BOCHNER Vs. PETTIS NORM: EXAMPLES and RESULTS 3

Contemp orary Mathematics

Volume 00, 0000

Bo chner vs. Pettis norm: examples and results

S.J. DILWORTH AND MARIA GIRARDI

Contemp. Math. 144 1993 69{80

Banach Spaces, Bor-Luh Lin and Wil liam B. Johnson, editors

Abstract. Our basic example shows that for an arbitrary in nite-dimen-

sional Banach space X, the Bo chner norm and the Pettis norm on L X are

not equivalent. Re nements of this example are then used to investigate

various mo des of sequential convergence in L X.

1. INTRODUCTION

Over the years, the Pettis integral along with the Pettis norm have grabb ed

the interest of many. In this note, we wish to clarify the di erences b etween the

Bo chner and the Pettis norms. We b egin our investigation by using Dvoretzky's

Theorem to construct, for an arbitrary in nite-dimensional Banach space, a se-

quence of Bo chner integrable functions whose Bo chner norms tend to in nity but

whose Pettis norms tend to zero. By re ning this example again working with

an arbitrary in nite-dimensional Banach space, we pro duce a Pettis integrable

function that is not Bo chner integrable and we show that the space of Pettis

integrable functions is not complete. Thus our basic example provides a uni-

ed constructiveway of seeing several known facts. The third section expresses

these results from a vector measure viewp oint. In the last section, with the aid

of these examples, we give a fairly thorough survey of the implications going

between various mo des of convergence for sequences of L X functions.

1991 Mathematics Subject Classi cation. 46E40, 28A20, 46G10.

This pap er is in nal form and no version of it will b e submitted for publication elsewhere.

0000 American Mathematical So ciety

0000-0000/00 $1.00 + $.25 p er page 1

DILWORTH and GIRARDI 2

2. EXAMPLES: L X vs. P X

1 1

Let X b e a Banach space with dual X and let ; ; b e the usual Leb esgue

measure space on [0; 1]. Let L X = L ; ;;X b e the space of X-valued

1 1

Bo chner integrable functions on with the usual Bo chner norm jjjj .

Bo chner

A quick review of the Pettis integral. Consider a weakly measurable func-

tion f : !Xsuch that x f 2 L R for all x 2 X . The Closed Graph

of X satisfying Theorem gives that for each E 2 there is an element x

is actually in X for each E 2 , x = x fd for each x 2 X .Ifx x

E E

then wesay that f is Pettis integrable with the Pettis integral of f over E b eing

the element x 2 X satisfying

x x = x fd

for each x 2 X . In this note we shall restrict our attention to Pettis integrable

functions that are strongly measurable. The space P X of equivalence classes

of all strongly measurable Pettis integrable functions forms a normed linear

space under the norm

k f k = sup j x f j d :

Pettis

x 2B X

Clearly if f 2 L X then f 2P X with

1 1

jjf jj jjf jj :

Pettis Bo chner

If X is nite-dimensional, then the Bo chner norm and the Pettis norm are equiva-

lent and so L X= PX. However, in any in nite-dimensional Banach space

1 1

X, the next example shows that the two norms are not equivalentonL X.

Sp eci cally, it constructs an essential ly bounded sequence of Bo chner-norm one

functions whose Pettis norms tend to 0.

EXAMPLE 1.

Let X b e an in nite-dimensional Banach space. Wenow construct a sequence

ff g of L X functions which tend to zero in the Pettis norm but not in the

n 1

Bo chner norm.

n n

Let fI : n =0;1;::: ;k =1;::: ;2 g b e the dyadic intervals on [0; 1], i.e.

k 1 k

I = ; :

n n

2 2

Fix n 2 N.We de ne f with the help of Dvoretzky's Theorem [D]. Find a 2 -

dimensional subspace E of X such that the Banach-Mazur distance b etween E

n n

2 2

and ` is at most 2. So there is an op erator T : ` ! E such that the norm

n n

2 2

n n

of T is at most 2 and the norm of the inverse of T is 1. Let fe : k =1;::: 2 g

n n k

BOCHNER vs. PETTIS NORM: EXAMPLES AND RESULTS 3

n n 2

b e the image under T of the standard unit vectors fu : k =1;::: 2 g of ` .

