On a half-discrete reverse Mulholland-type inequality and an extension

Abstract

By using the way of weight functions and the Hermite-Hadamard inequality, a half-discrete reverse Mulholland-type inequality with a best constant factor is given. The extension with multi-parameters, the equivalent forms as well as the relating homogeneous inequalities are also considered.

MSC:26D15.

1 Introduction

Assuming that $f,g\in L^2(R_{+})$, $\parallel f\parallel ={\{{\int }_0^{\mathrm{\infty }}{f}^2(x)dx\}}^{\frac{1}{2}}>0$, $\parallel g\parallel >0$, we have the following Hilbert integral inequality (cf. [1]):

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{f(x)g(y)}{x+y}dxdy<\pi \parallel f\parallel \parallel g\parallel ,$$

(1)

where the constant factor π is best possible. If $a={\{a_n\}}_{n=1}^{\mathrm{\infty }},b={\{b_n\}}_{n=1}^{\mathrm{\infty }}\in l^2$, $\parallel a\parallel ={\{{\sum }_{n=1}^{\mathrm{\infty }}{a}_n^2\}}^{\frac{1}{2}}>0$, $\parallel b\parallel >0$, then we still have the following discrete Hilbert inequality:

$$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{m+n}<\pi \parallel a\parallel \parallel b\parallel ,$$

(2)

with the same best constant factor π. Inequalities (1) and (2) are important in analysis and its applications (cf. [2–4]). Also we have the following Mulholland inequality with the same best constant factor π (cf. [1, 5]):

$$\sum _{m=2}^{\mathrm{\infty }}\sum _{n=2}^{\mathrm{\infty }}\frac{a_m{b}_n}{lnmn}<\pi {\left\{\sum _{m=2}^{\mathrm{\infty }}m{a}_m^2\sum _{n=2}^{\mathrm{\infty }}n{b}_n^2\right\}}^{\frac{1}{2}}.$$

(3)

In 1998, by introducing an independent parameter $\lambda \in (0,1]$, Yang [6] gave an extension of (1). For generalizing the results from [6], Yang [7] gave some best extensions of (1) and (2) as follows. If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, ${\lambda }_1+{\lambda }_2=\lambda$, $k_{\lambda }(x,y)$ is a non-negative homogeneous function of degree −λ, $k({\lambda }_1)={\int }_0^{\mathrm{\infty }}{k}_{\lambda }(t,1)t^{{\lambda }_1-1}dt\in R_{+}$, $\varphi (x)=x^{p(1-{\lambda }_1)-1}$, $\psi (x)=x^{q(1-{\lambda }_2)-1}$, $f(\ge 0)\in L_{p,\varphi }(R_{+})=\{f|{\parallel f\parallel }_{p,\varphi }:={\{{\int }_0^{\mathrm{\infty }}\varphi (x){|f(x)|}^{p}dx\}}^{\frac{1}{p}}<\mathrm{\infty }\}$, $g(\ge 0)\in L_{q,\psi }(R_{+})$, ${\parallel f\parallel }_{p,\varphi },{\parallel g\parallel }_{q,\psi }>0$, then

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{k}_{\lambda }(x,y)f(x)g(y)dxdy<k({\lambda }_1){\parallel f\parallel }_{p,\varphi }{\parallel g\parallel }_{q,\psi },$$

(4)

where the constant factor $k({\lambda }_1)$ is best possible. Moreover, if $k_{\lambda }(x,y)$ is finite and $k_{\lambda }(x,y)x^{{\lambda }_1-1}(k_{\lambda }(x,y)y^{{\lambda }_2-1})$ is strict decreasing for $x>0$ ($y>0$), then for $a_{m,}{b}_n\ge 0$, $a={\{a_m\}}_{m=1}^{\mathrm{\infty }}\in l_{p,\varphi }=\{a|{\parallel a\parallel }_{p,\varphi }:={\{{\sum }_{n=1}^{\mathrm{\infty }}\varphi (n){|a_n|}^p\}}^{\frac{1}{p}}<\mathrm{\infty }\}$, $b={\{b_n\}}_{n=1}^{\mathrm{\infty }}\in l_{q,\psi }$, ${\parallel a\parallel }_{p,\varphi },{\parallel b\parallel }_{q,\psi }>0$, we have

$$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{k}_{\lambda }(m,n)a_m{b}_n<k({\lambda }_1){\parallel a\parallel }_{p,\varphi }{\parallel b\parallel }_{q,\psi },$$

(5)

with the same best constant factor $k({\lambda }_1)$. Clearly, for $p=q=2$, $\lambda =1$, $k_1(x,y)=\frac{1}{x+y}$, ${\lambda }_1={\lambda }_2=\frac{1}{2}$, (4) reduces to (1), while (5) reduces to (2).

