1 Introduction
The Hilbert inequality [5] asserts that
holds for non-negative sequences \(a_{m}\) and \(b_{n}\), provided that \((\sum_{m=1}^{\infty }a_{m}^{p} )^{\frac{1}{p}}>0\) and \((\sum_{n=1}^{\infty }b_{n}^{q} )^{\frac{1}{q}}>0\). The parameters p and q appearing in (1) are mutually conjugate, i.e. \({\frac{1}{p}}+{\frac{1}{q}}=1\), where \(p>1\). In addition, the constant \(\frac{\pi }{\sin (\pi /p)}\) is the best possible in the sense that it can not be replaced with a smaller constant so that (1) still holds.
$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{ \pi }{\sin (\pi /p)} \Biggl( \sum_{m=1}^{\infty }a_{m}^{p} \Biggr)^{ \frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }b_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(1)
The Hilbert inequality is one of the most interesting inequalities in mathematical analysis. For a detailed review of the starting development of the Hilbert inequality the reader is referred to monograph [5]. The most important recent results regarding Hilbert-type inequalities are collected in monographs [4] and [7].
Advertisement
In 2006, Krnić and Pečarić [6], obtained the following generalization of classical Hilbert inequality.
Theorem 1
Let
\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), and let
\(2< s\leq 14\). Suppose that
\(\alpha _{1}\in [-\frac{1}{q},\frac{1}{q} )\), \(\alpha _{2}\in [-\frac{1}{p},\frac{1}{p} )\)
and
\(p\alpha _{2}+q \alpha _{1}=2-s\). If
\(\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1}a_{m}^{p}< \infty \)
and
\(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}b_{n}^{q}<\infty \), then
where the constant
\(B(1-p\alpha _{2}, p\alpha _{2}+s-1)\)
is the best possible.
$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{s}}< B(1-p \alpha _{2}, p\alpha _{2}+s-1) \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n ^{pq\alpha _{2}-1}b_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(2)
In the last few years, considerable attention is given to a class of Hilbert-type inequalities where the functions and sequences are replaced by certain integral or discrete operators. For example: in 2013, Azar [3] introduced a new Hilbert-type integral inequality including functions \(F(x)=\int _{0}^{x} f(t)\,dt\) and \(g(y)=\int _{0}^{y} g(t)\,dt\). For some related Hilbert-type inequalities where the functions and sequences are replaced by certain integral or discrete operators, the reader is referred to [1] and [2].
The main objective of this paper is to derive a discrete Hilbert-type inequality involving partial sums, similar to a result of Azar [3]. Such inequality is derived by virtue of inequality (2) and some well-known classical inequalities. As an application, we consider some particular settings.
Advertisement
2 Preliminaries and lemma
Recall that the Gamma function \(\varGamma (\theta )\) and the Beta function \(B ( \mu ,\nu ) \) are defined, respectively, by
and they satisfy the following relation
$$\begin{aligned}& \varGamma (\theta )= \int _{0}^{\infty }t^{\theta -1}e^{-t}\,dt, \quad \theta >0, \\& B ( \mu ,\nu ) = \int _{0}^{\infty }\frac{t^{\mu -1}}{ ( t+1 ) ^{\mu +\nu }}\,dt, \quad \mu ,\nu >0, \end{aligned}$$
$$ B ( \mu ,\nu ) =\frac{\varGamma (\mu )\varGamma (\nu )}{\varGamma (\mu +\nu )}. $$
By the definition of the Gamma function, the following equality holds:
$$ \frac{1}{ ( m+n ) ^{\lambda }}=\frac{1}{\varGamma ( \lambda ) } \int _{0}^{\infty }t^{\lambda -1}e^{- ( m+n ) t}\,dt. $$
(3)
To prove our main results we need the following lemma.
