Abstract

We introduce the notion of 𝜓-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.

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 ‖𝑥+𝑦‖2+‖𝑥−𝑦‖2=2‖𝑥‖2+‖𝑦‖2(𝑥,𝑦∈𝐻).(1.1) This implies that the parallelogram inequality ‖𝑥+𝑦‖2≤2‖𝑥‖2+‖𝑦‖2(𝑥,𝑦∈𝐻)(1.2) 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 a 𝑞-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 of 𝑞-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 on ℂ2. 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: ‖(𝑥,𝑦)‖𝑝=|𝑥|𝑝+||𝑦||𝑝1/𝑝||𝑦||if1≤𝑝<∞,max|𝑥|,if𝑝=∞.(1.3) 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 ‖(𝑥,𝑦)‖𝜓=⎧⎪⎨⎪⎩||𝑦||𝜓||𝑦|||𝑥|+||𝑦|||𝑥|+if(𝑥,𝑦)≠(0,0),0if(𝑥,𝑦)=(0,0).(1.4) 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. Let 𝑋 be a vector space and 𝜓∈Ψ2. Then a mapping ‖⋅‖∶𝑋→ℝ+ is called 𝜓-norm on 𝑋 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 all 𝑞 with 1≤𝑞<∞, any 𝜓𝑞-norm ‖⋅‖ is just a 𝑞-norm. Indeed, since the function 𝜓𝑞 takes the minimum at 𝑡=1/2 and 𝜓𝑞12=12𝑞+12𝑞1/𝑞=21/𝑞−1,(2.1) the condition (iii) of Definition 2.1 implies 1‖𝑥+𝑦‖≤𝜓𝑞(1/2)‖(‖𝑥‖,‖𝑦‖)‖𝜓𝑞=21−1/𝑞‖𝑥‖𝑞+‖𝑦‖𝑞1/𝑞.(2.2) Thus we have ‖𝑥+𝑦‖𝑞≤2𝑞−1(‖𝑥‖𝑞+‖𝑦‖𝑞) and so ‖⋅‖ becomes a 𝑞-norm.

If 𝜓=𝜓1, then the condition (iii) of Definition 2.1 is just a triangle inequality. Thus we suppose that 𝜓≠𝜓1.

Proposition 2.2. Let 𝑋 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]). Let 𝜓,𝜑∈Ψ2 and 𝜑≥𝜓. Put 𝑀=max0≤𝑡≤1𝜑(𝑡)𝜓(𝑡).(2.3) Then ‖⋅‖𝜓≤‖⋅‖𝜑≤𝑀‖⋅‖𝜓.(2.4)

Proof of Proposition 2.2. Let ‖⋅‖ be a norm on 𝑋 and 𝑥,𝑦∈𝑋. Since 𝜓≤𝜓1, we have by Lemma 2.3, ‖𝑥+𝑦‖≤‖𝑥‖+‖𝑦‖=‖(‖𝑥‖,‖𝑦‖)‖1≤max0≤𝑡≤11𝜓‖(𝑡)(‖𝑥‖,‖𝑦‖)‖𝜓=1min0≤𝑡≤1𝜓(𝑡)‖(‖𝑥‖,‖𝑦‖)‖𝜓.(2.5) Thus ‖⋅‖ is a 𝜓-norm on 𝑋.

We will show that every 𝜓-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 𝑋 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 that 𝐵𝑋 is convex. For every 𝑥,𝑦∈𝑋 such that 𝑥≠0,𝑦≠0, we have ‖‖‖𝑥+𝑦‖𝑥‖+‖𝑦‖‖‖‖=‖‖‖‖𝑥‖+‖𝑦‖‖𝑥‖𝑥‖𝑥‖+‖𝑦‖+‖𝑥‖‖𝑦‖𝑦‖𝑥‖+‖𝑦‖‖‖‖‖𝑦‖≤1.(2.6) This completes the proof.

Since every 𝜓1-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. Let 𝑥,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 ‖‖1−𝑡0𝑥+𝑡0𝑦‖‖≤1𝜓𝑡0‖‖1−𝑡0‖𝑥‖,𝑡0‖‖‖𝑦‖𝜓≤1𝜓𝑡0‖‖(1−𝑡0,𝑡0)‖‖𝜓=1,(2.7) which implies (1−𝑡0)𝑥+𝑡0𝑦∈𝐵𝑋.

Here we define the set 𝐴𝑛 for all 𝑛=1,2,…, by 𝐴0={0,1},𝐴𝑛=1−𝑡0𝑎+𝑡0𝑏∶𝑎,𝑏∈𝐴𝑛−1(𝑛=1,2,…).(2.8) Put ⋃𝐴=∞𝑛=0𝐴𝑛. 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 every 𝑥,𝑦∈𝐵𝑋, we have 𝑓(𝑥,𝑦,𝑡)∈𝐵𝑋 for all 𝑡∈𝐴.

Proof. Let 𝑥,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 =𝑓(𝑥,𝑦,𝑡)=(1−𝑡)𝑥+𝑡𝑦1−1−𝑡0𝑎+𝑡0𝑏𝑥+1−𝑡0𝑎+𝑡0𝑏𝑦=1−𝑡0((1−𝑎)𝑥+𝑎𝑦)+𝑡0=((1−𝑏)𝑥+𝑏𝑦)1−𝑡0𝑓(𝑥,𝑦,𝑎)+𝑡0𝑓(𝑥,𝑦,𝑏).(2.9) 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. Let 𝑋 be a vector space and 𝜓∈Ψ2 with 𝜓≠𝜓1. Then every 𝜓-norm on 𝑋 is a norm in the usual sense.

Proof. Let 𝑥,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, 𝛽𝑛𝑧=(1−𝜆)𝛽𝑛𝑥+𝜆𝛽𝑛𝑦=1−𝑟𝑛𝑥+𝑟𝑛𝜆𝛽𝑛𝑟𝑛𝑦=𝑓𝑥,𝜆𝛽𝑛𝑟𝑛𝑦,𝑟𝑛∈𝐵𝑋.(2.10) Since 𝛽𝑛‖𝑧‖=‖𝛽𝑛𝑧‖≤1, we get 𝑧∈𝐵𝑋. Thus 𝐵𝑋 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).