On a half-discrete Hilbert-type inequality similar to Mulholland’s inequality

Abstract

By using the way of weight functions and Hadamard’s inequality, a half-discrete Hilbert-type inequality similar to Mulholland’s inequality with a best constant factor is given. The extension with multi-parameters, the equivalent forms as well as the operator expressions 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 the 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 (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). By 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 with $k({\lambda }_1)={\int }_0^{\mathrm{\infty }}{k}_{\lambda }(t,1)t^{{\lambda }_1-1}dt\in R_{+}$, $\varphi (x)=x^{p(1-{\lambda }_1)-1}$, $\psi (x)=x^{q(1-{\lambda }_2)-1}$, $f(\ge 0)\in L_{p,\varphi }(R_{+})=\{f|{\parallel f\parallel }_{p,\varphi }:={\{{\int }_0^{\mathrm{\infty }}\varphi (x){|f(x)|}^{p}dx\}}^{\frac{1}{p}}<\mathrm{\infty }\}$, $g(\ge 0)\in L_{q,\psi }(R_{+})$, ${\parallel f\parallel }_{p,\varphi },{\parallel g\parallel }_{q,\psi }>0$, 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 the 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 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 about Hilbert-type inequalities are provided by [5, 8–16].

On the topic of half-discrete Hilbert-type inequalities with the general 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 the best possible. Moreover, Yang [17] gave an inequality with the particular kernel $\frac{1}{{(1+nx)}^{\lambda }}$ and an interval variable, and proved that the constant factor is the best possible. Recently, [18] and [19] gave the following half-discrete Hilbert inequality with the best constant factor π:

$${\int }_0^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{(x+n)}^{\lambda }}dx<\pi \parallel f\parallel \parallel a\parallel .$$

(6)

In this paper, by using the way of weight functions and Hadamard’s inequality, a half-discrete Hilbert-type inequality similar to (3) and (6) with the best constant factor is given as follows:

$${\int }_0^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{lne{(n+\frac{1}{2})}^x}dx<\pi \parallel f\parallel {\left\{\sum _{n=1}^{\mathrm{\infty }}(n+\frac{1}{2})a_n^2\right\}}^{\frac{1}{2}}.$$

(7)

Moreover, the best extension of (7) with multi-parameters, some equivalent forms as well as the operator expressions 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:

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

(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

$$\varpi (x)<\omega (n)=B(\frac{\lambda }{2},\frac{\lambda }{2}).$$

(10)

Proof Substitution of $t=xln(n+\alpha )$ in (8), by calculation, yields

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

Since, for fixed $x>0$ and in view of the conditions,

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

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

$$\begin{array}{rcl}\varpi (x)& <& x^{\frac{\lambda }{2}}{\int }_{\frac{1}{2}}^{\mathrm{\infty }}\frac{{ln}^{\frac{\lambda }{2}-1}(y+\alpha )}{(y+\alpha ){[1+xln(y+\alpha )]}^{\lambda }}dy\\ \stackrel{t=xln(y+\alpha )}{=}& {\int }_{xln(\frac{1}{2}+\alpha )}^{\mathrm{\infty }}\frac{t^{\frac{\lambda }{2}-1}}{{(1+t)}^{\lambda }}dt\le B(\frac{\lambda }{2},\frac{\lambda }{2}),\end{array}$$

namely, (10) follows. □

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

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

(11)

$$\begin{array}{c}{L}_1:={\left\{{\int }_0^{\mathrm{\infty }}\frac{x^{\frac{q\lambda }{2}-1}}{{[\varpi (x)]}^{q-1}}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{{ln}^{\lambda }e{(n+\alpha )}^x}\right]}^{q}dx\right\}}^{\frac{1}{q}}\hfill \\ \phantom{L_1}\le {\{B(\frac{\lambda }{2},\frac{\lambda }{2})\sum _{n=1}^{\mathrm{\infty }}{(n+\alpha )}^{q-1}{ln}^{q(1-\frac{\lambda }{2})-1}(n+\alpha )a_n^q\}}^{\frac{1}{q}}.\hfill \end{array}$$

(12)

Proof By Hölder’s inequality (cf. [20]) and (10), it follows

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

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

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

and (11) follows. Still by Hölder’s inequality, we have

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

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

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

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

3 Main results

We introduce two functions

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

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

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

$$\begin{array}{c}I:=\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{a_{n}f(x)dx}{{ln}^{\lambda }e{(n+\alpha )}^x}\hfill \\ \phantom{I}={\int }_0^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{a_{n}f(x)dx}{{ln}^{\lambda }e{(n+\alpha )}^x}<B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel f\parallel }_{p,\mathrm{\Phi }}{\parallel a\parallel }_{q,\mathrm{\Psi }},\hfill \end{array}$$

(13)

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

(14)

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

(15)

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

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

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

(16)

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

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

then $J^{p-1}={\parallel a\parallel }_{q,\mathrm{\Psi }}$. By (11), we find $J<\mathrm{\infty }$. If $J=0$, then (14) is trivially valid; if $J>0$, 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 ${[\varpi (x)]}^{1-q}>{[B(\frac{\lambda }{2},\frac{\lambda }{2})]}^{1-q}$, we have (15). By Hölder’s inequality, we find

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

(17)

