# A Reverse Hardy-Hilbert’s Inequality Involving One Partial Sum as the Terms of Double Series

Authors: Bicheng Yang; Shanhe Wu; Xingshou Huang
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, by constructing proper weight coefficients and utilizing the Euler-Maclaurin summation formula and the Abel partial summation formula, we establish reverse Hardy-Hilbert’s inequality involving one partial sum as the terms of double series. On the basis of the obtained inequality, the equivalent conditions of the best possible constant factor associated with several parameters are discussed. Finally, we illustrate that more reverse inequalities of Hardy-Hilbert type can be generated from the special cases of the present results.


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## Body

## 1. Introduction

The celebrated Hardy-Hilbert’s inequality asserts that ifp>1,1/p+1/q=1, am≥0,bn≥00<∑m=1∞amp<∞, and 0<∑n=1∞bnq<∞, then
(1)∑m=1∞∑n=1∞ambnm+n<πsinπ/p∑m=1∞amp1/p∑n=1∞bnq1/q,where the constant factor π/sinπ/p is best possible (see [1]).A generalization of inequality (Equation (1)) was posted by Krnić and Pečarić in [2], as follows:
(2)∑m=1∞∑n=1∞ambnm+nλ<Bλ1,λ2∑m=1∞mp1−λ1−1amp1/p∑n=1∞nq1−λ2−1bnq1/q,where λi∈0,2i=1,2,λ1+λ2=λ∈0,4, and also the constant factor Bλ1,λ2 given by the beta function is the best possible.In 2019, Adiyasuren et al. [3] put forwarded an analogous version of Hardy-Hilbert’s inequality containing the kernel 1/m+nλ and partial sums. Recently, Liu [4] and Yang [5] provided some extensions of Hardy-Hilbert type inequalities.In 2020, by introducing more parameters, a generalization of inequality (Equation (2)) was established by Yang et al. [6], which is a further generalization of inequality (Equation (1)), as follows:
(3)∑n=1∞∑m=1∞ambnmα+nβλ<1βkλλ21/p1αkλλ11/q×∑m=1∞mp1−αλ−λ2/p+λ1/q−1amp1/p∑n=1∞nq1−βλ−λ1/q+λ2/p−1bnq1/q,where 1/p+1/q=1,p>1,α,β∈0,1,λ∈0,6,λ1∈0,2/a∩0,λ,λ2∈0,2/β∩0,λ,kλλi≔Bλi,λ−λii=1,2.In 2021, Liao et al. [7] considered a variation of inequality (Equation (3)) and proposed the following inequality containing partial sums as the terms of series:
(4)∑m=1∞∑n=1∞ambnmα+nβλ<λ1βkλ+1λ2+11/p1αkλ+1λ11/q×∑m=1∞mp1−αλ∧1−1amp1/p∑n=1∞nq1−β1+λ∧2−1Bnq1/q,where 1/p+1/q=1,p>1,α,β∈0,1,λ∈0,5,λ1∈0,2/a∩0,λ+1,λ2∈0,2/β−1∩0,λ+1,λ^1≔λ−λ2/p+λ1/q,λ^2≔λ−λ1/q+λ2/p,kλ+1λi≔Bλi,λ+1−λii=1,2,Bn≔∑k=1nbk.Inspired by the work [4–7] mentioned above, in this paper, we establish reverse Hardy-Hilbert’s inequality which contains one partial sum as the terms of double series. Our method is mainly based on real analysis techniques and the applications of the Euler-Maclaurin summation formula and the Abel partial summation formula. Moreover, the equivalent conditions of the best possible constant factor associated with several parameters are discussed. As applications, we deal with some equivalent forms of the obtained inequality and illustrate that more reverse inequalities of Hardy-Hilbert type can be derived from the special cases of the current inequality.


