Skip to content

Advertisement

  • Research
  • Open Access

A more accurate reverse half-discrete Hilbert-type inequality

Journal of Inequalities and Applications20152015:85

https://doi.org/10.1186/s13660-015-0613-8

Received: 8 December 2014

Accepted: 25 February 2015

Published: 5 March 2015

Abstract

Using the way of weight functions and the idea of introducing parameters, and by means of Hermite-Hadamard’s inequality, a more accurate reverse half-discrete Hilbert-type inequality with the non-homogeneous kernel and a best constant factor is established. In addition, its best extension with parameters, the equivalent forms, as well as some particular cases are given.

Keywords

  • weight function
  • non-homogeneous kernel
  • reverse
  • equivalent form
  • Hermite-Hadamard’s inequality

MSC

  • 26D15
  • 47A07

1 Introduction

If \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(a_{n},b_{n}\geq0\), \(a=\{a_{n}\}_{n=1}^{\infty}\in l^{p}\), \(b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}\), \(\| a\|_{p}=\{\sum_{n=1}^{\infty}a_{n}^{p}\}^{1/p}>0\), and \(\| b\|_{q}=\{\sum_{n=1}^{\infty}b_{n}^{q}\}^{1/q}>0\), then we have the following famous discrete Hilbert-type inequality (cf. [1]):
$$ \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{\ln (m/n)}{m-n}a_{m}b_{n}< \biggl[\frac{\pi}{\sin(\pi/p)}\biggr]^{2}\| a\|_{p}\| b \|_{q}, $$
(1)
where the constant factor \([\pi/\sin(\pi/p)]^{2}\) is the best possible. Moreover the integral analogue of inequality (1) is given as follows (cf. [1]): If \(p>1\), \(\frac{1}{p}+\frac {1}{q}=1\), \(f(x),g(x)\geq0\), \(f\in L^{p}(0,\infty)\), \(g\in L^{q}(0,\infty)\), \(\| f\| _{p}=\{\int_{0}^{\infty}f^{p}(x)\,dx\}^{\frac{1}{p}}>0\), \(\| g\| _{q}=\{\int_{0}^{\infty}g^{q}(x)\,dx\}^{\frac{1}{q}}>0\), then
$$ \int_{0}^{\infty}\int_{0}^{\infty} \frac{\ln (x/y)}{x-y}f(x)g(y)\,dx\,dy< \biggl[\frac{\pi}{\sin(\pi/p)}\biggr]^{2}\| f \|_{p}\| g\|_{q}, $$
(2)
with the same best constant factor \([\pi/\sin(\pi/p)]^{2}\).
In 2006, Yang proved the following more accurate inequality of (1) (cf. [2]): If \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\frac {1}{2}\leq\alpha \leq1\), \(a_{n},b_{n}\geq0\), such that \(0<\| a\|_{p}<\infty\) and \(0<\| b\|_{q}<\infty\), then
$$ \sum_{n=0}^{\infty}\sum _{m=0}^{\infty}\frac{\ln(\frac{m+\alpha }{n+\alpha })}{m-n}a_{m}b_{n}< \biggl[\frac{\pi}{\sin(\pi/p)}\biggr]^{2}\| a\| _{p}\| b \|_{q}, $$
(3)
where the constant factor \([\pi/\sin(\pi/p)]^{2}\) is still the best possible. Inequalities (1)-(3) are important in mathematical analysis and its applications [3]. There are lots of improvements, generalizations, and applications of inequalities (1)-(3); for more details, refer to [412].
At present, the research on the half-discrete Hilbert-type inequalities has gradually become in focus. We find a few results on the half-discrete Hilbert-type inequalities with the non-homogeneous kernel, which were published early (cf. [1], Theorem 351 and [13]). Recently, Yang gave some half-discrete Hilbert-type inequalities (cf. [1417]). In 2011, Zhong proved a half-discrete Hilbert-type inequality with the non-homogeneous kernel as follows (cf. [18]): If \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\lambda\leq2\), \(a_{n},f(x)\geq0\), \(f(x)\) is a measurable function in \((0,\infty)\), such that \(0<\sum_{n=1}^{\infty}n^{p(1-\frac {\lambda}{2})-1}a_{n}^{p}<\infty\) and \(0<\int_{0}^{\infty}x^{q(1-\frac {\lambda}{2})-1}f^{q}(x)\,dx<\infty\), then
$$\begin{aligned} &\sum_{n=1}^{\infty}\int_{0}^{\infty} \frac{\ln(nx)}{(nx)^{\lambda }-1}f(x)a_{n}\,dx \\ &\quad< \biggl(\frac{\pi}{\lambda}\biggr)^{2}\Biggl\{ \sum _{n=1}^{\infty}n^{p(1-\frac {\lambda}{2})-1}a_{n}^{p} \Biggr\} ^{\frac{1}{p}}\biggl\{ \int_{0}^{\infty}x^{q(1-\frac{\lambda }{2})-1}f^{q}(x)\,dx \biggr\} ^{\frac{1}{q}}, \end{aligned}$$
(4)
where the constant factor \((\frac{\pi}{\lambda})^{2}\) is the best possible.
In this article, using the way of weight functions and the idea of introducing parameters, and by means of Hadamard’s inequality, we give a more accurate reverse inequality of (4) with a best constant factor as follows: For \(p<0\), \(\frac{1}{p}+\frac{1}{q}=1\), we have
$$\begin{aligned} &\sum_{n=1}^{\infty}a_{n}\int _{0}^{\infty}\frac{\ln[x(n-\frac {1}{2})]}{x^{\lambda}(n-\frac{1}{2})^{\lambda}-1}f(x)\,dx \\ &\quad>\biggl(\frac{\pi}{\lambda}\biggr)^{2}\biggl\{ \int _{0}^{\infty}x^{p(1-\frac{\lambda }{2})-1}f^{p}(x)\,dx\biggr\} ^{1/p}\Biggl\{ \sum_{n=1}^{\infty} \biggl(n-\frac{1}{2}\biggr)^{q(1-\frac{\lambda}{2})-1}a_{n}^{q} \Biggr\} ^{1/q}. \end{aligned}$$
(5)
The main objective of this paper is to consider its best extension with parameters, the equivalent forms, as well as some particular cases.

