Norm Attaining Bilinear Forms onL1[0, 1]

JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.

211, 295]300 Ž1997.

AY975461

Norm Attaining Bilinear Forms on L1w0, 1x Yun Sung Choi* Department of Mathematics, Pohang Uni¨ ersity of Science and Technology, Pohang, 790, Korea Submitted by Richard M. Aron Received March 11, 1996

We show that the set of norm attaining bilinear forms on L1 w0, 1x is not dense in the space of all continuous bilinear forms. Q 1997 Academic Press

For a real Banach space X we let BŽ X . denote the Banach space of continuous bilinear forms on X, endowed with the norm 5 w 5 s sup< BŽ x, y .<: 5 x 5 F 1, 5 y 5 F 14 . We say that w g BŽ X . attains its norm if there are x, y g X, 5 x 5 s 5 y 5 s 1 such that 5 w 5 s < w Ž x, y .<. We let Bn aŽ X . denote the set of all norm attaining ones. Motivated by the Bishop]Phelps theorem w3x that the set of norm attaining functions is dense in X *, we consider a natural question about the denseness of Bn aŽ X .. M. D. Acosta, F. Aguirre, and R. Paya ´ w1x first Ž B answered this question in the negative. They proved that n a G . is not dense in BŽ G ., where G is the Banach space used by W. Gowers w6x to show that l p Ž1 - p - `. fails Lindenstrauss’ property B w9x. In the positive direction, R. Aron, C. Finet, and E. Werner w2x showed that Bn aŽ X . is dense in BŽ X . whenever X satisfies either the Radon]Nikodym property or the so-called property Ž a . w10x. Moreover, the author and S. G. Kim w5x got the same result for a Banach space X with a monotone shrinking basis and the Dunford]Pettis property Že.g., c 0 .. In this note we characterize the norm attaining bilinear forms on L1w0, 1x, and show that Bn aŽ L1w0, 1x. is *Research supported by KOSEF Grant 941-0100-004-2 and the Basic Science Research Institute Program, Ministry of Education, BSRI-95-1401. E-mail address: mathchoi@euclid. postech.ac.kr. 295 0022-247Xr97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.

296

YUN SUNG CHOI

not dense in BŽ L1w0, 1x.. The author expresses his thanks to C. Boyd and Y. S. Shim for helpful conversations and the referees for kind suggestions concerning the style of this note. We begin by recalling a useful integral representation of continuous bilinear forms on L1w0, 1x which is deduced from the following natural isometric isomorphisms, where I stands for the unit interval w0, 1x:

ˆp L1 Ž I . . * , L1 Ž I = I . * , L` Ž I = I . . B Ž L1 Ž I . . , Ž L1 Ž I . m More precisely, to each w g BŽ L1 Ž I .. there corresponds a unique f g L`Ž I = I . such that 5 w 5 s 5 f 5 ` and

w Ž g , h. s

HI=I f Ž x, y . g Ž x . h Ž y . dx dy

for all g, h g L1 Ž I .. For f g L`Ž I = I . we set S f s  Ž x, y . g I = I : < f Ž x, y . < s 5 f 5 ` 4 , 5 5 Sq f s  Ž x, y . g I = I : f Ž x, y . s f ` 4 , and 5 5 Sy f s  Ž x, y . g I = I : f Ž x, y . s y f ` 4 . PROPOSITION 1. Let w g BŽ L1 Ž I .. be represented by f g L`Ž I = I . as explained abo¨ e. Then w attains its norm if and only if there is a measurable rectangle A = B in I = I with positi¨ e measure such that either < A = B < s <Ž A = B . l Sq < or < A = B < s <Ž A = B . l Sy < f f . Proof. Suppose that w g Bn aŽ L1 Ž I .. and let g, h g L1 Ž I . be such that 5 g 5 s 5 h5 s 1

and

5 w 5 s w Ž g , h. .

Ž 1.

Then we have 5 f 5` s

HI=I f Ž x, y . g Ž x . h Ž y . dx dy

F 5 f 5`

HI=I< g Ž x . < < h Ž y . < dx dy s 5 f 5

hence f Ž x, y . g Ž x . h Ž y . s 5 f 5 ` < g Ž x . < < h Ž y . <

`

Ž 2.

