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Research article

Some new Hardy-Hilbert-type inequalities with multiparameters

  • Received: 08 April 2021 Accepted: 11 October 2021 Published: 18 October 2021
  • MSC : 26D15

  • The purpose of this paper is to build some new Hardy-Hilbert-type inequalities with multiparameters and their equivalent forms and variants, which generalize some existing results. Similarly, the corresponding Hardy-Hilbert-type integral inequalities are also given.

    Citation: Limin Yang, Ruiyun Yang. Some new Hardy-Hilbert-type inequalities with multiparameters[J]. AIMS Mathematics, 2022, 7(1): 840-854. doi: 10.3934/math.2022050

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  • The purpose of this paper is to build some new Hardy-Hilbert-type inequalities with multiparameters and their equivalent forms and variants, which generalize some existing results. Similarly, the corresponding Hardy-Hilbert-type integral inequalities are also given.



    Let an,bn0,p>1, 1/p+1/q=1. If 0<n=1apn< and 0<n=1bqn<, then

    n=1m=1ambnm+n<πsin(π/p)[n=1apn]1/p[n=1bqn]1/q, (1.1)

    where the constant π/sin(π/p) is the best possible. The inequality (1.1) can be called as the well known Hardy-Hilbert's inequality [1]. An equivalent form of inequality (1.1) is presented as follows.

    n=1[m=1amm+n]p<[πsin(π/p)]pn=1apn, (1.2)

    where the constant [π/sin(π/p)]p is also the best possible. In connection with applications in analysis, their generalizations and variants have received considerable interest recent years [2,3,4,5,6,7,8,9,10,11,12]. By introducing some parameters, Yang [13] obtained a generalization of Hardy-Hilbert's integral inequality with a best constant factor that involves the beta function. In the paper [14], Das and sahoo considered a generalization of multiple Hardy-Hilbert's inequality with the best constant factor. Sroysang [15] established a generalization on the kinds of Hardy-Hilbert's integral inequality with the weight homogeneous function.

    By introducing a parameter, the following extension of (1.1) was obtained by Yang [3]. Let an,bn0,p>1, 1/p+1/q=1. If 0<n=1n(p1)(1λ)apn<,0<n=1n(q1)(1λ)bqn<, then

    n=1m=1ambnmλ+nλ<πλsin(π/p)[n=1n(p1)(1λ)apn]1/p[n=1n(q1)(1λ)bqn]1/q, (1.3)

    where the constat π/(λsin(π/p)) is the best possible.

    Nextly, Sun [8] gave an extension of (1.3) as follows. Let an,bn0, p>1, 1/p+1/q=1, 0<λ<min(p,q), c>0. If 0<n=1n(p1)(1λ)apn<, 0<n=1n(q1)(1λ)bqn<, then

    n=1m=1ambn(mλ+nλ)cλ<cλ,p[n=1n(p1)(1λ)apn]1/p[n=1n(q1)(1λ)bqn]1/q, (1.4)

    where the constant cλ,p=(1/c)B(λ/(cp),λ/(cq)) is the best possible, B(,) denotes the beta function.

    Finally, by introducing two parameters, Xu [9] presented another extension of (1.3) as follows. Let an,bn0,p>1, 1/p+1/q=1, 0<λ1<p, 0<λ2<q. If 0<n=1n(p1)(1λ1)apn<, 0<n=1n(q1)(1λ2)bqn<, then

    n=1m=1ambnmλ1+nλ2<πλ11/qλ21/psin(π/p)[n=1n(p1)(1λ1)apn]1/p[n=1n(q1)(1λ2)bqn]1/q, (1.5)

    where the constant π/(λ1/q1λ1/p2sin(π/p)) is the best possible.

    Recently, some Hilbert-type integral inequalities with quasi-homogeneous integral kernels and multiple functions were established [16]. By means of the technique of real analysis and the weight functions, Xin [17] obtained an equivalent statements of a Hilbert-type integral inequality with the nonhomogeneous kernel in the whole plane and the best constant factor related to the beta function. By using the weight function and the technique of real analysis, Liu [18,19] established some multi-parameter Hilbert-type integral inequalities with the hybrid kernel and the best constant factors, respectively. By using weight functions and introducing parameters, Chen and Yang [20] presented a reverse Hardy-Hilbert-type integral inequality involving one derivative function and the beta function. Based on the theory of operators, Hong et al. [21] obtained a necessary and sufficient condition and the best constant factor for the Hilbert-type multiple integral inequality with kernel. Liao et al. [22] investigated a new half-discrete Hilbert-type inequality involving the variable upper limit integral and partial sums and proved the equivalent conditions of the best possible constant factor related to several parameters.

    In the paper, motivated by the mentioned references above, we will obtain a Hardy-Hilbert-type inequality with multiparameters, which can be see as a new generalization of (1.3)–(1.5). And its equivalent form and variant are given. Furthermore, their integral forms are also presented.

    For convenience, we assume always that B(,) represents the beta function throughout the paper.

    Theorem 2.1. Let an,bn0, and p>1, 1/p+1/q=1, 0<λ1<p, 0<λ2<q, 0<cλ. If 0<n=1n(p1)(1cλ1λ)apn<, 0<n=1n(q1)(1cλ2λ)bqn<, then

    n=1m=1ambn(mλ1+nλ2)cλ<cλ,p[n=1n(p1)(1cλ1λ)apn]1/p[n=1n(q1)(1cλ2λ)bqn]1/q, (2.1)

    where the constant cλ,p=(1/λ1/q1λ1/p2)B(c/(λp),c/(λq)) is the best possible.

