In what follows, we assume that \(p < 0\ (0 < q < 1),\frac{1}{p} + \frac{1}{q} = 1,\xi,\eta \in [0,\frac{1}{2}], \mathrm{s} \in \mathrm{N} = \{ 1,2, \ldots \}, 0 < c_{1} \le \cdots \le c_{s}, 0 < \lambda _{i} < \lambda \le s,\lambda _{i} \le 1\ (i = 1,2)\), \(a_{m},b_{n} \ge 0\), such that
$$\begin{aligned} &0 < \sum_{m = 2}^{\infty } \frac{ \ln ^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1}(m - \xi )}{(m - \xi )^{1 - p}} a_{m}^{p} < \infty\quad \text{and}\\ & 0 < \sum _{n = 2}^{\infty } \frac{ \ln ^{q[1 - (\frac{\lambda _{2}}{p} + \frac{\lambda - \lambda _{1}}{q})] - 1}(n - \eta )}{(n - \eta )^{1 - p}} b_{n}^{q} < \infty. \end{aligned}$$
For \(\gamma = \lambda _{1},\lambda - \lambda _{2}\), we set
$$ k_{s}(\gamma ): = \int _{0}^{\infty } \frac{t^{\gamma - 1}}{\prod_{k = 1}^{s} (t^{\lambda /s} + c_{k})} \,dt. $$
By Example 1 of [23], it follows that
$$ k_{s}(\gamma ) = \frac{\pi s}{\lambda \sin (\frac{\pi s\gamma }{ \lambda } )}\sum_{k = 1}^{s} c_{k}^{\frac{s\gamma }{\lambda } - 1} \prod_{j = 1(j \ne k)}^{s} \frac{1}{c_{j} - c_{k}} \in \mathrm{R}_{ +} = (0,\infty ). $$
(5)
In particular, for \(s = 1\), we have
$$ k_{1}(\gamma ) = \int _{0}^{\infty } \frac{t^{\gamma - 1}}{t^{\lambda } + c_{1}} \,dt= \frac{\pi }{\lambda \sin (\frac{\pi \gamma }{\lambda } )}c_{1}^{\frac{\eta }{\lambda } - 1}; $$
for \(s = 2\), we have
$$ k_{2}(\gamma ) = \int _{0}^{\infty } \frac{t^{\gamma - 1}}{(t^{\lambda /2} + c_{1})(t^{\lambda /2} + c_{2})} \,dt= \frac{2\pi }{\lambda \sin (\frac{2 \pi \gamma }{\lambda } )} \bigl(c_{1}^{\frac{2\gamma }{\lambda } - 1} - c _{2}^{\frac{2\gamma }{\lambda } - 1} \bigr)\frac{1}{c_{2} - c_{1}}. $$
Lemma 1
Define the following weight coefficients:
$$\begin{aligned} &\omega _{s}(\lambda _{2},m): = \ln ^{\lambda - \lambda _{2}}(m - \xi ) \sum_{n = 2}^{\infty } \frac{\ln ^{\lambda _{2} - 1}(n - \eta )}{ \prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \frac{1}{n - \eta } \\ &\quad \bigl(m \in \mathrm{N}\backslash \{ 1\} \bigr), \end{aligned}$$
(6)
$$\begin{aligned} &\varpi _{s}(\lambda _{1},n): = \ln ^{\lambda - \lambda _{1}}(n - \eta ) \sum_{m = 2}^{\infty } \frac{\ln ^{\lambda _{1} - 1}(m - \xi )}{ \prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \frac{1}{m - \xi } \\ &\quad \bigl(n \in \mathrm{N}\backslash \{ 1\} \bigr). \end{aligned}$$
(7)
For
\(\lambda _{2} \le 1\), we have
