Another Aspect of Triangle Inequalitydownloads.hindawi.com/archive/2011/514184.pdf · The triangle inequality is one of the most fundamental inequalities in analysis and has been

International Scholarly Research NetworkISRN Mathematical AnalysisVolume 2011, Article ID 514184, 5 pagesdoi:10.5402/2011/514184

Research ArticleAnother Aspect of Triangle Inequality

Kichi-Suke Saito,1 Runling An,2 Hiroyasu Mizuguchi,3and Ken-Ichi Mitani4

1 Department of Mathematics, Faculty of Science, Niigata University, Niigata 950-2181, Japan2 Department of Mathematics, Taiyuan University of Technology, Taiyuan 030024, China3 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

Correspondence should be addressed to Kichi-Suke Saito, [email protected]

Received 18 February 2011; Accepted 14 March 2011

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

Copyright q 2011 Kichi-Suke Saito et al. This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

We introduce the notion of ψ-norm by considering the fact that an absolute normalized norm onC2 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 weshow 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 beenstudied by several authors. For example, Kato et al. in [1] showed a sharpened triangleinequality and its reverse one with n elements in a Banach space (see also [2–4]). Here weconsider another aspect of the classical triangle inequality ‖x + y‖ ≤ ‖x‖ + ‖y‖. For a HilbertspaceH, 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)

2 ISRN Mathematical Analysis

holds. Saitoh in [5] noted the inequality (1.2)may be more suitable than the classical triangleinequality and used the inequality (1.2) to the setting of a natural sum Hilbert space for twoarbitrary 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 amapping ‖ · ‖ 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 with1 ≤ 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 ofψ-norm by considering the fact that an absolute normalized norm on R

2 corresponds to acontinuous convex function ψ on the unit interval [0, 1] with some conditions (cf. [7]). Weshow that a ψ-norm is a norm in the usual sense.

We recall some properties of absolute normalized norms on C2. A norm ‖ · ‖ on C

2 iscalled absolute if ‖(x, y)‖ = ‖(|x|, |y|)‖ for all (x, y) ∈ C

2 and normalized if ‖(1, 0)‖ = ‖(0, 1)‖ = 1.The �p-norms ‖ · ‖p are such examples:

∥∥(x, y)

∥∥p =

⎧

⎪⎨

⎪⎩

(|x|p + ∣∣y∣∣p)1/p if 1 ≤ p <∞,

max{|x|, ∣∣y∣∣} if p = ∞.

(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 onthe unit interval [0, 1] satisfying max{1 − t, t} ≤ ψ(t) ≤ 1 for t with 0 ≤ t ≤ 1 (see [7, 8]). Thecorrespondence is given by the equation ψ(t) = ‖(1 − t, t)‖ψ . Indeed, for all ψ inΨ2, we definethe norm ‖ · ‖ψ as

∥∥(x, y)

∥∥ψ =

⎧

⎪⎪⎪⎨

⎪⎪⎪⎩

(|x| + ∣∣y∣∣)ψ( ∣

∣y∣∣

|x| + ∣∣y∣∣

)

if(

x, y)

/= (0, 0),

0 if(

x, y)

= (0, 0).

(1.4)

Then ‖ · ‖ψ ∈ AN2 and satisfies ψ(t) = ‖(1 − t, t)‖ψ . The functions which correspond to the

�p-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).

ISRN Mathematical Analysis 3

Note that for all q with 1 ≤ q < ∞, any ψq-norm ‖ · ‖ is just a q-norm. Indeed, since thefunction ψq takes the minimum at t = 1/2 and

ψq

(12

)

=((

12

)q

+(12

)q)1/q

= 21/q−1, (2.1)

the condition (iii) of Definition 2.1 implies

∥∥x + y

∥∥ ≤ 1

ψq(1/2)

∥∥(‖x‖,∥∥y∥∥)∥∥ψq = 21−1/q

(‖x‖q + ∥∥y∥∥q)1/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 theusual 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≤1

ϕ(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 byLemma 2.3,

∥∥x + y

∥∥ ≤ ‖x‖ + ∥∥y∥∥ =

∥∥(‖x‖,∥∥y∥∥)∥∥1

≤ max0≤t≤1

1ψ(t)

∥∥(‖x‖,∥∥y∥∥)∥∥ψ

=1

min0≤t≤1 ψ(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 thefollowing 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.