    Remark 2.1. The above inequality presented in (2.1) concretely produces some known Hardy-Hilbert type inequalities based on different settings of the parameters λ1,λ2 and λ.

    (A1) If λ1=λ2=λ=c, then inequality (2.1) reduces to the inequality (1.3) given in Yang [3].

    (A2) If λ1=λ2=λ, then inequality (2.1) develops into the inequality (1.4) presented in Sun [8].

    (A3) If λ=c, then inequality (2.1) converts into the inequality (1.5) given in Xu [9].

    The idea of proof of Theorem 2.1 comes similarly from [3] and [9]. To prove the Theorem 2.1, we need some lemmas in the same way, which are new generalizations of some lemmas given in [8,9].

    Lemma 2.1. Let p>1, 1/p+1/q=1, λ1,λ2>0, 0<cλ. Define the weight function ωλ1,λ2(y,p)=0(1/(xλ1+yλ2)c/λ)(yλ2/xλ1)c/(λp)xcλ1/λ1dx, y(0,) and weight coefficient ˜ωλ1,λ2(n,p)=m=1(1/(mλ1+nλ2)c/λ)(nλ2/mλ1)c/(λp)mcλ1/λ1. Then (a) ωλ1,λ2(y,p)=(1/λ1)B(c/(λp),c/(λq)), and (b) If 0<λ1<p, and nN, then ˜ωλ1,λ2(n,p)<ωλ1,λ2(n,p) holds.

    Proof. (a) Set t=xλ1/yλ2. Then

    ωλ1,λ2(y,p)=01(xλ1+yλ2)cλ(yλ2xλ1)c/(λp)xcλ1λ1dx=1λ101(1+t)cλtcλq1dt=1λ1B(cλp,cλq).

    (b) Since 1(cλ1/λ)+(cλ1/λp)>0, foxed λ1,λ2, and nN, the function f(x)=(1/(xλ1+nλ2)c/λ)(nλ2/xλ1)c/(λp)xcλ1/λ1 is strict deceasing in (0,). Then

    f(m)<mm1f(x)dxm=1f(m)<0f(x)dx.

    So we have ˜ωλ1,λ2(n,p)<ωλ1,λ2(n,p). This completes the proof of Lemma 2.1.

    Similar to Lemma 2.1, we can introduce the following lemma.

    Lemma 2.2. Let p>1, 1/p+1/q=1, λ1,λ2>0, 0<cλ. Define the weight function ˉωλ1,λ2(x,p)=0(1/(xλ1+yλ2)c/λ)(xλ1/yλ2)c/(λp)ycλ2/λ1dy, x(0,) and weight coefficient ˆωλ1,λ2(m,p)=n=1(1/(mλ1+nλ2)c/λ)(mλ1/nλ2)c/(λp)ncλ2/λ1. Then (a) ˉωλ1,λ2(x,p)=(1/λ2)B(c/(λp),c/(λq)), and (b) If 0<λ2<q, and mN, then ˆωλ1,λ2(m,p)<ˉωλ1,λ2(m,p) holds.

    Lemma 2.3. Let p>1, 1/p+1/q=1, λ1,λ2>0, 0<cλ, ε(0,q/(2p)). Then

    11x1+cλ1ελ1xλ101(1+u)cλucλ(1pεq)1dudx=O(1),ε0+. (2.2)

    Proof. It follows from ε(0,q/(2p)) that 1/pε/q>1/(2p). Then we have

    0<11x1+cλ1ελ1xλ101(1+u)cλucλ(1pεq)1dudx<11x1xλ101(1+u)cλucλ(1pεq)1dudx<11x1xλ101(1+u)cλuc2pλ1dudx<2pλc1x1cλ12pλdx=4p2λ2λ1c2,

    which implies (2.2). This finishes the proof of Lemma 2.3.

    Nextly, we will give the proof of Theorem 2.1.

    Proof. By using the Hölder's inequality and Lemmas 2.1 and 2.2, we can capture

    n=1m=1ambn(mλ1+nλ2)cλ=n=1m=1[am(mλ1+nλ2)cλp(mλ1nλ2)cλpqncλ2λ1pmcλ1λ1q][bn(mλ1+nλ2)cλq(nλ2mλ1)cλpqmcλ1λ1qncλ2λ1p][n=1m=1am(mλ1+nλ2)cλ(mλ1nλ2)cλqncλ2λ1mp(cλ1λ1)q]1/p[n=1m=1bn(mλ1+nλ2)cλ(nλ2mλ1)cλpmcλ1λ1nq(cλ2λ1)p]1/q=[n=1m=11(mλ1+nλ2)cλ(mλ1nλ2)cλqncλ2λ1m(p1)(1cλ1λ)apm]1/p×[n=1m=11(mλ1+nλ2)cλ(nλ2mλ1)cλpmcλ1λ1n(q1)(1cλ2λ)bqn]1/q=[m=1ˆωλ1,λ2(m,q)m(p1)(1cλ1λ)apm]1/p[n=1˜ωλ1,λ2(n,p)n(q1)(1cλ2λ)bqn]1/q<[m=1ˉωλ1,λ2(m,q)m(p1)(1cλ1λ)apm]1/p[n=1ωλ1,λ2(n,p)n(q1)(1cλ2λ)bqn]1/q=[1λ2B(cλp,cλq)]1/p[m=1m(p1)(1cλ1λ)apm]1/p[1λ1B(cλp,cλq)]1/q[n=1n(q1)(1cλ2λ)bqn]1/q=cλ,p[m=1m(p1)(1cλ1λ)apm]1/p[n=1n(q1)(1cλ2λ)bqn]1/q, (2.3)

