# Inequalities Similar to Hilbert's Inequality

Authors: Chang-Jian Zhao
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

In the present paper, we establish some new inequalities similar
to Hilbert’s type inequalities. Our results provide some new estimates to these
types of inequalities.


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## Body

## 1. Introduction

The well-known classical Hilbert’s double-series inequality can be stated as follows [1, page 253].Theorem A.
Ifp1,p2>1 such that 1/p1+1/p2≥1 and 0<λ=2-1/p1-1/p2=1/q1+1/q2≤1, where, as usual, q1 and q2 are the conjugate exponents of p1 and p2, respectively,  then
(1)∑m=1∞‍∑n=1∞‍ambn(m+n)λ≤K(∑m=1∞‍amp1)1/p1(∑n=1∞‍bnp2)1/p2,
where K=K(p1,p2) depends on p1 and p2 only.In recent years, several authors [1–18] have given considerable attention to Hilbert’s double-series inequality together with its integral version, inverse version, and various generalizations. In particular, Pachpatte [11] established an inequality similar to inequality (1) as follows.Theorem 1.
Letp>1 be constant and 1/p+1/q=1. If a(s) and b(t) are real-valued functions defined for {0,1,…,m} and {0,1,…,n}, respectively, and a(0)=b(0)=0. Moreover, define the operators ∇ by ∇u(t)=u(t)-u(t-1). Then,
(2)∑s=1m∑t=1n|a(s)||b(t)|qsp-1+ptq-1≤1pqm(p-1)/pn(q-1)/q×(∑s=1m‍(m-s+1)|∇a(s)|p)1/p×(∑t=1n‍(n-t+1)|∇b(t)|q)1/q.The first aim of this paper is to establish a new inequality similar to Hilbert’s type inequality. Our result provides new estimates to this type of inequality.Theorem 2.
Letp>1 be constants, and 1/p+1/q=1. For i=1,2, let ai(si,ti) be real-valued functions defined for (si,ti), where si=1,2,…,mi;  ti=1,2,…,ni, and let mi, ni be natural numbers. Let ai(0,ti)=ai(si,0)=0, and define the operators ∇1,∇2 by
(3)∇1υi(si,ti)=υi(si,ti)-υi(si-1,ti),∇2υi(si,ti)=υi(si,ti)-υi(si,ti-1).
Then,(4)∑s1=1m1∑t1=1n1(∑s2=1m2∑t2=1n2(|a1(s1,t1)|p+|a2(s2,t2)|q)000000000000000×(max{p(s1t1)p/q,q(s2t2)q/p}Γp,q(s1,t1,s2,t2)000000000000000000·max{p(s1t1)p/q,q(s2t2)q/p})-1∑s1=1m1)≤1pq(m1n1)1/q(m2n2)1/p×(∑s1=1m1∑t1=1n1(m1-s1+1)(n1-t1+1)×|∇2∇1a1(s1,t1)|p∑t1=1n1)1/p×(∑s2=1m2∑t2=1n2(m2-s2+1)(n2-t2+1)×|∇2∇1a2(s2,t2)|p∑t2=1n2)1/p,
where
(5)Γp,q(s1,t1,s2,t2)=pqS((∑ξ2=1s2‍∑η2=1t2‍|∇2∇1a2(ξ2,η2)|q)-1∑ξ1=1s1‍∑η1=1t1‍|∇2∇1a1(ξ1,η1)|p00000000000×(∑ξ2=1s2‍∑η2=1t2‍|∇2∇1a2(ξ2,η2)|q)-1),(6)S(h)=h1/(h-1)elogh1/(h-1),h≠1.Remark 3.
Inequality (4) is just a similar version of the following inequality established by Pachpatte [11]:
(7)∑s=1x∑t=1y(∑k=1z∑r=1w|a(s,t)||b(k,r)|q(st)p-1+p(kr)q-1)≤1pq(xy)1/q(zw)1/p(∑s=1x∑t=1y(x-s+1)00000000000000000000000×(y-t+1)00000000000000000000000×|∇2∇1a(s,t)|p∑r=1w)1/p×(∑k=1z∑r=1w(z-k+1)(w-r+1)0000000000×|∇2∇1b(k,r)|q∑r=1w)1/q.On the other hand, leta1(s1,t1) and a2(s2,t2) change to a1(s1) and a2(s2), respectively, and, with appropriate transformation, we have
