# On a Hardy-Hilbert-type inequality with parameters

## Abstract

By means of the way of weight coefficients and technique of real analysis, an extension of a Hardy-Hilbert-type inequality with parameters and a best possible constant factor is given. The equivalent forms, the operator expression with the norm, the reverses and some particular cases are also considered.

## Introduction

Suppose that $$p>1$$, $$\frac{1}{p}+\frac{1}{q}=1$$, $$f(x),g(y)\geq0$$, $$f\in L^{p}(\mathbf{R}_{+})$$, $$g\in L^{q}(\mathbf{R}_{+})$$, $$\|f\|_{p} =(\int_{0}^{\infty }f^{p}(x)\,dx)^{\frac{1}{p}}>0$$, $$\|g\|_{q}>0$$. We have the following Hardy-Hilbert’s integral inequality with the best possible constant factor $$\frac{\pi}{\sin(\pi/p)}$$ (cf. [1]):

$$\int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \frac{\pi }{\sin(\pi/p)}\|f\|_{p}\|g\|_{q}.$$
(1)

Assuming that $$a_{m},b_{n}\geq0$$,

$$a=\{a_{m}\}_{m=1}^{\infty}\in l^{p}=\Biggl\{ a;\|a\|_{p}=\Biggl(\sum_{m=1}^{\infty }|a_{m}|^{p} \Biggr)^{\frac{1}{p}}< \infty\Biggr\} ,$$

$$b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}$$, $$\|a\|_{p},\|b\|_{q}>0$$, we have the following Hardy-Hilbert’s inequality with the same best possible constant factor $$\frac{\pi}{\sin(\pi/p)}$$ (cf. [1]):

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\pi/p)}\|a \|_{p}\|b\|_{q}.$$
(2)

Hardy-Hilbert-type inequalities, specially (1) and (2), are basically important in mathematical analysis and its applications (cf. [17]).

If $$\mu_{i},\upsilon_{j}>0$$ ($$i,j\in\mathbf{N}$$),

$$U_{m}:=\sum_{i=1}^{m} \mu_{i},\qquad V_{n}:=\sum_{j=1}^{n} \upsilon_{j}\quad (m,n\in \mathbf{N}),$$
(3)

then we have the following inequality (cf. [1], Theorem 321, p.261):

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{\mu_{m}^{1/q}\upsilon _{n}^{1/p}a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\pi/p)}\|a\|_{p}\|b\|_{q}.$$
(4)

Replacing $$\mu_{m}^{1/q}a_{m}$$ and $$\upsilon_{n}^{1/p}b_{n}$$ by $$a_{m}$$ and $$b_{n}$$ in (4), respectively, we obtain the following equivalent form of (4):

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac {a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\frac{\pi}{p})} \Biggl( \sum_{m=1}^{\infty}\frac {a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty} \frac {b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}.$$
(5)

For $$\mu_{i}=\upsilon_{j}=1$$ ($$i,j\in\mathbf{N}$$), both (4) and (5) reduce to (2). We call (4) and (5) Hardy-Hilbert-type inequalities.

### Note

The authors of [1] (Theorem 321, p.261) did not prove that (4) is valid with the best possible constant factor.

In 1998, by introducing an independent parameter $$\lambda\in(0,1]$$, Yang [8] gave an extension of (1) for $$p=q=2$$. Following the methods of [8], Yang [5] gave some best extensions of (1) and (2) as follows.

If $$\lambda_{1},\lambda_{2}\in\mathbf{R}=(-\infty,\infty)$$, $$\lambda _{1}+\lambda_{2}=\lambda$$, $$k_{\lambda}(x,y)$$ is a nonnegative homogeneous function of degree −λ, with $$k(\lambda_{1})=\int_{0}^{\infty }k_{\lambda}(t,1)t^{\lambda_{1}-1}\,dt\in\mathbf{R}_{+}$$, $$\phi (x)=x^{p(1-\lambda_{1})-1}$$, $$\psi(x)=x^{q(1-\lambda _{2})-1}$$, $$f(x),g(y)\geq 0$$,

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

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

$$\int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda }(x,y)f(x)g(y) \,dx\,dy< k(\lambda _{1})\|f\|_{p,\phi}\|g\|_{q,\psi},$$
(6)

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 $$a_{m},b_{n}\geq0$$,

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

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

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}k_{\lambda }(m,n)a_{m}b_{n}< k( \lambda_{1})\|a\|_{p,\phi}\|b\|_{q,\psi},$$
(7)

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 (6) reduces to (1), while (7) reduces to (2). For $$0<\lambda_{1},\lambda _{2}\leq1$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, we set $$k_{\lambda }(x,y)=\frac{1}{(x+y)^{\lambda}}$$. Then, by (7), it follows that

$$\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac {a_{m}b_{n}}{(m+n)^{\lambda}}< B( \lambda_{1},\lambda_{2})\|a\|_{p,\phi}\|b \|_{q,\psi},$$
(8)

where the constant factor $$B(\lambda_{1},\lambda_{2})$$ is the best possible ($$B(u,v)$$ is the beta function). Some other results including multidimensional Hilbert-type inequalities are provided by [927].

In 2015, by adding a few conditions, Yang [28] gave an extension of (8) and (5) as follows:

\begin{aligned} &\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{(U_{m}+V_{n})^{\lambda}} \\ &\quad< B(\lambda_{1},\lambda_{2}) \Biggl( \sum _{m=1}^{\infty}\frac{U_{m}^{p(1-\lambda_{1})-1}a_{m}^{p}}{\mu_{m}^{p-1}} \Biggr) ^{\frac {1}{p}} \Biggl( \sum_{n=1}^{\infty}\frac{V_{n}^{q(1-\lambda_{2})-1}b_{n}^{q}}{ \upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}, \end{aligned}
(9)

where the constant factor $$B(\lambda_{1},\lambda_{2})$$ is the best possible. For $$\mu_{i}=\upsilon_{j}=1$$ ($$i,j\in\mathbf{N}$$), (9) reduces to (8); for $$\lambda=1$$, $$\lambda_{1}=\frac{1}{q}$$, $$\lambda _{2}=\frac{1}{p}$$, (9) reduces to (5).

In this paper, by using the way of weight coefficients and technique of real analysis, a Hardy-Hilbert-type inequality with parameters and a best possible constant factor is given, which is with the kernel $$\frac{(\min \{x,c_{1}y\})^{\alpha}}{(\max\{x,c_{1}y\})^{\lambda+\alpha}}$$ similar to (9). The extended inequalities, the equivalent forms, the operator expression with the norm, the reverses and some particular cases are also considered.

## Some lemmas

In the following, we agree on that $$\mu_{i},\upsilon_{j}>0$$ ($$i,j\in \mathbf{N}$$), $$U_{m}$$ and $$V_{n}$$ are defined by (3), $$p\neq0,1$$, $$\frac {1}{p}+\frac{1}{q}=1$$, $$a_{m},b_{n}\geq0$$ ($$m,n\in\mathbf{N}$$), $$\|a\|_{p,\Phi _{\lambda}}=(\sum_{m=1}^{\infty}\Phi_{\lambda}(m)a_{m}^{p})^{\frac {1}{p}}$$, $$\|b\|_{q,\Psi_{\lambda}}=(\sum_{n=1}^{\infty}\Psi_{\lambda }(n)b_{n}^{q})^{\frac{1}{q}}$$, where

$$\Phi_{\lambda}(m):=\frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu _{m}^{p-1}},\qquad \Psi _{\lambda}(n):= \frac{V_{n}^{q(1-\lambda_{2})-1}}{\upsilon_{n}^{q-1}}\quad(m,n\in\mathbf{N}).$$

### Lemma 1

If $$g(t)$$ (>0) is decreasing in $$\mathbf{R}_{+}$$ and strictly decreasing in $$[n_{0},\infty)\subset\mathbf{R}_{+}$$ ($$n_{0}\in \mathbf{N}$$), satisfying $$\int_{0}^{\infty}g(t)\,dt\in\mathbf{R}_{+}$$, then we have

$$\int_{1}^{\infty}g(t)\,dt< \sum _{n=1}^{\infty}g(n)< \int_{0}^{\infty}g(t) \,dt.$$
(10)

### Proof

Since, by the assumption, we have

\begin{aligned}& \int_{n}^{n+1}g(t)\,dt \leq g(n)\leq\int _{n-1}^{n}g(t)\,dt\quad (n=1,\ldots,n_{0}),\\& \int_{n_{0}+1}^{n_{0}+2}g(t)\,dt < g(n_{0}+1)< \int_{n_{0}}^{n_{0}+1}g(t)\,dt, \end{aligned}

it follows that

$$0< \int_{1}^{n_{0}+2}g(t)\,dt< \sum _{n=1}^{n_{0}+1}g(n)< \sum_{n=1}^{n_{0}+1} \int_{n-1}^{n}g(t)\,dt=\int _{0}^{n_{0}+1}g(t)\,dt< \infty.$$

