# A Generalization on Some New Types of Hardy-Hilbert’s Integral Inequalities

Authors: Banyat Sroysang
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

Sulaiman presented, in 2008, new kinds of Hardy-Hilbert’s integral
inequality in which the weight function is homogeneous. In this
paper, we present a generalization on the kinds of Hardy-Hilbert’s integral
inequality.


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## Body

## 1. Introduction and Preliminaries

For any two nonnegative measurable functionsf and g such that
(1)0<∫0∞‍f2(x)dx<∞,0<∫0∞‍g2(y)dy<∞,
we have the Hilbert’s integral inequality [1] that
(2)∬0∞f(x)g(y)x+ydxdy<π(∫0∞f2(x)dx∫0∞‍g2(y)dy)1/2.
The constant π is the best possible. In 1925, Hardy [2] extended the Hilbert’s integral inequality into the integral inequality as follows. If p>1, 1/p+1/q=1, and f,g≥0 such that
(3)0<∫0∞‍fp(x)dx<∞,0<∫0∞‍gq(y)dy<∞,
then we have the Hardy-Hilbert’s integral inequality that
(4)∬0∞f(x)g(y)x+ydxdy<πsin(π/p)(∫0∞fp(x)dx)1/p(∫0∞gq(y)dy)1/q.
The constant π/sin(π/p) is the best possible. Both the two inequalities are important in mathematical analysis and its applications [3].In 1938, Widder [4] studied on the Stieltjes Transform Sf(y)=∫0∞‍f(x)/(x+y)dx.Now, we recall the beta functionB as follows:
(5)B(p,q)=∫01‍tp-1(1-t)q-1dt,wherep,q>0.In 2001, Yang [5] extended the Hardy-Hilbert’s integral inequality into the following integral inequality. If p,q>0, λ>2-min{p,q}, 1/p+1/q=1, and f,g≥0 such that
(6)0<∫0∞‍x1-λfp(x)dx<∞,0<∫0∞‍y1-λgq(y)dy<∞,then we have(7)∬0∞f(x)g(y)(x+y)λdxdy<kλ(p)(∫0∞x1-λfp(x)dx)1/p(∫0∞y1-λgq(y)dy)1/q,
where kλ(p)=B(1+(λ-2)/p,1+(λ-2)/p). The constant kλ(p) is the best possible.We also recall that a nonnegative functionf(x,y) which is said to be homogeneous function of degree λ if f(tx,ty)=tλf(x,y) for all t>0. And we say that K(u,v) is increasing if K(1,t) and K(t,1) are increasing functions.In 2008, Sulaiman [6] gave new integral inequality similar to the Hardy-Hilbert’s integral inequality. If a,b>0, p>1, 1/p+1/q=1, 0<λ≤min{(1-b)p/q,(1-a)q/p}, K(u,v) is a positive increasing homogeneous function of degree λ, and f,g≥0 and
(8)F(x)=∫0x‍f(t)dt,G(x)=∫0x‍g(t)dt∀x>0,
then, for all T>0, we have
(9)∬0TF(u)G(v)K(u,v)dudv≤TαpK1pqK2q(∫0T‍(T-t)Fp-1(t)f(t)dt)1/p×(∫0T‍(T-t)Gq-1(t)g(t)dt)1/q,
where
(10)K1=∫01‍ta-1K(1,t)dt,K2=∫01‍tb-1K(t,1)dt.In this paper, we present a generalization of the integral inequality (9) and its applications. Next proposition will be used in the next section.Proposition 1 (see [6]).
Letg be a positive increasing function, and a,b>0. Then, for all x≥1, one has
(11)x-a∫0x‍ta-1g(t)dt≤∫01‍ta-1g(t)dt.