## 2. Some Lemmas

For convenience, let us first state the conditions (C1) below, which will be used repeatedly in what follows.(C1)0<p<1,q<0,1/p+1/q=1,α,β∈0,1,λ∈0,6,λ1∈0,2/α∩0,λ,λ2∈0,2/β∩0,λ,λ^1≔λ−λ2/p+λ1/q,λ^2≔λ−λ1/q+λ2/p,kλλi≔Bλi,λ−λii=1,2. Furthermore, for am≥0 and bn≥0, we define the partial sums Bn≔∑k=1nbkn∈N=1,2,⋯, which satisfies limn⟶∞Bn/etnβ=0 denoted as Bn=oetnβt>0;n⟶∞ and
(5)0<∑m=1∞mp1−αλ∧1−1amp<∞,0<∑n=1∞nq1−βλ∧2−1bnq<∞.Lemma 1 (see [8]).
(i)
Let−1idi/dtiψt>0,t∈m,∞m∈N and ψi∞=0 and i=0,1,2,3, and let Pit,Bii∈N denote, respectively, the Bernoulli functions and Bernoulli numbers of i-order. Then,(6)∫m∞P2q−1tψtdt=−εqB2q2qψm,0<εq<1,q=1,2,⋯.
In particular, forq=1 and B2=1/6, we have the inequality:
(7)−112ψm<∫m∞P1tψtdt<0.
Forq=2 and B4=−1/30, we have the inequality:
(8)0<∫m∞P3tψtdt<1120ψm.(ii)
Ifft>0∈C3m,∞,fi∞=0i=0,1,2,3, then we have the following Euler-Maclaurin summation formula:(9)∑k=m∞fk=∫m∞ftdt+12fm+∫m∞P1tf′tdt,(10)∫m∞P1tf′tdt=−112f′m+16∫m∞P3tf‴tdt.Lemma 2.
Fors∈0,6,s2∈0,2/β∩0,s,kss2=Bs2,s−s2, we define the following weight coefficient:
(11)ϖαs2,m≔mαs−s2∑n=1∞βnβs2−1mα+nβs,m∈N.
Then, the following inequalities hold:(12)0<kss21−O1mαs2<ϖαs2,m<kss2,m∈N,where O1/mαs2≔1/kss2∫01/mαus2−1/1+usdu>0.Proof.
For fixedm∈N, we define a function ψm,t as follows:
(13)ψm,t≔βtβs2−1mα+tβs,t>0.
Using equality (Equation (9)), we have the following:
(14)∑n=1∞ψm,n=∫1∞ψm,tdt+12ψm,1+∫1∞P1tψ′m,tdt=∫0∞ψm,tdt−ℏm,ℏm≔∫01ψm,tdt−12ψm,1−∫1∞P1tψ′m,tdt.
It is easy to observe that−1/2ψm,1=−β/2mα+1s. In addition, integration by parts, it follows that
(15)∫01ψm,tdt=β∫01tβs2−1mα+tβsdt=u=tβ∫01us2−1mα+usdu=1s2∫01dus2mα+us=1s2us2mα+us01+ss2∫01us2mα+us+1du=1s21mα+1s+ss2s2+1∫01dus2+1mα+us+1>1s21mα+1s+ss2s2+1us2+1mα+us+101+ss+1s2s2+1mα+1s+2∫01us2+1du=1s21mα+1s+λs2s2+11mα+1s+1+ss+1s2s2+1s2+21mα+1s+2,−ψ′m,t=−ββs2−1tβs2−2mα+tβs+β2stβ+βs2−2mα+tβs+1=−ββs2−1tβs2−2mα+tβs+β2smα+tβ−mαtβs2−2mα+tβs+1=ββs−βs2+1tβs2−2mα+tβs−β2smαtβs2−2mα+tβs+1.Further, for 0<s2≤2/β,0<β≤1,s2<s≤6, it is not difficult to verify that
(16)−1ididtitβs2−2mα+tβs>0,−1ididtitβs2−2mα+tβs+1>0i=0,1,2,3.
