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On a more accurate half-discrete Hilbert's inequality

Journal of Inequalities and Applications20122012:106

https://doi.org/10.1186/1029-242X-2012-106

• Accepted: 8 May 2012
• Published:

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.

Keywords

• weight coefficient
• parameter
• equivalent form
• reverse
• Hilbert's inequality

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. ):
$\sum _{n=1}^{\infty }{\sum _{m=1}^{\infty }\frac{{a}_{m}{b}_{n}}{m+n}<\pi \left(\sum _{m=1}^{\infty }{a}_{m}^{2}\sum _{n=1}^{\infty }{b}_{n}^{2}\right)}^{1/2},$
(1)
where the constant factor π is the best possible. The integral analogue of inequality (1) is given as follows (cf. ): 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.  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 . There are lots of improvements, generalizations, and applications of inequalities (1-3), for more details, refer to literatures .

We find a few results on the half-discrete Hilbert-type inequalities with the non-homogeneous kernel, which were published early ([, Th. 351], ). Recently, Yang  gave some half-discrete Hilbert-type inequalities. A half-discrete Hilbert's inequality with the homogeneous kernel was derived as follows : 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. ) and the weight coefficients as follows:
$B\left(u,v\right):=\underset{0}{\overset{\infty }{\int }}\frac{{t}^{u-1}dt}{{\left(1+t\right)}^{u+v}}=\underset{0}{\overset{1}{\int }}\frac{{t}^{u-1}+{t}^{v-1}}{{\left(1+t\right)}^{u+v}}dt\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left(u,v>0\right),$
(6)
$\omega \left(n\right):={\left(n-\beta \right)}^{{\lambda }_{2}\alpha }\underset{\gamma }{\overset{\infty }{\int }}\frac{{\left(x-\gamma \right)}^{{\lambda }_{1}\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\phantom{\rule{0.3em}{0ex}}\left(n\in \mathbf{N}\right),$
(7)
$\varpi \left(x\right):={\left(x-\gamma \right)}^{{\lambda }_{1}\alpha }\sum _{n=1}^{\infty }\frac{{\left(n-\beta \right)}^{{\lambda }_{2}\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left(x\in \left(\gamma ,\infty \right)\right).$
(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, ${\theta }_{\lambda }\left(x\right):=\frac{1}{B\left({\lambda }_{1},{\lambda }_{2}\right)}{\int }_{0}^{{\left(\frac{1-\beta }{x-\gamma }\right)}^{\alpha }}\frac{{u}^{{\lambda }_{2}-1}}{{\left(1+u\right)}^{\lambda }}du>0\phantom{\rule{0.3em}{0ex}}and\phantom{\rule{0.3em}{0ex}}{\theta }_{\lambda }\left(x\right)=O\left(\frac{1}{{\left(x-\gamma \right)}^{{\lambda }_{2}\alpha }}\right)\left(x\in \left(\gamma ,\infty \right)\right).$

