Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2008, Article ID 637397, 10 pages doi:10.1155/2008/637397
Research Article On a New Weighted Hilbert Inequality He Leping,1 Gao Xuemei,2, 3 and Gao Mingzhe2 1
Department of Mathematics and Applied Mathematics, College of Mathematics and Computer Science, Jishou University, Hunan Jishou 416000, China 2 Department of Mathematics and Computer Science, Normal College of Jishou University, Hunan Jishou 416000, China 3 Department of Mathematics, College of Mathematics and Computer Science, Hunan Normal University, Hunan Changsha 410081, China Correspondence should be addressed to Gao Mingzhe,
[email protected] Received 13 January 2008; Accepted 18 May 2008 Recommended by L´aszlo´ Losonczi It is shown that a weighted Hilbert inequality for double series can be established by introducing a proper weight function. Thus, a quite sharp result of the classical Hilbert inequality for double series is obtained. And a similar result for the Hilbert integral inequality is also proved. Some applications are considered. Copyright q 2008 He Leping 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.
1. Introduction Let {an } and {bn } be two sequences of real numbers. It is all known that the inequality
∞ ∞ am bn mn n1 m1
2 ≤ π2
∞ n1
a2n
∞
bn2
1.1
n1
∞ 2 2 is called Hilbert theorem for double series 1, where ∞ n1 an < ∞, n1 bn < ∞, and the constant factor π 2 in 1.1 is the best possible value. And the equality in 1.1 holds if and only if {an } or {bn } is a zero-sequence. The corresponding integral form of 1.1 is that ∞
∞ ∞ 2 fxgy 1.2 dx dy ≤ π 2 f 2 xdx g 2 xdx, xy 0 0 0 ∞ ∞ where 0 f 2 xdx < ∞, 0 g 2 xdx < ∞, and the constant factor π 2 in 1.2 is the best possible value. Recently, various improvements and extensions of 1.1 and 1.2 appear in
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a great deal of papers see 2–6, etc.. The aim of the present paper is to build some new inequalities by using the weight function method and the technique of analysis, and then to study some applications of them. First we give some lemmas. Lemma 1.1. Let n be a positive integer. Then ∞
π 1 dx 1 ln n .
√ n1 2 n 2 n x2 1 x 0 Proof. Let a, e, f be real numbers. Then e 1 2 dx x 1 2 C, f ln |e fx| − ln a x arctan
2 2 a a a x2 e fx e2 a2 f 2
1.3
1.4
where C is an arbitrary constant. This result is given in the paper see 7. Based on this indefinite integral it is easy to deduce that the equality 1.3 holds. √ √ √ √ Lemma 1.2. If fx 1/x nn/x1/2 1 − x/1 x − n/1 n and gx √ √ √ √ 1/x nn/x1/2 1 x/1 x − n/1 n, where n ∈ N, x ∈ 0, ∞, then 1 fx and gx are monotonously decreasing in interval 0, ∞; 2
∞
fxdx π − πωn,
0
∞
gxdx π πωn,
1.5
0
where the weight function ω is defined by
√ √
n n − 1 ln n . 1.6 − ωn √ n1 π n1 √ √ √ Proof. 1 At first, notice that 1 − x/1 x 1/1 x, hence we can write fx in the √ √ √ form of: fx f1 x f2 x, where f1 x 1/x n xn/1 n and f2 x n/x √ √ n1 x x. It is obvious that the functions f1 x and f2 x are monotonously decreasing in 0, ∞. So, fx is also monotonously decreasing in 0, ∞. In the next place, notice that √ √ √ 1 − n/1 n 1/1 n, therefore we can write gx in the form of: gx g1 x √ √ √ √ √ g2 x, where g1 x n/1 nx n x and g2 x n/x n1 x. It is clear that the functions g1 x and g2 x are monotonously decreasing in 0, ∞. So, gx is also monotonously decreasing in 0, ∞. 2 Below, we need only to compute the first integral, √
∞ ∞ 1/2 √ n 1 n x 1− dx fxdx √ √ − x n x 1 n 1 x 0 0 1
√ ∞ ∞ 1/2 1/2 √ 1 1 n n x n dx − √ √ dx xn x xn x 1 n 0 1 x 0
π 1
√ ∞ ∞
√ n 1 1 dt − −2 n dt
√ 1 n n t2 n t2 1 t 0 0
√ ∞ π − π −2 n
0
√ 1 nπ . dt − √ 1 n n t2 1 t 1.7
He Leping et al.
