Another Aspect of Triangle Inequality

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


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 This implies that the parallelogram inequality ISRN Mathematical Analysis 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 qnorm 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: iii x y q ≤ 2 q−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 R 2 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

1.3
Let AN 2 be the family of all absolute normalized norms on C 2 .It is well known that the set AN 2 is a one-to-one correspondence with the set Ψ 2 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 Ψ 2 , we define the norm Then • ψ ∈ AN 2 and satisfies ψ t 1 − t, t ψ .The functions which correspond to the

ψ-Norm
Definition 2.1.Let X be a vector space and ψ ∈ Ψ 2 .Then a mapping • : X → R is called ψ-norm on X if it satisfies the following conditions: Note that for all q with 1 ≤ q < ∞, any ψ q -norm • is just a q-norm.Indeed, since the function ψ q takes the minimum at t 1/2 and the condition iii of Definition 2.1 implies x , y ψ q 2 1−1/q x q y q 1/q .2.2 Thus we have x y q ≤ 2 q−1 x q y q and so • becomes a q-norm.If ψ ψ 1 , then the condition iii of Definition 2.1 is just a triangle inequality.Thus we suppose that ψ / ψ 1 .
Proposition 2.2.Let X be a vector space and ψ ∈ Ψ 2 with ψ / ψ 1 .Then every norm on X in the usual sense is a ψ-norm.
To do this, we need the following lemma given in 7 . Then

2.4
Proof of Proposition 2.2.Let • be a norm on X and x, y ∈ X.Since ψ ≤ ψ 1 , we have by Lemma 2.3, 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 B X {x ∈ X : x ≤ 1} is convex.

ISRN Mathematical Analysis
Proof.Suppose that B X is convex.For every x, y ∈ X such that x / 0, y / 0, we have

2.6
This completes the proof.
Lemma 2.5.Let • be a ψ-norm on X.Then, for every x, y ∈ B X we have 1 − t 0 x t 0 y ∈ B X .
Proof.Let x, y ∈ B X .We may assume that x / y and x, y / 0. From the definition of a ψ-norm and Lemma 1 in 8 , we have Here we define the set A n for all n 1, 2, . .., by We also define a function f by f x, y, t 1−t x ty for all x, y ∈ B X and all t ∈ 0, 1 .
Lemma 2.6.For every x, y ∈ B X , we have f x, y, t ∈ B X for all t ∈ A.
Proof.Let x, y ∈ B X .It is clear that f x, y, t ∈ B X for all t ∈ A 0 .We suppose that f x, y, t ∈ B X for all t ∈ A n−1 .Then, for all t ∈ A n , there exist a, b

2.9
Since f x, y, a and f x, y, b are in B X , we have from Lemma 2.5, f x, y, t ∈ B X for all t ∈ A n .Thus f x, y, t ∈ B X for all t ∈ A.
Theorem 2.7.Let X be a vector space and ψ ∈ Ψ 2 with ψ / ψ 1 .Then every ψ-norm on X is a norm in the usual sense.
Proof.Let x, y ∈ B X and λ with 0 < λ < 1.Let z 1 − λ x λy.Take a strictly decreasing sequence {r n } in A such that r n λ.For each n, we define β n 1−r n / 1−λ .Then 0 < β n < 1 and β n 1.Since 0 < λβ n /r n < 1, we have λβ n /r n y ∈ B X .By Lemma 2.6, ≤ 1, we get z ∈ B X .Thus B X is convex.By Lemma 2.4, • becomes a norm.This completes the proof.