(8)∑s1=1m1∑s2=1m2|a1(s1)|p+|a2(s2)|qΓp,q(s1,s2)·max{ps1p/q,qs2q/p}≤1pqm11/qm21/p×(∑τ1=1m1‍(m1-τ1+1)|∇a1(τ1)|p)1/p×(∑τ2=1m2‍(m2-τ2+1)|∇a2(τ2)|q)1/q,
where
(9)Γp,q(s1,s2)=pqS(∑τ1=1s1|∇a1(τ1)|p∑τ2=1s2|∇a2(τ2)|q),
and S(h) is as in (6). This is just a similar version of inequality (2) in Theorem 1.The integral analogue of inequality (1) in Theorem A is as follows [1, page 254].Theorem B.
Letp1, p2, q1, q2, and λ be as in Theorem A. If f∈Lp(0,∞) and g∈Lq(0,∞), then
(10)∬0∞f(x)g(x)(x+y)λdxdy≤K(∫0∞‍fp(x)dx)1/p(∫0∞‍gq(y)dy)1/q,
where K=K(p,q) depends on p and q only.In [11], Pachpatte also established a similar version of inequality (10) as follows.Theorem 4.
Letp>1 be constants, and 1/p+1/q=1. If f(s) and g(t) are real-valued continuous functions defined on [0,x) and [0,y), respectively, and let f(0)=g(0)=0. Then,
(11)∫0x‍∫0y‍|f(s)||g(t)|qsp-1+ptq-1dtds≤1pqx(p-1)/py(q-1)/q×(∫0x‍(x-s)|f′(s)|pds)1/p×(∫0y‍(y-t)|g′(t)|qdt)1/q.Another aim of this paper is to establish a new integral inequality similar to Hilbert’s type inequality.Theorem 5.
Letp>1, and 1/p+1/q=1. For i=1,2, let hi≥1,  fi(si,ti) be real-valued differentiable functions defined on [0,xi)×[0,yi), where xi∈(0,∞),yi∈(0,∞), and fi(0,ti)=fi(si,0)=0. As usual, partial derivatives of fi are denoted by D1fi,D2fi,D12fi=D21fi, and so forth. Let
(12)D12*fi(si,ti)=D2(hifihi-1(si,ti)·D1fi(si,ti)).
Then,(13)∫0x1∫0y1(∫0x2∫0y2|f1h1(s1,t1)|p+|f2h2(s2,t2)|qLp,q(s1,t1,s2,t2)max{p(s1t1)p/q,q(s2t2)q/p}ds1dt1)ds2dt2≤1pq(x1y1)1/q(x2y2)1/p(∫0x1∫0y1(x1-s1)(y1-t1)|D12*f1(s1,t1)|pdt1ds1)1/p×(∫0x2∫0y2(x2-s2)(y2-t2)|D12*f2(s2,t2)|qdt2ds2∫0y2(x2-s2)(y2-t2))1/q,where
(14)Lp,q(s1,t1,s2,t2)=pqS(∫0s1∫0t1|D12*f1(ξ1,η1)|pdη1dξ1∫0s2∫0t2|D12*f2(ξ2,η2)|qdη2dξ2‍),
and S(h) is as in (6).Remark 6.
Inequality (13) is just a similar version of the following inequality established by Pachpatte [11]:
(15)∫0x1‍∫0y1‍(∫0x2‍∫0y2‍|f1(s1,t1)||f2(s2,t2)|q(s1t1)p-1+p(s2t2)q-1ds1dt1)ds2dt2≤1pq(x1y1)1/q(x2y2)1/p×(∫0x1∫0y1(x1-s1)(y1-t1)×|D2D1f1(s1,t1)|pds1dt1∫‍)1/p×(∫0x2∫0y2(x2-s2)(y2-t2)×|D2D1f2(s2,t2)|qds2dt2∫0y2(x2-s2)(y2-t2))1/q.On the other hand, letf1(s1,t1) and f2(s2,t2) change to f1(s1) and f2(s2), respectively, and, with appropriate transformation, we have
(16)∫0x1∫0x2|f1h1(s1)|p+|f2h2(s2)|qLp,q(s1,s2)max{ps1p/q,qs2q/p}ds1dt1≤1pqx11/qx21/p(∫0x1(x1-s1)|f1′(s1)|pds1)1/p×(∫0x2(x2-s2)|f2′(s2)|qds2)1/q,
where
(17)Lp,q(s1,s2)=pqS(∫0s1(s1-σ1)|f1′(σ1)|pdσ1∫0s2(s2-σ2)|f2′(σ2)|qdσ2).This is just a similar version of inequality (11) in Theorem 4.


