A Hilbert Integral-Type Inequality with Parameters

Abstract

A Hilbert-type integral inequality with parameters 𝛼 and (𝛼,𝜆>0) can be established by introducing a nonhomogeneous kernel function. And the constant factor is proved to be the best possible. And then some important and especial results are enumerated. As applications, some equivalent forms are studied.

1. Introduction

Let 𝑓(𝑥),𝑔(𝑥)∈𝐿²(0,+∞). Then ^∞₀𝑓(𝑥)𝑔(𝑦)𝑥+𝑦𝑑𝑥𝑑𝑦≤𝜋{^∞₀𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑔²(𝑥)𝑑𝑥}^1/2, (1.1) where the constant factor 𝜋 is the best possible, and the equality in (1.1) holds if and only if 𝑓(𝑥)=0, or 𝑔(𝑥)=0.This is the famous Hilbert integral inequality (see [1, 2]). Owing to the importance of the Hilbert inequality in analysis and applications, some mathematicians have been studying them. Recently, various improvements and extensions of (1.1) appear in a great deal of papers (see [3–11], etc.). Specially, Gao and Hsu enumerated the research articles to more than 40 in the paper [6]. It is obvious that the integral kernel function of the left hand side of (1.1) is a homogeneous form of degree −1. In general, the Hilbert type integral inequality with a homogeneous kernel of degree −1 has been studied in the paper [1]. The purpose of the present paper is to establish a Hilbert integral type inequality with a non-homogeneous kernel. And the constant factor is proved to be the best possible. And then some important and especial results are enumerated and some equivalent forms are studied.

For convenience, we need to introduce the Catalan constant and define a real function.

The Catalan constant is defined by 1𝐺=2¹₀𝐾𝑑𝑘=^∞_𝑛=0(−1)^𝑛(2𝑛+1)², (1.2) where 𝐾 is the complete elliptic integral, namely, 𝐾=𝐾(𝑘)=₀^𝜋/2𝑑𝜃1−𝑘²sin²𝜃. (1.3) And the approximation of 𝐺 is that 𝐺=0,915965594⋯. (1.4) These results can be found in the paper [12, page 503].

Let 𝜆>0. Define a real function by𝑆(𝜆)=^∞_𝑛=0(−1)^𝑛1(2𝑛+1)^𝜆+1. (1.5)

In particular, 𝑆(1)=𝐺, where 𝐺 is the Catalan constant.

In order to prove our main results, we need the following lemmas.

Lemma 1.1. Let 𝑎 be a positive number and 𝑏>−1. Then ^∞₀𝑥^𝑏𝑒^−𝑎𝑥𝑑𝑥=Γ(𝑏+1)𝑎^𝑏+1. (1.6)

Proof. According to the definition of the gamma function, we obtain immediately (1.6). This result can be also found in the paper [12, page 226, formula (1053)].

Lemma 1.2. Let 𝛼,𝜆>0. Define a function by 2𝐶(𝛼,𝜆)=2(𝛼)^𝜆+1Γ(𝜆+1)𝑆(𝜆), (1.7) where Γ(𝑧) is the gamma function, and 𝑆(𝜆) is a function defined by (1.5). Then ^∞₀𝑘(1,𝑢)𝑢^𝛼/2−1𝑑𝑢=𝐶(𝛼,𝜆), (1.8) where 𝑘(1,𝑢)=|ln(1/𝑢)|^𝜆/(1+𝑢^𝛼).