By Equations (7)–(10), we obtain the following:
(17)ββs−βs2+1∫1∞P1ttβs2−2mα+tβsdt>−ββs−βs2+112mα+1s,−β2mαs∫1∞P1ttβs2−2mα+tβs+1dt=β2mαs12mα+1s+1−β2mαs6∫1∞P3ttβs2−2mα+tβs+1′′dt>β2mαs12mα+1s+1−β2mαs720tβs2−2mα+tβs+1′′t=1>β2mα+1−1s12mα+1s+1−β2mα+1s720s+1s+2β2mα+1s+3+βs+15−β−2βs2mα+1s+2+2−βs23−βs2mα+1s+1=β2s12mα+1s−β2s12mα+1s+1−β2s720s+1s+2β2mα+1s+2+βs+15−β−2βs2mα+1s+1+2−βs23−βs2mα+1s,and then we get the following:
(18)ℏm>1mα+1sℏ1+λmα+1s+1ℏ2+ss+1mα+1s+2ℏ3,where
(19)ℏ1≔1s2−β2−β−β2s212−β2s2−βs23−βs2720,ℏ2≔1s2s2+1−β212−β3s+15−β−2βs2720,ℏ3≔1s2s2+1s2+2−β4s+2720.
It is easy to see that(20)ℏ1≥1s2−β2−β−β2s212−sβ22−βs23−βs2720=ψs2720s2,where ψσσ∈0,2/β is defined by the following:
(21)ψσ≔720−420β+6sβ2σ+60β2+5sβ3σ2−sβ4σ3.
Further, we obtain that forβ∈0,1,s∈0,6,
(22)ψ′σ=−420β+6sβ2+260β2+5sβ3σ−3β4σ2≤−420β−6sβ2+260β2+5sβ32β=14sβ−180β<0.
Thus, it follows that(23)ℏ1≥ψs2720s2≥ψ2/β720s2=16s2>0.
Also, we find that fors2∈0,2/β,
(24)ℏ2>β26−β212−5s+1β2720=112−s+1140β2>0,ℏ3≥124−s+2720β3>00<s≤6.
Hence, we haveℏm>0. Setting t=mα/βu1/β, it follows that
(25)ϖαs2,m=mαs−s2∑n=1∞ψm,n<mαs−s2∫0∞ψm,tdt=βmαs−s2∫0∞tβs2−1mα+tβsdt=∫0∞us2−11+usdu=Bs2,s−s2=kss2.
On the other hand, by Equation (9), we have the following:
(26)∑n=1∞ψm,n=∫1∞ψm,tdt+12ψm,1+∫1∞P1tψ′m,tdt=∫1∞ψm,tdt+Hm,Hm≔12ψm,1+∫1∞P1tψ′m,tdt.
Note that1/2ψm,1=β/2mα+1s, and
(27)ψ′m,t=−ββs−βs2+1tβs2−2mα+tβs+β2smαtβs2−2mα+tβs+1.
Fors2∈0,2/β∩0,s,0<s≤6, by Equation (7), we find −ββs−βs2+1∫1∞P1ttβs2−2/mα+tβsdt>0, and
(28)β2mαs∫1∞P1ttβs2−2mα+tβs+1dt>−β2mαs12mα+1s+1>−β2s12mα+1s.
Hence, we have the following:(29)Hm>β2mα+1s−β2s12mα+1s≥β2mα+1s−6β12mα+1s=0,and then we obtain the following:
(30)ϖαλ2,m=mαs−s2∑n=1∞ψm,n>mαs−s2∫1∞ψm,tdt=mαs−s2∫0∞ψm,tdt−mαs−s2∫01ψm,tdt=kss21−O1mαs2>0,where O1/mαs2=1/kss2∫01/mαus2−1du/1+us, which satisfies the following:
(31)0<∫01/mαus2−1du1+us<∫01/mαus2−1du=1s2mαs2.
