Norm Attaining Multilinear Forms on 𝐿₁(𝝁)

Abstract

Given an arbitrary measure 𝜇, this study shows that the set of norm attaining multilinear forms is not dense in the space of all continuous multilinear forms on 𝐿₁(𝜇). However, we have the density if and only if 𝜇 is purely atomic. Furthermore, the study presents an example of a Banach space 𝑋 in which the set of norm attaining operators from 𝑋 into 𝑋^∗ is dense in the space of all bounded linear operators 𝐿(𝑋,𝑋^∗). In contrast, the set of norm attaining bilinear forms on 𝑋 is not dense in the space of continuous bilinear forms on 𝑋.

1. Introduction

The Bishop-Phelps theorem [1] asserts that the set of norm attaining linear functionals on a Banach space 𝑋 is dense in the dual space 𝑋^∗. Some authors have considered the question of the density of norm attaining multilinear forms. To present the problem more precisely, given real Banach spaces 𝑋₁,…,𝑋_𝑁, we denote by ℒ^𝑁(𝑋₁,…,𝑋_𝑁) the space of all continuous 𝑁-linear mappings from 𝑋₁×⋯×𝑋_𝑁 into the scaler field. We say that 𝜑∈ℒ^𝑁(𝑋₁,…,𝑋_𝑁) attains its norm if there is 𝑥_𝑖∈𝐵_𝑋𝑖 (the unit ball of 𝑋_𝑖) for 𝑖=1,2,…,𝑁, such that|𝜑(𝑥₁,…,𝑥_𝑁)|=‖𝜑‖=sup{|𝜑(𝑦₁,…,𝑦_𝑁)|∶(𝑦₁,…,𝑦_𝑁)∈𝐵_𝑋1×⋯×𝐵_𝑋𝑁}, (1.1)and we denote by 𝒜ℒ^𝑁(𝑋₁,…,𝑋_𝑁) the set of all norm attaining 𝑁-linear forms. In the case where 𝑋₁=⋯=𝑋_𝑁=𝑋, we write simply ℒ^𝑁(𝑋) and 𝒜ℒ^𝑁(𝑋).

Aron et al. [2] posed the question of when 𝒜ℒ^𝑁(𝑋) is dense in ℒ^𝑁(𝑋), and gave sufficient conditions for this density to hold. The first example of a Banach space 𝑋 such that 𝒜ℒ²(𝑋) is not dense in ℒ²(𝑋) was given in [3]. Shortly after, Choi [4] showed that 𝒜ℒ²(𝐿₁[0,1]) is not dense in ℒ²(𝐿₁[0,1]). For additional results on this problem, we refer the reader to [5–9].

In this paper, we give some improvements on the results in [10]. More concretely, it was shown in that study that given an arbitrary finite measure 𝜇, 𝒜ℒ²(𝐿₁(𝜇)) is dense in ℒ²(𝐿₁(𝜇)) if and only if 𝜇 is purely atomic. In this note, we extend the above result to an arbitrary measure. Namely, we proved that, given any arbitrary measure 𝜇, 𝒜ℒ^𝑁(𝐿₁(𝜇)) is dense in ℒ^𝑁(𝐿₁(𝜇)) if and only if 𝜇 is purely atomic. Also, we present a new example of a Banach space 𝑋 such that the set of norm attaining operators from 𝑋 into 𝑋^∗ is dense in the space of all bounded linear operators from 𝑋 into 𝑋^∗, but the set 𝒜ℒ²(𝑋) is not dense in ℒ²(𝑋). This can be shown by relating the main result in our work to the following theorem.

Theorem 1.1 (see [11, Theorem 1]). Given an arbitrary measure 𝜇 and a localizable measure 𝜈, the set of norm attaining operators from 𝐿₁(𝜇) into 𝐿_∞(𝜈) is dense in the space 𝐿(𝐿₁(𝜇),𝐿_∞(𝜈)).

