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]): Assuming that \(a_{m},b_{n}\geq0\),
\(b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}\), \(\|a\|_{p},\|b\|_{q}>0\), we have the following Hardy-Hilbert’s inequality with the same best possible constant factor \(\frac{\pi}{\sin(\pi/p)}\) (cf. [1]): Hardy-Hilbert-type inequalities, specially (1) and (2), are basically important in mathematical analysis and its applications (cf. [1‐7]).
$$ \int_{0}^{\infty}\int_{0}^{\infty} \frac{f(x)g(y)}{x+y}\,dx\,dy< \frac{\pi }{\sin(\pi/p)}\|f\|_{p}\|g\|_{q}. $$
(1)
$$ a=\{a_{m}\}_{m=1}^{\infty}\in l^{p}=\Biggl\{ a;\|a\|_{p}=\Biggl(\sum_{m=1}^{\infty }|a_{m}|^{p} \Biggr)^{\frac{1}{p}}< \infty\Biggr\} , $$
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{m+n}< \frac {\pi}{\sin(\pi/p)}\|a \|_{p}\|b\|_{q}. $$
(2)
If \(\mu_{i},\upsilon_{j}>0\) (\(i,j\in\mathbf{N}\)), then we have the following inequality (cf. [1], Theorem 321, p.261): 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): 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.
$$ U_{m}:=\sum_{i=1}^{m} \mu_{i},\qquad V_{n}:=\sum_{j=1}^{n} \upsilon_{j}\quad (m,n\in \mathbf{N}), $$
(3)
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{\mu_{m}^{1/q}\upsilon _{n}^{1/p}a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\pi/p)}\|a\|_{p}\|b\|_{q}. $$
(4)
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac {a_{m}b_{n}}{U_{m}+V_{n}}< \frac{\pi}{\sin(\frac{\pi}{p})} \Biggl( \sum_{m=1}^{\infty}\frac {a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty} \frac {b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}. $$
(5)
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Note
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 \(k(\lambda_{1})=\int_{0}^{\infty }k_{\lambda}(t,1)t^{\lambda_{1}-1}\,dt\in\mathbf{R}_{+}\), \(\phi (x)=x^{p(1-\lambda_{1})-1}\), \(\psi(x)=x^{q(1-\lambda _{2})-1}\), \(f(x),g(y)\geq 0\),
\(g\in L_{q,\psi}(\mathbf{R}_{+})\), \(\|f\|_{p,\phi},\|g\|_{q,\psi}>0\), then 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 \(a_{m},b_{n}\geq0\),
\(b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\psi}\), \(\|a\|_{p,\phi },\|b\|_{q,\psi }>0\), we have where the constant factor \(k(\lambda_{1})\) is still the best possible.
$$ f\in L_{p,\phi}(\mathbf{R}_{+})= \biggl\{ f;\|f \|_{p,\phi }:=\biggl(\int_{0}^{\infty} \phi(x)\bigl|f(x)\bigr|^{p}\,dx\biggr)^{\frac{1}{p}}< \infty \biggr\} , $$
$$ \int_{0}^{\infty}\int_{0}^{\infty}k_{\lambda }(x,y)f(x)g(y) \,dx\,dy< k(\lambda _{1})\|f\|_{p,\phi}\|g\|_{q,\psi}, $$
(6)
$$ a\in l_{p,\phi}= \Biggl\{ a;\|a\|_{p,\phi}:=\Biggl(\sum _{n=1}^{\infty}\phi (n)|a_{n}|^{p} \Biggr)^{\frac{1}{p}}< \infty \Biggr\} , $$
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}k_{\lambda }(m,n)a_{m}b_{n}< k( \lambda_{1})\|a\|_{p,\phi}\|b\|_{q,\psi}, $$
(7)
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 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].
$$ \sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac {a_{m}b_{n}}{(m+n)^{\lambda}}< B( \lambda_{1},\lambda_{2})\|a\|_{p,\phi}\|b \|_{q,\psi}, $$
(8)
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In 2015, by adding a few conditions, Yang [28] gave an extension of (8) and (5) as follows: 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).
