proof (i) First, take b E Asa. Then pCb) = Ilbll by Lemma 6.17, and so we also have web) = Ilbll from Proposition 6.45(v). Now write a = b+ic, with b, c E Asa. Then either Ilbll 2 Ilall/2 or Ilcll 2 Ilall/2, say Ilbll 2 Ilall/2. There exists f E SeA) such that If(b)1 = Ilbll, and then
If(a)12 = f(b)2 + f(c)2 2 f(b)2 = IIbl1 2 2 (1Iall/2)2. Hence w(a) 2: Ilall/2. (ii) Suppose that a E Asa. Then, by Lemma 6.20, f(a) E JR for each f E SeA), and so W(a) C R
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Conversely, suppose that W(a) c R Write a = b + ic, with b, c E Asa. Then, for each f E S(A), we have f(a) = feb) + if(c) with f(a), f(b), f(c) E R Thus f(c) = 0, and so W(c) = {O} and w(c) = o. By Proposition 6.45(v), c = 0, and so a = b E Asa. (iii) Suppose that a ~ o. Then, by Theorem 6.30, a = b*b for some b E A, and so f(a) = f(b*b) ~ 0 (J E SeA)). Hence W(a) C JR+. Conversely, suppose that W(a) C JR+. By (ii), a E Asa, and then it follows that a ~ 0 because Spa ~ W(a). (iv) By Proposition 6.45(v), p(a*a) ::; w(a*a) ::; Ila*all. But p(a*a) = Ila*all by Lemma 6.17, and so w(a*a) = Ila*all. The result follows because we have
SeA) = K A .
0
We now have all the ingredients to complete the proof of the famous noncommutative Gel'fandNaimark theorem. Theorem 6.47 (Gel'fandNaimark theorem) Let A be a C* algebra. Then there is a Hilbert space H and an isometric *isomorphism of A onto a C* subalgebra
ofl3(H). Proof We may suppose that A is unital. By Corollary 6.42, the C* algebra A is *semisimple. By Theorem 6.15, there is a Hilbert space H and a faithful, universal *representation T of A on H with IITII ::; l. By Proposition 6.46(iv), for each a E A, there exists f E SeA) such that f(a*a) = Ila11 2. Thus IITfal12 ~ II(Tfa)(7rf(l))II~ = f(a*a) = IIal1 2, and this implies that IITal1 ~ IITfal1 ~ Iiali. It follows that IITal1 = lIall, and so T is an isometry; its image is a C*subalgebra of l3(H). 0 6.7 Weakoperator convergence in l3(H). Let H be a Hilbert space. We shall later need to use the notion of weakoperator convergence in l3(H); the weakoperator topology, written wo, on l3(H) was introduced in Section 3.6. In the present case, it is the locally convex topology defined by the collection of seminorms {Px : x E H}, where
Px(T) = I(Tx,x)1
(T E l3(H)).
That this is exactly the weakoperator topology, as defined earlier, follows from the polarization identity of Proposition 2.46. The weakoperator topology is strictly weaker than the norm topology on l3(H) whenever H is infinitedimensional. For example, let H = £2, let (en)n> 1 be the standard orthonormal basis of H, and let Pn be the operator of projection onto the nth coordinate for each n E N. Thus, for any x = (Xn)n~l in H, we have Pn(x) = xnen and (Pnx,x) = Ix n l2 + 0 as n + 00. Thus Pn~O. However, IIPnl1 = 1 (n EN), so certainly Pn f+ 0 in (l3(H); 11·11). The following result is a slight variation of Proposition 3.24.
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282
Proposition 6.48 Let H be a Hzlbert space. .) (1
wo
Suppose that Tn 7 T. Then
wo
T~ 7 T*
.
(ii) Suppose that Tn~T and S E l3(H). Then STn~ST and TnS~TS. In particular, ifTnS = STn (n EN), then TS = ST. (iii) For any subset E of l3(H), its commutant EC zs closed under the weak0 operator convergence of sequences. Indeed, a commutant E C is always weakoperator closed. Let H be a Hilbert space, and take T E l3(H)sa. Recall that T :::: 0 (i.e. T is positive) if and only if (Tx, x) :::: 0 (x E H) (cf. Proposition 6.32). Then (l3(H)sa; ~) is a poset.
Theorem 6.49 Let H be a Hzlbert space, and let (Tn)n2:1 be an increasmg sequence in (l3(H)sa;~) with Tn ~ kI (n E N) for some k :::: o. Then (Tn)n2:1 is strongoperator convergent to T E l3(H)sa. Moreover, T = sUPn2:1 Tn. Proof First, we remark that the convergence of the sequence (Tn)n2:1 is equivalent to the convergence of (Tn  Tdn2:1' so that we may suppose that we have o ~ Tn ~ kI (n EN). For each x E H, the sequence ((Tnx, x) )n2:1 is increasing in [0, kllxl1 2 ], and so it converges in JR.+ to a limit in [0, kllxI1 2 ]. By the polarization identity, ((Tnx, y) )n2:1 converges to a limit, say 'ljJ(x, y) for each x, y E H. Clearly, 'ljJ is a sesquilinear form on H, and also IITnl1 =sup{I(Tnx,x)l: Ilxll = I} ~ k because each Tn is selfadjoint. Let x,y E H. Then I(Tnx,y)1 ~ kllxllllyll, and so I'ljJ (x, y)1 ~ kllxlillyll, i.e. 1/; is a bounded sesquilinear form. By Theorem 2.55, there is a unique T E l3(H) such that (Tx, y)
= 'ljJ(x, y) = n>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], 'proC·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 VidavPalmer theorem; see [31, Section 38] and [127, Theorem 9.5.9]. For more on the strongoperator and weakoperator 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 weakoperator topology on B(H). Let A be a C· algebra. Then every derivation from A into a Banach Abimodule 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 StoneCech 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 selfadjoint, 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 infinitedimensional Hilbert space H. Then 1'*1' is compact and selfadjoint, 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
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 strongoperator and weakoperator 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 nonempty 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 pointwisemonotone 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 nonempty, locally compact Hausdorff space, and set L = CO,IR (K), the space of continuous, realvalued functions that vanish at infinity, so that L C;;;; JRK. Then the set of (realvalued) Baire functions on K is defined to be the monotone class generated by L. A realvalued Baire function on K is simply a member of this set; a complexvalued function on K is a Baire function if and only if its real and imaginary parts are both realvalued Baire functions.
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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 realvalued Baire functions is strictly larger than Cffi.(K). For example, the characteristic function of any nonempty, open subset of II is the pointwise limit of an increasing sequence of continuous functions, and so a realvalued 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 nonempty set. A subset M of IR x is a boundedmonotone 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 boundedmonotone class containing E; it is the boundedmonotone class generated by E. For the remainder of this section, we take K to be a nonempty, locally compact Hausdorff space, and set L = Co,ffi.(K). We write Bffi.(K) and B(K)for the sets of bounded, realvalued 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 boundedmonotone class generated by L.
(ii) The set Bffi.(K)+ is the boundedmonotone class generated by L+. Proof (i) From the definition of the Baire functions, it is clear that Bffi.(K) is a boundedmonotone class on K that contains L. Now suppose that M is any such boundedmonotone class, and define E to be the set of all those realvalued 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 realvalued 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 nonempty, 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 boundedmonotone 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).
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The Borel functional calculus
A similar argument (first considering nonnegative 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 complexvalued functions, by combining real and imaginary parts, is now immediate. D
Corollary 7.3 Let K be a nonempty, 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) <;;; SPB(K)f· Since SPB(K)f is compact, f(K) <;;; SPB(K)f. Now take A E
(i) X
E
I:;
(ii) if E E I:, then X \ E E I:; (iii) if (En)n?1 is any sequence of sets in I:, then also
Un?1 En
I:. Let I: be a O'field on X. Then, from (i) and (ii), 0 E I:, so that I: is also closed under finite unions. Also, by (ii) and (iii), I: is closed under countable intersections. Clearly, P(X) is itself a O'field on X, and the intersection of any collection of O'fields is a O'field. Thus, given a family E in P(X), there is a smallest O'field on X that includes E; this is the O'field generated by E. Let I: be a O'field on a set X. A complex measure on I: is a function fJ : I: +
of sets in I: with U~1 E J = E. Now let (X; T) be a topological space. Then the Borel O'field on X is the O'field generated by T; it is denoted by B(X). A Borel subset of X is simply a member of B(X). A measure defined on B(X) is a Borel measure. For technical reasons, it is sometimes necessary to use a (potentially) smaller O'field on X than B(X): this is the Baire O'field on X. It is the O'field generated by the collection of all the compact, Gosubsets of X, and it is denoted by Ba(X). It.is important to note that, when X is a compact metric space (in particular, a compact subset of
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Consider a function I : X ., C. Then I is a Borel function (respectively, a Baire function) if and only if, for every open subset U of C, the set 11 (U) is a Borel set (respectively, a Baire set) in X. Clearly, every function in C(X) is a Baire function. Suppose that I is a Borel function. Then it is easy to see that E B(X) for each B in B(C), and so g 0 I is a Borel function whenever g : C ., C is a Borel function. It is not difficult to show that this notion of a Baire function on a locally compact Hausdorff space K is identical with that previously defined in terms of monotone classes on K. Note that, for us, a Baire subset of X is just a subset E such that XE is a Baire function; again, this is equivalent to the definition in terms of O"fields. The following easy remark gives some examples of Baire functions.
1 1 (B)
Proposition 7.4 Let K be a nonempty, compact metric space, let a E K, and let I E eOO(K) be continuous on K \ {a}. Then I E B(K). Proof Since eOO(K) = lineOO(K)+, it suffices to suppose that I E eOO(K)+. Set U = K \ {a}. Then there is an increasing sequence (h t ) in C~(K)+ with h t i Xu, and with supph t <:;;; U (i EN). So each htl is continuous on K, and h,f i xul, which is therefore a Baire function on K. But also X{a} is a Baire function, so that I = I(a)x{a} + Xu I is a Baire function. 0
7.2 The Daniell extension process. Throughout this section, K will be a nonempty, compact Hausdorff space, and we again set L = C~(K). (We require K to be compact, not just locally compact, to ensure that 1 E K; more general situations are indicated in the notes.) An integral on L is defined to be a reallinear functional 1 : L ., IR which is positive, in the sense that 1(f) 2': 0 whenever I E L +. For example, set K = il, and let 1(f) =
10
1
I(x) dx
(f E L),
the Riemann integral. Then 1 is an integral on L in our new sense.
Lemma 7.5 Let 1 be an integral on L. Then:
(i) 11(f)1 ::; 1(1) IIIK (f E L); (ii) il (ft) in L and
IE
L, and il either It
1I
or It
i I, then 1(f,) ., 1(f).
Proof (i) This follows from the positivity of 1 since IIIK ::; I ::; I/IK (f E L). (ii) By Dini's theorem, Corollary 5.38, I, ., I uniformly on K, so that 1(f,) ., 1(f) by (i). 0 The Daniell process will give a way to extend 1 to an integral on B~ (K). In the case of the above example, with K = IT, this would lead to the Lebesgue integral restricted to the bounded, Borel functions, rather than on the slightly
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more general space of all Lebesgue integrable functions. But it should be noted that, in general, the definition of an 'integrable function' would depend on the particular integral 1 being extended. The definition of B(K), in contrast, depends only on L, and provides a convenient common domain to which all integrals on L can be extended. This will be important for us in Theorem 7.16. We shall suppose, for the rest of this section, that we have a specified integral 1 on L = C]R(K).
The first extension step. Let U be the subset of JRK consisting of those f E £;0 for which there is some increasing sequence (ft) in L with ft i f. It is immediately clear that U :2 L and that U is closed under addition, multiplication by nonnegative scalars, and the lattice operations. It is also clear that f 9 E U+ whenever f, 9 E U+. Let fEU, and take (ft) in L with ft i f. Since 1 is positive, the sequence (J(ft) )t>l is increasing and bounded above in JR, and so limH(X) 1(ft) exists in R We would like to define 1 on U by setting 1(f) = lim 1(ft). t+(X)
(*)
We must first check that 1(f) is well defined.
Lemma 7.6 Let (ft) and (gJ) be increasing sequences in L, each uniformly bounded above, with limJ+(X) gJ ~ limt+(X) ft. Then lim 1(gJ) ~ lim 1(ft). J~CX>
1,+00
Proof Suppose first that h ELand that limH(X) ft :2: h. Then ft :2: ft 1\ hand ft 1\ hi h, so that limH(X) 1(ft) :2: limH(X) 1(ft 1\ h) = 1(h) by Lemma 7.5(ii) and the orderpreserving property of 1. Taking h = 9J' and then passing to the limit as j + 00, gives the result. D Corollary 7.7 (i) The mapping 1 is well defined on U by (*). (ii) 1 I L
= 1.
(iii) 1(f + g)
= 1(f) + 1(g) (f,g E U) and 1(>.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;
290
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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 <;;; U and that II L = I. Also, U is closed under bounded, decreasing limits, addition, multiplication by nonnegative scalars, and the lattice operations. The mapping 1 is additive and positivehomogeneous on U. Further, if f, 9 E U with f :::; g, then l(f) :::; [(g). We also have the following simple lemma. Lemma 7.9 Take 9 E U and h E U with 9 :::; h. Then leg) :::; J(h) and h  9 E U.
Proof Using Corollary 7.7, we have h  9 J(h) leg)
= h + (g)
E U and
= J(h) + J( g) = J(h  g)
~
o. o
This implies the result.
The second extension step. It remains the case that we are taking K to be a nonempty, compact Hausdorff space and that L = CJR(K). Let f E CiR(K). Then f is Isummable (or just sllmmable, when I is clear) provided that: sup{l(g) : 9 E U, 9 :::; f}
= inf{I(h)
: h E U, h ~ f}.
In considering this definition, it should be noted that, by Lemma 7.9, [(g) :::; J(h) when 9 :::; f :::; h in U. We denote the set of Isummable functions in IRK by L~(K). (We use the notation 'L~', rather than 'L~', because we are considering just bounded functions on K.) Let f E L~(K). Then we define J(f) to be the common value of the above sup and inf, i.e. we define J(f)
= sup{l(g) : 9
E
U, 9 :::; f}
= inf{J(h) : hE U, h ~ f}.
We now aim to show that BJR(K) <:::; L'fi'(K); it will then be the case that is the extension of I that we are seeking. Before proceeding, it is useful to note two alternative forms of the definition of summable functions.
J I BJR
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291
Lemma 7.10 Let f E fYr(K). Then:
(i) f is summable if and only if, for every c > 0, there are functions 9 E U and h E U with 9 :s: f :s: hand I(h) [(g) < C; (ii) f is summable if and only if there is an increasing sequence (g')'"21 in U and a decreasing sequence (h,),>l in U with g, :s: f :s: h, (i E N) and I (h,) [(g,) * 0 as 1, * 00; moreover~ in this case, for any such pair of sequences (g')'"21 and (h')'"21, we have IU) = lim [(g,) = lim I(h,) . 1,+00
,/+00
Proof This is an easy exercise.
D
Note that, in clause (ii) of Lemma 7.10 it is NOT claimed that the sequences (h')'"21 and (g')'"21 converge pointwise to f·
Corollary 7.11 With the above notation: (i) UU(U) <:;; Lif(K); (ii) II U = I; (iii) II (U) = Ii (iv) L = I.
Yi
Proof Let fEU. Then there exists (g,) in L <:;; (U) with g, i f and with [(g,) = I(g,) * IU), and we may take h, = f (i EN). Now use Lemma 7.1O(ii) to establish (i) and (ii); (iii) is similar, and then (iv) follows. D The next lemma is the easy part of what must be proved.
Lemma 7.12 The set Lif (K) is a real vector subspace and sublattice ofF. K , and a closed subalgebm of (fYr(K); I·I K ). Further, I is a positive linear functional on Lif(K) such that IIU)I:S: IfI K l(l) U E Lif(K)). Proof This follows easily from either of the alternative formulations of the definition of summability. That IIU)I :s: IflK 1(1) U E Lif(K)) follows as in the proof of Lemma 7.5(i). D Let E be a boundedmonotone class with L <:;; E <:;; fYr(K). Then a linear functional J on E has the monotoneconvergence property if JU,) * JU) whenever U,) in E with f, i f·
Theorem 7.13 (Monotoneconvergence theorem) Let K be a nonempty, compact Hausdorff space. Then the Banach space Lif(K) (with the norm I·I K ) is a boundedmonotone class, and the functional I on Lif(K) has the monotoneconvergence property. Proof Suppose that I E fYr(K) is such that f, i f for a sequence U,) in Lif(K). Note first that, since I is bounded, there is c > 0 with I, :s: I :s: c (i EN), and so (IU,)).>l is increasing with JU.) :::; cI(l) (i EN), and hence limt+oo JU.) exists. By ~onsidering the sequence U,  f)'"21, we may suppose that 1=0.