    which implies (2.1). Now we prove that the constant cλ,p is the best possible. Suppose that the constant cλ,p in inequality (2.1) is not the best possible, then there exists positive k<cλ,p such that k replaces with cλ,p in inequality (2.1) holds all the same. Especially, for ε(0,q/2p), let ˆan=(1/n)1+(cλ1ε/λ)+(p1)(1cλ1/λ)p and ˆbn=(1/n)1+(cλ2ε/λ)+(q1)(1cλ2/λ)q for nN, then we obtain

    n=1n(p1)(1cλ1λ)ˆapn=n=11n1+λ1ελandn=1n(q1)(1cλ2λ)ˆbqn=n=11n1+λ2ελ. (2.4)

    On the other hand, we have

    1σ=11x1+σdx<n=11n1+σ=1+n=21n1+σ<1+11x1+σdx=1+1σ(σ>0). (2.5)

    It follows from (2.4) and (2.5) that we observe

    [n=1n(p1)(1cλ1λ)ˆapn]1/p[n=1n(q1)(1cλ2λ)ˆbqn]1/q=[λcλ1ε+O(1)]1/p[λcλ2ε+O(1)]1/q=[λcλ1ε(1+o(1))]1/p[λcλ2ε(1+o(1))]1/p=λcλ1/p1λ1/q2ε(1+o(1)), (2.6)

    as ε tends to 0+. According to the assumption, we have

    m=1n=1ˆamˆbn(mλ1+nλ2)cλ=m=1n=11(mλ1+nλ2)cλ(1m)1+(cλ1ε/λ)+(p1)(1cλ1/λ)p(1n)1+(cλ2ε/λ)+(q1)(1cλ2/λ)q>111(xλ1+yλ2)cλ(1x)1+(cλ1ε/λ)+(p1)(1cλ1/λ)p(1y)1+(cλ2ε/λ)+(q1)(1cλ2/λ)qdxdy=1λ211x1+cλ1ελ1xλ11(1+u)cλucλ(1pεq)1dudx=1λ211x1+cλ1ελ01(1+u)cλucλ(1pεq)1dudx1λ211x1+cλ1ελ1xλ101(1+u)cλucλ(1pεq)1dudx,

    where in the third equality we set that u=yλ2/xλ1. At the same time, when ε tends to 0+, we have

    limε0+B(cλ(1q+εq),cλ(1pεq))=B(cλq,cλp)=B(cλp,cλq),01(1+u)cλucλ(1pεq)1du=B(cλ(1q+εq),cλ(1pεq))=B(cλp,cλq)+o(1),

    which implies that if ε tends to 0+, the following equations holds true

    1λ211x1+cλ1ελ01(1+u)cλucλ(1pεq)1dudx=λcλ1λ2ε[B(cλp,cλq)+o(1)].

    It follows from Lemma 2.3 that when ε tends to 0+, we get

    m=1n=1ˆamˆbn(mλ1+nλ2)cλ>λcλ1λ2ε[B(cλp,cλq)+o(1)]O(1)=λcλ1λ2εB(cλp,cλq)(1+o(1))=λcλ1/p1λ1/q2εcλ,p[1+o(1)]. (2.7)

    Due to the equations (2.6) and (2.7), we can obtain

    λcλ1/p1λ1/q2εcλ,p(1+o(1))<m=1n=1ˆamˆbn(mλ1+nλ2)cλ<λcλ1/p1λ1/q2εk(1+o(1)),

    as ε tends to 0+. We have cλ,p<k, in contradiction with supposition. So the constant cλ,p in inequality (2.1) is the best possible. This finishes the proof of Theorem 2.1.

    Theorem 2.2. Let am0, and p>1, 1/p+1/q=1, 0<λ1<p, 0<λ2<q, 0<cλ. cλ,p=(1/λ1/q1λ1/p2)B(c/(λp),c/(λq)). If 0<m=1m(p1)(1λ1)apm<, then

    0<n=1ncλ2λ1[m=1am(mλ1+nλ2)cλ]p<cpλ,pm=1m(p1)(1cλ1λ)apm, (2.8)

    where the constant cpλ,p is the best possible. Moreover, inequality (2.8) is equivalent to inequality (2.1).

    Proof. There exists k0N, such that km=1(am/(mλ1+nλ2)c/λ)>0 for k>k0. Setting bn(k)=ncλ2/λ1[km=1(am/(mλ1+nλ2)c/λ)]p1 for k>k0. Thanks to (2.1), then we get

    0<kn=1n(q1)(1cλ2λ)bqn(k)=kn=1ncλ2λ1[km=1am(mλ1+nλ2)cλ]p=kn=1km=1ambn(k)(mλ1+nλ2)cλ<cλ,p[kn=1n(p1)(1cλ1λ)apn]1/p[kn=1n(q1)(1cλ2λ)bqn(k)]1/q,

    which imply the following inequality

    kn=1n(q1)(1cλ2λ)bqn(k)<cpλ,pkn=1n(p1)(1cλ1λ)apn.

