We also set
$$ K(\lambda_{1}): = h_{\beta}^{1/p}( \lambda_{1})h_{\alpha}^{1/q}(\lambda_{1}) = 2B(\lambda_{1},\lambda_{2})\csc^{2/p}\beta \csc^{2/q}\alpha. $$
(16)
Theorem 1
Suppose that
\(p > 1 \), \(\lambda_{1},\lambda_{2} \le 1 \), \(a_{m},b_{n} \ge 0 \) (\(|m|,|n| \in \mathbb{N}\setminus \{ 1\} \)), and
$$0 < \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}} a_{m}^{p} < \infty,\qquad0 < \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} < \infty. $$
We have the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b]I&: = \sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n) a_{m}b_{n} \\&< K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}$$
(17)
$$\begin{aligned}& \begin{aligned}[b]J&: = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}}\\& < K( \lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{aligned} \end{aligned}$$
(18)
In particular, (i) for
\(\xi,\eta \in [ - \frac{1}{2},\frac{1}{2}]\) (\(\alpha = \beta = \frac{\pi}{2}\)), we have the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b] &\sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} \frac{a_{m}b_{n}}{\ln^{\lambda} (|m - \xi ||n - \eta |)} \\&\quad< 2B( \lambda_{1},\lambda_{2}) \Biggl[ \sum _{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}|m - \xi |}{|m - \xi |^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}|n - \eta |}{|n - \eta |^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}$$
(19)
$$\begin{aligned}& \begin{aligned}[b]&\Biggl\{ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}|n - \eta |}{|n - \eta |} \Biggl[ \sum_{|m| = 2}^{\infty} \frac{a_{m}}{\ln^{\lambda} (|m - \xi ||n - \eta |)} \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\&\quad< 2B(\lambda_{1}, \lambda_{2}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}|m - \xi |}{|m - \xi |^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{aligned} \end{aligned}$$
(20)
(ii) For
\(\alpha,\beta \in [\arccos \frac{1}{3},\pi - \arccos \frac{1}{3}]\) (\(\xi = \eta = 0\)), we have the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b]&\sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} \frac{a_{m}b_{n}}{\ln^{\lambda} [(|m| + m\cos \alpha )(|n| + n\cos \beta )]} \\&\quad< K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}(|m| + m\cos \alpha )}{(|m| + m\cos \alpha )^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &\qquad\times\Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}(|n| + n\cos \beta )}{(|n| + n\cos \beta )^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}$$
(21)
$$\begin{aligned}& \begin{aligned}[b]&\Biggl\{ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}(|n| + n\cos \beta )}{|n| + n\cos \beta} \Biggl[ \sum_{|m| = 2}^{\infty} \frac{a_{m}}{\ln^{\lambda} [(|m| + m\cos \alpha )(|n| + n\cos \beta )]} \Biggr]^{p} \Biggr\} ^{\frac{1}{p}} \\&\quad< K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}(|m| + m\cos \alpha )}{(|m| + m\cos \alpha )^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{aligned} \end{aligned}$$
(22)
Proof
By Hölder’s inequality with weight (cf. [22]) and (9) we find
$$\begin{aligned} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} ={}& \Biggl\{ \sum _{|m| = 2}^{\infty} k(m,n) \biggl[ \frac{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)}{\ln^{\frac{1 - \lambda_{2}}{p}}B_{\eta,\beta} (n)}a_{m} \biggr] \\ &\times\biggl[ \frac{\ln^{\frac{1 - \lambda_{2}}{p}}B_{\eta,\beta} (n)}{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)} \biggr] \Biggr\} ^{p} \\\le{}& \sum_{|m| = 2}^{\infty} k(m,n) \frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p}\\ &\times \Biggl[ \sum _{|m| = 2}^{\infty} k(m,n)\frac{\ln^{\frac{(1 - \lambda_{2})q}{p}}B_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)} \Biggr]^{p - 1} \\ ={}& \frac{(\varpi (\lambda_{1},n))^{p - 1}B_{\eta,\beta} (n)}{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}\sum_{|m| = 2}^{\infty} k(m,n) \frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p}.\end{aligned} $$
By (13) it follows that
$$\begin{aligned}[b] J &< k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n)\frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &= k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{|m| = 2}^{\infty} \sum _{|n| = 2}^{\infty} k(m,n)\frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\ &= k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{|m| = 2}^{\infty} \omega (\lambda_{2},m) \frac{n^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{aligned} $$
(23)
By (10) and (16) we have (18).
