DILWORTH and GIRARDI 2 2. XAMPLS: L 1 (X) vs. P 1 (X) Let X be a Banach space with dual X and let (; ;) be the usual Lebesgue measure space on [0; 1].

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1 Contemporary Mathematics Volume 00, 0000 Bochner vs. Pettis norm: examples and results S.J. DILWORTH AND MARIA GIRARDI Contemp. Math. 144 (1993) 69{80 Banach Spaces, Bor-Luh Lin and William B. Johnson, editors Abstract. Our basic example shows that for an arbitrary innite-dimensional Banach space X, the Bochner norm and the Pettis norm on L 1 (X) are not equivalent. Renements of this example are then used to investigate various modes of sequential convergence in L 1 (X). 1. INTRODUCTION Over the years, the Pettis integral along with the Pettis norm have grabbed the interest of many. In this note, we wish to clarify the dierences between the Bochner and the Pettis norms. We begin our investigation by using Dvoretzky's Theorem to construct, for an arbitrary innite-dimensional Banach space, a sequence of Bochner integrable functions whose Bochner norms tend to innity but whose Pettis norms tend to zero. By rening this example (again working with an arbitrary innite-dimensional Banach space), we produce a Pettis integrable function that is not Bochner integrable and we show that the space of Pettis integrable functions is not complete. Thus our basic example provides a uni- ed constructive way of seeing several known facts. The third section expresses these results from a vector measure viewpoint. In the last section, with the aid of these examples, we give a fairly thorough survey of the implications going between various modes of convergence for sequences of L 1 (X) functions Mathematics Subject Classication. 4640, 28A20, 46G10. This paper is in nal form and no version of it will be submitted for publication elsewhere. c0000 American Mathematical Society /00 $ $.25 per page 1

2 DILWORTH and GIRARDI 2 2. XAMPLS: L 1 (X) vs. P 1 (X) Let X be a Banach space with dual X and let (; ;) be the usual Lebesgue measure space on [0; 1]. Let L 1 (X) = L 1 (; ;;X) be the space of X-valued Bochner integrable functions on with the usual Bochner norm jjjj Bochner. A quick review of the Pettis integral. Consider a weakly measurable function f :!Xsuch that x f 2 L 1 (R) for all x 2 X. The Closed Graph Theorem gives R that for each 2 there is an element x of X satisfying x (x ) = x fd for each x 2 X. If x is actually in X for each 2, then we say that f is Pettis integrable with the Pettis integral of f over being 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 1 (X) of (equivalence classes of) all strongly measurable Pettis integrable functions forms a normed linear space under the norm k f k Pettis = sup j x (f) j d : x 2B(X ) Clearly if f 2 L 1 (X) then f 2P 1 (X) with jjfjj Pettis jjfjj Bochner : If X is nite-dimensional, then the Bochner norm and the Pettis norm are equivalent and so L 1 (X) = P 1 (X). However, in any innite-dimensional Banach space X, the next example shows that the two norms are not equivalent onl 1 (X). Specically, it constructs an essentially bounded sequence of Bochner-norm one functions whose Pettis norms tend to 0. XAMPL 1. Let X be an innite-dimensional Banach space. We now construct a sequence ff n g of L 1 (X) functions which tend to zero in the Pettis norm but not in the Bochner norm. Let fi n k : n =0;1;::: ;k =1;::: ;2n g be the dyadic intervals on [0; 1], i.e. I n k = k,1 2 n ; k 2 n : Fix n 2 N. We dene f n with the help of Dvoretzky's Theorem [D]. Find a 2 n - dimensional subspace n of X such that the Banach-Mazur distance between n and `2n 2 is at most 2. So there is an operator T n : `2n 2! n such that the norm of T n is at most 2 and the norm of the inverse of T n is 1. Let fe n k : k =1;:::2n g