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. ):
$f\left(n\right)<\underset{n-\frac{1}{2}}{\overset{n+\frac{1}{2}}{\int }}f\left(t\right)dt\phantom{\rule{0.3em}{0ex}}\left(n\in \mathbf{N}\right),$
(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. ), we find
$\begin{array}{c}\phantom{\rule{1em}{0ex}}{\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 \\ \phantom{\rule{1em}{0ex}}×\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 \\ \phantom{\rule{1em}{0ex}}×{\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)
$\begin{array}{c}\phantom{\rule{1em}{0ex}}{J}^{p}\le {k}_{{\lambda }_{1}}^{p-1}\left(\alpha \right)\sum _{n=1}^{\infty }\underset{\gamma }{\overset{\infty }{\int }}\frac{{f}^{p}\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{{\left(x-\gamma \right)}^{\left(1-{\lambda }_{1}\alpha \right)\left(p-1\right)}}{{\left(n-\beta \right)}^{1-{\lambda }_{2}\alpha }}dx\hfill \\ ={k}_{{\lambda }_{1}}^{p-1}\left(\alpha \right)\underset{\gamma }{\overset{\infty }{\int }}\sum _{n=1}^{\infty }\frac{{\left(n-\beta \right)}^{{\lambda }_{2}\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}{\left(x-\gamma \right)}^{{\lambda }_{1}\alpha +p\left(1-{\lambda }_{1}\alpha \right)-1}{f}^{p}\left(x\right)dx\hfill \\ ={k}_{{\lambda }_{1}}^{p-1}\left(\alpha \right)\underset{\gamma }{\overset{\infty }{\int }}\varpi \left(x\right){\left(x-\gamma \right)}^{p\left(1-{\lambda }_{1}\alpha \right)-1}{f}^{p}\left(x\right)dx.\hfill \end{array}$
(16)
Hence (13) is valid. Using Hölder's inequality again, we have
(17)
$\begin{array}{c}\phantom{\rule{1em}{0ex}}{L}_{1}^{q}\le \underset{\gamma }{\overset{\infty }{\int }}\sum _{n=1}^{\infty }\frac{1}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\frac{{\left(n-\beta \right)}^{\left(1-{\lambda }_{2}\alpha \right)\left(q-1\right)}}{{\left(x-\gamma \right)}^{1-{\lambda }_{1}\alpha }}{a}_{n}^{q}dx\hfill \\ =\sum _{n=1}^{\infty }\left[{\left(n-\beta \right)}^{{\lambda }_{2}\alpha }\underset{\gamma }{\overset{\infty }{\int }}\frac{{\left(x-\gamma \right)}^{{\lambda }_{1}\alpha -1}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx\right]{\left(n-\beta \right)}^{q\left(1-{\lambda }_{2}\alpha \right)-1}{a}_{n}^{q}\hfill \\ =\sum _{n=1}^{\infty }\omega \left(n\right){\left(n-\beta \right)}^{q\left(1-{\lambda }_{2}\alpha \right)-1}{a}_{n}^{q}={k}_{{\lambda }_{1}}\left(\alpha \right)\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-\lambda 2\alpha \right)-1}{a}_{n}^{q}.\hfill \end{array}$
(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 $\varphi \left(x\right):={\left(x-\gamma \right)}^{p\left(1-{\lambda }_{1}\alpha \right)-1}$, $\stackrel{̃}{\varphi }\left(x\right):=\left(1-{\theta }_{\lambda }\left(x\right)\right)\varphi \left(x\right)$, $\psi \left(n\right):={\left(n-\beta \right)}^{q\left(1-{\lambda }_{2}\alpha \right)-1}$,
$\begin{array}{c}\hfill {L}_{p,\varphi }\left(\gamma ,\infty \right):=\left\{f;{\parallel f\parallel }_{p,\varphi }={\left\{\underset{\gamma }{\overset{\infty }{\int }}\varphi \left(x\right)|f\left(x\right){|}^{p}dx\right\}}^{1/p}<\infty \right\},\\ \hfill \phantom{\rule{1em}{0ex}}{l}_{q,\psi }:=\left\{a=\left\{{a}_{n}\right\};\parallel a\parallel {,}_{\psi }={\left\{\sum _{n=1}^{\infty }\psi \left(n\right)|{a}_{n}{|}^{q}\right\}}^{1/q}<\infty \right\}\end{array}$
(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, λ1}, setting $\stackrel{˜}{a}={\left\{{\stackrel{˜}{a}}_{n}\right\}}_{n=1}^{\infty }$ and $\stackrel{̃}{f}\left(x\right)$as follows
${\stackrel{˜}{a}}_{n}={\left(n-\beta \right)}^{{\lambda }_{2}\alpha -\frac{\epsilon }{q}-1};\phantom{\rule{2.77695pt}{0ex}}\stackrel{̃}{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 }{\stackrel{˜}{a}}_{n}\underset{\gamma }{\overset{\infty }{\int }}\frac{\stackrel{̃}{f}\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx
(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
$\mathit{Ĩ}-\sum _{n=1}^{\infty }{\stackrel{˜}{a}}_{n}\underset{\gamma }{\overset{\infty }{\int }}\frac{\stackrel{̃}{f}\left(x\right)}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}dx>k{∥\stackrel{̃}{f}∥}_{p,\stackrel{̃}{\varphi }}{∥\stackrel{˜}{a}∥}_{q,\psi },$
(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
$\begin{array}{cc}\hfill {∥\stackrel{̃}{f}∥}_{p,\varphi }& ={\left\{\underset{\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{p\left(1-{\lambda }_{1}\alpha \right)-1}{\stackrel{̃}{f}}^{p}\left(x\right)dx\right\}}^{1/p}\hfill \\ ={\left\{\underset{1+\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{-1-\epsilon }dx\right\}}^{1/p}={\left(\frac{1}{\epsilon }\right)}^{1/p},\hfill \end{array}$
(24)
$\begin{array}{cc}\hfill {∥\stackrel{˜}{a}∥}_{q,\psi }^{q}& =\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-{\lambda }_{2}\alpha \right)-1}{\stackrel{˜}{a}}_{n}^{q}=\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{-1-\epsilon }\hfill \\ <{\left(1-\beta \right)}^{-1-\epsilon }+\underset{1}{\overset{\infty }{\int }}{\left(x-\beta \right)}^{-1-\epsilon }dx=\frac{\epsilon +1-\beta }{\epsilon {\left(1-\beta \right)}^{\epsilon +1}}.\hfill \end{array}$
(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}&