Using Hölder’s inequality again, we have
$$\begin{aligned}[b] I &= \sum_{|n| = 2}^{\infty} \Biggl[ \frac{(B_{\eta,\beta} (n))^{\frac{ - 1}{p}}}{\ln^{\frac{1}{p} - \lambda_{2}}B_{\eta,\beta} (n)}\sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr] \biggl[ \frac{\ln^{\frac{1}{p} - \lambda_{2}}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{\frac{ - 1}{p}}}b_{n} \biggr] \\&\le J \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} $$
(24)
and then by (18) we have (17).
On the other hand, assuming that (17) is valid, we set
$$b_{n}: = \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggl( \sum _{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p - 1},\quad|n| \in \mathbb{N}\setminus \{ 1\}, $$
and find
$$J = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}}. $$
By (23) it follows that \(J < \infty \). If \(J = 0 \), then (18) is trivially valid; if \(J > 0 \), then we have
$$\begin{gathered}\begin{aligned} 0 &< \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} = J^{p} = I \\ &< K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \\J = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}} < K( \lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{gathered} $$
Hence (18) is valid, which is equivalent to (17). □
Theorem 2
With regards to the assumptions of Theorem
1, the constant factor
\(K(\lambda_{1}) \)
in (17) and (18) is the best possible.
Proof
For \(0 < \varepsilon < q\lambda_{2} \), we set \(\tilde{\lambda}_{1} = \lambda_{1} + \frac{\varepsilon}{q}\) (>0), \(\tilde{\lambda}_{2} = \lambda_{2} - \frac{\varepsilon}{q}\) (\(\in (0,1)\)), and
$$\begin{gathered} \tilde{a}_{m}: = \frac{\ln^{\lambda_{1} - \frac{\varepsilon}{p} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} = \frac{\ln^{\tilde{\lambda}_{1} - \varepsilon - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)}\quad\bigl(|m| \in \mathbb{N}\setminus \{ 1\} \bigr), \\ \tilde{b}_{n}: = \frac{\ln^{\lambda_{2} - \frac{\varepsilon}{q} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} = \frac{\ln^{\tilde{\lambda}_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)}\quad\bigl(|n| \in \mathbb{N}\setminus \{ 1\} \bigr).\end{gathered} $$
By (15) and (10) we find
$$\begin{gathered}\begin{aligned} \tilde{I}_{1}&: = \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}\tilde{a}_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} \tilde{b}_{n}^{q} \Biggr]^{\frac{1}{q}} \\ &= \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \Biggr]^{\frac{1}{p}} \Biggl[ \sum _{|n| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggr]^{\frac{1}{q}} \\ &= \frac{1}{\varepsilon} \bigl(2\csc^{2}\alpha + o(1) \bigr)^{\frac{1}{p}}\bigl(2\csc^{2}\beta + \tilde{o}(1) \bigr)^{\frac{1}{q}}\quad\bigl(\varepsilon \to 0^{ +} \bigr),\end{aligned} \\\begin{aligned}\tilde{I}&: = \sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} \\ &= \sum_{|m| = 2}^{\infty} \sum_{|n| = 2}^{\infty} k(m,n) \frac{\ln^{\tilde{\lambda}_{1} - \varepsilon - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \frac{\ln^{\tilde{\lambda}_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \\ &= \sum_{|m| = 2}^{\infty} \omega (\tilde{ \lambda}_{2},m)\frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \ge k_{\beta} (\tilde{ \lambda}_{1})\sum_{|m| = 2}^{\infty} \bigl(1 - \theta (\tilde{\lambda}_{2},m)\bigr)\frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \\&= k_{\beta} (\tilde{\lambda}_{1}) \Biggl[ \sum _{|m| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} - \sum _{|m| = 2}^{\infty} \frac{O(\ln^{ - 1 - (\frac{\varepsilon}{p} + \lambda_{2})}A_{\xi,\alpha} (m))}{A_{\xi,\alpha} (m)} \Biggr] \\ &= \frac{1}{\varepsilon} k_{\beta} (\tilde{\lambda}_{1}) \bigl(2\csc^{2}\alpha + o(1) - \varepsilon O(1)\bigr).\end{aligned}\end{gathered} $$
If there exists a positive number \(k \le K(\lambda_{1}) \), such that (17) is still valid when replacing \(K(\lambda_{1}) \) by k, then in particular we have
$$\varepsilon \tilde{I} = \varepsilon \sum_{|n| = 2}^{\infty} \sum_{|m| = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} < \varepsilon k\tilde{I}_{1}. $$
We obtain from the above results that
$$k_{\beta} \biggl(\lambda_{1} + \frac{\varepsilon}{q} \biggr) \bigl(2\csc^{2}\alpha + o(1) - \varepsilon O(1) \bigr) < k \bigl(2 \csc^{2}\alpha + o(1) \bigr)^{\frac{1}{p}} \bigl(2\csc^{2} \beta + \tilde{o}(1) \bigr)^{\frac{1}{q}}, $$
and then
$$4B(\lambda_{1},\lambda_{2})\csc^{2}\beta \csc^{2}\alpha \le 2k\csc^{\frac{2}{p}}\alpha \csc^{\frac{2}{q}}\beta \quad\bigl(\varepsilon \to 0^{ +} \bigr), $$
namely, \(K(\lambda_{1}) = 2B(\lambda_{1},\lambda_{2})\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha \le k \). Hence, \(k = K(\lambda_{1}) \) is the best value of (17).
The constant factor \(K(\lambda_{1}) \) in (18) is still the best possible. Otherwise, we would reach a contradiction by (24) that the constant factor in (17) is not the best possible. □
Theorem 3
Suppose that
\(0 < p < 1\), \(\lambda_{1},\lambda_{2} \le 1\), \(a_{m},b_{n} \ge 0\) (\(|m|,|n| \in \mathbb{N}\setminus \{ 1\} \)), and
$$0 < \sum_{|m| = 2}^{\infty} \frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}} a_{m}^{p} < \infty,\qquad0 < \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} < \infty. $$
We have the following equivalent inequalities:
$$\begin{aligned}& \begin{aligned}[b]I &= \sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n) a_{m}b_{n} \\&> K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}\\&\quad\times \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}$$
(25)
$$\begin{aligned}& \begin{aligned}[b]J &= \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} \Biggr]^{\frac{1}{p}} \\&> K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}},\end{aligned} \end{aligned}$$
(26)
$$\begin{aligned}& \begin{aligned}[b]L&: = \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{q\lambda_{1} - 1}A_{\xi,\alpha} (m)}{(1 - \theta (\lambda_{2},m))^{q - 1}A_{\xi,\alpha} (m)} \Biggl( \sum_{|n| = 2}^{\infty} k(m,n)b_{n} \Biggr)^{q} \Biggr]^{\frac{1}{q}} \\&> K(\lambda_{1}) \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} \end{aligned}$$
(27)
where the constant factor
\(K(\lambda_{1}) \)
in (25), (26), and (27) is the best possible.