4 ISRN Mathematical Analysis

Proof. Suppose that BX is convex. For every x, y ∈ X such that x /= 0, y /= 0, we have

∥∥∥∥∥

x

‖x‖ + ∥∥y∥∥ +y

‖x‖ + ∥∥y∥∥

∥∥∥∥∥=

∥∥∥∥∥

‖x‖‖x‖ + ∥∥y∥∥

x

‖x‖ +

∥∥y∥∥

‖x‖ + ∥∥y∥∥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 t0with 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-normand Lemma 1 in [8], we have

∥∥(1 − t0)x + t0y

∥∥ ≤ 1

ψ(t0)

∥∥(

(1 − t0)‖x‖, t0∥∥y∥∥)∥∥ψ

≤ 1ψ(t0)

‖(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 − t0)a + t0b : a, b ∈ An−1} (n = 1, 2, . . .). (2.8)

PutA =⋃∞n=0An. It is clear thatA = [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) ∈ BXfor 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 + 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)

.

(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 normin the usual sense.

ISRN Mathematical Analysis 5

Proof. Let x, y ∈ BX and λ with 0 < λ < 1. Let z = (1 − λ)x + λy. Take a strictly decreasingsequence {rn} inA such that rn ↘ λ. For each n, we define βn = (1−rn)/(1−λ). Then 0 < βn < 1and βn ↗ 1. Since 0 < λβn/rn < 1, we have (λβn/rn)y ∈ BX . By Lemma 2.6,

βnz = (1 − λ)βnx + λβny

= (1 − rn)x + rnλβnrn

y

= f(

x,λβnrn

y, rn

)

∈ BX.

(2.10)

Since βn‖z‖ = ‖βnz‖ ≤ 1, we get z ∈ BX . Thus BX is convex. By Lemma 2.4, ‖ · ‖ becomes anorm. 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. RunlingAnwas supported by Programfor Top Young Academic Leaders of Higher Learning Institutions of Shanxi (TYAL) and agrant from National Foundation of China (No. 11001194).

References

[1] M. Kato, K.-S. Saito, and T. Tamura, “Sharp triangle inequality and its reverse in Banach spaces,”Mathematical Inequalities & Applications, vol. 10, no. 2, pp. 451–460, 2007.

[2] M. Fujii, M. Kato, K.-S. Saito, and T. Tamura, “Sharpmean triangle inequality,”Mathematical Inequalities& Applications, vol. 13, no. 4, pp. 743–752, 2010.

[3] K.-I. Mitani, K.-S. Saito, M. Kato, and T. Tamura, “On sharp triangle inequalities in Banach spaces,”Journal of Mathematical Analysis and Applications, vol. 336, no. 2, pp. 1178–1186, 2007.

[4] K.-I. Mitani and K.-S. Saito, “On sharp triangle inequalities in Banach spaces II,” Journal of Inequalitiesand Applications, vol. 2010, Article ID 323609, 17 pages, 2010.

[5] S. Saitoh, “Generalizations of the triangle inequality,” Journal of Inequalities in Pure and AppliedMathematics, vol. 4, no. 3, article 62, pp. 1–5, 2003.

[6] H. Belbachir, M. Mirzavaziri, and M. S. Moslehian, “q-norms are really norms,” The Australian Journalof Mathematical Analysis and Applications, vol. 3, no. 1, article 2, pp. 1–3, 2006.

[7] K.-S. Saito, M. Kato, and Y. Takahashi, “Von Neumann-Jordan constant of absolute normalized normson C2,” Journal of Mathematical Analysis and Applications, vol. 244, no. 2, pp. 515–532, 2000.

[8] Y. Takahashi, M. Kato, and K.-S. Saito, “Strict convexity of absolute norms on C2 and direct sums ofBanach spaces,” Journal of Inequalities and Applications, vol. 7, no. 2, pp. 179–186, 2002.

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Another Aspect of Triangle Inequalitydownloads.hindawi.com/archive/2011/514184.pdf · The triangle inequality is one of the most fundamental inequalities in analysis and has been