Some other results including the reverse Hilbert-type inequalities are provided by [8–16]. 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 in the inequalities are best possible. However, Yang [17] gave a result by introducing an interval variable and proved that the constant factor is best possible. Recently, Yang [18] gave a half-discrete Hilbert inequality with multi-parameters, and [19] gave the following half-discrete reverse Hilbert-type inequality with the best constant factor 4: For $0<p<1$, $\frac{1}{p}+\frac{1}{q}=1$, we have ${\theta }_1(x)\in (0,1)$, and

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}min\{x,n\}{a}_{n}dx\\ >4{\{{\int }_0^{\mathrm{\infty }}\frac{1-{\theta }_1(x)}{x^{1-\frac{3p}{2}}}{f}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{a_n^q}{n^{1-\frac{3q}{2}}}\right\}}^{\frac{1}{q}}.\end{array}$$

(6)

In this paper, by using the way of weight functions and the Hermite-Hadamard inequality, a half-discrete reverse Mulholland-type inequality similar to (6) is given as follows:

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{lne{(n+\frac{1}{2})}^x}dx\\ >\pi {\{{\int }_0^{\mathrm{\infty }}\frac{1-{\theta }_1(x)}{x^{1-\frac{p}{2}}}{f}^p(x)dx\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{{(n+\frac{1}{2})}^{q-1}{a}_n^q}{{ln}^{1-\frac{q}{2}}(n+\frac{1}{2})}\right\}}^{\frac{1}{q}}.\end{array}$$

(7)

Moreover, a best extension of (7) with multi-parameters, the equivalent forms and the relating homogeneous inequalities are considered.

2 Some lemmas

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

$$\omega (n):={ln}^{\frac{\lambda }{2}}(n+\alpha ){\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}{x}^{\frac{\lambda }{2}-1}dx,n\in \mathbf{N},$$

(8)

$$\varpi (x):=x^{\frac{\lambda }{2}}\sum _{n=1}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}(n+\alpha )}{(n+\alpha ){ln}^{\lambda }e{(n+\alpha )}^x},x\in (0,\mathrm{\infty }),$$

(9)

we have

$$B(\frac{\lambda }{2},\frac{\lambda }{2})(1-{\theta }_{\lambda }(x))<\varpi (x)<\omega (n)=B(\frac{\lambda }{2},\frac{\lambda }{2}),$$

(10)

where

$${\theta }_{\lambda }(x)=\frac{1}{B(\frac{\lambda }{2},\frac{\lambda }{2})}{\int }_0^{xln(1+\alpha )}\frac{t^{(\lambda /2)-1}}{{(1+t)}^{\lambda }}dt\in (0,1),$$

satisfying ${\theta }_{\lambda }(x)=O(x^{\frac{\lambda }{2}})$.

Proof Substituting of $t=xln(n+\alpha )$ in (8), by calculation, we have

$$\omega (n)={\int }_0^{\mathrm{\infty }}\frac{1}{{(1+t)}^{\lambda }}{t}^{\frac{\lambda }{2}-1}dt=B(\frac{\lambda }{2},\frac{\lambda }{2}).$$