## 2. Proof of Theorems

Proof of Theorem2.
From the hypotheses of Theorem2, we have
(18)|a1(s1,t1)|≤∑ξ1=1s1∑η1=1t1|∇2∇1a1(ξ1,η1)|,|a2(s2,t2)|≤∑ξ2=1s2∑η2=1t2|∇2∇1a2(ξ2,η2)|.
By using Hölder’s inequality and noticing the reverse Young’s inequality [19],
(19)s11/αs21/βS(s1s2)≥s1α+s2β,
for positive real numbers s1, s2 and 1/α+1/β=1, α>1, where S(h) is as in (6). Hence,
(20)|a1(s1,t1)|p+|a2(s2,t2)|qmax{p(s1t1)p/q,q(s2t2)q/p}≤1p∑ξ1=1s1∑η1=1t1|∇2∇1a1(ξ1,η1)|p+1q∑ξ2=1s2∑η2=1t2|∇2∇1a2(ξ2,η2)|q≤S(∑ξ1=1s1∑η1=1t1|∇2∇1a1(ξ1,η1)|p∑ξ2=1s2∑η2=1t2|∇2∇1a2(ξ2,η2)|q)×(∑ξ1=1s1∑η1=1t1|∇2∇1a1(ξ1,η1)|p)1/p×(∑ξ2=1s2∑η2=1t2|∇2∇1a2(ξ2,η2)|q)1/q.
Dividing both sides of (20) by
(21)Γp,q(s1,t1,s2,t2)=pqS(∑ξ1=1s1∑η1=1t1|∇2∇1a1(ξ1,η1)|p∑ξ2=1s2∑η2=1t2|∇2∇1a2(ξ2,η2)|q),
taking the sum of both sides of (20) over ti and si from 1 to mi and ni(i=1,2), respectively, and making use of Hölder’s inequality, we have
(22)∑s1=1m1∑t1=1n1(∑s2=1m2∑t2=1n2(|a1(s1,t1)|p+|a2(s2,t2)|q)000000000000000×(Γp,q(s1,t1,s2,t2)0000000000000000000·max{p(s1t1)p/q,q(s2t2)q/p})-1∑s2=1m2∑t2=1n2(|a1(s1,t1)|p+|a2(s2,t2)|q))≤1pq∑s1=1m1‍∑t1=1n1‍(∑ξ1=1s1‍∑η1=1t1‍|∇2∇1a1(ξ1,η1)|p)1/p×∑s2=1m2‍∑t2=1n2‍(∑ξ2=1s2‍∑η2=1t2‍|∇2∇1a2(ξ2,η2)|q)1/q.≤1pq(m1n1)1/q×(∑s1=1m1∑t1=1n1∑ξ1=1s1∑η1=1t1|∇2∇1a1(ξ1,η1)|p)1/p×(m2n2)1/p×(∑s2=1m2‍∑t2=1n2‍∑ξ2=1s2‍∑η2=1t2‍|∇2∇1a2(ξ2,η2)|q)1/q.=1pq(m1n1)1/q(m2n2)1/p×(∑ξ1=1m1∑η1=1n1(m1-ξ1+1)×(n1-η1+1)|∇2∇1a1(ξ1,η1)|p∑ξ1=1m1‍)1/p×(∑ξ2=1m2∑η2=1n2(m2-ξ2+1)(n2-η2+1)×|∇2∇1a2(ξ2,η2)|q∑ξ2=1m2∑η2=1n2(m2-ξ2+1)(n2-η2+1))1/q=1pq(m1n1)1/q(m2n2)1/p×(∑s1=1m1∑t1=1n1(m1-s1+1)(n1-t1+1)×|∇2∇1a1(s1,t1)|p∑t2=1n2(m2-s2+1)(n2-t2+1))1/p×(∑s2=1m2∑t2=1n2(m2-s2+1)(n2-t2+1)×|∇2∇1a2(s2,t2)|p∑t2=1n2(m2-s2+1)(n2-t2+1))1/p.
