In the following, we agree that \(\mu_{1},\upsilon_{1}\geq1\), \(\mu _{i},\upsilon_{j}>0\) (\(i,j\in\mathbf{N}\backslash\{1\}\)), \(U_{m}\) and \(V_{n}\) are defined by (4), \(-1<\alpha\leq1\), \(0<\lambda _{1},\lambda_{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(K_{\lambda }(m,n)\), and \(k_{\alpha}(\lambda_{1})\) are indicated by (7) and (9), \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(a_{m},b_{n}\geq0 \) (\(m,n\in\mathbf{N}\backslash\{1\}\)),
$$ \|a\|_{p,\Phi_{\lambda}}:=\Biggl(\sum_{m=2}^{\infty} \Phi_{\lambda }(m)a_{m}^{p}\Biggr)^{\frac{1}{p}},\qquad \|b\|_{q,\Psi_{\lambda}}:=\Biggl(\sum_{n=2}^{\infty } \Psi_{\lambda}(n)b_{n}^{q}\Biggr)^{\frac{1}{q}}, $$
where
$$\begin{aligned}& \Phi_{\lambda}(m) : =\biggl(\frac{U_{m}}{\mu_{m}}\biggr)^{p-1}(\ln U_{m})^{p(1-\lambda_{1})-1}, \\& \Psi_{\lambda}(n) : =\biggl(\frac{V_{n}}{\upsilon_{n}}\biggr)^{q-1}(\ln V_{n})^{q(1-\lambda_{2})-1}\quad \bigl(m,n\in\mathbf{N}\backslash\{1\}\bigr). \end{aligned}$$
Theorem 1
If
\(0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi _{\lambda}}<\infty\), then we have the following equivalent inequalities:
$$\begin{aligned}& I : =\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}K_{\lambda }(m,n)a_{m}b_{n}< k_{\alpha}( \lambda_{1})\|a\|_{p,\Phi_{\lambda }}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(28)
$$\begin{aligned}& J : = \Biggl[ \sum_{n=2}^{\infty} \frac{\upsilon_{n}}{V_{n}}(\ln V_{n})^{p\lambda_{2}-1}\Biggl(\sum _{m=2}^{\infty}K_{\lambda}(m,n)a_{m} \Biggr)^{p} \Biggr] ^{\frac{1}{p}}< k_{\alpha}( \lambda_{1})\|a\|_{p,\Phi_{\lambda}}. \end{aligned}$$
(29)
In particular, for
\(\lambda_{1}=\lambda_{2}=\frac{\lambda}{2}\in(0,1]\), the constant factor
\(k_{\alpha}(\lambda_{1})\)
in the above inequalities is expressed in the following form:
$$ k_{\alpha}\biggl(\frac{\lambda}{2}\biggr)=\frac{4}{\lambda(1-\alpha^{2})^{\frac {1}{2}}}\arctan \biggl( \frac{1-\alpha}{1+\alpha} \biggr) ^{\frac{1}{2}}. $$
Proof
By Hölder’s inequality with weight (cf. [32]) and (16), we have
$$\begin{aligned} \Biggl( \sum_{m=2}^{\infty}K_{\lambda}(m,n)a_{m} \Biggr) ^{p} =&\Biggl[ \sum_{m=2}^{\infty}K_{\lambda}(m,n) \biggl( \frac{U_{m}^{1/q}(\ln U_{m})^{(1-\lambda _{1})/q}}{(\ln V_{n})^{(1-\lambda_{2})/p}\mu_{m}^{1/q}}a_{m} \biggr) \biggl( \frac {(\ln V_{n})^{(1-\lambda_{2})/p}\mu_{m}^{1/q}}{U_{m}^{1/q}(\ln U_{m})^{(1-\lambda_{1})/q}} \biggr) \Biggr] ^{p} \\ \leq&\sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac{U_{m}^{p-1}(\ln U_{m})^{(1-\lambda_{1})p/q}}{(\ln V_{n})^{1-\lambda_{2}}\mu _{m}^{p/q}}a_{m}^{p} \\ &{}\times \Biggl[ \sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac{(\ln V_{n})^{(1-\lambda_{2})(q-1)}\mu_{m}}{U_{m}(\ln U_{m})^{1-\lambda _{1}}} \Biggr] ^{p-1} \\ =&\frac{(\varpi(\lambda_{1},n))^{p-1}V_{n}}{(\ln V_{n})^{p\lambda _{2}-1}\upsilon_{n}}\sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac{\upsilon _{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda_{1})(p-1)}a_{m}^{p}}{V_{n}(\ln V_{n})^{1-\lambda_{2}}\mu_{m}^{p-1}}. \end{aligned}$$