Since by the conditions and for fixed $x>0$

$$h(x,y):=\frac{{ln}^{\frac{\lambda }{2}-1}(y+\alpha )}{(y+\alpha ){ln}^{\lambda }e{(y+\alpha )}^x}=\frac{{ln}^{\frac{\lambda }{2}-1}(y+\alpha )}{(y+\alpha ){[1+xln(y+\alpha )]}^{\lambda }}$$

is strictly decreasing and strictly convex in $y\in (\frac{1}{2},\mathrm{\infty })$, then by the Hermite-Hadamard inequality (cf. [20]), we find

$$\begin{array}{c}\varpi (x)<x^{\frac{\lambda }{2}}{\int }_{\frac{1}{2}}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}(y+\alpha )}{(y+\alpha ){[1+xln(y+\alpha )]}^{\lambda }}dy\stackrel{t=xln(y+\alpha )}{=}{\int }_{xln(\frac{1}{2}+\alpha )}^{\mathrm{\infty }}\frac{t^{\frac{\lambda }{2}-1}}{{(1+t)}^{\lambda }}dt\le B(\frac{\lambda }{2},\frac{\lambda }{2}),\hfill \\ \begin{array}{rl}\varpi (x)& >x^{\frac{\lambda }{2}}{\int }_1^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}(y+\alpha )}{(y+\alpha ){[1+xln(y+\alpha )]}^{\lambda }}dy\\ \stackrel{t=xln(y+\alpha )}{=}{\int }_{xln(1+\alpha )}^{\mathrm{\infty }}\frac{t^{\frac{\lambda }{2}-1}dt}{{(1+t)}^{\lambda }}=B(\frac{\lambda }{2},\frac{\lambda }{2})(1-{\theta }_{\lambda }(x))>0,\end{array}\hfill \\ \begin{array}{r}0<{\theta }_{\lambda }(x):=\frac{1}{B(\frac{\lambda }{2},\frac{\lambda }{2})}{\int }_0^{xln(1+\alpha )}\frac{t^{\frac{\lambda }{2}-1}}{{(1+t)}^{\lambda }}dt<\frac{1}{B(\frac{\lambda }{2},\frac{\lambda }{2})}{\int }_0^{xln(1+\alpha )}{t}^{\frac{\lambda }{2}-1}dt=\frac{2{[xln(1+\alpha )]}^{\frac{\lambda }{2}}}{\lambda B(\frac{\lambda }{2},\frac{\lambda }{2})},\end{array}\hfill \end{array}$$

that is, (10) is valid. □

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

$$\begin{array}{rl}J& :={\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{{ln}^{\frac{p\lambda }{2}-1}(n+\alpha )}{n+\alpha }{\left[{\int }_0^{\mathrm{\infty }}\frac{f(x)}{{ln}^{\lambda }e{(n+\alpha )}^x}dx\right]}^p\right\}}^{\frac{1}{p}}\\ \ge {\left[B(\frac{\lambda }{2},\frac{\lambda }{2})\right]}^{\frac{1}{q}}{\{{\int }_0^{\mathrm{\infty }}\varpi (x)x^{p(1-\frac{\lambda }{2})-1}{f}^p(x)dx\}}^{\frac{1}{p}},\end{array}$$

(11)

$$\begin{array}{rl}{L}_1& :={\left\{{\int }_0^{\mathrm{\infty }}\frac{x^{\frac{q\lambda }{2}-1}}{{[\varpi (x)]}^{q-1}}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{ln}^{\lambda }e{(n+\alpha )}^x}\right]}^{q}dx\right\}}^{\frac{1}{q}}\\ \ge {\{B(\frac{\lambda }{2},\frac{\lambda }{2})\sum _{n=1}^{\mathrm{\infty }}{(n+\alpha )}^{q-1}{ln}^{q(1-\frac{\lambda }{2})-1}(n+\alpha )a_n^q\}}^{\frac{1}{q}}.\end{array}$$

(12)

