# On a half-discrete inequality with a generalized homogeneous kernel

## Abstract

By introducing a real number homogeneous kernel and estimating the weight function through the real function techniques, a half-discrete inequality with a best constant factor is established. In addition, the operator expressions, equivalent forms, reverse inequalities and some particular cases are given.

Mathematics Subject Classification (2000): 26D15.

## 1. Introduction

One hundred years ago, Hilbert proved the following classic inequality 

$\sum _{n}\sum _{m}\frac{{a}_{m}{b}_{n}}{m+n}\le \pi {\left(\sum _{n}{a}_{n}^{2}\right)}^{1/2}{\left(\sum _{n}{b}_{n}^{2}\right)}^{1/2}.$
(1.1)

During the past century, ever since the advent the inequality (1.1), numerous related results have been obtained. The inequality (1.1) may be classified into several types (discrete and integral etc.), being the following integral form:

If f, g are real functions such that $0<{\int }_{0}^{\infty }{f}^{2}\left(x\right)\phantom{\rule{2.77695pt}{0ex}}\text{d}x<\infty ,0<{\int }_{0}^{\infty }{g}^{2}\left(x\right)\phantom{\rule{2.77695pt}{0ex}}\text{d}x<\infty$, then we have 

$\underset{0}{\overset{\infty }{\int }}{\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{x+y}\text{d}x\text{d}y<\pi \left\{\underset{0}{\overset{\infty }{\int }}{f}^{2}\left(x\right)\text{d}x\underset{0}{\overset{\infty }{\int }}{g}^{2}\left(x\right)\text{d}x\right\}}^{\frac{1}{2}},$
(1.2)

where the constant factor π is the best possible. Inequality (1.2) had been generalized by Hardy-Riesz in 1925 as :

If p > 1, $\frac{1}{p}+\frac{1}{q}=1,f,g\ge 0$ such that $0<{\int }_{0}^{\infty }{f}^{p}\left(x\right)\text{d}x,{\int }_{0}^{\infty }{g}^{q}\left(x\right)\text{d}x<\infty$, then

$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{x+y}\text{d}x\text{d}y<\frac{\pi }{\text{sin}\left(\frac{\pi }{p}\right)}{\left\{\underset{0}{\overset{\infty }{\int }}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\underset{0}{\overset{\infty }{\int }}{g}^{q}\left(x\right)\text{d}x\right\}}^{1/q},$
(1.3)

where the constant factor $\frac{\pi }{\text{sin}\left(\frac{\pi }{p}\right)}$ is the best possible. Inequality (1.3) is named as Hardy-Hilbert's integral inequality, which is of great importance in analysis and its applications . Its generalization can be seen in .

Until now, we only studied the related inequalities with pure discrete or integral inequalities, but half-discrete inequality is very rare in the literature . Now we attempt investigation for it, lots of related results will appear in the coming future.

The main purpose of this article is to establish a half-discrete inequality with the mixed homogeneous kernel of real number degree. For example: If $0<{\int }_{0}^{\infty }{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x<\infty ,0<{\sum }_{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}<\infty$, then

$\begin{array}{c}\sum _{n=1}^{\infty }{a}_{n}\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\text{d}x\\ <\frac{\alpha }{{\lambda }_{1}{\lambda }_{2}}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q},\end{array}$
(1.4)

where α = λ1 + λ2, 0 <λ1 <α and the constant factor $\frac{\alpha }{{\lambda }_{1}{\lambda }_{2}}$ is the best possible. Meanwhile, the extended inequality, operator expressions, reverse inequality, and equivalent forms are given. We hope this work will expand our understanding of inequality and the scope of the study.

## 2. Lemmas

LEMMA 2.1. Let α, β and λ1 + λ2 = α - β, -β <λ1 <α, λ2 ≤ 1 - β, ${k}_{{\lambda }_{1}}:=\frac{1}{\alpha -{\lambda }_{1}}+\frac{1}{\beta +{\lambda }_{1}}$define the weight function and the weight coefficient as follows

$\omega \left(n\right):={n}^{{\lambda }_{2}}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\cdot {x}^{{\lambda }_{1}-1}dx,n\in {ℕ}_{+},$
(2.1)
$\varpi \left(x\right):={x}^{{\lambda }_{1}}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\cdot {n}^{{\lambda }_{2}-1},x\in \left(0,\infty \right),$
(2.2)

then

$0<{k}_{{\lambda }_{1}}\left(1-{\theta }_{\lambda }\left(x\right)\right)<\varpi \left(x\right)<\omega \left(n\right)={k}_{{\lambda }_{1}},$
(2.3)