(30)
Then by (18), we obtain
$$\begin{aligned} J \leq&\bigl(k_{\alpha}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{n=2}^{\infty }\sum _{m=2}^{\infty}K_{\lambda}(m,n)\frac{\upsilon_{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda_{1})(p-1)}a_{m}^{p}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}\mu_{m}^{p-1}} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{\alpha}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=2}^{\infty }\sum _{n=2}^{\infty}K_{\lambda}(m,n)\frac{\upsilon_{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda_{1})(p-1)}a_{m}^{p}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}\mu_{m}^{p-1}} \Biggr] ^{\frac{1}{p}} \\ =&\bigl(k_{\alpha}(\lambda_{1})\bigr)^{\frac{1}{q}} \Biggl[ \sum_{m=2}^{\infty }\omega(\lambda_{2},m) \frac{(\ln U_{m})^{p(1-\lambda_{1})-1}}{U_{m}^{1-p}\mu_{m}^{p-1}}a_{m}^{p} \Biggr] ^{\frac{1}{p}}. \end{aligned}$$
(31)
Hence, by (17), we have (29).
By Hölder’s inequality (cf. [32]), we have
$$\begin{aligned} I =&\sum_{n=2}^{\infty} \Biggl[ \biggl( \frac{\upsilon_{n}}{V_{n}}\biggr)^{1/p}(\ln V_{n})^{\lambda_{2}-\frac{1}{p}} \sum_{m=2}^{\infty}K_{\lambda }(m,n)a_{m} \Biggr] \\ &{}\times \biggl[ \biggl(\frac{\upsilon_{n}}{V_{n}}\biggr)^{-1/p}(\ln V_{n})^{\frac {1}{p}-\lambda_{2}}b_{n} \biggr] \leq J\|b \|_{q,\Psi_{\lambda}}, \end{aligned}$$
(32)
and then by (29), we have (28).
On the other hand, assuming that (29) is valid, setting
$$ b_{n}:=\frac{\upsilon_{n}}{V_{n}}(\ln V_{n})^{p\lambda_{2}-1} \Biggl( \sum_{m=2}^{\infty}K_{\lambda}(m,n)a_{m} \Biggr) ^{p-1},\quad n\in\mathbf{ N}\backslash\{1\}, $$
(33)
we find \(J^{p}=\|b\|_{q,\Psi_{\lambda}}^{q}\). If \(J=0\), then (29) is trivially valid; if \(J=\infty\), then by (31) and (17), it is impossible; if \(0< J<\infty\), then by (28), it follows that
$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q} = J^{p}=I< k_{\alpha}( \lambda _{1})\|a\|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(34)
$$\begin{aligned}& \|b\|_{q,\Psi_{\lambda}}^{q-1} = J< k_{\alpha}(\lambda _{1})\|a\|_{p,\Phi _{\lambda}}, \end{aligned}$$
(35)
namely, (29) follows, which is equivalent to (28). □
Theorem 2
With regards the assumptions of Theorem
1, if
\(U(\infty )=V(\infty)=\infty\), there exist
\(m_{0},n_{0}\in\mathbf{N}\backslash \{1\}\), such that
\(\mu_{m}\geq\mu_{m+1}\) (\(m\in \{m_{0},m_{0}+1,\ldots\}\)), \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\in \{n_{0},n_{0}+1,\ldots\}\)), then the constant factor
\(k_{\alpha }(\lambda_{1})\)
in (28) and (29) is the best possible.
