Abstract
By means of the way of weight coefficients and technique of real analysis, an extension of a Hardy-Hilbert-type inequality with parameters and a best possible constant factor is given. The equivalent forms, the operator expression with the norm, the reverses and some particular cases are also considered.
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1 Introduction
Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x),g(y)\geq0\), \(f\in L^{p}(\mathbf{R}_{+})\), \(g\in L^{q}(\mathbf{R}_{+})\), \(\|f\|_{p} =(\int_{0}^{\infty }f^{p}(x)\,dx)^{\frac{1}{p}}>0\), \(\|g\|_{q}>0\). We have the following Hardy-Hilbert’s integral inequality with the best possible constant factor \(\frac{\pi}{\sin(\pi/p)}\) (cf. [1]):
Assuming that \(a_{m},b_{n}\geq0\),
\(b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}\), \(\|a\|_{p},\|b\|_{q}>0\), we have the following Hardy-Hilbert’s inequality with the same best possible constant factor \(\frac{\pi}{\sin(\pi/p)}\) (cf. [1]):
Hardy-Hilbert-type inequalities, specially (1) and (2), are basically important in mathematical analysis and its applications (cf. [1–7]).
If \(\mu_{i},\upsilon_{j}>0\) (\(i,j\in\mathbf{N}\)),
then we have the following inequality (cf. [1], Theorem 321, p.261):
Replacing \(\mu_{m}^{1/q}a_{m}\) and \(\upsilon_{n}^{1/p}b_{n}\) by \(a_{m}\) and \(b_{n}\) in (4), respectively, we obtain the following equivalent form of (4):
For \(\mu_{i}=\upsilon_{j}=1\) (\(i,j\in\mathbf{N}\)), both (4) and (5) reduce to (2). We call (4) and (5) Hardy-Hilbert-type inequalities.
Note
The authors of [1] (Theorem 321, p.261) did not prove that (4) is valid with the best possible constant factor.
In 1998, by introducing an independent parameter \(\lambda\in(0,1]\), Yang [8] gave an extension of (1) for \(p=q=2\). Following the methods of [8], Yang [5] gave some best extensions of (1) and (2) as follows.
If \(\lambda_{1},\lambda_{2}\in\mathbf{R}=(-\infty,\infty)\), \(\lambda _{1}+\lambda_{2}=\lambda\), \(k_{\lambda}(x,y)\) is a nonnegative homogeneous function of degree −λ, with \(k(\lambda_{1})=\int_{0}^{\infty }k_{\lambda}(t,1)t^{\lambda_{1}-1}\,dt\in\mathbf{R}_{+}\), \(\phi (x)=x^{p(1-\lambda_{1})-1}\), \(\psi(x)=x^{q(1-\lambda _{2})-1}\), \(f(x),g(y)\geq 0\),
\(g\in L_{q,\psi}(\mathbf{R}_{+})\), \(\|f\|_{p,\phi},\|g\|_{q,\psi}>0\), then
where the constant factor \(k(\lambda_{1})\) is the best possible. Moreover, if \(k_{\lambda}(x,y)\) is finite and \(k_{\lambda}(x,y)x^{\lambda _{1}-1}(k_{\lambda}(x,y)y^{\lambda_{2}-1})\) is decreasing with respect to \(x>0\) (\(y>0\)), then for \(a_{m},b_{n}\geq0\),
\(b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\psi}\), \(\|a\|_{p,\phi },\|b\|_{q,\psi }>0\), we have
where the constant factor \(k(\lambda_{1})\) is still the best possible.
Clearly, for \(\lambda=1\), \(k_{1}(x,y)=\frac{1}{x+y}\), \(\lambda_{1}=\frac {1}{q}\), \(\lambda_{2}=\frac{1}{p}\), inequality (6) reduces to (1), while (7) reduces to (2). For \(0<\lambda_{1},\lambda _{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda\), we set \(k_{\lambda }(x,y)=\frac{1}{(x+y)^{\lambda}}\). Then, by (7), it follows that
where the constant factor \(B(\lambda_{1},\lambda_{2})\) is the best possible (\(B(u,v)\) is the beta function). Some other results including multidimensional Hilbert-type inequalities are provided by [9–27].
