We also set the following functions:
$$\begin{aligned} {\widetilde{\Phi }}_{\lambda }(m):= & {} \omega (\lambda _{2},m)\frac{(\ln U_{m})^{p(1-\lambda _{1})-1}}{U_{m}^{1-p}\mu _{m}^{p-1}}, \\ {\widetilde{\Psi }}_{\lambda }(n):= & {} \varpi (\lambda _{1},n)\frac{(\ln V_{n})^{q(1-\lambda _{2})-1}}{V_{n}^{1-q}\upsilon _{n}^{q-1}}\quad (m,n\in {\mathbf {N}}\backslash \{1\}). \end{aligned}$$
(23)
Theorem 3.1
If \(0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,\) then:
-
(i)
for \(p>1,\) we have the following equivalent inequalities:
$$\begin{aligned} I:= & {} \sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}b_{n}\le \Vert a\Vert _{p,{\widetilde{\Phi }}_{\lambda }}\Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}, \end{aligned}$$
(24)
$$\begin{aligned} J:= & {} \left\{ \sum _{n=2}^{\infty }\frac{\upsilon _{n}\ln ^{p\lambda _{2}-1}V_{n}}{(\varpi (\lambda _{1},n))^{p-1}V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p}\right\} ^{\frac{1}{p}} \\\le & {} \Vert a\Vert _{p,{\widetilde{\Phi }}_{\lambda }}; \end{aligned}$$
(25)
-
(ii)
for \(0<p<1\) (or \(p<0),\) we have the equivalent reverses of (24) and (25).
Proof
(i) By Hölder’s inequality with weight (cf. [11]) and (12), we have
$$\begin{aligned}&\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{ (\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p} \\&\quad =\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha } }{(\ln U_{m}V_{n})^{\lambda +\alpha }}\right. \\&\qquad \times \left. \left( \frac{U_{m}^{1/q}(\ln U_{m})^{(1-\lambda _{1})/q}}{ (\ln V_{n})^{(1-\lambda _{2})/p}\mu _{m}^{1/q}}a_{m}\right) \left( \frac{ (\ln V_{n})^{(1-\lambda _{2})/p}\mu _{m}^{1/q}}{U_{m}^{1/q}(\ln U_{m})^{(1-\lambda _{1})/q}}\right) \right] ^{p} \\&\quad \le \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{ (\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{U_{m}^{p-1}(\ln U_{m})^{(1-\lambda _{1})p/q}}{(\ln V_{n})^{1-\lambda _{2}}\mu _{m}^{p/q}}a_{m}^{p} \\&\qquad \times \left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{(\ln V_{n})^{(1-\lambda _{2})(q-1)}\mu _{m}}{U_{m}(\ln U_{m})^{1-\lambda _{1}}}\right] ^{p-1} \\&\quad =\frac{(\varpi (\lambda _{1},n))^{p-1}V_{n}}{(\ln V_{n})^{p\lambda _{2}-1}\upsilon _{n}} \\&\qquad \times \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{ (\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{\upsilon _{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda _{1})(p-1)}a_{m}^{p}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}\mu _{m}^{p-1}}. \end{aligned}$$
(26)
Hence by (11), we deduce that
$$\begin{aligned} J\le & {} \left[ \sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{ \upsilon _{n}U_{m}^{p-1}(\ln U_{m})^{(1-\lambda _{1})(p-1)}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}\mu _{m}^{p-1}}a_{m}^{p}\right] ^{\frac{1}{p}} \\= & {} \left[ \sum _{m=2}^{\infty }\sum _{n=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{ \upsilon _{n}(\ln U_{m})^{\lambda _{1}}}{V_{n}(\ln V_{n})^{1-\lambda _{2}}} \frac{(\ln U_{m})^{p(1-\lambda _{1})-1}}{U_{m}^{1-p}\mu _{m}^{p-1}}a_{m}^{p} \right] ^{\frac{1}{p}} \\= & {} \left[ \sum _{m=2}^{\infty }\omega (\lambda _{2},m)\frac{(\ln U_{m})^{p(1-\lambda _{1})-1}}{U_{m}^{1-p}\mu _{m}^{p-1}}a_{m}^{p}\right] ^{ \frac{1}{p}}, \end{aligned}$$
(27)
and then (25) follows.