Proof
For \(\varepsilon\in(0,p\lambda_{1})\), we set \(\tilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon}{p}\) (\(\in(0,1)\)), \(\tilde{ \lambda}_{2}=\lambda_{2}+\frac{\varepsilon}{p}\) (>0), \(\tilde {a}=\{\tilde{a}_{m}\}_{m=2}^{\infty}\), \(\tilde{b}=\{\tilde{b}_{n}\}_{n=2}^{\infty}\), where
$$ \begin{aligned} &\tilde{a}_{m} : =\frac{\mu_{m}}{U_{m}} \ln^{\tilde{\lambda}_{1}-1}U_{m}=\frac{\mu_{m}}{U_{m}}\ln^{\lambda_{1}-\frac{\varepsilon }{p}-1}U_{m}, \\ &\tilde{b}_{n} : =\frac{\upsilon_{n}}{V_{n}}\ln^{\tilde {\lambda}_{2}-\varepsilon-1}V_{n}= \frac{\upsilon_{n}}{V_{n}}\ln^{\lambda_{2}- \frac{\varepsilon}{q}-1}V_{n}. \end{aligned} $$
(36)
Then by (25), (26), and (22), we find
$$\begin{aligned}& \|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi_{\lambda}}= \Biggl( \sum _{m=2}^{\infty}\frac{\mu_{m}}{U_{m}\ln^{1+\varepsilon }U_{m}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=2}^{\infty} \frac{\upsilon_{n}}{V_{n}\ln^{1+\varepsilon}V_{n}} \Biggr) ^{\frac{1}{q}} \\& \hphantom{\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi_{\lambda}}} = \frac{1}{\varepsilon} \biggl( \frac{1}{\ln^{\varepsilon}U_{m_{0}}}+ \varepsilon O_{1}(1) \biggr) ^{\frac{1}{p}} \biggl( \frac{1}{\ln ^{\varepsilon }V_{n_{0}}}+\varepsilon O_{2}(1) \biggr) ^{\frac{1}{q}}, \\& \tilde{I} : =\sum_{n=2}^{\infty}\sum _{m=2}^{\infty}K_{\lambda }(m,n)\tilde{a}_{m}\tilde{b}_{n} \\& \hphantom{\tilde{I} } = \sum_{n=2}^{\infty} \Biggl( \sum_{m=2}^{\infty}K_{\lambda}(m,n) \frac {\mu _{m}\ln^{\tilde{\lambda}_{2}}V_{n}}{U_{m}\ln^{1-\tilde {\lambda}_{1}}U_{m}} \Biggr) \frac{\upsilon_{n}}{V_{n}\ln^{\varepsilon +1}V_{n}} \\& \hphantom{\tilde{I} } = \sum_{n=2}^{\infty} \frac{\varpi(\tilde{\lambda }_{1},n)\upsilon_{n}}{V_{n}\ln^{\varepsilon+1}V_{n}}\geq k_{\alpha}(\tilde{\lambda}_{1}) \sum_{n=2}^{\infty} \biggl( 1-O\biggl( \frac{1}{\ln^{\tilde{\lambda}_{1}}V_{n}}\biggr) \biggr) \frac{\upsilon_{n}}{V_{n}\ln^{\varepsilon +1}V_{n}} \\& \hphantom{\tilde{I} } = k_{\alpha}(\tilde{\lambda}_{1}) \Biggl[ \sum _{n=2}^{\infty}\frac {\upsilon_{n}}{V_{n}\ln^{\varepsilon+1}V_{n}}-\sum _{n=2}^{\infty }O \biggl( \frac{\upsilon_{n}}{V_{n}(\ln V_{n})^{(\frac{\varepsilon}{q}+\lambda _{1})+1}} \biggr) \Biggr] \\& \hphantom{\tilde{I} } = \frac{1}{\varepsilon}k_{\alpha}(\tilde{ \lambda}_{1}) \biggl[ \frac{1}{\ln^{\varepsilon}V_{n_{0}}}+\varepsilon \bigl(O_{2}(1)-O(1)\bigr) \biggr] . \end{aligned}$$
If there exists a positive constant \(K\leq k_{\alpha}(\lambda_{1})\), such that (28) is valid when replacing \(k_{\alpha}(\lambda_{1})\) by K, then, in particular, we have \(\varepsilon\tilde{I}<\varepsilon K\|\tilde{a}\|_{p,\Phi_{\lambda}}\|\tilde{b}\|_{q,\Psi_{\lambda}}\), namely