where

${\theta }_{\lambda }\left(x\right):=\frac{1}{{k}_{{\lambda }_{1}}}\underset{0}{\overset{1/x}{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\cdot {t}^{{\lambda }_{2}-1}\text{d}t=\left\{\begin{array}{cc}\hfill \frac{1}{{k}_{{\lambda }_{1}}}\left(\frac{1}{\beta +{\lambda }_{2}}+\frac{1-{x}^{\alpha -{\lambda }_{2}}}{\alpha -{\lambda }_{2}}\right),\hfill & \hfill x\in \left(0,1\right),\hfill \\ \hfill \frac{{x}^{-\left(\beta +{\lambda }_{2}\right)}}{\left(\beta +{\lambda }_{2}\right){k}_{{\lambda }_{1}}}=O\left(\frac{1}{{x}^{\beta +{\lambda }_{2}}}\right),\hfill & \hfill x\in \left[0,\infty \right).\hfill \end{array}\right\$

Proof. For fixed n, let $t=\frac{x}{n}$, substituting into ω(n) gives

$\begin{array}{ll}\hfill \omega \left(n\right)& =\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\cdot {t}^{{\lambda }_{1}-1}\text{d}t\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{1}{\int }}{t}^{\beta +{\lambda }_{1}-1}\text{d}t+\underset{1}{\overset{\infty }{\int }}{t}^{-\alpha +{\lambda }_{1}-1}\text{d}t\phantom{\rule{2em}{0ex}}\\ =\frac{1}{\alpha -{\lambda }_{1}}+\frac{1}{\beta +{\lambda }_{1}}={k}_{{\lambda }_{1}}.\phantom{\rule{2em}{0ex}}\end{array}$

In view of λ2 ≤ 1 - β, α - λ2 = β + λ1 > 0, for fixed x > 0, the function

$\frac{{\left(\text{min}\left\{x,y\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,y\right\}\right)}^{\alpha }}\cdot {y}^{{\lambda }_{2}-1}=\left\{\begin{array}{cc}\hfill {x}^{-\alpha }{y}^{\beta +{\lambda }_{2}-1},\hfill & \hfill 0

is monotonically decreasing with respect to y, then

$\begin{array}{ll}\hfill \varpi \left(x\right)& <{x}^{{\lambda }_{1}}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,y\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,y\right\}\right)}^{\alpha }}\cdot {y}^{{\lambda }_{2}-1}\text{d}y\phantom{\rule{1em}{0ex}}\left(t=y/x\right)\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\cdot {t}^{{\lambda }_{2}-1}\text{d}t=\underset{0}{\overset{1}{\int }}{t}^{\beta +{\lambda }_{2}-1}\text{d}t+\underset{1}{\overset{\infty }{\int }}{t}^{-\alpha +{\lambda }_{2}-1}\text{d}t\phantom{\rule{2em}{0ex}}\\ =\frac{1}{\beta +{\lambda }_{2}}+\frac{1}{\alpha -{\lambda }_{2}}=\frac{1}{\alpha -{\lambda }_{1}}+\frac{1}{\beta +{\lambda }_{1}}={k}_{{\lambda }_{1}}.\phantom{\rule{2em}{0ex}}\end{array}$
$\begin{array}{ll}\hfill \varpi \left(x\right)& >{x}^{{\lambda }_{1}}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,y\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,y\right\}\right)}^{\alpha }}\cdot {y}^{{\lambda }_{2}-1}\text{d}y\phantom{\rule{1em}{0ex}}\left(t=y/x\right)\phantom{\rule{2em}{0ex}}\\ =\underset{1/x}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\cdot {t}^{{\lambda }_{2}-1}\text{d}t\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}-\underset{0}{\overset{1/x}{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\cdot {t}^{{\lambda }_{2}-1}\text{d}t={k}_{{\lambda }_{1}}\left(1-{\theta }_{\lambda }\left(x\right)\right)>0.\phantom{\rule{2em}{0ex}}\end{array}$

where ${\theta }_{\lambda }\left(x\right)=\frac{1}{{k}_{{\lambda }_{1}}}{\int }_{0}^{1/x}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\cdot {t}^{{\lambda }_{2}-1}\text{d}t$. If x (0,1), then

${\theta }_{\lambda }\left(x\right)=\frac{1}{{k}_{{\lambda }_{1}}}\left(\underset{0}{\overset{1}{\int }}{t}^{\beta +{\lambda }_{2}-1}\text{d}t+\underset{1}{\overset{1/x}{\int }}{t}^{-\alpha +{\lambda }_{2}-1}\text{d}t\right)=\frac{1}{{k}_{{\lambda }_{1}}}\left[\frac{1}{\beta +{\lambda }_{2}}+\frac{1-{x}^{\alpha -{\lambda }_{2}}}{\alpha -{\lambda }_{2}}\right];$

if x [1,∞), then

${\theta }_{\lambda }\left(x\right)=\frac{1}{{k}_{{\lambda }_{1}}}\underset{0}{\overset{1/x}{\int }}{t}^{\beta +{\lambda }_{2}-1}\text{d}t=\frac{{x}^{-\left(\beta +{\lambda }_{2}\right)}}{\left(\beta +{\lambda }_{2}\right){k}_{{\lambda }_{1}}}=O\left(\frac{1}{{x}^{\beta +{\lambda }_{2}}}\right).$

Thus (2.3) is valid.

