# A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension

## Abstract

Using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel is obtained, and a best extension with two interval variables is given. The equivalent forms, the operator expressions, the reverses and some particular cases are considered.

2000 Mathematics Subject Classification: 26D15; 47A07.

## 1 Introduction

Assuming that p > 1, $\frac{1}{p}+\frac{1}{q}=1$, f (≥ 0) Lp (R+), g(≥ 0) Lq (R+), ${∥f∥}_{p}={\left\{{\int }_{0}^{\infty }{f}^{p}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}>0$, || g || q > 0, we have the following Hardy-Hilbert's integral inequality [1]:

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

where the constant factor $\frac{\pi }{sin\left(\pi \mathsf{\text{/}}p\right)}$ is the best possible. If a m , b n ≥ 0, $a={\left\{{a}_{m}\right\}}_{m=1}^{\infty }\in {l}^{p}$, $b={\left\{{b}_{n}\right\}}_{n=1}^{\infty }\in {l}^{q}$, $||a|{|}_{p}={\left\{{\sum }_{m=1}^{\infty }{a}_{m}^{p}\right\}}^{\frac{1}{p}}>0$, || b || q > 0, then we still have the following discrete Hardy-Hilbert's inequality with the same best constant factor $\frac{\pi }{sin\left(\pi \mathsf{\text{/}}p\right)}$:

$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{{a}_{m}{b}_{n}}{m+n}<\frac{\pi }{sin\left(\pi \mathsf{\text{/}}p\right)}||a{||}_{p}||b{||}_{q}.$
(2)

For p = q = 2, the above two inequalities reduce to the famous Hilbert's inequalities. Inequalities (1) and (2) are important in analysis and its applications [24].

In 1998, by introducing an independent parameter λ (0, 1], Yang [5] gave an extension of (1) for p = q = 2. Refinement and generalizing the results from [5], Yang [6] gave some best extensions of (1) and (2) as follows: If λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative homogeneous function of degree - λ satisfying for any x, y, t > 0, kλ(tx, ty) = t kλ (x, y), $k\left({\lambda }_{1}\right)={\int }_{0}^{\infty }{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{1}-1}\mathsf{\text{d}}t\in {R}_{+}$, $\varphi \left(x\right)={x}^{p\left(1-{\lambda }_{1}\right)-1}$, $\psi \left(x\right)={x}^{q\left(1-{\lambda }_{2}\right)-1}$, $f\left(\ge 0\right)\in {L}_{p,\varphi }\left({R}_{+}\right)=\left\{f|||f{||}_{p,\varphi }:={\left\{{\int }_{0}^{\infty }\varphi \left(x\right)|f\left(x{\right)|}^{p}\text{d}x\right\}}^{\frac{1}{p}}<\infty \right\}$, g(≥ 0) L q , ψ , || f || p , ϕ , || g || q , ψ > 0, then we have

$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(x,y\right)f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y
(3)

where the constant factor k1) is the best possible. Moreover, if kλ(x, y) is finite and ${k}_{\lambda }\left(x,y\right){x}^{{\lambda }_{1}-1}\left({k}_{\lambda }\left(x,y\right){y}^{{\lambda }_{2}-1}\right)$ is decreasing with respect to x > 0(y > 0), then for a m ,b n ≥ 0, $a={\left\{{a}_{m}\right\}}_{m=1}^{\infty }\in {l}_{p,\varphi }=\left\{a|||a|{|}_{p,\varphi }:={\left\{{\sum }_{n=1}^{\infty }\varphi \left(n\right)|{a}_{n}{|}^{p}\right\}}^{\frac{1}{p}}<\infty \right\}$, $b={\left\{{b}_{n}\right\}}_{n=1}^{\infty }\in {l}_{q,\psi }$, || a || p , ϕ , || b || q , Ψ > 0, we have

$\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }{k}_{\lambda }\left(m,n\right){a}_{m}{b}_{n}
(4)

with the best constant factor k1). Clearly, for λ = 1, ${k}_{1}\left(x,y\right)=\frac{1}{x+y}$, ${\lambda }_{1}=\frac{1}{q}$, ${\lambda }_{2}=\frac{1}{p}$ (3) reduces to (1), and (4) reduces to (2). Some other results about Hilbert-type inequalities are provided by [715].

On half-discrete Hilbert-type inequalities with the non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are the best possible. And, Yang [16] gave a result by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang [17] gave the following half-discrete Hilbert's inequality with the best constant factor B1, λ2)(λ, λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ):

$\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }\frac{{a}_{n}}{{\left(x+n\right)}^{\lambda }}\mathsf{\text{d}}x
(5)

In this article, using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel and a best constant factor is given as follows:

$\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sum _{n=1}^{\infty }{k}_{\lambda }\left(x,\phantom{\rule{2.77695pt}{0ex}}n\right){a}_{n}\mathsf{\text{d}}x
(6)

which is a generalization of (5). A best extension of (6) with two interval variables, some equivalent forms, the operator expressions, the reverses and some particular cases are considered.

## 2 Some lemmas

We set the following conditions:

Condition (i) v(y)(y [n0 - 1, ∞)) is strictly increasing with v(n0 - 1) ≥ 0 and for any fixed x (b, c), f(x, y) is decreasing for y (n0 - 1, ∞) and strictly decreasing in an interval of (n0 - 1, ∞).

Condition (ii) $v\left(y\right)\left(y\in \left[{n}_{0}-\frac{1}{2},\infty \right)\right)$ is strictly increasing with $v\left({n}_{0}-\frac{1}{2}\right)\ge 0$ and for any fixed x (b, c), f(x, y) is decreasing and strictly convex for $y\in \left({n}_{0}-\frac{1}{2},\infty \right)$.

Condition (iii) There exists a constant β ≥ 0, such that v(y)(y [n0 - β, ∞)) is strictly increasing with v(n0 - β) ≥ 0, and for any fixed x (b, c), f(x, y) is piecewise smooth satisfying

$R\left(x\right):=\underset{{n}_{0}-\beta }{\overset{{n}_{0}}{\int }}f\left(x,y\right)\mathsf{\text{d}}y-\frac{1}{2}f\left(x,{n}_{0}\right)-\underset{{n}_{0}}{\overset{\infty }{\int }}\rho \left(y\right){{f}^{\prime }}_{y}\left(x,y\right)\mathsf{\text{d}}y>0,$

where $\rho \left(y\right)\left(=y-\left[y\right]-\frac{1}{2}\right)$ is Bernoulli function of the first order.

