Volume 9 (2008), Issue 2, Article 41, 4 pp.
ON THE TRIANGLE INEQUALITY IN QUASI-BANACH SPACES
CONG WU AND YONGJIN LI
D EPARTMENT OF M ATHEMATICS
S UN YAT-S EN U NIVERSITY
G UANGZHOU , 510275
P. R. C HINA
congwu@hotmail.com
stslyj@mail.sysu.edu.cn
Received 11 May, 2007; accepted 10 June, 2008
Communicated by S.S. Dragomir
A BSTRACT. In this paper, we show the triangle inequality and its reverse inequality in quasiBanach spaces.
Key words and phrases: Triangle inequality, Quasi-Banach spaces.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
The triangle inequality is one of the most fundamental inequalities in analysis. The following
sharp triangle inequality was given earlier in H. Hudzik and T. R. Landes [2] and also found in
a recent paper of L. Maligranda [5].
Theorem 1.1. For all nonzero elements x, y in a normed linear space X with kxk ≥ kyk,
x
y kyk
kx + yk+ 2 − +
kxk kyk ≤ kxk + kyk
x
y kxk.
≤ kx + yk + 2 − +
kxk kyk We recall that a quasi-norm k · k defined on a vector space X (over a real or complex field K)
is a map X → R+ such that:
(i) kxk > 0 for x 6= 0;
(ii) kαxk = |α|kxk for α ∈ K, x ∈ X;
(iii) kx + yk ≤ C(kxk + kyk) for all x, y ∈ X, where C is a constant independent of x, y.
If k · k is a quasi-norm on X defining a complete metrizable topology, then X is called a
quasi-Banach space.
In the present paper we will present the triangle inequality in quasi-normed spaces.
157-07
2
C ONG W U AND YONGJIN L I
2. M AIN R ESULTS
Theorem 2.1. For all nonzero elements x, y in a quasi-Banach space X with kxk ≥ kyk
x
y kyk
kx + yk+C 2 − +
kxk kyk (2.1)
≤ C(kxk + kyk)
y
x
(2.2)
≤ kx + yk + 2C 2 − +
kxk kyk kxk,
where C ≥ 1.
Proof. Let kxk ≥ kyk. We first show the inequality (2.1).
x
y
x
x
kx + yk = kyk kxk + kyk + kxk kxk − kyk kxk x
y
x
x
+ C kxk
≤C
kyk
+
−
kyk
kxk kyk kxk
kxk x
y + C(kxk − kyk)
= Ckyk +
kxk kyk x
y + C(kxk + kyk − 2kyk)
= Ckyk +
kxk kyk x
y
= Ckyk kxk + kyk − 2 + C(kxk + kyk).
Since
x
y
y
y kx + yk = kxk
+
− kxk
− kyk
kxk kyk
kyk
kyk 1 x
y x
x ≥ kxk
+
− kxk
− kyk
C
kxk kyk kxk
kxk x
1
y
− (kxk − kyk)
= kxk +
C
kxk kyk x
1
y
= kxk +
kxk kyk + (kxk + kyk − 2kxk)
C
1 x
y = kxk
+
− 2 + (kxk + kyk).
C kxk kyk we have
x
y
C(kxk + kyk) ≤ Ckx + yk + 2C − kxk + kyk kxk
x
y
= kx + yk + (C − 1)kx + yk + 2C − +
kxk kyk kxk
x
y kxk
+
≤ kx + yk + (C − 1)C(kxk + kyk) + 2C − kxk kyk J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 41, 4 pp.
http://jipam.vu.edu.au/
T RIANGLE I NEQUALITY IN Q UASI -BANACH S PACES
3
x
y kxk
≤ kx + yk + (C − 1)C(2kxk) + 2C − +
kxk kyk x
y
2
= kx + yk + 2C − kxk + kyk kxk.
Thus the inequality (2.2) holds.
T. Aoki [1] and S. Rolewicz [6] characterized quasi-Banach spaces as follows:
Theorem 2.2 (Aoki-Rolewicz Theorem). Let X be a quasi-Banach space. Then there exists
0 < p ≤ 1 and an equivalent quasi-norm k| · k| on X that satisfies for every x, y ∈ X
k|x + yk|p ≤ k|xk|p + k|yk|p .
Idea of the proof. Let k · k be the original quasi-norm on X, denote by k = inf{K ≥ 1 :
for any x, y ∈ X, kx + yk ≤ K(kxk + kyk)} and p is such that 21/p = 2k. It is shown [3] that
the function k| · k| defined on X by:


! p1
n
n
 X
X 
p
k|xk| = inf
kxi k
:x=
xi


i=1
i=1
is an equivalent quasi-norm on X that satisfies the required inequality.
Next, we will prove the p-triangle inequality in quasi-Banach spaces.
Theorem 2.3. For all nonzero elements x, y in a quasi-Banach space X with kxk ≥ kyk,
p y p
p
p
p
p x
kx + yk + kxk + kyk − (kxk − kyk) − kyk +
kxk kyk ≤ kxkp + kykp
p x
y
+
≤ kx + ykp + kxkp + kykp + (kxk − kyk)p − kxkp kxk kyk ,
where 0 < p ≤ 1.
Proof. We have
p
x
y
x
x kx + yk = kyk
+
+ kxk
− kyk
kxk kyk
kxk
kxk p
p x
y
x
x
+ kxk
≤
kyk
+
−
kyk
kxk kyk kxk
kxk p
y p x
+ (kxk − kyk)p
= kyk +
kxk kyk p
y p x
+ kxkp
= kyk +
kxk kyk + kykp − (kxkp + kykp ) + (kxk − kyk)p .
p
Thus
p x
y ≤ kxkp + kykp
kx + yk + kxk + kyk − (kxk − kyk) − kyk +
kxk kyk p
p
p
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 41, 4 pp.
p
p
http://jipam.vu.edu.au/
4
C ONG W U AND YONGJIN L I
and
p
x
y
y y
kx + yk = kxk
+
− kxk
− kyk
kxk kyk
kyk
kyk p
p x
y
x
x
≥
kxk kxk + kyk − kxk kxk − kyk kxk p
y
x
p
+
= kxkp kxk kyk − (kxk − kyk)
p
y p x
= kxk +
kxk kyk + kxkp + kykp − (kxkp + kykp ) − (kxk − kyk)p .
p
Hence
p x
y .
kxk + kyk ≤ kx + yk + kxk + kyk + (kxk − kyk) − kxk +
kxk kyk This completes the proof.
p
p
p
p
p
p
p
R EFERENCES
[1] T. AOKI, Locally bounded linear topological spaces, Proc. Imp. Acad. Tokyo, 18 (1942), 588–594.
[2] H. HUDZIK AND T.R. LANDES, Characteristic of convexity of Köthe function spaces, Math. Ann.,
294 (1992), 117–124.
[3] N.J. KALTON, N.T. PECK AND J.W. ROBERTS, An F-Space Sampler, London Math. Soc. Lecture
Notes 89, Cambridge University Press, Cambridge, 1984.
[4] K.-I. MITANI, K.-S. SAITO, M.I. KATO AND T. TAMURA, On sharp triangle inequalities in Banach spaces, J. Math. Anal. Appl., 336 (2007), 1178–1186.
[5] L. MALIGRANDA, Simple norm inequalities, Amer. Math. Monthly, 113 (2006), 256–260.
[6] S. ROLEWICZ, On a certain class of linear metric spaces, Bull. Acad. Polon. Sci. Sér. Sci. Math.
Astrono. Phys., 5 (1957), 471–473.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 41, 4 pp.
http://jipam.vu.edu.au/