In what follows, α, β will be real numbers such that λ1 + λ2 = α - β, -β <λ1 <α, λ2 ≤ 1 - β.

LEMMA 2.2. Suppose that p > 0(p ≠ 1), $\frac{1}{p}+\frac{1}{q}=1$, a n ≥ 0, f(x) is a non-negative measurable function in (0, ∞), then

(a) if p > 1, then the following two inequalities hold:

$J:=\sum _{n=1}^{\infty }{n}^{p{\lambda }_{2}-1}{\left[\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\right]}^{p}\le {k}_{{\lambda }_{1}}^{p}\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x,$
(2.4)
$L:=\underset{0}{\overset{\infty }{\int }}{x}^{q{\lambda }_{1}-1}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q}\text{d}x\le {k}_{{\lambda }_{1}}^{q}\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q};$
(2.5)

(b) if 0 <p < 1, then we have

$J\ge {k}_{{\lambda }_{1}}^{p}\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x,$
(2.6)
$\stackrel{̃}{L}:=\underset{0}{\overset{\infty }{\int }}\frac{{x}^{q{\lambda }_{1}-1}}{{\left[1-{\theta }_{\lambda }\left(x\right)\right]}^{q-1}}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q}\text{d}x\le {k}_{{\lambda }_{1}}^{q}\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}.$
(2.7)

Proof. (a) Using Hölder's inequality with weight  and (2.3) gives

$\begin{array}{l}\phantom{\rule{1em}{0ex}}{\left[\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\right]}^{p}\phantom{\rule{2em}{0ex}}\\ ={\left\{\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\left[\frac{{x}^{\left(1-{\lambda }_{1}\right)/q}}{{n}^{\left(1-{\lambda }_{2}\right)/p}}f\left(x\right)\right]\left[\frac{{n}^{\left(1-{\lambda }_{2}\right)/p}}{{x}^{\left(1-{\lambda }_{1}\right)/q}}\right]\text{d}x\right\}}^{p}\phantom{\rule{2em}{0ex}}\\ \le \underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{x}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{{n}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)\text{d}x{\left[\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}\text{d}x\right]}^{p-1}\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{x}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{{n}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)\text{d}x{\left[{n}^{q\left(1-{\lambda }_{2}\right)-1}\omega \left(n\right)\right]}^{p-1}\phantom{\rule{2em}{0ex}}\\ ={n}^{-p{\lambda }_{2}+1}{k}_{{\lambda }_{1}}^{p-1}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{x}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{{n}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)\text{d}x.\phantom{\rule{2em}{0ex}}\end{array}$
$\begin{array}{ll}\hfill J& \le {k}_{{\lambda }_{1}}^{p-1}\sum _{n=1}^{\infty }\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{x}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{{n}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)\text{d}x\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{p-1}\underset{0}{\overset{\infty }{\int }}\left[{x}^{{\lambda }_{1}}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{n}^{{\lambda }_{2}-1}\right]{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{p-1}\underset{0}{\overset{\infty }{\int }}\varpi \left(x\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\le {k}_{{\lambda }_{1}}^{p}\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x.\phantom{\rule{2em}{0ex}}\end{array}$
(2.8)

Hence (2.4) is valid. By similar reasoning to the above it may be shown that

$\begin{array}{l}\phantom{\rule{1em}{0ex}}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q}\phantom{\rule{2em}{0ex}}\\ ={\left\{\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\left[\frac{{x}^{\left(1-{\lambda }_{1}\right)/q}}{{n}^{\left(1-{\lambda }_{2}\right)/p}}\right]\left[\frac{{n}^{\left(1-{\lambda }_{2}\right)/p}}{{x}^{\left(1-{\lambda }_{1}\right)/q}}{a}_{n}\right]\right\}}^{q}\phantom{\rule{2em}{0ex}}\\ \le {\left[\varpi \left(x\right){x}^{p\left(1-{\lambda }_{1}\right)-1}\right]}^{q-1}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}{a}_{n}^{q}\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{q-1}{x}^{-q{\lambda }_{1}+1}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}{a}_{n}^{q}.\phantom{\rule{2em}{0ex}}\end{array}$
(2.9)
$\begin{array}{ll}\hfill L& \le {k}_{{\lambda }_{1}}^{q-1}\underset{0}{\overset{\infty }{\int }}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}{a}_{n}^{q}\text{d}x\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{q-1}\sum _{n=1}^{\infty }\left[{n}^{{\lambda }_{2}}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{x}^{{\lambda }_{1}-1}\text{d}x\right]{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{q-1}\sum _{n=1}^{\infty }\omega \left(n\right){n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}={k}_{{\lambda }_{1}}^{q}\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}.\phantom{\rule{2em}{0ex}}\end{array}$

Thus (2.5) is valid.

