# On a Generalization of Hilbert-Type Integral Inequality

Authors: Sun Baoju
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

By introducing some parameters, we establish generalizations of the Hilbert-type inequality. As applications, the reverse and its equivalent form are considered.


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## Body

## 1. Introduction

Considerable attention has been given to Hilbert inequalities
and Hilbert-type inequalities by several authors including Gao and Yang [1],
Yang [2–4], Jichang and Debnath [5], Pachpatte [6], Zhao [7], Brnetić and Pečarić [8]. In 2007, Li et al.
[9] gave a new inequality similar to Hilbert inequality for integrals:Iff(x),g(x)≥0,  0<∫0∞f2(x)dx<∞,0<∫0∞g2(x)dx<∞, then one has(1.1)∬0∞|lnx−lny|x+y+|x−y|f(x)g(y)dxdy<4[∫0∞f2(x)dx∫0∞g2(x)dx]1/2, where constant factor 4 is the best possible.An equivalent
inequality is(1.2)∫0∞[∫0∞|lnx−lny|x+y+|x−y|f(x)dx]2dy<16∫0∞f2(x)dx.In this paper,
by introducing some parameters we generalize (1.1), (1.2), and we obtain the reverse
form for each of them. The equivalent forms are also considered.


## 2. Main Results

Lemma 2.1.
Suppose   thatλ>0,  p>1(1/p+1/q=1), define
weight functions wλ(x,q),  wλ(y,p),
respectively as
(2.1)wλ(x,q)=∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|(xy)λ/q1y1−λdy,wλ(y,p)=∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|(yx)λ/p1x1−λdx.
One
has
wλ(x,q)=wλ(y,p)=(1/2λ2)(p2+q2).Proof.
Lettingt=yλ/xλ,
we have
(2.2)wλ(x,q)=1λ2∫0∞|lnt|1+t+|1−t|⋅t−1/qdt=1λ2[∫01−lnt2⋅t−1/qdt+∫1∞lnt2t⋅t−1/qdt]=12λ2(p2+q2). By symmetry we have
(2.3)wλ(y,p)=12λ2(p2+q2). The lemma is proved.Lemma 2.2.
Letp>1(or0<p<1), 1/p+1/q=1, λ>0, and 0<ε<qλ/2p, setting
(2.4)J(ε)=∬1∞|lnx−lny|xλ+yλ+|xλ−yλ|x−[1+ε+(p−1)(1−λ)]/py−[1+ε+(q−1)(1−λ)]/qdxdy. Then for ε→0+,
one gets
(2.5)p2+q22λ2ε[1+o(1)]−O(1)<J(ε)<p2+q22λ2ε[1+o(1)].Proof.
Lettingt=yλ/xλ,
we have
(2.6)J(ε)=1λ2∫1∞1x1+ε[∫1/xλ∞|lnt|1+t+|1−t|⋅t1/p−1−ε/λqdt]dx=1λ2∫1∞1x1+ε[∫0∞|lnt|1+t+|1−t|⋅t1/p−1−ε/λqdt]dx−1λ2∫1∞1x1+ε[∫01/xλ|lnt|1+t+|1−t|⋅t1/p−1−ε/λqdt]dx=1λ2ε∫1∞|lnt|1+t+|1−t|⋅t1/p−1−ε/λqdt−1λ2∫1∞1x1+ε[∫01/xλ|lnt|1+t+|1−t|⋅t1/p−1−ε/λqdt]dx:=I1−I2. Now, observe that
(2.7)∫0∞|lnt|1+t+|1−t|⋅t1/p−1−ε/λqdt=p2+q22[1+o(1)],(ε→0+),0<∫01/xλ|lnt|1+t+|1−t|⋅t1/p−1−ε/λqdt=∫01/xλ−lnt2⋅t1/p−1−ε/λqdt<∫01/xλ−lnt2⋅t1/2p−1dt=pλx−λ/2plnx+2p2x−λ/2p, then
(2.8)I1=p2+q22λ2ε[1+o(1)],0<I2<1λ2[pλ∫1∞x−1−λ/2plnxdx+2p2∫1∞x−1−λ/2pdx]=8p3λ3. We get
(2.9)J(ε)>p2+q22λ2ε[1+o(1)]−O(1). On the other hand,
(2.10)J(ε)=∬1∞|lnx−lny|xλ+yλ+|xλ−yλ|x−[1+ε+(p−1)(1−λ)]/py−[1+ε+(q−1)(1−λ)]/qdxdy<∫1∞[∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|x−[1+ε+(p−1)(1−λ)]/py−[1+ε+(q−1)(1−λ)]/qdy]dx=p2+q22λ2ε[1+o(1)]. Hence, (2.5) is valid. The lemma is proved.Theorem 2.3.
