A more accurate half-discrete reverse Hilbert-type inequality with a non-homogeneous kernel

Abstract

By means of weight functions and the improved Euler-Maclaurin summation formula, a more accurate half-discrete reverse Hilbert-type inequality with the kernel $\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}$ and a best constant factor is given. Some equivalent forms, the dual forms as well as some related homogeneous cases, are also considered.

MSC:26D15.

1 Introduction

Assuming that $f,g\in L^2(R_{+})$, $\parallel f\parallel ={\{{\int }_0^{\mathrm{\infty }}{f}^2(x)dx\}}^{\frac{1}{2}}>0$, $\parallel g\parallel >0$, we have the following Hilbert integral inequality (cf. [1]):

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{f(x)g(y)}{x+y}dxdy<\pi \parallel f\parallel \parallel g\parallel ,$$

(1)

where the constant factor π is best possible. If $a={\{a_n\}}_{n=1}^{\mathrm{\infty }}$, $b={\{b_n\}}_{n=1}^{\mathrm{\infty }}\in l^2$, $\parallel a\parallel ={\{{\sum }_{n=1}^{\mathrm{\infty }}{a}_n^2\}}^{\frac{1}{2}}>0$, $\parallel b\parallel >0$, then we have the following analogous discrete Hilbert inequality:

$$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{m+n}<\pi \parallel a\parallel \parallel b\parallel ,$$

(2)

with the same best constant factor π. Inequalities (1) and (2) are important in analysis and its applications (cf. [2–4]).

In 1998, by introducing an independent parameter $\lambda \in (0,1]$, Yang [5] gave an extension of (1). For generalizing the results from [5], Yang [6] gave some best extensions of (1) and (2): If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, ${\lambda }_1+{\lambda }_2=\lambda$, $k_{\lambda }(x,y)$ is a non-negative homogeneous function of degree −λ satisfying $k({\lambda }_1)={\int }_0^{\mathrm{\infty }}{k}_{\lambda }(t,1)t^{{\lambda }_1-1}dt\in R_{+}$, $\varphi (x)=x^{p(1-{\lambda }_1)-1}$, $\psi (x)=x^{q(1-{\lambda }_2)-1}$, $f(\ge 0)\in L_{p,\varphi }(R_{+})=\{f|{\parallel f\parallel }_{p,\varphi }:={\{{\int }_0^{\mathrm{\infty }}\varphi (x){|f(x)|}^{p}dx\}}^{\frac{1}{p}}<\mathrm{\infty }\}$, $g(\ge 0)\in L_{q,\psi }(R_{+})$, and ${\parallel f\parallel }_{p,\varphi },{\parallel g\parallel }_{q,\psi }>0$, then

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{k}_{\lambda }(x,y)f(x)g(y)dxdy<k({\lambda }_1){\parallel f\parallel }_{p,\varphi }{\parallel g\parallel }_{q,\psi },$$

(3)

where the constant factor $k({\lambda }_1)$ is best possible. Moreover if the value of $k_{\lambda }(x,y)$ is finite and $k_{\lambda }(x,y)x^{{\lambda }_1-1}$ ($k_{\lambda }(x,y)y^{{\lambda }_2-1}$) is decreasing for $x>0$ ($y>0$), then for $a_m,b_n\ge 0$, $a={\{a_m\}}_{m=1}^{\mathrm{\infty }}\in l_{p,\varphi }=\{a|{\parallel a\parallel }_{p,\varphi }:={\{{\sum }_{n=1}^{\mathrm{\infty }}\varphi (n){|a_n|}^p\}}^{\frac{1}{p}}<\mathrm{\infty }\}$, and $b={\{b_n\}}_{n=1}^{\mathrm{\infty }}\in l_{q,\psi }$, ${\parallel a\parallel }_{p,\varphi },{\parallel b\parallel }_{q,\psi }>0$, we have

$$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{k}_{\lambda }(m,n)a_m{b}_n<k({\lambda }_1){\parallel a\parallel }_{p,\varphi }{\parallel b\parallel }_{q,\psi },$$

(4)

where the constant $k({\lambda }_1)$ is still best value. Clearly, for $p=q=2$, $\lambda =1$, $k_1(x,y)=\frac{1}{x+y}$, ${\lambda }_1={\lambda }_2=\frac{1}{2}$, (3) reduces to (1), while (4) reduces to (2). The reverses of (3) and (4) as well as the equivalent forms are also considered by [6].

Some other results about integral and discrete Hilbert-type inequalities can be found in [7–15]. On half-discrete Hilbert-type inequalities with the general non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are best possible. In 2005, Yang [16] gave a result with the kernel $\frac{1}{{(1+nx)}^{\lambda }}$ by introducing a variable and proved that the constant factor is best possible. Very recently, Yang [17] and [18] gave the following half-discrete reverse Hilbert inequality with best constant factor: For $0<p<1$, $\frac{1}{p}+\frac{1}{q}=1$, ${\lambda }_1>0$, $0<{\lambda }_2\le 1$, ${\lambda }_1+{\lambda }_2=\lambda$, ${\theta }_{\lambda }(x)=O(\frac{1}{x^{{\lambda }_2}})\in (0,1)$, $\tilde{\varphi }(x)=(1-{\theta }_{\lambda }(x))x^{p(1-{\lambda }_1)-1}$,

$${\int }_0^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{(x+n)}^{\lambda }}dx>B({\lambda }_1,{\lambda }_2){\parallel f\parallel }_{p,\tilde{\varphi }}{\parallel a\parallel }_{q,\psi }.$$

(5)

In this paper, by means of weight functions and the improved Euler-Maclaurin summation formula, a more accurate half-discrete reverse Hilbert-type inequality with the kernel $\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}$ similar to (5) and a best constant factor is given. Moreover, some equivalent forms, the dual forms as well as some relating homogeneous cases are also considered.

