# On Banach spaces with the approximate hyperplane series property

Yun Sung Choi Department of Mathematics, POSTECH, Pohang (790-784), Republic of Korea Sun Kwang Kim Department of Mathematics, Kyonggi University , Suwon 443-760, Republic of Korea Han Ju Lee Department of Mathematics Education, Dongguk University - Seoul, 100-715 Seoul, Republic of Korea  and  Miguel Martín Departamento de Análisis Matemático, Facultad de Ciencias, Universidad de Granada, E-18071 Granada, Spain
July 25th, 2014
###### Abstract.

We present a sufficient condition for a Banach space to have the approximate hyperplane series property (AHSP) which actually covers all known examples. We use this property to get a stability result to vector-valued spaces of integrable functions. On the other hand, the study of a possible Bishop-Phelps-Bollobás version of a classical result of V. Zizler leads to a new characterization of the AHSP for dual spaces in terms of -continuous operators and other related results.

###### Key words and phrases:
Banach space, approximation, norm-attaining operators, Bishop-Phelps-Bollobás theorem.
###### 2000 Mathematics Subject Classification:
Primary 46B20; Secondary 46B04, 46B22
The first author was supported by by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (No.2010-0008543 and No. 2013053914). The third author partially supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology (NRF-2012R1A1A1006869). The fourth author partially supported by Spanish MICINN and FEDER project no. MTM2012-31755 and Junta de Andalucía and FEDER grants FQM-185 and P09-FQM-4911.

## 1. Introduction

Given two (real or complex) Banach spaces and , we write for the Banach space of all bounded linear operators from into , endowed with the operator norm. We use the symbol and to denote, respectively, the closed unit ball and the unit sphere of . The topological dual of is denote by . An operator is said to be norm-attaining if there is such that . We write for the subset of consisting on all norm-attaining operators. After the celebrated Bishop-Phelps theorem, which says that the set of all norm-attaining linear functionals is dense in , it was a natural question to study whether the set of norm-attaining linear operators is dense in for all Banach spaces and . In 1963, Lindenstrauss [12] gave a negative answer to this question and showed, among many other things, that is dense in for all Banach spaces .

Bollobás sharpened the Bishop-Phelps theorem to which is now called the Bishop-Phelps-Bollobás theorem: given a Banach space , if and satisfy for some , then there exist and such that , and . See [7] for a sharper version of this result. In 2008, Acosta, Aron, García and Maestre [1] introduced the notion of Bishop-Phelps-Bollobás theorem for operators. Precisely, a pair of Banach spaces is said to have the Bishop-Phleps-Bollobás property (BPBp in short) if given , there is such that whenever with and satisfy , there exist and such that

 ∥S−T∥<ε,   ∥x−y∥<ε,   and    ∥S∥=∥Sy∥=1.

In the same paper [1], the authors characterize Banach spaces for which the pair () has the BPBp in terms of convex series and use the result to show that there is a reflexive Banach space such that does not have the BPBp (while, as commented above, is dense in for every ). Specifically, has the BPBp if and only if has the approximate hyperplane series property (AHSP): for every there exists such that for every sequence and for every convex series with

 ∥∥ ∥∥∞∑k=1αkyk∥∥ ∥∥>1−η(ε),

there exist , and satisfying

• ,

• and for all .

Let us remark that the definition of the AHSP does not change if we replace infinite sequences by finite (but of arbitrary length) sequences.

The following spaces are known to have the AHSP [1, 8]: finite dimensional spaces, uniformly convex spaces, spaces with property (see definition in section 2), and the so-called lush spaces (see definition in section 2). In particular, spaces, spaces, the disc algebra and , all have the AHSP.

In section 2 of the present paper, we introduce a sufficient condition for the AHSP, called the AHP (see Definition 2.1), which actually subsumes all previously known examples. We use the AHP to present new examples of spaces with the AHSP, namely, the spaces when is an arbitrary measure and is finite-dimensional, uniformly convex, lush or has property .