$$\begin{aligned}& k_{\alpha}\biggl(\lambda_{1}-\frac{\varepsilon}{p}\biggr) \biggl[ \frac{1}{\ln ^{\varepsilon}V_{n_{0}}}+\varepsilon\bigl(O_{2}(1)-O(1)\bigr) \biggr] \\& \quad < K \biggl( \frac{1}{\ln^{\varepsilon}U_{m_{0}}}+\varepsilon O_{1}(1) \biggr) ^{\frac{1}{p}} \biggl( \frac{1}{\ln^{\varepsilon }V_{n_{0}}}+\varepsilon O_{2}(1) \biggr) ^{\frac{1}{q}}. \end{aligned}$$
By (27), it follows that \(k_{\alpha}(\lambda_{1})\leq K\) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k_{\alpha}(\lambda_{1})\) is the best possible constant factor of (28).
The constant factor \(k_{\alpha}(\lambda_{1})\) in (29) is still the best possible. Otherwise, we would reach a contradiction by (32) that the constant factor in (28) is not the best possible. □
For \(p>1\), we find \(\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon _{n}}{V_{n}}(\ln V_{n})^{p\lambda_{2}-1}\) (\(n\in\mathbf{N}\backslash\{1\}\)) and define the following 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
$$ c=\{c_{n}\}_{n=2}^{\infty},\qquad c_{n}:= \sum_{m=2}^{\infty}K_{\lambda }(m,n)a_{m}, \quad n\in\mathbf{N}, $$
we can rewrite (29) as \(\|c\|_{p,\Psi_{\lambda }^{1-p}}< k_{\alpha }(\lambda_{1})\|a\|_{p,\Phi_{\lambda}}<\infty\), 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 \(Ta=c\in l_{p,\Psi_{\lambda}^{1-p}}\). Define the formal inner product of Ta and \(b=\{b_{n}\}_{n=2}^{\infty}\in l_{q,\Psi_{\lambda}}\) as follows:
$$ (Ta,b):=\sum_{n=2}^{\infty} \Biggl( \sum _{m=2}^{\infty}K_{\lambda }(m,n)a_{m} \Biggr) b_{n}. $$
(37)
Then we can rewrite (28) and (29) as follows:
$$\begin{aligned}& (Ta,b) < k_{\alpha}(\lambda_{1})\|a\|_{p,\Phi_{\lambda }}\|b \|_{q,\Psi _{\lambda}}, \end{aligned}$$
(38)
$$\begin{aligned}& \|Ta\|_{p,\Psi_{\lambda}^{1-p}} < k_{\alpha}(\lambda _{1})\|a \|_{p,\Phi _{\lambda}}. \end{aligned}$$
(39)
Define the norm of operator T as follows:
$$ \|T\|:=\sup_{a\, (\neq\theta)\in l_{p,\Phi_{\lambda}}}\frac {\|Ta\|_{p,\Psi _{\lambda}^{1-p}}}{\|a\|_{p,\Phi_{\lambda}}}. $$
Then by (39), it follows that \(\|T\|\leq k_{\alpha}(\lambda_{1})\). In view of Theorem 2, the constant factor in (39) is the best possible, we have
$$ \|T\|=k_{\alpha}(\lambda_{1})= \int_{0}^{1}\frac{t^{\lambda _{1}-1}+t^{\lambda_{2}-1}}{1+\alpha+(1-\alpha)t^{\lambda}}\,dt. $$
(40)
Remark 1
(i) For \(\lambda=1\), \(\lambda_{1}=\frac{1}{q}\), \(\lambda_{2}=\frac{1}{p}\), (28) reduces to