2. The Results

We begin by recalling the isometric classification of 𝐿₁-spaces and a technical lemma which deals with the density of norm attaining bilinear forms on arbitrary 𝑙₁-sums of Banach spaces in order to reduce the proof of our problem to the case where 𝜇 is a finite measure. Recall that if 𝜇 is an arbitrary measure, 𝐿₁(𝜇) can be decomposed in the form𝐿₁(𝜇)≅(⊕_𝑖∈𝐼𝐿₁(𝜇_𝑖))_ℓ1 (2.1)where 𝜇_𝑖 is a finite measure for all 𝑖∈𝐼 (see, e.g., [12, Appendix B]). On the other hand, if 𝜈 is a localizable measure we have that 𝐿_∞(𝜈)=𝐿₁(𝜈)^∗, and we get a set of finite measures {𝜈_𝑗∶𝑗∈𝐽} such that

𝐿_∞(𝜈)≅(⊕_𝑗∈𝐽𝐿_∞(𝜈_𝑗))_ℓ∞. (2.2)In what follows, we may assume without loss of generality that (Ω,𝒜,𝜇) is a finite measure space. The well-known representation of the space ℒ²(𝐿₁(𝜇)) is nothing but 𝐿_∞(𝜇⊗𝜇) “the space of all essential bounded measurable functions,” where 𝜇⊗𝜇 denotes the product measure on Ω×Ω. More concretely,ℒ²(𝐿₁(𝜇)≅𝐿(𝐿₁(𝜇),𝐿₁(𝜇)^∗)≅𝐿(𝐿₁(𝜇),𝐿_∞(𝜇))≅𝐿_∞(𝜇⊗𝜇); (2.3)see [12, Example 3.27]. In view of the above, we get the integral representation for the continuous bilinear form ℎ on ℒ²(𝐿₁(𝜇)) as follows:ℎ(𝑓,𝑔)=_Ω×Ωℎ(𝑥,𝑦)𝑓(𝑥)𝑔(𝑦)𝑑𝜇(𝑥)𝑑𝜇(𝑦), (2.4)for 𝑓,𝑔∈𝐿₁(𝜇), 𝑥,𝑦∈Ω, and ℎ∈𝐿_∞(𝜇⊗𝜇). Moreover, the application ℎ↦ℎ is linear isometric bijection from 𝐿_∞(𝜇⊗𝜇) onto ℒ²(𝐿₁(𝜇)); see [4].

To make the vision more comprehensive, we state the following technical lemmas that will be needed later. To simplify the notation, we consider the case 𝑁=2. The proof for the general case is exactly the same.

Lemma 2.1 (see [10, Lemma 2.1]). Let 𝜈 be an arbitrary nonzero finite measure and 𝜇=𝜈⊗𝑚, where 𝑚 denotes Lebesgue measure on 𝐼=[0,1]. Then 𝒜ℒ²(𝐿₁(𝜇)) is not dense in ℒ²(𝐿₁(𝜇)).

The other technical lemma deals with 𝑙₁-sums of Banach spaces. By 𝑌⊕₁𝑍 we denote the ℓ₁-sum of two Banach spaces 𝑌 and 𝑍, that is, ‖𝑦+𝑧‖=‖𝑦‖+‖𝑧‖ for arbitrary 𝑦∈𝑌, 𝑧∈𝑍.

Lemma 2.2 (see [10, Lemma 2.2]). Let 𝑌,𝑍 be Banach spaces and 𝑋=𝑌⊕₁𝑍. If 𝒜ℒ²(𝑋) is dense in ℒ²(𝑋), then 𝒜ℒ²(𝑌) is dense in ℒ²(𝑌).

Our first result of this paper is a characterization of those functions ℎ∈ℒ_∞(𝜇⊗𝜇), where ℎ its corresponding bilinear form in ℒ²(𝐿₁(𝜇)) that attains its norm (see [4]).