By Hölder’s inequality (cf. [11]), we have
$$\begin{aligned} I= & {} \sum _{n=2}^{\infty }\left[ \frac{(\ln V_{n})^{\lambda _{2}-\frac{1}{p} }\upsilon _{n}^{1/p}}{(\varpi (\lambda _{1},n))^{\frac{1}{q}}V_{n}^{\frac{1}{ p}}}\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] \\&\times \left[ (\varpi (\lambda _{1},n))^{\frac{1}{q}}\frac{(\ln V_{n})^{ \frac{1}{p}-\lambda _{2}}}{V_{n}^{\frac{-1}{p}}\upsilon _{n}^{\frac{1}{p}}} b_{n}\right] \le J\Vert b\Vert _{q,\widetilde{\Psi }_{\lambda }}. \end{aligned}$$
(28)
Then by (25), we derive (24).
On the other hand, assuming that (24) holds true, we set
$$\begin{aligned} b_{n}:=\frac{(\ln V_{n})^{p\lambda _{2}-1}\upsilon _{n}}{(\varpi (\lambda _{1},n))^{p-1}V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p-1},\quad n\in {\mathbf {N}}\backslash \{1\}. \end{aligned}$$
(29)
Then we obtain that \(J^{p}=\Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}^{q}.\) If \(J=0,\) then (25) is trivially valid; if \(J=\infty ,\) then by (27), (25) takes the form of equality (\(=\infty \)). Suppose that \(0<J<\infty .\) By (24), it follows that
$$\begin{aligned} \Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}^{q}= & {} J^{p}=I\le \Vert a\Vert _{p, {\widetilde{\Phi }}_{\lambda }}\Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}, \end{aligned}$$
(30)
$$\begin{aligned} \Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}^{q-1}= & {} J\le \Vert a\Vert _{p,{\widetilde{\Phi }}_{\lambda }}, \end{aligned}$$
(31)
and then (25) follows, which is equivalent to (24).
(ii) For \(0<p<1\) (or \(p<0),\) by the reverse Hölder inequality with weight (cf. [11]) and (12), we obtain the reverse of (26) (or (26)). Then by (11), we derive the reverse of (27), and thus the reverse of (25) follows. By Hölder’s inequality (cf. [11]), we obtain the reverse of (28) and then by the reverse of (25), we deduce the reverse of (24).
On the other hand, assuming that the reverse of (24) holds true, we set \(b_{n}\) as in (29). Then we obtain that \(J^{p}=\Vert b\Vert _{q,{\widetilde{\Psi }} _{\lambda }}^{q}.\) If \(J=\infty ,\) then the reverse of (25) is trivially valid; if \(J=0,\) then by the reverse of (27), (25) takes the form of equality (\(=0\)). Suppose that \(0<J<\infty .\) By the reverse of (24), it follows that the reverses of (30) and (31) are valid and then the reverse of (25) follows, which is equivalent to the reverse of (24). \(\square \)
Theorem 3.2
If \(0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})\) is defined as in (8), there exist \(m_{0},n_{0}\in {\mathbf {N}},\) such that \(\{\mu _{m}\}_{m=m_{0}}^{\infty }\) and \(\{\upsilon _{n}\}_{n=n_{0}}^{\infty }\) are decreasing, \(U_{\infty }=V_{\infty }=\infty \), then for \(p>1\), \(\Vert a\Vert _{p,\Phi _{\lambda }}\in {\mathbf {R}}_{+}\) and \(\Vert b\Vert _{q,\Psi _{\lambda }}\in {\mathbf {R}}_{+},\) we have the following equivalent inequalities:
$$\begin{aligned}&\sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}b_{n}<k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}\Vert b\Vert _{q,\Psi _{\lambda }}, \end{aligned}$$
(32)
$$\begin{aligned}&J_{1} :=\left\{ \sum _{n=2}^{\infty }\frac{\upsilon _{n}\ln ^{p\lambda _{2}-1}V_{n}}{V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p}\right\} ^{\frac{1}{p}} \\&\quad <k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}, \end{aligned}$$
(33)
where the constant factor \(k_{\alpha }(\lambda _{1})\) is the best possible.
Proof
Applying (13) and (14) in (24) and (25), since
$$\begin{aligned} (\omega (\lambda _{2},m))^{\frac{1}{p}}<(k_{\alpha }(\lambda _{1}))^{\frac{1 }{p}}\quad (p>1),\qquad (\varpi (\lambda _{1},n))^{\frac{1}{q}}<(k_{\alpha }(\lambda _{1}))^{\frac{1}{q}}\quad (q>1), \end{aligned}$$
and
$$\begin{aligned} \frac{1}{(k_{\alpha }(\lambda _{1}))^{p-1}}<\frac{1}{(\varpi (\lambda _{1},n))^{p-1}}\quad (p>1), \end{aligned}$$
by simplification, we obtain the equivalent inequalities (32) and (33).