Lemma 1 If λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree - λ in ${\mathsf{\text{R}}}_{+}^{2},u\left(x\right)\left(x\in \left(b,c\right),-\infty \le b and v(y)(y [n0, ∞), n0 N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, setting K(x, y) = kλ(u(x), v(y)), then we define weight functions ω(n) and ϖ(x) as follows:

$\omega \left(n\right):\phantom{\rule{0.5em}{0ex}}={\left[v\left(n\right)\right]}^{{\lambda }_{2}}\underset{b}{\overset{c}{\int }}K\left(x,n\right)\left[u\left(x{\right)\right]}^{{\lambda }_{1}-1}{u}^{\prime }\left(x\right)\text{d}x,n\ge {n}_{0}\left(n\in \mathbf{N}\right),$
(7)
$\varpi \left(x\right):\phantom{\rule{0.5em}{0ex}}={\left[u\left(x\right)\right]}^{{\lambda }_{1}}\sum _{n={n}_{0}}^{\infty }K\left(x,n\right)\left[v\left(n{\right)\right]}^{{\lambda }_{2}-1}{v}^{\prime }\left(n\right),x\in \left(b,c\right).$
(8)

It follows

$\omega \left(n\right)=k\left({\lambda }_{1}\right):\phantom{\rule{2.77695pt}{0ex}}=\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{1}-1}\mathsf{\text{d}}t.$
(9)

Moreover, setting $f\left(x,y\right):\phantom{\rule{0.5em}{0ex}}={\left[u\left(x\right)\right]}^{{\lambda }_{1}}K\left(x,y\right)\left[v\left(y{\right)\right]}^{{\lambda }_{2}-1}{v}^{\prime }\left(y\right)$, if k1) R+ and one of the above three conditions is fulfilled, then we still have

$\varpi \left(x\right)
(10)

Proof . Setting $t=\frac{u\left(x\right)}{v\left(n\right)}$ in (7), by calculation, we have (9).

1. (i)

If Condition (i) is fulfilled, then we have

$\begin{array}{l}\varpi \left(x\right)=\sum _{n={n}_{0}}^{\infty }f\left(x,n\right)<{\left[u\left(x\right)\right]}^{{\lambda }_{1}}\underset{{n}_{0}-1}{\overset{\infty }{\int }}K\left(x,y\right)\left[v\left(y{\right)\right]}^{{\lambda }_{2}-1}{v}^{\prime }\left(y\right)\text{d}y\\ \stackrel{t=u\left(x\right)\text{/}v\left(y\right)}{=}\phantom{\rule{0.5em}{0ex}}\underset{0}{\overset{\frac{u\left(x\right)}{v\left({n}_{0}-1\right)}}{\int }}{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{1}-1}\text{d}t\le k\left({\lambda }_{1}\right).\end{array}$
2. (ii)

If Condition (ii) is fulfilled, then by Hadamard's inequality [18], we have

$\begin{array}{c}\varpi \left(x\right)=\sum _{n={n}_{0}}^{\infty }f\left(x,n\right)<\underset{{n}_{0}-\frac{1}{2}}{\overset{\infty }{\int }}f\left(x,y\right)\mathsf{\text{d}}y\\ \stackrel{t=u\left(x\right)\mathsf{\text{/}}v\left(y\right)}{=}\underset{0}{\overset{\frac{u\left(x\right)}{v\left({n}_{0}-\frac{1}{2}\right)}}{\int }}{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{1}-1}\mathsf{\text{d}}t\le k\left({\lambda }_{1}\right).\end{array}$
3. (iii)

If Condition (iii) is fulfilled, then by Euler-Maclaurin summation formula [6], we have

$\begin{array}{ll}\hfill \varpi \left(x\right)& =\sum _{n={n}_{0}}^{\infty }f\left(x,n\right)\phantom{\rule{2em}{0ex}}\\ =\underset{{n}_{0}}{\overset{\infty }{\int }}f\left(x,y\right)\mathsf{\text{d}}y+\frac{1}{2}f\left(x,{n}_{0}\right)+\underset{{n}_{0}}{\overset{\infty }{\int }}\rho \left(y\right){{f}^{\prime }}_{y}\left(x,y\right)\mathsf{\text{d}}y\phantom{\rule{2em}{0ex}}\\ =\underset{{n}_{0}-\beta }{\overset{\infty }{\int }}f\left(x,y\right)\mathsf{\text{d}}y-R\left(x\right)\phantom{\rule{2em}{0ex}}\\ =\underset{0}{\overset{\frac{u\left(x\right)}{v\left({n}_{0}-\beta \right)}}{\int }}{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{1}-1}\mathsf{\text{d}}t-R\left(x\right)\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left({\lambda }_{1}\right)-R\left(x\right)

The lemma is proved.    ■

Lemma 2 Let the assumptions of Lemma 1 be fulfilled and additionally, p > 0(p ≠ 1), $\frac{1}{p}+\frac{1}{q}=1,$ a n ≥ 0, nn0(n N), f (x) is a non-negative measurable function in (b, c). Then, (i) for p > 1, we have the following inequalities:

$\begin{array}{ll}\hfill {J}_{1}:& ={\left\{\sum _{n={n}_{0}}^{\infty }\frac{{v}^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{1-p{\lambda }_{2}}}{\left[\underset{d}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^{p}\right\}}^{\frac{1}{p}}\phantom{\rule{2em}{0ex}}\\ \le {\left[k\left({\lambda }_{1}\right)\right]}^{\frac{1}{q}}{\left\{\underset{d}{\overset{c}{\int }}\varpi \left(x\right)\frac{{\left[u\left(x\right)\right]}^{p\left(1-{\lambda }_{1}\right)-1}}{{\left[{u}^{\prime }\left(x\right)\right]}^{p-1}}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}},\phantom{\rule{2em}{0ex}}\end{array}$
(11)
$\begin{array}{ll}\hfill {L}_{1}:& ={\left\{\underset{b}{\overset{c}{\int }}\frac{{\left[\varpi \left(x\right)\right]}^{1-q}{u}^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{1-q{\lambda }_{1}}}{\left[\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\right]}^{q}\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}{\left\{k\left({\lambda }_{1}\right)\sum _{n={n}_{0}}^{\infty }\frac{{\left[v\left(n\right)\right]}^{q\left(1-{\lambda }_{2}\right)-1}}{{\left[{v}^{\prime }\left(n\right)\right]}^{q-1}}{a}_{n}^{q}\right\}}^{\frac{1}{q}};\phantom{\rule{2em}{0ex}}\end{array}$
(12)

(ii) for 0 < p < 1, we have the reverses of (11) and (12).

