On a Reverse Half-Discrete Hilbert-Type Inequality with the General Homogeneous Kernel as Well as Multiple Lower Limit Function and Remainder Sum

1. Introduction

If $p>1,\frac{1}{p}+\frac{1}{q}=1,a_m,b_n\ge 0,0<{\displaystyle {\sum }_{m=1}^{\infty }{a}_m^p}<\infty and0<{\displaystyle {\sum }_{n=1}^{\infty }{b}_n^q}<\infty$ , then we have the following Hardy-Hilbert’s inequality with the best value $\frac{\pi }{\sin (\pi /p)}$ as follows (cf. [1], Theorem 315)：

$${\displaystyle {\displaystyle \sum _{m=1}^{\infty }}{\displaystyle \sum _{n=1}^{\infty }\frac{a_m{b}_n}{m+n}}}<\frac{\pi }{\sin (\pi /p)}{\left({\displaystyle \sum _{m=1}^{\infty }{a}_m^p}\right)}^{\frac{1}{p}}{\left({\displaystyle \sum _{n=1}^{\infty }{b}_n^q}\right)}^{\frac{1}{q}}.$$

(1)

In 2006, an extension of (1) was given by Krnic et al. [2] as the follows:

$${\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{a_m{b}_n}{{(m+n)}^{\lambda }}}}<B({\lambda }_1,{\lambda }_2){\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1-{\lambda }_1)-1}{a}_m^p}\right]}^{\frac{1}{p}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{q(1-{\lambda }_2)-1}{b}_n^q}\right]}^{\frac{1}{q}},$$

(2)

where, ${\lambda }_i\in (0,2](i=1,2),{\lambda }_1+{\lambda }_2=\lambda \in (0,4],$ the constant $B({\lambda }_1,{\lambda }_2)$ is the best value, and

$B(u,v):={\displaystyle {\int }_0^{\infty }\frac{t^{u-1}}{{(1+t)}^{u+v}}}dt(u,v>0)$ is the Beta function. For $p=q=2,$ ${\lambda }_1={\lambda }_2=\frac{\lambda }{2},$ (2) deduces to Yang’s inequality published in 2001 (cf. [3]). Some generalizations of (1) were cited by [4,5,6,7,8,9,10,11,12,13,14,15,16].

In 2016, by using the techniques of real analysis, Hong et al. [17] considered some equivalent conditions of the best value related to several parameters in the general form of (1). Some further results were published in [18,19,20,21,22,23]. In 2019, following [2], Adiyasuren et al. [24] gave an extended application of (3) involving two partial sums as follows:

For ${\lambda }_i\in (0,1]\cap (0,\lambda )$ $(\lambda \in (0,2];i=1,2),$ ${\lambda }_1+{\lambda }_2=\lambda ,$

$${\displaystyle \sum _{m=1}^{\infty }{\displaystyle \sum _{n=1}^{\infty }\frac{a_m{b}_n}{{(m+n)}^{\lambda }}}}<{\lambda }_1{\lambda }_{2}B({\lambda }_1,{\lambda }_2){\left({\displaystyle \sum _{m=1}^{\infty }{m}^{-p{\lambda }_1-1}{A}_m^p}\right)}^{\frac{1}{p}}{\left({\displaystyle \sum _{n=1}^{\infty }{n}^{-q{\lambda }_2-1}{B}_n^q}\right)}^{\frac{1}{q}},$$

(3)

where, ${\lambda }_1{\lambda }_{2}B({\lambda }_1,{\lambda }_2)$ is the best value, $a_m,b_n\ge 0,$ $A_m={\displaystyle {\sum }_{i=1}^m{a}_i}$ and $B_n={\displaystyle {\sum }_{k=1}^n{b}_k}$ $(m,n\in \mathrm{N}$   $=\{1,2,\dots \})$ , satisfying $A_m=o(e^{tm})$ , $B_n=o(e^{tn})$ $(t>0;m,n\to \infty ),$

$$0<{\displaystyle {\displaystyle \sum _{m=1}^{\infty }}{m}^{-p{\lambda }_1-1}{A}_m^p}<\infty and 0<{\displaystyle {\displaystyle \sum _{n=1}^{\infty }}{n}^{-q{\lambda }_2-1}{B}_n^q}<\infty .$$

In 2021, by using Hermite-Hadamard’s inequality, Huang et al. [26] proved a half-discrete Mulholland-type inequality involving one multiple upper limit function. Recently, Yang et al. [25] provided a more accurate extension of (3) with the kernel involving two partial sums. Some new extended applications on half-discrete Hardy-Hilbert’s inequalities with the particular kernels as (3) were published in [27,28,29,30]. In 2025-2026, by using Hardy’s integral inequality, a new Hilbert-type integral inequality with the general homogeneous kernel involving two derivative functions of higher-order, and an extended multidimensional half-discrete Hilbert-type inequality with the general homogeneous kernel involving one derivative function of m-order were provided in Yang and Liao et al. [31,32].

In this article, by means of the weight functions, the idea of introduced parameters and the reverse Hardy’s integral inequality, we obtain a new extended reverse Hardy-type inequality and then derive a reverse half-discrete Hilbert-type inequality with the general homogeneous kernel as $K_{\lambda }(m,y)$   $(\lambda \in \mathrm{R};$ $m\in \mathrm{N},$   $\mathrm{y}\in {\mathrm{R}}_{+}),$ as well as one multiple lower limit function and one remainder sum. The equivalent statements of the best value related to several parameters are considered. The equivalent forms and some particular examples are provided as applications..

2. Some Lemmas

In what follows, we assume that

(A1). $p<0(0<q<1),\frac{1}{p}+\frac{1}{q}=1,$ $\lambda ,{\lambda }_2\in \mathrm{R}=(-\infty ,\infty ),{\lambda }_1\in (0，1],$ ${\widehat{\lambda }}_1:=\frac{\lambda -{\lambda }_2}{p}+\frac{{\lambda }_1}{q},$ ${\widehat{\lambda }}_2:=\frac{\lambda -{\lambda }_1}{q}+\frac{{\lambda }_2}{p},$ $a_m,g(y)\ge 0$ $(m\in \mathrm{N},y\in {\mathrm{R}}_{+}=(0,\infty )),$ for $n\in {\mathrm{N}}_0=\mathrm{N}\cup \left\{0\right\},$ $0<{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}}{a}_m^p<\infty$ and $0<{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\widehat{\lambda }}_2)-1}}{g}^q(y)dy<\infty .$ We also assume that ${\hat{\mathrm{G}}}_0(y):=g(y)$ is continuous unless at finite points in ${\mathrm{R}}_{+}，$ and define multiple lower limit functions as follows: For $\mathrm{n}\in \mathrm{N},k=1,\dots ,n,$.