By the same way, we still have

$$0< \int_{n_{0}+2}^{\infty}g(t)\,dt\leq\sum _{n=n_{0}+2}^{\infty}g(n)\leq \int_{n_{0}+1}^{\infty}g(t) \,dt< \infty.$$

Hence, making plus for the above two inequalities, we have (10). □

### Example 1

For $$s\in\mathbf{N}$$, $$0< c_{1}\leq\cdots\leq c_{s}<\infty$$, $$\lambda_{1},\lambda_{2}>-\alpha$$, $$\lambda_{1}+\lambda _{2}=\lambda$$, we set

$$k_{\lambda}(x,y):=\prod_{k=1}^{s} \frac{(\min\{x,c_{k}y\})^{\frac {\alpha}{s}}}{(\max\{x,c_{k}y\})^{\frac{\lambda+\alpha}{s}}}\quad\bigl((x,y)\in\mathbf {R}_{+}^{2}= \mathbf{R}_{+}\times\mathbf{R}_{+}\bigr).$$

(a) We find

\begin{aligned} k_{s}(\lambda_{1}) :=&\int_{0}^{\infty}k_{\lambda}(1,u)t^{\lambda _{2}-1} \,du\overset{u=1/t}{=}\int_{0}^{\infty}k_{\lambda}(t,1)t^{\lambda _{1}-1} \,dt\\ =&\int_{0}^{\infty}\prod _{k=1}^{s}\frac{(\min\{t,c_{k}\})^{\frac {\alpha}{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda _{1}-1}\,dt\\ =&\int_{0}^{c_{1}}\prod _{k=1}^{s}\frac{(\min\{t,c_{k}\})^{\frac {\alpha}{s}}t^{\lambda_{1}-1}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}\,dt+\int _{c_{s}}^{\infty}\prod_{k=1}^{s} \frac{(\min\{t,c_{k}\})^{\frac{ \alpha}{s}}t^{\lambda_{1}-1}}{(\max\{t,c_{k}\})^{\frac{\lambda +\alpha}{s}}}\,dt \\ &{}+\sum_{i=1}^{s-1}\int _{c_{i}}^{c_{i+1}}\prod_{k=1}^{s} \frac{(\min \{t,c_{k}\})^{\frac{\alpha}{s}}t^{\lambda_{1}-1}}{(\max\{t,c_{k}\})^{ \frac{\lambda+\alpha}{s}}}\,dt \\ =&\prod_{k=1}^{s}\frac{1}{c_{k}^{(\lambda+\alpha)/s}}\int _{0}^{c_{1}}t^{\lambda_{1}+\alpha-1}\,dt+\prod _{k=1}^{s}c_{k}^{\alpha /s}\int _{c_{s}}^{\infty}t^{-\lambda_{2}-\alpha-1}\,dt \\ &{}+\sum_{i=1}^{s-1}\int _{c_{i}}^{c_{i+1}}\prod_{k=1}^{i} \frac {c_{k}^{\frac{\alpha}{s}}}{t^{\frac{\lambda+\alpha}{s}}}\prod_{k=i+1}^{s} \frac {t^{\frac{\alpha}{s}}}{c_{k}^{\frac{\lambda+\alpha}{s}}}t^{\lambda_{1}-1}\,dt \\ =&\frac{c_{1}^{\lambda_{1}+\alpha}}{\lambda_{1}+\alpha}\frac{1}{\prod_{k=1}^{s}c_{k}^{\frac{\lambda+\alpha}{s}}}+\frac{1}{(\lambda _{2}+\alpha)c_{s}^{\lambda_{2}+\alpha}}\prod _{k=1}^{s}c_{k}^{\frac {\alpha }{s}} \\ &{}+\sum_{i=1}^{s-1}\frac{\prod_{k=1}^{i}c_{k}^{\frac{\alpha}{s}}}{\prod_{k=i+1}^{s}c_{k}^{\frac{\lambda+\alpha}{s}}}\int_{c_{i}}^{c_{i+1}}t^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac {2i}{s})\alpha-1}\,dt. \end{aligned}

If $$\lambda_{1}-\frac{i\lambda}{s}+(1-\frac{2i}{s})\alpha\neq0$$, then

$$\int_{c_{i}}^{c_{i+1}}t^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac {2i}{s})\alpha-1}\,dt= \frac{c_{i+1}^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac {2i}{s})\alpha}-c_{i}^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac {2i}{s})\alpha}}{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac{2i}{s})\alpha};$$

if there exists $$i_{0}\in\{1,\ldots,s-1\}$$ such that $$\lambda_{1}-\frac{i_{0}\lambda}{s}+(1-\frac{2i_{0}}{s})\alpha=0$$, then we find

$$\int_{c_{i_{0}}}^{c_{i_{0}+1}}t^{\lambda_{1}-\frac{i_{0}\lambda }{s}+(1-\frac{2i_{0}}{s})\alpha-1}\,dt=\ln\biggl( \frac{c_{i_{0}+1}}{c_{i_{0}}}\biggr)=\lim_{i\rightarrow i_{0}}\int _{c_{i}}^{c_{i+1}}t^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac{2i}{s})\alpha-1}\,dt,$$

and we still indicate $$\ln(\frac{c_{i_{0}+1}}{c_{i_{0}}})$$ by the following formal expression:

$$\frac{c_{i_{0}+1}^{\lambda_{1}-\frac{i_{0}\lambda}{s}+(1-\frac {2i_{0}}{s})\alpha}-c_{i_{0}}^{\lambda_{1}-\frac{i_{0}\lambda}{s}+(1-\frac {2i_{0}}{s})\alpha}}{\lambda_{1}-\frac{i_{0}\lambda}{s}+(1-\frac {2i_{0}}{s})\alpha}.$$

Hence, we may set

\begin{aligned} k_{s}(\lambda_{1}) =&\frac{c_{1}^{\lambda_{1}+\alpha}}{\lambda _{1}+\alpha} \frac{1}{\prod_{k=1}^{s}c_{k}^{\frac{\lambda+\alpha }{s}}}+\frac{1}{(\lambda_{2}+\alpha)c_{s}^{\lambda_{2}+\alpha}}\prod _{k=1}^{s}c_{k}^{\frac{\alpha}{s}} \\ &{}+\sum_{i=1}^{s-1} \biggl[ \frac{c_{i+1}^{\lambda_{1}-\frac{i\lambda }{s}+(1-\frac{2i}{s})\alpha}-c_{i}^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac {2i}{s})\alpha}}{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac{2i}{s})\alpha }\frac{\prod_{k=1}^{i}c_{k}^{\frac{\alpha}{s}}}{\prod_{k=i+1}^{s}c_{k}^{\frac{ \lambda+\alpha}{s}}} \biggr] . \end{aligned}
(11)

In particular, (i) for $$s=1$$ (or $$c_{s}=\cdots=c_{1}$$), we have $$k_{\lambda }(x,y)=\frac{(\min\{x,c_{1}y\})^{\alpha}}{(\max\{x,c_{1}y\})^{\lambda +\alpha}}$$ and

$$k_{1}(\lambda_{1})=\frac{\lambda+2\alpha}{(\lambda_{1}+\alpha )(\lambda _{2}+\alpha)}\frac{1}{c_{1}^{\lambda_{2}}};$$
(12)

(ii) for $$s=2$$, we have $$k_{\lambda}(x,y)=\frac{(\min\{x,c_{1}y\}\min \{x,c_{2}y\})^{\alpha/2}}{(\max\{x,c_{1}y\}\max\{x,c_{2}y\} )^{(\lambda +\alpha)/2}}$$ and

$$k_{2}(\lambda_{1})= \biggl( \frac{c_{1}}{c_{2}} \biggr) ^{\frac{\alpha }{2}} \biggl[ \frac{c_{1}^{\lambda_{1}-\frac{\lambda}{2}}}{(\lambda _{1}+\alpha )c_{2}^{\frac{\lambda}{2}}}+\frac{1}{(\lambda_{2}+\alpha )c_{2}^{\lambda _{2}}}+ \frac{c_{2}^{\lambda_{1}-\frac{\lambda}{2}}-c_{1}^{\lambda _{1}-\frac{\lambda}{2}}}{(\lambda_{1}-\frac{\lambda}{2})c_{2}^{\frac {\lambda}{2}}} \biggr] ;$$
(13)

(iii) for $$\alpha=0$$, we have $$\lambda_{1},\lambda_{2}>0$$, $$k_{\lambda }(x,y)=\frac{1}{\prod_{k=1}^{s}(\max\{x,c_{k}y\})^{\frac{\lambda }{s}}}$$ and

\begin{aligned} k_{s}(\lambda_{1}) =&\widetilde{k}_{s}( \lambda_{1}):=\frac {c_{1}^{\lambda _{1}}}{\lambda_{1}}\frac{1}{\prod_{k=1}^{s}c_{k}^{\frac{\lambda}{s}}}+ \frac{1}{\lambda_{2}c_{s}^{\lambda_{2}}} +\sum_{i=1}^{s-1}\frac{c_{i+1}^{\lambda_{1}-\frac{i}{s}\lambda }-c_{i}^{\lambda_{1}-\frac{i}{s}\lambda}}{\lambda_{1}-\frac {i}{s}\lambda}\frac{1}{\prod_{k=i+1}^{s}c_{k}^{\frac{\lambda}{s}}}; \end{aligned}
(14)