Proof(i) By Hölder's inequality with weight [18] and (9), it follows

$\begin{array}{c}{\left[\underset{b}{\overset{c}{\int }}K\left(x,\phantom{\rule{0.5em}{0ex}}n\right)f\left(x\right)\text{d}x\right]}^{p}=\left\{\underset{b}{\overset{c}{\int }}K\left(x,n\right)\left[\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_{1}\right)\text{/}q}}{{\left[v\left(n\right)\right]}^{\left(1-{\lambda }_{2}\right)\text{/}p}}\frac{{\left[{v}^{\prime }\left(n\right)\right]}^{1\text{/}p}}{{\left[{u}^{\prime }\left(x\right)\right]}^{1\text{/}q}}f\left(x\right)\right]\\ \phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}{\left[\frac{{\left[v\left(n\right)\right]}^{\left(1-{\lambda }_{2}\right)\text{/}p}}{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_{1}\right)\text{/}q}}\frac{{\left[{u}^{\prime }\left(x\right)\right]}^{1\text{/}q}}{{\left[{v}^{\prime }\left(n\right)\right]}^{1\text{/}p}}\right]\text{d}x\right\}}^{p}\\ \le \phantom{\rule{0.5em}{0ex}}\underset{b}{\overset{c}{\int }}K\left(x,n\right)\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}{v}^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{1-{\lambda }_{2}}{\left[{u}^{\prime }\left(x\right)\right]}^{p-1}}{f}^{p}\left(x\right)\text{d}x\\ \phantom{\rule{0.5em}{0ex}}×\phantom{\rule{0.5em}{0ex}}{\left\{\underset{b}{\overset{c}{\int }}K\left(x,n\right)\frac{{\left[v\left(n\right)\right]}^{\left(1-{\lambda }_{2}\right)\left(q-1\right)}{u}^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{1-{\lambda }_{1}}{\left[{v}^{\prime }\left(n\right)\right]}^{q-1}}\text{d}x\right\}}^{p-1}\\ =\phantom{\rule{0.5em}{0ex}}{\left\{\frac{\omega \left(n\right)\left[v\left(n{\right)\right]}^{q\left(1-{\lambda }_{2}\right)-1}}{{\left[{v}^{\prime }\left(n\right)\right]}^{q-1}}\right\}}^{p-1}\underset{b}{\overset{c}{\int }}K\left(x,n\right)\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}{v}^{\prime }\left(n\right){f}^{p}\left(x\right)}{{\left[v\left(n\right)\right]}^{1-{\lambda }_{2}}{\left[{u}^{\prime }\left(x\right)\right]}^{p-1}}\text{d}x\\ =\phantom{\rule{0.5em}{0ex}}\frac{{\left[k\left({\lambda }_{1}\right)\right]}^{p-1}}{{\left[v\left(n\right)\right]}^{p{\lambda }_{2}-1}{v}^{\prime }\left(n\right)}\underset{b}{\overset{c}{\int }}K\left(x,n\right)\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}{v}^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{1-{\lambda }_{2}}{\left[{u}^{\prime }\left(x\right)\right]}^{p-1}}{f}^{p}\left(x\right)\text{d}x.\end{array}$

Then, by Lebesgue term-by-term integration theorem [19], we have

$\begin{array}{ll}\hfill {J}_{1}& \le {\left[k\left({\lambda }_{1}\right)\right]}^{\frac{1}{q}}{\left\{\sum _{n={n}_{0}}^{\infty }\underset{b}{\overset{c}{\int }}K\left(x,n\right)\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}{v}^{\prime }\left(n\right){f}^{p}\left(x\right)}{{\left[v\left(n\right)\right]}^{1-{\lambda }_{2}}{\left[{u}^{\prime }\left(x\right)\right]}^{p-1}}\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}\phantom{\rule{2em}{0ex}}\\ ={\left[k\left({\lambda }_{1}\right)\right]}^{\frac{1}{q}}{\left\{\underset{b}{\overset{c}{\int }}\sum _{n={n}_{0}}^{\infty }K\left(x,n\right)\frac{{\left[u\left(x\right)\right]}^{\left(1-{\lambda }_{1}\right)\left(p-1\right)}{v}^{\prime }\left(n\right){f}^{p}\left(x\right)}{{\left[v\left(n\right)\right]}^{1-{\lambda }_{2}}{\left[{u}^{\prime }\left(x\right)\right]}^{p-1}}\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}\phantom{\rule{2em}{0ex}}\\ ={\left[k\left({\lambda }_{1}\right)\right]}^{\frac{1}{q}}{\left\{\underset{b}{\overset{c}{\int }}\varpi \left(x\right)\frac{{\left[u\left(x\right)\right]}^{p\left(1-{\lambda }_{1}\right)-1}}{{\left[{u}^{\prime }\left(x\right)\right]}^{p-1}}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}},\phantom{\rule{2em}{0ex}}\end{array}$

and (11) follows.

Still by Hölder's inequality, we have

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

$\begin{array}{ll}\hfill {L}_{1}& \le {\left\{\underset{b}{\overset{c}{\int }}\sum _{n={n}_{0}}^{\infty }K\left(x,n\right)\frac{{u}^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{1-{\lambda }_{1}}}\frac{{\left[v\left(n\right)\right]}^{\left(q-1\right)\left(1-{\lambda }_{2}\right)}}{{\left[{v}^{\prime }\left(n\right)\right]}^{q-1}}{a}_{n}^{q}\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\\ ={\left\{\sum _{n={n}_{0}}^{\infty }\left[{\left[v\left(n\right)\right]}^{{\lambda }_{2}}\underset{b}{\overset{c}{\int }}K\left(x,n\right)\frac{{u}^{\prime }\left(x\right)\mathsf{\text{d}}x}{{\left[u\left(x\right)\right]}^{1-{\lambda }_{1}}}\right]\frac{{\left[v\left(n\right)\right]}^{q\left(1-{\lambda }_{2}\right)-1}}{{\left[{v}^{\prime }\left(n\right)\right]}^{q-1}}{a}_{n}^{q}\right\}}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\\ ={\left\{\sum _{n={n}_{0}}^{\infty }\omega \left(n\right)\frac{{\left[v\left(n\right)\right]}^{q\left(1-{\lambda }_{2}\right)-1}}{{\left[{v}^{\prime }\left(n\right)\right]}^{q-1}}{a}_{n}^{q}\right\}}^{\frac{1}{q}},\phantom{\rule{2em}{0ex}}\end{array}$

and then in view of (9), inequality (12) follows.

1. (ii)

By the reverse Hölder's inequality [18] and in the same way, for q < 0, we have the reverses of (11) and (12). ■

## 3 Main results

We set $\Phi \left(x\right):=\frac{{\left[u\left(x\right)\right]}^{p\left(1-{\lambda }_{1}\right)-1}}{{\left[{u}^{\prime }\left(x\right)\right]}^{p-1}}\left(x\in \left(b,c\right)\right)$, $\Psi \left(n\right):=\frac{{\left[v\left(n\right)\right]}^{q\left(1-{\lambda }_{2}\right)-1}}{{\left[{v}^{\prime }\left(n\right)\right]}^{q-1}}\left(n\ge {n}_{0},n\in \mathbf{N}\right)$, wherefrom ${\left[\Phi \left(x\right)\right]}^{1-q}=\frac{{u}^{\prime }\left(x\right)}{{\left[u\left(x\right)\right]}^{1-q{\lambda }_{1}}}$, ${\left[\Psi \left(n\right)\right]}^{1-p}=\frac{{v}^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{1-p{\lambda }_{2}}}$.