## 2. Main Results

Theorem 2.
Let0<a, b<1<p, 1/p+1/q=1, 0<λ≤min{(1-b)p/q,(1-a)q/p}, and let K(u,v) be positive increasing homogeneous function of degree λ, and f,g≥0 and
(12)F(x)=∫0x‍f(t)dt,G(x)=∫0x‍g(t)dt∀x>0,
and letψ be a function such that ψ(x)≥x for all x>0.
Then, for allT>0, one has
(13)∬0TF(u)G(v)ψ(K(u,v))dudv≤T1-λpK1pqK2q(∫0T‍(T-t)Fp-1(t)f(t)dt)1/p×(∫0T‍(T-t)Gq-1(t)g(t)dt)1/q,
where
(14)K1=∫01‍ta-1K(1,t)dt,K2=∫01‍tb-1K(t,1)dt.Proof.
LetT>0 and M=∬0TF(u)G(v)/ψ(K(u,v))dudv.
By the Hölder inequality, the assumption ofψ, and the Tonelli theorem, we have
(15)M=∬0TF(u)v(a-1)/pu(b-1)/qψ1/p(K(u,v))×G(v)u(b-1)/qv(a-1)/pψ1/q(K(u,v))dudv≤(∬0TFp(u)va-1u(b-1)p/qψ(K(u,v))dudv)1/p×(∬0TGq(v)ub-1v(a-1)q/pψ(K(u,v))dudv)1/q=(∫0T‍Fp(u)∫0T‍u(1-b)p/qva-1ψ(K(u,v))dvdu)1/p×(∫0T‍Gq(v)∫0T‍ub-1v(1-a)q/pψ(K(u,v))dudv)1/q≤(∫0T‍Fp(u)∫0T‍u(1-b)p/qva-1K(u,v)dvdu)1/p×(∫0T‍Gq(v)∫0T‍ub-1v(1-a)q/pK(u,v)dudv)1/q=(∫0T‍Fp(u)∫0T‍u(1-b)(p/q)+a-1(v/u)a-1uλK(1,v/u)dvdu)1/p×(∫0T‍Gq(v)∫0T‍v(1-a)(q/p)+b-1(u/v)b-1vλK(u/v,1)dudv)1/q.
Now, we putt=v/u and dt=dv/u for the first integral, and then we put t=u/v and dt=du/v for the second integral.
And, by Proposition1, one has
(16)M≤(∫0T‍ua+(1-b)(p/q)-λFp(u)∫0T/u‍ta-1K(1,t)dtdu)1/p×(∫0T‍vb+(1-a)(q/p)-λGq(v)∫0T/u‍tb-1K(t,1)dtdv)1/q=(Ta+(1-b)(p/q)-λ∫0T‍Fp(u)(uT)a+(1-b)(p/q)-λ×∫0T/u‍ta-1K(1,t)dtdu)1/p×(Tb+(1-a)(q/p)-λ∫0T‍Gq(v)(vT)b+(1-a)(q/p)-λ×∫0T/u‍tb-1K(t,1)dtdv)1/q≤(Ta+(1-b)(p/q)-λ∫0T‍Fp(u)(uT)a×∫0T/u‍ta-1K(1,t)dtdu)1/p×(Tb+(1-a)(q/p)-λ∫0T‍Gq(v)(vT)b×∫0T/u‍tb-1K(t,1)dtdv)1/q=T1-λ(∫0T‍Fp(u)(uT)a∫0T/u‍ta-1K(1,t)dtdu)1/p×(∫0T‍Gq(v)(vT)b∫0T/u‍tb-1K(t,1)dtdv)1/q≤T1-λ(∫0T‍Fp(u)∫01‍ta-1K(1,t)dtdu)1/p×(∫0T‍Gq(v)∫01‍tb-1K(t,1)dtdv)1/q=T1-λ(K1∫0T‍Fp(u)du)1/p×(K2∫0T‍Gq(v)dv)1/q.
Then, by the assumption, one has(17)M≤T1-λ(K1∫0T‍∫0u‍(Fp(t))′dtdu)1/p×(K2∫0T‍∫0v‍(Gq(t))′dtdv)1/q=T1-λ(pK1∫0T‍∫0u‍Fp-1(t)f(t)dtdu)1/p×(qK2∫0T‍∫0v‍Gq-1(t)f(t)dtdv)1/q=T1-λ(pK1∫0T‍Fp-1(t)f(t)∫tT‍dudt)1/p×(qK2∫0T‍Gq-1(t)f(t)∫tT‍dvdt)1/q=T1-λ(pK1∫0T‍(T-t)Fp-1(t)f(t)dt)1/p×(qK2∫0T‍(T-t)Gq-1(t)f(t)dt)1/q=T1-λpK1pqK2q×(∫0T‍(T-t)Fp-1(t)f(t)dt)1/p×(∫0T‍(T-t)Gq-1(t)g(t)dt)1/q.
This proof is completed.


## 3. Applications

Corollary 3.
Let0<a, b<1<p, 1/p+1/q=1 and 0<λ≤min{(1-b)p/q,(1-a)q/p}, and let K(u,v) be a positive increasing homogeneous function of degree λ, and f,g≥0 and
(18)F(x)=∫0x‍f(t)dt,G(x)=∫0x‍g(t)dt∀x>0.
Then, for allT>0, one has(19)(a)∬0TF(u)G(v)K(u,v)dudv≤T1-λpK1pqK2q×(∫0T‍(T-t)Fp-1(t)f(t)dt)1/p×(∫0T‍(T-t)Gq-1(t)g(t)dt)1/q,(20)(b)∬0TF(u)G(v)1+K(u,v)dudv≤T1-λpK1pqK2q×(∫0T‍(T-t)Fp-1(t)f(t)dt)1/p×(∫0T‍(T-t)Gq-1(t)g(t)dt)1/q,(21)(c)∬0TF(u)G(v)(1+K(u,v))K(u,v)dudv≤T1-λpK1pqK2q×(∫0T‍(T-t)Fp-1(t)f(t)dt)1/p×(∫0T‍(T-t)Gq-1(t)g(t)dt)1/q,(22)(d)∬0TF(u)G(v)eK(u,v)dudv≤T1-λpK1pqK2q(∫0T‍(T-t)Fp-1(t)f(t)dt)1/p×(∫0T‍(T-t)Gq-1(t)g(t)dt)1/q,
where
(23)K1=∫01‍ta-1K(1,t)dt,K2=∫01‍tb-1K(t,1)dt.Proof.
(a) This follows from Theorem2 where ψ(x)=x for all x.
(b) This follows from Theorem2 where ψ(x)=1+x for all x.
(c) This follows from Theorem2 where ψ(x)=x+x2 for all x.
(d) This follows from Theorem2 where ψ(x)=ex for all x.


## 4. Open Problem

In this section, we pose a question that is how to generalize the integral inequality (13) if ψ may not satisfy the property ψ(x)≥x for all x>0.



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*Source: 925464-2013-10-30.xml*