A new half-discrete Mulholland-type inequality with multi-parameters

Abstract

By means of weight functions and Hermite-Hadamard’s inequality, a new half-discrete Mulholland-type inequality with a best constant factor is given. A best extension with multi-parameters, some equivalent forms, the operator expressions as well as some particular cases are considered.

1 Introduction

Assuming that $f,g\in L^{2}(\mathbf{R}_{+})$, $\|f\| =\{\int_{0}^{\infty }f^{2}(x)\,dx\}^{\frac{1}{2}} >0$, $\|g\|>0$, we have the following Hilbert integral inequality (cf. [1]):

$$ \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \pi\|f\|\|g\|, $$

(1.1)

where the constant factor π is best possible. If $a=\{a_{m}\} _{m=1}^{\infty},b=\{b_{n}\}_{n=1}^{\infty}\in l^{2}$, $\|a\|=\{\sum_{m=1}^{\infty }a_{m}^{2}\}^{\frac{1}{2}}>0$, $\|b\|>0$, then we have the following discrete Hilbert inequality:

$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \pi \|a\|\|b\|, $$

(1.2)

with the same best constant factor π. Inequalities (1.1) and (1.2) are important in analysis and its applications (cf. [2, 3]). On the other hand, we have the following Mulholland inequality with the same best constant factor π (cf. [1, 4]):

$$ \sum_{m=2}^{\infty}\sum _{n=2}^{\infty}\frac{a_{m}b_{n}}{\ln mn}< \pi \Biggl\{ \sum _{m=2}^{\infty}ma_{m}^{2}\sum _{n=2}^{\infty }nb_{n}^{2} \Biggr\} ^{\frac{1}{2}}. $$

(1.3)

In 1998, by introducing an independent parameter $\lambda\in(0,1]$, Yang [5] gave an extension of (1.1). Generalizing the results from [5], Yang [3] gave some extensions of (1.1) and (1.2) as follows: If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $\lambda_{1}+\lambda _{2}=\lambda \in\mathbf{R}$, $k_{\lambda}(x,y)$ is a non-negative homogeneous function of degree −λ satisfying

$$k(\lambda_{1})=\int_{0}^{\infty}k_{\lambda }(t,1)t^{\lambda_{1}-1} \,dt\in\mathbf{R}_{+}, $$

$\phi(x)=x^{p(1-\lambda _{1})-1}$, $\psi(x)=x^{q(1-\lambda_{2})-1}$, $f(x),g(y)\geq0$,

$$ f\in L_{p,\phi}(\mathbf{R}_{+})= \biggl\{ f\Bigm|\|f \|_{p,\phi}:= \biggl\{ \int_{0}^{\infty} \phi(x) \bigl|f(x) \bigr|^{p}\,dx \biggr\} ^{\frac{1}{p}}< \infty \biggr\} , $$

$g\in L_{q,\psi}(\mathbf{R}_{+})$, and $\|f\|_{p,\phi}, \|g\|_{q,\psi }>0$, then

$$ \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda }(x,y)f(x)g(y) \,dx\,dy< k(\lambda _{1})\|f\|_{p,\phi}\|g\|_{q,\psi}, $$

(1.4)

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 decreasing for $x>0$ ($y>0$), then for $a_{m},b_{n}\geq0$,

$$ a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\phi}= \Biggl\{ a\Bigm|\|a\|_{p,\phi }:= \Biggl\{ \sum_{m=1}^{\infty} \phi(m)|a_{m}|^{p} \Biggr\} ^{\frac{1}{p}}< \infty \Biggr\} , $$

and $b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\psi}$, $\|a\|_{p,\phi },\|b\|_{q,\psi}>0$, we have

$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}k_{\lambda }(m,n)a_{m}b_{n}< k( \lambda_{1})\|a\|_{p,\phi}\|b\|_{q,\psi}, $$

(1.5)

where the constant factor $k(\lambda_{1})$ is still the best possible. Clearly, for $p=q=2$, $\lambda=1$, $k_{1}(x,y)=\frac{1}{x+y}$ and $\lambda _{1}=\lambda_{2}=\frac{1}{2}$, (1.4) reduces to (1.1), while (1.5) reduces to (1.2).