3 BOCHNR vs. PTTIS NORM: XAMPLS AND RSULTS 3 be the image under T n of the standard unit vectors fu n k : k =1;:::2n g of `2n 2. Dene f n :!Xby f n (!) = e n k 1 I n k (!) ; The sequence ff n g has the desired properties. Since X2 n 2 n X jjf n jj Bochner = jje n k jj X d = 2,n jje n k jj X and 1 jje n k jj X 2, I n k 1 jjf n jj Bochner 2 : As for the Pettis norm, x x 2 B(X ). Note that the restriction y n of x to n is an element in nof norm at most 1. Also, the adjoint T n of T n has norm at most 2. Thus jx (e n k)j = = X2 n jyn(t n u n k)j X2 n and so jx (f n )j d = giving that 2 n j(t n y n)(u n k)j jjt n y njj (`2n 2 ) jj I n k X X u n kjj`2n 2 2 p 2 n ; 2 n jx (e n k)j d = 2,n jx (e n k)j 22,n 2 ; jjf n jj Pettis 22,n 2 : Thus the Pettis norm of f n tends to zero. This dierence between the Bochner and Pettis norms is further emphasized by the following minor variation of xample 1. XAMPL 2. Fix 0 << 1 and put 2 n 2 n. Dene f ~ n :!Xby f ~ n = n f n where f n is as in xample 1. The Bochner norm of f ~ n tends to innity at the rate of n but the Pettis norm of f ~ n tends to zero. Avariation on this example gives a (strongly measurable) Pettis integrable function that is not Bochner integrable { working in any innite-dimensional Banach space. The usual way of constructing such a function uses the Dvoretsky- Rogers Theorem. Recall [DR] that in any innite-dimensional P Banach space X, there is an unconditionally convergent series n x n that is not absolutely convergent. The function f :!Xgiven by f() = x n ( n ) 1 n ();

4 DILWORTH and GIRARDI 4 (where f n g is a partition of into sets of strictly positive measure) is Pettis integrable but not Bochner integrable. Our example is in the same spirit. XAMPL 3. Let X be an innite-dimensional Banach space. Consider the above construction. Note that the collection fi n 2 g1 partitions [0; 1]. Let g n :!Xbe a normalized f n (from xample 1) supported on I n 2, i.e. Dene f :!Xby g n (!) = 2 n f n (2 n!, 1) 1 I n 2 (!) : f(!) = g n (!) 1 I n 2 (!): Clearly f is strongly measurable. But f is not Bochner integrable since for any N 2 N jjfjj d NX I n 2 jjg n jjx d NX 2,n 2 n = N : However f is Pettis integrable. First note that x f 2 L 1 (R) for all x 2 X since for a xed x 2 B(X )wehave (using computations from the rst example) and so I n 2 jx fj d = jx fj d = I n 2 I n 2 jx g n j d = jx fj d 2 jx f n j d 22,n 2 ; 2, n 2 = 2( p 2+1): Next x R 2. We know there is an element x of X satisfying that x (x ) = x fd for each x 2 X.To see that x is actually in X, consider the sequence fs N g of L 1 (X) functions on where s N (!) = NX g n (!) 1 I n 2 (!): Let x N be the Bochner integral of s N over. Viewed as a sequence in X, fx N g N converges in norm to x since for any x 2 B(X ) j(x N, x )(x )j = j = x s N d, n=n+1 I n 2 x fdj jx fj d 2 n=n+1 jx (s N,f)jd 2, n 2 : Thus x is actually in X, as needed. Janicka and Kalton [JK] showed that if X is innite-dimensional then P 1 (X)is never complete under the Pettis norm. The rst example provides a constructive way to see this.