Therefore, the two-sided inequalities in Equation (12) follow. This completes the proof of Lemma 2.Moreover, by the same way as above, fors∈0,6,s1∈0,2/α∩0,s,kss1=Bs1,s−s1, one can obtain the following inequalities for another weight coefficient:
(32)kss11−O1nβs1<ωβs1,n≔nβs−s1∑m=1∞αmαs1−1mα+nβs<kss1n∈N.Lemma 3.
Under the assumption (C1), we have the following reverse Hardy-Hilbert’s inequality:(33)I≔∑n=1∞∑m=1∞ambnmα+nβλ>1βkλλ21/p1αkλλ11/q×∑m=1∞1−O1mαλ2mp1−αλ∧1−1amp1/p∑n=1∞nq1−βλ∧2−1bnq1/q.Proof.
By using the reverse Hӧlder’s inequality [9], we obtain the following:
(34)I=∑n=1∞∑m=1∞1mα+nβλmα1−λ1/qβnβ−11/pnβ1−λ2/pαmα−11/qamnβ1−λ2/pαmα−11/qmα1−λ1/qβnβ−11/pbn≥∑m=1∞∑n=1∞βmα+nβλmα1−λ1p−1nβ−1ampnβ1−λ2αmα−1p−11/p∑n=1∞∑m=1∞αmα+nβλnβ1−λ2q−1mα−1bnqmα1−λ1βnβ−1q−11/q=1β∑m=1∞ϖαλ2,mmp1−αλ∧1−1amp1/p1α∑n=1∞ωβλ1,nnq1−βλ∧2−1bnq1/q.
Hence, by virtue of Equations (12) and (32), for s=λ,si=λii=1,2, we get inequality (Equation (33)). The proof of Lemma 3 is complete.Lemma 4.
Lett>0, then we have the inequality:
(35)∑n=1∞e−tnβbn≤t∑n=1∞e−tnβBn.Proof.
In view ofBne−tnβ=o1n⟶∞, applying the Abel partial summation formula, we find the following:
(36)∑n=1∞e−tnβbn=limn⟶∞Bne−tnβ+∑n=1∞Bne−tnβ−e−tn+1β=∑n=1∞Bne−tnβ−e−tn+1β.
Since1−e−t<tt>0, and for β∈0,1,
(37)e−tn+1β≥e−tnβ+1⇔etn+1β−nβ−1≤1⇔n+1β−nβ−1=αn+θnβ−1−1≤0θn∈0,1,we acquire that
(38)∑n=1∞e−tnβbn≤∑n=1∞Bne−tnβ−e−tnβ+1=1−e−t∑n=1∞Bne−tnβ≤t∑n=1∞Bne−tnβ,which implies inequality (Equation (35)). Lemma 4 is proved.


## 3. Main Results

Theorem 5.
Under the assumption (C1), we have the following reverse Hardy-Hilbert’s inequality with one partial sum as the terms of double series:(39)I1≔∑m=1∞∑n=1∞amBnmα+nβλ+1>1λ1βkλλ21/p1αkλλ11/q×∑m=1∞1−O1mαλ2mp1−αλ∧1−1amp1/p∑n=1∞nq1−βλ∧2−1bnq1/q.
In particular, forλ1+λ2=λ, we have the following:
(40)0<∑m=1∞mp1−αλ1−1amp<∞,0<∑n=1∞nq1−βλ2−1bnq<∞,and the following inequality:
(41)I1=∑m=1∞∑n=1∞amBnmα+nβλ+1>Bλ1,λ2β1/pα1/qλ×∑m=1∞1−O1mαλ2mp1−αλ1−1amp1/p∑n=1∞nq1−βλ2−1bnq1/q.Proof.
