On a more accurate multidimensional Mulholland-type inequality

Abstract

In this paper, by using the way of weight coefficients and technique of real analysis, a more accurate multidimensional discrete Mulholland-type inequality with the best possible constant factor is given, which is an extension of the Mulholland inequality. The equivalent form, the operator expression with the norm as well as a few particular cases are also considered.

MSC:26D15, 47A07.

1 Introduction

Suppose that $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f(x),g(y)\ge 0$, $f\in L^p({\mathbf{R}}_{+})$, $g\in L^q({\mathbf{R}}_{+})$, ${\parallel f\parallel }_p={\{{\int }_0^{\mathrm{\infty }}{f}^p(x)dx\}}^{\frac{1}{p}}>0$, ${\parallel g\parallel }_q>0$. We have the following Hardy-Hilbert integral inequality (cf. [1]):

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

(1)

where the constant factor $\frac{\pi }{sin(\pi /p)}$ is the best possible. Assuming that $a_m,b_n\ge 0$, $a={\{a_m\}}_{m=1}^{\mathrm{\infty }}\in l^p$, $b={\{b_n\}}_{n=1}^{\mathrm{\infty }}\in l^q$, ${\parallel a\parallel }_p={\{{\sum }_{m=1}^{\mathrm{\infty }}{a}_m^p\}}^{\frac{1}{p}}>0$, ${\parallel b\parallel }_q>0$, we have the following Hardy-Hilbert inequality with the same best constant $\frac{\pi }{sin(\pi /p)}$ (cf. [1]):

$$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{m+n}<\frac{\pi }{sin(\pi /p)}{\parallel a\parallel }_p{\parallel b\parallel }_q.$$

(2)

Inequalities (1) and (2) are important in analysis and its applications (cf. [1–6]). Also we have the following Mulholland inequality (cf. [1]):

$$\sum _{m=2}^{\mathrm{\infty }}\sum _{n=2}^{\mathrm{\infty }}\frac{a_m{b}_n}{lnmn}<\frac{\pi }{sin(\pi /p)}{\left\{\sum _{m=2}^{\mathrm{\infty }}\frac{a_m^p}{m^{1-p}}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=2}^{\mathrm{\infty }}\frac{b_n^q}{n^{1-q}}\right\}}^{\frac{1}{q}}.$$

(3)

In 1998, by introducing an independent parameter $\lambda \in (0,1]$, Yang [7] gave an extension of (1) for $p=q=2$. Yang [5] gave some extensions of (1) and (2) as follows: If ${\lambda }_1,{\lambda }_2,\lambda \in \mathbf{R}$, ${\lambda }_1+{\lambda }_2=\lambda$, $k_{\lambda }(x,y)$ is a non-negative homogeneous function of degree −λ, with

$$k({\lambda }_1)={\int }_0^{\mathrm{\infty }}{k}_{\lambda }(t,1)t^{{\lambda }_1-1}dt\in {\mathbf{R}}_{+},$$

$\varphi (x)=x^{p(1-{\lambda }_1)-1}$, $\psi (x)=x^{q(1-{\lambda }_2)-1}$, $f(x),g(y)\ge 0$,

$$f\in L_{p,\varphi }({\mathbf{R}}_{+})=\{f;{\parallel f\parallel }_{p,\varphi }:={\{{\int }_0^{\mathrm{\infty }}\varphi (x)|f(x){|}^{p}dx\}}^{\frac{1}{p}}<\mathrm{\infty }\},$$

$g\in L_{q,\psi }({\mathbf{R}}_{+})$, ${\parallel f\parallel }_{p,\varphi },{\parallel g\parallel }_{q,\psi }>0$, 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 with respect to $x>0$ ($y>0$), then for $a_m,b_n\ge 0$,

$$a\in l_{p,\varphi }=\{a;{\parallel a\parallel }_{p,\varphi }:={\{\sum _{n=1}^{\mathrm{\infty }}\varphi (n){|a_n|}^p\}}^{\frac{1}{p}}<\mathrm{\infty }\},$$

$b={\{b_n\}}_{n=1}^{\mathrm{\infty }}\in l_{q,\psi }$, ${\parallel a\parallel }_{p,\varphi },{\parallel b\parallel }_{q,\psi }>0$, it follows that

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

(5)

where the constant factor $k({\lambda }_1)$ is still the best possible.

Clearly, for $\lambda =1$, $k_1(x,y)=\frac{1}{x+y}$, ${\lambda }_1=\frac{1}{q}$, ${\lambda }_2=\frac{1}{p}$, inequality (3) reduces to (1), while (5) reduces to (2). Some other results including the multidimensional Hilbert-type integral inequalities are provided by [8–24].

About 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. However, Yang [25] gave a result with the kernel $\frac{1}{{(1+nx)}^{\lambda }}$ ($0<\lambda \le 2$) by introducing a variable and proved that the constant factor is the best possible. In 2011 Yang [26] gave a half-discrete Hardy-Hilbert inequality with the best possible constant factor. Zhong et al. [27–33] investigated several half-discrete Hilbert-type inequalities with particular kernels. Using the way of weight functions and the techniques of discrete and integral Hilbert-type inequalities with some additional conditions on the kernel, a half-discrete Hilbert-type inequality with a general homogeneous kernel of degree $-\lambda \in \mathbf{R}$ and a best constant factor $k({\lambda }_1)$ is obtained as follows:

$${\int }_0^{\mathrm{\infty }}f(x)\sum _{n=1}^{\mathrm{\infty }}{k}_{\lambda }(x,n)a_{n}dx<k({\lambda }_1){\parallel f\parallel }_{p,\varphi }{\parallel a\parallel }_{q,\psi }$$

(6)

(see Yang and Chen [34]). At the same time, a half-discrete Hilbert-type inequality with a general non-homogeneous kernel and the best constant factor is given by Yang [35].

In this paper, by using the way of weight coefficients and technique of real analysis, a more accurate multidimensional discrete Mulholland-type inequality with the best possible constant factor is given, which is an extension of (3). The equivalent form, the operator expression with the norm as well as a few particular cases are also considered.