${\widehat{G}}_k(y):={\displaystyle {\int }_y^{\infty }\left({\displaystyle {\int }_{t_{k-1}}^{\infty }\dots {\displaystyle {\int }_{t_1}^{\infty }g(t_0)d{t}_0\dots d{t}_{k-2}}}\right)}d{t}_{k-1}.$

By the following reverse Hardy’s integral inequality (cf. [1], Theorem 347):

$${\displaystyle {\int }_0^{\infty }{\mathrm{y}}^{-\mathrm{r}}}({\displaystyle {\int }_y^{\infty }f(x})dx{)}^{q}dy>{(\frac{q}{1-r})}^q{\displaystyle {\int }_0^{\infty }{y}^{q-r}{f}^q(y})dy,$$

(1)

where, $0<\mathrm{q}<1,\mathrm{r}<1,\mathrm{f}(\mathrm{y})\ge 0,0<{\displaystyle {\int }_0^{\infty }{\mathrm{y}}^{\mathrm{q}-\mathrm{r}}{\mathrm{f}}^{\mathrm{q}}(y)dy}<\infty$ . For ${\widehat{\lambda }}_2>0,\mathrm{k}=1,\dots ,\mathrm{n}$ , setting $\mathrm{f}(\mathrm{y})={\mathrm{G}^}_{\mathrm{k}-1}(y)$ , ${\displaystyle {\int }_y^{\infty }\mathrm{f}}(\mathrm{x})\mathrm{dx}={\mathrm{G}^}_{\mathrm{k}}(y)$ , and $\mathrm{r}=-\mathrm{q}({\widehat{\lambda }}_2+\mathrm{n}-\mathrm{k})+1(<1)$ in (1), we have

$${[{\displaystyle {\int }_0^{\infty }{\mathrm{y}}^{\mathrm{q}({\widehat{\lambda }}_2+n-k)-1}}{\widehat{G}}_k^q(y)dy]}^{\frac{1}{q}}\ge \frac{1}{{\widehat{\lambda }}_2+n-k}{[{\displaystyle {\int }_0^{\infty }{\mathrm{y}}^{\mathrm{q}({\widehat{\lambda }}_2+n-k+1)-1}{\widehat{G}}_{k-1}^q(y)dy}]}^{\frac{1}{q}}$$

(2)

where, for $\mathrm{k}=1，$ since $0<{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\widehat{\lambda }}_2)-1}}{g}^q(y)dy<\infty ,$ (1) keeps the form of a strict inequality. Substitution of $\mathrm{k}=1,\dots ,\mathrm{n}$ in (1), by simplifications, since ${\displaystyle {\prod }_{\mathrm{k}=1}^{\mathrm{n}}{({\widehat{\lambda }}_2+n-k)}^{-1}}$   $={\displaystyle {\prod }_{\mathrm{k}=0}^{\mathrm{n}-1}{({\widehat{\lambda }}_2+k)}^{-1}},$ we obtain the following reverse inequality:

${[{\displaystyle {\int }_0^{\infty }{\mathrm{y}}^{\mathrm{q}{\widehat{\lambda }}_2-1}}{\widehat{G}}_n^q(y)dy]}^{\frac{1}{q}}\ge$

$${\displaystyle \prod _{k=0}^{n-1}\frac{1}{{\widehat{\lambda }}_2+k}}{[{\displaystyle {\int }_0^{\infty }{\mathrm{y}}^{\mathrm{q}(\mathrm{n}+{\widehat{\lambda }}_2)-1}gq(y)dy}]}^{\frac{1}{q}}.$$

(3)

Note. If we denote ${\displaystyle {\prod }_{\mathrm{k}=0}^{-1}{(a+k)}^{-1}：=0(a>0)},$ then (3) keeps the form of an equality

for $\mathrm{n}=0$ ; for $\mathrm{n}\in \mathrm{N}，$ (3) keeps the form of a strict inequality.

Lemma 1

. If $a_m\ge 0,m\in {\mathrm{N},\mathrm{s},\mathrm{s}}_1,s_2\in {\mathrm{R},\mathrm{s}}_1+s_2={\mathrm{s},\mathrm{k}}_{\mathrm{s}}(x,y)(\ge 0)$  is a homogeneous function of -s-degree in   ${\mathrm{R}}_{+}^2,$  satisfying for any   $u,x,y\in {\mathrm{R}}_{+},$  ${\mathrm{k}}_{\mathrm{s}}(ux,uy)=u^{-s}{\mathrm{k}}_{\mathrm{s}}(x,y)$  , then we define the following weight functions:

$\omega (s_2,m):=m^{s_1}{\displaystyle \sum _{n=1}^{\infty }{k}_s}(m,n)n^{s_2-1}(m\in \mathrm{N}),$                      (4)

$\varpi (s_1,n):=n^{s_2}{\displaystyle \sum _{m=1}^{\infty }{k}_s}(m,n)m^{s_1-1}(n\in \mathrm{N}).$                       (5)

For  $0<$  $\varpi (s_1,n)<\infty (n\in \mathrm{N}),$  the following reverse inequality holds:

$$J_0:={\left[{\displaystyle \sum _{n=1}^{\infty }\frac{n^{p{s}_2-1}}{{(\varpi (s_1,n))}^{p-1}}{\left({\displaystyle \sum _{m=1}^{\infty }{k}_s(m,n)a_m}\right)}^p}\right]}^{\frac{1}{p}}\ge {\left[{\displaystyle \sum _{m=1}^{\infty }\omega (s_2,m)m^{p(1-s_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}.$$

(6)

Proof. If $J_0=\infty ,$ then (6) is naturally valid; if $J_0=0,$ then (6) keeps the form of an equality, or it is impossible that makes (6) valid for ${\displaystyle {\sum }_{m=1}^{\infty }\omega (s_2,m)m^{p(1-s_1)-1}}{a}_m^p>0$ . In the following, we suppose $0<J_0<\infty .$ For ${\mathrm{s}}_1+s_2=\mathrm{s},$ setting

$b_n:=\frac{n^{p{s}_2-1}}{{(\varpi (s_1,n))}^{p-1}}{\left({\displaystyle \sum _{m=1}^{\infty }{k}_s(m,n)a_m}\right)}^{p-1},n\in \mathrm{N},$

by the reverse H $\ddot{o}$ lder’s inequality (cf. [33]), we find

${\displaystyle {\displaystyle \sum _{n=1}^{\infty }}\varpi }(s{}_1,n)n{}_{q(1-s_2)-1}bnq=J_0^p=I{}_0:={\displaystyle {\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \sum _{m=1}^{\infty }{k}_s}}(m,n)a_m{b}_n$

$\begin{array}{l}={\displaystyle \sum _{n=1}^{\infty }{\displaystyle \sum _{m=1}^{\infty }{k}_s}}(m,n)\left[\frac{m^{(1-s_1)/q}}{n^{(1-s_2)/p}}{a}_m\right]\cdot \left[\frac{n^{(1-s_2)/p}}{m^{(1-s_1)/q}}{b}_n\right]\\ \ge {\left\{{\displaystyle \sum _{m=1}^{\infty }\left[{\displaystyle {\displaystyle \sum _{n=1}^{\infty }}{k}_s}(m,n)\frac{m^{(1-s_1)(p-1)}}{n^{1-s_2}}\right]}{a}_m^p\right\}}^{\frac{1}{p}}{\left\{{\displaystyle \sum _{n=1}^{\infty }\left[{\displaystyle {\displaystyle \sum _{m=1}^{\infty }}{k}_s}(m,n)\frac{n^{(1-s_2)(q-1)}}{m^{1-s_1}}\right]}{b}_n^q\right\}}^{\frac{1}{q}}\end{array}$

$={\left[{\displaystyle \sum _{m=1}^{\infty }\omega (s_2,m)}{m}^{p(1-s_1)-1}{a}_m^p\right]}^{\frac{1}{p}}$${\left[{\displaystyle \sum _{n=1}^{\infty }\varpi }(s{}_1,n)n{}_{q(1-s_2)-1}bnq\right]}^{\frac{1}{q}}$

$={\left[{\displaystyle \sum _{m=1}^{\infty }\omega (s_2,m)}{m}^{p(1-s_1)-1}{a}_m^p\right]}^{\frac{1}{p}}{J}_0^{p-1}.$

Dividing $J_0^{p-1}(\in (0,\infty ))$ in the above inequality, we obtain (6). The lemma is proved.

