# On a half-discrete Hilbert-type inequality similar to Mulholland’s inequality

## Abstract

By using the way of weight functions and Hadamard’s inequality, a half-discrete Hilbert-type inequality similar to Mulholland’s inequality with a best constant factor is given. The extension with multi-parameters, the equivalent forms as well as the operator expressions are also considered.

MSC:26D15.

## 1 Introduction

Assuming that $f,g\in {L}^{2}\left({R}_{+}\right)$, $\parallel f\parallel ={\left\{{\int }_{0}^{\mathrm{\infty }}{f}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{2}}>0$, $\parallel g\parallel >0$, we have the following Hilbert integral inequality (cf. [1]):

${\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{f\left(x\right)g\left(y\right)}{x+y}\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy<\pi \parallel f\parallel \parallel g\parallel ,$
(1)

where the constant factor π is the best possible. If $a={\left\{{a}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$, $b={\left\{{b}_{n}\right\}}_{n=1}^{\mathrm{\infty }}\in {l}^{2}$, $\parallel a\parallel ={\left\{{\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{2}\right\}}^{\frac{1}{2}}>0$, $\parallel b\parallel >0$, then we still have the following 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. [24]). Also we have the following Mulholland inequality with the same best constant factor (cf. [1, 5]):

$\sum _{m=2}^{\mathrm{\infty }}\sum _{n=2}^{\mathrm{\infty }}\frac{{a}_{m}{b}_{n}}{lnmn}<\pi {\left\{\sum _{m=2}^{\mathrm{\infty }}m{a}_{m}^{2}\sum _{n=2}^{\mathrm{\infty }}n{b}_{n}^{2}\right\}}^{\frac{1}{2}}.$
(3)

In 1998, by introducing an independent parameter $\lambda \in \left(0,1\right]$, Yang [6] gave an extension of (1). By generalizing the results from [6], Yang [7] gave some best extensions of (1) and (2) as follows: If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, ${\lambda }_{1}+{\lambda }_{2}=\lambda$, ${k}_{\lambda }\left(x,y\right)$ is a non-negative homogeneous function of degree −λ with $k\left({\lambda }_{1}\right)={\int }_{0}^{\mathrm{\infty }}{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{1}-1}\phantom{\rule{0.2em}{0ex}}dt\in {R}_{+}$, $\varphi \left(x\right)={x}^{p\left(1-{\lambda }_{1}\right)-1}$, $\psi \left(x\right)={x}^{q\left(1-{\lambda }_{2}\right)-1}$, $f\phantom{\rule{0.25em}{0ex}}\left(\ge 0\right)\in {L}_{p,\varphi }\left({R}_{+}\right)=\left\{f|{\parallel f\parallel }_{p,\varphi }:={\left\{{\int }_{0}^{\mathrm{\infty }}\varphi \left(x\right){|f\left(x\right)|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}<\mathrm{\infty }\right\}$, $g\phantom{\rule{0.25em}{0ex}}\left(\ge 0\right)\in {L}_{q,\psi }\left({R}_{+}\right)$, ${\parallel f\parallel }_{p,\varphi },{\parallel g\parallel }_{q,\psi }>0$, then

${\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}{k}_{\lambda }\left(x,y\right)f\left(x\right)g\left(y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy
(4)

where the constant factor $k\left({\lambda }_{1}\right)$ is the best possible. Moreover, if ${k}_{\lambda }\left(x,y\right)$ is finite and ${k}_{\lambda }\left(x,y\right){x}^{{\lambda }_{1}-1}$ (${k}_{\lambda }\left(x,y\right){y}^{{\lambda }_{2}-1}$) is decreasing for $x>0$ ($y>0$), then for ${a}_{m,}{b}_{n}\ge 0$, $a={\left\{{a}_{m}\right\}}_{m=1}^{\mathrm{\infty }}\in {l}_{p,\varphi }=\left\{a|{\parallel a\parallel }_{p,\varphi }:={\left\{{\sum }_{n=1}^{\mathrm{\infty }}\varphi \left(n\right){|{a}_{n}|}^{p}\right\}}^{\frac{1}{p}}<\mathrm{\infty }\right\}$, $b={\left\{{b}_{n}\right\}}_{n=1}^{\mathrm{\infty }}\in {l}_{q,\psi }$, ${\parallel a\parallel }_{p,\varphi },{\parallel b\parallel }_{q,\psi }>0$, we have