2 Some lemmas

Lemma 1 If $n_0\in \mathbf{N}$, $s>n_0$, $g_1(y)$ ($y\in [n_0,s)$), $g_2(y)$ ($y\in [s,\mathrm{\infty })$) are continuous decreasing functions satisfying $g_1(n_0)-g_1(s-0)+g_2(s)>0$, $g_2(\mathrm{\infty })=0$, define a function $g(y)$ as follows:

$$g(y):=\{\begin{array}{ll}{g}_1(y),& y\in [n_0,s),\\ g_2(y),& y\in [s,\mathrm{\infty }).\end{array}$$

(6)

Then there exists $\epsilon \in [0,1]$, such that

$$\begin{array}{c}\frac{-1}{8}[g_1(n_0)+\epsilon (g_2(s)-g_1(s-0))]\hfill \\ <{\int }_{n_0}^{\mathrm{\infty }}\rho (y)g(y)dy<\frac{1-\epsilon }{8}(g_2(s)-g_1(s-0)),\hfill \end{array}$$

(7)

where $\rho (y)=y-[y]-\frac{1}{2}$ is the Bernoulli function of the first order. In particular, for $g_1(y)=0$, $y\in [n_0,s)$, we have $g_2(s)>0$ and

$$\frac{-1}{8}{g}_2(s)<{\int }_s^{\mathrm{\infty }}\rho (y)g(y)dy<\frac{1}{8}{g}_2(s);$$

(8)

for $g_2(y)=0$, $y\in [s,\mathrm{\infty })$, if $g_1(s-0)\ge 0$, then it follows $g_1(n_0)>0$ and

$$\frac{-1}{8}{g}_1(n_0)<{\int }_{n_0}^s\rho (y)g_1(y)dy<0.$$

(9)

Proof Define a continuous decreasing function $\tilde{g}(y)$ as follows:

$$\tilde{g}(y):=\{\begin{array}{ll}{g}_1(y)+g_2(s)-g_1(s-0),& y\in [n_0,s),\\ g_2(y),& y\in [s,\mathrm{\infty }).\end{array}$$

Then it follows that

$$\begin{array}{c}{\int }_{n_0}^{\mathrm{\infty }}\rho (y)g(y)dy={\int }_{n_0}^s\rho (y)g(y)dy+{\int }_s^{\mathrm{\infty }}\rho (y)g(y)dy\hfill \\ \phantom{{\int }_{n_0}^{\mathrm{\infty }}\rho (y)g(y)dy}={\int }_{n_0}^s\rho (y)(\tilde{g}(y)-g_2(s)+g_1(s-0))dy+{\int }_s^{\mathrm{\infty }}\rho (y)\tilde{g}(y)dy\hfill \\ \phantom{{\int }_{n_0}^{\mathrm{\infty }}\rho (y)g(y)dy}={\int }_{n_0}^{\mathrm{\infty }}\rho (y)\tilde{g}(y)dy-(g_2(s)-g_1(s-0)){\int }_{n_0}^s\rho (y)dy,\hfill \\ {\int }_{n_0}^s\rho (y)dy={\int }_{n_0}^{[s]}\rho (y)dy+{\int }_{[s]}^s\rho (y)dy={\int }_{[s]}^s(y-[s]-\frac{1}{2})dy\hfill \\ \phantom{{\int }_{n_0}^s\rho (y)dy}=\frac{1}{8}[4{(s-[s]-\frac{1}{2})}^2-1]=\frac{\epsilon -1}{8}(\epsilon \in [0,1]).\hfill \end{array}$$

Since $\tilde{g}(n_0)=g_1(n_0)+g_2(s)-g_1(s-0)>0$, $\tilde{g}(y)$ is a non-constant continuous decreasing function with $\tilde{g}(\mathrm{\infty })=g_2(\mathrm{\infty })=0$, by the improved Euler-Maclaurin summation formula (cf. [6], Theorem 2.2.2), it follows that

$$\frac{-1}{8}(g_1(n_0)+g_2(s)-g_1(s-0))=\frac{-1}{8}\tilde{g}(n_0)<{\int }_{n_0}^{\mathrm{\infty }}\rho (y)\tilde{g}(y)dy<0,$$

and then in view of the above results and by simple calculation, we have (7). □

Lemma 2 If $0<\alpha +\beta \le 2$, $\gamma \in \mathbf{R}$, $\eta \le 1-\frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})$, and $\omega (n)$ and $\varpi (x)$ are weight functions given by

$$\begin{array}{c}\omega (n):={\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}\frac{{(n-\eta )}^{\frac{\alpha -\beta }{2}}}{{(x-\gamma )}^{1-\frac{\alpha -\beta }{2}}}dx,n\in \mathbf{N},\hfill \\ \varpi (x):=\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}\frac{{(x-\gamma )}^{\frac{\alpha -\beta }{2}}}{{(n-\eta )}^{1-\frac{\alpha -\beta }{2}}},x>\gamma ,\hfill \end{array}$$

(10)

then we have

$$0<\frac{4}{\alpha +\beta }(1-\theta (x))<\varpi (x)<\omega (n)=\frac{4}{\alpha +\beta },$$

(11)

$$\theta (x)=\{\begin{array}{ll}\frac{1}{2}{(1-\eta )}^{\frac{\alpha +\beta }{2}}{(x-\gamma )}^{\frac{\alpha +\beta }{2}},& 0<x-\gamma \le \frac{1}{1-\eta },\\ 1-\frac{1}{2}{(1-\eta )}^{-\frac{\alpha +\beta }{2}}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}},& x-\gamma >\frac{1}{1-\eta }.\end{array}$$

(12)

Proof Substituting $t=(x-\gamma )(n-\eta )$ in (10), and by simple calculation, we have

$$\omega (n)={\int }_0^{\mathrm{\infty }}\frac{{(min\{1,t\})}^{\beta }}{{(max\{1,t\})}^{\alpha }}{t}^{\frac{\alpha -\beta }{2}-1}dt={\int }_0^1{t}^{\beta +\frac{\alpha -\beta }{2}-1}dt+{\int }_1^{\mathrm{\infty }}{t}^{-\alpha +\frac{\alpha -\beta }{2}-1}dt=\frac{4}{\alpha +\beta }.$$

For fixed $x>\gamma$, we find

$$\begin{array}{c}h(x,y):={(x-\gamma )}^{\frac{\alpha -\beta }{2}}\frac{{(min\{1,(x-\gamma )(y-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(y-\eta )\})}^{\alpha }}{(y-\eta )}^{\frac{\alpha -\beta }{2}-1}\hfill \\ \phantom{h(x,y)}=\{\begin{array}{ll}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}{(y-\eta )}^{\frac{\alpha +\beta }{2}-1},& \eta <y<\eta +\frac{1}{x-\gamma },\\ {(x-\gamma )}^{-\frac{\alpha +\beta }{2}}{(y-\eta )}^{-\frac{\alpha +\beta }{2}-1},& y\ge \eta +\frac{1}{x-\gamma },\end{array}\hfill \\ h_y^{\prime }(x,y)=\{\begin{array}{ll}-(1-\frac{\alpha +\beta }{2}){(x-\gamma )}^{\frac{\alpha +\beta }{2}}{(y-\eta )}^{\frac{\alpha +\beta }{2}-2},& \eta <y<\eta +\frac{1}{x-\gamma },\\ -(\frac{\alpha +\beta }{2}+1){(x-\gamma )}^{-\frac{\alpha +\beta }{2}}{(y-\eta )}^{-\frac{\alpha +\beta }{2}-2},& y\ge \eta +\frac{1}{x-\gamma },\end{array}\hfill \\ {\int }_{\eta }^{\mathrm{\infty }}h(x,y)dy\stackrel{t=(x-\gamma )(y-\eta )}{=}{\int }_0^{\mathrm{\infty }}\frac{{(min\{1,t\})}^{\beta }}{{(max\{1,t\})}^{\alpha }}{t}^{\frac{\alpha -\beta }{2}-1}dt=\frac{4}{\alpha +\beta }.\hfill \end{array}$$