(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
$\begin{array}{cc}\hfill A:\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}& =\frac{1}{\alpha }\underset{1}{\overset{\infty }{\int }}{s}^{-1-\epsilon }\underset{{\left(\frac{1-\beta }{s}\right)}^{\alpha }}{\overset{1}{\int }}\frac{{t}^{{\lambda }_{2}-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds;\hfill \\ \hfill B:\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}& =\frac{1}{\alpha }\underset{1}{\overset{\infty }{\int }}{s}^{-1-\epsilon }\underset{1}{\overset{\infty }{\int }}\frac{{t}^{{\lambda }_{2}-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds=\frac{1}{\alpha \epsilon }\underset{1}{\overset{\infty }{\int }}\frac{{t}^{{\lambda }_{2}-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dt\hfill \\ \stackrel{u=\frac{1}{t}}{=}\frac{1}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{{u}^{{\lambda }_{1}+\frac{\epsilon }{q\alpha }-1}}{{\left(1+u\right)}^{\lambda }}du\ge \frac{1}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{{u}^{{\lambda }_{1}+\frac{\epsilon }{q\alpha }-1}}{{\left(1+u\right)}^{\lambda +\epsilon /\alpha }}du.\hfill \end{array}$
(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
$\begin{array}{cc}\hfill A& =\frac{1}{\alpha }\left[\underset{1-\beta }{\overset{\infty }{\int }}{s}^{-1-\epsilon }\underset{{\left(\frac{1-\beta }{s}\right)}^{\alpha }}{\overset{1}{\int }}\frac{{t}^{{\lambda }_{2}-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds-\underset{1-\beta }{\overset{1}{\int }}{s}^{-1-\epsilon }\underset{{\left(\frac{1-\beta }{s}\right)}^{\alpha }}{\overset{1}{\int }}\frac{{t}^{{\lambda }_{2}-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dtds\right]\hfill \\ =\frac{1}{\alpha }\underset{0}{\overset{1}{\int }}\frac{{t}^{{\lambda }_{2}-\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}\left[\underset{\left(1-\beta \right)/{t}^{\left(1/\alpha \right)}}{\overset{\infty }{\int }}{s}^{-1-\epsilon }ds\right]\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dt-O\left(1\right)\hfill \\ =\frac{{\left(1-\beta \right)}^{-\epsilon }}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{{t}^{{\lambda }_{2}+\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda }}dt-O\left(1\right)\ge \frac{1}{\alpha \epsilon }\underset{0}{\overset{1}{\int }}\frac{{t}^{{\lambda }_{2}+\frac{\epsilon }{q\alpha }-1}}{{\left(1+t\right)}^{\lambda +\epsilon /\alpha }}dt-O\left(1\right).\hfill \end{array}$
(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)
$\begin{array}{cc}\hfill \mathit{Ĩ}& >k{\left\{\underset{1+\gamma }{\overset{\infty }{\int }}\left[1-O\left(\frac{1}{{\left(x-\gamma \right)}^{{\lambda }_{2}\alpha }}\right)\right]\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{0.3em}{0ex}}{\left(x-\gamma \right)}^{-1-\epsilon }dx\right\}}^{1/p}{∥\stackrel{˜}{a}∥}_{\mathsf{\text{q,}}\psi }\hfill \\ =k\phantom{\rule{2.77695pt}{0ex}}{\left(\frac{1}{\epsilon }-\underset{1+\gamma }{\overset{\infty }{\int }}O\left(\frac{1}{{\left(x-\gamma \right)}^{{\lambda }_{2}\alpha +\epsilon +1}}\right)dx\right)}^{1/\mathsf{\text{p}}}{∥\stackrel{˜}{a}∥}_{q\mathsf{\text{,}}\psi }\hfill \\ >k\phantom{\rule{2.77695pt}{0ex}}{\left(\frac{1}{\epsilon }-O\left(1\right)\right)}^{1/p}{\left(\frac{\epsilon +1-\beta }{\epsilon {\left(1-\beta \right)}^{\epsilon +1}}\right)}^{1/q}\hfill \\ =\frac{k}{\epsilon }{\left(1-\epsilon O\left(1\right)\right)}^{1/p}{\left(\frac{\epsilon +1-\beta }{{\left(1-\beta \right)}^{\epsilon +1}}\right)}^{1/q}.\hfill \end{array}$
(30)
On the other hand, setting $t={\left(\frac{x-\gamma }{n-\beta }\right)}^{\alpha }$in , 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 = λ12, $\varphi \left(x\right)={\left(x-y\right)}^{p\left(1-{\lambda }_{1}\alpha \right)-1}$, $\psi \left(n\right)={\left(n-\beta \right)}^{q\left(1-{\lambda }_{2}\alpha \right)-1}$, 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:
$\begin{array}{cc}\hfill I\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}& :\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\sum _{n=1}^{\infty }{a}_{n}\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}dx\hfill \\ =\underset{\gamma }{\overset{\infty }{\int }}f\left(x\right)\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }\frac{{a}_{n}dx}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\varphi }\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi },\hfill \end{array}$
(32)
$J\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{\left\{\sum _{n=1}^{\infty }{\left(n\phantom{\rule{2.77695pt}{0ex}}-\beta \right)}^{p{\lambda }_{2}\alpha -1}\phantom{\rule{0.3em}{0ex}}{\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)dx}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^{p}\right\}}^{\frac{1}{p}}\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\varphi },$
(33)
$L={\left\{\underset{\gamma }{\overset{\infty }{\int }}{\left(x-\gamma \right)}^{q{\lambda }_{1}\alpha -1}{\left[\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left[{\left(x-\gamma \right)}^{\alpha }+{\left(n-\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^{q}dx\right\}}^{\frac{1}{q}}<{k}_{{\lambda }_{1}}\left(\alpha \right){‖a‖}_{q,\psi },$
(34)