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Take c
> 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 boundedmonotone class, and, further, IU) = limn>oo IUn) , so that the space L'fR(K) has the monotoneconvergence property. 0 Corollary 7.14 We have B'R,(K) <:;; L'fR(K). Proof By the theorem, L'fR(K) is a boundedmonotone class with L <:;; L'fR(K). But B'R,(K) is the smallest boundedmonotone class containing L, and so we have B'R,(K) <:;; L'fR(K). 0 For simplicity, we now write I for II B'R,(K). Thus I is a reallinear functional on B'R,(K). Theorem 7.15 The functional I is the unique extension of I to a positive linear
functional on B'R,(K) satisfying the monotoneconvergence property. ~roof
Certainly I is a positive linear functional on B'R,(K) that extends I, and I has the monotoneconvergence property. Now let J be any positive linear functional on B'R,(K) with J I L = I such that J has the monotoneconvergence property. Then J must agree with I on U and with [ on U, and hence J agrees with Ion U U (U). But then, since B'R,(K) <:;; L'fR(K), the definition of Isummability makes it clear that J = Ion the whole of B'R,(K). 0 We conclude by extending the above to the complexvalued case. An integral on C(K) is now defined to be a complexlinear functional I : C(K) ~
IU so that
+ ig)
f: B(K)
+ iI(g)
U
+ ig E
=
IU)
B'R,(K) EB iBlR(K)
~
=
B(K)) ,
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293
In a similar way, we can define LOO(K) to be the complexification of L';:(K); specifically, LOO(K) = {f + ig: f,g E LOO(K)}. It is easily seen that LOO(K) is a C*subalgebra of (1;OO(K); I·I K ). Notes An approach to the Daniell integral via measuretheoretic considerations is given in [110], and there is a more extensive account in [128, Section 6.1]; we have modified this theory to concentrate on bounded functions. One can run the extension process described above starting with any nonempty set S (rather than K) and any vector space L ~ ]Rs such that L is closed under V and 1\ (rather than just L = CII(K)) and any positive linear functional J on L such that JU,) > 0 whenever j, 1 O. For example, we could take K to be a nonempty, locally compact Hausdorff space, and L to be C o,II(K), the space of continuous, realvalued functions on K that vanish at infinity. The results in Section 7.1 still apply, but in Section 7.2 a problem arises because it is no longer true that 1 E L (in the case where K is not compact). In this case, it is natural to modify our approach by dealing with monotone classes, rather than boundedmonotone classes; in this modified approach, one finishes with the Lebesgue measurable functions. See [110, Section 9.4] and [117, Section 12]. The notation L'{'(K) is not standard. In the case where K = IT and the initial integral on CII(IT) is the Riemann integral, functions in L'{'(IT) are the 'bounded, Lebesguemeasurable functions' of measure theory; this gives some justification for the notation. Let K be a nonempty, compact Hausdorff space. Then we can define a 'measurable set' as a subset A of K such that XA E L'{'(K); we obtain a O'field of measurable subsets of K. A subset A ofIT is 'Lebesgue measurable' if and only if XA E L'{'(IT). Let K be a nonempty, compact Hausdorff space. The Baire functions of order a are the functions in C(K). Given a definition of the Baire class of order {3 for each (3 < n, the class of order n is the space of bounded functions on K which are pointwise limits of sequences of functions in the union of the earlier classes; the construction terminates at n = WI. The Baire functions on K are the members of this final class. Each Baire class is itself a Banach algebra which is a closed subalgebra of (B(K); I·I K ). Some properties of the C* algebra B( K) and of its character space are given in [53]. Exercise 7.1 Let K be a nonempty, compact, metrizable space. Show that the space U is precisely the set of bounded, lower semicontinuous, realvalued functions on K. Exercise 7.2 Let X be a nonempty, discrete space (so that X is a locally compact, metric space). Show that B(X) = P(X) and that Ba(X) consists of subsets Y of X such that either Y or X \ Y is countable. Thus, in the case where X is uncountable, B(X) =I Ba(X). Exercise 7.3 Let K be an uncountable, compact, metrizable space (such as K = IT). Prove that IKI = c, that IBII(K) I = c, and that 1l'nr'(K) I = 2', so that BII(K) <;; l'nr'(K). Exercise 7.4 Let C be the Cantor set, introduced in Exercise 1.7. Show that the characteristic function of every subset of C belongs to L'{' (IT). Deduce that IL'{' (IT) I = 2', and hence that BII(IT) <;; LR'(IT). Exercise 7.5 The following exercise requires a subset E of IT which is not Lebesgue measurable; such a set is 'constructed' in [143, Example 2.22], for example. (The construction of such a set E requires the axiom of choice.) Then XE E l'nr'(][), but XE f/ L'{'(][), and so LR'(][) <;; l'nr'(][).
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Exercise 7.6 This exercise requires some knowledge of the StoneCech compactification j'3N of the discrete space N; this compact space was introduced in Exercise 6.6. The Baire subsets of j'3N are determined by the space C(j'3N), and IC(j'3N)I = c, so that IBa(j'3N)I = c. However, each singleton in j'3N is a Borel set, and so IB(j'3N)I ;::: Ij'3N I = 2'. This shows that Ba(j'3N) <;;; B(j'3N). [In fact, IB(j'3N)I = 2'.J
The Borel functional calculus and the spectral theorem 7.3 Introduction to the spectral theorem. We shall now establish a Borel functional calculus for normal operators on a Hilbert space, en route to a spectral theorem. We first think what such a calculus might consist of. Let T be a normal operator on the finitedimensional Hilbert space en. Then the spectral theorem for T is another name for the diagonalization theorem of linear algebra: en has an orthonormal basis consisting of eigenvectors of T. The next stage of generality is for T to be a compact, normal operator on an infinitedimensional Hilbert space H. In this case, the spectrum Sp T of T consists of {a}, together with a finite or countable sequence of nonzero eigenvalues, each having a finitedimensional eigenspace. In the case where the sequence of eigenvalues (An) is infinite, An + a as n + 00. As we remarked on page 273, H has an orthonormal basis that consists of eigenvectors of T. These first two cases may be stated together as follows. Let T be a compact, normal operator on a Hilbert space H. For each eigenvalue A ofT, with eigenspace E(A) and corresponding orthogonal projection P).., the 'spectral theorem' for T is equivalent to the statement:
T = with convergence in (l3(H); T gives:
f ~ f(T) =
L {>. P).. : A E Sp T} ,
(*)
II . II). Further, the continuous functional calculus for
L {J(A) P).. : A E SpT},
C(SpT)
+
l3(H) ,
with convergence in the strongoperator, and hence weakoperator, topology on l3(H); see Section 4.6. Now suppose that T is an arbitrary normal operator in l3(H). Then T need not have any eigenvaluesand even when it does, Sp T may have many points that are not eigenvalues. There is still a good analogue of (*), but it is less elementary, in that the infinite (or finite) sum in (*) is replaced by a kind of integral with respect to a 'projectionvalued measure', so that, symbolically,
T =
r
A dE).. .
JSpT
In terms of this representation, the continuous functional calculus of Theorem 6.26 appears as:
The Borel functional calculus
f(T) =
295
r
f().) dE)..
(fEC(SpT)).
(**)
iSpT
However, a very important point about the spectral theorem is that we may make good sense of the above formula (**) for functions f that are more general than continuous functionsnamely for bounded Borel (or Baire) functions. In this case, the map f f+ f(T) becomes the Borel functional calculus (precise definitions will be given in the next section). We shall reverse the more usual order by developing the Borel functional calculus without assuming knowledge of integration theory (we shall prove what we need). Indeed, we shall extend the continuous functional calculus map to a certain *homomorphism from the commutative C *algebra B(Sp T) into B(H) (where B(SpT) is the algebra of bounded Borel functions on SpT). This extended homomorphism is precisely the Borel functional calculus. We shall give a few applications of this calculus and, finally, explain briefly and informally how the integral representation may be obtained from the Borel functional calculus. 7.4 The Borel functional calculus. Let H be a Hilbert space, and let T be a normal operator in B(H). Then, from Theorem 6.26, the continuous functional calculus BT is an isometric *isomorphism from C(Sp T) onto the commutative C*subalgebra C*(T) of B(H) generated by T. As has already been indicated, the Borel functional calculus is a certain extension of BT to a *homomorphism
f3T : B(Sp T)
>
B(H) .
Note that, in view of Corollary 6.19, f3T will be automatically continuous, with IIf3T(f) II :::; Ifl spT (f E B(SpT)). The homomorphism f3T will be uniquely determined, as an extension of BT , subject to one very natural extra condition involving monotone sequences.
Remarks: (i) Unlike the continuous functional calculus, the Borel functional calculus is definitely about homomorphisms into B(H), and not just into some abstract C* algebra. This arises because, although the continuous functional calculus maps into (indeed, onto) A = C*(T), the extended map f3T (in all interesting cases) takes values outside A. In fact, as will be seen, f3T maps into (but not, generally, onto) the algebra A ee, the bicommutant of A (which we know to be a commutative C*subalgebra of B(H)). (ii) Also, unlike the continuous functional calculus, the Borel functional calculus is not generally injective; this gives significance to clause (ii), below. Theorem 7.16 (Extension of homomorphisms) Let K be a nonempty, compact Hausdorff space, let H be a Hilbert space, and let B : C(K) > A be an isometric *isomorphism onto a C*subalgebra A of B(H). Then B can be extended to a *homomorphism f3 : B(K) > B(H), and the extension f3 is unique subject to the following monotoneconvergence property: if (f.) in BIR(K) , if
f
E
BIR(K) , and if f. i f,
then
f3(f.)~f3(f).
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Moreover: (i) (3(B(K)) <;;; Ace; (ii) (3(XU) =I 0 for each nonempty, open subset U of K ; (iii) SPB(H)(3(f) <;;; f(K) (f E B(K)) . Proof Again, we set L = CJR(K). By Corollary 6.31, 0 ILL orderpreserving, isometric, reallinear algebra isomorphism. For each x E H, define Ix : L f lR by
Ix(f) = (O(f)x,x)
f
Asa is an
(f E L).
Then Ix is an integral on L,~nd so, by Theorem 7.15, it has a unique extension to a positive linear functional Ix on BJR(K) that satisfies the monotoneconvergence property. Next, for each f E BJR(K), we define a mapping IJ f : H x H f C by the formula:
IJf(x, y) = For every
f
E
~ (Ix+y(f) Ix_y(f) + iIx+iy(f) 
iIxiy(f))
(x, y
E
H).
L, the polarization identity shows that
IJf(x,y) = (O(f)x,y)
(x,y
E
H).
In particular, IJ f is a bounded, selfadjoint, sesquilinear form on H such that IlJf(x,y)l:::; Ilfllllxllllyll (using the fact that 110(f)11 = IfI K )· Let M be the set of all those f E BJR(K) such that IJ f is a selfadjoint, sesquilinear form on H. We have just seen that M ::2 L, so that, to prove that M = BJR(K), it suffices to show that M is a boundedmonotone class. But this follows easily from the fact that if, say, f. l' f with f., f E M, then Iz(f.) f IAf) for every Z E H, and thus IJf,(x,y) f IJf(x,y) for all X,y E H. We shall now show that IJ f is a bounded form for every f E BJR (K). Since we know that IJ f is sesquilinear, it suffices to show that it is a bounded function on the set {(x, y) : x, y E H[lJ}. Indeed, take x, y E H[lJ. By Lemma 7.12, we have
IIIzl1 = I z (1) = IIzI12
(z E H), and so the formula for IJf shows that
IlJf(x, y)1
:::;
IflK
(11x11 2+ IIYI12) :::; 21flK
,
so that the boundedness of IJ f is proved. (We shall later see that the '2' is unnecessary, but that seems to be unclear at this point.) By Theorem 2.55, it follows that, for each f E BJR(K), there is a unique operator (3(f) E 8(H)sa such that IJf(x,y) = ((3(f)x,y) (x,y E H)or, more simply, (3(f) is unique subject to the condition that
((3(f)x, x) = Ix (f) It is then clear that the mapping (3 : BJR(K)
(x
E
H).
8(H)sa is a reallinear mapping and that, if either f. i f or f. ! f in BJR(K), then (3(f.)~(3(f). f
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The Borel functional calculus
We now extend (3 to be a complexlinear mapping from B(K) into 8(H) by combining the real and imaginary parts of functions in the obvious way. Trivially, this extended mapping satisfies (3(l) = (3(J)* (J E B(K)). To show that (3 is multiplicative, it suffices to consider just nonnegative functions f, 9 E B(K)+. So, for 9 E B(K)+, set
M(g) = {J
E
B(K)+ : (3(Jg) = (3(J)(3(g)}.
Then M(g) is a monotone class (using Proposition 6.48(iii)). Also, if 9 E L+, then M(g) :2 L+, so that M(g) = B(K)+. But clearly f E M(g) if and only if 9 E M(J), so that, for every 9 E B(K)+, we have M(g):2 L+, and so, again, M(g) = B(K)+. It has thus been shown that (3(Jg) = (3(1)(3(g) (1, 9 E B(K)+), and hence (3 is multiplicative. Combining real and imaginary parts, it then follows that (3 : B(K) + 8(H) is a *homomorphism. So (3 is automatically normdecreasing (whence it may be noted that 100f(x,y)1 :::; IflK Ilx1111Y11 (x,y E H), without the extra '2'). (i) To show that we have (3(BJR(K)) <:;;; A CC, we just note that, by the monotoneconvergence property of (3 and the fact that A cc is closed under weakoperator convergence of sequences, it is clear that {J E BJR(K) : (3(J) E A CC} is a boundedmonotone class containing L, and thus equal to the whole of BJR(K). The result follows immediately. (ii) Let U be a nonempty, open subset of K. Then, by Urysohn's lemma, there is a nonzero function h E L with 0 :::; h :::; 1 on K, while h(x) = 0 for all x E K \ U. Then 0 :::; h :::; Xu, so that (3(Xu) ::::: (3(h) = e(h) > 0 because e is isometric and orderpreserving. (iii) Let f E B(K). By Corollary 7.3, Sp B(K)f follows because SPI3(H)(3(J) <:;;; SPB(K)f.
=
f(K), and so the result D
Theorem 7.17 (Borel functional calculus) Let T be a normal operator on a Hilbert space H, and let eT : C(SpT) + 8(H) be the continuous functional
calculus for T. Then eT extends to a (normdecreasing) *homomorphism (3T: B(SpT)
+
{T,T*Yc
<:;;;
8(H).
Further, (3T has the monotoneconvergence property, and (3T is unique subject to satisfying this extra condition. Let U be a nonempty, open subset of Sp T. Then (3T(XU) =J: o. Proof This is a special case of Theorem 7.16.
D
Remark on terminology. Let K be a compact metric space (such as SpT). As explained above, the families of Baire and Borel subsets coincide, and the space B(K) of bounded Baire functions on K is the same as the space of bounded Borel functions on K. This is why Theorem 7.17 is almost always called the Borel functional calculus, although 'Baire' would be just as good.
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298
As with the continuous functional calculus, it is quite common to write f(T) as a notation for /3r(f) when f E B(Sp T). We shall use this notation in the applications to follow. Let f E B(SpT). By Theorem 7.16(iii), SPl3(H)f(T) <:;;; f(SpT), and so we have go
f
E
B(Sp T) whenever 9 E B(f(Sp T)).
Corollary 7.18 Let T be a normal operator on a Hilbert space H. Then
(gof)(T)=g(f(T))
(gEB(f(SpT)),JEB(SpT)).
(*)
Proof Equation (*) certainly holds whenever 9 is a polynomial in Z and Z and f E B(SpT). It follows by uniform approximation on f(SpT) that (*) holds whenever 9 E C(f(SpT)) and f E B(SpT) (and this is sufficient for later applications). The two maps 9 f+ g(f(T)) and 9 f+ (g 0 f)(T) from B(f(SpT)) to B(H) both have the monotoneconvergence property, and so the result follows 0 from the uniqueness clause in the theorem. 7.5
Applications of the Borel functional calculus.
Invariant subspaces. Let E be a Banach space, and let T E B(E). We have defined a proper invariant subspace for T in Section 4.6. As we explained, we are hoping to show that, when dim E ~ 2, many such T have such a proper invariant subspace. We have noted that this is true for operators T with an eigenvalue; also, by Corollary 4.97, it is true whenever Sp T is a disconnected subset of C Thus we are now mainly concerned with the case where Sp T is connected. We shall show, using the Borel functional calculus, that normal operators on Hilbert spaces have proper hyperinvariant subspaces. Recall that a closed subspace F of a Hilbert space H is a hyperinvariant subspace for T E B(H) if S(F) <:;;; F for each S E B(H) with ST = TS.
Theorem 7.19 Let H be a Hilbert space with dim H ~ 2, and let T be a normal operator on H with T tJ. CIH. Then there is a proper hyperinvariant subspace forT. Proof Suppose that Sp T has an isolated point. Then, by Corollary 6.28, T has an eigenvalue \, and, as in Section 4.6, the corresponding eigenspace, E(\), is a closed subspace of H such that E(\) is invariant for each S E B(H) with ST = TS. Assume that E(\) = H. Then T = AIH, a contradiction. Thus E(\) is a proper hyperinvariant subspace for T. We may suppose, therefore, that Sp T has no isolated points. (In fact, this implies that Sp T is uncountablebut the rest of the argument works just on the hypothesis that Sp T has at least two points.) In this case we may find an open subset U of C such that both Un K and K \ U are nonempty. Then P := f3T(XU) is a selfadjoint operator such that p 2 = P. By the last sentence of Theorem 7.17, P =1= 0 and P =1= I H .
