volume 7, issue 2, article 76, 2006.
Received 22 September, 2005;
accepted 05 January, 2006.
Communicated by:B. Yang
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Journal of Inequalities in Pure and Applied Mathematics
FOUR INEQUALITIES SIMILAR TO HARDY-HILBERT’S INTEGRAL INEQUALITY
W.T. SULAIMAN
College of Computer Sciences and Mathematics University of Mosul, Iraq.
EMail:waadsulaiman@hotmail.com
c
2000Victoria University ISSN (electronic): 1443-5756 285-05
Four Inequalities Similar To Hardy-Hilbert’s Integral
Inequality W.T. Sulaiman
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Abstract
Four new different types of inequalities similar to Hardy-Hilbert’s inequality are given.
2000 Mathematics Subject Classification:26D15.
Key words: Hardy-Hilbert integral inequality.
The author is grateful to the referee who read through the paper very carefully and correct many typing mistakes.
Contents
1 Introduction. . . 3 2 New Inequalities . . . 6
References
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Inequality W.T. Sulaiman
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1. Introduction
Suppose that f and g are real functions, such that 0 < R∞
0 f2(t)dt < ∞ and 0<R∞
0 g2(t)dt <∞, then (1.1)
Z ∞
0
Z ∞
0
f(x)g(y) x+y < π
Z ∞
0
f2(t)dt Z ∞
0
g2(t)dt 12
,
where π is best possible. If(an)and(bn)are sequences of real numbers such that0<P∞
n=1a2n<∞and0<P∞
n=1b2n <∞, then (1.2)
∞
X
n=1
∞
X
m=1
ambn m+n < π
∞
X
n=1
a2n
∞
X
n=1
b2n
!12 .
The inequalities (1.1) and (1.2) are called Hilbert’s inequalities. These inequal- ities play an important role in analysis (cf. [1, Chap. 9]). In their recent papers Hu [5] and Gao [3] gave two distinct improvements of (1.1) and Gao [4] gave a strengthened version of (1.2).
The following definitions are given:
ϕλ(r) = r+λ−2
r (r=p, q), kλ(p) = B(ϕλ(p), ϕλ(q)), andB is the beta function.
Recently, by introducing some parameters, Yang and Debnath [2] gave the following extensions:
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Inequality W.T. Sulaiman
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Theorem A. Iff, g≥0, p >1, 1p +1q = 1, λ >2−min{p, q},such that 0<
Z ∞
0
t1−λfp(t)dt <∞ and 0<
Z ∞
0
t1−λgq(t)dt <∞, then
(1.3) Z ∞
0
Z ∞
0
f(x)g(y)
(Ax+By)λdxdy
< kλ(p) Aϕλ(p)Bϕλ(q)
Z ∞
0
x1−λfp(x)dx
1pZ ∞
0
y1−λgq(y)dy 1q
, where the constant factor[kλ(p)/Aϕλ(p)Bϕλ(q)]is the best possible.
Theorem B. If f ≥0, p > 1, 1p + 1q = 1, λ > 2−min{p, q}, A, B > 0such that0<R∞
0 t1−λfp(t)dt <∞,then (1.4)
Z ∞
0
y(λ−1)(p−1) Z ∞
0
f(x) (Ax+By)λdx
!p
dy
<
kλ(p) Aϕλ(p)Bϕλ(q)
pZ ∞
0
x1−λfp(x)dx, where the constant factor
kλ(p)/Aϕλ(p)Bϕλ(q)p
is the best possible. The in- equalities (1.3) and (1.4) are equivalent.
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Inequality W.T. Sulaiman
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Theorem C. Ifan, bn >0 (n ∈N), p >1,1p +1q = 1,2−min{p, q}< λ <2, A, B >0such that
0<
∞
X
n=1
n1−λapn<∞ and 0<
∞
X
n=1
n1−λbqn <∞, then
(1.5)
∞
X
n=1
∞
X
m=1
ambn
(Am+Bn)λ
< kλ(p) Aϕλ(p)Bϕλ(q)
∞
X
n=1
n1−λapn
!p1 ∞ X
n=1
n1−λbqn
!1q , where the constant factor
kλ(p)/Aϕλ(p)Bϕλ(q)
is the best possible.
Theorem D. Ifan ≥ 0 (n ∈ N), p >1,1p + 1q = 1,2−min{p, q} < λ ≤ 2, A, B >0such that0<P∞
n=1n1−λapn <∞,then (1.6)
∞
X
n=1
n(λ−1)(p−1)
∞
X
m=1
am (Am+Bn)λ
!p
<
kλ(p) Aϕλ(p)Bϕλ(q)
p ∞
X
n=1
n1−λapn, where the constant factor
kλ(p)/Aϕλ(p)Bϕλ(q)p
is the best possible. The in- equalities (1.5) and (1.6) are equivalent.