Remark 1

. For $s_2\in (0,1],s_1+s_2=s,$$0<{\displaystyle {\sum }_{m=1}^{\infty }{m}^{p(1+s_2)-1}}{a}_m^p<\infty ,$  we set

$k_{\mathrm{s}}(x,y):=\left\{\begin{array}{c}0,0<x\le y\\ \frac{1}{x^s},x>y\end{array}\right.,k_{\mathrm{s}}(m,n)m^{s_1-1}=\left\{\begin{array}{c}0,m\le n,\\ \frac{1}{m^{s_2+1}},m\ge n+1\end{array}\right.(m,n\in \mathrm{N}),$

and  ${\displaystyle {\sum }_{m=1}^{\infty }{k}_s(m,n)}am={\displaystyle {\sum }_{m=n+1}^{\infty }\frac{1}{m^s}{a}_m}.$ 

In view of the decreasingness property of series, for   ${\mathrm{s}}_1+s_2={\mathrm{s},\mathrm{s}}_2\in (0,1],$  we obtain

$\begin{array}{l}0<\varpi (s_1,n)=n^{s_2}{\displaystyle {\displaystyle \sum _{m=n+1}^{\infty }}\frac{1}{m^{s_2+1}}}<n^{s_2}{\displaystyle {\int }_n^{\infty }{x}^{-s_2-1}}dx=\frac{1}{s_2}<\infty ,\\ \varpi (s_1,n)=n^{s_2}\left({\displaystyle {\displaystyle \sum _{m=n}^{\infty }}\frac{1}{m^{s_2+1}}}-\frac{1}{n^{s_2+1}}\right)>n^{s_2}{\displaystyle {\int }_n^{\infty }{x}^{-s_2-1}}dx-\frac{1}{n}=\frac{1}{s_2}(1-\frac{s_2}{n}),\\ \omega (s_2,m)=m^{-s_2}{\displaystyle {\displaystyle \sum _{n=1}^{m-1}}{n}^{s_2-1}}<m^{-s_2}{\displaystyle {\int }_0^m{x}^{s_2-1}}dx=\frac{1}{s_2}\end{array}$

(Note. We denote   ${\displaystyle {\sum }_{n=1}^0{n}^{s_2-1}}：=0$  ). Then by (6), for   $\mathrm{s}={0,-\mathrm{s}}_1=s_2\in (0,1],$  we obtain

$s_2^{\frac{1}{q}}{\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{p{s}_2-1}{(1-\frac{s_2}{n})}^{1-p}{\left({\displaystyle \sum _{m=n+1}^{\infty }{a}_m}\right)}^p}\right]}^{\frac{1}{p}}$   $\ge J_0$ $>{(\frac{1}{s_2})}^{\frac{1}{p}}{\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+s_2)-1}}{a}_m^p\right]}^{\frac{1}{p}},$

and then it follows that

$${\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{p{s}_2-1}{(1-\frac{s_2}{n})}^{1-p}{\left({\displaystyle \sum _{m=n+1}^{\infty }{a}_m}\right)}^p}\right]}^{\frac{1}{p}}>\frac{1}{s_2}{\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+s_2)-1}}{a}_m^p\right]}^{\frac{1}{p}}.$$

(7)

Setting  $\mathrm{r}=-{\mathrm{ps}}_2+1\in (1，1-p](s_2\in (0,1],p<0)$  in (7), equivalently, we find an extended reverse Hardy-type inequality as follows:

$${\left[{\displaystyle \sum _{n=1}^{\infty }{n}^{-r}{(1-\frac{1-r}{pn})}^{1-p}{\left({\displaystyle \sum _{m=n+1}^{\infty }{a}_m}\right)}^p}\right]}^{\frac{1}{p}}>\frac{p}{1-r}{\left({\displaystyle \sum _{m=1}^{\infty }{m}^{p-r}}{a}_m^p\right)}^{\frac{1}{p}}.$$

(8)

In particular, for   $\mathrm{r}=1-p(p<0)$  , we have

$${\displaystyle \sum _{n=1}^{\infty }{n}^{p-1}{(1-\frac{1}{n})}^{1-p}{\left({\displaystyle \sum _{m=n+1}^{\infty }{a}_m}\right)}^p}<{\displaystyle \sum _{m=1}^{\infty }{m}^{2p-1}}{a}_m^p.$$

(9)

For   ${\widehat{\lambda }}_1\in (0,1],$  $\mathrm{r}=-\mathrm{p}{\widehat{\lambda }}_1+1(\in (1,1-p]),$  ${\widehat{A}}_m^{(0)}：=a_m,{\widehat{A}}_m^{(1)}：={\displaystyle {\sum }_{k=m+1}^{\infty }{a}_k}$  in (8),we have

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p({\widehat{\lambda }}_1+1-i)-1}{(1-\frac{{\widehat{\lambda }}_1}{n})}^{i(1-p)}({\widehat{A}}_m^{(i)}}{)}^p\right]}^{\frac{1}{p}}\ge \frac{1}{{\widehat{\lambda }}_1^i}{\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}{a}_m^p}\right]}^{\frac{1}{p}},$$

(10)

where,   $i\in \{0,1\},$  for   $i=0，$  (10) keeps the form of an equality, and we don’t need   ${\widehat{\lambda }}_1\in (0,1]$  ; for   $i=1，$  (10) keeps the form of a strict inequality, and we need   ${\widehat{\lambda }}_1\in (0,1]$  in this case.

Lemma 2

. With regards to the assumption (A1), if $K_{\lambda }(x,y)(\ge 0)$is a homogeneous function of  $-\lambda -$  degree in   ${\mathrm{R}}_{+}^2$  , for any fixed   $y>0,K_{\lambda }(x,y)$  is decreasing in  $\mathrm{x}\in {\mathrm{R}}_{+}$  and strictly decreasing in an interval   $I\subset {\mathrm{R}}_{+}$  , such that

$$K_{\lambda }(\eta ):={\displaystyle {\int }_0^{\infty }{K}_{\lambda }(u,1)}{u}^{\eta -1}du\in {\mathrm{R}}_{+}(\eta ={\lambda }_1,\lambda -{\lambda }_2),$$

(11)

then we define the following weight functions:

$${\omega }_{\lambda }({\lambda }_1,y):=y^{\lambda -{\lambda }_1}{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}(m,y)m^{{\lambda }_1-1}(y\in {\mathrm{R}}_{+}),$$

(12)

$${\varpi }_{\lambda }({\lambda }_2,m):=m^{\lambda -{\lambda }_2}{\displaystyle {\int }_0^{\infty }{K}_{\lambda }}(m,y)y^{{\lambda }_2-1}dy(m\in \mathrm{N}),$$

(13)

We have the following inequality and expression:

$${\omega }_{\lambda }({\lambda }_1,y)<K_{\lambda }({\lambda }_1)(y\in {\mathrm{R}}_{+}^{\mathrm{n}}),$$

(14)

${\varpi }_{\lambda }({\lambda }_2,m)=K_{\lambda }(\lambda -{\lambda }_2)(m\in \mathrm{N}).$                           (15)

Proof. For ${\lambda }_1\le 1,$ in view of the decreasingness property of series, setting $\mathrm{u}=\frac{\mathrm{x}}{\mathrm{y}},$ we obtain

${\omega }_{\lambda }({\lambda }_1,y)<y^{\lambda -{\lambda }_1}{\displaystyle {\int }_0^{\infty }{K}_{\lambda }}(x,y)x^{{\lambda }_1-1}dx={\displaystyle {\int }_0^{\infty }{K}_{\lambda }}(u,1)u^{{\lambda }_1-1}du=K_{\lambda }({\lambda }_1),$ and then (14) follows. By (13), setting $u=\frac{m}{y},$ we find

${\varpi }_{\lambda }({\lambda }_2,m)=$${\displaystyle {\int }_0^{\infty }{K}_{\lambda }}(u,1)u^{(\lambda -{\lambda }_2)-1}du=K_{\lambda }(\lambda -{\lambda }_2),$ namely, (15) follows. The lemma is proved.

