# Another Aspect of Triangle Inequality

Authors: Kichi-Suke Saito; Runling An; Hiroyasu Mizuguchi; Ken-Ichi Mitani
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

We introduce the notion ofψ-norm by considering the fact that an absolute normalized norm on
ℂ2 corresponds to a continuous convex function
ψ on the unit interval
[0,1] with some conditions. This is a generalization of the notion of
q-norm introduced by  Belbachir et al. (2006). Then we show that a
ψ-norm is a norm in the usual sense.


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## Body

## 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 n elements in a Banach space (see also [2–4]). Here we consider another aspect of the classical triangle inequality ∥x+y∥≤∥x∥+∥y∥. For a Hilbert space H, we recall the parallelogram law
(1.1)‖x+y‖2+‖x-y‖2=2(‖x‖2+‖y‖2)(x,y∈H).
This implies that the parallelogram inequality
(1.2)‖x+y‖2≤2(‖x‖2+‖y‖2)(x,y∈H)
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 q-norm (1≤q<∞) in a vector space X over 𝕂(=ℝ or ℂ), where the definition of q-norm is a mapping ∥·∥ from X into ℝ+(={a∈ℝ:a≥0}) satisfying the following conditions: (i)
∥x∥=0⇔x=0,(ii)
∥αx∥=|α|∥x∥(x∈X,α∈𝕂),(iii)
∥x+y∥q≤2q-1(∥x∥q+∥y∥q)(x,y∈X).We easily show that every norm is aq-norm. Conversely, they proved that for all q with 1≤q<∞, every q-norm is a norm in the usual sense.In this paper, we generalize the notion ofq-norm, that is, we introduce the notion of ψ-norm by considering the fact that an absolute normalized norm on ℝ2 corresponds to a continuous convex function ψ on the unit interval [0,1] with some conditions (cf. [7]). We show that a ψ-norm is a norm in the usual sense.We recall some properties of absolute normalized norms onℂ2. A norm ∥·∥ on ℂ2 is called  absolute if ∥(x,y)∥=∥(|x|,|y|)∥ for all (x,y)∈ℂ2 and normalized if ∥(1,0)∥=∥(0,1)∥=1. The ℓp-norms ∥·∥p are such examples:
(1.3)‖(x,y)‖p={(|x|p+|y|p)1/pif1≤p<∞,max⁡⁡{|x|,|y|}ifp=∞.
Let AN2 be the family of all absolute normalized norms on ℂ2. It is well known that the set AN2 is a one-to-one correspondence with the set Ψ2 of all continuous convex functions ψ on the unit interval [0,1] satisfying max{1-t,t}≤ψ(t)≤1 for t with 0≤t≤1 (see [7, 8]). The correspondence is given by the equation ψ(t)=∥(1-t,t)∥ψ. Indeed, for all ψ in Ψ2, we define the norm ∥·∥ψ as
(1.4)‖(x,y)‖ψ={(|x|+|y|)ψ(|y||x|+|y|)if(x,y)≠(0,0),0if(x,y)=(0,0).
Then ∥·∥ψ∈AN2 and satisfies ψ(t)=∥(1-t,t)∥ψ. The functions which correspond to the ℓp-norms ∥·∥p on ℂ2 are ψp(t)={(1-t)p+tp}1/p if 1≤p<∞ and ψ∞(t)=max{1-t,t} if p=∞.


