The Generalizations of Hilbert's Inequality

Abstract

By introducing some parameters and estimating the weight functions, we establish a new Hilbert-type inequalities with best constant factors. The equivalent inequalities are also considered.

1. Introduction

If 𝑓,𝑔 are real functions such that 0<∫^∞₀𝑓²(𝑥)𝑑𝑥<∞ and 0<∫^∞₀𝑔²(𝑥)𝑑𝑥<∞, then we have (see [1])^∞₀^∞₀𝑓(𝑥)𝑔(𝑦)𝑥+𝑦𝑑𝑥𝑑𝑦<𝜋{^∞₀𝑓²(𝑥)𝑑𝑥^∞₀𝑔²(𝑥)𝑑𝑥}^1/2, (1.1)where the constant factor 𝜋 is the best possible. Inequality (1.1) is the well-known Hilbert's inequality. Inequality (1.1) had been generalized by Hardy-Riesz (see [2]) in 1925 as if 𝑓,𝑔 are real functions such that 0<∫^∞₀𝑓^𝑝(𝑥)𝑑𝑥<∞ and 0<∫^∞₀𝑔^𝑞(𝑥)𝑑𝑥<∞, then^∞₀^∞₀𝑓(𝑥)𝑔(𝑦)𝜋𝑥+𝑦𝑑𝑥𝑑𝑦<{sin(𝜋/𝑝)^∞₀𝑓^𝑝(𝑥)𝑑𝑥}^1/𝑝{^∞₀𝑔^𝑞(𝑥)𝑑𝑥}^1/𝑞, (1.2)where the constant factor 𝑐=𝜋/sin(𝜋/𝑝) is the best possible. When 𝑝=𝑞=2, (1.2) reduces to (1.1). Inequality (1.2) is named after Hardy-Hilbert's integral inequality, which is important in the analysis and its applications (see [3]), it has been studied and generalized in many directions by a number of mathematicians (see [4–8]).

Under the same condition of (1.2), we have Hardy-Hilbert's type inequality (see [1, Theorems 341 and 342]):^∞₀^∞₀𝑓(𝑥)𝑔(𝑦)max{𝑥,𝑦}𝑑𝑥𝑑𝑦<4{^∞₀𝑓²(𝑥)𝑑𝑥^∞₀𝑔²(𝑥)𝑑𝑥}^1/2,_∞0∞0ln𝑥−ln𝑦𝑥−𝑦𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦<𝜋²{^∞₀𝑓²(𝑥)𝑑𝑥^∞₀𝑔²(𝑥)𝑑𝑥}^1/2, (1.3)where the constant factors 4 and 𝜋² are both the best possible.

Recently, Li et al. [9], by introducing the function |ln𝑥−ln𝑦|/(𝑥+𝑦+|𝑥−𝑦|), establish new inequalities similar to Hilbert-type inequality for integrals.

Theorem 1.1. If 𝑓(𝑥),𝑔(𝑥)≥0,0<∫^∞₀𝑓²(𝑥)𝑑𝑥<∞,0<∫^∞₀𝑔²(𝑥)𝑑𝑥<∞. Then, one has^∞₀^∞₀|ln𝑥−ln𝑦|𝑥+𝑦+|𝑥−𝑦|𝑓(𝑥)𝑔(𝑦)d𝑥d𝑦<4(^∞₀𝑓²(𝑥)d𝑥)^1/2(^∞₀𝑔²(𝑥)d𝑥)^1/2, (1.4)where the constant factor 4 is the best possible.

In this paper, we give further analogs of Hilbert-type inequality and its applications. The main result unifies and generalizes the classical results as follows.

Assume that 𝑟>0,𝑠>−min{1,𝑟},𝑡⩾0,𝑝,𝑞>1, and 1/𝑝+1/𝑞=1. If 𝑓,𝑔⩾0, such that 0<∫^∞₀𝑓^𝑝(𝑥)𝑑𝑥<∞,0<∫^∞₀𝑔^𝑞(𝑥)𝑑𝑥<∞, then^∞₀^∞₀|ln𝑥−ln𝑦|^𝑡𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦<𝐾₁(^∞₀𝑓^𝑝(𝑥)𝑑𝑥)^1/𝑝(^∞₀𝑔^𝑞(𝑥)𝑑𝑥)^1/𝑞, (1.5)where the constant factor𝐾₁=^∞₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^−1/𝑞𝑑𝑢 (1.6)is the best possible.

