We define the following functions:
$$\begin{aligned} &\widetilde{\Phi}_{\lambda}(m) :=\omega(\lambda_{2},m) \frac{[\ln (U_{m}-\alpha)]^{p(1-\lambda_{1})-1}}{(U_{m}-\alpha)^{1-p}\mu _{m+1}^{p-1}}, \\ &\widetilde{\Psi}_{\lambda}(n) :=\varpi(\lambda_{1},n) \frac{[\ln (V_{n}-\beta)]^{q(1-\lambda_{2})-1}}{(V_{n}-\beta)^{1-q}\upsilon _{n+1}^{q-1}}\quad\bigl(m,n\in\mathbf{N}\backslash\{1\}\bigr). \end{aligned}$$
(22)
Theorem 1
We have the following equivalent inequalities:
$$\begin{aligned}& I :=\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln ^{\lambda}[(U_{m}-\alpha)(V_{n}-\beta)]}\leq\|a\|_{p,\widetilde {\Phi}_{\lambda}} \|b\|_{q,\widetilde{\Psi}_{\lambda}}, \end{aligned}$$
(23)
$$\begin{aligned}& J := \Biggl\{ \sum_{n=2}^{\infty} \frac{\upsilon_{n+1}\ln^{p\lambda _{2}-1}(V_{n}-\beta)}{(\varpi(\lambda_{1},n))^{p-1}(V_{n}-\beta )} \Biggl[ \sum_{m=2}^{\infty} \frac{a_{m}}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} \leq\|a \|_{p,\widetilde{\Phi}_{\lambda}}. \end{aligned}$$
(24)
Proof
By Hölder’s inequality (cf. [15]) and (13), we find
$$\begin{aligned} & \Biggl[ \sum_{m=2}^{\infty} \frac{a_{m}}{\ln^{\lambda}[(U_{m}-\alpha )(V_{n}-\beta)]} \Biggr] ^{p} \\ &\quad= \Biggl[ \sum _{m=2}^{\infty}\frac{1}{\ln ^{\lambda}[(U_{m}-\alpha)(V_{n}-\beta)]} \\ &\qquad{}\times \biggl( \frac{(U_{m}-\alpha)^{1/q}\ln ^{(1-\lambda_{1})/q}(U_{m}-\alpha)}{\mu_{m+1}^{1/q}\ln^{(1-\lambda _{2})/p}(V_{n}-\beta)}a_{m} \biggr) \biggl( \frac{\mu_{m+1}^{1/q}\ln^{(1-\lambda _{2})/p}(V_{n}-\beta)}{(U_{m}-\alpha)^{1/q}\ln ^{(1-\lambda_{1})/q}(U_{m}-\alpha)} \biggr) \Biggr] ^{p} \\ &\quad\leq\sum_{m=2}^{\infty}\frac{1}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]} \frac{(U_{m}-\alpha)^{p-1}\ln ^{(1-\lambda _{1})p/q}(U_{m}-\alpha)}{\mu_{m+1}^{p/q}\ln^{1-\lambda _{2}}(V_{n}-\beta)}a_{m}^{p} \\ &\qquad{}\times \Biggl[ \sum_{m=2}^{\infty} \frac{\mu_{m+1}}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]}\frac{\ln^{(1-\lambda _{2})(q-1)}(V_{n}-\beta)}{(U_{m}-\alpha)\ln ^{1-\lambda_{1}}(U_{m}-\alpha)} \Biggr] ^{p-1} \\ &\quad=\frac{(\varpi(\lambda_{1},n))^{p-1}(V_{n}-\beta)}{\upsilon _{n+1}\ln ^{p\lambda_{2}-1}(V_{n}-\beta)} \\ &\qquad{}\times\sum_{m=2}^{\infty}\frac{\upsilon_{n+1}(U_{m}-\alpha )^{p-1}\ln ^{(1-\lambda_{1})(p-1)}(U_{m}-\alpha)}{\mu_{m+1}^{p-1}(V_{n}-\beta )\ln ^{\lambda}[(U_{m}-\alpha)(V_{n}-\beta)]\ln^{1-\lambda _{2}}(V_{n}-\beta)}a_{m}^{p}. \end{aligned}$$
(25)
Then by (12) we obtain
$$\begin{aligned} J &\leq \Biggl[ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{\upsilon _{n+1}(U_{m}-\alpha)^{p-1}a_{m}^{p}}{\ln^{\lambda}[(U_{m}-\alpha )(V_{n}-\beta)]}\frac{\ln^{(1-\lambda_{1})(p-1)}(U_{m}-\alpha)}{\mu _{m+1}^{p-1}(V_{n}-\beta)\ln^{1-\lambda_{2}}(V_{n}-\beta)} \Biggr] ^{ \frac{1}{p}} \\ &= \Biggl[ \sum_{m=2}^{\infty}\sum _{n=2}^{\infty}\frac{\upsilon _{n+1}(U_{m}-\alpha)^{p-1}a_{m}^{p}}{\ln^{\lambda}[(U_{m}-\alpha )(V_{n}-\beta)]}\frac{\ln^{(1-\lambda_{1})(p-1)}(U_{m}-\alpha)}{\mu _{m+1}^{p-1}(V_{n}-\beta)\ln^{1-\lambda_{2}}(V_{n}-\beta)} \Biggr] ^{ \frac{1}{p}} \\ &= \Biggl[ \sum_{m=2}^{\infty}\omega( \lambda_{2},m)\frac{\ln ^{p(1-\lambda _{1})-1}(U_{m}-\alpha)}{(U_{m}-\alpha)^{1-p}\mu_{m+1}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}, \end{aligned}$$
