Introduction to Banach Spaces and Algebras

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 _ -2-dt -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 Hahn-Banach 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 (Hahn-Banach 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 one-step 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 'one-step' 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 non-empty 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 one-step 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 (Hahn-Banach 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.

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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 real-linear 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 real-linear 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 Hahn-Banach 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 real-linear 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 non-negative. 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 f---t 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 f---7 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 norm-preserving 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 Banach-space 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 norm-preserving extension, say f E E*. The functional f has the required properties. 0

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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 finite-dimensional. 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 norm-preserving 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 vector-valued 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 real-valued, continuous functions e 1 , . .. ,ek on the interval [a, b] such that: 0 ::; e) ::; 1 on [a, b], such that suppe) ~ I) (j = 1, ... ,k), and such that L:~=1 e) (x) = 1 for all x E [a,b]. (Thus we have constructed a so-called continuous partition of unity subordinate to the open covering {It, h ... , Id of [a, b].) Now define k

fo(x)

=

L e)(x)f(x))

(x

[a, b]).

E

)=1

Then fo E C~[a, b] and, for every x E [a, b], we have

Ilf(x) - fo(x)11 =

lit e) (x)(J(x) -

f(x)))II::; teleX) Ilf(x) - f(x))11 <

)=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 Hahn-Banach theorem, E* separates the points of E, it follows that, for every I E Gi, the element I is well defined in E, and is the unique element of E which satisfies (*). It is clear that the map

J:

I

f-7

ib I , Gi E, -4

is a linear mapping from Gi into E. To see that this mapping is also continuous, let I E Gi, and then, using Corollary 3.4(ii), choose 1jJ E E* with 111jJ11 = 1 and

(i I) Ilib III· b

1jJ

Clearly,

Ili

b

It =

=

ib 1jJ I :S ib III(x)11 dx :S (b - a) III[a,bl

J:

0

Thus the map I f-7 I is continuous on Gi and, more specifically, satisfies Thus this map extends uniquely to a continuous clause (ii) for each I E linear mapping from Gda, b] into E; we also write I f-7 I for this extension. That properties (*), (i), and (ii) hold for all I E GE[a,b] all follow byextension from Gi. The uniqueness of I, subject to (*), is immediate from the fact that E* separates the points of E. 0

Gl

J:

J:

We now use Theorem 3.10 to define some E-valued path integrals. Let 'Y: [a, b] - 4 C be a contour, and suppose that F : b] - 4 E is continuous, where E is a Banach space. Define

1

F(z) dz =

lb

F('Y(t)) 'Y'(t) dt.

This definition has all the properties that you would expect by analogy with the case of complex-valued functions (cf. Section 1.11).

113

Banach spaces

Let U be a non-empty, open subset of C, and let E be a Banach space. A function J : U - t E is an analytic (or holomorphic) E-valued function if

J'(z) = lim J(w) - J(z) w ..... z

W -

Z

exists in E (with convergence in the norm topology) for each z E U; J is weakly analytic (or weakly holomorphic) if A 0 J : U - t C is analytic in the usual sense for each A in E*. It can be shown that each weakly analytic function is already analytic, but this will not be important for us because the only property of an analytic function that we shall use is that it is weakly analytic. We now give vector-valued forms of Cauchy's integral theorem (and formula). Theorem 3.11 Let U be a non-empty, open subset oj C, let E be a complex Banach space, and let J be an analytic E-valued Junction on U. Let '"'( be a

contour in U such that n( '"'(; z) = 0 Jor every z E C \ U. Then: (i)

1

J(z) dz = 0 ;

(ii) Jor every w E U \ b]' we have 1 n('"'(; w)J(w) = -2. 1fl

1

J(z) - dz.

'Y Z -

W

Proof To prove that the equations in (i) and (ii) hold, we apply A E E* to each side of each equation, and use Theorem 3.10 and the standard Cauchy's integral theorem and integral formula, Theorem 1.32, to see that the action of A on each side of both of the equations gives equality. The result follows because E* separates the points of E. D The following result will be used later in our account of spectral theory. Theorem 3.12 (Liouville's theorem for vector-valued functions) Let E be a complex normed space, and let J : C - t E be a Junction which is weakly an-

alytic and bounded. Then J is constant. ,Proof For every A E E*, the function A 0 J is a complex-valued entire function on Co Since J is bounded, each function A 0 J is also bounded, and so is constant by the classical Liouville's theorem, Proposition 1.39. Then, for Zl, Z2 E
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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 spaces-usually 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 complex-valued analytic functions on a non-empty, open subset U of e (or e n see Section 8.4). The topology on O(U) is the standard one-that of local uniform convergence. Indeed, we may write U = U~=l K n , with each Kn compact and Kn C int Kn+l for each n E fiI, and then define

Banach spaces

115

Pn(f) = sup{IJ(z)1 : z E Kn}

(n E N)

for J E C(U), so that each Pn is a semi norm on C(U). The family {Pn : n E N} defines the topology of local uniform convergence on C(U), and hence on O(U). The next result is analogous to the one where E and Fare normed spaces. Lemma 3.13 Let (E; P) and (F; Q) be locally convex spaces, and let T : E be a linear map. Then the following are equivalent: (a) T is continuous on Ei (b) T is continuous at OEi (c) for each q E Q, there are seminorms PI, ... ,Pn E P on E and M such that q(Tx) :'S Mmax{PI(x), ... ,Pn(x)} for all x E E.

-+

F

> 0 D

Corollary 3.14 (Hahn-Banach theorem for locally convex spaces) Let E be a locally convex space, let F be a subspace, and let g be a continuous linear functional on F. Then there is a continuous linear functional J on E such that f I F = g. In particular, Jor any x -1= 0 in E, there is some continuous linear functional J on E wzth J ( x) -1= o. Proof This is immediate from Corollary 3.3.

D

Corollary 3.15 Let E be a locally convex (in particular, a normed) space. Then a linear Junctional), on E is continuous if and only zJ ker). is closed in E.

Remark: the 'if' part of this result does NOT extend to linear mappings that are not into the scalar field. Proof The scalar field of E is OC. Of course, it is clear that the vector subspace ker). is closed whenever). is continuous. Conversely, let a linear map). : E -+ OC have closed kernel, say Z. We claim that ). is continuous. We may suppose that). -1= 0, since otherwise the result is trivial. Then Z -1= E, and so we may choose Xo E E \ Z. Since Z is assumed to be closed, there is then a continuous seminorm P on E such that, setting V = {x E E : p(x) < I}, we . have (xo + V) n Z = 0. Assume towards a contradiction that )'(V) is an unbounded subset of oc. Then, for any JL E OC, there is some x E V with 1).(x)1 > IJLI. But now

). ().tX) x) = JL

and

P

().tX) x) < 1 ,

so that JLx/),(x) E V, and therefore JL E ).(V). Thus, in this case, )'(V) = II(. But then, in particular, there is some v E V with ).(v) = -).(xo), and so with ).(xo + v) = 0, it follows that Xo + v E (xo + V) n Z, contrary to the choice of V. Hence )'(V) is a bounded subset of II(. By the usual argument, there is a constant M > 0 such that 1).(x)1 :'S Mp(x) (x E E), and the continuity of). is proved. 0

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Introduction to Banach Spaces and Algebras

The following result is proved by a slight modification of the proof of Theorem 2.22. Theorem 3.16 Let E be a metrizable locally c~nvex space. Then there is a unique Frechet space structure on the c~mpletio~E of E such that E is isometrically embedded as a dense subspace of E. Thus E is a Frechet-space completzon ofE. 0

3.4 Paired vector spaces. We wish to discuss, mainly, the weak topology on a Banach space and the weak-* topology on its dual. It is useful to set these in a slightly wider context. Let E and F be linear spaces over IK. A pairing of (E, F) is a specified nondegenerate, bilinear form (x, y) f---> (x, y) on ExF. [Here, 'non-degenerate' means: for each x 1= 0 in E, there is some y E F such that (x, y) 1= 0, and, for each y 1= 0 in F, there is some x E E such that (x, y) 1= o. Unlike the finite-dimensional case, the two parts of this condition are not equivalent.] Examples 3.17 (i) Let E be a IK-linear space, and let F be its algebraic dual space, with (x,!) = f(x) for x E X and f E F. Then (., .) is a pairing of (E,F). (ii) Let E be a locally convex space (in particular a normed space), and set F = E*, the 'continuous dual'; the pairing is as in (i). The proof that this pairing is non-degenerate uses the Hahn-Banach theorem. (iii) Let F be a normed space, and take E = F* to be its dual space. Then (F*, F) is a non-degenerate pairing; this pairing is not, in general, a special case of (ii). However, the pairing (F*, F**) is a special case of (ii). (iv) If (E, F) is any pairing, then there is always the 'reverse pairing', in which the bilinear form remains the same, but the roles of E and F are interchanged. 0 Let us use the above notation (x,1) f---> (x,!) for the pamng between a normed space E and its dual space E*, where x E E and f E E*, and also the notation (j, A) f---> (j, A) for the pairing between E* and E**. Then the canonical embedding J : x -+ X, E -+ E**, described on page 108, takes the form

(j,Jx)

=

(x,!)

(x E E, f E E*).

