http://jipam.vu.edu.au/
Volume 7, Issue 2, Article 61, 2006
SOME EXTENSIONS OF HILBERT’S TYPE INEQUALITY AND ITS APPLICATIONS
YONGJIN LI AND BING HE DEPARTMENT OFMATHEMATICS
SUNYAT-SENUNIVERSITY
GUANGZHOU, 510275 P. R. CHINA
stslyj@mail.sysu.edu.cn hzs314@163.com
Received 28 November, 2005; accepted 19 January, 2006 Communicated by B. Yang
ABSTRACT. By introducing parametersλandµ, we give a generalization of the Hilbert’s type integral inequality. As applications, we give its equivalent form.
Key words and phrases: Hilbert’s integral inequality, Weight function.
2000 Mathematics Subject Classification. 26D15.
1. INTRODUCTION
Iff, g ≥0, p >1, 1p + 1q = 1,0<R∞
0 fp(x)dx <∞and0<R∞
0 gq(x)dx <∞, then
(1.1)
Z ∞ 0
Z ∞ 0
f(x)g(y)
x+y dxdy < π sin(π/p)
Z ∞ 0
fp(x)dx
1p Z ∞ 0
gq(x)dx 1q
,
(1.2)
Z ∞ 0
Z ∞ 0
f(x) x+ydx
p
dy <
π sin(π/p)
pZ ∞ 0
fp(x)dx,
where the constant factor sin(π/p)π is the best possible. Inequality (1.1) is known as Hardy- Hilbert’s integral inequality (see [1]); it is important in analysis and its applications (see [4]).
Under the same condition of (1.1), we have the Hardy-Hilbert type inequality similar to (1.1):
(1.3)
Z ∞ 0
Z ∞ 0
f(x)g(y)
max{x, y}dxdy < pq Z ∞
0
fp(x)dx
1pZ ∞ 0
gq(x)dx 1q
,
ISSN (electronic): 1443-5756 c
2006 Victoria University. All rights reserved.
The authors are grateful to the referee for the careful reading of the manuscript and for several useful remarks. The work was partially supported by the Foundation of Sun Yat-sen University Advanced Research Centre.
342-05
(1.4)
Z ∞ 0
Z ∞ 0
f(x) max{x, y}dx
p
dy <(pq)p Z ∞
0
fp(x)dx,
where the constant factorpqis the best possible. The corresponding inequality for series is:
(1.5)
∞
X
n=1
∞
X
m=1
ambn
max{m, n} < pq
∞
X
n=1
apn
!1p ∞ X
n=1
bqn
!1q ,
where the constant factorpq is also the best possible. In particular, whenp =q = 2, we have the well-known Hilbert type inequality:
(1.6)
Z ∞ 0
Z ∞ 0
f(x)g(y)
max{x, y}dxdy <4 Z ∞
0
f2(x)dx
12 Z ∞ 0
g2(x)dx 12
.
In recent years, Kuang (see [3]) gave a strengthened form as:
(1.7)
∞
X
n=1
∞
X
m=1
ambn max{m, n} <
( ∞ X
n=1
[pq−G(p, n)]apn
)1p ( ∞ X
n=1
[pq−G(q, n)]bqn )1q
,
whereG(r, n) = r+1/3r−4/3(2n+1)1/r >0 (r=p, q).
Yang (see [5, 8]) gave: forλ >2−min{p, q}
(1.8) Z ∞
0
Z ∞ 0
f(x)g(y)
max{xλ, yλ}dxdy < pqλ
(p+λ−2)(q+λ−2)
× Z ∞
0
x(p−1)(2−λ)−1
fp(x)dx
1pZ ∞ 0
x(q−1)(2−λ)−1
gq(x)dx 1q
and (1.9)
Z ∞ 0
Z ∞ 0
f(x)g(y)
max{xλ, yλ}dxdy < pqλ
(p+λ−2)(q+λ−2)
× Z ∞
0
x1−λfp(x)dx
1pZ ∞ 0
x1−λgq(x)dx 1q
.