3. Main Results

Theorem 1

. With regards to the assumption (A1), if for $i=1,$

$$p(1-{\lambda }_1)\le \lambda -\lambda {}_1-{\lambda }_2<-p{\lambda }_1,$$

(16)

and for  $n\in {\mathrm{N}}_0,\lambda -{\lambda }_1-{\lambda }_2>-q{\lambda }_2,$  then for   $i\in \{0,1\},n\in {\mathrm{N}}_0$  , we have

$I_i:={\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)m^{2{\widehat{\lambda }}_1-i}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}{y}^{2{\widehat{\lambda }}_2-1}{\widehat{G}}_n(y)dy$  $>\frac{1}{{\widehat{\lambda }}_1^i}{K}_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1){\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\widehat{\lambda }}_2+k}}$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\widehat{\lambda }}_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}.$$

(17)

In particular, for   ${\lambda }_1+\lambda {}_2=\lambda ,$  in view of the assumption, we have   ${\lambda }_2>0(n\in {\mathrm{N}}_0),$ 

$0<{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p<\infty ,$  $0<{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\lambda }_2)-1}}{g}^q(y)dy<\infty ,$ 

and the following inequality:

${\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)m^{2{\lambda }_1-i}{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}{y}^{2{\lambda }_2-1}{\widehat{G}}_n(y)dy$  $>\frac{1}{{\lambda }_1^i}{K}_{\lambda }({\lambda }_1){\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\lambda }_2+k}}$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\lambda }_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}.$$

(18)

Proof. In view of (16), for $i=1,$ we have ${\widehat{\lambda }}_1=\frac{\lambda -{\lambda }_1-{\lambda }_2}{p}+{\lambda }_1\in (0,1]；$ for $n\in {\mathrm{N}}_0,$ since $\lambda -{\lambda }_1-{\lambda }_2>-q{\lambda }_2,$ we have ${\widehat{\lambda }}_2=\frac{\lambda -{\lambda }_1-{\lambda }_2}{q}+{\lambda }_2>0$ .

By the reverse H $\ddot{o}$ lder’s inequality (cf. [33]), it follows that

$\begin{array}{l}{I}_i={\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)\left[\frac{m^{(1-{\lambda }_1)/q}}{y^{(1-{\lambda }_2)/p}}{m}^{2{\widehat{\lambda }}_1-i}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}\right]\cdot \left[\frac{y^{(1-{\lambda }_2)/p}}{m^{(1-{\lambda }_1)/q}}{y}^{2{\widehat{\lambda }}_2-1}{\widehat{G}}_n(y)\right]dy\\ \ge {\left\{{\displaystyle {\displaystyle \sum _{m=1}^{\infty }}{\displaystyle {\int }_0^{\infty }{K}_{\lambda }}}(m,y)\frac{m^{(1-{\lambda }_1)(p-1)}}{y^{1-{\lambda }_2}}dy{[m^{2{\widehat{\lambda }}_1-i}{m}^{-i}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}]}^p\right\}}^{\frac{1}{p}}\end{array}$

$\times {\left\{{\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)\frac{y^{(1-{\lambda }_2)(q-1)}}{m^{1-{\lambda }_1}}{(y^{2{\widehat{\lambda }}_2-1}{\widehat{G}}_n(y))}^{q}dy\right\}}^{\frac{1}{q}}$

$\begin{array}{l}={\left\{{\displaystyle \sum _{m=1}^{\infty }\left[m^{\lambda -{\lambda }_2}{\displaystyle {\int }_0^{\infty }{K}_{\lambda }}(m,y)y^{{\lambda }_2-1}dy\right]}{m}^{p(1-{\widehat{\lambda }}_1)-1}{[m^{2{\widehat{\lambda }}_1-i}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}]}^p\right\}}^{\frac{1}{p}}\\ \times {\left\{{\displaystyle {\int }_0^{\infty }\left[y^{\lambda -{\lambda }_1}{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }(m,y)m^{{\lambda }_1-1}}\right]}{y}^{q(1-{\widehat{\lambda }}_2)-1}{(y^{2{\widehat{\lambda }}_2-1}{\widehat{G}}_n(y))}^{q}dy\right\}}^{\frac{1}{q}}\end{array}$

$\begin{array}{l}={\left[{\displaystyle \sum _{m=1}^{\infty }{\varpi }_{\lambda }({\lambda }_2,m)}{m}^{p({\widehat{\lambda }}_1+1-i)-1}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{i(1-p)}{({\widehat{A}}_m^{(i)})}^p\right]}^{\frac{1}{p}}\\ \times {\left[{\displaystyle {\int }_0^{\infty }{\omega }_{\lambda }({\lambda }_1,}y)y^{q{\widehat{\lambda }}_2-1}{\widehat{G}}_n^q(y)dy\right]}^{\frac{1}{q}}.\end{array}$

Then by (14) and (15), we have

$I_i\ge K_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p({\widehat{\lambda }}_1+1-i)-1}}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{i(1-p)}{({\widehat{A}}_m^{(i)})}^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q{\widehat{\lambda }}_2-1}}{\widehat{G}}_n^q(y)dy\right]}^{\frac{1}{q}},$$

(19)

where, for $i=n=0,$ in view of the assumption, (19) keeps the form of a strict inequality, and then (17) follows. For $i=1$ or $n\in \mathrm{N},$ by (3) and the strict form of (10), we still have (17).

The theorem is proved.

Theorem 2

. With regards to the assumption of Theorem 1, if there exist constants${\delta }_0,M_0>0,$such that the following inequality

$$K_{\lambda }(u,1)\le M_0{u}^{-{\delta }_0-{\lambda }_1}(u\in (0,\infty ))$$

(20)

holds, then for  ${\lambda }_1+\lambda {}_2=\lambda ,$  the constant factor in (17) is the best value.

Proof. We need to show that the constant factor in (18) is the best value for $i\in \{0,1\}.$ For any $0<\epsilon <\min \left\{\right|p|{\lambda }_1,\frac{q}{2}{\delta }_0\},$ we set

${\tilde{A}}_m^{(0)}:={\tilde{a}}_m=m^{-{\lambda }_1-\frac{\epsilon }{p}-1},{\tilde{A}}_m^{(1)}:={\displaystyle {\displaystyle \sum _{k=m+1}^{\infty }}{\tilde{a}}_k}={\displaystyle {\displaystyle \sum _{k=m+1}^{\infty }}{k}^{-{\lambda }_1-\frac{\epsilon }{p}-1}}(m\in \mathrm{N}),and$${\tilde{G}}_0(y)=\tilde{g}(y):=\left\{\begin{array}{c}0,0<y<1,\\ y^{-n-{\lambda }_2-\frac{\epsilon }{q}},y\ge 1\end{array}\right..$

