On a more accurate half-discrete Hilbert's inequality

Abstract

By using the way of weight coefficients and the idea of introducing parameters and by means of Hadamard's inequality, we give a more accurate half-discrete Hilbert's inequality with a best constant factor. We also consider its best extension with parameters, the equivalent forms, the operator expressions as well as some reverses.

2000 Mathematics Subject Classification: 26D15; 47A07.

1 Introduction

If a _n , b _n ≥ 0, $0<{\sum }_{n=1}^{\infty }{a}_n^2<\infty$ and $0<{\sum }_{n=1}^{\infty }{b}_n^2<\infty$, then we have the following well-known Hilbert's inequality (cf. [1]):

$$\sum _{n=1}^{\infty }{\sum _{m=1}^{\infty }\frac{a_m{b}_n}{m+n}<\pi \left(\sum _{m=1}^{\infty }{a}_m^2\sum _{n=1}^{\infty }{b}_n^2\right)}^{1/2},$$

(1)

where the constant factor π is the best possible. The integral analogue of inequality (1) is given as follows (cf. [2]): If $0<{\int }_0^{\infty }{f}^2\left(x\right)dx<\infty$ and $0<{\int }_0^{\infty }{g}^2\left(x\right)dx<\infty$, then

$$\underset{0}{\overset{\infty }{\int }}{\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{x+y}dxdy<\pi \left(\underset{0}{\overset{\infty }{\int }}{f}^2\left(x\right)dx\underset{0}{\overset{\infty }{\int }}{g}^2\left(x\right)dx\right)}^{1/2},$$

(2)

where the constant factor π is the best possible. We named inequality (2) as Hilbert's integral inequality. Hardy et al. [3] proved the following more accurate Hilbert's inequality:

$$\sum _{n=1}^{\infty }{\sum _{m=1}^{\infty }\frac{a_m{b}_n}{m+n-1}<\pi \left(\sum _{m=1}^{\infty }{a}_m^2\sum _{n=1}^{\infty }{b}_n^2\right)}^{1/2},$$

(3)

where the constant factor π is still the best possible. Inequalities (1)-(3) are important in analysis and its applications [4]. There are lots of improvements, generalizations, and applications of inequalities (1-3), for more details, refer to literatures [5–18].

We find a few results on the half-discrete Hilbert-type inequalities with the non-homogeneous kernel, which were published early ([[3], Th. 351], [19]). Recently, Yang [20–22] gave some half-discrete Hilbert-type inequalities. A half-discrete Hilbert's inequality with the homogeneous kernel was derived as follows [20]: If $0<{\int }_0^{\infty }{f}^2\left(x\right)dx<\infty$ and $0<{\sum }_{n=1}^{\infty }{a}_n^2<\infty$, then

$$\sum _{n=1}^{\infty }{a}_n\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)}{x+n}dx<\pi {\left(\sum _{n=1}^{\infty }{a}_n^2\underset{0}{\overset{\infty }{\int }}{f}^2\left(x\right)dx\right)}^{1/2},$$

(4)

where the constant factor π is the best possible.

In this article, by using the way of weight coefficients and the idea of introducing parameters and by means of Hadamard's inequality, we give a more accurate inequality of (4) with a best constant factor as follows:

$$\sum _{n=1}^{\infty }{a}_n\underset{-\frac{1}{2}}{\overset{\infty }{\int }}\frac{f\left(x\right)}{x+n}dx<\pi {\left(\sum _{n=1}^{\infty }{a}_n^2\underset{-\frac{1}{2}}{\overset{\infty }{\int }}{f}^2\left(x\right)dx\right)}^{1/2}.$$

(5)

We also consider its best extension with parameters, the equivalent forms, the operator expressions as well as some reverses.

2 Some lemmas

Lemma 1 Suppose 0 < α ≤ 1, $0\le \beta \le \frac{1}{2}$, γ ∈ (-∞, ∞), λ₁ > 0, 0 < λ₂α ≤ 1, λ = λ₁ + λ₂. Define the beta function (cf. [18]) and the weight coefficients as follows:

$$B\left(u,v\right):=\underset{0}{\overset{\infty }{\int }}\frac{t^{u-1}dt}{{\left(1+t\right)}^{u+v}}=\underset{0}{\overset{1}{\int }}\frac{t^{u-1}+t^{v-1}}{{\left(1+t\right)}^{u+v}}dt\left(u,v>0\right),$$

(6)

$$\omega \left(n\right):={\left(n-\beta \right)}^{{\lambda }_2\alpha }\underset{\gamma }{\overset{\infty }{\int }}\frac{{\left(x-\gamma \right)}^{{\lambda }_1\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\left(n\in \mathbf{N}\right),$$

(7)

$$\varpi \left(x\right):={\left(x-\gamma \right)}^{{\lambda }_1\alpha }\sum _{n=1}^{\infty }\frac{{\left(n-\beta \right)}^{{\lambda }_2\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\left(x\in \left(\gamma ,\infty \right)\right).$$

(8)

Setting $k_{{\lambda }_1}\left(\alpha \right):=\frac{1}{\alpha }B\left({\lambda }_1,{\lambda }_2\right)$, we have the following inequalities:

$$0<k_{{\lambda }_1}\left(\alpha \right)\left(1-{\theta }_{\lambda }\left(x\right)\right)<\varpi \left(x\right)<\omega \left(n\right)=k_{{\lambda }_1}\left(\alpha \right),$$

(9)

where, ${\theta }_{\lambda }\left(x\right):=\frac{1}{B\left({\lambda }_1,{\lambda }_2\right)}{\int }_0^{{\left(\frac{1-\beta }{x-\gamma }\right)}^{\alpha }}\frac{u^{{\lambda }_2-1}}{{\left(1+u\right)}^{\lambda }}du>0and{\theta }_{\lambda }\left(x\right)=O\left(\frac{1}{{\left(x-\gamma \right)}^{{\lambda }_2\alpha }}\right)\left(x\in \left(\gamma ,\infty \right)\right).$