5 BOCHNR vs. PTTIS NORM: XAMPLS AND RSULTS 5 XAMPL 4. Taking some care in our construction in xample 1, we may arrange for f n g to form a nite dimensional decomposition (FDD) of some subspace of X. To see this, rst choose a basic sequence fx n g in X. Take a blocking ff n g of the basis so that each subspace F n is of large enough dimension to nd (using the nite-dimensional version of Dvoretzky's Theorem) a 2 n -dimensional subspace n of F n such that the Banach-Mazur distance between n and `2n 2 is at most 2. Then f n g forms a FDD. Keeping with the notation from xample 1, dene h n :!Xby h n = Clearly each h n is in P 1 (X). Also, fh n g is Cauchy in the Pettis norm since jjh n, h n+j jj Pettis k=n nx f k : jjf k jj Pettis 2 k=n 2, k 2 : To show that h n cannot converge to an elementinp 1 (X), we need the following two lemmas whose proofs are given shortly. lemma 1. If the sequence ff n g of functions in P 1 (X) converges to f 2P 1 (X) and each f n is essentially valued in some subspace Y of X, then f is also essentially valued in Y. lemma 2. Let ff n g be a sequence of P 1 (X) functions that converges in Pettis norm to f 2P 1 (X). Furthermore, suppose that each f n is essentially valued in some subspace Y of X and that T : Y! X is a bounded linear operator. Then Tf n converges in the Pettis norm to Tf. P Suppose that h n converges in the Pettis norm to h 2P 1 (X). Let Y n and let P n : Y! X be the natural projection from Y onto n. Note that each h n,thus also h, are essentially valued in Y. Applying the second lemma to the operator P n and noting that P n h m f n for m n, wehave that P n h f n. Thus for each n 2 N 1 jjp n h(!)jjx 2 for almost all!. This contradicts the fact that for almost all! X N lim jjh(!), P n h(!)jjx = 0 : N!1 Thus h n cannot converge to an element inp 1 (X). Now for the proofs of the lemmas. proof of lemma 1. Let ff n g be a sequence of P 1 (X) functions that converges in the Pettis norm to f 2P 1 (X). Furthermore, suppose that each f n is essentially valued in some subspace Y of X and that f is essentially valued in some subspace with Y. Since f and each f n are strongly measurable, we may assume that is separable.

6 DILWORTH and GIRARDI 6 Since is separable, there is a sequence fzm g of functionals from such that an element z from is in Y if and only if zm(z) = 0 for each m. Since ff n g converges in the Pettis norm to f, for each zm the sequence fzm f ng n converges to zm f in L 1(R) -norm. But for each m, note that zm f n is zero almost everywhere and so zm f is also zero almost everywhere. Thus f is also essentially valued in Y. proof of lemma 2. Let ff n g be a sequence of P 1 (X) functions which converges in Pettis norm to f 2 P 1 (X). Furthermore, suppose that each f n is essentially valued in some subspace Y of X. Lemma 1 gives that f is also essentially valued in Y. Let T : Y! X be a bounded linear operator. Note that for a xed x 2 X wemay extend T x to an element inx without increasing its norm. Note that Tf (and likewise each Tf n ) is actually in P 1 (X). To see this, rst observe that for a xed x 2 X (viewing T x as an element inx of the same norm) jx (Tf)jd = j(t x )(f)j d jjt x jj jjfjj Pettis : and so x (Tf) 2 L 1. Next, let x be the Pettis integral of f over 2. Note that x 2 Y and for each x 2 X x (Tx ) = (T x )(x ) = (T x )(f) d = x (Tf)d ; and so T (x ) is the Pettis integral of Tf over 2. Thus Tf is indeed in P 1 (X). Since for x 2 B(X ) jx (Tf n,tf)jd = j(t x )(f n, f)j d jjt jj jjf n, fjj Pettis ; the sequence Tf n converges in the Pettis norm to Tf. 3. VCTOR MASURS Let's look at our examples from a vector measure viewpoint. APettis integrable function f gives rise to a vector measure G f :!X where G f () is the Pettis integral of f over. The Pettis norm of f is the semivariation of G f.if f2 L 1 (X), then the Bochner norm of f is the variation of G f. xample 2 shows that for any innite-dimensional Banach space X, there is a sequence of X-valued measures whose variations tend to innity but whose semivariations tend to zero. On the other hand, xample 3 gives a X-valued measure of bounded semivariation but innite variation. For f 2P(X), the corresponding measure G f is -continuous and has a relatively compact range. In fact [DU], the completion of P 1 (X) is (isometrically)