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{k}_{\lambda }\left(m,n\right){a}_{m}{b}_{n}
(5)

with the same best constant factor $k\left({\lambda }_{1}\right)$. Clearly, for $p=q=2$, $\lambda =1$, ${k}_{1}\left(x,y\right)=\frac{1}{x+y}$, ${\lambda }_{1}={\lambda }_{2}=\frac{1}{2}$, (4) reduces to (1), while (5) reduces to (2). Some other results about Hilbert-type inequalities are provided by [5, 816].

On the topic of 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 in the inequalities are the best possible. Moreover, Yang [17] gave an inequality with the particular kernel $\frac{1}{{\left(1+nx\right)}^{\lambda }}$ and an interval variable, and proved that the constant factor is the best possible. Recently, [18] and [19] gave the following half-discrete Hilbert inequality with the best constant factor π:

${\int }_{0}^{\mathrm{\infty }}f\left(x\right)\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{n}}{{\left(x+n\right)}^{\lambda }}\phantom{\rule{0.2em}{0ex}}dx<\pi \parallel f\parallel \parallel a\parallel .$
(6)

In this paper, by using the way of weight functions and Hadamard’s inequality, a half-discrete Hilbert-type inequality similar to (3) and (6) with the best constant factor is given as follows:

${\int }_{0}^{\mathrm{\infty }}f\left(x\right)\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{n}}{lne{\left(n+\frac{1}{2}\right)}^{x}}\phantom{\rule{0.2em}{0ex}}dx<\pi \parallel f\parallel {\left\{\sum _{n=1}^{\mathrm{\infty }}\left(n+\frac{1}{2}\right){a}_{n}^{2}\right\}}^{\frac{1}{2}}.$
(7)

Moreover, the best extension of (7) with multi-parameters, some equivalent forms as well as the operator expressions are considered.

## 2 Some lemmas

Lemma 1 If $0<\lambda \le 2$, $\alpha \ge \frac{1}{2}$, setting weight functions $\omega \left(n\right)$ and $\varpi \left(x\right)$ as follows:

$\omega \left(n\right):={ln}^{\frac{\lambda }{2}}\left(n+\alpha \right){\int }_{0}^{\mathrm{\infty }}\frac{{x}^{\frac{\lambda }{2}-1}}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\phantom{\rule{0.2em}{0ex}}dx,\phantom{\rule{1em}{0ex}}n\in \mathbf{N},$
(8)
$\varpi \left(x\right):={x}^{\frac{\lambda }{2}}\sum _{n=1}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}\left(n+\alpha \right)}{\left(n+\alpha \right){ln}^{\lambda }e{\left(n+\alpha \right)}^{x}},\phantom{\rule{1em}{0ex}}x\in \left(0,\mathrm{\infty }\right),$
(9)

we have

$\varpi \left(x\right)<\omega \left(n\right)=B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right).$
(10)

Proof Substitution of $t=xln\left(n+\alpha \right)$ in (8), by calculation, yields

$\omega \left(n\right)={\int }_{0}^{\mathrm{\infty }}\frac{1}{{\left(1+t\right)}^{\lambda }}{t}^{\frac{\lambda }{2}-1}\phantom{\rule{0.2em}{0ex}}dt=B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right).$

Since, for fixed $x>0$ and in view of the conditions,

$\begin{array}{rcl}h\left(x,y\right)& :=& \frac{{ln}^{\frac{\lambda }{2}-1}\left(y+\alpha \right)}{\left(y+\alpha \right){ln}^{\lambda }e{\left(y+\alpha \right)}^{x}}\\ =& \frac{{ln}^{\frac{\lambda }{2}-1}\left(y+\alpha \right)}{\left(y+\alpha \right){\left[1+xln\left(y+\alpha \right)\right]}^{\lambda }}\end{array}$

is decreasing and strictly convex for $y\in \left(\frac{1}{2},\mathrm{\infty }\right)$, then by Hadamard’s inequality (cf. [20]), we find