2 Some lemmas

Lemma 1 If ${(-1)}^i{h}^{(i)}(t)>0$ ($t>0$; $i=1,2$), then for $b>0$, $0<\alpha \le 1$,

$${(-1)}^i\frac{d^i}{d{x}^i}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)>0(x>1;i=1,2).$$

(7)

Proof We find

$$\begin{array}{c}\frac{d}{dx}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)=\frac{1}{x}{h}^{\prime }\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right){(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }-1}{ln}^{\alpha -1}x<0,\hfill \\ \frac{d^2}{d{x}^2}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)=\frac{d}{dx}\left[\frac{1}{x}{h}^{\prime }\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right){(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }-1}{ln}^{\alpha -1}x\right]\hfill \\ \phantom{\frac{d^2}{d{x}^2}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)}=-\frac{1}{x^2}{h}^{\prime }\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right){(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }-1}{ln}^{\alpha -1}x\hfill \\ \phantom{\frac{d^2}{d{x}^2}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)=}+\frac{1}{x^2}{h}^{″}\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right){(b+{ln}^{\alpha }x)}^{\frac{2}{\alpha }-2}{ln}^{2\alpha -2}x\hfill \\ \phantom{\frac{d^2}{d{x}^2}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)=}+\alpha (\frac{1}{\alpha }-1)\frac{1}{x^2}{h}^{\prime }\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right){(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }-2}{ln}^{2\alpha -2}x\hfill \\ \phantom{\frac{d^2}{d{x}^2}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)=}+(\alpha -1)\frac{1}{x^2}{h}^{\prime }\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right){(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }-1}{ln}^{\alpha -2}x\hfill \\ \phantom{\frac{d^2}{d{x}^2}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)}=[-h^{\prime }\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)(b+{ln}^{\alpha }x)lnx\hfill \\ \phantom{\frac{d^2}{d{x}^2}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)=}+h^{″}\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right){(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}{ln}^{\alpha }x\hfill \\ \phantom{\frac{d^2}{d{x}^2}h\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)=}+b(\alpha -1)h^{\prime }\left({(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }}\right)]\frac{1}{x^2}{(b+{ln}^{\alpha }x)}^{\frac{1}{\alpha }-2}{ln}^{\alpha -2}x>0.\hfill \end{array}$$

Then we have (7). □

If $i_0,j_0\in \mathbf{N}$ (N is the set of positive integers), $\alpha ,\beta >0$, we set

$${\parallel x\parallel }_{\alpha }:={\left(\sum _{k=1}^{i_0}{|x_k|}^{\alpha }\right)}^{\frac{1}{\alpha }}(x=(x_1,\dots ,x_{i_0})\in {\mathbf{R}}^{i_0}),$$

(8)

$${\parallel y\parallel }_{\beta }:={\left(\sum _{k=1}^{j_0}{|y_k|}^{\beta }\right)}^{\frac{1}{\beta }}(y=(y_1,\dots ,y_{j_0})\in {\mathbf{R}}^{j_0}).$$

(9)

Lemma 2 If $s\in \mathbf{N}$, $\gamma ,M>0$, $\mathrm{\Psi }(u)$ is a non-negative measurable function in $(0,1]$, and

$$D_M:=\{x\in {\mathbf{R}}_{+}^s;\sum _{i=1}^s{x}_i^{\gamma }\le M^{\gamma }\}=\{x;\sum _{i=1}^s{\left(\frac{x_i}{M}\right)}^{\gamma }\le 1\},$$

then we have (cf. [36])

$$\begin{array}{c}\int \cdots {\int }_{D_M}\mathrm{\Psi }\left(\sum _{i=1}^s{\left(\frac{x_i}{M}\right)}^{\gamma }\right)d{x}_1\cdots d{x}_s\hfill \\ =\frac{M^s{\mathrm{\Gamma }}^s(\frac{1}{\gamma })}{{\gamma }^s\mathrm{\Gamma }(\frac{s}{\gamma })}{\int }_0^1\mathrm{\Psi }(u)u^{\frac{s}{\gamma }-1}du.\hfill \end{array}$$

(10)

Lemma 3 If $s\in \mathbf{N}$, $\gamma >0$, $\epsilon >0$, $d=(d_1,\dots ,d_s)\in {[\frac{1}{2},1]}^s$, then

$$\begin{array}{rcl}{A}_s(\epsilon )& :=& \sum _m{\parallel ln(m+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^s(m_i+d_i)}\\ =& \frac{{\mathrm{\Gamma }}^s(\frac{1}{\gamma })}{\epsilon s^{\epsilon /\gamma }{\gamma }^{s-1}\mathrm{\Gamma }(\frac{s}{\gamma })}+O(1)(\epsilon \to 0^{+}).\end{array}$$

(11)

Proof For $M>s^{1/\gamma }$, we set

$$\mathrm{\Psi }(u)=\{\begin{array}{ll}0,& 0<u<\frac{s}{M^{\gamma }},\\ {(M{u}^{1/\gamma })}^{-s-\epsilon },& \frac{s}{M^{\gamma }}\le u\le 1.\end{array}$$

Then by the decreasing property and (10), it follows that

$$\begin{array}{rcl}{A}_s(\epsilon )& \ge & {\int }_{\{x\in {\mathbf{R}}_{+}^s;x_i\ge e-d_i\}}{\parallel ln(x+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{dx}{{\prod }_{i=1}^s(x_i+d_i)}\\ \stackrel{u_i=ln(x_i+d_i)}{=}& {\int }_{\{u\in {\mathbf{R}}_{+}^s;u_i\ge 1\}}{\parallel u\parallel }_{\gamma }^{-s-\epsilon }du\\ =& \underset{M\to \mathrm{\infty }}{lim}\int \cdots {\int }_{D_M}\mathrm{\Psi }\left(\sum _{i=1}^s{\left(\frac{x_i}{M}\right)}^{\gamma }\right)d{x}_1\cdots d{x}_s\\ =& \underset{M\to \mathrm{\infty }}{lim}\frac{M^s{\mathrm{\Gamma }}^s(\frac{1}{\gamma })}{{\gamma }^s\mathrm{\Gamma }(\frac{s}{\gamma })}{\int }_{s/M^{\gamma }}^1{\left(M{u}^{1/\gamma }\right)}^{-s-\epsilon }{u}^{\frac{s}{\gamma }-1}du=\frac{{\mathrm{\Gamma }}^s(\frac{1}{\gamma })}{\epsilon s^{\epsilon /\gamma }{\gamma }^{s-1}\mathrm{\Gamma }(\frac{s}{\gamma })}.\end{array}$$

In the following, by mathematical induction we prove that, for any $s\in \mathbf{N}$,

$$A_s(\epsilon )\le O_s(1)+\frac{{\mathrm{\Gamma }}^s(\frac{1}{\gamma })}{\epsilon s^{\epsilon /\gamma }{\gamma }^{s-1}\mathrm{\Gamma }(\frac{s}{\gamma })}(\epsilon \to 0^{+}).$$