By the Euler-Maclaurin summation formula (cf. [6]), it follows that

$$\begin{array}{c}\varpi (x)=\sum _{n=1}^{\mathrm{\infty }}h(x,n)={\int }_1^{\mathrm{\infty }}h(x,y)dy+\frac{1}{2}h(x,1)+{\int }_1^{\mathrm{\infty }}\rho (y)h_y^{\prime }(x,y)dy\hfill \\ \phantom{\varpi (x)}={\int }_{\eta }^{\mathrm{\infty }}h(x,y)dy-R(x)=\frac{4}{\alpha +\beta }-R(x),\hfill \\ R(x):={\int }_{\eta }^{1}h(x,y)dy-\frac{1}{2}h(x,1)-{\int }_1^{\mathrm{\infty }}\rho (y)h_y^{\prime }(x,y)dy.\hfill \end{array}$$

(13)

• (i) For $0<x-\gamma <\frac{1}{1-\eta }$, we obtain $-\frac{1}{2}h(x,1)=-\frac{1}{2}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}{(1-\eta )}^{\frac{\alpha +\beta }{2}-1}$, and

  $${\int }_{\eta }^{1}h(x,y)dy={(x-\gamma )}^{\frac{\alpha +\beta }{2}}{\int }_{\eta }^1{(y-\eta )}^{\frac{\alpha +\beta }{2}-1}dy=\frac{2{(1-\eta )}^{\frac{\alpha +\beta }{2}}}{\alpha +\beta }{(x-\gamma )}^{\frac{\alpha +\beta }{2}}.$$

Setting $g(y):=-h_y^{\prime }(x,y)$, wherefrom $g_1(y)=(1-\frac{\alpha +\beta }{2}){(x-\gamma )}^{\frac{\alpha +\beta }{2}}{(y-\eta )}^{\frac{\alpha +\beta }{2}-2}$, $g_2(y)=(\frac{\alpha +\beta }{2}+1){(x-\gamma )}^{-\frac{\alpha +\beta }{2}}{(y-\eta )}^{-\frac{\alpha +\beta }{2}-2}$ and

$$\begin{array}{c}{g}_2(\eta +\frac{1}{x-\gamma })-g_1((\eta +\frac{1}{x-\gamma })-0)\hfill \\ =(\frac{\alpha +\beta }{2}+1){(x-\gamma )}^2-(1-\frac{\alpha +\beta }{2}){(x-\gamma )}^2\hfill \\ =(\alpha +\beta ){(x-\gamma )}^2>0,\hfill \end{array}$$

then by (7), we find

$$\begin{array}{rcl}-{\int }_1^{\mathrm{\infty }}\rho (y)h_y^{\prime }(x,y)dy& =& {\int }_1^{\mathrm{\infty }}\rho (y)g(y)dy\\ >& \frac{-1}{8}[g_1(1)+g_2(\eta +\frac{1}{x-\gamma })-g_1((\eta +\frac{1}{x-\gamma })-0)]\\ =& \frac{-1}{8}[(1-\frac{\alpha +\beta }{2}){(x-\gamma )}^{\frac{\alpha +\beta }{2}}{(1-\eta )}^{\frac{\alpha +\beta }{2}-2}+(\alpha +\beta ){(x-\gamma )}^2]\\ >& \frac{-1}{8}[(1-\frac{\alpha +\beta }{2}){(1-\eta )}^{\frac{\alpha +\beta }{2}-2}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}\\ +(\alpha +\beta ){(1-\eta )}^{\frac{\alpha +\beta }{2}-2}{(x-\gamma )}^{\frac{\alpha +\beta }{2}-2}{(x-\gamma )}^2]\\ =& \frac{-1}{8}(1+\frac{\alpha +\beta }{2}){(1-\eta )}^{\frac{\alpha +\beta }{2}-2}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}.\end{array}$$

In view of (11) and the above results, since for $\eta \le 1-\frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})$, namely $1-\eta \ge \frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})$, it follows that

$$\begin{array}{rcl}R(x)& >& \frac{2}{\alpha +\beta }{(1-\eta )}^{\frac{\alpha +\beta }{2}}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}-\frac{1}{2}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}{(1-\eta )}^{\frac{\alpha +\beta }{2}-1}\\ -\frac{1}{8}(1+\frac{\alpha +\beta }{2}){(1-\eta )}^{\frac{\alpha +\beta }{2}-2}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}\\ =& [\frac{2{(1-\eta )}^2}{\alpha +\beta }-\frac{(1-\eta )}{2}-\frac{2+\alpha +\beta }{16}]\frac{{(x-\gamma )}^{\frac{\alpha +\beta }{2}}}{{(1-\eta )}^{2-\frac{\alpha +\beta }{2}}}\ge 0.\end{array}$$

• (ii) For $x-\gamma \ge \frac{1}{1-\eta }$, we obtain $-\frac{1}{2}h(x,1)=-\frac{1}{2}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}{(1-\eta )}^{-\frac{\alpha +\beta }{2}-1}$, and

  $$\begin{array}{rcl}{\int }_{\eta }^{1}h(x,y)dy& =& {\int }_{\eta }^{\eta +\frac{1}{x-\gamma }}\frac{{(x-\gamma )}^{\frac{\alpha +\beta }{2}}}{{(y-\eta )}^{1-\frac{\alpha +\beta }{2}}}dy+{\int }_{\eta +\frac{1}{x-\gamma }}^1\frac{{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}}{{(y-\eta )}^{\frac{\alpha +\beta }{2}+1}}dy\\ =& \frac{4}{\alpha +\beta }-\frac{2}{\alpha +\beta }{(1-\eta )}^{-\frac{\alpha +\beta }{2}}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}\\ \ge & \frac{4{(1-\eta )}^{-\frac{\alpha +\beta }{2}}}{\alpha +\beta }{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}-\frac{2{(1-\eta )}^{-\frac{\alpha +\beta }{2}}}{\alpha +\beta }{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}\\ =& \frac{2}{\alpha +\beta }{(1-\eta )}^{-\frac{\alpha +\beta }{2}}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}.\end{array}$$