3
By Lemma 1.1, we obtain the first integral of 1.5 at once after some simple computations and simplifications. Similarly, the second integral of 1.5 can be gotten. Lemma 1.3. Let {an } be a sequence of real numbers, and let cx be a real function and 1−cncm ≥ 2 0 n, m ∈ N. If ∞ n1 an < ∞, then ∞ ∞ ∞ ∞ am an am an
1 − cn cm . m n m1 n1 m n m1 n1
1.8
Proof. It is obvious that ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ am an
am an am an am an 1 − cn cm − cn cm. mn m n m1 n1 m n mn m1 n1 m1 n1 m1 n1
1.9
We need only to show that
Let hm
∞
k1
∞ ∞ ∞ ∞ am an am an cn cm. m n m n m1 n1 m1 n1
1.10
ak /m k. Then
∞ ∞ am an cm m n m1 n1
∞ ∞ m1
n1
an mn
am cm
∞ ∞ m1
k1
ak mk
am cm
∞
hmam cm
m1
∞
hnan cn
n1
∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ak am am an am an an cn an cn cn cn. n k n m n m m n n1 k1 n1 m1 n1 m1 m1 n1 1.11 ∞ Lemma 1.4. Let α be a real number, and let fx and cx be two real functions, and α f 2 xdx < ∞ and 1 − cx − α cy − α ≥ 0, where x, y ∈ α, ∞ × α, ∞. Then ∞ ∞ fxfy fxfy
dx dy 1 − cx − α cy − α dx dy. 1.12 α x y − 2α α x y − 2α Its proof is similar to that of Lemma 1.3. Hence, it is omitted. 2. Main results 2 Theorem 2.1. Let {an } and {bn } be two sequences of real numbers. If ∞ n1 an < ∞ and , then 2 2 2 2 4 ∞ ∞ ∞ ∞ ∞ ∞ am bn 4 2 2 2 2 ≤ π an − ωnan bn − ωnbn , mn m1 n1 n1 n1 n1 n1 2.1 where the weight function ωn is defined by 1.6.
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Journal of Inequalities and Applications
Proof. Let cx be a real function and it satisfies condition 1 − cn cm ≥ 0 n, m ∈ N. First, we suppose that bn an . By Lemma 1.3 and then applying Cauchy’s inequality we have
∞ ∞ am an m n m1 n1
2
∞ ∞ am an
1 − cn cm m n m1 n1
2
1/2 1/4
1/2 1/4 2 ∞ ∞ am 1 − cn cm an 1 − cn cm m n m n1/2
m1 n1
m n1/2
n
m
≤ J1 J2 , 2.2 where 1/2 ∞ ∞
a2m m J1 1 − cn cm , m n n m1 n1
J2
∞ ∞ m1 n1
1/2
a2n n 1 − cn cm . mn m 2.3
It is easy to deduce that 1/2 1/2 ∞ ∞ ∞ ∞
2 a2m m n 1 1 − cn cm 1 − cm cn an . J1 mn n mn m m1 n1 n1 m1 2.4 √ √ √ √ √ √ Let cx x/1 x. It is obvious that 1 − x/1 x y/1 y ≥ 0 for x ≥ 0 and √ √ √ √ y ≥ 0. Consider the function fx 1/x nn/x1/2 1 − x/1 x − n/1 n. By Lemma 1.2, the function fx is monotonously decreasing in 0, ∞. Hereby, we have J1
∞ ∞ n1
√ √ 1/2 n 1 m n 1− a2n √ √ m n m 1 m 1 n m1
∞ √
√ 1/2 ∞ 1 n x n 1− ≤ dx a2n . √ √ − x n x n 1 x 1 0 n1
2.5
Using 1.5, we can obtain immediately J1 ≤ π
∞
a2n − π
n1
∞
ωna2n .
2.6
ωna2n .
2.7
n1
Similarly, we have J2 ≤ π
∞ n1
a2n π
∞ n1
He Leping et al.
5
It follows from 2.2, 2.6, and 2.7 that
∞ ∞ am an m n m1 n1
2 ≤π
2
∞
2 −
a2n
n1
2
∞
ωna2n
2.8
,
n1
where the weight function ωn is defined by 1.6. Next, consider the case for bn / an . We can apply Schwarz’s inequality to estimate the left-hand side of 2.1 as follows:
∞ ∞ am bn m n m1 n1
4
1 ∞ 0
≤
am t
m1
1 0
m−1/2
∞
∞
2 2 n−1/2
bn t
dt
n1
m−1/2
am t
2 1 2 2 ∞ n−1/2 dt bn t dt 0
m1
∞ ∞ am an m n m1 n1
2.9
n1
2
∞ ∞ bm bn m n m1 n1
2 .
And then by using the inequality 2.8, the inequality 2.1 follows from 2.9 at once. It is obvious that the inequality 2.1 is a refinement of 1.1. Below, we give an extension of 1.2. Theorem 2.2. Let α be a real ∞ ∞number, x ≥ α and y ≥ α, and let fx and gx be two real functions, and α f 2 xdx < ∞ and α g 2 xdx < ∞. Then ∞ α
fxgx dx dy x y − 2α
4 ≤π
4
∞
2 f xdx 2
−
α
×
∞
∞
2 ωxf xdx 2
α
2 g 2 xdx
α
−
∞
2.10
2 2 ωxg xdx
,
α
where the weight function ω is defined by
ωx
⎧ ⎪ 0, ⎪ ⎨ x−α ⎪ ⎪ ⎩ − 1x−α
x α, √ √ x − α lnx − α x−α − , x > α. √ π1 x − α 1 x−α
Specially, when α 0, it is a refinement of 1.2.