Proof. Putting $u={\left(\frac{x-\gamma }{n-\beta }\right)}^{\alpha }$ in (7), we have

$$\omega \left(n\right)=\frac{1}{\alpha }\underset{0}{\overset{\infty }{\int }}\frac{u^{{\lambda }_1-1}}{{\left(1+u\right)}^{\lambda }}du=\frac{1}{\alpha }B\left({\lambda }_1,{\lambda }_2\right)=k_{{\lambda }_1}\left(\alpha \right).$$

(10)

For fixed x ∈ (γ, ∞), setting

$$f\left(t\right):=\frac{{\left(x-\gamma \right)}^{{\lambda }_1\alpha }{\left(t-\beta \right)}^{{\lambda }_2\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(t-\beta \right)}^{\alpha }\right]}^{\lambda }}\left(t\in \left(\beta ,\infty \right)\right),$$

(11)

in view of the conditions, we find f' (t) < 0 and f" (t) > 0. By the following Hadamard's inequality (cf. [18]):

$$f\left(n\right)<\underset{n-\frac{1}{2}}{\overset{n+\frac{1}{2}}{\int }}f\left(t\right)dt\left(n\in \mathbf{N}\right),$$

(12)

and putting $u={\left(\frac{t-\beta }{x-\gamma }\right)}^{\alpha }$, it follows

$$\begin{array}{cc}\hfill \varpi \left(x\right)& =\sum _{n=1}^{\infty }f\left(n\right)<\sum _{n=1}^{\infty }\underset{n-\frac{1}{2}}{\overset{n+\frac{1}{2}}{\int }}f\left(t\right)dt=\underset{\frac{1}{2}}{\overset{\infty }{\int }}f\left(t\right)dt\hfill \\ <\underset{\beta }{\overset{\infty }{\int }}f\left(t\right)dt=\frac{1}{\alpha }\underset{0}{\overset{\infty }{\int }}\frac{u^{{\lambda }_2-1}}{{\left(1+u\right)}^{\lambda }}du=\frac{1}{\alpha }B\left({\lambda }_2,{\lambda }_1\right)=k_{{\lambda }_1}\left(\alpha \right),\hfill \\ \hfill \varpi \left(x\right)& =\sum _{n=1}^{\infty }f\left(n\right)>\underset{1}{\overset{\infty }{\int }}f\left(t\right)dt=\underset{\beta }{\overset{\infty }{\int }}f\left(t\right)dt-\underset{\beta }{\overset{1}{\int }}f\left(t\right)dt\hfill \\ =k_{{\lambda }_1}\left(\alpha \right)-\frac{1}{\alpha }\underset{0}{\overset{{\left(\frac{1-\beta }{x-\gamma }\right)}^{\alpha }}{\int }}\frac{u^{{\lambda }_2-1}}{{\left(1+u\right)}^{\lambda }}du=k_{{\lambda }_1}\left(\alpha \right)\left(1-{\theta }_{\lambda }\left(x\right)\right)>0,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill 0<{\theta }_{\lambda }\left(x\right)& =\frac{1}{B\left({\lambda }_1,{\lambda }_2\right)}\underset{0}{\overset{{\left(\frac{1-\beta }{x-\gamma }\right)}^{\alpha }}{\int }}\frac{u^{{\lambda }_2-1}}{{\left(1+u\right)}^{\lambda }}du\hfill \\ <\frac{1}{B\left({\lambda }_1,{\lambda }_2\right)}\underset{0}{\overset{{\left(\frac{1-\beta }{x-\gamma }\right)}^{\alpha }}{\int }}{u^{{\lambda }_2}}^{-1}du=\frac{{\left(1-\beta \right)}^{{\lambda }_2\alpha }}{{\lambda }_{2}B\left({\lambda }_1,{\lambda }_2\right)}\frac{1}{{\left(x-\gamma \right)}^{{\lambda }_2\alpha }}.\hfill \end{array}$$

Hence, we prove that (9) is valid.

Lemma 2 Suppose that $\frac{1}{p}+\frac{1}{q}=1\left(p\ne 0,1\right)$, 0 < α ≤ 1, $0\le \beta \le \frac{1}{2}$, γ ∈ (-∞ + ∞), λ₁ > 0, 0 < λ₂α ≤ 1, λ = λ₁ + λ₂, α _n ≥ 0, f(x) ≥ 0 is a real measurable function in (γ,∞), then (i) for p > 1, we have the following

$$\begin{array}{cc}\hfill J& :={\left\{\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{p{\lambda }_2\alpha -1}{\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\right]}^p\right\}}^{1/p}\hfill \\ \le {\left(k_{{\lambda }_1}\left(\alpha \right)\right)}^{1/q}{\left\{\underset{\gamma }{\overset{\infty }{\int }}\varpi \left(x\right){\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}{f}^p\left(x\right)dx\right\}}^{1/p},\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill L_1& :={\left\{\underset{\gamma }{\overset{\infty }{\int }}\frac{{\left(x-\gamma \right)}^{q{\lambda }_1\alpha -1}}{{\varpi }^{q-1}\left(x\right)}{\left[\sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^{q}dx\right\}}^{1/q}\hfill \\ <{\left\{k_{{\lambda }_1}\left(\alpha \right)\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}{a}_n^q\right\}}^{1/q},\hfill \end{array}$$

(14)

where ϖ(x) and ω(n) are indicated by (7) and (8).

(ii) for p < 1(p ≠ 0), we have the reverses of (13) and (14).

