Abstract and Applied Analysis

Volume 2014 (2014), Article ID 479208, 15 pages

http://dx.doi.org/10.1155/2014/479208

Research Article

On the Bishop-Phelps-Bollobás Property for Numerical Radius

Sun Kwang Kim,¹ Han Ju Lee,² and Miguel Martín³

¹Department of Mathematics, Kyonggi University, Suwon 443-760, Republic of Korea

²Department of Mathematics Education, Dongguk University-Seoul, Seoul 100-715, Republic of Korea

³Departamento de Análisis Mátematico, Facultad de Ciencias, Universidad de Granada, E-18071 Granada, Spain

Received 30 December 2013; Accepted 14 February 2014; Published 6 April 2014

Academic Editor: Manuel Maestre

Copyright © 2014 Sun Kwang Kim et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We study the Bishop-Phelps-Bollobás property for numerical radius (in short, BPBp-nu) and find sufficient conditions for Banach spaces to ensure the BPBp-nu. Among other results, we show that -spaces have this property for every measure μ. On the other hand, we show that every infinite-dimensional separable Banach space can be renormed to fail the BPBp-nu. In particular, this shows that the Radon-Nikodým property (even reflexivity) is not enough to get BPBp-nu.

1. Introduction

Let be a (real or complex) Banach space and its dual space. The unit sphere of will be denoted by . We write for the space of all bounded linear operators on . For , its numerical radius is defined by

where . It is clear that is a seminorm on . We refer the reader to the monographs [1, 2] for background. An operator attains its numerical radius if there exists such that .

In this paper we will discuss the density of numerical radius attaining operators, actually on a stronger property called Bishop-Phelps-Bollobás property for numerical radius. Let us present first a short account on the known results about numerical radius attaining operators. Motivated by the study of norm attaining operators initiated by J. Lindenstrauss in the 1960s, Sims [3] asked in 1972 whether the numerical radius attaining operators are dense in the space of all bounded linear operators on a Banach space. Berg and Sims [4] gave a positive answer for uniformly convex spaces and Cardassi showed that the answer is positive for , , (where is a metrizable compact), , and uniformly smooth spaces [5–7]. Acosta showed that the numerical radius attaining operators are dense in for every compact Hausdorff space [8]. Acosta and Payá showed that numerical radius attaining operators are dense in if has the Radon-Nikodým property [9]. On the other hand, Payá [10] showed in 1992 that there is a Banach space such that the numerical radius attaining operators are not dense in , which gave a negative answer to Sims’ question. Some authors also paid attention to the study of denseness of numerical radius attaining nonlinear mappings [11–14].

Motivated by the work [15] of Acosta et al. on the Bishop-Phelps-Bollobás property for operators, Guirao and Kozhushkina [16] introduced very recently the notion of Bishop-Phelps-Bollobás property for numerical radius.

Definition 1 (see [16]). A Banach space is said to have the Bishop-Phelps-Bollobás property for numerical radius (in short, BPBp-nu) if, for every , there exists such that, whenever and satisfy and , there exit and such that

Notice that if a Banach space has the BPBp-nu, then the numerical radius attaining operators are dense in . One of the main results of this paper is to show that the converse result is no longer true (Section 5).

It is shown in [16] that the real or complex spaces and have the BPBp-nu. This result has been extended to the real space by Falcó [17]. Aviles et al. [18] give sufficient conditions on a compact space for the real space to have the BPBp-nu which, in particular, include all metrizable compact spaces.

The content of this paper is the following. First, we introduce in Section 2 a modulus of the BPBp-nu analogous to the one introduced in [19] for the Bishop-Phelps-Bollobás property for the operator norm, and we will use it as a tool in the rest of the paper. As easy applications, we prove that finite-dimensional spaces always have the BPBp-nu and that a reflexive space has the BPBp-nu if and only if its dual does. Next, Section 3 is devoted to prove that Banach spaces which are both uniformly convex and uniformly smooth satisfy a weaker version of the BPBp-nu and to discuss such weaker version. In particular, it is shown that spaces have the BPBp-nu for every measure when , . We show in Section 4 that, given any measure , the real or complex space has the BPBp-nu. Finally, we prove in Section 5 that every separable infinite-dimensional Banach space can be equivalently renormed to fail the BPBp-nu (actually, to fail the weaker version). In particular, this shows that reflexivity (or even superreflexivity) is not enough for the BPBp-nu, while the Radon-Nikodým property was known to be sufficient for the density of numerical radius attaining operators.