Theorem 1 Suppose that λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree - λ¸ in ${\mathsf{\text{R}}}_{+}^{2},u\left(x\right)\left(x\in \left(b,c\right),-\infty \le b and v(y)(y [n0, ∞), n0 N are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, ϖ(x) < k1) R+(x (b, c)). If $p>1,\frac{1}{p}+\frac{1}{q}=1$ , f ( x ), a n ≥ 0, f L p Φ ( b , c ), $a={\left\{{a}_{n}\right\}}_{n={n}_{0}}^{\infty }\in {l}_{q,\Psi },$ || f || p ,Φ > 0 and || a || q ,Ψ > 0, then we have the following equivalent inequalities:

$\begin{array}{ll}\hfill I:& =\sum _{n={n}_{0}}^{\infty }{a}_{n}\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x=\underset{b}{\overset{c}{\int }}f\left(x\right)\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\mathsf{\text{d}}x\phantom{\rule{2em}{0ex}}\\
(13)
$J:={\left\{\sum _{n={n}_{0}}^{\infty }{\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^{p}\right\}}^{\frac{1}{p}}
(14)
$L:={\left\{\underset{b}{\overset{c}{\int }}{\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\right]}^{q}\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}
(15)

Moreover, if $\frac{{v}^{\prime }\left(y\right)}{v\left(y\right)}\left(y\ge {n}_{0}\right)$ is decreasing and there exist constants δ < λ1 and M > 0, such that, then the constant factor k1) in the above inequalities is the best possible.

Proof By Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for ϖ(x) < k1) R+, we have (14). By Hölder's inequality, we have

$I=\sum _{n={n}_{0}}^{\infty }\left[{\Psi }^{{\frac{-1}{q}}^{}}\left(n\right)\underset{b}{\overset{c}{\int }}K\left(x,\phantom{\rule{2.77695pt}{0ex}}n\right)f\left(x\right)\mathsf{\text{d}}x\right]\left[{\Psi }^{\frac{1}{q}}\left(n\right){a}_{n}\right]\le J||a{||}_{q,\Psi }.$
(16)

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

${a}_{n}:={\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{b}{\overset{c}{\int }}K\left(x,\phantom{\rule{2.77695pt}{0ex}}n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^{p-1},n\ge {n}_{0},$

then Jp-1 = || a || q . By (11), we find J < ∞. If J = 0, then (14) is naturally valid; if J > 0, then by (13), we have

$||a|{|}_{q,\Psi }^{q}={J}^{p}=I

and we have (14), which is equivalent to (13).

In view of (12), for [ϖ(x) ]1-q> [k1)]1-q, we have (15). By Hölder's in equality, we find

$I=\underset{b}{\overset{c}{\int }}\left[{\Phi }^{\frac{1}{p}}\left(x\right)f\left(x\right)\right]\left[{\Phi }^{\frac{-1}{p}}\left(x\right)\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\right]\mathsf{\text{d}}x\le \phantom{\rule{2.77695pt}{0ex}}||f{||}_{p,\Phi }L.$
(17)

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

$f\left(x\right):={\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\right]}^{q-1},x\in \left(b,c\right),$

then Lq-1 = || f || p Φ . By (12), we find L < ∞. If L = 0, then (15) is naturally valid; if L > 0, then by (13), we have

$||f|{|}_{p,\Phi }^{p}={L}^{q}=I

and we have (15) which is equivalent to (13).

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

There exists an unified constant d (b, c), satisfying u(d) = 1. For 0 < ε < p1 - δ), setting $\stackrel{̃}{f}\left(x\right)=0$, x (b, d); $\stackrel{̃}{f}\left(x\right)={\left[u\left(x\right)\right]}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}{u}^{\prime }\left(x\right)$, x (d, c), and ${ã}_{n}={\left[v\left(n\right)\right]}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}{v}^{\prime }\left(n\right)$, nn0, if there exists a positive number k(≤ k1)), such that (13) is valid as we replace k1) by k, then in particular, we find

$\begin{array}{ll}\hfill Ĩ:& =\sum _{n={n}_{0}}^{\infty }\underset{b}{\overset{c}{\int }}K\left(x,n\right){ã}_{n}\stackrel{̃}{f}\left(x\right)\mathsf{\text{d}}x
(18)
$\begin{array}{l}\stackrel{˜}{I}=\sum _{n={n}_{0}}^{\infty }{\left[v\left(n\right)\right]}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}{v}^{\prime }\left(n\right)\underset{d}{\overset{c}{\int }}K\left(x,n\right)\left[u\left(x{\right)\right]}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}{u}^{\prime }\left(x\right)\text{d}x\\ \stackrel{t=u\left(x\right)\text{/}v\left(n\right)}{=}\sum _{n={n}_{0}}^{\infty }{\left[v\left(n\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(n\right)\underset{1/v\left(n\right)}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,\phantom{\rule{0.5em}{0ex}}1\right){t}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}\text{d}t\\ =k\left({\lambda }_{1}-\frac{\epsilon }{p}\right)\sum _{n={n}_{0}}^{\infty }{\left[v\left(n\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(n\right)-A\left(\epsilon \right)\\ >\phantom{\rule{0.5em}{0ex}}k\left({\lambda }_{1}-\frac{\epsilon }{p}\right)\underset{{n}_{0}}{\overset{\infty }{\int }}{\left[v\left(y\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(y\right)\text{d}y-A\left(\epsilon \right)\\ =\phantom{\rule{0.5em}{0ex}}\frac{1}{\epsilon }k\left({\lambda }_{1}-\frac{\epsilon }{p}\right){\left[v\left({n}_{0}\right)\right]}^{-\epsilon }-A\left(\epsilon \right),\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}A\left(\epsilon \right):=\sum _{n={n}_{0}}^{\infty }{\left[v\left(n\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(n\right)\underset{0}{\overset{1/v\left(n\right)}{\int }}{k}_{\lambda }\left(\mathrm{t,}1\right){t}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}\text{d}t.\end{array}$
(19)

For ${k}_{\lambda }\left(t,1\right)\le M\left(\frac{1}{{t}^{\delta }}\right)\left(\delta <{\lambda }_{1};t\in \left(0,1∕v\left({n}_{0}\right)\right]\right)$, we find

$\begin{array}{ll}\hfill 0

namely A(ε) = O(1)(ε → 0+). Hence, by (18) and (19), it follows

$k\left({\lambda }_{1}-\frac{\epsilon }{p}\right){\left[v\left({n}_{0}\right)\right]}^{-\epsilon }-\epsilon O\left(\mathsf{\text{1}}\right)
(20)

By Fatou Lemma [19], we have $k\left({\lambda }_{1}\right)\le {\underset{}{\text{lim}}}_{\epsilon \to {0}^{+}}k\left({\lambda }_{1}-\frac{\epsilon }{p}\right)$, then by (20), it follows k1) ≤ k(ε → 0+). Hence, k = k1) is the best value of (12).