2 Some lemmas

Lemma 1

If \(0<\lambda_{1}<1\), \(\lambda_{1}+\lambda_{2}=1\), then we have the following expression for the Beta function (cf. [1]):
$$ \int_{0}^{\infty}\frac{\ln u}{u-1}u^{\lambda_{1}-1}\,du= \bigl[B(\lambda _{1},\lambda_{2})\bigr]^{2}=\biggl[ \frac{\pi}{\sin(\pi\lambda_{1})}\biggr]^{2}. $$
(6)

Lemma 2

Suppose that \(\lambda>0\), \(0<\sigma<\lambda\leq1\), \(\beta _{1}\in(-\infty,\infty)\), \(0\leq\beta_{2}\leq\frac{1}{2}\), \(\delta\in \{-1,1\}\). We define the weight functions \(\omega_{\sigma}(n)\) and \(\widetilde{\omega}_{\sigma}(x)\) as follows:
$$\begin{aligned}& \omega_{\sigma}(n):=(n-\beta_{2})^{\sigma}\int _{\beta_{1}}^{\infty }\frac{(x-\beta_{1})^{\delta\sigma-1}\ln[(x-\beta_{1})^{\delta }(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\,dx\quad(n\in \mathbf{N}), \end{aligned}$$
(7)
$$\begin{aligned}& \widetilde{\omega}_{\sigma}(x):=(x-\beta_{1})^{\delta\sigma }\sum _{n=1}^{\infty}\frac{(n-\beta_{2})^{\sigma-1}\ln[(x-\beta _{1})^{\delta}(n-\beta_{2})]}{[(x-\beta_{1})^{\delta}(n-\beta _{2})]^{\lambda}-1}\quad\bigl(x\in( \beta_{1},\infty)\bigr). \end{aligned}$$
(8)
Setting \(k_{\lambda}(\sigma):=\frac{1}{\lambda^{2}}[B(\frac{\sigma}{ \lambda},1-\frac{\sigma}{\lambda})]^{2}=[\frac{\pi}{\lambda\sin (\frac{\pi\sigma}{\lambda})}]^{2}\), we have the following inequalities:
$$\begin{aligned}& 0< k_{\lambda}(\sigma) \bigl(1-\theta_{\lambda}(x)\bigr)<\widetilde{ \omega}_{\sigma }(x)<\omega_{\sigma}(n)=k_{\lambda}(\sigma), \end{aligned}$$
(9)
$$\begin{aligned}& \begin{aligned}[b] 0 &< \theta_{\lambda}(x):=\frac{1}{\lambda^{2}k_{\lambda}(\sigma)}\int _{0}^{[(x-\beta_{1})^{\delta}(1-\beta_{2})]^{\lambda}}\frac{\ln v}{v-1}v^{\frac{\sigma}{\lambda}-1}\,dv\\ &=O\bigl((x-\beta_{1})^{\frac{\delta\sigma}{2}}\bigr) \quad\bigl(x\in( \beta_{1},\infty)\bigr). \end{aligned} \end{aligned}$$
(10)

Proof

Putting \(u=[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}\) in (7), by Lemma 1, we find
$$ \omega_{\sigma}(n)=\frac{1}{\lambda^{2}}\int_{0}^{\infty} \frac{\ln u}{u-1}u^{\frac{\sigma}{\lambda}-1}\,du=k_{\lambda}(\sigma). $$
(11)
For fixed \(x\in(\beta_{1},\infty)\), setting
$$ f(t):=\frac{(x-\beta_{1})^{\delta\sigma}\ln[(x-\beta_{1})^{\delta }(t-\beta_{2})]}{[(x-\beta_{1})^{\delta}(t-\beta_{2})]^{\lambda}-1} (t-\beta_{2})^{\sigma-1}\quad \bigl(t\in(\beta_{2},\infty)\bigr), $$
(12)
in view of \(0<\sigma<\lambda\leq1\), we find \([\frac{\ln u}{u^{\lambda}-1}]^{\prime}<0\), \([\frac{\ln u}{u^{\lambda}-1}]^{\prime\prime}>0\) (\(u>0\)) (cf. [19], Example 2.2.1) and then \(f^{\prime}(t)<0\), and \(f^{\prime \prime}(t)>0\). By the following Hermite-Hadamard inequality (cf. [20]):
$$ f(n)< \int_{n-\frac{1}{2}}^{n+\frac{1}{2}}f(t)\,dt\quad(n\in\mathbf{N}), $$
(13)
and putting \(v=[(x-\beta_{1})^{\delta}(t-\beta_{2})]^{\lambda}\), it follows that
$$\begin{aligned}& \begin{aligned}[b] \widetilde{\omega}_{\sigma}(x) &=\sum_{n=1}^{\infty}f(n)< \sum_{n=1}^{\infty }\int_{n-\frac{1}{2}}^{n+\frac{1}{2}}f(t)\,dt= \int_{\frac{1}{2}}^{\infty }f(t)\,dt\\ &\leq\int_{\beta_{2}}^{\infty}f(t)\,dt=\frac{1}{\lambda^{2}}\int_{0}^{\infty}\frac{\ln v}{v-1}v^{\frac{\sigma}{\lambda}-1}\,dv=k_{\lambda}( \sigma), \end{aligned}\\& \begin{aligned}[b] \widetilde{\omega}_{\sigma}(x) &=\sum_{n=1}^{\infty}f(n)> \int_{1}^{\infty }f(t)\,dt=\int_{\beta_{2}}^{\infty}f(t)\,dt- \int_{\beta_{2}}^{1}f(t)\,dt\\ &=k_{\lambda}(\sigma)-\frac{1}{\lambda^{2}}\int_{0}^{[(x-\beta _{1})^{\delta}(1-\beta_{2})]^{\lambda}} \frac{\ln v}{v-1}v^{\frac {\sigma}{\lambda}-1}\,dv\\ &=k_{\lambda}(\sigma) \bigl(1-\theta_{\lambda}(x)\bigr)>0, \end{aligned} \end{aligned}$$
where
$$ 0< \theta_{\lambda}(x):=\frac{1}{\lambda^{2}k_{\lambda}(\sigma)}\int _{0}^{[(x-\beta_{1})^{\delta}(1-\beta_{2})]^{\lambda}}\frac{\ln v}{v-1}v^{\frac{\sigma}{\lambda}-1}\,dv\quad \bigl(x\in(\beta_{1},\infty)\bigr). $$
Since \(\lim_{v\rightarrow0^{+}}\frac{\ln v}{v-1}v^{\frac{\sigma}{2\lambda}}=\lim_{v\rightarrow\infty}\frac{\ln v}{v-1}v^{\frac {\sigma}{2\lambda}}=0\) and \(\frac{\ln v}{v-1}v^{\frac{\sigma}{2\lambda}}| _{v=1}=1\), in view of the bounded properties of continuous function, there exists a constant \(M>0\), such that \(0<\frac{\ln v}{v-1}v^{\frac{\sigma }{2\lambda}}\leq M\) (\(v\in(0,\infty)\)). For \(x\in(\beta_{1},\infty)\), we have
$$\begin{aligned} 0 < &\int_{0}^{[(x-\beta_{1})^{\delta}(1-\beta_{2})]^{\lambda}}\frac {\ln v}{v-1}v^{\frac{\sigma}{\lambda}-1}\,dv \\ =&\int_{0}^{[(x-\beta_{1})^{\delta }(1-\beta_{2})]^{\lambda}}\biggl(\frac{\ln v}{v-1}v^{\frac{\sigma }{2\lambda}} \biggr)v^{\frac{\sigma}{2\lambda}-1}\,dv \\ \leq&M\int_{0}^{[(x-\beta_{1})^{\delta}(1-\beta_{2})]^{\lambda }}v^{\frac{\sigma}{2\lambda}-1}\,dv \\ =&\frac{2M\lambda}{\sigma}(1-\beta_{2})^{\frac{\sigma}{2}}(x- \beta_{1})^{\frac{\delta\sigma}{2}}. \end{aligned}$$
(14)
Hence we proved that (9) and (10) are valid. □