Let us introduce some notations for later use. The -dimensional space with the norm is denoted by . Given a family of Banach spaces, (resp., ) is the Banach space consisting of all sequences such that each is in and (resp., ) equipped with the norm (resp., ).

2. Modulus of the Bishop-Phelps-Bollobás for Numerical Radius

Analogously to what is done in [19] for the BPBp for the operator norm, we introduce here a modulus to quantify the Bishop-Phelps-Bollobás property for numerical radius.

Notation 1. Let be a Banach space. Consider the set

which is closed in with respect to the following metric:

The modulus of the Bishop-Phelps-Bollobás property for numerical radius is the function defined by

for every . Equivalently, is the supremum of those scalars such that, whenever and satisfy and , there exist and such that

It is immediate that a Banach space has the BPBp-nu if and only if for every . By construction, if a function is valid in the definition of the BPBp-nu, then .

An immediate consequence of the compactness of the unit ball of a finite-dimensional space is the following result. It was previously known to A. Guirao (private communication).

Proposition 2. Let be a finite-dimensional Banach space. Then has the Bishop-Phelps-Bollobás property for numerical radius.

Proof. Let . Then is a norm-closed subspace of . Hence is a finite-dimensional space with two norms:

where is the class of in the quotient space . Hence there is a constant such that

Suppose that does not have the BPBp-nu. Then, there is such that . That is, there are sequences and with such that

By compactness, we may assume that for some and . Hence there exists a sequence in such that . Observe that for every .

By compactness again, we may assume that converges to . This implies that , and , that is, for all . This is a contradiction with the fact that

We may also give the following easy result concerning duality.

Proposition 3. Let be a reflexive space. Then

for every . In particular, has the BPBp-nu if and only if has the BPBp-nu.

We will use that for all , where denotes the adjoint operator of . This result can be found in [1], but it is obvious if is reflexive.

Proof. By reflexivity, it is enough to show that . Let be fixed. If , there is nothing to prove. Otherwise, consider . Suppose that and satisfy

By considering , we may find and such that

Then and satisfy

This implies that . We finish by just taking supremum on .

We do not know whether the result above is valid in the nonreflexive case.

3. Spaces Which Are Both Uniformly Convex and Uniformly Smooth

For a Banach space which is both uniformly convex and uniformly smooth, we get a property which is weaker than BPBp-nu. This result was known to A. Guirao (private communication).

Proposition 4. Let be a uniformly convex and uniformly smooth Banach space. Then, given , there exists such that, whenever with and satisfy , there exist and such that

Proof. Notice that the uniform smoothness of is equivalent to the uniform convexity of . Let and be the moduli of convexity and , respectively. Given , consider

Consider with and satisfying . Define by

for all , where is the scalar satisfying and . Now, choose and such that , , and

Now we define a sequence in inductively. Indeed, suppose that we have a defined sequence for and let

Then choose and such that and :

Notice that, for all , we have

This implies that is a Cauchy sequence and assume that it converges to . Then we have

We will show that both sequences and are Cauchy. From the definition, we have

In summary, we have

Hence

This means that

for all . So and are Cauchy. Let and . Then and . Hence, and

Let , , and . Then we have , , and . Notice that

Therefore

This completes the proof.

Let us discuss a little bit about the equivalence between the property in the result above and the BPBp-nu. For convenience, let us introduce the following definition.

Definition 5. A Banach space has the weak Bishop-Phelps-Bollobás property for the numerical radius (in short weak-BPBp-nu); if given , there exists such that whenever with and satisfy , there exist and such that

Notice that the only difference between this concept and the BPBp-nu is the normalization of the operator by the numerical radius. Of course, if the numerical radius and the operator norm are equivalent, these two properties are the same. This equivalence is measured by the so-called numerical index of the Banach space, as follows. For a Banach space , the numerical index of is defined by

It is clear that and for all . The value means that equals the usual operator norm. This is the case of and , among many others. On the other hand, if and only if the numerical radius is equivalent to the norm of . We refer the reader to [20] for more information and background.

The following result is immediate. We include a proof for the sake of completeness.

Proposition 6. Let be a Banach space with . Then, has the BPBp-nu if and only if has the weak-BPBp-nu.