At the same time, Yang (see [6, 7]) considered the refinement of other types of Hilbert’s in- equalities.
In this paper, we give a generalization of Hilbert’s type inequality and an improvement as:
Z ∞ 0
Z ∞ 0
f(xλ)g(yµ)
max{x, y}dxdy < pq λ1pµ1q
Z ∞ 0
xλ1−1fp(x)dx
1pZ ∞ 0
xµ1−1gq(x)dx 1q
,
whereλ >0andµ >0.
2. MAINRESULTS
Lemma 2.1. Supposer >1,1s+ 1r = 1, λ, µ, ε > 0.Then
(2.1)
Z ∞ 1
x−ε(1s+λrµ)−1 Z x−λ1
0
1
max{1, t}t−1−µεr dtdx=O(1) (ε→o+).
Proof. There existn ∈ Nwhich is large enough, such that1 + −1−µεr > 0forε ∈ (0,µn1 ]and x≥1, we have
Z x−λ1 0
1
max{1, t}t−1−µεr dt= Z x−λ1
0
t−1−µεr dt= 1 1 + −1−µεr
x−λ11+−1−µεr
.
Since fora≥1the functiong(y) = ya1y (y ∈(0,∞))is decreasing, we find 1
1 + −1−µεr
x−1λ1+−1−µεr
≤ 1
1 + −1−1/nr
x−λ11+−1−1/nr
,
so
0<
Z ∞ 1
x−ε(1s+λrµ)−1 Z x−1λ
0
1
max{1, t}t−1−µr dtdx
<
Z ∞ 1
x−1 1 1 + −1−1/nr
x−λ11+−1−1/nr
= 1 λ
1 1 + −1−1/nr
!2
.
Hence relation (2.1) is valid. The lemma is proved.
Now we study the following inequality:
Theorem 2.2. Supposef(x), g(x)≥0, p >1, 1p + 1q = 1,λ >0, µ >0and 0<
Z ∞ 0
xλ1−1fp(x)dx <∞, 0<
Z ∞ 0
xµ1−1gq(x)dx <∞.
Then (2.2)
Z ∞ 0
Z ∞ 0
f(xλ)g(yµ)
max{x, y}dxdy < pq λ1pµ1q
Z ∞ 0
xλ1−1fp(x)dx
1p Z ∞ 0
xµ1−1gq(x)dx 1q
,
where the constant factor pq
λ1pµ1q
is the best possible forλ =µ.
Proof. By Hölder’s inequality, we have Z ∞
0
Z ∞ 0
f(xλ)g(yµ) max{x, y}dxdy
= 1 λ
1 µ
Z ∞ 0
Z ∞ 0
h
x1λ−1f(x) i h
yµ1−1g(y) i
maxn
x1λ, y1µo dxdy
= 1 λ
1 µ
Z ∞ 0
Z ∞ 0
xλ1−1f(x)
maxn
xλ1, yµ1o1p
x(1−λ1)p/q y1−µ1
!1p x1λ yµ1
!pq1
×
yµ1−1g(y)
maxn
x1λ, y1µo1q
y(1−µ1) x(1−λ1)p/q
!1p y1µ xλ1
!pq1
dxdy
≤ 1 λ
1 µ
Z ∞
0
Z ∞ 0
xp(1λ−1) fp(x) maxn
xλ1, yµ1o
x(1−λ1)p/q y1−µ1
x1λ yµ1
!1q dxdy
1 p
×
Z ∞
0
Z ∞ 0
yq(1µ−1) gq(y) maxn
xλ1, yµ1o
y(1−µ1)q/p x1−λ1
yµ1 x1λ
!1p dxdy
1 q
.