2. Some Lemmas

Our results will be based on the following results. In the following lemmas, assume that 𝑟>0, 𝑠>−min{1,𝑟}, 𝑡⩾0,𝑝,𝑞>1, and 1/𝑝+1/𝑞=1.

Lemma 2.1. Define the following weight functions:𝜔₁(𝑥)=∫^∞₀|ln𝑥−ln𝑦|^𝑡(𝑥𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑦)^1/𝑞𝜔𝑑𝑦,𝑥>0,₂(𝑦)=∫^∞₀|ln𝑥−ln𝑦|^𝑡(𝑦𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑥)^1/𝑝𝑑𝑥,𝑦>0. (2.1)Then, 𝜔₁=𝜔₂=𝐾₁, where 𝐾₁ is defined by (1.1).

Proof. For 𝜔₁(𝑥), let 𝑢=𝑦/𝑥, and we have𝜔₁=^∞₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^−1/𝑞𝑑𝑢=𝐾₁. (2.2)
For 𝜔₂(𝑦), first let 𝑣=𝑥/𝑦, and then let 𝑢=1/𝑣. Thus, we have𝜔₂=^∞₀|ln𝑣|^𝑡𝑣1+𝑟𝑣+𝑠|𝑣−1|^−1/𝑝𝑑𝑣=^∞₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^−1/𝑞𝑑𝑢=𝐾₁. (2.3)Hence, the lemma is proved.

Lemma 2.2. Assume that 𝜀>0, then^∞₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^{−(1+𝜀)/𝑞}𝑑𝑢=𝐾₁+𝑜(1)(𝜀→0⁺). (2.4)

Proof. Sincelim_𝑢→0𝑢^1/2𝑝|ln𝑢|^𝑡=0,lim_𝑢→+∞𝑢^1/2𝑝|ln𝑢|^𝑡=+∞, (2.5)it follows that there exists 𝛿₁∈(0,1), such that if 𝑢∈(0,𝛿₁), then 𝑢^−1/2𝑝>|ln𝑢|^𝑡. Moreover, there exists 𝛿₂∈(1,+∞), such that if 𝑢∈(𝛿₂,+∞), then 𝑢^1/2𝑞>|ln𝑢|^𝑡. Using the expression of 𝐾₁, if 𝑠⩽0, then|^∞₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^{−(1+𝜀)/𝑞}𝑑𝑢−𝐾₁|=|^∞₀|ln𝑢|^𝑡1+𝑟𝑢+𝑠|1−𝑢|(𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞)𝑑𝑢|⩽(^𝛿1₀+^𝛿2_𝛿1+^∞_𝛿2)|ln𝑢|^𝑡1+𝑟𝑢+𝑠|1−𝑢||𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞<|𝑑𝑢_𝛿10|ln𝑢|^𝑡1+𝑠(𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞)𝑑𝑢+^𝛿2_𝛿1|ln𝑢|^𝑡1+𝑟𝑢+𝑠|1−𝑢||𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞+|𝑑𝑢_∞𝛿2|ln𝑢|^𝑡(𝑟+𝑠)𝑢(𝑢^−1/𝑞−𝑢^{−(1+𝜀)/𝑞}<)𝑑𝑢_𝛿10𝑢^−1/2𝑝1+𝑠(𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞)𝑑𝑢+^𝛿2_𝛿1|ln𝑢|^𝑡1+𝑟𝑢+𝑠|1−𝑢||𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞|𝑑𝑢+^∞_𝛿2𝑢^1/2𝑞(𝑟+𝑠)𝑢(𝑢^−1/𝑞−𝑢^{−(1+𝜀)/𝑞}=1)𝑑𝑢(11+𝑠𝛿1−1/2𝑝−(1+𝜀)/𝑞^{11−1/2𝑝−(1+𝜀)/𝑞}−1𝛿1−1/2𝑝−1/𝑞^{11−1/2𝑝−1/𝑞})+^𝛿2_𝛿1|ln𝑢|^𝑡1+𝑟𝑢+𝑠|1−𝑢||𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞1|𝑑𝑢+𝑟+𝑠(2𝑞𝛿₂^−1/2𝑞−2𝑞𝛿1+2𝜀^{2−(1+2𝜀)/2𝑞})=𝑜(1)(𝜀→0⁺). (2.6)Therefore, Lemma 2.2 is proved for 𝑠⩽0. If 𝑠>0, then we can replace the right-hand side of the first strict inequality above with^𝛿1₀|ln𝑢|^𝑡1(𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞)𝑑𝑢+^𝛿2_𝛿1|ln𝑢|^𝑡1+𝑟𝑢+𝑠|1−𝑢||𝑢^{−(1+𝜀)/𝑞}−𝑢^−1/𝑞+|𝑑𝑢_∞𝛿2|ln𝑢|^𝑡𝑟𝑢(𝑢^−1/𝑞−𝑢^{−(1+𝜀)/𝑞})𝑑𝑢. (2.7)By the same way, we can show that the lemma is valid for 𝑠>0. Hence, the lemma is proved.