Proposition 2.3. Let (Ω,𝒜,𝜇) be a finite measure space, fixed ℎ∈𝐿_∞(𝜇⊗𝜇), and let ℎ be its corresponding bilinear form as defined in (2.4).
(1) There exist sets 𝐴,𝐵∈𝒜 with 𝜇(𝐴)>0, 𝜇(𝐵)>0 and a scalar 𝑡 with |𝑡|=1 such that𝑡ℎ(𝑥,𝑦)=‖ℎ‖_∞ (2.5)for [𝜇⊗𝜇]-almost every (𝑥,𝑦)∈𝐴×𝐵.(2)There are sets 𝐴,𝐵 like in (1) and measurable functions 𝜑,𝜓 on Ω such that|𝜑(𝑤)|=|𝜓(𝑤)|=1, (2.6)where 𝑤∈Ω and 𝜑(𝑥)𝜓(𝑦)ℎ(𝑥,𝑦)=‖ℎ‖_∞, for [𝜇⊗𝜇]-almost every (𝑥,𝑦)∈𝐴×𝐵.(3) The bilinear form ℎ∈ℒ²(𝐿₁(𝜇)) corresponding to ℎ∈𝐿_∞(𝜇⊗𝜇) attains its norm.Then(1)⇒(2)⟺(3). (2.7)Moreover, in the real case all three statements are equivalent.

Proof. (1)⇒(2) is clear, just take 𝜑=𝑡 and 𝜓=1.
For (2)⇒(3), just consider the functions 𝑓=𝜑𝜒_𝐴/𝜇(𝐴),𝑔=𝜓𝜒_𝐴/𝜇(𝐵) where 𝑓,𝑔 are in the unit sphere of 𝐿₁(𝜇), 𝜒_𝐴,𝜒_𝐵 denote the characteristic functions on 𝐴 and 𝐵, respectively, and1ℎ(𝑓,𝑔)=𝜇(𝐴)𝜇(𝐵)_𝐴×𝐵1ℎ(𝑥,𝑦)𝜑(𝑥)𝜓(𝑦)𝑑𝜇(𝑥)𝑑𝜇(𝑦)=𝜇(𝐴)𝜇(𝐵)_𝐴×𝐵‖ℎ‖_∞𝑑(𝜇⊗𝜇)=‖ℎ‖_∞. (2.8)
(3)⇒(2) Let 𝑓,𝑔∈𝐿₁(𝜇) be such that ‖𝑓‖₁=‖𝑔‖₁=1 and ℎ(𝑓,𝑔)=‖ℎ‖_∞. Take𝐴={𝑥∈Ω∶𝑓(𝑥)≠0},𝐵={𝑦∈Ω∶𝑔(𝑦)≠0} (2.9)to be two measurable sets in Ω with 𝜇(𝐴)>0,𝜇(𝐵)>0, and write 𝑓,𝑔 in the forms 𝑓=𝜑|𝑓|,𝑔=𝜓|𝑔| where 𝜑,𝜓 are measurable functions on Ω with |𝜑|=1,|𝜓|=1, then we have‖ℎ‖_∞=ℎ(𝑓,𝑔)=_𝐴×𝐵ℎ(𝑥,𝑦)𝜑(𝑥)|𝑓(𝑥)|𝜓(𝑦)|𝑔(𝑦)|𝑑𝜇(𝑥)𝑑𝜇(𝑦)≤‖ℎ‖_∞‖𝑓‖₁‖𝑔‖₁=‖ℎ‖_∞, (2.10)from which we conclude thatℎ(𝑥,𝑦)𝜑(𝑥)𝜓(𝑦)=‖ℎ‖_∞ (2.11)for [𝜇⊗𝜇]-almost every (𝑥,𝑦)∈𝐴×𝐵.
In the real case, the functions 𝜑,𝜓 have only the values ±1, then we can choose measurable subsets 𝐴₀⊆𝐴 and 𝐵₀⊆𝐵 such that 𝜇(𝐴₀)𝜇(𝐵₀)>0, where 𝜑,𝜓 are constants on 𝐴₀,𝐵₀, respectively. If 𝑡=±1 is the product of these constants, then we have clearly 𝑡ℎ(𝑥,𝑦)=‖ℎ‖_∞ for [𝜇⊗𝜇]-almost every (𝑥,𝑦)∈𝐴₀×𝐵₀, so we get that (3)⇒(1), as required.

In the special case ℎ=𝜒_𝐸, the characteristic function of a measurable set 𝐸∈𝒜×𝒜, we have the following result.