$$ \omega _{s}(\lambda _{2},m) < k_{s}(\lambda - \lambda _{2}) \quad \bigl(m \in \mathrm{N}\backslash \{ 1\} \bigr); $$
(8)
for
\(\lambda _{1} \le 1\), we have
$$ k_{s}(\lambda _{1}) \bigl(1 - \theta _{s}( \lambda _{1},n) \bigr) < \varpi _{s}(\lambda _{1},n) < k_{s}(\lambda _{1})\quad \bigl(n \in \mathrm{N}\backslash \{ 1\} \bigr), $$
(9)
where
\(\theta _{s}(\lambda _{1})\)
is indicated by
$$ \theta _{s}(\lambda _{1},n): = \frac{1}{k_{s}(\lambda _{1})} \int _{0}^{\frac{ \ln (2 - \xi )}{\ln (n - \eta )}} \frac{u^{\lambda _{1} - 1}}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c_{k})} \,du= O \biggl( \frac{1}{\ln ^{\lambda _{1}}(n - \eta )} \biggr) \in (0,1). $$
(10)
Proof
Since for \(0 < \lambda _{2} \le 1,0 < \lambda \le s,y > \frac{3}{2}\), we find that
$$\begin{aligned} &( - 1)^{i}\frac{d^{i}}{dy^{i}}\ln ^{\lambda _{2} - 1}(y - \eta ) \ge 0, \qquad ( - 1)^{i}\frac{d^{i}}{dx^{i}}\frac{1}{y - \eta } > 0\quad \text{and} \\ &( - 1)^{i}\frac{d^{i}}{dy^{i}}\frac{1}{\prod_{k = 1}^{s} [\ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}( y - \eta )]} > 0\quad (i = 1,2). \end{aligned}$$
It follows that
$$ ( - 1)^{i}\frac{d^{i}}{dy^{i}}\frac{\ln ^{\lambda _{2} - 1}(y - \eta )}{ \prod_{k = 1}^{s} [\ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}( y - \eta )]}\frac{1}{y - \eta } > 0 \quad (i = 0,1,2). $$
By Hermite–Hadamard’s inequality (cf. [24]), we find
$$\begin{aligned} \omega _{s}(\lambda _{2},m) &< \ln ^{\lambda - \lambda _{2}}(m - \xi ) \int _{\frac{3}{2}}^{\infty } \frac{\ln ^{\lambda _{2} - 1}(y - \eta )}{ \prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(y - \eta )]} \frac{1}{y - \eta } \,dy \\ & = \ln ^{\lambda - \lambda _{2}}(m - \xi ) \int _{\frac{3}{2}}^{\infty } \frac{ \ln ^{\lambda _{2} - \lambda - 1}(y - \eta )}{\prod_{k = 1}^{s} \{ [ \frac{ \ln (m - \xi )}{\ln (y - \eta )}]^{\lambda /s} + c_{k}\}} \frac{1}{y - \eta } \,dy. \end{aligned}$$
Setting \(u = \frac{\ln (m - \xi )}{\ln (y - \eta )}\), it follows that \(du = \frac{ - \ln (m - \xi )}{\ln ^{2}(y - \eta )}\frac{1}{y - \eta } \,dy\) and
$$ \omega _{s}(\lambda _{2},m) < \int _{0}^{\frac{\ln (m - \xi )}{\ln ( \frac{3}{2} - \eta )}} \frac{u^{(\lambda - \lambda _{2}) - 1}}{ \prod_{k = 1}^{s} ( u^{\lambda /s} + c_{k})} \,du < \int _{0}^{\infty } \frac{u ^{(\lambda - \lambda _{2}) - 1}}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c _{k})} \,du = k_{s}(\lambda - \lambda _{2}), $$
namely (8) follows.