Proof. The necessity is clear. For the converse, assume that we have satisfying the conditions of the weak-BPBp-nu for all . If with and satisfy for , then there exist and such that

As by the above, let . Then we have

Finally, we have

An obvious change of parameters finishes the proof.

We do not know whether the hypothesis of can be omitted in the above result.

Putting together Propositions 4 and 6, we get the following.

Corollary 7. Let be a uniformly convex and uniformly smooth Banach space with . Then has the BPBp-nu.

Let us comment that every complex Banach space satisfies , so the above corollary automatically applies in the complex case. In the real case, this is no longer true, as the numerical index of a Hilbert space of dimension greater than or equal to two is . On the other hand, it is proved in [21] that real spaces have nonzero numerical index for every measure when . Therefore, we have the following examples.

Example 8. (a) Complex Banach spaces which are uniformly smooth and uniformly convex satisfy the BPBp-nu.

(b) In particular, for every measure , the complex spaces have the BPBp-nu for .

(c) For every measure , the real spaces have the BPBp-nu for , .

Note Added in Revision. Very recently, H. J. Lee, M. Martín, and J. Merí have proved that Proposition 6 can be extended to some Banach spaces with numerical index zero as, for instance, real Hilbert spaces. Hence, they have shown that Hilbert spaces have the BPBp-nu. These results will appear elsewhere.

4. L ₁ Spaces

In this section, we will show that has the BPBp-nu for every measure . In the proof, we are dealing with complex integrable functions since the real case is followed easily by applying the same proof. Our main result here is the following.

Theorem 9. Let be a measure. Then has the Bishop-Phelps-Bollobás property for numerical radius. More precisely, given , there exists (which does not depend on ) such that whenever with and satisfy , then there exist , such that

As a first step, we have to start dealing with finite regular positive Borel measures, for which a representation theorem for operators exists.

Proposition 10. Let be a finite regular positive Borel measure on a compact Hausdorff space . Then has the Bishop-Phelps-Bollobás property for numerical radius. More precisely, given , there is (which is independent of the measure ) such that if a norm-one element in and satisfy , then there exist an operator , such that

To prove this proposition, we need some background on representation of operators on Lebesgue spaces on finite regular positive Borel measures and several preliminary lemmas.

Let be a finite regular positive Borel measure on a compact Hausdorff space . If is a complex-valued Borel measure on the product space , then define their marginal measures on () as follows: and , where and are Borel measurable subsets of .

Let be the complex Banach lattice of measures consisting of all complex-valued Borel measures on the product space such that are absolutely continuous with respect to for , endowed with the norm

Each defines a bounded linear operator from to itself by

where and . Iwanik [22] showed that the mapping is a lattice isometric isomorphism from onto . Even though he showed this for the real case, it can be easily generalized to the complex case. For details, see [22, Theorem 1] and [23, IV Theorem 1.5(ii), Corollary 2].

We will also use that, given an arbitrary measure , every satisfies [24] (that is, the space has numerical index ).

Lemma 11 (see [15, Lemma 3.3]). Let be a sequence of complex numbers with for every , and let such that, for a convex series , . Then for every , the set satisfies the estimate

From now on, will be a finite regular positive Borel measure on the compact Hausdorff space .

Lemma 12. Suppose that there exist a nonnegative simple function and a function such that

Then there exist a nonnegative simple function and a function such that

Proof. Let for some such that for all and , and 's are mutually disjoint. By the assumption, we have

and letting

we have by Lemma 11

For each , we have

This implies that

Define for all ,

and on and elsewhere. Then it is clear that , , and . Finally we will show that . Notice first that

Hence

Lemma 13 (see [25, Lemma 3.3]). Suppose that is a norm-one element in for some and there is a nonnegative simple function such that is a norm-one element of and for some . Then there exist a norm-one bounded linear operator for some and a nonnegative simple function in such that , , and for all .

Lemma 14. Suppose that is a norm-one operator, , where for all and are mutually disjoint Borel subsets of , is a norm-one nonnegative simple function, and is an element of such that

for some and

for all x in the support of .

Then there exist a nonnegative simple function , a function , and an operator in such that

Proof. Since

we have

Let

Then from Lemma 11 we have . Let , where for all . Then

Note that there is a Borel measurable function on such that and for all . Let

Define two measures and as follows:

for every Borel subset of . It is clear that

Since for all , we have

for all , and we deduce that for all .