Proof. (i) By (7)-(9) and Hölder's inequality (cf. [18]), we find

$$\begin{array}{c}{\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\right]}^p=\left\{\underset{\gamma }{\overset{\infty }{\int }}\frac{1}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right.\hfill \\ \times \left[\frac{{\left(x-\gamma \right)}^{\left(1-{\lambda }_1\alpha \right)/q}}{{\left(n-\beta \right)}^{\left(1-{\lambda }_2\alpha \right)/q}}f\left(x\right)\right]{\left[\frac{{\left(n-\beta \right)}^{\left(1-{\lambda }_2\alpha \right)/q}}{{\left(x-\gamma \right)}^{\left(1-{\lambda }_1\alpha \right)/q}}dx\right]}^p\hfill \\ \le \underset{\gamma }{\overset{\infty }{\int }}\frac{1}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{{\left(x-\gamma \right)}^{\left(1-{\lambda }_1\alpha \right)\left(\rho -1\right)}}{{\left(n-\beta \right)}^{1-{\lambda }_2\alpha }}{f}^p\left(x\right)dx\hfill \\ \times {\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{1}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{{\left(n-\beta \right)}^{\left(1-{\lambda }_2\alpha \right)\left(q-1\right)}}{{\left(x-\gamma \right)}^{1-{\lambda }_1\alpha }}dx\right]}^{p-1}\hfill \\ =\underset{\gamma }{\overset{\infty }{\int }}\frac{f^p\left(x\right){\left(x-\gamma \right)}^{\left(1-{\lambda }_1\alpha \right)\left(p-1\right)}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{dx}{{\left(n-\beta \right)}^{1-{\lambda }_2\alpha }}{\left[{\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}\omega \left(n\right)\right]}^{p-1}\hfill \\ ={\left(n-\beta \right)}^{1-p{\lambda }_2\alpha }{k}_{{\lambda }_1}^{p-1}\left(\alpha \right)\underset{\gamma }{\overset{\infty }{\int }}\frac{f^p\left(x\right){\left(x-\gamma \right)}^{\left(1-{\lambda }_1\alpha \right)\left(\rho -1\right)}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{1}{{\left(n-\beta \right)}^{1-{\lambda }_2\alpha }}dx,\hfill \end{array}$$

(15)

$$\begin{array}{c}{J}^p\le k_{{\lambda }_1}^{p-1}\left(\alpha \right)\sum _{n=1}^{\infty }\underset{\gamma }{\overset{\infty }{\int }}\frac{f^p\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{{\left(x-\gamma \right)}^{\left(1-{\lambda }_1\alpha \right)\left(p-1\right)}}{{\left(n-\beta \right)}^{1-{\lambda }_2\alpha }}dx\hfill \\ =k_{{\lambda }_1}^{p-1}\left(\alpha \right)\underset{\gamma }{\overset{\infty }{\int }}\sum _{n=1}^{\infty }\frac{{\left(n-\beta \right)}^{{\lambda }_2\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}{\left(x-\gamma \right)}^{{\lambda }_1\alpha +p\left(1-{\lambda }_1\alpha \right)-1}{f}^p\left(x\right)dx\hfill \\ =k_{{\lambda }_1}^{p-1}\left(\alpha \right)\underset{\gamma }{\overset{\infty }{\int }}\varpi \left(x\right){\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}{f}^p\left(x\right)dx.\hfill \end{array}$$

(16)

Hence (13) is valid. Using Hölder's inequality again, we have

$$\begin{array}{c}{\left[\sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^q=\left\{\sum _{n=1}^{\infty }\frac{1}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right.\hfill \\ {\left.\times \left[\frac{{\left(x-\gamma \right)}^{\left(1-{\lambda }_1\alpha \right)/q}}{{\left(n-\beta \right)}^{\left(1-{\lambda }_2\alpha \right)/q}}\right]\left[\frac{{\left(n-\beta \right)}^{\left(1-{\lambda }_2\alpha \right)/q}}{{\left(x-\gamma \right)}^{\left(1-{\lambda }_1\alpha \right)/q}}{a}_n\right]\right\}}^q\hfill \\ \le {\left[\varpi \left(x\right){\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}\right]}^{q-1}\sum _{n=1}^{\infty }\frac{a_n^q}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{{\left(n-\beta \right)}^{\left(1-{\lambda }_2\alpha \right)q/p}}{{\left(x-\gamma \right)}^{1-{\lambda }_1\alpha }}\hfill \\ ={\varpi }^{q-1}\left(x\right){\left(x-\gamma \right)}^{1-q{\lambda }_1\alpha }\sum _{n=1}^{\infty }\frac{{\left(n-\beta \right)}^{\left(1-{\lambda }_2\alpha \right)\left(q-1\right)|}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{1}{{\left(x-\gamma \right)}^{1-{\lambda }_1\alpha }}{a}_n^q,\hfill \end{array}$$

(17)

$$\begin{array}{c}{L}_1^q\le \underset{\gamma }{\overset{\infty }{\int }}\sum _{n=1}^{\infty }\frac{1}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{{\left(n-\beta \right)}^{\left(1-{\lambda }_2\alpha \right)\left(q-1\right)}}{{\left(x-\gamma \right)}^{1-{\lambda }_1\alpha }}{a}_n^{q}dx\hfill \\ =\sum _{n=1}^{\infty }\left[{\left(n-\beta \right)}^{{\lambda }_2\alpha }\underset{\gamma }{\overset{\infty }{\int }}\frac{{\left(x-\gamma \right)}^{{\lambda }_1\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\right]{\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}{a}_n^q\hfill \\ =\sum _{n=1}^{\infty }\omega \left(n\right){\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}{a}_n^q=k_{{\lambda }_1}\left(\alpha \right)\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-\lambda 2\alpha \right)-1}{a}_n^q.\hfill \end{array}$$

(18)

Hence (14) is valid.

(ii) For 0 < p < 1(q < 0) or p < 0(0 < q < 1), using the reverse Hölder's inequality and in the same way, we have the reverses of (13) and (14).

