Theorem 1
Suppose that
\(0 < p < 1,\frac{1}{p} + \frac{1}{q} = 1\),
$$ k(\lambda_{1}): = k_{\beta}^{1/p}( \lambda_{1})k_{\alpha}^{1/q}(\lambda_{1}) = \frac{2\pi^{2}\csc^{2/p}\beta \csc^{2/q}\alpha}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}}. $$
(16)
If
\(a_{m},b_{n} \ge 0\ ( \vert m \vert , \vert n \vert \in \mathbf{N}\backslash \{ 1\} )\), satisfy
$$0 < \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}} a_{m}^{p} < \infty,\qquad 0 < \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} < \infty, $$
then for
$$\begin{aligned} \theta (\lambda_{2},m) &= \biggl[\frac{\lambda}{\pi} \sin \biggl( \frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln [(2 + \eta )(1 + \cos \beta )]}{\ln A_{\xi,\alpha} (m)}} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du \\ &= O\biggl(\frac{1}{\ln^{\lambda_{2}/2}A_{\xi,\alpha} (m)}\biggr) \in (0,1), \end{aligned}$$
we obtain the following equivalent reverse Mulholland-type inequalities:
$$\begin{aligned} I: ={}& \sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln (\ln A_{\xi,\alpha} (m)/\ln A_{\eta,\beta} (n))}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} A_{\eta,\beta} (n)} a_{m}b_{n} \\ &\quad> \frac{2\pi^{2}\csc^{2/p}\beta \csc^{2/q}\alpha}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \\ &\qquad{}\times \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &\qquad{}\times \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned}$$
(17)
$$\begin{aligned} J_{1}: ={}& \Biggl\{ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln (\ln A_{\xi,\alpha} (m)/\ln A_{\eta,\beta} (n))}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} A_{\eta,\beta} (n)}a_{m} \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\ >{}& \frac{2\pi^{2}\csc^{2/p}\beta \csc^{2/q}\alpha}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}, \end{aligned}$$
(18)
$$\begin{aligned} J_{2}: = {}&\Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{q\lambda_{1} - 1}A_{\xi,\alpha} (m)}{(1 - \theta (\lambda_{2},m))^{q - 1}A_{\xi,\alpha} (m)} \Biggl( \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln (\ln A_{\xi,\alpha} (m)/\ln A_{\eta,\beta} (n))}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} A_{\eta,\beta} (n)}b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} \\ >{}& \frac{2\pi^{2}\csc^{2/p}\beta \csc^{2/q}\alpha}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(19)
Particularly, (i) for
\(\alpha = \beta = \frac{\pi}{2},\xi,\eta \in [0,\frac{1}{2}]\), setting
$$\begin{aligned} \theta_{1}(\lambda_{2},m)&: = \biggl[\frac{\lambda}{\pi} \sin \biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln (2 + \eta )}{\ln \vert m - \xi \vert }} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du \\ &= O\biggl(\frac{1}{\ln^{\lambda_{2}/2} \vert m - \xi \vert }\biggr) \in (0,1), \end{aligned}$$
we have the following equivalent reverse Mulholland-type inequalities:
$$\begin{aligned} &\sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln ( \vert m - \xi \vert / \vert n - \eta \vert )}{\ln^{\lambda} \vert m - \xi \vert - \ln^{\lambda} \vert n - \eta \vert } a_{m}b_{n} \\ &\quad > \frac{2\pi^{2}}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \\ &\qquad{}\times \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta_{1}(\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1} \vert m - \xi \vert }{ \vert m - \xi \vert ^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &\qquad{}\times\Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1} \vert n - \eta \vert }{ \vert n - \eta \vert ^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned}$$
(20)