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Inequality W.T. Sulaiman
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2. New Inequalities
The aim of this paper is to give the following results:
Theorem 2.1. Let lnf, lng be convex for nonnegative functionsf andg such that f(0) = g(0) = 0, f(∞) = g(∞) = ∞, f0(s) ≥ 0, g0(s) ≥ 0, s ∈ {xp, yq}.Letλ >max{p, q}, p >1, 1p + 1q = 1. Let
0<
Z ∞
0
t−p2/q2[f(tp)]2−λ+p/q
[f0(t)]pq dt <∞, 0<
Z ∞
0
t−q2/p2[g(tq)]2−λ+q/p
[g0(t)]qp dt <∞, then we have
(2.1) Z ∞
0
Z ∞
0
f(xy)g(xy)
(f(xp) +g(yq))λdxdy
≤ 1
√p
p√q
qBp1 (p, λ−p)B1q (q, λ−q)
× Z ∞
0
t−p2/q2[f(tp)]2−λ+p/q [f0(t)]pq dt
!p1
× Z ∞
0
t−q2/p2[g(tq)]2−λ+q/p [g0(t)]qp dt
!1q .
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Proof. Sincelnf is convex andxy≤ xpp + yqq,then f(xy) =elnf(xy) ≤elnf
xp p+yqq
≤elnfp(xp)+lnf(yqq ) =fp1(xp)f1q(yq).
Therefore, we have Z ∞
0
Z ∞
0
f(xy)g(xy)
(f(xp) +g(yq))λdxdy
≤ Z ∞
0
Z ∞
0
fp1(xp)g1q(yq)[g0(yq)]
1 p
[f0(xp)]1q y
q−1 p
x
p−1 q
(f(xp) +g(yq))λp
f1q(yq)g1p(xp)[f0(xp)]
1q
[g0(yq)]
1 p
x
p−1 q
y
q−1 p
(f(xp) +g(yq))λq
dxdy
≤ Z ∞
0
Z ∞
0
f(xp)gp/q(yq)g0(yq)yq−1
x(p−1)p/q[f0(xp)]pq (f(xp) +g(yq))λdxdy
!p1
× Z ∞
0
Z ∞
0
f(yq)gq/p(xp)f0(xp)xp−1
y(q−1)q/p[g0(yq)]pq (f(xp) +g(yq))λdxdy
!1q
=M1pN1q, say.
Then
M = 1 q
Z ∞
0
x(1−p)p/q[f(xp)]2−λ+p/q [f0(x)]pq dx
Z ∞
0
g(yq) f(xp)
pq
g0(yq)qyf(xq−1p)
1 + fg(y(xqp))
λ dy
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Inequality W.T. Sulaiman
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= 1 q
Z ∞
0
x−p2/q2[f(xp)]2−λ+p/q [f0(x)]pq dx
Z ∞
0
up/q (1 +u)λdu
= 1
qB(p, λ−p) Z ∞
0
x−p2/q2[f(xp)]2−λ+p/q [f0(x)]pq dx.
Similarly,
N = 1
pB(q, λ−q) Z ∞
0
y−q2/p2[g(yq)]2−λ+q/p [g0(y)]qp dy.
Therefore Z ∞
0
Z ∞
0
f(xy)g(xy)
(f(xp) +g(yq))λdxdy
≤ 1
√q
p√p
qB1p(p, λ−p)B1q (q, λ−q)
× Z ∞
0
tt−p2/q2[f(tp)]2−λ+p/q [f0(t)]pq dt
!1p
× Z ∞
0
t−q2/p2[g(tq)]2−λ+q/p [g0(t)]qp dt
!1q .
Theorem 2.2. Let f, g, hbe nonnegative functions, h(x, y)is homogeneous of ordern such thath(0,1) =h(1,0) = 0, h(∞,1) = h(1,∞) =∞. Letp > 1,
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Inequality W.T. Sulaiman
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1
p + 1q = 1,
0<1 +µ−r < λ, r∈n
p q,qpo
, hx(x, y)≥0, hy(x, y)≥0, wherehx =dh/dx,0<R∞
0 tp/qfp(t)dt <∞,0<R∞
0 tq/pgq(t)dt <∞,then (2.2)
Z ∞
0
Z ∞
0
f(x)g(y)hµ(x, y) (1 +h(x, y))λ dxdy
≤ 1 nB1p
1 +µ−p
q, λ−1−µ+p q
B1q
1 +µ−q
p, λ−1−µ−q p
× Z ∞
0
tq−1p fp(t)dt
1pZ ∞
0
tp−1q gq(t)dt 1q
. Proof. Observe that
Z ∞
0
Z ∞
0
f(x)g(y)hµ(x, y) (1 +h(x, y))λ dxdy
≤ Z ∞
0
Z ∞
0
fp(x)hµ(x, y)hy(x, y) hp/qx (x, y) (1 +h(x, y))λdxdy
!1p
× Z ∞
0
Z ∞
0
gq(y)hµ(x, y)hx(x, y) hq/py (x, y) (1 +h(x, y))λdxdy
!1q
=M1pN1q, say.