3.5 Weak and weak-* topologies. It is important to know that, associated with any pairing, there is a weak topology. Let (E, F) be a pairing. Then the mappings

Pf:xf--->l(x,!)I,

E-+lR.,

as f runs through F, form a point-separating family of seminorms on E. The associated locally convex topology on E is called the weak topology associated

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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-*-to-weak-*-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 a-closed. Conversely, let F be a closed subspace (with F -I- E), and let Xo E E \ F. Then, using the Hahn-Banach 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 (Banach-Alaoglu 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, conjugate-linear map from H onto the Banach space dual of H that was given in Corollary 2.54. Then the adjoint of T is strictly J-1T* 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 strong-operator 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 strong-operator 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 weak-operator 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 so-continuous if and only zf it is wo-contmuous. Proof Clearly, if f is wo-continuous, then it is so-continuous. Suppose that f is so-continuous. 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 wo-continuous. 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 Hahn-Banach 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 non-separable. A Banach space E is quasi-reflexive if E** / E is finite-dimensional. In fact there are non-reflexive, quasi-reflexive spaces. For example, let J be the James space of Exercise 2.14. Then J** / J is even one-dimensional. Further, J** is isometrically isomorphic to J-but still J is non-reflexive. For a discussion of J, see [2, Section 3.4] and [47, Example 4.1.45]. For a more sophisticated approach to the integration of Banach-space-valued functions, involving, for example, the Bochner integral, the Pettis integral, and the Bartle integral, see [58, Chapter II]. The fact that a weakly analytic Banach-space-valued function is already analytic is proved in [144, Theorem 3.31]. We have mentioned the strong-operator and weak-operator topology on B(E, F). In fact, there are several other useful topologies that can be defined on B(E, F); see [63, Chapter VI, Section 1], for example. Exercise 3.1 Let E be a separable normed space. Prove the Hahn-Banach theorem for E without using the Zorn's-lemma part of the argument. In fact, one can arrange to use the main part of the proof of Theorem 3.1 count ably many times to extend a given functional to a dense linear subspace of E, and then extend the functional to the whole space by continuity. Exercise 3.2 Let X be the space (£nlR, where 1 ::;: p ::;: 00, so that X is just the plane IR X 1R, with the norm II· lip' Let Y be a one-dimensional subspace of X, so that Y is a line through the origin, and let g be a linear functional on Y. Of course there is a norm-preserving extension of g to a linear functional f on X. Discuss whether or not such a functional f is uniquely defined for various values of p. Exercise 3.3 Let L be the left shift operator on £It', as in Exercise 2.11. Prove that there is a continuous linear functional f on £It' such that f(Lx) = f(x) and liminfx n n~oo

for all x =

(Xn )n2:1 E

::;:

f(x)::;: lim sup Xn n~=

CIt'. (Such a functional is called a Banach limit.)

122

Introduction to Banach Spaces and Algebras Here is a hint. For n E N, define

fn(x)

=

!(Xl n

+ ... + Xn)

(x

E

£llf'),

and set Y = {x E £llf' : limn~oo fn(x) exists}. Set p(x) x E £llf', and apply the Hahn-Banach theorem.

limsuPn~oo

fn(x) for

Exercise 3.4 For f,g E C(ll), define

d(j,g) =

11 . If~tL-:-g(t),I"

dt.

Show that d is a metric on C(ll), and that C(ll) is a topological vector space with respect to the metric topology. Show that the only continuous linear functional on this space is the zero functional. Exercise 3.5 Let E = co. Then E* = £1 and E** = £00 (see Example 2.18). Show, directly, that the linear functional f on £ 1 , defined by

00 f(Xl,X2,X3"")=Lx n

((Xn)E£I)

n=l

is continuous (and so weakly continuous), but that it is not weak-* continuous. By considering the subspace ker f, show that there is no analogue of Theorem 3.20, with 'weak' replaced by 'weak-*'. Exercise 3.6 Let E be a Banach space with dual space E*. Set B = (E*)[!]. Prove that B is metrizable (in the weak-* topology) if and only if E is separable. Exercise 3.7 (i) Let E be a Banach space. Show that the dual space E* is reflexive if and only if E is reflexive. (ii) Show that the closed unit ball of a Banach space E is weakly compact if and only if E is reflexive. (iii) Let E and F be Banach spaces. Show that the closed unit ball of B(E, F) is compact in the weak-operator topology if and only if F is a reflexive space.

Separation theorems We now give some geometric versions of the Hahn-Banach theorem. 3.7 Separation of convex sets. In what follows, E is a locally convex space over JR., except where otherwise stated. (The results may also be applied to a complex space by the usual device of restricting multiplication to real scalars.) Lemma 3.25 Let C be a convex, open neighbourhood of OE in E. For every x E E, define J-lc(X) = inf{a > 0: x E aC}.

Then J-lc is a sub linear functional on E and C = {x

E

E : J-lc(x) < I}.

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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 non-empty, convex subsets of E with An B = 0. (i) Suppose that A is open. Then there exist f E E* and ~ E lR such that

f(x) <

~

::; 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 non-empty, 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 Hahn-Banach 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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Introduction to Banach Spaces and Algebras

elementary to see) f(A) is open. We may thus take of the interval f(A) to establish (i).

~

as the right-hand 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, real-linear 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 non-negative. 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 wo-closed if and only if it is so-closed. 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 Krein-Milman 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 non-empty 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 non-empty. Theorem 3.31 (Krein-Milman theorem) Let E be a real locally convex space, and let K be a non-empty, 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 non-empty (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 non-empty, 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 Krein-Milman 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 Krein-Milman 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 Krein-Milman 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 Radon-Nikodjm property of Banach spaces. Indeed, a dual Banach space has the Krein-Milman property if and only if it has the RadonNikodym property. This is further related to the important Bishop-Phelps 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 Go-set 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 Banach-space 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 (Banach-Steinhaus 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:xf--tT(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 f--t 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 k-dense 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 k-dense 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 k-dense 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 Ilk-I(TxO - y)11 ::; 1, so that there is an element Xl E E[R] such that IITxI - k-l(y - Txo)11 ::; k, i.e.

Ily -

(Txo

+ kTxI)1I

::; k 2 .

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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 i-dense 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 non-empty, 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 2/3-dense 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. T-1 E B(F, E). Proof Clearly, T- I : F --* E is linear. Let K be the constant specified in the theorem. For each y E F, there is a unique element x E E with Tx = y. By the theorem, Ilxll :s; K lIyll, and so IIT-I(y)11 :::; K Ilyli. This shows that T- I is bounded, and hence continuous. D

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Corollary 3.42 Let E be a vector space that is a Banach space with respect to two norms, II ·11 and III . III· Suppose that there zs a constant C > 0 such that Illxlll :::; C Ilxll (x E E). Then 11·11 and 111·111 are equivalent. 0

Recall from the definition immediately before Proposition 2.23 that an operator T E [3(E, F) is bounded below if and only if there is a constant c> 0 such that IITxl1 2': cllxll (x E E). Corollary 3.43 Let E and F be Banach spaces, and let T E [3(E, F). Then the following assertions are equzvalent: (a) T is injectzve and has closed range; (b) T is bounded below. Proof We may suppose that E f= {O} since the result is trivial in the case where E = {O}. (a) =}(b) Since imT is closed in F, it is a Banach space, and now T defines a linear homeomorphism of E onto im T. By Corollary 3.41, T has a continuous inverse, say S : im T ---+ E. Of course x = (ST) (x) (x E E), so that

Ilxll :::; IISllllTxl1 f= {O}, necessarily IISII f= 0, and so

(x E E).

Since E (b) holds with (b) =}(a) This was proved in Proposition 2.23(i).

c = IISII- 1 . o

Let E be a Banach space, and let S, T E [3(E). Then 5 and T are said to be similar if there is an invertible operator U E [3(E) such that 5U = UT. Proposition 3.44 Let E be a Banach space, and let 5, T E [3(E) be similar operators. Then 5 is bounded below if and only if T is bounded below. Proof Take U E [3(E) with U invertible such that 5U = UT, and set V = U- 1 , so that V5 = TV. Suppose that T is bounded below, so that there exists c > 0 such that IITyl1 2': c Ilyll (y E E). Let x E E, and set y = U.T. Then

cIlxll :::; c1lUIIIIVxli :::; 1lUIIIITVxli = 1lUIII1V5xli :::; 1lUIIIIVIII15xli , and so 5 is bounded below. The result follows. Let 5 and T be non-empty sets, and let j : 5 graph of f is the set

o ---+

T be a function. Then the

Gr(J) = {(8,j(S)) E 5 x T: s E 5}. In the case where 5 and T are topological spaces and j is a continuous map, it is clear that Gr(J) is a closed subset of 5 x T; in general, the converse is not true. In the case where E and F are vector spaces and T : E ---+ F is a linear map, it is also clear that Gr(T) is a vector subspace of the vector space E x F.

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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 non-trivial point to be proved is that T is continuous whenever Gr(T) is a closed subspace of Ex F. The space E x F is a Banach space, for example for the norm defined by

II(x, y)11 = Ilxll + Ilyll

(x E E, y E F)

(see Exercise 2.5). Since Gr(T) is closed in E x F, the subspace Gr(T) is itself a Banach space. Let 7fE and 7fF be the coordinate projections of E x F onto E and F, respectively; each of 7fE and 7fF is a continuous linear mapping. However, 7fE I Gr(T) maps Gr(T) bijectively onto E, and so, by Banach's isomorphism theorem, Corollary 3.41, 7fE I Gr(T) : Gr(T) ----> E is a linear homeomorphism. But then T =

7fF

0

(7fE

I Gr(T))-1

is continuous, as required.

D

3.12 Quotient norms. We now discuss quotient norms. Let (E;p) be a seminormed space, let M be a subspace of E, and let 7fM : E ----> ElM be the quotient mapping. For every ~ EEl M, define PM(O = inf{p(x) : x E E, 7fM(X) =

o.

Then it is easy to check that PM is a seminorm on ElM; it is the quotient seminorm. The quotient mapping 7f M is an open mapping and is 'seminorm decreasing' . Lemma 3.46 Let (E;p) be a seminormed space, and let M be a subspace of E. Then PM is a norm if and only if M is closed in E. Proof Suppose that M is a closed subspace of E, and let ~ E ElM with o. Then there is a sequence (Xnk:':l in E with 7fM(X n ) = ~ (n E N) and such that p(xn) ----> o. But then Xl - Xn E M (n EN), whilst Xl - Xn ----> Xl. Since M is closed, it follows that Xl E M, and thus ~ = 7fM(Xt} = o. Thus PM is a norm on ElM. Conversely, suppose that PM is a norm on ElM. Let X E M, so that there is a sequence (Xn)n:::>:l in M with Xn ----> x. But then, writing ~ = 7fM(X), we have PM(~) ::; p(x - Xn) ----> 0 for each n E N, so that PM(~) = 0, and hence ~ = 0, i.e. X E M. Thus the subspace M is closed. D PM(~) =

,

In particular, for a seminormed space (E;p), set Eo = {x E E : p(x) = O}, so that Eo is a subspace which is the closure of {O} in (EiP). Then the quotient seminorm on E I Eo is a norm.