In view of the fact that(42)1mα+nβλ+1=1Γλ+1∫0∞tλ+1−1e−mα+nβtdt,by using inequality (Equation (35)), it follows that
(43)I1=1Γλ+1∑m=1∞∑n=1∞amBn∫0∞tλ+1−1e−mα+nβtdt=1Γλ+1∫0∞tλ−1∑m=1∞e−mαtamt∑n=1∞e−nβtBndt≥1Γλ+1∫0∞tλ−1∑m=1∞e−mαtam∑n=1∞e−nβtbndt=1Γλ+1∑m=1∞∑n=1∞ambn∫0∞tλ−1e−mα+nβtdt=ΓλΓλ+1∑m=1∞∑n=1∞ambnmα+nβλ=1λI.
Then by Equation (33), we deduce inequality (Equation (39)). The proof of Theorem 5 is complete.Remark 6.
Fors=λ+1∈1,6,λ∈0,5,β=1,s1=λ^1∈0,2/a∩0,λ+1, in Equation (32), we have the following:
(44)kλ+1λ~11−O1nβλ~1<ω1λ~1,n=nλ−λ~1+1∑m=1∞αmαλ~1−1mα+nλ+1<kλ+1λ~1n∈N.Theorem 7.
Under the assumption (C1), ifλ1+λ2=λ∈0,5, then for β=1, λ1∈0,2/α∩0,λ,λ2∈0,1∩0,λ, the constant factor1/λ1/βkλλ21/p1/αkλλ11/q in Equation (39) is the best possible.Proof.
We may divide two cases ofλ2∈0,1∩0,λ andλ2=1<λ to prove that the constant factor 1/λα1/qBλ1,λ2 in Equation (41) (for β=1) is the best possible.
Case (i)λ2∈0,1∩0,λ. For any 0<ε<minpλ1,q1−λ2, we set the following:
(45)a~m≔mαλ1−ε/p−1,b~n≔nλ2−ε/q−1m,n∈N.
Since0<λ2−ε/q<1, by the decreasingness property of series, we have the following:
(46)B~n≔∑k=1nb~k=∑k=1nkλ2−ε/q−1<∫0ntλ2−ε/q−1dt=1λ2−ε/qnλ2−ε/qn∈N.
If there exists a constantM, M≥1/λα1/qBλ1,λ2, such that Equation (41) for β=1 is valid when 1/λα1/qBλ1,λ2 is replaced by M; then, in particular, by substitution of am=a~m,bn=b~n, and Bn=B~n in Equation (41) for β=1, we have the following:
(47)I~1≔∑n=1∞∑m=1∞1mα+nλ+1a~mB~n>M∑m=1∞1−O1mαλ2mp1−αλ1−1a~mp1/p∑n=1∞nq1−λ2−1b~nq1/q.
By using inequality (Equation (47)) and the decreasingness property of series, we obtain the following:
(48)I~1>M∑m=1∞1−O1mαλ2mp1−αλ1−1mpαλ1−αε−p1/p∑n=1∞nq1−λ2−1nqλ2−ε−q1/q=M∑m=1∞m−αε−1−∑m=1∞O1mαλ2+ε+11p1+∑n=2∞n−ε−11q>M∫1∞x−αε−1dx−O11p1+∫1∞y−ε−1dy1q=Mε1α−εO11pε+11q.
By utilizing inequality (Equation (44)) and setting λ^1=λ1−ε/p∈0,2/a∩0,λ,λ^2=λ−λ^1, we find the following:
(49)I~2<1αλ2−ε/q∑n=1∞nλ2+ε/p+1∑m=1∞αm+nλ+1mαλ1−ε/p−1n−ε−1=1αλ2−ε/q∑n=1∞ω1λ~1,nn−ε−1<1αλ2−ε/qkλ+1λ~11+∑n=2∞n−ε−1<Γλ~1Γλ~2+1αλ2−ε/qΓλ+11+∫1∞y−ε−1dy=λ~2Γλ~1Γλ~2εαλ2−ε/qλΓλε+1.