De ne f : !Xby

f ! = e 1 ! ;

n I

k=1

The sequence ff g has the desired prop erties.

Since

n n

2 2

X X

n n n

jjf jj = jje jj d = 2 jje jj

n Bo chner X X

k k

k =1 k =1

and 1 jje jj 2,

1 jjf jj 2 :

n Bo chner

As for the Pettis norm, x x 2 B X . Note that the restriction y of x to

E is an elementin E of norm at most 1. Also, the adjoint T of T has norm

n n

at most 2. Thus

n n

2 2

X X

n n

jx e j = jy T u j

k n k

k =1 k =1

n n

2 2

X X

n n

2 ; = jT y u j jjT y jj u jj jj 2

2 2

n n k n n k

` `

2 2

k =1 k =1

and so

n n

Z Z

2 2

X X

n n n

jx f j d = ; jx e j d = 2 jx e j 22

k k

k =1 k =1

giving that

: jjf jj 22

n Pettis

Thus the Pettis norm of f tends to zero.

This di erence b etween the Bo chner and Pettis norms is further emphasized

by the following minor variation of Example 1.

EXAMPLE 2.

~ ~

Fix 0 < < and put 2 . De ne f : !X by f = f where

n n n n n

f is as in Example 1. The Bo chner norm of f tends to in nity at the rate of

n n

but the Pettis norm of f tends to zero.

n n

Avariation on this example gives a strongly measurable Pettis integrable

function that is not Bo chner integrable { working in any in nite-dimensional

Banach space. The usual way of constructing such a function uses the Dvoretsky-

Rogers Theorem. Recall [DR] that in any in nite-dimensional Banach space

X, there is an unconditionally convergent series x that is not absolutely

convergent. The function f : !Xgiven by

f = 1 ;

n=1

DILWORTH and GIRARDI 4

where fE g is a partition of into sets of strictly p ositive measure is Pettis

integrable but not Bo chner integrable. Our example is in the same spirit.

EXAMPLE 3.

Let X b e an in nite-dimensional Banach space. Consider the ab ove construc-

n 1

tion. Note that the collection fI g partitions [0; 1]. Let g : ! X be a

2 n=1

normalized f from Example 1 supp orted on I , i.e.

n n

g ! = 2 f 2 ! 1 1 ! :

n n I

De ne f : !Xby

g ! 1 ! : f! =

n I

n=1

Clearly f is strongly measurable.

But f is not Bo chner integrable since for any N 2 N

Z Z

N N

X X

n n

jjf jj d jjg jj d 2 2 = N:

n X

n=1 n=1

However f is Pettis integrable. First note that x f 2 L R for all x 2 X since

for a xed x 2 B X wehave using computations from the rst example

Z Z Z

jx f j d = jx g j d = jx f j d 22 ;

n n

I I

2 2

and so

Z Z

1 1

X X

jx f j d = jx f j d 2 2 = 2 2+1 :

n=1 n=1

Next x E 2 . We know there is an element x of X satisfying that

xfd for each x 2 X .To see that x is actually in X, consider x x =

E E

the sequence fs g of L X functions on where

N 1

s ! = g ! 1 !:

N n I

n=1

Let x b e the Bo chner integral of s over E . Viewed as a sequence in X ,

fx g converges in norm to x since for any x 2 B X

E E

Z Z Z

jx x x j = j x s d x fdj jx s fjd

N N

E E

1 1

X X

= jx f j d 2 2 :

n=N +1 n=N +1

Thus x is actually in X, as needed.

Janicka and Kalton [JK] showed that if X is in nite-dimensional then P Xis

never complete under the Pettis norm. The rst example provides a constructive

way to see this.

BOCHNER vs. PETTIS NORM: EXAMPLES AND RESULTS 5

EXAMPLE 4.