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Notation: In the case where the subspace M is a closed subspace of a seminormed space E, the norm PM is more usually written in a norm notation, for example, as II . II E/ M or just II . II· It is called the quotient norm on ElM.

Lemma 3.47 Let E and F be normed spaces, and let T E B(E, F). Further, set M = ker T, let 7r M : E ----> ElM be the quotient mapping, and let ElM be given the quotient norm. Then there is a unique linear mapping T M : ElM such that TM 07rM = T. Moreover, TM continuous with IITMII:::; IITII.

is

---->

F

injective, im TM

= im T, and TM zs

Proof The existence and uniqueness of the linear map TM with TM 07rM = T, with T M injective, and with im TM = im T is all elementary algebra (assumed to be well known). We write II·II E/M for the quotient norm. Let ~ E ElM. Then, for every x E E such that 7r M (x) = ~, it follows that IITM(~)II

= IITxl1 :::; IITllllxll·

Hence, taking the infimum over all such x, we see that IITM(~)II Thus TM is continuous with IITM I :::; IITII.

:::;

IITIIII~IIE/M. D

Let E and F be normed spaces, and let T E B(E, F). Then we can regard T (E) as a normed space by 'transferring the norm from E I ker T', since the latter space is linearly isomorphic to T(E). In general, this norm is not equivalent to the relative norm from F. We now show that when we take appropriate quotients of Banach spaces, we 'remain within the category'.

Theorem 3.48 (Completeness of quotients) Let E be a Banach space, and let M be a closed subspace of E. Then (ElM; II·II E/M) is a Banach space. Proof The definition of II·II E / M makes it clear that, for any R > 1 (e.g. R = 2 will suffice), we have 7rM(E[RJ) :2 (EIM)[lJ. Hence, by the open mapping lemma, Lemma 3.38(ii), (ElM; II·IIE/M) is complete. D For a direct proof of the above result, see Exercise 3.13.

Corollary 3.49 Let E and F be Banach spaces, let T E B(E, F), let M = ker T, and suppose that im T is closed in F. Then the induced map T M : ElM ----> F is a linear homeomorphism between ElM and im T. Proof By Lemma 3.47, TM is a continuous, injective linear mapping of ElM onto im T. But ElM is complete by Theorem 3.48, and also im T is complete because it is a closed subspace of F. The result now follows from Banach's isomorphism theorem, Corollary 3.41. D

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Banach spaces

Corollary 3.50 (Universal property of £ I) Every separable Banach space is linearly homeomorphic to a quotient space of £ I . Proof Let F be a separable Banach space, and choose a countable, dense subset {en: n E N} of the closed unit ball F[ll' For x = (Xnk;~l E £ I (real or complex according to whether F is a real or complex space), define CXJ

Tx= Lxnen . n=l Note that the series converges in F because L~=l Ilxnenll :S Ilxlll and F is assumed to be complete. Evidently T E B( £ I , F). Let B be the closed unit ball of £ I. Then the construction makes it clear that {en: n E N} ~ T(B) ~ F[ll' so that T(B) is dense in F[ll' Now the open mapping lemma, Lemma 3.38, shows that, in fact, T( £ I) = F. By Corollary 3.49, T induces a linear homeomorphism between £ I I ker T and F. 0 3.13 The separating spaces and a stability theorem. A null sequence in a normed space E is a sequence (Xn)n>l in E such that lim n ---+ CXJ Xn = O. Let E and F be Banach spaces, and let T E £(E, F). To apply the above closed graph theorem, we must show that Gr(T) is closed in E x F. It is easily seen that, to do this, it suffices to prove the following: for each null sequence (X n )n21 in E such that TX n ----+ y in F, it is necessarily the case that y = O. The point is that we can suppose in advance that the sequence (Tx n )n21 converges to some element of F. Indeed, we can reformulate the closed graph theorem in the terminology of a separating space; this reformulation will be important in Part II. Let E and F be normed spaces, and let T : E ----+ F be a linear map. Then the separating space, 6(T), of T is defined to be the set of all y E F such that , TX n ----+ Y for some null sequence (Xn)n>l in E. Equivalently, y E 6(T) if and only if, for each n E N, there exists x E E with Ilxll < lin and IITx - yll < lin. The following theorem is immediate from the closed graph theorem. Theorem 3.51 Let E and F be Banach spaces, and let T : E ----+ F be a linear map. Then 6 (T) is a closed subspace of F, and T is continuous if and only if

6(T) = {O}.

0

Corollary 3.52 Let E be a vector space that is a Banach space with respect to two norms, 11·11 and 111·111. Suppose that y = 0 whenever Xn ----+ 0 in (E; 11·11) and Xn ----+ Y in (E; 111·111). Then 11·11 and 111·111 are equivalent. Proof By the closed graph theorem, the identity map ~ : (E; 11·11) ----+ (E; 111·111) is continuous, and so there is a constant C > 0 such that Illxlll :s: C Ilxll (x E E). By Corollary 3.42, II ·11 and 111·111 are equivalent. 0

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Proposition 3.53 Let E and F be Banach spaces, let T : E --+ F be a linear map, and take Q : F --+ FIS(T) to be the quotient map. Then S(QT) = {O}, QT is continuous, and T-1(S(T)) is a closed subspace of E. Proof Let y E F be such that Qy E S(QT). Then there is a null sequence (X n )n>l with QTx n --+ Qy as n --+ 00. Thus there exists (Yn)n>l in SeT) such that TX n - Y - Yn --+ 0 in F as n --+ 00. For each n E N, there exists Zn E E with Ilznll < lin and IITzn - Ynll < lin. But now (xn - Zn)n21 is a null sequence in E such that T(xn - zn) --+ Y as n --+ 00, and so Y E SeT) and Qy = O. Hence S(QT) = {O}, and so QT is continuous. We have T-1(S(T)) = kerQT, and so T-1(S(T)) is closed in E. 0 Proposition 3.54 Let E and F be Banach spaces, and let T : E linear map.

--+

F be a

(i) Suppose that E1 zs a Banach space and that R E B( E 1, E). Then we have S(TR) <:;;; 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 n-1 IISnTUnYnl1 ::::: n + 1 +

L

IITUJYJII

(n::::: 2).

(*)

J=l

We now define Xn = 2::;:n UJYJ in E for n E N; each series converges because IIUJYJII ::; 2- J (j EN). Now Xl = U1Y1 + ... + UnYn + Xn+1 (n EN), and so n-1

IISnTUnYnl1 ::;

IITxIil + L

IITUJYJII

+ II S nTx n+111

.

(**)

J=l

But IISnTxn+111 ::; 2::;:n+11ISnTUJIIIIYnll ::; 1, and so, from (*) and (**), it follows that n ::; IITx111 for each n E N, a contradiction. Thus there exists N E N such that SnT R1 ... Rn is continuous for each n:::::N. For n E N, set 6 n = 6(TR1··· R n ), and let Qn : F -+ F/6 n be the quotient map. By Proposition 3.54(i), 6 n +1 s: 6 n (n EN), and so (6(TR1··· R n ))n2 1 is a nest in F. By the above result with Sn = Qn+1, there exists N E N such that Qn+1TRl··· Rn is continuous for each n ::::: N. By Proposition 3.54(ii), 6(TRl··· Rn) s: kerQn+! = 6 n+! (n::::: N). Thus 6 n = 6N (n ::::: N).

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Introduction to Banach Spaces and Algebras

(ii) For n E N, define Un = 8 1 ... 8 n T - TR1 ... R n , so that Un+1 = UnRn+1 - 8 1 " , 8n(TRn+1 - 8 n+1T). By hypothesis, U 1 is continuous and T Rn+1 - 8 n+1T is continuous for each n E N, and so, by an immediate induction, each Un is continuous. Thus 6(8 1 ···8n T)

= 6(TR1'" Rn)

By Proposition 3.54(ii), 8 1 " , 8 n 6(T) 8 1 ···8n 6(T)

(n E N).

= 6(81 ... 8 n T) , and so

= 6(TR1 ... Rn)

(n E N).

The result now follows from (i).

0

3.14 Open mapping theorem for Frechet spaces. In our later study of homomorphisms between algebras of analytic functions, we shall need to know that the open mapping theorem, proved in Section 3.11 for Banach spaces (see Theorem 3.40), is also valid for Frechet spaces. It does, in fact, work for even more general complete, metrizable topological vector spaces, but we shall restrict ourselves to the Frechet case. Although, in retrospect, it may seem that the proof is not dissimilar from the proof for Banach spaces, it will be seen that the details are a good deal more fiddly. Let E and F be Frechet spaces, and let T : E ----> F be a linear map such that T is a bijection. In the case where both T and T- 1 are continuous, we again say that T is a linear homeomorphism and that the spaces E and F are linearly homeomorphic. Let E be a metrizable, locally convex space, with its topology given by an increasing sequence, say (Pn)n21, of seminorms. Then, as before, we may suppose that the associated complete metric d is specified by the formula d(x, y)

=

L

n21

1 Pn(x-y) 2n . 1 + Pn(x _ y)

(x,YEE).

The metric is translation-invariant: that is, d(x + z, y + z) = d(x, y), and, in particular, d(x, y) = d(x - y, 0) for all x, y, z E E. Suppose that the topology of E is defined by a metric d, and let a E E. For r > 0, we again write B(a;r)

= {x

E

E: d(x, a)

< r}

for the open ball of radius r around a. It is clear that B(a; r) = a + E[r] , where now E[r] = {x E E: d(x,O) :::; r}. Most of the work for the open mapping theorem for Frechet spaces is contained in the following lemma.

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Banach spaces

Lemma 3.56 Let (E; d) be a Frechet space, let (F; d) be a metrizable, locally convex space, and let T : E ----+ F be a continuous linear mappzng. Suppose that, for every r > 0, there exists some P == per) > 0 with F[p] ~ T(B(O;r)). Then: (i) B(O; p) ~ T(E[a]) for every a > r; (ii) T is a surjective, open mapping; (iii) F is complete.