(iv) for $$\alpha=-\lambda$$, we have $$\lambda<\lambda_{1},\lambda_{2}<0$$, $$k_{\lambda}(x,y)=\frac{1}{\prod_{k=1}^{s}(\min\{x,c_{k}y\})^{\frac{\lambda}{s}}}$$ and

\begin{aligned} k_{s}(\lambda_{1}) =&\widehat{k}_{s}( \lambda_{1}):=\frac {c_{1}^{-\lambda _{2}}}{(-\lambda_{2})}+\frac{1}{(-\lambda_{1})c_{s}^{-\lambda_{1}}}\prod _{k=1}^{s}c_{k}^{\frac{-\lambda}{s}} +\sum_{i=1}^{s-1} \Biggl( \frac{c_{i+1}^{\lambda_{1}-\frac {s-i}{s}\lambda }-c_{i}^{\lambda_{1}-\frac{s-i}{s}\lambda}}{\lambda_{1}-\frac{s-i}{s} \lambda}\prod_{k=1}^{i}c_{k}^{\frac{-\lambda}{s}} \Biggr) ; \end{aligned}
(15)

(v) for $$\lambda=0$$, we have $$\lambda_{2}=-\lambda_{1}$$, $$|\lambda _{1}|<\alpha$$ ($$\alpha>0$$),

$$k_{0}(x,y)=\prod_{k=1}^{s} \biggl( \frac{\min\{x,c_{k}y\}}{\max\{ x,c_{k}y\}}\biggr) ^{\frac{\alpha}{s}},$$

and

\begin{aligned} k_{s}(\lambda_{1}) =&k_{s}^{(0)}( \lambda_{1}):=\frac{c_{1}^{\lambda _{1}+\alpha}}{a+\lambda_{1}}\frac{1}{\prod_{k=1}^{s}c_{k}^{\frac {\alpha}{s}}}+\frac{c_{s}^{\lambda_{1}-\alpha}}{a-\lambda_{1}} \prod_{k=1}^{s}c_{k}^{\frac{\alpha}{s}} \\ &+\sum_{i=1}^{s-1} \biggl[ \frac{c_{i+1}^{\lambda_{1}+(1-\frac {2i}{s})\alpha }-c_{i}^{\lambda_{1}+(1-\frac{2i}{s})\alpha}}{\lambda_{1}+(1-\frac {2i}{s})\alpha}\frac{\prod_{k=1}^{i}c_{k}^{\frac{\alpha}{s}}}{\prod_{k=i+1}^{s}c_{k}^{\frac{\alpha}{s}}} \biggr] . \end{aligned}
(16)

(b) Since we find

\begin{aligned} k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}&=\frac{1}{y^{1-\lambda _{2}}}\prod _{k=1}^{s}\frac{(\min\{c_{k}^{-1}x,y\})^{\frac{\alpha }{s}}}{c_{k}^{\frac{\lambda}{s}}(\max\{c_{k}^{-1}x,y\})^{\frac{\lambda+\alpha}{s}}}\\ &=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1}{y^{1-\lambda_{2}-\alpha}}\prod_{k=1}^{s}\frac{1}{c_{k}^{\frac {\lambda}{s}}(c_{k}^{-1}x)^{\frac{\lambda+\alpha}{s}}},& 0< y\leq c_{s}^{-1}x,\\ \frac{1}{y^{1+\lambda_{1}+\alpha-\frac{i}{s}(\lambda+2\alpha)}}\frac {\prod_{k=i+1}^{s}(c_{k}^{-1}x)^{\frac{\alpha}{s}}}{\prod_{k=1}^{s}c_{k}^{\frac{\lambda}{s}}\prod_{k=1}^{i}(c_{k}^{-1}x)^{\frac{\lambda+\alpha }{s}}},& c_{i+1}^{-1}x< y\leq c_{i}^{-1}x \ (i=1,\ldots,s-1), \\ \frac{1}{y^{1+\lambda_{1}+\alpha}}\prod_{k=1}^{s}\frac{(c_{k}^{-1}x)^{ \frac{\alpha}{s}}}{c_{k}^{\frac{\lambda}{s}}(y)^{\frac{\lambda +\alpha}{s}}},& c_{1}^{-1}x< y< \infty,\end{array}\displaystyle \right . \end{aligned}

then for $$\lambda_{2}\leq1-\alpha$$ ($$\lambda_{1}>-\alpha$$), $$k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}$$ is decreasing for $$y>0$$ and strictly decreasing for the large enough variable y. By the same way, since

\begin{aligned} k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}&=\frac{1}{x^{1-\lambda _{1}}}\prod _{k=1}^{s}\frac{(\min\{x,c_{k}y\})^{\frac{\alpha}{s}}}{(\max \{x,c_{k}y\})^{\frac{\lambda+\alpha}{s}}}\\ &=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1}{x^{1-\lambda_{1}-\alpha}}\prod_{k=1}^{s}\frac {1}{(c_{k}y)^{\frac{\lambda+\alpha}{s}}},& 0< x\leq c_{1}y, \\ \frac{1}{x^{1-\lambda_{1}-\alpha+\frac{i}{s}(\lambda+2\alpha)}}\frac {\prod_{k=1}^{i}(c_{k}y)^{\frac{\alpha}{s}}}{\prod_{k=i+1}^{s}(c_{k}y)^{ \frac{\lambda+\alpha}{s}}},& c_{i}y< x\leq c_{i+1}y\ (i=1,\ldots,s-1), \\ \frac{1}{x^{1+\lambda_{2}+\alpha}}\prod_{k=1}^{s}(c_{k}y)^{\frac {\alpha}{s}},& c_{s}y< x< \infty,\end{array}\displaystyle \right . \end{aligned}

then for $$\lambda_{1}\leq1-\alpha$$ ($$\lambda_{2}>-\alpha$$), $$k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}$$ is decreasing for $$x>0$$ and strictly decreasing for the large enough variable x.

In view of (a) and (b), for $$-\alpha<\lambda_{1},\lambda_{2}\leq 1-\alpha$$, $$\lambda_{1}+\lambda_{2}=\lambda$$, $$k_{\lambda}(x,y)\frac {1}{y^{1-\lambda _{2}}}$$ ($$k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}$$) is decreasing for $$y>0$$ ($$x>0$$) and strictly decreasing for the large enough variable $$y^{{}} (x)$$ satisfying $$k_{s}(\lambda_{1})\in\mathbf{R}_{+}$$.

### Lemma 2

If $$s\in\mathbf{N}$$, $$0< c_{1}\leq\cdots \leq c_{s}$$, $$-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha$$, $$\lambda _{1}+\lambda_{2}=\lambda$$, $$k_{s}(\lambda_{1})$$ is indicated by (11), define the following weight coefficients:

\begin{aligned}& \omega(\lambda_{2},m) :=\sum_{n=1}^{\infty} \prod_{k=1}^{s}\frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}} \frac{U_{m}^{\lambda_{1}}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}},\quad m\in\mathbf{N}, \end{aligned}
(17)
\begin{aligned}& \varpi(\lambda_{1},n) :=\sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}} \frac{V_{n}^{\lambda_{2}}\mu_{m}}{U_{m}^{1-\lambda_{1}}},\quad n\in\mathbf{N}. \end{aligned}
(18)

Then we have the following inequalities:

\begin{aligned}& \omega(\lambda_{2},m) < k_{s}(\lambda_{1}) \quad(- \alpha< \lambda_{2}\leq 1-\alpha,\lambda_{1}>-\alpha;m\in \mathbf{N}), \end{aligned}
(19)
\begin{aligned}& \varpi(\lambda_{1},n) < k_{s}(\lambda_{1}) \quad (-\alpha< \lambda_{1}\leq 1-\alpha,\lambda_{2}>-\alpha;n\in \mathbf{N}). \end{aligned}
(20)

### Proof

We set $$\mu(t):=\mu_{m}$$, $$t\in(m-1,m]$$ ($$m\in\mathbf{N}$$); $$\upsilon(t):=\upsilon_{n}$$, $$t\in(n-1,n]$$ ($$n\in\mathbf{N}$$),

$$U(x):=\int_{0}^{x}\mu(t)\,dt\quad(x\geq0),\qquad V(y):= \int_{0}^{y}\upsilon (t)\,dt\quad(y\geq 0).$$
(21)