Then, we have(50)λ~2Γλ~1Γλ~2αλ2−ε/qλΓλε+1>εI~>M1/α−εO11/pε+11/q.
Lettingε⟶0+ along with the continuity of the gamma function, we deduce that 1/λα1/qBλ1,λ2≥M. Hence, M=1/λα1/qBλ1,λ2 is the best possible constant factor of Equation (41) (for β=1).
Case (ii)λ2=1<λ. For 0<ε<1, replacing λ by λ−ε in Equation (41) and taking λ1=λ−1,λ2=1−ε, and β=1, we obtain the following:
(51)∑m=1∞∑n=1∞amBnmα+nλ−ε+1>Bλ−1,1−εα1/qλ−εKp,ε∑n=1∞nqε−1bnq1/q,where Kp,ε≔∑m=1∞1−O1/mα1−εmq1−αλ−1−1amp1/p.
Ifλ1=1, then we have the following:
(52)nqε−1≤n−1=nq1−λ2−1,∑n=1∞nqε−1bnq≤∑n=1∞nq1−λ2−1bnq<∞.It follows that limε⟶0+∑n=1∞nqε−1bnq=∑n=1∞nq1−1−1bnq,, and similarly, limε⟶0+Kp,ε=Kp,0.
If there exists a constant factorM, M≥1/α1/qλBλ−1,1=1/α1/qλλ−1, such that (Equation (41)) for β=1,λ2=1,λ1=λ−1 is valid when we replace 1/α1/qλλ−1 by M, namely,
(53)∑m=1∞∑n=1∞amBnmα+nλ+1>M⋅Kp,0∑n=1∞n−1bnq1/q,
then, by applying Fatou’s lemma [10] and Equation (53), we deduce that
(54)limε⟶0+∑m=1∞∑n=1∞amBn/mα+nλ−ε+1Kp,ε∑n=1∞nqε−1bnq−1/q≥∑m=1∞∑n=1∞amBn/mα+nλ+1Kp,0∑n=1∞n−1bnq−1/q>M.
By the property of limitation, there exists a constantδ0∈0,1, such that for any δ∈0,δ0,
(55)∑m=1∞∑n=1∞amBn/mα+nλ−δ+1Kq,δ∑n=1∞nq1−1−δ−1bnq−1/q>M,namely,
(56)∑m=1∞∑n=1∞amBnmα+nλ−δ+1>M⋅Kp,δ∑n=1∞nq1−1−δ−1bnq1/q.
By Case (i), if the constant factor1/α1/qλ−δBλ−1,1−δ (Equation (51)) ε=δ is the best possible, we have 1/α1/qλ−δBλ−1,1−δ≥M. Then, for δ⟶0+, we have the following:
(57)1α1/qλλ−1=1α1/qλBλ−1,1≥M,which implies that M=1/α1/qλλ−1 is the best possible factor of Equation (41) (for β=λ1=1,λ2=λ−1). This completes the proof of Theorem 7.Theorem 8.
Under the assumption (C1), if the constant factor1/λ1/βkλλ21/p1/αkλλ11/q in Equation (39) is the best possible, then for
(58)λ−λ1−λ2∈−λ1p,2α−λ1p∩λ−λ1−2βp,λ−λ1p⊃0,we have λ1+λ2=λ..Proof.
Forλ^1=λ−λ2/p+λ1/q,λ^2=λ−λ1/q+λ2/p, we have the following:
(59)λ^1+λ^2=λ−λ2p+λ1q+λ−λ1q+λ2p=λ.
Ifλ−λ1−λ2∈−λ1p,2/α−λ1p⊃0, then 0<λ^1=λ−λ2/p+λ1/q≤2/α; if
(60)λ−λ1−λ2∈λ−λ1−2βp,λ−λ1p⊃0,then 0<λ^2≤2/β.