$\begin{array}{rcl}\varpi \left(x\right)& <& {x}^{\frac{\lambda }{2}}{\int }_{\frac{1}{2}}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}\left(y+\alpha \right)}{\left(y+\alpha \right){\left[1+xln\left(y+\alpha \right)\right]}^{\lambda }}\phantom{\rule{0.2em}{0ex}}dy\\ \stackrel{t=xln\left(y+\alpha \right)}{=}& {\int }_{xln\left(\frac{1}{2}+\alpha \right)}^{\mathrm{\infty }}\frac{{t}^{\frac{\lambda }{2}-1}}{{\left(1+t\right)}^{\lambda }}\phantom{\rule{0.2em}{0ex}}dt\le B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right),\end{array}$

namely, (10) follows. □

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

$\begin{array}{c}J:={\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{{ln}^{\frac{p\lambda }{2}-1}\left(n+\alpha \right)}{n+\alpha }{\left[{\int }_{0}^{\mathrm{\infty }}\frac{f\left(x\right)}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\phantom{\rule{0.2em}{0ex}}dx\right]}^{p}\right\}}^{\frac{1}{p}}\hfill \\ \phantom{J}\le {\left[B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\right]}^{\frac{1}{q}}{\left\{{\int }_{0}^{\mathrm{\infty }}\varpi \left(x\right){x}^{p\left(1-\frac{\lambda }{2}\right)-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}},\hfill \end{array}$
(11)
$\begin{array}{c}{L}_{1}:={\left\{{\int }_{0}^{\mathrm{\infty }}\frac{{x}^{\frac{q\lambda }{2}-1}}{{\left[\varpi \left(x\right)\right]}^{q-1}}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{n}}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\right]}^{q}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{q}}\hfill \\ \phantom{{L}_{1}}\le {\left\{B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\sum _{n=1}^{\mathrm{\infty }}{\left(n+\alpha \right)}^{q-1}{ln}^{q\left(1-\frac{\lambda }{2}\right)-1}\left(n+\alpha \right){a}_{n}^{q}\right\}}^{\frac{1}{q}}.\hfill \end{array}$
(12)

Proof By Hölder’s inequality (cf. [20]) and (10), it follows

$\begin{array}{c}{\left[{\int }_{0}^{\mathrm{\infty }}\frac{f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\right]}^{p}\hfill \\ \phantom{\rule{1em}{0ex}}={\left\{{\int }_{0}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\left[\frac{{x}^{\left(1-\frac{\lambda }{2}\right)/q}}{{ln}^{\left(1-\frac{\lambda }{2}\right)/p}\left(n+\alpha \right)}\frac{f\left(x\right)}{{\left(n+\alpha \right)}^{\frac{1}{p}}}\right]\left[\frac{{ln}^{\left(1-\frac{\lambda }{2}\right)/p}\left(n+\alpha \right)}{{x}^{\left(1-\frac{\lambda }{2}\right)/q}}{\left(n+\alpha \right)}^{\frac{1}{p}}\right]\phantom{\rule{0.2em}{0ex}}dx\right\}}^{p}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\int }_{0}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}\left(n+\alpha \right)}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\frac{{x}^{\left(1-\frac{\lambda }{2}\right)\left(p-1\right)}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{n+\alpha }{\left\{{\int }_{0}^{\mathrm{\infty }}\frac{{\left(n+\alpha \right)}^{q-1}}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\frac{{ln}^{\left(1-\frac{\lambda }{2}\right)\left(q-1\right)}\left(n+\alpha \right)}{{x}^{1-\frac{\lambda }{2}}}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{p-1}\hfill \\ \phantom{\rule{1em}{0ex}}={\left\{\frac{\omega \left(n\right){\left(n+\alpha \right)}^{q-1}}{{ln}^{q\left(\frac{\lambda }{2}-1\right)+1}\left(n+\alpha \right)}\right\}}^{p-1}{\int }_{0}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}\left(n+\alpha \right)}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\frac{{x}^{\left(1-\frac{\lambda }{2}\right)\left(p-1\right)}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{n+\alpha }\hfill \\ \phantom{\rule{1em}{0ex}}={\left[B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\right]}^{p-1}\frac{n+\alpha }{{ln}^{\frac{p\lambda }{2}-1}\left(n+\alpha \right)}{\int }_{0}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}\left(n+\alpha \right)}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\frac{{x}^{\left(1-\frac{\lambda }{2}\right)\left(p-1\right)}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{n+\alpha }.\hfill \end{array}$

