integration in banach spaces

8/12/2019 Integration in Banach Spaces

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Integration in Banach spaces

When integrating a continuous function f : [a, b] E, where E is a Banachspace, it usually suffices to use the Riemann integral. We shall be concernedfrequently with E-valued functions defined on some abstract measure space(typically, a probability space), and in this context the notions of continuityand Riemann integral make no sense. For this reason we start this first lecturewith generalising the Lebesgue integral to the E-valued setting.

1.1 Banach spaces

Throughout this lecture, E is a Banach space over the scalar field K, whichmay be either Ror C unless otherwise stated. The norm of an element x Eis denoted byxE, or, if no confusion can arise, byx. We write

BE= {x E: x 1}

for the closed unit ball ofE.TheBanach space dualofEis the vector space E of all continuous linear

mappings fromEto K. This space is a Banach space with respect to the norm

x

E

:= supx1 |x, x

|.

Here, x, x := x(x) denotes the duality pairing of the elements x Eand x E. We shall simply write x instead ofxE if no confusioncan arise. The elements ofE are often called (linear) functionalson E. TheHahn-Banach separation theoremguarantees an ample supply of functionalsonE: for every convex closed set C Eand convex compact setKEsuchthat C K= there exist x E and real numbers a < b such that

Rex, x a < b Rey, x


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2 1 Integration in Banach spaces

for all x C and y K. As is well-known, from this one derives the Hahn-Banach extension theorem: ifFis a closed subspace ofE, then for all y F

there exists an x E such that x|F = y and x = y. This easilyimplies that for all x Ewe have

x= supx1

|x, x|.

A linear subspaceF ofE is called normingfor a subset S ofE if for all

x Swe have x= supxFx1

|x, x|.

A subspace of E which is norming for E is simply called norming. Thefollowing lemma will be used frequently.

Lemma 1.1.IfE0 is a separable subspace ofE andF is a linear subspace ofE which is norming for E0, thenF contains a sequence of unit vectors that

is norming forE0.

Proof. Choose a dense sequence (xn)n=1 inE0 and choose a sequence of unit

vectors (xn)n=1 inF such that|xn, x

n| (1 n)xn for all n 1, where

the numbers 0 < n 1 satisfy limn n = 0. The sequence (xn)n=1 is

norming for E0. To see this, fix an arbitrary x E0 and let > 0. Pickn0 1 such that 0 < n0 andx xn0 . Then,

(1 )x (1 n0)x (1 n0)xn0 + (1 n0)

|xn0 , xn0

| + |x, xn0| + 2.

Since >0 was arbitrary it follows that x supn1 |x, xn|.

A linear subspaceFofE is said toseparate the pointsof a subsetSofEiffor every pairx, y Swithx =y there exists anx Fwith x, x =y, x.Clearly, norming subspaces separate points, but the converse need not be true.

Lemma 1.2.IfE0 is a separable subspace ofE andF is a linear subspace ofE which separates the points ofE0, thenFcontains a sequence that separates

the points ofE0.

Proof. By the Hahn-Banach theorem, for each x E0 \ {0} there exists avector x(x) F such that x, x(x) = 0. Defining

Vx := {y E0\ {0} : y, x(x) = 0}

we obtain an open cover{Vx}xE0\{0} ofE0 \ {0}. Since every open cover of aseparable metric space admits a countable subcover it follows that there existsa sequence (xn)n=1 inE0 \ {0}such that{Vxn}

n=1 coversE0 \ {0}. Then the

sequence{x(xn)}n=1 separates the points ofE0: indeed, every x E0\ {0}belongs to some Vxn , which means thatx, x

(xn) = 0.


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1.2 The Pettis measurability theorem 3

1.2 The Pettis measurability theorem

We begin with a discussion of weak and strong measurability of E-valuedfunctions. The main result in this direction is the Pettis measurability the-orem which states, roughly speaking, that an E-valued function is stronglymeasurable if and only if it is weakly measurable and takes its values in aseparable subspace ofE.

1.2.1 Strong measurability

Throughout this section (A,A) denotes a measurable space, that is, A is aset and A is a -algebra in A, that is, a collection of subsets ofA with thefollowing properties:

1. A A;2. B A implies B A;3. B1 A, B2 A, . . . imply

n=1 Bn A.

The first property guarantees that Ais non-empty, the second expresses thatA is closed under taking complements, and the third that Ais closed undertaking countable unions.

TheBorel-algebraof a topological space T, notation B(T), is the smallest-algebra containing all open subsets of T. The sets in B(T) are the Borel

setsofT.

Definition 1.3.A functionf :A Tis calledA-measurableiff1(B) Afor allB B(T).

The collection of allB B(T) satisfying f1(B) Ais easily seen to bea-algebra. As a consequence, f is A-measurable if and only iff1(U) Afor all open sets U inT.