It is easy to observe thatλ^i<λi=1,2, and then, by Equation (58), it follows that λ^1∈0,2/α∩0,λ, λ^2∈0,2/β∩0,λ, and 1/λBλ^1,λ^2∈R+=0,∞. By Equation (41), we have the following:
(61)I1=∑m=1∞∑n=1∞amBnmα+nβλ+1>Bλ^1,λ^2β1/pα1/qλ×∑m=1∞1−O1mαλ∧2mp1−αλ∧1−1amp1/p∑n=1∞nq1−βλ∧2−1bnq1/q.
If the constant factor1/λ1/βkλλ21/p1/αkλλ11/q in Equation (39) is the best possible, then by Equation (61), we have the inequality:
(62)1λ1βkλλ21/p1αkλλ11/q≥Bλ^1,λ^2β1/pα1/qλ,that is
(63)kλλ21/pkλλ11/q≥Bλ^1,λ^2.
By the reverse Hӧlder’s inequality [9], we obtain the following:
(64)Bλ^1,λ^2=kλλ^1=kλλ−λ2p+λ1q=∫0∞11+uλuλ−λ2/p+λ1/q−1du=∫0∞11+uλuλ−λ2−1/puλ1−1/qdu≥∫0∞11+uλuλ−λ2−1du1/p∫0∞11+uλuλ1−1du1/q=∫0∞11+vλvλ2−1dv1/p∫0∞11+uλuλ1−1du1/q=kλλ21/pkλλ11/q.
In light of Equation (63), we deduce that Equation (64) keeps the form of equality. Thus, there exist constants A and B, such that they are not both zero and they satisfy Auλ−λ2−1=Buλ1−1a.e. in R+ (see [9]). Assuming that A≠0, we have uλ−λ2−λ1=B/Aa.e. in R+, and hence, λ−λ2−λ1=0, namely, λ1+λ2=λ. Theorem 8 is proved.


## 4. Equivalent Forms and Some Particular Inequalities

Theorem 9.
Under the assumption (C1), we have the following inequality which is equivalent to Equation (39):
(65)J1≔∑m=1∞mqαλ∧1−11−O1/mαλ2q−1∑n=1∞Bnmα+nβλ+1q1/q>1λ1βkλλ21/p1αkλλ11/q∑n=1∞nq1−βλ∧2−1bnq1/q.
In particular, forλ1+λ2=λ, we have the following:
(66)0<∑m=1∞mp1−αλ1−1amp<∞,0<∑n=1∞nq1−βλ2−1bnq<∞,and the following inequality which is equivalent to Equation (41):
(67)∑m=1∞mqαλ1−11−O1/mαλ2q−1∑n=1∞Bnmα+nβλ+1q1/q>1α1/qλBλ1,λ2∑n=1∞nq1−βλ2−1bnq1/q.Proof.
Suppose that Equation (65) is valid. By the reverse Hӧlder’s inequality [9], we have the following:
(68)I1=∑m=1∞1−O1mαλ21/pm1/q−αλ∧1amm−1/q+αλ∧11−O1/mαλ21/p∑n=1∞Bnmα+nβλ+1≥∑m=1∞mp1−αλ∧1−1amp1/pJ1.
Then by Equation (65), we obtain Equation (39). On the other hand, assuming that Equation (39) is valid, we set the following:
(69)am≔mqαλ∧1−11−O1/mαλ2q−1∑n=1∞Bnmα+nβλ+1q−1,m∈N.
Then, it follows thatJ1=∑m=1∞1−O1/mαλ2mp1−αλ∧1−1amp1/q.
IfJ1=∞, then Equation (65) is naturally valid; if J1=0, then it is impossible that makes Equation (65) valid, namely, J1>0. Suppose that 0<J1<∞. By Equation (39), we have the following:
(70)∑m=1∞1−O1mαλ2mp1−αλ∧1−1amp=J1q=I1>1λ1βkλλ21/p1αkλλ11/qJ1q−1∑n=1∞nq1−βλ∧2−1bnq1/q,J1>1λ1βkλλ21/p1αkλλ11/q∑n=1∞nq1−βλ∧2−1bnq1/q,namely, Equation (65) follows, which is equivalent to Equation (39). Theorem 9 is proved.Remark 10.
