volume 4, issue 1, article 16, 2003.
Received 31 October, 2002;
accepted 8 January, 2003.
Communicated by:P.S. Bullen
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Journal of Inequalities in Pure and Applied Mathematics
HILBERT-PACHPATTE TYPE INTEGRAL INEQUALITIES AND THEIR IMPROVEMENT
S.S. DRAGOMIR AND YOUNG-HO KIM
School of Computer Science and Mathematics Victoria University of Technology
PO Box 14428 , Melbourne City MC Victoria 8001, Australia.
EMail:sever.dragomir@vu.edu.au
URL:http://rgmia.vu.edu.au/SSDragomirWeb.html Department of Applied Mathematics
Changwon National University Changwon 641-773, Korea.
EMail:yhkim@sarim.changwon.ac.kr
c
2000Victoria University ISSN (electronic): 1443-5756 114-02
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Abstract
In this paper, we obtain an extension of multivariable integral inequality of Hilbert-Pachpatte type. By specializing the upper estimate functions in the hy- pothesis and the parameters, we obtain many special cases.
2000 Mathematics Subject Classification:26D15.
Key words: Hilbert’s inequality, Hilbert-Pachpatte type inequality, Hölder’s inequality, Jensen inequality.
The authors would like to thank Professor P.S. Bullen, University of British Columbia, Canada, for the careful reading of the manuscript which led to a considerable im- provement in the presentation of this paper.
Contents
1 Intoduction. . . 3 2 Main Results . . . 5 3 The Various Inequalities. . . 13
References
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1. Intoduction
Hilbert’s double series theorem [3, p. 226] was proved first by Hilbert in his lectures on integral equations. The determination of the constant, the integral analogue, the extension, other proofs of the whole or of parts of the theorems and generalizations in different directions have been given by several authors (cf. [3, Chap. 9]). Specifically, in [10] – [14] the author has established some new inequalities similar to Hilbert’s double-series inequality and its integral analogue which we believe will serve as a model for further investigation. Re- cently, G.D. Handley, J.J. Koliha and J.E. Peˇcari´c [2] established a new class of related integral inequalities from which the results of Pachpatte [12] – [14] are obtained by specializing the parameters and the functions Φi.A representative sample is the following.
Theorem 1.1 (Handley, Koliha and Peˇcari´c [2, Theorem 3.1]). Let ui ∈ Cmi([0, xi])fori∈I.If
u(ki i)(si) ≤
Z si
0
(si−τi)mi−ki−1Φi(τi)dτi, si ∈[0, xi], i∈I,
then
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) Pn
i=1ωis(αi i+1)/(qiωi) ds1· · ·dsn
≤U
n
Y
i=1
x
1 qi
i n
Y
i=1
Z xi
0
(xi −si)βi+1Φi(si)pidsi 1
pi ,
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whereU = 1 .Qn
i=1[(αi+ 1)
1
qi(βi+ 1)
1 pi].
The purpose of the present paper is to derive an extension of the inequality given in Theorem1.1. In addition, we obtain some new inequalities as Hilbert- Pachpatte type inequalities, these inequalities improve the results obtained by Handley, Koliha and Peˇcari´c [2].
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2. Main Results
In what follows we denote byRthe set of real numbers;R+denotes the interval [0,∞).The symbolsN,Zhave their usual meaning. The following notation and hypotheses will be used throughout the paper:
I ={1, ..., n} n∈N mi, i∈I mi ∈N
ki, i∈I ki ∈ {0,1, . . . , mi−1}
xi, i∈I xi ∈R, xi >0
pi, qi, i∈I pi, qi ∈R, pi, qi >0, p1
i +q1
i = 1
p, q 1p =Pn
i=1
1 pi
, 1q =Pn i=1
1 qi
ai, bi, i∈I ai, bi ∈R+, ai+bi = 1 ωi, i∈I ωi ∈R, ωi >0, Pn
i=1ωi = Ωn αi, i∈I αi = (ai+biqi)(mi−ki−1) βi, i∈I βi =ai(mi−ki−1)
ui, i∈I ui ∈Cm0i([0, xi]) for some m0i ≥mi Φi, i∈I Φi ∈C1([0, xi]), Φi ≥mi.