Proof (i) Let r > 0, let p = per) be as specified in the hypotheses, and then take a > r. For each x E E, it is clear that B(Tx;p) ~ T(B(x;r)), where

p = per). Let y E B(O; p). We shall show that y = Tx for some x E E[a]. Indeed, let (rn)n~l be a sequence in jR+. with rl = r and a = L~=l rn. Take PI = p and, for each n ::::: 2, take Pn with 0 < Pn s: p(rn) and such that Pn ----+ 0 as n ----+ 00. Evidently, for every x E E and n E N, the set T(B(x; rn)) is dense in B(Tx; Pn). We shall next define, inductively, a sequence (.Tn)n~O in E with Xo = 0 and such that: (a) Xn E B(Xn-l; rn) (n::::: 1), and (b) TX n E B(y; Pn+l) (n::::: 0). For n = 0: clause (b) holds since y E B(O;p), and so 0 = Txo E B(Y;PI)' For n = 1: y E F[Pl] ~ T(B(O; rt)), and so there is some Xl E B(xo; rt) with TXI E B(y; P2). Now take n ::::: 2, and assume that we have chosen Xo, ... ,Xn-l such that (a) and (b) hold. We have B(Txn-l; Pn) ~ T(B(xn-l; rn)), and B(Txn-l; Pn) contains y because TXn-1 E B(y; Pn). Thus there is an element Xn E B(Xn-l; rn) with TX n E B(y; Pn+1)' This continues the inductive definition. We have seen that there is a sequence (Xn)n~l that satisfies (a) and (b). For n, k ::::: 1, we have d(xn, Xn+k) s: rn+l + ... + rn+k, so that (Xn)n>l is a Cauchy sequence in E, and thus limn->oo Xn exists, say x = limn->oo Xn. We have d(x,O) s: L~=l rn = a, i.e. x E E[a]' Also, since T is continuous, TX n ----+ Tx in F. However, d(Txn' y) s: Pn+1 (n EN). Hence TX n ----+ y, and so Tx = y, which implies that y E T(E[a]). This proves (i). (ii) It is immediate that T is surjective. Now let U be a non-empty, open subset of E, let y E T(U), and choose some x E U with Tx = y. Next, choose rS > 0 such that x + E[28] ~ U. Set P = p(rS). It follows from (i) that B(O;p) ~ T(E[28]), so that T(U) :2 T(x

+ E[28]) :2 y + B(O; p).

Thus T(U) is open in F. This proves that T is an open mapping. (iii) Let F be the completi(~n of F (see Theorem 3.16). For every r > 0, there exists some P > 0 such that F[p] ~ T(B(O; r)), where the latter closure is now taken in F. Clause (ii) applied to the map T : E ----+ F shows that im T = F. However, imT ~ F, and so F = F. Thus F is complete. 0

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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.

141

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 Hahn-Banach 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 'norm-closed'; 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 norm-closed.

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 Hahn-Banach 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 E-L = {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 > IITII-E.

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 finite-dimensional.

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 Arzela-Ascoli 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 .

145

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 Hahn-Banach theorem that every 9 E M* has a norm-preserving 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 Hahn-Banach 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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T* = 7r* aTo"; by Corollary 3.63(ii), the map To" is a linear homeomorphism, and, by Proposition 3.68, 7r* is isometric. Thus IIT*fll = IITofll;::: II(To)-lllllfll

(f E F*),

and so IIT*fll ;:::cllfll (fEF*),wherec= II(To")-111 = liTo-III. Corollary 3.70 Let E and F be Banach spaces, and let T following assertions are equivalent:

E

0

B(E, F). Then the

(a) imT = F; (b) T* is injective with norm-closed range; (c) there is some c > a such that IIT*fll ;::: cllfll (f E F*); (d) T* is injective with weak-*-closed range. Proof The equivalence of (b) and (c) is immediate from Corollary 3.43, and the equivalence of (a) and (c) comes from Theorem 3.69. Clearly, (d) implies (b). The proof will now be completed by showing that (a) implies that im T* is weak-*-closed. We know (from Lemma 3.61(i)) that (kerT)~ is the weak-*closure of im T*, so that it will suffice to show that (ker T) ~ s;.; im T* . Indeed, let 7r : E ---4 E / ker T be the quotient map, and let To : E / ker T ---4 F be the unique linear homeomorphism such that T = To a 7r. Then

7r* : (E / ker T)*

---4

E*

is an isometric mapping with im 7r* = (ker T)~, and To" is a linear homeomorphism of F * onto (E / ker T) * with T* = 7r* a To" . Now let g E (kerT)~. Then there is a unique 'Y E (E/kerT)* such that 7r*("() = g, and then there is a unique f E F* with To"(f) = 'Y. It follows that g = 7r*("() = (7r* a To") (f) = T*(f) E imT*, so that (ker T)~ s;.; im T*. The proof is complete. Corollary 3.71 Let E and F be Banach spaces, and let T following assertions are equivalent:

0 E

B( E, F). Then the

(a) imT is closed in F;

(b) im T* is weak-*-closed in E* ; (c) im T* is norm-closed in E* . Proof (a) =}(b) Set Y = imT, a closed subspace of F, and let J : Y ---4 F be the inclusion map. Define TI : E ---4 Y by setting Tix = Tx (x E E), so that TI E B( E, Y) with im TI = Y. By Corollary 3.70, the dual operator Ti is injective, with weak-*-closed range. However, T = J a T 1 , and so T* = Ti a J*. But we know that J* : F * ---4 Y* is surjective, so im T* = im Ti and (b) is proved.

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Banach spaces

(b) :=;. (c) This is trivial. (c):=;.(a) Now set Y = imT, and let J : Y ----+ F be the inclusion map. Define Tl : E ----+ Y by setting T1x = Tx (x E E), so that Tl E 13(E, Y) with im Tl = im T and J 0 Tl = T. Now im Tl = Y, so that Ti : Y* ----+ E* is injective. Also, T* = Ti 0 J* and J* : F * ----+ Y* is surjective, so that im Ti = im T*, which is norm-closed in E*. It follows from Corollary 3.70 that im Tl = Y, which is closed in F. Since im T = im T 1 , the proof is complete. 0

3.17 Short exact sequences. The main results of this section may be neatly summarized in the language of exact sequences. Let X, Y, and Z be vector spaces (over the same field OC), and let S : X ----+ Y and T : Y ----+ Z be linear maps. Then the sequence

X~Y~Z is said to be a complex if im S ~ ker T, and exact at Y if im S = ker T. We now write 0 for the vector space {a}. The only linear maps O---+X or X ---+0 are necessarily the zero map, which is usually not written in a diagram, and so a complex

o---+X~Y is exact at X if and only if S is injective; similarly, a complex

Y~Z---+o is exact at Z if and only if T is surjective. A short sequence of vector spaces and linear maps is a sequence

o ---+

S

T

X ---+ Y ---+ Z ---+ 0;

the sequence is exact if it is exact at X, Y, and Z, i.e. it is such that S is injective, im S = ker T, and T is surjective. In particular, it necessarily follows . that, in this case, we have Z ~ Y / im S as vector spaces. Note also that a short sequence, as above, is a complex if and only if TS = O. We now consider complexes of Banach spaces and continuous linear maps, often written £: O---+X~Y~Z---+O. (The symbol £ is just a way of giving a name to the sequence). We note that by 'exactness' of £, we shall continue to mean the purely algebraic notion just explained. We remark that, if £ is a short exact sequence of Banach spaces and continuous linear maps, then necessarily Sand T have closed ranges, being equal, respectively, to ker T and Z. Also, using the open mapping theorem, we see that the induced linear isomorphism Z ~ Y / im S is also a homeomorphism.

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Given E, there is the naturally induced dual complex E*, given by

E* : 0

f--

X*

?

Y*

Z

Z*

f--

0.

Note that, in the case where imS ~ kerT, so that TS = 0, it follows that S*T* = (T S)* = 0* = 0, so that im T* ~ ker S*. Also, if im T* ~ ker S*, then 0= S*T* = (TS)*, so that also TS = 0 and therefore imS ~ kerT. Propositions 3.67 and 3.68 may be stated in terms of particular short exact sequences. Indeed, let E be a Banach space, and let M be a closed subspace. We now write J : M -+ E for the inclusion map and 7r : E -+ E / M for the quotient map. Then we have the 'canonical' short exact sequence: EM: 0 ---->

M

J ---->

E

7r ---->

E/ M

---->

o.

The dual sequence is:

E~I: 0

f--

M*

J:-

E* ~ (E/M)*

f--

O.

From Propositions 3.67 and 3.68, we know that: (i) J* is surjective; (ii) ker J* = M l.. = im 7r*; (iii) 7r* is injective. Thus EM is exact and, even more strongly, 7r* is isometric and J* is a 'quotient map', in the sense that the norm of M* is precisely the quotient norm on E* / M l... Theorem 3.72 Let E be a complex of Banach spaces and continuous linear

maps, as above. Then E is exact if and only if E* is exact. Proof Suppose that E is exact. Then we shall prove that E* is also exact. (i) Exactness at X*. The map S is injective, so that, by Lemma 3.61(ii), im S* is weak-*-dense in X*. But also S has closed range, and so, by Corollary 3.70, (a) =} (d), imS* is weak-*-closed. Hence imS* = X*. (ii) Exactness at Z*. The map T is surjective, so that T* is injective (with weak-*-closed range) by Corollary 3.70, (a) =} (d). (iii) Exactness at Y*. We know that im T* ~ ker S*. As noted in the proof of exactness at Z*, we know that im T* is weak-*-closed. Thus, since we have (im T*) T = ker T = im S, this result follows because

imT* = (imT*)Tl.. = (imS)l.. = kerS*. For the converse, we suppose that E* is exact, and prove that E is exact. (i) Exactness at X. Since kerS = (imS*)T = (X*)T = {O}, E is exact at X. (ii) Exactness at Z. Since T* is injective and has closed range, we have imT = Z, by Corollary 3.70. (iii) Exactness at Y. We know that im S ~ ker T, and also that imS = (im S)l..T = (ker S*) T = (im T*) T = ker T . But im S* = X*, so that, by Corollary 3.71, im S is closed. Thus im S and the proof is complete.