We claim that for all . Indeed, if , then

So we have

This proves our claim.

We also claim that, for each , there exists a Borel subset of such that

for all . Indeed, set . Then

This shows that . This proves our second claim.

Now, we define by if and if , and we write . It is clear that , , and for all .

Finally, we define the measure

where . It is easy to see that on and elsewhere. Note that

If , then and

Hence, for all , we have

So, we have, for all ,

This gives that . Note also that, for all ,

Hence we get , which implies that . Finally,

We are now ready to present the proof of the main result in the case of finite regular positive Borel measures.

Proof of Proposition 10. Let , , and for some . Suppose that with and that there is an and such that and . Then there is an isometric isomorphism from onto itself such that and there is a scalar number in such that . Then letting , , and , we have

Since , by Lemma 13, there exists a norm-one bounded operator and a nonnegative simple function such that , , and for all . Then

Notice also that

By Lemma 12 there are a nonnegative simple function and a function such that

So we have

By Lemma 14, there exist and and an operator such that

So we have

Let , , and ; then we have

This completes the proof.

Finally, we may give the proof of the main result in full generality.

Proof of Theorem 9. Notice that the Kakutani representation theorem (see [26] for a reference) says that, for every -finite measure , the space is isometrically isomorphic to for some positive Borel regular measure on a compact Hausdorff space. Then, by Proposition 10, there is a universal function which gives the BPBp-nu for for every -finite measure .

Fix . Suppose that with and satisfy

Choose a sequence in such that and let be the closed linear span of

As is separable, there is a dense subset of and let , where is the support of . Then the measure is -finite. Let

be a closed subspace of . It is clear that and is isometrically isomorphic to . So has the BPBp-nu with .

Now, write , consider , , and observe that and . Hence, there exist and such that

Finally consider the operator given by

We have (and so ). Indeed,

for all and . Let and . Then . Moreover, we have

This completes the proof.

5. Examples of Spaces Failing the Bishop-Phelps-Bollobás Property for Numerical Radius

Our goal here is to prove that the density of numerical radius attaining operators does not imply the BPBp-nu. Actually, we will show that, among separable spaces, there is no isomorphic property implying the BPBp-nu other than finite-dimensionality.

We need to relate the BPBp-nu to the Bishop-Phelps-Bollobás property for operators which, as mentioned in the introduction, was introduced in [15]. A pair of Banach spaces has the Bishop-Phelps-Bollobás property for operators (in short, BPBp); if given there exists such that, given with and such that , then there exist and satisfying

We refer the reader to [15, 19, 25] and references therein for more information and background. Among the interesting results on the BPBp, we emphasize that a pair when is finite-dimensional does not necessarily have the BPBp. For instance, if is a strictly convex space which is not uniformly convex, then the pair fails to have the BPBp (this is contained in [15]; see [19, Section  3]).

The next result relates the BPBp-nu to the BPBp for operators in a particular case. We will deduce our example from it.

Theorem 15. If has the BPBp-, then the pair has the BPBp for operators.

Before proving this proposition, we will use it to get the main examples of this section. The first example shows that the density of numerical radius attaining operators does not imply the BPBp-nu.

Example 16. There is a reflexive space (and so numerical radius attaining operators on it are dense) which fails to have the BPBp-nu. Indeed, let be a reflexive separable space which is not superreflexive and we may suppose that is strictly convex. Observe that cannot be uniformly convex since it is not superreflexive. Now, is reflexive, but the pair fails the BPBp since is strictly convex but not uniformly convex [19, Corollary 3.3]. Therefore, Theorem 15 gives us that does not have the BPBp-nu.

The example above can be extended to get the result that every infinite-dimensional separable Banach space can be renormed to fail the BPBp-nu. This follows from the fact that every infinite-dimensional separable Banach space can be renormed to be strictly convex but not uniformly convex (this result can be proved “by hand”; an alternative categorical argument for it can be found in [27] and references therein). With a little more of effort, we may get the main result of the section.

Theorem 17. Every infinite-dimensional separable Banach space can be renormed to fail the weak-BPBp-nu (and so, in particular, to fail the BPBp-nu).

We need the following result which is surely well known. As we have not found a reference, we include a nice and easy proof kindly given to us by Vladimir Kadets. We recall that, given a Banach space , the set of all equivalent norms on can be viewed as a metric space using the Banach-Mazur distance.

