NORM ATTAINING BILINEAR FORMS ON L 1 ( μ )

Given a finite measure μ, we show that the set of norm attaining bilinear forms is dense in the space of all continuous bilinear forms on L1(μ) if and only if μ is purely atomic.


Introduction. A classical result of Bishop and Phelps
asserts that the set of norm attaining linear functionals on a Banach space is dense in the dual space.Very recently some attention has been paid to the question if the Bishop-Phelps theorem still holds for multilinear forms.To pose the problem more precisely, given a real or complex Banach space X and a natural number N, let us denote by ᏸ N (X) the space of all continuous N-linear forms on X and let us say that ϕ ∈ ᏸ N (X) attains its norm if there are x 1 ,...,x N ∈ B X (the closed unit ball of X) such that ϕ x 1 ,...,x N = ϕ := sup ϕ y 1 ,...,y N : y 1 ,...,y N ∈ B X . (1.1) We denote by Ꮽᏸ N (X) the set of norm attaining continuous N-linear forms on X.
The question is whether Ꮽᏸ N (X) is dense in ᏸ N (X) or not.Unlike the linear case, the answer to this question is negative, an example of a Banach space X such that Ꮽᏸ 2 (X) is not dense in ᏸ 2 (X) was recently exhibited by Acosta, Aguirre, and Payá [1].
In [4], Choi gave a more striking counterexample by showing that Ꮽᏸ ).In the positive direction, Aron, Finet, and Werner [2] showed that Ꮽᏸ N (X) is dense in ᏸ N (X) whenever X satisfies either the Radon-Nikodým property or the so-called property (α).Choi and Kim [5] obtained the same result for a Banach space X with a monotone shrinking basis and the Dunford-Pettis property (e.g., c 0 ).Jiménez and Payá [7] gave, for each N, an example of a Banach space X such that Ꮽᏸ N (X) is dense in ᏸ N (X) but Ꮽᏸ N+1 (X) is not dense in ᏸ N+1 (X), actually X is the canonical predual of a suitable Lorentz sequence space.
In this paper, we discuss the denseness of norm attaining multilinear forms on the space L 1 (µ), where µ is an arbitrary finite measure.We show that Ꮽᏸ N (L 1 (µ)) is dense in ᏸ N (L 1 (µ)) (for all N, or just for N = 2) if and only if µ is purely atomic.Half of this characterization follows from the main result in [2], since L 1 (µ) satisfies the Radon-Nikodým property if µ is purely atomic.For the converse, we first extend Choi's example [4] to show that Ꮽᏸ 2 (L 1 (µ)) is not dense in ᏸ 2 (L 1 (µ)) where µ is the product measure on an arbitrary product of copies of the unit interval.Then the result follows from the isometric classification of L 1 -spaces (cf.[8]) through an elementary lemma which deals with the denseness of Ꮽᏸ 2 (X) when X = Y ⊕ 1 Z is the l 1 -sum of two Banach spaces.

Results.
In what follows (Ω, Ꮽ,µ) will be a finite measure space.Let us start by recalling that the Banach space ᏸ 2 (L 1 (µ)) of all continuous bilinear forms on L 1 (µ) is isometrically isomorphic to L ∞ (µ ⊗ µ), where µ ⊗ µ denotes the product measure on Ω × Ω.More precisely, the bilinear form ϕ which corresponds to a function for every f ,g ∈ L 1 (µ) (see [6]).Choi [4], has shown that Ꮽᏸ ).This result can be extended in the following way.
Lemma 2.1.Let ν be an arbitrary nonzero finite measure and µ = ν ⊗ m, where m denotes Lebesgue measure on Proof.Let U be the set where ν is defined, so that µ works on Ω = I ×U and µ ⊗µ lives on the set Ω × Ω = I × U × I × U .We want a function h ∈ L ∞ (µ ⊗ µ) such that the corresponding bilinear form cannot be approximated by norm attaining bilinear forms.Actually h will be the characteristic function χ T of a suitable measurable set T ⊆ Ω × Ω with positive measure.The same argument used by Choi [4, Theorem 3] shows that the bilinear form corresponding to χ T belongs to the closure of Ꮽᏸ 2 (L 1 (µ)) only if there are measurable sets E,F ⊆ Ω with µ(E) > 0, µ(F ) > 0, such that [µ ⊗ µ] (E × F \ T )= 0. Therefore, we are left with finding a measurable set By [4, Lemma 2] there exists a set S ⊆ I × I with the analogous property.More concretely, S is a measurable set in I ×I, with [m⊗m](S) > 0, such that [m⊗m](A×B \S) for any pair A, B of measurable subsets of I with m(A) > 0 and m(B) > 0. To get our set T , we modify S in the obvious way, namely we define Clearly, T is a measurable set in Ω × Ω, with positive measure.Let E,F ⊆ Ω be measurable sets in Ω with µ(E) > 0, µ(F ) > 0, write H = E × F \ T and assume that [µ ⊗ µ](H) = 0 to get a contradiction.For u, v ∈ U, let us consider the section and note that where (2.4) By Fubini's theorem (or the definition of the product measure) we have for [ν ⊗ν]-almost every (u, v) ∈ U ×U .The property satisfied by S then implies that for [ν ⊗ ν]-almost every (u, v) ∈ U × U and by applying to E × F the same argument used with H, we get which is the required contradiction.