On Hilbert-Pachpatte Multiple Integral Inequalities
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Research Article On Hilbert-Pachpatte Multiple Integral Inequalities Changjian Zhao,1 Lian-ying Chen,1 and Wing-Sum Cheung2 1
Department of Mathematics, College of Science, China Jiliang University, Hangzhou 310018, China 2 Department of Mathematics, The University of Hong Kong, Pokfulam Road, Hong Kong, China Correspondence should be addressed to Changjian Zhao, [email protected] Received 11 March 2010; Revised 16 July 2010; Accepted 28 July 2010 Academic Editor: N. Govil Copyright q 2010 Changjian Zhao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We establish some multiple integral Hilbert-Pachpatte-type inequalities. As applications, we get some inverse forms of Pachpatte’s inequalities which were established in 1998.
1. Introduction In 1934, Hilbert 1 established the following well-known integral inequality. If f ∈ Lp 0, ∞, g ∈ Lp 0, ∞, f, g ≥ 0, p > 1 and 1/p 1/q 1, then
∞ ∞ 0
0
fxgx π dx dy ≤ xy sin π/p
∞ 0
f p xdx
1/p ∞
1/q g q xdx
,
1.1
0
where π/ sinπ/p is the best value. In recent years, considerable attention has been given to various extensions and improvements of the Hilbert inequality form different viewpoints 2–10. In particular, Pachpatte 11 proved some inequalities similar to Hilbert’s integral inequalities in 1998. In this paper, we establish some new multiple integral Hilbert-Pachpatte-type inequalities.
2
Journal of Inequalities and Applications
2. Main Results Theorem 2.1. Let hi ≥ 1, let fi σi ∈ C1 xi , 0, 0, ∞, i 1, . . . , n, where xi are positive real 0 numbers, and define Fi si si fi σi dσi , for si ∈ xi , 0. Then for 1/αi 1/βi 1, 0 < βi < 1 and n i1 1/αi 1/α, 0
···
x1
0 xn
n
hi i1 Fi si
1/α ds1 · · · dsn ≥ n α i1 1/αi −si
n
−xi
1/αi
0
1/βi
βi hi −1 . si −xi F i si fi si dsi
hi
xi
i1
2.1
Proof. From the hypotheses and in view of inverse Holder integral inequality see 12, it is ¨ easy to observe that n
Fihi si
i1
n
0 hi si
i1
≥
n
Fihi −1 σi fi σi dσi
hi −si
1/αi
0
Fihi −1 σi fi σi
si
i1
βi
2.2
1/βi dσi
,
si ∈ xi , 0, i 1, . . . , n.
Let us note the following means inequality: n
i m1/α ≥ i
n 1 α mi α i1 i
i1
1/α ,
2.3
m > 0.
We obtain that n
hi i1 Fi si
1/α ≥ n α i1 1/αi −si
n
0 hi si
i1
Fihi −1 σi fi σi
βi
1/βi dσi
.
2.4
Integrating both sides of 2.4 over si from xi i 1, 2, . . . , n to 0 and using the special case of inverse Holder integral inequality, we observe that ¨ 0 x1
···
n
0
hi i1 Fi si
1/α ds1 · · · dsn n α i1 1/αi −si 1/βi 0 0 n
βi
hi −1 Fi σi fi σi dσi ≥ hi dsi xn
i1
≥
n
xi
si
hi −xi 1/αi
xi
i1
n
i1
The proof is complete.
0 0
−xi
1/αi
0 hi
xi
si
Fihi −1 σi
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