Finally, we start section 3 by showing that there is no BPB version of the theorem of Zizler [16] which states that the set of operators between two Banach spaces whose adjoints attain their norm is dense in the space of all operators. Then, we exploit this idea getting several results: a new characterization of the AHSP for dual spaces in terms of -continuous operators, a characterization of pairs having the AHSP (a concept recently introduced in [2], see its definition in section 3), and a proof that has the AHSP for every locally compact Hausdorff space and every uniformly smooth space , thereby extending a result in [2] to the vector-valued case.

Acknowledgement: This research was initiated during the visit of the first, third and fourth authors to the Department of Mathematical Sciences, Kent State University in 2012. They would like to thank the department and specially Richard Aron for their hospitality and fruitful discussions about this research topic.

## 2. A sufficient condition for the AHSP

We devote this section to study a sufficient condition for the AHSP, which actually covers all known examples, and which will be useful to provide new examples of the form . We need some notation. Let be a Banach space. A face of is a non-empty subset of the form

 F(x∗):={x∈SX:Rex∗(x)=1}

for a suitable attaining its norm. A subset is said to be norming if

 ∥x∥=sup{|x∗(x)|:x∗∈C}(x∈X)

and it is said to be rounded if for every with .

###### Definition 2.1.

A Banach space is said to have the approximate hyperplane property (AHP in short) if there exist a function and a norming subset of for satisfying the following:
Given , there is a function such that if and satisfy , then .

Observe that by a routine argument, we may suppose the set to be rounded. It is also straightforward to prove that it is enough to check the property just for a dense subset of .

As announced in the introduction, we show that the AHP implies the AHSP. Actually, more can be said.

###### Proposition 2.2.

Let be a Banach space. Suppose that there is a function such that for every finite-dimensional subspace of , there exists a subspace of containing and having the AHP with the function . Then, has the AHSP. In particular, the AHP implies the AHSP.

###### Proof.

Let and write . Consider a finite convex combination of elements of satisfying . By hypothesis, there is a subspace of with the AHP containing the subspace spanned by ; we denote by the norming subset of given by the AHP of , which we may and do suppose that it is rounded. Then, there exists such that

 Rex∗(n∑j=1αjxj)=Ren∑j=1αjx∗(xj)>1−δ1(ε)2.

Setting

 A={j:1⩽j⩽n, Rex∗(xj)>1−δ1(ε)},

we can get easily that

 ∑j∈Aαj>1−δ1(ε)⩾1−ε

(see the proof of [11, Proposition 2.1]). By the AHP of , for all . Letting be a Hahn-Banach extension of to the whole of , then we get for , as desired. ∎

Examples of spaces with AHP appeared already (without this name) in the literature. Actually, the usual way to provide examples of Banach spaces with the AHSP has been to prove that they have the AHP. Let us discuss the main examples.

###### Proposition 2.3.

Every uniformly convex Banach space has the AHP. Besides, here is the whole dual unit sphere and is the identity.

This result appeared in [8, Lemma 13] (without this notation) and also follows easily from [2, Lemma 2.1].

Next, it is shown in [1, Lemma 3.4] that every finite-dimensional Banach space has the AHP.

###### Proposition 2.4.

Every finite-dimensional Banach space has the AHP. Besides, here is the whole dual unit sphere but is not, in general, equal to the identity.

It follows from the result above, Proposition 2.2 and the results in [4, §4], that property AHP is not stable by infinite -, - or -sums.

###### Example 2.5.

AHP is not stable by infinite -, - or -sums.  Indeed, a sequence of finite-dimensional spaces is presented in [4, §4] such that its -, - and -sums fail the AHSP. Now, all the ’s have the AHP (Proposition 2.4) and the sums fail it by Proposition 2.2.

The next family of examples we present here is the one of lush spaces. A Banach space is said to be lush [6] if for every and every , there is a slice with such that and the distance from to the absolutely convex hull of is smaller than . We refer to [5, 6, 10, 11] for information about lush spaces. Among lush spaces we may find spaces, spaces and their isometric preduals, the disk algebra, , and finite-codimensional subspaces of .

###### Proposition 2.6.

Every separable lush space has the AHP. Besides, the function does not depend on the particular lush space, is not the whole dual unit sphere and is the identity.