=

ker T, 0

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Banach spaces

Notes The theorems given above are, with the Hahn-Banach theorem, the great pillars of functional analysis. They are expounded, sometimes in more general form, in all the books on functional analysis; see, for example, [63, Chapter II] and [144, Chapter 2]. We proved the Tietze extension theorem in Theorem 3.39. One could ask if the extension go of fa can be 'chosen linearly'. This is answered in certain cases by a theorem of Borsuk. Let K be an infinite, compact, metric space, and let F be a closed subset of K. Then there is a linear map T : C(F) - t C(K) with (Tf) I F = f (j E C(F» and such that 111'11 = 1 and T(XF) = XK. This is related to Milutin's theorem, which states that C(K) is linearly homeomorphic to C(ll) for each uncountable, compact metric space K. See [2, Section 4.4]) for these results. There are many generalizations of the open mapping and closed graph theorems. First, there are versions for more general topological vector spaces than Fn§chet spaces; see [144]. Secondly, one can replace the vector spaces by various so-called topological groups. Thirdly, it is sometimes not necessary for, say, the graph of a linear map to be closed, but only that it be an analytic topological space; an analytic space is defined to be the continuous image of complete and separable metric space. For this latter generalization, see [47, Appendix A]. The separating space and the stability lemma are discussed in some detail in [47, Section 5.2]; some history is given there. The results have many generalizations. Let E and F be Banach spaces, and let l' E B(E, F). Our dual operator 1'* is often called the adjoint of T; see [144, Chapter 4]. It is denoted by T' in some texts, including [47]. The notions of a complex and of an exact sequence arise in many different contexts in mathematics, and are often formulated as holding in a particular 'category'. (This terminology has no connection with that of 'Baire category', in Section 1.7.) Exact sequences arise in algebraic topology, in group theory, and in pure algebra, for example, as well as in Banach space theory. Classic texts on homological algebra that use this language are [119, 120, 164]; for applications within functional analysis, see [92, 93], for example. These notions are developed substantially within the cohomology theory of Banach algebras [47, Section 2.8]. Complexes and exact sequences also arise in the theory of analytic function of several complex variables; we shall make a brief mention of Dolbeault cohomology groups in Section 8.5. lI!#

Exercise 3.13 Let E be a Banach space, and let M be a closed subspace of E. Give a direct proof that (ElM; II·IIEIM) is a Banach space (cf. Theorem 3.48). Indeed, let (Xn + M)n>l be a Cauchy sequence in ElM. Define inductively a subsequence (xnkh2:l of (X n )n2:l such that IIXnk - xnk+l + < 2- k for each k E 1':1. Then inductively choose a sequence (Ykh2: l in E such that, for each k E 1':1, we have Yk E x nk + M and IIYk - Yk+lll < Tk+l . Using Exercise 2.4, show that (Ykh>l has a limit, say Y E E, and then, using Exercise 1.4, that (Xn + M)n2:1 converges to -Y + M.

Mil

Exercise 3.14 Let (Xn)n2:l be a sequence in s, and take p > 1 with conjugate index q. Suppose that the series L~=l XnYn converges for each sequence (Yn)n2:l in £q. Deduce from the uniform boundedness theorem that (Xn)n2:l belongs to £P. Exercise 3.15 Let E be the Banach space co. For n 2': 1, define

Pn

:

(x)

f-4

(Xl, ... ,Xn,O,O, ... ),

E

-t

E,

so that Pn is 'projection onto the first n coordinates'. Note that (Pn )n2:1 is a sequence in B(E) such that Pnx - t X = lEX (X E E), but that IIIE - Pnll = 1 (n E 1':1). This shows

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Introduction to Banach Spaces and Algebras

that we cannot conclude that theorem, Theorem 3.36.

limn~=

'1' in B(E, F) in the Banach-Steinhaus

'l'n

Exercise 3.16 (i) Let E be the normed space (coo; 11·11=). For x = (Xnk:::1 E E, define Tx = (Xnln)n~l' Check that '1' : E --> E is a continuous linear isomorphism, but that '1' is not a linear homeomorphism. Why is this not a contradiction of Banach's isomorphism theorem, Corollary 3.41?

(ii) Let E

= (C l(IT); I· In) and F = (C(IT); I· In)' Define D:ff--tj',

E-->F.

Show that D has a closed graph, but that D is not continuous. Why is this not a contradiction of the closed graph theorem, Theorem 3.45? (iii) Let (E; 11·11) be an infinite-dimensional Banach space, and take (e", : ex E A) to be basis for E, as in Example 1.5(i). We may suppose that Ileall = 1 (ex E A). For x = L: Aaea E E (the sum is finite and uniquely defines x), set Illxlll = L: IAal. Then (E; 111·111) is a normed space. Show that the identity map t : (E; 11·11) --> (E; 111·111) has a closed graph. Clearly, the projection maps 7r(3 : L:A",ea f--t A(3 are continuous on (E; 111·111) for each (3 E A. Assume towards a contradiction that infinitely many of these maps are continuous on (E; 11·11), say 7r(3n is continuous for each n E N. Set En = ker7r(3" (n EN). Then each En is closed in (E; 11·11). Also, for each x E E, we have 7r(3" (x) = 0 for all but finitely many n E N, and so U::-'=l En = E. By the Baire category theorem, Corollary 1.22, there exists no E N such that int Eno =1= 0. But now Eno = E because Eno is a subspace of E. This is a contradiction. Deduce that t: (E; 11·11) --> (E; 111·111) is not continuous. Why is this not a contradiction of Corollary 3.52? Exercise 3.17 Let (Dn)n~l be the Dirichlet kernel of Section 2.17. Recall from page 92 that IF(Dn)lz = 1 (n E Z) and from Exercise 2.35 that IIDnill ~ logn as n --> 00. Deduce from Banach's isomorphism theorem that A(Z) <;; 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) finite-dimensional linear subspace G of F such that F = 'I'(E) EVG. Define S: (x + M, y) f--t '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 infinite-dimensional Banach space. (i) Use the fact that every vector space has a basis to show that there are discontinuous linear functionals on E. (ii) Let f be a discontinuous linear functional on E. Fix a E IC with a #- 1 and Xo E E with f(xo) = 1, and then set

Illxlll", = Ilx -

af(x)xoll

(x E E).

Show that 111·111", is a complete norm on E. (Hint: By Proposition 3.8(i), there is a closed subspace G of E such that E = ICxo EEl G; the projections onto ICxo and G are continuous.) Now take a,(3 E IC\ {I} with a #- (3. Show that 111·111", and 111·111,6 are not equivalent. Exercise 3.24 In the text, we deduced Banach's isomorphism theorem for Banach spaces, Corollary 3.41, from the open mapping theorem, Theorem 3.40, and we deduced the closed graph theorem, Theorem 3.45, from Banach's isomorphism theorem. In fact these three theorems are essentially equivalent. Indeed, first deduce Banach's isomorphism theorem from the closed graph theorem by considering the graph of the inverse mapping. Then deduce the open mapping theorem by using Lemma 3.47. Deduce an isomorphism theorem and a closed graph theorem for Frechet spaces from the open mapping theorem, Theorem 3.58. Exercise 3.25 Let E and F be Banach spaces, and take l' E B(E, F). (i) Prove that the map 1'** E B(E**, F**) is surjective if and only if Tis surjective, and that 1'** is injective with closed range if and only if l' is injective with closed range.

(ii) Take F = Co and E = F* = £ 1, so that E* = F** = £00. As in Proposition 3.22, there is a projection P : E** --+ E such that E** = E EEl G, where G = ker P <;;; 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 vector-space 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 two-sided 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 non-empty 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 'two-sided 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

157

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 algebra-norm 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 non-zero algebra and 11·11 is a given algebra-norm 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 multiplicatively-convex 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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Introduction to Banach Spaces and Algebras

It is easy to show (see Lemma 4.8, below) that any normed algebra with an

identity may be given an equivalent algebra-norm 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, complex-valued functions on a non-empty, 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 non-empty, locally compact Hausdorff space, and let A = (Co(K); I·I K ) be the Banach space of all complex-valued, 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 non-empty, compact Hausdorff space K is a subalgebra A of C(K) that is closed in the norm-topology, 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 non-empty 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 'infinite-dimensional generalization' of the previous example.

Example 4.6 For each complex Banach space E, the algebra B(E) of all bounded linear operators on E, equipped with the operator norm and composition as

Introduction to Banach Spaces and Algebras

160

product, is another important example of a unital Banach algebra. This algebra is a subalgebra of £(E). It is, of course, a non-commutative algebra (provided that dim E > 1). A central role will be played by certain closed subalgebras of B(H), where H is a Hilbert space. Let Xo E E and fo E E*. Then, as in Example 2.17, the rank-one operator

Xo ® fo : x

f---+

fo(x)xo,

E

--t

E,

belongs to B(E). The space of continuous finite-rank operators on E is F(E). For each T E B(E), Xo E E, and fo E E*, we have

To (xo®fo) =Txo®fo

and

(xo®fo)

0

T=xo®T*(fo),

(*)

and so we see that F(E) is an ideal in B(E). We claim that F(E) is the minimum (with respect to inclusion) non-zero ideal in B(E). Indeed, let J be a non-zero ideal in B(E), and take S E J with S =I- O. Then there exist non-zero elements Xo and Yo in E with Sxo = Yo. Choose fo E E* with fo(Yo) = 1, take T = Xl ® h, where Xl E E and h E E*, and set RI = Xl ® fo and R2 = Xo ® h· Then R I , R2 E B(E), and

R ISR2x = h(x)RIYO = h(X)XI = Tx

(x E E),

so that T = R I SR 2 E J. It follows that F(E) <;;;; 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 infinite-dimensional. The closure A(E) of F(E) in (B(E); 11·11) is the collection of approximable operators; this collection is the minimum non-zero, 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 non-zero, 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

161

Banach algebras

Example 4.7 Let U be a non-empty, open subset of C, and let O(U) be the space of all complex-valued, 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 non-unital 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 Banach-algebra completion of A. (By Theorem 2.22, it is the unique 0 completion of A as a Banach space.)

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Introduction to Banach Spaces and Algebras

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

a-l=I+~)I-a)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 a-I = 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 < lilia-III, then b E G(A). Proof Let a E G(A). Then

b=a-(a-b)=a(1-a- 1 (a-b))

(bEA).