Proof By the reverse Hölder inequality (cf. [20]) and (10), it follows that

$$\begin{array}{r}{\left[{\int }_0^{\mathrm{\infty }}\frac{f(x)dx}{{ln}^{\lambda }e{(n+\alpha )}^x}\right]}^p\\ =\{{\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}\left[\frac{x^{(1-\frac{\lambda }{2})/q}}{{ln}^{(1-\frac{\lambda }{2})/p}(n+\alpha )}\frac{f(x)}{{(n+\alpha )}^{\frac{1}{p}}}\right]\\ {\cdot \left[\frac{{ln}^{(1-\frac{\lambda }{2})/p}(n+\alpha )}{x^{(1-\frac{\lambda }{2})/q}}{(n+\alpha )}^{\frac{1}{p}}\right]dx\}}^p\\ \ge {\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}\frac{x^{(1-\frac{\lambda }{2})(p-1)}}{(n+\alpha ){ln}^{1-\frac{\lambda }{2}}(n+\alpha )}{f}^p(x)dx\\ \cdot {\left\{{\int }_0^{\mathrm{\infty }}\frac{{(n+\alpha )}^{q-1}}{{ln}^{\lambda }e{(n+\alpha )}^x}\frac{{ln}^{(1-\frac{\lambda }{2})(q-1)}(n+\alpha )}{x^{1-\frac{\lambda }{2}}}dx\right\}}^{p-1}\\ ={\{\omega (n){(n+\alpha )}^{q-1}{ln}^{q(1-\frac{\lambda }{2})-1}(n+\alpha )\}}^{p-1}\\ \cdot {\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}\frac{x^{(1-\frac{\lambda }{2})(p-1)}}{(n+\alpha ){ln}^{1-\frac{\lambda }{2}}(n+\alpha )}{f}^p(x)dx\\ ={\left[B(\frac{\lambda }{2},\frac{\lambda }{2})\right]}^{p-1}(n+\alpha ){ln}^{1-\frac{p\lambda }{2}}(n+\alpha )\\ \cdot {\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}\frac{x^{(1-\frac{\lambda }{2})(p-1)}}{(n+\alpha ){ln}^{1-\frac{\lambda }{2}}(n+\alpha )}{f}^p(x)dx.\end{array}$$

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

$$\begin{array}{rl}J& \ge {\left[B(\frac{\lambda }{2},\frac{\lambda }{2})\right]}^{\frac{1}{q}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}\frac{x^{(1-\frac{\lambda }{2})(p-1)}{f}^p(x)dx}{(n+\alpha ){ln}^{1-\frac{\lambda }{2}}(n+\alpha )}\right\}}^{\frac{1}{p}}\\ ={\left[B(\frac{\lambda }{2},\frac{\lambda }{2})\right]}^{\frac{1}{q}}{\left\{{\int }_0^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{x^{\frac{\lambda }{2}}}{{ln}^{\lambda }e{((n+\alpha ))}^x}\frac{x^{p(1-\frac{\lambda }{2})-1}{f}^p(x)dx}{(n+\alpha ){ln}^{1-\frac{\lambda }{2}}(n+\alpha )}\right\}}^{\frac{1}{p}}\\ ={\left[B(\frac{\lambda }{2},\frac{\lambda }{2})\right]}^{\frac{1}{q}}{\{{\int }_0^{\mathrm{\infty }}\varpi (x)x^{p(1-\frac{\lambda }{2})-1}{f}^p(x)dx\}}^{\frac{1}{p}},\end{array}$$

and (11) follows. Still, by the reverse Hölder inequality, we have

$$\begin{array}{r}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{ln}^{\lambda }e{(n+\alpha )}^x}\right]}^q\\ ={\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}[\frac{x^{(1-\frac{\lambda }{2})/q}}{{ln}^{(1-\frac{\lambda }{2})/p}(n+\alpha )}\cdot \frac{1}{{(n+\alpha )}^{\frac{1}{p}}}]\left[\frac{{ln}^{(1-\frac{\lambda }{2})/p}(n+\alpha )}{x^{(1-\frac{\lambda }{2})/q}}{(n+\alpha )}^{\frac{1}{p}}{a}_n\right]\right\}}^q\\ \le {\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}\frac{x^{(1-\frac{\lambda }{2})(p-1)}}{(n+\alpha ){ln}^{1-\frac{\lambda }{2}}(n+\alpha )}\right\}}^{q-1}\\ \cdot \sum _{n=1}^{\mathrm{\infty }}\frac{{(n+\alpha )}^{q-1}}{{ln}^{\lambda }e{(n+\alpha )}^x}\frac{{ln}^{(1-\frac{\lambda }{2})(q-1)}(n+\alpha )}{x^{1-\frac{\lambda }{2}}}{a}_n^q\\ =\frac{{[\varpi (x)]}^{q-1}}{x^{\frac{q\lambda }{2}-1}}\sum _{n=1}^{\mathrm{\infty }}\frac{{(n+\alpha )}^{q-1}}{{ln}^{\lambda }e{(n+\alpha )}^x}{x}^{\frac{\lambda }{2}-1}{ln}^{(1-\frac{\lambda }{2})(q-1)}(n+\alpha )a_n^q.\end{array}$$