Lemma 2
Let
\(a_{m}>0\), \(a_{m}\in \ell ^{1}\), \(A_{m}=\sum_{k=1}^{m} a_{k}\), then for
\(t>0\), we have
$$ \sum_{m=1}^{\infty }e^{-tm}a_{m} \leq t\sum_{m=1}^{\infty }e^{-tm}A _{m}. $$
(4)
Proof
Using Abel’s summation by parts formula and the inequality \(1-\frac{1}{e ^{t}}\leq t\), we have
The lemma is proved. □
$$\begin{aligned} \sum_{m=1}^{\infty }e^{-tm}a_{m} =&\lim_{m\to \infty }A_{m}e^{-t(m+1)}+ \sum _{m=1}^{\infty }A_{m}\bigl(e^{-tm}-e^{-t(m+1)} \bigr) \\ =& \biggl(1-\frac{1}{e^{t}} \biggr)\sum_{m=1}^{\infty }e^{-tm}A_{m} \\ \leq &t\sum_{m=1}^{\infty }e^{-tm}A_{m}. \end{aligned}$$
3 Main results
Theorem 3
Let
\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\lambda >0\), \(a_{m}, b_{n}>0\), \(a _{m}, b_{n}\in \ell ^{1}\), define
\(A_{m}=\sum_{k=1}^{m}a_{k}\), \(B_{n}= \sum_{k=1}^{n}b_{k}\). If
\(\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1}A_{m} ^{p}<\infty \)
and
\(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q}< \infty \), then
where
\(\alpha _{1}\in [-\frac{1}{q},0 )\), \(\alpha _{2}\in [-\frac{1}{p},0 )\)
and
\(p\alpha _{2}+q\alpha _{1}=-\lambda \). In addition, the constant
\(C=pq\alpha _{1}\alpha _{2}B(-p\alpha _{2}, -q\alpha _{1})\)
is the best possible in (5).
$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< C \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr) ^{\frac{1}{q}}, $$
(5)
Proof
Using (3), the left-hand side of inequality (5) can be expressed in the following form:
Now, by applying inequality (4) and equality (3) to the previous equality, we have
Moreover, the last double series represents the left-hand side of the Hilbert-type inequality (2) for \(s=2+\lambda \), that is, we have the inequality
so by (7) we get
Now we shall prove that the constant factor is the best possible. Assuming that the constant C is not the best possible, then there exists a positive constant K such that \(K< C\) and (5) still remains valid if C is replaced by K. Further, consider the \(\tilde{a}_{m}=m^{-q\alpha _{1}-1-\frac{\varepsilon }{p}}\) and \(\tilde{b}_{n}=n^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}\), where \(\varepsilon >0\) is sufficiently small number. Then, we have
and similarly
Inserting the above sequences in (5), the right-hand side of (5) becomes
Now, let us estimate the left-hand side of inequality (5). Namely, by inserting the above defined sequences \(\tilde{a}_{m}\) and \(\tilde{b}_{n}\) in the left-hand side of inequality (5), we get the inequality
It follows from inequalities (8) and (9) that
Now, letting \(\varepsilon \to 0+\), relation (10) yields a contradiction with the assumption \(K< C\). So the constant C, in inequality (5) is the best possible. □
$$\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }} =&\frac{1}{\varGamma (\lambda )}\sum _{m=1}^{\infty }\sum_{n=1} ^{\infty }a_{m}b_{n} \biggl( \int _{0}^{\infty }t^{\lambda -1}e^{-(m+n)t}\,dt \biggr) \\ =&\frac{1}{\varGamma (\lambda )} \int _{0}^{\infty }t^{\lambda -1} \Biggl(\sum _{m=1}^{\infty }e^{-tm}a_{m} \Biggr) \Biggl(\sum_{n=1}^{\infty }e^{-tn}b _{n} \Biggr)\,dt. \end{aligned}$$
(6)
$$\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }} \leq &\frac{1}{\varGamma (\lambda )} \int _{0}^{\infty }t^{ \lambda +1} \Biggl(\sum _{m=1}^{\infty }e^{-tm}A_{m} \Biggr) \Biggl(\sum_{n=1} ^{\infty }e^{-tn}B_{n} \Biggr)\,dt \\ =&\frac{1}{\varGamma (\lambda )}\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }A_{m}B_{n} \biggl( \int _{0}^{\infty }t^{\lambda +1}e^{-(m+n)t}\,dt \biggr) \\ =&\frac{\varGamma (\lambda +2)}{\varGamma (\lambda )}\sum_{m=1}^{\infty } \sum _{n=1}^{\infty }\frac{A_{m}B_{n}}{(m+n)^{\lambda +2}}. \end{aligned}$$