Proof
By the reverse Hölder inequality with weight (cf. [22]), and (9), we find
$$\begin{aligned} \Biggl( \sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p} ={}& \Biggl\{ \sum _{|m| = 2}^{\infty} k(m,n) \biggl[ \frac{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)}{\ln^{\frac{1 - \lambda_{2}}{p}}B_{\eta,\beta} (n)}a_{m} \biggr] \\ &\times\biggl[ \frac{\ln^{\frac{1 - \lambda_{2}}{p}}B_{\eta,\beta} (n)}{(A_{\xi,\alpha} (m))^{\frac{1}{q}}\ln^{\frac{1 - \lambda_{1}}{q}}A_{\xi,\alpha} (m)} \biggr] \Biggr\} ^{p} \\\ge{}&\sum_{|m| = 2}^{\infty} k(m,n) \frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p}\\ &\times \Biggl[ \sum _{|m| = 2}^{\infty} k(m,n)\frac{\ln^{\frac{(1 - \lambda_{2})q}{p}}B_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)} \Biggr]^{p - 1} \\ ={}& \frac{(\varpi (\lambda_{1},n))^{p - 1}B_{\eta,\beta} (n)}{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}\sum_{|m| = 2}^{\infty} k(m,n) \frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p}.\end{aligned} $$
Since \(p - 1 < 0 \), by (13) it follows that
$$\begin{aligned}[b] J &> k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n)\frac{(A_{\xi,\alpha} (m))^{\frac{p}{q}}\ln^{\frac{(1 - \lambda_{1})p}{q}}A_{\xi,\alpha} (m)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)}a_{m}^{p} \Biggr]^{\frac{1}{p}} \\&= k_{\alpha}^{1/q}(\lambda_{1}) \Biggl[ \sum _{|m| = 2}^{\infty} \omega (\lambda_{2},m) \frac{n^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{aligned} $$
(28)
By (10) and (16) we have (26).
Using the reverse Hölder’s inequality again, we have
$$\begin{aligned}[b] I &= \sum_{|n| = 2}^{\infty} \Biggl[ \frac{\ln^{\lambda_{2} - \frac{1}{p}}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{\frac{1}{p}}}\sum_{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr] \biggl[ \frac{\ln^{\frac{1}{p} - \lambda_{2}}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{\frac{ - 1}{p}}}b_{n} \biggr] \\&\ge J \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{q}},\end{aligned} $$
(29)
and then by using (26) we have (25).
On the other hand, assuming that (25) is valid, we set
$$b_{n}: = \frac{\ln^{p\lambda_{2} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggl( \sum _{|m| = 2}^{\infty} k(m,n)a_{m} \Biggr)^{p - 1},\quad|n| \in \mathbb{N}\setminus \{ 1\}, $$
and find
$$J = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}}. $$
By (28) it follows that \(J > 0 \). If \(J = \infty \), then (26) is trivially valid; if \(0 < J < \infty \), then we have
$$\begin{gathered} \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q}\\ \quad = J^{p} = I \\ \quad> K(\lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{q}},\\J = \Biggl[ \sum_{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{p}} > K( \lambda_{1}) \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}.\end{gathered} $$
Hence (26) is valid, which is equivalent to (25).