In 2015, by adding a few conditions, Yang [28] gave an extension of (8) and (5) as follows:
where the constant factor \(B(\lambda_{1},\lambda_{2})\) is the best possible. For \(\mu_{i}=\upsilon_{j}=1\) (\(i,j\in\mathbf{N}\)), (9) reduces to (8); for \(\lambda=1\), \(\lambda_{1}=\frac{1}{q}\), \(\lambda _{2}=\frac{1}{p}\), (9) reduces to (5).
In this paper, by using the way of weight coefficients and technique of real analysis, a Hardy-Hilbert-type inequality with parameters and a best possible constant factor is given, which is with the kernel \(\frac{(\min \{x,c_{1}y\})^{\alpha}}{(\max\{x,c_{1}y\})^{\lambda+\alpha}}\) similar to (9). The extended inequalities, the equivalent forms, the operator expression with the norm, the reverses and some particular cases are also considered.
2 Some lemmas
In the following, we agree on that \(\mu_{i},\upsilon_{j}>0\) (\(i,j\in \mathbf{N}\)), \(U_{m}\) and \(V_{n}\) are defined by (3), \(p\neq0,1\), \(\frac {1}{p}+\frac{1}{q}=1\), \(a_{m},b_{n}\geq0\) (\(m,n\in\mathbf{N}\)), \(\|a\|_{p,\Phi _{\lambda}}=(\sum_{m=1}^{\infty}\Phi_{\lambda}(m)a_{m}^{p})^{\frac {1}{p}}\), \(\|b\|_{q,\Psi_{\lambda}}=(\sum_{n=1}^{\infty}\Psi_{\lambda }(n)b_{n}^{q})^{\frac{1}{q}}\), where
Lemma 1
If \(g(t)\) (>0) is decreasing in \(\mathbf{R}_{+}\) and strictly decreasing in \([n_{0},\infty)\subset\mathbf{R}_{+}\) (\(n_{0}\in \mathbf{N}\)), satisfying \(\int_{0}^{\infty}g(t)\,dt\in\mathbf{R}_{+}\), then we have
Proof
Since, by the assumption, we have
it follows that
By the same way, we still have
Hence, making plus for the above two inequalities, we have (10). □
Example 1
For \(s\in\mathbf{N}\), \(0< c_{1}\leq\cdots\leq c_{s}<\infty\), \(\lambda_{1},\lambda_{2}>-\alpha\), \(\lambda_{1}+\lambda _{2}=\lambda\), we set
(a) We find
If \(\lambda_{1}-\frac{i\lambda}{s}+(1-\frac{2i}{s})\alpha\neq0\), then
if there exists \(i_{0}\in\{1,\ldots,s-1\}\) such that \(\lambda_{1}-\frac{i_{0}\lambda}{s}+(1-\frac{2i_{0}}{s})\alpha=0\), then we find
and we still indicate \(\ln(\frac{c_{i_{0}+1}}{c_{i_{0}}})\) by the following formal expression:
Hence, we may set
In particular, (i) for \(s=1\) (or \(c_{s}=\cdots=c_{1}\)), we have \(k_{\lambda }(x,y)=\frac{(\min\{x,c_{1}y\})^{\alpha}}{(\max\{x,c_{1}y\})^{\lambda +\alpha}}\) and
(ii) for \(s=2\), we have \(k_{\lambda}(x,y)=\frac{(\min\{x,c_{1}y\}\min \{x,c_{2}y\})^{\alpha/2}}{(\max\{x,c_{1}y\}\max\{x,c_{2}y\} )^{(\lambda +\alpha)/2}}\) and
(iii) for \(\alpha=0\), we have \(\lambda_{1},\lambda_{2}>0\), \(k_{\lambda }(x,y)=\frac{1}{\prod_{k=1}^{s}(\max\{x,c_{k}y\})^{\frac{\lambda }{s}}}\) and
(iv) for \(\alpha=-\lambda\), we have \(\lambda<\lambda_{1},\lambda_{2}<0\), \(k_{\lambda}(x,y)=\frac{1}{\prod_{k=1}^{s}(\min\{x,c_{k}y\})^{\frac{\lambda}{s}}}\) and
(v) for \(\lambda=0\), we have \(\lambda_{2}=-\lambda_{1}\), \(|\lambda _{1}|<\alpha\) (\(\alpha>0\)),
and
(b) Since we find
then for \(\lambda_{2}\leq1-\alpha \) (\(\lambda_{1}>-\alpha\)), \(k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}\) is decreasing for \(y>0\) and strictly decreasing for the large enough variable y. By the same way, since
then for \(\lambda_{1}\leq1-\alpha\) (\(\lambda_{2}>-\alpha\)), \(k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}\) is decreasing for \(x>0\) and strictly decreasing for the large enough variable x.