Since for $y\ge 1$, $y-\eta \ge \frac{1}{x-\gamma }$, by the improved Euler-Maclaurin summation formula (cf. [6]), it follows that

$$\begin{array}{rcl}-{\int }_1^{\mathrm{\infty }}\rho (y)h_y^{\prime }(x,y)dy& =& (\frac{\alpha +\beta }{2}+1){(x-\gamma )}^{-\frac{\alpha +\beta }{2}}{\int }_1^{\mathrm{\infty }}\rho (y){(y-\eta )}^{-\frac{\alpha +\beta }{2}-2}dy\\ >& -\frac{1}{8}(\frac{\alpha +\beta }{2}+1){(x-\gamma )}^{-\frac{\alpha +\beta }{2}}{(1-\eta )}^{-\frac{\alpha +\beta }{2}-2}.\end{array}$$

In view of (13) and the above results, for $1-\eta \ge \frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})$, we find

$$\begin{array}{rcl}R(x)& >& \frac{2}{\alpha +\beta }{(1-\eta )}^{-\frac{\alpha +\beta }{2}}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}-\frac{1}{2}{(1-\eta )}^{-\frac{\alpha +\beta }{2}-1}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}\\ -\frac{1}{8}(\frac{\alpha +\beta }{2}+1){(1-\eta )}^{-\frac{\alpha +\beta }{2}-2}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}\\ =& [\frac{2{(1-\eta )}^2}{\alpha +\beta }-\frac{1-\eta }{2}-\frac{2+\alpha +\beta }{16}]\frac{{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}}{{(1-\eta )}^{2+\frac{\alpha +\beta }{2}}}\ge 0.\end{array}$$

Hence for $x>\gamma$, we have $R(x)>0$, and then $\varpi (x)<\omega (n)=\frac{4}{\alpha +\beta }$.

On the other-hand, since $h(x,y)$ is decreasing with respect to $y>\eta$, we find

$$\begin{array}{rcl}\varpi (x)& >& {\int }_1^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(y-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(y-\eta )\})}^{\alpha }}\frac{{(x-\gamma )}^{\frac{\alpha -\beta }{2}}}{{(y-\eta )}^{1-\frac{\alpha -\beta }{2}}}dy\\ \stackrel{t=(x-\gamma )(y-\eta )}{=}& {\int }_{(1-\eta )(x-\gamma )}^{\mathrm{\infty }}\frac{{(min\{1,t\})}^{\beta }}{{(max\{1,t\})}^{\alpha }}{t}^{\frac{\alpha -\beta }{2}-1}dt=\frac{4}{\alpha +\beta }(1-\theta (x)),\end{array}$$

where $\theta (x):=\frac{\alpha +\beta }{4}{\int }_0^{(1-\eta )(x-\gamma )}\frac{{(min\{1,t\})}^{\beta }}{{(max\{1,t\})}^{\alpha }}{t}^{\frac{\alpha -\beta }{2}-1}dt\in (0,1)$.

• (i) For $0<x-\gamma \le \frac{1}{1-\eta }$, we obtain

  $$\theta (x)=\frac{\alpha +\beta }{4}{\int }_0^{(1-\eta )(x-\gamma )}{t}^{\beta +\frac{\alpha -\beta }{2}-1}dt=\frac{1}{2}{(1-\eta )}^{\frac{\alpha +\beta }{2}}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}.$$

• (ii) For $x-\gamma >\frac{1}{1-\eta }$, it follows that

  $$\begin{array}{rcl}\theta (x)& =& \frac{\alpha +\beta }{4}[{\int }_0^1{t}^{\beta +\frac{\alpha -\beta }{2}-1}dt+{\int }_1^{(1-\eta )(x-\gamma )}{t}^{-\alpha +\frac{\alpha -\beta }{2}-1}dt]\\ =& 1-\frac{1}{2}{(1-\eta )}^{-\frac{\alpha +\beta }{2}}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}.\end{array}$$

Hence we have (11) and (12). □

Lemma 3 Let the assumptions of Lemma  2 be fulfilled and additionally, let $0<p<1$ or $p<0$, $\frac{1}{p}+\frac{1}{q}=1$, $a_n\ge 0$, $n\in \mathbf{N}$, $f(x)$ be a non-negative measurable function in $(\gamma ,\mathrm{\infty })$. Then we have the following inequalities:

$$\begin{array}{c}J:={\left\{\sum _{n=1}^{\mathrm{\infty }}{(n-\beta )}^{\frac{p(\alpha -\beta )}{2}-1}{[{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}f(x)dx]}^p\right\}}^{\frac{1}{p}}\hfill \\ \phantom{J}\ge {\left(\frac{4}{\alpha +\beta }\right)}^{\frac{1}{q}}{\{{\int }_{\gamma }^{\mathrm{\infty }}\varpi (x){(x-\gamma )}^{p(1-\frac{\alpha -\beta }{2})-1}{f}^p(x)dx\}}^{\frac{1}{p}},\hfill \end{array}$$

(14)

$$\begin{array}{c}{L}_1:={\left\{{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(x-\alpha )}^{\frac{q(\alpha -\beta )}{2}-1}}{{[\varpi (x)]}^{q-1}}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }{a}_n}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}\right]}^{q}dx\right\}}^{\frac{1}{q}}\hfill \\ \phantom{L_1}\ge {\left\{\frac{4}{\alpha +\beta }\sum _{n=1}^{\mathrm{\infty }}{(n-\eta )}^{q(1-\frac{\alpha -\beta }{2})-1}{a}_n^q\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(15)

Proof For $0<p<1$, setting $k(x,n):=\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}$, by the reverse Hölder inequality (cf. [19]) and (11), it follows that

$$\begin{array}{c}{[{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}f(x)dx]}^p\hfill \\ ={\{{\int }_{\gamma }^{\mathrm{\infty }}k(x,n)[\frac{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})/q}}{{(n-\eta )}^{(1-\frac{\alpha -\beta }{2})/p}}f(x)]\left[\frac{{(n-\gamma )}^{(1-\frac{\alpha -\beta }{2})/p}}{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})/q}}\right]dx\}}^p\hfill \\ \ge {\int }_{\gamma }^{\mathrm{\infty }}k(x,n)\frac{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})(p-1)}}{{(n-\eta )}^{1-\frac{\alpha -\beta }{2}}}{f}^p(x)dx\hfill \\ \times {\{{\int }_{\gamma }^{\mathrm{\infty }}k(x,n)\frac{{(n-\eta )}^{(1-\frac{\alpha -\beta }{2})(q-1)}}{{(x-\gamma )}^{1-\frac{\alpha -\beta }{2}}}dx\}}^{p-1}\hfill \\ ={\{\omega (n){(n-\eta )}^{q(1-\frac{\alpha -\beta }{2})-1}\}}^{p-1}{\int }_{\gamma }^{\mathrm{\infty }}k(x,n)\frac{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})(p-1)}}{{(n-\eta )}^{1-\frac{\alpha -\beta }{2}}}{f}^p(x)dx\hfill \\ ={\left(\frac{4}{\alpha +\beta }\right)}^{p-1}{(n-\eta )}^{1-\frac{p(\alpha -\beta )}{2}}{\int }_{\gamma }^{\mathrm{\infty }}k(x,n)\frac{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})(p-1)}}{{(n-\eta )}^{1-\frac{\alpha -\beta }{2}}}{f}^p(x)dx.\hfill \end{array}$$