Lemma 3

Suppose that \(\frac{1}{p}+\frac{1}{q}=1\) (\(p\neq 0,1\)), \(0<\sigma<\lambda\leq1\), \(\beta_{1}\in(-\infty,\infty)\), \(0\leq \beta_{2}\leq\frac{1}{2}\), \(\delta\in\{-1,1\}\), \(a_{n}\geq0\), \(f(x)\) is a non-negative real measurable function in \((\beta_{1},\infty)\). Then

(i) for \(p>1\), we have the following inequalities:
$$\begin{aligned}& \begin{aligned}[b] J &:=\Biggl\{ \sum_{n=1}^{\infty}(n- \beta_{2})^{p\sigma-1}\biggl[\int_{\beta _{1}}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]}{[(x-\beta _{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}f(x)\,dx\biggr]^{p}\Biggr\} ^{\frac{1}{p}} \\ &\leq\bigl[k_{\lambda}(\sigma)\bigr]^{\frac{1}{q}}\biggl\{ \int _{\beta_{1}}^{\infty }\widetilde{\omega}_{\sigma}(x) (x-\beta_{1})^{p(1-\delta\sigma )-1}f^{p}(x)\,dx\biggr\} ^{\frac{1}{p}}, \end{aligned} \end{aligned}$$
(15)
$$\begin{aligned}& \begin{aligned}[b] L_{1} &:=\Biggl\{ \int_{\beta_{1}}^{\infty} \frac{(x-\beta_{1})^{q\delta \sigma -1}}{\widetilde{\omega}_{\sigma}^{q-1}(x)}\Biggl[\sum_{n=1}^{\infty} \frac{\ln [(x-\beta_{1})^{\delta}(n-\beta_{2})]}{[(x-\beta_{1})^{\delta }(n-\beta _{2})]^{\lambda}-1}a_{n}\Biggr]^{q}\,dx\Biggr\} ^{\frac{1}{q}}\\ &\leq\Biggl\{ k_{\lambda}(\sigma)\sum_{n=1}^{\infty}(n- \beta _{2})^{q(1-\sigma )-1}a_{n}^{q}\Biggr\} ^{\frac{1}{q}}, \end{aligned} \end{aligned}$$
(16)
where \(\omega_{\sigma}(n)\) and \(\widetilde{\omega}_{\sigma}(x)\) are defined by (7) and (8);

(ii) for \(0< p<1\) or \(p<0\), we have the reverses of (15) and (16).

Proof

(i) By (7)-(10) and the Hölder inequality [20], we have
$$\begin{aligned} &\biggl[\int_{\beta_{1}}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta }(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda }-1}f(x)\,dx \biggr]^{p} \\ &\quad=\biggl\{ \int_{\beta_{1}}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta }(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1} \biggl[\frac{(x-\beta_{1})^{(1-\delta\sigma)/q}}{(n-\beta _{2})^{(1-\sigma)/p}}f(x)\biggr] \biggl[\frac{(n-\beta_{2})^{(1-\sigma )/p}}{(x-\beta _{1})^{(1-\delta\sigma)/q}} \biggr]\,dx\biggr\} ^{p} \\ &\quad\leq\int_{\beta_{1}}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta }(n-\beta_{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1} \frac{(x-\beta_{1})^{(1-\delta\sigma)(p-1)}}{(n-\beta _{2})^{1-\sigma}}f^{p}(x)\,dx \\ &\qquad{}\times\biggl[\int_{\beta_{1}}^{\infty}\frac{\ln[(x-\beta _{1})^{\delta}(n-\beta_{2})]}{[(x-\beta_{1})^{\delta}(n-\beta _{2})]^{\lambda}-1} \frac{(n-\beta_{2})^{(1-\sigma)(q-1)}}{(x-\beta _{1})^{1-\delta\sigma}}\,dx\biggr]^{p-1} \\ &\quad=\int_{\beta_{1}}^{\infty}\frac{f^{p}(x)\ln[(x-\beta_{1})^{\delta }(n-\beta_{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1} \frac{(x-\beta_{1})^{(1-\delta\sigma)(p-1)}}{(n-\beta _{2})^{1-\sigma}}\,dx \bigl[(n-\beta_{2})^{q(1-\sigma)-1}\omega_{\sigma}(n) \bigr]^{p-1} \\ &\quad=\frac{k_{\lambda}^{p-1}(\sigma)}{(n-\beta_{2})^{p\sigma-1}} \int_{\beta_{1}}^{\infty}\frac{f^{p}(x)\ln[(x-\beta _{1})^{\delta}(n-\beta_{2})]}{[(x-\beta_{1})^{\delta}(n-\beta _{2})]^{\lambda}-1} \frac{(x-\beta_{1})^{(1-\delta\sigma )(p-1)}}{(n-\beta _{2})^{1-\sigma}}\,dx. \end{aligned}$$
(17)
By the Lebesgue term by term integration theorem [21], (17), and (9), we obtain
$$\begin{aligned} J^{p} \leq&k_{\lambda}^{p-1}(\sigma)\sum _{n=1}^{\infty}\int_{\beta _{1}}^{\infty} \frac{f^{p}(x)\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\frac{(x-\beta _{1})^{(1-\delta\sigma)(p-1)}}{(n-\beta_{2})^{1-\sigma}}\,dx \\ =&k_{\lambda}^{p-1}(\sigma)\int_{\beta_{1}}^{\infty} \sum_{n=1}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta}(n-\beta_{2})]}{[(x-\beta _{1})^{\delta }(n-\beta_{2})]^{\lambda}-1} \frac{(x-\beta_{1})^{(1-\delta\sigma )(p-1)}}{(n-\beta_{2})^{1-\sigma}}f^{p}(x)\,dx \\ =&k_{\lambda}^{p-1}(\sigma)\int_{\beta_{1}}^{\infty} \widetilde{\omega}_{\sigma}(x) (x-\beta_{1})^{p(1-\delta\sigma)-1}f^{p}(x)\,dx. \end{aligned}$$
(18)
Hence (15) is valid. Using the Hölder inequality, in view of the Lebesgue term by term integration theorem and (9) again, we have
$$\begin{aligned} &\Biggl[\sum_{n=1}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}a_{n}\Biggr]^{q} \\ &\quad=\Biggl\{ \sum_{n=1}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\frac{(x-\beta _{1})^{(1-\delta\sigma)/q}}{(n-\beta_{2})^{(1-\sigma)/p}}\frac {(n-\beta _{2})^{(1-\sigma)/p}a_{n}}{(x-\beta_{1})^{(1-\delta\sigma)/q}}\Biggr\} ^{q} \\ &\quad\leq\bigl[\widetilde{\omega}_{\sigma}(x) (x-\beta_{1})^{p(1-\delta\sigma )-1} \bigr]^{q-1}\sum_{n=1}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1} \frac {(n-\beta _{2})^{(1-\sigma)(q-1)}a_{n}^{q}}{(x-\beta_{1})^{1-\delta\sigma}} \\ &\quad=\widetilde{\omega}_{\sigma}^{q-1}(x) (x-\beta_{1})^{1-q\delta\sigma} \sum_{n=1}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1} \frac {(n-\beta _{2})^{(1-\sigma)(q-1)}}{(x-\beta_{1})^{1-\delta\sigma}}a_{n}^{q}, \end{aligned}$$
(19)
and therefore
$$\begin{aligned} L_{1}^{q} \leq&\int_{\beta_{1}}^{\infty} \sum_{n=1}^{\infty}\frac {\ln [(x-\beta_{1})^{\delta}(n-\beta_{2})]}{[(x-\beta_{1})^{\delta }(n-\beta _{2})]^{\lambda}-1} \frac{(n-\beta_{2})^{(1-\delta\sigma )(q-1)}}{(x-\beta _{1})^{1-\delta\sigma}}a_{n}^{q}\,dx \\ =&\sum_{n=1}^{\infty}\biggl[(n- \beta_{2})^{\sigma}\int_{\beta _{1}}^{\infty}\frac{(x-\beta_{1})^{\delta\sigma-1}\ln[(x-\beta_{1})^{\delta }(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\,dx\biggr] (n-\beta_{2})^{q(1-\sigma)-1}a_{n}^{q} \\ =&\sum_{n=1}^{\infty}\omega_{\sigma}(n) (n-\beta_{2})^{q(1-\sigma )-1}a_{n}^{q}=k_{\lambda}( \sigma)\sum_{n=1}^{\infty}(n-\beta _{2})^{q(1-\sigma)-1}a_{n}^{q}. \end{aligned}$$
(20)
Hence (16) is valid.