By the same way as above, under the assumption (C1), ifp<0,0<q<1, we can obtain the following equivalent inequalities:
(71)∑m=1∞∑n=1∞amBnmα+nβλ+1>1λ1βkλλ21/p1αkλλ11/q×∑m=1∞mp1−αλ∧1−1amp1/p∑n=1∞1−O1nβλ1nq1−βλ∧2−1bnq1/q,(72)∑m=1∞mqαλ∧1−1∑n=1∞Bnmα+nβλ+1q1/q>1λ1βkλλ21/p1αkλλ11/q∑n=1∞1−O1nβλ1nq1−βλ∧2−1bnq1/q.Theorem 11.
Under the assumption (C1), ifλ1+λ2=λ∈0,5, then for β=1, λ1∈0,2/α∩0,λ, and λ2∈0,1∩0,λ, the constant factor 1/λ1/βkλλ21/p1/αkλλ11/q in Equation (65) is the best possible. On the other hand, if the same constant factor in Equation (65) is the best possible, then for
(73)λ−λ1−λ2∈−λ1p,2α−λ1p∩λ−λ1−2βp,λ−λ1p⊃0,we have λ1+λ2=λ..Proof.
Ifλ1+λ2=λ, then by Theorem 7, the constant factor 1/α1/qλBλ1,λ2 in Equation (41) (for β=1) is the best possible. Then by Equation (68) (for λ1+λ2=λ,β=1), the constant factor in Equation (65) is still the best possible.
On the other hand, if the same constant factor,1/λ1/βkλλ21/p1/αkλλ11/q, in Equation (65) is the best possible, then by the equivalency of Equations (65) and (39), in view of I=Jq (in the proof of Theorem 9), it follows that the same constant factor in Equation (39) is the best possible. By using Theorem 7, we obtain λ1+λ2=λ. The proof of Theorem 11 is complete.Below, we illustrate that some reverse Hardy-Hilbert type inequalities can be derived from the special cases of our main results stated in Theorems5 and 9.Remark 12.
(i) Choosingβ=1,λ=1,λ1=1/r,λ2=1/sr>1,1/r+1/s=1 in Equations (41) and (67), respectively, we acquire the following inequalities with the best possible constant factor π/α1/qsinπ/r:
(74)∑m=1∞∑n=1∞amBnmα+n2>πα1/qsinπ/r∑m=1∞1−O1mα/smp1−α/r−1amp1/p∑n=1∞nq/r−1bnq1/q,∑m=1∞mqα/r−11−O1/mα/sq−1∑n=1∞Bnmα+n2q1/q>πα1/qsinπ/r∑n=1∞nq/r−1bnq1/q.
In particular, forr=s=2, we deduce from Equation (68) and Equation (71) that
(75)∑m=1∞∑n=1∞amBnmα+n2>πα1/q∑m=1∞1−O1mα/2mp1−α/2−1amp1/p∑n=1∞nq/2−1bnq1/q,∑m=1∞mqα/2−11−O1/mα/2q−1∑n=1∞Bnmα+n2q1/q>πα1/q∑n=1∞nq/2−1bnq1/q.
(ii) Puttingβ=1,λ=3,λ1=2,λ2=1 in Equations (41) and (67), respectively, we obtain the following inequalities with the best possible constant factor 1/6α1/q(76)∑m=1∞∑n=1∞amBnmα+n4>16α1/q∑m=1∞1−O1mαmp1−2α−1amp1/p∑n=1∞n−1bnq1/q,∑m=1∞m2qα−11−O1/mαq−1∑n=1∞Bnmα+n4q1/q>16α1/q∑n=1∞n−1bnq1/q.



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