Proof. It is easy to deduce that ^∞₀𝑘(1,𝑢)𝑢^𝛼/2−1𝑑𝑢=¹₀(ln(1/𝑢))^𝜆1+𝑢^𝛼𝑢^𝛼/2−1𝑑𝑢+^∞₁(ln𝑢)^𝜆1+𝑢^𝛼𝑢^𝛼/2−1=𝑑𝑢₁₀(ln(1/𝑢))^𝜆1+𝑢^𝛼𝑢^(𝛼/2)−1𝑑𝑢+¹₀(ln(1/𝑣))^𝜆1+𝑣^𝛼𝑢^(𝛼/2)−1𝑑𝑢=2¹₀(ln(1/𝑢))^𝜆1+𝑢^𝛼𝑢^(𝛼/2)−1𝑑𝑢=2^∞₀𝑡^𝜆𝑒^{−(𝛼/2)𝑡}1+𝑒^−𝛼𝑡𝑑𝑡. (1.9) Expanding 1/(1+𝑒^−𝛼𝑡) into power series of 𝑒^−𝛼𝑡and then using Lemma 1.1, we have ^∞₀𝑡^𝜆𝑒^{−(𝛼/2)𝑡}1+𝑒^−𝛼𝑡𝑑𝑡=^∞₀𝑡^𝜆𝑒^{−(𝛼/2)𝑡}(1−𝑒^−𝛼𝑡+𝑒^−2𝛼𝑡+𝑒^−3𝛼𝑡−𝑒^−4𝛼𝑡=+⋯)𝑑𝑡_∞0(𝑡^𝜆𝑒^{−(𝛼/2)𝑡}−𝑡^𝜆𝑒^{−(3/2)𝛼𝑡}+𝑡^𝜆𝑒^{−(5/2)𝛼𝑡}−𝑡^𝜆𝑒^{−(7/2)𝛼𝑡}2+⋯)𝑑𝑡=(𝛼)^𝜆+1Γ(𝜆+1)^∞_𝑛=0(−1)^𝑛1(2𝑛+1)^𝜆+12=(𝛼)^𝜆+1Γ(𝜆+1)𝑆(𝜆). (1.10)
It follows from (1.9), (1.10), and (1.7) that the equality (1.8) holds.

2. Main Results

In the section we will formulate our main results.

Theorem 2.1. Let 𝑓and 𝑔 be two real functions, and let 𝛼 and 𝜆 be arbitrary two positive numbers. If ∫^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥<+∞ and ∫^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥<+∞, then ^∞₀|ln𝑥𝑦|^𝜆𝑓(𝑥)𝑔(𝑦)1+(𝑥𝑦)^𝛼𝑑𝑥𝑑𝑦≤𝐶(𝛼,𝜆){^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥}^1/2, (2.1) where 𝐶(𝛼,𝜆) is defined by (1.7). And the constant factor 𝐶(𝛼,𝜆) in (2.1) is the best possible. And the equality in (2.1) holds if and only if 𝑓(𝑥)=0, or 𝑔(𝑥)=0.