1. (b)

Similarly, using the reverse Hölder's inequality with weight  and (2.3) gives (2.6) and (2.7).

LEMMA 2.3. Suppose that 0 <q < 1, $\frac{1}{p}+\frac{1}{q}=1$, a n ≥ 0, f(x) is a non-negative measurable function in (0, ∞), then (Let J, L be as in Lemma 2.2)

$J\le {k}_{{\lambda }_{1}}^{p}\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x,$
(2.10)
$L\ge {k}_{{\lambda }_{1}}^{q}\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}.$
(2.11)

Proof. Applying Hölder's inequality  and (2.3), where p < 0 gives

$\begin{array}{l}\phantom{\rule{1em}{0ex}}{\left[\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\right]}^{p}\phantom{\rule{2em}{0ex}}\\ ={\left\{\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\left[\frac{{x}^{\left(1-{\lambda }_{1}\right)/q}}{{n}^{\left(1-{\lambda }_{2}\right)/p}}f\left(x\right)\right]\left[\frac{{n}^{\left(1-{\lambda }_{2}\right)/p}}{{x}^{\left(1-{\lambda }_{1}\right)/q}}\right]\text{d}x\right\}}^{p}\phantom{\rule{2em}{0ex}}\\ \le \underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{x}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{{n}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)\text{d}x{\left[\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}\text{d}x\right]}^{p-1}\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{min}\left\{x,n\right\}\right)}^{\alpha }}\frac{{x}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{{n}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)\text{d}x{\left[{n}^{q\left(1-{\lambda }_{2}\right)-1}\omega \left(n\right)\right]}^{p-1}\phantom{\rule{2em}{0ex}}\\ ={n}^{-p{\lambda }_{2}+1}{k}_{{\lambda }_{1}}^{p-1}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{x}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{{n}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)\text{d}x.\phantom{\rule{2em}{0ex}}\end{array}$
$\begin{array}{ll}\hfill J& \le {k}_{{\lambda }_{1}}^{p-1}\sum _{n=1}^{\infty }\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{x}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}}{{n}^{1-{\lambda }_{2}}}{f}^{p}\left(x\right)\text{d}x\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{p-1}\underset{0}{\overset{\infty }{\int }}\left[{x}^{{\lambda }_{1}}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{n}^{{\lambda }_{2}-1}\right]{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{p-1}\underset{0}{\overset{\infty }{\int }}\varpi \left(x\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\le {k}_{{\lambda }_{1}}^{p}\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\phantom{\rule{2em}{0ex}}\end{array}$
(2.12)

Hence (2.10) is valid. By similar reasoning to the above, in view of 0 <q < 1, it may be shown that

$\begin{array}{l}\phantom{\rule{1em}{0ex}}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q}\phantom{\rule{2em}{0ex}}\\ ={\left\{\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\left[\frac{{x}^{\left(1-{\lambda }_{1}\right)/q}}{{n}^{\left(1-{\lambda }_{2}\right)/p}}\right]\left[\frac{{n}^{\left(1-{\lambda }_{2}\right)/p}}{{x}^{\left(1-{\lambda }_{1}\right)/q}}{a}_{n}\right]\right\}}^{q}\phantom{\rule{2em}{0ex}}\\ \ge {\left[\varpi \left(x\right){x}^{p\left(1-{\lambda }_{1}\right)-1}\right]}^{q-1}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}{a}_{n}^{q}\phantom{\rule{2em}{0ex}}\\ \ge {k}_{{\lambda }_{1}}^{q-1}{x}^{-q{\lambda }_{1}+1}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}{a}_{n}^{q}.\phantom{\rule{2em}{0ex}}\end{array}$
(2.13)
$\begin{array}{ll}\hfill L& \ge {k}_{{\lambda }_{1}}^{q-1}\underset{0}{\overset{\infty }{\int }}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}{a}_{n}^{q}\text{d}x\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{q-1}\sum _{n=1}^{\infty }\left[{n}^{{\lambda }_{2}}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{x}^{{\lambda }_{1}-1}\text{d}x\right]{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\phantom{\rule{2em}{0ex}}\\ ={k}_{{\lambda }_{1}}^{q-1}\sum _{n=1}^{\infty }\omega \left(n\right){n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}={k}_{{\lambda }_{1}}^{q}\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}.\phantom{\rule{2em}{0ex}}\end{array}$

Thus (2.11) is valid.