7 BOCHNR vs. PTTIS NORM: XAMPLS AND RSULTS 7 the space K(X) of all -continuous measures G :!Xwhose range is relatively compact, equipped with the semivariation norm. Thus xample 4 gives a measure in K(X) which is not representable by a (strongly measurable) Pettis integrable function. 4. CONVRGNC We now restrict our attention to sequences of L 1 (X) functions which converge (in the Bochner norm, in the Pettis norm, or weakly in L 1 (X)) to functions actually in L 1 (X). We have the following obvious relations: (1) Bochner-norm convergence implies weak and Pettis-norm convergence; (2) weak convergence implies neither Bochner-norm nor Pettis-norm convergence (e.g. consider the Rademacher functions in L 1 (R)); (3) if X is nite dimensional then the Pettis norm and the Bochner norm are equivalent. Our examples help to clarify the other implications. If X is innite-dimensional, then there is a sequence of functions which converges in Pettis norm but whose Bochner norms tend to innity (xample 2). Thus this sequence does not converge in the Bochner norm or weakly. What if we require the sequence to be bounded (sup n jjf n jj L1 (X) < 1), or uniformly integrable, or even essentially bounded (sup n jjf n jj L 1(X) < 1)? Again, if X is innite-dimensional then xample 1 gives a sequence of essentially bounded functions which converges in Pettis norm but not in the Bochner norm. In certain situations we can pass from Pettis-norm convergence R to R weak convergence. If f n! f in the Pettis norm, then clearly g(f n) d! g(f) d for each simple function g 2 L 1 (X ). As a rst step towards weak convergence, when can we at least conclude that f n! f in the (L 1 (X);L 1 (X ))-topology? Aword of caution, the simple functions P need not be dense in L 1 (X ). For example, the L 1 (`2) function g() = 1 e n1 I n 2, where fe n g are the standard unit vectors of `2, is at least p 2 from any simple function in 2 L 1(`2). Our next example illustrates the fact that for any innite-dimensional Banach space there is a bounded sequence in L 1 (X) that converges in the Pettis norm but not in the (L 1 (X);L 1 (X ))-topology. XAMPL 5. Consider the sequence fg n g from xample 3 where g n (!) = 2 n f n (2 n!, 1) 1 I n 2 (!) : Since jjf n jj L1 (X) 2, the Bochner norm of each g n is at most 2. To examine the Pettis norm, x x 2 B(X ). Computations in xample 1 show that jx g n j d = I n 2 jx g n j d = jx f n j d 22,n 2

8 and so DILWORTH and GIRARDI 8 jjg n jj Pettis 22,n 2 : Thus g n! 0 in the Pettis norm. As for the (L 1 (X);L 1 (X ))-topology, nd y n k 2 B(X ) such that y n k (en k )= jje n k jj X. Note that fi n 2 : n =1;2;::: g partitions [0; 1] and that fi 2n j : j =2 n + 1;::: ;2 n+1 g partitions the interval I n 2. Dene g :!X by g(!) = y n k 1 I 2n 2n +k (!) : Clearly g 2 L 1 (X ). But g(g n ) d = = Ik n X2 n g(g n ) d = I 2n I 2n 2n +k y n k (2 n e n k) d 2n +k g(g n ) d 2 n 2,2n 1 = 1 : Thus fg n g does not converge in the (L 1 (X);L 1 (X ))-topology. However, a uniformly integrable Pettis-norm convergent sequence does converge in the (L 1 (X);L 1 (X ))-topology. To see why, consider a uniformly R integrable sequence ff n g such that g(f n) d! 0 for each simple function g 2 L 1 (X ). For an arbitrary g 2 L 1 (X ), note that there is a sequence of simple functions which converges to g in measure. So we can approximate g (in L 1 (X R ) norm) by a simple function ~g on a subset 2 of full enough measure that c jjf n jj d is small for each n and the measure of c is also small. Note that g(f n ) d jjgjjx jjf njjx d + jjg,~gjjx jjf njjx d + ~g(f n ) d : c The rst two integrals on the left-hand side are small for each n. The last integral converges to zero as n!1.thus R g(f n) d! 0, as needed. A Pettis-norm convergent sequence need not be uniformly integrable (xample 2). However, if a sequence converges in the (L 1 (X);L 1 (X ))-topology then it is necessarily uniformly integrable. To see why this is so, recall the usual proof [DU p. 104] that a weakly convergent sequence is uniformly integrable. If a subset K of L 1 (X) is not uniformly integrable, then we can nd (with the help of Rosenthal's Lemma) a sequence ff n g in K and a disjoint sequence f n g in such that n jjf n jj d > 2 and [ j6=n j jjf n jj d <