(ii) For \(0< p<1\) (\(q<0\)) or \(p<0\) (\(0< q<1\)), using the reverse Hölder inequality (cf. [20]) and in the same way, we obtain the reverses of (15) and (16). □

Definition 1

As the assumptions of Lemma 2 and Lemma 3, we define \(\phi(x):=(x-\beta_{1})^{p(1-\delta\sigma)-1}\), \(\widetilde{\phi}(x):=(1-\theta_{\lambda}(x))\phi(x)\), \(\psi(n):=(n-\beta _{2})^{q(1-\sigma)-1}\), and the following sets:
$$\begin{aligned}& L_{p,\phi}(\beta_{1},\infty):=\biggl\{ f;\| f \|_{p,\phi}=\biggl[\int_{\beta _{1}}^{\infty} \phi(x)\bigl|f(x)\bigr|^{p}\,dx\biggr]^{1/p}< \infty\biggr\} , \\& l_{q,\psi}:=\Biggl\{ a=\{a_{n}\};\| a\|_{q,\psi}= \Biggl[\sum_{n=1}^{\infty }\psi (n)|a_{n}|^{q}\Biggr]^{1/q}<\infty\Biggr\} . \end{aligned}$$

Note

If \(p>1\), then \(L_{p,\phi}(\beta_{1},\infty)\) and \(l_{q,\psi}\) are normed spaces; if \(0< p<1\) or \(p<0\), then both \(L_{p,\phi }(\beta_{1},\infty)\) and \(l_{q,\psi}\) are not normed spaces, but we still use the formal symbols in the following.

3 Main results and applications

Theorem 1

Suppose that \(p<0\), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\sigma <\lambda\leq1\), \(\beta_{1}\in(-\infty,+\infty)\), \(0\leq\beta _{2}\leq \frac{1}{2}\), \(\delta\in\{-1,1\}\), \(f(x),a_{n}\geq0\), satisfying \(f\in L_{p,\phi}(\beta_{1},\infty)\), \(a=\{a_{n}\}_{n=1}^{\infty}\in l_{q,\psi }\), \(\| f\|_{p,\phi}>0\), \(\| a\|_{q,\psi}>0\). Then we have the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b] I&:=\sum_{n=1}^{\infty}a_{n}\int _{\beta_{1}}^{\infty}\frac{\ln [(x-\beta _{1})^{\delta}(n-\beta_{2})]}{[(x-\beta_{1})^{\delta}(n-\beta _{2})]^{\lambda}-1}f(x)\,dx \\ &=\int_{\beta_{1}}^{\infty}f(x)\sum _{n=1}^{\infty}\frac{\ln [(x-\beta _{1})^{\delta}(n-\beta_{2})]a_{n}}{[(x-\beta_{1})^{\delta}(n-\beta _{2})]^{\lambda}-1}\,dx>k_{\lambda}( \sigma)\| f\|_{p,\phi }\| a\|_{q,\psi}, \end{aligned} \end{aligned}$$
(21)
$$\begin{aligned}& J=\Biggl\{ \sum_{n=1}^{\infty}(n- \beta_{2})^{p\sigma-1}\biggl[\int_{\beta _{1}}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta_{2})]f(x)}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\,dx\biggr]^{p}\Biggr\} ^{\frac {1}{p}}>k_{\lambda}( \sigma)\| f\|_{p,\phi}, \end{aligned}$$
(22)
$$\begin{aligned}& L:=\Biggl\{ \int_{\beta_{1}}^{\infty}(x-\beta_{1})^{q\delta\sigma -1} \Biggl[\sum_{n=1}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]a_{n}}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\Biggr]^{q}\,dx\Biggr\} ^{\frac{1}{q}}>k_{\lambda}(\sigma)\| a \|_{q,\psi}, \end{aligned}$$
(23)
where the constant factor \(k_{\lambda}(\sigma)=[\frac{\pi}{\lambda \sin(\frac{\pi\sigma}{\lambda})}]^{2}\) is the best possible.