Letp>1, 1/p+1/q=1, λ>0,  f(x)≥0,g(y)≥0. If 0<∫0∞x(p−1)(1−λ)fp(x)dx<∞,  0<∫0∞y(q−1)(1−λ)gq(y)dy<∞,  then one has
(2.11)∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)g(y)dxdy<p2+q22λ2(∫0∞x(p−1)(1−λ)fp(x)dx)1/p(∫0∞y(q−1)(1−λ)gq(y)dy)1/q,where constant factor (p2+q2)/2λ2 is the best possible. In particular, for λ=1, inequality (2.11) reduces to
(2.12)∬0∞|lnx−lny|x+y+|x−y|f(x)g(y)dxdy<p2+q22(∫0∞fp(x)dx)1/p(∫0∞gq(y)dy)1/q.Proof.
Applying Hölder’s inequality and Lemma2.1, we have
(2.13)∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)g(y)dxdy=∬0∞[(|lnx−lny|xλ+yλ+|xλ−yλ|)1/p(xy)λ/pqx(1−λ)/qy(1−λ)/pf(x)]×[(|lnx−lny|xλ+yλ+|xλ−yλ|)1/q(yx)λ/pqy(1−λ)/px(1−λ)/qf(y)]dxdy≤{∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|(xy)λ/qx(p/q)(1−λ)y1−λfp(x)dxdy}1/p×{∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|(yx)λ/py(q/p)(1−λ)x1−λgq(y)dxdy}1/q≤p2+q22λ2(∫0∞x(p−1)(1−λ)fp(x)dx)1/p(∫0∞y(q−1)(1−λ)gq(y)dy)1/q.
If (2.13)
takes the form of equality, then there exist constants A and B,
such that they are not all zero, and
(2.14)A|lnx−lny|xλ+yλ+|xλ−yλ|(xy)λ/qx(p/q)(1−λ)y1−λfp(x)=B|lnx−lny|xλ+yλ+|xλ−yλ|(yx)λ/py(q/p)(1−λ)x1−λgq(y), a.e.  (x,y) in (0,∞)×(0,∞).
It follows that there exists a constant C, such that
(2.15)Ax⋅x(p−1)(1−λ)fp(x)=By⋅y(q−1)(1−λ)gq(y)=C,a.e.(x,y)in(0,∞)×(0,∞). Without lose of generality, suppose A≠0,
then we have
(2.16)x(p−1)(1−λ)fp(x)=1Ax, which contradicts the fact that 0<∫0∞x(p−1)(1−λ)fp(x)dx<∞, hence (2.13) takes the form of strict
inequality, so we obtain (2.11).
Assume that the
constant factor(p2+q2)/2λ2 in (2.11) is not the best possible, then there
exists a positive number k (with k<(p2+q2)/2λ2) such that (2.11)
is still valid if one replaces (p2+q2)/2λ2 by k.
In particular, for 0<ε<qλ/2p, setting f˜ and g˜ as f˜(x)=g˜(x)=0 for x∈(0,1),  f˜(x)=x−[1+ε+(p−1)(1−λ)]/p,  g˜(x)=x−[1+ε+(q−1)(1−λ)]/q for x∈[1,∞), then we have
(2.17)k(∫0∞x(p−1)(1−λ)f˜p(x)dx)1/p(∫0∞y(q−1)(1−λ)g˜q(y)dy)1/q=kε. By using Lemma 2.2, we find
(2.18)∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|f˜(x)g˜(y)dxdy=∬1∞|lnx−lny|xλ+yλ+|xλ−yλ|x−[1+ε+(p−1)(1−λ)]/py−[1+ε+(q−1)(1−λ)]/qdxdy>p2+q22λ2ε[1+o(1)]−O(1). Therefore, we get
(2.19)p2+q22λ2ε[1+o(1)]−O(1)<kε or
(2.20)p2+q22λ2[1+o(1)]−εO(1)<k. For ε→0+, it
follows that (p2+q2)/2λ2≤k. This contradicts the fact that k<(p2+q2)/2λ2. Hence, the constant factor in (2.11) is the best
possible. Theorem 2.3 is proved.Theorem 2.4.
Let0<p<1,  1/p+1/q=1,  λ>0,  f(x)≥0,g(y)≥0. If 0<∫0∞x(p−1)(1−λ)fp(x)dx<∞,  0<∫0∞y(q−1)(1−λ)gq(y)dy<∞,  then one has
(2.21)∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)g(y)dxdy>p2+q22λ2(∫0∞x(p−1)(1−λ)fp(x)dx)1/p(∫0∞y(q−1)(1−λ)gq(y)dy)1/q, where the constant factor (p2+q2)/2λ2 is the best possible. In particular, for λ=1,  the inequality reduces to
