# Norm Attaining Multilinear Forms onL1(μ)

Authors: Yousef Saleh
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

Given an arbitrary measureμ, 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.


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## Body

## 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,…,XN,
we denote by ℒN(X1,…,XN) the space of all continuous N-linear mappings
from X1×⋯×XN into the scaler field. We say that φ∈ℒN(X1,…,XN) attains its norm if there is xi∈BXi (the unit ball of Xi) for i=1,2,…,N,
such that(1.1)|φ(x1,…,xN)|=∥φ∥=sup{|φ(y1,…,yN)|:(y1,…,yN)∈BX1×⋯×BXN},and we denote by 𝒜ℒN(X1,…,XN) the set of all norm attaining N-linear forms. In the case where X1=⋯=XN=X,
we write simply ℒN(X) and 𝒜ℒN(X).Aron et al. [2] posed the
question of when 𝒜ℒN(X) is dense in ℒN(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 measureμ, 𝒜ℒN(L1(μ)) is dense in ℒN(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 measureμ 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 ofL1-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 form(2.1)L1(μ)≅(⊕i∈IL1(μi))ℓ1where μi 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 {νj:j∈J} such that(2.2)L∞(ν)≅(⊕j∈JL∞(νj))ℓ∞.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,(2.3)ℒ2(L1(μ)≅L(L1(μ),L1(μ)*)≅L(L1(μ),L∞(μ))≅L∞(μ⊗μ);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:(2.4)h^(f,g)=∫Ω×Ωh(x,y)f(x)g(y)dμ(x)dμ(y),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 caseN=2.
The proof for the general case is exactly the same.Lemma 2.1 (see [10, Lemma 2.1]).
Letν 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 withl1-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]).
LetY,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 functionsh∈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 setsA,B∈𝒜 with μ(A)>0, μ(B)>0 and a scalar t with |t|=1 such that(2.5)th(x,y)=∥h∥∞for [μ⊗μ]-almost every (x,y)∈A×B.(2)
There are setsA,B like in (1) and measurable functions φ,ψ on Ω such that(2.6)|φ(w)|=|ψ(w)|=1,where w∈Ω and φ(x)ψ(y)h(x,y)=∥h∥∞,
for [μ⊗μ]-almost every (x,y)∈A×B.(3)
The bilinear formh^∈ℒ2(L1(μ)) corresponding to h∈L∞(μ⊗μ) attains its norm.
(2.7)Then(1)⇒(2)⟺(3).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=φχA/μ(A),g=ψχA/μ(B) where f,g  are in the unit sphere of L1(μ), χA,χB denote the characteristic functions on A and B, respectively, and(2.8)h^(f,g)=1μ(A)μ(B)∫A×Bh(x,y)φ(x)ψ(y)dμ(x)dμ(y)=1μ(A)μ(B)∫A×B∥h∥∞d(μ⊗μ)=∥h∥∞.
(3)⇒(2) Let f,g∈L1(μ) be such that ∥f∥1=∥g∥1=1 and h^(f,g)=∥h∥∞. Take(2.9)A={x∈Ω:f(x)≠0},B={y∈Ω:g(y)≠0}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(2.10)∥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∥∞,from which we conclude that(2.11)h(x,y)φ(x)ψ(y)=∥h∥∞for [μ⊗μ]-almost every (x,y)∈A×B.
In the real case, the functionsφ,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 caseh=χE,
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 subsetsA,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 rectangleA×B is contained in the set E.Proof.
(1)⇒(2).
This is trivial.
(2)⇒(3).
Let h∈L∞(μ⊗μ) be such that ∥χE−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 that(2.12)φ(x)ψ(y)h(x,y)=∥h∥∞,then(2.13)|h(x,y)|=∥h∥∞>12,for [μ⊗μ]-almost every (x,y)∈A×B.
Hence(2.14)|χE(x,y)|≥|h(x,y)|−|h(x,y)−χE(x,y)|>12−∥h−χE∥∞>0.for [μ⊗μ]-almost every (x,y)∈A×B, from which we get that χE=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 χE=1=∥χE∥∞, for [μ⊗μ]-almost every (x,y)∈A×B, then the function f=χE 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 forL1-spaces. Indeed, if ν is a finite measure, we may write(2.15)L1(ν)=(⊕iIXi)ℓ1,where each space Xi 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 measureμ,
the following statements are equivalents.
(1)
μ is purely atomic.(2)
𝒜ℒN(L1(μ)) is dense in ℒN(L1(μ)) for any number N.(3)
𝒜ℒN(L1(μ)) is dense in ℒN(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∈IXi)ℓ1 for each i∈I,
where Xi is 1-dimensional or of the form L1[0,1]Λi for appropriate nonempty set Λi (see [13, Theorem 14]). It follows then from
Lemma 2.2 that 𝒜ℒ2(Xi) is dense in ℒ2(Xi) for all i∈I.
But in view of Remark 2.5, none of the spaces Xi are of the form L1[0,1]Λi.
Then all Xi 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(μi))ℓ1,
where μi is a finite measure for all i∈I.
So by Lemma 2.2, we get that 𝒜ℒ2(L1(μi)) is dense in ℒ2(L1(μi)),
and this proves that μi is purely atomic for each i∈I,
which clearly means that μ is purely atomic.Remark 2.7.
Let
us mention the relation between theℒ2(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(2.16)[T(x)](y)=φ(x,y)(x,y∈X).The bilinear form φ attains its norm if and only if the operator T attains its norm at a point x∈BX,
that is, T(x) also attains its norm as a functional on X,
therefore, T∈NA(X,X*) whenever φ∈𝒜ℒN(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.
Ifμ 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)).



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