$$\begin{aligned} &\Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1} \vert n - \eta \vert }{ \vert n - \eta \vert } \Biggl( \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln ( \vert m - \xi \vert / \vert n - \eta \vert )}{\ln^{\lambda} \vert m - \xi \vert - \ln^{\lambda} \vert n - \eta \vert }a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} \\ &\quad > \frac{2\pi^{2}}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta_{1}(\lambda_{2},m)\bigr) \frac{\ln^{p(1 - \lambda_{1}) - 1} \vert m - \xi \vert }{ \vert m - \xi \vert ^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}, \end{aligned}$$
(21)
$$\begin{aligned} &\Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{q\lambda_{1} - 1} \vert m - \xi \vert }{(1 - \theta_{1}(\lambda_{2},m))^{q - 1} \vert m - \xi \vert } \Biggl( \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln ( \vert m - \xi \vert / \vert n - \eta \vert )a_{m}}{\ln^{\lambda} \vert m - \xi \vert - \ln^{\lambda} \vert n - \eta \vert }b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} \\ &\quad > \frac{2\pi^{2}}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1} \vert n - \eta \vert }{ \vert n - \eta \vert ^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(22)
(ii) For
\(\xi = \eta = 0,\alpha,\beta \in [\arccos \frac{1}{3},\frac{\pi}{ 2}]\), setting
$$\begin{aligned} \theta_{2}(\lambda_{2},m): ={}& \biggl[\frac{\lambda}{\pi} \sin \biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln 2(1 + \cos \beta )}{\ln ( \vert m \vert + m\cos \alpha )}} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du \\ ={}& O\biggl(\frac{1}{\ln^{\lambda_{2}/2}A_{\xi,\alpha} (m)}\biggr) \in (0,1), \end{aligned}$$
we have the following equivalent reverse Mulholland-type inequalities:
$$\begin{aligned} &\sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln [\ln ( \vert m \vert + m\cos \alpha )/\ln ( \vert n \vert + n\cos \beta )]}{\ln^{\lambda} ( \vert m \vert + m\cos \alpha ) - \ln^{\lambda} ( \vert n \vert + n\cos \beta )} a_{m}b_{n} \\ &\quad > \frac{2\pi^{2}\csc^{2/p}\beta \csc^{2/q}\alpha}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \\ &\qquad{}\times \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta_{2}(\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}( \vert m \vert + m\cos \alpha )}{( \vert m \vert + m\cos \alpha )^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &\qquad{}\times\Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}( \vert n \vert + n\cos \beta )}{( \vert n \vert + n\cos \beta )^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \end{aligned}$$
(23)
$$\begin{aligned} &\Biggl\{ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}( \vert n \vert + n\cos \beta )}{ \vert n \vert + n\cos \beta} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln [\ln ( \vert m \vert + m\cos \alpha )/\ln ( \vert n \vert + n\cos \beta )]}{\ln^{\lambda} ( \vert m \vert + m\cos \alpha ) - \ln^{\lambda} ( \vert n \vert + n\cos \beta )}a_{m} \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\ &\quad> \frac{2\pi^{2}\csc^{2/p}\beta \csc^{2/q}\alpha}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta_{2}(\lambda_{2},m)\bigr) \frac{\ln^{p(1 - \lambda_{1}) - 1}( \vert m \vert + m\cos \alpha )}{( \vert m \vert + m\cos \alpha )^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}, \end{aligned}$$
(24)
$$\begin{aligned} &\Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{q\lambda_{1} - 1}( \vert m \vert + m\cos \alpha )}{(1 - \theta_{2}(\lambda_{2},m))^{q - 1}( \vert m \vert + m\cos \alpha )} \\ &\qquad{}\times \Biggl( \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln [\ln ( \vert m \vert + m\cos \alpha )/\ln ( \vert n \vert + n\cos \beta )]}{\ln^{\lambda} ( \vert m \vert + m\cos \alpha ) - \ln^{\lambda} ( \vert n \vert + n\cos \beta )}b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} \\ &\quad > \frac{2\pi^{2}\csc^{2/p}\beta \csc^{2/q}\alpha}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}( \vert n \vert + n\cos \beta )}{( \vert n \vert + n\cos \beta )^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(25)
Proof
Applying the reverse Hölder inequality with weight (see [17]) and (8), we find