Corollary 2.4. Let (Ω,𝒜,𝜇) be a finite measure space, let 𝐸∈𝒜×𝒜 be a measurable set with (𝜇⊗𝜇)(𝐸)>0, and consider the following bilinear form 𝜒_𝐸 corresponding to the characteristic function of 𝐸. The following statements are equivalent:
(1)𝜒_𝐸∈𝒜ℒ²(𝐿₁(𝜇));(2)𝜒_𝐸∈𝒜ℒ²(𝐿₁(𝜇));(3) There exist subsets 𝐴,𝐵∈𝒜 with 𝜇(𝐴)𝜇(𝐵)>0 such that [𝜇⊗𝜇]((𝐴×𝐵)∩𝐸)=𝜇(𝐴)𝜇(𝐵).
Note that we can say that the measurable rectangle 𝐴×𝐵 is contained in the set 𝐸.

Proof. (1)⇒(2). This is trivial.
(2)⇒(3). Let ℎ∈𝐿_∞(𝜇⊗𝜇) be such that ‖𝜒_𝐸−ℎ‖_∞<1/2, and ℎ∈𝒜ℒ²(𝐿₁(𝜇)), then it is clear that ‖ℎ‖_∞>1/2. From the implication (3)⇒(2) of Proposition 2.3, we have two measurable sets 𝐴,𝐵∈𝒜 with 𝜇(𝐴)𝜇(𝐵)>0, and measurable functions 𝜑,𝜓 on Ω with |𝜑(𝑥)|=|𝜓(𝑦)|=1, such that𝜑(𝑥)𝜓(𝑦)ℎ(𝑥,𝑦)=‖ℎ‖_∞, (2.12)then|ℎ(𝑥,𝑦)|=‖ℎ‖_∞>12, (2.13)for [𝜇⊗𝜇]-almost every (𝑥,𝑦)∈𝐴×𝐵. Hence|𝜒_𝐸(𝑥,𝑦)|≥|ℎ(𝑥,𝑦)|−|ℎ(𝑥,𝑦)−𝜒_𝐸1(𝑥,𝑦)|>2−‖ℎ−𝜒_𝐸‖_∞>0. (2.14)for [𝜇⊗𝜇]-almost every (𝑥,𝑦)∈𝐴×𝐵, from which we get that 𝜒_𝐸=1, for [𝜇⊗𝜇]-almost every (𝑥,𝑦)∈𝐴×𝐵, which means that (3) holds.
(3)⇒(1). If 𝐴,𝐵 are the sets that satisfy the conditions of the statement (3), then we may clearly see that the function 𝜒_𝐸=1=‖𝜒_𝐸‖_∞, for [𝜇⊗𝜇]-almost every (𝑥,𝑦)∈𝐴×𝐵, then the function 𝑓=𝜒_𝐸 verifies the statement (1) of Proposition 2.3 including the case 𝑡=1.

Remark 2.5. Let us point out the following consequence of the representation theory for 𝐿₁-spaces. Indeed, if 𝜈 is a finite measure, we may write𝐿₁(𝜈)=(⊕_𝑖𝐼𝑋_𝑖)_ℓ1, (2.15)where each space 𝑋_𝑖 is either 1-dimensional or of the form 𝐿₁([0,1]^Λ) and Λ is a finite or infinite set. For each coordinate interval, we consider the Lebesgue measure on the Borel subsets of [0,1] and [0,1]^Λ provided with the product measure on the Borel 𝜎-algebra (see [13]).

We are now ready to provide the main result.

Theorem 2.6. Given an arbitrary measure 𝜇, the following statements are equivalents.
(1)𝜇 is purely atomic.(2)𝒜ℒ^𝑁(𝐿₁(𝜇)) is dense in ℒ^𝑁(𝐿₁(𝜇)) for any number 𝑁.(3)𝒜ℒ^𝑁(𝐿₁(𝜇)) is dense in ℒ^𝑁(𝐿₁(𝜇)) for any number 𝑁≥2.(4)𝒜ℒ²(𝐿₁(𝜇)) is dense in ℒ²(𝐿₁(𝜇)).