Then by the Lebesgue term-by-term integration theorem (cf. [21]), we have

$\begin{array}{rcl}J& \le & {\left[B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\right]}^{\frac{1}{q}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}\left(n+\alpha \right)}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\frac{{x}^{\left(1-\frac{\lambda }{2}\right)\left(p-1\right)}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{n+\alpha }\right\}}^{\frac{1}{p}}\\ =& {\left[B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\right]}^{\frac{1}{q}}{\left\{{\int }_{0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}\left(n+\alpha \right)}{{ln}^{\lambda }e{\left(\left(n+\alpha \right)\right)}^{x}}\frac{{x}^{\left(1-\frac{\lambda }{2}\right)\left(p-1\right)}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{n+\alpha }\right\}}^{\frac{1}{p}}\\ =& {\left[B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\right]}^{\frac{1}{q}}{\left\{{\int }_{0}^{\mathrm{\infty }}\varpi \left(x\right){x}^{p\left(1-\frac{\lambda }{2}\right)-1}{f}^{p}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}},\end{array}$

and (11) follows. Still by Hölder’s inequality, we have

$\begin{array}{c}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{n}}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\right]}^{q}\hfill \\ \phantom{\rule{1em}{0ex}}={\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\left[\frac{{x}^{\left(1-\frac{\lambda }{2}\right)/q}}{{ln}^{\left(1-\frac{\lambda }{2}\right)/p}\left(n+\alpha \right)}\frac{1}{{\left(n+\alpha \right)}^{\frac{1}{p}}}\right]\left[\frac{{ln}^{\left(1-\frac{\lambda }{2}\right)/p}\left(n+\alpha \right)}{{x}^{\left(1-\frac{\lambda }{2}\right)/q}}{\left(n+\alpha \right)}^{\frac{1}{p}}{a}_{n}\right]\right\}}^{q}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}\left(n+\alpha \right)}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\frac{{x}^{\left(1-\frac{\lambda }{2}\right)\left(p-1\right)}}{\left(n+\alpha \right)}\right\}}^{q-1}\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(n+\alpha \right)}^{q-1}}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\frac{{ln}^{\left(1-\frac{\lambda }{2}\right)\left(q-1\right)}\left(n+\alpha \right)}{{x}^{1-\frac{\lambda }{2}}}{a}_{n}^{q}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{\left[\varpi \left(x\right)\right]}^{q-1}}{{x}^{\frac{q\lambda }{2}-1}}\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(n+\alpha \right)}^{q-1}}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}{x}^{\frac{\lambda }{2}-1}{ln}^{\left(1-\frac{\lambda }{2}\right)\left(q-1\right)}\left(n+\alpha \right){a}_{n}^{q}.\hfill \end{array}$

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

$\begin{array}{rcl}{L}_{1}& \le & {\left\{{\int }_{0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(n+\alpha \right)}^{q-1}}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}{x}^{\frac{\lambda }{2}-1}{ln}^{\left(1-\frac{\lambda }{2}\right)\left(q-1\right)}\left(n+\alpha \right){a}_{n}^{q}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{q}}\\ =& {\left\{\sum _{n=1}^{\mathrm{\infty }}\left[{ln}^{\frac{\lambda }{2}}\left(n+\alpha \right){\int }_{0}^{\mathrm{\infty }}\frac{{x}^{\frac{\lambda }{2}-1}\phantom{\rule{0.2em}{0ex}}dx}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\right]{\left(n+\alpha \right)}^{q-1}{ln}^{q\left(1-\frac{\lambda }{2}\right)-1}\left(n+\alpha \right){a}_{n}^{q}\right\}}^{\frac{1}{q}}\\ =& {\left\{\sum _{n=1}^{\mathrm{\infty }}\omega \left(n\right){\left(n+\alpha \right)}^{q-1}{ln}^{q\left(1-\frac{\lambda }{2}\right)-1}\left(n+\alpha \right){a}_{n}^{q}\right\}}^{\frac{1}{q}},\end{array}$