In the same way, for \(\lambda _{1} \le 1\), by Hermite–Hadamard’s inequality, we find
$$ \varpi _{s}(\lambda _{1},n) < \ln ^{\lambda - \lambda _{1}}(n - \eta ) \int _{\frac{3}{2}}^{\infty } \frac{\ln ^{\lambda _{1} - 1}(x - \xi )}{ \prod_{k = 1}^{s} [ \ln ^{\lambda /s}(x - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \frac{1}{x - \xi } \,dx. $$
Setting \(u = \frac{\ln (x - \xi )}{\ln (n - \eta )}\), it follows that
$$ \varpi _{s}(\lambda _{1},n) < \int _{\frac{\ln (\frac{3}{2} - \xi )}{\ln (n - \eta )}}^{\infty } \frac{u ^{\lambda _{1} - 1}}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c_{k})} \,du \le \int _{0}^{\infty } \frac{u^{\lambda _{1} - 1}}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c_{k})} \,du = k_{s}(\lambda _{i}). $$
By the decreasing property, we also find
$$\begin{aligned} &\varpi _{s}(\lambda _{1},n) > \ln ^{\lambda - \lambda _{1}}(n - \eta ) \int _{2}^{\infty } \frac{\ln ^{\lambda _{1} - 1}(x - \xi )}{\prod_{k = 1} ^{s} [ \ln ^{\lambda /s}(x - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \frac{1}{x - \xi } \,dx \\ &\phantom{\varpi _{s}(\lambda _{1},n) }= \int _{\frac{\ln (2 - \xi )}{\ln (n - \eta )}}^{\infty } \frac{u^{ \lambda _{1} - 1}}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c_{k})} \,du = k _{s}(\lambda _{1}) \bigl[1 - \theta _{s}( \lambda _{1},n) \bigr] > 0, \\ &0 < \theta _{s}(\lambda _{1},n) \le \frac{1}{k_{s}(\lambda _{1})} \int _{0}^{\frac{\ln (2 - \xi )}{\ln (n - \eta )}} \frac{u^{\lambda _{1} - 1}}{c _{1}^{s}} \,du = \frac{1}{\lambda _{1}k_{s}(\lambda _{1})c_{1}^{s}} \biggl[\frac{ \ln (2 - \xi )}{\ln (n - \eta )} \biggr]^{\lambda _{1}}. \end{aligned}$$
Hence, (9) and (10) follow. □
Lemma 2
We have the following inequality:
$$\begin{aligned} I: ={}& \sum_{n = 2}^{\infty } \sum _{m = 2}^{\infty } \frac{a_{m}b_{n}}{ \prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \\ >{}& k_{s}^{\frac{1}{p}}(\lambda - \lambda _{2})k_{s}^{\frac{1}{q}}( \lambda _{1}) \Biggl\{ \sum_{m = 2}^{\infty } \frac{ \ln ^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1}(m - \xi )}{(m - \xi )^{1 - p}} a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &{}\times \Biggl\{ \sum_{n=2}^{\infty} \bigl(1-\theta_{s}(\lambda_{1},n) \bigr) \frac{\ln^{q[1 - (\frac{\lambda - \lambda_{1}}{q} + \frac{\lambda_{2}}{p})] - 1}(n - \eta )}{(n - \eta)^{1-q}}b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
(11)
Proof
By reverse Hölder’s inequality (cf. [24]), we obtain
$$\begin{aligned} I: ={}& \sum_{n = 2}^{\infty } \sum _{m = 2}^{\infty } \frac{1}{\prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \biggl[ \frac{\ln ^{(\lambda _{2} - 1)p}(n - \xi )}{(n - \eta )^{1/p}}\frac{ \ln ^{(1 - \lambda _{1})/q}(m - \xi )}{(m - \xi )^{ - 1/q}}a_{m} \biggr] \\ &{}\times \biggl[\frac{\ln ^{(\lambda _{1} - 1)/q}(m - \xi )}{(m - \xi )^{1/q}}\frac{ \ln ^{(1 - \lambda _{2})/p}(n - \eta )}{(n - \eta )^{ - 1/p}}b_{n} \biggr] \\ \ge{}& \Biggl\{ \sum_{m = 2}^{\infty } \Biggl[\ln ^{\lambda _{1}}(m - \xi )\sum_{n = 2} ^{\infty } \frac{\ln ^{\lambda _{2} - 1}(n - \eta )}{\prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \frac{1}{n - \eta } \Biggr]\\ &{}\times\frac{\ln ^{p(1 - \lambda _{1}) - 1}(m - \xi )}{(m - \xi )^{1 - p}}a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &{}\times \Biggl\{ \sum_{n = 2}^{\infty } \Biggl[ \ln ^{\lambda _{2}}(n - \eta )\sum_{m = 2}^{\infty } \frac{\ln ^{\lambda _{1} - 1}(m - \xi )}{\prod_{k = 1} ^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \frac{1}{m - \xi } \Biggr]\\ &{}\times\frac{\ln ^{q(1 - \lambda _{2}) - 1}(n - \eta )}{(n - \eta )^{1 - q}}b_{n}^{q} \Biggr\} ^{\frac{1}{q}} \\ ={}& \Biggl\{ \sum_{m = 2}^{\infty } \omega _{s} (\lambda _{2},m)\frac{ \ln ^{p[1 - (\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q})] - 1}(m - \xi )}{(m - \xi )^{1 - p}}a_{m}^{p} \Biggr\} ^{\frac{1}{p}} \\ &{}\times\Biggl\{ \sum_{n = 2}^{\infty } \varpi _{s}(\lambda _{1},n) \frac{ \ln ^{q[1 - (\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p})] - 1}(n - \eta )}{(n - \eta )^{1 - q}}b_{n}^{q} \Biggr\} ^{\frac{1}{q}}. \end{aligned}$$
Then, by (8) and (9), we have (11). □
Remark 1
By (11), for \(\lambda _{1} + \lambda _{2} = \lambda \), we find
$$\begin{aligned} &\omega _{s}(\lambda _{2},m) = \ln ^{\lambda _{1}}(m - \xi )\sum_{n = 2} ^{\infty } \frac{\ln ^{\lambda _{2} - 1}(n - \eta )}{\prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \frac{1}{n - \eta } \quad \bigl(m \in \mathrm{N}\backslash \{ 1\} \bigr), \\ &0 < \sum_{m = 2}^{\infty } \frac{\ln ^{p(1 - \lambda _{1}) - 1}(m - \xi )}{(m - \xi )^{1 - p}} a_{m}^{p} < \infty,\qquad 0 < \sum _{n = 2}^{ \infty } \frac{\ln ^{q(1 - \lambda _{2}) - 1}(n - \eta )}{(n - \eta )^{1 - p}} b_{n}^{q} < \infty, \end{aligned}$$
and the following inequality:
$$\begin{aligned} &\sum_{n = 2}^{\infty } \sum _{m = 2}^{\infty } \frac{a_{m}b_{n}}{ \prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \\ &\quad > k_{s}(\lambda _{1}) \Biggl[\sum _{m = 2}^{\infty } \frac{ \ln ^{p(1 - \lambda _{1}) - 1}(m - \xi )}{(m - \xi )^{1 - p}} a_{m}^{p} \Biggr]^{ \frac{1}{p}} \\ &\qquad{}\times\Biggl[\sum_{n = 2}^{\infty } \bigl(1 - \theta _{s}(\lambda _{1},n) \bigr) \frac{ \ln ^{q(1 - \lambda _{2}) - 1}(n - \eta )}{(n - \eta )^{1 - q}} b_{n} ^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(12)
In particular, for \(\xi = \eta = 0\), we have \(\tilde{\theta }_{s}( \lambda _{1},n) = O(\frac{1}{\ln ^{\lambda _{1}}n}) \in (0,1)\), and
$$\begin{aligned} &\sum_{n = 2}^{\infty } \sum _{m = 2}^{\infty } \frac{a_{m}b_{n}}{ \prod_{k = 1}^{s} ( \ln ^{\lambda /s}m + c_{k}\ln ^{\lambda /s}n)} \\ &\quad > k_{s}(\lambda _{1}) \Biggl[\sum _{m = 2}^{\infty } \frac{ \ln ^{p(1 - \lambda _{1}) - 1}m}{m^{1 - p}} a_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[ \sum_{n = 2}^{\infty } \bigl(1 - \tilde{\theta }_{s}(\lambda _{1},n) \bigr) \frac{ \ln ^{q(1 - \lambda _{2}) - 1}n}{n^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(13)
Hence, (12) is a more accurate extension of (13).