Taking some care in our construction in Example 1, wemay arrange for fE g

to form a nite dimensional decomp osition FDD of some subspace of X. To

see this, rst cho ose a basic sequence fx g in X. Take a blo cking fF g of the

n n

basis so that each subspace F is of large enough dimension to nd using the

nite-dimensional version of Dvoretzky's Theorem a 2 -dimensional subspace

E of F such that the Banach-Mazur distance b etween E and ` is at most

n n n

2. Then fE g forms a FDD.

Keeping with the notation from Example 1, de ne h : !Xby

h = f :

n k

k=1

Clearly each h is in P X. Also, fh g is Cauchy in the Pettis norm since

n 1 n

1 1

X X

: jjh h jj jjf jj 2 2

n n+j Pettis k Pettis

k =n k =n

To show that h cannot converge to an elementin P X, we need the following

n 1

two lemmas whose pro ofs are given shortly.

lemma 1. If the sequence ff g of functions in P X converges to f 2P X

n 1 1

and each f is essentially valued in some subspace Y of X, then f is also essentially

valued in Y.

lemma 2. Let ff g b e a sequence of P X functions that converges in Pettis

n 1

norm to f 2P X. Furthermore, supp ose that each f is essentially valued in

1 n

some subspace Y of X and that T : Y ! X is a b ounded linear op erator. Then

Tf converges in the Pettis norm to Tf.

Supp ose that h converges in the Pettis norm to h 2P X. Let Y E

n 1 n

and let P : Y ! X b e the natural pro jection from Y onto E . Note that each

n n

h ,thus also h, are essentially valued in Y. Applying the second lemma to the

op erator P and noting that P h f for m n,wehave that P h f .

n n m n n n

Thus for each n 2 N

1 jjP h! jj 2

n X

for almost all ! . This contradicts the fact that for almost all !

lim jjh! P h! jj = 0 :

n X

N !1

n=1

Thus h cannot converge to an elementin P X.

n 1

Now for the pro ofs of the lemmas.

proof of lemma 1. Let ff g b e a sequence of P X functions that con-

n 1

verges in the Pettis norm to f 2P X. Furthermore, supp ose that each f is

1 n

essentially valued in some subspace Y of X and that f is essentially valued in

some subspace Z with Y Z. Since f and each f are strongly measurable, we

may assume that Z is separable.

DILWORTH and GIRARDI 6

Since Z is separable, there is a sequence fz g of functionals from Z such

that an element z from Z is in Y if and only if z z = 0 for each m.

Since ff g converges in the Pettis norm to f , for each z the sequence

fz f g converges to z f in L R -norm. But for each m, note that z f

n n 1 n

m m m

is zero almost everywhere and so z f is also zero almost everywhere. Thus f is

also essentially valued in Y.

proof of lemma 2. Let ff g b e a sequence of P X functions which con-

n 1

verges in Pettis norm to f 2PX. Furthermore, supp ose that each f is

1 n

essentially valued in some subspace Y of X. Lemma 1 gives that f is also essen-

tially valued in Y. Let T : Y ! X b e a b ounded linear op erator. Note that for

a xed x 2 X wemay extend T x to an elementin X without increasing its

norm.

Note that Tf and likewise each Tf is actually in P X. To see this, rst

n 1

observe that for a xed x 2 X viewing T x as an elementin X of the same

norm

Z Z

jx Tfj d = jT x f j d jjT x jj jjf jj :

Pettis

and so x Tf 2 L . Next, let x b e the Pettis integral of f over E 2 . Note

1 E

that x 2 Y and for each x 2 X

Z Z

x Tx = T x x = T x f d = x Tf d ;

E E

and so T x is the Pettis integral of Tf over E 2 . Thus Tf is indeed in

P X.

Since for x 2 B X

Z Z

jx Tf Tfj d = jT x f f j d jjT jj jjf f jj ;

n n n Pettis

the sequence Tf converges in the Pettis norm to Tf.

3. VECTOR MEASURES

Let's lo ok at our examples from a vector measure viewp oint.

APettis integrable function f gives rise to a vector measure G : ! X

where G E is the Pettis integral of f over E . The Pettis norm of f is the

semivariation of G .Iff2LX, then the Bo chner norm of f is the variation

f 1

of G .

Example 2 shows that for any in nite-dimensional Banach space X, there is

a sequence of X-valued measures whose variations tend to in nity but whose

semivariations tend to zero. On the other hand, Example 3 gives a X-valued

measure of b ounded semivariation but in nite variation.