$$\begin{aligned} &\sum_{m=1}^{\infty}\sum _{n=1}^{\infty}\frac{a_{m}b_{n}}{(U_{m}+V_{n})^{\lambda}} \\ &\quad< B(\lambda_{1},\lambda_{2}) \Biggl( \sum _{m=1}^{\infty}\frac{U_{m}^{p(1-\lambda_{1})-1}a_{m}^{p}}{\mu_{m}^{p-1}} \Biggr) ^{\frac {1}{p}} \Biggl( \sum_{n=1}^{\infty}\frac{V_{n}^{q(1-\lambda_{2})-1}b_{n}^{q}}{ \upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}, \end{aligned}$$
(9)
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 it follows that By the same way, we still have Hence, making plus for the above two inequalities, we have (10). □
$$\begin{aligned}& \int_{n}^{n+1}g(t)\,dt \leq g(n)\leq\int _{n-1}^{n}g(t)\,dt\quad (n=1,\ldots,n_{0}),\\& \int_{n_{0}+1}^{n_{0}+2}g(t)\,dt < g(n_{0}+1)< \int_{n_{0}}^{n_{0}+1}g(t)\,dt, \end{aligned}$$
$$ 0< \int_{1}^{n_{0}+2}g(t)\,dt< \sum _{n=1}^{n_{0}+1}g(n)< \sum_{n=1}^{n_{0}+1} \int_{n-1}^{n}g(t)\,dt=\int _{0}^{n_{0}+1}g(t)\,dt< \infty. $$
$$ 0< \int_{n_{0}+2}^{\infty}g(t)\,dt\leq\sum _{n=n_{0}+2}^{\infty}g(n)\leq \int_{n_{0}+1}^{\infty}g(t) \,dt< \infty. $$
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 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 and we still indicate \(\ln(\frac{c_{i_{0}+1}}{c_{i_{0}}})\) by the following formal expression:
$$ \int_{c_{i}}^{c_{i+1}}t^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac {2i}{s})\alpha-1}\,dt= \frac{c_{i+1}^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac {2i}{s})\alpha}-c_{i}^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac {2i}{s})\alpha}}{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac{2i}{s})\alpha}; $$
$$ \int_{c_{i_{0}}}^{c_{i_{0}+1}}t^{\lambda_{1}-\frac{i_{0}\lambda }{s}+(1-\frac{2i_{0}}{s})\alpha-1}\,dt=\ln\biggl( \frac{c_{i_{0}+1}}{c_{i_{0}}}\biggr)=\lim_{i\rightarrow i_{0}}\int _{c_{i}}^{c_{i+1}}t^{\lambda_{1}-\frac{i\lambda}{s}+(1-\frac{2i}{s})\alpha-1}\,dt, $$
$$ \frac{c_{i_{0}+1}^{\lambda_{1}-\frac{i_{0}\lambda}{s}+(1-\frac {2i_{0}}{s})\alpha}-c_{i_{0}}^{\lambda_{1}-\frac{i_{0}\lambda}{s}+(1-\frac {2i_{0}}{s})\alpha}}{\lambda_{1}-\frac{i_{0}\lambda}{s}+(1-\frac {2i_{0}}{s})\alpha}. $$
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 \(\lambda_{2}=-\lambda_{1}\), \(|\lambda _{1}|<\alpha\) (\(\alpha>0\)), and
$$ k_{0}(x,y)=\prod_{k=1}^{s} \biggl( \frac{\min\{x,c_{k}y\}}{\max\{ x,c_{k}y\}}\biggr) ^{\frac{\alpha}{s}}, $$
$$\begin{aligned} k_{s}(\lambda_{1}) =&k_{s}^{(0)}( \lambda_{1}):=\frac{c_{1}^{\lambda _{1}+\alpha}}{a+\lambda_{1}}\frac{1}{\prod_{k=1}^{s}c_{k}^{\frac {\alpha}{s}}}+\frac{c_{s}^{\lambda_{1}-\alpha}}{a-\lambda_{1}} \prod_{k=1}^{s}c_{k}^{\frac{\alpha}{s}} \\ &+\sum_{i=1}^{s-1} \biggl[ \frac{c_{i+1}^{\lambda_{1}+(1-\frac {2i}{s})\alpha }-c_{i}^{\lambda_{1}+(1-\frac{2i}{s})\alpha}}{\lambda_{1}+(1-\frac {2i}{s})\alpha}\frac{\prod_{k=1}^{i}c_{k}^{\frac{\alpha}{s}}}{\prod_{k=i+1}^{s}c_{k}^{\frac{\alpha}{s}}} \biggr] . \end{aligned}$$
(16)
(b) Since we find 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 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.