By integration, we find that for $n\in \mathrm{N},$

${\tilde{G}}_n(y):={\displaystyle {\int }_y^{\infty }\left({\displaystyle {\int }_{t_{n-1}}^{\infty }\dots {\displaystyle {\int }_{t_1}^{\infty }\tilde{g}(t_0)d{t}_0\dots d{t}_{n-2}}}\right)}d{t}_{n-1}$

$$={\displaystyle \prod _{k=0}^{n-1}{({\lambda }_2+\frac{\epsilon }{q}+k)}^{-1}}\left\{\begin{array}{c}{p}_{n-1}(y),0<y<1,\\ y^{-{\lambda }_2-\frac{\epsilon }{q}},y\ge 1\end{array}\right.,$$

(21)

where, $p_{n-1}(y)$ is a polynomial 0f $n-1$ -degree with positive coefficients in $(0,1)$ , and $p_{n-1}(1^{-})=1.$ If we denote $p_{-1}(y):=0,$ and ${\displaystyle {\prod }_{k=0}^{-1}{(a+k)}^{-1}：=1}(a>0)$ , then (21) satisfies for $n\in {\mathrm{N}}_0.$ In view of the decreasingness property of series, we find

${\tilde{A}}_m^{(1)}<{\displaystyle {\int }_m^{\infty }{x}^{-{\lambda }_1-\frac{\epsilon }{p}-1}}dx=\frac{1}{{\lambda }_1+\frac{\epsilon }{p}}{m}^{-{\lambda }_1-\frac{\epsilon }{p}},$

and then for $i\in \{0,1\},$ it follows that $m^{{\lambda }_1-\frac{\epsilon }{q}-i}{\tilde{A}}_m^{(i)}\le \frac{1}{{({\lambda }_1+\frac{\epsilon }{p})}^i}{m}^{-\epsilon -1}(m\in \mathrm{N}).$ If there exists a positive constant $M,$ with

$$M\ge \frac{1}{{\lambda }_1^i}{K}_{\lambda }({\lambda }_1){\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\lambda }_2+k}}，$$

(22)

such that (18) is valid when we replace the constant factor by $M,$ then in particular, we have

${\tilde{I}}_i:={\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)m^{2{\lambda }_1-i}{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}{\tilde{A}}_m^{(i)}{y}^{2{\lambda }_2-1}{\tilde{G}}_n(y)dy$

$>M{\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{\tilde{a}}_m^p\right]}^{\frac{1}{p}}$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\lambda }_2)-1}}{\tilde{g}}^q(y)dy\right]}^{\frac{1}{q}}$

$=M{\left(1+{\displaystyle \sum _{m=2}^{\infty }{m}^{-\epsilon -1}}\right)}^{\frac{1}{p}}$${\left({\displaystyle {\int }_1^{\infty }{y}^{-\epsilon -1}}dy\right)}^{\frac{1}{q}}$

$>M{\left(1+{\displaystyle {\int }_1^{\infty }{x}^{-\epsilon -1}dx}\right)}^{\frac{1}{p}}{\left({\displaystyle {\int }_1^{\infty }{y}^{-\epsilon -1}}dy\right)}^{\frac{1}{q}}=\frac{M}{\epsilon }{(\epsilon +1)}^{\frac{1}{p}}.$

Since we find

$\underset{x\to 0^{+}}{\lim }\frac{{(1-{\lambda }_{1}x)}^{-\frac{1}{q}}-1}{x}=\frac{{\lambda }_1}{q}\underset{x\to 0^{+}}{\lim }{(1-{\lambda }_{1}x)}^{-\frac{1}{q}-1}=\frac{{\lambda }_1}{q},$

There exists a positive constant $M_1,$ such that

${(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}\le {(1-\frac{{\lambda }_1}{m})}^{-\frac{1}{q}}\le 1+\frac{M_1}{m}(m\in \mathrm{N},\mathrm{i}\in \{0,1\}).$

There exists a positive constant $M_2,$ such that $p_{n-1}(y)\le M_2(y\in (0,1)).$ By (20), we have

$\begin{array}{l}A:={\displaystyle {\int }_0^1{K}_{\lambda }(m,y)y^{2{\lambda }_2-1}}{p}_{n-1}(y)dy={\displaystyle {\int }_0^1{K}_{\lambda }(\frac{m}{y},1)y^{-\lambda +2{\lambda }_2-1}}{p}_{n-1}(y)dy\\ \le M_0{M}_2{\displaystyle {\int }_0^1{(\frac{m}{y})}^{-{\delta }_0-{\lambda }_1}{y}^{-\lambda +2{\lambda }_2-1}}dy=\frac{M_0{M}_2}{{\delta }_0+{\lambda }_2}\frac{1}{m^{{\delta }_0+{\lambda }_1}}.\end{array}$

In view of the above results, setting $u=\frac{m}{y},$ we obtain that

${\tilde{I}}_i\le {\displaystyle {\displaystyle \prod _{k=0}^{n-1}}{({\lambda }_2+\frac{\epsilon }{q}+k)}^{-1}}{\displaystyle {\displaystyle \sum _{m=1}^{\infty }}{m}^{2{\lambda }_1-i}(1+\frac{M_1}{m}){\tilde{A}}_m^{(i)}}\left({\displaystyle {\int }_0^{\infty }{K}_{\lambda }}(m,y)y^{{\lambda }_2-\frac{\epsilon }{q}-1}dy+A\right)$

$\begin{array}{l}={\displaystyle {\displaystyle \prod _{k=0}^{n-1}}{({\lambda }_2+\frac{\epsilon }{q}+k)}^{-1}}{\displaystyle {\displaystyle \sum _{m=1}^{\infty }}(1+\frac{M_1}{m})m^{{\lambda }_1-\frac{\epsilon }{q}-i}{\tilde{A}}_m^{(i)}}(K_{\lambda }({\lambda }_1+\frac{\epsilon }{q})+m^{-{\lambda }_1-\frac{\epsilon }{q}}\mathrm{A})\\ \le {\displaystyle \prod _{k=0}^{n-1}{({\lambda }_2+\frac{\epsilon }{q}+k)}^{-1}\frac{K_{\lambda }({\lambda }_1+\frac{\epsilon }{q})}{{({\lambda }_1+\frac{\epsilon }{p})}^i}}\\ \times {\displaystyle {\displaystyle \sum _{m=1}^{\infty }}(\frac{1}{m^{1+\epsilon }}+\frac{M_1}{m^{2+\epsilon }})}(1+K_{\lambda }-1({\lambda }_1+\frac{\epsilon }{q})\frac{M_0{M}_2}{{\delta }_0+{\lambda }_2}{m}^{-{\delta }_0-2{\lambda }_1-\frac{\epsilon }{q}})\\ ={\displaystyle \prod _{k=0}^{n-1}{({\lambda }_2+\frac{\epsilon }{q}+k)}^{-1}\frac{K_{\lambda }({\lambda }_1+\frac{\epsilon }{q})}{{({\lambda }_1+\frac{\epsilon }{p})}^i}\left({\displaystyle \sum _{m=2}^{\infty }\frac{1}{m^{1+\epsilon }}+O(1)}\right)}\end{array}$

$\begin{array}{l}<{\displaystyle \prod _{k=0}^{n-1}{({\lambda }_2+\frac{\epsilon }{q}+k)}^{-1}\frac{K_{\lambda }({\lambda }_1+\frac{\epsilon }{q})}{{({\lambda }_1+\frac{\epsilon }{p})}^i}}\left({\displaystyle {\int }_1^{\infty }{x}^{-\epsilon -1}dx}+O(1)\right)\\ ={\displaystyle \prod _{k=0}^{n-1}{({\lambda }_2+\frac{\epsilon }{q}+k)}^{-1}\frac{K_{\lambda }({\lambda }_1+\frac{\epsilon }{q})}{\epsilon {({\lambda }_1+\frac{\epsilon }{p})}^i}}\left(1+\epsilon O(1)\right).\end{array}$