Proof. (1)⇒(2). If 𝜇 is purely atomic, then 𝐿₁(𝜇) has the Radon-Nikodym property, and (2) follows from [2, Theorem 1].
(2)⇒(3). This is trivial.
(3)⇒(4). This follows from [8, Proposition 2.1].
(4)⇒(1). Given an arbitrary nonempty set Λ, consider the product [0,1]^Λ of so many copies of [0,1] as indicated by Λ with product measure. We have clearly 𝜇=𝜈⊗𝑚, where 𝜈 is an arbitrary nonzero finite measure and 𝑚 denotes the Lebesgue measure on [0,1]. Then it follows form Lemma 2.1 that 𝒜ℒ²(𝐿₁[0,1]^Λ) is not dense in ℒ²(𝐿₁[0,1]^Λ). Indeed, if 𝜇 is a finite measure satisfying statement (4) of the above theorem, then by Remark 2.5, 𝐿₁(𝜇)≅(⊕_𝑖∈𝐼𝑋_𝑖)_ℓ1 for each 𝑖∈𝐼, where 𝑋_𝑖 is 1-dimensional or of the form 𝐿₁[0,1]^Λ𝑖 for appropriate nonempty set Λ_𝑖 (see [13, Theorem 14]). It follows then from Lemma 2.2 that 𝒜ℒ²(𝑋_𝑖) is dense in ℒ²(𝑋_𝑖) for all 𝑖∈𝐼. But in view of Remark 2.5, none of the spaces 𝑋_𝑖 are of the form 𝐿₁[0,1]^Λ𝑖. Then all 𝑋_𝑖 are 1-dimensional, and then 𝐿₁(𝜇)≅ℓ₁(𝐼), which means that 𝜇 is purely atomic. Finally, if 𝜇 is not necessarily a finite measure satisfying (4) of our theorem, we recall that 𝐿₁(𝜇)≅(⊕_𝑖∈𝐼𝐿₁(𝜇_𝑖))_ℓ1, where 𝜇_𝑖 is a finite measure for all 𝑖∈𝐼. So by Lemma 2.2, we get that 𝒜ℒ²(𝐿₁(𝜇_𝑖)) is dense in ℒ²(𝐿₁(𝜇_𝑖)), and this proves that 𝜇_𝑖 is purely atomic for each 𝑖∈𝐼, which clearly means that 𝜇 is purely atomic.

Remark 2.7. Let us mention the relation between the ℒ²(𝑋), the space of all continuous bilinear forms on the Banach space 𝑋, and 𝐿(𝑋,𝑋^∗), the space of all bounded linear operators from 𝑋 into 𝑋^∗, to see that just consider the canonical identification of ℒ²(𝑋) with 𝐿(𝑋,𝑋^∗). The operator 𝑇∈𝐿(𝑋,𝑋^∗) corresponding to a bilinear form 𝜑∈ℒ²(𝑋) is given by[𝑇(𝑥)](𝑦)=𝜑(𝑥,𝑦)(𝑥,𝑦∈𝑋). (2.16)The bilinear form 𝜑 attains its norm if and only if the operator 𝑇 attains its norm at a point 𝑥∈𝐵_𝑋, that is, 𝑇(𝑥) also attains its norm as a functional on 𝑋, therefore, 𝑇∈𝑁𝐴(𝑋,𝑋^∗) whenever 𝜑∈𝒜ℒ^𝑁(𝑋), but the converse is not true (see [4, 14, 15]). Connecting our main result in this paper with Theorem 1.1, we get a new example of a Banach space 𝑋 such that the set of norm attaining bounded linear operators from 𝑋 into 𝑋^∗ is dense in the space of all bounded linear operators from 𝑋 into 𝑋^∗, but 𝒜ℒ²(𝑋) is not dense in ℒ²(𝑋).

Therefore, the following result is inevitable.

Corollary 2.8. If 𝜇 is a localizable and not purely atomic measure, then the set of norm attaining bounded linear operators from 𝐿₁(𝜇) into 𝐿_∞(𝜇) is dense in the space 𝐿(𝐿₁(𝜇),𝐿_∞(𝜇)) but 𝒜ℒ²(𝐿₁(𝜇)) is not dense in ℒ²(𝐿₁(𝜇)).