Then, by (3), it follows that $$U(m)=U_{m}$$, $$V(n)=V_{n}$$ ($$m,n\in \mathbf{N}$$). For $$x\in(m-1,m]$$, $$U^{\prime}(x)=\mu(x)=\mu_{m}$$ ($$m\in\mathbf{N}$$); for $$y\in(n-1,n]$$, $$V^{\prime}(y)=\upsilon(y)=\upsilon_{n}$$ ($$n\in\mathbf {N}$$). Since $$V(y)$$ is strictly increasing in $$(n-1,n]$$, $$-\alpha<\lambda _{2}\leq1-\alpha$$, $$\lambda_{1}>-\alpha$$, in view of Lemma 1 and Example 1, we find

\begin{aligned} \omega(\lambda_{2},m) =&\sum_{n=1}^{\infty} \int_{n-1}^{n}\prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}\frac{U_{m}^{\lambda _{1}}}{V_{n}^{1-\lambda_{2}}}V^{\prime}(y)\,dy \\ < &\sum_{n=1}^{\infty}\int_{n-1}^{n} \prod_{k=1}^{s}\frac{(\min \{U_{m},c_{k}V(y)\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V(y)\} )^{\frac{\lambda+\alpha}{s}}} \frac{U_{m}^{\lambda_{1}}V^{\prime }(y)}{V^{1-\lambda _{2}}(y)}\,dy. \end{aligned}

Setting $$t=\frac{V(y)}{U_{m}}$$, we obtain $$V^{\prime}(y)\,dy=U_{m}\,dt$$ and

\begin{aligned} \omega(\lambda_{2},m) < &\sum_{n=1}^{\infty} \int_{\frac {V(n-1)}{U_{m}}}^{\frac{V(n)}{U_{m}}}\prod_{k=1}^{s} \frac{(\min\{1,c_{k}t\})^{\frac {\alpha}{s}}}{(\max\{1,c_{k}t\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda _{2}-1}\,dt \\ =&\int_{0}^{\frac{V(\infty)}{U_{m}}}\prod _{k=1}^{s}\frac{(\min \{1,c_{k}t\})^{\frac{\alpha}{s}}}{(\max\{1,c_{k}t\})^{\frac{\lambda +\alpha}{s}}}t^{\lambda_{2}-1}\,dt \\ \leq&\int_{0}^{\infty}\prod _{k=1}^{s}\frac{(\min\{1,c_{k}t\})^{\frac {\alpha}{s}}}{(\max\{1,c_{k}t\})^{\frac{\lambda+\alpha }{s}}}t^{\lambda _{2}-1} \,dt=k_{s}(\lambda_{1}). \end{aligned}

Since $$U(x)$$ is strictly increasing in $$(m-1,m]$$, $$-\alpha<\lambda _{1}\leq 1-\alpha$$, $$\lambda_{2}>-\alpha$$, by the same way, we have

\begin{aligned} \varpi(\lambda_{1},n) =&\sum_{m=1}^{\infty} \int_{m-1}^{m}\prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}\frac{V_{n}^{\lambda _{2}}U^{\prime}(x)}{U_{m}^{1-\lambda_{1}}}\,dx \\ < &\sum_{m=1}^{\infty}\int_{m-1}^{m} \prod_{k=1}^{s}\frac{(\min \{U(x),c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U(x),c_{k}V_{n}\} )^{\frac{\lambda+\alpha}{s}}} \frac{V_{n}^{\lambda_{2}}U^{\prime }(x)}{U^{1-\lambda _{1}}(x)}\,dx \\ \overset{t=U(x)/V_{n}}{=}&\sum_{m=1}^{\infty} \int_{\frac {U(m-1)}{V_{n}}}^{\frac{U(m)}{V_{n}}}\prod_{k=1}^{s} \frac{(\min\{t,c_{k}\})^{\frac {\alpha}{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda_{1}-1}\,dt \\ =&\int_{0}^{\frac{U(\infty)}{V_{n}}}\prod _{k=1}^{s}\frac{(\min \{t,c_{k}\})^{\frac{\alpha}{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda +\alpha }{s}}}t^{\lambda_{1}-1}\,dt\leq k_{s}(\lambda_{1}). \end{aligned}

Hence, we have (19) and (20). □

### Lemma 3

If $$s\in\mathbf{N}$$, $$0< c_{1}\leq\cdots\leq c_{s}$$, $$-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha$$, $$\lambda _{1}+\lambda_{2}=\lambda$$, $$k_{s}(\lambda_{1})$$ is indicated by (11), $$m_{0},n_{0}\in\mathbf{N}$$, $$\mu_{m}\geq\mu_{m+1}$$ ($$m\in \{m_{0},m_{0}+1,\ldots\}$$), $$\upsilon_{n}\geq\upsilon_{n+1}$$ ($$n\in \{n_{0},n_{0}+1,\ldots\}$$), $$U(\infty)=V(\infty)=\infty$$, then (i) for $$m,n\in\mathbf{N}$$, we have

\begin{aligned}& k_{s}(\lambda_{1}) \bigl(1-\theta(\lambda_{2},m) \bigr) < \omega(\lambda _{2},m) \quad(-\alpha< \lambda_{2}\leq1- \alpha,\lambda_{1}>-\alpha), \end{aligned}
(22)
\begin{aligned}& k_{s}(\lambda_{1}) \bigl(1-\vartheta( \lambda_{1},n)\bigr) < \varpi(\lambda _{1},n)\quad (-\alpha< \lambda_{1}\leq1-\alpha,\lambda_{2}>-\alpha), \end{aligned}
(23)

where

\begin{aligned}& \theta(\lambda_{2},m) :=\frac{1}{k_{s}(\lambda_{1})}\int_{0}^{\frac {U_{m_{0}}}{V_{n}}} \prod_{k=1}^{s}\frac{(\min\{t,c_{k}\})^{\frac{\alpha }{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda_{1}-1} \,dt =O\biggl(\frac{1}{U_{m}^{\lambda_{2}+\alpha}}\biggr)\in(0,1), \\& \vartheta(\lambda_{1},n) :=\frac{1}{k_{s}(\lambda_{1})}\int _{0}^{\frac{U_{m_{0}}}{V_{n}}}\prod_{k=1}^{s} \frac{(\min\{t,c_{k}\})^{\frac{\alpha }{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda_{1}-1}\,dt =O\biggl(\frac{1}{V_{n}^{\lambda_{1}+\alpha}}\biggr)\in(0,1); \end{aligned}

(ii) for any $$b>0$$, we have

\begin{aligned}& \sum_{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+b}}= \frac{1}{b} \biggl( \frac {1}{U_{m_{0}}^{b}}+bO(1) \biggr) , \end{aligned}
(24)
\begin{aligned}& \sum_{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{1+b}}= \frac{1}{b} \biggl( \frac{1}{V_{n_{0}}^{b}}+b\widetilde{O}(1) \biggr) . \end{aligned}
(25)

### Proof

Since $$\upsilon_{n}\geq\upsilon_{n+1}$$ ($$n\geq n_{0}$$), $$-\alpha<\lambda_{2}\leq1-\alpha$$, $$\lambda_{1}>-\alpha$$ and $$V(\infty)=\infty$$, by Lemma 1, we have

\begin{aligned} \omega(\lambda_{2},m) \geq&\sum_{n=n_{0}}^{\infty} \prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \frac{U_{m}^{\lambda _{1}}}{V_{n}^{1-\lambda_{2}}}\upsilon_{n+1} \\ =&\sum_{n=n_{0}}^{\infty}\int_{n}^{n+1} \prod_{k=1}^{s}\frac{(\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\} )^{\frac{\lambda+\alpha}{s}}} \frac{U_{m}^{\lambda_{1}}V^{\prime}(y)}{V_{n}^{1-\lambda_{2}}}\,dy \\ >&\sum_{n=n_{0}}^{\infty}\int_{n}^{n+1} \prod_{k=1}^{s}\frac{(\{U_{m},c_{k}V(y)\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V(y)\} )^{\frac{\lambda+\alpha}{s}}} \frac{U_{m}^{\lambda_{1}}V^{\prime }(y)}{V^{1-\lambda _{2}}(y)}\,dy \\ =&\sum_{n=n_{0}}^{\infty}\int_{\frac{V(n)}{U_{m}}}^{\frac {V(n+1)}{U_{m}}} \prod_{k=1}^{s}\frac{(\min\{1,c_{k}t\})^{\frac{\alpha}{s}}}{(\max \{1,c_{k}t\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda_{2}-1} \,dt \\ =&\int_{\frac{V_{n_{0}}}{U_{m}}}^{\infty}\prod _{k=1}^{s}\frac{(\min \{1,c_{k}t\})^{\frac{\alpha}{s}}t^{\lambda_{2}-1}}{(\max\{1,c_{k}t\} )^{\frac{\lambda+\alpha}{s}}}\,dt=k_{s}( \lambda_{1}) \bigl(1-\theta(\lambda _{2},m)\bigr). \end{aligned}

For $$U_{m}>c_{s}V_{n_{0}}$$, we obtain $$c_{k}t\leq c_{s}t\leq c_{s}\frac{ V_{n_{0}}}{U_{m}}<1$$ ($$t\in(0,\frac{V_{n_{0}}}{U_{m}}]$$; $$k=1,\ldots,s$$) and

$$\theta(\lambda_{2},m)=\frac{\prod_{k=1}^{s}c_{k}}{k_{s}(\lambda_{1})} \int _{0}^{\frac{V_{n_{0}}}{U_{m}}}t^{\lambda_{2}+\alpha-1}\,dt= \frac{\prod_{k=1}^{s}c_{k}}{(\lambda_{2}+\alpha)k_{s}(\lambda_{1})} \biggl( \frac{V_{n_{0}}}{U_{m}} \biggr) ^{\lambda_{2}+\alpha},$$

and then $$\theta(\lambda_{2},m)=O(\frac{1}{U_{m}^{\lambda_{2}+\alpha}})$$. Hence we have (22).