(26)
namely, (24) follows. By Hölder’s inequality (cf. [15]), we find
$$\begin{aligned} I =&\sum_{n=2}^{\infty} \Biggl[ \frac{\upsilon_{n+1}^{1/p}\ln^{\lambda _{2}-\frac{1}{p}}(V_{n}-\beta)}{(\varpi(\lambda_{1},n))^{\frac{1}{q}}(V_{n}-\beta)^{1/p}}\sum_{m=1}^{\infty} \frac{a_{m}}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]} \Biggr] \\ &{}\times \biggl[ \bigl(\varpi(\lambda_{1},n)\bigr)^{\frac{1}{q}} \frac{\ln^{\frac {1}{p}-\lambda_{2}}(V_{n}-\beta)}{(V_{n}-\beta)^{-1/p}\upsilon _{n+1}^{1/p}}b_{n} \biggr] \leq J\|b\|_{q,\widetilde{\Psi}_{\lambda}}. \end{aligned}$$
(27)
Then by (24), (23) follows.
On the other hand, suppose that (23) is valid. We set
$$\begin{aligned} b_{n} :=&\frac{\upsilon_{n+1}\ln^{p\lambda_{2}-1}(V_{n}-\beta )}{(\varpi (\lambda_{1},n))^{p-1}(V_{n}-\beta)} \Biggl[ \sum _{m=2}^{\infty}\frac {a_{m}}{\ln^{\lambda}[(U_{m}-\alpha)(V_{n}-\beta)]} \Biggr] ^{p-1},\quad n \in\mathbf{N}\backslash\{1\}. \end{aligned}$$
(28)
Then we have \(J^{p}=\|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q}\). If \(J=0\), then (24) is trivially valid; if \(J=\infty\), then in view of (26), (24) takes the form of an equality. Suppose that \(0< J<\infty\). By (23), we obtain
$$\begin{aligned}& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q} =J^{p}=I\leq\|a\|_{p,\widetilde{\Phi}_{\lambda}} \|b\|_{q,\widetilde{\Psi}_{\lambda}}, \end{aligned}$$
(29)
$$\begin{aligned}& \|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q-1} =J\leq \|a\|_{p,\widetilde{\Phi}_{\lambda}}, \end{aligned}$$
(30)
namely, (24) follows, which is equivalent to (23). □
Theorem 2
Assuming that
\(\{\mu_{m}\}_{m=1}^{\infty}\)
and
\(\{\upsilon _{n}\}_{n=1}^{\infty}\)
are decreasing, \(U(\infty)=V(\infty)=\infty\), \(0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi_{\lambda }}<\infty\), we have the following equivalent inequalities:
$$\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]}< B(\lambda_{1}, \lambda _{2})\|a\|_{p,\Phi _{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(31)
$$\begin{aligned}& \begin{aligned}[b] J_{1} :={}& \Biggl\{ \sum_{n=2}^{\infty} \frac{\upsilon_{n+1}\ln ^{p\lambda _{2}-1}(V_{n}-\beta)}{V_{n}-\beta} \\ &{} \times \Biggl[ \sum_{m=2}^{\infty} \frac{a_{m}}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}} < B( \lambda_{1},\lambda_{2})\|a\|_{p,\Phi_{\lambda}}, \end{aligned} \end{aligned}$$
(32)
where the constant factor
\(B(\lambda_{1},\lambda_{2})\)
is the best possible.