## 3. Main results

THEOREM 3.1. If p > 1, $\frac{1}{p}+\frac{1}{q}=1$, a n ≥ 0, f(x) ≥ 0 such that$0<{\int }_{0}^{\infty }{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x<\infty$and$0<{\sum }_{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}<\infty$, then we have the following equivalent inequalities

$\begin{array}{ll}\hfill I:& =\sum _{n=1}^{\infty }{a}_{n}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x=\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\text{d}x\phantom{\rule{2em}{0ex}}\\ <{k}_{{\lambda }_{1}}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\sum _{n=1}^{p}{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q},\phantom{\rule{2em}{0ex}}\end{array}$
(3.1)
$J=\sum _{n=1}^{\infty }{n}^{p{\lambda }_{2}-1}{\left[\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\right]}^{p}<{k}_{{\lambda }_{1}}^{p}\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x,$
(3.2)
$L=\underset{0}{\overset{\infty }{\int }}{x}^{q{\lambda }_{1}-1}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q}\text{d}x<{k}_{{\lambda }_{1}}^{q}\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q},$
(3.3)

where the constant factor${k}_{{\lambda }_{1}}=\frac{1}{\alpha -{\lambda }_{1}}+\frac{1}{\beta +{\lambda }_{1}},{k}_{{\lambda }_{1}}^{p},{k}_{{\lambda }_{1}}^{q}$, are the best possible.

Proof. Using Lebesgue term-by-term integration theorem, there are two forms of I of (3.1). In view of $0<{\int }_{0}^{\infty }{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x<\infty$, (2.8) takes the strict inequality, thus (3.1) is valid. On one hand, using Hölder's inequality  gives

$I=\sum _{n=1}^{\infty }\left[{n}^{{\lambda }_{2}-\frac{1}{p}}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\right]\left[{n}^{\frac{1}{p}-{\lambda }_{2}}{a}_{n}\right]\le {J}^{1/p}={\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q}.$
(3.4)

By (3.2), (3.1) is valid. On the other hand, suppose that (3.1) is valid. Let

${a}_{n}:={n}^{p{\lambda }_{2}-1}{\left[\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\right]}^{p-1},n\in {ℕ}_{+},$

then from (3.1), it follows

$\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}=J=I\le {k}_{{\lambda }_{1}}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q}.$
(3.5)

By (2.8) and the conditions, it follows that J < ∞. If J = 0, then (3.2) is naturally valid. If J > 0, in view of the conditions of (3.1), then (3.5) takes the strict inequality, and

${J}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/p}<{k}_{{\lambda }_{1}}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}.$

Hence (3.2) is valid, which is equivalent to (3.1).

On one hand, in view of the conditions, (2.9) takes the strict inequality, thus (3.3) is valid. Using Hölder's inequality  gives

$\begin{array}{ll}\hfill I& =\underset{0}{\overset{\infty }{\int }}\sum _{n=1}^{\infty }{a}_{n}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{\infty }{\int }}\left[{x}^{\frac{1}{q}-{\lambda }_{1}}f\left(x\right)\right]\left[{x}^{{\lambda }_{1}-\frac{1}{q}}{a}_{n}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\right]\text{d}x\phantom{\rule{2em}{0ex}}\\ \ge {\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\right\}}^{1/p}{L}^{1/q}.\phantom{\rule{2em}{0ex}}\end{array}$
(3.6)

By (3.3), (3.1) is valid. On the other hand, suppose that (3.1) is valid. Let

$f\left(x\right):={x}^{q{\lambda }_{1}-1}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q-1},x\in \left(0,\infty \right).$

Applying (3.1) gives

$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)=L=I\\ \le {k}_{{\lambda }_{1}}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q}.\end{array}$
(3.7)

By (2.9) and the conditions, it follows that L < ∞. If L = 0, then (3.3) is naturally valid. If L > 0, in view of the conditions of (3.1), then (3.7) takes the strict inequality, and

${L}^{1/q}={\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\right\}}^{1/q}<{k}_{{\lambda }_{1}}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q}.$

Hence (3.3) is valid, which is equivalent to (3.1). Thus (3.1), (3.2), and (3.3) are equivalent to each other.