By the equivalence, the constant factor k1) in (14) and (15) is the best possible, otherwise we can imply a contradiction by (16) and (17) that the constant factor in (13) is not the best possible. ■

Remark 1 (i) Define a half-discrete Hilbert's operator T : ${L}_{p,\Phi }\left(b,c\right)\to {l}_{p,{\Psi }^{1-p}}$ as: for f L p Φ (b, c), we define $Tf\in {l}_{p,{\Psi }^{1-p}}$, satisfying

$Tf\left(n\right)=\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)dx,\phantom{\rule{1em}{0ex}}n\ge {n}_{0}.$

Then, by (14), it follows $||Tf|{|}_{p.{\Psi }^{1-p}}\le k\left({\lambda }_{1}\right)||f|{|}_{p,\Phi }$ and then T is a bounded operator with || T || ≤ k1). Since, by Theorem 1, the constant factor in (14) is the best possible, we have || T || = k1).

1. (ii)

Define a half-discrete Hilbert's operator $\stackrel{̃}{T}:{l}_{q,\Psi }\to {L}_{q,{\Phi }^{1-q}}\left(b,c\right)$ as: for a l q Ψ , we define $\stackrel{̃}{T}a\in {L}_{q,{\Phi }^{1-q}}\left(b,c\right)$, satisfying

$\stackrel{̃}{T}a\left(x\right)=\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n},x\in \left(b,c\right).$

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

In the following theorem, for 0 < p < 1, or p < 0, we still use the formal symbols of $||f|{|}_{p,\stackrel{̃}{\Phi }}$ and ||a|| q ,Ψ and so on. ■

Theorem 2 Suppose that λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degreein ${R}_{+}^{2}$, u(x)(x (b, c), -∞ ≤ b < c ≤ ∞) and v(y)(y [n0, ∞), n0 N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, k1) R+, θλ(x) (0, 1), k1)(1 - θλ(x)) < ϖ(x) < k1)(x (b, c)). If 0 < p < 1, $\frac{1}{p}+\frac{1}{q}=1$, f(x), a n ≥ 0, $\stackrel{̃}{\Phi }\left(x\right):=\left(\mathsf{\text{1}}\phantom{\rule{2.77695pt}{0ex}}-\phantom{\rule{2.77695pt}{0ex}}{\theta }_{\lambda }\left(x\right)\right)\Phi \left(x\right)\left(x\in \left(b,c\right)\right)$, $0<\phantom{\rule{2.77695pt}{0ex}}||f|{|}_{q,\stackrel{̃}{\Phi }}<\infty$ and 0 < ||a|| q ,Ψ < ∞. Then, we have the following equivalent inequalities:

$\begin{array}{ll}\hfill I\phantom{\rule{1em}{0ex}}:& =\phantom{\rule{1em}{0ex}}\sum _{n={n}_{0}}^{\infty }\underset{b}{\overset{c}{\int }}K\left(x,n\right){a}_{n}f\left(x\right)\mathsf{\text{d}}x=\underset{b}{\overset{c}{\int }}\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}f\left(x\right)\mathsf{\text{d}}x\phantom{\rule{2em}{0ex}}\\ >\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}k\left({\lambda }_{1}\right)||f|{|}_{p,\stackrel{̃}{\Phi }}||a|{|}_{q,\Psi },\phantom{\rule{2em}{0ex}}\end{array}$
(21)
$J\phantom{\rule{1em}{0ex}}:=\phantom{\rule{1em}{0ex}}{\left\{\sum _{n={n}_{0}}^{\infty }{\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^{p}\right\}}^{\frac{1}{p}}>k\left({\lambda }_{1}\right)||f|{|}_{p,\stackrel{̃}{\Phi }},$
(22)
$\stackrel{̃}{L}\phantom{\rule{1em}{0ex}}:=\phantom{\rule{1em}{0ex}}{\left\{\underset{b}{\overset{c}{\int }}{\left[\stackrel{̃}{\Phi }\left(x\right)\right]}^{1-q}{\left[\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\right]}^{q}\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}>k\left({\lambda }_{1}\right)||a|{|}_{q,\Psi }.$
(23)

Moreover, if $\frac{{v}^{\prime }\left(y\right)}{v\left(y\right)}\left(y\ge {n}_{0}\right)$ is decreasing and there exist constants δ, δ0 > 0, such that ${\theta }_{\lambda }\left(x\right)=O\left(\frac{1}{{\left[u\left(x\right)\right]}^{\delta }}\right)\left(x\in \left(d,c\right)\right)$ and k1 - δ0) R+, then the constant factor k1) in the above inequalities is the best possible.

Proof. In view of (9) and the reverse of (11), for ϖ(x) > k1)(1 - θλ(x)), we have (22). By the reverse Hölder's inequality, we have

$I=\sum _{n={n}_{0}}^{\infty }\left[{\Psi }^{\frac{-1}{q}}\left(n\right)\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]\phantom{\rule{2.77695pt}{0ex}}\left[{\Psi }^{\frac{1}{q}}\left(n\right){a}_{n}\right]\ge J||a|{|}_{q,\Psi }.$
(24)

Then, by (22), we have (21). On the other hand, assuming that (21) is valid, setting a n as Theorem 1, then Jp-1 = ||a|| q Ψ. By the reverse of (11), we find J > 0. If J = ∞, then (24) is naturally valid; if J < ∞, then by (21), we have

$||a|{|}_{q,\Psi }^{q}={J}^{p}=I>k\left({\lambda }_{1}\right)||f|{|}_{p,\stackrel{̃}{\Phi }}||a|{|}_{q,\Psi },||a|{|}_{q,\Psi }^{q-1}=J>k\left({\lambda }_{1}\right)||f|{|}_{p,\stackrel{̃}{\Phi }},$

and we have (22) which is equivalent to (21).