(12)

For $s=1$, by the Hermite-Hadamard inequality (cf. [37]), it follows that

$$\begin{array}{rcl}{A}_1(\epsilon )& =& \sum _{m_1=1}^2\frac{{ln}^{-1-\epsilon }(m_1+d_1)}{m_1+d_1}+\sum _{m_1=3}^{\mathrm{\infty }}\frac{{ln}^{-1-\epsilon }(m_1+d_1)}{m_1+d_1}\\ \le & O_1(1)+{\int }_{\frac{5}{2}}^{\mathrm{\infty }}\frac{{ln}^{-1-\epsilon }(x+d_1)dx}{x+d_1}\le O_1(1)+{\int }_{e-d_1}^{\mathrm{\infty }}\frac{{ln}^{-1-\epsilon }(x+d_1)dx}{x+d_1}\\ \stackrel{u=ln(x+d_1)}{=}& O_1(1)+{\int }_1^{\mathrm{\infty }}{u}^{-1-\epsilon }du=O_1(1)+\frac{1}{\epsilon },\end{array}$$

and then (12) is valid. Assuming that (12) is valid for $s-1\in \mathbf{N}$, then for s, we set

$$\begin{array}{rcl}{A}_s(\epsilon )& =& \sum _{\{m\in {\mathbf{N}}^s;\mathrm{\exists }{i}_0,m_{i_0}=1,2\}}{\parallel ln(m+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^s(m_i+d_i)}\\ +\sum _{\{m\in {\mathbf{N}}^s;m_i\ge 3\}}{\parallel ln(m+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^s(m_i+d_i)}.\end{array}$$

There exist constants $a,b\in {\mathbf{R}}_{+}$, such that

$$\begin{array}{c}\sum _{\{m\in {\mathbf{N}}^s;\mathrm{\exists }{i}_0,m_{i_0}=1,2\}}{\parallel ln(m+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^s(m_i+d_i)}\hfill \\ \le a+b\sum _{\{m\in {\mathbf{N}}^{s-1};m_i\ge 1\}}{\parallel ln(m+d)\parallel }_{\gamma }^{-(s-1)-(1+\epsilon )}\frac{1}{{\prod }_{i=1}^{s-1}(m_i+d_i)}.\hfill \end{array}$$

By the assumption of mathematical induction for $s-1$, we find

$$\begin{array}{c}\sum _{\{m\in {\mathbf{N}}^{s-1};m_i\ge 1\}}{\parallel ln(m+d)\parallel }_{\gamma }^{-(s-1)-(1+\epsilon )}\frac{1}{{\prod }_{i=1}^{s-1}(m_i+d_i)}\hfill \\ \le O_{s-1}(1)+\frac{{\mathrm{\Gamma }}^{s-1}(\frac{1}{\gamma })}{(1+\epsilon ){(s-1)}^{(1+\epsilon )/\gamma }{\gamma }^{s-2}\mathrm{\Gamma }(\frac{s-1}{\gamma })},\hfill \end{array}$$

and then

$$\sum _{\{m\in {\mathbf{N}}^s;\mathrm{\exists }{i}_0,m_{i_0}=1,2\}}{\parallel ln(m+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^s(m_i+d_i)}\le O_s(1).$$

By Lemma 1 and the Hermite-Hadamard inequality (cf. [37]), we obtain

$$\begin{array}{c}\sum _{\{m\in {\mathbf{N}}^s;m_i\ge 3\}}{\parallel ln(m+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^s(m_i+d_i)}\hfill \\ \le {\int }_{\{x\in {\mathbf{R}}_{+}^s;x_i\ge \frac{5}{2}\}}{\parallel ln(x+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^s(x_i+d_i)}dx\hfill \\ \le {\int }_{\{x\in {\mathbf{R}}_{+}^s;x_i\ge e-d_i\}}{\parallel ln(x+d)\parallel }_{\gamma }^{-s-\epsilon }\frac{1}{{\prod }_{i=1}^s(x_i+d_i)}dx\hfill \\ \stackrel{u_i=ln(x_i+d_i)}{=}{\int }_{\{u\in {\mathbf{R}}_{+}^s;u_i\ge 1\}}{\parallel u\parallel }_{\gamma }^{-s-\epsilon }du=\frac{{\mathrm{\Gamma }}^s(\frac{1}{\gamma })}{\epsilon s^{\epsilon /\gamma }{\gamma }^{s-1}\mathrm{\Gamma }(\frac{s}{\gamma })}.\hfill \end{array}$$

Hence we prove that (12) is valid for $s\in \mathbf{N}$. Therefore, we have (11). □

Lemma 4 If C is the set of complex numbers and ${\mathbf{C}}_{\mathrm{\infty }}=\mathbf{C}\cup \{\mathrm{\infty }\}$, $z_k\in \mathbf{C}\mathrm{\setminus }\{z|Rez\ge 0,Imz=0\}$ ($k=1,2,\dots ,n$) are different points, the function $f(z)$ is analytic in ${\mathbf{C}}_{\mathrm{\infty }}$ except for $z_i$ ($i=1,2,\dots ,n$), and $z=\mathrm{\infty }$ is a zero point of $f(z)$ whose order is not less than 1, then for $\alpha \in \mathbf{R}$, we have

$${\int }_0^{\mathrm{\infty }}f(x)x^{\alpha -1}dx=\frac{2\pi i}{1-e^{2\pi \alpha i}}\sum _{k=1}^{n}Res[f(z)z^{\alpha -1},z_k],$$

(13)

where $0<Imlnz=argz<2\pi$. In particular, if $z_k$ ($k=1,\dots ,n$) are all poles of order 1, setting ${\phi }_k(z)=(z-z_k)f(z)$ (${\phi }_k(z_k)\ne 0$), then

$${\int }_0^{\mathrm{\infty }}f(x)x^{\alpha -1}dx=\frac{\pi }{sin\pi \alpha }\sum _{k=1}^n{(-z_k)}^{\alpha -1}{\phi }_k(z_k).$$

(14)

Proof By [[38], p.118], we have (13). We find

$$\begin{array}{rcl}1-e^{2\pi \alpha i}& =& 1-cos2\pi \alpha -isin2\pi \alpha \\ =& -2isin\pi \alpha (cos\pi \alpha +isin\pi \alpha )=-2i{e}^{i\pi \alpha }sin\pi \alpha .\end{array}$$