Proof. We may apply Hardy’s technique and Cauchy-Schwarz’s inequality to estimate the left-hand side of (2.1) as follows: ^∞₀|ln𝑥𝑦|^𝜆𝑓(𝑥)𝑔(𝑦)1+(𝑥𝑦)^𝛼=𝑑𝑥𝑑𝑦_∞0|ln𝑥𝑦|^𝜆/2𝑓(𝑥)(1+(𝑥𝑦)^𝛼)^1/2(𝑥𝑦)^(2−𝛼)/4|ln𝑥𝑦|^𝜆/2𝑓(𝑥)(1+(𝑥𝑦)^𝛼)^1/2(𝑦𝑥)^(2−𝛼)/4𝑔(𝑦)𝑑𝑥𝑑𝑦≤{^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼(𝑥𝑦)^(2−𝛼)/2𝑓²(𝑥)𝑑𝑥𝑑𝑦}^1/2{^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼(𝑦𝑥)^(2−𝛼)/2𝑔²(𝑦)𝑑𝑥𝑑𝑦}^1/2={^∞₀𝜔(𝛼,𝜆,𝑥)𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝜔(𝛼,𝜆,𝑥)𝑔²(𝑥)𝑑𝑥}^1/2, (2.2) where 𝜔(𝛼,𝜆,𝑥)=∫^∞₀(|ln𝑥𝑦|^𝜆/(1+(𝑥𝑦)^𝛼))(𝑥/𝑦)^(2−𝛼)/2𝑑𝑦.
By substituting 𝑢 for 𝑥𝑦, it is easy to deduce that 𝜔(𝛼,𝜆,𝑥)=^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼(𝑥𝑦)^(2−𝛼)/2𝑑𝑦=𝑥^1−𝛼∞₀|ln𝑢|^𝜆1+𝑢^𝛼(1𝑢)^(2−𝛼)/2𝑑𝑢=𝑥^1−𝛼∞₀|ln(1/𝑢)|^𝜆1+𝑢^𝛼𝑢^(𝛼/2)−1𝑑𝑢. (2.3) Based on (1.8), we obtain 𝜔(𝛼,𝜆,𝑥)=𝑥^1−𝛼𝐶(𝛼,𝜆). (2.4) It is known from (2.2) and (2.4) that the inequality (2.1) is valid.
If 𝑓(𝑥)=0, or 𝑔(𝑥)=0, then the equality in (2.1) obviously holds. If 𝑓(𝑥)≠0 and 𝑔(𝑥)≠0, then 0<^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥<+∞,0<^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥<+∞. (2.5)
If (2.2) takes the form of the equality, then there exists a pair of non-zero constants 𝑐₁ and 𝑐₂ such that 𝑐₁|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑓²𝑥(𝑥)(𝑦)^1−𝛼/2=𝑐₂|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑔²𝑦(𝑦)(𝑥)^1−𝛼/2a.e.on(0,+∞)×(0,+∞). (2.6) Then we have 𝑐₁𝑥^2−𝛼𝑓²(𝑥)=𝑐₂𝑦^2−𝛼𝑔²(𝑦)=𝐶₀.(constant)a.e.on(0,+∞)×(0,+∞). (2.7) Without losing the generality, we suppose that 𝑐₁≠0; then ^∞₀𝑥^1−𝛼𝑓²𝐶(𝑥)𝑑𝑥=⁰𝑐₁^∞₀𝑥⁻¹𝑑𝑥. (2.8) This contradicts that 0<∫^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥<+∞. Hence it is impossible to take the equality in (2.2). It shows that it is also impossible to take the equality in (2.1).
It remains to need only to show that 𝐶(𝛼,𝜆) in (2.1) is the best possible. Below we will apply Yang’s technique (see [13]) to verify our assertion.
For all 𝑛∈𝑁, define two functions by 𝑓_𝑛𝑥(𝑥)=^{(𝛼/2)−1+1/2𝑛}𝑔𝑥∈(0,1),0𝑥∈[1,∞),_𝑛𝑥(𝑥)=0𝑥∈(0,1],_{(𝛼/2)−1−1/2𝑛}𝑥∈(1,∞). (2.9) Then we have (¹₀𝑥^1−𝛼𝑓²_𝑛(𝑥)𝑑𝑥)^1/2=(^∞₁𝑥^1−𝛼𝑔²_𝑛(𝑥)𝑑𝑥)^1/2=√𝑛. (2.10)
Let 0≤𝑘≤𝐶(𝛼,𝜆) such that the inequality (2.1) is still valid when C(𝛼,𝜆) is replaced by 𝑘. Specially, we have 1𝑛^∞₀|ln𝑥𝑦|^𝜆𝑓_𝑛(𝑥)𝑔_𝑛(𝑦)1+(𝑥𝑦)^𝛼1𝑑𝑥𝑑𝑦≤𝑘(𝑛){^∞₀𝑥^1−𝛼𝑓²_𝑛(𝑥)𝑑𝑥}^1/2{^∞₀𝑥^1−𝛼𝑔²_𝑛(𝑥)𝑑𝑥}^1/2=𝑘. (2.11) Let 𝑘(1,𝑥𝑦)=|ln𝑥𝑦|^𝜆/(1+(𝑥𝑦)^𝛼). By Fubini’s theorem, we have 1𝑘≥𝑛^∞₀𝑘(1,𝑥𝑦)𝑓_𝑛(𝑥)𝑔_𝑛=1(𝑦)𝑑𝑥𝑑𝑦𝑛^∞₁𝑦^{(𝛼/2)−1−1/2𝑛}(¹₀𝑘(1,𝑥𝑦)𝑥^{(𝛼/2)−1+1/2𝑛}=1𝑑𝑥)𝑑𝑦𝑛^∞₁𝑦^{−1−1/𝑛}(^𝑦₀𝑘(1,𝑢)𝑢^{(𝛼/2)−1+1/2𝑛}=1𝑑𝑢)𝑑𝑦𝑛{^∞₁𝑦^{−1−1/𝑛}(¹₀𝑘(1,𝑢)𝑢^{(𝛼/2)−1+1/2𝑛}+𝑑𝑢)𝑑𝑦_∞1𝑦^{−1−1/𝑛}(^𝑦₁𝑘(1,𝑢)𝑢^{(𝛼/2)−1+1/2𝑛}=1𝑑𝑢)𝑑𝑦}𝑛{^∞₁𝑛(¹₀𝑘(1,𝑢)𝑢^{(𝛼/2)−1+1/2𝑛}+𝑑𝑢)_∞1𝑘(1,𝑢)𝑢^{(𝛼/2)−1+1/2𝑛}(^∞_𝑢𝑦^{−1−1/𝑛}=𝑑𝑦)𝑑𝑢}₁₀𝑘(1,𝑢)𝑢^{(𝛼/2)−1+1/2𝑛}𝑑𝑢+^∞₁𝑘(1,𝑢)𝑢^{(𝛼/2)−1−1/2𝑛}𝑑𝑢. (2.12) By Fatou’s lemma and (1.8), we have 𝑘≥lim_𝑛→∞¹₀𝑘(1,𝑢)𝑢^{𝛼/2−1+1/2𝑛}𝑑𝑢+lim_𝑛→∞^∞₁𝑘(1,𝑢)𝑢^{𝛼/2−1−1/2𝑛}≥𝑑𝑢₁₀lim_𝑛→∞𝑘(1,𝑢)𝑢^{𝛼/2−1+1/2𝑛}𝑑𝑢+^∞₁lim_𝑛→∞𝑘(1,𝑢)𝑢^{𝛼/2−1−1/2𝑛}=𝑑𝑢₁₀𝑘(1,𝑢)𝑢^𝛼/2−1𝑑𝑢+^∞₁𝑘(1,𝑢)𝑢^𝛼/2−1𝑑𝑢=^∞₀𝑘(1,𝑢)𝑢^𝛼/2−1𝑑𝑢=𝐶(𝛼,𝜆). (2.13)
It follows that 𝑘=𝐶(𝛼,𝜆) in (2.1) is the best possible. Thus the proof of theorem is completed.