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Let F = P(H). Then F is a closed subspace of H, with F =1= {O} and F =1= H. We have P E {T, T*} CC. Now suppose that S E B(H) with ST = TS. Then, by Fuglede's theorem, Theorem 6.35, ST* = T* S, and so S E {T, T*} C. Thus SP = PS, and so S(F) <;;; F. This shows that F is a proper hyperinvariant subspace for T. D
Polar decomposition. In Corollary 6.40, we gave a polar decomposition of an invertible element in a unital C * algebra. We shall now give a similar decomposition of an arbitrary normal operator. Theorem 7.20 (Polar decomposition) Let H be a Hilbert space, and let T be a normal operator on H. Then there exist R E B(H)+ and U E U(B(H)) such that T = RU, and R, U, and T are pairwise commuting. Proof Define bounded Borel functions rand u on Sp T by
r(A) = IAI,
U(A) = AlIAI (A
=1=
0),
u(O) = 1
for A E C. Notice that r E C(SpT) and that u E B(K) by Proposition 7.4. Let R = r(T) and U = u(T). Then R E B(H)+ and U E U(B(H)). Since Z = ru, we have T = RU = UR. Further, RT = R 2U = RUR = TR and UT = U RU = TU, and so R, U, and T are pairwise commuting. D
Unitary operators as exponentials. In Section 4.15, we defined e a = expa for an element a of a unital Banach algebra A. Proposition 7.21 Let H be a Hilbert space, and let U E U(B(H)). Then there exists Q E B(H)sa such that U = eiQ . Proof Since U is unitary, Sp U <;;; 'JI'. There exists a bounded, realvalued function f on 'JI' with eij(z) = Z (z E 'JI') and such that f is continuous on 'JI' save at Z = 1. By Proposition 7.4, f E BJR.('JI'). Set Q = f(U) E B(H)sa. Then e iQ = U by Corollary 7.18. D
The group of units of B(H). Let A be a unital Banach algebra. Then Go(A) is the principal component of the group of units of A; by Theorem 4.105(ii), Go(A) consists of the finite products of exponentials. Theorem 7.22 Let H be a Hilbert space. Then the group of all invertible operators in B(H) is connected, and every invertible operator is a product of two exponentials. Proof Let T E G(H). By Corollary 6.40, T = RU, where R = (TT*)1/2 is an invertible positive operator and U is unitary. Since Sp R c ~+., we can apply Proposition 6.33 to see that R = e S for some S E B(H)sa. Also, by the previous application, U = e iQ for some Q E B(H)sa. Thus T = e S e iQ , a product of two exponentials. The mapping t f+ e tS e itQ , IT + B(H), is a path in G(H) from IH to T, so that G(H) is connected. D
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Introduction to Banach Spaces and Algebras
Let H be a Hilbert
7.6 The spectral theorem for normal operators. space, and let T be a normal operator in B(H); as above, (3T : B(SpT)
>
A = {T} cc
is the Borel functional calculus for T. For every Borel subset E of Sp T, the characteristic function XE is an idempotent element of BIR(SpT), so that
P(E)
:= (3T(XE)
is a selfadjoint, and hence orthogonal, projection contained in A. Evidently P(0) = 0 and P(SpT) = I H ; also, P(U) # 0 for each nonempty, open subset of U of SpT. Let (Ekh>l be a sequence of pairwisedisjoint Borel subsets of SpT, and set E = U Ek~ Then the partial sums of the series 2::r=1 XE k increase pointwisemonotonically to the Borel set XE. By Theorem 6.49 and the fact that (3T satisfies the monotoneconvergence property, (2::~=1 P(Ek) )n21 is an increasing sequence of projections converging in the strongoperator topology of B(H) to
P(E) = P
(U
Ek) = sup
k=l
nEN
t
P(Ek).
k=l
(For m # n, P(Em ) ~ P(En ) because XEm . XEn = 0, so that P(Em)P(En) = 0.) The correspondence E f+ P(E) is then a kind of 'projectionvalued Borel measure' on Sp T. We shall not formalize this concept, but shall instead note that it follows immediately from the remarks just made that: for every x E H, the mapping
E
f+
JLx(E)
:=
(P(E)x, x),
B(SpT)
>
1R,
is a nonnegative, 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 ). dPx .
(** )
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)
(***) .
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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 functionalanalytic, 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 nontrivial, closed invariant subspace. (Here, H is an infinitedimensional, 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 <;;; SpT. Then l' has a proper invariant subspace. A generalization of this theorem is given in [14J. Exercise 7.7 Let H be a Hilbert space, and let l' be a normal operator in B(H). (i) Use the polarization identity to show that, for each x, y E H, there exists a nonnegative, Borel measure /lx,y on on SpT such that
(f3r(g)x,y)
=
r
gd/lx,y
(x,y E H)
iSpT
for every g E B(Sp T). (ii) Show that, for each E > 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 'halfopen' square in C. (iii) Show that the linear span of the orthogonal projections is II . IIdense 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 2n 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 'severalvariable holomorphic functional calculus' for finitely many elements of a unital (commutative) Banach algebra. The singlevariable 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 severalvariable 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 purposesand 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 nonempty, 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 realdifferentiable 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),
(*)
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Introduction to Banach Spaces and Algebras
where E( z) > E( a) = 0 as Z > a in U. The space of realdifferentiable 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 realdifferentiable functions whose partial derivatives are not continuous. The CauchyRiemann criterion for analyticity becomes: fED 1 (U) is analytic (i.e. complexdifferentiable) 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 twodimensional Lebesgue measure) of a compact plane set K is denoted by rn(K). Theorem 8.1 (CauchyGreen theorem for a rectangle) Let R be a closed rectangle in
hR f(z) dz = 2i f
l
8f
dx dy.
Remarks. 1) Under the slightly stronger hypothesis that f has continuous partial derivatives of jax and of joy on R, this is a special case of the usual Green's theorem in the plane; this latter version would actually be sufficient for our purposes. 2) The theorem is given as stated since it seems pleasant to include the usual Cauchy theorem for a rectangle (where the assumed condition is 8 f = 0 on R) as a special case. Proof For every subrectangle S of R, define
w(S) = hsf(Z)dZ2i ffs8fdXdY. Let Iw(R)1 = k 2: 0; we need to show that k = O. We first quadrisect R into four subrectangles, say R i , ... , R 4 . Clearly, we have w(R) = L~=i w(RJ ), so we can choose one of the quarterrectanglescall it R(l)with Iw(R(1))1 2: kj4. Now iterate the quadrisection process, obtaining a decreasing sequence of closed rectangles
R = R(O) :2 R(i) :2 R(2) :2 ... :2 R(n) :2 ... , with Iw(R(n))1 2: kj4n (n EN). Clearly, nn>i R(n) is a singleton, say {a}, where a E R. Since f is realdifferentiable at a, we may suppose that (*) holds; for n E N, define En = SUp{IE(Z)1 : z E R(n)}, so that En > 0 as n > 00.
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Introduction to several complex variables
As in the usual proof of Cauchy's theorem for a rectangle, we do simple direct calculations to show that:
r (J(a) JaR(n)
+ (z 
a)f)J(a)) dz = 0,
r (z  a) dz = 2i m(R(n») JaR(n)
=
2ibc/4n ,
say, where R has sides of length band c. It follows that
Iw(R(n»)1 ::; 2 Jr r 18 J(z)  8 J(a)1 dx dy JWn)
+ 2cn (b + c)(b 2 + C2 )1/2 4n
(n
E
N).
Fix ry > 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 twodimensional Lebesgue measure, in the sense that
JJKlzlr dxdy
<
00
for each compact subset K of C; to see this, use polar coordinates to evaluate the double integral over a disc, centre 0, that contains K. Corollary 8.2 Let R be a compact rectangle in C (with the sides oj R parallel to the axes). Let JED l(U), where U is an open neighbourhood oj R, and suppose that (8 J) IRE C(R). Let a E int R. Then
J(a)=~
r
J(z)
2m JaR z  a
dz~Jrr 7r
8 J (z)dxdy. JR z  a
Proof For every sub rectangle S of R, define
w(S)
=
r J(z) dZ2iJrr 8J(z) dxdy. Jasza Js za
For some 6 E (0,1]' the point a is at the centre of a square, say R a, with sides, of length 6, parallel to those of R and which is contained in the interior
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Introduction to Banach Spaces and Algebras
of R. By extending the sides of Ro to meet those of R, we divide R into nine subrectangles, Ro and Rk (k = I, ... ,8), of which only Ro contains the point a.
~:
I R, I I
R8
IR
Then, by applying the above theorem to each of RI, ... , Rs for the function z f> f(z)/(z  a), we see that w(RJ ) = 0 for j = 1, ... ,8, so that w(R) = w(Ro) for all sufficiently small S > o. U sing the analyticity of the function z f> 1/ (z  a) on C \ {a}, the local integrability of z f> 1/ z, and the continuity of f on R, it is simple to estimate that
a
flo a (!~~)
I
Iflo ~~;
dXdyl =
dXdyl
~ MIS,
say, where MI is a constant independent of S. Since f satisfies (*) on page 305, we estimate that, for each z E oRo, we have
 f(a) I ~ Jof(a)J +I(:2  a) af(a) 1+ Jc:(z)J ~ M Ifez)za za
2 ,
say, where M2 is a constant independent of S. Thus
r
lim I fez)  f(a) dzl 0>0 JaR" z a and so
.1
~
lim 4M2 S = 0, 0>0
1
fez) . f(a) . hm  dz = hm  dz = 27rlf(a). 0>0 aR" z  a 0>0 aR" z  a Hence
w(R) = w(Ro) =
r
fez) dz JaR" z  a
2iJrJR"r a(f(Z) ) dxdy z a
>
27rif(a)
o
as S > 0, from which the result follows. Corollary 8.3 Let fED on Co Then
I
f (w) = 
(q,
~ 7r
with supp f compact and with
JJIJII.2r af (z) dx dy Z 
(w
E
af
continuous
q .
W
Proof Let w E C, and take a rectangle R such that supp f U { w} <:;;; int R. Since fez) = 0 (z E oR), Corollary 8.2 immediately gives the required result. 0
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Introduction to several complex variables
8.2 Existence of smooth functions. Before going further, it is useful to describe briefly the existence of smooth functions (and these will be very much needed in the next chapter). Throughout this section, n is fixed in N. A complexvalued function f on a nonempty, open subset U of C n is called smooth provided that it has continuous partial derivatives of all orders with respect to the underlying real coordinates. Using the alaz and alOi notations described at the start of this chapter~but now using all the coordinates Zl, ... , Zn ~we are requiring that all the partial derivatives ar+s f Iar z) asz) exist and are continuous. The space of smooth functions on U is denoted by C=(U); clearly C=(U) is a subalgebra of C l(U). By wellknown 'advanced calculus techniques', it follows that each of these partial derivatives is itself a smooth function, and that the various partial differentiation operations are mutually commuting. We shall sketch a few elementary facts about the existence of smooth functions. We successively define functions 0:, (3, "( : lR + H by setting:
o:(x) =
{~l/X
if x> 0, if x ::; 0,
(3(x)
o:(x)
=
(xElR),
and then let
"((x) = (3(x
+ 2) (3(2 
x)
(x
E
lR).
Then,,( is smooth on lR, "((lR) <:;; H, "((x) = 1 (Ixl ::; 1), and "((x) = 0 (Ixl ::::: 2). Now, for z E cn, define B(z) = "((llzl!), where 11·11 is the Euclidean norm on cn. Then B is smooth. B(z) = 1 (lizil ::; 1), and B(z) = 0 (11zll ::::: 2). Let f be a (real, complex or, more generally, vector)valued function on a topological space X. Recall that the support of f, denoted by supp f, is the closure in X of {x EX: f(x) =I a}. Lemma 8.4 Let U be an open neighbourhood of a nonempty, compact subset K of C n . Then there is a compactly supported, smooth function X on C n such that X I K = 1, supp XC U, and X(C n ) <:;; H. Proof By an elementary compactness argument, we can find and 1'1 ... ,rk > 0 such that
UB(z); 1')) )=1
K
k
k
K C
Zl ... ,Zk E
and
UB(z); 21')) C U, )=1
where B( w; 1') denotes the Euclidean open ball with centre wand radius l' in C n . For j = 1, ... ,k, set B) (z) = B( (z  z) )11')) (z E C n ). Then each B) is smooth on cn, B)(C n ) <:;; II, B)(z) = 1 for z E B(z);r)), and B)(z) = 0 for z tf. B(z);2r)). It is easily checked that the smooth function X := 1(IB1)(IB2) ... (IBk) has the required properties. 0 We shall now state, without proof, the following more elaborate existence theorem for smooth partitions of unity.
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Introduction to Banach Spaces and Algebras
Theorem 8.5 Let U be a nonempty, open subset of cn, and let {Un}nEA be an open covering of U. Then there is a family {en}nEA of smooth functions on U such that: (i) en(U) <;;; II (0: E A); (ii) on each compact subset of U, all but finitely many of the functions en are identically zero;
(iii) suppen
C
Un (0:
(iv) LnEA en(z)
=
E
1 (z
A); E
U).
0
It follows from (ii) that, for each z E U, en(z)
I
0 for only finitely many
0: E A, and so the sum in (iv) is well defined. The family {en}nEA is called a smooth partition of unity subordinate to the covering {Un}nEA. Lemma 8.6 (Inhomogeneous CauchyRiemann equations, with compact support) Let f E COO(q be such that supp f is compact, and set g(w)
1 = 
J~ 1Ft2
7r
Then 9 E COO(q and 8g =
fez) 
z
dxdy
W
(W
Eq.
f on C
Proof Let wEe We use the change of variable z expression for 9 as g(W)
=
~ 7r
f+
z
+W
to write the
Jr r fez + w) dxdy. l1Ft2
Z
Then an easily justified 'differentiation under the integral sign' using the local integrability of 1/ z shows that 9 is smooth and that 8g(w) =
_~ 7r
and so 8 g( w) 8.3.
Jrr
l1Ft2
8f(z+w) Z dx dy ,
= f (w) by reversing the change of variables and applying Corollary 0
We remark that, on the open set C \ supp f, we have 8 9 = 0, so that 9 is analytic on that set. It is generally not possible to choose 9 to have compact support. Corollary 8.7 Let U be an open neighbourhood of a nonempty, compact subset K of C, and let f E COO(U). Then, for each open subset V with compact closure and such that K eVe V c U, there exists 9 E COO(V) with 8g = f on V. Suppose that the function f depends also on some additional complex variables, say on (Wi, ... ,Wk), and that f is smooth in (Z,Wi, ... ,Wk) and analytic in certain of the variables wJ • Then 9 may be chosen also to be smooth in (z, Wi, ... , Wk) and analytic in the corresponding variables Wj.
311
Introduction to several complex variables
Proof By Lemma 8.4, there is a smooth function e on C with compact support suppe C U and with e(z) = 1 (z E V). Then ef (defined to be a on C\suppf), is a compactly supported, smooth function on C. If we also allow extra variables (WI, ... ,Wk), as in the second part of the statement, then we define
1 g(Z,WI, ... ,Wk) =  7f
J1
dc;d1] e(z)f(Z,WI, ... ,Wk);
]R2
.., 
Z
(z
E
q,
where (= c; +i1] E C. By Lemma 8.6, 8g = e· f on C, so that 8g = f on V. The statements about smoothness and analyticity follow by differentiation under the integral sign. D 8.3 CauchyGreen formula. Although not strictly needed, we should like to complete the 'CauchyGreen' approach that we have been following by giving a version of Cauchy's integral formula that applies to realdifferentiable functions; it does, of course, also provide a proof of the more familiar Cauchy integral formula of the windingnumber variety. We shall use the notion of a contour as introduced in Section
1.11.
Theorem 8.8 (CauchyGreen formula) Let U be a nonempty, open subset of C, and let"( be a contour in U such that nh; w) = 0 (w E C\ U). Let fED leU) with 8 f E C(U). Then
nh;w)f(w)=~l 2m
Proof Let
W
E U \
'Y
fez)
z W
dZ~Jrr 7f
Ju
n h ;z)8 f (z)dxdy
z W
(wEU\b]).
b]' and define K =
bl U {z
E C : nh; z) =1= O} U {w}.
Then K is a compact subset of U. By Lemma 8.4, there is a compactly supported, smooth function e, with e == 1 on a neighbourhood of K and with supp e c U. Then, since f = e . f on a neighbourhood of K and since nh; z) = 0 on U \ K, we may replace f bye· f in the theorem. Thus, without loss of generality, we may suppose that f itself satisfies the condition that supp feU. Now, for every ( E bl and W E U \ b]' by applying Corollary 8.3 to f, we have fee)  few) = r 8 fez) dxdy (w 7f((w) Ju z( zw
1 Jr
=
_~Jr{ 7f
Ju
(_1___1_)
8f(Z) dxdy. (z()(zw)
Next, we integrate round ['Yl with respect to (. By an elementary argument (and recalling that l/(z  () and l/(z  w) are locally integrable with respect
312
Introduction to Banach Spaces and Algebras
to m), we may use Fubini's theorem to interchange the path and area integrals, obtaining:
~ 2m
1 'Y
J(()  J(w) d( = ( 
W
! Je r (~1~) 8 J(z) dxdy 7r JU 27rz 'Y Z  ( z  w
= !Jer n(r;z)8J(z) dxdy. 7r Ju z w Thus
n(r;w)J(w)=~l 2m
'Y
J(z) dz!Jer n(r;z)8 J (z)dxdy, z w W 7r Ju
z
o
as required. We may immediately deduce a form of Cauchy's integral formula.