In particular, since $f(z)z^{\alpha -1}=\frac{1}{z-z_k}({\phi }_k(z)z^{\alpha -1})$, it is obvious that

$$Res[f(z)z^{\alpha -1},-a_k]=z_k{}^{\alpha -1}{\phi }_k(z_k)=-e^{i\pi \alpha }{(-z_k)}^{\alpha -1}{\phi }_k(z_k).$$

Then by (13), we obtain (14). □

Example 1 For $s\in \mathbf{N}$, we set

$$k_{\lambda }(x,y)=\prod _{k=1}^s\frac{1}{(x^{\lambda /s}+c_k{y}^{\lambda /s})}(0<c_1<\cdots <c_s,0<\lambda \le s).$$

For $0<{\lambda }_1\le i_0$, $0<{\lambda }_2\le j_0$, ${\lambda }_1+{\lambda }_2=\lambda$, by (14), we find

$$\begin{array}{rcl}{k}_s({\lambda }_1)& :=& {\int }_0^{\mathrm{\infty }}\prod _{k=1}^s\frac{1}{t^{\lambda /s}+c_k}{t}^{{\lambda }_1-1}dt\\ \stackrel{u=t^{\lambda /s}}{=}& \frac{s}{\lambda }{\int }_0^{\mathrm{\infty }}\prod _{k=1}^s\frac{1}{u+c_k}{u}^{\frac{s{\lambda }_1}{\lambda }-1}du\\ =& \frac{\pi s}{\lambda sin(\frac{\pi s{\lambda }_1}{\lambda })}\sum _{k=1}^s{c}_k^{\frac{s{\lambda }_1}{\lambda }-1}\prod _{j=1(j\ne k)}^s\frac{1}{c_j-c_k}\in {\mathbf{R}}_{+}.\end{array}$$

(15)

In particular, for $s=1$, we obtain

$$k_1({\lambda }_1)=\frac{1}{\lambda }{\int }_0^{\mathrm{\infty }}\frac{u^{({\lambda }_1/\lambda )-1}}{u+c_1}du=\frac{\pi }{\lambda sin(\frac{\pi {\lambda }_1}{\lambda })}{c}_1^{\frac{{\lambda }_1}{\lambda }-1}.$$

Definition 1 For $s\in \mathbf{N}$, $0<\alpha ,\beta \le 1$, $0<c_1<\cdots <c_s$, $0<\lambda \le s$, $0<{\lambda }_1\le i_0$, $0<{\lambda }_2\le j_0$, ${\lambda }_1+{\lambda }_2=\lambda$, $\tau =({\tau }_1,\dots ,{\tau }_{i_0})\in {[\frac{1}{2},1]}^{i_0}$, $\sigma =({\sigma }_1,\dots ,{\sigma }_{j_0})\in {[\frac{1}{2},1]}^{j_0}$, $ln(m+\tau )=(ln(m_1+{\tau }_1),\dots ,ln(m_{i_0}+{\tau }_{i_0}))\in {\mathbf{R}}_{+}^{i_0}$, $ln(n+\sigma )=(ln(n_1+{\sigma }_1),\dots ,ln(n_{j_0}+{\sigma }_{j_0}))\in {\mathbf{R}}_{+}^{j_0}$, we define two weight coefficients $w_{\lambda }({\lambda }_2,n)$ and $W_{\lambda }({\lambda }_1,m)$ as follows:

$$\begin{array}{r}{w}_{\lambda }({\lambda }_2,n):=\sum _m\frac{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\lambda }_2}{\parallel ln(m+\tau )\parallel }_{\alpha }^{{\lambda }_1-i_0}}{{\prod }_{k=1}^s[{\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]{\prod }_{i=1}^{i_0}(m_i+{\tau }_i)},\\ W_{\lambda }({\lambda }_1,m):=\sum _n\frac{{\parallel ln(m+\tau )\parallel }_{\alpha }^{{\lambda }_1}{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\lambda }_2-j_0}}{{\prod }_{k=1}^s[{\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]{\prod }_{j=1}^{j_0}(n_j+{\sigma }_j)},\end{array}$$

(16)

where ${\sum }_m={\sum }_{m_{i_0}=1}^{\mathrm{\infty }}\cdots {\sum }_{m_1=1}^{\mathrm{\infty }}$ and ${\sum }_n={\sum }_{n_{j_0}=1}^{\mathrm{\infty }}\cdots {\sum }_{n_1=1}^{\mathrm{\infty }}$.

Lemma 5 Let the assumptions as in Definition  1 be fulfilled. Then:

• (i) we have

  $$w_{\lambda }({\lambda }_2,n)<K_2(n\in {\mathbf{N}}^{j_0}),$$

  (17)

$$W_{\lambda }({\lambda }_1,m)<K_1(m\in {\mathbf{N}}^{i_0}),$$

(18)

where

$$K_1:=\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta })}{{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta })}{k}_s({\lambda }_1),K_2:=\frac{{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{{\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}{k}_s({\lambda }_1),$$

(19)

and $k_s({\lambda }_1)$ is indicated by (15);

• (ii) for $p>1$, $0<\epsilon <pmin\{{\lambda }_1,1-{\lambda }_2\}$, setting ${\tilde{\lambda }}_1={\lambda }_1-\frac{\epsilon }{p}$, ${\tilde{\lambda }}_2={\lambda }_2+\frac{\epsilon }{p}$, we have

  $$0<{\tilde{K}}_2(1-{\tilde{\theta }}_{\lambda }(n))<w_{\lambda }({\tilde{\lambda }}_2,n),$$

  (20)

where

$$\begin{array}{c}{\tilde{\theta }}_{\lambda }(n):=\frac{1}{k_s({\tilde{\lambda }}_1)}{\int }_0^{i_0^{\lambda /(\alpha s)}/{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}}\frac{v^{\frac{s{\lambda }_1}{\lambda }-1}}{{\prod }_{k=1}^s(v+c_k)}dv\hfill \\ \phantom{{\tilde{\theta }}_{\lambda }(n)}=O\left(\frac{1}{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\tilde{\lambda }}_1}}\right),\hfill \end{array}$$

(21)

$${\tilde{K}}_2=\frac{{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })k_s({\tilde{\lambda }}_1)}{{\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}\in {\mathbf{R}}_{+}.$$

(22)

