Open Access

Denseness of Numerical Radius Attaining Holomorphic Functions

Journal of Inequalities and Applications20092009:981453

DOI: 10.1155/2009/981453

Received: 21 June 2009

Accepted: 6 October 2009

Published: 8 October 2009

Abstract

We study the density of numerical radius attaining holomorphic functions on certain Banach spaces using the Lindenstrauss method. In particular, it is shown that if a complex Banach space is locally uniformly convex, then the set of all numerical attaining elements of is dense in .

1. Introduction

Let be a complex Banach space and its dual space. We consider the topological subspace of the product space , equipped with norm and weak- topology on the unit ball of and its dual unit ball , respectively. It is easy to see that is a closed subspace of .

For two complex Banach spaces and , denote by the Banach space of all bounded continuous functions from to with sup norm . We are interested in the following two subspaces of :
(1.1)

We denote by either or . When , we write instead of . A nonzero function is said to be a strong peak function at if whenever there is a sequence in with , the sequence converges to . The corresponding point is said to be a strong peak point of . It is easy to see that is a strong peak point of if and only if is a strong peak point of . By the maximum modulus theorem, it is easy to see that if is a strong peak point of and , then is contained in the unit sphere of .

Harris [1] introduced the notion of numerical radius of holomorphic function . More precisely, for each , . An element is said [2] to be numerical radius attaining if there is such that .

Acosta and Kim [2] showed that if is a complex Banach space with the Radon-Nikodým property, then the set of all numerical radius attaining elements in is dense. In this paper, we show that if is a locally uniformly convex space or locally uniformly -convex, order continuous, sequence space, then the set of all numerical radius attaining elements in is dense.

We need the notion of numerical boundary. The subset of is said [3] to be a numerical boundary of if for every , . For more properties of numerical boundaries, see [35].

2. Main Results

The following is an application of the numerical boundary to the density of numerical radius attaining holomorphic functions. Similar application of the norming subset to the density of norm attaining holomorphic functions is given in [4]. We use the Lindenstrauss method [6].

Theorem 2.1.

Suppose that is a Banach space and there is a numerical boundary of such that for every , is a strong peak point of . Then the set of numerical radius attaining elements in is dense.

Proof.

We may assume that and for each , where each is a strong peak function in . Notice that if and , then for any and attains its numerical radius. Hence we have only to show that if and , then there is such that attains its numerical radius and

Let with and with be given. We choose a monotonically decreasing sequence of positive numbers so that
(2.1)
We next choose inductively sequences , in satisfying
(2.2)
(2.3)
(2.4)
(2.5)
where each is chosen in to satisfy and each is for some positive integer . Having chosen these sequences, we verify that the following hold:
(2.6)
(2.7)
(2.8)
(2.9)
Assertion (2.6) is easy by using induction on . By (2.3) and (2.4),
(2.10)
so the relation (2.7) is proved. Therefore (2.8) is an immediate consequence (2.1) and (2.7). For , by the triangle inequality, (2.3) and (2.6), we have
(2.11)
Hence by (2.4) and (2.7),
(2.12)
so that
(2.13)

and this proves (2.9). Let be the limit of in the norm topology. By (2.1) and (2.6), holds. The relations (2.5) and (2.9) mean that the sequence converges to a point , say and by (2.3), we have , where is a weak- limit point of in . Then it is easy to see that . Hence attains its numerical radius. This concludes the proof.

Recall that a Banach space is said to be locally uniformly convex if and there is a sequence in satisfying , then .

Corollary 2.2.

Let be a locally uniformly convex Banach space. Then the set of numerical radius attaining elements in is dense.

Proof.

Let and notice that every element in is a strong peak point for . Indeed, if , choose so that . Set for . Then and . If for some sequence in , then . Since for every , and as . By Theorem 2.1, we get the desired result.

It was shown in [7] that if a Banach sequence space is locally uniformly -convex and order continuous, then the set of all strong peak points for is dense in . Therefore, the set of all strong peak points for is dense in . For the definition of a Banach sequence space and order continuity, see [8, 9]. For the characterization of local uniform -convexity in function spaces, see [7, 10].

Corollary 2.3.

Suppose that is a locally uniformly -convex order continuous Banach sequence space. Then the set of numerical radius attaining elements in is dense.

Proof.

Let . Then by [11, Theorem  2.5], and the remark above the Corollary 2.3, is a numerical boundary of . Hence the proof is complete by Theorem 2.1.

Declarations

Acknowledgment

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2009-0069100).

Authors’ Affiliations

(1)
Department of Mathematics Education, Dongguk University

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Copyright

© Han Ju Lee. 2009

This article is published under license to BioMed Central Ltd. 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.