Corollary 8.9 Let U be a nonempty, open subset oj C, and let, be a contour m U such that n(r;w) = 0 (w E C \ U). Let J E Dl(U) with 8J E C(U) be such that J is analytic on each component oj {z E U : '11,(,; z) I o}. Then
n(r; w)J(w) = _1 27ri
1 'Y
J(z) d z W Z
(WEU\[r]).
Proof For every z E U \ [r], either '11,(,; z) = 0 or 8 J(z) = o. The integrand in the area integral of Theorem 8.8 is thus zero almost everywhere. 0
Notes Theorem 8.5 is also valid with any finitedimensional, differentiable manifold in place of an open subset of en. A proof of this theorem may be found in many introductions to differentiable manifolds, or to distribution theory; a good reference is [33, Theorem 10.1]. The CauchyGreen formula is sometimes called the 'CauchyPompeiu formula'. Exercise 8.1 Let U be a nonempty, open subset of]Rn. Show that C leU) <;;; D leU). Exercise 8.2 Let U be a nonempty, open subset of e, let 'Y be a contour in U such that nCr, w) = 0 (w E e \ U), and let F be a closed subset of U \ b] such that F has no limit point in U. (i) Show that nCr, () = 0 for all but finitely many ( E F. (ii) (Residue theorem) Let 1 E O(U \ F). Show that
!
'Y
I(z) dz = 27ri
L
Res(f, ()nCr, ()
(EF
where Res(f, () is the residue of the function
1 at (.
,
313
Introduction to several complex variables
(iii) (Argument principle) Suppose, further, that f is nonconstant on U, has a pole at each point of F, and that Z(J) c U \ b]. Show that n(g 0
/"
0)
=
L mj(()nb, ()  L mj(()n(/" (), (EZ
(EF
where Z = Z(f) and mj(() denotes the multiplicity of fat ( for ( of the pole of fat ( for ( E F.
E
Z and the order
Functions of several variables 8.4 Holomorphic functions. We now consider analytic functions of several complex variables. We shall work in the space en, where n E No A generic point of en is
Z = (Zl,'" ,zn)
=
(Xl
+ iYI,'"
,Xn + iYn).
When we denote a point of en by a, say, we understand that a = (al,"" an) as an element of en. Suppose that 1 ::; m ::; n. Then we shall write Zm for the coordinate projection onto the mth coordinate and 7rm ,n for the projection of en onto the first m coordinates, so that
Zm
(ZI"",Zn)
f+
Zm,
7r
m,n (ZI, ... ,Zm, ... ,zn)
f+
(ZI"",Zm)'
Let a E en and let I = ('l"""n) E (JR+e)n, so that II, ... ,ln the open polydisc, with centre a and polyradius I, is
6.(a;l) = {z
E
> O. Then
en: IZk akl < 'k (k = 1, ... ,n)}.
The closed polydisc is the closure of 6.(a; I), namely
6.(a;l)
= {z
E
en: IZk akl::; 'k (k
= 1, ... ,n)}.
Evidently an open polydisc is a product of open discs, whilst a closed polydisc is a product of closed discs, and is a compact subset of en. In the case of a closed polydisc, we may allow Ik = 0 for some (or even for all) k. We shall also sometimes refer to the open polyball, defined for a E en and I> 0 by
B(a;l)
=
{z
E
en:
IZI 
al1 2 + ... + IZn 
an l 2 <
12}.
Let U be a nonempty, open subset of en. Then the space of those functions > e such that all the partial derivatives af lax) and af lay) exist on U for k = 1, ... , n is denoted by P D I (U), generalizing the onedimensional situation. It is clear that P D 1 (U) is an algebra of functions on U.
f: U
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Introduction to Banach Spaces and Algebras
For a function
JE
P D 1 (U), it is again very convenient to set
i!!L),
ok! == oj == ~(oJ _ OZk 2 OXk 0Yk
ak! == oj == ~ (OJ Ozk 2 OXk
+i
oj ) 0Yk
for k = 1, ... , n. The operators Ok and ak are derivations from P D 1 (U) into
Cu. A function J E P D 1 (U) is realdifferentiable if it is differentiable at each point of U as a map into JR2n; here, J is realdifferentiable at a E U if and only if there is a function c : U ~ C such that n
J(z) = J(a)
+L
((Zk  ak)okJ(a)
+ (Zk
 (lk)akJ(a))
+ c(z) liz 
all
(*)
k=1
for z E U, where c(z) ~ c(a) = 0 as z ~ a in U and 11·11 denotes the Euclidean norm on JR 2n. The space of realdifferentiable functions in P D 1 (U) is denoted by Dl(U); clearly, Dl(U) is a subalgebra of C(U). A function J E Dl(U) is analytic on U if it is complexdifferentiable at each point of U, in the sense that, for each a E U, there exist continuous functions L 1 , ... ,Ln : U ~ C such that n
J(z) = J(a) + L(Zk  ak)Lk(z)
(z
E
U).
k=1
It is easily seen that the values of the functions L 1 , . .. ,Ln at the point a are uniquely determined and that, for an analytic function J and k = 1, ... ,17" ak! = 0 and ok! coincides with the ordinary kth complex partial derivative, defined as a limit, so that an analytic function on U is analytic 'in each variable separately'. A classical result of Hartogs gives the remarkable converse of this statement; we shall give only a weaker form of that resultunder the unnecessary additional assumption of the continuity of Jin Theorem 8.11, (c)=?(a). The space of analytic functions on U is denoted by O(U), extending the notation of page 114. Clearly, O(U) is a unital subalgebra of C 1 (U). The composition of two analytic functions is analytic, with the usual 'chain rule' giving the derivative of the composition. For example, J E O(U) is invertible if and only if J(z) I 0 (z E U). Lemma 8.10 (Cauchy's integral formula for a polydisc) Let U be a nonempty, open subset oj en, and let J E C(U) be analytic in each variable separately.
Then, Jor every closed polydisc .6.(a; r) c U, we have J(Z) = (~) n 2m
Jor Z E .6. (a; r).
1 IA1all=rl
... 1 IAna"l=r n
J(A) dAl ... dAn (AI  zd··· (An  Zn)
315
Introduction to several complex variables
Proof Since f is analytic in each variable separately, repeated application of Cauchy's integral formula in one variable establishes the required formula as an iterated integral. However, the integrand is continuous, and hence integrable, on the domain of integration (a product of circles), with respect to the product measure. The result then follows from Fubini's theorem. D The domain of integration in the above Cauchy integral is sometimes called the distinguished boundary of ~(a; r). As a product of n circles, it is a set ofreal dimension n; it is (for n > 1) a much smaller set than the topological boundary of ~(a; r), which has real dimension 2n  l. We now need some notation involving powerseries expansions. Let X = (Xl"",Xn)' where we are thinking of Xl"",Xn as 'indeterminates', and take I = (i l , ... , in) E z+n. Then we set Xl
= X?l1
.. ·X?" n ,
etc' .,
also, III = L~=l ik and I! = i l !i2! ... in!. Then the standard formal power series f = f(Xl, ... ,Xn ) E In := C[[Xl, ... ,Xn]] may be written, using the contracted notation, as f(X)
=
LaIX I
=
L
{a(?l, ... ,?,,)X~'" ·X~n :
(i l , ... ,in) E z+n} ,
I
where aI = a
L
lallrI
<
00
and
f(z) =
L
al(z  a)l
(z E ~(a; r)).
316
Introduction to Banach Spaces and Algebras
We have the following fundamental equivalences. Theorem 8.11 Let U be a nonempty, open subset of en, and let f : U be a function. Then the following assertions are equivalent:
~
e
(a) f is analytic on U, so that f E O(U);
(b) f is realdifferentiable on U and satisfies the CauchyRiemann equations Elkf = 0 on U for k = 1, ... , n; (c) fEe (U) and f is analytic in each variable separately;
(d) f has a powerseries expansion about every point of u. Proof In fact, we shall just sketch the proof of this theorem. First, note that the implications (a) =}(b) =}(c) are all straightforward. Also, that (d) =}(a) follows easily by uniform convergence results. (c) =}(d). Let a E U. By termwise integration of a multivariable geometric series, we obtain the powerseries representation f(z) = I:I O:I(Z  a)I of f on ~(a; r), where
O:I
=
( l)nl 27fi 1)I1 a ll=rl
1 ...
f(A)dAI···dA n IAnanl=rn (AI  al)""+l ... (An  an)"n+ l
o
for I = (i l , ... , in) E z+n.
The following four corollaries follow easily, essentially as in the onevariable case. Throughout, U is a nonempty, open subset of en. Corollary 8.12 Let f E O(U). Then f has complex partial derivatives of all orders, each being an analytic function on U. The coefficients in the powerseries expansion f(z) = I:I O:I(Z  a)I of f about a E U are aI =
..!.. I!
ailif
azI (a)
(I E z+n).
o
Corollary 8.13 (Maximum modulus theorem) Let f E O(U), and take a E U. Suppose that U is connected and that If (z) I :::; If (a) I for all z in some neighbourhood of a. Then f is constant on U. 0 Corollary 8.14 (Uniqueness theorem) Let f,g E O(U). Suppose that U is connected and f(z) = g(z) for all z in some nonempty, open subset of U. Then f =g. 0 Corollary 8.15 Let (fn)n'21 be a sequence in O(U) such that fn on every compact subset of U. Then f E O(U).
~
f uniformly 0
317
Introduction to several complex variables
So far, the theory of analytic functions of several complex variables looks very much like the onedimensional theory. However, there are striking differences between the two theories. We shall now show one way in which the nvariable theory (for n ~ 2) starts to differ from the onevariable case. In the next proof, we shall use the following easily checked topological fact. Let U be a connected, open neighbourhood of the frontier 811 of an open polydisc 11 c en. Then U n 11 is also connected.
Theorem 8.16 Let U be a connected, open neighbourhood oj 811 Jor an open polydisc 11 c en, where n ~ 2. Then, Jor every J E O(U), there zs a unique FE 0(11) such that F(z) = J(z) (z E Un 11). Proof We may suppose that 11 z = (Zl, ... , zn) E 11, define
F(z)
= _1
1
27ri 1(I=rn
11(0; r) for somer E
(~+.)n.
For each
J(Zl, ... , Znl, () Zn d( . (

Clearly, F E C(I1) and F is analytic in each variable separately. By Theorem 8.11, (c) =} (a), we have F E 0(11). By the compactness of 811, there exists c > 0 such that Z E U whenever Z E 11 with r1  c < [Zl[ < '1'1. For such a z, the onevariable Cauchy integral formula gives F(z) = J(z). But then F and J agree on a nonempty, open subset of Unl1, and so, by Corollary 8.14 and the above topological remark, they agree on all of U n 11. 0 It follows from this theorem that, for n ~ 2, an analytic function of n variables cannot have an isolated singularity; neither can it have an isolated zero, since an isolated zero of J would be an isolated singularity for 1/J. Again, let U be a nonempty, open subset of en. We have already remarked that O(U) is an algebra of functions on U. It is clear that O(U) always contains some unbounded function, so that, by Example 4.20, O(U) cannot be given any Banachalgebra norm. There is, however, a very natural complete, metrizable topology on O(U) that we shall now describe. It specifies the topology of local uniform convergence (or uniform convergence on compacta), as on page 114. We can again write U = Um >l K m , where (Km)m~l is a sequence of compact subsets of U and Km C int Km+l (m EN). In particular, for each compact subset K of U, we have K <;;; Km for all sufficiently large m. For mEN, we define
Pm(f)
=
sup{[J(z)[ : z E Km}
(f
E
O(U)).
Then (Pm)m>l is a sequence of submultiplicative semi norms that defines a metrizable locally ;onvex topology on O(U); using Corollary 8.15, the topology is easily seen to be complete, so that O(U) is a F'rechet algebra, as in the case where n = 1.
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Introduction to Banach Spaces and Algebras
8.5 Differential forms. We shall need a few ideas about smooth differential forms. Since we shall need such forms only on open subsets of en (rather than on more general manifolds), our approach can be quite simpleminded. The main point of the formalism to be introduced lies in the definition of the exterior which will provide an algebraically efficient means of studying derivative integrability conditions for the inhomogeneous CauchyRiemann equations. It will be seen that an equation between forms is just a concise way of writing a certain set of partial differential equations. Again, we fix n E N, and let U be a nonempty, open subset of en. A mapping JL : U + em is given by an mtuple (JL1, ... , JLm) of complexvalued functions on U; we shall write JL = (JL1, ... , JLm) as a shorthand for the function specified by
a,
JL(z) = (JL1(Z), ... ,JLm(z))
(z
E U).
We say that JL is smooth (respectively, analytic) if each of JL1 ... , JL m is smooth (respectively, analytic) on U. We shall now try to introduce our ideas about forms. Let I E C=(U). Then we have seen in Theorem 8.11 that I is analytic if and only if ad = 0 (k = 1, ... , n) on U. With this in mind, we let "[l(U) be a direct sum of n copies of C=(U), and define a mapping
a: I
f4
(ad, ... ,anI),
c=(U)
+ "[ 1 (U).
a
Evidently, is a complexlinear mapping and kera = O(U). For k = 1, ... , n, we now write dZk
= (0, ... ,0,1,0, ... ,0)
1
E [
(U),
where '1', the constant function of value 1, is in the kth place. We note that dZk is just a notation for something else, but, of course, it is chosen for a particular 1 reason. Thus we can regard [ (U) as a 'free C=(U)module'; however is not a C = (U)module homomorphism. With this notation, a generic element of"[ 1 (U) appears as W = 91 dZ l + ... + 9n dz n ,
a
where 91, ... ,9n E C=(U). For _
aI
I
n
=
E C=(U), we have _
LOkI dZk
=
k=l
n 01 L 0
k=l
dZk;
Zk
this suggestive formula reveals one reason for the choice of the notation dZk for a certain vector in "[ 1 (U). 0 Set [ (U) = C=(U), and then note that we have the sequence
o
4
O(U)
0 4 [
(U)
aI (U),
4 [
where the second arrow is inclusion, and the sequence is an exact sequence, in the language of Section 3.17, for every open subset U of en.
319
Introduction to several complex variables
We remark that we are really telling only half the story: there is also a mapping 0 based on the operators ojoZk. However, we shall only use the part of the story that we are describing; the full treatment would be very similar. We shall find that we need to consider the socalled inhomogeneous CauchyRiemann equations: given w E £ 1(U), find an element f E £0 (U) with f = w. Notice that this formulation just gives a concise way of saying: given elements gl, ... ,gn E COO(U), find f E COO(U) with ad = gk (k = 1, ... ,n). Clearly, there is at least the following necessary conditions on the gk for this to hold: ogk oge (k,£= 1, ... ,n). Oze Ozk That this is so follows at once from the fact that, for f E COO(U), we have
a
o2f
o2f
OzeOZk
Ozk Oze
(k,£= 1, ... ,n).
(*)
(It should be clear that these relations for the 'symbolic partial derivatives' oj OZk and ojOzk, &c., follow at once from the corresponding wellknown properties of the actual partial derivatives.) We shall also be forced to consider a sequence of higherdegree analogues of these inhomogeneous CauchyRiemann equations. These are handled very effectively by the formalism of exterior algebra, to which we shall give a brief introduction. It should be realized that, for our present purposes, the point of the algebra is just so that we can handle efficiently the consequences of the above equations (*). Fix n EN, and let X = {e1, ... , en} be a finite set of n distinct elements. For p = 1, ... , n, let Sp be the set of all ordered ptuples of elements of X such that the ptuple has suffixes in strictly increasing order, so that Sp = {e'l 1\ e'2 1\ ... 1\ e,p : 1 ::; i1
< i2 < ... < ip ::; n} ,
where we have written e'l 1\ e'2 1\ ... 1\ e,p instead of (e'l , e'2' ... , e,p). Evidently the number of elements in Sp is ISpl= (;)
=~
Of course we identify Sl with X itself. We also set So = {1}, where 1 is an additional symbol. Write S = U;=o Sp, so that lSI = 2n. Now let A == A(Cn ) be the free complex vector space on S as a basis, so that dim A = 2n; let AP == AP(C n ) be the subspace of A spanned by Sp, so that dimAP = ISpl, and set n
A= L:EBAP. p=o A1(C n ) ~ cn.
Note that It is also sometimes useful to define AP(Cn ) to be {O} for each p > n, and we shall do this.
320
Introduction to Banach Spaces and Algebras
For n E N, we write 6 n for the group of permutations of n distinct elements, so that 16 n l = n!. We shall now define a product operation, denoted by /\, on A, making A into a finitedimensional, complex (associative) algebra. The product /\ is often called the exterior product. First, we define the product of any finite sequence (e", e'2' ... ,e,p) of elements of X, as follows. (i) Set e" /\ e'2 /\ ... /\ e,,, = 0 whenever ir = is for some 1 ~ r < s ~ p. (ii) In the case where all of il, ... , ip are distinct, let a E 6 p be the unique permutation with io(l) < ... < ia(p) , so that e,"(,) /\ ... /\ e,"(p) ESp. Then we define e" /\ e'2 /\ ... /\ e,,, = c:( a )e'"(l) /\ ... /\ e,"(p) , where c:( a) (= ± 1) is the sign of the permutation a. (Thus our new definition coincides with the earlier symbolism in the case where (e", e'2' ... , e,,,) ESp.) Now suppose that a = e" /\ et2 /\ ... /\ e,p E Sp and b = eJ, /\ eJ2 /\ ... /\ eJ'I E Sq for some p, q E N. Then we set a /\
b = e" /\ e'2 /\ ... /\ e,p /\ eJ, /\ eJ2 /\ ... /\ eJ,/ '
as defined by (i) or (ii), above. The basis element 1 of S is to act as a multiplicative identity for A(C n ). Finally, we extend the product to all elements in A by bilinearity. It is now simple to verify that A(Cn ) is a complex algebra with an identity. Note also that
a/\b=(l) pq b/\a
(aEAP,bEAq).