Proof By Lemma 1, the Hermite-Hadamard inequality (cf. [37]), (10), and (15), it follows that

$$\begin{array}{c}{w}_{\lambda }({\lambda }_2,n)\hfill \\ <{\int }_{{(\frac{1}{2},\mathrm{\infty })}^{i_0}}\frac{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\lambda }_2}{\parallel ln(x+\tau )\parallel }_{\alpha }^{{\lambda }_1-i_0}dx}{{\prod }_{k=1}^s[{\parallel ln(x+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]{\prod }_{i=1}^{i_0}(x_i+{\tau }_i)}\hfill \\ \stackrel{u_i=ln(x_i+{\tau }_i)}{=}{\int }_{\{u\in {\mathbf{R}}_{+}^{i_0};u_i>ln(\frac{1}{2}+{\tau }_i)\}}\frac{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\lambda }_2}{\parallel u\parallel }_{\alpha }^{{\lambda }_1-i_0}}{{\prod }_{k=1}^s[{\parallel u\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}du\hfill \\ \le {\int }_{{\mathbf{R}}_{+}^{i_0}}\frac{{\parallel n-\sigma \parallel }_{\beta }^{{\lambda }_2}{\parallel u\parallel }_{\alpha }^{{\lambda }_1-i_0}}{{\prod }_{k=1}^s[{\parallel u\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}du\hfill \\ =\underset{M\to \mathrm{\infty }}{lim}{\int }_{{\mathbf{D}}_M}\frac{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\lambda }_2}{M}^{{\lambda }_1-i_0}{[{\sum }_{i=1}^{j_0}{(\frac{u_i}{M})}^{\alpha }]}^{({\lambda }_1-i_0)/\alpha }}{{\prod }_{k=1}^s\{M^{\frac{\lambda }{s}}{[{\sum }_{i=1}^{i_0}{(\frac{u_i}{M})}^{\alpha }]}^{\frac{\lambda }{\alpha s}}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\frac{\lambda }{s}}\}}du\hfill \\ =\underset{M\to \mathrm{\infty }}{lim}\frac{M^{i_0}{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{{\alpha }^{i_0}\mathrm{\Gamma }(\frac{i_0}{\alpha })}{\int }_0^1\frac{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\lambda }_2}{M}^{{\lambda }_1-i_0}{t}^{({\lambda }_1-i_0)/\alpha }{t}^{\frac{i_0}{\alpha }-1}}{{\prod }_{k=1}^s(M^{\frac{\lambda }{s}}{t}^{\frac{\lambda }{\alpha s}}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\frac{\lambda }{s}})}dt\hfill \\ =\underset{M\to \mathrm{\infty }}{lim}\frac{M^{{\lambda }_1}{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{{\alpha }^{i_0}\mathrm{\Gamma }(\frac{i_0}{\alpha })}{\int }_0^1\frac{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\lambda }_2}{t}^{\frac{{\lambda }_1}{\alpha }-1}}{{\prod }_{k=1}^s(M^{\frac{\lambda }{s}}{t}^{\frac{\lambda }{\alpha s}}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\frac{\lambda }{s}})}dt\hfill \\ \stackrel{t={\parallel ln(n+\sigma )\parallel }_{\beta }^{\alpha }{M}^{-\alpha }{v}^{\alpha s/\lambda }}{=}\frac{s{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{\lambda {\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}{\int }_0^{\mathrm{\infty }}\frac{v^{\frac{s{\lambda }_1}{\lambda }-1}}{{\prod }_{k=1}^s(v+c_k)}dv\hfill \\ =\frac{{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{{\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}{k}_s({\lambda }_1)=K_2.\hfill \end{array}$$

Hence, we have (17). In the same way, we have (18).

By the decreasing property and (10), similarly to the proof of (11), we find

$$\begin{array}{c}{w}_{\lambda }({\tilde{\lambda }}_2,n)\hfill \\ >{\int }_{\{x\in {\mathbf{R}}_{+}^{i_0};x_i\ge e-{\tau }_i\}}\frac{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\tilde{\lambda }}_2}{\parallel ln(x+\tau )\parallel }_{\alpha }^{{\tilde{\lambda }}_1-i_0}dx}{{\prod }_{k=1}^s[{\parallel ln(x+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]{\prod }_{i=1}^{i_0}(x_i+{\tau }_i)}\hfill \\ \stackrel{u_i=ln(x_i+{\tau }_i)}{=}{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\tilde{\lambda }}_2}{\int }_{\{u\in {\mathbf{R}}_{+}^{i_0};u_i\ge 1\}}\frac{{\parallel u\parallel }_{\alpha }^{{\tilde{\lambda }}_1-i_0}du}{{\prod }_{k=1}^s[{\parallel u\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}\hfill \\ ={\parallel ln(n+\sigma )\parallel }_{\beta }^{{\tilde{\lambda }}_2}\underset{M\to \mathrm{\infty }}{lim}\int \cdots {\int }_{D_M}\frac{{\{{\sum }_{i=1}^{i_0}{(\frac{u_i}{M})}^{\alpha }\}}^{\frac{{\tilde{\lambda }}_1-i_0}{\alpha }}{M}^{{\tilde{\lambda }}_1-i_0}d{u}_1\cdots d{u}_{i_0}}{{\prod }_{k=1}^s[{\{{\sum }_{i=1}^{i_0}{(\frac{u_i}{M})}^{\alpha }\}}^{\frac{\lambda }{\alpha s}}{M}^{\frac{\lambda }{s}}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}\hfill \\ =\frac{M^{i_0}{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{{\alpha }^{i_0}\mathrm{\Gamma }(\frac{i_0}{\alpha })}{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\tilde{\lambda }}_2}\underset{M\to \mathrm{\infty }}{lim}{\int }_{\frac{i_0}{M^{\alpha }}}^1\frac{t^{\frac{{\tilde{\lambda }}_1-i_0}{\alpha }}{M}^{{\tilde{\lambda }}_1-i_0}{t}^{\frac{i_0}{\alpha }-1}}{{\prod }_{k=1}^s[t^{\frac{\lambda }{\alpha s}}{M}^{\frac{\lambda }{s}}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}dt\hfill \\ =\frac{s{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{\lambda {\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}{\int }_{\frac{i_0^{\lambda /(\alpha s)}}{{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}}}^{\mathrm{\infty }}\frac{v^{\frac{s{\tilde{\lambda }}_1}{\lambda }-1}dv}{{\prod }_{k=1}^s(v+c_k)}={\tilde{K}}_2(1-{\tilde{\theta }}_{\lambda }(n))>0,\hfill \\ 0<{\tilde{\theta }}_{\lambda }(n)=\frac{s}{\lambda k_s({\tilde{\lambda }}_1)}{\int }_0^{i_0^{\lambda /(\alpha s)}/{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}}\frac{v^{\frac{s{\lambda }_1}{\lambda }-1}}{{\prod }_{k=1}^s(v+c_k)}dv\hfill \\ \phantom{0}\le \frac{s}{\lambda k_s({\tilde{\lambda }}_1){\prod }_{k=1}^s{c}_k}{\int }_0^{i_0^{\lambda /(\alpha s)}/{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}}{v}^{\frac{s{\tilde{\lambda }}_1}{\lambda }-1}dv\hfill \\ \phantom{0}=\frac{1}{{\tilde{\lambda }}_1{k}_s({\tilde{\lambda }}_1){\prod }_{k=1}^s{c}_k}\frac{i_0^{{\tilde{\lambda }}_1/\alpha }}{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\tilde{\lambda }}_1}}.\hfill \end{array}$$