For any 0 <ε <2, suppose that $\stackrel{̃}{f}\left(x\right)=0,x\in \left(0,1\right);\stackrel{̃}{f}\left(x\right)={x}^{{\lambda }_{1}-\frac{\epsilon }{p}-1},x\in \left[1,\infty \right)$ and ${\stackrel{̃}{a}}_{n}={n}^{{\lambda }_{2}-\frac{\epsilon }{q}-1},n\in {ℕ}_{+}$. Assuming there exists a positive number k with $k\le {k}_{{\lambda }_{1}}$, such that (3.1) is still valid by changing ${k}_{{\lambda }_{1}}$ to k. In particular, on one hand,

$\begin{array}{ll}\hfill \stackrel{̃}{I}& =\sum _{n=1}^{\infty }{\stackrel{̃}{a}}_{n}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\stackrel{̃}{f}\left(x\right)\text{d}x\phantom{\rule{2em}{0ex}}\\
(3.8)

On the other hand, by monotonicity and Fubini theorem, it follows that

$\begin{array}{ll}\hfill \stackrel{̃}{I}& =\underset{1}{\overset{\infty }{\int }}{x}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}\left[\sum _{n=1}^{\infty }{n}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\right]\text{d}x\phantom{\rule{2em}{0ex}}\\ \ge \underset{1}{\overset{\infty }{\int }}{x}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}\left[\underset{1}{\overset{\infty }{\int }}{y}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{x,y\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,y\right\}\right)}^{\alpha }}\text{d}y\right]\text{d}x,\left(t=y/x\right)\phantom{\rule{2em}{0ex}}\\ =\underset{1}{\overset{\infty }{\int }}{x}^{-\epsilon -1}\left[\underset{1/x}{\overset{\infty }{\int }}{t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{x,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,t\right\}\right)}^{\alpha }}\text{d}t\right]\text{d}x.\phantom{\rule{2em}{0ex}}\end{array}$
$\begin{array}{l}=\underset{1}{\overset{\infty }{\int }}{x}^{-\epsilon -1}\left[\underset{1/x}{\overset{1}{\int }}{t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\text{d}t\right]\text{d}x+\frac{1}{\epsilon }\underset{1}{\overset{\infty }{\int }}{t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\text{d}t\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{1}{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\left(\underset{1/t}{\overset{\infty }{\int }}{x}^{-\epsilon -1}\text{d}x\right){t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\text{d}t+\frac{1}{\epsilon }\underset{1}{\overset{\infty }{\int }}{t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\text{d}t\phantom{\rule{2em}{0ex}}\\ =\frac{1}{\epsilon }\underset{0}{\overset{1}{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}{t}^{{\lambda }_{2}+\frac{\epsilon }{p}-1}\text{d}t+\frac{1}{\epsilon }\underset{1}{\overset{\infty }{\int }}{t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\text{d}t.\phantom{\rule{2em}{0ex}}\end{array}$
(3.9)

Applying (3.8) and (3.9) gives

$\underset{0}{\overset{1}{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}{t}^{{\lambda }_{2}+\frac{\epsilon }{p}-1}\text{d}t+\underset{1}{\overset{\infty }{\int }}{t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\text{d}t

Using Fatou theorem gives

$\begin{array}{ll}\hfill {k}_{{\lambda }_{1}}& =\underset{0}{\overset{1}{\int }}\underset{\epsilon \to {0}^{+}}{\text{lim}}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}{t}^{{\lambda }_{2}+\frac{\epsilon }{p}-1}\text{d}t+\underset{1}{\overset{\infty }{\int }}\underset{\epsilon \to {0}^{+}}{\text{lim}}{t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\text{d}t\phantom{\rule{2em}{0ex}}\\ \le \frac{\text{lim}}{\epsilon \to {0}^{+}}\left[\underset{0}{\overset{1}{\int }}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}{t}^{{\lambda }_{2}+\frac{\epsilon }{p}-1}\text{d}t+\underset{1}{\overset{\infty }{\int }}{t}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\text{d}t\right]\phantom{\rule{2em}{0ex}}\\ \le \frac{\text{lim}}{\epsilon \to {0}^{+}}k{\left(\epsilon +1\right)}^{1/q}=k.\phantom{\rule{2em}{0ex}}\end{array}$

Hence $k={k}_{{\lambda }_{1}}$ is the best constant factor of (3.1). It is obvious that the constant factor in (3.2) (or (3.3)) is the best possible. Otherwise, by (3.4) (or (3.6)), we may get a contradiction that the constant factor in (3.1) is not the best possible. This completes the proof.

Remark 1. Let $\Phi \left(x\right)={x}^{p\left(1-{\lambda }_{1}\right)-1}$, x (0,∞) and $\Psi \left(n\right)={n}^{q\left(1-{\lambda }_{2}\right)-1}$, n +, then ${\left[\Phi \left(x\right)\right]}^{1-q}={x}^{q{\lambda }_{1}-1},{\left[\Psi \left(n\right)\right]}^{1-p}={n}^{p{\lambda }_{2}-1}$.