Lemma 3 As the assumptions of Lemmas 1 and 2, we set $\varphi \left(x\right):={\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}$, $\tilde{\varphi }\left(x\right):=\left(1-{\theta }_{\lambda }\left(x\right)\right)\varphi \left(x\right)$, $\psi \left(n\right):={\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}$,

$$\begin{array}{c}\hfill L_{p,\varphi }\left(\gamma ,\infty \right):=\left\{f;{\parallel f\parallel }_{p,\varphi }={\left\{\underset{\gamma }{\overset{\infty }{\int }}\varphi \left(x\right)|f\left(x\right){|}^{p}dx\right\}}^{1/p}<\infty \right\},\\ \hfill l_{q,\psi }:=\left\{a=\left\{a_n\right\};\parallel a\parallel {,}_{\psi }={\left\{\sum _{n=1}^{\infty }\psi \left(n\right)|a_n{|}^q\right\}}^{1/q}<\infty \right\}\end{array}$$

(Note. if p > 1, then L _p , _ϕ (γ, ∞) and l _q , _ψ are normal spaces; if 0 < p < 1 or p < 0, then both L _p , _ϕ (γ, ∞) and l _q,ψ are not normal spaces, but we still use the formal symbols in the following.) For 0 < ε < min{1, λ₁pα}, setting $\tilde{a}={\left\{{\tilde{a}}_n\right\}}_{n=1}^{\infty }$ and $\tilde{f}\left(x\right)$as follows

$${\tilde{a}}_n={\left(n-\beta \right)}^{{\lambda }_2\alpha -\frac{\epsilon }{q}-1};\tilde{f}\left(x\right)=\left\{\begin{array}{cc}0,\hfill & \hfill x\in \left(\gamma ,1+\gamma \right),\hfill \\ \hfill {\left(x-\gamma \right)}^{{\lambda }_1\alpha -\frac{\epsilon }{p}-1},\hfill & \hfill x\in \left[1+\gamma ,\infty \right),\hfill \end{array}\right.$$

(19)

(i) if p > 1, there exists a constant k > 0, such that

$$\mathit{Ĩ}:=\sum _{n=1}^{\infty }{\tilde{a}}_n\underset{\gamma }{\overset{\infty }{\int }}\frac{\tilde{f}\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx<k{∥\tilde{f}∥}_{p,\varphi }{∥\tilde{a}∥}_{q,\psi },$$

(20)

then it follows

$$k{\left(\frac{\epsilon +1-\beta }{{\left(1-\beta \right)}^{\epsilon +1}}\right)}^{1/q}\ge \frac{1}{\alpha }B\left({\lambda }_2+\frac{\epsilon }{p\alpha },{\lambda }_1+\frac{\epsilon }{p\alpha }\right)-\epsilon O\left(1\right);$$

(21)

(ii) if 0 < p < 1, there exists a constant k > 0, such that

$$\mathit{Ĩ}-\sum _{n=1}^{\infty }{\tilde{a}}_n\underset{\gamma }{\overset{\infty }{\int }}\frac{\tilde{f}\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx>k{∥\tilde{f}∥}_{p,\tilde{\varphi }}{∥\tilde{a}∥}_{q,\psi },$$

(22)

then it follows

$$k{\left(1-\epsilon O\left(1\right)\right)}^{1/p}<\frac{1}{\alpha }{\left(\frac{\epsilon +1-\beta }{{\left(1-\beta \right)}^{\epsilon +1}}\right)}^{1/p}B\left({\lambda }_1-\frac{\epsilon }{p\alpha },{\lambda }_2+\frac{\epsilon }{p\alpha }\right).$$

(23)

Proof. we obtain

$$\begin{array}{cc}\hfill {∥\tilde{f}∥}_{p,\varphi }& ={\left\{\underset{\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}{\tilde{f}}^p\left(x\right)dx\right\}}^{1/p}\hfill \\ ={\left\{\underset{1+\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{-1-\epsilon }dx\right\}}^{1/p}={\left(\frac{1}{\epsilon }\right)}^{1/p},\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill {∥\tilde{a}∥}_{q,\psi }^q& =\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}{\tilde{a}}_n^q=\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{-1-\epsilon }\hfill \\ <{\left(1-\beta \right)}^{-1-\epsilon }+\underset{1}{\overset{\infty }{\int }}{\left(x-\beta \right)}^{-1-\epsilon }dx=\frac{\epsilon +1-\beta }{\epsilon {\left(1-\beta \right)}^{\epsilon +1}}.\hfill \end{array}$$

(25)

(i) For p > 1, then q > 1, ${\lambda }_2\alpha -\frac{\epsilon }{q}-1<0$, by (20), (24), and (25), we find

$$\begin{array}{cc}\hfill \mathit{Ĩ}& <k{\left(\frac{1}{\epsilon }\right)}^{1/p}{\left[\frac{\epsilon +1-\beta }{\epsilon {\left(1-\beta \right)}^{\epsilon +1}}\right]}^{1/q}=\frac{k}{\epsilon }{\left[\frac{\epsilon +1-\beta }{{\left(1-\beta \right)}^{\epsilon +1}}\right]}^{1/q},\hfill \\ \hfill \mathit{Ĩ}& =\underset{1+\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{{\lambda }_1\alpha -\frac{\epsilon }{p}-1}\left(\sum _{n=1}^{\infty }\frac{{\left(n-\beta \right)}^{{\lambda }_2\alpha -\frac{\epsilon }{q}-1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right)dx\hfill \\ \ge \underset{1+\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{{\lambda }_1\alpha -\frac{\epsilon }{p}-1}\left(\underset{1}{\overset{\infty }{\int }}\frac{{\left(y-\beta \right)}^{{\lambda }_2\alpha -\frac{\epsilon }{q}-1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(y-\beta \right)}^{\alpha }\right]}^{\lambda }}dy\right)dx.\hfill \end{array}$$

(26)

Setting s = x - γ, $t={\left(\frac{y-\beta }{x-\gamma }\right)}^{\alpha }$ in the above integral, we have

$$\mathit{Ĩ}\ge \frac{1}{\alpha }\underset{1}{\overset{\infty }{\int }}{s}^{-1-\epsilon }\left[\underset{{\left(\frac{1-\beta }{s}\right)}^{\epsilon }}{\overset{\infty }{\int }}{t}^{{\lambda }_2-\frac{\epsilon }{q\alpha }-1}\frac{1}{{\left(1+t\right)}^{\lambda }}dt\right]ds=A+B,$$