A proof of this result is contained in the proof of [11, Proposition 2.1.c]. We are going to comment this proof here for the sake of completeness and also in order to extend the result to some non-separable spaces. Indeed, let be a separable lush space. By [10, Theorem 4.3] and [3, Corollary 3.5], there exists a rounded subset of norming for such that

 (1) BX=¯¯¯¯¯¯¯¯¯¯¯¯¯¯aconv(F(x∗))

for every , where denotes the absolutely closed convex hull. With this in mind, one may follows the proof of [11, Proposition 2.1.c] to get Proposition 2.6 with a function (which is independent of ), the norming set and equals to the identity.

Therefore, the key ingredient of the proof is to get the existence of a norming set such that (1) holds for every element of . Another family of spaces for which this happens is the one of almost-CL-spaces. A Banach space is said to be an almost-CL-space if is the closed absolutely convex hull of every maximal convex subset of . We refer the reader to [13] and references therein for more information and background. Almost-CL-spaces are lush, but the converse is not true [6]. The main examples of almost-CL-spaces are -spaces and -spaces. It is easy to show (see [13, §2]) that if is an almost-CL-space, then there is a rounded subset of norming for such that (1) holds for every element of . By the comments above, almost-CL-spaces have the AHP.

###### Proposition 2.7.

Every almost-CL-space has the AHP. Besides, the function does not depend on the particular almost-CL-space, is not the whole dual unit sphere and is the identity.

We may particularize this result to the main examples of almost-CL-spaces.

###### Corollary 2.8.

All -spaces and all -spaces have the AHP. Besides, the function does not depend on the particular space, is not the whole dual unit sphere and is the identity.

We do not know if this result extends to general non-separable lush spaces, but a reduction to the separable case of lushness property allows to get this weaker version.

###### Proposition 2.9.

There exists a function such that for every lush space and every separable subspace of , there is a (separable) subspace of containing and having the AHP with the function .

###### Proof.

Let the universal function provided in Proposition 2.6. Let be a lush space and a separable subspace of . We use [5, Theorem 4.2] to get a separable subspace of containing which is lush. Now, Proposition 2.6 gives that has the AHP with the function , as required. ∎

The last family of examples of spaces with the AHSP is given by property . A Banach space has property if there are two sets , and a constant such that the following conditions hold:

1. , .

2. if .

3. for every .

This property was introduced by J. Lindenstrauss [12] in his study of norm-attaining operators. We refer to [14] and references therein for more information and background. It is known that if has property , then has the BPBp for every Banach space [1, Theorem 2.2]. In particular, has the BPBp and so has the AHSP. Actually, property implies AHP.

###### Proposition 2.10.

Property implies property AHP. Besides, the function only depends on the constant of the definition of property , is not the whole dual unit sphere, and the function is the identity.

###### Proof.

Suppose has property with constant and consider the sets , given in the definition of the property. We write , which is a norming set for , and for , we consider such that

 δ(1+ρ)+2ρ√2δ1−(1−δ)ρ+ρ√2δ<ε.

Now, we fix and consider any such that . We write

 a=1−ρ1−(1−δ)ρ+ρ√2δ∈(0,1] and b=x∗i(x0).

Then and so . Consider the vector

 x=ax0+(1−ab)xi∈X.

Observe that, clearly, and that

 ∥x0−x∥ ⩽(1−a)∥x0∥+|1−ab|∥xi∥ =(1−a)+(1−aReb)+a|Im b|<1−a+(1−a(1−δ))+a√2δ =δ(1+ρ)+2ρ√2δ1−(1−δ)ρ+ρ√2δ<ε.

It remains to show that for which it suffices to check that for every . Indeed, fix , and observe that

 |x∗j(x)| ⩽a|x∗j(x0)|+(1−ab)|x∗j(xi)|⩽a+|1−ab|ρ

It is now time to present the main result of the section, namely the lifting property of the AHP from to , which we will use to get a lifting property of the AHSP. Recall that is the space of all strongly measurable functions with

 ∥f∥1=∫Ω∥f(ω)∥dμ<∞.

That is, is the completion of the space of all simple functions with support of finite measure.

###### Theorem 2.11.

Let be a measure space and let be a Banach space. Suppose that there exists a function such that for every separable subspace of , there exists a subspace of containing and having AHP with function . Then, there exists a function such that every separable subspace of is contained in a subspace of which has the AHP with the function . Moreover, if has the AHP, then has the AHP.