1. (i)

A half-discrete Hilbert's operator $T:{L}_{p,\Phi }\left(0,\infty \right)\to {l}_{p,{\Psi }^{1-p}}$ is defined by:

$Tf\left(n\right)=\underset{0}{\overset{\infty }{\int }}{h}_{\lambda }\left(x,n\right)f\left(x\right)\text{d}x,n\ge 1.$

where $f\in {L}_{p,\Phi }\left(0,\infty \right),Tf\in {l}_{p,{\Psi }^{1-p}},{h}_{\lambda }\left(x,n\right)=\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }},\lambda =:\alpha -\beta$. Then by (3.2), it follows that: ${∥Tf∥}_{p,{\Psi }^{1-p}}\le {k}_{{\lambda }_{1}}{∥f∥}_{p,\Phi }$, i.e., T is a bounded operator with $∥T∥={k}_{{\lambda }_{1}}$. Since the constant factor in (3.2) is the best possible, we have $∥T∥={k}_{{\lambda }_{1}}$.

1. (ii)

Similarly, another half-discrete Hilbert's operator $\stackrel{̃}{T}:{l}_{q,\Psi }\to {L}_{q,{\Phi }^{1-q}}\left(0,\infty \right)$ is defined by:

$\stackrel{̃}{T}a\left(x\right)=\sum _{1}^{\infty }{h}_{\lambda }\left(x,n\right){a}_{n},x\in \left(0,\infty \right).$

where $a\in {l}_{q,\Psi },\stackrel{̃}{T}a\in {L}_{q,{\Phi }^{1-q}}\left(0,\infty \right)$. Then by (3.3), it follows that: ${∥\stackrel{̃}{T}a∥}_{q,{\Phi }^{1-q}}\le {k}_{{\lambda }_{1}}{∥a∥}_{q,\Psi }$. In another word, T is a bounded operator with $∥\stackrel{̃}{T}∥\le {k}_{{\lambda }_{1}}$. Since the constant factor in (3.3) is the best possible, we obtain $∥\stackrel{̃}{T}∥\le {k}_{{\lambda }_{1}}$.

THEOREM 3.2. If 0 <p < 1, $\frac{1}{p}+\frac{1}{q}=1,{\theta }_{\lambda }\left(x\right):=\frac{1}{{k}_{{\lambda }_{1}}}{\int }_{0}^{1/x}\frac{{\left(\text{min}\left\{1,t\right\}\right)}^{\beta }}{{\left(\text{max}\left\{1,t\right\}\right)}^{\alpha }}\cdot {t}^{{\lambda }_{2}-1}\left(x\in \left(0,\infty \right)\right)$, a n ≥ 0, f(x) ≥ 0 such that$0<{\int }_{0}^{\infty }\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x<\infty$and$0<{\sum }_{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}<\infty$, then we have the following equivalent inequalities (Let I, J be as in Theorem 3.1)

$I>{k}_{{\lambda }_{1}}{\left\{\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q},$
(3.10)
$J>{k}_{{\lambda }_{1}}^{p}\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x,$
(3.11)
$\stackrel{̃}{L}=\underset{0}{\overset{\infty }{\int }}\frac{{x}^{q{\lambda }_{1}-1}}{{\left[1-{\theta }_{\lambda }\left(x\right)\right]}^{q-1}}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q}\text{d}x<{k}_{{\lambda }_{1}}^{q}\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q},$
(3.12)

where the constant factor ${k}_{{\lambda }_{1}}=\frac{1}{\alpha -{\lambda }_{1}}+\frac{1}{\beta +{\lambda }_{1}},{k}_{{\lambda }_{1}}^{p},{k}_{{\lambda }_{1}}^{q}$ are the best possible.

Proof. Similar to (2.8), by the reverse Hölder's inequality , (2.3) and the conditions, we have

$J\ge {k}_{{\lambda }_{1}}^{p-1}\underset{0}{\overset{\infty }{\int }}\varpi \left(x\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x>{k}_{{\lambda }_{1}}^{p}\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x,$
(3.13)

thus (2.11) is valid. On one hand, by the reverse Hölder's inequality , we obtain the reverse form of (3.4) as follows

$I=\sum _{n=1}^{\infty }\left[{n}^{{\lambda }_{2}-\frac{1}{p}}\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\right]\left[{n}^{\frac{1}{p}-{\lambda }_{2}}{a}_{n}\right]\ge {J}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q}.$
(3.14)

by (3.11), (3.10) is valid. On the other hand, suppose that (3.10) is valid. Let

${a}_{n}={n}^{p{\lambda }_{2}-1}{\left[\underset{0}{\overset{\infty }{\int }}\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}f\left(x\right)\text{d}x\right]}^{p-1},n\in {ℕ}_{+}.$

Applying (3.10) gives

$\begin{array}{c}\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}=J=I\\ \ge {k}_{{\lambda }_{1}}{\left\{\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q},\end{array}$
(3.15)

By (3.13) and the conditions, it follows that J > 0. If J = ∞, then (3.11) is naturally valid. If J < ∞, in view of the conditions and (3.10), then (3.15) takes the strict inequality, and

${J}^{1/p}={\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q}>{k}_{{\lambda }_{1}}{\left\{\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}.$

Hence (3.11) is valid, which is equivalent to (3.10).