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

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

then $L^{q-1}={\parallel f\parallel }_{p,\mathrm{\Phi }}$. By (12), we find $L<\mathrm{\infty }$. If $L=0$, then (15) is trivially valid; if $L>0$, 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 $0<\epsilon <\frac{p\lambda }{2}$, setting $\tilde{f}(x)=x^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}$, $x\in (0,1)$; $\tilde{f}(x)=0$, $x\in [1,\mathrm{\infty })$, 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 ($\le B(\frac{\lambda }{2},\frac{\lambda }{2})$) such that (13) is valid as we replace $B(\frac{\lambda }{2},\frac{\lambda }{2})$ with k, then, in particular, it follows

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

(18)

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

$$A(\epsilon ):=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+\alpha ){ln}^{\epsilon +1}(n+\alpha )}{\int }_{ln(n+\alpha )}^{\mathrm{\infty }}\frac{1}{{(t+1)}^{\lambda }}{t}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}dt.$$

(19)

We find

$$\begin{array}{rcl}0& <& A(\epsilon )\le \sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+\alpha ){ln}^{\epsilon +1}(n+\alpha )}{\int }_{ln(n+\alpha )}^{\mathrm{\infty }}\frac{1}{t^{\lambda }}{t}^{\frac{\lambda }{2}+\frac{\epsilon }{p}-1}dt\\ =& \frac{1}{\frac{\lambda }{2}-\frac{\epsilon }{p}}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+\alpha ){ln}^{\frac{\lambda }{2}+\frac{\epsilon }{q}+1}(n+\alpha )}<\mathrm{\infty },\end{array}$$

and then $A(\epsilon )=O(1)$ ($\epsilon \to 0^{+}$). Hence by (18) and (19), it follows

$$\begin{array}{c}\frac{1}{{ln}^{\epsilon }(1+\alpha )}B(\frac{\lambda }{2}+\frac{\epsilon }{p},\frac{\lambda }{2}-\frac{\epsilon }{p})-\epsilon O(1)\hfill \\ <k{\{\frac{\epsilon }{(1+\alpha ){ln}^{\epsilon +1}(1+\alpha )}+\frac{1}{{ln}^{\epsilon }(1+\alpha )}\}}^{\frac{1}{q}},\hfill \end{array}$$

(20)

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

By equivalence, the constant factor $B(\frac{\lambda }{2},\frac{\lambda }{2})$ 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 the first type half-discrete Hilbert-type operator $T_1:L_{p,\mathrm{\Phi }}(R_{+})\to l_{p,{\mathrm{\Psi }}^{1-p}}$ as follows: For $f\in L_{p,\mathrm{\Phi }}(R_{+})$, we define $T_{1}f\in l_{p,{\mathrm{\Psi }}^{1-p}}$, satisfying

$$T_{1}f(n)={\int }_0^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}f(x)dx,n\in \mathbf{N}.$$

Then by (14) it follows ${\parallel T_{1}f\parallel }_{p.{\mathrm{\Psi }}^{1-p}}\le B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel f\parallel }_{p,\mathrm{\Phi }}$, and then $T_1$ is a bounded operator with $\parallel T_1\parallel \le B(\frac{\lambda }{2},\frac{\lambda }{2})$. Since by Theorem 1 the constant factor in (14) is the best possible, we have $\parallel T_1\parallel =B(\frac{\lambda }{2},\frac{\lambda }{2})$.

• (ii) Define the second type half-discrete Hilbert-type operator $T_2:l_{q,\mathrm{\Psi }}\to L_{q,{\mathrm{\Phi }}^{1-q}}(R_{+})$ as follows: For $a\in l_{q,\mathrm{\Psi }}$, we define $T_{2}a\in L_{q,{\mathrm{\Phi }}^{1-q}}(R_{+})$, satisfying

  $$T_{2}a(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{ln}^{\lambda }e{(n+\alpha )}^x}{a}_n,x\in (0,\mathrm{\infty }).$$

Then by (15) it follows ${\parallel T_{2}a\parallel }_{q,{\mathrm{\Phi }}^{1-q}}\le B(\frac{\lambda }{2},\frac{\lambda }{2}){\parallel a\parallel }_{q,\mathrm{\Psi }}$, and then $T_2$ is a bounded operator with $\parallel T_2\parallel \le B(\frac{\lambda }{2},\frac{\lambda }{2})$. Since by Theorem 1 the constant factor in (15) is the best possible, we have $\parallel T_2\parallel =B(\frac{\lambda }{2},\frac{\lambda }{2})$.

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

$${\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n+\frac{1}{2}}{\left[{\int }_0^{\mathrm{\infty }}\frac{f(x)}{lne{(n+\frac{1}{2})}^x}dx\right]}^2\right\}}^{\frac{1}{2}}<\pi \parallel f\parallel ,$$

(21)

$${\left\{{\int }_0^{\mathrm{\infty }}{\left[\sum _{n=1}^{\mathrm{\infty }}\frac{a_n}{lne{(n+\frac{1}{2})}^x}\right]}^{2}dx\right\}}^{\frac{1}{2}}<\pi {\left\{\sum _{n=1}^{\mathrm{\infty }}(n+\frac{1}{2})a_n^2\right\}}^{\frac{1}{2}}.$$

(22)