Some other results about Hilbert-type inequalities can be found in [6–13]. On 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 are best possible. In 2005, Yang [14] gave a result with the kernel $\frac {1}{(1+nx)^{\lambda}}$ by introducing a variable and proved that the constant factor is best possible. Recently, Wang and Yang [15] gave a more accurate reverse half-discrete Hilbert-type inequality, and Yang [16] provided the following half-discrete Hilbert inequality with best constant factor:

$$ \int_{0}^{\infty}f(x)\sum _{n=1}^{\infty}\frac{a_{n}}{x+n}\,dx< \pi\|f\|\|a\|. $$

(1.6)

In this paper, by means of weight functions and Hermite-Hadamard’s inequality, a new half-discrete Mulholland-type inequality similar to (1.3) and (1.6) with a best possible constant factor is given as follows:

$$ \int_{1}^{\infty}f(x)\sum _{n=2}^{\infty}\frac{a_{n}}{1+\ln x\ln n}\,dx< \pi \Biggl\{ \int _{1}^{\infty}xf^{2}(x)\,dx\sum _{n=2}^{\infty }na_{n}^{2} \Biggr\} ^{\frac{1}{2}}. $$

(1.7)

Moreover, a best extension of (1.7) with multi-parameters, some equivalent forms, the operator expressions as well as some particular cases are considered.

2 Some lemmas

Lemma 2.1

If $0<\sigma<\lambda$ ($\sigma\leq1$), $\alpha>0$, $\beta\geq \frac{2}{3}$, $\delta\in\{-1,1\}$, the weight functions $\omega(n)$ and $\varpi(x)$ are defined by

$$\begin{aligned}& \omega(n) :=(\ln\beta n)^{\sigma}\int_{\frac{1}{\alpha}}^{\infty} \frac{(\ln\alpha x)^{\delta\sigma-1}}{x(1+\ln^{\delta}\alpha x\ln \beta n)^{\lambda}}\,dx,\quad n\in\mathbf{N}\backslash\{1 \}, \end{aligned}$$

(2.1)

$$\begin{aligned}& \varpi(x) :=(\ln\alpha x)^{\delta\sigma}\sum_{n=2}^{\infty} \frac {(\ln \beta n)^{\sigma-1}}{n(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} ,\quad x\in \biggl(\frac{1}{\alpha},\infty \biggr), \end{aligned}$$

(2.2)

then we have

$$ \varpi(x)< \omega(n)=B(\sigma,\lambda-\sigma). $$

(2.3)

Proof

Substituting $t=\ln^{\delta}\alpha x\ln\beta n$ in (2.1), and by a simple calculation, for $\delta\in\{-1,1\}$, we have

$$ \omega(n)=\int_{0}^{\infty}\frac{1}{(1+t)^{\lambda}}t^{\sigma -1} \,dt=B(\sigma,\lambda-\sigma). $$

For fixed $x>\frac{1}{\alpha}$, in view of the conditions, it is easy to find that

$$ h(x,y):=\frac{(\ln\beta y)^{\sigma-1}}{y(1+\ln^{\delta}\alpha x\ln \beta y)^{\lambda}}=\frac{1}{y(1+\ln^{\delta}\alpha x\ln\beta y)^{\lambda }(\ln\beta y)^{1-\sigma}} $$

is decreasing and strictly convex with $h_{y}^{\prime}(x,y)<0$ and $h_{y^{2}}^{\prime\prime}(x,y)>0$, for $y\in(\frac{3}{2},\infty)$. Hence by the Hermite-Hadamard inequality (cf. [17]), we find

$$\begin{aligned} &\varpi(x)< (\ln\alpha x)^{\delta\sigma}\int_{\frac{3}{2}}^{\infty } \frac{1}{y(1+\ln^{\delta}\alpha x\ln\beta y)^{\lambda}(\ln\beta y)^{1-\sigma}}\,dy \\ &\hphantom{a}\overset{t=\ln^{\delta}\alpha x\ln\beta y}{=}\int _{\ln^{\delta }\alpha x\ln(\frac{3}{2}\beta)}^{\infty} \frac{t^{\sigma-1}}{(1+t)^{\lambda }}\,dt\leq B(\sigma, \lambda-\sigma), \end{aligned}$$

and then (2.3) follows. □

Lemma 2.2

Let the assumptions of Lemma  2.1 be fulfilled and, additionally, let $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $a_{n}\geq0$, $n\in \mathbf{N}\backslash\{1\}$, $f(x)$ is a non-negative measurable function in $(\frac{1}{\alpha},\infty)$. Then we have the following inequalities:

$$\begin{aligned}& \begin{aligned}[b] J &:= \Biggl\{ \sum_{n=2}^{\infty} \frac{1}{n}(\ln\beta n)^{p\sigma -1} \biggl[ \int_{\frac{1}{\alpha}}^{\infty} \frac{f(x)}{(1+\ln^{\delta}\alpha x\ln \beta n)^{\lambda}}\,dx \biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\ &\leq \bigl[ B(\sigma,\lambda-\sigma) \bigr] ^{\frac{1}{q}} \biggl\{ \int _{\frac{1}{\alpha}}^{\infty}\varpi(x)x^{p-1}(\ln\alpha x)^{p(1-\delta \sigma)-1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}}, \end{aligned} \end{aligned}$$