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Here theui are given functions of sufficient smoothness, and theΦi are subject to choice. The coefficients pi, qi are conjugate Hölder exponents to be used in applications of Hölder’s inequality, and the coefficients ai, bi will be used in exponents to factorize integrands. The coefficients ωi will act as weights in applications of the geometric-arithmetic mean inequality. The coefficients αi
andβi arise naturally in the derivation of the inequalities. Our main results are given in the following theorems.
Theorem 2.1. Letui ∈Cmi([0, xi])fori∈I.If
(2.1)
u(ki i)(si) ≤
Z si
0
(si−τi)mi−ki−1Φi(τi)dτi, si ∈[0, xi], i∈I, then
(2.2) Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis(αi i+1)/(qiωi)iΩn ds1· · ·dsn
≤V
n
Y
i=1
x
1 qi
i n
Y
i=1
Z xi
0
(xi −si)βi+1Φi(si)pidsi pi1
,
where
(2.3) V = 1
Qn i=1
h
(αi+ 1)qi1(βi+ 1)pi1i. Proof. Factorize the integrand on the right side of (2.1) as
(si−τi)(ai/qi+bi)(mi−ki−1)×(si−τi)(ai/pi)(mi−ki−1)Φi(τi)
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and apply Hölder’s inequality [9, p.106]. Then
u(ki i)(si) ≤
Z si
0
(si−τi)(ai+biqi)(mi−ki−1)dτi
qi1
× Z si
0
(si−τi)ai(mi−ki−1)Φi(τi)pidτi
pi1
= s(αi i+1)/qi (αi+ 1)qi1
Z si
0
(si−τi)βiΦi(τi)pidτi pi1
.
Using the inequality of means [9, p. 15]
n
Y
i=1
swii
!Ωn1
≤ 1
Ωn n
X
i=1
wisri
!1r
forr >0,we deduce that
n
Y
i=1
swiir≤
"
1 Ωn
n
X
i=1
wisri
#Ωn
forr >0.According to above inequality, we have
n
Y
i=1
u(ki i)(si)
≤ 1 Qn
i=1(αi+ 1)qi1
"
1 Ωn
n
X
i=1
ωis(αi i+1)/(qiωi)
#Ωn
×
n
Y
i=1
Z si
0
(si −τi)βiΦi(τi)pidτi 1
pi
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for r = (αi + 1)/qiωi. In the following estimate we apply Hölder’s inequality and, at the end, change the order of integration:
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)si
h 1
Ωn
Pn
i=1ωis(αi i+1)/(qiωi)iΩn ds1· · ·dsn
≤ 1
Qn
i=1(αi+ 1)qi1
n
Y
i=1
"
Z xi
0
Z si
0
(si−τi)βiΦi(τi)pidτi, pi1
dsi
#
≤ 1
Qn
i=1(αi+ 1)qi1
n
Y
i=1
x
1 qi
i
Z xi
0
Z si
0
(si−τi)βiΦi(τi)pidτi,
dsi
pi1
= 1
Qn
i=1[(αi+ 1)qi1(βi+ 1)pi1]
n
Y
i=1
x
1 qi
i n
Y
i=1
Z xi
0
(xi−si)βi+1Φi(si)pidsi 1
pi .
This proves the theorem.
Remark 2.1. In Theorem2.1, settingΩn = 1, we have Theorem1.1.
Corollary 2.2. Under the assumptions of Theorem2.1, ifr >0, we have
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)si
h 1
Ωn
Pn
i=1ωis(αi i+1)/(qiωi)iΩn ds1· · ·dsn
≤pr·p1 V
n
Y
i=1
x
1 qi
i
" n X
i=1
1 pi
Z xi
0
(xi−si)βi+1Φispiidsi r#r·p1
,
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whereV is defined by (2.3).
Proof. By the inequality of means, for anyAi ≥0andr >0,we obtain
n
Y
i=1
A
1 pi
i ≤
"
p
n
X
i=1
1 piAri
#r·p1 .
The corollary then follows from the preceding theorem.