Proof
Applying (14) and (15) in (23) and (24), we have the equivalent inequalities (31) and (32).
For \(\varepsilon\in(0,p\lambda_{1})\), we set \(\widetilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon}{p}\) (\(\in(0,1)\)), \(\widetilde{\lambda} _{2}=\lambda_{2}+\frac{\varepsilon}{p}\) (>0), and
$$ \widetilde{a}_{m} :=\frac{\mu_{m+1}}{U_{m}-\alpha}\ln^{\widetilde{\lambda}_{1}-1}(U_{m}- \alpha),\qquad \widetilde{b}_{n} :=\frac{\upsilon_{n+1}}{V_{n}-\beta}\ln^{\widetilde {\lambda}_{2}-\varepsilon-1}(V_{n}- \beta). $$
(33)
Then by (20), (21), and (17), we obtain
$$\begin{aligned}& \begin{aligned}[b] &\|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\Psi _{\lambda }}\\ &\quad= \Biggl[ \sum_{m=2}^{\infty}\frac{\mu_{m+1}}{(U_{m}-\alpha)\ln ^{1+\varepsilon}(U_{m}-\alpha)} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=2}^{\infty} \frac{\upsilon_{n+1}}{(V_{n}-\beta )\ln ^{1+\varepsilon}(V_{n}-\beta)} \Biggr] ^{\frac{1}{q}} \\ &\quad=\frac{1}{\varepsilon} \biggl[ \frac{1}{\ln^{\varepsilon}(1+\mu _{2}-\alpha)}+\varepsilon O(1) \biggr] ^{\frac{1}{p}} \biggl[ \frac{1}{\ln ^{\varepsilon}(1+\upsilon_{2}-\beta)}+\varepsilon\widetilde {O}(1) \biggr] ^{\frac{1}{q}}, \end{aligned}\\& \begin{aligned}[b] \widetilde{I} :={}&\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac {\widetilde{a}_{m}\widetilde{b}_{n}}{\ln^{\lambda}[(U_{m}-\alpha)(V_{n}-\beta)]} \\ ={}&\sum_{n=2}^{\infty} \Biggl[ \sum _{m=2}^{\infty}\frac{1}{\ln^{\lambda }[(U_{m}-\alpha)(V_{n}-\beta)]}\frac{\mu_{m+1}\ln^{\widetilde {\lambda}_{2}}(V_{n}-\beta)}{(U_{m}-\alpha)\ln^{1-\widetilde{\lambda}_{1}}(U_{m}-\alpha)} \Biggr] \\ &{}\times\frac{\upsilon_{n+1}}{(V_{n}-\beta)\ln^{\varepsilon +1}(V_{n}-\beta)} =\sum_{n=2}^{\infty} \varpi(\widetilde{\lambda}_{1},n)\frac{\upsilon _{n+1}}{(V_{n}-\beta)\ln^{\varepsilon+1}(V_{n}-\beta)} \\ \geq{}&B(\widetilde{\lambda}_{1},\widetilde{\lambda}_{2}) \Biggl[ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{(V_{n}-\beta)\ln ^{\varepsilon +1}(V_{n}-\beta)}\\ &{}- \sum_{n=2}^{\infty}O\biggl( \frac{\upsilon_{n+1}}{(V_{n}-\beta)\ln ^{\lambda_{1}+\frac{\varepsilon}{q}+1}(V_{n}-\beta)}\biggr) \Biggr]\\ ={}&\frac{1}{\varepsilon}B(\widetilde{\lambda}_{1},\widetilde{\lambda }_{2}) \biggl[ \frac{1}{\ln^{\varepsilon}(1+\upsilon_{2}-\beta )}+\varepsilon \bigl(\widetilde{O}(1)-O(1)\bigr) \biggr]. \end{aligned} \end{aligned}$$
If there exists a positive constant \(K\leq B(\lambda_{1},\lambda_{2})\), such that (31) is valid when replacing \(B(\lambda_{1},\lambda_{2})\) by K, then in particular, we have \(\varepsilon\widetilde {I}<\varepsilon K\|\widetilde{a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\Psi _{\lambda }}\), namely
$$\begin{aligned} &B\biggl(\lambda_{1}-\frac{\varepsilon}{p},\lambda_{2}+ \frac{\varepsilon }{p}\biggr) \biggl[ \frac{1}{\ln^{\varepsilon}(1+\upsilon_{2}-\beta )}+\varepsilon \bigl(\widetilde{O}(1)-O(1)\bigr) \biggr] \\ &\quad< K \biggl[ \frac{1}{\ln^{\varepsilon}(1+\mu_{2}-\alpha )}+\varepsilon O(1) \biggr] ^{\frac{1}{p}} \biggl[ \frac{1}{\ln^{\varepsilon}(1+\upsilon _{2}-\beta)}+\varepsilon\widetilde{O}(1) \biggr] ^{\frac{1}{q}}. \end{aligned}$$
It follows that \(B(\lambda_{1},\lambda_{2})\leq K(\varepsilon \rightarrow 0^{+})\). Hence, \(K=B(\lambda_{1},\lambda_{2})\) is the best possible constant factor of (31).