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Then
M =
Z ∞
0
fp(x)dx Z ∞
0
hµ(x, y)hy(x, y)
hp/qx (x, y) (1 +h(x, y))λdy.
Lety=xv,dy=xdvand hence hy(x, y) = dh(x, xv)
dy =xndh(1, v)
dy =xn−1dh(1, v)
dv =xn−1hv(1, v), hx(x, y) = dh(x, xv)
dx = d
dxxnh(1, v) =nxn−1h(1, v), therefore
M = 1
np/q Z ∞
0
xp/q−1fp(x)dx Z ∞
0
[xnh(1, v)]µ−p/qxnhv(1, v) (1 +xnh(1, v))λ dv
= 1
np/qB
1 +µ−p
q, λ−1−µ+p q
Z ∞
0
xp/q−1fp(x)dx.
Similarly,
N = 1
nq/pB
1 +µ− q
p, λ−1−µ+ q p
Z ∞
0
yq/p−1gq(y)dy.
This implies Z ∞
0
Z ∞
0
f(x)g(y)hµ(x, y) (1 +h(x, y))λ dxdy
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≤ 1 nB1p
1 +µ−p
q, λ−1−µ+p q
B1q
1 +µ− q
p, λ−1−µ+q p
× Z ∞
0
tp/q−1fp(t)dt
1pZ ∞
0
tq/p−1gq(t)dt 1q
.
The following lemma is needed for the coming result.
Lemma 2.3. Lets ≥1,0<1 +µ≤min{α, λ}and define f(s) = s−α
Z s
0
tµ (1 +t)λdt, thenf(s)≤f(1).
Proof. We have
f0(s) =s−α sµ (1 +s)λ +
Z s
0
tµ
(1 +t)λdt(−α)s−α−1
≤ sµ−α
(1 +s)λ − αs−α−1 (1 +s)λ
Z s
0
tµdt
= sµ−α (1 +s)λ
1− α 1 +µ
≤0.
This shows thatf is nonincreasing and hencef(s)≤f(1).
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Theorem 2.4. Letf, g, F, Gbe nonnegative functions such thatF(s) =Rs
0 f(t)dt, G(s) = Rs
0 g(t)dt, let, p > 1, 1p + 1q = 1, (1−λ/2)r ≤ λ/2 +α ≤ 2α, r ∈n
p q,qpo
, 0<
Z x
0
(x−t)t(1−λ/2)p/q−λ/2−α
Fp−1(t)f(t)dt < ∞, 0<
Z x
0
(x−t)t(1−λ/2)q/p−λ/2−α
Gq−1(t)g(t)dt <∞, then
(2.3) Z x
0
Z x
0
F(s)G(t)
(s+t)λ dsdt≤
√p
p√q q 2 xαB
λ 2,λ
2
× Z x
0
(x−t)t(1−λ/2)p/q−λ/2−αFp−1(t)f(t)dt 1p
× Z x
0
(x−t)t(1−λ/2)q/p−λ/2−α
Gq−1(t)g(t)dt 1q
. Proof. Observe that
Z x
0
Z x
0
F(s)G(t) (s+t)λ dsdt
= Z x
0
Z x
0
F(s)
t1/p s1/q
λ/2−1
(s+t)λ/p ·
G(t)
s1/q t1/p
λ/2−1
(s+t)λ/q dsdt
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≤ Z x
0
Z x
0
Fp(s)tλ/2−1
(s+t)λs(λ/2−1)p/qdsdt
!1p Z x
0
Z x
0
Gq(t)sλ/2−1
(s+t)λt(λ/2−1)q/pdsdt
!1q
=M1pN1q, say.
Then
M =
Z x
0
s(1−λ/2)p/q−λ/2
Fp(s)ds Z x
0 t s
λ/2−1 1
s
1 + stλ dt
= Z x
0
s(1−λ/2)p/q−λ/2
Fp(s)dsx s
αx s
−αZ x/s
0
uλ/2−1 (1 +u)λdu
≤ Z x
0
s(1−λ/2)p/q−λ/2Fp(s)dsx s
αZ 1
0
uλ/2−1 (1 +u)λdu
= xα 2 B
λ 2,λ
2 Z x
0
s(1−λ/2)p/q−λ/2−α
Fp(s)ds, by virtue of the lemma.