Proof

By the Lebesgue term by term integration theorem [21], we find that there are two expressions of I in (21). By (9), the reverse of (15) and \(0<\| f\|_{p,\phi}<\infty\), we have (22). By the reverse Hölder inequality, we find
$$\begin{aligned} I =&\sum_{n=1}^{\infty}\biggl[(n- \beta_{2})^{\sigma-\frac{1}{p}}\int_{\beta _{1}}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta_{2})]f(x)}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\,dx\biggr] \bigl[(n-\beta _{2})^{\frac{1}{p}-\sigma}a_{n} \bigr] \\ \geq&J\Biggl\{ \sum_{n=1}^{\infty}[(n- \beta_{2})^{q(1-\sigma )-1}a_{n}^{q}\Biggr\} ^{1/q}=J\| a\|_{q,\psi}. \end{aligned}$$
(24)
Then by (22), (21) is valid. On the other hand, assuming that (21) is valid, setting
$$ a_{n}:=(n-\beta_{2})^{p\sigma-1}\biggl[\int _{\beta_{1}}^{\infty}\frac{\ln [(x-\beta_{1})^{\delta}(n-\beta_{2})]}{[(x-\beta_{1})^{\delta }(n-\beta _{2})]^{\lambda}-1}f(x)\,dx\quad \biggr]^{p-1}\quad(n\in\mathbf{N}), $$
(25)
then by (21), we have
$$ \| a\|_{q,\psi}^{q}=\sum_{n=1}^{\infty}(n- \beta _{2})^{q(1-\sigma )-1}a_{n}^{q}=J^{p}=I \geq k_{\lambda}(\sigma)\| f\|_{p,\phi }\| a\|_{q,\psi}. $$
(26)
By (9), the reverse of (15), and \(0<\| f\|_{p,\phi }<\infty\), it follows that \(J>0\). If \(J=\infty\), then (22) is trivially valid; if \(J<\infty\), then \(0<\| a\|_{q,\psi }=J^{p-1}<\infty\). Thus the conditions of applying (21) are fulfilled and by (21), (26) takes a strict sign inequality. Thus we find
$$ J=\| a\|_{q,\psi}^{q-1}>k_{\lambda}(\sigma)\| f\| _{p,\phi}. $$
(27)
Hence, (22) is valid, which is equivalent to (21).
By (9), the reverse of (16) and \(0<\| a\|_{q,\psi }<\infty\), we obtain (23). By the reverse Hölder inequality again, we have
$$\begin{aligned} I =&\int_{\beta_{1}}^{\infty}\Biggl[(x-\beta_{1})^{\delta \sigma-\frac{1}{q}} \sum_{n=1}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta }(n-\beta _{2})]a_{n}}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\Biggr] \bigl[(x-\beta_{1})^{\frac{1}{q}-\delta\sigma}f(x)\bigr]\,dx \\ \geq&L\biggl\{ \int_{\beta_{1}}^{\infty}(x- \beta_{1})^{p(1-\delta\sigma )-1}f^{p}(x)\,dx\biggr\} ^{1/p}=L\| f\|_{p,\phi}. \end{aligned}$$
(28)
Hence (21) is valid by using (23). On the other hand, assuming that (21) is valid, setting
$$ f(x):=(x-\beta_{1})^{q\delta\sigma-1}\Biggl[\sum _{n=1}^{\infty}\frac{\ln [(x-\beta_{1})^{\delta}(n-\beta_{2})]a_{n}}{[(x-\beta_{1})^{\delta }(n-\beta _{2})]^{\lambda}-1}\Biggr]^{q-1}\quad \bigl(x\in(\beta_{1},\infty)\bigr), $$
(29)
then by (21), we find
$$ \| f\|_{p,\phi}^{p}=\int_{\beta_{1}}^{\infty}(x- \beta _{1})^{p(1-\delta\sigma)-1}f^{p}(x)\,dx=L^{q}=I\geq k_{\lambda}(\sigma )\| f\|_{p,\phi}\| a\|_{q,\psi}. $$
(30)
By (9), the reverse of (16) and \(0<\| a\|_{q,\psi }<\infty\), it follows that \(L>0\). If \(L=\infty\), then (23) is trivially valid; if \(L<\infty\), then \(0<\| f\|_{p,\phi }=L^{q-1}<\infty\), i.e. the conditions of applying (21) are fulfilled and by (30), we still have
$$ \| f\|_{p,\phi}^{p}=L^{q}=I>k_{\lambda}(\sigma) \| f\| _{p,\phi}\| a\|_{q,\psi}, \quad\textit{i.e. } L=\| f\| _{p,\phi }^{p-1}>k_{\lambda}(\sigma)\| a\|_{q,\psi}. $$
Hence (23) is valid, which is equivalent to (21).

It follows that (21), (22), and (23) are equivalent.

We prove that the constant factor in (21) is the best possible. For \(0<\varepsilon<q\sigma\), we set \(E_{\delta}:=\{x|0<(x-\beta _{1})^{\delta }<1\}\), \(\widetilde{a}=\{\widetilde{a}_{n}\}_{n=1}^{\infty}\), and \(\widetilde{f}(x)\) as follows:
$$ \widetilde{a}_{n}=(n-\beta_{2})^{\sigma-\frac{\varepsilon}{q}-1};\qquad \widetilde{f}(x)= \left \{ \begin{array}{@{}l@{\quad}l} (x-\beta_{1})^{\delta(\sigma+\frac{\varepsilon}{p})-1}, & x\in E_{\delta},\\ 0,& x\in(\beta_{1},\infty)\backslash E_{\delta}.\end{array} \right . $$
(31)
If there exists a positive number \(k\geq k_{\lambda}(\sigma)\), such that (21) is still valid as we replace \(k_{\lambda}(\sigma)\) by k, then in particular, on substitution of \(\widetilde{a}\) and \(\widetilde {f}(x)\), we have
$$ \widetilde{I}:=\sum_{n=1}^{\infty} \widetilde{a}_{n}\int_{\beta _{1}}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]}{[(x-\beta _{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\widetilde{f}(x)\,dx>k\| \widetilde{f}\|_{p,\phi}\| \widetilde{a}\|_{q,\psi}. $$
(32)
For \(p<0\), \(0< q<1\), setting \(\widetilde{\sigma}=\sigma-\frac{\varepsilon }{q}\), we find by Lemma 2 that
$$\begin{aligned} \widetilde{I} =&\int_{E_{\delta}}(x-\beta_{1})^{\delta\varepsilon -1}(x- \beta_{1})^{\delta(\sigma-\frac{\varepsilon}{q})} \sum_{n=1}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}(n-\beta _{2})^{(\sigma-\frac{\varepsilon}{q})-1}\,dx \\ =&\int_{E_{\delta}}(x-\beta_{1})^{\delta\varepsilon-1}\widetilde{ \omega }_{\widetilde{\sigma}}(x)\,dx< k_{\lambda}(\widetilde{\sigma})\int _{E_{\delta }}(x-\beta_{1})^{\delta\varepsilon-1}\,dx \\ =&\frac{1}{\varepsilon}k_{\lambda}\biggl(\sigma-\frac{\varepsilon}{q}\biggr). \end{aligned}$$
(33)
Setting \(u=(x-\beta_{1})^{\delta}\), by calculation we obtain
$$\begin{aligned}& \begin{aligned}[b] \|\widetilde{f}\|_{p,\phi} &=\biggl\{ \int_{\beta_{1}}^{\infty }(x- \beta _{1})^{p(1-\delta\sigma)-1}\widetilde{f}^{p}(x)\,dx\biggr\} ^{1/p} \\ &=\biggl\{ \int_{E_{\delta}}(x-\beta_{1})^{-1+\delta\varepsilon }\,dx \biggr\} ^{1/p}=\biggl\{ \int_{0}^{1}u^{-1+\varepsilon}\,du \biggr\} ^{1/p}=\biggl(\frac{1}{\varepsilon}\biggr)^{1/p}, \end{aligned} \end{aligned}$$
(34)
$$\begin{aligned}& \begin{aligned}[b] \|\widetilde{a}\|_{q,\psi}^{q} &=\sum _{n=1}^{\infty}(n-\beta _{2})^{q(1-\sigma)-1} \widetilde{a}_{n}^{q}=\sum_{n=1}^{\infty}(n- \beta _{2})^{-1-\varepsilon} \\ &>\int_{1}^{\infty}(x-\beta_{2})^{-1-\varepsilon}\,dx= \frac {1}{\varepsilon (1-\beta_{2})^{\varepsilon}}. \end{aligned} \end{aligned}$$
(35)
In view of (32), (33) and \(0< q<1\), it follows that
$$ k_{\lambda}\biggl(\sigma-\frac{\varepsilon}{q}\biggr)>\varepsilon \widetilde{I}>\varepsilon k\|\widetilde{f}\|_{p,\phi}\| \widetilde {a}\| _{q,\psi}> k\biggl[\frac{1}{(1-\beta_{2})^{\varepsilon}}\biggr]^{1/q}. $$
(36)
For \(\varepsilon\rightarrow0^{+}\) in (36), we have \(k_{\lambda }(\sigma)\geq k\). Hence \(k=k_{\lambda}(\sigma)\) is the best value of (21). We confirm that the constant factor \(k_{\lambda}(\sigma)\) in (22) ((23)) is the best possible. Otherwise we can get the contradiction by (24) ((28)) that the constant factor in (21) is not the best possible. □