$$\begin{aligned} k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}&=\frac{1}{y^{1-\lambda _{2}}}\prod _{k=1}^{s}\frac{(\min\{c_{k}^{-1}x,y\})^{\frac{\alpha }{s}}}{c_{k}^{\frac{\lambda}{s}}(\max\{c_{k}^{-1}x,y\})^{\frac{\lambda+\alpha}{s}}}\\ &=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1}{y^{1-\lambda_{2}-\alpha}}\prod_{k=1}^{s}\frac{1}{c_{k}^{\frac {\lambda}{s}}(c_{k}^{-1}x)^{\frac{\lambda+\alpha}{s}}},& 0< y\leq c_{s}^{-1}x,\\ \frac{1}{y^{1+\lambda_{1}+\alpha-\frac{i}{s}(\lambda+2\alpha)}}\frac {\prod_{k=i+1}^{s}(c_{k}^{-1}x)^{\frac{\alpha}{s}}}{\prod_{k=1}^{s}c_{k}^{\frac{\lambda}{s}}\prod_{k=1}^{i}(c_{k}^{-1}x)^{\frac{\lambda+\alpha }{s}}},& c_{i+1}^{-1}x< y\leq c_{i}^{-1}x \ (i=1,\ldots,s-1), \\ \frac{1}{y^{1+\lambda_{1}+\alpha}}\prod_{k=1}^{s}\frac{(c_{k}^{-1}x)^{ \frac{\alpha}{s}}}{c_{k}^{\frac{\lambda}{s}}(y)^{\frac{\lambda +\alpha}{s}}},& c_{1}^{-1}x< y< \infty,\end{array}\displaystyle \right . \end{aligned}$$
$$\begin{aligned} k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}&=\frac{1}{x^{1-\lambda _{1}}}\prod _{k=1}^{s}\frac{(\min\{x,c_{k}y\})^{\frac{\alpha}{s}}}{(\max \{x,c_{k}y\})^{\frac{\lambda+\alpha}{s}}}\\ &=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1}{x^{1-\lambda_{1}-\alpha}}\prod_{k=1}^{s}\frac {1}{(c_{k}y)^{\frac{\lambda+\alpha}{s}}},& 0< x\leq c_{1}y, \\ \frac{1}{x^{1-\lambda_{1}-\alpha+\frac{i}{s}(\lambda+2\alpha)}}\frac {\prod_{k=1}^{i}(c_{k}y)^{\frac{\alpha}{s}}}{\prod_{k=i+1}^{s}(c_{k}y)^{ \frac{\lambda+\alpha}{s}}},& c_{i}y< x\leq c_{i+1}y\ (i=1,\ldots,s-1), \\ \frac{1}{x^{1+\lambda_{2}+\alpha}}\prod_{k=1}^{s}(c_{k}y)^{\frac {\alpha}{s}},& c_{s}y< x< \infty,\end{array}\displaystyle \right . \end{aligned}$$
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:
Then we have the following inequalities:
$$\begin{aligned}& \omega(\lambda_{2},m) :=\sum_{n=1}^{\infty} \prod_{k=1}^{s}\frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}} \frac{U_{m}^{\lambda_{1}}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}},\quad m\in\mathbf{N}, \end{aligned}$$
(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)
$$\begin{aligned}& \omega(\lambda_{2},m) < k_{s}(\lambda_{1}) \quad(- \alpha< \lambda_{2}\leq 1-\alpha,\lambda_{1}>-\alpha;m\in \mathbf{N}), \end{aligned}$$
(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 \(\mu(t):=\mu_{m}\), \(t\in(m-1,m]\) (\(m\in\mathbf{N}\)); \(\upsilon(t):=\upsilon_{n}\), \(t\in(n-1,n]\) (\(n\in\mathbf{N}\)), 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 Setting \(t=\frac{V(y)}{U_{m}}\), we obtain \(V^{\prime}(y)\,dy=U_{m}\,dt\) and
$$ U(x):=\int_{0}^{x}\mu(t)\,dt\quad(x\geq0),\qquad V(y):= \int_{0}^{y}\upsilon (t)\,dt\quad(y\geq 0). $$
(21)
$$\begin{aligned} \omega(\lambda_{2},m) =&\sum_{n=1}^{\infty} \int_{n-1}^{n}\prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}\frac{U_{m}^{\lambda _{1}}}{V_{n}^{1-\lambda_{2}}}V^{\prime}(y)\,dy \\ < &\sum_{n=1}^{\infty}\int_{n-1}^{n} \prod_{k=1}^{s}\frac{(\min \{U_{m},c_{k}V(y)\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V(y)\} )^{\frac{\lambda+\alpha}{s}}} \frac{U_{m}^{\lambda_{1}}V^{\prime }(y)}{V^{1-\lambda _{2}}(y)}\,dy. \end{aligned}$$