When T1 and T2 are topological spaces, a function g : T1 T2 is Borelmeasurable if g1(B) B(T1) for all B B(T2), that is, if g is B(T1)-measurable. Note that if f : A T1 is A-measurable and g : T1 T2 isBorel measurable, then the composition g f : A T2 is A-measurable.By the above observation, every continuous function g : T1 T2 is Borelmeasurable.

It is a matter of experience that the notion ofA-measurability does notlead to a satisfactory theory from the point of view of vector-valued analysis.Indeed, the problem is that this definition does not provide the means forapproximation arguments. It is for this reason that we shall introduce nextanother notion of measurability. We shall restrict ourselves to Banach space-valued functions, although some of the results proved below can be generalisedto functions with values in metric spaces.

Let E be a Banach space and (A,A) a measurable space. A function

f : A E is called A-simple if it is of the form f =N

n=11Anxn withAn A and xn E for all 1 n N. Here 1A denotes the indicator

functionof the set A, that is, 1A() = 1 if A and 1A() = 0 ifA.


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Definition 1.4.A function f : A E is strongly A-measurable if thereexists a sequence ofA-simple functionsfn: A Esuch thatlimn fn= fpointwise onA.

In order to be able to characterise strong A-measurability of E-valuedfunctions we introduce some terminology. A function f : A E is calledseparably valued if there exists a separable closed subspace E0 E suchthat f() E0 for all A, and weakly A-measurable if the functionsf, x: A K,f, x() :=f(), x, are A-measurable for all x E.

Theorem 1.5 (Pettis measurability theorem, first version). Let(A,A)be a measurable space and letF be a norming subspace ofE. For a function

f :A E the following assertions are equivalent:

(1)f is stronglyA-measurable;(2)f is separably valued andf, x isA-measurable for all x E;(3)f is separably valued andf, x isA-measurable for all x F.

Proof.(1)(2): Let (fn)n=1 be a sequence ofA-simple functions convergingto fpointwise and let E0 be the closed subspace spanned by the countablymany values taken by these functions. Then E0 is separable and f takes itsvalues in E0. Furthermore, eachf, xis A-measurable, being the pointwise

limit of theA

-measurable functions fn, x

.(2)(3): This implication is trivial.(3)(1): Using Lemma1.1,choose a sequence (xn)

n=1 of unit vectors in

Fthat is norming for a separable closed subspace E0 ofEwhere f takes itsvalues. By the A-measurability of the functions f, xn, for each x E0 thereal-valued function

f() x= supn1

|f() x, xn|

is A-measurable. Let (xn)n=1 be a dense sequence in E0.Define the functions sn : E0 {x1, . . . , xn} as follows. For each y E0

let k(n, y) be the least integer 1 k n with the property that

y xk= min1jn y xj

and put sn(y) := xk(n,y). Notice that

limn

sn(y) y= 0 y E0

since (xn)n=1 is dense in E0. Now define fn: A Eby

fn() :=sn(f()), A.

For all 1 k n we have


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1.2 The Pettis measurability theorem 5

{ A : fn() =xk}

=

A : f() xk= min1jn

f() xj

A : f() xl> min1jn

f() xj for l = 1, . . . , k 1

.

Note that the sets on the right hand side are in A. Hence eachfnis A-simple,and for all A we have

limn

fn() f()= limn

sn(f()) f()= 0.

Corollary 1.6.The pointwise limit of a sequence of stronglyA-measurablefunctions is stronglyA-measurable.

Proof. Each function fn takes its values in a separable subspace ofE. Thenftakes its values in the closed linear span of these spaces, which is separable.The measurability of the functions f, x follows by noting that each f, xis the pointwise limit of the measurable functions fn, x.

Corollary 1.7.If an E-valued function f is stronglyA-measurable and :E Fis continuous, whereFis another Banach space, then fis stronglyA-measurable.

Proof. Choose simple functions fn converging to fpointwise. Then fn

fpointwise and the result follows from the previous corollary.

Proposition 1.8. For a function f : A E, the following assertions areequivalent:

(1)f is stronglyA-measurable;(2)fis separably valued and for all B B(E) we havef1(B) A.

Proof. (1)(2): Letfbe strongly A-measurable. Thenfis separably-valued.To prove thatf1(B) A for allB B(E) it suffices to show that f1(U)A for all open sets U.

LetUbe open and choose a sequence ofA-simple functionsfn convergingpointwise tof. Forr >0 let Ur= {x U : d(x, U)> r}, whereUdenotesthe complement ofU. Then f1n (Ur) A for all n 1, by the definition of

anA

-simple function. Sincef1(U) =

m1

n1

kn

f1k (U 1m

)

(the inclusion being a consequence of the fact that U is open) it followsthat also f1(U) A.

