Abstract
By introducing independent parameters, and applying weight coefficients and the technique of real analysis, we give a new extension of a Hardy-Hilbert-type inequality with a best possible constant factor. Furthermore, the equivalent forms, the operator expressions, and the reverses are considered.
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1 Introduction
If \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(a_{n},b_{n}\geq 0\), \(0<\sum_{n=1}^{\infty }a_{n}^{p}<\infty \) and \(0<\sum_{n=1}^{\infty }b_{n}^{q}<\infty \), then we have the Hardy-Hilbert inequality as follows (cf. [1]):
where the constant factor \(\frac{\pi }{\sin (\pi /p)}\) is the best possible. We also have the following Hardy-Hilbert-type inequality (cf. [2]):
where the constant factor \([ \frac{\pi }{\sin (\pi /p)} ] ^{2}\) is still the best possible. In 2008, by introducing some parameters, Yang gave an extension of inequality (2) (cf. [3]): If \(0<\lambda _{1},\lambda _{2}\leq 1\), \(\lambda _{1}+\lambda _{2}=\lambda\), \(a_{n},b_{n}\geq 0\), \(0<\sum_{n=1}^{\infty }n^{p(1-\lambda _{1})-1}a_{n}^{p}<\infty\), and \(0<\sum_{n=1}^{\infty }n^{q(1-\lambda _{2})-1}b_{n}^{q}<\infty \), then the following inequality holds:
where the constant factor \([ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} ] ^{2}\) is the best possible. There are lots of improvements, generalizations, and applications of inequality (2) ([3–11]). For more details, Yang gives a summary of introducing independent parameters ([12, 13]).
In this article, by introducing independent parameters, and applying weight coefficients and the technique of real analysis, we give a new extension of (2) with a best possible constant factor. Furthermore, the equivalent forms, the operator expressions, and the reverses are considered.
2 Some lemmas
We agree on the following assumptions in this paper: \(p\neq 0\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\lambda >0\), \(0<\lambda _{i}\leq 1\) (\({i=1,2}\)), \(\lambda _{1}+\lambda _{2}=\lambda \), \(k_{\lambda }(\lambda _{2})=k_{\lambda }(\lambda _{1})= [ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} ] ^{2}\), \(\{\mu _{m}\}_{m=1}^{\infty }\) and \(\{\nu _{n}\}_{n=1}^{\infty }\) are positive sequences, \(U_{m}=\sum_{i=1}^{m}\mu _{i}\), \(V_{n}=\sum_{i=1}^{n}\nu _{i}\), and \(a_{n},b_{n}\geq 0\) (\(m,n\in \mathbf{N}=\{1,2,\ldots \}\)),
Lemma 1
Define the weight coefficients as follows:
We have the following inequalities:
Proof
Putting \(\mu (t):=\mu _{m}\), \(t\in (m-1,m]\) (\(m=1,2,\ldots\)), \(\nu (t):=\nu _{n}\), \(t\in (n-1,n]\) (\(n=1,2,\ldots \)),
Then we have \(U(m)=U_{m}\), \(V(n)=V_{n}\) (\(m,n\in \mathbf{N}\)). \(U^{\prime} (x)=\mu (x)=\mu _{m}\) when \(x\in (m-1,m]\); \(V^{\prime}(y)=\nu (y)=\nu _{n}\) when \(y\in (n-1,n]\). Since the function \(V(y)\) (\(y>0\)) is strictly increasing and \(f(x)=\frac{\ln (m/x)}{m^{\lambda }-x^{\lambda }}\) (\(x>0\)) is strictly decreasing (cf. [4], Example 2.2.1), in view of \(1-\lambda _{2}\geq 0\), we have
Putting \(u=\frac{V^{\lambda }(t)}{U_{m}^{\lambda }}\) in the above integral, and in view of the fact that (cf. [2])
it follows that
Hence we prove that (6) is valid. In the same way, we can prove that (7) is valid too. □
Lemma 2
Suppose that \(\{\mu _{m}\}_{m=1}^{\infty }\) and \(\{\nu _{n}\}_{n=1}^{\infty }\) are decreasing sequences, and \(U(\infty )=V(\infty )=\infty \), then we have the following inequalities:
where \(\theta _{1}(\lambda _{2},m)=O(\frac{1}{U_{m}^{\lambda _{2}/2}})\in (0,1)\) and \(\theta _{2}(\lambda _{1},n)=O(\frac{1}{V_{n}^{\lambda _{1}/2}})\in (0,1)\). Moreover, we get
Proof
By the decreasing property of \(\{\nu _{n}\}_{n=1}^{\infty }\), and in view of \(1-\lambda _{2}\geq 0\), \(V(\infty )=\infty \), we find
where
In virtue of
it is obvious that \(\theta _{1}(\lambda _{2},m)=O(\frac{1}{U_{m}^{\lambda _{2}/2}})\). Hence (8) is valid. In the same way, we can prove that (9) is valid too. Moreover, we have
Then we have (10). In the same way, we have (11). □
Remark 1
Taking \(\varepsilon =a>0\), we write by (10) and (11) that
3 Equivalent forms and operator expressions
Theorem 1
Suppose that \(p>1\), then we have the following equivalent inequalities:
Proof
By Hölder’s inequality with weight (cf. [14]), we find
By (7), it follows that
Combining (8) and (15), we have (13).