where the constant factor ${k}_{{\lambda }_{1}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\frac{1}{\alpha }B\left({\lambda }_{1},\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{\lambda }_{2}\right)$ is the best possible.

Proof. By Lebesgue term-by-term integration theorem , 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
$\begin{array}{cc}\hfill I\phantom{\rule{0.3em}{0ex}}& =\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }\left[{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{{\lambda }_{2}\alpha -\frac{1}{p}}\phantom{\rule{0.3em}{0ex}}\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)dx}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\right]\phantom{\rule{0.3em}{0ex}}\left[{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\frac{1}{p}-{\lambda }_{2}\alpha }\phantom{\rule{0.3em}{0ex}}{a}_{n}\right]\phantom{\rule{0.3em}{0ex}}\hfill \\ \le \phantom{\rule{0.3em}{0ex}}J\phantom{\rule{0.3em}{0ex}}{\left\{\sum _{n=1}^{\infty }{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{q\left(1\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{\lambda }_{2}\alpha \right)-1}\phantom{\rule{0.3em}{0ex}}{a}_{n}^{q}\right\}}^{1/q}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}J{∥a∥}_{q,\psi }.\hfill \end{array}$
(35)
Hence (32) is valid by (33). On the other hand, setting
${a}_{n}\phantom{\rule{0.3em}{0ex}}:\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{p{\lambda }_{2}\alpha -1}\phantom{\rule{0.3em}{0ex}}{\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\phantom{\rule{0.3em}{0ex}}dx\right]}^{p-1}\phantom{\rule{0.3em}{0ex}}\left(n\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\text{N}\right),$
(36)
then we have
${∥a∥}_{q,\psi }^{q}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }{\left(n\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{q\left(1-{\lambda }_{2}\alpha \right)-1}\phantom{\rule{0.3em}{0ex}}{a}_{n}^{q}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{J}^{p}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}I.$
(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
(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
$\begin{array}{cc}\hfill I\phantom{\rule{0.3em}{0ex}}& =\phantom{\rule{0.3em}{0ex}}\underset{\gamma }{\overset{\infty }{\int }}\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{{\lambda }_{1}\alpha -\frac{1}{q}}\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\right]\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\frac{1}{q}-{\lambda }_{1}\alpha }\phantom{\rule{0.3em}{0ex}}f\left(x\right)\right]\phantom{\rule{0.3em}{0ex}}dx\hfill \\ \le \phantom{\rule{0.3em}{0ex}}L\phantom{\rule{0.3em}{0ex}}{\left\{\underset{\gamma }{\overset{\infty }{\int }}{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{p\left(1-{\lambda }_{1}\alpha \right)-1}\phantom{\rule{0.3em}{0ex}}{f}^{p}\left(x\right)dx\right\}}^{1/p}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}L\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\varphi }.\hfill \end{array}$
(39)
Hence (32) is valid by using (34). Assuming that (32) is valid, setting
$f\left(x\right)\phantom{\rule{0.3em}{0ex}}:={\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{2.77695pt}{0ex}}\gamma \right)}^{q{\lambda }_{1}\alpha -1}{\left[\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)\right]}^{\lambda }}\right]}^{q-1}\phantom{\rule{0.3em}{0ex}}\left(x\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}\left(\gamma ,\phantom{\rule{0.3em}{0ex}}\infty \right)\right),$
(40)
then we find
${∥f∥}_{p,\varphi }^{p}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\underset{\gamma }{\overset{\infty }{\int }}{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{p\left(1-{\lambda }_{1}\alpha \right)-1}{f}^{p}\left(x\right)dx\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{L}^{q}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}I.$
(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
${‖f‖}_{p,\varphi }^{p}={L}^{q}=I<{k}_{{\lambda }_{1}}\left(\alpha \right){‖f‖}_{p,\varphi }{‖a‖}_{q,\psi },\text{i}.\text{e}.L={‖f‖}_{p,\varphi }^{p-1}<{k}_{{\lambda }_{1}}\left(\alpha \right){‖a‖}_{q,\psi }.$