and then in view of (10), inequality (12) follows. □

## 3 Main results

We introduce two functions

$\mathrm{\Phi }\left(x\right):={x}^{p\left(1-\frac{\lambda }{2}\right)-1}\phantom{\rule{1em}{0ex}}\left(x>0\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\Psi }\left(n\right):={\left(n+\alpha \right)}^{q-1}{ln}^{q\left(1-\frac{\lambda }{2}\right)-1}\left(n+\alpha \right)\phantom{\rule{1em}{0ex}}\left(n\in \mathbf{N}\right),$

wherefrom, ${\left[\mathrm{\Phi }\left(x\right)\right]}^{1-q}={x}^{\frac{q\lambda }{2}-1}$, and ${\left[\mathrm{\Psi }\left(n\right)\right]}^{1-p}=\frac{{ln}^{\frac{p\lambda }{2}-1}\left(n+\alpha \right)}{n+\alpha }$.

Theorem 1 If  $0<\lambda \le 2$, $\alpha \ge \frac{1}{2}$, $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f\left(x\right),{a}_{n}\ge 0$, $f\in {L}_{p,\mathrm{\Phi }}\left({R}_{+}\right)$, $a={\left\{{a}_{n}\right\}}_{n=1}^{\mathrm{\infty }}\in {l}_{q,\mathrm{\Psi }}$, ${\parallel f\parallel }_{p,\mathrm{\Phi }}>0$ and ${\parallel a\parallel }_{q,\mathrm{\Psi }}>0$, then we have the following equivalent inequalities:

$\begin{array}{c}I:=\sum _{n=1}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{{a}_{n}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\hfill \\ \phantom{I}={\int }_{0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{n}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}
(13)
$\begin{array}{c}J={\left\{\sum _{n=1}^{\mathrm{\infty }}{\left[\mathrm{\Psi }\left(n\right)\right]}^{1-p}{\left[{\int }_{0}^{\mathrm{\infty }}\frac{f\left(x\right)}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\phantom{\rule{0.2em}{0ex}}dx\right]}^{p}\right\}}^{\frac{1}{p}}\hfill \\ \phantom{J}
(14)
$\begin{array}{c}L:={\left\{{\int }_{0}^{\mathrm{\infty }}{\left[\mathrm{\Phi }\left(x\right)\right]}^{1-q}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{n}}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}\right]}^{q}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{q}}\hfill \\ \phantom{L}
(15)

where the constant $B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$ is the best possible in the above inequalities.

Proof By the Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for $\varpi \left(x\right), we have (14). By Hölder’s inequality, we have

$I=\sum _{n=1}^{\mathrm{\infty }}\left[{\mathrm{\Psi }}^{\frac{-1}{q}}\left(n\right){\int }_{0}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right]\left[{\mathrm{\Psi }}^{\frac{1}{q}}\left(n\right){a}_{n}\right]\le J{\parallel a\parallel }_{q,\mathrm{\Psi }}.$
(16)

Then by (14), we have (13). On the other hand, assuming that (13) is valid, setting

${a}_{n}:={\left[\mathrm{\Psi }\left(n\right)\right]}^{1-p}{\left[{\int }_{0}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right]}^{p-1},\phantom{\rule{1em}{0ex}}n\in \mathbf{N},$

then ${J}^{p-1}={\parallel a\parallel }_{q,\mathrm{\Psi }}$. By (11), we find $J<\mathrm{\infty }$. If $J=0$, then (14) is trivially valid; if $J>0$, then by (13), we have

$\begin{array}{c}{\parallel a\parallel }_{q,\mathrm{\Psi }}^{q}={J}^{p}=I

that is, (14) is equivalent to (13). In view of (12), for ${\left[\varpi \left(x\right)\right]}^{1-q}>{\left[B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\right]}^{1-q}$, we have (15). By Hölder’s inequality, we find

$I={\int }_{0}^{\mathrm{\infty }}\left[{\mathrm{\Phi }}^{\frac{1}{p}}\left(x\right)f\left(x\right)\right]\left[{\mathrm{\Phi }}^{\frac{-1}{p}}\left(x\right)\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}{a}_{n}\right]\phantom{\rule{0.2em}{0ex}}dx\le {\parallel f\parallel }_{p,\mathrm{\Phi }}L.$
(17)