For f 2PX, the corresp onding measure G is -continuous and has a rel-

atively compact range. In fact [DU], the completion of P X is isometrically 1

BOCHNER vs. PETTIS NORM: EXAMPLES AND RESULTS 7

the space K Xofall -continuous measures G :!Xwhose range is rela-

tively compact, equipp ed with the semivariation norm. Thus Example 4 gives a

measure in K X which is not representable by a strongly measurable Pettis

integrable function.

4. CONVERGENCE

Wenow restrict our attention to sequences of L X functions which converge

in the Bo chner norm, in the Pettis norm, or weakly in L X to functions

actually in L X. Wehave the following obvious relations:

1 Bo chner-norm convergence implies weak and Pettis-norm convergence;

2 weak convergence implies neither Bo chner-norm nor Pettis-norm conver-

gence e.g. consider the Rademacher functions in L R;

3 if X is nite dimensional then the Pettis norm and the Bo chner norm are

equivalent.

Our examples help to clarify the other implications.

If X is in nite-dimensional, then there is a sequence of functions which con-

verges in Pettis norm but whose Bo chner norms tend to in nity Example 2.

Thus this sequence do es not converge in the Bo chner norm or weakly.

What if we require the sequence to b e b ounded sup jjf jj < 1, or uni-

L X

formly integrable, or even essentially b ounded sup jjf jj < 1? Again, if

L X

X is in nite-dimensional then Example 1 gives a sequence of essentially b ounded

functions which converges in Pettis norm but not in the Bo chner norm.

In certain situations we can pass from Pettis-norm convergence to weak con-

R R

g f d g f d ! vergence. If f ! f in the Pettis norm, then clearly

n n

for each simple function g 2 L X . As a rst step towards weak convergence,

when can we at least conclude that f ! f in the L X;L X -top ology?

n 1 1

Aword of caution, the simple functions need not b e dense in L X . For ex-

ample, the L ` function g = e 1 , where fe g are the standard

1 2 n I n

n=1

from any simple function in L ` . Our next unit vectors of ` , is at least

1 2 2

example illustrates the fact that for any in nite-dimensional Banach space there

is a b ounded sequence in L X that converges in the Pettis norm but not in the

L X;L X -top ology.

1 1

EXAMPLE 5.

Consider the sequence fg g from Example 3 where

n n

g ! = 2 f 2 ! 1 1 ! :

n n I

Since jjf jj 2, the Bo chner norm of each g is at most 2.

n n

L X

To examine the Pettis norm, x x 2 B X . Computations in Example 1

show that

Z Z Z

jx g j d = jx g j d = jx f j d 22

n n n

I 2

DILWORTH and GIRARDI 8

and so

jjg jj 22 :

n Pettis

Thus g ! 0 in the Pettis norm.

n n n

As for the L X;L X -top ology, nd y 2 B X such that y e =

1 1

k k k

n n n 2n

: j =2 + jje jj . Note that fI : n =1;2;::: g partitions [0; 1] and that fI

n+1 n

1;::: ;2 g partitions the interval I . De ne g : !X by

2 1

g! = y 1 ! :

2 +k

n=1

k =1

Clearly g 2 L X . But

Z Z Z

g g d = g g d = g g d

n n n

n 2n

I I

2 +k

k =1

n n

2 2

X X

n n n n 2n

= y 2 e d 2 2 1 = 1 :

k k

2 +k

k =1 k =1

Thus fg g do es not converge in the L X;L X -top ology.

n 1 1

However, a uniformly integrable Pettis-norm convergent sequence do es con-

verge in the L X;L X -top ology. To see why, consider a uniformly

1 1

g f d ! 0 for each simple function integrable sequence ff g such that

n n

g 2 L X . For an arbitrary g 2 L X , note that there is a sequence of

1 1

simple functions which converges to g in measure. So we can approximate g in

L X norm by a simple functiong ~ on a subset E 2 of full enough measure

jjf jj d is small for each n and the measure of E is also small. Note that

that

Z Z Z Z

g f d jjg jj jjf jj d + jjg g~jj jjf jj d + g~f d :

n X n X X n X n

E E E

The rst twointegrals on the left-hand side are small for each n. The last integral

g f d ! 0, as needed. converges to zero as n !1.Thus

APettis-norm convergent sequence need not b e uniformly integrable Exam-

ple 2. However, if a sequence converges in the L X;L X -top ology then

1 1

it is necessarily uniformly integrable. To see why this is so, recall the usual pro of