Define the weight functionϕ(x),ψ(y)as ϕ(x) :=
Z ∞ 0
1 maxn
xλ1, yµ1o · x(1−1λ)p/q y1−1µ
xλ1 yµ1
!1q
dy, x∈(0,∞),
ψ(y) :=
Z ∞ 0
1 maxn
xλ1, yµ1o · y(1−µ1)q/p x1−λ1
yµ1 x1λ
!1p
dx, y∈(0,∞),
then above inequality yields Z ∞
0
Z ∞ 0
f(xλ)g(yµ) max{x, y}dxdy
≤ 1 λ
1 µ
Z ∞ 0
ϕ(x)xp(1λ−1)fp(x)dx
1pZ ∞ 0
ψ(y)yq(1µ−1)gq(y)dy 1q
.
For fixedx, lety1µ =xλ1t, we have
ϕ(x) := µx(p−1)(1−1λ) Z ∞
0
1
max{1, t}tp1−1dt
=µpqx(p−1)(1−λ1)
=µpqx(p−1)(1−λ1).
By the same token,ψ(y) =λpqy(q−1)(1−µ1), thus (2.3)
Z ∞ 0
Z ∞ 0
f(xλ)g(yµ)
max{x, y}dxdy ≤ pq λ1pµ1q
Z ∞ 0
xλ1−1fp(x)dx
p1 Z ∞ 0
xµ1−1gq(x)dx 1q
.
If (2.3) takes the form of the equality, then there exist constantscandd, such that Kuang (see [2])
cxp(λ1−1)fp(x)
max{xλ1, yµ1} · x(1−1λ)p/q y1−1µ
xλ1 y1µ
!1q
=d yq(1µ−1)gq(y) maxn
xλ1, yµ1o · y(1−µ1)q/p x1−λ1
yµ1 x1λ
!1p
a.e. on(0,∞)×(0,∞).
Then we have
cx1λfp(x) = dyµ1gq(y) a.e. on(0,∞)×(0,∞).
Hence we have
cxλ1fp(x) =dyµ1gq(y) = constant a.e. on(0,∞)×(0,∞), which contradicts the facts that
0<
Z ∞ 0
x1λfp(x)dx <∞ and 0<
Z ∞ 0
y1µgq(x)dx <∞.
Hence (2.3) takes the form of strict inequality. So we have (2.2).
For 0 < ε < 12, settingfε(x) = x
−ε−1/λ
p , forx ∈ [1,∞); fε(x) = 0, forx ∈ (0,1), and gε(y) = y
−ε−1/µ
q , for y ∈ [1,∞);gε(y) = 0, for y ∈ (0,1). Assume that the constant factor
pq
λ1pµ1q in (2.2) is not the best possible, then there exists a positive numberK with K < pq
λ1pµ1q, such that (2.2) is valid by changing pq
λ1pµ1q
toK. We have Z ∞
0
Z ∞ 0
f(xλ)g(yµ)
max{x, y}dxdy < K Z ∞
0
x1λ−1fp(x)dx
1pZ ∞ 0
x1µ−1gq(x)dx 1q
= K ε . Since
Z ∞ 0
1 max{1, t}t
−1−εµ
q dt =pq+o(1) (ε→0+),
settingyµ1 =xλ1t, by (2.1), we find Z ∞
0
Z ∞ 0
f(xλ)g(yµ) max{x, y}dxdy
= 1 λ
1 µ
Z ∞ 0
Z ∞ 0
h
x1λ−1f(x)i h
yµ1−1g(y)i maxn
x1λ, y1µo dxdy
= 1 λ
1 µ
Z ∞ 1
Z ∞ 1
x(λ1−1)+
−ε−1 λ
p y(1µ−1)+
−ε−1 µ q
max n
x1λ, y1µ
o dxdy
= 1 λ
1 µ
Z ∞ 1
Z ∞ x−1λ
x(λ1−1)+
−ε−1 λ
p (tµxµλ)(µ1−1)+
−ε−1 µ q
maxn
xλ1, txλ1o xµλµtµ−1dxdt
= 1 λ
Z ∞ 1
x−ε(1p+λqµ)−1 Z ∞
x−λ1
1
max{1, t}t−1−εµq dtdx
= 1 λ
Z ∞ 1
x−ε(1p+λqµ)−1
Z ∞
0
1
max{1, t}t−1−εµq dtdx− Z x−λ1
0
1
max{1, t}t−1−εµq dtdx
= 1 λε
"
pq
1
p + λqµ +o(1)
# .
Since forε >0small enough, we have 1 λ
"
pq
1
p +λqµ +o(1)
#
< K.