Lemma 2.3. Assume that 𝜀>0, then^∞₁𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^{−(1+𝜀)/𝑞}𝑑𝑢=𝑂(1)(𝜀→0⁺). (2.8)

Proof. Using 𝛿₁ introduced in the proof of Lemma 2.2, if 𝑠⩽0, then0<^∞₁𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^{−(1+𝜀)/𝑞}=𝑑𝑢_∞1𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠(1−𝑢)^{−(1+𝜀)/𝑞}𝑑𝑢<^∞₁𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}𝑑𝑢=(^1/𝛿1₁+^∞_1/𝛿1)𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}=𝑑𝑢_1/𝛿11𝑥^−1−𝜀𝑑𝑥(^𝛿1₀+^𝑥−1_𝛿1)|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}𝑑𝑢+^∞_1/𝛿1𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}=𝑑𝑢_1/𝛿11𝑥^−1−𝜀𝑑𝑥^𝛿1₀|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}𝑑𝑢+^1/𝛿1₁𝑥^−1−𝜀𝑑𝑥^𝑥−1_𝛿1|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}+𝑑𝑢_∞1/𝛿1𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}𝑑𝑢<^1/𝛿1₁𝑥^−1−𝜀𝑑𝑥^𝛿1₀𝑢^−1/2𝑝𝑢1+𝑠^{−(1+𝜀)/𝑞}𝑑𝑢+^1/𝛿1₁𝑥^−1−𝜀𝑑𝑥^𝑥−1_𝛿1|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}+𝑑𝑢_∞1/𝛿1𝑥^−1−𝜀𝑑𝑥^𝑥−1₀𝑢^−1/2𝑝𝑢1+𝑠^{−(1+𝜀)/𝑞}=𝑑𝑢(1−𝛿₁)𝛿₁^{1−1/2𝑝−(1+𝜀)/𝑞}+𝜀(1+𝑠)(1−1/2𝑝−(1+𝜀)/𝑞)^1/𝛿1₁𝑥^−1−𝜀𝑑𝑥^𝑥−1_𝛿1|ln𝑢|^𝑡𝑢1+𝑠^{−(1+𝜀)/𝑞}+𝑑𝑢2𝑝𝛿₁^{(1+2𝜀)/2𝑝}(1+2𝜀)(1+𝑠)(1−1/2𝑝−(1+𝜀)/𝑞)=𝑂(1)(𝜀→0⁺). (2.9)Therefore, Lemma 2.3 is proved for 𝑠⩽0. If 𝑠>0, then we can replace the right-hand side of the second strict inequality mentioned above with^∞₁𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡1𝑢^{−(1+𝜀)/𝑞}𝑑𝑢. (2.10)By the same way, we can show that the lemma is valid for 𝑠>0. Therefore, the lemma is proved.

3. Integral Case

In this section, we will state our main results.

Proposition 3.1. Assume that 𝑟>0,𝑠>−min{1,𝑟},𝑡⩾0,𝑝,𝑞>1, and 1/𝑝+1/𝑞=1. If 𝑓,𝑔⩾0, such that 0<∫^∞₀𝑓^𝑝(𝑥)𝑑𝑥<∞,0<∫^∞₀𝑔^𝑞(𝑥)𝑑𝑥<∞, then^∞₀^∞₀|ln𝑥−ln𝑦|^𝑡𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦<𝐾₁(^∞₀𝑓^𝑝(𝑥)𝑑𝑥)^1/𝑝(^∞₀𝑔^𝑞(𝑥)𝑑𝑥)^1/𝑞, (3.1)where the constant factor𝐾₁=^∞₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^−1/𝑞𝑑𝑢 (3.2)is the best possible.