    Letting k tends to , we have 0<n=1n(p1)(1cλ1λ)apn<. Then we obtain

    0<n=1ncλ2λ1[m=1am(mλ1+nλ2)cλ]p<cpλ,pm=1m(p1)(1cλ1λ)apm.

    Therefore, inequality (2.8) holds from inequality (2.1).

    On the other hand, if inequality (2.8) holds, by applying the Hölder's inequality, we have

    n=1m=1ambn(mλ1+nλ2)cλ=n=1ncλ2λ1pm=1am(mλ1+nλ2)cλ[n1cλ2λpbn]<[n=1ncλ2λ1[m=1am(mλ1+nλ2)cλ]p]1/p[n=1n(q1)(1cλ2λ)bqn]1/q<cλ,p[n=1n(p1)(1cλ1λ)apn]1/p[n=1n(q1)(1cλ2λ)bqn]1/q,

    which means that inequality (2.1) holds. Since the constant cλ,p in inequality (2.1) is the best possible, and inequality (2.1) is the equivalent to inequality (2.8), the constant cλ,p in inequality (2.8) is also the best possible. This completes the proof of Theorem 2.2.

    Now we present a variant of Theorem 2.1 as follows.

    Theorem 2.3. Let an,bn0, and p>1, 1/p+1/q=1, 0<λ1<p, 0<λ2<q, 0<cλ. If 0<n=2(lnnλ1)(p1)(1cλ)np1apn<, and 0<n=2(lnnλ2)(q1)(1cλ)nq1bqn<, then

    n=2m=2ambn(lnmλ1nλ2)cλ<cλ,p[n=2(lnnλ1)(p1)(1cλ)np1apn]1/p[n=2(lnnλ2)(q1)(1cλ)nq1bqn]1/q, (2.9)

    where the constant cλ,p=(1/λ1/q1λ1/p2)B(c/(λp),c/(λq)) is the best possible.

    Remark 2.2. The above inequality presented in (2.9) concretely produces some known Hardy-Hilbert type inequalities based on different settings of the parameters λ1,λ2 and λ.

    (B1) If λ1=λ2=λ=c=1, Theorem 2.3 reduces to Theorem 2.1 given by Yang [4].

    (B2) If λ=c, then Theorem 2.3 is converted into Theorem 4 obtained by Xu [9].

    (B3) If λ1=λ2=λ=c=1 and am,bn are replaced by am/m,bn/n, respectively, then the inequality (2.9) is developed into the Muholland's inequality presented in [10].

    When am,bn are replaced by am/m,bn/n in inequality (2.9), respectively, it is easy to obtain that inequality (2.9) is equivalent to the following new variant of Muholland's inequality.

    n=2m=2ambnmn(lnmλ1nλ2)cλ<cλ,p[n=2(lnnλ1)(p1)(1cλ)n1apn]1/p[n=2(lnnλ2)(q1)(1cλ)n1bqn]1/q,

    where 0<n=2(lnnλ1)(p1)(1cλ)n1apn<, 0<n=2(lnnλ2)(q1)(1cλ)n1bqn<, and the constant cλ,p=(1/λ1/q1λ1/p2)B(c/(λp),c/(λq)) is also the best possible.

    The method of proof of Theorem 2.3 comes from [4,9]. To prove Theorem 2.3, we need the following two lemmas, which are new generalizations of Lemma 3 and Lemma 4 in [9].

    Lemma 2.4. Let p>1, 1/p+1/q=1, λ1,λ2>0,0<cλ. Define the weight function ωλ1,λ2(y,p)=1(1/(lnxλ1yλ2)cλ)(lnyλ2/lnxλ1)cλp(lnxλ1)cλ1x1dx, y(0,), and weight coefficient ~ωλ1,λ2(n,p)=m=2(1/(lnmλ1nλ2)cλ)(lnnλ2/lnmλ1)cλp(lnmλ1)cλ1m1. Then (a) ωλ1,λ2(y,p)=(1/λ1)B(c/(λp),c/(λq)), and (b) If nN and n2, then ~ωλ1,λ2(n,p)<ωλ1,λ2(n,p) holds.

    Proof. (a) Set t=lnxλ1/lnyλ2. Then

    ωλ1,λ2(y,p)=11(lnxλ1yλ2)cλ(lnyλ2lnxλ1)cλp(lnxλ1)cλ1x1dx=1λ101(1+t)cλtcλq1dt=1λ1B(cλp,cλq).

    (b) For fixed λ1,λ2, and 2nN, the function f(x)=(1/(lnxλ1nλ2)cλ)(lnnλ2/lnxλ1)cλs(lnxλ1)cλ1x1 is strict deceasing in (1,). Then f(m)<mm1f(x)dx. Furthermore, we can obtain m=1f(m)<0f(x)dx. So we have ~ωλ1,λ2(n,s)<ωλ1,λ2(n,s).

    Similar to Lemma 2.4, we can get the following lemma.