In view of (a) and (b), for \(-\alpha<\lambda_{1},\lambda_{2}\leq 1-\alpha\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(k_{\lambda}(x,y)\frac {1}{y^{1-\lambda _{2}}}\) (\(k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}\)) is decreasing for \(y>0\) (\(x>0\)) and strictly decreasing for the large enough variable \(y^{{}} (x)\) satisfying \(k_{s}(\lambda_{1})\in\mathbf{R}_{+}\).
Lemma 2
If \(s\in\mathbf{N}\), \(0< c_{1}\leq\cdots \leq c_{s}\), \(-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha\), \(\lambda _{1}+\lambda_{2}=\lambda\), \(k_{s}(\lambda_{1})\) is indicated by (11), define the following weight coefficients:
Then we have the following inequalities:
Proof
We set \(\mu(t):=\mu_{m}\), \(t\in(m-1,m]\) (\(m\in\mathbf{N}\)); \(\upsilon(t):=\upsilon_{n}\), \(t\in(n-1,n]\) (\(n\in\mathbf{N}\)),
Then, by (3), it follows that \(U(m)=U_{m}\), \(V(n)=V_{n}\) (\(m,n\in \mathbf{N}\)). For \(x\in(m-1,m]\), \(U^{\prime}(x)=\mu(x)=\mu_{m}\) (\(m\in\mathbf{N}\)); for \(y\in(n-1,n]\), \(V^{\prime}(y)=\upsilon(y)=\upsilon_{n}\) (\(n\in\mathbf {N}\)). Since \(V(y)\) is strictly increasing in \((n-1,n]\), \(-\alpha<\lambda _{2}\leq1-\alpha\), \(\lambda_{1}>-\alpha\), in view of Lemma 1 and Example 1, we find
Setting \(t=\frac{V(y)}{U_{m}}\), we obtain \(V^{\prime}(y)\,dy=U_{m}\,dt\) and
Since \(U(x)\) is strictly increasing in \((m-1,m]\), \(-\alpha<\lambda _{1}\leq 1-\alpha\), \(\lambda_{2}>-\alpha\), by the same way, we have
Hence, we have (19) and (20). □
Lemma 3
If \(s\in\mathbf{N}\), \(0< c_{1}\leq\cdots\leq c_{s}\), \(-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha\), \(\lambda _{1}+\lambda_{2}=\lambda\), \(k_{s}(\lambda_{1})\) is indicated by (11), \(m_{0},n_{0}\in\mathbf{N}\), \(\mu_{m}\geq\mu_{m+1}\) (\(m\in \{m_{0},m_{0}+1,\ldots\}\)), \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\in \{n_{0},n_{0}+1,\ldots\}\)), \(U(\infty)=V(\infty)=\infty\), then (i) for \(m,n\in\mathbf{N}\), we have
where
(ii) for any \(b>0\), we have
Proof
Since \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\geq n_{0}\)), \(-\alpha<\lambda_{2}\leq1-\alpha\), \(\lambda_{1}>-\alpha\) and \(V(\infty)=\infty\), by Lemma 1, we have
For \(U_{m}>c_{s}V_{n_{0}}\), we obtain \(c_{k}t\leq c_{s}t\leq c_{s}\frac{ V_{n_{0}}}{U_{m}}<1\) (\(t\in(0,\frac{V_{n_{0}}}{U_{m}}]\); \(k=1,\ldots,s\)) and
and then \(\theta(\lambda_{2},m)=O(\frac{1}{U_{m}^{\lambda_{2}+\alpha}})\). Hence we have (22).