Hence, we have (20) and (21). □

3 Main results and operator expressions

Setting $\mathrm{\Phi }(m):={\prod }_{i=1}^{i_0}{(m_i+{\tau }_i)}^{p-1}{\parallel ln(m+\tau )\parallel }_{\alpha }^{p(i_0-{\lambda }_1)-i_0}$ ($m\in {\mathbf{N}}^{i_0}$) and $\mathrm{\Psi }(n):={\prod }_{j=1}^{j_0}{(n_j+{\sigma }_j)}^{q-1}{\parallel ln(n+\sigma )\parallel }_{\beta }^{q(j_0-{\lambda }_2)-j_0}$ ($n\in {\mathbf{N}}^{j_0}$), wherefrom

$${[\mathrm{\Psi }(n)]}^{1-p}=\prod _{j=1}^{j_0}{(n_j+{\sigma }_j)}^{-1}{\parallel ln(n+\sigma )\parallel }_{\beta }^{p{\lambda }_2-j_0},$$

we have the following.

Theorem 1 If $s\in \mathbf{N}$, $0<\alpha ,\beta \le 1$, $0<c_1<\cdots <c_s$, $0<\lambda \le s$, $0<{\lambda }_1\le i_0$, $0<{\lambda }_2\le j_0$, ${\lambda }_1+{\lambda }_2=\lambda$, $\tau \in {[\frac{1}{2},1]}^{i_0}$, $\sigma \in {[\frac{1}{2},1]}^{j_0}$, then for $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $a_m,b_n\ge 0$, $0<{\parallel a\parallel }_{p,\mathrm{\Phi }},{\parallel b\parallel }_{q,\mathrm{\Psi }}<\mathrm{\infty }$, we have

$$\begin{array}{rcl}I& :=& \sum _n\sum _m\frac{a_m{b}_n}{{\prod }_{k=1}^s[{\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}\\ <& K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}{\parallel b\parallel }_{q,\mathrm{\Psi }},\end{array}$$

(23)

where the constant factor

$$K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}={\left[\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta })}{{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta })}\right]}^{\frac{1}{p}}{\left[\frac{{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{{\beta }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}\right]}^{\frac{1}{q}}{k}_s({\lambda }_1)$$

(24)

is the best possible ($k_s({\lambda }_1)$ is indicated by (15)).

Proof By the Hölder inequality (cf. [37]), we have

$$\begin{array}{rcl}I& =& \sum _n\sum _m\frac{1}{{\prod }_{k=1}^s[{\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}\\ \times \left[\frac{{\parallel ln(m+\tau )\parallel }_{\alpha }^{(i_0-{\lambda }_1)/q}}{{\parallel ln(n+\sigma )\parallel }_{\beta }^{(j_0-{\lambda }_2)/p}}\frac{{\prod }_{i=1}^{i_0}{(m_i+{\tau }_i)}^{1/q}}{{\prod }_{j=1}^{j_0}{(n_j+{\sigma }_j)}^{1/p}}{a}_m\right]\\ \times \left[\frac{{\parallel ln(n+\sigma )\parallel }_{\beta }^{(j_0-{\lambda }_2)/p}}{{\parallel ln(m+\tau )\parallel }_{\alpha }^{(i_0-{\lambda }_1)/q}}\frac{{\prod }_{j=1}^{j_0}{(n_j+{\sigma }_j)}^{1/p}}{{\prod }_{i=1}^{i_0}{(m_i+{\tau }_i)}^{1/q}}{b}_n\right]\\ \le & {\{\sum _m{W}_{\lambda }({\lambda }_1,m)\prod _{i=1}^{i_0}{(m_i+{\tau }_i)}^{p-1}{\parallel ln(m+\tau )\parallel }_{\alpha }^{p(i_0-{\lambda }_1)-i_0}{a}_m^p\}}^{\frac{1}{p}}\\ \times {\{\sum _n{w}_{\lambda }({\lambda }_2,n)\prod _{j=1}^{j_0}{(n_j+{\sigma }_j)}^{q-1}{\parallel ln(n+\sigma )\parallel }_{\beta }^{q(j_0-{\lambda }_2)-j_0}{b}_n^q\}}^{\frac{1}{q}}.\end{array}$$

Then by (17) and (18), we have (23).

For $0<\epsilon <pmin\{{\lambda }_1,1-{\lambda }_2\}$, ${\tilde{\lambda }}_1={\lambda }_1-\frac{\epsilon }{p}$, ${\tilde{\lambda }}_2={\lambda }_2+\frac{\epsilon }{p}$, we set

$$\begin{array}{c}{\tilde{a}}_m={\parallel ln(m+\tau )\parallel }_{\alpha }^{-i_0+{\lambda }_1-\frac{\epsilon }{p}}\frac{1}{{\prod }_{i=1}^{i_0}(m_i+{\tau }_i)},\hfill \\ {\tilde{b}}_n={\parallel ln(n+\sigma )\parallel }_{\beta }^{-j_0+{\lambda }_2-\frac{\epsilon }{q}}\frac{1}{{\prod }_{j=1}^{j_0}(n_j+{\sigma }_j)}(m\in {\mathbf{N}}^{i_0},n\in {\mathbf{N}}^{j_0}).\hfill \end{array}$$