(2.4)

$$\begin{aligned}& \begin{aligned}[b] L_{1} &:= \Biggl\{ \int_{\frac{1}{\alpha}}^{\infty} \frac{(\ln\alpha x)^{q\delta\alpha-1}}{x[\varpi(x)]^{q-1}} \Biggl[ \sum_{n=2}^{\infty} \frac{a_{n}}{(1+\ln ^{\delta}\alpha x\ln\beta n)^{\lambda}} \Biggr] ^{q}\,dx \Biggr\} ^{\frac {1}{q}} \\ &\leq \Biggl\{ B(\sigma,\lambda-\sigma)\sum_{n=2}^{\infty }n^{q-1}( \ln \beta n)^{q(1-\sigma)-1}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned} \end{aligned}$$

(2.5)

Proof

By Hölder’s inequality (cf. [17]) and (2.3), it follows that

$$\begin{aligned} & \biggl[ \int_{\frac{1}{\alpha}}^{\infty}\frac{f(x)\,dx}{(1+\ln^{\delta }\alpha x\ln\beta n)^{\lambda}} \biggr]^{p} \\ &\quad= \biggl\{ \int_{\frac{1}{\alpha}}^{\infty}\frac{1}{(1+\ln^{\delta }\alpha x\ln\beta n)^{\lambda}} \biggl[ \frac{(\ln\alpha x)^{(1-\delta \sigma)/q}}{(\ln\beta n)^{(1-\sigma)/p}}\frac{x^{\frac {1}{q}}f(x)}{n^{\frac{1}{p}}} \biggr] \\ &\qquad{}\times \biggl[ \frac{(\ln\beta n)^{(1-\sigma)/p}}{(\ln\alpha x)^{(1-\delta\sigma)/q}}\frac{n^{\frac{1}{p}}}{x^{\frac{1}{q}}} \biggr]\,dx \biggr\} ^{p}\leq\int_{\frac{1}{\alpha}}^{\infty} \frac{x^{p-1}(\ln \alpha x)^{(1-\delta\sigma)(p-1)}}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} \frac{f^{p}(x)\,dx}{n(\ln\beta n)^{1-\sigma}} \\ &\qquad{}\times \biggl\{ \int_{\frac{1}{\alpha}}^{\infty} \frac{n^{q-1}}{(1+\ln ^{\delta}\alpha x\ln\beta n)^{\lambda}} \frac{(\ln\beta n)^{(1-\sigma )(q-1)}}{x(\ln\alpha x)^{1-\delta\sigma}}\,dx \biggr\} ^{p-1} \\ &\quad= \bigl\{ \omega(n)n^{q-1}(\ln\beta n)^{q(1-\sigma)-1} \bigr\} ^{p-1}\int_{\frac{1}{\alpha}}^{\infty}\frac{x^{p-1}(\ln\alpha x)^{(1-\delta\sigma)(p-1)}}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda }} \frac{f^{p}(x)\,dx}{n(\ln\beta n)^{1-\sigma}} \\ &\quad= \bigl[ B(\sigma,\lambda-\sigma) \bigr] ^{p-1}n(\ln\beta n)^{1-p\sigma }\int_{\frac{1}{\alpha}}^{\infty}\frac{x^{p-1}(\ln\alpha x)^{(1-\delta \sigma)(p-1)}}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} \frac{f^{p}(x)\,dx}{n(\ln\beta n)^{1-\sigma}}. \end{aligned}$$

Then by Lebesgue term-by-term integration theorem (cf. [18]), we have

$$\begin{aligned} J \leq& \bigl[ B(\sigma,\lambda-\sigma) \bigr] ^{\frac{1}{q}} \Biggl\{ \sum _{n=2}^{\infty}\int_{\frac{1}{\alpha}}^{\infty} \frac{x^{p-1}(\ln \alpha x)^{(1-\delta\sigma)(p-1)}}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}}\frac{f^{p}(x)\,dx}{n(\ln\beta n)^{1-\sigma}} \Biggr\} ^{\frac{1}{p}} \\ =& \bigl[ B(\sigma,\lambda-\sigma) \bigr] ^{\frac{1}{q}} \Biggl\{ \int _{\frac{1}{\alpha}}^{\infty}\sum_{n=2}^{\infty} \frac{x^{p-1}(\ln\alpha x)^{(1-\delta\sigma)(p-1)}}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda }}\frac{f^{p}(x)\,dx}{n(\ln\beta n)^{1-\sigma}} \Biggr\} ^{\frac{1}{p}} \\ =& \bigl[ B(\sigma,\lambda-\sigma) \bigr] ^{\frac{1}{q}} \biggl\{ \int _{\frac{1}{\alpha}}^{\infty}\varpi(x)x^{p-1}(\ln\alpha x)^{p(1-\delta \sigma)-1}f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}}, \end{aligned}$$

hence, (2.4) follows.