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

If there exists a positive number $k\phantom{\rule{0.3em}{0ex}}\le {k}_{{\lambda }_{1}}\left(\alpha \right)$, such that (32) is still valid as we replace ${k}_{{\lambda }_{1}}\left(\alpha \right)$, by k, then in particular, (20) is valid (${\stackrel{˜}{a}}_{n},\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\stackrel{̃}{f}\left(x\right)$ are taken as (19)). Then we have (21). For ε → 0+ in (21), we have $k\phantom{\rule{0.3em}{0ex}}\ge \phantom{\rule{0.3em}{0ex}}\frac{1}{\alpha }B\left({\lambda }_{2},\phantom{\rule{0.3em}{0ex}}{\lambda }_{1}\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right).$ Hence, $k\phantom{\rule{0.3em}{0ex}}={k}_{{\lambda }_{1}}\left(\alpha \right)$ 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
$T\phantom{\rule{0.3em}{0ex}}f\left(n\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left(n\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}\text{N}\right).$
Then by (33), it follows ${∥T\phantom{\rule{0.3em}{0ex}}f∥}_{p,{\psi }^{1-p}}\phantom{\rule{0.3em}{0ex}}\le \phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\varphi }\phantom{\rule{0.3em}{0ex}},\phantom{\rule{0.3em}{0ex}}$ i.e. T is the bounded operator with $∥T∥\phantom{\rule{0.3em}{0ex}}\le \phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right).$ Since the constant factor ${k}_{{\lambda }_{1}}\left(\alpha \right)$ in (33) is the best possible, we have $∥T∥\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right).$
1. (ii)
Define a half-discrete Hilbert's operator $\stackrel{̃}{T}:{l}_{q,\psi }\phantom{\rule{0.3em}{0ex}}\to \phantom{\rule{0.3em}{0ex}}{L}_{q,{\varphi }^{1-q}}\left(\gamma ,\phantom{\rule{0.3em}{0ex}}\infty \right)$ in the following way: For a l q , ψ , we define $\stackrel{̃}{T}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}a\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{L}_{q,{\varphi }^{1-q}}$, satisfying
$\stackrel{̃}{T}a\left(x\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left(x\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}\left(\gamma ,\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\infty \right)\right).$