Now suppose that lib - all < lilia-III. Then Ila-l(a - b)11 ::::; Ila-Illlia - bll < 1, and so it follows from Lemma 4.10 that 1 - a-I(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

164

Introduction to Banach Spaces and Algebras

Corollary 4.12 Let A be a unital Banach algebra. Then the mapping

a

f---+

a-I,

G(A) ----+ G(A) ,

is a homeomorphism of G(A) onto itself. Proof Set 8(a) = a-I (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,

lib-III :::; Ilb- 1 - a-III

+ lIa-111 =

Ilb-1(a - b)a-111

+ Ila-111

1

:::; 211b-lll

+ Ila-111,

whence lib-III:::; 211a- 111 and Ilb- 1 - a-III:::; Ilb-11l11a - blllla-111 :::; 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 ----+ a-I as n ----+ 00, so that certainly sUPnEN Ila;;-lll < 00. Conversely, suppose that (a;;-l)n>l is bounded, say Ila;;-lll :::; K (n EN). Choose N E N such that Klla - aN11 < 1. Then

a=aN+(a-aN)=aN(l+aj/(a-aN)) .

Since Ilai\/(a - aN)11 :::; Klla - aN11 < 1, a is invertible by Lemma 4.10. [Remark: if a ~ G(A), then it follows that no subsequence of the above sequence (1Ia;;-lllk:::l can be bounded, so that even Ila;;-lll ----+ 00.] (ii) Let a E 8G(A). Then, since G(A) is an open subset of A, it follows that a ~ G(A), but that there is a sequence (an)n>l in G(A) such that an ----+ a. By (i), we may suppose that Ila;;-lll----+ 00 as n ----+ 00. Define bn = a;;-l/lla;;-lll (n EN). Then IIbnll = 1 (n E N) and IIbnall = Ila;;-l(a - an)

+ llllla;;-lll-l

so that bna ----+ O. Similarly, ab n ----+ O.

::::: Iia - anll

+ Ila;;-lll-l

----+ 0

as

n ----+

00,

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 non-empty (since lA E G(A)). Then also ] n G(A) = 0, so that] i- A. Let M be a maximal (left) ideal in A. Then M is also a (left) ideal in A, and clearly M ~ M i- A. By the maximality of M, we have M = M, i.e. Mis closed. 0 The fact that maximal ideals in Banach algebras with identities are closed does not extend to Frechet algebras; for this, see Example 4.53, below. Recall that, if E is a Banach space, then an operator T E B(E) is said to be invertible if and only if there is some S E B(E) such that ST = TS = Ie, where IE is the identity operator on E (see Section 2.5). It is clear that this is precisely the same as T being an invertible element of the Banach algebra B(E). The following theorem summarizes some special cases of Lemma 4.10 and of Corollaries 4.11 and 4.12. We shall write G(E) for the group of invertible, bounded linear operators on E.

Theorem 4.15 Let E be a Banach space. (i) Let T E B(E) with IIIe - Til < 1. Then T E G(E). (ii) The set G(E) is an open subset of B(E). (iii) The map T f-+ T- 1 is a homeomorphism of G(E) onto itself.

0

4.5 The spectrum. We note that, in this section, it is important that all algebras are over the complex field, C. Let A be any complex algebra with an identity, and let a E A. Then the spectrum of a, denoted by SPA (a) (or usually just Sp a) is defined as: SPA (a) =

{A

E

C: Al - art G(A)}.

We remark that, in the trivial case where A = {O}, the zero algebra, then I = 0 and the unique element of A is invertible, so that Sp A(O) = 0. In the case where A is a complex algebra without an identity and a E A, we define Sp A a = Sp A+ a . We have the obvious remark that, in this case, 0 E Sp a.

166

Introduction to Banach Spaces and Algebras

Let A be an algebra. Then an idempotent in A is an element p such that p2 = p. Of course 0 is an idempotent, and the identity of a unital algebra is an idempotent; these are the trivial idempotents. But there may be other, non-trivial, idempotents. Suppose that A has an identity 1 and that p is an idempotent in A. Then 1 - p is also an idempotent. Here is a simple remark. Proposition 4.16 Let A be an algebra.

(i) Let p be an idempotent in A. Then Spp <;;;; {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- >')l-p. Then (>'l-p)b = b(>'l-p) = (>. - >.2)1. Thus, if >. -=I- 0,1, necessarily>. Sp p.

rt

(ii) It is sufficient to note that 1 + ba is invertible if and only if 1 + ab is invertible. For this, suppose that c(l + ab) = (1 + ab)c = 1. Then (1 - bca)(l + ba) = 1 + ba - bca - bcaba = 1 + ba - bca - b(l - c)a = 1,

o

and also (1 + ba)(l - bca) = 1.

An element a in an algebra A is nilpotent if there exists n E N such that an

= o. For example, the matrix T =

(~ ~)

is such that T2 = 0, and so T is nilpotent in the algebra M 2. It is clear that Sp a <;;;; {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 an-I) 1+-+···+-- . >. >.n-I

An element a in an algebra A is said to be quasi-nilpotent if either Sp a = 0 or Sp a = {O}. We shall write Q(A) for the set of quasi-nilpotent elements of A; in general, Q(A) is not a vector subspace of A. Every nilpotent element is quasi-nilpotent; we shall soon see some quasi-nilpotent, non-nilpotent elements of Banach algebras. Let A and B be two complex algebras with identities, and let 8 : A ----> B be a unital homomorphism. Then clearly 8(G(A)) <;;;; 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 non-zero.

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 Iiall, we have Al - a = A(1 - A-Ia). But IIA-Iall = IAI-Illall < 1, and so, by Lemma 4.10, Al - a E G(A), i.e. A tJ- Spa. Thus Spa C;;; {A E A, of the closed subset A \ G(A) of A. So Spa is a bounded and closed-i.e. compact-subset of rc. It remains to show that Sp a =I- 0. For each A E
R(A) - R(J-L) = -(A - J-L)R(A) R(J-L)

(A, J-L E
(*)

so that R is an analytic A-valued 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 - A-Ia)-lll

---->

0

as

IAI

----> 00

since 1 - A-la ----> 1 as IAI ----> 00 and inversion is continuous. Assume towards a contradiction that Spa = 0. Then R is an A-valued entire function with R(A) ----> 0 as IAI ----> 00. By Liouville's theorem for vector-valued functions, Theorem 3.12, we see that R(A) = 0 (A E q, and in particular a-I = 0, which is a contradiction. Hence Spa =I- 0. D Thus, for a quasi-nilpotent 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'fand-Mazur theorem) Let A be a normed division algebra over C Then A ~ C Proof Let a E A. By Corollary 4.18, Spa =I- 0, and so there exists an element, say J-L, in Spa. We have J-Ll - a tJ- G(A). But then, since A is a division algebra, J-Ll - a = o. The injective mapping A f---+ Al is thus an isomorphism of C onto A (and this map is clearly an isometry if A is unital). D

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Introduction to Banach Spaces and Algebras

Example 4.20 Let A be an algebra of complex-valued functions on some set X (with the algebraic operations defined pointwise on X). Suppose that A can be given some Banach-algebra norm, I . II. Then every function in A is bounded. Indeed, IJ(x)1 :'S IIJII for all J E A and x E X; this is immediate from Theorem 4.17 since it is clear that J(x) ESp J (x EX). 0 Note that, for a normed algebra A, it may be that the spectrum of an element of A is neither bounded nor closed in C. For example, let A = O(q be the algebra of all entire functions on C, so that A is a normed algebra for the uniform norm I·I~. Then

G(A) = {J E A : Z(J) = 0}

and

Sp J

= J(q

(J E A) ,

as is immediately checked. Now set exp(z) = e Z (z E q, so that exp E A. Then SPA(exp) = exp(q = C \ {O}. Proposition 4.21 Let A be a Banach algebra, let a E A, and let U be an open neighbourhood oj Sp a in C. Then there exists 6 > 0 such that Sp b c U whenever lib - all < 6. Proof We may suppose that A is unital. Assume to the contrary that the result is false. Then there is a sequence (bn)n~l in A with limn->oo bn = a and such that there exists An E Sp bn \ U for each n E N. Since (b n )n>l and hence (An)n>l - is bounded, so is (A n )n>l, - has a convergent subsequence. Thus we may suppose that An -> IL, say. Clearly, IL 1- U because U is open. But then AnI - bn -> ILl - a E G(A) as n -> 00. Since G(A) is open, AnI - bn E G(A) for sufficiently large n E N, contrary to the fact that An E Spb n for each n E N. 0 Hence the result holds. Lemma 4.22 (Spectral mapping property for polynomials) Let A be a complex algebra wzth an identity, let a E A, and let p be a complex polynomial. Then

Spp(a)

=

{p(A) : A E Spa}.

Proof We may suppose that degp = n 2': 1 because the result is trivial for a constant polynomial. For every A E C, simple algebra shows that p(A)I- p(a) = (AI- a)b for some bE A such that ab = ba. It is immediate that p(A) E Spp(a) whenever A E Spa. Conversely, let IL E C, and suppose that AI, ... , An are the roots (not necessarily distinct) of p( A) - IL in C. Thus

p(A) - IL

=

C(A - Ad ... (A - An)

(A

E

q,

where C is a non-zero constant. It follows that

p( a) - ILl = C( a - All) ... (a - An 1) EA. But then, in the case where ~ E Sp p( a), it follows that Ak E Sp a for at least one k = 1, ... , n, and hence ~ = p(Ak) E p(Sp a). 0

169

Banach algebras

Let A be a Banach algebra, and let a PA(a) (or just pea)) of a to be

A. We define the spectral radius

E

PA(a) = sup{I.\1 :.\

E

Spa}.

From Theorem 4.17, it is immediate that pea) ::; Iiali. For example, pea) = 0 if and only if a is a quasi-nilpotent element of A. We now give the famous Beurling-Gel'fand spectral radius formula for an element in a Banach algebra.