    Lemma 2.5. Let p>1, 1/p+1/q=1, λ1,λ2>0,0<cλ. Define the weight function ¯ωλ1,λ2(x,p)=1(1/(lnxλ1yλ2)cλ)(lnxλ1/lnyλ2)cλp(lnyλ2)cλ1y1dy, x(0,), and weight coefficient ^ωλ1,λ2(m,p)=n=2(1/(lnmλ1nλ2)cλ)(lnmλ1/lnnλ2)cλp(lnnλ2)cλ1n1. Then (a) ¯ωλ1,λ2(x,p)=(1/λ2)B(c/(λp),c/(λq)), and (b) If mN, and m2 then ^ωλ1,λ2(m,p)<¯ωλ1,λ2(m,p) holds.

    Nextly, we will give the proof of Theorem 2.1.

    Proof. By using the Hölder's inequality and Lemmas 2.4 and 2.5, we have

    n=2m=2ambn(lnmλ1nλ2)cλ=n=2m=2[am(lnmλ1nλ2)cλp(lnmλ1lnnλ2)cλpq(lnnλ2)1p(cλ1)(lnmλ1)1q(1cλ)m1qn1p]×[bn(lnmλ1nλ2)cλq(lnnλ2lnmλ1)cλpq(lnnλ2)1p(1cλ)(lnmλ1)1q(cλ1)n1pm1q][n=2m=2apm(lnmλ1nλ2)cλ(lnmλ1lnnλ2)cλq(lnnλ2)cλ1(lnmλ1)pq(1cλ)mp1n]1/p×[n=2m=2bqn(lnmλ1nλ2)cλ(lnnλ2lnmλ1)cλp(lnnλ2)qp(1cλ)(lnmλ1)cλ1nq1m]1/q=[n=2m=21(lnmλ1nλ2)cλ(lnmλ1lnnλ2)cλq(lnnλ2)cλ1n1(lnmλ1)(p1)(1cλ)mp1apm]1/p×[n=2m=21(lnmλ1nλ2)cλ(lnnλ2lnmλ1)cλp(lnmλ1)cλ1m1(lnnλ2)(q1)(1cλ)nq1bqn]1/q=[m=2^ωλ1,λ2(m,q)(lnmλ1)(p1)(1cλ)mp1apm]1/p[n=2~ωλ1,λ2(n,p)(lnnλ2)(q1)(1cλ)nq1bqn]1/q<[m=2¯ωλ1,λ2(m,q)(lnmλ1)(p1)(1cλ)mp1apm]1/p[n=2ωλ1,λ2(n,p)(lnnλ2)(q1)(1cλ)nq1bqn]1/q=cλ,p[m=2(lnmλ1)(p1)(1cλ)mp1apm]1/p[n=2(lnnλ2)(q1)(1cλ)nq1bqn]1/q,

    which implies (2.9). Now we prove that the constant cλ,p is the best possible. Suppose that the constant cλ,p in inequality (2.9) is not the best possible, then there exists positive k<cλ,p, such that k replaces with cλ,p, inequality (2.9) holds all the same. Especially, for ε(0,q/(2p)), set ˆan=1/(n(lnnλ1)(1c/λ)/p+(1+cε/λ)/p)) and ˆbn=1/(n(lnnλ2)(1c/λ)/p+(1+cε/λ)/q) for nN, then we have

    n=2(lnnλ1)(p1)(1cλ)np1ˆapn=n=21n(lnnλ1)1+cελandn=2(lnnλ2)(q1)(1cλ)nq1ˆbqn=n=21n(lnnλ2)1+cελ.

    As σ tends to 0+, we have

    1λ1+σσ=e1x(lnxλ)1+σdx<n=21n(lnnλ)1+σ=12(ln2λ)1+σ+13(ln3λ)1+σ+n=41n(lnnλ)1+σ<O(1)+e1x(lnxλ)1+σdx=O(1)+1λ1+σσ,

    which implies that as ε tends to 0+, we obtain

    [n=2(lnnλ1)(p1)(1cλ)np1ˆapn]1/p[n=2(lnnλ2)(q1)(1cλ)nq1ˆbqn]1/q=[λcλ11+cε/λε+O(1)]1/p[λcλ21+cε/λε+O(1)]1/q=λcλ(1+cε/λ)/p1λ(1+cε/λ)/q2ε(1+o(1)). (2.10)

    From the assumptions, we have

    n=2m=2ˆamˆbnln(mλ1nλ2)cλ=n=2m=21ln(mλ1nλ2)cλ1m(lnmλ1)1q(1cλ)+1+cελp1n(lnnλ2)1p(1cλ)+1+cελq>ee1ln(xλ1yλ2)cλ1x(lnxλ1)1q(1cλ)+1+cελp1y(lnyλ2)1p(1cλ)+1+cελqdxdy=1λ2e1x(lnxλ1)1+cελλ2lnxλ11(1+u)cλucλ(1pεq)1dudx=1λ2e1x(lnxλ1)1+cελ01(1+u)cλucλ(1pεq)1dudx1λ2e1x(lnxλ1)1+cελλ2lnxλ101(1+u)cλucλ(1pεq)1dudx,

    where in the third equality we set that u=lnyλ2/lnxλ1. Since

    1λ2e1x(lnxλ1)1+cελ01(1+u)cλucλ(1pεq)1dudx=1λ2{λcλ11+cελε[B(cλp,cλq)+o(1)]}=λcλ11+cελλ2ε[B(cλp,cλq)+o(1)],ε0+,

    which implies

    0<e1x(lnxλ1)1+cελλ2lnxλ101(1+u)cλucλ(1pεq)1dudx<e1x(lnxλ1)1+cελλ2lnxλ10ucλ(1pεq)1dudx=qp2λ2λcλ(1pεq)2c2(1+ε)(qpε)λ1cλp(1+ε)=O(1),ε0+.