###### Proof.

Let be a separable subspace of . Then it is contained in , where is a separable subspace of . So we are done if we assume that the whole space has the AHP and prove that does. Let be the subset of norming for and and the functions, given by the definition of the AHP. First, observe that the set of those elements in of the form

 N∑i=1x∗iχAi

where is a disjoint family of measurable subsets with and for all , is norming for . Now, fix . For , define

 ΥL1(μ,X),ε(ϕ)=N∑i=1ΥX,ε/2(x∗i)χAi

and observe that this definition does not depend on the particular decomposition of . Write . Consider and a simple function with such that (as simple functions are dense in , it is enough tho check the property for them). Then has the form where is a disjoint family of measurable subsets with and for every . Besides, considering a finer partition if needed, we may and do suppose that .

Next, let . Then, by the AHP of , for each , there is such that . Hence, by setting , we have

 ΥX,ε/2(x∗i)(yi)=∥xi∥=∥yi∥and% ∥yi−xi∥<ε/2∥xi∥(i∈E).

By the assumption, we have

 1−δ1(ε)

where . Hence

 ∑i∈ERex∗i(xi)μ(Ai)>1−δ1(ε)δ(ε/2)=1−ε4.

In particular, and is not empty. Finally, let in . Then

 ΥL1(μ,X),ε(ϕ)(f)=1β∑i∈EΥX,ε/2(x∗i)(yi)μ(Ai)=1β∑i∈E∥xi∥μ(Ai)=1=∥f∥

and

 ∥g−f∥ ⩽∑i∈E∥∥∥yiβ−xi∥∥∥μ(Ai)+∑i∈Ec∥xi∥μ(Ai) ⩽∑i∈E∥∥∥yiβ−yi∥∥∥μ(Ai)+∑i∈E∥xi−yi∥μ(Ai)+∑i∈Ec∥xi∥μ(Ai) ⩽∑i∈Eε2∥xi∥μ(Ai)+2(1−β)<ε.\qed

We particularize the above result to the known examples of spaces with the AHP. This generalizes [8, Theorem 14], where the result was only given for uniformly convex ’s.

###### Corollary 2.12.

Let be a measure space and let be a Banach space. Then has the AHP (and so the AHSP), provided either of the following holds:

1. is finite-dimensional.

2. is uniformly convex.

3. is lush and separable.

4. is an almost-CL-space.

5. has property .

For non-separable lush spaces we have the following result.

###### Corollary 2.13.

Let be a measure space and let be a (non-separable) lush space. Then every separable subspace of is contained in a separable subspace of with the AHP and the function in the definition of the AHP does not depend on subspaces. In particular, has the AHSP.

Let us observe that finite dimensionality, uniform convexity and property do not pass from to if is non-trivial. Whether is lush for every lush space is not known to the best of our knowledge.

## 3. On a possible extension of a theorem by Zizler

It is proved in [1, Example 6.3] that the classical Lindenstrauss theorem, proving the density of the set of those operators acting between two arbitrary Banach spaces whose second adjoint attain the norm, has no Bishop-Phelps-Bollobás counterpart. We may wonder if the result by Zizler providing the density of the set of operators whose first adjoint attains the norm has a Bishop-Phelps-Bollobás counterpart. More concretely, we may ask about the validity of the following Bishop-Phelps-Bollobás version of Zizler’s result:

Given a pair of Banach spaces, is there a function such that for every , whenever with and satisfy , then there exist with and such that , and ?

The following easy example shows that this is not always possible.

###### Example 3.1.

Let be a smooth reflexive space whose dual is not super-reflexive and let . Then the question above has a negative answer for . Indeed, since and are reflexive, the above question is equivalent to whether has the BPBp. Since is strictly convex, this would imply to be uniformly convex by [4, Corollary 3.3]. This is impossible since is not super-reflexive.

We observe that it is immediate that if is a reflexive space, then for every Banach space the question of whether a pair satisfies the BPB version of Zizler result is equivalent to the question of whether has the BPBp. (This follows because every operator from to is automatically --continuous.)

Next, we investigate the Bishop-Phelps-Bollobás version of Zizler result when the range space is . We have the following result, whose proof is based on [1, Theorem 4.1].