Theorem 4.23 (Spectral radius formula) Let A be a Banach algebra, and let a E A. Then PA(a) = lim Ilanll l / n = inf Ilanll l / n . n--->oo nEN

[Remark. It is elementary to prove the existence of the limit limn--->oo Ila n III/n and to show that it is equal to infnEN Ila n III/n just by using the obvious fact that Ilaffi+nli ::; Ilaffilillanil (m,n EN). However, the existence of this limit emerges anyway in the proof to follow.] Proof We mula would Let .\ E that pea) ::; Now let

may suppose that A is unital (it being clear that the proposed foryield the same value for any two equivalent norms on A). Sp a. Then, as an easy application of Lemma 4.22, .\n E Sp an, so Ilanll l / n (n EN). Thus pea) ::; infnEN Ilanll l / n .

g(z) = (1 - za)-l

(Izl

< 1Ip(a))

(where we define 9 on all of p(a), so that liz ~ Spa, and hence 1 - za is invertible in A. Thus the element g(z) is always well defined. By Corollary 4.12, 9 is continuous, and so, for each R > p(a), we have

MI/R(g)

:=

sup{llg(z)11 : Izl = 11R} <

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 pea), we have

IA(an)1 =

.1

1 1-2 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 algebra-norms on A that are equivalent to II . II. Let a E A. Then

PA(a) = inf{lllalll :

111·111 E

E}.

Proof Certainly, by Theorem 4.17, PA(a) ::::: inf{lllalll : 111·111 E E}. Now take M > PA(a). We have PA(M-1a) < 1, and so it follows from Theorem 4.23 that S := {M-na n : n E N} is bounded, and this set is certainly closed under multiplication. By Lemma 4.8, there is some III . III E E such that Illslll ::::: 1 for every s E S. In particular, Illalll ::::: M. The result follows. 0 Corollary 4.26 Let A be a Banach algebra, and let a, b E A with ab

p(ab) ::::: p(a)p(b) and p(a

+ b)

::::: p(a)

= ba. Then

+ p(b).

Proof Fix E > 0, and set u = aj(p(a) +E) and v = aj(p(b) +E). Since uv = vu, the set S = {uJ,vk,uJv k : j,k E N} is a bounded semigroup in (A,·). By Lemma 4.8, there is an algebra-norm III . Ilion A that is equivalent to II . II such that Illalll < p(a) + E and Illblll < p(b) + E. But now

p(a + b) ::::: Ilia + bill::::: Illalll + Illblll < p(a) + p(b) + 2E. This holds true for each E > 0, and so p(a + b) ::::: p(a) + p(b). Similarly, we have p(ab) ::::: p(a)p(b).

o

171

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 hyper-invariant 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 non-zero, closed subspace, and E(A) is a hyper-invariant subspace for T; each proper, closed subspace of E(A) is an invariant subspace for T. In particular, suppose that E is a finite-dimensional (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 upper-triangular. 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 finite-dimensional (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 non-empty, proper subset of Sp T. For example, an operator T E B(E) is quasi-nilpotent 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

172

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 Pi-IE' 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 non-zero 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(T-AI)xnll:::::IA-Anl+-----+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 self-adjoint. 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 self-adjoint 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 self-adjoint 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 quasi-nilpotent 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 self-adjoint or unitary operator. Let T be a bounded operator on a non-zero 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 self-adjoint. 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 = {Z-1 : z E SpU}, so in fact SpU ~ {z: Izl = I}. Now assume towards a contradiction that Sp U = 0, and consider the operator

T = i(U + IH)(U - I H )-1 (noting that U - IH is invertible since 1

rf- Sp U by hypothesis). Then

T* = -i(U* + IH)(U* - I H )-1 = -i(U* = -i(IH + U)(IH - U)-1 = T, so that T is self-adjoint, and clearly U elementary from (i) that Sp U = giving the result.

{

+

X - .i : X -1

x

(T

E

+ IH)U((U* + iIH )(T

- I H )U)-1

- iIH )-1. Then it is

Sp T } =I- 0 ,

o

175

Banach algebras

4.8 The spectrum of a compact operator. Let E be a non-zero Banach space, and consider an operator T E K(E), so that T is a compact operator. Compact operators were introduced in Example 2.17. Our aim is to describe Sp T. As in Section 4.6, E(A)is the eigenspace corresponding to an eigenvalue A and Spp(T) is the point spectrum of T. Lemma 4.32 Let E be a Banach space, let T E K(E). Take A E SpT with Ai- o. Then:

(i) E(A) is finite-dimensional; (ii) im(T - AI) is closed in E; (iii) either there exists x E E with x f E E* with f i- 0 such that T* f = Af·

i- 0

such that Tx

= AX, or there exists

Proof (i) As above, E(A) is an invariant subspace for T. Set S = T I E(A). Then S E K(E(A)). For X E E(A), we have S(X/A) = x, and so S is a surjection. By Proposition 3.64, E(A) is finite-dimensional.

(ii) By Proposition 3.8(i), E(A) is complemented in E, and so there is a closed subspace G of E such that E = E(A) EEl G. Define

S :x

f----+

AX - Tx ,

G

---*

E,

so that S E B(G,E), S is injective, and imS = im(T - AI). We shall show that S is bounded below. Assume to the contrary that S is not bounded below. Then there exists a sequence (Xn)n>l in Se with SX n ---* O. Since T is compact, we may suppose that (TXn)n>1 is convergent in E, say TX n ---* Xo as n ---* 00. It follows that AX n ---* Xo, so that Xo E G and, further, SXo = limn->(X) ASX n = O. Since S is injective, Xo = O. But Ilxnll = 1 (n EN), and so Ilxoll = IAI > 0, a contradiction. By Proposition 2.23(i), im(T - AI) is closed in E. (iii) Assume that T - AI is an injection and that im(T - AI) is dense in E. By (ii), T - AI is a surjection, and hence a bijection. By Banach's isomorphism theorem, Corollary 3.41, T -AI E G(E), a contradiction ofthe fact that A E Sp T. Suppose that T - AI is not an injection. Then there exists X E E with x i- 0 such that Tx = AX. Suppose that im(T - AI) is not dense in E. Then there exists f E E* with f i- 0 such that f 0 (T - AI)(X) = 0 for all X E E, and this implies that T* f = Af. 0 Lemma 4.33 Let E be a Banach space, and let T E K(E). Suppose that (An)n~l is a sequence of distinct complex numbers and that (Xn)n~l is a sequence of nonzero vectors in E such that TX n = AnXn (n EN). Then An ---* 0 as n ---* 00. Proof We first note that {xn : n E N} is linearly independent. For assume inductively that {Xl, ... ,xn } is linearly independent, and then assume that Xn+l

= aixi + ... + anxn

Introduction to Banach Spaces and Algebras

176 for some al, ... ,an E C. Then

0= (An+1I - T)(Xn+l) = al(A n+1 - AI)XI

+ ... + an(An+1

- An)Xn,

and so al = ... = an = 0 because An+1-AJ -I- 0 for j = 1, ... , n, a contradiction. Thus {Xl, ... , Xn+1} is linearly independent, and the induction continues. Assume to the contrary that An f+ O. Then, by passing to subsequences, we may suppose that IAnl 2: E for some E > O. For n E N, set

Ln = lin{xI"'" x n }, a finite-dimensional, and hence closed, vector subspace of E. Since {xn : n E N} is linearly independent, Ln <;; 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

Ln-d·

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 infinite-dimensional. Each non-zero A E Sp T is an isolated point of Sp T and an ezgenvalue, and the corresponding eigenspace E(A) is finite-dimensional. Proof In the case where E is infinite-dimensional, T ~ G(E), for otherwise E K( E), and so 0 E Sp T. Since Sp T is a non-empty, compact subset of C, the conclusion will follow from the fact that every non-zero point of Sp T is isolated in Sp T. Take A E Sp T with A -I- O. Assume to the contrary that A is not isolated. Then there is a sequence (An)n2':1 of distinct points of Sp T with An ----+ A as n ----+ 00. By Lemma 4.33, only a finite number of the maps AnI - T fail to be injective. By Lemma 4.32(iii), only a finite number of the maps AnI* - T* are injective. However, by Schauder's theorem, Theorem 3.65, T* is compact, and so, by Lemma 4.33 applied to T*, we obtain a contradiction. Thus every non-zero point of Sp T is isolated. We have A E SpT. By Theorem 4.28, A E SPapT, and so there exists (Xn)n2':1 in SE such that (T - AI)X n ----+ 0 as n ----+ 00. Since T is compact, we may suppose that limn--->oo TX n = Y E E. But now limn--->oo AXn = Y with Ilyll = IAI > 0, so that y -I- O. Clearly, limn--->oo ATxn = Ty, and so Ty = )..y. Thus).. E SpT is an eigenvalue. By Lemma 4.32, E()") is finite-dimensional. D

Ie

a

177

Banach algebras

We shall now prove the following striking result of Lomonosov on hyperinvariant subspaces for compact operators. Theorem 4.35 (Lomonosov's theorem) Let E be an infinite-dimensional Banach space, and let T E K(E) with T '" o. Then T has a proper hyper-invariant subspace. Proof We first remark that we can suppose that T is quasi-nilpotent. For otherwise there exists A E Sp T with A '" 0; by Theorem 4.34, A is an eigenvalue, and the eigenspace E(A) is finite-dimensional, and hence a proper hyper-invariant su bspace for T. Let 21 = {T} C = {S E 8(E) : ST = T S}, so that 21 is a unital subalgebra of 8(E). Assume towards a contradiction that there is no proper, closed subspace of E which is invariant for each S E 21. For x E E, set 21x = {Sx : S E 21}. The assumption implies that, for each x '" 0, we have 21x = E, for otherwise 21x '" E would be a proper invariant subspace of E for each S E 21. For y E E, we write B(y) for the open ball B(y; 1). We choose a vector y E E such that the compact, convex set C := T(B(y)) does not contain the zero vector OE, say r := inf{llzll : z E C} > o. For each x E C, choose Ax E 21 with Axx E B(y), and then take an open neighbourhood Ux of x such that Ax(Ux ) ~ B(y). Since C is compact, there are finitely many operators AI, ... ,An E 21 such that n

C ~

U A;I(B(y)). )=1

We have Ty E C, and so there exists jl E {1, ... ,n} with A)ITy E B(y). But then T A)l Ty = A)l T2y E C, and so there exists j2 E {I, ... , n} with A)2A)IT2y E B(y). Continuing in this way, we obtain, for each mEN, vectors Ym

Set k

= A)m ... A)l Tmy

= max{IIA 1 11, ... , IIAnll}. 1

E

B(y) .

Then r ~ IIYml1 ~ k m IITmllllyl1 (m EN), and

= lim rl/m ~ lim k IIT m ll l / m = kp(T) = 0, rn----+CX)

m~CX)

the required contradiction.

