A more accurate half-discrete 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 Hilbert-type inequality with a non-homogeneous kernel and a best constant factor is given. A best extension, some equivalent forms, the operator expressions as well as some particular cases are also considered.

MSC:26D15, 47A07.

1 Introduction

Assuming that $f,g\in L^2({\mathbf{R}}_{+})$, $\parallel f\parallel ={\{{\int }_0^{\mathrm{\infty }}{f}^2(x)dx\}}^{\frac{1}{2}}>0$, $\parallel g\parallel >0$, we have the following Hilbert’s 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’s 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) as follows. 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 $k({\lambda }_1)={\int }_0^{\mathrm{\infty }}{k}_{\lambda }(t,1)t^{{\lambda }_1-1}dt\in {\mathbf{R}}_{+}$, $\varphi (x)=x^{p(1-{\lambda }_1)-1}$, $\psi (x)=x^{q(1-{\lambda }_2)-1}$, $f(\ge 0)\in L_{p,\varphi }({\mathbf{R}}_{+})=\{f|{\parallel f\parallel }_{p,\varphi }:={\{{\int }_0^{\mathrm{\infty }}\varphi (x){|f(x)|}^{p}dx\}}^{\frac{1}{p}}<\mathrm{\infty }\}$, $g(\ge 0)\in L_{q,\psi }({\mathbf{R}}_{+})$, 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 $k_{\lambda }(x,y)$ 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 the 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).

Some other results about integral and discrete Hilbert-type inequalities can be found in [7–16]. 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 [17] 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 [18] and [19] gave the following half-discrete Hilbert’s inequality with the best constant factor:

$${\int }_0^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{(x+n)}^{\lambda }}dx<\pi \parallel f\parallel \parallel a\parallel ;$$

(5)

Chen [20] and Yang [21] gave two more accurate half-discrete Mulholland’s inequalities by using Hadamard’s inequality.

In this paper, by means of weight functions and the improved Euler-Maclaurin summation formula, a more accurate half-discrete Hilbert-type inequality with a non-homogeneous kernel and a best constant factor is given as follows. For $0<\alpha +\beta \le 2$, $\gamma \in \mathbf{R}$, $\eta \le 1-\frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})$,

(6)

Moreover, a best extension of (6), some equivalent forms, the operator expressions as well as some particular inequalities are 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 decreasing continuous 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:

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

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

(7)

where $\rho (y)=y-[y]-\frac{1}{2}$ is a 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 decreasing continuous function $\tilde{g}(y)$ as

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

Then it follows

Since $\tilde{g}(n_0)=g_1(n_0)+g_2(s)-g_1(s-0)>0$, $\tilde{g}(y)$ is a non-constant decreasing continuous 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

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

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

(10)

(11)

then we have

$$\varpi (x)<\omega (n)=\frac{4}{\alpha +\beta }.$$

(12)

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

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

For fixed $x>\gamma$, we find

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

(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 $g(y):=-h_y^{\mathrm{\prime }}(x,y)$, 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

then by (7), we find

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

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

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

• (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

$$\begin{array}{rcl}-{\int }_1^{\mathrm{\infty }}\rho (y)h_y^{\mathrm{\prime }}(x,y)dy& =& (\frac{\alpha +\beta }{2}+1){(x-\gamma )}^{-\frac{\alpha +\beta }{2}}{\int }_1^{\mathrm{\infty }}\rho (y){(y-\eta )}^{-\frac{\alpha +\beta }{2}-2}dy\\ >& -\frac{1}{12}(\frac{\alpha +\beta }{2}+1){(x-\gamma )}^{-\frac{\alpha +\beta }{2}}{(1-\eta )}^{-\frac{\alpha +\beta }{2}-2}.\end{array}$$

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

$$\begin{array}{rcl}R(x)& >& \frac{2}{\alpha +\beta }{(1-\eta )}^{-\frac{\alpha +\beta }{2}}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}-\frac{1}{2}{(1-\eta )}^{-\frac{\alpha +\beta }{2}-1}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}\\ -\frac{1}{12}(\frac{\alpha +\beta }{2}+1){(1-\eta )}^{-\frac{\alpha +\beta }{2}-2}{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}\\ >& [\frac{2{(1-\eta )}^2}{\alpha +\beta }-\frac{1-\eta }{2}-\frac{2+\alpha +\beta }{16}]\frac{{(x-\gamma )}^{-\frac{\alpha +\beta }{2}}}{{(1-\eta )}^{2+\frac{\alpha +\beta }{2}}}\ge 0.\end{array}$$

Hence, for $x>\gamma$, we have $R(x)>0$, and then (12) follows. □

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

(14)

(15)