Based on Theorem 2.1, we may build some important and interesting inequalities.

Theorem 2.2. Let 𝑛 be a nonnegative integer and 𝛼>0. If ∫^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥<+∞ and ∫^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥<+∞, then ^∞₀(ln𝑥𝑦)^2𝑛𝑓(𝑥)𝑔(𝑦)1+(𝑥𝑦)^𝛼𝑑𝑥𝑑𝑦≤𝐶(𝛼,2𝑛){^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥}^1/2, (2.14) where 𝐶(𝛼,2𝑛)=(𝜋/𝛼)^2𝑛+1𝐸_𝑛,the𝐸_𝑛′𝑠 are the Euler numbers, namely, 𝐸₀=1,𝐸₁=1,𝐸₂=5,𝐸₃=61,𝐸₄=1385, and so forth, and the constant factor 𝐶(𝛼,2𝑛)in (2.14) is the best possible. And the equality in (2.14) holds if and only if 𝑓(𝑥)=0, or 𝑔(𝑥)=0.

Proof. We need only to verify the constant factor 𝐶(𝛼,2𝑛) in (2.14). When 𝜆=2𝑛, it is known from (1.10) that 2𝐶(𝛼,2𝑛)=2(𝛼)^2𝑛+1Γ(2𝑛+1)^∞_𝑘=0(−1)^𝑘1(2𝑘+1)^2𝑛+1. (2.15)
It is known from the paper [14] that ^∞_𝑘=0(−1)^𝑘1(2𝑘+1)^2𝑛+1=𝜋^2𝑛+12^2𝑛+2𝐸(2𝑛)!^𝑛, (2.16) where the 𝐸_𝑛′𝑠 are the Euler numbers, namely, 𝐸₀=1,𝐸₁=1,𝐸₂=5,𝐸₃=61,𝐸₄=1385, and so forth.
It follows that the constant factor 𝐶(𝛼,2𝑛) in (2.14) is correct.