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

$$\begin{array}{ll}{L}_1& \ge {\{{\int }_0^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{(n+\alpha )}^{q-1}}{{ln}^{\lambda }e{(n+\alpha )}^x}{x}^{\frac{\lambda }{2}-1}{ln}^{(1-\frac{\lambda }{2})(q-1)}(n+\alpha )a_n^{q}dx\}}^{\frac{1}{q}}\\ ={\{\sum _{n=1}^{\mathrm{\infty }}[{ln}^{\frac{\lambda }{2}}(n+\alpha ){\int }_0^{\mathrm{\infty }}\frac{x^{\frac{\lambda }{2}-1}dx}{{ln}^{\lambda }e{(n+\alpha )}^x}]{(n+\alpha )}^{q-1}{ln}^{q(1-\frac{\lambda }{2})-1}(n+\alpha )a_n^q\}}^{\frac{1}{q}}\\ ={\{\sum _{n=1}^{\mathrm{\infty }}\omega (n){(n+\alpha )}^{q-1}{ln}^{q(1-\frac{\lambda }{2})-1}(n+\alpha )a_n^q\}}^{\frac{1}{q}},\end{array}$$

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

3 Main results

In this paper, for $0<p<1$ ($q<0$), we still use the normal expressions ${\parallel f\parallel }_{p,\mathrm{\Phi }}$ and ${\parallel a\parallel }_{q,\mathrm{\Psi }}$. We also introduce two functions

$$\begin{array}{c}\mathrm{\Phi }(x):=(1-{\theta }_{\lambda }(x))x^{p(1-\frac{\lambda }{2})-1}(x>0)\text{and}\hfill \\ \mathrm{\Psi }(n):={(n+\alpha )}^{q-1}{ln}^{q(1-\frac{\lambda }{2})-1}(n+\alpha )(n\in \mathbf{N}),\hfill \end{array}$$

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

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

$$\begin{array}{rl}I& :=\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_0^{\mathrm{\infty }}\frac{f(x)}{{ln}^{\lambda }e{(n+\alpha )}^x}dx\\ ={\int }_0^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{ln}^{\lambda }e{(n+\alpha )}^x}dx>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel f\parallel }_{p,\mathrm{\Phi }}{\parallel a\parallel }_{q,\mathrm{\Psi }},\end{array}$$

(13)

$$J={\left\{\sum _{n=1}^{\mathrm{\infty }}{[\mathrm{\Psi }(n)]}^{1-p}{\left[{\int }_0^{\mathrm{\infty }}\frac{f(x)}{{ln}^{\lambda }e{(n+\alpha )}^x}dx\right]}^p\right\}}^{\frac{1}{p}}>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel f\parallel }_{p,\mathrm{\Phi }},$$

(14)

$$L:={\left\{{\int }_0^{\mathrm{\infty }}{[\mathrm{\Phi }(x)]}^{1-q}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{ln}^{\lambda }e{(n+\alpha )}^x}\right]}^{q}dx\right\}}^{\frac{1}{q}}>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel a\parallel }_{q,\mathrm{\Psi }},$$

(15)

where the constant $B(\frac{\lambda }{2},\frac{\lambda }{2})$ is best possible.