By the same way, since $$\mu_{m}\geq\mu_{m+1}$$ ($$m\geq m_{0}$$), $$-\alpha <\lambda_{1}\leq1-\alpha$$, $$\lambda_{2}>-\alpha$$ and $$U(\infty )=\infty$$, we have

\begin{aligned} \varpi(\lambda_{1},n) \geq&\sum_{m=m_{0}}^{\infty} \prod_{k=1}^{s}\frac {(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \frac{V_{n}^{\lambda _{2}}\mu_{m+1}}{U_{m}^{1-\lambda_{1}}}\\ =&\sum_{m=m_{0}}^{\infty}\int_{m}^{m+1} \prod_{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}} \frac{V_{n}^{\lambda_{2}}U^{\prime}(x)}{U_{m}^{1-\lambda_{1}}}\,dx \\ >&\sum_{m=m_{0}}^{\infty}\int_{m}^{m+1} \prod_{k=1}^{s}\frac{(\min \{U(x),c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U(x),c_{k}V_{n}\} )^{\frac{\lambda+\alpha}{s}}} \frac{V_{n}^{\lambda_{2}}U^{\prime }(x)}{U^{1-\lambda _{1}}(x)}\,dx \\ \overset{t=U(x)/V_{n}}{=}&\sum_{m=m_{0}}^{\infty} \int_{\frac {U(m)}{V_{n}}}^{\frac{U(m+1)}{V_{n}}}\prod_{k=1}^{s} \frac{(\min\{t,c_{k}\})^{\frac{\alpha}{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda _{1}-1}\,dt \\ =&\int_{\frac{U_{m_{0}}}{V_{n}}}^{\infty}\prod _{k=1}^{s}\frac{(\min \{t,c_{k}\})^{\frac{\alpha}{s}}t^{\lambda_{1}-1}}{(\max\{t,c_{k}\})^{ \frac{\lambda+\alpha}{s}}}\,dt=k_{s}( \lambda_{1}) \bigl(1-\vartheta(\lambda _{1},n)\bigr). \end{aligned}

For $$V_{n}>c_{1}^{-1}U_{m_{0}}$$, we obtain $$t\leq\frac {U_{m_{0}}}{V_{n}}< c_{1}\leq c_{k}$$ ($$t\in(0,\frac{U_{m_{0}}}{V_{n}}]$$; $$k=1,\ldots,s$$) and

$$\vartheta(\lambda_{1},n)=\frac{\int_{0}^{\frac{U_{m_{0}}}{V_{n}}}t^{\lambda_{1}+\alpha-1}\,dt}{k_{s}(\lambda_{1})\prod_{k=1}^{s}c_{k}^{ \frac{\lambda+\alpha}{s}}}=\frac{(\lambda_{1}+\alpha)^{-1}}{k_{s}(\lambda_{1})\prod_{k=1}^{s}c_{k}^{\frac{\lambda+\alpha }{s}}} \biggl( \frac{U_{m_{0}}}{V_{n}} \biggr) ^{\lambda_{1}+\alpha}.$$

Hence, we have (23).

For $$b>0$$, we find

\begin{aligned}& \begin{aligned}[b] \sum_{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+b}}&=\sum _{m=1}^{m_{0}}\frac {\mu _{m}}{U_{m}^{1+b}}+\sum _{m=m_{0}+1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+b}}\\ &=\sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+b}}+ \sum_{m=m_{0}+1}^{\infty }\int_{m-1}^{m} \frac{U^{\prime}(x)}{U_{m}^{1+b}}\,dx \\ &< \sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+b}}+ \sum_{m=m_{0}+1}^{\infty }\int_{m-1}^{m} \frac{U^{\prime}(x)}{U^{1+b}(x)}\,dx\\ &=\sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+b}}+ \int_{m_{0}}^{\infty} \frac{dU(x)}{U^{1+b}(x)}=\sum _{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+b}}+ \frac{1}{bU_{m_{0}}^{b}} \\ &=\frac{1}{b} \Biggl( \frac{1}{U_{m_{0}}^{b}}+b\sum _{m=1}^{m_{0}}\frac {\mu _{m}}{U_{m}^{1+b}} \Biggr) , \end{aligned}\\& \begin{aligned}[b] \sum_{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+b}} &\geq \sum_{m=m_{0}}^{\infty}\frac{\mu_{m+1}}{U_{m}^{1+b}}=\sum _{m=m_{0}}^{ \infty}\int_{m}^{m+1} \frac{U^{\prime}(x)}{U_{m}^{1+b}}\,dx \\ &>\sum_{m=m_{0}}^{\infty}\int_{m}^{m+1} \frac{U^{\prime }(x)\,dx}{U^{1+b}(x)}=\int_{m_{0}}^{\infty} \frac{dU(x)}{U^{1+b}(x)}=\frac{1}{bU_{m_{0}}^{b}}. \end{aligned} \end{aligned}

Hence we have (24). By the same way, we still have (25). □

### Note

For example, $$\mu_{m}=\frac{1}{m^{\sigma}}$$, $$\upsilon _{n}=\frac{1}{n^{\sigma}}$$ ($$0\leq\sigma\leq1$$; $$m,n\in\mathbf{N}$$) satisfy the conditions of Lemma 3 ($$m_{0}=n_{0}=1$$).

## Main results and operator expressions

### Theorem 1

If $$s\in\mathbf{N}$$, $$0< c_{1}\leq\cdots \leq c_{s}$$, $$-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha$$, $$\lambda _{1}+\lambda_{2}=\lambda$$, $$k_{s}(\lambda_{1})$$ is indicated by (11), then for $$p>1$$, $$0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi_{\lambda }}<\infty$$, we have the following equivalent inequalities:

\begin{aligned}& I:=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}b_{n}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}< k_{s}(\lambda _{1})\|a \|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}
(26)
\begin{aligned}& J:= \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac{(\min\{ U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda +\alpha }{s}}} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< k_{s}(\lambda _{1}) \|a\|_{p,\Phi _{\lambda}}. \end{aligned}
(27)

In particular, for $$s=1$$ (or $$c_{s}=\cdots=c_{1}$$), we have the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(\min \{U_{m},c_{1}V_{n}\})^{\alpha}a_{m}b_{n}}{(\max \{U_{m},c_{1}V_{n}\})^{\lambda+\alpha}}< k_{1}( \lambda _{1})\|a\|_{p,\Phi _{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}
(28)
\begin{aligned}& \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \frac{(\min\{U_{m},c_{1}V_{n}\})^{\alpha }a_{m}}{(\max\{U_{m},c_{1}V_{n}\})^{\lambda+\alpha}} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< k_{1}( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}, \end{aligned}
(29)

where $$k_{1}(\lambda_{1})$$ is indicated by (12).

### Proof

By Hölder’s inequality with weight (cf. [29]), we have

\begin{aligned} &\Biggl[ \sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{ U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}a_{m} \Biggr] ^{p} \\ &\quad= \Biggl[ \sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}} \biggl( \frac{U_{m}^{\frac{1-\lambda_{1}}{q} }a_{m}}{V_{n}^{\frac{1-\lambda_{2}}{p}}\mu_{m}^{\frac{1}{q}}} \biggr) \biggl( \frac{V_{n}^{\frac{1-\lambda_{2}}{p}}\mu_{m}^{\frac {1}{q}}}{U_{m}^{\frac{1-\lambda_{1}}{q}}} \biggr) \Biggr] ^{p} \\ &\quad\leq\sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{ U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}} \biggl( \frac{U_{m}^{(1-\lambda_{1})p/q}}{V_{n}^{1-\lambda_{2}}\mu _{m}^{p/q}}a_{m}^{p} \biggr) \\ &\qquad{}\times \Biggl[ \sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}} \frac{V_{n}^{(1-\lambda_{2})(q-1)}\mu _{m}}{U_{m}^{1-\lambda_{1}}} \Biggr] ^{p-1} \\ &\quad=\frac{V_{n}^{1-p\lambda_{2}}}{(\varpi(\lambda _{1},n))^{1-p}\upsilon _{n}}\sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}\frac{U_{m}^{(1-\lambda_{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda _{2}}\mu _{m}^{p-1}}a_{m}^{p}. \end{aligned}
(30)

In view of (20), we find

\begin{aligned} J \leq&\bigl(k_{s}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}\frac{U_{m}^{(1-\lambda_{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{s}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty}\prod_{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}\frac{U_{m}^{(1-\lambda_{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{s}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=1}^{\infty}\omega (\lambda_{2},m) \frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}
(31)

Then, by (19), we have (27).