By Hölder’s inequality again, we have

$$\begin{aligned} & \Biggl[ \sum_{n=2}^{\infty}\frac{a_{n}}{(1+\ln^{\delta}\alpha x\ln \beta n)^{\lambda}} \Biggr] ^{q} \\ &\quad= \Biggl\{ \sum_{n=2}^{\infty} \frac{1}{(1+\ln^{\delta}\alpha x\ln \beta n)^{\lambda}} \biggl[ \frac{(\ln\alpha x)^{(1-\delta\sigma)/q}}{(\ln \beta n)^{(1-\sigma)/p}}\frac{x^{\frac{1}{q}}}{n^{\frac{1}{p}}} \biggr] \\ &\qquad{}\times \biggl[ \frac{(\ln\beta n)^{(1-\sigma)/p}}{(\ln\alpha x)^{(1-\delta\sigma)/q}}\frac{n^{\frac{1}{p}}a_{n}}{x^{\frac {1}{q}}} \biggr] \Biggr\} ^{q}\leq \Biggl\{ \sum_{n=2}^{\infty} \frac{x^{p-1}(\ln\alpha x)^{(1-\delta\sigma)(p-1)}}{n(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}(\ln\beta n)^{1-\sigma}} \Biggr\} ^{q-1} \\ &\qquad{}\times\sum_{n=2}^{\infty} \frac{n^{q-1}}{(1+\ln^{\delta}\alpha x\ln \beta n)^{\lambda}} \frac{(\ln\beta n)^{(1-\sigma)(q-1)}}{x(\ln\alpha x)^{1-\delta\sigma}}a_{n}^{q} \\ &\quad=\frac{x[\varpi(x)]^{q-1}}{(\ln\alpha x)^{q\delta\sigma-1}}\sum_{n=2}^{\infty} \frac{n^{q-1}}{x(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}}(\ln\alpha x)^{\delta\sigma-1}(\ln\beta n)^{(1-\sigma )(q-1)}a_{n}^{q}. \end{aligned}$$

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

$$\begin{aligned} L_{1}&\leq \Biggl\{ \int_{\frac{1}{\alpha}}^{\infty}\sum _{n=2}^{\infty} \frac{n^{q-1}}{x(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}}(\ln \alpha x)^{\delta\sigma-1}(\ln\beta n)^{(1-\sigma)(q-1)}a_{n}^{q} \,dx \Biggr\} ^{\frac{1}{q}} \\ &= \Biggl\{ \sum_{n=2}^{\infty} \biggl[ (\ln \beta n)^{\sigma}\int_{\frac {1}{\alpha}}^{\infty} \frac{(\ln\alpha x)^{\delta\sigma-1}\,dx}{x(1+\ln ^{\delta}\alpha x\ln\beta n)^{\lambda}} \biggr] n^{q-1}(\ln\beta n)^{q(1-\sigma)-1}a_{n}^{q} \Biggr\} ^{\frac{1}{q}} \\ &= \Biggl\{ \sum_{n=2}^{\infty} \omega(n)n^{q-1}(\ln\beta n)^{q(1-\sigma )-1}a_{n}^{q} \Biggr\} ^{\frac{1}{q}}, \end{aligned}$$

and in view of (2.3), inequality (2.5) follows. □

3 Main results

We introduce the functions

$$\begin{aligned}& \Phi_{\delta}(x) :=x^{p-1}(\ln\alpha x)^{p(1-\delta\sigma )-1}\quad \biggl(x>\frac{1}{\alpha} \biggr), \\& \Psi(n) :=n^{q-1}(\ln\beta n)^{q(1-\sigma)-1}\quad \bigl(n\in\mathbf {N} \backslash \{1\} \bigr), \end{aligned}$$

wherefrom $[\Phi_{\delta}(x)]^{1-q}=\frac{1}{x}(\ln\alpha x)^{q\delta \sigma-1}$, and $[\Psi(n)]^{1-p}=\frac{1}{n}(\ln\beta n)^{p\sigma-1}$.