$$\begin{aligned} &\Biggl( \sum_{ \vert m \vert = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} \\ &\quad= \Biggl\{ \sum_{ \vert m \vert = 2}^{\infty} k(m,n) \biggl[ \frac{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)}{\ln^{\frac{1 - \lambda_{2}}{p}}A_{\eta,\beta} (n)}a_{m} \biggr] \biggl[ \frac{\ln^{\frac{1 - \lambda_{2}}{p}}A_{\eta,\beta} (n)}{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)} \biggr] \Biggr\} ^{p} \\ &\quad\ge \sum_{ \vert m \vert = 2}^{\infty} k(m,n) \frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}a_{m}^{p} \\ &\qquad{}\times\Biggl[ \sum _{ \vert m \vert = 2}^{\infty} k(m,n)\frac{\ln^{\frac{(1 - \lambda_{2})q}{p}}A_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)} \Biggr]^{p - 1} \\ &\quad= \frac{(\varpi (\lambda_{1},n))^{p - 1}A_{\eta,\beta} (n)}{\ln^{p\lambda_{2} - 1}A_{\eta,\beta} (n)}\sum_{ \vert m \vert = 2}^{\infty} k(m,n) \frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{A_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}a_{m}^{p}. \end{aligned}$$
Then since \(0 < p < 1\), by (13) this yields
$$\begin{aligned} J &> k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} k(m,n)\frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{A_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &= k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert m \vert = 2}^{\infty} \sum _{ \vert n \vert = 2}^{\infty} k(m,n)\frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{A_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}A_{\eta,\beta} (n)}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &= k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{ \vert m \vert = 2}^{\infty} \omega (\lambda_{2},m) \frac{n^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}. \end{aligned}$$
(26)
Combining (9) and (16), we obtain (18).
Using the reverse Hölders inequality again, we obtain
$$\begin{aligned} I &= \sum_{ \vert n \vert = 2}^{\infty} \Biggl[ \frac{(A_{\eta,\beta} (n))^{\frac{ - 1}{p}}}{\ln^{\frac{1}{p} - \lambda_{2}}A_{\eta,\beta} (n)}\sum_{ \vert m \vert = 2}^{\infty} k(m,n)a_{m} \Biggr] \biggl[ \frac{\ln^{\frac{1}{p} - \lambda_{2}}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{\frac{ - 1}{p}}}b_{n} \biggr] \\ &\ge J_{1} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(27)
Then by (18) we obtain (17).
On the other-hand, assuming that (17) is valid, letting
$$b_{n}: = \frac{\ln^{p\lambda_{2} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \Biggl( \sum _{ \vert m \vert = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p - 1}, \quad \vert n \vert \in \mathbf{N}\backslash \{ 1\}, $$
we find
$$J_{1} = \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}}. $$
By (26) it follows that \(J_{1} > 0\). If \(J_{1} = \infty\), then (19) is trivially valid; if \(J_{1} < \infty\), then by (17) we have
$$\begin{aligned} &\sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q}\\ &\quad = J_{1}^{p} = I \\ &\quad> k(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \\ &J_{1} = \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}} > k( \lambda_{1}) \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}. \end{aligned}$$
Thus (18) is valid, which is equivalent to (17).
We further prove that (19) is equivalent to (17). Using the reverse Hölders inequality, we have
$$\begin{aligned} I={}& \sum_{ \vert m \vert = 2}^{\infty} \biggl[ \bigl(1 - \theta (\lambda_{2},m)\bigr)^{\frac{1}{p}}\frac{\ln^{\frac{1}{q} - \lambda_{1}}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{\frac{ - 1}{q}}}a_{m} \biggr] \\ &{}\times \Biggl[ \frac{n^{\frac{ - 1}{q} + \lambda_{1}}A_{\xi,\alpha} (m)}{(1 - \theta (\lambda_{2},m))^{\frac{1}{p}}(A_{\xi,\alpha} (m))^{\frac{1}{q}}}\sum_{ \vert n \vert = 2}^{\infty} k(m,n)b_{n} \Biggr] \\ \ge{}& \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}J_{2}, \end{aligned}$$
(28)
and then (19) is valid by (17).