D

4.9 Spectrum relative to a subalgebra. about spectra relative to closed subalgebras. For a Banach algebra A and a E A, let

There are a few useful results

RA(a) = C \ SPAa.

Recall that RA(a) is an open subset of the complex plane C and that RA(a) :2 {A E C :

IAI > PA(a)} :2

The set R A (a) is called the resolvent set of a.

{A

E

C : 1>'1 > lIall}·

178

Introduction to Banach Spaces and Algebras

Theorem 4.36 Let A be a unital Banach algebra, let B be a closed subalgebra of A with 1A E B, and let a E B. Then SPAa t;;; SPBa and RB(a) is a relatively open-and-closed subset of RA (a). Proof Clearly, RB(a) = {A E RA(a) : (AlA - a)-l E B} t;;; RA(a), so that Sp A a t;;; Sp Ba. Also, RB (a) is certainly open relative to RA (a) since it is even an open subset of Co By Corollary 4.12, the map A f---> (AlA - a)-l is a continuous mapping from R A (a) into A. Since B is closed in A, it follows that R B (a) is closed relative to RA(a). 0

We now need an elementary topological remark. Let K be a non-empty, compact subset of C. Then C \ K is open, and each of its components is also open. The set of these components is either finite or countably infinite; precisely one of these components is unbounded. Corollary 4.37 Let A be a unital Banach algebra, let B be a closed subalgebra of A with 1A E B, and let a E B.

(i) Let U be a component of RA(a). Then either U is also a component of RB(a) or Un RB(a)

=

0.

(ii) The open sets RA(a) and RB(a) have the same unbounded component with respect to C. (iii) 8SPBa t;;; 8SPAa. Proof (i) This is immediate from Theorem 4.36.

(ii) Let U be the unbounded component of RA(a). Then Un RB(a) ;;2 {A E C :

IAI > Ilall}·

By (i), U must be a component of RB(a), and it is then clearly the unbounded component. (iii) Let A E 8SPBa = SPBa n RB(a). Assume towards a contradiction that RA(a). Then). E RB(a) because RB(a) is closed relative to RA(a). But this contradicts the fact that A E Sp Ba, and therefore)' E SPA a. Since also A E RB(a) t;;; RA(a), we have A E 8SPAa. 0 ). E

Remark: By Corollary 4.37(i),(ii), the compact set SPBa is the union of SPA a with certain bounded components of RA(a). In particular, if RA(a) is connected, then Sp Ba = SPA a. We state a special case of this explicitly. Corollary 4.38 Let A be a unital Banach algebra, let B be a closed subalgebra of A with 1A E B, and let a E B. Suppose that SPA a C JR. Then SPBa = SPAa. 0

Moreover, if B is taken to be as small as possible, then all the 'holes' in SPA a are filled in. This is the content of the next corollary.

179

Banach algebras

Let K be a non-empty, compact subset of C. Then the polynomial hull of K is defined to be the set

K=

{z

E

C : Ip(z)1 :::; IplK for all polynomials p};

here, I· IK is the uniform norm on K. We say that K is polynomially convex if K = K. It is clear that, for any compact K <:;;; 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 f-L tj. SPA(a)' Then (f-L1 - a)-l E A(a). It follows from the definition of A(a) that (f-L1 - a)-l may be approximated by polynomials in a. In particular, there is some complex polynomial, say p, such that

IIp(a) (f-LI - 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 non-empty, 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 non-empty, 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 bi-commutant 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 inverse-closed if 1A E B and if a-I 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 mverse-closed 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 bi-commutant subalgebra if C = CCC. By the last lemma, it is clear that C is bi-commutant if and only if C = Xc for some subset X of A. By Lemma 4.41(i), such a subalgebra is always inverse-closed, 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 non-zero 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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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, infinite-dimensional 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'fand-Mazur 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,

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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 quasi-nilpotent 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 )th-position and 0 elsewhere. Show that the centre 3 (Mn) of Mn consists of the multiples of the identity matrix and that lin{E11 , ... , Enn} is a maximal commutative subalgebra of Mn. Exercise 4.9 Let H be the Hilbert space g 2, and consider the left and right shift operators, Land R, of Exercise 2.11 as elements of B(H). Show that RL = I H , but that LR i= I H. Calculate the spectra, sets of eigenvalues, and approximate point spectra of Land R. Exercise 4.10 Fix p 2: 1, and set E = gPo For n E N, let en be the characteristic function of {n}, as in Section 2.3. Let (Wn)n~l be a sequence in (0,1], and define T E B(E) by requiring that Ten = wnen+l (n EN). Then T is a weighted right shift operator on E. It is clear that IITII S 1 and that n{1,n(E) = {O} : n EN}. Show that Sp T contains 0 and that T has no eigenvalues. For ( E 'f, define U( E B(E) by requiring that U(e n = (ne n (n EN). Show that U( E G(E) and that (TU( = U(T, so that (T is similar to T. Deduce that SpT and SPapT are invariant with respect to rotation about the origin, and that Sp T is a disc (possibly degenerate) centred at the origin. Express the spectral radius of T in terms of the numbers W n . Operators of the above type can be used as examples of operators which do or do not belong to many different classes; see Exercise 5.10 and [114, Section 1.6].

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Banach algebras

Commutative Banach algebras 4.11 Characters and ideals. Throughout this section, A will usually be a non-zero, commutative Banach algebra, but initially A will denote a more general algebra. Let A be a unital algebra. We shall write MA for the set of all maximal ideals of A. In the case where A is commutative, the quotient of A by a maximal ideal is a field. Let A be a unital algebra, and let B be a simple algebra. Then ker e is a maximal ideal in A whenever A ----> B is an epimorphism; in particular, ker 'P is a maximal ideal in A for each character 'P on A.

e:

Theorem 4.46 Let A be a commutative, unital Banach algebra. Then cI> A and the mapping 'P f---4 ker'P is a bijection from cI> A onto MA.

-I- 0

Proof Certainly ker 'P is a maximal ideal in A for each 'P E cI> A. Suppose that 'P,1/) E cI> A with ker'P = ker 1/J. Then, for each a E A, we have a - 'P(a)1 E ker'P = ker1/J, and so 1/J(a - 'P(a)l) = 0, i.e. 1/J(a) = 'P(a). Thus 'P = 1/J, and so the specified map is an injection. It remains to show that every maximal ideal is the kernel of a character on A. But, for each M E MA, the quotient AIM is both a field and a Banach algebra in its quotient norm. By the Gel'fand-Mazur theorem, Theorem 4.19, AIM ~ C, so that the quotient homomorphism Q : A ----> AIM 'is' a character, with ker Q = M. Thus the specified map is a surjection. Since MA -I- 0, we have cI>A -I- 0. D Let A be any Banach space, taken with the zero product, so that, in this product, ab = 0 (a, bE A). Then A is a commutative Banach algebra without an identity. Clearly, cI> A = 0; every vector subspace of co dimension 1 is a maximal ideal in A. Corollary 4.47 Let A be a commutative, unital Banach algebra, and let a

(i) a E G(A) if and only if'P(a) (ii) Spa (iii) p(a) (iii) a

E

= =

-I- 0 for

E

A.

each 'P E cI>A.

{'P(a) : 'P E cI>A}.

sup{I'P(a)1 : 'P

E cI>A}.

Q(A) if and only if'P(a)

= 0 for each 'P

E cI>A.

Proof (i) By elementary algebra, a E G(A) if and only if Aa = A (since A is commutative). But Aa is an ideal in A, and so it is proper if and only if it is included in some maximal ideal. Thus a E G(A) if and only if a does not belong to any maximal ideal of A. Hence, by the theorem, a E G(A) if and only if 'P(a) -I- 0 for each 'P E cI>A. D (ii), (iii), and (iv) These are now very simple deductions. We now give another proof of a more general version of Corollary 4.26.

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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 non-empty, compact Hausdorff space, so that, as before, A is a commutative, unital Banach algebra. For every x E K there is the character cx, defined by

cxU) = J(x) U E C(K)) , i.e.

Cx

is 'evaluation at x', with corresponding maximal ideal Mx = {J E C(K) : J(x) = O}.

We shall now see that these are the only characters (equivalently, maximal ideals) of A. In fact, let M be a maximal ideal of C(K), and assume towards a contradiction that, for every x E K, there is a function gx E M such that gx(x) i- o. By continuity, gx i- 0 on an open neighbourhood, say U(x), of x. Since K is compact, there are finitely many points, say X1, ... ,X n , such that K = U~l U(x,). But then n

g:= L ,=1

n

Igx,1 2

=

Lgx,gx, EM ,=1

and g(x) > 0 for all x E K, which immediately implies that g E G(C(K)). Thus M is not proper, contrary to hypothesis. Hence there exists x E K such that M ~ NIx. Since Jl1 is a maximal ideal, necessarily M = Mx. 0 Example 4.50 Let A = A(~), the disc algebra. Once again, for each z E ~, there is the corresponding evaluation character cz, defined as before. Again, we shall see that these are all the characters on A, but the proof is quite different. Let cp be a character on A. Then, setting z = cp(Z), where Z is the coordinate functional, we have Izl ::; IZIb. = 1, so that z E ~. Then it is clear that, for each polynomial p, we have cp(p) = p(z). But, as remarked above, it is a consequence of Fejer's theorem that the set of polynomials is dense in A(~) (this shows that A is polynomially generated by Z), and cp is continuous, so that cpU) = J(z) U E A), and hence cp = Cz. 0

Banach algebras

187

Example 4.51 Let K be a non-empty, compact subset of
~

= 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

=

~ (~akf3n-k) 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 I-t am; these are the coefficient functionals. Further, the maps m

00

Pm: LanX n n=O

I-t

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 algebra-norm. 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) ----+ 0 and n I-t e- na for a > 1 are weight functions on Z+. For such a weight function w on Z+, we set

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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 quasi-nilpotent element, and indeed each element 2::::=0 anxn of j! 1(Z+, w) with ao = 0 is quasi-nilpotent. In particular, we obtain commutative Banach algebras A for which Q(A) "I {O}. 0

Example 4.53 Let U be a non-empty, 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 Banach-space 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 I----t cp(a), 1> A ----> C, for a E A, continuous.

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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