Using Hölder’s inequality again, we have
and then we have (12) by using (13). On the other hand, assuming that (12) is valid, setting
then we find \(J= [ \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} ] ^{1/p}\). By (15), it follows that \(J<\infty \). If \(J=0\), then (13) is trivially valid. If \(0< J<\infty \), then we have
Hence (13) is valid, which is equivalent to (12). □
Theorem 2
Suppose that \(p>1\), \(\{\mu _{m}\}_{m=1}^{\infty }\) and \(\{\nu _{n}\}_{n=1}^{\infty }\) are decreasing positive sequences, and \(U(\infty )=V(\infty )=\infty \), then the constant factor \(k_{\lambda }(\lambda _{1})= [ \frac{\pi }{\lambda \sin (\lambda _{1}\pi /\lambda )} ] ^{2}\) is the best possible in (12) and (13).
Proof
For \(0<\varepsilon <p\lambda _{1}\), we set \(\widetilde{\lambda }_{1}=\lambda _{1}-\frac{\varepsilon }{p}\) (\(\in (0,1)\)), \(\widetilde{\lambda }_{2}=\lambda _{2}+\frac{\varepsilon }{p}\) (>0), \(\widetilde{a}_{m}=U_{m}^{\widetilde{\lambda }_{1}-1}\mu _{m}\), \(\widetilde{b}_{n}=V_{n}^{\widetilde{\lambda }_{2}-\varepsilon -1}\nu _{n}\). By (10), (11), and (9), in view of Remark 1, we find
If there exists a positive number \(K\leq k_{\lambda }(\lambda _{1})\), such that (12) is still valid when replacing \(k_{\lambda }(\lambda _{1})\) by K, then, in particular, we have
We obtain from the above results
and then it follows that \(k_{\lambda }(\lambda _{1})\leq K\) (for \(\varepsilon \rightarrow 0^{+}\)). Hence \(K=k_{\lambda }(\lambda _{1})\) is the best value of (12).
We conform that the constant factor \(k_{\lambda }(\lambda _{1})\) in (13) is the best possible. Otherwise we can get a contradiction by (16): that the constant factor in (12) is not the best value. □
For \(p>1\), setting
then it follows that \([ \psi (n) ] ^{1-p}=\frac{\nu _{n}}{V_{n}^{1-p\lambda _{2}}}\), and we define the real weighted normed function spaces as follows:
For \(a=\{a_{m}\}_{m=1}^{\infty }\in l_{p,\varphi }\), putting \(h_{n}:=\sum_{m=1}^{\infty }\frac{\ln (U_{m}/V_{n})}{U_{m}^{\lambda }-V_{n}^{\lambda }}a_{m}\), \(h=\{h_{n}\}_{n=1}^{\infty }\), then it follows by (13) that \(\Vert h\Vert _{p,\psi ^{1-p}}< k_{\lambda }(\lambda _{1})\Vert a\Vert _{p,\varphi }\), and \(h\in l_{p,\psi ^{1-p}}\).
Definition 1
Define a Hardy-Hilbert-type operator \(T: l_{p,\varphi }\rightarrow l_{p,\psi ^{1-p}}\) as follows: For \({a_{m}\geq 0}\), \(a=\{a_{m}\}_{m=1}^{\infty }\in l_{p,\varphi }\), there exists a unique representation \(Ta=h\in l_{p,\psi ^{1-p}}\). We define the following formal inner product of Ta and \(b=\{b_{n}\}_{n=1}^{\infty }\in l_{q,\psi }\) (\(b_{n}\geq 0\)) as follows:
Hence (12) and (13) may be rewritten in terms of the following operator expressions:
It follows that the operator T is bounded with
Since the constant factor \(k_{\lambda }(\lambda _{1})\) in (19) is the best possible, we have
4 Some reverses
We set \(\widetilde{\varphi }(m):=(1-\theta _{1}(\lambda _{2},m))\frac{U_{m}^{p(1-\lambda _{1})-1}}{\mu _{m}^{p-1}}\), \(\widetilde{\psi }(n):=(1-\theta _{2}(\lambda _{1},m))\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}\) (\(n,m\in \mathbf{N}\)). For \(0< p<1\) or \(p<0\), we still use the formal symbol of the norm in this part for convenience.
Theorem 3
Suppose that \(0< p<1\), \(\{\mu _{m}\}_{m=1}^{\infty }\) and \(\{\nu _{n}\}_{n=1}^{\infty }\) are decreasing positive sequences, and \(U(\infty )=V(\infty )=\infty \), then we have the following equivalent inequalities:
where the constant factor \([ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} ] ^{2}\) is the best possible.