By Hölder’s inequality (cf. [29]), we have

\begin{aligned} I =&\sum_{n=1}^{\infty} \Biggl[ \frac{\upsilon_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\lambda_{2}}}\sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}}{(\max\{ U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr] \biggl( \frac{V_{n}^{\frac {1}{p}-\lambda _{2}}}{\upsilon_{n}^{\frac{1}{p}}}b_{n} \biggr) \\ \leq&J\|b\|_{q,\Psi_{\lambda}}. \end{aligned}
(32)

Then, by (27), we have (26).

On the other hand, assuming that (26) is valid, we set

$$b_{n}:=\frac{\upsilon_{n}}{V_{n}^{1-p\lambda_{2}}} \Biggl[ \sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}a_{m}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}} \Biggr] ^{p-1},\quad n\in\mathbf{N}.$$

Then we find $$J^{p}=\|b\|_{q,\Psi_{\lambda}}^{q}$$. If $$J=0$$, then (27) is trivially valid; if $$J=\infty$$, then, by (31) and (19), it is impossible. Suppose that $$0< J<\infty$$. By (26), it follows that

\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q} =J^{p}=I< k_{s}( \lambda _{1})\|a\|_{p,\Phi _{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \\& \|b\|_{q,\Psi_{\lambda}}^{q-1} =J< k_{s}(\lambda_{1}) \|a\|_{p,\Phi _{\lambda}}, \end{aligned}

and then (27) follows, which is equivalent to (26). □

### Theorem 2

With the assumptions of Theorem  1, if $$m_{0},n_{0}\in \mathbf{N}$$, $$\mu_{m}\geq\mu_{m+1}$$ ($$m\in\{m_{0},m_{0}+1,\ldots\}$$), $$\upsilon_{n}\geq\upsilon_{n+1}$$ ($$n\in\{n_{0},n_{0}+1,\ldots \}$$), $$U(\infty)=V(\infty)=\infty$$, then the constant factor $$k_{s}(\lambda_{1})$$ in (26) and (27) is the best possible.

### Proof

For $$\varepsilon\in(0,p(\lambda_{1}+\alpha))$$, we set $$\widetilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon}{p}$$ ($${\in} (-\alpha ,1-\alpha)$$), $$\widetilde{\lambda}_{2}=\lambda_{2}+\frac{\varepsilon }{p}$$ ($${>}-\alpha$$), and $$\widetilde{a}=\{\widetilde{a}_{m}\}_{m=1}^{\infty}$$, $$\widetilde{b}=\{\widetilde{b}_{n}\}_{n=1}^{\infty}$$,

$$\widetilde{a}_{m}:=U_{m}^{\widetilde{\lambda}_{1}-1}\mu _{m}=U_{m}^{\lambda _{1}-\frac{\varepsilon}{p}-1}\mu_{m},\qquad\widetilde {b}_{n}=V_{n}^{\widetilde{\lambda}_{2}-\varepsilon-1}\upsilon_{n}=V_{n}^{\lambda_{2}-\frac{\varepsilon}{q}-1} \upsilon_{n}.$$
(33)

Then, by (24), (25) and (23), we have

\begin{aligned}& \begin{aligned}[b] \|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\Psi _{\lambda }}&= \Biggl( \sum _{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}}\\ &=\frac{1}{\varepsilon} \biggl( \frac{1}{U_{m_{0}}^{\varepsilon }}+\varepsilon O(1) \biggr) ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I} &:=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \widetilde{a}_{m}\widetilde{b}_{n} \\ &=\sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}}\frac{V_{n}^{\widetilde{\lambda}_{2}}\mu _{m}}{U_{m}^{1-\widetilde{\lambda}_{1}}} \Biggr] \frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\sum_{n=1}^{\infty}\varpi(\widetilde{ \lambda}_{1},n)\frac {\upsilon_{n}}{V_{n}^{\varepsilon+1}}\geq k_{s}(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty} \bigl(1-\vartheta(\widetilde{\lambda}_{1},n)\bigr)\frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=k_{s}(\widetilde{\lambda}_{1}) \Biggl( \sum _{n=1}^{\infty}\frac {\upsilon _{n}}{V_{n}^{\varepsilon+1}}-\sum _{n=1}^{\infty}O\biggl(\frac{\upsilon _{n}}{V_{n}^{\frac{\varepsilon}{q}+\lambda_{1}+\alpha+1}}\biggr) \Biggr) \\ &=\frac{1}{\varepsilon}k_{s}(\widetilde{\lambda}_{1}) \biggl[ \frac{1}{ V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl(\widetilde{O}(1)-O(1)\bigr) \biggr] . \end{aligned} \end{aligned}

If there exists a positive constant $$K\leq k_{s}(\lambda_{1})$$ such that (26) is valid when replacing $$k_{s}(\lambda_{1})$$ with K, then, in particular, we have $$\varepsilon\widetilde{I}<\varepsilon K\|\widetilde {a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\Psi_{\lambda}}$$, namely

\begin{aligned} &k_{s}(\widetilde{\lambda}_{1}) \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon }}+\varepsilon\bigl(\widetilde{O}(1)-O(1)\bigr) \biggr] < K \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{ \frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon \widetilde{O}(1) \biggr) ^{\frac{1}{q}}. \end{aligned}

It follows that $$k_{s}(\lambda_{1})\leq K(\varepsilon\rightarrow0^{+})$$. Hence, $$K=k_{s}(\lambda_{1})$$ is the best possible constant factor of (26).

The constant factor $$k_{s}(\lambda_{1})$$ in (27) is still the best possible. Otherwise, we would reach a contradiction by (32) that the constant factor in (26) is not the best possible. □

### Remark 1

Inequality (26) is an extension of Hardy-Hilbert-type inequality (28) with parameters and a best possible constant factor.

For $$p>1$$, we find $$\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon_{n}}{V_{n}^{1-p\lambda_{2}}}$$ and define the following normed spaces:

\begin{aligned}& l_{p,\Phi_{\lambda}} :=\bigl\{ a=\{a_{m}\}_{m=1}^{\infty}; \|a\|_{p,\Phi _{\lambda}}< \infty\bigr\} , \\& l_{q,\Psi_{\lambda}} :=\bigl\{ b=\{b_{n}\}_{n=1}^{\infty}; \|b\|_{q,\Psi _{\lambda}}< \infty\bigr\} , \\& l_{p,\Psi_{\lambda}^{1-p}} :=\bigl\{ c=\{c_{n}\}_{n=1}^{\infty }; \|c\|_{p,\Psi _{\lambda}^{1-p}}< \infty\bigr\} . \end{aligned}

Assuming that $$a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}$$, setting

$$c=\{c_{n}\}_{n=1}^{\infty},\qquad c_{n}:=\sum _{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}a_{m},\quad n\in \mathbf{N},$$

we can rewrite (27) as follows:

$$\|c\|_{p,\Psi_{\lambda}^{1-p}}< k_{s}(\lambda_{1})\|a \|_{p,\Phi _{\lambda }}< \infty,$$

namely $$c\in l_{p,\Psi_{\lambda}^{1-p}}$$.

### Definition 1

Define a Hardy-Hilbert-type operator $$T:l_{p,\Phi _{\lambda}}\rightarrow l_{p,\Psi_{\lambda}^{1-p}}$$ as follows: For any $$a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}$$, there exists a unique representation $$Ta=c\in l_{p,\Psi_{\lambda}^{1-p}}$$. Define the formal inner product of Ta and $$b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\Psi _{\lambda}}$$ as follows:

$$(Ta,b):=\sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}a_{m} \Biggr] b_{n}.$$
(34)

Then we can rewrite (26) and (27) as follows:

\begin{aligned}& (Ta,b) < k_{s}(\lambda_{1})\|a\|_{p,\Phi_{\lambda}}\|b \|_{q,\Psi _{\lambda}}, \end{aligned}
(35)
\begin{aligned}& \|Ta\|_{p,\Psi_{\lambda}^{1-p}} < k_{s}(\lambda_{1})\|a \|_{p,\Phi _{\lambda}}. \end{aligned}
(36)

Define the norm of operator T as follows:

$$\|T\|:=\sup_{a(\neq\theta)\in l_{p,\Phi_{\lambda}}}\frac {\|Ta\|_{p,\Psi _{\lambda}^{1-p}}}{\|a\|_{p,\Phi_{\lambda}}}.$$
(37)

Then, by (36), we find $$\|T\|\leq k_{s}(\lambda_{1})$$. Since by Theorem 2 the constant factor in (36) is the best possible, we have

$$\|T\|=k_{s}(\lambda_{1}).$$
(38)

## Some reverses

In the following, we also set

\begin{aligned}& \widetilde{\Phi}_{\lambda}(m) :=\bigl(1-\theta(\lambda_{2},m) \bigr)\frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu_{m}^{p-1}}, \\& \widetilde{\Psi}_{\lambda}(n) :=\bigl(1-\vartheta(\lambda_{1},n) \bigr)\frac{ V_{n}^{q(1-\lambda_{2})-1}}{\upsilon_{n}^{q-1}}\quad(m,n\in\mathbf{N}). \end{aligned}

For $$0< p<1$$ or $$p<0$$, we still use the formal symbols of $$\|a\|_{p,\Phi _{\lambda}}$$, $$\|b\|_{q,\Psi_{\lambda}}$$, $$\|a\|_{p,\widetilde{\Phi} _{\lambda}}$$ and $$\|b\|_{q,\widetilde{\Psi}_{\lambda}}$$.