Then by (34), it follows ${∥\stackrel{̃}{T}a∥}_{q,{\varphi }^{1-q}}\phantom{\rule{0.3em}{0ex}}\le \phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi },$ i.e. $\stackrel{̃}{T}$ is the bounded operator with $∥\stackrel{̃}{T}∥\phantom{\rule{0.3em}{0ex}}\le \phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right).$ Since the constant factor ${k}_{{\lambda }_{1}}\left(\alpha \right)$ in (34) is the best possible, we have $∥\stackrel{̃}{T}∥\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right).$

Theorem 2 Suppose that 0 < p < 1, $\frac{1}{p}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\frac{1}{q}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}1$, 0 < α ≤ 1, $0\le \beta \le \frac{1}{2}$, γ (-∞, + ∞), λ1 > 0, 0 < λ2α ≤ 1, λ = λ1 + λ2, $\psi \left(n\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{q\left(1-{\lambda }_{2}\alpha \right)-1}$, $\stackrel{̃}{\varphi }\left(x\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\left(1-{\theta }_{\lambda }\left(x\right)\right)\phantom{\rule{0.3em}{0ex}}{\left(x-\gamma \right)}^{p\left(1-{\lambda }_{1}\alpha \right)-1}\left({\theta }_{\lambda }\left(x\right)=\frac{1}{B\left({\lambda }_{1},\phantom{\rule{0.3em}{0ex}}{\lambda }_{2}\right)}\phantom{\rule{0.3em}{0ex}}{\int }_{0}^{{\left(\frac{1-\beta }{x-\gamma }\right)}^{\alpha }}\frac{{u}^{{\lambda }_{2}-1}}{{\left(1+u\right)}^{\lambda }}du\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}\left(0,1\right)\right)$, f(x), a n ≥ 0, such that $f\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}{L}_{p,\stackrel{̃}{\varphi }}\left(\gamma ,\phantom{\rule{0.3em}{0ex}}\infty \right)$, $a\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{\left\{{a}_{n}\right\}}_{n=1}^{\infty }\phantom{\rule{0.3em}{0ex}}\in \phantom{\rule{0.3em}{0ex}}{l}_{q,\psi }$, ${∥f∥}_{p,\stackrel{̃}{\varphi }}\phantom{\rule{0.3em}{0ex}}>\phantom{\rule{0.3em}{0ex}}0$, ||a|| q , ψ > 0, then we have the following equivalent inequalities:
$\begin{array}{cc}\hfill I\phantom{\rule{0.3em}{0ex}}& =\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }{a}_{n}\phantom{\rule{0.3em}{0ex}}\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}dx\hfill \\ =\phantom{\rule{0.3em}{0ex}}\underset{\gamma }{\overset{\infty }{\int }}f\left(x\right)\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }\frac{{a}_{n}dx}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\phantom{\rule{0.3em}{0ex}}>\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\stackrel{̃}{\varphi }}\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi },\hfill \end{array}$
(42)
$\begin{array}{cc}\hfill J\phantom{\rule{0.3em}{0ex}}& =\phantom{\rule{0.3em}{0ex}}{\left\{\sum _{n=1}^{\infty }{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{p{\lambda }_{2}\alpha -1}\phantom{\rule{0.3em}{0ex}}{\left[\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\phantom{\rule{0.3em}{0ex}}dx\right]}^{p}\right\}}^{1/p}\hfill \\ >\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\stackrel{̃}{\varphi }},\hfill \end{array}$
(43)
$\begin{array}{cc}\hfill \stackrel{̃}{L}\phantom{\rule{0.3em}{0ex}}& :\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{\left\{\underset{\gamma }{\overset{\infty }{\int }}\frac{{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{q{\lambda }_{1}\alpha -1}}{{\left[1\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{\theta }_{\lambda }\phantom{\rule{0.3em}{0ex}}\left(x\right)\right]}^{q-1}}\phantom{\rule{0.3em}{0ex}}{\left[\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left[{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^{q}\phantom{\rule{0.3em}{0ex}}dx\right\}}^{1/q}\hfill \\ >\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi },\hfill \end{array}$
(44)