Now again take U to be a nonempty, open subset of C n . For p = 0,1, ... ,n, we define a (smooth, complexconjugate) pform on U to be a smooth mapping, say W : U + AP(C n ); in this case, the degree of the form is p. Thus, for p :::: 1, a pform may be written uniquely as:
L
W=
g[ e" /\ e'2 /\ ... /\ e,1' '
1$" <···
where I = (i 1 , ... , ip) and each g[ E COO(U). In accordance with the notation introduced before, we write' dZk' in place of 'ek'. Then the standard expression for a pform becomes:
L
W=
g[ dz" /\ dZ'2 /\ ... /\ dz,,, .
1$" <···<',,$n
For
f
COO(U) and was above, we define
E
f· w =
L
(f . g[) dz" /\ dZ'2 /\ ... /\ dz,,, .
1$" <··<',,$n p
p
[
Let [ (U)
(U) be the vector space of pforms on U. For convenience, we also set p > n. It should be clear that, for p = 0 and p = 1, these
= {O} for each
321
Introduction to several complex variables
definitions are the same as those discussed more informally at the beginning of this section, and that in fact £p(U) is a CCXl(U)module for the above module operation. Note also that we are identifying, for example, dZ 1 /\ dZ 2 E A2(C n ) with the constant 2  form z f> 1 . dz 1 /\ dz 2 , U + A2 (C n ). From this viewpoint, we are using the notation
{dz'l /\ dZ'2 /\ ... /\ dz",: 1 ~ i1
< ... < 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 complexlinear 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 rewriting 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 pform w (for p ? 0) on U is 8closed if and only ifaw = 0, and, secondly, that the pform 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 aclosed pforms on U (for p ? 0) and of all 8exact 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 CauchyRiemann equations and their higherdegree 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 nonempty, 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 complexlinear 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 pform 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 Iform 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 formsi.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 nonempty, 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 welldefined 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 nonempty, 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=ghonVnW. 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 severalvariable theory and the onevariable theory; for an attractive discussion of 10 key differences, see [111, Section 0.3]. For example, a nonempty, open set U in en is a domain of holomorphy if there do not exist nonempty, open sets V and W, with W connected, such that V <;;; W n U, such that W g U, and such that, for each f E O(U), there exists F E O(W) with F I V = f I V. Roughly, U is not a domain of holomorphy if there is a strictly larger open set such that each f E O( U) can be extended to be analytic on the larger set. Certainly, every nonempty, open set in e is a domain of holomorphy. In en where n 2: 2, some sets are indeed domains of holomorphy: this is trivially the case for polydiscs, and it is true for the polyball B(O; 1), but this is not so obvious. However,
325
Introduction to several complex variables
set Ur = {(Z1,Z2) E e 2 : IZJI < r (j = 1,2)} and U = U1 \ U 1/ 2. Then it is easy to see that U is not a domain of holomorphy. There are several deep theorems that characterize domains of holomorphy, for example involving 'pseudoconvex domains'. A compact subset K of U is holomorphically convex if, for each Z E U \ K, there exists IE O(U) with II(z)1 > 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 nonempty, open set in en. Then the Frechet algebra O(U) is functionally continuous, and the character space
Zn = g(Z1, ... , Zn~1). Exercise 8.5 Let U be a nonempty open set in en, and take p, q 2:
a(w 1\ T)) = (aw)
1\ T)
+ (l)P w 1\ (aT))
(w E lP(u),
T)
o.
Show that
E lq(U)).
Exercise 8.6 Complete the details of the following sketch that there is a subset U of
e 2 such that 1{ 1 (U) '" {O}. Set U = e 2\ {(O,O)}, an open set in e 2, and write r2 1
= IZ112 + IZ212. Define win
£ (U) by setting
w =
a ( Z1Z2r2 )
(Z1 '" 0) ,
w
=
a( z2r Z1
To show that w is well defined, note that (z2/z1r2)
Z1Z2 '" O. It is immediate that w is aclosed.
2
)
(Z2 '" 0) .
+ (Zl/Z2r2)
1/z1z2 when
326
Introduction to Banach Spaces and Algebras
Assume towards a contradiction that w is aexact, so that there exists f E COO(U) with f = w. Define
a
g(Z1,Z2)
= z1/(Z1,Z2) 
Z2 2" r
((Z1,Z2)
E
U).
Then 9 E COO(U) and, for Z1 f= 0, we have (a g)/ Z1 = a(g/ zd = a f  w = o. It is easy to see that 9 extends to be an analytic function on all of (:2. However, for each Z2 E (: with Z2 f= 0, we have g(0,Z2) = 1/z2. This is a contradiction. In fact, it can be shown that H 1 (U) is not a finitedimensional space. On the other hand, H 1(:n \ {(O, ... , O)}) = {O} for n :0:: 3. Exercise 8.7 Set U = {(z,w) E (:2:
Izl < 1,
1/2
< Iwl < I},
V = {(z,w) E (:2:
Izl < 1/2, Iwl < I}.
Show that each analytic function on U U V has a unique extension to an analytic function on the polydisc b.(0; (1, 1», so that U U V is not a domain of holomorphy.
Polynomial convexity
en,
Our current aim is to prove that, for suitably shaped open subsets U of we have H P (l (U») = {O} for each p ~ 1. In the first subsection below, we shall prove this when U is a polydisc; later, in Corollary 8.35, we shall extend this result to polynomial polyhedra.
8.6 The DolbeauIt complex of a polydisc. Again, throughout this section, n E N will be fixed. We begin with a preliminary proposition.
en
Proposition 8.22 Let G be a nonempty, open subset of with G = U:=1 K m , where Km is compact and Km C int Km+l for mEN. Suppose also that: (i) for each mEN, each open neighbourhood U of K m , and each wE ZP(U) pl
(where p ~ 1), there are an open set V with Km C V <:;; U and'f/ E [; (V) with 8'f/ = w on V; (ii) for each mEN, each open nezghbourhood U of K m , and each f E O(U), there is a sequence (hkh>1 in O(G) such that hk > f uniformly on some neighbourhood of Km. Then HP(G) = {O} for all p ~ 1. Proof Take W E ZP(G), where p :0:: 1. The cases p differently. Case p ~ 2. For any given mEN, hypothesis (i) on a neighbourhood of Km with 8'f/m = w. Now take neighbourhood of Km+1 with a'f/:"+1 = w. Then, near
8('f/:"+1  'f/m) = w  w =
:0:: 2 and p = 1 proceed gives a (p  1) form 'f/m 'f/:"+1 to be defined on a K m , we have
o.
So, again by (i), there is a (p  2) form, say ~m' defined on a neighbourhood of Km with 8~m = 17:"+1  17m near Km· Indeed, if we multiply ~m by a smooth
327
Introduction to several complex variables
function of compact support in G that is identically equal to 1 near Km (such a p2 smooth function exists by Lemma 8.4), we may even suppose that ~m E £ (G), with 8 ~m = '17~+ 1  '17m near K m· Take 'l7m+1 = '17~+1 8 ~m on a neighbourhood of K m+1. Then 'l7m+1 = '17m near Km and 8 '17m + 1 = 8'17~+1 = w near K m+1. In this way, we define inductively a sequence ('17m )m> 1 of smooth (p  1)forms, with each '17m defined on a neighbourhood of K m , satisfying 8 '17m = w on the neighbourhood, and also with 'l7m+1 = '17m on a neighbourhood of Km. p1 We may then define '17 E £ (G) by '17 I Km = '17m I Km (m EN). We now have 8'17 = won G, and so the case where p 2': 2 is proved. Case p = 1. We start similarly, with, for each mEN, a smooth function (i.e. with a Oform), say fm' satisfying 8 fm = w on a neighbourhood of K m , and then take a smooth function gm+l such that 8 gm+1 = w on a neighbourhood of K m + 1 ; these smooth functions exist by the case p = 1 of hypothesis (i). Then 8(gm+1  fm) = 0 near K m , so that gm+l  fm is an analytic function on a neighbourhood of Km. By hypothesis (ii), there is, for each mEN, a function hm+1 E O( G) such that
l(gm+1  fm  hm+d(z)1 < T m for z in some neighbourhood of Km. Now set fm+1 = gm+1  hm+1 near K m+1. Then 8fm+1 = 8g m+1 = w near Km+1' and Ifm+1(Z)  fm(z)1 < 2 m for z in some neighbourhood of Km. It is clear that the sequence (fm)m?l that we obtain converges locally uniformly on G, say to a function f. Fix N EN. For each z E K N, we have
f(z) = fN(Z)
+ m+oo lim (Jm(Z)
 fN(Z)) ,
with uniform convergence on some neighbourhood of K N . But each fm  fN for m 2': N is analytic near KN, so that f(z) = fN(Z) + gN(Z), say, near K N , where gN is analytic. But then f E COO(G), while 8 f = 8 fN = w near K N · This holds for each N E N, so that 8 f = w on G. 0 Theorem 8.23 (The DolbeaultGrothendieck lemma) Let ll. be a closed polydisc in en, let U be an open neighbourhood of ll., and let w E ZP(U), where p 2': 1. Then there exzst an open neighbourhood V of ll. with V <:;;; U and a form p1 '17 E £ (V) such that a '17 = w on v. Proof We may write w =
L WJ dZ
J1
1\ dZJ2 1\ ... 1\ dZJr> '
J
where 1 ::; j1 < ... < jp ::; n, J = (jl, ... ,jp), and each WJ E COO(U). Let v == v(w) be the greatest integer such that the Iform az" occurs explicitly in the expression for w, so that 0 ::; v ::; n. The proof is by induction on v.
328
Introduction to Banach Spaces and Algebras
Case 1: v(w) = o. Necessarily, w = 0, and so we may take V = U and T} = o. Case 2: v(w) ;::: 1, and we make the inductive hypothesis that the result holds for any form w' with v(w') < v := v(w). Then we may write w = pl
az
v 1\
a
+ (3,
p
where a E [ (U), (3 E [ (U), and v(a), v((3) :S v  l. Now 0 = = + (3, and, from the definition of it follows v 1\ that, if rp is any coefficient function of either a or (3, then akrp = 0 on U for k = v + 1, ... , n, so that rp is analytic in (Zv+l, ... , zn). For each such rp, we may find (by Corollary 8.7, which applies because ~ is a polydisc) an open neighbourhood W of ~ with W <;;; U and '¢ E COO(W), with '¢ analytic in (zv+1, ... , zn) and v'¢ = rp on W. Thus, for each coefficient function rp of a, we can find a corresponding ,¢, and we may suppose that all these functions are defined on a common set W by taking a finite intersection of such sets. Next, define a new (p  1) form 'Y on W by replacing each coefficient function rp of a by its corresponding '¢. Then a'Y = dz v 1\ a + J, say, where J E [PeW) and v( J) :S v  1. (We remark that no terms in azv+l, ... ,azn are introduced in this process because each '¢ is analytic in (Zv+l, ... , zn).)
aw az aa a
a,
a

p


2
Let E = wfh = (3J E [ (W). Then VeE) :S vI and oE = owa 'Y = o. By the induction hypothesis, there is an open neighbourhood V of ~ with pl V <;;; Wand a form E [ (V) such that = E on V. Thus, on V, we have 

e
oe
= o'Y + E = ob + e). We can thus take
= 'Y + e E
pl
[ (V). We have proved the result for the current value of v, and so the induction continues. 0
W
Corollary 8.24 Let
~
T}
be an open polydisc in
en.
Then
1iP(~) =
{O} for all
p;:::l.
Proof We simply take (Km)m>l to be an increasing sequence of compact polydiscs with Km C int K m+l (~E N) and U:=l Km = ~. Hypothesis (i) of Proposition 8.22 is supplied by the DolbeaultGrothendieck lemma, whilst hypothesis (ii) follows by considering the powerseries expansion of f in a neighbourhood of Km and letting (hk)k?l be the sequence of Taylor polynomials (see Lemma 8.11(d)). 0 We shall now give another application of Proposition 8.22: it provides an improvement to Lemma 8.6 and its corollary. We remark that, for an open subset U of the plane, it is trivial that 1iP(U) = {O} for p ;::: 2; the only question is for p = 1. Theorem 8.25 Let U be a nonempty, open subset of the complex plane C. Then
1i leU) = {O}. Proof We again apply Proposition 8.22. In this case, hypothesis (i) is immediate from Corollary 8.7, whilst hypothesis (ii) follows from Runge's theorem, Theorem 4.83. 0
329
Introduction to several complex variables
Corollary 8.26 Let V and W be open subsets of e with V n W I= 0. Then, for every f E O(V n W), there are g E O(V) and h E O(W) with f = g  h on VnW. Proof This follows at once from Theorem 8.25 and Corollary 8.21.
D
8.7 The Cousin problem. This section is not strictly necessary for the main results that we are aiming for, and so some details will be passed over quickly. It does, however, indicate a reason for the importance of the Dolbeault cohomology group 'H l(U). Take n E N, and let U be a nonempty, open set in en. Let us consider an open covering {UaJaEA of U. By Cousin data for this covering, we mean that, whenever a, (3 E A are such that Ua n U{3 I= 0, there is a specified function h a ,{3 E O(Ua n U(3), and that these functions satisfy the following conditions:
+ h(3,a = 0 whenever Ua n U(3 I= 0; h a ,(3 + h(3" + h"a = 0 whenever Ua n U(3 n U,
(i) h a ,(3
(ii) I= 0. The first Cousin problem for the given set of data is to find functions ha E O(Ua ) for each a E A such that ha  h(3 = h a ,(3 E O(Ua n U(3) whenever Ua n U(3 I= 0. The open set U is a Cousin domain if the first Cousin problem is solvable in U for each open covering {Ua}aEA of U and each set of Cousin data for the covering. Theorem 8.27 Let U be a nonempty, open set in en. Then U is a Cousin
domain if and only if'Hl(U)
= {a}.
Proof Let {Ua}aEA and {ha}aEA be as specified above. By Theorem 8.5, there are functions () a E Coo (Ua), defined for each a E A, that form a smooth partition of unity subordinate to the covering {Ua}aEA. For each a E A, define
fa =
2: {(),ha"
:I
E
A}
on
Ua ,
where h a " = 0 when Ua n U, = 0. It follows from Theorem 8.5 that, on each compact subset of Ua, fa is given by a finite sum, and so each fa is a smooth function on Ua' Further, in the case where a, (3 E A and Ua n U(3 I= 0, we have
fa  f!3 =
2: {(),(ha"
 h(3,,) : I E A} =
(2: {(), : I
E A}) ha,(3
= ha,(3,
and so the Cousin problem is solvable with smooth functions (for any nonempty, open set U in en). Now suppose that 'H l(U) = {a}. Then the proof that U is a Cousin domain is a slight extension of that of Corollary 8.21: we use the fact that 'H l(U) = {O} to modify the functions fa so that they become analytic. Conversely, suppose that U is a Cousin domain, and take w to be a Iform such that 73 w = O.
Introduction to Banach Spaces and Algebras
330 Let
{~a}aEA
en such that
be a family of open polydiscs in
:
U{~a a
E
A} = U.
By Corollary 8.24, for each a E A, there exists fa E COO(Ua ) such that a fa = w. For each a, (3 E A such that Ua n U{3 of. 0, define h a ,{3 E COO(Ua n U(3) by h a ,{3 = fa  f{3; on Ua n U{3, we have a h a ,{3 = a fa  a f{3 = 0, and so, in fact, h a ,{3 E O(Ua n U(3). It is now clear that we have Cousin data for the covering (~a : a E A). Since U is a Cousin domain, there exist ha E O(Ua ) for each a E A such that ha  h{3 = fa  f{3 whenever Ua n U{3 of. 0. We define f(z) = fa(z)  ha(z) (z E ~a) for each a E A. Clearly, f is well defined, f E coo(U), and a f = w. Thus w is aexact. It follows that 1t I(U) = {O}. 0
8.8 Joint spectra. Before discussing some more general domains in en than polydiscs, we return briefly to Banach algebra theory, in order to motivate the direction that we shall take. This section also includes some simple results that will be needed in proving (and even in stating) results on the severalvariable functional calculus. Let A be an algebra, and take n E N. In this section we shall write An for the nfold Cartesian product of A with itself. We have discussed the spectrum of an element a in A in Section 4.5; we shall now give an nvariable version of the spectrum. We shall restrict ourselves to commutative Banach algebras. Let A be a commutative, unital Banach algebra, and consider an element a = (al,"" an) E An. Then the joint spectrum SPA a = SPA (al,"" an) of a is the subset of en defined by: SPAa = {('P(aI), 00. ,'P(an ))
:
'P
E <J>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 nonempty, 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)(Ak1ak)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 nonempty, compact subset K of C. Now suppose that K is a nonempty, 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 nonempty, 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 nonempty, open subset of that p 2': 0 and that HP+1(U x D) = {O}. Take
U'" = {z
E
en . Suppose
U : 1
and define
P:Zf+(z,
U",.UxD.
Then, for each wE ZP(U",), there exists n E ZP(U x D) such that w
= p*n.