Lemma 2.3. Let γ1 > 0 and γ2 < −1. Let ωi > 0, Pn
i=1ωi = Ωn and let si >0, i= 1, . . . , nbe real numbers. Then
n
Y
i=1
sωiiγ1γ2 ≥
"
1 Ωn
n
X
i=1
ωis−γi 2
#−γ1Ωn
.
Proof. By the inequality of means, for anyγ1 >0andγ2 <−1,we have
n
Y
i=1
sωiiγ1γ2 ≥
"
1 Ωn
n
X
i=1
ωisi
#γ1γ2Ωn
.
Using the fact that x−γ12 is concave and using the Jensen inequality, we have that
"
1 Ωn
n
X
i=1
ωisi
#γ1γ2Ωn
=
"
1 Ωn
n
X
i=1
ωif(s−γi 2)
#γ1γ2Ωn
≥
"
f 1
Ωn
n
X
i=1
ωis−γi 2
!#γ1γ2Ωn
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=
1 Ωn
n
X
i=1
ωis−γi 2
!−γ1
2
γ1γ2Ωn
=
"
1 Ωn
n
X
i=1
ωis−γi 2
#−γ1Ωn
.
The proof of the lemma is complete.
Theorem 2.4. Under the assumptions of Theorem2.1, ifγ2 <−1,then
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis−γi 2i−(αi+1)Ωn/γ2qiωi ds1· · ·dsn
≤V
n
Y
i=1
x
1 qi
i n
Y
i=1
Z xi
0
(xi−si)βi+1Φi(si)pidsi
pi1 ,
whereV is given by (2.3).
Proof. Using the inequality of Lemma2.3, for anyγ1 >0andγ2 <−1,we get
n
Y
i=1
sωiiγ1 ≤
"
1 Ωn
n
X
i=1
ωis−γi 2
#−γ1Ωn
γ2
.
According to above inequality, we deduce that
n
Y
i=1
u(ki i)(si)
≤ 1 Qn
i=1(αi+ 1)qi1
"
1 Ωn
n
X
i=1
ωis−γi 2
#−W1
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×
n
Y
i=1
"
Z (si) 0
(si−τi)βiΦi(τi)pidτi
#1
pi
,
where W1 = (αi+ 1)Ωn/γ2qiωi. The proof of the theorem then follows from the preceding Theorem2.1.
Corollary 2.5. Under the assumptions of Theorem2.4, ifr >0, we have
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis−γi 2i−(αi+1)Ωn/γ2qiωi ds1· · ·dsn
≤pr·p1 V
n
Y
i=1
x
1 qi
i
" n X
i=1
1 pi
Z xi
0
(xi−si)βi+1Φi(si)pidsi r#r·p1
,
whereV is given by (2.3).
Proof. By the inequality of means, for anyAi ≥0andr >0,we obtain
n
Y
i=1
A
1 pi
i ≤
"
p
n
X
i=1
1 piAri
#r·p1 .
The corollary then follows from the preceding Theorem2.4.
In the following section we discuss some choice of the functionsΦi.
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3. The Various Inequalities
Theorem 3.1. Letui ∈Cmi([0, xi])be such thatu(j)i (0) = 0forj ∈ {0, . . . , mi− 1}, i∈I.Then
(3.1) Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis(αi i+1)/(qiωi)
iΩn ds1· · ·dsn
≤V1
n
Y
i=1
x
1 qi
i n
Y
i=1
Z xi
0
(xi−si)βi+1
u(mi i)(si)
pi
dsi pi1
,
where
(3.2) V1 = 1
Qn i=1
h
(mi−ki−1)!(αi+ 1)qi1(βi + 1)pi1 i.
Proof. Inequality (3.1) is proved when we set
Φi(si) =
u(mi i)(si) (mi−ki−1)!
in Theorem2.1.
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Corollary 3.2. Under the assumptions of Theorem3.1, ifr >0, we have
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis(αi i+1)/(qiωi)iΩn ds1· · ·dsn
≤pr·p1 V1
n
Y
i=1
x
1 qi
i
" n X
i=1
1 pi
Z xi
0
(xi−si)βi+1
u(mi i)(si)
pi
dsi r#r·p1
,
whereV1 is given by (3.2).