By the same way, since \(\mu_{m}\geq\mu_{m+1}\) (\(m\geq m_{0}\)), \(-\alpha <\lambda_{1}\leq1-\alpha\), \(\lambda_{2}>-\alpha\) and \(U(\infty )=\infty\), we have
For \(V_{n}>c_{1}^{-1}U_{m_{0}}\), we obtain \(t\leq\frac {U_{m_{0}}}{V_{n}}< c_{1}\leq c_{k}\) (\(t\in(0,\frac{U_{m_{0}}}{V_{n}}]\); \(k=1,\ldots,s\)) and
Hence, we have (23).
For \(b>0\), we find
Hence we have (24). By the same way, we still have (25). □
Note
For example, \(\mu_{m}=\frac{1}{m^{\sigma}}\), \(\upsilon _{n}=\frac{1}{n^{\sigma}}\) (\(0\leq\sigma\leq1\); \(m,n\in\mathbf{N}\)) satisfy the conditions of Lemma 3 (\(m_{0}=n_{0}=1\)).
3 Main results and operator expressions
Theorem 1
If \(s\in\mathbf{N}\), \(0< c_{1}\leq\cdots \leq c_{s}\), \(-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha\), \(\lambda _{1}+\lambda_{2}=\lambda\), \(k_{s}(\lambda_{1})\) is indicated by (11), then for \(p>1\), \(0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi_{\lambda }}<\infty\), we have the following equivalent inequalities:
In particular, for \(s=1\) (or \(c_{s}=\cdots=c_{1}\)), we have the following equivalent inequalities:
where \(k_{1}(\lambda_{1})\) is indicated by (12).
Proof
By Hölder’s inequality with weight (cf. [29]), we have
In view of (20), we find
By Hölder’s inequality (cf. [29]), we have
On the other hand, assuming that (26) is valid, we set
Then we find \(J^{p}=\|b\|_{q,\Psi_{\lambda}}^{q}\). If \(J=0\), then (27) is trivially valid; if \(J=\infty\), then, by (31) and (19), it is impossible. Suppose that \(0< J<\infty\). By (26), it follows that
and then (27) follows, which is equivalent to (26). □
Theorem 2
With the assumptions of Theorem 1, if \(m_{0},n_{0}\in \mathbf{N}\), \(\mu_{m}\geq\mu_{m+1}\) (\(m\in\{m_{0},m_{0}+1,\ldots\}\)), \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\in\{n_{0},n_{0}+1,\ldots \}\)), \(U(\infty)=V(\infty)=\infty\), then the constant factor \(k_{s}(\lambda_{1})\) in (26) and (27) is the best possible.
Proof
For \(\varepsilon\in(0,p(\lambda_{1}+\alpha))\), we set \(\widetilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon}{p}\) (\({\in} (-\alpha ,1-\alpha)\)), \(\widetilde{\lambda}_{2}=\lambda_{2}+\frac{\varepsilon }{p}\) (\({>}-\alpha\)), and \(\widetilde{a}=\{\widetilde{a}_{m}\}_{m=1}^{\infty}\), \(\widetilde{b}=\{\widetilde{b}_{n}\}_{n=1}^{\infty}\),
Then, by (24), (25) and (23), we have
If there exists a positive constant \(K\leq k_{s}(\lambda_{1})\) such that (26) is valid when replacing \(k_{s}(\lambda_{1})\) with K, then, in particular, we have \(\varepsilon\widetilde{I}<\varepsilon K\|\widetilde {a}\|_{p,\Phi_{\lambda}}\|\widetilde{b}\|_{q,\Psi_{\lambda}}\), namely
It follows that \(k_{s}(\lambda_{1})\leq K(\varepsilon\rightarrow0^{+})\). Hence, \(K=k_{s}(\lambda_{1})\) is the best possible constant factor of (26).
