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.

1 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. [1–7]).

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 [9–27].

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.

2 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 ($m_{0}=n_{0}=1$).

3 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)

4 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.