Theorem 3.3. Under the assumptions of Theorem3.1, ifγ2 <−1,then
(3.3) Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis−γi 2i−(αi+1)Ωn/γ2qiωi ds1· · ·dsn
≤V1
n
Y
i=1
x
1 qi
i n
Y
i=1
Z xi
0
(xi−si)βi+1
u(mi i)(si)
pi
dsi 1
pi ,
whereV1 is given by (3.2).
Proof. Inequality (3.3) is proved when we set
Φi(si) =
u(mi i)(si) (mi−ki−1)!
in Theorem2.4.
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Corollary 3.4. Under the assumptions of Theorem3.3, ifr >0, we have
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis−γi 2
i−(αi+1)Ωn/γ2qiωi ds1· · ·dsn
≤pr·p1 V1 n
Y
i=1
x
1 qi
i
" n X
i=1
1 pi
Z xi
0
(xi−si)βi+1
u(mi i)(si)
pi
dsi
r#r·p1 .
We discuss a number of special cases of Theorem 3.1. Similar examples apply also to Corollary3.2, Theorem3.3and Corollary3.4.
Example 3.1. Ifai = 0andbi = 1fori∈I,then Theorem3.1becomes
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis(qi imi−qiki−qi+1)/(qiωi)iΩn ds1· · ·dsn
≤V2
n
Y
i=1
x
1 qi
i n
Y
i=1
Z xi
0
(xi−si)
u(mi i)(si)
pi
dsi 1
pi ,
where
V2 = 1
Qn i=1
h
(mi−ki−1)!(qimi−qiki−qi+ 1)qi1i.
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Example 3.2. If ai = 0, bi = 1, qi =n, pi = n/(n−1), mi =mandki = k fori∈I,then
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis(nm−nk−n+1)/(nωi) i
iΩn ds1· · ·dsn
≤
√n
x1· · ·xn (m−k−1)!n
(nm−nk−n+ 1)
×
n
Y
i=1
Z xi
0
(xi−si)
u(m)i (si)
n n−1 dsi
n−1n .
For q = p = n = 2 andωi = n1 this is [12, Theorem 1]. Settingq = p = 2, k = 0, n= 1andωi = n1, we recover the result of [14].
Example 3.3. Ifai = 0andbi = 1fori∈I,then Theorem3.1becomes
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis(mi i−ki)/(qiωi)iΩn ds1· · ·dsn
≤V3
n
Y
i=1
x
1 qi
i n
Y
i=1
Z xi
0
(xi−si)mi−ki
u(mi i)(si)
pi
dsi pi1
,
where
V3 = 1
Qn i=1
(mi−ki)!.
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Example 3.4. If ai = 1, bi = 0, qi =n, pi = n/(n−1), mi =mandki = k fori∈I.Then (3.1) becomes
Z x1
0
· · · Z xn
0
Qn i=1
u(ki i)(si) h 1
Ωn
Pn
i=1ωis(m−k)/(nωi i)iΩn ds1· · ·dsn
≤
√n
x1· · ·xn
(m−k)!n n
Y
i=1
Z xi
0
(xi−si)m−k
u(m)i (si)
n/(n−1)
dsi (n−1)n
.
Example 3.5. Letp1, p2 ∈R+.If we setn = 2, ω1 = p1
1, ω2 = p1
2, mi = 1and ki = 0fori = 1,2in Theorem3.1, then by our assumptionsq1 =p1/(p1−1), q2 =p2/(p2−1),and we obtain
Z x1
0
Z x2
0
|u1(s1)| |u2(s2)|
h 1 p1p2Ω2
p2s(p11−1)+p1s(p22−1)iΩ2 ds1ds2
≤x(p11−1)/p1x(p2 2−1)/p2 Z x1
0
(x1−s1)|u01(s1)|p1 ds1
p1
1
× Z x2
0
(x2−s2)|u02(s2)|p2 ds2
p1
2
. If we set ω1 +ω2 = 1 in Example 3.5, then we have [13, Theorem 2]. (The values ofai andbiare irrelevant.)
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Hilbert-Pachpatte Type Integral Inequalities and their
Improvement S.S. Dragomir and Young-Ho Kim
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J. Ineq. Pure and Appl. Math. 4(1) Art. 16, 2003
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