Lemma 18. Let be an infinite-dimensional separable Banach space. Then the set of equivalent norms on which are strictly convex and are not (locally) uniformly convex is dense in the set of all equivalent norms on (with respect to the Banach-Mazur distance).

Proof. Fix and such that . For a fixed , denote

Evidently, for every . Fix a sequence of norm-one functionals separating the points of , and denote

Then, is a strictly convex norm on , , and for all . Finally, write

Then, is a strictly convex norm on and

We will finish the proof by showing that is not uniformly convex (actually, it is not locally uniformly convex). Indeed, for each we select with and consider . Then, , , and . At the same time, , so and . Consequently,

but , which means the absence of local uniform convexity at .

Proof of Theorem 17. Let be an infinite-dimensional separable Banach space. Take a closed subspace of of codimension two. By [28, Proposition 2], the map carrying every equivalent norm on to its numerical index is continuous and so, the set of values of the numerical index of up to reforming is a nontrivial interval [28, Theorem 9]. Then Lemma 18 allows us to find an equivalent norm on in such a way that is strictly convex and is not uniformly convex, and . Now, the space is an equivalent renorming of which does not have the BPBp- (indeed, otherwise, the pair would have the BPBp for the operator norm and so, would be uniformly convex by [19, Corollary 3.3], a contradiction.) Moreover, as

(see [20, Proposition 2], for instance), also fails the weak-BPBp-nu by Proposition 6.

To finish the section with the promised proof of Theorem 15, we first see the following stability result.

Lemma 19. Let or . If has the Bishop-Phelps-Bollobás property for numerical radius with a function , then each Banach space has the Bishop-Phelps-Bollobás property for numerical radius with . That is, for all .

Proof. Let and be the natural projections, and let and be the natural embeddings.

Assume that an operator and a pair satisfy that

We define an operator and by

then clearly we see that

From the assumption, there exist and a pair such that

Since this clearly shows that

we only need to show that .

We first show the case of sum. Since

for every , we have

This shows that and for every . So and . This and the fact that imply that

So we get that . Hence, we have

and so we get .

We next show the case of sum. The proof is almost the same as that of the case. However, for the sake of completeness, we provide it here.

Since for every , we have

which shows and for every . Since this implies , we get that . Hence, we have

and so .

Proof of Theorem 15. Note that as . Fix and choose such that . Let .

Suppose that with and satisfy

For any measurable subset , let

with the norm . Then it is easy to see that is isometrically isomorphic to a complemented subspace of . Let be the restriction defined by for all and let be the extension defined by if and otherwise. It is clear that and for all . Notice also that is isometrically isomorphic to .

Let and . Then

and define the operator by for every . Then,

Since is -finite, . Let be a function such that , and choose such that . Define the operator by

and observe that . Indeed,

It is immediate that

By Lemma 19, has the BPBp-nu with the function . Therefore, there exist , , and such that

Claim 1. We claim that .

Otherwise,

We deduce that

Since , we get that

a contradiction. This proves the claim.

We define the operator by for every and for every . Then we have

Indeed, from Claim 1, we have

On the other hand, we see that

for every . Therefore, . Also,

Claim 2. There exists such that , , for every , and .

Indeed, write , where and . We have that

This implies that

Therefore, equals on the support of . As , we also have that equals on the support of . Changing the values of by the ones of on , we may and do suppose that on the whole .

We also have . Indeed,

So we have

and .

Finally define the operator by

for . It is clear that

Notice also that

for all . Hence we have

On the other hand, let be defined by for every . Then, we have

where is the family of measurable subsets of .

Hence, for any simple function , where is a family of disjoint measurable subsets with strictly positive measure, we have

Since , is an isometric isomorphism, so for each there exists a sequence of norm-one simple functions such that converges to . Therefore,

On the other hand, we see that

Therefore, which proves Claim 2.

Finally, set for a suitable and define the operator by

for every . Then, we have

for every , so . Also,

so attains its norm at , and

We also have that, for any ,

Hence .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors thank Gilles Godefroy and Rafael Payá for fruitful conversations about the content of this paper and Vladimir Kadets for providing Lemma 18. They also appreciate anonymous referees for careful reading and fruitful suggestions about revision. The second author was 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 third author was partially supported by Spanish MICINN and FEDER project no. MTM2012-31755 and by Junta de Andalucía and FEDER Grants FQM-185 and P09-FQM-4911.

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