This completes the proof.Proof of Theorem5.
From the hypotheses of  Theorem5, we obtain for i=1,2:
(23)fihi(si,ti)=fihi(si,ti)-fihi(0,ti)-fihi(si,0)+fihi(0,0)=∫0si‍D1fihi(ξi,ti)dξi-∫0si‍D1fihi(ξi,0)dξi=∫0si‍(D1fihi(ξi,ti)-D1fihi(ξi,0))dξi=∫0si‍∫0ti‍D2(hifihi-1(ξi,ηi)·D1fi(ξi,ηi))dηidξi=∫0si‍∫0ti‍D12*fi(ξi,ηi)dηidξi.
From (23), Hölder’s integral inequality and in view of the reverse Young’s inequality (19), we have
(24)|f1h1(s1,t1)|p+|f2h2(s2,t2)|qmax{p(s1t1)p/q,q(s2t2)q/p}≤1p∫0s1∫0t1|D12*f1(ξ1,η1)|pdη1dξ1+1q∫0s2∫0t2|D12*f2(ξ2,η2)|qdη2dξ2.≤S(∫0s1∫0t1|D12*f1(ξ1,η1)|pdη1dξ1∫0s2∫0t2|D12*f2(ξ2,η2)|qdη2dξ2)×(∫0s1∫0t1|D12*f1(ξ1,η1)|pdη1dξ1)1/p×(∫0s2∫0t2|D12*f2(ξ2,η2)|qdη2dξ2)1/q.
Integrating both sides of (24) over si and ti from 1 to xi and yi (i=1,2), respectively, and by using Hölder’s integral inequality, we arrive at(25)∫0x1∫0y1(∫0x2∫0y2|f1h1(s1,t1)|p+|f2h2(s2,t2)|qLp,q(s1,t1,s2,t2)max{p(s1t1)p/q,q(s2t2)q/p}ds1dt1)ds2dt2≤1pq∫0x1∫0y1(∫0s1∫0t1|D12*f1(ξ1,η1)|pdη1dξ1)1/pds1dt1×∫0x2∫0y2(∫0s2∫0t2|D12*f2(ξ2,η2)|qdη2dξ2)1/qds2dt2.≤1pq(x1y1)1/q(∫0x1∫0y1(∫0s1∫0t1|D12*f1(ξ1,η1)|pdη1dξ1)dt1ds1∫)1/p×(x2y2)1/p(∫0x2∫0y2(∫0s2∫0t2|D12*f2(ξ2,η2)|pdη2dξ2)dt2ds2)1/p=1pq(x1y1)1/q(x2y2)1/p(∫0x1∫0y1(x1-s1)(y1-t1)|D12*f1(s1,t1)|pdt1ds1)1/p×(∫0x2∫0y2(x2-s2)(y2-t2)|D12*f2(s2,t2)|qdt2ds2)1/q.
This completes the proof.



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*Source: 861948-2013-09-10.xml*