On the other-hand, assuming that (17) is valid, we set
$$a_{m}: = \frac{\ln^{q\lambda_{1} - 1}A_{\xi,\alpha} (m)}{(1 - \theta (\lambda_{2},m))^{q - 1}A_{\xi,\alpha} (m)} \Biggl( \sum _{ \vert n \vert = 2}^{\infty} \frac{\ln (\ln A_{\xi,\alpha} (m)/\ln A_{\eta,\beta} (n))}{\ln^{\lambda} A_{\xi,\alpha} (m) - \ln^{\lambda} A_{\eta,\beta} (n)}b_{n} \Biggr)^{q - 1},\quad m \in \mathbf{N}\backslash \{ 1\}, $$
and find
$$J_{2} = \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{q}}. $$
If \(J_{2} = 0\), then (19) is impossible, so that \(J_{2} > 0\). If \(J_{2} = \infty\), then (19) is trivially valid; if \(J_{2} < \infty\), then by (17) we have
$$\begin{aligned} &\sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta ( \lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \\ &\quad= J_{2}^{q} = I \\ &\quad> k(\lambda_{1}) \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}, \\ &\Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta ( \lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{q}} \\ &\quad= J_{2} \\ &\quad > k(\lambda_{1}) \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
Thus (19) is valid, which is equivalent to (17).
Hence, inequalities (17), (18), and (19) are equivalent. □
Theorem 2
Under the assumptions in Theorem 1,
$$k(\lambda_{1}) = \frac{2\pi^{2}\csc^{2/p}\beta \csc^{2/q}\alpha}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} $$
is the best possible constant factor in (17), (18), and (19).
Proof
For \(0 < \varepsilon < \min \{ p\lambda_{1},p(1 - \lambda_{2})\}\), we set \(\tilde{\lambda}_{1} = \lambda_{1} - \frac{\varepsilon}{p}( \in (0,1)),\tilde{\lambda}_{2} = \lambda_{2} + \frac{\varepsilon}{p} ( \in (0,1))\), and
$$\begin{aligned} &\tilde{a}_{m}: = \frac{\ln^{\lambda_{1} - \frac{\varepsilon}{p} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} = \frac{\ln^{\tilde{\lambda}_{1} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)}\quad \bigl( \vert m \vert \in \mathbf{N}\backslash \{ 1\} \bigr), \\ &\tilde{b}_{n}: = \frac{\ln^{\lambda_{2} - \frac{\varepsilon}{q} - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} = \frac{\ln^{\tilde{\lambda}_{2} - \varepsilon - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)}\quad \bigl( \vert n \vert \in \mathbf{N}\backslash \{ 1\} \bigr). \end{aligned}$$
By (15) and (13) we find
$$\begin{aligned} &\tilde{I}_{2}: = \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}} \tilde{a}_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum _{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}A_{\eta,\beta} (n)}{(A_{\eta,\beta} (n))^{1 - q}} \tilde{b}_{n}^{q} \Biggr]^{\frac{1}{q}} \\ &\phantom{\tilde{I}_{2}:}= \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} - \sum_{ \vert m \vert = 2}^{\infty} \frac{O(\ln^{ - 1 - (\frac{\lambda_{2}}{2} + \varepsilon )}A_{\xi,\alpha} (m))}{A_{\xi,\alpha} (m)} \Biggr]^{\frac{1}{p}} \Biggl[ \sum _{ \vert n \vert = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \Biggr]^{\frac{1}{q}} \\ &\phantom{\tilde{I}_{2}:}= \frac{1}{\varepsilon} \bigl(2\csc^{2}\alpha + o(1) - \varepsilon O(1) \bigr)^{\frac{1}{p}}\bigl(2\csc^{2}\beta + \tilde{o}(1) \bigr)^{\frac{1}{q}}\quad\bigl(\varepsilon \to 0^{ +} \bigr), \\ &\tilde{I} = \sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} = \sum_{ \vert m \vert = 2}^{\infty} \sum_{ \vert n \vert = 