For \(\varepsilon \in (0,q(\alpha +\lambda _{2})),\) we set \({\widetilde{\lambda }}_{1}=\lambda _{1}+\frac{\varepsilon }{q}\,(>-\alpha ),{\widetilde{\lambda }}_{2}=\lambda _{2}-\frac{\varepsilon }{q}\,(\in (-\alpha ,1)),\) and \( {\widetilde{a}}=\{{\widetilde{a}}_{m}\}_{m=2}^{\infty },\) \({\widetilde{b}}=\{ {\widetilde{b}}_{n}\}_{n=2}^{\infty },\)
$$\begin{aligned} {\widetilde{a}}_{m}:= & {} \frac{\mu _{m}}{U_{m}}\ln ^{\widetilde{\lambda } _{1}-\varepsilon -1}U_{m}=\frac{\mu _{m}}{U_{m}}\ln ^{\lambda _{1}-\frac{\varepsilon }{p}-1}U_{m}, \\ {\widetilde{b}}_{n}= & {} \frac{\upsilon _{n}}{V_{n}}\ln ^{{\widetilde{\lambda }} _{2}-1}V_{n}=\frac{\upsilon _{n}}{V_{n}}\ln ^{\lambda _{2}-\frac{\varepsilon }{q}-1}V_{n}. \end{aligned}$$
(34)
Then by (20), (21) and (19), we have
$$\begin{aligned}&\Vert {\widetilde{a}}\Vert _{p,\Phi _{\lambda }}\Vert {\widetilde{b}}\Vert _{q,\Psi _{\lambda }}=\left( \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{1+\varepsilon }U_{m} }\right) ^{\frac{1}{p}}\left( \sum _{n=2}^{\infty }\frac{\upsilon _{n}}{ V_{n}\ln ^{1+\varepsilon }V_{n}}\right) ^{\frac{1}{q}} \\&\quad =\frac{1}{\varepsilon }\left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}} +\varepsilon O_{1}(1)\right] ^{\frac{1}{p}}\left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}}+\varepsilon O_{2}(1)\right] ^{\frac{1}{q}},\\&{\widetilde{I}} :=\sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\widetilde{ a}_{m}{\widetilde{b}}_{n} \\&\quad =\sum _{m=2}^{\infty }\left[ \sum _{n=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}\frac{ \upsilon _{n}\ln ^{{\widetilde{\lambda }}_{1}}U_{m}}{V_{n}\ln ^{1-{\widetilde{\lambda }}_{2}}V_{n}}\right] \frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m}}\\&\quad =\sum _{m=2}^{\infty }\omega ({\widetilde{\lambda }}_{2},m)\frac{\mu _{m}}{ U_{m}\ln ^{\varepsilon +1}U_{m}} \\&\quad \ge k_{\alpha }({\widetilde{\lambda }}_{1})\sum _{m=2}^{\infty }\left( 1-O\left( \frac{1 }{\ln ^{{\widetilde{\lambda }}_{2}+\alpha }U_{m}}\right) \right) \frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m}} \\&\quad = k_{\alpha }({\widetilde{\lambda }}_{1})\left[ \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{\varepsilon +1}U_{m}}-\sum _{m=2}^{\infty }O\left( \frac{\mu _{m}}{U_{m}(\ln U_{m})^{\left( \frac{\varepsilon }{p}+\alpha +\lambda _{2}\right) +1}} \right) \right] \\&\quad =\frac{1}{\varepsilon }k_{\alpha }\left( \lambda _{1}+\frac{\varepsilon }{q}\right) \left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon (O_{1}(1)-O(1)) \right] . \end{aligned}$$
If there exists a positive constant \(K\le k_{\alpha }(\lambda _{1}),\) such that (32) is valid when replacing \(k_{\alpha }(\lambda _{1})\) by K, then in particular, we have \(\varepsilon {\widetilde{I}}<\varepsilon K\Vert {\widetilde{a}}\Vert _{p,\Phi _{\lambda }}\Vert {\widetilde{b}}\Vert _{q,\Psi _{\lambda }},\) namely
$$\begin{aligned}&k_{\alpha }\left( \lambda _{1}+\frac{\varepsilon }{q}\right) \left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon (O_{1}(1)-O(1))\right] \\&\quad <K\left[ \frac{1}{\ln ^{\varepsilon }U_{m_{0}+1}}+\varepsilon O_{1}(1) \right] ^{\frac{1}{p}}\left[ \frac{1}{\ln ^{\varepsilon }V_{n_{0}+1}} +\varepsilon O_{2}(1)\right] ^{\frac{1}{q}}. \end{aligned}$$
(35)
In view of (22), it follows that \(k_{\alpha }(\lambda _{1})\le K\,(\varepsilon \rightarrow 0^{+}).\) Hence, \(K=k_{\alpha }(\lambda _{1})\) is the best possible constant factor of (32).