$$\begin{aligned} \omega(\lambda_{2},m) < &\sum_{n=1}^{\infty} \int_{\frac {V(n-1)}{U_{m}}}^{\frac{V(n)}{U_{m}}}\prod_{k=1}^{s} \frac{(\min\{1,c_{k}t\})^{\frac {\alpha}{s}}}{(\max\{1,c_{k}t\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda _{2}-1}\,dt \\ =&\int_{0}^{\frac{V(\infty)}{U_{m}}}\prod _{k=1}^{s}\frac{(\min \{1,c_{k}t\})^{\frac{\alpha}{s}}}{(\max\{1,c_{k}t\})^{\frac{\lambda +\alpha}{s}}}t^{\lambda_{2}-1}\,dt \\ \leq&\int_{0}^{\infty}\prod _{k=1}^{s}\frac{(\min\{1,c_{k}t\})^{\frac {\alpha}{s}}}{(\max\{1,c_{k}t\})^{\frac{\lambda+\alpha }{s}}}t^{\lambda _{2}-1} \,dt=k_{s}(\lambda_{1}). \end{aligned}$$
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 Hence, we have (19) and (20). □
$$\begin{aligned} \varpi(\lambda_{1},n) =&\sum_{m=1}^{\infty} \int_{m-1}^{m}\prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}\frac{V_{n}^{\lambda _{2}}U^{\prime}(x)}{U_{m}^{1-\lambda_{1}}}\,dx \\ < &\sum_{m=1}^{\infty}\int_{m-1}^{m} \prod_{k=1}^{s}\frac{(\min \{U(x),c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U(x),c_{k}V_{n}\} )^{\frac{\lambda+\alpha}{s}}} \frac{V_{n}^{\lambda_{2}}U^{\prime }(x)}{U^{1-\lambda _{1}}(x)}\,dx \\ \overset{t=U(x)/V_{n}}{=}&\sum_{m=1}^{\infty} \int_{\frac {U(m-1)}{V_{n}}}^{\frac{U(m)}{V_{n}}}\prod_{k=1}^{s} \frac{(\min\{t,c_{k}\})^{\frac {\alpha}{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda_{1}-1}\,dt \\ =&\int_{0}^{\frac{U(\infty)}{V_{n}}}\prod _{k=1}^{s}\frac{(\min \{t,c_{k}\})^{\frac{\alpha}{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda +\alpha }{s}}}t^{\lambda_{1}-1}\,dt\leq k_{s}(\lambda_{1}). \end{aligned}$$
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
where
$$\begin{aligned}& k_{s}(\lambda_{1}) \bigl(1-\theta(\lambda_{2},m) \bigr) < \omega(\lambda _{2},m) \quad(-\alpha< \lambda_{2}\leq1- \alpha,\lambda_{1}>-\alpha), \end{aligned}$$
(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)
$$\begin{aligned}& \theta(\lambda_{2},m) :=\frac{1}{k_{s}(\lambda_{1})}\int_{0}^{\frac {U_{m_{0}}}{V_{n}}} \prod_{k=1}^{s}\frac{(\min\{t,c_{k}\})^{\frac{\alpha }{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda_{1}-1} \,dt =O\biggl(\frac{1}{U_{m}^{\lambda_{2}+\alpha}}\biggr)\in(0,1), \\& \vartheta(\lambda_{1},n) :=\frac{1}{k_{s}(\lambda_{1})}\int _{0}^{\frac{U_{m_{0}}}{V_{n}}}\prod_{k=1}^{s} \frac{(\min\{t,c_{k}\})^{\frac{\alpha }{s}}}{(\max\{t,c_{k}\})^{\frac{\lambda+\alpha}{s}}}t^{\lambda_{1}-1}\,dt =O\biggl(\frac{1}{V_{n}^{\lambda_{1}+\alpha}}\biggr)\in(0,1); \end{aligned}$$
(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 and then \(\theta(\lambda_{2},m)=O(\frac{1}{U_{m}^{\lambda_{2}+\alpha}})\). Hence we have (22).
$$ \theta(\lambda_{2},m)=\frac{\prod_{k=1}^{s}c_{k}}{k_{s}(\lambda_{1})} \int _{0}^{\frac{V_{n_{0}}}{U_{m}}}t^{\lambda_{2}+\alpha-1}\,dt= \frac{\prod_{k=1}^{s}c_{k}}{(\lambda_{2}+\alpha)k_{s}(\lambda_{1})} \biggl( \frac{V_{n_{0}}}{U_{m}} \biggr) ^{\lambda_{2}+\alpha}, $$
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 Hence, we have (23).