By the reverse Hölder inequality with weight (cf. [22]), and (9) we find
$$\begin{aligned} \Biggl( \sum_{|n| = 2}^{\infty} k(m,n)b_{n} \Biggr)^{q} \le {}&\Biggl[ \sum _{|n| = 2}^{\infty} k(m,n)\frac{\ln^{(1 - \lambda_{1})p/q}A_{\xi,\alpha} (m)}{B_{\eta,\beta} (n)\ln^{1 - \lambda_{2}}B_{\eta,\beta} (n)} \Biggr]^{q - 1} \\ &\times \sum_{|n| = 2}^{\infty} k(m,n) \frac{(B_{\eta,\beta} (n))^{q/p}\ln^{(1 - \lambda_{2})q/p}B_{\eta,\beta} (n)}{\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)} b_{n}^{q} \\={}& \frac{(\omega (\lambda_{2},m))^{q - 1}A_{\xi,\alpha} (m)}{\ln^{q\lambda_{1} - 1}A_{\xi,\alpha} (m)}\sum_{|n| = 2}^{\infty} k(m,n) \frac{(B_{\eta,\beta} (n))^{\frac{q}{p}}\ln^{\frac{(1 - \lambda_{2})q}{p}}B_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)}b_{n}^{q}.\end{aligned} $$
Since \(q < 0 \), by (10) it follows that
$$\begin{aligned} L &> k_{\beta}^{1/p}(\lambda_{1}) \Biggl[ \sum _{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n)\frac{(B_{\eta,\beta} (n))^{\frac{q}{p}}\ln^{\frac{(1 - \lambda_{2})q}{p}}B_{\eta,\beta} (n)}{A_{\xi,\alpha} (m)\ln^{1 - \lambda_{1}}A_{\xi,\alpha} (m)}b_{n}^{q} \Biggr]^{\frac{1}{p}} \\ &= k_{\beta}^{1/p}(\lambda_{1}) \Biggl[ \sum _{|n| = 2}^{\infty} \varpi (\lambda_{1},n) \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}b_{n}^{q} \Biggr]^{\frac{1}{p}}.\end{aligned} $$
By (13) and (16) we have (27).
In the same way, we find
$$\begin{aligned}[b] I = {}&\sum_{|m| = 2}^{\infty} \biggl[ \bigl( 1 - \theta (\lambda_{2},m) \bigr)^{\frac{1}{p}}\frac{\ln^{\frac{1}{q} - \lambda_{1}}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{ - 1/q}}a_{m} \biggr]\\&\times \Biggl[ \frac{\ln^{\frac{ - 1}{q} + \lambda_{1}}A_{\xi,\alpha} (m)}{ ( 1 - \theta (\lambda_{2},m) )^{\frac{1}{p}}(A_{\xi,\alpha} (m))^{1/q}}\sum_{|n| = 2}^{\infty} k(m,n) b_{n} \Biggr] \\\ge{}& \Biggl[ \sum_{|m| = 2}^{\infty} \bigl( 1 - \theta (\lambda_{2},m) \bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}}a_{m}^{p} \Biggr]^{\frac{1}{p}}L,\end{aligned} $$
(30)
and then we can prove that (27) and (25) are equivalent.
For \(0 < \varepsilon < \min \{ p\lambda_{1},|q|\lambda_{1}\} \), we set \(\tilde{\lambda}_{1} = \lambda_{1} - \frac{\varepsilon}{p}\) (\(\in (0,1)\)), \(\tilde{\lambda}_{2} = \lambda_{2} + \frac{\varepsilon}{p}\) (>0), and
$$\begin{gathered} \tilde{a}_{m}: = \frac{\ln^{\lambda_{1} - \frac{\varepsilon}{p} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} = \frac{\ln^{\tilde{\lambda}_{1} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)}\quad\bigl(|m| \in \mathbb{N}\setminus \{ 1\} \bigr), \\ \tilde{b}_{n}: = \frac{\ln^{\lambda_{2} - \frac{\varepsilon}{q} - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} = \frac{\ln^{\tilde{\lambda}_{2} - \varepsilon - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)}\quad\bigl(|n| \in \mathbb{N}\setminus \{ 1\} \bigr).\end{gathered} $$
By (15) and (13) we find