###### Proposition 3.2.

Given a Banach space , has the AHSP if and only if there is a function such that for every , whenever with and satisfy , then there exist with and such that , and .

We restate this result as the following corollary.

###### Corollary 3.3.

Let be a dual space. Then the pair has the BPBp if and only if there is a function such that for every , whenever a --continuous and satisfy , then there exist a --continuous and such that , and (that is, in the definition of BPBp we may restrict ourselves to --continuous operators.)

###### Proof of Proposition 3.2.

This proof is based on the one of [1, Theorem 4.1]. But for the sake of completeness we give details. For convenience, let be the basis of and be the basis of .

Suppose that has the AHSP with a function . We fix with and satisfying . Since the set of finite convex sum of is dense in , we may assume that . Moreover, by composing with an appropriate --continuous isometry, we may assume that for all . Since

 1−η(ε)<∥T∗0(y∗0)∥=∥∥ ∥∥n∑i=1αiT∗0(e∗i)∥∥ ∥∥,

we apply the definition of AHSP to find , , and such that

 ∑i∈Aαi>1−ε,∥T∗(e∗i)−x∗i∥<εandx∗∗(x∗i)=1(i∈A).

Define and by

 S(x)=∑i∈Ax∗i(x)ei+∑i∈N∖AT∗(e∗i)(x)ei  (x∈X),andy∗=∑i∈Aαie∗i∑i∈Aαi.

We can see that for every and for every . Therefore, we get that and . Moreover, , and so completes our proof.

For the converse, given , choose such that

 0<√2(1−ρ)<ε/2,0<ε′<ε/2andε′1−ρ<ε/2.

Consider a finite convex series and a sequence satisfying . Write and define the operator by

 S0(x)=n∑i=1x∗i(x)ei(x∈X)

which clearly satisfies . Since , by hypothesis there exist with and such that

 ∥S∗(z∗)∥=1,∥S−S0∥<ε′,and∥z∗0−z∗∥<ε′.

It then follows that

 n∑i=1(αi−Rez∗(i))<∥z∗0−z∗∥<ε′,

and so . Set . Since

 1−ε′

we get . Hence,

 ∑i∈Aαi ⩾∑i∈ARez∗(i)−∥z∗0−z∗∥ ⩾1−ε′1−ρ−ε′>1−ε.

On the other hand, choose so that

 1=x∗∗S∗(z∗)=∞∑i=1x∗∗(z∗(i)S∗(e∗i)).

Set for every and observe that for each . Since

 ∣∣∣1−z∗(i)|z∗(i)|∣∣∣<√2(1−ρ)<ε/2

for every , we get

 ∥x∗i−y∗i∥ =∥∥∥z∗(i)|z∗(i)|S∗(e∗i)−S∗0(e∗i)∥∥∥ ⩽∥∥∥z∗(i)|z∗(i)|S∗(e∗i)−S∗(e∗i)∥∥∥+∥S∗(e∗i)−S∗0(e∗i)∥ <ε/2+ε/2=ε.

Finally, , the set , the sequence and the functional complete our proof. ∎

Very recently, another “flavor” of the approximate hyperplane series property has been studied, namely the AHSP for a pair . This property was introduced in [2] to characterize those Banach spaces such that has the BPBp for bilinear forms.

###### Definition 3.4 ([2]).

Let be a Banach space. We say that the pair has the approximate hyperplane series property for dual pairs (AHSP) if for every there exists such that for every convex series and for every sequence and with

 Re∞∑n=1αkx∗k(x0)>1−η(ε)

there exist a subset , a subset and satisfying

• ,

• , for all , and for all .

It is clear that by assuming the condition above just for finite sequences and , an equivalent property is obtained. As we already remarked, it is shown in [2] that has the BPBp for bilinear forms if and only if has the AHSP. It follows directly from the definition that if a pair has the AHSP, then has the AHSP. However the converse is not true, since has the AHSP but the pair does not have the AHSP since does not have the BPBp for bilinear forms [9]. Moreover, it is known that no pair of the form has the AHSP in the infinite dimensional case [2]. On the other hand, the pair has the AHSP for the following spaces: finite dimensional , uniformly smooth , ,
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