    From the above inequalities, we can obtain

    n=2m=2ˆamˆbn(lnmλ1nλ2)cλ>λcλ11+cε/λλ2ε[B(cλp,cλq)+o(1)]O(1)=λcλ11+cε/λλ2εB(cλp,cλq)(1+o(1))=λcλ1/p+cε/λ1λ1/q2εcλ,p(1+o(1)),ε0+. (2.11)

    According to (2.10) and (2.11), then we give

    λcλ1/p+cε/λ1λ1/q2εcλ,p(1+o(1))<n=2m=2ˆamˆbn(lnmλ1nλ2)cλ<λcλ(1+cε/λ)/p1λ(1+cε/λ)/q2εk(1+o(1)),ε0+,

    which impies cλ,p<k, in contradiction with supposition. So the constant cλ,p in inequality (2.9) is the best possible. This completes the proof of Theorem 2.3.

    Theorem 2.4. Let an,bn0, and p>1, 1/p+1/q=1, 0<λ1<p, 0<λ2<q, 0<cλ, cλ,p=(1/λ1/q1λ1/p2)B(c/(λp),c/(λq)). If 0<n=2(lnnλ1)(p1)(1cλ)np1apn<, then

    0<n=21n(lnnλ2)1cλ[m=2am(lnmλ1nλ2)cλ]p<cpλ,pm=2(lnmλ1)(p1)(1cλ)mp1apm, (2.12)

    where the constant cpλ,p is the best possible. And inequality (2.12) is equivalent to inequality (2.9).

    Proof. There exists k0N such that km=2am/(lnmλ1nλ2)cλ>0 for k>k0. Setting bn(k)=1/(n(lnnλ2)1cλ)[km=2am/(lnmλ1nλ2)cλ]p1 for k>k0, then

    0<kn=2(lnnλ2)(q1)(1cλ)nq1bqn(k)=kn=21n(lnnλ2)1cλ[km=2am(lnmλ1nλ2)cλ]p=kn=2km=2ambn(k)(lnmλ1nλ2)cλ<cλ,p[km=2(lnmλ1)(p1)(1cλ)mp1apm]1/p[kn=2(lnnλ2)(q1)(1cλ)nq1bqn(k)]1/q,

    which implies that the following inequality holds

    kn=2(lnnλ2)(q1)(1cλ)nq1bqn(k)<cpλ,pkm=2(lnmλ1)(p1)(1cλ)mp1apm.

    As k tends to , we have

    0<n=21n(lnnλ2)1cλ[m=2am(lnmλ1nλ2)cλ]p<cpλ,pm=2(lnmλ1)(p1)(1cλ)mp1apm,

    which implies inequality (2.12). If inequality (2.12) holds, by applying the Hölder's inequality, then

    n=2m=2ambn(lnmλ1nλ2)cλ=n=2[1n1p(lnnλ2)1p(1cλ)m=2am(lnmλ1nλ2)cλ][n1p(lnnλ2)1p(1cλ)bn]<[n=21n(lnnλ2)1cλ[m=2am(lnmλ1nλ2)cλ]p]1/p[n=2(lnnλ2)(q1)(1λ)nq1bqn]1/q.

    Thanks to the inequality (2.12), this means the inequality (2.9) holds. Since the inequalities (2.9) and (2.12) are equivalent, then cpλ,p is also the best possible. This completes the proof of Theorem 2.4.

    In this section, we will give new double integral forms of double series inequalities.

    Theorem 3.1. Let f(x),g(x)0, p>1, 1/p+1/q=1, λ1,λ2,λ,c>0. cλ,p=(1/(λ1/q1λ1/p2))B(c/(λp), c/(λq)). If 0<0x(p1)(1cλ1λ)fp(x)dx<,0<0y(q1)(1cλ2λ)gq(y)dy<, then

    00f(x)g(y)(xλ1+yλ2)cλdxdy<cλ,p[0x(p1)(1cλ1λ)fp(x)dx]1/p[0y(q1)(1cλ2λ)gq(y)dy]1/q, (3.1)
    0ycλ2λ1[0f(x)(xλ1+yλ2)cλdx]pdy<cpλ,p0x(p1)(1cλ1λ)fp(x)dx, (3.2)

    where the constant factors in the inequalities (3.1) and (3.2) are also the best possible. Moreover, the inequalities (3.1) and (3.2) are equivalent.