(2.22)∬0∞|lnx−lny|x+y+|x−y|f(x)g(y)dxdy>p2+q22(∫0∞fp(x)dx)1/p(∫0∞gq(y)dy)1/q.Proof.
Applying reverse Hölder’s inequality and the same arguments as before, we have (2.21).
If the
constant factor(p2+q2)/2λ2 in (2.21) is not the best possible, then there
exists a positive number h (with h>(p2+q2)/2λ2), such that (2.21)
is still valid if one replaces (p2+q2)/2λ2 by h.
In particular, for 0<ε<qλ/2p, setting f˜ and g˜ as in Theorem 2.3, we have
(2.23)h(∫0∞x(p−1)(1−λ)f˜p(x)dx)1/p(∫0∞y(q−1)(1−λ)g˜q(y)dy)1/q=hε. By using Lemma 2.2, we find
(2.24)∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|f˜(x)g˜(y)dxdy=∬1∞|lnx−lny|xλ+yλ+|xλ−yλ|x−[1+ε+(p−1)(1−λ)]/py−[1+ε+(q−1)(1−λ)]/qdxdy<p2+q22λ2ε[I+o(1)]. Therefore, we get
(2.25)p2+q22λ2ε[1+o(1)]>hεorp2+q22λ2[1+o(1)]>h for ε→0+, and it
follows that (p2+q2)/2λ2≥h. This contradicts the fact that h>(p2+q2)/2λ2. Hence, the constant factor  in (2.21) is the best
possible. Theorem 2.4 is proved.Theorem 2.5.
Ifp>1,  1/p+1/q=1,  λ>0, f(x)≥0,  0<∫0∞x(p−1)(1−λ)fp(x)dx<∞, then one has
(2.26)∫0∞yλ−1[∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)dx]pdy<(p2+q22λ2)p∫0∞x(p−1)(1−λ)fp(x)dx,where the constant factor ((p2+q2)/2λ2)p is the best possible. Inequality (2.26) is
equivalent to (2.11).Proof.
Setting
(2.27)g(y)=yλ−1[∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)dx]p−1, then by (2.11), we find
(2.28)∫0∞y(q−1)(1−λ)gq(y)dy=∫0∞yλ−1[∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)dx]pdy=∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)g(y)dxdy≤p2+q22λ2(∫0∞x(p−1)(1−λ)fp(x)dx)1/p(∫0∞yλ−1[∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)dx]pdy)1/q. Hence, we obtain
(2.29)∫0∞y(q−1)(1−λ)gq(y)dy≤(p2+q22λ2)p∫0∞x(p−1)(1−λ)fp(x)dx, Thus, by (2.11), both (2.28) and (2.29) keep the form of strict
inequalities, then we have (2.26).
Applying Hölder’s inequality, we have
(2.30)∬0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)g(y)dxdy=∫0∞[y(λ−1)/p∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)dx][y(1−λ)/pg(y)]dy≤{∫0∞yλ−1[∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)dx]pdy}1/p×{∫0∞y(q−1)(1−λ)gq(y)dy}1/q. Therefore, by (2.26) we have (2.11). It follows that inequality
(2.26) is equivalent to (2.11), and the
constant factors in (2.26) are the
best possible. The theorem is proved.Theorem 2.6.
If0<p<1,  1/p+1/q=1,  λ>0,  f(x)≥0,  0<∫0∞x(p−1)(1−λ)fp(x)dx<∞, then one has
(2.31)∫0∞yλ−1[∫0∞|lnx−lny|xλ+yλ+|xλ−yλ|f(x)dx]pdy>(p2+q22λ2)p∫0∞x(p−1)(1−λ)fp(x)dx, where the constant factor ((p2+q2)/2λ2)p is the best possible. Inequality (2.31) is equivalent to (2.21).The proof of
Theorem2.6 is similar to that of Theorem 2.5, so we omit it.



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