Proof By the Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for $\varpi (x)>B(\frac{\lambda }{2},\frac{\lambda }{2})(1-{\theta }_{\lambda }(x))$, we have (14). By the reverse Hölder inequality, we have

$$I=\sum _{n=1}^{\mathrm{\infty }}[{\mathrm{\Psi }}^{\frac{-1}{q}}(n){\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}f(x)dx][{\mathrm{\Psi }}^{\frac{1}{q}}(n)a_n]\ge J{\parallel a\parallel }_{q,\mathrm{\Psi }}.$$

(16)

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

$$a_n:={[\mathrm{\Psi }(n)]}^{1-p}{[{\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}f(x)dx]}^{p-1},n\in \mathbf{N},$$

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

$$\begin{array}{c}{\parallel a\parallel }_{q,\mathrm{\Psi }}^q=J^p=I>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel f\parallel }_{p,\mathrm{\Phi }}{\parallel a\parallel }_{q,\mathrm{\Psi }},\mathit{\text{i.e.}},\hfill \\ {\parallel a\parallel }_{q,\mathrm{\Psi }}^{q-1}=J>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel f\parallel }_{p,\mathrm{\Phi }},\hfill \end{array}$$

that is, (14) is equivalent to (13). In view of (12), for

$${[\varpi (x)]}^{1-q}>{\left[B(\frac{\lambda }{2},\frac{\lambda }{2})(1-{\theta }_{\lambda }(x))\right]}^{1-q},$$

we have (15). By the reverse Hölder inequality, we find

$$I={\int }_0^{\mathrm{\infty }}[{\mathrm{\Phi }}^{\frac{1}{p}}(x)f(x)][{\mathrm{\Phi }}^{\frac{-1}{p}}(x)\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}{a}_n]dx\ge {\parallel f\parallel }_{p,\mathrm{\Phi }}L.$$

(17)

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

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

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

$$\begin{array}{c}{\parallel f\parallel }_{p,\mathrm{\Phi }}^p=L^q=I>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel f\parallel }_{p,\mathrm{\Phi }}{\parallel a\parallel }_{q,\mathrm{\Psi }},\mathit{\text{i.e.}},\hfill \\ {\parallel f\parallel }_{p,\mathrm{\Phi }}^{p-1}=L>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel a\parallel }_{q,\mathrm{\Psi }},\hfill \end{array}$$

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

For $0<\epsilon <\frac{p\lambda }{2}$, set $\tilde{f}(x)=x^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}$, $x\in (0,1)$; $\tilde{f}(x)=0$, $x\in [1,\mathrm{\infty })$, and

$${\tilde{a}}_n=\frac{1}{n+\alpha }{ln}^{\frac{\lambda }{2}-\frac{\epsilon }{q}-1}(n+\alpha ),n\in \mathbf{N}.$$

If there exists a positive number k ($\ge B(\frac{\lambda }{2},\frac{\lambda }{2})$) such that (13) is valid when replacing $B(\frac{\lambda }{2},\frac{\lambda }{2})$ with k, then, in particular, it follows that

$$\begin{array}{rl}\tilde{I}:=& \sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}{\tilde{a}}_n\tilde{f}(x)dx>k{\parallel \tilde{f}\parallel }_{p,\mathrm{\Phi }}{\parallel \tilde{a}\parallel }_{q,\mathrm{\Psi }}\\ =& k{\left\{{\int }_0^1(1-O\left(x^{\frac{\lambda }{2}}\right))\frac{dx}{x^{-\epsilon +1}}\right\}}^{\frac{1}{p}}{\{\frac{1}{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n+\alpha ){ln}^{\epsilon +1}(n+\alpha )}\}}^{\frac{1}{q}}\\ >& k{\{\frac{1}{\epsilon }-O(1)\}}^{\frac{1}{p}}{\{\frac{1}{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+{\int }_1^{\mathrm{\infty }}\frac{dx}{(x+\alpha ){ln}^{\epsilon +1}(x+\alpha )}\}}^{\frac{1}{q}}\\ =& \frac{k}{\epsilon }{\{1-\epsilon O(1)\}}^{\frac{1}{p}}{\{\frac{\epsilon }{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+\frac{1}{{ln}^{\epsilon }(1+\alpha )}\}}^{\frac{1}{q}},\end{array}$$