Similarly to (28), we can still obtain that
$$\begin{aligned} I\le J_{1}\Vert b\Vert _{q,\Psi _{\lambda }}. \end{aligned}$$
(36)
Hence, we can prove that the constant factor \(k_{\alpha }(\lambda _{1})\) in (33) is the best possible. Otherwise, we would reach a contradiction by (36) that the constant factor in (32) is not the best possible.
For \(p>1,\)
$$\begin{aligned} \Psi _{\lambda }^{1-p}(n)=\frac{\upsilon _{n}}{V_{n}}(\ln V_{n})^{p\lambda _{2}-1}\quad (n\in {\mathbf {N}}\backslash \{1\}), \end{aligned}$$
we define the following normed spaces:
$$\begin{aligned} l_{p,\Phi _{\lambda }}:= & {} \{a=\{a_{m}\}_{m=2}^{\infty };\Vert a\Vert _{p,\Phi _{\lambda }}<\infty \}, \\ l_{q,\Psi _{\lambda }}:= & {} \{b=\{b_{n}\}_{n=2}^{\infty };\Vert b\Vert _{q,\Psi _{\lambda }}<\infty \}, \\ l_{p,\Psi _{\lambda }^{1-p}}:= & {} \{c=\{c_{n}\}_{n=2}^{\infty };\Vert c\Vert _{p,\Psi _{\lambda }^{1-p}}<\infty \}. \end{aligned}$$
Assuming that \(a=\{a_{m}\}_{m=2}^{\infty }\in l_{p,\Phi _{\lambda }},\) setting
$$\begin{aligned} c=\{c_{n}\}_{n=2}^{\infty },\quad c_{n}:=\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m},\ n\in {\mathbf {N}}\backslash \{1\}, \end{aligned}$$
we can rewrite (33) as follows:
$$\begin{aligned}\Vert c\Vert _{p,\Psi _{\lambda }^{1-p}}<k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}<\infty ,\end{aligned}$$
namely, \(c\in l_{p,\Psi _{\lambda }^{1-p}}.\)
Definition 3.3
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:
$$\begin{aligned} (Ta,b):=\sum _{n=2}^{\infty }\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m} \right] b_{n}. \end{aligned}$$
(37)
Then we can rewrite (32) and (33) as bellow:
$$\begin{aligned} (Ta,b)< & {} k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}\Vert b\Vert _{q,\Psi _{\lambda }}, \end{aligned}$$
(38)
$$\begin{aligned} \Vert Ta\Vert _{p,\Psi _{\lambda }^{1-p}}< & {} k_{\alpha }(\lambda _{1})\Vert a\Vert _{p,\Phi _{\lambda }}. \end{aligned}$$
(39)
Define the norm of operator T as follows:
$$\begin{aligned} \Vert T\Vert :=\sup _{a(\ne \theta )\in l_{p,\Phi _{\lambda }}}\frac{\Vert Ta\Vert _{p,\Psi _{\lambda }^{1-p}}}{\Vert a\Vert _{p,\Phi _{\lambda }}}. \end{aligned}$$
Then by (39), we derive that \(\Vert T\Vert \le k_{\alpha }(\lambda _{1}).\) Since the constant factor in (39) is the best possible, we have
$$\begin{aligned} \Vert T\Vert =k_{\alpha }(\lambda _{1})=\sum _{i=1}^{2}\frac{1}{\lambda _{i}+\alpha } F(\lambda +\alpha ,\lambda _{i}+\alpha ,\lambda _{i}+\alpha +1,-1). \end{aligned}$$
(40)