Theorem 2

Suppose that \(0< p<1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\sigma <\lambda\leq1\), \(\beta_{1}\in(-\infty,+\infty)\), \(0\leq\beta _{2}\leq \frac{1}{2}\), \(\delta\in\{-1,1\}\), \(f(x),a_{n}\geq0\), satisfying \(f\in L_{p,\widetilde{\phi}}(\beta_{1},\infty)\), \(a=\{a_{n}\} _{n=1}^{\infty }\in l_{q,\psi}\), \(\| f\|_{p,\widetilde{\phi}}>0\), \(\| a\| _{q,\psi}>0\). Then we have the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b] I &=\sum_{n=1}^{\infty}a_{n}\int _{\beta_{1}}^{\infty}\frac{\ln [(x-\beta _{1})^{\delta}(n-\beta_{2})]}{[(x-\beta_{1})^{\delta}(n-\beta _{2})]^{\lambda}-1}f(x)\,dx \\ &=\int_{\beta_{1}}^{\infty}f(x)\sum _{n=1}^{\infty}\frac{\ln [(x-\beta _{1})^{\delta}(n-\beta_{2})]a_{n}}{[(x-\beta_{1})^{\delta}(n-\beta _{2})]^{\lambda}-1}\,dx>k_{\lambda}( \sigma)\| f\|_{p,\widetilde{ \phi}}\| a\|_{q,\psi}, \end{aligned} \end{aligned}$$
(37)
$$\begin{aligned}& J=\Biggl\{ \sum_{n=1}^{\infty}(n- \beta_{2})^{p\sigma-1}\biggl[\int_{\beta _{1}}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta_{2})]f(x)}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\,dx\biggr]^{p}\Biggr\} ^{\frac {1}{p}}>k_{\lambda}( \sigma)\| f\|_{p,\widetilde{\phi}}, \end{aligned}$$
(38)
$$\begin{aligned}& \begin{aligned}[b] \widetilde{L} &:=\Biggl\{ \int_{\beta_{1}}^{\infty} \frac{(x-\beta _{1})^{q\delta \sigma-1}}{[(1-\theta_{\lambda}(x)]^{q-1}}\Biggl[\sum_{n=1}^{\infty} \frac {\ln [(x-\beta_{1})^{\delta}(n-\beta_{2})]a_{n}}{[(x-\beta_{1})^{\delta }(n-\beta_{2})]^{\lambda}-1}\Biggr]^{q}\,dx\Biggr\} ^{\frac{1}{q}} \\ &>k_{\lambda}(\sigma)\| a\|_{q,\psi}, \end{aligned} \end{aligned}$$
(39)
where the constant factor \(k_{\lambda}(\sigma)=[\frac{\pi}{\lambda \sin(\frac{\pi\sigma}{\lambda})}]^{2}\) is the best possible.