Lemma 3
For
\(0 < \varepsilon < q\lambda _{2}\), we have
$$ L: = \sum_{n = 2}^{\infty } O \biggl( \frac{1}{ \ln ^{\lambda _{1} + \varepsilon + 1}(n - \eta )} \biggr)\frac{1}{n - \eta } = O(1). $$
(14)
Proof
There exist constants \(m,M > 0\) such that
$$ 0 < m\sum_{n = 2}^{\infty } \frac{1}{\ln ^{\lambda _{1} + \varepsilon + 1}(n - \eta )} \frac{1}{n - \eta } \le L \le M \Biggl[\frac{(2 - \eta )^{ - 1}}{ \ln ^{\lambda _{1} + \varepsilon + 1}(2 - \eta )} + \sum _{n = 3}^{ \infty } \frac{1}{\ln ^{\lambda _{1} + \varepsilon + 1}(n - \eta )} \frac{1}{n - \eta } \Biggr]. $$
By Hermite–Hadamard’s inequality, it follows that
$$\begin{aligned} 0 &< L \le M \biggl[\frac{(2 - \eta )^{ - 1}}{\ln ^{\lambda _{1} + \varepsilon + 1}(2 - \eta )} + \int _{\frac{5}{2}}^{\infty } \frac{1}{ \ln ^{\lambda _{1} + \varepsilon + 1}(y - \eta )} \frac{1}{y - \eta } \,dy \biggr] \\ & = M \biggl[\frac{(2 - \eta )^{ - 1}}{\ln ^{\lambda _{1} + \varepsilon + 1}(2 - \eta )} + \frac{1}{\lambda _{1} + \varepsilon } \ln ^{ - \lambda _{1} - \varepsilon } \biggl( \frac{5}{2} - \eta \biggr) \biggr] \\ &\le M \biggl[\frac{(2 - \eta )^{ - 1}}{\ln ^{\lambda _{1} + q\lambda _{2} + 1}(2 - \eta )} + \frac{1}{\lambda _{1}}\ln ^{ - \lambda _{1} - q\lambda _{2}} \biggl( \frac{5}{2} - \eta \biggr) \biggr] < \infty. \end{aligned}$$
Hence, (14) follows. □
Lemma 4
The constant factor
\(k_{s}(\lambda _{1})\)
in (12) is the best possible.
Proof
For \(0 < \varepsilon < q\lambda {}_{2}\), we set
$$ \tilde{a}_{m}: = \frac{\ln ^{\lambda _{1} - \frac{\varepsilon }{p} - 1}(m - \xi )}{m - \xi },\qquad\tilde{b}_{n}: = \frac{ \ln ^{\lambda _{2} - \frac{\varepsilon }{q} - 1}(n - \eta )}{n - \eta }\quad \bigl(m,n \in \mathrm{N}\backslash \{ 1\} \bigr). $$
If there exists a constant \(M \ge k_{s}(\lambda _{1})\) such that (12) is valid when replacing \(k_{s}(\lambda _{1})\) by M, then, in particular, we have
$$\begin{aligned} \tilde{I}&: = \sum_{n = 2}^{\infty } \sum _{m = 2}^{\infty } \frac{ \tilde{a}_{m}\tilde{b}_{n}}{\prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c_{k}\ln ^{\lambda /s}(n - \eta )]} \\ &> M \Biggl[\sum_{m = 2}^{\infty } \frac{\ln ^{p(1 - \lambda _{1}) - 1}(m - \xi )}{(m - \xi )^{1 - p}} \tilde{a}_{m}^{p} \Biggr]^{\frac{1}{p}} \Biggl[\sum_{n = 2}^{\infty } \bigl(1 - \theta _{s}(\lambda _{1},n) \bigr)\frac{ \ln ^{q(1 - \lambda _{2}) - 1}(n - \eta )}{(n - \eta )^{1 - p}} \tilde{b}_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