Proof Note first that p : U'" . U x D is an analytic mapping and that p* was defined in Theorem 8.19. Set A = U'" x D and B = {(z, w) E U x D : w #
= (w 
E
A n B).
Then clearly T/ E Zp(AnB), and so, by Theorem 8.20, there are aclosed pforms Ct and (3 on A and B, respectively, such that T/ = Ct  (3 on A n B. Hence, on An B, we have
w(z)  (w 
=
It follows that there is a welldefined form
nez, w)
(w 
n E p £ (U
x D) defined by:
= {w(Z)  (w 
((z,w)
E E
A), B).
Since w, Ct, and (3 are each aclosed on their respective domains and the map (z, w) f+ W 
334
Introduction to Banach Spaces and Algebras
Corollary 8.33 Let U be a nonempty, open subset oj en.
(i) Suppose that H 1(U x D) = {a}. Then, Jar each f E O(U'P)' there exists O(U x D) such that J(z) = F (z, 'P(z)) (z E U'P)' (ii) Suppose that HP(U x D) = {a} (p;::: 1). Then HP(U'P) = {a} (p;::: 1).
FE
Proof (i) This is the case p = 0 of Oka's lemma. (ii) Fix p ;::: 1, and take W E ZP(U'P)' Since HP+I(U x D) = {O}, it follows from Lemma 8.32 that there exists D E ZP(U x D) with W = JL*D. But also pl HP(U x D) = {O}, so that there is a form 8 E [; (U x D) such that D = B8.
Setting e = JL*8 E lPI(U'P)' we then have (using Theorem 8.19(iv)):
ae = a(JL*8) = JL*(a8) = JL*D = w.
o
Thus HP(U'P) = {O}. We can now extend Oka's lemma as follows.
Corollary 8.34 Let U be a nonempty, open subset oj en, and suppose that HP(U x Dk) = {O} (p;::: 1). Take 'PI, ... , 'Pk E O(U), set
U'Pl"",'Pk = {z E U: l'Pr(z)1 < 1
(r = 1, ... , k)},
and define JL: z
f>
(Z,'Pl(Z), ... ,'Pk(Z)) ,
U'Pl"",'Pk
*
U
X
Dk.
Then: (i) Jor each p;::: 0 and wE ZP(U'Pl, ",'Pk)' there exists D E ZP(U X Dk) such that W = JL*D ; (ii) Jar each f E O(U'Pl,. ','Pk)' there exists FE O(U X Dk) such that J = JL* F, so that J(z) = F (z, 'Pl(Z), ... , 'Pk(Z)) = (F
0
JL)(z)
(z E U'Pl"",'Pk);
(iii) HP(U'Pl, ",'Pk) = {O} (p;::: 1). Proof We shall indicate the proof of just clause (i); clauses (ii) and (iii) will then follow in the same way as Corollary 8.33 followed from Oka's lemma. The proof of (i) is by induction on k. The case where k = 1 is just Lemma 8.32. Thus, take k ;::: 2, and assume that the result is true for sets defined by smaller numbers of the functions 'PJ' Define ILo: (z, w)
f>
(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 nonempty 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 (OkaWeil 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 powerseries 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 <:;;: ReP <:;;: U. The result therefore follows from the case already considered. 0 The following theorem uses some notation from Corollary 8.37. Theorem 8.39 Let Q be a polynomial polyhedron in en, let PI, ... ,Pk be polynomials, and set P = Qp1,. ,p<' Take F E O(Q X Dk) such that /1*F = O. Then
there are HI, ... ,Hk
E
O( Q
X
Dk) such that
k
F(z, WI,"" Wk) =
L Hr(z,
WI,""
Wk)(W r  Pr(Z))
((Z, W)
E
Q
X
Dk).
r=l
Proof The proof is again by induction on k. Case k = 1 (Note that this case makes no use of the vanishing cohomology of£(Q x Dk)). We have, say, FE O(Q x D) with
F(z,p(z)) = 0
(z
E
Qp == {z
E
Q: Ip(z)1 < I}).
We must find HE O(QxD) with F(z,w) = H(z,w)(wp(z)) ((z,w) E QxD). Any such function H (if it exists at all) is clearly unique on each nonempty,
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Introduction to several complex variables
open subset of Q x D, so that it is sufficient to define H on a neighbourhood of each point of Q x D. Thus, let (zo, wo) E Q x D, and suppose that Wo = p(zo)since, otherwise, the definition of H is trivial. Now the mapping 8: (z,w) ~ (z,w  p(z)) is an analytic isomorphism oH:n onto itself, with 8(zo, wo) = (zo, 0), and hence F 0 8 1 is analytic on a neighbourhood of (zo, 0) and (F 0 8 1 )(z, w) = 0 whenever W = o. Thus, considering the powerseries expansion of F 0 e l about (zo,O), we have that, say, F(e l (z,w)) = wG(z,w) in some open neighbourhood, say W, of (zo,O), where G E O(W). Thus, on the neighbourhood 8 1 (W) of (zo, wo), we have F(z,w) = H(z,w)(w  p(z)), where H = Go e E O(8 1(W)). This proves the result for the case where k = l.
Case k 2: 2, and, inductively, assume that the result is true for all sets defined by smaller numbers of the functions cpJ. We have F E O(Q X Dk), with F(Z,Pl(Z), ... ,Pk(Z)) = 0 (z E QPl, ... ,Pk). But QPl, ... ,Pk = (QPl)P2, ... ,Pk' so that, defining G E O(QPl x D k l ) by
G(z, W2, . .. ,Wk) = F(z, PI (z), W2, ... ,Wk)
((z, w)
E
QPl X D k l ) ,
we have G(Z,P2(Z), ... ,Pk(Z)) = 0 (z E (QPl)P2, . .,Pk). Then the induction hypothesis gives functions, say G 2 , •.. ,Gk E O(QPl x D k  l ), with k
G(z, W2, ... , Wk) =
L Gr(z, W2,···, Wk)(W r  Pr(z))
((z, w)
E
QPl X D k l ).
r=2 By Corollary 8.33(i), there are functions H 2, ... , Hk in O(Q x Dk) such that = H r (Z,Pl(Z),W2, ... ,Wk) on QPl x Dkl. Thus the function
G r (Z,W2, ... ,Wk)
k
F(z, WI,··· ,wd 
L
Hr(z, WI,···, Wk)(Wr  Pr(z))
r=2 is analytic on Q x Dk and vanishes whenever WI = PI (z). By the case k = 1 of the present theorem, proved above, this function has the form
H l (z,Wl, ... ,Wk)(WIPl(Z)) for some HI E O(Q X Dk). Thus the result holds for the present value of k, and so the induction continues. D Notes Theorem 8.23 and Corollary 8.24 are proved in [86, I, Section D] and [96, Theorem 2.3.3], for example. For an extension of this result and of Corollary 8.35, see [111, Section 6.3]. We have discussed the first Cousin problem in Section 8.7. For a more extensive discussion, see [111, Chapter 6] and [133, Chapter VI, Section 4]; in these sources, the second, or multiplicative, Cousin problem is also described. For example, it is shown in [111, Theorem 6.1.4] that a domain of holomorphy is a Cousin domain. For some applications of the Cousin problems, see [153, Section 2.1].
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Introduction to Banach Spaces and Algebras
A recent book devoted to the study of polynomial convexity in several complex variables is that of Stout [153], where it becomes clear that the subject is complex and that it is hard to tell when sets in en are polynomially convex. For example, it is known that the union of three pairwisedisjoint, closed balls in en is always polynomially convex [153, Theorem 1.6.20], but it is apparently not known whether or not the union of four pairwisedisjoint, closed balls in en is necessarily polynomially convex. Let K be a nonempty, compact subspace of en. The character space of R( K) is discussed in Exercise 8.12, below. However, the character space of O(K) need not be a subspace of any space em [91]. Even if
(ii) Let K = B(O; 1) be the closure of the unit polyball B(O; 1) in en. Show that lo(P(K)) = r(P(K)) = 8K, the topological boundary of K. Exercise 8.10 For r > 0, let Kr = {(z, w) E e 2 : zw = 1, Izl = r}, a set homeomorphic to the circle. Show that each Kr is polynomially convex and that Kl n Kl/2 = 0, but that the polynomial hull of Kl U Kl/2 is the annulus
{(z,w) E e 2 : zw = 1,1::;
Iwl::; 2} .
Exercise 8.11 Let K J = {(z, w) E e 2 : z = jill, Iwl ::; 2} for j = 1,2. Then each of Kl and K2 is a compact, convex subset ofe 2 , and Kl nK2 = {(O,O)}. However, their union, K = Kl U K 2, is not polynomially convex. To see this, consider the map ~:(f>(,I/(),
e\{0}>e 2
,
so that ~() E Kl when 1(1 = 1 and ~() E K2 when 1(1 = v'2. Take A to be the annulus {( E e : 1 ::; 1(1 ::; v'2}. Show that A ~ R, but that A g K, and so K is not polynomially convex. Exercise 8.12 Let K be a nonempty, compact subset of en. The rational hull of K consists of the points z E en such that If(z)1 ::; IflK (f E R(K)), and K is rationally convex if it is equal to its rational hull. (i) Show that
9
The holomorphic functional calculus in several variables
In this final chapter, we shall extend Theorem 4.89 to analytic functions of several complex variables. This severalvariable result has many applications and is certainly the single most powerful method in the study of general commutative Banach algebras; some applications will be given in the second part of the chapter.
The main theorem Preliminary version. Again, we fix n E N in this section. Let A be a commutative, unital Banach algebra, let al, ... , an E A, and write a = (al, ... ,an) E An, in the notation of Section 8.8. Given an open neighbourhood U of SPA a in en, we wish to construct a continuous, unital homomorphism O(U) > A such that (again writing Zk for the kth coordinate projection, now restricted to U): 9.1
e:
(i) e(zJ) = aJ (j = 1, ... ,n); (ii) the spectral mapping property holds, i.e. for every
f
E O(U) and
we have
)).
Such a map is a 11Olomorphic functional calculus map. The question of the uniqueness of the holomorphic functional calculus is more complicated than it was in the onevariable theorem. The standard statement asserts a uniqueness, not for each individual but for the whole family of homomorphisms, as a ranges over all ordered ntuples of elements of A, for all natural numbers n E N and, for each a, with U ranging over all open neighbourhoods of SPAa, subject to the appropriate conditions (i), and to certain compatibility conditions between the different homomorphisms. (In the most usual formulation, the mention of the neighbourhoods U is suppressed by expressing the result in terms of algebras of functiongerms; see the notes for details.) The spectral mapping condition (ii) is not, in the classical result, required as a condition for uniqueness: it is an important additional property that is satisfied by each of the unique family of homomorphisms. There is a much nicer uniqueness formulation, due to W. R. Zame; this formulation asserts the uniqueness of each separate e : O(U) > A, subject to both clauses (i) and (ii), above, holding; i.e. given e, subject to (i) and (ii), then this e must be just the appropriate member of the unique family of homomorphisms that appears in the standard result. The only problem with Zame's formulation is
e,
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Introduction to Banach Spaces and Algebras
that (at least so far) its proof requires a great deal of complex analysis in several variablesit would take a second course just to obtain that single improvement in the theorem (although the course would include a lot of good basic complex analysis along the way). In this book, we shall, with regret, dispense with proving any uniqueness statement, except in the simple special case of the next result that deals with polynomially convex neighbourhoods of the joint spectrum. Note that, in the proof of this result, we shall use most of the preliminary theorems on several complex variables that were established in the previous chapter. Lemma 9.1 (Preliminary functional calculus theorem). Let A be a commutative, unital Banach algebra, and let a = (al, . .. ,an) E An. Let U be an open, polynomially convex neighbourhood of Sp a. Then:
(i) there is a unique continuous, unital homomorphism 8 a such that 8 a (ZJ) = a J (j = 1, ... , n);
==
8~ : O(U) + A
(ii) for each f E O(U) and each
f (
Proof Notice first that the uniqueness of 8 a , subject to the condition that 8 a (ZJ) = aJ (j = 1 ... , n), follows at once from the density of the polynomials in O(U) given by the OkaWeil theorem, Corollary 8.38, and the hypothesis of the continuity of the homomorphism. Likewise, the spectral mapping property (ii) also holds for the same reason because we have already established this for polynomials in Lemma 8.28(iv). It thus remains to prove the existence of 8 a . This is broken into several cases. Case (i) U is an open polydisc. By the compactness of Sp a, we may choose an open polydisc 6. = 6.(w; r) such that Sp a c 6. <;;; 6. c U. For each f E O(U) define 8 a (J) to be
( 2~i)n
1 ···1 _ IZ1 W11 T 1
IZn Wn ITn
f(z)(ztlad l ... (Zn 1  a n)l dz l
···
dzn ·
It is clear that 8 a is a continuous linear mapping. Also, as in the onevariable case, _ ZJkJ( ZJ  a J )ld ZJ = a JkJ (j=I, ... ,n)
1
whenever k J
E
1z.,wJ IT J Z+, and so we have
8 a (Zk1 1
n) ... Zk n
= a kl 1 ... ankn
for every monomial Z~1 ··.Z~n. Hence, by linearity, Sa(P) = p(all ... ,an ) for every polynomial p. In particular, therefore, Sa acts as a unital algebra homomorphism on the algebra qX] of polynomials. So, by continuity, Sa : O(U) + A is also a unital algebra homomorphism. This completes the proof of case (i).
341
Holomorphic functional calculus
[Note: It is also possible to prove case (i) by showing that it makes good sense to substitute al, ... , an, respectively, for Zl,"" Zn in the powerseries expansion for f on ~; the convergence question is settled by using the spectral radius formula from Theorem 4.23.] Case (ii) U is an open polynomial polyhedron. We may suppose that U is the set P = ~Pl' ... 'Pk' in the notation of Corollary 8.35, where ~ is an open polydisc in <en and Pl, ... ,Pk are polynomials. Take r E {1, ... ,k}, and set br =Pr(al, ... ,a n ). Then Spbr = {Pr(z): Z E Spa}. Since Spa we have Spb r
C
C
P = {z E ~: IPr(z)1
D = {z E <e:
Sp (a, b)
=
Izi < I},
< 1 (r
1, ... ,k)},
=
and so, by Lemma 8.28(ii), we also have
Sp (al, ... , an, bl , ... , bk )
c P
x Dk <;;; ~
X
Dk.
Now ~x Dk is an open polydisc in <e n+k , so that, by case (i), there is a continuous, unital homomorphism, say B = 8(a,b) : O(~ x Dk) '> 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* <;;; ker B. By Theorem 8.39, ker fJ* is generated as an ideal by the functions VVr  Pr : (z,w) ~ Wr  Pr(z),
~
X
Dk
'>
<e,
for r = 1, ... , k. However, we do have B(VVrPr) =brPr(al, ... ,an)=O
(r=I, ... ,k),
and so indeed ker fJ* <;;; ker B. Thus the existence of 'ljJ is proved. For j = 1, ... , n, we have 7j;(ZJ) = (7j; 0 fJ*)(ZJ) = B(ZJ) = aJ • So, setting Sa = 7j;, case (ii) is established.
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Introduction to Banach Spaces and Algebras
Case (iii) U is an arbitrary open, polynomially convex neighbourhood of Spa. Let K = Spa. Then R C U and, by Lemma 8.31(ii), there is a polynomial polyhedron P with K ~ ReP ~ U. Let R : O(U) > 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 (ArensCalderon 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
((ZI, ... , zm+n) E U x V).
Show that the function h is analytic on the neighbourhood U x V of Sp c, where c = (al, ... ,a m ,bl, ... ,bn ) E Am+n, and that 8 c = 8 a 8b. Exercise 9.2 Let K be a nonempty, compact subset of en such that K is rationally convex, in the sense of Exercise 8.12. Show that O(K) = R(K). Exercise 9.3 Formulate and prove a version of the composition theorem, Theorem 9.4, in which the second function of the composition is a function of m variables, rather than just one variable. Exercise 9.4 Let Kl
= {(z, w) E e 2: 1/2 ::; IzI2 + Iwl 2
L = B(O; 1) = {(z,w) E e 2
:
::;
I}, K
IzI2 + Iw12::;
= Kl U {(O, O)}, and
I}.
Set A = C(K), al = ZI K, and a2 = Z2 K, so that SPA(al,a2) = K. Take U to be an open neighbourhood of K. Construct two distinct continuous, unital homomorphisms from O(U) into A, both taking Zl and Z2 to al and a2, respectively. 1
1
345
Holomorphic functional calculus
Indeed, the first homomorphism is 8(al,a2)' For the second, use the following theorem: each function which is analytic on a neighbourhood of K 1 is the restriction of a function which is analytic on a neighbourhood of L (cf. Exercise 8.7).