The constant factor \(k_{s}(\lambda_{1})\) in (27) is still the best possible. Otherwise, we would reach a contradiction by (32) that the constant factor in (26) is not the best possible. □
Remark 1
Inequality (26) is an extension of Hardy-Hilbert-type inequality (28) with parameters and a best possible constant factor.
For \(p>1\), we find \(\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon_{n}}{V_{n}^{1-p\lambda_{2}}}\) and define the following normed spaces:
Assuming that \(a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}\), setting
we can rewrite (27) as follows:
namely \(c\in l_{p,\Psi_{\lambda}^{1-p}}\).
Definition 1
Define a Hardy-Hilbert-type operator \(T:l_{p,\Phi _{\lambda}}\rightarrow l_{p,\Psi_{\lambda}^{1-p}}\) as follows: For any \(a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}\), there exists a unique representation \(Ta=c\in l_{p,\Psi_{\lambda}^{1-p}}\). Define the formal inner product of Ta and \(b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\Psi _{\lambda}}\) as follows:
Then we can rewrite (26) and (27) as follows:
Define the norm of operator T as follows:
Then, by (36), we find \(\|T\|\leq k_{s}(\lambda_{1})\). Since by Theorem 2 the constant factor in (36) is the best possible, we have
4 Some reverses
In the following, we also set
For \(0< p<1\) or \(p<0\), we still use the formal symbols of \(\|a\|_{p,\Phi _{\lambda}}\), \(\|b\|_{q,\Psi_{\lambda}}\), \(\|a\|_{p,\widetilde{\Phi} _{\lambda}}\) and \(\|b\|_{q,\widetilde{\Psi}_{\lambda}}\).
Theorem 3
If \(s\in\mathbf{N}\), \(0< c_{1}\leq\cdots\leq c_{s}\), \(-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha\), \(\lambda _{1}+\lambda_{2}=\lambda\), \(k_{s}(\lambda_{1})\) is indicated by (11), \(m_{0},n_{0}\in\mathbf{N}\), \(\mu_{m}\geq\mu_{m+1}\) (\(m\in \{m_{0},m_{0}+1,\ldots\}\)), \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\in \{n_{0},n_{0}+1,\ldots\}\)), \(U(\infty)=V(\infty)=\infty\), then for \(0< p<1\), \(0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi_{\lambda }}<\infty \), we have the following equivalent inequalities with the best possible constant factor \(k_{s}(\lambda_{1})\):
Proof
By the reverse Hölder’s inequality (cf. [29]) and (20), we have the reverses of (30), (31) and (32). Then, by (22), we have (40). By (40) and the reverse of (32), we have (39).
On the other hand, assuming that (39) is valid, we set \(b_{n}\) as in Theorem 1. Then we find \(J^{p}=\|b\|_{q,\Psi_{\lambda}}^{q}\). If \(J=\infty \), then (40) is trivially valid; if \(J=0\), then, by reverse of (31) and (22), it is impossible. Suppose that \(0< J<\infty\). By (39), it follows that
and then (40) follows, which is equivalent to (39).
For \(\varepsilon\in(0,p(\lambda_{1}+\alpha))\), we set \(\widetilde{\lambda}_{1}\), \(\widetilde{\lambda}_{2}\), \(\widetilde{a}_{m}\) and \(\widetilde{b}_{n}\) as (33). Then, by (24), (25) and (20), we find
If there exists a constant \(K\geq k_{s}(\lambda_{1})\) such that (39) is valid when replacing \(k_{s}(\lambda_{1})\) with K, then, in particular, we have \(\varepsilon\widetilde{I}>\varepsilon K\|\widetilde{a}\|_{p,\widetilde{\Phi}_{\lambda}}\|\widetilde{b}\|_{q,\Psi_{\lambda}}\), namely
It follows that \(k_{s}(\lambda_{1})\geq K\) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k_{s}(\lambda_{1})\) is the best possible constant factor of (39).