Then by the Lebesgue term by term integration theorem (cf. [20]), we have

$$\begin{array}{rcl}J& \ge & {\left(\frac{4}{\alpha +\beta }\right)}^{\frac{1}{q}}{\{\sum _{n=1}^{\mathrm{\infty }}{\int }_{\gamma }^{\mathrm{\infty }}k(x,n)\frac{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})(p-1)}}{{(n-\eta )}^{1-\frac{\alpha -\beta }{2}}}{f}^p(x)dx\}}^{\frac{1}{p}}\\ =& {\left(\frac{4}{\alpha +\beta }\right)}^{\frac{1}{q}}{\{{\int }_{\gamma }^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k(x,n)\frac{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})(p-1)}}{{(n-\eta )}^{1-\frac{\alpha -\beta }{2}}}{f}^p(x)dx\}}^{\frac{1}{p}}\\ =& {\left(\frac{4}{\alpha +\beta }\right)}^{\frac{1}{q}}{\{{\int }_{\gamma }^{\mathrm{\infty }}\varpi (x){(x-\gamma )}^{p(1-\frac{\alpha -\beta }{2})-1}{f}^p(x)dx\}}^{\frac{1}{p}},\end{array}$$

and then (14) follows. By the reverse Hölder inequality, for $q<0$, we have

$$\begin{array}{rcl}{[\sum _{n=1}^{\mathrm{\infty }}k(x,n)a_n]}^q& =& {\{\sum _{n=1}^{\mathrm{\infty }}k(x,n)\left[\frac{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})/q}}{{(n-\eta )}^{(1-\frac{\alpha -\beta }{2})/p}}\right]\left[\frac{{(n-\eta )}^{(1-\frac{\alpha -\beta }{2})/p}{a}_n}{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})/q}}\right]\}}^q\\ \le & {\{\sum _{n=1}^{\mathrm{\infty }}k(x,n)\frac{{(x-\gamma )}^{(1-\frac{\alpha -\beta }{2})(p-1)}}{{(n-\eta )}^{1-\frac{\alpha -\beta }{2}}}\}}^{q-1}\sum _{n=1}^{\mathrm{\infty }}k(x,n)\frac{{(n-\eta )}^{(1-\frac{\alpha -\beta }{2})(q-1)}}{{(x-\gamma )}^{1-\frac{\alpha -\beta }{2}}}{a}_n^q\\ =& \frac{{[\varpi (x)]}^{q-1}}{{(x-\gamma )}^{\frac{q(\alpha -\beta )}{2}-1}}\sum _{n=1}^{\mathrm{\infty }}k(x,n)\frac{{(n-\eta )}^{(1-\frac{\alpha -\beta }{2})(q-1)}}{{(x-\gamma )}^{1-\frac{\alpha -\beta }{2}}}{a}_n^q.\end{array}$$

By the Lebesgue term by term integration theorem, we have

$$\begin{array}{rcl}{L}_1& \ge & {\{{\int }_{\gamma }^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}k(x,n)\frac{{(n-\eta )}^{(1-\frac{\alpha -\beta }{2})(q-1)}}{{(x-\gamma )}^{1-\frac{\alpha -\beta }{2}}}{a}_n^{q}dx\}}^{\frac{1}{q}}\\ =& {\{\sum _{n=1}^{\mathrm{\infty }}{\int }_{\gamma }^{\mathrm{\infty }}k(x,n)\frac{{(n-\eta )}^{(1-\frac{\alpha -\beta }{2})(q-1)}}{{(x-\gamma )}^{1-\frac{\alpha -\beta }{2}}}{a}_n^{q}dx\}}^{\frac{1}{q}}\\ =& {\{\sum _{n=1}^{\mathrm{\infty }}\omega (n){(n-\eta )}^{q(1-\frac{\alpha -\beta }{2})-1}{a}_n^q\}}^{\frac{1}{q}},\end{array}$$

and in view of (11), inequality (15) follows. For $p<0$, by the same way we still have (14) and (15). □

Lemma 4 Let the assumptions of Lemma  2 be fulfilled and additionally, let $0<p<1$, $\frac{1}{p}+\frac{1}{q}=1$, $0<\epsilon <\frac{p}{2}(\alpha +\beta )$. Setting $\tilde{f}(x)={(x-\gamma )}^{\frac{\alpha -\beta }{2}+\frac{\epsilon }{p}-1}$, $x\in (\gamma ,\gamma +1)$; $\tilde{f}(x)=0$, $x\in [\gamma +1,\mathrm{\infty })$, and ${\tilde{a}}_n={(n-\eta )}^{\frac{\alpha -\beta }{2}-\frac{\epsilon }{q}-1}$, $n\in \mathbf{N}$, then we have

$$\begin{array}{c}\tilde{I}:=\sum _{n=1}^{\mathrm{\infty }}{\tilde{a}}_n{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}\tilde{f}(x)dx\hfill \\ \phantom{\tilde{I}}<\frac{1}{\epsilon }\frac{(\alpha +\beta )}{{(\frac{\alpha +\beta }{2})}^2-{(\frac{\epsilon }{p})}^2}[\frac{\epsilon }{{(1-\eta )}^{\epsilon +1}}+\frac{1}{{(1-\eta )}^{\epsilon }}],\hfill \end{array}$$

(16)

$$\begin{array}{c}\tilde{H}:={\{{\int }_{\gamma }^{\mathrm{\infty }}{(x-\gamma )}^{p(1-\frac{\alpha -\beta }{2})-1}{\tilde{f}}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{(n-\eta )}^{q(1-\frac{\alpha -\beta }{2})-1}{\tilde{a}}_n^q\right\}}^{\frac{1}{q}}\hfill \\ \phantom{\tilde{H}}>\frac{1}{\epsilon }{(1-\epsilon O(1))}^{\frac{1}{p}}{\{\frac{\epsilon }{{(1-\eta )}^{\epsilon +1}}+\frac{1}{{(1-\eta )}^{\epsilon }}\}}^{\frac{1}{q}}.\hfill \end{array}$$