As
Fp(s) =p Z s
0
Fp−1(u)f(u)du, then
M = p 2xαB
λ 2,λ
2 Z x
0
s(1−λ/2)p/q−λ/2−αds Z s
0
Fp−1(u)f(u)du
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= p 2xαB
λ 2,λ
2 Z x
0
Z s
0
u(1−λ/2)p/q−λ/2−α
Fp−1(u)f(u)dsdu
= p 2xαB
λ 2,λ
2 Z x
0
(x−s)s(1−λ/2)p/q−λ/2−α
Fp−1(s)f(s)ds.
Similarly,
N = q 2xαB
λ 2,λ
2 Z x
0
(x−t)t(1−λ/2)q/p−λ/2−α
Gq−1(t)g(t)dt.
Therefore, we have Z x
0
Z x
0
F(s)G(t) (s+t)λ dsdt
≤
√p
p√q q 2 xαB
λ 2,λ
2
Z x
0
(x−t)t(1−λ/2)p/q−λ/2−α
Fp−1(t)f(t)dt 1p
× Z x
0
(x−t)t(1−λ/2)q/p−λ/2−α
Gq−1(t)g(t)dt 1q
.
Theorem 2.5. Letf, gbe nonnegative functions,f is submultiplicative andgis concave nonincreasing, f0(x), g0(y) ≥ 0, f(0) = g(0) = 0, f(∞) = g(∞) =
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∞, p >1, 1p + 1q = 1,0< a+ 1 < λ,0< b+ 1 < λ,
0<
Z ∞
0
[f(x)]µp−bp/q[g(x)]1+a−λ
[f0(x)]pq dx <∞, 0<
Z ∞
0
[g(y)]µq−aq/p[f(y)]1+b−λ
[g0(y)]qp dy <∞, then
(2.4) Z ∞
0
Z ∞
0
fµ(xy) gλ/2(xy)dxdy
≤2λ/2B1p(a+ 1, λ−a−1)B1q (b+ 1, λ−b−1)
× Z ∞
0
[f(t)]µp[g(t)]1+a−λ−bp/q [g0(t)]pq dt
!1p
× Z ∞
0
[f(t)]µq[g(t)]1+b−λ−aq/p [g0(t)]qp dt
!1q .
Proof. Since√
xy ≤ x+y2 , then g(xy) = g
(√ xy)2
≥(g(√
xy))2 ≥
g
x+y 2
2
≥
g(x) +g(y) 2
2
,
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and hence Z ∞
0
Z ∞
0
fµ(xy) gλ/2(xy)dxdy
≤2λ/2 Z ∞
0
Z ∞
0
fµ(x)fµ(y)
(g(x) +g(y))λdxdy
= 2λ/2 Z ∞
0
Z ∞
0
fµ(x)[g(y)]a/p
[g(x)]b/q
[g0(y)]1/p [g0(x)]1/q
(g(x) +g(y))λp
·
fµ(y)[g(x)]b/q
[g(y)]a/p · [g0(x)]1/q
[g0(y)]1/p
(g(x) +g(y))λq
≤2λ/2 Z ∞
0
Z ∞
0
fµp(x)ga(y)g0(y)
[g(x)]bp/q[g0(x)]pq (g(x) +g(y))λdxdy
!1p
× Z ∞
0
Z ∞
0
fµq(y)gb(x)g0(x)
[g(y)]aq/p[g0(y)]qp(g(x) +g(y))λdxdy
!1q
= 2λ/2M1pN1q, say.
Then
M =
Z ∞
0
[f(x)]µp[g(x)]1+a−λ−bp/q [g0(x)]pq dx
Z ∞
0
g(y) g(x)
a g0(y) g(x)
1 + g(y)g(x)λdy
=B(a+ 1, λ−a−1) Z ∞
0
[f(x)]ηp[g(x)]1+a−λ−bp/q [g0(x)]pq dx.
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Similarly, we can show that
N =B(b+ 1, λ−b−1) Z ∞
0
[f(y)]µq[g(y)]1+b−λ−aq/p [g0(y)]qp dy.
The result follows.
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[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cam- bridge Univ. Press. Cambridge, 1952.
[2] B. YANGANDL. DEBNATH, On the extended Hardy-Hilbert’s inequality, J. Math. Anal. Appl., 272 (2002), 187–199.
[3] M. GAO, TAN LI AND L. DEBNATH, Some improvements on Hilbert’s integral inequality, J. Math. Anal. Appl., 229 (1999), 682–689.
[4] M. GAO, A note on Hilbert double series theorem, Hunan Ann. Math., 12(1- 2) (1992), 142–147.
[5] HU KE, On Hilbert’s inequality, Chinese Ann. of Math., 13B(1) (1992), 35–39.
[6] W. RUDIN, Principles of Mathematical Analysis, Third Edition, McGraw- Hill Book Co. 1976.