Proof
By the reverse Hölder inequality with weight (cf. [14]) and (7), we obtain the reverse forms of (14) and (15). It follows that (22) is valid by (8). Using the reverse Hölder inequality (cf. [14]), we find
Hence (21) is valid by using (22). Setting
then we have \(J= [ \sum_{n=1}^{\infty }\frac{V_{n}^{q(1-\lambda _{2})-1}}{\nu _{n}^{q-1}}b_{n}^{q} ] ^{1/p}\). Assume that (21) is valid. By the reverse of (15), it follows that \(J>0\). If \(J=\infty \), then (22) is trivially valid. If \(0< J<\infty \), then we find
Hence (22) is valid, which is equivalent to (21).
For \(0<\varepsilon <p\lambda _{1}\), we set \(\widetilde{\lambda }_{1}=\lambda _{1}-\frac{\varepsilon }{p}\) (\(\in (0,1)\)), \(\widetilde{\lambda }_{2}=\lambda _{2}+\frac{\varepsilon }{p}\) (>0), \(\widetilde{a}_{m}=U_{m}^{\widetilde{\lambda }_{1}-1}\mu _{m}\), \(\widetilde{b}_{n}=V_{n}^{\widetilde{\lambda }_{2}-\varepsilon -1}\nu _{n}\). By (10), (11), and (7), in view of Remark 1, we find
If there exists a positive number \(K\geq k_{\lambda }(\lambda _{1})\), such that (21) is still valid when replacing \(k_{\lambda }(\lambda _{1})\) by K, then in particular, we have
We obtain from the above results that
and then \(k_{\lambda }(\lambda _{1})\geq K\) (for \(\varepsilon \rightarrow 0^{+}\)). Hence \(K=k_{\lambda }(\lambda _{1})\) is the best value of (21).
We conform that the constant factor \(k_{\lambda }(\lambda _{1})\) in (22) is the best possible. Otherwise we can get a contradiction by (23): that the constant factor in (21) is not the best value. □
Theorem 4
Suppose that \(p<0\), \(\{\mu _{m}\}_{m=1}^{\infty }\) and \(\{\nu _{n}\}_{n=1}^{\infty }\) are decreasing positive sequences, and \(U(\infty )=V(\infty )=\infty \), then we have the following equivalent inequalities:
where the constant factor \([ \frac{\pi }{\lambda \sin (\pi \lambda _{1}/\lambda )} ] ^{2}\) is the best possible.
Proof
Using the same way of obtaining (14) and (15), by the reverse Hölder inequality with weight and (9), we have
then we obtain (25) by (6). Using the reverse Hölder inequality, we have
Hence (24) is valid by (25). Assuming that (24) is valid, setting
we find
It follows that \(J_{1}>0\) by (26). If \(J_{1}=\infty \), then (25) is trivially valid. If \(0< J_{1}<\infty \), then we find
Hence (25) is valid, which is equivalent to (24).
For \(0<\varepsilon <q\lambda _{2}\), we set \(\widetilde{\lambda }_{1}=\lambda _{1}+\frac{\varepsilon }{q}\) (>0), \(\widetilde{\lambda }_{2}=\lambda _{2}-\frac{\varepsilon }{q}\) (\(\in (0,1)\)), \(\widetilde{a}_{m}=U_{m}^{\widetilde{\lambda }_{1}-\varepsilon -1}\mu _{m}\), \(\widetilde{b}_{n}=V_{n}^{\widetilde{\lambda }_{2}-1}\nu _{n}\). By (10), (11), and (6), in view of Remark 1, we have
If there exists a positive number \(K\geq k_{\lambda }(\lambda _{1})\), such that (24) is still valid as we replace \(k_{\lambda }(\lambda _{1})\) by K, then, in particular, we have
From the above results, we have
It follows that \(k_{\lambda }(\lambda _{1})\geq K\) (for \(\varepsilon \rightarrow 0^{+}\)). Hence \(K=k_{\lambda }(\lambda _{1})\) is the best value of (24). We conform that the constant factor \(k_{\lambda }(\lambda _{1})\) in (25) is the best possible. Otherwise we can get a contradiction by (27): that the constant factor in (24) is not the best value. □
Remark 2
For \(\mu _{i}=\nu _{i}=1\) (\(i=1,2,\ldots \)), (12) reduces to (3); for \(\lambda =1\), \(\lambda _{1}=\frac{1}{q}\), \(\lambda _{2}=\frac{1}{p}\), it follows by (12) that
for \(\lambda =1\), \(\lambda _{1}=\frac{1}{p}\), \(\lambda _{2}=\frac{1}{q}\), (12) reduces to the dual form of (28) as follows:
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Acknowledgements
This work is supported by the 2013 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2013KJCX0140).
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QH carried out the mathematical studies, sequenced alignment, drafted the manuscript, and performed the numerical analysis. The author read and approved the final manuscript.
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Huang, Q. A new extension of a Hardy-Hilbert-type inequality. J Inequal Appl 2015, 397 (2015). https://doi.org/10.1186/s13660-015-0918-7
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DOI: https://doi.org/10.1186/s13660-015-0918-7