In particular, based on Theorem 2.2, the following results are gotten.

Corollary 2.3. If ∫^∞₀𝑓²(𝑥)𝑑𝑥<+∞ and ∫^∞₀𝑔²(𝑥)𝑑𝑥<+∞, then ^∞₀𝑓(𝑥)𝑔(𝑦)1+𝑥𝑦𝑑𝑥𝑑𝑦≤𝜋{^∞₀𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑔²(𝑥)𝑑𝑥}^1/2, (2.17) where the constant factor 𝜋 in (2.17) is the best possible. And the equality in (2.17) holds if and only if 𝑓(𝑥)=0, or 𝑔(𝑥)=0.

Corollary 2.4. If ∫^∞₀𝑓²(𝑥)𝑑𝑥<+∞ and ∫^∞₀𝑔²(𝑥)𝑑𝑥<+∞, then ^∞₀(ln𝑥𝑦)²𝑓(𝑥)𝑔(𝑦)1+𝑥𝑦𝑑𝑥𝑑𝑦≤𝜋³{^∞₀𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑔²(𝑥)𝑑𝑥}^1/2, (2.18) where the constant factor 𝜋³ in (2.18) is the best possible. And the equality in (2.18) holds if and only if 𝑓(𝑥)=0, or 𝑔(𝑥)=0.

Theorem 2.5. Let 𝑓and 𝑔 be two real functions, and 𝛼 a positive number. If ∫^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥<+∞ and ∫^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥<+∞, then ^∞₀|ln𝑥𝑦|𝑓(𝑥)𝑔(𝑦)1+(𝑥𝑦)^𝛼𝑑𝑥𝑑𝑦≤8𝐺𝛼²{^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥}^1/2, (2.19) where 𝐺 is the Catalan constant, that is, 𝐺=0,915965594⋯. And the constant factor 8𝐺/𝛼² in (2.19) is the best possible. And the equality in (2.19) holds if and only if 𝑓(𝑥)=0, or 𝑔(𝑥)=0.

Proof. We need only to verify the constant factor in (2.19).  For case 𝜆=1, based on (1.7), (1.5) and (1.2), it is easy to deduce that the constant factor is that 𝐶(𝛼,1)=8𝐺/𝛼².

In particular, when 𝛼=𝜆=1, the following result is obtained．

Corollary 2.6. Let 𝑓 and 𝑔 be two real functions. If ∫^∞₀𝑓²(𝑥)𝑑𝑥<+∞ and ∫^∞₀𝑔²(𝑥)𝑑𝑥<+∞, then ^∞₀|ln𝑥𝑦|𝑓(𝑥)𝑔(𝑦)1+𝑥𝑦𝑑𝑥𝑑𝑦≤8𝐺{^∞₀𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑔²(𝑥)𝑑𝑥}^1/2, (2.20) where 𝐺 is the Catalan constant, that is, 𝐺=0,915965594⋯. And the constant factor 8𝐺 in (2.20) is the best possible. And the equality in (2.20) holds if and only if𝑓(𝑥)=0, or 𝑔(𝑥)=0.

A great deal of the Hilbert type inequalities can be established provided that the parameters are properly selected．

3. Some Equivalent Forms

As applications, we will build some new inequalities.

Theorem 3.1. Let 𝑓 be a real function, and let 𝛼 and 𝜆 be arbitrary two positive numbers. If ∫^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥<+∞, then ^∞₀𝑦^𝛼−1{^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑑𝑥}²𝑑𝑦≤(𝐶(𝛼,𝜆))^2∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥, (3.1) where 𝐶(𝛼,𝜆) is defined by (1.7) and the constant factor (𝐶(𝛼,𝜆))² in (3.1) is the best possible. And the equality in (3.1) holds if and only if 𝑓(𝑥)=0. And the inequality (3.1) is equivalent to (2.1).