In view of (10) and (14), we obtain
$$\begin{aligned} \tilde{I} > {}&M \Biggl\{ \sum_{m = 2}^{\infty } \frac{\ln ^{p(1 - \lambda _{1}) - 1}(m - \xi )}{(m - \xi )^{1 - p}} \frac{ \ln ^{p\lambda _{1} - \varepsilon - p}(m - \xi )}{(m - \xi )^{p}} \Biggr\} ^{ \frac{1}{p}} \\ &{}\times \Biggl\{ \sum_{n = 2}^{\infty } \bigl(1 - \theta _{s}(\lambda _{1},n) \bigr)\frac{ \ln ^{q(1 - \lambda _{2}) - 1}(n - \eta )}{(n - \eta )^{1 - q}} \frac{ \ln ^{q\lambda _{2} - \varepsilon - q}(n - \eta )}{(n - \eta )^{q}} \Biggr\} ^{\frac{1}{q}} \\ ={}& M \Biggl[\frac{\ln ^{ - \varepsilon - 1}(2 - \xi )}{2 - \xi } + \sum_{m = 3} ^{\infty } \frac{\ln ^{ - \varepsilon - 1}(m - \xi )}{m - \xi } \Biggr]^{ \frac{1}{p}}\\ &{}\times \Biggl[\sum _{n = 2}^{\infty } \frac{\ln ^{ - \varepsilon - 1}(n - \eta )}{n - \eta } - \sum _{n = 2}^{\infty } O \biggl(\frac{1}{\ln ^{\lambda _{1}}(n - \eta )} \biggr) \frac{\ln ^{ - \varepsilon - 1}(n - \eta )}{n - \eta } \Biggr]^{\frac{1}{q}} \\ >&{} M \biggl[\frac{\ln ^{ - \varepsilon - 1}(2 - \xi )}{2 - \xi } + \int _{2} ^{\infty } \frac{\ln ^{ - \varepsilon - 1}(x - \xi )}{x - \xi } \,dx \biggr]^{ \frac{1}{p}} \\ &{}\times\Biggl[ \int _{2}^{\infty } \frac{\ln ^{ - \varepsilon - 1}(y - \eta )}{y - \eta } \,dy - \sum _{n = 2}^{\infty } O \biggl(\frac{1}{ \ln ^{\lambda _{1} + \varepsilon + 1}(n - \eta )} \biggr)\frac{1}{n - \eta } \Biggr]^{ \frac{1}{q}} \\ ={}& \frac{M}{\varepsilon } \biggl[\frac{\varepsilon \ln ^{ - \varepsilon - 1}(2 - \xi )}{2 - \xi } + \ln ^{ - \varepsilon } (2 - \xi ) \biggr]^{\frac{1}{p}} \bigl[ \ln ^{ - \varepsilon } (2 - \eta ) - \varepsilon O(1) \bigr]^{\frac{1}{q}}. \end{aligned}$$
By (8), setting \(\hat{\lambda }_{2} = \lambda _{2} - \frac{\varepsilon }{q} \in (0,\lambda )(\hat{\lambda }_{2} \le 1,\hat{\lambda }_{1} = \lambda _{1} + \frac{\varepsilon }{q})\), we find
$$\begin{aligned} \tilde{I}={}& \sum_{m = 2}^{\infty } \Biggl\{ \ln ^{(\lambda _{1} + \frac{\varepsilon }{q})}(m - \xi )\sum_{n = 2} ^{\infty } \frac{1}{\prod_{k = 1}^{s} [ \ln ^{\lambda /s}(m - \xi ) + c _{k}\ln ^{\lambda /s}(n - \eta )]} \frac{ \ln ^{(\lambda _{2} - \frac{\varepsilon }{q}) - 1}(n - \eta )}{n - \eta } \Biggr\} \\ &{}\times\frac{\ln ^{ - \varepsilon - 1}(m - \xi )}{m - \xi } \\ ={}& \sum_{m = 2}^{\infty } \omega _{s} ( \hat{\lambda }_{2},m)\frac{ \ln ^{ - \varepsilon - 1}(m - \xi )}{m - \xi } \le k_{s}(\hat{ \lambda }_{1}) \Biggl[\frac{\ln ^{ - 1 - \varepsilon } (2 - \xi )}{2 - \xi } + \sum _{m = 3}^{\infty } \frac{\ln ^{ - 1 - \varepsilon } (m - \xi )}{m - \xi } \Biggr] \\ \le{}& k_{s}(\hat{\lambda }_{1}) \biggl[ \frac{\ln ^{ - 1 - \varepsilon } (2 - \xi )}{2 - \xi } + \int _{2}^{\infty } \frac{\ln ^{ - 1 - \varepsilon } (x - \xi )}{x - \xi } \,dx \biggr] \\ ={}& \frac{1}{\varepsilon } k_{s}(\hat{\lambda }_{1}) \biggl[ \frac{\varepsilon \ln ^{ - 1 - \varepsilon } (2 - \xi )}{2 - \xi } + \ln ^{ - \varepsilon } (2 - \xi ) \biggr]. \end{aligned}$$
Then we have
$$\begin{aligned} &k_{s}(\hat{\lambda }_{1}) \biggl[\frac{\varepsilon \ln ^{ - 1 - \varepsilon } (2 - \xi )}{2 - \xi } + \ln ^{ - \varepsilon } (2 - \xi ) \biggr] \\ &\quad \ge \varepsilon \tilde{I} > M \biggl[\frac{\varepsilon \ln ^{ - \varepsilon - 1}(2 - \xi )}{2 - \xi } + \ln ^{ - \varepsilon } (2 - \xi ) \biggr]^{ \frac{1}{p}} \bigl[\ln ^{ - \varepsilon } (2 - \eta ) - \varepsilon O(1) \bigr]^{ \frac{1}{q}}. \end{aligned}$$
For \(\varepsilon \to 0^{ +} \), we find \(k_{s}(\lambda _{1}) \ge M\). Hence, \(M = k_{s}(\lambda _{1})\) is the best possible constant factor of (12).
Setting \(\tilde{\lambda }_{1}: = \frac{\lambda - \lambda _{2}}{p} + \frac{ \lambda _{1}}{q},\tilde{\lambda }_{2}: = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}\), we find
$$ \tilde{\lambda }_{1} + \tilde{\lambda }_{2} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} + \frac{\lambda - \lambda _{1}}{q} + \frac{ \lambda _{2}}{p} = \frac{\lambda }{p} + \frac{\lambda }{q} = \lambda, $$
and we can rewrite (11) as follows:
$$\begin{aligned} I >{}& k_{s}^{\frac{1}{p}}(\lambda - \lambda _{2})k_{s}^{\frac{1}{q}}( \lambda _{1}) \Biggl[\sum_{m = 2}^{\infty } \frac{ \ln ^{p(1 - \tilde{\lambda }_{1}) - 1}(m - \xi )}{(m - \xi )^{1 - p}} a _{m}^{p} \Biggr]^{\frac{1}{p}} \\ &\times{}\Biggl[ \sum_{n = 2}^{\infty } \bigl(1 - \theta _{s}(\lambda _{1},n) \bigr)\frac{\ln ^{q(1 - \tilde{\lambda }_{2}) - 1}(n - \eta )}{(n - \eta )^{1 - q}} b_{n}^{q} \Biggr]^{\frac{1}{q}}. \end{aligned}$$
(15)
□
Lemma 5
If
\(\lambda \in (\lambda _{1} + (1 - q)\lambda _{2},(1 - p)\lambda _{1} + \lambda _{2})\), the constant factor
\(k_{s}^{ \frac{1}{p}}(\lambda - \lambda _{2})k_{s}^{\frac{1}{q}}(\lambda _{1})\)
in (15) is the best possible, then we have
\(\lambda = \lambda _{1} + \lambda _{2}\).