9 BOCHNR vs. PTTIS NORM: XAMPLS AND RSULTS 9 for some >0. For each n we can nd l n 2 L 1 (X ) [L 1 (X)] of norm one and supported on n such that R n l n (f n ) > 2. Dene l:!x by l() = l n ()1 n () : Clearly, l is a norm one element ofl 1 (X )[L 1 (X)]. But R l(f n) 9 0 since where F n l(f n ) d = = [ m6=n m, and so l(f n ) d n l n (f n ) d + l n (f n ) d n 2, jjf n jj d F n 2, = : F n l(f n ) d ;, F n jjljjx jj(f n)jjx d So if g n! g in the (L 1 (X);L 1 (X ))-topology then the set fg n, gg, and thus also the set fg n g, are uniformly integrable. To pass from (L 1 (X);L 1 (X ))-convergence to weak convergence, recall [DU] that the dual of L 1 (X) is (isometrically) L 1 (X ) if and only if X has the Radon- Nikodym property (RNP). Thus we have: Fact. If X has the RNP, then a uniformly integrable Pettis-norm convergent sequence ofl 1 (X)functions also converges weakly. As noted above, the uniform integrability condition is necessary. Our next proposition shows that it is also necessary that X have the RNP. The proof, which uses Stegall's Factorization Theorem, may be pulled out of a result of Ghoussoub and P. Saab [GS] characterizing weakly compact sets in L 1 (X). Proposition. If X fails the RNP, then there is an essentially bounded sequence ofl 1 (X)functions that is Pettis-norm convergent but not weakly convergent. 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 bounded sequence fg n g of L 1 (Y) functions such that g n! 0 in the Pettis-norm on L 1 (Y) but g n 9 0weakly in L 1 (Y). This is sucient. Let = f,1; 1g N be the Cantor group with Haar measure. Let f n k : k = 1;::: ;2 n g be the standard n-th partition of. Thus 0 1 = and n k = n+1 2k,1 [ n+1 2k and ( n k )=2,n. We will work with our underlying measure space being the Cantor group endowed with its Haar measure instead of our usual Lebesgue measure space on [0; 1].

10 DILWORTH and GIRARDI 10 Consider the set C() of real-valued continuous function on as a subspace of L 1 (R). Let f1 g[fh n k : n =0;1;2;::: and k =1;::: ;2n gbe the usual Haar basis of C(), where h n k :!Ris given by h n k = 1 n+1 2k,1, 1 n+1 2k Let fe n k : n =0;1;2;::: and k =1;::: ;2n g be an enumeration (lexicographically) of the usual `1 basis and H : `1! L 1 be the Haar operator that takes e n k to h n k. By Stegall's Factorization Theorem [S, Theorem 4], H factors through Y, i.e. there are bounded linear operators R: `1! Y and S : Y! L 1 such that H = SR. Let ~ R be the natural extension of R to a bounded linear operator from L 1 (`1) tol 1 (Y). Consider the sequence ff m g of L 1 (`1) functions given by f m () = 1 m mx : h n k()e n k : Let g m = R(fm ~ ). The sequence fg m g of L 1 (Y) functions does the job. Since sup m jjf m (!)jj`1 = 1 for -a.e.!, wehave that fg m g is essentially bounded. To examine the Pettis norm of f m, consider the continuous convex function : `1! R where the image of x =( n)2`1'`1(again, k lexicographically ordered) is (x ) = jx (f m )j d = 1 m mx h n k()x (e n k) d : Since the image of f m is supported on fe n k : n =1;::: ;2m and k =1;::: ;2 n g, we consider the natural restriction map r m : `1! `2(22m,1) 1 `1 given by r m (( n k : n =0;1;2;::: and k =1;::: ;2 n )) = ( n k : n =1;::: ;2 m and k =1;::: ;2 n ) : Let K be the unit ball of r m (`1). The maximum of over K is attained at some extreme point x 0 of K. Being an extreme point, x 0 =( n k ) satises jn k j =1 for each admissible n and k. So jjf m jj Pettis = sup x 2B(`1) = (x 0) = 1 m (x ) = sup mx x 2K X2 n (x ) n k hn k() d : Since P 2 n n k hn k behaves like the nth -Rademacher function, Khintchine's inequality gives that jjf m jj Pettis decays like m, 1 2.