(17)

Proof We find

$$\begin{array}{rcl}\tilde{I}& =& \sum _{n=1}^{\mathrm{\infty }}{(n-\eta )}^{\frac{\alpha -\beta }{2}-\frac{\epsilon }{q}-1}{\int }_{\gamma }^{\gamma +1}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}{(x-\gamma )}^{\frac{\alpha -\beta }{2}+\frac{\epsilon }{p}-1}dx\\ <& \sum _{n=1}^{\mathrm{\infty }}{(n-\eta )}^{\frac{\alpha -\beta }{2}-\frac{\epsilon }{q}-1}{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}{(x-\gamma )}^{\frac{\alpha -\beta }{2}+\frac{\epsilon }{p}-1}dx\\ =& \frac{\alpha +\beta }{{(\frac{\alpha +\beta }{2})}^2-{(\frac{\epsilon }{p})}^2}[\frac{1}{{(1-\eta )}^{\epsilon +1}}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{(n-\eta )}^{\epsilon +1}}]\\ <& \frac{\alpha +\beta }{{(\frac{\alpha +\beta }{2})}^2-{(\frac{\epsilon }{p})}^2}[\frac{1}{{(1-\eta )}^{\epsilon +1}}+{\int }_1^{\mathrm{\infty }}\frac{dy}{{(y-\eta )}^{\epsilon +1}}]\\ =& \frac{1}{\epsilon }\frac{(\alpha +\beta )}{{(\frac{\alpha +\beta }{2})}^2-{(\frac{\epsilon }{p})}^2}[\frac{\epsilon }{{(1-\eta )}^{\epsilon +1}}+\frac{1}{{(1-\eta )}^{\epsilon }}],\end{array}$$

and then (16) is valid. We obtain

$$\begin{array}{rcl}\tilde{H}& =& {\left\{{\int }_{\gamma }^{\gamma +1}[1-\frac{1}{2}{(1-\eta )}^{\frac{\alpha +\beta }{2}}{(x-\gamma )}^{\frac{\alpha +\beta }{2}}]{(x-\gamma )}^{\epsilon -1}dx\right\}}^{\frac{1}{p}}\\ \times {\{\frac{1}{{(1-\eta )}^{\epsilon +1}}+\sum _{n=2}^{\mathrm{\infty }}\frac{1}{{(n-\eta )}^{\epsilon +1}}\}}^{\frac{1}{q}}\\ >& {(\frac{1}{\epsilon }-O(1))}^{\frac{1}{p}}{\{\frac{1}{{(1-\eta )}^{\epsilon +1}}+{\int }_1^{\mathrm{\infty }}\frac{dy}{{(y-\eta )}^{\epsilon +1}}\}}^{\frac{1}{q}}\\ =& \frac{1}{\epsilon }{(1-\epsilon O(1))}^{\frac{1}{p}}{\{\frac{\epsilon }{{(1-\eta )}^{\epsilon +1}}+\frac{1}{{(1-\eta )}^{\epsilon }}\}}^{\frac{1}{q}},\end{array}$$

and so (17) is valid. □

3 Main results

We introduce the functions

$$\begin{array}{r}\mathrm{\Phi }(x):={(x-\gamma )}^{p(1-\frac{\alpha -\beta }{2})-1},\tilde{\mathrm{\Phi }}(x)=(1-\theta (x))\mathrm{\Phi }(x)(x\in (\gamma ,\mathrm{\infty })),\\ \mathrm{\Psi }(n):={(n-\eta )}^{q(1-\frac{\alpha -\beta }{2})-1}(n\in \mathbf{N}),\end{array}$$

wherefrom ${[\mathrm{\Phi }(x)]}^{1-q}={(x-\gamma )}^{q\frac{\alpha -\beta }{2}-1}$, ${[\tilde{\mathrm{\Phi }}(x)]}^{1-q}={(1-\theta (x))}^{1-q}{(x-\gamma )}^{q\frac{\alpha -\beta }{2}-1}$ and ${[\mathrm{\Psi }(n)]}^{1-p}={(n-\eta )}^{p\frac{\alpha -\beta }{2}-1}$.

Theorem 1 If $0<\alpha +\beta \le 2$, $\gamma \in \mathbf{R}$, $\eta \le 1-\frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})$, $0<p<1$, $\frac{1}{p}+\frac{1}{q}=1$, $f(x),a_n\ge 0$, $f\in L_{p,\tilde{\mathrm{\Phi }}}(\gamma ,\mathrm{\infty })$, $a={\{a_n\}}_{n=1}^{\mathrm{\infty }}\in l_{q,\mathrm{\Psi }}$, ${\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}>0$ and ${\parallel a\parallel }_{q,\mathrm{\Psi }}>0$, then we have the following equivalent inequalities:

$$\begin{array}{c}I:=\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}f(x)dx\hfill \\ \phantom{I}={\int }_{\gamma }^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }{a}_n}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}dx>\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}{\parallel a\parallel }_{q,\mathrm{\Psi }},\hfill \end{array}$$

(18)

$$J={\left\{\sum _{n=1}^{\mathrm{\infty }}{[\mathrm{\Psi }(n)]}^{1-p}{\left[{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }f(x)}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}dx\right]}^p\right\}}^{\frac{1}{p}}>\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}},$$

(19)

$$L:={\left\{{\int }_{\gamma }^{\mathrm{\infty }}{[\tilde{\mathrm{\Phi }}(x)]}^{1-q}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }{a}_n}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}\right]}^{q}dx\right\}}^{\frac{1}{q}}>\frac{4}{\alpha +\beta }{\parallel a\parallel }_{q,\mathrm{\Psi }},$$

(20)

where the constant $\frac{4}{\alpha +\beta }$ is the best possible in the above inequalities.