$$ \vartheta(\lambda_{1},n)=\frac{\int_{0}^{\frac{U_{m_{0}}}{V_{n}}}t^{\lambda_{1}+\alpha-1}\,dt}{k_{s}(\lambda_{1})\prod_{k=1}^{s}c_{k}^{ \frac{\lambda+\alpha}{s}}}=\frac{(\lambda_{1}+\alpha)^{-1}}{k_{s}(\lambda_{1})\prod_{k=1}^{s}c_{k}^{\frac{\lambda+\alpha }{s}}} \biggl( \frac{U_{m_{0}}}{V_{n}} \biggr) ^{\lambda_{1}+\alpha}. $$
For \(b>0\), we find Hence we have (24). By the same way, we still have (25). □
$$\begin{aligned}& \begin{aligned}[b] \sum_{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+b}}&=\sum _{m=1}^{m_{0}}\frac {\mu _{m}}{U_{m}^{1+b}}+\sum _{m=m_{0}+1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+b}}\\ &=\sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+b}}+ \sum_{m=m_{0}+1}^{\infty }\int_{m-1}^{m} \frac{U^{\prime}(x)}{U_{m}^{1+b}}\,dx \\ &< \sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+b}}+ \sum_{m=m_{0}+1}^{\infty }\int_{m-1}^{m} \frac{U^{\prime}(x)}{U^{1+b}(x)}\,dx\\ &=\sum_{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+b}}+ \int_{m_{0}}^{\infty} \frac{dU(x)}{U^{1+b}(x)}=\sum _{m=1}^{m_{0}}\frac{\mu_{m}}{U_{m}^{1+b}}+ \frac{1}{bU_{m_{0}}^{b}} \\ &=\frac{1}{b} \Biggl( \frac{1}{U_{m_{0}}^{b}}+b\sum _{m=1}^{m_{0}}\frac {\mu _{m}}{U_{m}^{1+b}} \Biggr) , \end{aligned}\\& \begin{aligned}[b] \sum_{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+b}} &\geq \sum_{m=m_{0}}^{\infty}\frac{\mu_{m+1}}{U_{m}^{1+b}}=\sum _{m=m_{0}}^{ \infty}\int_{m}^{m+1} \frac{U^{\prime}(x)}{U_{m}^{1+b}}\,dx \\ &>\sum_{m=m_{0}}^{\infty}\int_{m}^{m+1} \frac{U^{\prime }(x)\,dx}{U^{1+b}(x)}=\int_{m_{0}}^{\infty} \frac{dU(x)}{U^{1+b}(x)}=\frac{1}{bU_{m_{0}}^{b}}. \end{aligned} \end{aligned}$$
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:
where
\(k_{1}(\lambda_{1})\)
is indicated by (12).
$$\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{(\min \{U_{m},c_{1}V_{n}\})^{\alpha}a_{m}b_{n}}{(\max \{U_{m},c_{1}V_{n}\})^{\lambda+\alpha}}< k_{1}( \lambda _{1})\|a\|_{p,\Phi _{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(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)
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 Then, by (19), we have (27).
$$\begin{aligned} J \leq&\bigl(k_{s}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}\frac{U_{m}^{(1-\lambda_{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{s}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty}\prod_{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}}\frac{U_{m}^{(1-\lambda_{1})(p-1)}\upsilon_{n}}{V_{n}^{1-\lambda_{2}}\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{s}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=1}^{\infty}\omega (\lambda_{2},m) \frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu _{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$
(31)
By Hölder’s inequality (cf. [29]), we have Then, by (27), we have (26).
$$\begin{aligned} I =&\sum_{n=1}^{\infty} \Biggl[ \frac{\upsilon_{n}^{\frac {1}{p}}}{V_{n}^{\frac{1}{p}-\lambda_{2}}}\sum_{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}}{(\max\{ U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr] \biggl( \frac{V_{n}^{\frac {1}{p}-\lambda _{2}}}{\upsilon_{n}^{\frac{1}{p}}}b_{n} \biggr) \\ \leq&J\|b\|_{q,\Psi_{\lambda}}. \end{aligned}$$
(32)
On the other hand, assuming that (26) is valid, we set 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 and then (27) follows, which is equivalent to (26). □
$$ b_{n}:=\frac{\upsilon_{n}}{V_{n}^{1-p\lambda_{2}}} \Biggl[ \sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\} )^{\frac{\alpha}{s}}a_{m}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha }{s}}} \Biggr] ^{p-1},\quad n\in\mathbf{N}. $$
$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q} =J^{p}=I< k_{s}( \lambda _{1})\|a\|_{p,\Phi _{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \\& \|b\|_{q,\Psi_{\lambda}}^{q-1} =J< k_{s}(\lambda_{1}) \|a\|_{p,\Phi _{\lambda}}, \end{aligned}$$