Proof Setting $k(x,n):=\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}$, by Hölder’s inequality (cf. [22]) and (12), it follows

Then by the Lebesgue term-by-term integration theorem (cf. [23]), we have

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

Hence, (14) follows. By Hölder’s inequality again, we have

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

By the Lebesgue term-by-term integration theorem, we have

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

and in view of (12), inequality (15) follows. □

Lemma 4 Let the assumptions of Lemma 2 be fulfilled and, additionally, let $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $0<\epsilon <\frac{p}{2}(\alpha +\beta )$. 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

(16)

(17)

Proof We find

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

and then (16) is valid. We obtain

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

and so (17) is valid. □

3 Main results

We introduce the functions

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

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

Theorem 5 If $0<\alpha +\beta \le 2$, $\gamma \in \mathbf{R}$, $\eta \le 1-\frac{\alpha +\beta }{8}(1+\sqrt{3+\frac{4}{\alpha +\beta }})$, $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f(x),a_n\ge 0$, $f\in L_{p,\mathrm{\Phi }}(\gamma ,\mathrm{\infty })$, $a={\{a_n\}}_{n=1}^{\mathrm{\infty }}\in l_{q,\mathrm{\Psi }}$, ${\parallel f\parallel }_{p,\mathrm{\Phi }}>0$ and ${\parallel a\parallel }_{q,\mathrm{\Psi }}>0$, then we have the following equivalent inequalities:

(18)

(19)

(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 the Lebesgue term-by-term integration theorem. By (14) and (12), we have (19). By Hölder’s inequality, we have

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

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

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

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

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

therefore ${\parallel a\parallel }_{q,\mathrm{\Psi }}^{q-1}=J<\frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\mathrm{\Phi }}$; that is, (19) is equivalent to (18). On the other hand, by (12) we have ${[\varpi (x)]}^{1-q}>{(\frac{4}{\alpha +\beta })}^{1-q}$. Then in view of (15), we have (20). By Hölder’s inequality, we find

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

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

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

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

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

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

If there exists a positive number k ($\le \frac{4}{\alpha +\beta }$) 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

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

and $\frac{4}{\alpha +\beta }\le k$ ($\epsilon \to 0^{+}$). 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. □

Remark 1 (i) Define the first type half-discrete Hilbert-type operator $T_1:L_{p,\mathrm{\Phi }}(\gamma ,\mathrm{\infty })\to l_{p,{\mathrm{\Psi }}^{1-p}}$ as follows. For $f\in L_{p,\mathrm{\Phi }}(\gamma ,\mathrm{\infty })$, we define $T_{1}f\in l_{p,{\mathrm{\Psi }}^{1-p}}$ by

$$T_{1}f(n)={\int }_{\gamma }^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}f(x)dx,n\in \mathbf{N}.$$

Then by (19), ${\parallel T_{1}f\parallel }_{p,{\mathrm{\Psi }}^{1-p}}\le \frac{4}{\alpha +\beta }{\parallel f\parallel }_{p,\mathrm{\Phi }}$ and so $T_1$ is a bounded operator with $\parallel T_1\parallel \le \frac{4}{\alpha +\beta }$. Since by Theorem 5 the constant factor in (19) is best possible, we have $\parallel T_1\parallel =\frac{4}{\alpha +\beta }$.

• (ii) Define the second type half-discrete Hilbert-type operator $T_2:l_{q,\mathrm{\Psi }}\to L_{q,{\mathrm{\Phi }}^{1-q}}(\gamma ,\mathrm{\infty })$ as follows. For $a\in l_{q,\mathrm{\Psi }}$, we define $T_{2}a\in L_{q,{\mathrm{\Phi }}^{1-q}}(\gamma ,\mathrm{\infty })$ by

  $$T_{2}a(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(min\{1,(x-\gamma )(n-\eta )\})}^{\beta }}{{(max\{1,(x-\gamma )(n-\eta )\})}^{\alpha }}{a}_n,x\in (\gamma ,\mathrm{\infty }).$$

Then by (20), ${\parallel T_{2}a\parallel }_{q,{\mathrm{\Phi }}^{1-q}}\le \frac{4}{\alpha +\beta }{\parallel a\parallel }_{q,\mathrm{\Psi }}$ and so $T_2$ is a bounded operator with $\parallel T_2\parallel \le \frac{4}{\alpha +\beta }$. Since by Theorem 5 the constant factor in (20) is best possible, we have $\parallel T_2\parallel =\frac{4}{\alpha +\beta }$.

Remark 2 (i) For $p=q=2$, (18) reduces to (6). Since we find

then for $\eta =\gamma =0$ in (18), we have the following inequality:

(21)

Hence, (18) is a more accurate inequality of (21).

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

  

  (22)

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

(23)

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

(24)