On one hand, similar to (2.9), by the reverse Hölder's inequality , (2.3) and the conditions, in view of q < 0, we have

$\begin{array}{l}\phantom{\rule{1em}{0ex}}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q}\phantom{\rule{2em}{0ex}}\\ \le {\left[\omega \left(x\right){x}^{p\left(1-{\lambda }_{1}\right)-1}\right]}^{q-1}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}{a}_{n}^{q}\phantom{\rule{2em}{0ex}}\\ <{k}_{{\lambda }_{1}}^{q-1}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{-q{\lambda }_{1}+1}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\frac{{n}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}}{{x}^{1-{\lambda }_{1}}}{a}_{n}^{q}.\phantom{\rule{2em}{0ex}}\end{array}$
(3.16)

Similarly, we get (3.12). Applying the reverse Hölder's inequality  gives

$\begin{array}{c}I=\underset{0}{\overset{\infty }{\int }}\left[{\left(1-{\theta }_{\lambda }\left(x\right)\right)}^{1/p}{x}^{\frac{1}{q}-{\lambda }_{1}}f\left(x\right)\right]\left[\frac{{x}^{{\lambda }_{1}-\frac{1}{q}}}{{\left(1-{\theta }_{\lambda }\left(x\right)\right)}^{1/p}}\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}\cdot {a}_{n}\right]\text{d}x\\ \ge {\left\{\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\right\}}^{1/p}{\stackrel{̃}{L}}^{1/q}.\end{array}$
(3.17)

By (3.12), (3.10) is valid. On the other hand, suppose that (3.10) is valid. Let

$f\left(x\right)=\frac{{x}^{q{\lambda }_{1}-1}}{{\left(1-{\theta }_{\lambda }\left(x\right)\right)}^{q-1}}{\left[\sum _{n=1}^{\infty }\frac{{\left(\text{min}\left\{x,n\right\}\right)}^{\beta }}{{\left(\text{max}\left\{x,n\right\}\right)}^{\alpha }}{a}_{n}\right]}^{q-1},x\in \left(0,\infty \right),$

applying (3.10) gives

$\begin{array}{c}\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)=\stackrel{̃}{L}=I\\ \le {k}_{{\lambda }_{1}}{\left\{\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q}.\end{array}$
(3.18)

By (3.16) and the conditions, it follows that $\stackrel{̃}{L}<\infty$. If $\stackrel{̃}{L}=0$, then (3.12) is naturally valid. If $\stackrel{̃}{L}>0$, in view of the conditions of (3.10), then (3.18) takes the strict inequality, and

${\stackrel{̃}{L}}^{1/q}={\left\{\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\text{d}x\right\}}^{1/q}>{k}_{{\lambda }_{1}}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{1/q}.$

In view of q < 0, hence (3.12) is valid, which is equivalent to (3.10). Thus (3.10), (3.11) and (3.12) are equivalent to each other.

For any 0 <ε <p(λ1 + β), suppose that $\stackrel{̃}{f}\left(x\right)=0,x\in \left(0,1\right);\stackrel{̃}{f}\left(x\right)={x}^{{\lambda }_{1}-\frac{\epsilon }{p}-1},x\in \left[1,\infty \right)$ and ${\stackrel{̃}{a}}_{n}={n}^{{\lambda }_{2}-\frac{\epsilon }{q}-1},n\in {ℕ}_{+}$. Assuming there exists a positive number K with $K\ge {k}_{{\lambda }_{1}}$, such that (3.10) is still valid by changing ${k}_{{\lambda }_{1}}$ to K. In particular, on one hand,

$\begin{array}{ll}\hfill \stackrel{̃}{I}& >K{\left\{\underset{0}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{p\left(1-{\lambda }_{1}\right)-1}{\stackrel{̃}{f}}^{p}\left(x\right)\text{d}x\right\}}^{1/p}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-{\lambda }_{2}\right)-1}{\stackrel{̃}{a}}_{n}^{q}\right\}}^{1/q}\phantom{\rule{2em}{0ex}}\\ =K{\left(\underset{1}{\overset{\infty }{\int }}\left(1-{\theta }_{\lambda }\left(x\right)\right){x}^{-1-\epsilon }\text{d}x\right)}^{1/p}{\left(1+\sum _{n=1}^{\infty }{n}^{-1-\epsilon }\right)}^{1/q}\phantom{\rule{2em}{0ex}}\\ =K{\left(\underset{1}{\overset{\infty }{\int }}\left({x}^{-1-\epsilon }-O\left(\frac{1}{{x}^{\beta +{\lambda }_{2}}}\right)\right)\right)}^{}\end{array}$