Proof. Using Hölder's inequality, we have^∞₀^∞₀|ln𝑥−ln𝑦|^𝑡=𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦_∞0∞0|ln𝑥−ln𝑦|^𝑡𝑥𝑥+𝑟𝑦+𝑠|𝑥−𝑦|[(𝑦)^1/𝑝𝑞𝑦𝑓(𝑥)][(𝑥)^1/𝑝𝑞𝑔(𝑦)]𝑑𝑥𝑑𝑦⩽(^∞₀^∞₀|ln𝑥−ln𝑦|^𝑡(𝑥𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑦)^1/𝑞𝑓^𝑝(𝑥)𝑑𝑥𝑑𝑦)^1/𝑝×(^∞₀^∞₀|ln𝑥−ln𝑦|^𝑡(𝑦𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑥)^1/𝑝𝑔^𝑞(𝑦)𝑑𝑥𝑑𝑦)^1/𝑞=(^∞₀𝜔₁(𝑥)𝑓^𝑝(𝑥)𝑑𝑥)^1/𝑝(^∞₀𝜔₂(𝑦)𝑔^𝑞(𝑦)𝑑𝑦)^1/𝑞=𝐾₁(^∞₀𝑓^𝑝(𝑥)𝑑𝑥)^1/𝑝(^∞₀𝑔^𝑞(𝑥)𝑑𝑥)^1/𝑞. (3.3)
If (3.3) takes the form of equality, then there are constants 𝑎 and 𝑏, such that they are not all zero, and𝑎|ln𝑥−ln𝑦|^𝑡(𝑥𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑦)^1/𝑞𝑓^𝑝(𝑥)=𝑏|ln𝑥−ln𝑦|^𝑡(𝑦𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑥)^1/𝑝𝑔^𝑞(𝑦),a.e.in(0,∞)×(0,∞). (3.4)Hence,𝑎𝑥𝑓^𝑝(𝑥)=𝑏𝑦𝑔^𝑞(𝑦),a.e.in(0,∞)×(0,∞). (3.5)Therefore, there is a constant 𝑐, such that𝑎𝑥𝑓^𝑝(𝑥)=𝑏𝑦𝑔^𝑞(𝑦)=𝑐,a.e.in(0,∞)×(0,∞). (3.6) We claim that 𝑎=0. In fact, if 𝑎≠0, then 𝑓^𝑝(𝑥)=𝑐/𝑎𝑥, a.e., in (0,∞), which contradicts the fact that 0<∫^∞₀𝑓^𝑝(𝑥)𝑑𝑥<∞. In the same way, we claim that 𝑏=0. This is a contradiction. Hence, by (3.3), we have (3.1).
If the constant factor 𝐾₁ in (3.1) is not the best possible factor, then there exists a positive constant 𝐻 (with 𝐻<𝐾₁), such that (3.1) is still valid if we replace 𝐾₁ with 𝐻. For 𝜀>0 sufficiently small, construct the following functions:𝑓_𝜀𝑥(𝑥)=0,𝑥∈(0,1),_{−(1+𝜀)/𝑝}𝑔,𝑥∈(0,+∞),_𝜀𝑥(𝑥)=0,𝑥∈(0,1),_{−(1+𝜀)/𝑞},𝑥∈(0,+∞), (3.7)Thus, we obtain𝐻(^∞₀𝑓^𝑝_𝜀(𝑥)𝑑𝑥)^1/𝑝(^∞₀𝑔^𝑞_𝜀(𝑥)𝑑𝑥)^1/𝑞=𝐻𝜀,𝐼∶=^∞₀^∞₀|ln𝑥−ln𝑦|^𝑡𝑓𝑥+𝑟𝑦+𝑠|𝑥−𝑦|^𝜀(𝑥)𝑔_𝜀=(𝑦)𝑑𝑥𝑑𝑦_∞1∞1|ln𝑥−ln𝑦|^𝑡𝑥𝑥+𝑟𝑦+𝑠|𝑥−𝑦|^{−(1+𝜀)/𝑝}𝑦^{−(1+𝜀)/𝑞}=𝑑𝑥𝑑𝑦_∞1𝑥^{−(1+𝜀)/𝑝}𝑑𝑥^∞₁|ln𝑥−ln𝑦|^𝑡𝑦𝑥+𝑟𝑦+𝑠|𝑥−𝑦|^{−(1+𝜀)/𝑞}𝑑𝑦. (3.8)Setting 𝑢=𝑦/𝑥, we have𝐼=^∞₁𝑥^−1−𝜀𝑑𝑥^∞_𝑥−1|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^{−(1+𝜀)/𝑞}𝑑𝑢=^∞₁𝑥^−1−𝜀𝑑𝑥^∞₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^{−(1+𝜀)/𝑞}𝑑𝑢−^∞₁𝑥^−1−𝜀𝑑𝑥^𝑥−1₀|ln𝑢|^𝑡𝑢1+𝑟𝑢+𝑠|1−𝑢|^{−(1+𝜀)/𝑞}=1𝑑𝑢𝜀(𝐾₁+𝑜(1))+𝑂(1)(𝜀→0⁺). (3.9)Since𝐻𝐼<𝜀, (3.10)it follows that𝐾₁+𝑜(1)<𝐻(𝜀→0⁺). (3.11)Hence,𝐾₁⩽𝐻. (3.12)This contradicts the fact that 𝐻<𝐾₁. So the constant factor 𝐾₁ in (3.1) is the best possible. Then Proposition 3.1 is proved.