Based on the above results, we have

${\displaystyle \prod _{k=0}^{n-1}{({\lambda }_2+\frac{\epsilon }{q}+k)}^{-1}\frac{K_{\lambda }({\lambda }_1+\frac{\epsilon }{q})}{{({\lambda }_1+\frac{\epsilon }{p})}^i}}\left(1+\epsilon O(1)\right)$  $<\epsilon {\tilde{I}}_i$

$$<M(\epsilon +1{)}^{\frac{1}{p}}.$$

(23)

In view of Levi theorem (cf. [34]), we have

${\displaystyle {\int }_0^1{K}_{\lambda }(u,1)u^{{\lambda }_1+\frac{\epsilon }{q}-1}}du={\displaystyle {\int }_0^1{K}_{\lambda }(u,1)u^{{\lambda }_1-1}}du+o_1(1)(\epsilon \to 0^{+}).$ Since by (20) and the assumption, we have

$0\le K_{\lambda }(u,1)u^{{\lambda }_1+\frac{\epsilon }{q}-1}\le M_0{u}^{-{\delta }_0-{\lambda }_1}{u}^{{\lambda }_1+\frac{{\delta }_0}{2}-1}=M_0{u}^{-\frac{{\delta }_0}{2}-1}(u\in [1,\infty )),$

and ${\displaystyle {\int }_1^{\infty }{M}_0}{u}^{-\frac{{\delta }_0}{2}-1}=\frac{2M_0}{{\delta }_0}\in {\mathrm{R}}_{+}$ , by Lebesgue dominate convergence theorem (cf. [34]), we have

${\displaystyle {\int }_1^{\infty }{K}_{\lambda }(u,1)u^{{\lambda }_1+\frac{\epsilon }{q}-1}}du={\displaystyle {\int }_1^{\infty }{K}_{\lambda }(u,1)u^{{\lambda }_1-1}}du+o_2(1)(\epsilon \to 0^{+}).$

Hence, we find

$\begin{array}{l}{K}_{\lambda }({\lambda }_1+\frac{\epsilon }{q})={\displaystyle {\int }_0^1{K}_{\lambda }(u,1)u^{{\lambda }_1+\frac{\epsilon }{q}-1}}du+{\displaystyle {\int }_1^{\infty }{K}_{\lambda }(u,1)u^{{\lambda }_1+\frac{\epsilon }{q}-1}}du\\ ={\displaystyle {\int }_0^1{K}_{\lambda }(u,1)u^{{\lambda }_1-1}}du+{\displaystyle {\int }_1^{\infty }{K}_{\lambda }(u,1)u^{{\lambda }_1-1}}du+o_1(1)+o_2(1)\\ =K_{\lambda }({\lambda }_1)+o(1)(\epsilon \to 0^{+}).\end{array}$

For $\epsilon \to 0^{+}$ , in view of (23) and the above results, we have $\frac{K_{\lambda }({\lambda }_1)}{{\lambda }_1^i}{\displaystyle {\prod }_{k=0}^{n-1}\frac{1}{{\lambda }_2+k}}$   $\ge M.$ Then by (22), we observe that $M=\frac{K_{\lambda }({\lambda }_1)}{{\lambda }_1^i}{\displaystyle {\prod }_{k=0}^{n-1}\frac{1}{{\lambda }_2+k}}$ is the best value in (18) (namely, for ${\lambda }_1+\lambda {}_2=\lambda$ in (17)).

The theorem is proved.

Theorem 3

. With regards to the assumption of Theorem 1, suppose that (16) is valid for $i\in \{0,1\}$; for  $n\in {\mathrm{N}}_0,$  $\lambda -{\lambda }_1-{\lambda }_2>-q{\lambda }_2,$  and   $K_{\lambda }({\widehat{\lambda }}_1)<\infty$  . If the constant factor in (17) is the best value, then we have  ${\lambda }_1+\lambda {}_2=\lambda$ .

Proof. In view of the assumptions, we have ${\widehat{\lambda }}_1+\widehat{\lambda }{}_2=\lambda$ with ${\widehat{\lambda }}_1\in (0,1],{\widehat{\lambda }}_2>0$ (for $i\in \{0,1\},$   $n\in {\mathrm{N}}_0$ ). By the reverse H $\ddot{o}$ lder’s inequality (cf. [33]), it follows that

$\begin{array}{l}{K}_{\lambda }({\widehat{\lambda }}_1)={\displaystyle {\int }_0^{\infty }{K}_{\lambda }(u,1)u^{{\widehat{\lambda }}_1-1}}du\\ ={\displaystyle {\int }_0^{\infty }{K}_{\lambda }(u,1)(u^{\frac{\lambda -{\lambda }_2-1}{p}})(u^{\frac{{\lambda }_1-1}{q}})}du\\ \ge {\left[{\displaystyle {\int }_0^{\infty }{K}_{\lambda }(u,1)u^{\lambda -{\lambda }_2-1}}du\right]}^{\frac{1}{p}}{\left[{\displaystyle {\int }_0^{\infty }{K}_{\lambda }(u,1)u^{{\lambda }_1-1}}du\right]}^{\frac{1}{q}}\end{array}$

$$=K_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)>0,$$

(24)

and then $K_{\lambda }({\widehat{\lambda }}_1)\in {\mathrm{R}}_{+}$ . Since the constant factor in (17) is the best value, comparing with the constant factors in (17) and (18) (for ${\lambda }_1={\widehat{\lambda }}_1,$   ${\lambda }_2={\widehat{\lambda }}_2$ ), we have

$\frac{1}{{\widehat{\lambda }}_1^i}{K}_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1){\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\widehat{\lambda }}_2+k}}$$\ge \frac{1}{{\widehat{\lambda }}_1^i}{K}_{\lambda }({\widehat{\lambda }}_1){\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\widehat{\lambda }}_2+k}},$

namely, $K_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)\ge K_{\lambda }({\widehat{\lambda }}_1),$ which follows that (24) keeps the form of an equality.

The necessary and sufficient condition for taking an equal sign is that there exist constants A and B, such that they are not both zero and (cf. [33]) $A{u}^{\lambda -{\lambda }_2-1}=B{u}^{{\lambda }_1-1}$ a,e, in ${\mathrm{R}}_{+}$ . Assuming that $A\ne 0$ , $u^{\lambda -{\lambda }_1-{\lambda }_2}=\frac{B}{A}$ a,e, in ${\mathrm{R}}_{+}$ , it means that $\lambda -{\lambda }_1-{\lambda }_2=0,$ and then ${\lambda }_1+{\lambda }_2=\lambda .$ The theorem is proved.

Remark 2

. (i) For $i=n=0$in (17), we have

${\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)m^{2{\widehat{\lambda }}_1}{a}_m{y}^{2{\widehat{\lambda }}_2-1}g(y)dy$  $>K_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q{\widehat{\lambda }}_2-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}.$$

(25)

In particular, for   ${\lambda }_1+{\lambda }_2=\lambda ，$  we have the following inequality with the best value:

${\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)m^{2{\lambda }_1}{a}_m{y}^{2{\lambda }_2-1}g(y)dy$  $>K_{\lambda }({\lambda }_1)$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q{\lambda }_2-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}.$$

(26)

Hence, (17) (resp. (18)) is an extended application of (25) (resp. (26)).