[DU p. 104] that a weakly convergent sequence is uniformly integrable. If a

subset K of L X is not uniformly integrable, then we can nd with the help

of Rosenthal's Lemma a sequence ff g in K and a disjoint sequence fE g in

n n

such that

Z Z

jjf jj d > 2 and jjf jj d <

n n

E [ E

n j 6=n j

BOCHNER vs. PETTIS NORM: EXAMPLES AND RESULTS 9

for some >0. For each n we can nd l 2 L X [ L X] of norm one

n 1 1

and supp orted on E such that l f > 2 . De ne l : !X by

n n n

l = l 1 :

n E

n=1

Clearly, l is a norm one elementof L X [L X] . But l f 9 0 since

1 1 n

Z Z Z

l f d = l f d + l f d ;

n n n n

E F

n n

where F = [ E , and so

n m6=n m

Z Z Z

l f d l f d jjl jj jjf jj d

n n n X n X

E F

n n

2 jjf jj d

2 = :

So if g ! g in the L X;L X -top ology then the set fg g g, and thus

n 1 1 n

also the set fg g, are uniformly integrable.

To pass from L X;L X -convergence to weak convergence, recall [DU]

1 1

that the dual of L X is isometrically L X if and only if X has the Radon-

1 1

Nikod ym prop erty RNP. Thus wehave:

Fact. If X has the RNP, then a uniformly integrable Pettis-norm convergent

sequenceofL Xfunctions also converges weakly.

As noted ab ove, the uniform integrability condition is necessary. Our next

prop osition shows that it is also necessary that X have the RNP. The pro of,

which uses Stegall's Factorization Theorem, may b e pulled out of a result of

Ghoussoub and P. Saab [GS] characterizing weakly compact sets in L X.

Proposition. If X fails the RNP, then there is an essential ly bounded se-

quenceofL Xfunctions that is Pettis-norm convergent but not weakly conver-

gent.

Proof.

Since X fails the RNP, there is a separable subspace Y of X such that Y is

not separable. We shall construct an essentially b ounded sequence fg g of L Y

n 1

functions such that g ! 0 in the Pettis-norm on L Y but g 9 0weakly in

n 1 n

L Y. This is sucient.

N n

Let = f1; 1g b e the Cantor group with Haar measure . Let f : k =

n 0 n

1;::: ;2 g b e the standard n-th partition of . Thus = and =

n+1 n+1

n n

[ and = 2 . We will work with our underlying measure

2k 1 2k

space b eing the Cantor group endowed with its Haar measure instead of our

usual Leb esgue measure space on [0; 1].

DILWORTH and GIRARDI 10

Consider the set C of real-valued continuous function on as a subspace

n n

of L R. Let f1 g[fh : n =0;1;2;::: and k =1;::: ;2 g b e the usual Haar

basis of C , where h :!Ris given by

n+1 n+1

: 1 h = 1

2k 2k1

n n

Let fe : n =0;1;2;::: and k =1;::: ;2 g b e an enumeration lexicographi-

cally of the usual ` basis and H : ` ! L b e the Haar op erator that takes e

1 1 1

to h .

By Stegall's Factorization Theorem [S, Theorem 4], H factors through Y,

i.e. there are b ounded linear op erators R : ` ! Y and S : Y ! L such that

1 1

H = SR. Let R b e the natural extension of R to a b ounded linear op erator from

L ` toL Y.

1 1 1

Consider the sequence ff g of L ` functions given by

m 1 1

m 2

X X

n n

f = h e :

k k

n=1

k =1

Let g = R f . The sequence fg g of L Y functions do es the job.

m m m 1

Since sup jjf ! jj = 1 for -a.e. ! ,wehave that fg g is essentially

m ` m

m 1

b ounded.

