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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 1 not equivalent. Re nements of this example are then used to investigate various mo des of sequential convergence in L X. 1 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. 1 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. c 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 1 of X satisfying Theorem gives that for each E 2 there is an element x E R is actually in X for each E 2 , x = x fd for each x 2 X .Ifx x E 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 E Z x x = x fd E E 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 1 of all strongly measurable Pettis integrable functions forms a normed linear space under the norm Z 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. 1 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 n k 1 k I = ; : n n k 2 2 n Fix n 2 N.We de ne f with the help of Dvoretzky's Theorem [D]. Find a 2 - n dimensional subspace E of X such that the Banach-Mazur distance b etween E n n 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 n 2 b e the image under T of the standard unit vectors fu : k =1;::: 2 g of ` . n 2 k De ne f : !Xby n n 2 X n n f ! = e 1 ! ; n I k k k=1 The sequence ff g has the desired prop erties. n Since n n Z 2 2 X X n n n jjf jj = jje jj d = 2 jje jj n Bo chner X X k k n I k k =1 k =1 n and 1 jje jj 2, X k 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 n E is an elementin E of norm at most 1. Also, the adjoint T of T has norm n n n n at most 2. Thus n n 2 2 X X n n jx e j = jy T u j n k n k k =1 k =1 n n 2 2 X X p n n n 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 n 2 jx f j d = ; jx e j d = 2 jx e j 22 n k k n I k k =1 k =1 giving that n 2 : jjf jj 22 n Pettis Thus the Pettis norm of f tends to zero. n This di erence b etween the Bo chner and Pettis norms is further emphasized by the following minor variation of Example 1. EXAMPLE 2. 1 n ~ ~ Fix 0 < < and put 2 . De ne f : !X by f = f where n n n n n 2 ~ 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 P X, there is an unconditionally convergent series x that is not absolutely n n convergent. The function f : !Xgiven by 1 X x n f = 1 ; E n E n n=1 DILWORTH and GIRARDI 4 where fE g is a partition of into sets of strictly p ositive measure is Pettis n 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 n 2 n=1 n normalized f from Example 1 supp orted on I , i.e. n 2 n n n g ! = 2 f 2 ! 1 1 ! : n n I 2 De ne f : !Xby 1 X n g ! 1 ! : f! = n I 2 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 I 2 n=1 n=1 However f is Pettis integrable. First note that x f 2 L R for all x 2 X since 1 for a xed x 2 B X wehave using computations from the rst example Z Z Z n 2 jx f j d = jx g j d = jx f j d 22 ; n n n n I I 2 2 and so Z Z 1 1 X X p n 2 jx f j d = jx f j d 2 2 = 2 2+1 : n I 2 n=1 n=1 Next x E 2 . We know there is an element x of X satisfying that E R xfd for each x 2 X .To see that x is actually in X, consider x x = E E E the sequence fs g of L X functions on where N 1 N X n s ! = g ! 1 !: N n I 2 n=1 N Let x b e the Bo chner integral of s over E . Viewed as a sequence in X , N E N fx g converges in norm to x since for any x 2 B X N E E Z Z Z N jx x x j = j x s d x fdj jx s fjd N N E E E E Z 1 1 X X n 2 = jx f j d 2 2 : n I 2 n=N +1 n=N +1 Thus x is actually in X, as needed.
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