(27)

where

$$\begin{array}{cc}\hfill A:& =\frac{1}{\alpha }\underset{1}{\overset{\infty }{\int }}{s}^{-1-\epsilon }\underset{{\left(\frac{1-\beta }{s}\right)}^{\alpha }}{\overset{1}{\int }}\frac{t^{{\lambda }_2-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds;\hfill \\ \hfill B:& =\frac{1}{\alpha }\underset{1}{\overset{\infty }{\int }}{s}^{-1-\epsilon }\underset{1}{\overset{\infty }{\int }}\frac{t^{{\lambda }_2-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds=\frac{1}{\alpha \epsilon }\underset{1}{\overset{\infty }{\int }}\frac{t^{{\lambda }_2-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dt\hfill \\ \stackrel{u=\frac{1}{t}}{=}\frac{1}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{u^{{\lambda }_1+\frac{\epsilon }{q\alpha }-1}}{{\left(1+u\right)}^{\lambda }}du\ge \frac{1}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{u^{{\lambda }_1+\frac{\epsilon }{q\alpha }-1}}{{\left(1+u\right)}^{\lambda +\epsilon /\alpha }}du.\hfill \end{array}$$

(28)

Since

$$0<\frac{1}{\alpha }\underset{1-\beta }{\overset{1}{\int }}{s}^{-1-\epsilon }\underset{\left(\frac{1-\beta }{s}\right)}{\overset{1}{\int }}\frac{t^{{\lambda }_2-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds\le \frac{1}{\alpha }\underset{1-\beta }{\overset{1}{\int }}\underset{{\left(1-\beta \right)}^{\alpha }}{\overset{1}{\int }}{s}^{-2}\frac{t^{{\lambda }_2-\frac{1}{\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds<\infty ,$$

then by Fubini's theorem, we have

$$\begin{array}{cc}\hfill A& =\frac{1}{\alpha }\left[\underset{1-\beta }{\overset{\infty }{\int }}{s}^{-1-\epsilon }\underset{{\left(\frac{1-\beta }{s}\right)}^{\alpha }}{\overset{1}{\int }}\frac{t^{{\lambda }_2-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds-\underset{1-\beta }{\overset{1}{\int }}{s}^{-1-\epsilon }\underset{{\left(\frac{1-\beta }{s}\right)}^{\alpha }}{\overset{1}{\int }}\frac{t^{{\lambda }_2-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds\right]\hfill \\ =\frac{1}{\alpha }\underset{0}{\overset{1}{\int }}\frac{t^{{\lambda }_2-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}\left[\underset{\left(1-\beta \right)/t^{\left(1/\alpha \right)}}{\overset{\infty }{\int }}{s}^{-1-\epsilon }ds\right]dt-O\left(1\right)\hfill \\ =\frac{{\left(1-\beta \right)}^{-\epsilon }}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{t^{{\lambda }_2+\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dt-O\left(1\right)\ge \frac{1}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{t^{{\lambda }_2+\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda +\epsilon /\alpha }}dt-O\left(1\right).\hfill \end{array}$$

(29)

In view of (28) and (29) and (6), it follows that

$$A+B\ge \frac{1}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{t^{{\lambda }_1+\frac{\epsilon }{q\alpha }-1}+t^{{\lambda }_2+\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda +\epsilon /\alpha }}dt-O\left(1\right)=\frac{1}{\alpha \epsilon }B\left({\lambda }_2+\frac{\epsilon }{p\alpha },{\lambda }_1+\frac{\epsilon }{q\alpha }\right)-O\left(1\right).$$

Then by (26) and (27), (21) is valid.

(ii) For 0 < p < 1, by (22) and (25), we find (notice that q < 0)

$$\begin{array}{cc}\hfill \mathit{Ĩ}& >k{\left\{\underset{1+\gamma }{\overset{\infty }{\int }}\left[1-O\left(\frac{1}{{\left(x-\gamma \right)}^{{\lambda }_2\alpha }}\right)\right]{\left(x-\gamma \right)}^{-1-\epsilon }dx\right\}}^{1/p}{∥\tilde{a}∥}_{\mathsf{\text{q,}}\psi }\hfill \\ =k{\left(\frac{1}{\epsilon }-\underset{1+\gamma }{\overset{\infty }{\int }}O\left(\frac{1}{{\left(x-\gamma \right)}^{{\lambda }_2\alpha +\epsilon +1}}\right)dx\right)}^{1/\mathsf{\text{p}}}{∥\tilde{a}∥}_{q\mathsf{\text{,}}\psi }\hfill \\ >k{\left(\frac{1}{\epsilon }-O\left(1\right)\right)}^{1/p}{\left(\frac{\epsilon +1-\beta }{\epsilon {\left(1-\beta \right)}^{\epsilon +1}}\right)}^{1/q}\hfill \\ =\frac{k}{\epsilon }{\left(1-\epsilon O\left(1\right)\right)}^{1/p}{\left(\frac{\epsilon +1-\beta }{{\left(1-\beta \right)}^{\epsilon +1}}\right)}^{1/q}.\hfill \end{array}$$

(30)

On the other hand, setting $t={\left(\frac{x-\gamma }{n-\beta }\right)}^{\alpha }$in $\mathit{ Ĩ}$, we have

$$\begin{array}{cc}\hfill \mathit{Ĩ}& =\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{-1-\epsilon }\frac{1}{\alpha }\underset{1/{\left(n-\beta \right)}^{\alpha }}{\overset{\infty }{\int }}\frac{t^{{\lambda }_1-\frac{\epsilon }{p\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dt\hfill \\ \le \sum _{n=1}^{\infty }{\left(n-\beta \right)}^{-1-\epsilon }\frac{1}{\alpha }\underset{0}{\overset{\infty }{\int }}\frac{t^{{\lambda }_1-\frac{\epsilon }{p\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dt\hfill \\ <\frac{\epsilon +1-\beta }{\alpha \epsilon {\left(1-\beta \right)}^{\epsilon +1}}B\left({\lambda }_1-\frac{\epsilon }{p\alpha },{\lambda }_2+\frac{\epsilon }{p\alpha }\right).\hfill \end{array}$$

(31)

In virtue of (30) and (31), (23) is valid.