2}^{\infty} k(m,n) \frac{\ln^{\tilde{\lambda}_{1} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \frac{\ln^{\tilde{\lambda}_{2} - \varepsilon - 1}A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \\ &\phantom{\tilde{I} =}= \sum_{ \vert n \vert = 2}^{\infty} \varpi (\tilde{ \lambda}_{1},n)\frac{\ln^{ - 1 - \varepsilon} A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} < k_{\alpha} (\tilde{ \lambda}_{1})\sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\eta,\beta} (n)}{A_{\eta,\beta} (n)} \\ &\phantom{\tilde{I} =}= \frac{1}{\varepsilon} k_{\alpha} (\tilde{\lambda}_{1}) \bigl(2\csc^{2}\beta + o(1)\bigr). \end{aligned}$$
If there exists a positive number \(k \ge k(\lambda_{1})\) such that (17) is still valid when replacing \(k(\lambda_{1})\) by k, then, in particular, we have
$$\varepsilon \tilde{I} = \varepsilon \sum_{ \vert m \vert = 2}^{\infty} \sum_{ \vert n \vert = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} > \varepsilon k\tilde{I}_{2}. $$
We obtain from the previous results that
$$\begin{aligned} &k_{\beta} \biggl(\lambda_{1} + \frac{\varepsilon}{ q}\biggr) \bigl(2\csc^{2}\alpha + o(1)\bigr) \\ &\quad > k\bigl(2\csc^{2}\alpha + o(1) - \varepsilon O(1) \bigr)^{\frac{1}{p}}\bigl(2\csc^{2}\beta + \tilde{o}(1) \bigr)^{\frac{1}{q}}, \end{aligned}$$
and then
$$\frac{4\pi^{2}}{\lambda^{2}\sin^{2}(\frac{\pi \lambda_{1}}{\lambda} )}\csc^{2}\beta \csc^{2}\alpha \ge 2k \csc^{\frac{2}{p}}\alpha \csc^{\frac{2}{q}}\beta \quad \bigl(\varepsilon \to 0^{ +} \bigr), $$
namely, \(k(\lambda_{1}) = \frac{2\pi^{2}}{\lambda^{2}\sin^{2}(\frac{\pi \lambda_{1}}{\lambda} )}\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha \ge k\). Hence, \(k = k(\lambda_{1})\) is the best possible constant factor of (17).
The constant factor \(k(\lambda_{1})\) in (18) and (19) is still the best possible. Otherwise, we would reach a contradiction by (27) and (28) that the constant factor in (17) is not the best possible. □
Remark 2
(i) For \(\xi = \eta = 0\) in (20), setting
$$\tilde{\theta}_{1}(\lambda_{2},m): = \biggl[ \frac{\lambda}{\pi} \sin \biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln 2}{\ln \vert m \vert }} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du = O\biggl(\frac{1}{\ln^{\lambda_{2}/2} \vert m \vert }\biggr) \in (0,1), $$
we have the following new inequality:
$$\begin{aligned} &\sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln (\ln \vert m \vert /\ln \vert n \vert )}{\ln^{\lambda} \vert m \vert - \ln^{\lambda} \vert n \vert } a_{m}b_{n} \\ &\quad > \frac{2\pi^{2}}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \\ &\qquad{}\times \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \tilde{\theta}_{1}(\lambda_{2},m)\bigr) \frac{\ln^{p(1 - \lambda_{1}) - 1} \vert m \vert }{ \vert m \vert ^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1} \vert n \vert }{ \vert n \vert ^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(29)
It follows that (20) is an extension of (29). In particular, for \(\lambda = 1,\lambda_{1} = \lambda_{2} = \frac{1}{2}\), setting
$$\tilde{\theta}_{1}(m): = \frac{1}{\pi^{2}} \int_{0}^{\frac{\ln 2}{\ln \vert m \vert }} \frac{\ln u}{u - 1} u^{\frac{ - 1}{2}} \,du = O\biggl(\frac{1}{\ln^{1/4} \vert m \vert }\biggr) \in (0,1), $$
we have the following simple reverse Mulholland-type inequality in the whole plane:
$$\begin{aligned} &\sum_{ \vert n \vert = 2}^{\infty} \sum _{ \vert m \vert = 2}^{\infty} \frac{\ln (\ln \vert m \vert /\ln \vert n \vert )}{\ln ( \vert m \vert / \vert n \vert )} a_{m}b_{n} \\ &\quad> 2\pi^{2} \Biggl[ \sum_{ \vert m \vert = 2}^{\infty} \bigl(1 - \tilde{\theta}_{1}(m)\bigr)\frac{\ln^{\frac{p}{2} - 1} \vert m \vert }{ \vert m \vert ^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl( \sum_{ \vert n \vert = 2}^{\infty} \frac{\ln^{\frac{q}{2} - 1} \vert n \vert }{ \vert n \vert ^{1 - q}}b_{n}^{q} \Biggr)^{\frac{1}{q}}. \end{aligned}$$
(30)
(ii) If \(a_{ - m} = a_{m},b_{ - n} = b_{n}\ (m,n \in \mathbf{N}\backslash \{ 1\} )\), for \(m \in \mathbf{N}\backslash \{ 1\}\), setting
$$\begin{aligned} &\stackrel{\frown}{\theta}_{1}(\lambda_{2},m): = \biggl[ \frac{\lambda}{\pi} \sin \biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln (2 + \eta )}{\ln (m - \xi )}} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du = O\biggl(\frac{1}{\ln^{\lambda_{2}/2}(m - \xi )}\biggr) \in (0,1), \\ &\stackrel{\smile}{\theta}_{1}(\lambda_{2},m): = \biggl[ \frac{\lambda}{\pi} \sin \biggl(\frac{\pi \lambda_{1}}{\lambda} \biggr)\biggr]^{2} \int_{0}^{\frac{\ln (2 + \eta )}{\ln (m + \xi )}} \frac{\ln u}{u^{\lambda} - 1} u^{\lambda_{2} - 1} \,du = O\biggl(\frac{1}{\ln^{\lambda_{2}/2}(m + \xi )}\biggr) \in (0,1), \end{aligned}$$
(20) reduces to
$$\begin{aligned} &\sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \biggl\{ \frac{\ln [\ln (m - \xi )/\ln (n - \eta )]}{\ln^{\lambda} (m - \xi ) - \ln^{\lambda} (n - \eta )} + \frac{\ln [\ln (m - \xi )/\ln (n + \eta )]}{\ln^{\lambda} (m - \xi ) - \ln^{\lambda} (n + \eta )} \\ &\qquad{} + \frac{\ln [\ln (m + \xi )/\ln (n - \eta )]}{\ln^{\lambda} (m + \xi ) - \ln^{\lambda} (n - \eta )} + \frac{\ln [\ln (m + \xi )/\ln (n + \eta )]}{\ln^{\lambda} (m + \xi ) - \ln^{\lambda} (n + \eta )} \biggr\} a_{m}b_{n} \\ &\quad> \frac{2\pi^{2}}{[\lambda \sin (\frac{\pi \lambda_{1}}{\lambda} )]^{2}} \Biggl\{ \sum_{m = 2}^{\infty} \biggl[ \bigl(1 - \stackrel{\frown}{\theta}_{1}( \lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}(m - \xi )}{(m - \xi )^{1 - p}} \\ &\qquad{}+ \bigl(1 - \stackrel{\smile}{ \theta}_{1}(\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}(m + \xi )}{(m + \xi )^{1 - p}} \biggr]a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &\qquad{}\times \Biggl\{ \sum_{n = 2}^{\infty} \biggl[ \frac{\ln^{q(1 - \lambda_{2}) - 1}(n - \eta )}{(n - \eta )^{1 - q}} + \frac{\ln^{q(1 - \lambda_{2}) - 1}(n + \eta )}{(n + \eta )^{1 - q}} \biggr]b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(31)
In particular, for \(\xi = \eta = 0,\lambda = 1,\lambda_{1} = \lambda_{2} = \frac{1}{2}\), setting
$$\stackrel{\frown}{\theta}_{1}(m): = \frac{1}{\pi^{2}} \int_{0}^{\frac{\ln 2}{\ln m}} \frac{\ln u}{u - 1} u^{ - \frac{1}{2}} \,du = O\biggl(\frac{1}{\ln^{1/4}m}\biggr) \in (0,1), $$
we have the following simple reverse Mulholland-type inequality:
$$\begin{aligned} &\sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} \frac{\ln (\ln m/\ln n)}{\ln (m/n)} a_{m}b_{n} \\ &\quad > \pi^{2} \Biggl[ \sum_{m = 2}^{\infty} \bigl(1 - \stackrel{\frown}{\theta}_{1}(m)\bigr) \frac{\ln^{\frac{p}{2} - 1}m}{m^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl( \sum_{n = 2}^{\infty} \frac{\ln^{\frac{q}{2} - 1}n}{n^{1 - q}}b_{n}^{q} \Biggr)^{\frac{1}{q}}. \end{aligned}$$
(32)