2.11
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Journal of Inequalities and Applications
Proof. Let cx be a real function, and 1 − cx − α cy − α ≥ 0x, y ∈ α, ∞ × α, ∞. Firstly, we suppose that f g. Using Lemma 1.4 and then applying Cauchy’s inequality we have ∞ α
fxfy dx dy x y − 2α
2
∞ α
∞ α
fxfy
1 − cx − α cy − α dx dy x y − 2α
2
1/2
fx 1 − cx − α cy − α x − α 1/4 y−α x y − 2α1/2
2
1/2
fy 1 − cx − α cy − α y − α 1/4 × dx dy x−α x y − 2α1/2 ≤ J1 J2 , 2.12 where J1
∞ α
J2
∞ α
f 2 x 1 − cx − α cy − α x − α 1/2 dx dy, x y − 2α y−α
f 2 y 1 − cx − α cy − α y − α 1/2 dx dy. x y − 2α x−α
2.13
In the first place, we consider J1 : J1
∞ ∞ α
α
∞ ∞ α
π
0
∞
1 − cx − α cy − α x − α 1/2 dy f 2 xdx x − α y − α y−α ∞ ∞ 1/2
cy − α − cx − α x − α 1/2 1 1 2 du f xdx dy f 2 xdx 1u u x − α y − α y − α α α
f xdx 2
∞
α
ωxf xdx , 2
α
2.14
where 1 ωx π
∞ α
1 x − α 1/2
cy − α − cx − α dy. x − α y − α y − α
We need only to compute the weight function ω.
2.15
He Leping et al.
7 √ √ √ Let us select still cx x/1 x. Then cy − α − cx − α y − α/1 y − α − √ √ x − α/1 x − α. Using 1.4, we have √
1 ωx π 1 π
∞ α
∞ α
1 − π
√
√
y−α 1 x − α 1/2 x−α − dy √ √ x − α y − α y − α 1 y−α 1 x−α
√
y−α 1 x − α 1/2 dy √ x − α y − α y − α 1 y−α
∞ α
√ x − α 1/2 x−α 1 dy √ x − α y − α y − α 1 x−α ∞
2 √ x − α π
0
du − x − α u2
∞ 0
2.16
√ 1 x−α du −
√ 2 x − α u 1 u 1 x−α
√ √ x−α x−α x − α lnx − α . − − √ 1x−α π1 x − α 1 x−α
Notice that limx→α ωx 0. Hence, the function defined by 2.11 is just. So,we attain
J1 π
∞
f xdx 2
∞
α
ωxf xdx .
2.17
2
α
Similarly, we have
J2 π
∞
f xdx − 2
∞
α
ωxf xdx .
2.18
2
α
We obtain from 2.12, 2.17, and 2.18 that ∞ α
fxfy dx dy x y − 2α
2 ≤π
2
∞
2 f xdx 2
α
where the weight function ω is defined by 2.11.
−
∞ α
2 ωxf xdx 2
,
2.19
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Journal of Inequalities and Applications
Secondly, consider the case for f / g. We can apply Schwarz’s inequality to estimate the left-hand side of 2.10 as follows: ∞ α
fxgx dx dy x y − 2α
∞ ∞ 0
∞ ∞ fxe
∞ dx
gye
−u y−α
fxe
−u x−α
2 dx
∞ ∞
2 2 −u y−α du gye dy du
α
0
∞ ∞
α
∞ ∞
2 −uxy−2α fxfydx dy e du gxgydx dy e−uxy−2α du
α
∞ α
2 2 dy du
α
∞ ∞
0
−u x−α
α
0
2 2 −uxy−2α fxgydx dy e du
α
0
≤
4
0
fxfy dx dy x y − 2α
∞ α
α
gxgy dx dy x y − 2α
2 . 2.20
It follows from 2.19 and 2.20 that the inequality 2.10 is valid. 3. Applications As applications, we will give some new refinements of Hardy-Littlewood’s theorem and Widder’s theorem below. Let fx ∈ L2 0, 1 and fx / 0 for all x. Define a sequence {an } by an
1
n 0, 1, 2, . . . .
xn fxdx,
3.1
0
Hardy-Littlewood 1 proved that ∞
a2n < π
n0
1
f 2 xdx,
3.2
0
where π is the best constant that the inequality 3.2 keeps valid. Theorem 3.1. With the assumptions as the above-mentioned, define a sequence {an } by an 1 n−1/2 x fxdx, n 1, 2, . . .. Then 0 ∞
2 a2n