Proof The two expressions for I in (18) follow from Lebesgue’s term by term integration theorem. By (14) and (11), we have (19). By the reverse Hölder inequality, we have

$$I=\sum _{n=1}^{\mathrm{\infty }}[{\mathrm{\Psi }}^{\frac{-1}{q}}(n){\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }f(x)}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}dx][{\mathrm{\Psi }}^{\frac{1}{q}}(n)a_n]\ge J{\parallel a\parallel }_{q,\mathrm{\Psi }}.$$

Then by (19), we have (18). On the other-hand, assume that (18) is valid. Setting

$$a_n:={[\mathrm{\Psi }(n)]}^{1-p}{\left[{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }f(x)}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}dx\right]}^{p-1},n\in \mathbf{N},$$

it follows that $J^{p-1}={\parallel a\parallel }_{q,\mathrm{\Psi }}$. By (14), we find $J>0$. If $J=\mathrm{\infty }$, then (19) is trivially valid; if $J<\mathrm{\infty }$, then by (18), we have

$${\parallel a\parallel }_{q,\mathrm{\Psi }}^q=J^{q(p-1)}=J^p=I>\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}{\parallel a\parallel }_{q,\mathrm{\Psi }},$$

therefore ${\parallel a\parallel }_{q,\mathrm{\Psi }}^{q-1}=J>\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}$, that is, (19) is equivalent to (18). On the other-hand, by (11) we have ${[\varpi (x)]}^{1-q}>{(1-\theta (x))}^{1-q}{(\frac{4}{\alpha +\beta })}^{1-q}$. Then in view of (15), we have (20). By the Hölder inequality, we find

$$I={\int }_{\gamma }^{\mathrm{\infty }}[{\tilde{\mathrm{\Phi }}}^{\frac{1}{p}}(x)f(x)][{\tilde{\mathrm{\Phi }}}^{\frac{-1}{p}}(x)\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }{a}_n}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}]dx\ge {\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}L.$$

Then by (20), we have (18). On the other-hand, assume that (18) is valid. Setting

$$f(x):={[\tilde{\mathrm{\Phi }}(x)]}^{1-q}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }{a}_n}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}\right]}^{q-1},x\in (\gamma ,\mathrm{\infty }),$$

then $L^{q-1}={\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}$. By (15), we find $L>0$. If $L=\mathrm{\infty }$, then (20) is trivially valid; if $L<\mathrm{\infty }$, then by (18), we have

$${\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}^p=L^{p(q-1)}=I>\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}{\parallel a\parallel }_{q,\mathrm{\Psi }},$$

therefore ${\parallel f\parallel }_{p,\tilde{\mathrm{\Phi }}}^{p-1}=L>\frac{4}{\alpha +\beta }{\parallel a\parallel }_{q,\mathrm{\Psi }}$, that is, (20) is equivalent to (18). Hence, (18), (19), and (20) are equivalent.

If there exists a positive number k ($\ge \frac{4}{\alpha +\beta }$), such that (18) is valid as we replace $\frac{4}{\alpha +\beta }$ with k, then in particular, it follows that $\tilde{I}>k\tilde{H}$. In view of (16) and (17), we have

$$\frac{(\alpha +\beta )}{{(\frac{\alpha +\beta }{2})}^2-{(\frac{\epsilon }{p})}^2}[\frac{\epsilon }{{(1-\eta )}^{\epsilon +1}}+\frac{1}{{(1-\eta )}^{\epsilon }}]>k{(1-\epsilon O(1))}^{\frac{1}{p}}{\{\frac{\epsilon }{{(1-\eta )}^{\epsilon +1}}+\frac{1}{{(1-\eta )}^{\epsilon }}\}}^{\frac{1}{q}},$$

and $\frac{4}{\alpha +\beta }\ge k$ ($\epsilon \to 0^{+}$). Hence $k=\frac{4}{\alpha +\beta }$ is the best value of (18).

By the equivalence of the inequalities, the constant factor $\frac{4}{\alpha +\beta }$ in (19) and (20) is the best possible. □

For $p<0$, we have the dual forms of (18), (19), and (20) as follows:

Theorem 2 If $0<\alpha +\beta \le 2$, $\gamma \in \mathbf{R}$, $\eta \le 1-\frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})$, $p<0$, $\frac{1}{p}+\frac{1}{q}=1$, $f(x),a_n\ge 0$, $f\in L_{p,\mathrm{\Phi }}(\gamma ,\mathrm{\infty })$, $a={\{a_n\}}_{n=1}^{\mathrm{\infty }}\in l_{q,\mathrm{\Psi }}$, ${\parallel f\parallel }_{p,\mathrm{\Phi }}>0$ and ${\parallel a\parallel }_{q,\mathrm{\Psi }}>0$, then we have the following equivalent inequalities:

$$\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}f(x)dx>\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\mathrm{\Phi }}{\parallel a\parallel }_{q,\mathrm{\Psi }},$$

(21)

$${\left\{\sum _{n=1}^{\mathrm{\infty }}{[\mathrm{\Psi }(n)]}^{1-p}{\left[{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }f(x)}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}dx\right]}^p\right\}}^{\frac{1}{p}}>\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\mathrm{\Phi }},$$

(22)

$${\left\{{\int }_{\gamma }^{\mathrm{\infty }}{[\mathrm{\Phi }(x)]}^{1-q}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }{a}_n}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}\right]}^{q}dx\right\}}^{\frac{1}{q}}>\frac{4}{\alpha +\beta }{\parallel a\parallel }_{q,\mathrm{\Psi }},$$

(23)

where the constant $\frac{4}{\alpha +\beta }$ is the best possible in the above inequalities.

Proof By means of Lemma 3 and the same way, we can prove that (21), (22), and (23) are valid and equivalent. For $0<\epsilon \frac{|p|}{2}(\alpha +\beta )$, setting $\tilde{f}(x)$ and ${\tilde{a}}_n$ as Lemma 4, if there exists a positive number k ($\ge \frac{4}{\alpha +\beta }$), such that (21) is valid as we replace $\frac{4}{\alpha +\beta }$ with k, then in particular, by (16), it follows that

$$\begin{array}{c}\frac{\alpha +\beta }{{(\frac{\alpha +\beta }{2})}^2-{(\frac{\epsilon }{p})}^2}[\frac{\epsilon }{{(1-\eta )}^{\epsilon +1}}+\frac{1}{{(1-\eta )}^{\epsilon }}]\hfill \\ >\epsilon \tilde{I}>\epsilon k{\left\{{\int }_{\gamma }^{\gamma +1}{(x-\gamma )}^{\epsilon -1}dx\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{(n-\eta )}^{\epsilon +1}}\right\}}^{\frac{1}{q}}\hfill \\ >\epsilon k{\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}{\left\{{\int }_1^{\mathrm{\infty }}\frac{dy}{{(y-\eta )}^{\epsilon +1}}\right\}}^{\frac{1}{q}}=k{\left\{\frac{1}{{(1-\eta )}^{\epsilon }}\right\}}^{\frac{1}{q}},\hfill \end{array}$$

and $\frac{4}{\alpha +\beta }\ge k$ ($\epsilon \to 0^{+}$). Hence $k=\frac{4}{\alpha +\beta }$ is the best value of (21). By the equivalence of the inequalities, the constant factor $\frac{4}{\alpha +\beta }$ in (21) and (22) is the best possible. □