(ii) For   $i=1$  in (18),   ${\lambda }_1\in (0,1],{\lambda }_2>0(n\in {\mathrm{N}}_0)$  , setting   ${\widehat{A}}_m={\widehat{A}}_m^{(1)}={\displaystyle {\sum }_{k=\mathrm{m}+1}^{\infty }{a}_k}$  , we have the following inequality with the best value:

${\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)m^{2{\lambda }_1-1}{(1-\frac{{\lambda }_1}{m})}^{-\frac{1}{q}}{\widehat{A}}_m{y}^{2{\lambda }_2-1}{\widehat{G}}_n(y)dy$  $>\frac{1}{{\lambda }_1}{K}_{\lambda }({\lambda }_1){\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\lambda }_2+k}}$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\lambda }_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}.$$

(27)

(iii) For   $n=0$  in(17),   ${\widehat{\lambda }}_1\in (0,1](i=1)$  , replacing   $g(y)$  by   $y^{1-2{\widehat{\lambda }}_2}g(y)$  , for

$0<{\displaystyle {\int }_0^{\infty }{y}^{q(1-{\widehat{\lambda }}_2)-1}}{g}^q(y)dy<\infty ,$  we have

$I_i^{(0)}:={\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)m^{2{\widehat{\lambda }}_1-i}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}g(y)dy$   $>\frac{1}{{\widehat{\lambda }}_1^i}{K}_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)$ $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(1-{\widehat{\lambda }}_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}.$$

(28)

In particular, for   ${\lambda }_1+\lambda {}_2=\lambda ,$  we have the following inequality with the best value:.

${\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}}(m,y)m^{2{\lambda }_1-i}{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}g(y)dy$  $>\frac{1}{{\lambda }_1^i}{K}_{\lambda }({\lambda }_1)$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(1-{\lambda }_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}.$$

(29)

Example 1

. For$0<{\lambda }_1,{\lambda }_2<\lambda ,{\lambda }_1\le 1,{\lambda }_1+{\lambda }_2=\lambda$ ,

$K_{\lambda }(x,y)=\frac{1}{{(x+y)}^{\lambda }},\frac{\ln (x/y)}{x^{\lambda }-y^{\lambda }},\frac{1}{{(\max \{x,y\})}^{\lambda }}(x,y\in {\mathrm{R}}_{+}^2),$ 

We observe that in the above any kernel, for fixed$y>0,K_{\lambda }(x,y)$is decreasing in $\mathrm{x}\in {\mathrm{R}}_{+}$  and strictly decreasing in an interval $I\subset {\mathrm{R}}_{+}$  , and (cf. [4])

$K_{\lambda }({\lambda }_1)=B({\lambda }_1,\lambda {}_2),{[\frac{\pi }{\lambda \sin (\frac{\pi {\lambda }_1}{\lambda })}]}^2,\frac{\lambda }{{\lambda }_1{\lambda }_2}\in {\mathrm{R}}_{+}.$ 

Since   $K_{\lambda }(u,1)$  is continuous in   ${\mathrm{R}}_{+}$  , and for any   $\mathrm{b}\in (0,\lambda ),\mathrm{K}{}_{\lambda }(u,1)u^b\to 0(u\to 0^{+},\infty )$  , there exists a constant   ${\mathrm{M}}_{\mathrm{b}}>0$  , such that   $\mathrm{K}{}_{\lambda }(u,1)u^b\le M_b(u\in {\mathrm{R}}_{+}),$  and then  $\mathrm{K}{}_{\lambda }(u,1)\le M_b{u}^{-b}$  $(u\in {\mathrm{R}}_{+}).$  Hence, there exists   ${\delta }_0=\frac{{\lambda }_2}{2}\in (0,{\lambda }_2),$  satisfying   $b={\lambda }_1+{\delta }_0\in (0,\lambda ),M_0=M_b$  , that makes (20) valid. By (18), we have the following extended half-discrete Hilbert-type inequalities:

${\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{m^{2{\lambda }_1-i}}{{(m+y)}^{\lambda }}{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}}}{\widehat{A}}_m^{(i)}{y}^{2{\lambda }_2-1}{\widehat{G}}_n(y)dy$  $>\frac{1}{{\lambda }_1^i}B({\lambda }_1,{\lambda }_2){\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\lambda }_2+k}}$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\lambda }_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}，$$

(30)

${\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{\ln (m/y)}{m^{\lambda }-y^{\lambda }}}}{m}^{2{\lambda }_1-i}{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}{y}^{2{\lambda }_2-1}{\widehat{G}}_n(y)dy$  $>\frac{1}{{\lambda }_1^i}{[\frac{\pi }{\lambda \sin (\frac{\pi {\lambda }_1}{\lambda })}]}^2{\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\lambda }_2+k}}$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\lambda }_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}，$$

(31)

${\displaystyle {\int }_0^{\infty }{\displaystyle \sum _{m=1}^{\infty }\frac{m^{2\lambda {}_1-i}}{{(\max \{m,y\})}^{\lambda }}}}{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}{y}^{2{\lambda }_2-1}{\widehat{G}}_n(y)dy$  $>\frac{\lambda }{{\lambda }_1^{\mathrm{i}+1}{\lambda }_2}{\displaystyle {\displaystyle \prod _{k=0}^{n-1}}\frac{1}{{\lambda }_2+k}}$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(n+{\lambda }_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}，$$

(32)

where, the constant factors in the above inequalities are the best value.

4. Equivalent Forms

Theorem 4.

With regards to the assumptions (A1) (for n=0), if for$i=1,$

$p(1-{\lambda }_1)\le \lambda -\lambda {}_1-{\lambda }_2<-p{\lambda }_1,$ 

then for  $i\in \{0,1\}$  , we havewe have the following reverse inequality equivalent to (28):

$J_i:={\left\{{\displaystyle {\int }_0^{\infty }{y}^{p{\widehat{\lambda }}_2-1}}{\left[{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}(m,y)m^{2{\widehat{\lambda }}_1-i}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}\right]}^{p}dy\right\}}^{\frac{1}{p}}$  $>\frac{1}{{\widehat{\lambda }}_1^i}{K}_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)$

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}.$$

(33)

In particular, for   ${\lambda }_1+{\lambda }_2=\lambda$  , we have the following reverse inequality equivalent to (29):

${\left\{{\displaystyle {\int }_0^{\infty }{y}^{p{\lambda }_2-1}}{\left[{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}(m,y)m^{2{\lambda }_1-i}{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}\right]}^{p}dy\right\}}^{\frac{1}{p}}$  $>\frac{1}{{\lambda }_1^i}{K}_{\lambda }({\lambda }_1)$

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}.$$

(34)

Proof. Assuming that (33) is valid, by the reverse H $\ddot{o}$ lder’s inequality (cf. [33]), we have

$I_i^{(0)}={\displaystyle {\int }_0^{\infty }\left[y^{\frac{-1}{p}+{\widehat{\lambda }}_2}{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }(m,y)m^{2{\widehat{\lambda }}_1-i}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}}\right]}(y^{-{\widehat{\lambda }}_2+\frac{1}{p}}g(y))dy$

$$\ge J_i{[{\displaystyle {\int }_0^{\infty }{y}^{q(1-{\widehat{\lambda }}_2)-1}}{g}^q(y)dy]}^{\frac{1}{q}}.$$

(35)

Then by (33), we have (28). On the other hand, assuming that (28) is valid, we set