Similarly, we can obtain
$$ I\leq J_{1}\|b\|_{q,\Psi_{\lambda}}. $$
(34)
Hence, we can prove that the constant factor \(B(\lambda_{1},\lambda_{2})\) in (32) is the best possible. Otherwise, we would reach a contradiction by (34) that the constant factor in (31) is not the best possible. □
We find \(\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon_{n+1}}{V_{n}-\beta }\ln ^{p\lambda_{2}-1}(V_{n}-\beta)\), and we define the following weighted normed spaces:
$$\begin{aligned}& l_{p,\Phi_{\lambda}} :=\bigl\{ a=\{a_{m}\}_{m=2}^{\infty}; \|a\|_{p,\Phi _{\lambda}}< \infty\bigr\} , \\& l_{q,\Psi_{\lambda}} :=\bigl\{ b=\{b_{n}\}_{n=2}^{\infty}; \|b\|_{q,\Psi _{\lambda}}< \infty\bigr\} , \\& l_{p,\Psi_{\lambda}^{1-p}} :=\bigl\{ c=\{c_{n}\}_{n=2}^{\infty }; \|c\|_{p,\Psi _{\lambda}^{1-p}}< \infty\bigr\} . \end{aligned}$$
Assuming that \(a=\{a_{m}\}_{m=2}^{\infty}\in l_{p,\Phi_{\lambda}}\), setting
$$ c=\{c_{n}\}_{n=2}^{\infty},\qquad c_{n}:=\sum _{m=2}^{\infty}\frac{a_{m}}{\ln ^{\lambda}[(U_{m}-\alpha)(V_{n}-\beta)]},\quad n\in\mathbf{N} \backslash\{ 1\}, $$
we can rewrite (32) as follows:
$$ \|c\|_{p,\Psi_{\lambda}^{1-p}}< B(\lambda_{1},\lambda _{2})\|a \|_{p,\Phi _{\lambda}}< \infty, $$
namely, \(c\in l_{p,\Psi_{\lambda}^{1-p}}\).
Definition 1
Define a Hardy-Mulholland-type operator \(T:l_{p,\Phi _{\lambda}}\rightarrow l_{p,\Psi_{\lambda}^{1-p}}\) as follows: For any \(a=\{a_{m}\}_{m=2}^{\infty}\in l_{p,\Phi_{\lambda}}\), there exists a unique representation \(Ta=c\in l_{p,\Psi_{\lambda}^{1-p}}\). We set the formal inner product of Ta and \(b=\{b_{n}\}_{n=2}^{\infty}\in l_{q,\Psi _{\lambda}}\) as follows:
$$ (Ta,b):=\sum_{n=2}^{\infty} \Biggl[ \sum _{m=2}^{\infty}\frac{a_{m}}{\ln ^{\lambda}[(U_{m}-\alpha)(V_{n}-\beta)]} \Biggr] b_{n}. $$
(35)
Then we can rewrite (31) and (32) as follows:
$$\begin{aligned}& (Ta,b) < B(\lambda_{1},\lambda_{2})\|a\|_{p,\Phi_{\lambda }} \|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(36)
$$\begin{aligned}& \|Ta\|_{p,\Psi_{\lambda}^{1-p}} < B(\lambda_{1},\lambda _{2})\|a \|_{p,\Phi_{\lambda}}. \end{aligned}$$
(37)
We set the norm of operator T as follows:
$$ \|T\|:=\sup_{a(\neq\theta)\in l_{p,\Phi_{\lambda}}}\frac {\|Ta\|_{p,\Psi _{\lambda}^{1-p}}}{\|a\|_{p,\Phi_{\lambda}}}. $$
By (37), we find \(\|T\|\leq B(\lambda_{1},\lambda_{2})\). Since the constant factor in (37) is the best possible, it follows that \(\|T\|=B(\lambda_{1},\lambda_{2})\).