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 \(\widetilde{a}=\{\widetilde{a}_{m}\}_{m=1}^{\infty}\), \(\widetilde{b}=\{\widetilde{b}_{n}\}_{n=1}^{\infty}\), Then, by (24), (25) and (23), we have
$$ \widetilde{a}_{m}:=U_{m}^{\widetilde{\lambda}_{1}-1}\mu _{m}=U_{m}^{\lambda _{1}-\frac{\varepsilon}{p}-1}\mu_{m},\qquad\widetilde {b}_{n}=V_{n}^{\widetilde{\lambda}_{2}-\varepsilon-1}\upsilon_{n}=V_{n}^{\lambda_{2}-\frac{\varepsilon}{q}-1} \upsilon_{n}. $$
(33)
$$\begin{aligned}& \begin{aligned}[b] \|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\Psi _{\lambda }}&= \Biggl( \sum _{m=1}^{\infty}\frac{\mu_{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}}\\ &=\frac{1}{\varepsilon} \biggl( \frac{1}{U_{m_{0}}^{\varepsilon }}+\varepsilon O(1) \biggr) ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I} &:=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \widetilde{a}_{m}\widetilde{b}_{n} \\ &=\sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}}\frac{V_{n}^{\widetilde{\lambda}_{2}}\mu _{m}}{U_{m}^{1-\widetilde{\lambda}_{1}}} \Biggr] \frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\sum_{n=1}^{\infty}\varpi(\widetilde{ \lambda}_{1},n)\frac {\upsilon_{n}}{V_{n}^{\varepsilon+1}}\geq k_{s}(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty} \bigl(1-\vartheta(\widetilde{\lambda}_{1},n)\bigr)\frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=k_{s}(\widetilde{\lambda}_{1}) \Biggl( \sum _{n=1}^{\infty}\frac {\upsilon _{n}}{V_{n}^{\varepsilon+1}}-\sum _{n=1}^{\infty}O\biggl(\frac{\upsilon _{n}}{V_{n}^{\frac{\varepsilon}{q}+\lambda_{1}+\alpha+1}}\biggr) \Biggr) \\ &=\frac{1}{\varepsilon}k_{s}(\widetilde{\lambda}_{1}) \biggl[ \frac{1}{ V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl(\widetilde{O}(1)-O(1)\bigr) \biggr] . \end{aligned} \end{aligned}$$
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 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).
$$\begin{aligned} &k_{s}(\widetilde{\lambda}_{1}) \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon }}+\varepsilon\bigl(\widetilde{O}(1)-O(1)\bigr) \biggr] < K \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{ \frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon \widetilde{O}(1) \biggr) ^{\frac{1}{q}}. \end{aligned}$$
Remark 1
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: Assuming that \(a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}\), setting we can rewrite (27) as follows: namely \(c\in l_{p,\Psi_{\lambda}^{1-p}}\).
$$\begin{aligned}& l_{p,\Phi_{\lambda}} :=\bigl\{ a=\{a_{m}\}_{m=1}^{\infty}; \|a\|_{p,\Phi _{\lambda}}< \infty\bigr\} , \\& l_{q,\Psi_{\lambda}} :=\bigl\{ b=\{b_{n}\}_{n=1}^{\infty}; \|b\|_{q,\Psi _{\lambda}}< \infty\bigr\} , \\& l_{p,\Psi_{\lambda}^{1-p}} :=\bigl\{ c=\{c_{n}\}_{n=1}^{\infty }; \|c\|_{p,\Psi _{\lambda}^{1-p}}< \infty\bigr\} . \end{aligned}$$
$$ c=\{c_{n}\}_{n=1}^{\infty},\qquad c_{n}:=\sum _{m=1}^{\infty}\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}a_{m},\quad n\in \mathbf{N}, $$
$$ \|c\|_{p,\Psi_{\lambda}^{1-p}}< k_{s}(\lambda_{1})\|a \|_{p,\Phi _{\lambda }}< \infty, $$
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: 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
$$ \|T\|:=\sup_{a(\neq\theta)\in l_{p,\Phi_{\lambda}}}\frac {\|Ta\|_{p,\Psi _{\lambda}^{1-p}}}{\|a\|_{p,\Phi_{\lambda}}}. $$
(37)
$$ \|T\|=k_{s}(\lambda_{1}). $$
(38)
4 Some reverses
In the following, we also set 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 \(\|b\|_{q,\widetilde{\Psi}_{\lambda}}\).
$$\begin{aligned}& \widetilde{\Phi}_{\lambda}(m) :=\bigl(1-\theta(\lambda_{2},m) \bigr)\frac{U_{m}^{p(1-\lambda_{1})-1}}{\mu_{m}^{p-1}}, \\& \widetilde{\Psi}_{\lambda}(n) :=\bigl(1-\vartheta(\lambda_{1},n) \bigr)\frac{ V_{n}^{q(1-\lambda_{2})-1}}{\upsilon_{n}^{q-1}}\quad(m,n\in\mathbf{N}). \end{aligned}$$
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 and then (40) follows, which is equivalent to (39).