NORM ATTAINING BILINEAR FORMS

297

for almost every Ž x, y . g I = I. If we replace g and h by yg and yh, Ž1. and Ž2. remain unchanged, so we may assume that the set A s  x g I: g Ž x . ) 04 has positive measure. If the set  y g I: hŽ y . ) 04 also has positive measure, we take this set for B and Ž2. implies that f Ž x, y . s 5 f 5 ` < < < for almost every Ž x, y . g A = B, that is, <Ž A = B . l Sq f s A=B. < Otherwise we take B s  y g I: hŽ y . - 04 and we get <Ž A = B . l Sy f s < A = B <. The ‘‘if’’ part is even easier. Assume without loss of generality that there is a measurable rectangle A = B in I = I with positive measure such that < A = B < s <Ž A = B . l Sq < f . Then

w Ž x Ar< A < , x Br< B < . s

HI=I f Ž x, y . Ž x

A

Ž x . r< A < .Ž xB Ž y . r< B < . dx dy

s 5 f 5` s 5 w 5 . LEMMA 2. There is a measurable subset S of I = I with positi¨ e measure such that for any measurable rectangle A = B in I = I with positi¨ e measure, <Ž A = B . l S < - < A = B <. Proof. Let C be a Cantor-type subset of I with positive measure, and let S s  Ž x, y . g I = I : < x y y < g C 4 . Being the inverse image of C under a continuous function, S is closed in I = I, hence compact. Moreover, it is easy to check that < S x < s < C < for every x g I, where S x s  y g I: Ž x, y . g S4 , so < S < s < C < ) 0 Žnote that we use < ? < to denote the Lebesgue measure both in R and R 2 .. Let A = B be a measurable rectangle in I = I with positive measure. To prove <Ž A = B . _ S < ) 0 we use the following CLAIM. If A, B ; R are measurable sets with positi¨ e measure, there exist a nonempty open inter¨ al H and d ) 0 such that <Ž A q t . l B < ) d for e¨ ery t g H. Note the following consequence: H ; B y A hence B y A has nonempty interior. This is a generalized version of a classical result due to Steinhaus Žsee w4, pp. 137]140x, for example.. To prove the above claim we will simply check that the usual proof of the Steinhaus Theorem actually gives our slightly stronger statement. For the sake of completeness we go into the details. Proof of the Claim. We may clearly assume that the sets A, B are bounded, just take bounded subsets with positive measure. Fix 0 - r - 1 and let G be a bounded open set with A ; G and < A < ) r < G <. Since G can

298

YUN SUNG CHOI

be written as a countable union of disjoint intervals, at least one of these intervals Žcall it J . will satisfy < A l J < ) r < J <. In the same way we find a bounded interval K such that < B l K < ) r < K <. We may clearly assume without loss of generality that < J < F < K < and then, by subdividing K if necessary, that < K < F 2 < J <. Moreover, up to translation we may also assume that inf J s inf K s 0. Then, for t G 0 we have Ž A l J . q t ; w0, < K < q t x, so <Ž Ž A l J . q t . j Ž B l K . < F < K < q t while < Ž A l J . q t < q < B l K < G 32 r < K < . Therefore, by choosing r ) 2r3 and 0 - 2 d - Ž3 rr2 y 1.< K < we get <Ž A q t . l B < G <Ž Ž A l J . q t . l Ž B l K . < G d for every t g H [ w0, d x, as claimed. Back to the proof of our lemma, let H and d be given by the claim and take a nonempty open subinterval L ; H such that L l Ž C j y C . s B. Then the set T s Ž y y t, y .: t g L, y g Ž A q t . l B4 is contained in Ž A = B . _ S and it is easy to check that T has positive measure. Indeed, under the measure-preserving mapping Ž u, ¨ . ª Ž ¨ y u, ¨ . T becomes the set Tˆ s Ž t, y .: t g L, y g Ž A q t . l B4 , so < T < s < Tˆ < s THEOREM 3.

HL< Ž A q t . l B < dt G d < L < ) 0.

Bn aŽ L1 Ž I .. is not dense in BŽ L1 Ž I ...

Proof. Let S be the measurable set given by the above lemma and c g BŽ L1 Ž I .. the bilinear form corresponding to xS g L`Ž I = I .. If w g Bn aŽ L1 Ž I .. is represented by f g L`Ž I = I ., then Proposition 1 gives us a measurable rectangle A = B with positive measure such that < f < s 5 f 5 ` almost everywhere in A = B. By the above lemma the set Ž A = B . _ S has positive measure and < f y xS < s 5 f 5 ` almost everywhere in this set, so 5 f y xS 5 ` G 5 f 5 ` , that is, 5 w y c 5 G 5 w 5 G 1 y 5 w y c 5, hence 5 w y c 5 G 1r2. Remark 4. It is worth noting that the above theorem remains true for a complex space L1 Ž I ., due to the following simple observation. If a complex-valued function f g L`Ž I = I . corresponds to a norm attaining bilinear form on the complex space L1 Ž I ., then the real-valued function < f <