Proof. First, we assume that the inequality (2.1) is valid. Define a function 𝑔(𝑦) by 𝑔(𝑦)=𝑦^𝛼−1∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑑𝑥,𝑦∈(0,+∞). (3.2) By using (2.1), we have ^∞₀𝑦^𝛼−1{^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑑𝑥}²=𝑑𝑦_∞0|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦≤𝐶(𝛼,𝜆){^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑦^1−𝛼𝑔²(𝑦)𝑑𝑦}^1/2=𝐶(𝛼,𝜆){^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑦^𝛼−1(^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑑𝑥)²𝑑𝑦}^1/2. (3.3) It follows from (3.3) that the inequality (3.1) is valid after some simplifications.
On the other hand, assume that the inequality (3.1) keeps valid; by applying in turn Cauchy’s inequality and (3.1), we have ^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼=𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦_∞0𝑦^(𝛼−1)/2{^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑑𝑥}𝑦^(𝛼−1)/2𝑔(𝑦)𝑑𝑦≤{^∞₀𝑦^𝛼−1(^∞₀|ln𝑥𝑦|^𝜆1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑑𝑥)²𝑑𝑦}^1/2{^∞₀𝑦^1−𝛼𝑔²(𝑦)𝑑𝑦}^1/2≤{(𝐶(𝛼,𝜆))^2∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑦^1−𝛼𝑔²(𝑦)𝑑𝑦}^1/2=𝐶(𝛼,𝜆){^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑦^1−𝛼𝑔²(𝑦)𝑑𝑦}^1/2. (3.4) Therefore, the inequality (3.1) is equivalent to (2.1).
If the constant factor (𝐶(𝛼,𝜆))² in (3.1) is not the best possible, then it is known from (3.4) that the constant factor 𝐶(𝛼,𝜆) in (2.1) is also not the best possible. This is a contradiction. The theorem is proved.

Theorem 3.2. Let 𝑛 be a nonnegative integer and 𝛼>0. If ∫^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥<+∞, then ^∞₀𝑦^𝛼−1{^∞₀(ln𝑥𝑦)^2𝑛1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑑𝑥}²𝑑𝑦≤(𝐶(𝛼,2𝑛))^2∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥, (3.5) where 𝐶(𝛼,2𝑛)=(𝜋/𝛼)^2𝑛+1𝐸_𝑛, the 𝐸_𝑛′𝑠 are the Euler numbers,namely, 𝐸₀=1,𝐸₁=1,𝐸₂=5,𝐸₃=61,𝐸₄=1385, and so forth, and the constant factor 𝐶(𝛼,2𝑛) is the best possible. And the equality in (3.5) holds if and only if 𝑓(𝑥)=0.  And the inequality (3.5) is equivalent to (2.14).

Theorem 3.3. Let 𝑓and 𝑔 be two real functions, and let 𝛼 be a positive numbers. If ∫^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥<+∞ and ∫^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥<+∞, then ^∞₀𝑦^𝛼−1{^∞₀|ln𝑥𝑦|1+(𝑥𝑦)^𝛼𝑓(𝑥)𝑑𝑥}²𝑑𝑦≤8𝐺𝛼²{^∞₀𝑥^1−𝛼𝑓²(𝑥)𝑑𝑥}^1/2{^∞₀𝑥^1−𝛼𝑔²(𝑥)𝑑𝑥}^1/2, (3.6) where 𝐺 is the Catalan constant, that is, 𝐺=0,915965594⋯. And the constant factor 8𝐺/𝛼² in (3.6) is the best possible. And the equality in (3.6) holds if and only if 𝑓(𝑥)=0. And the inequality (3.6) is equivalent to (2.19).

The proofs of Theorems 3.2 and 3.3 are similar to one of Theorem 3.1; they are omitted here.

Similarly, we can establish also some new inequalities which they are, respectively, equivalent to the inequalities (2.17), (2.18) and (2.20). These are omitted here.

Acknowledgments

Authors would like to express their thanks to the referees for their valuable comments and suggestions. The research is supported by Scientific Research Fund of Hunan Provincial Education Department (09C789).