Proposition 3.2. Under the same assumptions of Proposition 3.1, one has^∞₀[^∞₀|ln𝑥−ln𝑦|^𝑡𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑑𝑥]^𝑝𝑑𝑦<𝐾^𝑝₁^∞₀𝑓^𝑝(𝑥)𝑑𝑥, (3.13)where the constant factor 𝐾^𝑝₁ is the best possible. Inequalities (3.1) and (3.13) are equivalent.

Proof. Setting𝑔(𝑦)=[^∞₀|ln𝑥−ln𝑦|^𝑡𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑑𝑥]^𝑝−1,𝑦∈(0,∞), (3.14)by (3.1), we have^∞₀𝑔^𝑞(𝑦)𝑑𝑦=^∞₀[^∞₀|ln𝑥−ln𝑦|^𝑡𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑑𝑥]^𝑝=𝑑𝑦_∞0∞0|ln𝑥−ln𝑦|^𝑡𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦<𝐾₁(^∞₀𝑓^𝑝(𝑥)𝑑𝑥)^1/𝑝(^∞₀𝑔^𝑞(𝑥)𝑑𝑥)^1/𝑞. (3.15)As a result,(^∞₀𝑔^𝑞(𝑥)𝑑𝑥)^1/𝑝<𝐾₁(^∞₀𝑓^𝑝(𝑥)𝑑𝑥)^1/𝑝. (3.16)Hence,^∞₀𝑔^𝑞(𝑥)𝑑𝑥<𝐾^𝑝₁^∞₀𝑓^𝑝(𝑥)𝑑𝑥. (3.17)Then, we have (3.13).
Conversely, by Hölder's inequality, we have^∞₀^∞₀|ln𝑥−ln𝑦|^𝑡=𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦_∞0[^∞₀|ln𝑥−ln𝑦|^𝑡𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑑𝑥]𝑔(𝑦)𝑑𝑦⩽(^∞₀[^∞₀|ln𝑥−ln𝑦|^𝑡𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝑓(𝑥)𝑑𝑥]^𝑝𝑑𝑦)^1/𝑝(^∞₀𝑔^𝑞(𝑥)𝑑𝑥)^1/𝑞<𝐾₁(^∞₀𝑓^𝑝(𝑥)𝑑𝑥)^1/𝑝(^∞₀𝑔^𝑞(𝑥)𝑑𝑥)^1/𝑞. (3.18)Then, by (3.13), we have (3.1). Hence inequalities (3.1) and (3.13) are equivalent.
If the constant factor 𝐾^𝑝₁ in (3.13) is not the best possible, then by (3.18) we can get a contradiction that the constant factor 𝐾₁ in (3.1) is not the best possible. Hence Proposition 3.2 is proved.