Proof

By (9), the reverse of (15) and \(0<\| f\| _{p,\widetilde{\phi}}<\infty\), we have (38). Using the reverse Hölder inequality, we obtain the reverse form of (24) as follows:
$$ I\geq J\| a\|_{q,\psi}. $$
(40)
Then by (38), (37) is valid.
On the other hand, if (37) is valid, setting \(a_{n}\) as (25), then (26) still holds with \(0< p<1\). By (37), we have
$$ \| a\|_{q,\psi}^{q}=\sum_{n=1}^{\infty}(n- \beta _{2})^{q(1-\sigma )-1}a_{n}^{q}=J^{p}=I \geq k_{\lambda}(\sigma)\| f\| _{p,\widetilde{\phi}}\| a\|_{q,\psi}. $$
(41)
Then by (9), the reverse of (18) and \(0<\| f\|_{p,\widetilde{\phi}}<\infty\), it follows that
$$ J=\Biggl\{ \sum_{n=1}^{\infty}(n- \beta_{2})^{q(1-\sigma)-1}a_{n}^{q}\Biggr\} ^{1/p}>0. $$
If \(J=\infty\), then (38) is trivially valid; if \(J<\infty\), then \(0<\| a\|_{q,\psi}=J^{p-1}<\infty\), i.e. the conditions of applying (37) are fulfilled and by (41), we still have
$$ \| a\|_{q,\psi}^{q}=J^{p}=I>k_{\lambda}(\sigma) \| f\| _{p,\widetilde{\phi}}\| a\|_{q,\psi}, \quad\textit{i.e. } J=\| a\| _{q,\psi}^{q-1}>k_{\lambda}(\sigma)\| f\|_{p,\widetilde{\phi}}. $$
Hence (38) is valid, which is equivalent to (37).
By the reverse of (16), in view of \(\widetilde{\omega}_{\sigma }(x)>k_{\lambda}(\sigma)(1-\theta_{\lambda}(x))\) and \(q<0\), we have
$$ \widetilde{L}>k_{\lambda}^{\frac{q-1}{q}}(\sigma)L_{1}\geq k_{\lambda }^{\frac{1}{p}}(\sigma)\Biggl\{ k_{\lambda}(\sigma)\sum _{n=1}^{\infty}(n-\beta _{2})^{q(1-\sigma)-1}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}=k_{\lambda}(\sigma )\| a\|_{q,\psi}, $$
then (39) is valid. By the reverse Hölder inequality again, we have
$$\begin{aligned} I =&\int_{\beta_{1}}^{\infty}\Biggl[\frac{(x-\beta_{1})^{\delta \sigma-\frac{1}{q}}}{(1-\theta_{\lambda}(x))^{\frac{1}{p}}}\sum_{n=1}^{\infty}\frac{\ln[(x-\beta_{1})^{\delta}(n-\beta_{2})]}{ [(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}a_{n} \Biggr] \\ &{}\times\bigl[\bigl(1-\theta_{\lambda}(x)\bigr)^{\frac{1}{p}}(x- \beta_{1})^{ \frac{1}{q}-\delta\sigma}f(x)\bigr]\,dx\geq\widetilde{L}\| f \|_{p,\widetilde{\phi}}. \end{aligned}$$
(42)
Hence (37) is valid by (39). On the other hand, if (37) is valid, setting
$$ f(x)=\frac{(x-\beta_{1})^{q\delta\sigma-1}}{[1-\theta_{\lambda }(x)]^{q-1}}\Biggl[\sum_{n=1}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta }(n-\beta _{2})]a_{n}}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1}\Biggr]^{q-1}\quad\bigl(x\in( \beta_{1},\infty)\bigr), $$
then by the reverse of (16) and \(0<\| a\|_{q,\psi}<\infty\), it follows that
$$ \widetilde{L}=\biggl\{ \int_{\beta_{1}}^{\infty}\bigl[1-\theta _{\lambda }(x)\bigr]^{\frac{1}{p}}(x-\beta_{1})^{p(1-\delta\sigma)-1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{q}}=\| f\|_{p,\widetilde{\phi}}^{p-1}>0. $$
If \(\widetilde{L}=\infty\), then (39) is trivially valid; if \(\widetilde{L}<\infty\), then \(0<\| f\|_{p,\widetilde{\phi}}=\widetilde{L}^{q-1}<\infty\), i.e. the conditions of applying (37) are fulfilled and we have
$$ \| f\|_{p,\widetilde{\phi}}^{p}=\widetilde{L}^{q}=I>k_{\lambda }( \sigma)\| f\|_{p,\widetilde{\phi}}\| a\|_{q,\psi },\quad\textit{i.e. } \widetilde{L}=\| f\|_{p,\widetilde{\phi }}^{p-1}>k_{\lambda }(\sigma)\| a \|_{q,\psi}. $$
Hence (39) is valid, which is equivalent to (37). It follows that (37), (38), and (39) are equivalent.
If there exists a positive number \(K\geq k_{\lambda}(\sigma)\), such that (37) is still valid as we replace \(k_{\lambda}(\sigma)\) by K, then in particular, we have
$$ \widetilde{I}=\sum_{n=1}^{\infty} \widetilde{a}_{n}\int_{\beta _{1}}^{\infty } \frac{\ln[(x-\beta_{1})^{\delta}(n-\beta_{2})]}{[(x-\beta _{1})^{\delta }(n-\beta_{2})]^{\lambda}-1}\widetilde{f}(x)\,dx>K\| \widetilde {f}\| _{p,\widetilde{\phi}}\| \widetilde{a}\|_{q,\psi}, $$
(43)
where \(\widetilde{a}=\{\widetilde{a}_{n}\}_{n=1}^{\infty}\) and \(\widetilde{f}\) are taken as (31) (\(0<\varepsilon<p(\lambda-\sigma)\)). We find
$$\begin{aligned} \|\widetilde{f}\|_{p,\widetilde{\phi}} =&\biggl\{ \int_{E_{\delta }} \bigl[1-O\bigl((x-\beta_{1})^{\frac{\delta\widetilde{\sigma}}{2}}\bigr)\bigr](x-\beta _{1})^{\delta\varepsilon-1}\,dx\biggr\} ^{1/p} \\ =&\biggl(\frac{1}{\varepsilon}-O(1)\biggr)^{1/p}. \end{aligned}$$
Since by (35) and setting \(u=[(x-\beta_{1})^{\delta}(n-\beta _{2})]^{\lambda}\), it follows that
$$\begin{aligned} \widetilde{I} &=\sum_{n=1}^{\infty}(n- \beta_{2})^{\sigma-\frac{\varepsilon}{q}-1}\int_{E_{\delta}} \frac{\ln[(x-\beta_{1})^{\delta }(n-\beta_{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1} (x-\beta_{1})^{\delta(\sigma+\frac{\varepsilon}{p})-1}\,dx \\ &\leq\sum_{n=1}^{\infty}(n- \beta_{2})^{\sigma-\frac{\varepsilon }{q}-1}\int_{\beta_{1}}^{\infty} \frac{\ln[(x-\beta_{1})^{\delta }(n-\beta _{2})]}{[(x-\beta_{1})^{\delta}(n-\beta_{2})]^{\lambda}-1} (x-\beta _{1})^{\delta(\sigma+\frac{\varepsilon}{p})-1}\,dx \\ &=\frac{1}{\lambda^{2}}\sum_{n=1}^{\infty}(n- \beta _{2})^{-\varepsilon -1}\int_{0}^{\infty} \frac{\ln u}{u-1}u^{\frac{\sigma}{\lambda}+\frac{ \varepsilon}{p\lambda}-1}\,du \\ &\leq\biggl[\frac{1}{\lambda}\sin\frac{\pi}{\lambda}\biggl(\sigma+ \frac{\varepsilon}{p}\biggr)\biggr]^{2}\frac{\varepsilon+1-\beta_{2}}{\varepsilon (1-\beta _{2})^{\varepsilon+1}}, \end{aligned}$$
(44)
by (35), (43), and (44), we have (notice that \(q<0\))
$$\begin{aligned} &\biggl[\frac{1}{\lambda}\sin\frac{\pi}{\lambda}\biggl(\sigma+ \frac {\varepsilon}{p}\biggr)\biggr]^{2}\frac{\varepsilon+1-\beta_{2}}{(1-\beta_{2})^{\varepsilon+1}} \\ &\quad\geq\varepsilon\widetilde{I}>\varepsilon K\biggl(\frac{1}{\varepsilon}-O(1) \biggr)^{1/p}\biggl(\frac{\varepsilon+1-\beta_{2}}{\varepsilon(1-\beta _{2})^{\varepsilon+1}}\biggr)^{1/q} \\ &\quad=K\bigl(1-\varepsilon O(1)\bigr)^{1/p}\biggl(\frac{\varepsilon+1-\beta_{2}}{(1-\beta _{2})^{\varepsilon+1}} \biggr)^{1/q}. \end{aligned}$$
(45)
For \(\varepsilon\rightarrow0^{+}\) in (45), we obtain \(k_{\lambda }(\sigma)=[\frac{1}{\lambda}\sin( \frac{\pi\sigma}{\lambda })]^{2}\geq K\). Hence \(k_{\lambda}(\sigma)=K\) is the best value of (37). We confirm that the constant factor \(k_{\lambda}(\sigma)\) in (38) ((39)) is the best possible. Otherwise we can get the contradiction by (40) ((42)) that the constant factor in (37) is not the best possible. □

Remark 1

(i) For \(\beta_{1}=\beta_{2}=0\), \(\sigma=\frac{\lambda }{2}\), \(\delta=1\) in (21), we have the reverse of (4). In particular, for \(\lambda=1\), \(p=q=2 \) in the reverse of (4), we have
$$ \sum_{n=1}^{\infty}a_{n}\int _{0}^{\infty}\frac{\ln (nx)}{nx-1}f(x)\,dx>\pi ^{2}\Biggl\{ \sum_{n=1}^{\infty}a_{n}^{2} \int_{0}^{\infty}f^{2}(x)\,dx\Biggr\} ^{\frac {1}{2}}. $$
(46)
(ii) For \(\beta_{1}=0\), \(\beta_{2}=\frac{1}{2}\), \(\sigma=\frac{\lambda }{2}\), \(\delta=1\) in (21), and (22), it follows from (5) that
$$\begin{aligned} &\sum_{n=1}^{\infty}x^{\frac{q\lambda}{2}-1}\biggl\{ \int_{0}^{\infty}\frac {\ln [x(n-\frac{1}{2})]}{[x(n-\frac{1}{2})]^{\lambda}-1}a_{n} \biggr\} ^{q}\,dx >\biggl(\frac{\pi}{\lambda}\biggr)^{2q}\sum _{n=1}^{\infty}\biggl(n-\frac{1}{2} \biggr)^{q(1- \frac{\lambda}{2})-1}a_{n}^{q}. \end{aligned}$$
(47)
In particular, for \(\lambda=1\), \(p=q=2\) in (5), we obtain
$$\begin{aligned} &\sum_{n=1}^{\infty}a_{n}\int _{0}^{\infty}\frac{\ln[x(n-\frac {1}{2})]}{x(n-\frac{1}{2})-1}f(x)\,dx >\pi^{2}\Biggl(\sum_{n=1}^{\infty}a_{n}^{2} \int_{0}^{\infty}f^{2}(x)\,dx \Biggr)^{1/2}, \end{aligned}$$
(48)
which is a more accurate inequality than (46).