Remark (i) Since we find

$$\underset{0<\alpha +\beta \le 2}{min}\{1-\frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})\}=\frac{3-\sqrt{5}}{4}={0.19}^{+}>0,$$

then for $\eta =\gamma =0$ in (18), we have

$${\theta }_0(x)=\{\begin{array}{ll}\frac{1}{2}{x}^{\frac{\alpha +\beta }{2}},& 0<x\le 1,\\ 1-\frac{1}{2}{x}^{-\frac{\alpha +\beta }{2}},& x>1,\end{array}$$

and the following inequality:

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_0^{\mathrm{\infty }}\frac{{(min\{1,xn\})}^{\beta }}{{(max\{1,xn\})}^{\alpha }}f(x)dx\hfill \\ >\frac{4}{\alpha +\beta }{\{{\int }_0^{\mathrm{\infty }}(1-{\theta }_0(x))x^{p(1-\frac{\alpha -\beta }{2})-1}{f}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q(1-\frac{\alpha -\beta }{2})-1}{a}_n^q\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(24)

Hence (18) is a more accurate inequality of (24).

• (ii) For $\beta =0$ in (18), we have $0<\alpha \le 2$, $\gamma \in \mathbf{R}$, $\eta \le 1-\frac{\alpha }{8}(1+\sqrt{3+\frac{4}{\alpha }})$,

  $${\theta }_1(x)=\{\begin{array}{ll}\frac{1}{2}{(x-\gamma )}^{\frac{\alpha }{2}},& 0<x-\gamma \le \frac{1}{1-\eta },\\ 1-\frac{1}{2}{(x-\gamma )}^{-\frac{\alpha }{2}},& x-\gamma >\frac{1}{1-\eta }\end{array}$$

and the following inequality:

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_{\gamma }^{\mathrm{\infty }}\frac{f(x)dx}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}\hfill \\ >\frac{4}{\alpha }{\{{\int }_{\gamma }^{\mathrm{\infty }}(1-{\theta }_1(x)){(x-\gamma )}^{p(1-\frac{\alpha }{2})-1}{f}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{(n-\eta )}^{q(1-\frac{\alpha }{2})-1}{a}_n^q\right\}}^{\frac{1}{q}};\hfill \end{array}$$

(25)

for $\alpha =0$ in (18), we have $0<\beta \le 2$, $\gamma \in \mathbf{R}$, $\eta \le 1-\frac{\beta }{8}(1+\sqrt{3+\frac{4}{\beta }})$,

$${\theta }_2(x)=\{\begin{array}{ll}\frac{1}{2}{(x-\gamma )}^{\frac{\beta }{2}},& 0<x-\gamma \le \frac{1}{1-\eta },\\ 1-\frac{1}{2}{(x-\gamma )}^{-\frac{\beta }{2}},& x-\gamma >\frac{1}{1-\eta }\end{array}$$

and the following inequality:

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_{\gamma }^{\mathrm{\infty }}{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }f(x)dx\hfill \\ >\frac{4}{\beta }{\{{\int }_{\gamma }^{\mathrm{\infty }}(1-{\theta }_2(x)){(x-\gamma )}^{p(1+\frac{\beta }{2})-1}{f}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{(n-\eta )}^{q(1+\frac{\beta }{2})-1}{a}_n^q\right\}}^{\frac{1}{q}};\hfill \end{array}$$

(26)

for $\beta =\alpha =\lambda$ in (18), we have $0<\lambda \le 1$, $\gamma \in \mathbf{R}$, $\eta \le 1-\frac{\lambda }{4}(1+\sqrt{3+\frac{2}{\lambda }})$,

$${\theta }_3(x)=\{\begin{array}{ll}\frac{1}{2}{(x-\gamma )}^{\lambda },& 0<x-\gamma \le \frac{1}{1-\eta },\\ 1-\frac{1}{2}{(x-\gamma )}^{-\lambda },& x-\gamma >\frac{1}{1-\eta }\end{array}$$

and the following inequality:

$$\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_{\gamma }^{\mathrm{\infty }}{\left[\frac{min\{1,(x-\gamma )(n-\eta )\}}{max\{1,(x-\gamma )(n-\eta )\}}\right]}^{\lambda }f(x)dx\hfill \\ >\frac{2}{\lambda }{\{{\int }_{\gamma }^{\mathrm{\infty }}(1-{\theta }_3(x)){(x-\gamma )}^{p-1}{f}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{(n-\eta )}^{q-1}{a}_n^q\right\}}^{\frac{1}{q}}.\hfill \end{array}$$

(27)

• (iii) Setting $y=\frac{1}{x-\gamma }+\gamma$, $g(y)={(y-\gamma )}^{\alpha -\beta -2}f(\frac{1}{y-\gamma }+\gamma )$, $\phi (y)={(y-\gamma )}^{p(1-\frac{\alpha -\beta }{2})-1}$ and $\tilde{\phi }(y)=(1-\theta (\frac{1}{y-\gamma }+\gamma ))\phi (y)$ in (18), by simplification, we obtain the following inequality with the homogeneous kernel:

  $$\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{y-\gamma ,n-\eta \})}^{\beta }}{{(max\{y-\gamma ,n-\eta \})}^{\alpha }}g(y)dy>\frac{4}{\alpha +\beta }{\parallel g\parallel }_{p,\tilde{\phi }}{\parallel a\parallel }_{q,\mathrm{\Psi }}.$$

  (28)

It is evident that (28) is equivalent to (18), and then the same constant factor $\frac{4}{\alpha +\beta }$ in (28) is still the best possible. In the same way, we can find the following inequalities equivalent to (28) with the same best possible constant factor $\frac{4}{\alpha +\beta }$:

$${\left\{\sum _{n=1}^{\mathrm{\infty }}{[\mathrm{\Psi }(n)]}^{1-p}{\left[{\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{y-\gamma ,n-\eta \})}^{\beta }g(y)}{{(max\{y-\gamma ,n-\eta \})}^{\alpha }}dy\right]}^p\right\}}^{\frac{1}{p}}>\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\tilde{\phi }},$$

(29)

$${\left\{{\int }_{\gamma }^{\mathrm{\infty }}{[\tilde{\phi }(y)]}^{1-q}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{y-\gamma ,n-\eta \})}^{\beta }{a}_n}{{(max\{y-\gamma ,n-\eta \})}^{\alpha }}\right]}^{q}dy\right\}}^{\frac{1}{q}}>\frac{4}{\alpha +\beta }{\parallel a\parallel }_{q,\mathrm{\Psi }}.$$

(30)

• (iv) Applying the same way in Theorem 2, we still can obtain some particular dual forms as (i) and (ii) and some equivalent inequalities similar to (28), (29), and (30).