Remark 3.3. In (3.1), let 𝑟=1,𝑠=0,𝑡=0,𝑝=𝑞=2, and we have Hilbert's integral inequality^∞₀^∞₀𝑓(𝑥)𝑔(𝑦)𝑥+𝑦𝑑𝑥𝑑𝑦<𝜋{^∞₀𝑓²(𝑥)𝑑𝑥^∞₀𝑔²(𝑥)𝑑𝑥}^1/2. (3.19)Let 𝑟=1,𝑠=1,𝑡=0,𝑝=𝑞=2; we have Hardy-Hilbert's classical inequality^∞₀^∞₀𝑓(𝑥)𝑔(𝑦)2max{𝑥,𝑦}𝑑𝑥𝑑𝑦<2{^∞₀𝑓²(𝑥)𝑑𝑥^∞₀𝑔²(𝑥)𝑑𝑥}^1/2. (3.20)Let 𝑟=1,𝑠=1/3,𝑡=0,𝑝=𝑞=2, and we can combine the above two inequalities as follows:^∞₀^∞₀𝑓(𝑥)𝑔(𝑦)(2/3)(𝑥+𝑦+max{𝑥,𝑦})𝑑𝑥𝑑𝑦<𝐾₁{^∞₀𝑓²(𝑥)𝑑𝑥^∞₀𝑔²(𝑥)𝑑𝑥}^1/2. (3.21)Let 𝑟=1,𝑠=−1/3,𝑡=0, and we can get the following inequality:^∞₀^∞₀𝑓(𝑥)𝑔(𝑦)(2/3)(𝑥+𝑦+min{𝑥,𝑦})𝑑𝑥𝑑𝑦<𝐾₁{^∞₀𝑓²(𝑥)𝑑𝑥^∞₀𝑔²(𝑥)𝑑𝑥}^1/2. (3.22)

4. Discrete Case

We also give results for the discrete case.

Proposition 4.1. Assume that 𝑟>0,𝑠∈(−min{1,𝑟},min{1,𝑟}],𝑝,𝑞>1 and 1/𝑝+1/𝑞=1. If 𝑎_𝑚,𝑏_𝑛⩾0, such that0<^∞_𝑚=1𝑎^𝑝_𝑚<∞,0<^∞_𝑛=1𝑏^𝑞_𝑛<∞, (4.1)then one has^∞∑^∞_𝑛=1∑_𝑚=1𝑎_𝑚𝑏_𝑛𝑚+𝑟𝑛+𝑠|𝑚−𝑛|<𝐾₁(∑^∞_𝑚=1𝑎^𝑝_𝑚)^1/𝑝(∑^∞_𝑛=1𝑏^𝑞_𝑛)^1/𝑞,_∞𝑛∑₌₁[∑^∞_𝑚=1𝑎_𝑚]𝑚+𝑟𝑛+𝑠|𝑚−𝑛|^𝑝<𝐾∑^𝑝₁^∞_𝑚=1𝑎^𝑝_𝑚, (4.2)(4.3)where the constant factor𝐾₁=^∞₀1𝑢1+𝑟𝑢+𝑠|1−𝑢|^−1/𝑞𝑑𝑢 (4.4)is the best possible, and inequalities (4.2) and (4.3) are equivalent.