Remark 1
(i) For \(\alpha=\beta=0\) in (31) and (32), setting
$$ \varphi_{\lambda}(m):=\frac{(\ln U_{m})^{p(1-\lambda_{1})-1}}{U_{m}^{1-p}\mu_{m+1}^{p-1}},\qquad\psi_{\lambda}(n):= \frac{(\ln V_{n})^{q(1-\lambda_{2})-1}}{V_{n}^{1-q}\upsilon_{n+1}^{q-1}} \quad \bigl(m,n\in \mathbf{N}\backslash\{1\}\bigr), $$
we have the following equivalent Hardy-Mulholland-type inequalities:
$$\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }(U_{m}V_{n})}< B(\lambda_{1}, \lambda_{2})\|a\|_{p,\varphi_{\lambda }}\|b\|_{q,\psi_{\lambda}}, \end{aligned}$$
(38)
$$\begin{aligned}& \Biggl\{ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}} \ln^{p\lambda _{2}-1}V_{n} \Biggl[ \sum_{m=2}^{\infty} \frac{a_{m}}{\ln^{\lambda }(U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< B( \lambda_{1},\lambda _{2})\|a\|_{p,\varphi_{\lambda}}. \end{aligned}$$
(39)
For \(\lambda=1\), \(\lambda_{1}=\frac{1}{q}\), \(\lambda_{2}=\frac{1}{p}\) in (38) and (39), we have the following equivalent inequality:
$$\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln (U_{m}V_{n})}< \frac{\pi}{\sin(\frac{\pi}{p})} \Biggl[ \sum_{m=2}^{\infty}\biggl( \frac{U_{m}}{\mu_{m+1}}\biggr)^{p-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum_{n=2}^{\infty}\biggl( \frac{V_{n}}{\upsilon_{n+1}}\biggr)^{q-1}b_{n}^{q} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
(40)
$$\begin{aligned}& \Biggl\{ \sum_{n=2}^{\infty}\frac{\upsilon_{n+1}}{V_{n}} \Biggl[ \sum_{m=2}^{\infty}\frac{a_{m}}{\ln(U_{m}V_{n})} \Biggr] ^{p} \Biggr\} ^{\frac{1}{p}}< \frac{\pi}{\sin(\frac{\pi}{p})} \Biggl[ \sum _{m=2}^{\infty }\biggl(\frac{U_{m}}{\mu_{m+1}} \biggr)^{p-1}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$
(41)
Hence, (38) is an extension of (40), and (31) is a more accurate inequality of (38) (for \(0<\alpha\leq\frac {\mu_{2}}{2}\), \(0<\beta\leq\frac{\upsilon_{2}}{2}\)).
(ii) For \(\mu_{i}=\upsilon_{j}=1 \) (\(i,j\in\mathbf{N}\)), \(\lambda =1\), \(\lambda _{1}=\frac{1}{q}\), \(\lambda_{2}=\frac{1}{p}\) in (31), we reduce our case to the following inequality: For \(\alpha,\beta\leq\frac{1}{2}\),
$$\begin{aligned} &\sum_{m=2}^{\infty}\sum _{n=2}^{\infty}\frac{a_{m}b_{n}}{\ln [(m-\alpha )(n-\beta)]} \\ &\quad< \frac{\pi}{\sin(\pi/p)} \Biggl[ \sum_{m=2}^{\infty} \frac {a_{m}^{p}}{(m-\alpha)^{1-p}} \Biggr] ^{\frac{1}{p}} \Biggl[ \sum _{n=2}^{\infty}\frac {b_{n}^{q}}{(n-\beta)^{1-q}} \Biggr] ^{\frac{1}{q}}, \end{aligned}$$
(42)
Hence, (42) is a more accurate inequality of (3) (for \(0<\alpha,\beta\leq\frac{1}{2}\)).