3 Main results

Theorem 1 Suppose that $p>1,\frac{1}{p}+\frac{1}{q}=1$, 0 < α ≤ 1, $0\le \beta \le \frac{1}{2}$, γ ∈ (-∞, +∞), λ₁ > 0, 0 < λ₂α ≤ 1, λ = λ₁+λ₂, $\varphi \left(x\right)={\left(x-y\right)}^{p\left(1-{\lambda }_1\alpha \right)-1}$, $\psi \left(n\right)={\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}$, f(x), a _n ≥ 0, such that f ∈ L _p , _ϕ (γ, ∞),$a={\left\{a_n\right\}}_{n=1}^{\infty }\in l_{q,\psi }$, ||f|| _p , _ϕ > 0, ||a|| _q , _ψ > 0, then we have the following equivalent inequalities:

$$\begin{array}{cc}\hfill I& :=\sum _{n=1}^{\infty }{a}_n\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\hfill \\ =\underset{\gamma }{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }\frac{a_{n}dx}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}<k_{{\lambda }_1}\left(\alpha \right){∥f∥}_{p,\varphi }{∥a∥}_{q,\psi },\hfill \end{array}$$

(32)

$$J={\left\{\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{p{\lambda }_2\alpha -1}{\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)dx}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^p\right\}}^{\frac{1}{p}}<k_{{\lambda }_1}\left(\alpha \right){∥f∥}_{p,\varphi },$$

(33)

$$L={\left\{{\displaystyle \underset{\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{q{\lambda }_1\alpha -1}}{\left[{\displaystyle \sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}}\right]}^{q}dx\right\}}^{\frac{1}{q}}<k_{{\lambda }_1}\left(\alpha \right){\Vert a\Vert }_{q,\psi },$$

(34)

where the constant factor $k_{{\lambda }_1}\left(\alpha \right)=\frac{1}{\alpha }B\left({\lambda }_1,{\lambda }_2\right)$ is the best possible.

Proof. By Lebesgue term-by-term integration theorem [23], we find that there are two expressions of I in (32). By (9), (13) and 0 < ||f|| _p , _ϕ < ∞, we have (33). By Hö lder's inequality, we find

$$\begin{array}{cc}\hfill I& =\sum _{n=1}^{\infty }\left[{\left(n-\beta \right)}^{{\lambda }_2\alpha -\frac{1}{p}}\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)dx}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]\left[{\left(n-\beta \right)}^{\frac{1}{p}-{\lambda }_2\alpha }{a}_n\right]\hfill \\ \le J{\left\{\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}{a}_n^q\right\}}^{1/q}=J{∥a∥}_{q,\psi }.\hfill \end{array}$$

(35)

Hence (32) is valid by (33). On the other hand, setting

$$a_n:={\left(n-\beta \right)}^{p{\lambda }_2\alpha -1}{\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\right]}^{p-1}\left(n\in \text{N}\right),$$

(36)

then we have

$${∥a∥}_{q,\psi }^q=\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}{a}_n^q=J^p=I.$$

(37)

By (9), (13) and 0 < ||f|| _p , _ϕ < ∞, it follows that J < ∞. If J = 0, then (33) is trivially valid. If J > 0, then 0 < ||a|| _q , _ψ = J^p-1 < ∞. Assuming that (32) is valid, we have

$${∥a∥}_{q,\psi }^q=J^p=I<k_{{\lambda }_1}\left(\alpha \right){∥f∥}_{p,\varphi }{∥a∥}_{q,\psi },\mathsf{\text{i}}\mathsf{\text{.e}}\mathsf{\text{.}}J\mathsf{\text{ = }}{∥a∥}_{q,\psi }^{q-1}<k_{{\lambda }_1}\left(\alpha \right){∥f∥}_{p,\varphi }.$$

(38)

Hence (33) is valid, which is equivalent to (32).

By (14) and (9), we obtain (34). By Hö lder's inequality again, we have

$$\begin{array}{cc}\hfill I& =\underset{\gamma }{\overset{\infty }{\int }}\left[{\left(x-\gamma \right)}^{{\lambda }_1\alpha -\frac{1}{q}}\sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]\left[{\left(x-\gamma \right)}^{\frac{1}{q}-{\lambda }_1\alpha }f\left(x\right)\right]dx\hfill \\ \le L{\left\{\underset{\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}{f}^p\left(x\right)dx\right\}}^{1/p}=L{∥f∥}_{p,\varphi }.\hfill \end{array}$$

(39)

Hence (32) is valid by using (34). Assuming that (32) is valid, setting

$$f\left(x\right):={\left(x-\gamma \right)}^{q{\lambda }_1\alpha -1}{\left[\sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+\left(n-\beta \right)\right]}^{\lambda }}\right]}^{q-1}\left(x\in \left(\gamma ,\infty \right)\right),$$

(40)

then we find

$${∥f∥}_{p,\varphi }^p=\underset{\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}{f}^p\left(x\right)dx=L^q=I.$$

(41)

By (14) and (9), it follows that L < ∞. If L = 0, then (34) is trivially valid; if L > 0, i.e. 0 < ||f|| _p , _ϕ < ∞, then by (32), we have

$${\Vert f\Vert }_{p,\varphi }^p=L^q=I<k_{{\lambda }_1}\left(\alpha \right){\Vert f\Vert }_{p,\varphi }{\Vert a\Vert }_{q,\psi },\text{i}.\text{e}.L={\Vert f\Vert }_{p,\varphi }^{p-1}<k_{{\lambda }_1}\left(\alpha \right){\Vert a\Vert }_{q,\psi }.$$

Hence (34) is valid, which is equivalent to (32). It follows that (32), (33), and (34) are equivalent.