where the constant factor ${k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\frac{1}{\alpha }B\left({\lambda }_{1},\phantom{\rule{0.3em}{0ex}}{\lambda }_{2}\right)$ is the best possible.

Proof. By (9) and the reverse of (13) and $0\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\stackrel{̃}{\varphi }}\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}\infty ,$ we have (43). Using the reverse Hö lder's inequality, we obtain the reverse form of (36) as follows
$I\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\ge \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}J\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi }.$
(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 $0\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi }\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{J}^{p-1}\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}\infty ,$ and we have
${‖a‖}_{q,\psi }^{q}={J}^{p}=I>{k}_{{\lambda }_{1}}\left(\alpha \right){‖f‖}_{p,\stackrel{˜}{\varphi }}{‖a‖}_{q,\psi },\text{i}.\text{e}.J={‖a‖}_{,\psi }^{q-1}>{k}_{{\lambda }_{1}}\left(\alpha \right){‖f‖}_{,\stackrel{˜}{\varphi }},$

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

By the reverse of (14), in view of $\varpi \left(x\right)\phantom{\rule{0.3em}{0ex}}>\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}\left(1\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{\theta }_{\lambda }\phantom{\rule{0.3em}{0ex}}\left(x\right)\right)$ and q < 0, we have
$\stackrel{̃}{L}\phantom{\rule{0.3em}{0ex}}>\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}^{\frac{q-1}{q}}\phantom{\rule{0.3em}{0ex}}{L}_{1}\phantom{\rule{0.3em}{0ex}}\ge \phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}^{\frac{q-1}{q}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{\left\{{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{q\left(1\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{\lambda }_{2}\alpha \right)-1}\phantom{\rule{0.3em}{0ex}}{a}_{n}^{q}\right\}}^{1/q}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi },$
then (44) is valid. By the reverse Hö lder's inequality again, we have
$\begin{array}{cc}\hfill I& =\underset{\gamma }{\overset{\infty }{\int }}\left[\frac{{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{\gamma 1\alpha -\frac{1}{q}}}{{\left(1\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{\theta }_{\lambda }\left(x\right)\right)}^{\frac{1}{q}}}\phantom{\rule{0.3em}{0ex}}\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left[{\left(x-\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\right]\hfill \\ \phantom{\rule{1em}{0ex}}×\phantom{\rule{0.3em}{0ex}}\left[{\left(1\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{\theta }_{\lambda }\left(x\right)\right)}^{\frac{1}{p}}\phantom{\rule{0.3em}{0ex}}{\left(x\phantom{\rule{0.3em}{0ex}}-\gamma \right)}^{\frac{1}{q}-{\lambda }_{1}\alpha }\phantom{\rule{0.3em}{0ex}}f\left(x\right)\right]\phantom{\rule{0.3em}{0ex}}dx\hfill \\ \ge \phantom{\rule{0.3em}{0ex}}\stackrel{̃}{L}\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\stackrel{̃}{\varphi }}.\hfill \end{array}$
(46)
Hence (42) is valid by (44). On the other hand, if (42) is valid, setting
$f\left(x\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\frac{{\left(x\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \right)}^{q{\lambda }_{1}\alpha -1}}{{\left[1\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}{\theta }_{\lambda }\left(x\right)\right]}^{q-1}}\phantom{\rule{0.3em}{0ex}}{\left[\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left[{\left(x-\gamma \right)}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{\left(n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta \right)}^{\alpha }\right]}^{\lambda }}\right]}^{q-1}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\left(x\in \left(\gamma ,\infty \right)\right),$
then ${∥f∥}_{p,\stackrel{̃}{\varphi }}^{p}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{\int }_{\gamma }^{\infty }\left[1-{\theta }_{\lambda }\left(x\right)\right]\phantom{\rule{0.3em}{0ex}}{\left(x-\gamma \right)}^{p\left(1-{\lambda }_{1}\alpha \right)-1}\phantom{\rule{0.3em}{0ex}}{f}^{p}\left(x\right)dx={\stackrel{̃}{L}}^{q}=I.$ By the reverse of (14), it follows that $\stackrel{̃}{L}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}>\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}0$. If $\stackrel{̃}{L}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\infty$, then (44) is trivially valid; if $0\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\stackrel{̃}{L}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\infty$, then by (42), we have
${∥f∥}_{p,\stackrel{̃}{\varphi }}^{p}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{\stackrel{̃}{L}}^{q}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}I\phantom{\rule{0.3em}{0ex}}>\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\stackrel{̃}{\varphi }}\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi },\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{i}.\text{e}.\phantom{\rule{1em}{0ex}}\stackrel{̃}{L},=\phantom{\rule{0.3em}{0ex}}{∥f∥}_{p,\stackrel{̃}{\varphi }}^{p-1}\phantom{\rule{0.3em}{0ex}}>\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right)\phantom{\rule{0.3em}{0ex}}{∥a∥}_{q,\psi }.$

