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Sun Kwang Kim, Han Ju Lee, Miguel Martín, "On the Bishop-Phelps-Bollobás Property for Numerical Radius", Abstract and Applied Analysis, vol. 2014, Article ID 479208, 15 pages, 2014. https://doi.org/10.1155/2014/479208
On the Bishop-Phelps-Bollobás Property for Numerical Radius
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.
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  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  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 . Acosta and Payá showed that numerical radius attaining operators are dense in if has the Radon-Nikodým property . On the other hand, Payá  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  of Acosta et al. on the Bishop-Phelps-Bollobás property for operators, Guirao and Kozhushkina  introduced very recently the notion of Bishop-Phelps-Bollobás property for numerical radius.
Definition 1 (see ). 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  that the real or complex spaces and have the BPBp-nu. This result has been extended to the real space by Falcó . Aviles et al.  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  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  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 , 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  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.
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  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. L1 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  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  (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
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