(18)

$$\begin{array}{rl}\tilde{I}& =\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n+\alpha }{ln}^{\frac{\lambda }{2}-\frac{\epsilon }{q}-1}(n+\alpha ){\int }_0^1\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}{x}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}dx\\ \stackrel{t=xln(n+\alpha )}{=}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+\alpha ){ln}^{\epsilon +1}(n+\alpha )}{\int }_0^{ln(n+\alpha )}\frac{1}{{(t+1)}^{\lambda }}{t}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}dt\\ \le B(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p})[\frac{1}{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n+\alpha ){ln}^{\epsilon +1}(n+\alpha )}]\\ <B(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p})[\frac{1}{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+{\int }_1^{\mathrm{\infty }}\frac{1}{(y+\alpha ){ln}^{\epsilon +1}(y+\alpha )}dy]\\ =\frac{1}{\epsilon }B(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p})[\frac{\epsilon }{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+\frac{1}{{ln}^{\epsilon }(1+\alpha )}].\end{array}$$

(19)

Hence by (18) and (19) it follows that

$$\begin{array}{r}B(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p})[\frac{\epsilon }{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+\frac{1}{{ln}^{\epsilon }(1+\alpha )}]\\ >k{\{1-\epsilon O(1)\}}^{\frac{1}{p}}{\{\frac{\epsilon }{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+\frac{1}{{ln}^{\epsilon }(1+\alpha )}\}}^{\frac{1}{q}},\end{array}$$

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

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

Remark 1 (i) For $\lambda =1$, ${\lambda }_1={\lambda }_2=\frac{1}{2}$, $\alpha =\frac{1}{2}$ in (13), (14) and (15), we have (7) and the following equivalent inequalities:

$$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}\frac{{ln}^{\frac{p}{2}-1}(n+\frac{1}{2})}{n+\frac{1}{2}}{\left[{\int }_0^{\mathrm{\infty }}\frac{f(x)}{lne{(n+\frac{1}{2})}^x}dx\right]}^p\\ >{\pi }^p{\int }_0^{\mathrm{\infty }}\frac{1-{\theta }_1(x)}{x^{1-\frac{p}{2}}}{f}^p(x)dx,\end{array}$$

(20)

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{{(1-{\theta }_1(x))}^{1-q}}{x^{1-\frac{q}{2}}}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{lne{(n+\frac{1}{2})}^x}\right]}^{q}dx\\ <{\pi }^q\sum _{n=1}^{\mathrm{\infty }}\frac{{(n+\frac{1}{2})}^{q-1}}{{ln}^{1-\frac{q}{2}}(n+\frac{1}{2})}{a}_n^q.\end{array}$$

(21)

(ii) Setting $x=\frac{1}{lny}$, $g(y):=\frac{1}{y}{(lny)}^{\lambda -2}f(\frac{1}{lny})$ and

$$\varphi (y):=(1-{\theta }_{\lambda }\left(\frac{1}{lny}\right))y^{p-1}{(lny)}^{p(1-\frac{\lambda }{2})-1}(y\in (1,\mathrm{\infty }))$$

in (13), by simplifications, we find the following inequality with the homogeneous kernel:

$$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}{a}_n{\int }_1^{\mathrm{\infty }}\frac{g(y)}{{ln}^{\lambda }y(n+\alpha )}dy\\ ={\int }_1^{\mathrm{\infty }}g(y)\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{ln}^{\lambda }y(n+\alpha )}dx>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel g\parallel }_{p,\varphi }{\parallel a\parallel }_{q,\mathrm{\Psi }}.\end{array}$$

(22)

It is evident that (22) is equivalent to (13), and then the constant factor $B(\frac{\lambda }{2},\frac{\lambda }{2})$ in (22) is still best possible. In the same way as in (14) and (15), we have the following inequalities equivalent to (13) with the best constant factor $B(\frac{\lambda }{2},\frac{\lambda }{2})$:

$${\left\{\sum _{n=1}^{\mathrm{\infty }}{[\mathrm{\Psi }(n)]}^{1-p}{\left[{\int }_1^{\mathrm{\infty }}\frac{g(y)}{{ln}^{\lambda }y(n+\alpha )}dy\right]}^p\right\}}^{\frac{1}{p}}>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel g\parallel }_{p,\varphi },$$

(23)

$${\left\{{\int }_1^{\mathrm{\infty }}{[\varphi (y)]}^{1-q}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{ln}^{\lambda }y(n+\alpha )}\right]}^{q}dy\right\}}^{\frac{1}{q}}>B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel a\parallel }_{q,\mathrm{\Psi }}.$$

(24)