Theorem 3.1

If $0<\sigma<\lambda$ ($\sigma\leq1$), $\alpha>0$, $\beta \geq\frac{2}{3}$, $\delta\in\{-1,1\}$, $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f(x)$, $a_{n}\geq0$, $f\in L_{p,\Phi}(\frac{1}{\alpha},\infty)$, $a=\{a_{n}\}_{n=2}^{\infty}\in l_{q,\Psi}$, $\|f\|_{p,\Phi_{\delta}}>0$, and $\|a\|_{q,\Psi}>0$, then we have the following equivalent inequalities:

$$\begin{aligned}& \begin{aligned}[b] I &:=\sum_{n=2}^{\infty}a_{n} \int_{\frac{1}{\alpha}}^{\infty}\frac {f(x)\,dx}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} \\ &=\int_{\frac{1}{\alpha}}^{\infty}f(x)\sum _{n=2}^{\infty}\frac {a_{n}\,dx}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}}< B(\sigma,\lambda -\sigma )\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, \end{aligned} \end{aligned}$$

(3.1)

$$\begin{aligned}& J= \Biggl\{ \sum_{n=2}^{\infty} \bigl[\Psi(n) \bigr]^{1-p} \biggl[ \int_{\frac {1}{\alpha}}^{\infty} \frac{f(x)\,dx}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda }} \biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< B( \sigma,\lambda-\sigma )\|f\|_{p,\Phi _{\delta}}, \end{aligned}$$

(3.2)

$$\begin{aligned}& L:= \Biggl\{ \int_{\frac{1}{\alpha}}^{\infty} \bigl[ \Phi_{\delta }(x) \bigr]^{1-q} \Biggl[ \sum _{n=2}^{\infty}\frac{a_{n}}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} \Biggr] ^{q} \,dx \Biggr\} ^{\frac{1}{q}}< B(\sigma,\lambda -\sigma)\|a\|_{q,\Psi}, \end{aligned}$$

(3.3)

where the constant $B(\sigma,\lambda-\sigma)$ is the best possible in the above inequalities.

Proof

The two expressions for I in (3.1) follow from Lebesgue’s term-by-term integration theorem. By (2.4) and (2.3), we have (3.2). By Hölder’s inequality, we have

$$ I=\sum_{n=2}^{\infty} \biggl[ \Psi^{\frac{-1}{q}}(n)\int_{\frac {1}{\alpha}}^{\infty} \frac{f(x)\,dx}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda }} \biggr] \bigl[\Psi^{\frac{1}{q}}(n)a_{n} \bigr]\leq J\|a\|_{q,\Psi}. $$

(3.4)

Then by (3.2), we have (3.1).

On the other hand, assuming that (3.1) is valid, we set

$$ a_{n}:= \bigl[\Psi(n) \bigr]^{1-p} \biggl[ \int _{\frac{1}{\alpha}}^{\infty}\frac {f(x)\,dx}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} \biggr] ^{p-1},\quad n\in \mathbf{N}\backslash\{1\}. $$

It follows that $J^{p-1}=\|a\|_{q,\Psi}$. By (2.4), we find $J<\infty$. If $J=0$, then (3.2) is trivially valid; if $J>0$, then by (3.1), we have

$$ \|a\|_{q,\Psi}^{q}=J^{q(p-1)}=J^{p}=I< B( \sigma,\lambda-\sigma )\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, $$

namely, $\|a\|_{q,\Psi}^{q-1}=J< B(\sigma,\lambda-\sigma )\|f\|_{p,\Phi _{\delta}}$. That is, (3.2) is equivalent to (3.1).

By (2.3) we have $[\varpi(x) ]^{1-q}>[B(\sigma,\lambda-\sigma )]^{1-q}$. Then in view of (2.5), we have (3.3). By Hölder’s inequality, we find

$$ I=\int_{\frac{1}{\alpha}}^{\infty} \bigl[\Phi_{\delta}^{\frac {1}{p}}(x)f(x) \bigr] \Biggl[ \Phi_{\delta}^{\frac{-1}{p}}(x)\sum _{n=2}^{\infty}\frac {a_{n}}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} \Biggr]\,dx\leq \|f \|_{p,\Phi_{\delta}}L. $$

(3.5)

Then by (3.3), we have (3.1).

On the other hand, assume that (3.1) is valid. Setting

$$ f(x):= \bigl[\Phi_{\delta}(x) \bigr]^{1-q} \Biggl[ \sum _{n=2}^{\infty}\frac{a_{n}}{ (1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} \Biggr] ^{q-1}, \quad x \in \biggl(\frac{1}{\alpha},\infty \biggr), $$

then $L^{q-1}=\|f\|_{p,\Phi_{\delta}}$. By (2.5), we find $L<\infty$. If $L=0$, then (3.3) is trivially valid; if $L>0$, then by (3.1), we have

$$ \|f\|_{p,\Phi_{\delta}}^{p}=L^{p(q-1)}=L^{q}=I< B( \sigma,\lambda -\sigma )\|f\|_{p,\Phi_{\delta}}\|a\|_{q,\Psi}, $$

therefore $\|f\|_{p,\Phi_{\delta}}^{p-1}=L< B(\sigma,\lambda-\sigma )\|a\|_{q,\Psi}$, that is, (3.3) is equivalent to (3.1). Hence, inequalities (3.1), (3.2), and (3.3) are equivalent.