Applications of the functional calculus 9.3 Shilov's idempotent theorem. As a first application of the holomorphic functional calculus, we shall prove the famous Shilov's idempotent theorem. In fact, the statement of this theorem makes no mention of analytic functions; they are simply used as a tool in its proof. In 1955, Shilov devised the functional calculus (in a weaker version than the one that we have given, and with a much more difficult proof) precisely to prove the idempotent theorem. It remains just possible that a quite different proof might be found, but this has evaded us for more than 50 years. As before, the characteristic function of a set K is denoted by XK. Theorem 9.5 (Shilov's idempotent theorem) Let A be a commutative, unital Banach algebra, and let K be subset of if? A which is both open and closed in if? A. Then there is a unique idempotent eK E A such that eK = XK· Proof We first deal with the question of uniqueness. Thus, let e and f be two idempotents in A with = [; we must show that e = f. Since A is commutative, we have
e
(e 
/)3
= e3

3e 2f
+ 3ef2 
f3 = e  3ef
+ 3ef 
f =e f ,
and so (e  /)(1  (e  /)2) = O. But, for every r.p E if? A, we have r.p(e) = r.p(f), and so r.p(1  (e  /)2) = 1. By Corollary 4.47(i), 1  (e  /)2 E G(A), and so e  f = O. Now we give the proof of the existence of e K. Clearly, if K = 0, then e K = 0, and, if K = if? A, then e K = 1. So, we may suppose that 0 =f=. K =f=. if? A; we write L = if? A \ K, and note that then if? A = K u L expresses the compact Hausdorff space if? A as a disjoint union of two nonempty, closed subsets. By the definition of the Gel'fand topology, it is elementary to find finitely many elements, say aI, ... ,an, of A such that a(K) and a(L) are disjoint, nonempty, compact subsets of en. Of course, SPAa = a(if?A) = a(K) u a(L). Now let V and W be disjoint, open neighbourhoods ofa(K) and a(L), respectively, in en, and set U = V u W. Then U is an open neighbourhood of Sp Aa. Take 8 a : O(U) + A to be the functional calculus homomorphism given by Theorem 9.3. Define h = Xv E O(U), so that h 2 = h in O(U), and define eK = 8 a (h), so that e'k = eK, and hence eK is an idempotent in A. Also, for each r.p E if? A, the spectral mapping property gives
r.p(eK) = r.p(8 a (h)) = h(r.p(al), ... , r.p(a n )) = Thus
eK
= XK, and so the result is proved.
{~
(r.p E K) , (r.pEL).
o
346
Introduction to Banach Spaces and Algebras
9.4 An implicit function theorem. The proof of Shilov's idempotent theorem amounted to solving the equation X 2  X = 0 in a commutative Banach algebra A, with the solution having a specified Gel'fand transform. There is a considerable extension of this idea; this gives the result usually known as the implicit function theorem for Banach algebras. The statement of this theorem is as follows.
Theorem 9.6 Let A be a commutative, unital Banach algebra. Let 9 take a1, ... , an E A, and set L:
= {(a1('P), ... , an('P),g('P)) : 'P
E
C(cI>A),
E cI>A},
a compact subset of c n+1. Suppose that F is an analytic function on a neighbourhood of L: such that:
(i) F(Zl, ... , Zn+1) = 0 ((Zl, ... , zn+d E L:); (ii) 8F/8zn + 1 =1= 0 on L:. !hen there is a unique element bE A such that both F(a1, ... ,an ,b) = 0 and b= g.
0
The proof of the above result is rather complicated in detail, and we shall just give a special case, which still has interesting applications. Theorem 9.7 (Implicit function theorem) Let A be a commutative, unital Banach algebra. Let 9 E C (cI> A), and take a EA. Let f be an analytic function on an open neighbourhood of g( cI> A) in C, and suppose that:
(i) f(g('P)) = a('P) ('P E cI>A); (ii) f'(z) =1= 0 (z E g(cI>A)). Then there is a unique element b E A such that both f (b) = a and b = g. Proof Step 1: There is a jinzte subset {a1, . .. ,am} of A such that, whenever 'P1, 'P2 E cI> A have the property that ak ('Pd = ak ('P2) (k = 1, ... , m), then necessarily g( 'Pd = g( 'P2). Indeed, let 'Po E cI>A. Then f(g('Po)) = a('Po) and !'(g('Po)) =1= o. By the inverse function theorem, Proposition 1.36, there are neighbourhoods Vo and Wo of g( 'Po) and a( 'Po), respectively, in C such that f : Vo > Wo is bijective, with analytic inverse f 1 : Wo > Vo. Now choose an open neighbourhood U = U('Po) of 'Po in cI> A such that both g( 'P) E Vo and a( 'P) E Wo when 'P E U. Since fog = a on cI> A, it follows that, for each 'P E U, the equation f (z) = a( 'P) has the solution z = g('P), and that this solution is the unique solution in Vo· Next, set W = U{U('Po) x U('Po) : 'Po E cI>A}. Then W is an open neighbourhood of the diagonal in cI> A xcI> A. If ('Pb 'P2) E W and ifa( 'Pd = a( 'P2), then, first, ('P1, 'P2) E U( 'Po) x U( 'Po) for some 'Po E cI> A, and
347
Holomorphic functional calculus
also g('Pd and g('P2) are both solutions of the equation J(z) = a('Pd = a('P2) with z in the corresponding Vo. It follows that g('Pd = g('P2). Suppose now that ('P~, 'Pg) E (cI> A X cI> A) \ W, so that 'P~ =1= 'Pg. Then there is an element C E A with 2('P~) =1= 2('Pg), and hence also 2('P1) =1= 2('P2) for all ('PI, 'P2) in a neighbourhood of ('P~ , 'Pg). Using the compactness of (cI> A X cI> A) \ W, we can thus find finitely many elements, say a2, ... ,am, of A such that, for each 'P1,'P2 E cI>A with the property that ak('Pd = ad'P2) (k = 2, ... ,m), we have ('P1,'P2) E W. We set al = a. We then see that g('Pt) = g('P2) whenever 'Pl,'P2 E cI>A and ak('Pt) = ak('P2) (k = 1, ... , m). This proves Step 1. Write a = (al,'" ,am) and a = (a1,'" ,am) : cI>A + em. Then, from Step 1, there is a unique function G : SPAa + e such that Goa = g. Elementary topology then shows that G is continuous. Step 2 : G extends to an analytic Junction on a neighbourhood oj Sp Aa.
Write ~ == SPA a. For each z = (Zl,"" zm) E ~, there is some 'P E cI> 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 welldefined 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 openandclosed 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 ArensRoyden 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 openandclosed subsets of A. The ArensRoyden 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) and that g(expA) <:;; expC. Thus 9 certainly induces a group homomorphism G(A)/expA > G(C)/expC.
g:
Theorem 9.9 (ArensRoyden) Let A be a commutative, unital Banach algebra. Then the homomorphism
g: G(A)/ exp 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(
A). Define W
E A}.
Then c > 0, and, for each g E C with Ig  fl
n
g
= Lakak+n. k=1
Define a: A
~
t
== SPA(al, ... ,a2n) by
a(
c 2n
= (al(
E
A),
by n
G(ZI, ... ,Z2n)
=
LZkZk+n.
(*)
k=1
Then G E coo(C2n) and G(z) :/= 0 on ~, so that it is also the case that G(z) :/= 0 on some open neighbourhood U of ~ in C 2n. By the ArensCalderon lemma, Lemma 9.2, there is a finite subset, say {a2n+l, . .. ,am}, of A and a polynomial polyhedron P such that
SPB(al, ... ,am) C P ~ U
X
cm 2n ,
where B = A( aI, ... ,am). By a slight 'abuse of notation', we now regard G as being defined on P by the equation (*), above. Then G E COO(P) and G(z) :/= 0 for all Z E P. Also, we have Go(3 = g, where (3: A  t SPA(al, ... ,am) is defined by (3('1') = (al(
E
1
£ (P) by
w(z)
=
G(z)1(8G)(z)
(z E P).
aw
A simple calculation shows that = O. By Corollary 8.35, 'H 1 (P) = {O}, and so there is some 'lj; E COO(P) such that w = a'lj;. Set h = exp( 'lj;)G E COO(P), so that Gh 1 = e'I/J on P and 8h = exp('lj;) (8G  G8'lj;) = 0,
which shows that h
E
O(P).
350
Introduction to Banach Spaces and Algebras
Now, applying a functional calculus map O(P) a
+
A, define
= h(al, ... ,am)'
Then, by the spectral mapping property (in fact only Lemma 9.1 is needed),
a = h 0 (3, so that
9 . aI = (G 0 (3)(h 0 (3)1 = e,po{3 E exp C.
Thus fa 1 = (f g1 ) (ga 1) E exp C, and the proof is complete.
o
9.6 The local maximum modulus theorem. The final application of the functional calculus in several variables that we shall offer is Rossi's local maximum modulus theorem. Let A be a natural Banach function algebra on a compact space K. In Section 4.13, we defined the notion of a peak set for A. We now say that a subset E of K is a local peak set for A if there exists f E A and a neighbourhood U of E such that f(x)
= 1 (x
E E)
If(x)1 < 1
and
(x E U \ E);
the function f peaks locally on E. A point Xo E K such that {xo} is a local peak set is a local peak point. We make a preliminary topological remark. Let mEN, and let U1 and U2 be nonempty, open sets in em. Suppose that L is a compact subset of U U V. Then we claim that there exist compact sets Ll and L2 such that L1 CUI, L2 C U2 , and L = Ll U L 2 . Indeed, for j = 1,2, take compact subsets K),n of U) for n E N such that K),n C int K),n+l (n E N) and U{ K),n : n E N} = U). Then there exists no EN such that L C intK 1,no UintK2 ,no' Set L) = LnK),nn for J = 1,2. Theorem 9.10 (Rossi's local peak set theorem) Let A be a natural Banach function algebra on a compact space K, and let E be a local peak set for A. Then E is a peak set for A. Proof Choose f E A and a neighbourhood U of E such that f(x) = 1 (x E E) and If(x)1 < 1 (x E U \ E). Set h = f  I, so that Z(h) n U = E and Reh(x) < 0 (x E U \ E), and hence Reh(x) :::; 0 (x E U). Since E is compact, we may suppose that there are finitely many basic open sets U1 , ... , Ur of the form Uk
= {x
E
K : If)(x)1 < 1 (j
where 1 = no < ... < nr and that
= nkl + 1, ... , nk)} (k = 1, ... , r),
12, ... , fnr
{x E K: If)(x)1 < 2 (j = nkl
E
A, such that E <;;; U~=1 Uk, and such
+ l, ... ,nk)} <;;; U
(k = l, ... ,r).
351
Holomorphic functional calculus
Set n = n r , and then define Vk ={ZEe n :lz]I<1 (j=nkl+1, ... ,nk)}
(k=1, ... ,r),
an open set in en. Next, set V = U~=l Vk and
W
{z E en: Rezl < O} U (en \ V),
=
so that V and Ware open sets in en. Clearly, V n W ~ {z E en : Rezl < O}. We recall that SPA (fl,'''' In) = {f(x) : X E K}, where f(x)
= (Il(x), ... ,In(x))
(x E K).
We claim that f(E) ~ V \ Wand f(K \ E) ~ W, and hence that V U W is an open neighbourhood of SPA (fl,"" In). Indeed, first, suppose that x E K \ U. Then, for each k = 1, ... , r, there exists j E {nkl + 1, ... , nd such that II](x)1 ~ 2, and so f(x) tf V k. Hence f(x) E en \ V ~ W. Next, suppose that x E U \ E. Then, by the choice of I, we have f(x) E {z E en: Rezl < O} ~ W. Thus I(K\E) ~ W. Finally, suppose that x E E. Then x E Uk for some k E {1, ... , r}, and then f(x) E Vk ~ V, and certainly f(x) 1 W. This establishes our claim. By the ArensCalderon lemma, Lemma 9.2, there exist In+l,.'" 1m E A and a polynomial polyhedron P ~ em such that L:= SPA(fl, ... ,1m) c P and 7rn ,m(P) ~ V U W. Set P v = 7r~,~(V) and Pw = 7r~,~(W), so that P v and P w are open sets in em with P v U Pw ~ P::::> Land
Pv n Pw ~ {w E
em: Rewl < O}.
By the preliminary remark, there are compact sets Ll and L2 with Ll C Pw, with L2 c Pv , and with Ll U L2 = L. Set Lo = {(h(x), ... , Im(x)) : x E E}. Since f(E) C V, we see that Lo is a compact subset of L n P v , and so we may suppose that Lo ~ L 2. We can define a branch of the analytic function log on {WI E e : Re WI < O}, and then W ~ (log wI) jWl is a welldefined analytic function on Pv n Pw. By Corollary 8.36, there exist G E O(Pv ) and H E O(Pw) such that
(G  H)( w)
log WI
=  
WI
(w
E
Pv n Pw ) .
Define F(w)
=
{WI exp(wlG(w)) exp(wlH(w))
(WEPv ), (WEPw).
Then F is well defined and analytic on P v U P w . Set g = F(h, ... , 1m) E A. If x E E, then f(x) E V and h(x) = 0, and so g(x) = 0. If x E K \ E, then f(x) E Wand so g(x) 10. Hence Z(g) = E.
352
Introduction to Banach Spaces and Algebras
The function 1/ F is bounded on compact subsets of Pw , and so there exists Ml > 0 such that Re(l/F(w)) ::; Ml (w ELI)' Define
0/'( )
=
exp( WI G (W)) 1 = ~ wiGJ+l(W)
0/ W
6 J=o
WI
(
J
(
1)1
+ .
wE
P)
v ,
so that 1jJ E O( P v ). Thus there exists M2 > 0 such that Re 1jJ( w) ::; M2 (w E L2)' Suppose that W E P v \ Lo and that (WI, ... , Wn ) = f(x) for some x E K. Then x
F(w)
1
=
WI
+ 1jJ(w)
(w E Pv \ Lo),
we have Re(l/F(w))::; M2 (w E L2 \Lo). Set M = max{M1 ,M2 } + 1. Then Re(l/F(w)) < M (w E L \ L o), and so Re(1/g(x)) < M (x E K \ E). Suppose that ( E e with I(  EI < E. Then ( = E(l + re iO ) for some r < 1 and E [0,27rJ, and so
e
Re(l/()
1
= . E
1 1 + 2r cos
1
e + r2
>. 4E
This implies that the set g(K) omits the open disc {( E e : IE  (I < E} in e, where E = 1/4M, and hence that the function E(cl g)1 E A peaks exactly on the set E. Thus E is a peak set, as required. 0 Theorem 9.11 (Local maximum modulus theorem) Let A be a natural Banach
function algebra on a nonempty, compact Hausdorff space K, and let U be a nonempty, open set in K. Then Iflu
=
Ifl(r(A)nu)u8u
(f
E
A).
Proof Let E = {x E U: If(x)1 = Iflu}' We must show that either EnaU =f. 0 or that En r(A) n U =f. 0. Suppose that E n aU = 0. We may suppose that there exists x E U with f(x) = Iflu = 1. Set F = {y E U : f(y) = I} and 9 = (1 + 1)/2. Then F =f. 0, g(y) = 1 (y E F), and Ig(y)1 < 1 (y E U \ F). By Theorem 9.10, F is a peak set for A, and so F n r( A) =f. 0. Since F ~ E c U, the result follows. 0 Recall that it is a conjecture of Gel'fand that the only natural uniform algebra on IT is C(IT). Corollary 9.12 Let A be a natural Banach function algebra on the closed unit interval IT. Then r(A) = IT.
353
Holomorphic functional calculus
Proof Assume towards a contradiction that f(A) C;; lI. Then there exist a, bE II with a < b such that (a, b) n f(A) = 0. Clearly, there exists a function f E A with f(a) = f(b) = 0, but with If(c)1 = Ifl[a,b] > for some c E (a, b). This contradicts Theorem 9.11. D
°
Notes For the applications that we have given, and other applications, of the functional calculus, see [4], [31, Section 21], [32, Chapitre I, Section 4], [47, Section 2.4], [79, Chapter III, Section 6], [96, Chapter III], and [126, Section 3.5]. In particular, Theorem 9.6 is proved in [79, Chapter III, Theorem 6.1] and [126, Theorem 3.5.12], and the ArensRoyden theorem, Theorem 9.9 is proved in [4, Chapter 15] and [79, Chapter III, Theorem 7.2]. Let A be a commutative, unital Banach algebra, and let C = C( A). For n E N, set An = Mn(A) and Cn = Mn(C). Then an extension ofthe ArensRoyden theorem states that G(An)/Go(An) ~ G(Cn)/GO(Cn ). Here Go(An) and Go(Cn ) are the principal components of the identity in G(An) and G(Cn ), respectively. These results explore the relationship between properties of a Banach algebra and of its character space; for further results along these lines, see [132, 157]. The original proof of Theorem 9.10 is due to Rossi; a simplified version is in [79, Chapter III, Theorem 8.1] and [86, Chapter I, Theorem 19]. A more general form is [152, Theorem 9.3]. Further extensions are given in [9]. Exercise 9.5 Let A be a natural Banach function algebra on A. (i) In the case where A has an identity, A is compact and nonempty. Deduce the converse from Shilov's idempotent theorem. (ii) Suppose that A is totally disconnected. Prove that A is dense in C( A). (iii) Let K be a compact, totally disconnected subset of en, where n E N. Prove that O(K) = C(K). Is it true that necessarily R(K) = C(K)? (iv) Let K be a compact, polynomially convex, totally disconnected subset of en. Prove that P(K) = C(K). (A compact, totally disconnected subset of en is not necessarily polynomially convex; see [79, Theorem 2.5].) Exercise 9.6 Let A be a commutative, unital Banach algebra, and let ao, .. . ,an E A. Suppose that g E C(A) satisfies the equation 'L.;=o aJgJ = 0 in C(A), whilst the function 'L.;=1 jaJg J 1 does not vanish on A. Prove that there is a unique element
b E A with
'L.;=o a b1 J
= 0 in A and with
b=
g.