$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q} =J^{p}=I>k_{s}( \lambda_{1})\|a\|_{p,\widetilde{\Phi}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \\& \|b\|_{q,\Psi_{\lambda}}^{q-1} =J>k_{s}(\lambda _{1})\|a\|_{p,\widetilde{\Phi}_{\lambda}}, \end{aligned}$$
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 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 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).
$$\begin{aligned}& \begin{aligned}[b] \|a\|_{p,\widetilde{\Phi}_{\lambda}}\|b\|_{q,\Psi_{\lambda}}&= \Biggl[ \sum _{m=1}^{\infty}\bigl(1-\theta(\lambda_{2},m) \bigr)\frac{\mu_{m}}{U_{m}^{1+\varepsilon}} \Biggr] ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}}\\ &= \Biggl( \sum_{m=1}^{\infty} \frac{\mu_{m}}{U_{m}^{1+\varepsilon}}-\sum_{m=1}^{\infty}O \biggl(\frac{\mu_{m}}{U_{m}^{1+\lambda_{2}+\alpha +\varepsilon}}\biggr) \Biggr) ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr) ^{\frac{1}{q}} \\ &=\frac{1}{\varepsilon} \biggl[ \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon \bigl(O(1)-O_{1}(1)\bigr) \biggr] ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I}&=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}}\widetilde{a}_{m}\widetilde{b}_{n}\\ &=\sum_{n=1}^{\infty} \Biggl[ \sum _{m=1}^{\infty}\prod_{k=1}^{s} \frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}}\frac{V_{n}^{\widetilde{\lambda}_{2}}\mu _{m}}{U_{m}^{1-\widetilde{\lambda}_{1}}} \Biggr] \frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\sum_{n=1}^{\infty}\varpi(\widetilde{ \lambda}_{1},n)\frac {\upsilon_{n}}{V_{n}^{\varepsilon+1}}\leq k_{s}(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty} \frac{\upsilon_{n}}{V_{n}^{\varepsilon+1}} \\ &=\frac{1}{\varepsilon}k_{s}(\widetilde{\lambda}_{1}) \biggl( \frac{1}{ V_{n_{0}}^{\varepsilon}}+\varepsilon\widetilde{O}(1) \biggr) . \end{aligned} \end{aligned}$$
$$\begin{aligned} &k_{s}(\widetilde{\lambda}_{1}) \biggl( \frac{1}{V_{n_{0}}^{\varepsilon }}+\varepsilon\widetilde{O}(1) \biggr) >K \biggl[ \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon \bigl(O(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{p}} \biggl( \frac{1}{V_{n_{0}}^{\varepsilon }}+\varepsilon \widetilde{O}(1) \biggr) ^{\frac{1}{q}}. \end{aligned}$$
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 Then, by (19), we have (44).
$$\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)
By the reverse Hölder’s inequality (cf. [29]), we have Then, by (42), we have (41).
$$\begin{aligned} I&=\sum_{n=1}^{\infty}\frac{V_{n}^{\lambda_{2}-\frac{1}{p}}\upsilon _{n}^{1/p}}{(1-\vartheta(\lambda_{1},n))^{1/q}} \Biggl[ \sum_{m=1}^{\infty }\prod _{k=1}^{s}\frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha }{s}}a_{m}}{(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr]\biggl[ \bigl(1-\vartheta(\lambda_{1},n)\bigr)^{\frac{1}{q}} \frac {V_{n}^{\frac{1}{p}-\lambda_{2}}}{\upsilon_{n}^{1/p}}b_{n} \biggr] \\ & \geq J_{1}\|b \|_{q,\widetilde{\Psi}_{\lambda}}. \end{aligned}$$
(44)
On the other hand, assuming that (41) is valid, we set \(b_{n}\) as follows: 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 and then (42) follows, which is equivalent to (41).
$$ b_{n}:=\frac{V_{n}^{p\lambda_{2}-1}\upsilon_{n}}{(1-\vartheta(\lambda _{1},n))^{p-1}} \Biggl[ \sum_{m=1}^{\infty} \prod_{k=1}^{s}\frac{(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}a_{m}}{(\max\{ U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \Biggr] ^{p-1},\quad n\in\mathbf{N}. $$
$$\begin{aligned}& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q} =J_{1}^{p}=I>k_{s}( \lambda _{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\widetilde{\Psi}_{\lambda}},\\& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q-1} =J_{1}>k_{s}( \lambda _{1})\|a\|_{p,\Phi_{\lambda}}, \end{aligned}$$
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 Then, by (24), (25) and (19), we have
$$ \widetilde{a}_{m}:=U_{m}^{\widetilde{\lambda}_{1}-1-\varepsilon}\mu _{m}=U_{m}^{\lambda_{1}-\frac{\varepsilon}{p}-1}\mu_{m}, \qquad\widetilde{b} _{n}=V_{n}^{\widetilde{\lambda}_{2}-1}\upsilon_{n}=V_{n}^{\lambda _{2}-\frac{\varepsilon}{q}-1} \upsilon_{n}. $$