To examine the Pettis norm of f , consider the continuous convex function

: ` ! R where the image of x = 2 ` '` again, lexicographically

1 1

ordered is

Z Z

m 2

X X

n n

x = jx f j d = h x e d :

k k

n=1

k =1

n m n

Since the image of f is supp orted on fe : n =1;::: ;2 and k =1;::: ;2 g,

22 1

we consider the natural restriction map r : ` ! ` ` given by

m 1 1

n n

r : n =0;1;2;::: and k =1;::: ;2 =

n m n

: n =1;::: ;2 and k =1;::: ;2 :

Let K b e the unit ball of r ` . The maximum of over K is attained at

m 1

n n

some extreme p oint x of K . Being an extreme p oint, x = satis es j j =1

0 0

k k

for each admissible n and k .So

jjf jj = sup x = sup x

m Pettis

x 2K

x 2B `

m 2

X X

n n

h d : = x =

k k 0

n=1

k =1

n n th

Since h b ehaves like the n -Rademacher function, Khintchine's in-

k k

k =1

equality gives that jjf jj decays like m .

m Pettis

BOCHNER vs. PETTIS NORM: EXAMPLES AND RESULTS 11

Now for the Pettis norm of g , note that if y 2 Y then

Z Z

j y g j d = jy Rf j d

m m

= jR y f j d jjR jj jjy jj jjf jj :

m m Pettis

Thus g ! 0 in the Pettis norm on L Y.

m 1

To see that g 9 0weakly in L Y, it suces to show that Hf 9 0weakly

m 1 m

in L L where H is the natural extension of H to an op erator from L ` to

1 1 1 1

L L . But since

1 1

2 m

X X

n n

h h ; Hf =

k k

n=1

k =1

wemay view Hf as a function from into C and thus we only need to

show that Hf 9 0weakly in L C . To see this, just consider the linear

m 1

functional l 2 [L C ] given by

l f = [f!]! d where f 2 L C

and note that

m 2

X X

n n

l Hf = h ! h ! d = 1 :

k k

n=1

k =1

Thus the sequence fg g do es all it is supp osed to do.

We close with one last example illustrating the ab ove prop osition in the case

that X is ` .

EXAMPLE 6. Once again working on the Leb esgue measure space on [0; 1],

consider the sequence fg g of L ` functions where

n 1 1

r e g =

k k n

k =1

where fe g are the standard unit vectors of ` and fr g are the Rademacher

k 1 k

functions.

The sequence is uniformly integrable, indeed it is even essentially b ounded by

1. As for the Pettis norm of a g , xx 2` ,say x = 2` . Since

n 1 j 1

Z Z

jx g j d = j r j d ;

n k k

k =1

Khintchine's inequality shows us that the Pettis norm of g b ehaves like .

Thus g ! 0 in the Pettis norm and so also in the L X;L X -top ology.

n 1 1

However, fg g do es not converge weakly. For consider the vector measure

G :!` given by

GE = E\[r =1] : j

DILWORTH and GIRARDI 12

Clearly G is a -continuous vector measure of b ounded variation. Thus G gives

rise to a functional l 2 [L ` ] given by

1 1

l f = fdG for f 2 L ` ;

1 1

where we view G as taking values in ` . Since

n n n

X X X

1 1 1 1

l g = l e r = l e 1 e 1 = ; 0=

n k k k k

[r =1] [r =1]

k k

n n n 2

k=1 k=1 k=1

we see that fg g do es not converge weakly.

Note that the ab ove linear functional l : L ` ! R is continuous for the

1 1

Bo chner norm but not for the Pettis norm.

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[D].A.Dvoretzky, Some results on convex bodies and Banach spaces, Pro c. Internat. Symp.

on Linear Spaces, Jerusalem, 1961, pp. 123{160.

[DR].A.Dvoretzky and C. A. Rogers, Absolute and unconditional convergence in normed

spaces, Pro c. Nat. Acad. Sci. USA 36 1950, 192{197.

[GS]. N. Ghoussoub and P. Saab, Weak Compactness in Spaces of Bochner Integrable Func-

tions and the Radon-Nikod ym property,Paci c J. Math. 110 1984, no. 1, 65{70.

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University of South Carolina, Mathematics Department, Columbia, SC 29208