Remark 2

For \(\delta=-1\), \(\mu=\lambda-\sigma\) (>0) in Theorem 1, setting \(\varphi(x):=(x-\beta_{1})^{p(1-\mu)-1}\), and \(F(x):=(x-\beta_{1})^{\lambda}f(x)\), we have the following equivalent inequalities with the homogeneous kernel and the best possible constant factor \(k_{\lambda}(\sigma)\):
$$\begin{aligned}& \begin{aligned}[b] &\sum_{n=1}^{\infty}a_{n}\int _{\beta_{1}}^{\infty}\frac{\ln [(n-\beta _{2})/(x-\beta_{1})]}{(n-\beta_{2})^{\lambda}-(x-\beta _{1})^{\lambda}}F(x)\,dx \\ &\quad=\int_{\beta_{1}}^{\infty}F(x)\sum _{n=1}^{\infty}\frac{\ln [(n-\beta _{2})/(x-\beta_{1})]}{(n-\beta_{2})^{\lambda}-(x-\beta _{1})^{\lambda}}a_{n}\,dx>k_{\lambda}( \sigma)\| F\|_{p,\varphi}\| a\| _{q,\psi }, \end{aligned} \end{aligned}$$
(49)
$$\begin{aligned}& \Biggl\{ \sum_{n=1}^{\infty}(n- \beta_{2})^{p\sigma-1}\biggl[\int_{\beta _{1}}^{\infty}\frac{\ln[(n-\beta_{2})/(x-\beta_{1})]F(x)}{(n-\beta_{2})^{\lambda }-(x-\beta_{1})^{\lambda}}\,dx\biggr]^{p}\Biggr\} ^{\frac{1}{p}}>k_{\lambda}( \sigma )\| F\|_{p,\varphi}, \end{aligned}$$
(50)
$$\begin{aligned}& \Biggl\{ \int_{\beta_{1}}^{\infty}(x-\beta_{1})^{-q\sigma-1} \Biggl[\sum_{n=1}^{\infty }\frac{\ln[(n-\beta_{2})/(x-\beta_{1})]a_{n}}{(n-\beta_{2})^{\lambda }-(x-\beta_{1})^{\lambda}} \Biggr]^{q}\,dx\Biggr\} ^{\frac{1}{q}}>k_{\lambda}(\sigma )\| a \|_{q,\psi}. \end{aligned}$$
(51)

In the same way, for \(\delta=-1\), \(\mu=\lambda-\sigma\) (>0) in Theorem 2, setting \(\varphi(x)=(x-\beta_{1})^{p(1-\mu)-1}\), and \(F(x)=(x-\beta_{1})^{\lambda}f(x)\), we still can find some new equivalent reverse inequalities with the best possible constant factor.

Declarations

Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 61370186) and 2013 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2013KJCX0140).

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Mathematics, Guangdong University of Education, Guangzhou, P.R. China

References

  1. Hardy, GH, Littlewood, JE, Pólya, G: Inequalities. Cambridge University Press, Cambridge (1952) MATHGoogle Scholar
  2. Yang, B: On a more accurate Hardy-Hilbert’s type inequality and its applications. Acta Math. Sin. 49(2), 363-368 (2006) MATHGoogle Scholar
  3. Mitrinović, DS, Pečarić, JE, Fink, AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic, Boston (1991) View ArticleMATHGoogle Scholar
  4. Pachpatte, BG: On some new inequalities similar to Hilbert’s inequality. J. Math. Anal. Appl. 226, 166-179 (1998) View ArticleMATHMathSciNetGoogle Scholar
  5. Gao, M, Yang, B: On the extended Hilbert’s inequality. Proc. Am. Math. Soc. 126(3), 751-759 (1998) View ArticleMATHMathSciNetGoogle Scholar
  6. Hu, K: On Hardy-Littlewood-Pólya inequality. Acta Math. Sci. 20, 684-687 (2000) MATHGoogle Scholar
  7. Lu, Z: Some new generalizations of Hilbert’s integral inequalities. Indian J. Pure Appl. Math. 33(5), 691-704 (2002) MathSciNetGoogle Scholar
  8. Kuang, J, Debnath, L: On new generalizations of Hilbert’s inequality and their applications. J. Math. Anal. Appl. 245, 248-265 (2000) View ArticleMATHMathSciNetGoogle Scholar
  9. Sulaiman, WT: On Hardy-Hilbert’s integral inequality. J. Inequal. Pure Appl. Math. 5(2), 25 (2004) MathSciNetGoogle Scholar
  10. Zhong, W, Yang, B: A reverse Hilbert’s type integral inequality with some parameters and the equivalent forms. Pure Appl. Math. 24(2), 401-407 (2008) MathSciNetGoogle Scholar
  11. Yang, B: On a Hilbert-type operator with a symmetric homogeneous kernel of -1-order and applications. J. Inequal. Appl. 2007, Article ID 47812 (2007) View ArticleGoogle Scholar
  12. Zhong, J, Yang, B: On a relation to four basic Hilbert-type integral inequalities. Appl. Math. Sci. 4(19), 923-930 (2010) MATHMathSciNetGoogle Scholar
  13. Yang, B: A mixed Hilbert-type inequality with a best constant factor. Int. J. Pure Appl. Math. 20(3), 319-328 (2005) MATHMathSciNetGoogle Scholar
  14. Yang, B: A half-discrete Hilbert’s inequality. J. Guangdong Univ. Educ. 31(3), 1-8 (2011) Google Scholar
  15. Xie, Z, Zeng, Z: A new half-discrete Hilbert’s inequality with the homogeneous kernel of degree \(-4\mu\). J. Zhanjiang Norm. Coll. 32(6), 13-19 (2011) Google Scholar
  16. Yang, B: A new half-discrete Mulholland-type inequality with parameters. Ann. Funct. Anal. 3(1), 142-150 (2012) View ArticleMATHMathSciNetGoogle Scholar
  17. Chen, Q, Yang, B: On a more accurate half-discrete Mulholland’s inequality and an extension. J. Inequal. Appl. 2012, Article ID 70 (2012). doi:10.1186/1029-242X-2012-70 View ArticleGoogle Scholar
  18. Zhong, W: A mixed Hilbert-type inequality and its equivalent forms. J. Guangdong Univ. Educ. 31(5), 18-22 (2011) MATHGoogle Scholar
  19. Yang, B: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing (2009) Google Scholar
  20. Kuang, J: Applied Inequalities. Shandong Science Technic Press, Jinan (2010) Google Scholar
  21. Kuang, J: Real and Functional Analysis. Higher Education Press, Beijing (2002) Google Scholar

Copyright

© Wang and Yang; licensee Springer. 2015

Advertisement