For $0<\varepsilon<p(\lambda-\sigma)$, setting $E_{\delta}:=\{ x;x>\frac{1}{\alpha},\ln^{\delta}\alpha x\in(0,1)\}$,

$$ \widetilde{f}(x)=\frac{1}{x}(\ln\alpha x)^{\delta(\sigma+\frac{\varepsilon}{p})-1},\quad x\in E_{\delta};\qquad \widetilde{f}(x)=0,\quad \biggl\{ x;x>\frac{1}{ \alpha} \biggr\} \backslash E_{\delta}, $$

and $\widetilde{a}_{n}=\frac{1}{n}(\ln\beta n)^{\sigma-\frac {\varepsilon}{q}-1}$, $n\in\mathbf{N}\backslash\{1\}$, if there exists a positive number k ($\leq B(\sigma,\lambda-\sigma)$), such that (3.1) is valid when replacing $B(\sigma,\lambda-\sigma)$ with k, then in particular, for $\delta=\pm1$, setting $u=\ln^{\delta}\alpha x$, it follows that

$$\begin{aligned} &\int_{E_{\delta}}\frac{dx}{x(\ln\alpha x)^{-\delta\varepsilon+1}}=\int _{0}^{1}\frac{|\delta|u^{\delta-1}}{u^{-\varepsilon+\delta}}\,du= \frac{1}{\varepsilon}, \\ &\widetilde{I} :=\sum_{n=2}^{\infty} \int_{\frac{1}{\alpha}}^{\infty} \frac{1}{(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}} \widetilde {a}_{n}\widetilde{f}(x)\,dx< k\|\widetilde{f} \|_{p,\Phi_{\delta}}\|\widetilde{a} \|_{q,\Psi} \\ &\hphantom{\widetilde{I}}=k \biggl\{ \int_{E_{\delta}}\frac{dx}{x(\ln\alpha x)^{-\delta \varepsilon +1}} \biggr\} ^{\frac{1}{p}} \Biggl\{ \frac{1}{2(\ln2\beta)^{\varepsilon +1}}+\sum _{n=3}^{\infty}\frac{1}{n(\ln\beta n)^{\varepsilon+1}} \Biggr\} ^{\frac{1}{q}} \\ &\hphantom{\widetilde{I}}< k \biggl(\frac{1}{\varepsilon} \biggr)^{\frac{1}{p}} \biggl\{ \frac{1}{2(\ln2\beta )^{\varepsilon+1}}+ \int_{2}^{\infty}\frac{1}{x(\ln\beta x)^{\varepsilon +1}} \,dx \biggr\} ^{\frac{1}{q}} \\ &\hphantom{\widetilde{I}}=\frac{k}{\varepsilon} \biggl\{ \frac{\varepsilon}{2(\ln2\beta )^{\varepsilon+1}}+\frac{1}{(\ln2\beta)^{\varepsilon}} \biggr\} ^{\frac{1}{q}}, \end{aligned}$$

(3.6)

$$\begin{aligned} &\widetilde{I} =\sum_{n=2}^{\infty} \frac{1}{n}(\ln\beta n)^{\sigma -\frac{\varepsilon}{q}-1}\int_{E_{\delta}} \frac{(\ln\alpha x)^{\delta (\sigma+\frac{\varepsilon}{p})-1}}{x(1+\ln^{\delta}\alpha x\ln\beta n)^{\lambda}}\,dx \\ &\overset{t=\ln^{\delta}\alpha x\ln\beta n}{=}\sum _{n=2}^{\infty }\frac{1}{n(\ln\beta n)^{\varepsilon+1}}\int_{0}^{\ln\beta n} \frac{1}{(1+t)^{\lambda}}t^{\sigma+\frac{\varepsilon}{p}-1}\,dt \\ &\hphantom{\widetilde{I}}=B \biggl( \sigma+\frac{\varepsilon}{p},\lambda-\sigma-\frac {\varepsilon }{p} \biggr) \sum_{n=2}^{\infty}\frac{1}{n(\ln\beta n)^{\varepsilon +1}}-A( \varepsilon) \\ &\hphantom{\widetilde{I}}>B \biggl( \sigma+\frac{\varepsilon}{p},\lambda-\sigma-\frac {\varepsilon }{p} \biggr) \int_{2}^{\infty}\frac{1}{y(\ln\beta y)^{\varepsilon+1}}\,dy-A(\varepsilon) \\ &\hphantom{\widetilde{I}}=\frac{1}{\varepsilon(\ln2\beta)^{\varepsilon}}B \biggl( \sigma +\frac{\varepsilon}{p},\lambda-\sigma- \frac{\varepsilon}{p} \biggr) -A(\varepsilon), \\ &A(\varepsilon) :=\sum_{n=2}^{\infty} \frac{1}{n(\ln\beta n)^{\varepsilon +1}}\int_{\ln\beta n}^{\infty}\frac{1}{(t+1)^{\lambda}}t^{\sigma +\frac{\varepsilon}{p}-1} \,dt. \end{aligned}$$