## 2.ψ-Norm

Definition 2.1.
LetX be a vector space and ψ∈Ψ2. Then a mapping ∥·∥:X→ℝ+ is called  ψ-norm  onX if it satisfies the following conditions: (i)
∥x∥=0⇔x=0,(ii)
∥αx∥=|α|∥x∥(x∈X,α∈𝕂),(iii)
∥x+y∥≤(1/min0≤t≤1ψ(t))∥(∥x∥,∥y∥)∥ψ(x,y∈X).Note that for allq with 1≤q<∞, any ψq-norm ∥·∥ is just a q-norm. Indeed, since the function ψq takes the minimum at t=1/2 and
(2.1)ψq(12)=((12)q+(12)q)1/q=21/q-1,
the condition (iii) of Definition 2.1 implies
(2.2)‖x+y‖≤1ψq(1/2)‖(‖x‖,‖y‖)‖ψq=21-1/q(‖x‖q+‖y‖q)1/q.
Thus we have ∥x+y∥q≤2q-1(∥x∥q+∥y∥q) and so ∥·∥ becomes a q-norm.Ifψ=ψ1, then the condition (iii) of Definition 2.1 is just a triangle inequality. Thus we suppose that ψ≠ψ1.Proposition 2.2.
LetX be a vector space and ψ∈Ψ2 with ψ≠ψ1. Then every norm on X in the usual sense is a ψ-norm.To do this, we need the following lemma given in [7].Lemma 2.3 (see [7]).
Letψ,φ∈Ψ2 and φ≥ψ. Put
(2.3)M=max⁡0≤t≤1φ(t)ψ(t).
Then
(2.4)‖⋅‖ψ≤‖⋅‖φ≤M‖⋅‖ψ.Proof of Proposition2.2.
Let∥·∥ be a norm on X and x,y∈X. Since ψ≤ψ1, we have by Lemma 2.3,
(2.5)‖x+y‖≤‖x‖+‖y‖=‖(‖x‖,‖y‖)‖1≤max⁡0≤t≤11ψ(t)‖(‖x‖,‖y‖)‖ψ=1min⁡0≤t≤1ψ(t)‖(‖x‖,‖y‖)‖ψ.
Thus ∥·∥ is a ψ-norm on X.We will show that everyψ-norm is a norm in the usual sense. To do this, we need the following lemma given in [6].Lemma 2.4 (see [6]).
LetX be a vector space. Let ∥·∥:X→ℝ+ be a mapping satisfying the conditions (i) and (ii) in Definition 2.1. Then ∥·∥ is a norm if and only if the set BX={x∈X:∥x∥≤1} is convex.Proof.
Suppose thatBX is convex. For every x,y∈X such that x≠0,y≠0, we have
(2.6)‖x‖x‖+‖y‖+y‖x‖+‖y‖‖=‖‖x‖‖x‖+‖y‖x‖x‖+‖y‖‖x‖+‖y‖y‖y‖‖≤1.
This completes the proof.Since everyψ1-norm is just a usual norm, we suppose that ψ∈Ψ2 with ψ≠ψ1. Put t0 with 0<t0<1 such that min0≤t≤1ψ(t)=ψ(t0). Then we have the following lemma.Lemma 2.5.
Let∥·∥ be a ψ-norm on X. Then, for every x,y∈BX we have (1-t0)x+t0y∈BX.Proof.
Letx,y∈BX. We may assume that x≠y and x,y≠0. From the definition of a ψ-norm and Lemma  1 in [8], we have
(2.7)‖(1-t0)x+t0y‖≤1ψ(t0)‖((1-t0)‖x‖,t0‖y‖)‖ψ≤1ψ(t0)‖(1-t0,t0)‖ψ=1,
which implies (1-t0)x+t0y∈BX.Here we define the setAn for all n=1,2,…, by
(2.8)A0={0,1},An={(1-t0)a+t0b:a,b∈An-1}(n=1,2,…).
Put A=⋃n=0∞An. It is clear that A¯=[0,1]. We also define a function f by f(x,y,t)=(1-t)x+ty for all x,y∈BX and all t∈[0,1].Lemma 2.6.
For everyx,y∈BX, we have f(x,y,t)∈BX for all t∈A.Proof.
Letx,y∈BX. It is clear that f(x,y,t)∈BX for all t∈A0. We suppose that f(x,y,t)∈BX for all t∈An-1. Then, for all t∈An, there exist a,b∈An-1 such that t=(1-t0)a+t0b. Hence
(2.9)f(x,y,t)=(1-t)x+ty=(1-((1-t0)a+t0b))x+((1-t0)a+t0b)y=(1-t0)((1-a)x+ay)+t0((1-b)x+by)=(1-t0)f(x,y,a)+t0f(x,y,b).
Since f(x,y,a) and f(x,y,b) are in BX, we have from Lemma 2.5, f(x,y,t)∈BX for all t∈An. Thus f(x,y,t)∈BX for all t∈A.Theorem 2.7.
LetX be a vector space and ψ∈Ψ2 with ψ≠ψ1. Then every ψ-norm on X is a norm in the usual sense.Proof.
Letx,y∈BX and λ with 0<λ<1. Let z=(1-λ)x+λy. Take a strictly decreasing sequence {rn} in A such that rn↘λ. For each n, we define βn=(1-rn)/(1-λ). Then 0<βn<1 and βn↗1. Since 0<λβn/rn<1, we have (λβn/rn)y∈BX. By Lemma 2.6,
(2.10)βnz=(1-λ)βnx+λβny=(1-rn)x+rnλβnrny=f(x,λβnrny,rn)∈BX.
Since βn∥z∥=∥βnz∥≤1, we get z∈BX. Thus BX is convex. By Lemma 2.4, ∥·∥ becomes a norm. This completes the proof.



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*Source: 514184-2011-04-14.xml*