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Any two-dimensional Normed space is a generalized Day-James space

Journal of Inequalities and Applications20112011:2

https://doi.org/10.1186/1029-242X-2011-2

Received: 11 February 2011

Accepted: 15 June 2011

Published: 15 June 2011

Abstract

It is proved that any two-dimensional normed space is isometrically isomorphic to a generalized Day-James space ψ - φ , introduced by W. Nilsrakoo and S. Saejung.

Keywords

Normed spaceDay-James spaceBirkhoff orthogonality

1991 Mathematics Subject Classification 46B20

The Day-James space p - q is defined for 1 ≤ p, q ≤ ∞ as the space 2 endowed with the norm

where x = (x1, x2). James [1] considered the space p - q with 1/p + 1/q = 1 as an example of a two-dimensional normed space where Birkhoff orthogonality is symmetric. Recall that if x and y are vectors in a normed space then x is said to be Birkhoff orthogonal to y, (x B y), if ||x +λy|| ≥||x|| for every scalar λ[2]. Birkhoff orthogonality coincides with usual orthogonality in inner product spaces. In arbitrary normed spaces Birkhoff orthogonality is in general not symmetric (e.g., in 2 with ||·||), and it is symmetric in a normed space of three or more dimension if and only if the norm is induced by an inner product. This last significant property was obtained in gradual stages by Birkhoff [2], James [1, 3], and Day [4]. The first reference related to the symmetry of Birkhoff orthogonality in two-dimensional spaces seems to be Radon [5] in 1916. He considered plane convex curves with conjugate diameters (as in ellipses) in order to solve certain variational problems.

The procedure that James used to get two-dimensional normed spaces where Birkhoff orthogonality is symmetric was extended by Day [4] in the following way. Let (X, ||·|| X ) be a two-dimensional normed space and let u, v X be such that ||u|| X = ||v|| X = 1, u B v, and v B u (see Lemma below). Then, taking a coordinate system where u = (1, 0) and v = (0, 1) and defining

one gets that in the space (X, ||·|| X,X *) Birkhoff orthogonality is symmetric. Moreover, Day also proved that surprisingly the norm of any two-dimensional space where Birkhoff orthogonality is symmetric can be constructed in the above way.

A norm on 2 is called absolute if ||(x1, x2)|| = ||(|x1|, |x2|)|| for any (x1, x2) 2. Following Nilsrakoo and Saejung [6] let AN2 be the family of all absolute and normalized (i.e., ||(1, 0)|| = ||(0, 1)|| = 1) norms on 2. Examples of norms in AN 2 are p norms. Bonsall and Duncan [7] showed that there is a one-to-one correspondence between AN2 and the family Ψ2 of all continuous and convex functions ψ : [0, 1] such that ψ(0) = ψ(1) = 1 and max{1-t, t} ≤ ψ(t) 1 (0 ≤ t ≤ 1). The correspondence is given by ψ(t) = ||(1-t, t)|| for ||·|| in AN2, and by

for ψ in Ψ2.

In [6] the family of norms ||·|| p,q of Day-James spaces p - q is extended to the family N2 of norms defined in 2 as

for ψ, φ Ψ2. The space 2 endowed with the above norm is called an ψ - φ space.

The purpose of this paper is to show that any two-dimensional normed space is isometrically isomorphic to an ψ - φ space. To this end we shall use the following lemma due to Day [8]. The nice proof we reproduce here is taken from the PhD Thesis of del Río [9], and is based on explicitly developing the idea underlying one of the two proofs given by Day.

Lemma 1[8]. Let (X, ||·||) be a two-dimensional normed space. Then, there exist u, v X such that ||u|| = ||v|| = 1, u B v, and v B u.

Proof. Let e, be linearly independent, and for x X let (x1, x2) 2 be the coordinates of x in the basis . Let S = {x X : ||x|| = 1}, and for x S consider the linear functional f x : y X f x (y) = x2y1 - x1y2. Then it is immediate to see that f x attains the norm in y S (i.e., |x2y1 - x1y2| ≥ |x2z1-x1z2|, for all ) if and only if y B x. Therefore if u, v S are such that |u2v1- u1v2| = max(x, y)S×S|x2y1 - x1y2| then u B v and v B u.    □

Theorem 2 For any two-dimensional normed space (X, ||·|| X ) there exist ψ, φ Ψ2such that (X, ||·|| X ) is isometrically isomorphic to (2, ||·|| ψ, φ ).

Proof. By Lemma 1 we can take u, v X such that ||u|| = ||v|| = 1, u B v, and v B u. Then u and v are linearly independent and (X, ||·|| X ) is isometrically isomorphic to (2, ||·||2), where || (x1, x2) ||2 := ||x1u + x2v|| X . Defining ψ(t) = || (1 - t)u + tv|| X , φ(t) = || (1 - t)u - tv|| X , (0 ≤ t ≤ 1), one trivially has that ψ, φ Ψ2 and || (x1, x2) ||2 = || (x1, x2) || ψ, φ for all (x1, x2) 2.    □

Declarations

Acknowledgements

Research partially supported by MICINN (Spain) and FEDER (UE) grant MTM2008-05460, and by Junta de Extremadura grant GR10060 (partially financed with FEDER).

Authors’ Affiliations

(1)
Department of Mathematics, University of Extremadura, Badajoz, Spain

References

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Copyright

© Alonso; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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