(3.7)

We find

$$\begin{aligned} 0 < &A(\varepsilon)\leq\sum_{n=2}^{\infty} \frac{1}{n(\ln\beta n)^{\varepsilon+1}}\int_{\ln\beta n}^{\infty}\frac{1}{t^{\lambda}}t^{\sigma+\frac{\varepsilon}{p}-1} \,dt \\ =&\frac{1}{\frac{\lambda}{2}-\frac{\varepsilon}{p}}\sum_{n=2}^{\infty}\frac{1}{n(\ln\beta n)^{\sigma+\frac{\varepsilon}{q}+1}}< \infty, \end{aligned}$$

and so $A(\varepsilon)=O(1)(\varepsilon\rightarrow0^{+})$. Hence by (3.6) and (3.7), it follows that

$$ \frac{1}{(\ln2\beta)^{\varepsilon}}B \biggl( \sigma+\frac{\varepsilon }{p},\lambda-\sigma- \frac{\varepsilon}{p} \biggr) -\varepsilon O(1)< k \biggl\{ \frac{\varepsilon}{2(\ln2\beta)^{\varepsilon+1}}+ \frac{1}{(\ln 2\beta )^{\varepsilon}} \biggr\} ^{\frac{1}{q}}, $$

and $B(\sigma,\lambda-\sigma)\leq k(\varepsilon\rightarrow0^{+})$. Hence $k=B(\sigma,\lambda-\sigma)$ is the best value of (3.1).

By the equivalence of the inequalities, the constant factor $B(\sigma ,\lambda-\sigma)$ in (3.2) ((3.3)) is the best possible. Otherwise, we would reach the contradiction by (3.4) ((3.5)) that the constant factor in (3.1) is not the best possible. □

Remark 3.2

(i) Define the first type half-discrete Hilbert-type operator $T_{1}:L_{p,\Phi_{\delta}}(\frac{1}{\alpha},\infty )\rightarrow l_{p,\Psi^{1-p}}$ as follows: For $f\in L_{p,\Phi_{\delta}}(\frac{1}{ \alpha},\infty)$, we define $T_{1}f \in l_{p,\Psi^{1-p}}$ by

$$ T_{1}f(n)=\int_{\frac{1}{\alpha}}^{\infty} \frac{1}{(1+\ln^{\delta }\alpha x\ln\beta n)^{\lambda}}f(x)\,dx,\quad n\in\mathbf{N}\backslash\{1\}. $$

Then by (3.2), $\|T_{1}f\|_{p,\Psi^{1-p}}\leq B(\sigma,\lambda -\sigma)\|f\|_{p,\Phi_{\delta}}$ and so $T_{1}$ is a bounded operator with $\|T_{1}\|\leq B(\sigma,\lambda-\sigma)$. Since by Theorem 3.1, the constant factor in (3.2) is best possible, we have $\|T_{1}\|=B(\sigma ,\lambda-\sigma)$.

(ii) Define the second type half-discrete Hilbert-type operator $T_{2}:l_{q,\Psi}\rightarrow L_{q,\Phi_{\delta}^{1-q}}(\frac {1}{\alpha},\infty)$ as follows: For $a\in l_{q,\Psi}$, we define $T_{2}a \in L_{q,\Phi_{\delta}^{1-q}}(\frac{1}{\alpha},\infty)$ by

$$ T_{2}a(x)=\sum_{n=2}^{\infty} \frac{1}{(1+\ln^{\delta}\alpha x\ln \beta n)^{\lambda}}a_{n},\quad x\in \biggl(\frac{1}{\alpha},\infty \biggr). $$

Then by (3.3), $\|T_{2}a\|_{q,\Phi_{\delta}^{1-q}}\leq B(\sigma ,\lambda-\sigma)\|a\|_{q,\Psi}$ and so $T_{2}$ is a bounded operator with $\|T_{2}\|\leq B(\sigma,\lambda-\sigma)$. Since by Theorem 3.1, the constant factor in (3.3) is best possible, we have $\|T_{2}\|=B(\sigma ,\lambda-\sigma)$.