(2)(1): By assumption, f is A-measurable, and therefore f, x is A-measurable for all x E. The result now follows from the Pettis measura-bility theorem.

Thus if E is separable, then an E-valued function f is strongly A-measurable if and only if it is A-measurable.


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1.2.2 Strong -measurability

So far, we have considered measurability properties of E-valued functionsdefined on a measurable space (A,A). Next we consider functions defined ona -finite measure space (A,A, ), that is, is a non-negative measure on ameasurable space (A,A) and there exist sets A(1) A(2) . . . in A with(A(n))< for all n 1 and A =

n=1 A

(n).A -simplefunction with values in Eis a function of the form

f=Nn=1

1Anxn,

where xn Eand the sets An A satisfy (An)< .We say that a property holds -almost everywhereif there exists a -null

set N A such that the property holds on the complement N ofN.

Definition 1.9.A function f : A E is strongly -measurable if thereexists a sequence(fn)n1 of-simple functions converging to f -almost ev-erywhere.

Using the -finiteness of it is easy to see that every strongly A-measurable function is strongly -measurable. Indeed, if f is strongly A-

measurable and limn fn = fpointwise with each fn an A-simple func-tions, then also limn1A(n)fn = f pointwise, where A =

n=1 A

(n) asbefore, and each 1A(n)fn is -simple. The next proposition shows that in theconverse direction, every strongly -measurable function is equal -almosteverywhere to a strongly A-measurable function.

Let us call two functions which agree -almost everywhere -versionsofeach other.

Proposition 1.10.For a function f : A E the following assertions areequivalent:

(1)f is strongly-measurable;(2)f has a-version which is stronglyA-measurable.

Proof. (1)(2): Suppose that fn foutside the null set N A, with eachfn -simple. Then we have limn1Nfn= 1Nfpointwise on A, and sincethe functions 1Nfn are A-simple, 1Nf is strongly A-measurable. It followsthat 1Nfis a strongly A-measurable-version off.

(2)(1): Letfbe a strongly A-measurable-version offand let N Abe a null set such that f =f on N. If ( fn)n=1 is a sequence ofA-simplefunctions converging pointwise tof, then limnfn= fon N, which meansthat limnfn= f -almost everywhere.

WriteA =n=1 A

(n) withA(1) A(2) A and(A(n))< for all

n 1. The functions fn:= 1A(n)fn are -simple and we have limn fn= f-almost everywhere.


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1.3 The Bochner integral 7

We say that f is -separably valuedif there exists a closed separable sub-space E0 of E such that f() E0 for -almost all A, and weakly -measurableiff, xis -measurable for all x E.

Theorem 1.11 (Pettis measurability theorem, second version). Let(A,A, ) be a -finite measure space and letF be a norming subspace ofE.For a functionf :A E the following assertions are equivalent:

(1)f is strongly-measurable;

(2)f is -separably valued andf, x

is-measurable for all x

E

;(3)f is -separably valued andf, x is-measurable for all x F.

Proof. The implication (1)(2) follows the corresponding implication in The-orem1.5 combined with Proposition1.10,and (2)(3) is trivial. The impli-cation (3)(1) is proved in the same way as the corresponding implication

in Theorem1.5,observing that this time the functions fn have -versionsfnthat are A-simple. If we write A=

n=1 A

(n) as before with each A(n) of fi-

nite-measure, the functions 1A(n)fnare -simple and converge tof -almost

everywhere.

By combining Proposition1.10with Corollaries1.6 and 1.7 we obtain:

Corollary 1.12.The -almost everywhere limit of a sequence of strongly-measurableE-valued functions is strongly-measurable.

Corollary 1.13.If an E-valued function f is strongly -measurable and :E Fis continuous, whereFis another Banach space, then fis strongly-measurable.

The following result will be applied frequently.

Corollary 1.14.If f and g are strongly -measurable E-valued functionswhich satisfyf, x=g, x -almost everywhere for everyx F, whereFis subspace ofE separating the points ofE. Thenf=g -almost everywhere.

Proof. Both f and g take values in a separable closed subspace E0 -almost

everywhere, say outside the -null set N. Since E0 is separable, by Lemma1.2some countable family of elements (xn)

n=1 in F separates the points of

E0. Sincef, xn= g, xnoutside a -null set Nn, we conclude that f and g

agree outside the -null set Nn=1 Nn.

1.3 The Bochner integral

The Bochner integral is the natural generalisation of the familiar Lebesgueintegral to the vector-valued setting.

Throughout this section, (A,A, ) is a -finite measure space.


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1.3.1 The Bochner integral

Definition 1.15. A function f : A E is -Bochner integrable if thereexists a sequence of -simple functions fn : A E such that the followingtwo conditions are met:

(1) limn fn= f -almost everywhere;

(2) limn

A

fn f d= 0.