Exercise 9.7 Let A be a natural Banach function algebra on a nonempty, compact Hausdorff space K, and let f, g E A. (i) Set M(f,g) = {x E K: If(x)1 :0:: Ig(x)I}. Let E be a component of M(f,g) such that En Z(g) = 0. Prove that M(f,g) n r(A) 1= 0. (ii) Let g E A. Suppose that there is a set E c:::; K such that g(x) = 1 (x E E) and Ig(x)1 > 1 (x E K \ E). Prove that E is a peak set for A. This is a local minimum modulus theorem. This exercise is taken from [9].
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I ndex of terms
Abelsummable, 65 absolutely convergent power series, 210, 315 absolutely convex, 34 absolutely summable, 38 absorbing, 34 abstract harmonic analysis, 95 Alexander's subbase lemma, 27 algebra, 155 commutative, 1.56 division, 156 identity, 156 Opposite, 232 primitive, 231 radical, 194, 232 semisimple, 194, 232 simple, 156 unital, 156 analytic function, 31, 314 Avalued, 249 Evalued, 113 weakly, 113 analytic mapping, 318 analytic multifunction, 257 analytic semigroup, 225 analytic space, 149 analytically closed, 257 analytically closure, 257 annihilator, 141 approximate identity, 97, 210 approximation property, 56 Arens' theorem, 183, 184, 207 ArensCalderon lemma, 342 ArensRoyden theorem, 348, 353 argument principle, 313 ArgyosHaydon theorem, 182 ArzelaAscoli theorem, 22 Aupetit's lemma, 254 Aupetit's theorem, 257 automorphism, 157 axiom of choice, 8, 14 BadeCurtis theorem, 246 Baire category theorem, 22, 23, 128 Baire function, 285, 288, 293 realvalued, 285 balanced, 34 Banach algebra, 157 amenable, 239
strongly decomposable, 248 Banach *algebra, 260 Banach Amodule, 229 Banach algebra of power series, 187, 206 Banach extension, 183 Banach function algebra, 189 natural, 190 Banach limit, 121 Banach operator algebra, 160 Banach sequence algebra, 192 Banach sequence space, 47 Banach space, 35 Banach's isomorphism theorem, 131, 150, 151 BanachAlaoglu theorem, 118 BanachSteinhaus theorem, 128 Basener's theorem, 208 Bergman space, 83 Bernstein polynomial, 64, 69 Bernstein's approximation theorem, 64 Bessel's inequality, 79 bicommutant, 180 bidual,46 biholomorphic, 325 bilinear mapping, 59, 129 bimodule, 229 Banach,229 dual, 229 Bishop's theorem, 207 BishopPhelps theorem, 127 Borel function, 288 Borsuk's theorem, 149 boundary, 202 Choquet, 208 closed, 202 Shilov, 203 bounded below, 132 bounded operator, 43 Calkin algebra, 261 canonical embedding, 46, 108 Cantor's intersection lemma, 20 Cantor's set, 29 Cantor's theorem, 15, 16 cardinal, 15 Carleson's corona theorem, 209 Carleson's theorem, 96
364 Cauchy's integral formula, 31, 113, 307, 312, 314 Cauchy's integral theorem, 113 Cauchy's theorem, 31, 306 CauchyGreen formula, 311 CauchyGreen theorem, 306 CauchyRiemann criterion, 306 CauchyRiemann equations, inhomogeneous, 310, 318, 319 CauchySchwarz inequality, 70, 263 Cech cohomology group, 225 centre, 180 chain, 10 character, 181, 207 evaluation, 189 character space, 189 Choquet's theorem, 208 closed graph theorem, 133, 151 codimension, 34 Cole's theorem, 208 commutant, 180 complemented, 110 completely regular, 283 completion, 17, 36, 50 complex 147 dual, 148 exact, 147 condItion (F), 48 conjugate index, 39 conjugatelinear map, 42 continuum hypothesis, 15, 247 contour, 30, 311 convergent, absolutely, 56 convergent, unconditionally, 57 convex, 34 convolution product, 159, 195, 196 coordinate functional, 7 coordinatewise convergence, 187 countable, 15 Cousin data, 329 Cousin domain, 329 Cousin problem, first, 329, 337 Cousin problem, second, 337 cyclic vector, 227 C • algebra, 269 C' property, 77 C' subalgebra, 269 Dales' theorem, 247 de la Vallee Poussin kernel, 97 dense range problem, 255 derivation, 229 inner, 229 point, 230 on A, 230 diameter, 20 dimension, 33 Dini's lemma, 249 Dini's theorem, 250
Index of terms Dirichlet kernel, 92, 97, 150 Dirichlet problem, 65, 68 disc algebra, 61, 158, 186 distinguished boundary, 315 Dixon's example, 258 DixonEsterle theorem, 183 Dolbeault cohomology group, 322 Dolbeault complex, 322 DolbeaultGrothendieck lemma, 327 domain of holomorphy, 324 dual Banach space, 45 dual group, 207 dual norm, 45 dual space, 45 Eidelheit's theorem, 243 eigenspace, 171 eigenvalue, 171 approximate, 172 eigenvector, 171 entire function, 31 envelope of holomorphy, 325 equicontinuous, 22 equipotent, 15 equivalent norms, 44 Esterle's theorem, 246, 247 exterior algebra, 319 exterior derivative, 318, 321 exterior product, 320, 321 extreme point, 34, 125 extreme subset, 125 Fejer kernel, 94 Fejer's theorem, 94 Feldman's example, 248 field, 156 ordered, 206 Ford's square root lemma, 262 Fourier coefficient, 66 generalized, 80 Fourier inversion formula, 104 Fourier series, 63, 85 absolutely convergent, 192 Fourier transform, 86, 89, 95, 100, 103, 104, 197, 200, 207 fractional integral semigroup, 225 Frechet algebra, 157, 208 Frechet space, 114 Fuglede's theorem, 277 functional calculus, 215 Borel, 295, 297 continuous, 272 holomorphic, 215, 224, 339, 343 functionally continuous, 182, 325 fundamental theorem of Banach algebras, 166 Gaussian semigroup, 225
365
Index of terms Gel'fand representation theorem, 191 Gel'fand topology, 188 Gel'fand transform, 191 Gel'fand's conjecture, 207, 352 Gel'fandMazur theorem, 167 Gel'fandNaimark theorem, 270, 281 commutative, 272 germs, 212 GNSconstruction, 266 Goldstine's theorem, 125 graph, 132 group algebra, 159 Haar measure, 206 HahnBanach theorem, 106, 107, 115, 121 harmonic, 65 HarrisKadison theorem, 235 Hartogs theorem, 314 HartogsRosenthal theorem, 207 Hausdorff metric, 29 HausdorffYoung theorem, 96 Hermite function, 98 Hermite polynomial, 98 Hilbert space, 71, 73 Hilbertian sum, 267 Holder's inequality, 40 holomorphically convex, 325 homomorphism, 157 hyperinvariant subspace, 171, 298 ideal, 156 left, right 156 maximal, 12, 156, 185 modular, 239 prime, 240 primitIve, 231 proper, 156 quotient, 231 separating, 244 idempotent, 166 trivial, 166 implicit functIOn theorem, 325, 346 independent, 16 index group, 224 infimum, 10 inner product, 70 inner product space, 70 integral, 53, 288, 292 invariant subspace, 171, 298 proper, 171 invariant subspace problem, 301 inverse, 162 inverse function theorem, 32 invertible, 162 left, right, 162 involution, 260 hermitian, 268 isometric, 260 symmetric, 268
Isometric, 17 isometrically isomorphic, 44 isometry, 17, 44 isomorphism, 157 Jacobson radical, 193, 232 Jacobson's density theorem, 236 James space, 59, 121, 209 Johnson's theorem, 239, 243, 246, 257, 262 generalized, 255 JohnsonSinclair theorem, 245 KahaneKatznelson theorem, 96 KahaneZelazko theorem, 222 Kaplansky's question, 247 KelleyVaught theorem, 265 Kolmogorov's theorem, 96 KreinMilman property, 127 KreinMilman theorem, 125, 127 LaceyThiele theorem, 96 Laplace transform, 105, 211 lattice, 10 Laurent series, 210 LCA group, 207 Lebesgue constants, 97 Lebesgue spaces, 52 left Amodule, 226 Banach, 227 lexicographic ordering, 9 linear homeomorphism, 44, 138 linear map, 42 linearly dependent, independent, 11 linearly homeomorphic, 44, 138 Liouville's theorem, 32, 113 realpart, 32 Lipschitz functions, 58 Lipschitz algebra, 209 LMC algebra, 157 local maximum modulus theorem, 352 local minimum modulus theorem, 353 local uniform convergence, 115, 317 locally compact group, 206 locally convex space, 114 metrizable, 114 Lomonosov's theorem, 177 Losert's theorem, 239 lower semicontinuous, 18 main bounded ness theorem, 246 maximal, maximum, 10 maximum modulus theorem, 31, 316 maximum principle for the spectral radius, 253 maximum set, 202 meagre, nonmeagre, comeagre, 22 measure, 287 Borel, 287
366 complex, 287 Mergelyan's theorem, 207 merom orphic function, 210 metric space, 16 complete, 16 Michael's problem, 183 Mi!utin's theorem, 149 mmimal, minimum, 10 Minkowski functional, 127 Minkowski's inequality, 40, 56 MittagLeffler theorem, 23 module homomorphism, 227 module isomorphism, 227 module 227 cyclic, 227 irreducible, 227 separating, 244 monomorphism, 157 monotone class, 285 bounded, 286 monotoneconvergence property, 291, 295 monotoneconvergence theorem, 291 Morera's theorem, 31 Miiller's example, 258 multiplicative semigroup, 155 nest, 136 stabilizes, 136 lllipotent, 166 norm, 34 submultiplicative, 157 normpreserving extension, 109 normal, 261 normed algebra, 157 normed space, 34 nowhere dense, 22 null sequence, 135 numerical radius, 280 numerical range, 280 ntransitive, 236 Oka's lemma, 333 OkaCartan theory, 344 OkaWei! theorem, 336 open mapping lemma, 129 open mapping theorem, 31, 131, 140, 151 operator norm, 44 operator, 43 adjoint, 76 approximable, 47, 160 bounded, 43 bounded below, 52 compact, 46, 160, 175 diagonal, 58 dual, 119 finiterank, 47, 160 invertible, 51, 78, 165 left, right, 51 normal, 76, 173
Index of terms positive, 77, 275 rankone, 47 selfadjoint, 76 spectral, 301 unitary, 76, 173, 299 ordinal, 15 limit, 15 orthogonal, 73 orthogonal family, 73 orthogonal projection, 75 orthonormal basis, 79 sequential, 79 orthonormal sequence, 79 orthonormal subset, 79 pairing, 116 nondegenerate, 116 parallelogram rule, 71 Parseval's equation, 80 partial ordering, 8 partially ordered set, 9 partition of unity, continuous, 111 partition of unity, smooth, 310 path integral, 30 closed, 30 track,30 peak point, 202 local,350 peak set, 202 local,350 peakpoint conjecture, 208 PhillipsSobczyk theorem, 121 Plancherel formula, 104 generalized, 104 Poincare's theorem, 325 point spectrum, 171 pointwise monotone limit, 285 POisson integral, 67 Poisson kernel, 67, 91 polar decomposition, 278, 299 polarization identity, 71 general,71 polyball, 313 polydisc, open, closed, 313 polynomial hull, 179, 331 polynomial polyhedron, 333 polynomially convex, 179,331, 332 polynomially generated, 179, 331 Pontryagin duality theorem, 207 poset, 9 positive, 275 positive linear functional, 263, 288 principal component, 223 principal square root, 221, 262 principle of mathematical induction, 12 principle of transfinite induction, 13 product space, 8, 26 product topology, 26 projection, 171
367
Index of terms Pythagoras's theorem, 73 pform, 320
aclosed, 322 degree, 320 quasinilpotent, 166, 233 quotient algebra, 161 quotient module, 227 quotIent norm, 134 quotient semi norm, 133 radical, 193 Jacobson, 193, 232, 239 prime, 240 strong, 239 RadonNlkodym property, 127 Ransford's proof, 257 Ransford's theorem, 258 rational hull, 338 rationally convex, 338 realdifferentiable function, 305, 314 reflexive, 109 representation, 226 faithful, 226 leftregular, 228 normed,226 continuous, 227 standard, 231 residue theorem, 312 resolvent set, 177 RiemannLebesgue lemma, 86, 96, 198, 207 Riesz's representation theorem, 75 Riesz's lemma, 55 RieszFischer theorem, 89 Ringrose's theorem, 283 Rossi's local peak set theorem, 350 Runde's theorem, 239 Runge's theorem, 212, 213 RussoDye theorem, 284 Sakai's theorem, 283 Schauder's theorem, 144 Schwarz's lemma, 32 selfadjoint, 190, 261 semiinner product, 70 semiinner product space, 70 semigroup, 159 semigroup algebra, 159 weighted, 159, 188 seminorm, 34 seminormed space, 34 separable, 19 separating ideal, 244 separating module, 244 separating space, 135 separation theorem, 123 sesquilinear form, 75 bounded,75
pOSItive, 75 selfadjoint, 75 shIft, left, right, 58 Shilov's idempotent theorem, 345 short exact sequence, 147 afield, 287 Baire, 287 Borel,287 simple functIOn, 52 SingerWermer theorem, 230, 239 smooth function, 309 smooth mapping, 318 spectral radius, 169 spectral radius formula, 169 spectral theorem, 294 spectral threecircles theorem, 252 spectrum, 165, 171 approximate point spectrum, 172 compression, 182 joint, 330 surjectivity, 182 stability theorem, 137 *algebra, 260 *closed, 260 *homomorphism, 260 *ideal, 260 *radical, 264 *representation, 261 faithful, 267 universal, 267 *semisimple, 264 *subalgebra, 260 state, 263 pure, 284 state space, 263 StoneWeierstrass theorem, 60 complex, 61 StoneCech compactification, 272, 283, 294 strongoperator topology, 119 sub algebra, 60, 157 bicommutant, 181 inverseclosed, 180 sublattice, 10 sublinear functional, 106 subspace topology, 26 sum mabie function, 290 support, 309 supremum, 10, 15 surrounds, 30 Swiss cheese, 209 Thomas's theorem, 239 Tietze extension theorem, 130 Titchmarsh's theorem, 206 topological algebra, 182 topological divisor, 183 topological group, 223 topological space, 18 normal,23
368 topological vector space, 121 topology, 18 base, 25 subbase, 25 weakest, 25 totally bounded, 20 totally ordered, 9 trace, 83 trigonometric polynomial, 62 Tychonoff's theorem, 27 uncountable, 15 uniform algebra, 158 uniform boundedness theorem, 128 uniform norm, 36 unique complete norm, 241 uniquenessofnorm theorem, 243 unit, 162 unitary, 261 upper semicontinuous, 18 Urysohn's lemma, 24 vector space, 33 Vesentini's theorem, 257 VidavPalmer theorem, 283 Volterra algebra, 196 von Neumann algebra, 283
Index of terms weak spectral maximum principle, 250 weak topology, 116, 117 weak* topology, 117 weak*continuous, 118 weakoperator topology, 120, 281 weakly bounded, 118 weakly compact, ll8 Wedderburn's principal theorem, 240 Wedderburn's theorem, 237 WedderburnArt in theorem, 240 Weierstrass' polynomial approximation theorem, 62 trigonometnc polynomials, 62 weight function, 187, 196 well ordered, 12 wellordering principle, 12, 14 Wiener algebra, 192 Wiener's theorem, 193 winding number, 30 Woodin's theorem, 247 Zame's theorem, 339, 344 Zemanek's theorem, 251 ZermeloFraenkel axioms, 16 zero set, 23 Zorn's lemma, 10
Index of symbols
a· E, 226 a . 1;, 226 [a, b], 7, 235 A EB.l B, 73 A(~), 61, 95, 186, 260 A(JR), 198 A(Z), 86, 89, 197 A ·1;,227 A"", B, 157 A+, 161 A+,263 ii, 192 A.l, A.l.l, 73 A O P,232 Apos,275 Asa, 261 AC,16 A(E, F), 47, 56 A(E), 47, 160 bv, 209 B(a;r), 16, 138,313 B(K), BJJl.(K), 286 BP(U),322 B(E,F),44 B(E, F; G), 59 B(E), 45, 159 B(H),261 B(X), Ba(X), 287
co,38 co(Z), 38 cardinals No, Nl, 2 ND , 15 cony A, 34 C(K), 37, 60, 158 CJJl.(K),37 CoCK), 37, 158, 260 C oo(K),37 C(U),31 Cl(U),305 C~X), CJJl.(X), C(X)+, 18 C (X), 58 Cb(X), C~(X), 37 C(li),42 CR(li),29 CI(li),83 Ck(li), 208
Coo (li), 208, 246 C(T), 65, 82 Co(JR),37 C[a, b], 37 CE [a, b], 110 C*(a),272 CH, 16, 247 C, 7 C X ,9 C
XT, 7 c, 15 Card, 15 diam(E),20 dimE,33 dZk, 318 dZtt /\ dZ'2 /\ ... /\ dz,,,, 321 DI(U), 306, 314 DN,91 6" 39 ~, 7, 61 ~(a; r), 313 8,318 aj, 8J, 305 akj, akj, 314
exC, 34, 125 expA,220 e" /\ e'2 /\ ... /\ e,l" 319
E(>"), 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 9780199206544
1111111 III
9 780199 206544