In view of (9) and the reverse of (12), for [ϖ(x)]1-q> [k1)(1 - θ λ (x))]1-q(q < 0), we have (23). By the reverse Hölder's inequality, we have

$I=\underset{b}{\overset{c}{\int }}\left[{\stackrel{̃}{\Phi }}^{\frac{1}{p}}\left(x\right)f\left(x\right)\right]\left[{\stackrel{̃}{\Phi }}^{\frac{-1}{p}}\left(x\right)\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\right]\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{d}}x\ge \phantom{\rule{2.77695pt}{0ex}}||f|{|}_{p,\stackrel{̃}{\Phi }}\stackrel{̃}{L}.$
(25)

Then, by (23), we have (21). On the other hand, assuming that (21) is valid, setting

$f\left(x\right):={\left[\stackrel{̃}{\Phi }\left(x\right)\right]}^{1-q}{\left[\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\right]}^{q-1},\phantom{\rule{1em}{0ex}}x\in \left(b,c\right),$

then ${\stackrel{̃}{L}}^{q-1}=\phantom{\rule{2.77695pt}{0ex}}||f|{|}_{p,\stackrel{̃}{\Phi }}$. By the reverse of (12), we find $\stackrel{̃}{L}>0$. If $\stackrel{̃}{L}=\infty$, then (23) is naturally valid; if $\stackrel{̃}{L}<\infty$, then by (21), we have

$||f|{|}_{p,\stackrel{̃}{\Phi }}^{p}={\stackrel{̃}{L}}^{q}=I>k\left({\lambda }_{1}\right)||f|{|}_{p,\stackrel{̃}{\Phi }}||a|{|}_{q,\Psi },||f|{|}_{p,\stackrel{̃}{\Phi }}^{p-1}=\stackrel{̃}{L}>k\left({\lambda }_{1}\right)||a|{|}_{q,\Psi },$

and we have (23) which is equivalent to (21).    ■

Hence, inequalities (21), (22) and (23) are equivalent.

For 0 < ε < pδ0, setting $\stackrel{̃}{f}\left(x\right)$ and ${ã}_{n}$ as Theorem 1, if there exists a positive number k(≥ k1)), such that (21) is still valid as we replace k1) by k, then in particular, for q < 0, in view of (9) and the conditions, we have

$\begin{array}{ll}\hfill Ĩ:& =\underset{b}{\overset{c}{\int }}\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){ã}_{n}\stackrel{̃}{f}\left(x\right)\mathsf{\text{d}}x>k||\stackrel{̃}{f}|{|}_{p,\stackrel{̃}{\Phi }}||ã|{|}_{q,\Psi }\phantom{\rule{2em}{0ex}}\\ =k{\left\{\underset{d}{\overset{c}{\int }}\left(1-O\left(\frac{1}{{\left[u\left(x\right)\right]}^{\delta }}\right)\right)\phantom{\rule{2.77695pt}{0ex}}\frac{{u}^{\prime }\left(x\right)\mathsf{\text{d}}x}{{\left[u\left(x\right)\right]}^{\epsilon +1}}\right\}}^{\frac{1}{p}}{\left\{\sum _{n={n}_{0}}^{\infty }\frac{{v}^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{\epsilon +1}}\right\}}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\\ =k{\left\{\frac{1}{\epsilon }-O\left(1\right)\right\}}^{\frac{1}{p}}{\left\{\frac{{v}^{\prime }\left({n}_{0}\right)}{{\left[v\left({n}_{0}\right)\right]}^{\epsilon +1}}+\sum _{n={n}_{0}+1}^{\infty }\frac{{v}^{\prime }\left(n\right)}{{\left[v\left(n\right)\right]}^{\epsilon +1}}\right\}}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\\ >k{\left\{\frac{1}{\epsilon }-O\left(1\right)\right\}}^{\frac{1}{p}}{\left\{\frac{{v}^{\prime }\left({n}_{0}\right)}{{\left[v\left({n}_{0}\right)\right]}^{\epsilon +1}}+\underset{{n}_{0}}{\overset{\infty }{\int }}\frac{{v}^{\prime }\left(y\right)}{{\left[v\left(y\right)\right]}^{+1}}\mathsf{\text{d}}y\right\}}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\\ =\frac{k}{\epsilon }{\left\{1-\epsilon O\left(1\right)\right\}}^{\frac{1}{p}}{\left\{\epsilon \frac{{v}^{\prime }\left({n}_{0}\right)}{{\left[v\left({n}_{0}\right)\right]}^{\epsilon +1}}+{\left[v\left({n}_{0}\right)\right]}^{-\epsilon }\right\}}^{\frac{1}{q}},\phantom{\rule{2em}{0ex}}\end{array}$
(26)
$\begin{array}{ll}\hfill Ĩ\phantom{\rule{1em}{0ex}}& =\phantom{\rule{1em}{0ex}}\sum _{n={n}_{0}}^{\infty }{\left[v\left(n\right)\right]}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}{v}^{\prime }\left(n\right)\underset{d}{\overset{c}{\int }}K\left(x,\phantom{\rule{2.77695pt}{0ex}}n\right){\left[u\left(x\right)\right]}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}{u}^{\prime }\left(x\right)\mathsf{\text{d}}x\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}\sum _{n={n}_{0}}^{\infty }{\left[v\left(n\right)\right]}^{{\lambda }_{2}-\frac{\epsilon }{q}-1}{v}^{\prime }\left(n\right)\underset{b}{\overset{c}{\int }}K\left(x,\phantom{\rule{2.77695pt}{0ex}}n\right)\phantom{\rule{2.77695pt}{0ex}}{\left[u\left(x\right)\right]}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}{u}^{\prime }\left(x\right)\mathsf{\text{d}}x\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\stackrel{t=u\left(x\right)/v\left(n\right)}{=}\sum _{n={n}_{0}}^{\infty }{\left[v\left(n\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(n\right)\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,\phantom{\rule{2.77695pt}{0ex}}1\right){t}^{\left({\lambda }_{1}-\frac{\epsilon }{p}\right)-1}\mathsf{\text{d}}t\phantom{\rule{2em}{0ex}}\\ \le \phantom{\rule{1em}{0ex}}k\left({\lambda }_{1}-\frac{\epsilon }{p}\right)\left[\frac{{v}^{\prime }\left({n}_{0}\right)}{{\left[v\left({n}_{0}\right)\right]}^{\epsilon +1}}+\underset{{n}_{0}}{\overset{\infty }{\int }}{\left[v\left(y\right)\right]}^{-\epsilon -1}{v}^{\prime }\left(y\right)\mathsf{\text{d}}y\right]\phantom{\rule{2em}{0ex}}\\ =\phantom{\rule{1em}{0ex}}\frac{1}{\epsilon }k\left({\lambda }_{1}-\frac{\epsilon }{p}\right)\left[\epsilon \frac{{v}^{\prime }\left({n}_{0}\right)}{{\left[v\left({n}_{0}\right)\right]}^{\epsilon +1}}+{\left[v\left({n}_{0}\right)\right]}^{-\epsilon }\right].\phantom{\rule{2em}{0ex}}\end{array}$
(27)

Since we have ${k}_{\lambda }\left(t,\phantom{\rule{2.77695pt}{0ex}}1\right){t}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}\le {k}_{\lambda }\left(t,\phantom{\rule{2.77695pt}{0ex}}1\right){t}^{{\lambda }_{1}-{\delta }_{0}-1}$, t (0, 1] and