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

If there exists a positive number $k\phantom{\rule{0.3em}{0ex}}\ge \phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right),$ 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 $k\phantom{\rule{0.3em}{0ex}}\le \phantom{\rule{0.3em}{0ex}}\frac{1}{\alpha }B\left({\lambda }_{1},\phantom{\rule{0.3em}{0ex}}{\lambda }_{2}\right)\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\phantom{\rule{0.3em}{0ex}}\left(\alpha \right).$ Hence $k\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}{k}_{{\lambda }_{1}}\left(\alpha \right)$ 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
$\sum _{n=1}^{\infty }{a}_{n}\phantom{\rule{0.3em}{0ex}}\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left({x}^{\alpha }\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}{n}^{\alpha }\right)}^{1/\alpha }}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}\frac{1}{\alpha }B\left(\frac{1}{{q}^{\alpha }},\phantom{\rule{0.3em}{0ex}}\frac{1}{{p}^{\alpha }}\right)\phantom{\rule{0.3em}{0ex}}{\left\{\underset{0}{\overset{\infty }{\int }}{f}^{p}\left(x\right)dx\right\}}^{1/p}\phantom{\rule{0.3em}{0ex}}{\left\{\sum _{n=1}^{\infty }{a}_{n}^{q}\right\}}^{1/q}.$
(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
$\sum _{n=1}^{\infty }{a}_{n}\phantom{\rule{0.3em}{0ex}}\underset{\gamma }{\overset{\infty }{\int }}\frac{f\left(x\right)}{x\phantom{\rule{0.3em}{0ex}}+\phantom{\rule{0.3em}{0ex}}n\phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\gamma \phantom{\rule{0.3em}{0ex}}-\phantom{\rule{0.3em}{0ex}}\beta }dx\phantom{\rule{0.3em}{0ex}}<\phantom{\rule{0.3em}{0ex}}\frac{\pi }{sin\left(\pi /p\right)}\phantom{\rule{0.3em}{0ex}}{\left\{\underset{\gamma }{\overset{\infty }{\int }}{f}^{p}\left(x\right)dx\right\}}^{1/p}\phantom{\rule{0.3em}{0ex}}{\left\{\sum _{n=1}^{\infty }{a}_{n}^{q}\right\}}^{1/q}.$
(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.

Declarations

Acknowledgements

This study was supported by the Emphases Natural Science Foundation of Guangdong Institution, Higher Learning, College and University (No. 05Z026), and Guangdong Natural Science Foundation (No. 7004344).

Authors’ Affiliations

(1)
Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong, 510303, People's Republic of China

References 