    Proof. By mean of Hölder's inequality and Lemmas 2.1 and 2.2, we have

    00f(x)g(y)(xλ1+yλ2)cλdxdy=00[f(x)(xλ1+yλ2)cλp(xλ1yλ2)cλpqycλ2λ1pxcλ1λ1q][g(y)(xλ1+yλ2)cλq(yλ2xλ1)cλpqxcλ1λ1qycλ2λ1p]dxdy[00fp(x)(xλ1+yλ2)cλ(xλ1yλ2)cλqycλ2λ1xp(cλ1λ1)/qdxdy]1/p[00gq(y)(xλ1+yλ2)cλ(yλ2xλ1)cλpxcλ1λ1yq(cλ2λ1)/pdxdy]1/q=[0dx01(xλ1+yλ2)cλ(xλ1yλ2)cλqycλ2λ1x(p1)(1cλ1λ)fp(x)dy]1/p×[0dy01(xλ1+nλ2)cλ(yλ2xλ1)cλpxcλ1λ1y(q1)(1cλ2λ)gq(y)dx]1/q=[0ˉωλ1,λ2(x,q)x(p1)(1cλ1λ)fp(x)dx]1/p[0ωλ1,λ2(y,p)y(q1)(1cλ2λ)gq(y)dy]1/q=[1λ2B(cλp,cλq)]1/p[0x(p1)(1cλ1λ)fp(x)dx]1/p[1λ1B(cλp,cλq)]1q[0y(q1)(1cλ2λ)gq(y)dy]1/q=cλ,p[0x(p1)(1cλ1λ)fp(x)dx]1/p[0y(q1)(1cλ2λ)gq(y)dy]1/q.

    According to the hypotheses, it is easy to see that the equality in the above the second inequality is not possible. Now we prove that the constant cλ,p is the best possible. If the constant cλ,p is not the best possible, then there exists k<cλ,p such that k replaces with cλ,p, inequality (3.1) holds all the same. Especially, for ε(0,q/(2p)), we set

    ˜f(x)={1x1+cλ1ελ+(p1)(1cλ1λ)p,x[0,1),0,x[1,),and˜g(x)={1x1+cλ2ελ+(q1)(1cλ2λ)q,x[0,1),0,x[1,).

    Based on the course of the proof of Theorem 2.1, we have

    λcλ1/p1λ1/q2εcλ,p(1+o(1))<00˜f(x)˜g(y)(xλ1+yλ2)cλdxdy<λcλ1/p1λ1/q2εk(1+o(1)),

    as ε tends to 0+. We have cλ,p<k, in contradiction with supposition. So the constant cλ,p in the inequality (3.1) is the best possible.

    There exists t0>0 such that T0f(x)/(xλ1+yλ2)cλdx>0 for T>t0, setting g(y,T)=ycλ2λ1[T0f(x)/(xλ1+yλ2)cλdx]p1, y(0,). Thanks to the inequality (3.1), thus we get

    T0y(q1)(1cλ2λ)gq(y,T)dy=T0ycλ2λ1[T0f(x)(xλ1+yλ2)cλdx]pdy=T0T0f(x)g(y,T)(xλ1+yλ2)cλdxdy<cλ,p[T0x(p1)(1cλ1λ)fp(x)dx]1/p[T0y(q1)(1cλ2λ)gq(y,T)dy]1/q,

    which implies that as T tends to , then we have

    0ycλ2λ1[0f(x)(xλ1+yλ2)cλdx]pdy<cpλ,p0x(p1)(1cλ1λ)fp(x)dx,

    which means the inequality (3.2) holds. If inequality (3.2) holds, by using the Hölder's inequality, then

    00f(x)g(y)(xλ1+yλ2)cλdxdy=0[ycλ2λ1p0f(x)(xλ1+yλ2)cλdx][y1cλ2λpg(y)]dy<[0ycλ2λ1(0f(x)(xλ1+yλ2)cλdx)pdy]1/p[0x(p1)(1cλ1λ)fp(x)dx]1/q.

    which implies that the inequality (3.1) holds. Therefore, inequalities (3.1) and (3.2) are equivalent. It is easy to see that the constant cpλ,p in the inequality (3.2) is also the best possible.

    Now we present a new variant of Theorem 3.1 as follows.

    Theorem 3.2. Let f(x),g(x)0, p>1, 1/p+1/q=1, λ1,λ2,λ,c>0. cλ,p=(1/(λ1/q1λ1/p2))B(c/(λp), c/(λq)). If 0<1(lnxλ1)(p1)(1cλ)xp1fp(x)dx<, and 0< 1(lnyλ2)(q1)(1cλ)yq1gq(y)dy<, then

    11f(x)g(y)(lnxλ1yλ2)cλdxdy<cλ,p[1(lnxλ1)(p1)(1cλ)xp1fp(x)dx]1/p×[1(lnyλ2)(q1)(1cλ)yq1gq(y)dy]1/q, (3.3)
    11y(lnyλ2)1cλ[1f(x)(lnxλ1yλ2)cλdx]pdy<cpλ,p1(lnxλ1)(p1)(1cλ)xp1fp(x)dx, (3.4)

    where the constant factors in the inequalities (3.3) and (3.4) are also the best possible. Furthermore, inequalities (3.3) and (3.4) are equivalent.