Remark 3.3

For $p=q=2$, $\lambda=1$, $\sigma=\frac{1}{2}$, $\delta=1$ in (3.1), (3.2), and (3.3), (i) if $\alpha=\beta=1$, then we have (1.7) and the following equivalent inequalities:

$$\begin{aligned}& \sum_{n=2}^{\infty}\frac{1}{n} \biggl( \int_{1}^{\infty}\frac {f(x)}{1+\ln x\ln n}\,dx \biggr) ^{2}< \pi^{2}\int_{1}^{\infty}xf^{2}(x) \,dx, \end{aligned}$$

(3.8)

$$\begin{aligned}& \int_{1}^{\infty}\frac{1}{x} \Biggl( \sum _{n=2}^{\infty}\frac {a_{n}}{1+\ln x\ln n} \Biggr) ^{2}\,dx< \pi^{2}\sum_{n=2}^{\infty}na_{n}^{2}; \end{aligned}$$

(3.9)

(ii) if $\alpha=\beta=\frac{2}{3}$, then we have the following equivalent inequalities:

$$\begin{aligned}& \int_{\frac{3}{2}}^{\infty}\sum_{n=2}^{\infty} \frac{a_{n}f(x)\,dx}{1+\ln \frac{2}{3}x\ln\frac{2}{3}n}< \pi \Biggl\{ \int_{\frac{3}{2}}^{\infty }xf^{2}(x) \,dx\sum_{n=2}^{\infty}na_{n}^{2} \Biggr\} ^{\frac{1}{2}}, \end{aligned}$$

(3.10)

$$\begin{aligned}& \sum_{n=2}^{\infty}\frac{1}{n} \biggl( \int_{\frac{3}{2}}^{\infty}\frac {f(x)}{1+\ln\frac{2}{3}x\ln\frac{2}{3}n}\,dx \biggr) ^{2}< \pi^{2}\int_{\frac {3}{2}}^{\infty}xf^{2}(x) \,dx, \end{aligned}$$

(3.11)

$$\begin{aligned}& \int_{\frac{3}{2}}^{\infty}\frac{1}{x} \Biggl( \sum _{n=2}^{\infty}\frac {a_{n}}{1+\ln\frac{2}{3}x\ln\frac{2}{3}n} \Biggr) ^{2}\,dx< \pi ^{2}\sum_{n=2}^{\infty}na_{n}^{2}. \end{aligned}$$

(3.12)

Remark 3.4

For $\delta=-1$ in (3.1), (3.2), and (3.3), setting $F(x)=\ln^{\lambda}(\alpha x)f(x)$, $\mu=\lambda-\sigma$ (>0), and $\Phi(x):=x^{p-1}(\ln\alpha x)^{p(1-\mu)-1}$, we have the following new equivalent inequalities with the same best possible constant factor $B(\sigma,\mu)$:

$$\begin{aligned} &\sum_{n=2}^{\infty}a_{n} \int_{\frac{1}{\alpha}}^{\infty}\frac {F(x)\,dx}{\ln ^{\lambda}(\alpha\beta nx)} =\int _{\frac{1}{\alpha}}^{\infty }F(x)\sum_{n=2}^{\infty} \frac{a_{n}}{\ln^{\lambda}(\alpha\beta nx)}\,dx \\ &\hphantom{\sum_{n=2}^{\infty}a_{n}\int_{\frac{1}{\alpha}}^{\infty}\frac {F(x)\,dx}{\ln ^{\lambda}(\alpha\beta nx)}}< B(\sigma,\mu)\|F\|_{p,\Phi}\|a\|_{q,\Psi}, \end{aligned}$$

(3.13)

$$\begin{aligned} & \Biggl\{ \sum_{n=2}^{\infty} \bigl[\Psi(n) \bigr]^{1-p} \biggl[ \int_{\frac {1}{\alpha}}^{\infty} \frac{F(x)\,dx}{\ln^{\lambda}(\alpha\beta nx)} \biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< B(\sigma, \mu) \|F\|_{p,\Phi}, \end{aligned}$$

(3.14)

$$\begin{aligned} & \Biggl\{ \int_{\frac{1}{\alpha}}^{\infty} \bigl[ \Phi(x) \bigr]^{1-q} \Biggl[ \sum_{n=2}^{\infty} \frac{a_{n}}{\ln^{\lambda}(\alpha\beta nx)} \Biggr] ^{q}\,dx \Biggr\} ^{\frac{1}{q}}< B( \sigma, \mu)\|a\|_{q,\Psi}. \end{aligned}$$

(3.15)