$\underset{0}{\overset{1}{\int }}{k}_{\lambda }\left(t,\phantom{\rule{2.77695pt}{0ex}}1\right){t}^{{\lambda }_{1}-{\delta }_{0}-1}\mathsf{\text{d}}t\le k\left({\lambda }_{1}-{\delta }_{0}\right)<\infty ,$

then by Lebesgue control convergence theorem [19], it follows

$\begin{array}{ll}\hfill k\left({\lambda }_{1}-\frac{\epsilon }{p}\right)& \le \underset{1}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,1\right){t}^{{\lambda }_{1}-1}\mathsf{\text{d}}t+\underset{0}{\overset{1}{\int }}{k}_{\lambda }\left(t,\phantom{\rule{2.77695pt}{0ex}}1\right){t}^{{\lambda }_{1}-\frac{\epsilon }{p}-1}\mathsf{\text{d}}t\phantom{\rule{2em}{0ex}}\\ =k\left({\lambda }_{1}\right)+o\left(1\right)\left(\epsilon \to {0}^{+}\right).\phantom{\rule{2em}{0ex}}\end{array}$

By (26) and (27), we have

$\left(k\left({\lambda }_{1}\right)+o\left(1\right)\right)\left[\epsilon \frac{{v}^{\prime }\left({n}_{0}\right)}{{\left[v\left({n}_{0}\right)\right]}^{\epsilon +1}}+{\left[v\left({n}_{0}\right)\right]}^{-\epsilon }\right]>k{\left\{1-\epsilon O\left(1\right)\right\}}^{\frac{1}{p}}{\left[\epsilon \frac{{v}^{\prime }\left({n}_{0}\right)}{{\left[v\left({n}_{0}\right)\right]}^{\epsilon +1}}+{\left[v\left({n}_{0}\right)\right]}^{-\epsilon }\right]}^{\frac{1}{q}},$

and then k1) ≥ k(ε → 0+). Hence, k = k1) is the best value of (21).

By the equivalence, the constant factor k1) in (22) and (23) is the best possible, otherwise we can imply a contradiction by (24) and (25) that the constant factor in (21) is not the best possible.    ■

In the same way, for p < 0, we also have the following theorem.

Theorem 3 Suppose that λ1, λ2 R, λ1 + λ2 = λ, kλ(x, y) is a non-negative finite homogeneous function of degree¸ in ${R}_{+}^{2}$, u(x)(x (b, c), -∞ ≤ b < c ≤ ∞) and v(y)(y [n0, ∞), n0 N) are strictly increasing differential functions with u(b+) = 0, v(n0) > 0, u(c-) = v(∞) = ∞, ϖ(x) < k1) R + (x (b, c)). If p < 0, $\frac{1}{p}+\frac{1}{q}=1$, f(x), a n ≥ 0, 0 < || f || p Φ < ∞ and 0 < ||a|| q ,Ψ < ∞. Then, we have the following equivalent inequalities:

$\begin{array}{c}I:=\sum _{n={n}_{0}}^{\infty }\underset{b}{\overset{c}{\int }}K\left(x,n\right){a}_{n}f\left(x\right)\mathsf{\text{d}}x=\underset{b}{\overset{c}{\int }}\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}f\left(x\right)\mathsf{\text{d}}x\\ \phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}>k\left({\lambda }_{1}\right)||f|{|}_{p,\Phi }||a|{|}_{q,\Psi },\end{array}$
(28)
$J:={\left\{\sum _{n={n}_{0}}^{\infty }{\left[\Psi \left(n\right)\right]}^{1-p}{\left[\underset{b}{\overset{c}{\int }}K\left(x,n\right)f\left(x\right)\mathsf{\text{d}}x\right]}^{p}\right\}}^{\frac{1}{p}}>k\left({\lambda }_{1}\right)||f|{|}_{p,\Phi },$
(29)
$\stackrel{̃}{L}:={\left\{\underset{b}{\overset{c}{\int }}{\left[\Phi \left(x\right)\right]}^{1-q}{\left[\sum _{n={n}_{0}}^{\infty }K\left(x,n\right){a}_{n}\right]}^{q}\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}>k\left({\lambda }_{1}\right)||a|{|}_{q,\Psi }.$
(30)

Moreover, if $\frac{{v}^{\prime }\left(y\right)}{v\left(y\right)}\left(y\ge {n}_{0}\right)$ is decreasing and there exists constant δ0 > 0, such that k1 + δ0) R+, then the constant factor k1) in the above inequalities is the best possible.

Remark 2 (i) For n0 = 1, b = 0, c = ∞, u(x) = v(x) = x, if

$\varpi \left(x\right)={x}^{{\lambda }_{1}}\sum _{n=1}^{\infty }{k}_{\lambda }\left(x,n\right){n}^{{\lambda }_{2}-1}

then (13) reduces to (6). In particular, for ${k}_{\lambda }\left(x,n\right)=\frac{1}{{\left(x+n\right)}^{\lambda }}\left(\lambda ={\lambda }_{1}+{\lambda }_{2},{\lambda }_{1}>0,0<{\lambda }_{2}\le 1\right)$,(6) reduces to (5).

1. (ii)

For n0 = 1, b = 0, c = ∞, u(x) = v(x) = xα(α > 0), ${k}_{\lambda }\left(x,y\right)=\frac{1}{{\left(max\left\{x,y\right\}\right)}^{\lambda }}\left(\lambda ,{\lambda }_{1}>0,0<\alpha {\lambda }_{2}\le 1\right)$, since

$f\left(x,y\right)=\frac{\alpha {x}^{\alpha {\lambda }_{1}}{y}^{\alpha {\lambda }_{2}-1}}{{\left(max\left\{{x}^{\alpha },{y}^{\alpha }\right\}\right)}^{\lambda }}=\left\{\begin{array}{c}\hfill \alpha {x}^{-\alpha {\lambda }_{2}}{y}^{\alpha {\lambda }_{2}-1},y\le x\hfill \\ \hfill \alpha {x}^{\alpha {\lambda }_{1}}{y}^{-\alpha {\lambda }_{1}-1},y>x\hfill \end{array}\right\$

is decreasing for y (0, ∞) and strictly decreasing in an interval of (0, ∞), then by Condition (i), it follows

$\begin{array}{l}\varpi \left(x\right)<\alpha {x}^{\alpha {\lambda }_{1}}\underset{0}{\overset{\infty }{\int }}\frac{1}{{\left(max\left\{{x}^{\alpha },{y}^{\alpha }\right\}\right)}^{\lambda }}{y}^{\alpha \left({\lambda }_{2}-1\right)}{y}^{\alpha -1}\text{d}y\\ \phantom{\rule{0.5em}{0ex}}\stackrel{t=\left(y/x{\right)}^{\alpha }}{=}\underset{0}{\overset{\infty }{\int }}\frac{{t}^{{\lambda }_{2}-1}}{{\left(max\left\{1,t\right\}\right)}^{\lambda }}\text{d}t=\frac{\lambda }{{\lambda }_{1}{\lambda }_{2}}=k\left({\lambda }_{1}\right).\end{array}$