NORM ATTAINING BILINEAR FORMS

299

corresponds to a norm attaining bilinear form on the real space L1 Ž I ., hence Proposition 1 tells us that < f < s 5 f 5 ` almost everywhere in a measurable rectangle with positive measure. This is just what we need in the proof of the above theorem. Let us mention an interesting open problem. There is a natural isometric isomorphism between BŽ X . and the space L Ž X, X *. of bounded operators of X into X *, under which every w g BŽ X . is associated with T g L Ž X, X *. by T Ž x .Ž y . s w Ž x, y .. It is clear that T attains its norm whenever w does, but the converse is not true in general. Actually, we do not know any example X such that Bn aŽ X . is not dense in BŽ X . but the set NAŽ X, X *. of norm attaining operators is dense in L Ž X, X *.. Since L`Ž I . doesn’t have the Radon]Nikodym property, every T g Ž L L1 Ž I ., L`Ž I .. is not representable. Nevertheless, to each T g L Ž L1 Ž I ., L`Ž I .. there corresponds a unique f g L`Ž I = I . such that 5 T 5 s 5 f 5 ` and T Ž g . Ž y. s

HI f Ž x, y . g Ž x . dx

for all g g L1 Ž I . and y g I. Note that L`Ž I, L`Ž I .. is not isomorphic to L`Ž I = I .. However, we don’t know if NAŽ L1 Ž I ., L`Ž I .. is dense in L Ž L1 Ž I ., L`Ž I ... As results related with this, J. Lindenstrauss w9x showed that for some Banach space Y, NAŽ L1 Ž I ., Y . is not dense in L Ž L1 Ž I ., Y . due to the lack of extreme points in the closed unit ball of L1 Ž I ., and in fact, W. Schachermayer w11x showed that C w0, 1x is such a space Y. On the other hand, J. Johnson and J. Wolfe w8x proved that for any Banach space Y every compact operator T of L1 Ž I . into Y can be approximated by finite rank norm attaining operators. For a strictly convex Banach space Y, J. J. Uhl w12x proved that NAŽ L1 Ž I ., Y . is dense in L Ž L1 Ž I ., Y . if and only if Y has the Radon]Nikodym property, and next A. Iwanik w7x proved that NAŽ L1 Ž I ., L1 Ž I .. is dense in L Ž L1 Ž I ., L1 Ž I ...

REFERENCES 1. M. D. Acosta, F. Aguirre, and R. Paya, ´ There is no bilinear Bishop]Phelps theorem, Israel J. Math. 93 Ž1996., 221]227. 2. R. Aron, C. Finet, and E. Werner, Norm-attaining n-linear forms and the Radon] Nikodym property, in ‘‘Proc. 2nd Conf. on Function Spaces’’ ŽK. Jarosz, Ed.., pp. 19]28, Lecture Notes in Pure and Appl. Math., Dekker, New York, 1995. 3. E. Bishop and R. R. Phelps, A proof that every Banach space is subreflexive, Bull. Amer. Math. Soc. 67 Ž1961., 97]98.

300

YUN SUNG CHOI

4. S. B. Chae, ‘‘Lebesgue Integration,’’ Dekker, New York, 1980. 5. Y. S. Choi and S. G. Kim, Norm or numerical radius attaining multilinear mappings and polynomials, J. London Math. Soc. 54, 2 Ž1996., 135]147. 6. W. Gowers, Symmetric block bases of sequences in large average growth, Israel J. Math. 69 Ž1990., 129]149. 7. A. Iwanik, Norm attaining operators on Lebesgue spaces, Pacific J. Math. 83, Ž1979., 381]386. 8. J. Johnson and J. Wolfe, Norm attaining operators, Studia Math. 65 Ž1979., 7]19. 9. J. Lindenstrauss, On operators which attain their norm, Israel J. Math 1 Ž1963., 139]148. 10. W. Schachermayer, Norm attaining operators and renormings of Banach spaces, Israel J. Math. 44 Ž1983., 201]212. 11. W. Schachermayer, Norm attaining operators on some classical Banach spaces, Pacific J. Math. 105 Ž1983., 427]438. 12. J. J. Uhl, Norm attaining operators on L1 w0, 1x and the Radon]Nikodym property, Pacific J. Math. 63 Ž1976., 293]300.