### Theorem 3

If $$s\in\mathbf{N}$$, $$0< c_{1}\leq\cdots\leq c_{s}$$, $$-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha$$, $$\lambda _{1}+\lambda_{2}=\lambda$$, $$k_{s}(\lambda_{1})$$ is indicated by (11), $$m_{0},n_{0}\in\mathbf{N}$$, $$\mu_{m}\geq\mu_{m+1}$$ ($$m\in \{m_{0},m_{0}+1,\ldots\}$$), $$\upsilon_{n}\geq\upsilon_{n+1}$$ ($$n\in \{n_{0},n_{0}+1,\ldots\}$$), $$U(\infty)=V(\infty)=\infty$$, then for $$0< p<1$$, $$0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi_{\lambda }}<\infty$$, we have the following equivalent inequalities with the best possible constant factor $$k_{s}(\lambda_{1})$$:

\begin{aligned}& \begin{aligned}[b] I &=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}b_{n}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}\\ &>k_{s}(\lambda _{1})\|a \|_{p,\widetilde{\Phi}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned} \end{aligned}
(39)
\begin{aligned}& \begin{aligned}[b] J &= \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}}{(\max\{ U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\ &>k_{s}(\lambda_{1})\|a\|_{p,\widetilde{\Phi}_{\lambda}}. \end{aligned} \end{aligned}
(40)

### Proof

By the reverse Hölder’s inequality (cf. [29]) and (20), we have the reverses of (30), (31) and (32). Then, by (22), we have (40). By (40) and the reverse of (32), we have (39).

On the other hand, assuming that (39) is valid, we set $$b_{n}$$ as in Theorem 1. Then we find $$J^{p}=\|b\|_{q,\Psi_{\lambda}}^{q}$$. If $$J=\infty$$, then (40) is trivially valid; if $$J=0$$, then, by reverse of (31) and (22), it is impossible. Suppose that $$0< J<\infty$$. By (39), it follows that

\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q} =J^{p}=I>k_{s}( \lambda_{1})\|a\|_{p,\widetilde{\Phi}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \\& \|b\|_{q,\Psi_{\lambda}}^{q-1} =J>k_{s}(\lambda _{1})\|a\|_{p,\widetilde{\Phi}_{\lambda}}, \end{aligned}

and then (40) follows, which is equivalent to (39).

For $$\varepsilon\in(0,p(\lambda_{1}+\alpha))$$, we set $$\widetilde{\lambda}_{1}$$, $$\widetilde{\lambda}_{2}$$, $$\widetilde{a}_{m}$$ and $$\widetilde{b}_{n}$$ as (33). Then, by (24), (25) and (20), we find

\begin{aligned}& \begin{aligned}[b] \|a\|_{p,\widetilde{\Phi}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}&= \Biggl[ \sum _{m=1}^{\infty}\bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\mu_{m}}{U_{m}^{1+\varepsilon}} \Biggr] ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}}\\ &= \Biggl( \sum_{m=1}^{\infty} \frac{\mu_{m}}{U_{m}^{1+\varepsilon}}-\sum_{m=1}^{\infty}O \biggl(\frac{\mu_{m}}{U_{m}^{1+\lambda_{2}+\alpha +\varepsilon}}\biggr) \Biggr) ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}} \\ &=\frac{1}{\varepsilon} \biggl[ \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon \bigl(O(1)-O_{1}(1)\bigr) \biggr] ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I}&=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}\widetilde{a}_{m}\widetilde{b}_{n}\\ &=\sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}}\frac{V_{n}^{\widetilde{\lambda}_{2}}\mu _{m}}{U_{m}^{1-\widetilde{\lambda}_{1}}} \Biggr] \frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\sum_{n=1}^{\infty}\varpi(\widetilde{ \lambda}_{1},n)\frac {\upsilon_{n}}{V_{n}^{\varepsilon+1}}\leq k_{s}(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\frac{1}{\varepsilon}k_{s}(\widetilde{\lambda}_{1}) \biggl( \frac{1}{ V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) . \end{aligned} \end{aligned}

If there exists a constant $$K\geq k_{s}(\lambda_{1})$$ such that (39) is valid when replacing $$k_{s}(\lambda_{1})$$ with K, then, in particular, we have $$\varepsilon\widetilde{I}>\varepsilon K\|\widetilde{a}\|_{p,\widetilde{\Phi}_{\lambda}}\|\widetilde{b}\|_{q,\Psi_{\lambda}}$$, namely

\begin{aligned} &k_{s}(\widetilde{\lambda}_{1}) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon }}+\varepsilon\widetilde{O}(1) \biggr) >K \biggl[ \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon \bigl(O(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon }}+\varepsilon \widetilde{O}(1) \biggr) ^{\frac{1}{q}}. \end{aligned}

It follows that $$k_{s}(\lambda_{1})\geq K$$ ($$\varepsilon\rightarrow0^{+}$$). Hence, $$K=k_{s}(\lambda_{1})$$ is the best possible constant factor of (39).

The constant factor $$k_{s}(\lambda_{1})$$ in (40) is still the best possible. Otherwise, we would reach a contradiction by the reverse of (32) that the constant factor in (39) is not the best possible. □

### Theorem 4

With the assumptions of Theorem  3, if $$p<0$$, then we have the following equivalent inequalities with the best possible constant factor $$k_{s}(\lambda_{1})$$:

\begin{aligned}& I =\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}b_{n}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}>k_{s}(\lambda _{1})\|a \|_{p,\Phi_{\lambda}}\|b\|_{q,\widetilde{\Psi}_{\lambda}}, \end{aligned}
(41)
\begin{aligned}& \begin{aligned}[b] J_{1} &:= \Biggl\{ \sum_{n=1}^{\infty} \frac{V_{n}^{p\lambda _{2}-1}\upsilon _{n}}{(1-\vartheta(\lambda_{1},n))^{p-1}} \Biggl[ \sum_{m=1}^{\infty } \prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha }{s}}a_{m}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \\ &>k_{s}(\lambda_{1})\|a\|_{p,\Phi_{\lambda}}. \end{aligned} \end{aligned}
(42)

### Proof

By the reverse Hölder’s inequality with weight (cf. [29]), since $$p<0$$, by (23), we have

\begin{aligned} & \Biggl[ \sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}}{(\max\{ U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr] ^{p} \\ &\quad= \Biggl[ \sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}} \biggl( \frac{U_{m}^{(1-\lambda_{1})/q}}{V_{n}^{(1-\lambda_{2})/p}\mu_{m}^{1/q}}a_{m} \biggr) \biggl( \frac{V_{n}^{(1-\lambda_{2})/p}\mu_{m}^{1/q}}{U_{m}^{(1-\lambda _{1})/q}} \biggr) \Biggr] ^{p} \\ &\quad\leq\sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{ U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}\frac{U_{m}^{(1-\lambda_{1})p/q}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p/q}} a_{m}^{p} \\ &\qquad{}\times \Biggl[ \sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}} \frac{V_{n}^{(1-\lambda_{2})(q-1)}\mu _{m}}{U_{m}^{1-\lambda_{1}}} \Biggr] ^{p-1} \\ &\quad=\frac{V_{n}^{1-p\lambda_{2}}}{(\varpi(\lambda_{1},n))^{1-p}}\sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}} \frac{U_{m}^{(1-\lambda_{1})(p-1)}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p-1}}a_{m}^{p} \\ &\quad\leq\frac{(k_{s}(\lambda_{1}))^{p-1}V_{n}^{1-p\lambda_{2}}}{(1-\vartheta(\lambda_{1},n))^{1-p}\upsilon_{n}}\sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha }{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \frac {U_{m}^{(1-\lambda _{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu_{m}^{p-1}}a_{m}^{p}, \\ &J_{1} \geq\bigl(k_{s}(\lambda_{1}) \bigr)^{\frac{1}{q}} \Biggl\{ \sum_{n=1}^{\infty } \sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}\frac{U_{m}^{(1-\lambda_{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu _{m}^{p-1}}a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &\hphantom{J_{1}}=\bigl(k_{s}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl\{ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty}\prod_{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}\frac{U_{m}^{(1-\lambda_{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu _{m}^{p-1}}a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &\hphantom{J_{1}}=\bigl(k_{s}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl\{ \sum_{m=1}^{\infty}\omega (\lambda_{2},m) \frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu_{m}^{p-1}}a_{m}^{p} \Biggr\} ^{\frac{1}{p}}. \end{aligned}
(43)

Then, by (19), we have (44).