$$\begin{gathered}\begin{aligned} \tilde{I}_{2}&: = \Biggl[ \sum_{|m| = 2}^{\infty} \bigl(1 - \theta (\lambda_{2},m)\bigr)\frac{\ln^{p(1 - \lambda_{1}) - 1}A_{\xi,\alpha} (m)}{(A_{\xi,\alpha} (m))^{1 - p}} \tilde{a}_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum _{|n| = 2}^{\infty} \frac{\ln^{q(1 - \lambda_{2}) - 1}B_{\eta,\beta} (n)}{(B_{\eta,\beta} (n))^{1 - q}}\tilde{b}_{n}^{q} \Biggr]^{\frac{1}{q}} \\ &= \Biggl[ \sum_{|m| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} - \sum_{|m| = 2}^{\infty} \frac{O(\ln^{ - 1 - (\lambda_{2} + \varepsilon )}A_{\xi,\alpha} (m))}{A_{\xi,\alpha} (m)} \Biggr]^{\frac{1}{p}} \Biggl[ \sum _{|n| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \Biggr]^{\frac{1}{q}} \\ &= \frac{1}{\varepsilon} \bigl(2\csc^{2}\alpha + o(1) - \varepsilon O(1) \bigr)^{\frac{1}{p}}\bigl(2\csc^{2}\beta + \tilde{o}(1) \bigr)^{\frac{1}{q}}\quad\bigl(\varepsilon \to 0^{ +} \bigr),\end{aligned} \\\begin{aligned}\tilde{I} &= \sum_{|n| = 2}^{\infty} \sum _{|m| = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} = \sum_{|m| = 2}^{\infty} \sum_{|n| = 2}^{\infty} k(m,n) \frac{\ln^{\tilde{\lambda}_{1} - 1}A_{\xi,\alpha} (m)}{A_{\xi,\alpha} (m)} \frac{\ln^{\tilde{\lambda}_{2} - \varepsilon - 1}B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \\ &= \sum_{|n| = 2}^{\infty} \varpi (\tilde{ \lambda}_{1},n)\frac{\ln^{ - 1 - \varepsilon} B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \le k_{\alpha} (\tilde{ \lambda}_{1})\sum_{|n| = 2}^{\infty} \frac{\ln^{ - 1 - \varepsilon} B_{\eta,\beta} (n)}{B_{\eta,\beta} (n)} \\ &= \frac{1}{\varepsilon} k_{\alpha} (\tilde{\lambda}_{1}) \bigl(2\csc^{2}\beta + o(1)\bigr).\end{aligned}\end{gathered} $$
If there exists a positive number \(k \ge K(\lambda_{1}) \), such that (25) is still valid when replacing \(K(\lambda_{1}) \) by k, then in particular we have
$$\varepsilon \tilde{I} = \varepsilon \sum_{|m| = 2}^{\infty} \sum_{|n| = 2}^{\infty} k(m,n) \tilde{a}_{m} \tilde{b}_{n} > \varepsilon k\tilde{I}_{2}. $$
We obtain from the above results that
$$k_{\beta} \biggl(\lambda_{1} + \frac{\varepsilon}{q} \biggr) \bigl(2\csc^{2}\alpha + o(1) \bigr) > k \bigl(2\csc^{2} \alpha + o(1) - \varepsilon O(1) \bigr)^{\frac{1}{p}} \bigl(2\csc^{2} \beta + \tilde{o}(1) \bigr)^{\frac{1}{q}}, $$
and then
$$4B(\lambda_{1},\lambda_{2})\csc^{2}\beta \csc^{2}\alpha \ge 2k\csc^{\frac{2}{p}}\alpha \csc^{\frac{2}{q}}\beta\quad \bigl(\varepsilon \to 0^{ +} \bigr), $$
namely, \(K(\lambda_{1}) = 2B(\lambda_{1},\lambda_{2})\csc^{\frac{2}{p}}\beta \csc^{\frac{2}{q}}\alpha \ge k \). Hence, \(k = K(\lambda_{1}) \) is the best value of (25).
The constant factor \(K(\lambda_{1}) \) in (26) is still the best possible. Otherwise, we would reach a contradiction by (30) that the constant factor in (25) is not the best possible.
In the same way, by (30) we can proved that the constant factor \(K(\lambda_{1}) \) in (27) is still the best possible. □