    Proof. By applying the Hölder's inequality and Lemmas 2.4 and 2.5, we have

    11f(x)g(y)(lnxλ1yλ2)cλdxdy=11[f(x)(lnxλ1yλ2)cλp(lnxλ1lnyλ2)cλpq(lnyλ2)1p(cλ1)(lnxλ1)1q(1cλ)x1qy1p]×[g(y)(lnxλ1yλ2)cλq(lnyλ2lnxλ1)cλpq(lnyλ2)1p(1cλ)(lnxλ1)1q(cλ1)y1px1q]dxdy[11fp(x)(lnxλ1yλ2)cλ(lnxλ1lnyλ2)cλq(lnyλ2)cλ1(lnxλ1)pq(1cλ)xp1ydxdy]1/p×[11gq(y)(lnxλ1yλ2)cλ(lnyλ2lnxλ1)cλp(lnyλ2)qp(1cλ)(lnxλ1)cλ1yq1xdxdy]1/q=[1dx11(lnxλ1yλ2)cλ(lnxλ1lnyλ2)cλq(lnyλ2)cλ1y1(lnxλ1)(p1)(1cλ)xp1fp(x)dy]1/p×[1dy11(lnxλ1yλ2)cλ(lnyλ2lnxλ1)cλp(lnxλ1)cλ1x1(lnyλ2)(q1)(1cλ)yq1gq(y)dx]1/q=[1¯ωλ1,λ2(x,q)(lnxλ1)(p1)(1cλ)xp1fp(x)dx]1/p[1ωλ1,λ2(y,p)(lnyλ2)(q1)(1cλ)yq1gq(y)dy]1/q=cλ,p[1(lnxλ1)(p1)(1cλ)xp1fp(x)dx]1/p[1(lnyλ2)(q1)(1cλ)yq1gq(y)dy]1/q,

    which implies the inequality (3.3). According to the hypotheses, it is easy to see that the equality in the above the second inequality is not possible. Now we prove that the constant cλ,p is the best possible. Suppose that the constant cλ,p in inequality (3.3) is not the best possible, then exists positive k<cλ,p such that when k replaces with cλ,p, the inequality (3.3) holds all the same. Especially, for ε(0,q/(2p)), we set

    ˆf(x)={1x(lnxλ1)1q(1cλ)+1+cελp,x[0,1),0,x[1,),andˆg(x)={1x(lnxλ2)1p(1cλ)+1+cελq,x[0,1),0,x[1,).

    Due to the course of the proof of Theorem 2.3, we have

    λcλ1/p+cε/λ1λ1/q2εcλ,p(1+o(1))<11ˆf(x)ˆg(y)(lnxλ1yλ2)cλdxdy<λcλ(1+cε/λ)/p1λ(1+cε/λ)/q2εk(1+o(1)),

    as ε tends to 0+. We have cλ,p<k, in contradiction with supposition. So the constant cλ,p in inequality (3.3) is the best possible.

    There exists t0>0 such that T1f(x)/(lnxλ1yλ2)cλdx>0 for T>t0. Setting g(y,T)=(1/(y(lnyλ2)1c/λ))[T1f(x)/(lnxλ1yλ2)cλdx]p1 for T>t0, then

    T1(lnyλ2)(q1)(1cλ)yq1gq(y,T)dy=T1T1f(x)g(y,T)(lnxλ1yλ2)cλdxdy<cλ,p[T1(lnxλ1)(p1)(1cλ)xp1fp(x)dx]1/p[T1(lnyλ2)(q1)(1cλ)yq1gq(y,T)dy]1/q,

    which implies that as T tends to , then we have

    11y(lnyλ2)1cλ[1f(x)(lnxλ1xλ2)cλdx]pdy<cpλ,p1(lnxλ1)(p1)(1cλ)xp1fp(x)dx,

    which implies the inequality (3.4). If inequality (3.4) holds, by Hölder's inequality, then

    11f(x)g(y)(lnxλ1yλ2)cλdxdy=1[1y1p(lnyλ2)1p(1cλ)1f(x)(lnxλ1yλ2)cλdx][y1p(lnyλ2)1p(1cλ)g(y)]dy<[11y(lnyλ2)1cλ(1f(x)(lnxλ1yλ2)cλdx)pdy]1/p[1(lnyλ2)(q1)(1λ)yq1gq(y)dy]1/q,

    which means the inequality (3.3) holds. Since the inequalities (3.3) and (3.4) are equivalent, then cpλ,p is also the best possible. The proof of Theorem 3.2 is completed.

    Remark 3.1. It is easy to see that inequality (3.3) is equivalent to the following new Mullholand's inequality associated integral.

    11f(x)g(y)xy(lnxλ1yλ2)cλdxdy<cλ,p[1(lnxλ1)(p1)(1cλ)x1fp(x)dx]1/p×[1(lnyλ2)(q1)(1cλ)y1gq(y)dy]1/q,

    where 0<1(lnxλ1)(p1)(1cλ)x1fp(x)dx<, 0<1(lnyλ2)(q1)(1cλ)y1gq(y)dy <, and the constant factor cλ,p is also the best possible.

    Remark 3.2. In the paper [16], by using the quasi-homogeneous integral kernels, Cao {et al.} obtained some Hilbert-type integral inequalities involving multiple functions with the best constant factors. In this paper, we not only give some Hilbert-type integral inequalities with multiparameters, but also obtain some Hilbert-type inequalities with multiparameters for double series. Therefore, our results are different from Cao et al. [16].

    In this paper, we have established a new Hardy-Hilbert-type inequality with multiparameters. Furthermore, its equivalent forms and variants, which generalize some existing results, have been also presented. Finally, the corresponding Hardy-Hilbert-type integral inequalities haven been obtained.

    The authors declare that there is no conflict of interests.



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