The constant factor \(k_{s}(\lambda_{1})\) in (40) is still the best possible. Otherwise, we would reach a contradiction by the reverse of (32) that the constant factor in (39) is not the best possible. □
Theorem 4
With the assumptions of Theorem 3, if \(p<0\), then we have the following equivalent inequalities with the best possible constant factor \(k_{s}(\lambda_{1})\):
Proof
By the reverse Hölder’s inequality with weight (cf. [29]), since \(p<0\), by (23), we have
By the reverse Hölder’s inequality (cf. [29]), we have
On the other hand, assuming that (41) is valid, we set \(b_{n}\) as follows:
Then we find \(J_{1}^{p}=\|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q}\). If \(J_{1}=\infty\), then (42) is trivially valid; if \(J_{1}=0\), then by (43) and (19) it is impossible. Suppose that \(0< J_{1}<\infty \). By (41), it follows that
and then (42) follows, which is equivalent to (41).
For \(\varepsilon\in(0,q(\lambda_{2}+\alpha))\), we set \(\widetilde{\lambda}_{1}=\lambda_{1}+\frac{\varepsilon}{q}\) (\({>}-\alpha\)), \(\widetilde{ \lambda}_{2}=\lambda_{2}-\frac{\varepsilon}{q}\) (\({\in}(-\alpha ,1-\alpha)\)), and
Then, by (24), (25) and (19), we have
If there exists a constant \(K\geq k_{s}(\lambda_{1})\) such that (41) is valid when replacing \(k_{s}(\lambda_{1})\) with K, then, in particular, we have \(\varepsilon\widetilde{I}>\varepsilon K\|\widetilde {a}\|_{p,\Phi _{\lambda}}\|\widetilde{b}\|_{q,\widetilde{\Psi}_{\lambda}}\), namely
It follows that \(k_{s}(\lambda_{1})\geq K\) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k_{s}(\lambda_{1})\) is the best possible constant factor of (41).
The constant factor \(k_{s}(\lambda_{1})\) in (42) is still the best possible. Otherwise, we would reach a contradiction by (44) that the constant factor in (41) is not the best possible. □
Remark 2
(i) For \(\alpha=0\), \(0<\lambda _{1},\lambda_{2}\leq1\) in (26) and (27), we have the following equivalent inequalities:
where \(\widetilde{k}_{s}(\lambda_{1})\) is indicated by (14);
(ii) for \(\alpha=-\lambda\), \(-1\leq\lambda_{1},\lambda_{2}<0\) in (26) and (27), we have the following equivalent inequalities:
where \(\widehat{k}_{s}(\lambda_{1})\) is indicated by (15);
(iii) for \(\lambda=0\), \(\lambda_{2}=-\lambda_{1}\), in (26) and (27), we have the following equivalent inequalities:
where \(k_{s}^{(0)}(\lambda_{1})\) is indicated by (16) (\(|\lambda _{1}|<\alpha\), \(0<\alpha\leq\frac{1}{2}\); \(|\lambda_{1}|<1-\alpha\), \(\frac {1}{2}<\alpha\leq1\)).
By Theorem 2, the constant factors in the above inequalities are all the best possible. We still can obtain some particular reverse inequalities with the best possible constant factors by Theorem 3 and Theorem 4.
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Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 61370186), and the Science and Technology Planning Project of Guangzhou (No. 2014J4100032, No. 201510010203). We are grateful for their help.
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BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. QC participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
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Yang, B., Chen, Q. On a Hardy-Hilbert-type inequality with parameters. J Inequal Appl 2015, 339 (2015). https://doi.org/10.1186/s13660-015-0861-7
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DOI: https://doi.org/10.1186/s13660-015-0861-7