If there exists a positive number $k\le k_{{\lambda }_1}\left(\alpha \right)$, such that (32) is still valid as we replace $k_{{\lambda }_1}\left(\alpha \right)$, by k, then in particular, (20) is valid (${\tilde{a}}_n,\tilde{f}\left(x\right)$ are taken as (19)). Then we have (21). For ε → 0⁺ in (21), we have $k\ge \frac{1}{\alpha }B\left({\lambda }_2,{\lambda }_1\right)=k_{{\lambda }_1}\left(\alpha \right).$ Hence, $k=k_{{\lambda }_1}\left(\alpha \right)$ is the best value of (32). We conform that the constant factor $k_{{\lambda }_1}\left(\alpha \right)$ in (33) [(34)] is the best possible, otherwise we can get a contradiction by (35) [(39)] that the constant factor in (32) is not the best possible.

Remark 1 (i) Define a half-discrete Hilbert's operator $T:L_{p,\varphi }\left(\gamma ,\infty \right)\to l_{p,{\psi }^{1-p}}$ as follows: For f ∈ L _p , _ϕ (γ, ∞), we define $Tf\in l_{p,{\psi }^{1-p}}$, satisfying

$$Tf\left(n\right)=\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\left(n\in \text{N}\right).$$

Then by (33), it follows ${∥Tf∥}_{p,{\psi }^{1-p}}\le k_{{\lambda }_1}\left(\alpha \right){∥f∥}_{p,\varphi },$ i.e. T is the bounded operator with $∥T∥\le k_{{\lambda }_1}\left(\alpha \right).$ Since the constant factor $k_{{\lambda }_1}\left(\alpha \right)$ in (33) is the best possible, we have $∥T∥=k_{{\lambda }_1}\left(\alpha \right).$

• (ii) Define a half-discrete Hilbert's operator $\tilde{T}:l_{q,\psi }\to L_{q,{\varphi }^{1-q}}\left(\gamma ,\infty \right)$ in the following way: For a ∈ l _q , _ψ , we define $\tilde{T}a\in L_{q,{\varphi }^{1-q}}$, satisfying

  $$\tilde{T}a\left(x\right)=\sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\left(x\in \left(\gamma ,\infty \right)\right).$$

Then by (34), it follows ${∥\tilde{T}a∥}_{q,{\varphi }^{1-q}}\le k_{{\lambda }_1}\left(\alpha \right){∥a∥}_{q,\psi },$ i.e. $\tilde{T}$ is the bounded operator with $∥\tilde{T}∥\le k_{{\lambda }_1}\left(\alpha \right).$ Since the constant factor $k_{{\lambda }_1}\left(\alpha \right)$ in (34) is the best possible, we have $∥\tilde{T}∥=k_{{\lambda }_1}\left(\alpha \right).$

Theorem 2 Suppose that 0 < p < 1, $\frac{1}{p}+\frac{1}{q}=1$, 0 < α ≤ 1, $0\le \beta \le \frac{1}{2}$, γ ∈ (-∞, + ∞), λ₁ > 0, 0 < λ₂α ≤ 1, λ = λ₁ + λ₂, $\psi \left(n\right)={\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}$, $\tilde{\varphi }\left(x\right)=\left(1-{\theta }_{\lambda }\left(x\right)\right){\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}\left({\theta }_{\lambda }\left(x\right)=\frac{1}{B\left({\lambda }_1,{\lambda }_2\right)}{\int }_0^{{\left(\frac{1-\beta }{x-\gamma }\right)}^{\alpha }}\frac{u^{{\lambda }_2-1}}{{\left(1+u\right)}^{\lambda }}du\in \left(0,1\right)\right)$, f(x), a _n ≥ 0, such that $f\in L_{p,\tilde{\varphi }}\left(\gamma ,\infty \right)$, $a={\left\{a_n\right\}}_{n=1}^{\infty }\in l_{q,\psi }$, ${∥f∥}_{p,\tilde{\varphi }}>0$, ||a|| _q , _ψ > 0, then we have the following equivalent inequalities:

$$\begin{array}{cc}\hfill I& =\sum _{n=1}^{\infty }{a}_n\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\hfill \\ =\underset{\gamma }{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }\frac{a_{n}dx}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}>k_{{\lambda }_1}\left(\alpha \right){∥f∥}_{p,\tilde{\varphi }}{∥a∥}_{q,\psi },\hfill \end{array}$$

(42)

$$\begin{array}{cc}\hfill J& ={\left\{\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{p{\lambda }_2\alpha -1}{\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\right]}^p\right\}}^{1/p}\hfill \\ >k_{{\lambda }_1}\left(\alpha \right){∥f∥}_{p,\tilde{\varphi }},\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill \tilde{L}& :={\left\{\underset{\gamma }{\overset{\infty }{\int }}\frac{{\left(x-\gamma \right)}^{q{\lambda }_1\alpha -1}}{{\left[1-{\theta }_{\lambda }\left(x\right)\right]}^{q-1}}{\left[\sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^{q}dx\right\}}^{1/q}\hfill \\ >k_{{\lambda }_1}\left(\alpha \right){∥a∥}_{q,\psi },\hfill \end{array}$$

(44)

where the constant factor $k_{{\lambda }_1}\left(\alpha \right)=\frac{1}{\alpha }B\left({\lambda }_1,{\lambda }_2\right)$ is the best possible.

Proof. By (9) and the reverse of (13) and $0<{∥f∥}_{p,\tilde{\varphi }}<\infty ,$ we have (43). Using the reverse Hö lder's inequality, we obtain the reverse form of (36) as follows

$$I\ge J{∥a∥}_{q,\psi }.$$

(45)

Then by (43), (42) is valid.