It is obvious thatλ1pµ1q ≤ λp+µq (i.e. λ1 1 1
p+λqµ ≤ 1
λ1pµ1q
)by Young’s inequality. Consider the case of taking the form of the equality for Young’s inequality, we getµ1q =λ
p−1
p ,i.e. λ=µ, Then 1
λ
"
pq
1
p +λqµ +o(1)
#
= pq λ1pµ1q
+o(1) < K.
Thus we get pq
λ1pµ1q
≤ K, which contradicts the hypothesis. Hence the constant factor pq
λ1pµ1q in
(2.2) is best possible forλ =µ.
Remark 2.3. Forλ=µ, inequality (2.2) becomes
(2.4)
Z ∞ 0
Z ∞ 0
f(xλ)g(yλ)
max{x, y}dxdy < pq λ
Z ∞ 0
xλ1−1fp(x)dx
p1 Z ∞ 0
xλ1−1gq(x)dx 1q
,
where the constant factor pqλ is the best possible.
Theorem 2.4. Supposef ≥0, p >1, λ > 0and0<R∞
0 x1λ−1fp(x)dx <∞. Then (2.5)
Z ∞ 0
Z ∞ 0
f(xλ) max{x, y}dx
p
dy < 1 λ(pq)p
Z ∞ 0
x1λ−1f(x)pdx,
where the constant factor λ1(pq)pis the best possible. Inequality (2.5) is equivalent to (2.4).
Proof. Settingg(y)as
Z ∞
0
xλ1−1f(x) maxn
xλ1, yλ1odx
p−1
>0, y∈(0,∞).
then by (2.4), we find λ−2
Z ∞ 0
y1λ−1gq(y)dy=λp−1 Z ∞
0
Z ∞ 0
f(xλ) max{x, y}dx
p
dy
= Z ∞
0
Z ∞ 0
f(xλ)g(yλ) max{x, y}dxdy
≤ pq λ
Z ∞ 0
xλ1−1fp(x)dx
1pZ ∞ 0
y1λ−1gq(y)dy 1q
. (2.6)
Hence we obtain
(2.7) 0<
Z ∞ 0
yλ1−1gq(y)dy≤λp(pq)p Z ∞
0
x1λ−1fp(x)dx <∞.
By (2.4), both (2.6) and (2.7) take the form of strict inequality, so we have (2.5).
On the other hand, suppose that (2.5) is valid. By Hölder’s inequality, we find Z ∞
0
Z ∞ 0
f(xλ)g(yλ) max{x, y}dxdy
= Z ∞
0
Z ∞ 0
f(xλ) max{x, y}dx
g(yλ)dy
≤ Z ∞
0
Z ∞ 0
f(xλ) max{x, y}dx
p
dy
1pZ ∞ 0
gq(yλ)dy 1q
. (2.8)
Then by (2.5), we have (2.4). Thus (2.4) and (2.5) are equivalent.
If the constant λ1(pq)p in (2.5) is not the best possible, by (2.8), we may get a contradiction that the constant factor in (2.4) is not the best possible. Thus we complete the proof of the
theorem.
Remark 2.5.
(i) For λ = µ = 1, (2.2) and (2.5) reduce respectively to (1.3) and (1.4). It follows that (2.2) is a new extension of (1.6) and (1.3) with some parameters and the equivalent form (2.4) is a new extension of (1.4).
(ii) It is amazing that (2.4) and (1.9) are different, although both of them are the extensions of (1.6) with(p, q)−parameter and the best constant factor.
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