By the reverse Hölder’s inequality (cf. [29]), we have

\begin{aligned} I&=\sum_{n=1}^{\infty}\frac{V_{n}^{\lambda_{2}-\frac{1}{p}}\upsilon _{n}^{1/p}}{(1-\vartheta(\lambda_{1},n))^{1/q}} \Biggl[ \sum_{m=1}^{\infty }\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha }{s}}a_{m}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr]\biggl[ \bigl(1-\vartheta(\lambda_{1},n)\bigr)^{\frac{1}{q}} \frac {V_{n}^{\frac{1}{p}-\lambda_{2}}}{\upsilon_{n}^{1/p}}b_{n} \biggr] \\ & \geq J_{1}\|b \|_{q,\widetilde{\Psi}_{\lambda}}. \end{aligned}
(44)

Then, by (42), we have (41).

On the other hand, assuming that (41) is valid, we set $$b_{n}$$ as follows:

$$b_{n}:=\frac{V_{n}^{p\lambda_{2}-1}\upsilon_{n}}{(1-\vartheta(\lambda _{1},n))^{p-1}} \Biggl[ \sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}}{(\max\{ U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr] ^{p-1},\quad n\in\mathbf{N}.$$

Then we find $$J_{1}^{p}=\|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q}$$. If $$J_{1}=\infty$$, then (42) is trivially valid; if $$J_{1}=0$$, then by (43) and (19) it is impossible. Suppose that $$0< J_{1}<\infty$$. By (41), it follows that

\begin{aligned}& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q} =J_{1}^{p}=I>k_{s}( \lambda _{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\widetilde{\Psi}_{\lambda}},\\& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q-1} =J_{1}>k_{s}( \lambda _{1})\|a\|_{p,\Phi_{\lambda}}, \end{aligned}

and then (42) follows, which is equivalent to (41).

For $$\varepsilon\in(0,q(\lambda_{2}+\alpha))$$, we set $$\widetilde{\lambda}_{1}=\lambda_{1}+\frac{\varepsilon}{q}$$ ($${>}-\alpha$$), $$\widetilde{ \lambda}_{2}=\lambda_{2}-\frac{\varepsilon}{q}$$ ($${\in}(-\alpha ,1-\alpha)$$), and

$$\widetilde{a}_{m}:=U_{m}^{\widetilde{\lambda}_{1}-1-\varepsilon}\mu _{m}=U_{m}^{\lambda_{1}-\frac{\varepsilon}{p}-1}\mu_{m}, \qquad\widetilde{b} _{n}=V_{n}^{\widetilde{\lambda}_{2}-1}\upsilon_{n}=V_{n}^{\lambda _{2}-\frac{\varepsilon}{q}-1} \upsilon_{n}.$$

Then, by (24), (25) and (19), we have

\begin{aligned}& \begin{aligned}[b] \|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\widetilde {\Psi}_{\lambda}}&= \Biggl( \sum _{m=1}^{\infty}\frac{\mu _{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty} \bigl(1-\vartheta(\lambda _{1},n)\bigr)\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr] ^{\frac{1}{q}} \\ &= \Biggl( \sum_{m=1}^{\infty} \frac{\mu_{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}}-\sum _{n=1}^{\infty}O\biggl(\frac{\upsilon_{n}}{V_{n}^{1+\lambda_{1}+\alpha+\varepsilon}}\biggr) \Biggr) ^{\frac{1}{q}} \\ &=\frac{1}{\varepsilon} \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{p}} \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl( \widetilde{O}(1)-O_{1}(1)\bigr) \biggr] ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I} &=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod _{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \widetilde{a}_{m}\widetilde{b}_{n} \\ &=\sum_{m=1}^{\infty} \Biggl[ \sum _{n=1}^{\infty}\prod_{k=1}^{s} \frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}}\frac{U_{m}^{\widetilde{\lambda }_{1}}\upsilon _{n}}{V_{n}^{1-\widetilde{\lambda}_{2}}} \Biggr] \frac{\mu_{m}}{U_{m}^{1+\varepsilon}} \\ &=\sum_{m=1}^{\infty}\omega(\widetilde{ \lambda}_{2},m)\frac{\mu _{m}}{U_{m}^{1+\varepsilon}}\leq k_{s}(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty} \frac{\mu_{m}}{U_{m}^{1+\varepsilon}} \\ &=\frac{1}{\varepsilon}k_{s}(\widetilde{\lambda}_{1}) \biggl( \frac{1}{ U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) . \end{aligned} \end{aligned}

If there exists a constant $$K\geq k_{s}(\lambda_{1})$$ such that (41) is valid when replacing $$k_{s}(\lambda_{1})$$ with K, then, in particular, we have $$\varepsilon\widetilde{I}>\varepsilon K\|\widetilde {a}\|_{p,\Phi _{\lambda}}\|\widetilde{b}\|_{q,\widetilde{\Psi}_{\lambda}}$$, namely

\begin{aligned} &k_{s}(\widetilde{\lambda}_{1}) \biggl( \frac{1}{U_{m_{0}}^{\varepsilon }}+\varepsilon O(1) \biggr) \\ &\quad>K \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{ \frac{1}{p}} \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl(\widetilde{O}(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{q}}. \end{aligned}

It follows that $$k_{s}(\lambda_{1})\geq K$$ ($$\varepsilon\rightarrow0^{+}$$). Hence, $$K=k_{s}(\lambda_{1})$$ is the best possible constant factor of (41).

The constant factor $$k_{s}(\lambda_{1})$$ in (42) is still the best possible. Otherwise, we would reach a contradiction by (44) that the constant factor in (41) is not the best possible. □

### Remark 2

(i) For $$\alpha=0$$, $$0<\lambda _{1},\lambda_{2}\leq1$$ in (26) and (27), we have the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\prod_{k=1}^{s}(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda }{s}}}< \widetilde{k}_{s}( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}
(45)
\begin{aligned}& \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \frac{a_{m}}{\prod_{k=1}^{s}(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda}{s}}} \Biggr] ^{p} \Biggr\} ^{\frac {1}{p}}< \widetilde{k}_{s}(\lambda_{1})\|a\|_{p,\Phi_{\lambda}}, \end{aligned}
(46)

where $$\widetilde{k}_{s}(\lambda_{1})$$ is indicated by (14);

(ii) for $$\alpha=-\lambda$$, $$-1\leq\lambda_{1},\lambda_{2}<0$$ in (26) and (27), we have the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\prod_{k=1}^{s}(\min\{U_{m},c_{k}V_{n}\})^{\frac{\lambda }{s}}}< \widehat{k}_{s}( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}
(47)
\begin{aligned}& \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1-p\lambda _{2}}} \Biggl[ \sum_{m=1}^{\infty} \frac{a_{m}}{\prod_{k=1}^{s}(\min \{U_{m},c_{k}V_{n}\})^{\frac{\lambda}{s}}} \Biggr] ^{p} \Biggr\} ^{\frac {1}{p}}< \widehat{k}_{s}(\lambda_{1})\|a\|_{p,\Phi_{\lambda}}, \end{aligned}
(48)

where $$\widehat{k}_{s}(\lambda_{1})$$ is indicated by (15);

(iii) for $$\lambda=0$$, $$\lambda_{2}=-\lambda_{1}$$, in (26) and (27), we have the following equivalent inequalities:

\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod_{k=1}^{s} \biggl( \frac{\min \{U_{m},c_{k}V_{n}\}}{\max\{U_{m},c_{k}V_{n}\}} \biggr) ^{\frac{\alpha }{s}}a_{m}b_{n}< k_{s}^{(0)}( \lambda_{1})\|a\|_{p,\Phi_{\lambda }}\|b\|_{q,\Psi _{\lambda}}, \end{aligned}
(49)
\begin{aligned}& \Biggl\{ \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1+p\lambda _{1}}} \Biggl[ \sum_{m=1}^{\infty} \prod_{k=1}^{s} \biggl( \frac{\min \{U_{m},c_{k}V_{n}\}}{\max\{U_{m},c_{k}V_{n}\}} \biggr) ^{\frac{\alpha }{s}}a_{m} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< k_{s}^{(0)}(\lambda _{1})\|a \|_{p,\Phi_{\lambda}}, \end{aligned}
(50)

where $$k_{s}^{(0)}(\lambda_{1})$$ is indicated by (16) ($$|\lambda _{1}|<\alpha$$, $$0<\alpha\leq\frac{1}{2}$$; $$|\lambda_{1}|<1-\alpha$$, $$\frac {1}{2}<\alpha\leq1$$).

By Theorem 2, the constant factors in the above inequalities are all the best possible. We still can obtain some particular reverse inequalities with the best possible constant factors by Theorem 3 and Theorem 4.

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## Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 61370186), and the Science and Technology Planning Project of Guangzhou (No. 2014J4100032, No. 201510010203). We are grateful for their help.

## Author information

Authors

### Corresponding author

Correspondence to Bicheng Yang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. QC participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.

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Reprints and Permissions

Yang, B., Chen, Q. On a Hardy-Hilbert-type inequality with parameters. J Inequal Appl 2015, 339 (2015). https://doi.org/10.1186/s13660-015-0861-7

• Accepted:

• Published:

• DOI: https://doi.org/10.1186/s13660-015-0861-7

• 26D15
• 47A07

### Keywords

• Hardy-Hilbert-type inequality
• weight coefficient
• equivalent form
• reverse
• operator