Proof
For \(\lambda _{1} + (1 - q)\lambda _{2} < \lambda \le \lambda _{1} + \lambda _{2}\), we obtain
$$\begin{aligned} &\tilde{\lambda }_{1} \ge \frac{\lambda - \lambda _{2}}{p} + \frac{ \lambda - \lambda _{2}}{q} = \lambda - \lambda _{2} > 0, \qquad \tilde{\lambda }_{1} = \frac{\lambda }{p} - \frac{\lambda _{2}}{p} + \frac{ \lambda _{1}}{q} < \lambda, \\ &\quad 0 < \tilde{\lambda }_{1} < \lambda,0 < \tilde{\lambda }_{2} = \lambda - \tilde{\lambda }_{1} < \lambda; \end{aligned}$$
for \(\lambda _{1} + \lambda _{2} < \lambda < (1 - p)\lambda _{1} + \lambda _{2}\), we still obtain
$$\begin{aligned} &\tilde{\lambda }_{2} \ge \frac{\lambda _{2}}{q} + \frac{\lambda _{2}}{p} = \lambda _{2} > 0,\qquad \tilde{\lambda }_{2} = \frac{ \lambda }{q} - \frac{\lambda _{1}}{q} + \frac{\lambda _{2}}{p} < \lambda , \\ &\quad 0 < \tilde{\lambda }_{2} < \lambda,0 < \tilde{\lambda }_{1} = \lambda - \tilde{\lambda }_{2} < \lambda. \end{aligned}$$
Hence, we have \(\tilde{\lambda }_{i} \in (0,\lambda )\ (i = 1,2)\), and then \(k _{s}(\tilde{\lambda }_{1}) \in \mathrm{R}_{ +} \).
If the constant factor \(k_{s}^{\frac{1}{p}}(\lambda - \lambda _{2})k _{s}^{\frac{1}{q}}(\lambda _{1})\) in (15) is the best possible, then, in view of (12), the unique best possible constant factor must be the form of \(k_{s}(\tilde{\lambda }_{1})\), namely
$$ k_{s}(\tilde{\lambda }_{1})= k_{s}^{\frac{1}{p}}( \lambda - \lambda _{2})k_{s}^{\frac{1}{q}}(\lambda _{1}). $$
By reverse Hölder’s inequality, we find
$$\begin{aligned} k_{s}(\tilde{\lambda }_{1})& = k_{\lambda } \biggl(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} \biggr) \\ &= \int _{0}^{\infty } \frac{1}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c _{k})} u^{\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} - 1} \,du = \int _{0}^{\infty } \frac{1}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c _{k})} \bigl(u^{\frac{\lambda - \lambda _{2} - 1}{p}} \bigr) \bigl(u^{\frac{\lambda _{1} - 1}{q}} \bigr)\,du \\ &\ge \biggl( \int _{0}^{\infty } \frac{1}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c_{k})} u^{\lambda - \lambda _{2} - 1} \,du \biggr)^{\frac{1}{p}} \biggl( \int _{0}^{ \infty } \frac{1}{\prod_{k = 1}^{s} ( u^{\lambda /s} + c_{k})} u^{ \lambda _{1} - 1} \,du \biggr)^{\frac{1}{q}} \\ &= k_{s}^{\frac{1}{p}}(\lambda - \lambda _{2})k_{s}^{\frac{1}{q}}( \lambda _{1}). \end{aligned}$$
(16)
We conclude that (16) keeps the form of equality if and only if there exist constants A and B such that they are not all zero and (cf. [24])
$$ Au^{\lambda - \lambda _{2} - 1} = Bu^{\lambda _{1} - 1}\quad \text{a.e. in } \mathrm{R}_{ +} = (0,\infty ). $$
Assuming that \(A \ne 0\) (otherwise, \(B = A = 0\)), it follows that \(u^{\lambda - \lambda _{2} - \lambda _{1}} = \frac{A}{B}\) a.e. in \(\mathrm{R}_{ +} \), and then \(\lambda - \lambda _{2} - \lambda _{1} = 0\), namely \(\lambda = \lambda _{1} + \lambda _{2}\). □