$$\begin{aligned}& \begin{aligned}[b] \|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\widetilde {\Psi}_{\lambda}}&= \Biggl( \sum _{m=1}^{\infty}\frac{\mu _{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl[ \sum_{n=1}^{\infty} \bigl(1-\vartheta(\lambda _{1},n)\bigr)\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}} \Biggr] ^{\frac{1}{q}} \\ &= \Biggl( \sum_{m=1}^{\infty} \frac{\mu_{m}}{U_{m}^{1+\varepsilon }} \Biggr) ^{\frac{1}{p}} \Biggl( \sum _{n=1}^{\infty}\frac{\upsilon_{n}}{V_{n}^{1+\varepsilon}}-\sum _{n=1}^{\infty}O\biggl(\frac{\upsilon_{n}}{V_{n}^{1+\lambda_{1}+\alpha+\varepsilon}}\biggr) \Biggr) ^{\frac{1}{q}} \\ &=\frac{1}{\varepsilon} \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{\frac{1}{p}} \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl( \widetilde{O}(1)-O_{1}(1)\bigr) \biggr] ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I} &=\sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod _{k=1}^{s} \frac{(\min\{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max \{U_{m},c_{k}V_{n}\})^{\frac{\lambda+\alpha}{s}}} \widetilde{a}_{m}\widetilde{b}_{n} \\ &=\sum_{m=1}^{\infty} \Biggl[ \sum _{n=1}^{\infty}\prod_{k=1}^{s} \frac {(\min \{U_{m},c_{k}V_{n}\})^{\frac{\alpha}{s}}}{(\max\{U_{m},c_{k}V_{n}\})^{ \frac{\lambda+\alpha}{s}}}\frac{U_{m}^{\widetilde{\lambda }_{1}}\upsilon _{n}}{V_{n}^{1-\widetilde{\lambda}_{2}}} \Biggr] \frac{\mu_{m}}{U_{m}^{1+\varepsilon}} \\ &=\sum_{m=1}^{\infty}\omega(\widetilde{ \lambda}_{2},m)\frac{\mu _{m}}{U_{m}^{1+\varepsilon}}\leq k_{s}(\widetilde{ \lambda}_{1})\sum_{n=1}^{\infty} \frac{\mu_{m}}{U_{m}^{1+\varepsilon}} \\ &=\frac{1}{\varepsilon}k_{s}(\widetilde{\lambda}_{1}) \biggl( \frac{1}{ U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) . \end{aligned} \end{aligned}$$
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 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).
$$\begin{aligned} &k_{s}(\widetilde{\lambda}_{1}) \biggl( \frac{1}{U_{m_{0}}^{\varepsilon }}+\varepsilon O(1) \biggr) \\ &\quad>K \biggl( \frac{1}{U_{m_{0}}^{\varepsilon}}+\varepsilon O(1) \biggr) ^{ \frac{1}{p}} \biggl[ \frac{1}{V_{n_{0}}^{\varepsilon}}+\varepsilon\bigl(\widetilde{O}(1)-O_{1}(1) \bigr) \biggr] ^{\frac{1}{q}}. \end{aligned}$$
Remark 2
(i) For \(\alpha=0\), \(0<\lambda _{1},\lambda_{2}\leq1\) in (26) and (27), we have the following equivalent inequalities:
where \(\widetilde{k}_{s}(\lambda_{1})\) is indicated by (14);
$$\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\prod_{k=1}^{s}(\max\{U_{m},c_{k}V_{n}\})^{\frac{\lambda }{s}}}< \widetilde{k}_{s}( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(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)
(ii) for \(\alpha=-\lambda\), \(-1\leq\lambda_{1},\lambda_{2}<0\) in (26) and (27), we have the following equivalent inequalities:
where \(\widehat{k}_{s}(\lambda_{1})\) is indicated by (15);
$$\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\frac{a_{m}b_{n}}{\prod_{k=1}^{s}(\min\{U_{m},c_{k}V_{n}\})^{\frac{\lambda }{s}}}< \widehat{k}_{s}( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(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)
(iii) for \(\lambda=0\), \(\lambda_{2}=-\lambda_{1}\), in (26) and (27), we have the following equivalent inequalities:
where \(k_{s}^{(0)}(\lambda_{1})\) is indicated by (16) (\(|\lambda _{1}|<\alpha\), \(0<\alpha\leq\frac{1}{2}\); \(|\lambda_{1}|<1-\alpha\), \(\frac {1}{2}<\alpha\leq1\)).
$$\begin{aligned}& \sum_{n=1}^{\infty}\sum _{m=1}^{\infty}\prod_{k=1}^{s} \biggl( \frac{\min \{U_{m},c_{k}V_{n}\}}{\max\{U_{m},c_{k}V_{n}\}} \biggr) ^{\frac{\alpha }{s}}a_{m}b_{n}< k_{s}^{(0)}( \lambda_{1})\|a\|_{p,\Phi_{\lambda }}\|b\|_{q,\Psi _{\lambda}}, \end{aligned}$$
(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)
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.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. QC participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.