Proof. Define the following weight functions:𝜛₁(𝑚)=^∞_𝑛=11(𝑚𝑚+𝑟𝑛+𝑠|𝑚−𝑛|𝑛)^1/𝑞𝜛,𝑚=1,2,…,₂(𝑛)=^∞_𝑚=11(𝑛𝑚+𝑟𝑛+𝑠|𝑚−𝑛|𝑚)^1/𝑝,𝑛=1,2,…, (4.5)then,𝜛₁(𝑚)<𝜔₁(𝑚)=𝐾₁,𝜛₂(𝑛)<𝜔₂(𝑛)=𝐾₁, (4.6)where 𝜔₁,𝜔₂ are defined in Lemma 2.1. By Hölder's inequality, we have^∞∞_𝑛=1𝑚=1𝑎_𝑚𝑏_𝑛=𝑚+𝑟𝑛+𝑠|𝑚−𝑛|_{∞∞𝑛=1𝑚=1}1𝑚𝑚+𝑟𝑛+𝑠|𝑚−𝑛|[(𝑛)^1/𝑝𝑞𝑎_𝑚𝑛][(𝑚)^1/𝑝𝑞𝑏_𝑛]≤(^∞∞_𝑛=1𝑚=11(𝑚𝑚+𝑟𝑛+𝑠|𝑚−𝑛|𝑛)^1/𝑞𝑎^𝑝_𝑚)^1/𝑝(^∞∞_𝑛=1𝑚=11(𝑛𝑚+𝑟𝑛+𝑠|𝑚−𝑛|𝑚)^1/𝑝𝑏^𝑞_𝑛)^1/𝑞=(^∞_𝑚=1𝜛₁(𝑚)𝑎^𝑝_𝑚)^1/𝑝(^∞_𝑛=1𝜛₂(𝑛)𝑏^𝑞_𝑛)^1/𝑞<𝐾₁(^∞_𝑚=1𝑎^𝑝_𝑚)^1/𝑝(^∞_𝑛=1𝑏^𝑞_𝑛)^1/𝑞. (4.7)The last strict inequality holds because both series {𝑎_𝑚} and {𝑏_𝑛} have positive terms. Thus, we have (4.2).
If the constant factor 𝐾₁ in (4.2) is not the best possible, then there exists a positive constant 𝐻 (with 𝐻<𝐾₁), such that (4.2) is still valid if we replace 𝐾₁ by 𝐻. For 𝜀>0 small enough, construct series^0𝑥0𝑎^02𝑑𝑐_𝑚=𝑚^{−(1+𝜀)/𝑝},𝑏^{0𝑥002𝑑𝑐}_𝑛=𝑛^{−(1+𝜀)/𝑞}. (4.8)Then, we have𝑎𝑏^∞∞_𝑛=1𝑚=1^{0𝑥002𝑑𝑐}_𝑚^{0𝑥002𝑑𝑐}_𝑛>𝑚+𝑟𝑛+𝑠|𝑚−𝑛|^∞₀^∞₀𝑓_𝜀(𝑥)𝑔_𝜀(𝑦)𝑑𝑥𝑑𝑦=1𝑥+𝑟𝑦+𝑠|𝑥−𝑦|𝜀(𝐾₁(+𝑜(1))+𝑂(1),_{∞𝑚=10𝑥002𝑑𝑐}𝑎^𝑝_𝑚)^1/𝑝(𝑏^∞_𝑛=1^{0𝑥002𝑑𝑐𝑞}_𝑛)^1/𝑞=^∞_𝑛=1𝑛^−1−𝜀<1+^∞₁𝑥^−1−𝜀1𝑑𝑥=1+𝜀, (4.9)(4.10)where 𝑓_𝜀(𝑥) and 𝑔_𝜀(𝑥) are defined in the proof of Proposition 3.1. Since1𝜀(𝐾₁1+𝑜(1))+𝑂(1)<𝐻(1+𝜀), (4.11)it follows that𝐾₁+𝑜(1)<𝐻. (4.12)Hence,𝐾₁⩽𝐻. (4.13)This contradicts with the fact that 𝐻<𝐾₁. So the constant factor 𝐾₁ in (4.2) is the best possible one.
Setting𝑏_𝑛=[^∞_𝑚=1𝑎_𝑚]𝑚+𝑟𝑛+𝑠|𝑚−𝑛|^𝑝−1,𝑛=1,2,…, (4.14)we find^∞_𝑛=1[^∞_𝑚=1𝑎_𝑚]𝑚+𝑟𝑛+𝑠|𝑚−𝑛|^𝑝=^∞_𝑛=1𝑏^𝑞_𝑛=^∞∞_𝑛=1𝑚=1𝑎_𝑚𝑏_𝑛𝑚+𝑟𝑛+𝑠|𝑚−𝑛|. (4.15)By the same argument used in the proof of Proposition 3.2, we can show that (4.3) is valid, the constant factor 𝐾^𝑝₁ in (4.3) is the best possible one and inequalities (4.2) and (4.3) are equivalent.

Remark 4.2. Proposition 4.1 is the corresponding series form of Propositions 3.1 and 3.2 for 𝑠∈(−min{1,𝑟},min{1,𝑟}],𝑡=0, and it is also a generalization of Hilbert's inequality. Here, we restrict the constants 𝑠,𝑡 so that we can use the monotony of functions to obtain (4.6) and (4.9).

Acknowledgment

This work was partially supported by the National Natural Science Foundation of China (Grant no. 10871213).