On the other hand, if (42) is valid, setting a _n as (36), then (37) still holds with 0 < p < 1. By (42), it follows that J > 0. If J = ∞, then (43) is trivially valid; if J < ∞, then $0<{∥a∥}_{q,\psi }=J^{p-1}<\infty ,$ and we have

$${\Vert a\Vert }_{q,\psi }^q=J^p=I>k_{{\lambda }_1}\left(\alpha \right){\Vert f\Vert }_{p,\tilde{\varphi }}{\Vert a\Vert }_{q,\psi },\text{i}.\text{e}.J={\Vert a\Vert }_{,\psi }^{q-1}>k_{{\lambda }_1}\left(\alpha \right){\Vert f\Vert }_{,\tilde{\varphi }},$$

Hence (43) is valid, which is equivalent to (42).

By the reverse of (14), in view of $\varpi \left(x\right)>k_{{\lambda }_1}\left(\alpha \right)\left(1-{\theta }_{\lambda }\left(x\right)\right)$ and q < 0, we have

$$\tilde{L}>k_{{\lambda }_1}^{\frac{q-1}{q}}{L}_1\ge k_{{\lambda }_1}^{\frac{q-1}{q}}\left(\alpha \right){\left\{k_{{\lambda }_1}\left(\alpha \right)\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-{\lambda }_2\alpha \right)-1}{a}_n^q\right\}}^{1/q}=k_{{\lambda }_1}\left(\alpha \right){∥a∥}_{q,\psi },$$

then (44) is valid. By the reverse Hö lder's inequality again, we have

$$\begin{array}{cc}\hfill I& =\underset{\gamma }{\overset{\infty }{\int }}\left[\frac{{\left(x-\gamma \right)}^{\gamma 1\alpha -\frac{1}{q}}}{{\left(1-{\theta }_{\lambda }\left(x\right)\right)}^{\frac{1}{q}}}\sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]\hfill \\ \times \left[{\left(1-{\theta }_{\lambda }\left(x\right)\right)}^{\frac{1}{p}}{\left(x-\gamma \right)}^{\frac{1}{q}-{\lambda }_1\alpha }f\left(x\right)\right]dx\hfill \\ \ge \tilde{L}{∥f∥}_{p,\tilde{\varphi }}.\hfill \end{array}$$

(46)

Hence (42) is valid by (44). On the other hand, if (42) is valid, setting

$$f\left(x\right)=\frac{{\left(x-\gamma \right)}^{q{\lambda }_1\alpha -1}}{{\left[1-{\theta }_{\lambda }\left(x\right)\right]}^{q-1}}{\left[\sum _{n=1}^{\infty }\frac{a_n}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^{q-1}\left(x\in \left(\gamma ,\infty \right)\right),$$

then ${∥f∥}_{p,\tilde{\varphi }}^p={\int }_{\gamma }^{\infty }\left[1-{\theta }_{\lambda }\left(x\right)\right]{\left(x-\gamma \right)}^{p\left(1-{\lambda }_1\alpha \right)-1}{f}^p\left(x\right)dx={\tilde{L}}^q=I.$ By the reverse of (14), it follows that $\tilde{L}>0$. If $\tilde{L}=\infty$, then (44) is trivially valid; if $0<\tilde{L}<\infty$, then by (42), we have

$${∥f∥}_{p,\tilde{\varphi }}^p={\tilde{L}}^q=I>k_{{\lambda }_1}\left(\alpha \right){∥f∥}_{p,\tilde{\varphi }}{∥a∥}_{q,\psi },\text{i}.\text{e}.\tilde{L},={∥f∥}_{p,\tilde{\varphi }}^{p-1}>k_{{\lambda }_1}\left(\alpha \right){∥a∥}_{q,\psi }.$$

Hence (44) is valid, which is equivalent to (42). It follows that (42), (43), and (44) are equivalent.

If there exists a positive number $k\ge k_{{\lambda }_1}\left(\alpha \right),$ such that (42) is still valid as we replace $k_{{\lambda }_1}\left(\alpha \right)$ by k, then in particular, (22) is valid. Hence we have (23). For ε → 0⁺ in (23), we obtain $k\le \frac{1}{\alpha }B\left({\lambda }_1,{\lambda }_2\right)=k_{{\lambda }_1}\left(\alpha \right).$ Hence $k=k_{{\lambda }_1}\left(\alpha \right)$ is the best value of (42). We conform that the constant factor $k_{{\lambda }_1}\left(\alpha \right)$ in (43) [(44)] is the best possible, otherwise we can get a contradiction by (45) [(46)] that the constant factor in (42) is not the best possible.

In the same way, for p < 0, we also have the following result:

Theorem 3 If the assumption of p > 1 in Theorem 1 is replaced by p < 0, then the reverses of (32), (33), and (34) are valid and equivalent. Moreover, the same constant factor is the best possible.

Remark 2 (i) For β = γ = 0, ${\lambda }_1=\frac{1}{q\alpha }$, ${\lambda }_2=\frac{1}{p\alpha }$ in (32), it follows

$$\sum _{n=1}^{\infty }{a}_n\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left(x^{\alpha }+n^{\alpha }\right)}^{1/\alpha }}dx<\frac{1}{\alpha }B\left(\frac{1}{q^{\alpha }},\frac{1}{p^{\alpha }}\right){\left\{\underset{0}{\overset{\infty }{\int }}{f}^p\left(x\right)dx\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{a}_n^q\right\}}^{1/q}.$$

(47)

In particular, for α = 1, p = q = 2, (47) reduces to (4). (ii) For λ = α = 1, ${\lambda }_1=\frac{1}{q}$, ${\lambda }_2=\frac{1}{p}$ in (32), it follows

$$\sum _{n=1}^{\infty }{a}_n\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{x+n-\gamma -\beta }dx<\frac{\pi }{sin\left(\pi /p\right)}{\left\{\underset{\gamma }{\overset{\infty }{\int }}{f}^p\left(x\right)dx\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{a}_n^q\right\}}^{1/q}.$$

(48)

In particular, for $\gamma =-\frac{1}{2}$, $\beta =\frac{1}{2}$, p = q = 2 in (48), we obtain (5). Hence, inequality (32) is the best extension of (4) and (5) with parameters.