$$ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln U_{m}V_{n}+\alpha|\ln\frac{U_{m}}{V_{n}}|}< k_{\alpha} \biggl(\frac {1}{q}\biggr) \Biggl( \sum_{m=2}^{\infty} \frac{a_{m}^{p}}{\mu_{m}^{p-1}} \Biggr) ^{\frac {1}{p}} \Biggl( \sum _{n=2}^{\infty}\frac{b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}. $$
(41)
In particular, for \(\alpha=0\), we have the following simple Hardy-Mulholland-type inequality:
$$ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln U_{m}V_{n}}< \frac{\pi}{\sin(\pi/p)} \Biggl( \sum_{m=2}^{\infty} \frac {a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum _{n=2}^{\infty}\frac {b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}, $$
(42)
which is an extension of (3); for \(\alpha=1\), we have another simple Hardy-Mulholland-type inequality:
$$ \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\max\{\ln U_{m},\ln V_{n}\}}< pq \Biggl( \sum _{m=2}^{\infty}\frac{a_{m}^{p}}{\mu _{m}^{p-1}} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{n=2}^{\infty} \frac {b_{n}^{q}}{\upsilon_{n}^{q-1}} \Biggr) ^{\frac{1}{q}}. $$
(43)
Hence, inequality (28) is a relation between (42) and (43).
(ii) For \(\alpha=1\) in (28) and (29), in view of (9), we have the following equivalent Hardy-Mulholland-type inequalities with parameters:
$$\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\max\{\ln ^{\lambda}U_{m},\ln^{\lambda}V_{n}\}}< \frac{\lambda}{\lambda _{1}\lambda _{2}}\|a \|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(44)
$$\begin{aligned}& \Biggl[ \sum_{n=2}^{\infty}\frac{\upsilon_{n}}{V_{n}}( \ln V_{n})^{p\lambda_{2}-1} \Biggl( \sum_{m=2}^{\infty} \frac{a_{m}}{\max\{ \ln ^{\lambda}U_{m},\ln^{\lambda}V_{n}\}} \Biggr) ^{p} \Biggr] ^{\frac{1}{p}} \\& \quad < \frac{\lambda}{\lambda_{1}\lambda_{2}}\|a\|_{p,\Phi_{\lambda}}. \end{aligned}$$
(45)
(iii) For \(\alpha=0\) in (28) and (29), in view of (11), we have another equivalent Hardy-Mulholland-type inequalities with parameters:
$$\begin{aligned}& \sum_{n=2}^{\infty}\sum _{m=2}^{\infty}\frac{a_{m}b_{n}}{\ln^{\lambda }U_{m}+\ln^{\lambda}V_{n}}< \frac{\pi}{\lambda\sin(\frac{\pi \lambda_{1}}{\lambda})}\|a \|_{p,\Phi_{\lambda}}\|b\|_{q,\Psi_{\lambda}}, \end{aligned}$$
(46)
$$\begin{aligned}& \Biggl[ \sum_{n=2}^{\infty}\frac{\upsilon_{n}}{V_{n}}( \ln V_{n})^{p\lambda_{2}-1} \Biggl( \sum_{m=2}^{\infty} \frac{a_{m}}{\ln ^{\lambda}U_{m}+\ln^{\lambda}V_{n}} \Biggr) ^{p} \Biggr] ^{\frac{1}{p}} \\& \quad < \frac{\pi}{\lambda\sin(\frac{\pi\lambda_{1}}{\lambda})}\|a\|_{p,\Phi_{\lambda}}. \end{aligned}$$
(47)
In view of Theorem 2, the constant factors in the above inequalities are all the best possible. Inequality (28) is also a relation between the two Hardy-Mulholland-type inequalities (46) and (44) with parameters.