Then by (11) and (20)-(22), we obtain

$$\begin{array}{c}{\parallel \tilde{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \tilde{b}\parallel }_{q,\mathrm{\Psi }}={\left\{\sum _m\prod _{i=1}^{i_0}{(m_i+{\tau }_i)}^{p-1}{\parallel ln(m+\tau )\parallel }_{\alpha }^{p(i_0-{\lambda }_1)-i_0}{\tilde{a}}_m^p\right\}}^{\frac{1}{p}}\hfill \\ \phantom{{\parallel \tilde{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \tilde{b}\parallel }_{q,\mathrm{\Psi }}=}\times {\left\{\sum _n\prod _{j=1}^{j_0}{(n_j+{\sigma }_j)}^{q-1}{\parallel ln(n+\sigma )\parallel }_{\beta }^{q(j_0-{\lambda }_2)-j_0}{\tilde{b}}_n^q\right\}}^{\frac{1}{q}}\hfill \\ \phantom{{\parallel \tilde{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \tilde{b}\parallel }_{q,\mathrm{\Psi }}}={\left\{\sum _m{\parallel ln(m+\tau )\parallel }_{\alpha }^{-i_0-\epsilon }\frac{1}{{\prod }_{i=1}^{i_0}(m_i+{\tau }_i)}\right\}}^{\frac{1}{p}}\hfill \\ \phantom{{\parallel \tilde{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \tilde{b}\parallel }_{q,\mathrm{\Psi }}=}\times {\left\{\sum _n{\parallel ln(n+\sigma )\parallel }_{\beta }^{-j_0-\epsilon }\frac{1}{{\prod }_{j=1}^{j_0}(n_j+{\sigma }_j)}\right\}}^{\frac{1}{q}}\hfill \\ \phantom{{\parallel \tilde{a}\parallel }_{p,\mathrm{\Phi }}{\parallel \tilde{b}\parallel }_{q,\mathrm{\Psi }}}=\frac{1}{\epsilon }{[\frac{{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{i_0^{\epsilon /\alpha }{\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}+\epsilon O(1)]}^{\frac{1}{p}}{[\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta })}{j_0^{\epsilon /\beta }{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta })}+\epsilon \tilde{O}(1)]}^{\frac{1}{q}},\hfill \end{array}$$

(25)

$$\begin{array}{c}\tilde{I}:=\sum _n\left[\sum _m\frac{{\tilde{a}}_m}{{\prod }_{k=1}^s({\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s})}\right]{\tilde{b}}_n\hfill \\ \phantom{\tilde{I}}=\sum _n{w}_{\lambda }({\tilde{\lambda }}_2,n){\parallel ln(n+\sigma )\parallel }_{\beta }^{-j_0-\epsilon }\frac{1}{{\prod }_{j=1}^{j_0}(n_j+{\sigma }_j)}\hfill \\ \phantom{\tilde{I}}>{\tilde{K}}_2\sum _n(1-O\left(\frac{1}{{\parallel ln(n+\sigma )\parallel }_{\beta }^{{\tilde{\lambda }}_1}}\right)){\parallel ln(n+\sigma )\parallel }_{\beta }^{-j_0-\epsilon }\frac{1}{{\prod }_{j=1}^{j_0}(n_j+{\sigma }_j)}\hfill \\ \phantom{\tilde{I}}={\tilde{K}}_2[\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta })}{\epsilon j_0^{\epsilon /\beta }{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta })}+\tilde{O}(1)-O(1)].\hfill \end{array}$$

(26)

If there exists a constant $K\le K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}$, such that (23) is valid when replacing $K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}$ by K, then we have

$$\begin{array}{c}(K_2+o(1))[\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta })}{j_0^{\epsilon /\beta }{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta })}+\epsilon \tilde{O}(1)-\epsilon O(1)]\hfill \\ <\epsilon \tilde{I}<\epsilon K{\parallel \tilde{a}\parallel }_{p,\phi }{\parallel \tilde{b}\parallel }_{q,\psi }=K{[\frac{{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{i_0^{\epsilon /\alpha }{\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}+\epsilon O(1)]}^{\frac{1}{p}}{[\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta })}{j_0^{\epsilon /\beta }{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta })}+\epsilon \tilde{O}(1)]}^{\frac{1}{q}}.\hfill \end{array}$$

For $\epsilon \to 0^{+}$, we find

$$\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta }){\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })k_s({\lambda }_1)}{{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta }){\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}\le K{\left[\frac{{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{{\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}\right]}^{\frac{1}{p}}{\left[\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta })}{{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta })}\right]}^{\frac{1}{q}},$$

and then $K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}\le K$. Hence, $K=K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}$ is the best possible constant factor of (23). □

Theorem 2 With the assumptions of Theorem  1, for $0<{\parallel a\parallel }_{p,\mathrm{\Phi }}<\mathrm{\infty }$, we have the following inequality with the best constant factor $K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}$:

$$\begin{array}{rcl}J& :=& {\left\{\sum _n{[\mathrm{\Psi }(n)]}^{1-p}{\left(\sum _m\frac{a_m}{{\prod }_{k=1}^s[{\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}\right)}^p\right\}}^{\frac{1}{p}}\\ <& K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }},\end{array}$$

(27)

which is equivalent to (23).

Proof We set $b_n$ as follows:

$$b_n:={[\mathrm{\Psi }(n)]}^{1-p}{\left(\sum _m\frac{a_m}{{\prod }_{k=1}^s({\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s})}\right)}^{p-1}.$$

Then it follows that $J^p={\parallel b\parallel }_{q,\mathrm{\Psi }}^q$. If $J=0$, then (27) is trivially valid, since $0<{\parallel a\parallel }_{p,\mathrm{\Phi }}<\mathrm{\infty }$; if $J=\mathrm{\infty }$, then it is a contradiction since the right hand side of (27) is finite. Suppose that $0<J<\mathrm{\infty }$. Then by (23), we find

$${\parallel b\parallel }_{q,\mathrm{\Psi }}^q=J^p=I<K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}{\parallel b\parallel }_{q,\mathrm{\Psi }},$$

namely, ${\parallel b\parallel }_{q,\mathrm{\Psi }}^{q-1}=J<K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}$, and then (27) follows.

On the other hand, assuming that (27) is valid, by the Hölder inequality, we have

$$\begin{array}{rcl}I& =& \sum _n{(\mathrm{\Psi }(n))}^{\frac{-1}{q}}\left[\sum _m\frac{a_m}{{\prod }_{k=1}^s({\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s})}\right]\\ \times \left[{(\mathrm{\Psi }(n))}^{\frac{1}{q}}{b}_n\right]\le J{\parallel b\parallel }_{q,\mathrm{\Psi }}.\end{array}$$

(28)

Then by (27), we have (23). Hence (27) and (23) are equivalent.

