`ISRN Mathematical AnalysisVolumeΒ 2011, Article IDΒ 514184, 5 pageshttp://dx.doi.org/10.5402/2011/514184`
Research Article

## Another Aspect of Triangle Inequality

1Department of Mathematics, Faculty of Science, Niigata University, Niigata 950-2181, Japan
2Department of Mathematics, Taiyuan University of Technology, Taiyuan 030024, China
3Department of Mathematics and Information Science, Graduate School of Science and Technology, Niigata University, Niigata 950-2181, Japan
4Department of Systems Engineering, Okayama Prefectural University, Soja, Okayama 719-1197, Japan

Received 18 February 2011; Accepted 14 March 2011

Academic Editors: Y.Β Dai and B.Β Djafari-Rouhani

Copyright Β© 2011 Kichi-Suke Saito 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 introduce the notion of -norm by considering the fact that an absolute normalized norm on corresponds to a continuous convex function on the unit interval with some conditions. This is a generalization of the notion of -norm introduced by Belbachir et al. (2006). Then we show that a -norm is a norm in the usual sense.

#### 1. Introduction

The triangle inequality is one of the most fundamental inequalities in analysis and has been studied by several authors. For example, Kato et al. in [1] showed a sharpened triangle inequality and its reverse one with elements in a Banach space (see also [2β4]). Here we consider another aspect of the classical triangle inequality . For a Hilbert space , we recall the parallelogram law This implies that the parallelogram inequality holds. Saitoh in [5] noted the inequality (1.2) may be more suitable than the classical triangle inequality and used the inequality (1.2) to the setting of a natural sum Hilbert space for two arbitrary Hilbert spaces. Motivated by this, Belbachir et al. [6] introduced the notion of -norm () in a vector space over or , where the definition of -norm is a mapping from into satisfying the following conditions: (i), (ii), (iii).

We easily show that every norm is a -norm. Conversely, they proved that for all with , every -norm is a norm in the usual sense.

In this paper, we generalize the notion of -norm, that is, we introduce the notion of -norm by considering the fact that an absolute normalized norm on corresponds to a continuous convex function on the unit interval with some conditions (cf. [7]). We show that a -norm is a norm in the usual sense.

We recall some properties of absolute normalized norms on . A norm on is called absolute if for all and normalized if . The -norms are such examples: Let be the family of all absolute normalized norms on . It is well known that the set is a one-to-one correspondence with the set of all continuous convex functions on the unit interval satisfying for with (see [7, 8]). The correspondence is given by the equation . Indeed, for all in , we define the norm as Then and satisfies . The functions which correspond to the -norms on are if and if .

#### 2. π-Norm

Definition 2.1. Let be a vector space and . Then a mapping is called -norm on if it satisfies the following conditions: (i),(ii),(iii).

Note that for all with , any -norm is just a -norm. Indeed, since the function takes the minimum at and the condition (iii) of Definition 2.1 implies Thus we have and so becomes a -norm.

If , then the condition (iii) of Definition 2.1 is just a triangle inequality. Thus we suppose that .

Proposition 2.2. Let be a vector space and with . Then every norm on in the usual sense is a -norm.

To do this, we need the following lemma given in [7].

Lemma 2.3 (see [7]). Let and . Put Then

Proof of Proposition 2.2. Let be a norm on and . Since , we have by Lemma 2.3, Thus is a -norm on .

We will show that every -norm is a norm in the usual sense. To do this, we need the following lemma given in [6].

Lemma 2.4 (see [6]). Let be a vector space. Let be a mapping satisfying the conditions (i) and (ii) in Definition 2.1. Then is a norm if and only if the set is convex.

Proof. Suppose that is convex. For every such that , we have This completes the proof.

Since every -norm is just a usual norm, we suppose that with . Put with such that . Then we have the following lemma.

Lemma 2.5. Let be a -norm on . Then, for every we have .

Proof. Let . We may assume that and . From the definition of a -norm and Lemmaββ1 in [8], we have which implies .

Here we define the set for all , by Put . It is clear that . We also define a function by for all and all .

Lemma 2.6. For every , we have for all .

Proof. Let . It is clear that for all . We suppose that for all . Then, for all , there exist such that . Hence Since and are in , we have from Lemma 2.5, for all . Thus for all .

Theorem 2.7. Let be a vector space and with . Then every -norm on is a norm in the usual sense.

Proof. Let and with . Let . Take a strictly decreasing sequence in such that . For each , we define . Then and . Since , we have . By Lemma 2.6, Since , we get . Thus is convex. By Lemma 2.4, becomes a norm. This completes the proof.

#### Acknowledgments

Kichi-Suke Saito was supported in part by Grants-in-Aid for Scientific Research (No. 20540158), Japan Society for the Promotion of Science. Runling An was supported by Program for Top Young Academic Leaders of Higher Learning Institutions of Shanxi (TYAL) and a grant from National Foundation of China (No. 11001194).

#### References

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