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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.

References

  1. 1.

    Hardy, GH, Littlewood, JE, Pólya, G: Inequalities. Cambridge University Press, Cambridge (1934)

    Google Scholar 

  2. 2.

    Mitrinović, DS, Pečarić, JE, Fink, AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic, Boston (1991)

    Google Scholar 

  3. 3.

    Yang, BC: Hilbert-Type Integral Inequalities. Bentham Science Publishers Ltd., Sharjah (2009)

    Google Scholar 

  4. 4.

    Yang, BC: Discrete Hilbert-Type Inequalities. Bentham Science Publishers Ltd., Sharjah (2011)

    Google Scholar 

  5. 5.

    Yang, BC: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing (2009)

    Google Scholar 

  6. 6.

    Yang, BC: Two Types of Multiple Half-Discrete Hilbert-Type Inequalities. Lambert Academic Publishing, Saarbrücken (2012)

    Google Scholar 

  7. 7.

    Persson, LE, et al.: Commutators of Hardy operators in vanishing Morrey spaces. In: 9th International Conference on Mathematical Problems in Engineering, Aerospace and Sciences (ICNPAA 2012). AIP Conference Proceedings, vol. 1493, pp. 859-866 (2012). doi:10.1063/1.4765588

    Google Scholar 

  8. 8.

    Yang, BC: On Hilbert’s integral inequality. J. Math. Anal. Appl. 220, 778-785 (1998)

    MATH  MathSciNet  Article  Google Scholar 

  9. 9.

    Yang, BC, Brnetić, I, Krnić, M, Pečarić, JE: Generalization of Hilbert and Hardy-Hilbert integral inequalities. Math. Inequal. Appl. 8(2), 259-272 (2005)

    MATH  MathSciNet  Google Scholar 

  10. 10.

    Krnić, M, Pečarić, JE: Hilbert’s inequalities and their reverses. Publ. Math. (Debr.) 67(3-4), 315-331 (2005)

    MATH  Google Scholar 

  11. 11.

    Yang, BC, Rassias, TM: On the way of weight coefficient and research for Hilbert-type inequalities. Math. Inequal. Appl. 6(4), 625-658 (2003)

    MATH  MathSciNet  Google Scholar 

  12. 12.

    Yang, BC, Rassias, TM: On a Hilbert-type integral inequality in the subinterval and its operator expression. Banach J. Math. Anal. 4(2), 100-110 (2010)

    MATH  MathSciNet  Article  Google Scholar 

  13. 13.

    Azar, L: On some extensions of Hardy-Hilbert’s inequality and applications. J. Inequal. Appl. 2009, Article ID 546829 (2009)

    MathSciNet  Google Scholar 

  14. 14.

    Arpad, B, Choonghong, O: Best constant for certain multilinear integral operator. J. Inequal. Appl. 2006, Article ID 28582 (2006)

    Google Scholar 

  15. 15.

    Kuang, JC, Debnath, L: On Hilbert’s type inequalities on the weighted Orlicz spaces. Pac. J. Appl. Math. 1(1), 95-103 (2007)

    MATH  MathSciNet  Google Scholar 

  16. 16.

    Zhong, WY: The Hilbert-type integral inequality with a homogeneous kernel of Lambda-degree. J. Inequal. Appl. 2008, Article ID 917392 (2008)

    Article  Google Scholar 

  17. 17.

    Hong, Y: On Hardy-Hilbert integral inequalities with some parameters. J. Inequal. Pure Appl. Math. 6(4), 92 (2005)

    MathSciNet  Google Scholar 

  18. 18.

    Zhong, WY, Yang, BC: On multiple Hardy-Hilbert’s integral inequality with kernel. J. Inequal. Appl. 2007, Article ID 27962 (2007). doi:10.1155/2007/27962

    MathSciNet  Article  Google Scholar 

  19. 19.

    Yang, BC, Krnić, M: On the norm of a multi-dimensional Hilbert-type operator. Sarajevo J. Math. 7(20), 223-243 (2011)

    MathSciNet  Google Scholar 

  20. 20.

    Krnić, M, Pečarić, JE, Vuković, P: On some higher-dimensional Hilbert’s and Hardy-Hilbert’s type integral inequalities with parameters. Math. Inequal. Appl. 11, 701-716 (2008)

    MATH  MathSciNet  Google Scholar 

  21. 21.

    Krnić, M, Vuković, P: On a multidimensional version of the Hilbert-type inequality. Anal. Math. 38, 291-303 (2012)

    MATH  MathSciNet  Article  Google Scholar 

  22. 22.

    Rassias, TM, Yang, BC: On half-discrete Hilbert’s inequality. Appl. Math. Comput. 220, 75-93 (2013)

    MathSciNet  Article  Google Scholar 

  23. 23.

    Rassias, TM, Yang, BC: A multidimensional half-discrete Hilbert-type inequality and the Riemann zeta function. Appl. Math. Comput. 225, 263-277 (2013)

    MathSciNet  Article  Google Scholar 

  24. 24.

    Rassias, TM, Yang, BC: On a multidimensional half - discrete Hilbert - type inequality related to the hyperbolic cotangent function. Appl. Math. Comput. 242, 800-813 (2014)

    MathSciNet  Article  Google Scholar 

  25. 25.

    Rassias, TM, Yang, BC: On a multidimensional Hilbert-type integral inequality associated to the gamma function. Appl. Math. Comput. 249, 408-418 (2014)

    MathSciNet  Article  Google Scholar 

  26. 26.

    Li, YJ, He, B: On inequalities of Hilbert’s type. Bull. Aust. Math. Soc. 76(1), 1-13 (2007)

    MATH  Article  Google Scholar 

  27. 27.

    Yang, BC: On a more accurate multidimensional Hilbert-type inequality with parameters. Math. Inequal. Appl. 18(2), 429-441 (2015)

    MathSciNet  Google Scholar 

  28. 28.

    Yang, BC: An extension of a Hardy-Hilbert-type inequality. J. Guangdong Univ. Educ. 35(3), 1-8 (2015)

    Google Scholar 

  29. 29.

    Kuang, JC: Applied Inequalities. Shangdong Science Technic Press, Jinan (2004)

    Google Scholar 

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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.

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Correspondence to Bicheng Yang.

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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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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

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MSC

  • 26D15
  • 47A07

Keywords

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