Then by (15), we have (13). On the other hand, assuming that (13) is valid, setting

$f\left(x\right):={\left[\mathrm{\Phi }\left(x\right)\right]}^{1-q}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}{a}_{n}\right]}^{q-1},\phantom{\rule{1em}{0ex}}x\in \left(0,\mathrm{\infty }\right),$

then ${L}^{q-1}={\parallel f\parallel }_{p,\mathrm{\Phi }}$. By (12), we find $L<\mathrm{\infty }$. If $L=0$, then (15) is trivially valid; if $L>0$, then by (13), we have

$\begin{array}{c}{\parallel f\parallel }_{p,\mathrm{\Phi }}^{p}={L}^{q}=I

that is, (15) is equivalent to (13). Hence, inequalities (13), (14) and (15) are equivalent.

For $0<\epsilon <\frac{p\lambda }{2}$, setting $\stackrel{˜}{f}\left(x\right)={x}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}$, $x\in \left(0,1\right)$; $\stackrel{˜}{f}\left(x\right)=0$, $x\in \left[1,\mathrm{\infty }\right)$, and ${\stackrel{˜}{a}}_{n}=\frac{1}{n+\alpha }{ln}^{\frac{\lambda }{2}-\frac{\epsilon }{q}-1}\left(n+\alpha \right)$, $n\in \mathbf{N}$, if there exists a positive number k ($\le B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$) such that (13) is valid as we replace $B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$ with k, then, in particular, it follows

$\begin{array}{c}\stackrel{˜}{I}:=\sum _{n=1}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}{\stackrel{˜}{a}}_{n}\stackrel{˜}{f}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx
(18)
$\begin{array}{rcl}\stackrel{˜}{I}& =& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n+\alpha }{ln}^{\frac{\lambda }{2}-\frac{\epsilon }{q}-1}\left(n+\alpha \right){\int }_{0}^{1}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}{x}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}\phantom{\rule{0.2em}{0ex}}dx\\ \stackrel{t=xln\left(n+\alpha \right)}{=}& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n+\alpha \right){ln}^{\epsilon +1}\left(n+\alpha \right)}{\int }_{0}^{ln\left(n+\alpha \right)}\frac{1}{{\left(t+1\right)}^{\lambda }}{t}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}\phantom{\rule{0.2em}{0ex}}dt\\ =& B\left(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p}\right)\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n+\alpha \right){ln}^{\epsilon +1}\left(n+\alpha \right)}-A\left(\epsilon \right)\\ >& B\left(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p}\right){\int }_{1}^{\mathrm{\infty }}\frac{1}{\left(y+\alpha \right){ln}^{\epsilon +1}\left(y+\alpha \right)}\phantom{\rule{0.2em}{0ex}}dy-A\left(\epsilon \right)\\ =& \frac{1}{\epsilon {ln}^{\epsilon }\left(1+\alpha \right)}B\left(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p}\right)-A\left(\epsilon \right),\end{array}$
$A\left(\epsilon \right):=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n+\alpha \right){ln}^{\epsilon +1}\left(n+\alpha \right)}{\int }_{ln\left(n+\alpha \right)}^{\mathrm{\infty }}\frac{1}{{\left(t+1\right)}^{\lambda }}{t}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}\phantom{\rule{0.2em}{0ex}}dt.$
(19)

We find

$\begin{array}{rcl}0& <& A\left(\epsilon \right)\le \sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n+\alpha \right){ln}^{\epsilon +1}\left(n+\alpha \right)}{\int }_{ln\left(n+\alpha \right)}^{\mathrm{\infty }}\frac{1}{{t}^{\lambda }}{t}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}\phantom{\rule{0.2em}{0ex}}dt\\ =& \frac{1}{\frac{\lambda }{2}-\frac{\epsilon }{p}}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n+\alpha \right){ln}^{\frac{\lambda }{2}+\frac{\epsilon }{q}+1}\left(n+\alpha \right)}<\mathrm{\infty },\end{array}$

and then $A\left(\epsilon \right)=O\left(1\right)$ ($\epsilon \to {0}^{+}$). Hence by (18) and (19), it follows

$\begin{array}{c}\frac{1}{{ln}^{\epsilon }\left(1+\alpha \right)}B\left(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p}\right)-\epsilon O\left(1\right)\hfill \\ \phantom{\rule{1em}{0ex}}
(20)

and $B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)\le k$ ($\epsilon \to {0}^{+}$). Hence $k=B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$ is the best value of (13).