11 BOCHNR vs. PTTIS NORM: XAMPLS AND RSULTS 11 Now for the Pettis norm of g m, note that if y 2 Y then jy g m j d = = jy (Rf m )j d j(r y )(f m )j d jjr jj jjy jj jjf m jj Pettis : Thus g m! 0 in the Pettis norm on L 1 (Y). To see that g m 9 0weakly in L 1 (Y), it suces to show that Hfm ~ 9 0weakly in L 1 (L 1 ) where H ~ is the natural extension of H to an operator from L1 (`1) to L 1 (L 1 ). But since ~Hf m () = 1 m mx h n k()h n k ; we may view Hfm ~ as a function from into C() and thus we only need to show that Hfm ~ 9 0weakly in L 1 (C()). To see this, just consider the linear functional l 2 [L 1 (C())] given by and note that l(f) = l( ~ Hfm ) = 1 m [f(!)](!) d where f 2 L 1 (C()) mx h n k(!)h n k(!) d = 1 : Thus the sequence fg m g does all it is supposed to do. We close with one last example illustrating the above proposition in the case that X is `1. XAMPL 6. Once again working on the Lebesgue measure space on [0; 1], consider the sequence fg n g of L 1 (`1) functions where g n () = 1 n nx r k ()e k where fe k g are the standard unit vectors of `1 and fr k g are the Rademacher functions. The sequence is uniformly integrable, indeed it is even essentially bounded by 1. As for the Pettis norm of a g n,xx 2(`1),sayx =( j )2`1. Since jx g n j d = 1 n j nx k r k j d ; Khintchine's inequality shows us that the Pettis norm of g n behaves like 1 p n. Thus g n! 0 in the Pettis norm and so also in the (L 1 (X);L 1 (X ))-topology. However, fg n g does not converge weakly. For consider the vector measure G:!`1given by G() = (( \[r j =1])) :

12 DILWORTH and GIRARDI 12 Clearly G is a -continuous vector measure of bounded variation. Thus G gives rise to a functional l 2 [L 1 (`1)] given by l(f) = fdg for f 2 L 1 (`1) ; where we view G as taking values in `1. Since l(g n ) = 1 n nx l(e k r k ) = 1 n nx l(e k 1 [rk =1], e k 1 [rk =,1]) = 1 n nx 1 2,0 = 1 2 ; we see that fg n g does not converge weakly. Note that the above linear functional l: L 1 (`1)! R is continuous for the Bochner norm but not for the Pettis norm. References [DU]. J. Diestel and J. J. Uhl, Jr., Vector Measures, Math. Surveys, no. 15, Amer. Math. Soc., Providence, R.I., [D]. A. Dvoretzky, Some results on convex bodies and Banach spaces, Proc. Internat. Symp. on Linear Spaces, Jerusalem, 1961, pp. 123{160. [DR]. A. Dvoretzky and C. A. Rogers, Absolute and unconditional convergence in normed spaces, Proc. Nat. Acad. Sci. USA 36 (1950), 192{197. [GS]. N. Ghoussoub and P. Saab, Weak Compactness in Spaces of Bochner Integrable Functions and the Radon-Nikodym property, Pacic J. Math. 110 (1984), no. 1, 65{70. [KJ]. L. Janicka and N.J. Kalton, Vector Measures of Innite Variation, Bulletin de L'Acad- emie Polonaise Des Sciences XXV (1977), no. 3, 239{241. [P]. B.J. Pettis, On integration in vector spaces, Trans. Amer. Math. Soc. 44 (1938), 277{ 304. [S]. Charles Stegall, The Radon-Nikodym property in conjugate Banach spaces, Trans. Amer. Math. Soc. 206 (1975), 213{223. University of South Carolina, Mathematics Department, Columbia, SC 29208