By the equivalency, the constant factor $K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}$ in (27) is the best possible. Otherwise, we would reach a contradiction by (28) that the constant factor $K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}$ in (23) is not the best possible. □

For $p>1$, we define two real weight normal discrete spaces $l_{p,\phi }$ and $l_{q,\psi }$ as follows:

$$\begin{array}{c}{l}_{p,\phi }:=\{a=\{a_m\};{\parallel a\parallel }_{p,\mathrm{\Phi }}={\{\sum _m\mathrm{\Phi }(m)a_m^p\}}^{\frac{1}{p}}<\mathrm{\infty }\},\hfill \\ l_{q,\psi }:=\{b=\{b_n\};{\parallel b\parallel }_{q,\mathrm{\Psi }}={\{\sum _n\mathrm{\Psi }(n)b_n^q\}}^{\frac{1}{q}}<\mathrm{\infty }\}.\hfill \end{array}$$

With the assumptions of Theorem 2, in view of $J<K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}$, we have the following definition.

Definition 2 Define a multidimensional Hilbert-type operator $T:l_{p,\mathrm{\Phi }}\to l_{p,{\mathrm{\Psi }}^{1-p}}$ as follows: For $a\in l_{p,\mathrm{\Phi }}$, there exists an unique representation $Ta\in l_{p,{\mathrm{\Psi }}^{1-p}}$, satisfying for $n\in {\mathbf{N}}^{j_0}$,

$$(Ta)(n):=\sum _m\frac{a_m}{{\prod }_{k=1}^s[{\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}.$$

(29)

For $b\in l_{q,\mathrm{\Psi }}$, we define the following formal inner product of Ta and b as follows:

$$(Ta,b):=\sum _n\sum _m\frac{a_m{b}_n}{{\prod }_{k=1}^s[{\parallel ln(m+\tau )\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+\sigma )\parallel }_{\beta }^{\lambda /s}]}.$$

(30)

Then by Theorem 1 and Theorem 2, for $0<{\parallel a\parallel }_{p,\phi },{\parallel b\parallel }_{q,\psi }<\mathrm{\infty }$, we have the following equivalent inequalities:

$$(Ta,b)<K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}{\parallel b\parallel }_{q,\mathrm{\Psi }},$$

(31)

$${\parallel Ta\parallel }_{p,{\mathrm{\Psi }}^{1-p}}<K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,\mathrm{\Phi }}.$$

(32)

It follows that T is bounded since

$$\parallel T\parallel :=\underset{a(\ne \theta )\in l_{p,\mathrm{\Phi }}}{sup}\frac{{\parallel Ta\parallel }_{p,{\mathrm{\Psi }}^{1-p}}}{{\parallel a\parallel }_{p,\mathrm{\Phi }}}\le K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}.$$

(33)

Since the constant factor $K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}$ in (32) is the best possible, we have:

Corollary 1 With the assumptions of Theorem  2, T is defined by Definition  2, it follows that

$$\parallel T\parallel =K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}={\left[\frac{{\mathrm{\Gamma }}^{j_0}(\frac{1}{\beta })}{{\beta }^{j_0-1}\mathrm{\Gamma }(\frac{j_0}{\beta })}\right]}^{\frac{1}{p}}{\left[\frac{{\mathrm{\Gamma }}^{i_0}(\frac{1}{\alpha })}{{\alpha }^{i_0-1}\mathrm{\Gamma }(\frac{i_0}{\alpha })}\right]}^{\frac{1}{q}}{k}_s({\lambda }_1).$$

(34)

Remark 1 (i) Setting ${\mathrm{\Phi }}_1(m):={\prod }_{i=1}^{i_0}{(m_i+1)}^{p-1}{\parallel ln(m+1)\parallel }_{\alpha }^{p(i_0-{\lambda }_1)-i_0}$ ($m\in {\mathbf{N}}^{i_0}$) and ${\mathrm{\Psi }}_1(n):={\prod }_{j=1}^{j_0}{(n_j+1)}^{q-1}{\parallel ln(n+1)\parallel }_{\beta }^{q(j_0-{\lambda }_2)-j_0}$ ($n\in {\mathbf{N}}^{j_0}$), then putting $\tau =\sigma =1$ in (23) and (27), we have the following equivalent inequalities with the best constant factor $K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}$:

$$\sum _n\sum _m\frac{a_m{b}_n}{{\prod }_{k=1}^s[{\parallel ln(m+1)\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+1)\parallel }_{\beta }^{\lambda /s}]}<K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,{\mathrm{\Phi }}_1}{\parallel b\parallel }_{q,{\mathrm{\Psi }}_1},$$

(35)

$$\begin{array}{c}{\left\{\sum _n{[{\mathrm{\Psi }}_1(n)]}^{1-p}{\left(\sum _m\frac{a_m}{{\prod }_{k=1}^s[{\parallel ln(m+1)\parallel }_{\alpha }^{\lambda /s}+c_k{\parallel ln(n+1)\parallel }_{\beta }^{\lambda /s}]}\right)}^p\right\}}^{\frac{1}{p}}\hfill \\ <K_1^{\frac{1}{p}}{K}_2^{\frac{1}{q}}{\parallel a\parallel }_{p,{\mathrm{\Phi }}_1}.\hfill \end{array}$$

(36)

Hence, (23) and (27) are more accurate inequalities than (35) and (36).

• (ii) Putting $i_0=j_0=1$, $\lambda =s$, ${\varphi }_1(m):={(m+1)}^{p-1}{ln}^{p(1-{\lambda }_1)-1}(m+1)$ ($m\in \mathbf{N}$) and ${\psi }_1(n):={(n+1)}^{q-1}{ln}^{q(1-{\lambda }_2)-1}(n+1)$ ($n\in \mathbf{N}$), in (32), we have the following new inequality:

  $$\begin{array}{c}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{{\prod }_{k=1}^{s}ln(m+1){(n+1)}^{c_k}}\hfill \\ <\frac{\pi }{sin(\pi {\lambda }_1)}\sum _{k=1}^s\prod _{j=1(j\ne k)}^s\frac{c_k^{{\lambda }_1-1}}{c_j-c_k}{\parallel a\parallel }_{p,{\varphi }_1}{\parallel b\parallel }_{q,{\psi }_1}.\hfill \end{array}$$

  (37)

In particular, for $s=c_k=1$, ${\lambda }_1=\frac{1}{q}$, ${\lambda }_2=\frac{1}{p}$ in (37), we can deduce (4). Hence, (23) is an extension of (4).