By equivalence, the constant factor $B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$ in (14) and (15) is the best possible. Otherwise, we can imply a contradiction by (16) and (17) that the constant factor in (13) is not the best possible. □

Remark 1 (i) Define the first type half-discrete Hilbert-type operator ${T}_{1}:{L}_{p,\mathrm{\Phi }}\left({R}_{+}\right)\to {l}_{p,{\mathrm{\Psi }}^{1-p}}$ as follows: For $f\in {L}_{p,\mathrm{\Phi }}\left({R}_{+}\right)$, we define ${T}_{1}f\in {l}_{p,{\mathrm{\Psi }}^{1-p}}$, satisfying

${T}_{1}f\left(n\right)={\int }_{0}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx,\phantom{\rule{1em}{0ex}}n\in \mathbf{N}.$

Then by (14) it follows ${\parallel {T}_{1}f\parallel }_{p.{\mathrm{\Psi }}^{1-p}}\le B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right){\parallel f\parallel }_{p,\mathrm{\Phi }}$, and then ${T}_{1}$ is a bounded operator with $\parallel {T}_{1}\parallel \le B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$. Since by Theorem 1 the constant factor in (14) is the best possible, we have $\parallel {T}_{1}\parallel =B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$.

1. (ii)

Define the second type half-discrete Hilbert-type operator ${T}_{2}:{l}_{q,\mathrm{\Psi }}\to {L}_{q,{\mathrm{\Phi }}^{1-q}}\left({R}_{+}\right)$ as follows: For $a\in {l}_{q,\mathrm{\Psi }}$, we define ${T}_{2}a\in {L}_{q,{\mathrm{\Phi }}^{1-q}}\left({R}_{+}\right)$, satisfying

${T}_{2}a\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{\left(n+\alpha \right)}^{x}}{a}_{n},\phantom{\rule{1em}{0ex}}x\in \left(0,\mathrm{\infty }\right).$

Then by (15) it follows ${\parallel {T}_{2}a\parallel }_{q,{\mathrm{\Phi }}^{1-q}}\le B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right){\parallel a\parallel }_{q,\mathrm{\Psi }}$, and then ${T}_{2}$ is a bounded operator with $\parallel {T}_{2}\parallel \le B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$. Since by Theorem 1 the constant factor in (15) is the best possible, we have $\parallel {T}_{2}\parallel =B\left(\frac{\lambda }{2},\frac{\lambda }{2}\right)$.

Remark 2 For $p=q=2$, $\lambda =1$, ${\lambda }_{1}={\lambda }_{2}=\frac{1}{2}$, $\alpha =\frac{1}{2}$ in (13), (14) and (15), we have (7) and the following equivalent inequalities:

${\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n+\frac{1}{2}}{\left[{\int }_{0}^{\mathrm{\infty }}\frac{f\left(x\right)}{lne{\left(n+\frac{1}{2}\right)}^{x}}\phantom{\rule{0.2em}{0ex}}dx\right]}^{2}\right\}}^{\frac{1}{2}}<\pi \parallel f\parallel ,$
(21)
${\left\{{\int }_{0}^{\mathrm{\infty }}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{{a}_{n}}{lne{\left(n+\frac{1}{2}\right)}^{x}}\right]}^{2}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{2}}<\pi {\left\{\sum _{n=1}^{\mathrm{\infty }}\left(n+\frac{1}{2}\right){a}_{n}^{2}\right\}}^{\frac{1}{2}}.$
(22)

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## Acknowledgements

This work is supported by 2012 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2012KJCX0079).

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Correspondence to Bicheng Yang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

ZH wrote and reformed the article. BY conceived of the study, and participated in its design and coordination. All authors read and approved the final manuscript.

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Huang, Z., Yang, B. On a half-discrete Hilbert-type inequality similar to Mulholland’s inequality. J Inequal Appl 2013, 290 (2013). https://doi.org/10.1186/1029-242X-2013-290