$g(y):=y^{p{\widehat{\lambda }}_2-1}{\left[{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}(m,y)m^{2{\widehat{\lambda }}_1-i}{(1-\frac{{\widehat{\lambda }}_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}\right]}^{p-1},y\in {\mathrm{R}}_{+}.$ If $J_i=\infty ,$ then (33) is naturally valid; if $J_i=0,$ then it is impossible that makes (33) valid. In the following, we suppose that $0<J_i<\infty .$ By (28), we have

${\displaystyle {\int }_0^{\infty }{y}^{q(1-{\widehat{\lambda }}_2)-1}}{g}^q(y)dy=J_i^p=I_i^{(0)}$  $>\frac{1}{{\widehat{\lambda }}_1^i}{K}_{\lambda }(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)$  $\times {\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}$

$${\left[{\displaystyle {\int }_0^{\infty }{y}^{q(1-{\widehat{\lambda }}_2)-1}}{g}^q(y)dy\right]}^{\frac{1}{q}}，$$

(36)

$J_i={[{\displaystyle {\int }_0^{\infty }{y}^{q(1-{\widehat{\lambda }}_2)-1}}{g}^q(y)dy]}^{\frac{1}{p}}$$>\frac{1}{{\widehat{\lambda }}_1^i}{K}_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}},$ namely, (33) follows, which is equivalent to (28). The theorem is proved.

Theorem 5.

With regards to the assumption of Theorem 4, if${\lambda }_1+{\lambda }_2=\lambda$, then the constant factor in (33) is the best value. On the other hand, if (16) is valid for   $i\in \{0,1\}$  , and the constant factor in (33) is the best value, then we have   ${\lambda }_1+{\lambda }_2=\lambda$ .

Proof. If the constant factor in (33) is not the best value, then by (35), for ${\lambda }_1+{\lambda }_2=\lambda$ , we would reach a contradiction that the constant factor in (29) is not the best possible. Hence, for ${\lambda }_1+{\lambda }_2=\lambda$ , the constant factor in (33) is the best value. On the other hand, if the same constant factor in (33) is the best value, then in view of the assumption and (36) (for $I_i^{(0)}=J_i^p$ ), we can show that the constant factor in (28) ( resp. (17) for $n=0$ ) is the best value, and then by Theorem 3 with the assumption, we have ${\lambda }_1+{\lambda }_2=\lambda$ . The theorem is proved.

Remark 3

. (i) For $i=0$in (33), we have the following reverse inequality equivalent to (25):

${\left[{\displaystyle {\int }_0^{\infty }{y}^{p{\widehat{\lambda }}_2-1}}{\left({\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}(m,y)m^{2{\widehat{\lambda }}_1}{a}_m\right)}^{p}dy\right]}^{\frac{1}{p}}$  $>K_{\lambda }^{\frac{1}{p}}(\lambda -{\lambda }_2)K_{\lambda }^{\frac{1}{q}}({\lambda }_1)$

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\widehat{\lambda }}_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}.$$

(37)

In particular, for  ${\lambda }_1+{\lambda }_2=\lambda$  in (37), we have the following inequality with the best value equivalent to (26):

${\left[{\displaystyle {\int }_0^{\infty }{y}^{p{\lambda }_2-1}}{\left({\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}(m,y)m^{2{\lambda }_1}{a}_m\right)}^{p}dy\right]}^{\frac{1}{p}}$  $>K_{\lambda }({\lambda }_1)$

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}.$$

(38)

(ii) For  $i=1$  in (34),   ${\lambda }_1\in (0,1],{\lambda }_2>0$  , setting  ${\widehat{A}}_m={\widehat{A}}_m^{(1)}={\displaystyle {\sum }_{k=\mathrm{m}+1}^{\infty }{a}_k}$  , we have the following inequality with the best value equivalent to (27):

${\left\{{\displaystyle {\int }_0^{\infty }{y}^{p{\lambda }_2-1}}{\left[{\displaystyle \sum _{m=1}^{\infty }{K}_{\lambda }}(m,y)m^{2{\lambda }_1-1}{(1-\frac{{\lambda }_1}{m})}^{-\frac{1}{q}}{\widehat{A}}_m\right]}^{p}dy\right\}}^{\frac{1}{p}}$  $>\frac{1}{{\lambda }_1}{K}_{\lambda }({\lambda }_1)$

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}}.$$

(39)

Example 2

. For $0<{\lambda }_1,{\lambda }_2<\lambda ,{\lambda }_1\le 1,{\lambda }_1+{\lambda }_2=\lambda$ ,

$K_{\lambda }(x,y)=\frac{1}{{(x+y)}^{\lambda }},\frac{\ln (x/y)}{x^{\lambda }-y^{\lambda }},\frac{1}{{(\max \{x,y\})}^{\lambda }}(x,y\in {\mathrm{R}}_{+}^2),$ 

by Example 1 and (33), we have the following extended reverse half-discrete Hilbert-type inequalities with one remainder sum equivalent to (30)-(32) (for   $n=0$  ):

${\left\{{\displaystyle {\int }_0^{\infty }{y}^{p\lambda {}_2-1}}{\left[{\displaystyle \sum _{m=1}^{\infty }{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}\frac{m^{2{\lambda }_1-i}{\widehat{A}}_m^{(i)}}{{(m+y)}^{\lambda }}}\right]}^{p}dy\right\}}^{\frac{1}{p}}$  $>\frac{1}{{\lambda }_1^i}B({\lambda }_1,{\lambda }_2)$

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}},$$

(40)

${\left\{{\displaystyle {\int }_0^{\infty }{y}^{p{\lambda }_2-1}}{\left[{\displaystyle \sum _{m=1}^{\infty }\frac{\ln (m/y)}{m^{\lambda }-y^{\lambda }}{m}^{2{\lambda }_1-i}{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}{\widehat{A}}_m^{(i)}}\right]}^{p}dy\right\}}^{\frac{1}{p}}$  $>\frac{1}{{\lambda }_1^i}{[\frac{\pi }{\lambda \sin (\frac{\pi {\lambda }_1}{\lambda })}]}^2$

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}},$$

(41)

${\left\{{\displaystyle {\int }_0^{\infty }{y}^{p{\lambda }_2-1}}{\left[{\displaystyle \sum _{m=1}^{\infty }{(1-\frac{{\lambda }_1}{m})}^{-\frac{i}{q}}\frac{m^{2{\lambda }_1-i}{\widehat{A}}_m^{(i)}}{{(\max \{m,y\})}^{\lambda }}}\right]}^{p}dy\right\}}^{\frac{1}{p}},$  $>\frac{\lambda }{{\lambda }_1^{1+i}{\lambda }_2}$

$${\left[{\displaystyle \sum _{m=1}^{\infty }{m}^{p(1+{\lambda }_1)-1}}{a}_m^p\right]}^{\frac{1}{p}},$$

(42)

where, the constant factors in the above inequalities are the best value.

5. Conclusions

In this article, by means of the weight functions, the idea of introduced parameters, the techniques of real analysis and the well known reverse Hardy’s integral inequality, we obtain a new extended reverse Hardy-type inequality in Lemma 1 and Remark 1, and then derive an extended reverse half-discrete Hilbert -type inequality with the general homogeneous kernel as $K_{\lambda }(m,y)(\lambda \in {\mathrm{R}}_{+};m\in \mathrm{N},\mathrm{y}\in {\mathrm{R}}_{+})$ , as well as one multiple lower limit function and one remainder sum in Theorem 1. The equivalent statements of the best value related to several parameters are considered in Theorem 2-3, the equivalent forms are obtained in Theorem 4-5, and some particular examples are provided in Remark 2-3 and Example 1-2.