Since for $\delta =\frac{{\lambda }_{1}}{2}<{\lambda }_{1}$, ${k}_{\lambda }\left(t,1\right)=1\le \frac{1}{{t}^{\delta }}\left(t\in \left(0,1\right]\right)$, then by (13), we have the following inequality with the best constant factor $\frac{\lambda }{\alpha {\lambda }_{1}{\lambda }_{2}}$:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\infty }{a}_{n}\underset{0}{\overset{\infty }{\int }}\frac{1}{{\left(max\left\{{x}^{\alpha },{n}^{\alpha }\right\}\right)}^{\lambda }}f\left(x\right)\mathsf{\text{d}}x\\ <\frac{\lambda }{\alpha {\lambda }_{1}{\lambda }_{2}}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1-\alpha {\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\infty }{n}^{q\left(1-\alpha {\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}}.\end{array}$
(31)
1. (iii)

For n0 = 1, b = β, c = ∞, $u\left(x\right)=v\left(x\right)=x-\beta \left(0\le \beta \le \frac{1}{2}\right)$, ${k}_{\lambda }\left(x,y\right)=\frac{ln\left(x∕y\right)}{{x}^{\lambda }-{y}^{\lambda }}\left(\lambda ,{\lambda }_{1}>0,0<{\lambda }_{2}\le 1\right)$, since for any fixed x (β, ∞),

$f\left(x,y\right)=\left(x-\beta {\right)}^{{\lambda }_{1}}\frac{ln\left[\left(x-\beta \right)/\left(y-\beta \right)\right]}{{\left(x-\beta \right)}^{\lambda }-{\left(y-\beta \right)}^{\lambda }}{\left(y-\beta \right)}^{{\lambda }_{2}-1}$

is decreasing and strictly convex for $y\in \left(\frac{1}{2},\infty \right)$, then by Condition (ii), it follows

$\begin{array}{l}\varpi \left(x\right)\phantom{\rule{0.5em}{0ex}}<\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}{\left(x-\beta \right)}^{{\lambda }_{1}}\underset{\frac{1}{2}}{\overset{\infty }{\int }}\frac{ln\left[\left(x-\beta \right)/\left(y-\beta \right)\right]}{{\left(x-\beta \right)}^{\lambda }-{\left(y-\beta \right)}^{\lambda }}{\left(y-\beta \right)}^{{\lambda }_{2}-1}\text{d}y\\ \phantom{\rule{0.5em}{0ex}}\stackrel{t=\left[\left(y-\beta \right)/\left(x-\beta {\right)\right]}^{\lambda }}{=}\frac{1}{{\lambda }^{2}}\underset{{\left[\frac{\frac{1}{2}-\beta }{x-\beta }\right]}^{\lambda }}{\overset{\infty }{\int }}\frac{lnt}{t-1}{t}^{\left({\lambda }_{2}/\lambda \right)-1}\text{d}t\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\le \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\frac{1}{{\lambda }^{2}}\underset{0}{\overset{\infty }{\int }}\frac{lnt}{t-1}{t}^{\left({\lambda }_{2}/\lambda \right)-1}\text{d}t={\left[\frac{\pi }{\lambda sin\left(\frac{\pi {\lambda }_{1}}{\lambda }\right)}\right]}^{2}.\end{array}$

Since for $\delta =\frac{{\lambda }_{1}}{2}<{\lambda }_{1}$, ${k}_{\lambda }\left(t,1\right)=\frac{lnt}{{t}^{\lambda }-1}\le \frac{M}{{t}^{\delta }}\left(t\in \left(0,1\right]\right)$, then by (13), we have the following inequality with the best constant factor ${\left[\frac{\pi }{\lambda sin\left(\frac{\pi {\lambda }_{1}}{\lambda }\right)}\right]}^{2}$:

$\begin{array}{c}\sum _{n=1}^{\infty }{a}_{n}\underset{\beta }{\overset{\infty }{\int }}\frac{ln\left[\left(x-\beta \right)∕\left(n-\beta \right)\right]}{{\left(x-\beta \right)}^{\lambda }-{\left(n-\beta \right)}^{\lambda }}f\left(x\right)\mathsf{\text{d}}x<{\left[\frac{\pi }{\lambda sin\left(\frac{\pi {\lambda }_{1}}{\lambda }\right)}\right]}^{2}\\ ×{\left\{\underset{\beta }{\overset{\infty }{\int }}{\left(x-\beta \right)}^{p\left(1-{\lambda }_{1}\right)-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\infty }{\left(n-\beta \right)}^{q\left(1-{\lambda }_{2}\right)-1}{a}_{n}^{q}\right\}}^{\frac{1}{q}}.\end{array}$
(32)
1. (iv)

For n0 = 1, b = 1 - β = γ, c = ∞, u(x) = v(x) = (x - γ), $\gamma \le 1-\frac{1}{8}\left[\lambda +\sqrt{\lambda \left(3\lambda +4\right)}\right]$, ${k}_{\lambda }\left(x,y\right)=\frac{1}{{x}^{\lambda }+{y}^{\lambda }}\left(0<\lambda \le 4\right)$, ${\lambda }_{1}={\lambda }_{2}=\frac{\lambda }{2}$, we have

$\begin{array}{ll}\hfill k\left(\frac{\lambda }{2}\right)& =\underset{0}{\overset{\infty }{\int }}{k}_{\lambda }\left(t,1\right){t}^{\frac{\lambda }{2}-1}\mathsf{\text{d}}t=\frac{2}{\lambda }\underset{0}{\overset{\infty }{\int }}\frac{1}{{t}^{\lambda }+1}\mathsf{\text{d}}{t}^{\frac{\lambda }{2}}\phantom{\rule{2em}{0ex}}\\ =\frac{2}{\lambda }arctan{t}^{\frac{\lambda }{2}}{|}_{0}^{\infty }=\frac{\pi }{\lambda }\in {R}_{+}\phantom{\rule{2em}{0ex}}\end{array}$

and

$f\left(x,y\right)=\frac{{\left(x-\gamma \right)}^{\frac{\lambda }{2}}{\left(y-\gamma \right)}^{\frac{\lambda }{2}-1}}{{\left(x-\gamma \right)}^{\lambda }+{\left(y-\gamma \right)}^{\lambda }}\left(x,y\in \left(\gamma ,\infty \right)\right).$

Hence, v(y)(y [γ, ∞)) is strictly increasing with v(1 - β) = v(γ) = 0, and for any fixed x (γ, ∞), f(x, y) is smooth with

${{f}^{\prime }}_{y}\left(x,y\right)=-\left(1+\frac{\lambda }{2}\right)\frac{{\left(x-}^{}}{}$