Note thatfis strongly -measurable. The functions fnf are-measurableby Corollary1.13.

It follows trivially from the definitions that every -simple function is -Bochner integrable. For f=

Nn=11Anxn we put

A

f d:=

Nn=1

(An)xn.

It is routine to check that this definition is independent of the representationoff. Iff is-Bochner integrable, the limit

A

f d := limn

A

fn d

exists in Eand is called the Bochner integraloffwith respect to . It is routineto check that this definition is independent of the approximating sequence(fn)n=1.

Iff is-Bochner integrable andg is a-version off, theng is -Bochnerintegrable and the Bochner integrals of f and g agree. In particular, in thedefinition of the Bochner integral the function f need not be everywheredefined; it suffices that f be -almost everywhere defined.

Iff is-Bochner integrable, then for all x E we have the identityA

f d, x

=

A

f, x d.

For

-simple functions this is trivial, and the general case follows by approx-imating fwith -simple functions.

Proposition 1.16.A strongly -measurable function f : A E is -Bochner integrable if and only if

A

f d < ,

and in this case we have A

f d

A

f d.


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Proof. First assume that f is -Bochner integrable. If the -simple functionsfn satisfy the two assumptions of Definition 1.15, then for large enough n weobtain

A

f d

A

f fn d +

A

fn d 0 such that

|Rex, x Rev, x| v V .

The Hahn-Banach separation theorem asserts that ifV is convex and x V,then there exists a functional x E which strictly separates x from V.

For x E, let

m(x) := inf{Ref(), x: A},

M(x) := sup{Ref(), x: A},


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allowing these values to be and , respectively. Then, since (A) = 1,

Re

A

f d, x

=

A

Ref, x d [m(x), M(x)].

This shows thatA

f d cannot be strictly separated from the convex setconv{f() : A}by functionals in E. Therefore the conclusion follows byan application of the Hahn-Banach separation theorem.

As a rule of thumb, results from the theory of Lebesgue integration carryover to the Bochner integral as long as there are no non-negativity assumptionsinvolved. For example, there are no analogues of the Fatou lemma and themonotone convergence theorem, but we do have the following analogue of thedominated convergence theorem:

Proposition 1.18 (Dominated convergence theorem). Let fn : A Ebe a sequence of functions, each of which is -Bochner integrable. Assume

that there exist a function f : A E and a -Bochner integrable functiong: A K such that:

(1) limn fn= f -almost everywhere;(2)fn |g| -almost everywhere.

Then f is-Bochner integrable and we have

limn

A

fn f d= 0.

In particular we have

limn

A

fn d=

A

f d.

Proof. We have fn f 2|g| -almost everywhere, and therefore the resultfollows from the scalar dominated convergence theorem.

It is immediate from the definition of the Bochner integral that iff :A E is -Bochner integrable and T is a bounded linear operator from E intoanother Banach space F, then T f :A F is -Bochner integrable and

T

A

f d=

A

T f d.

This identity has a useful extension to a suitable class of unbounded oper-ators. A linear operator T, defined on a linear subspace D(T) ofEand takingvalues in another Banach space F, is said to be closedif its graph

G(T) := {(x , T x) : x D(T)}

is a closed subspace of E F. If T is closed, then D(T) is a Banach spacewith respect to the graph norm


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xD(T):= x + T x

andTis a bounded operator from D(T) toE.The closed graph theoremasserts that ifT : E F is a closed operator

with domain D(T) = E, then Tis bounded.

Theorem 1.19 (Hille). Let f : A E be -Bochner integrable and let Tbe a closed linear operator with domainD(T) inEtaking values in a BanachspaceF. Assume thatftakes its values inD(T) -almost everywhere and the

-almost everywhere defined function T f : A F is -Bochner integrable.Then

A

f d D(T) and

T

A

f d=

A

T f d.

Proof. We begin with a simple observation which is a consequence of Propo-sition1.16and the fact that the coordinate mappings commute with Bochnerintegrals: ifE1 and E2 are Banach spaces and f1 : A E1 and f2 : A E2are -Bochner integrable, then f = (f1, f2) : A E1 E2 is -Bochnerintegrable and

A

f d=

A

f1 d,

A

f2 d

.

Turning to the proof of the proposition, by the preceding observation thefunction g : A E F, g() : = (f(), T f()), is -Bochner integrable.Moreover, since g takes its values in the graph G(T), we have

A

g() d()G(T). On the other hand,

A

g() d() =

A

f() d(),

A

T f() d()

.

The result follows by combining these facts.

We finish this section with a result on integration of E-valued functionswhich may fail to be Bochner integrable.

Theorem 1.20 (Pettis). Let(A,A, ) be a finite measure space and let1

integration in banach spaces

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