(7)
$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{A_{m}B_{n}}{(m+n)^{ \lambda +2}}< B(1-p\alpha _{2}, p\alpha _{2}+\lambda +1) \Biggl(\sum _{m=1} ^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< C \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr) ^{\frac{1}{q}}. $$
$$ \tilde{A}_{m}=\sum_{k=1}^{m} \tilde{a}_{k}=\sum_{k=1}^{m}m^{-q\alpha _{1}-1-\frac{\varepsilon }{p}} \leq \int _{0}^{m}x^{-q\alpha _{1}-1-\frac{ \varepsilon }{p}}\,dx= \frac{m^{-q\alpha _{1}-\frac{\varepsilon }{p}}}{-q \alpha _{1}-\frac{\varepsilon }{p}}, $$
$$ \tilde{B}_{n}=\sum_{k=1}^{n} \tilde{b}_{k}\leq \frac{n^{-p\alpha _{2}-\frac{ \varepsilon }{q}}}{-p\alpha _{2}-\frac{\varepsilon }{q}}. $$
$$\begin{aligned} &K \Biggl(\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1} \tilde{A}_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty }n^{pq\alpha _{2}-1}\tilde{B} _{n}^{q} \Biggr)^{\frac{1}{q}} \\ & \quad \leq K \Biggl(\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1} \frac{m^{-pq\alpha _{1}-\varepsilon }}{(-q\alpha _{1}-\frac{\varepsilon }{p})^{p}} \Biggr) ^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty }n^{pq\alpha _{2}-1}\frac{n ^{-pq\alpha _{2}-\varepsilon }}{(-p\alpha _{2}-\frac{\varepsilon }{q})^{q}} \Biggr) ^{\frac{1}{q}} \\ & \quad =\frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \Biggl(\sum_{m=1}^{\infty }m^{-1-\varepsilon } \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{-1-\varepsilon } \Biggr) ^{\frac{1}{q}} \\ & \quad =\frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \Biggl(1+\sum_{m=2}^{\infty }m^{-1-\varepsilon } \Biggr) \\ & \quad \leq \frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \biggl(1+ \int _{1}^{\infty }x^{-1-\varepsilon }\,dx \biggr) \\ & \quad =\frac{K(1+\varepsilon )}{\varepsilon (-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{\varepsilon }{q})}. \end{aligned}$$
(8)
$$\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{\tilde{a}_{m}\tilde{b} _{n}}{(m+n)^{\lambda }} &=\sum _{m=1}^{\infty }\sum_{n=1}^{\infty } \frac{m ^{-q\alpha _{1}-1-\frac{\varepsilon }{p}}n^{-p\alpha _{2}-1-\frac{ \varepsilon }{q}}}{(m+n)^{\lambda }} \\ &\geq \int _{1}^{\infty } \int _{1}^{\infty }\frac{x^{-q\alpha _{1}-1-\frac{ \varepsilon }{p}}y^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(x+y)^{ \lambda }}\,dx \,dy \\ &= \int _{1}^{\infty }x^{-1-\varepsilon } \int _{1/x}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du \,dx \\ &= \int _{1}^{\infty }x^{-1-\varepsilon } \biggl( \int _{0}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du- \int _{0}^{1/x}\frac{u^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{ \lambda }}\,du \biggr)\,dx \\ &\geq \int _{1}^{\infty }x^{-1-\varepsilon } \biggl( \int _{0}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du- \int _{0}^{1/x}u^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}\,du \biggr)\,dx \\ &=\frac{1}{\varepsilon }\cdot B \biggl(-p\alpha _{2}- \frac{\varepsilon }{q},-q\alpha _{1}+\frac{\varepsilon }{q} \biggr)- \frac{1}{ (p \alpha _{2}+\frac{\varepsilon }{q} ) (p\alpha _{2}-\frac{ \varepsilon }{p} )}. \end{aligned}$$
(9)
$$ B \biggl(-p\alpha _{2}-\frac{\varepsilon }{q},-q\alpha _{1}+\frac{\varepsilon }{q} \biggr)-\frac{\varepsilon }{ (p\alpha _{2}+\frac{\varepsilon }{q} ) (p\alpha _{2}-\frac{\varepsilon }{p} )}\leq \frac{K(1+ \varepsilon )}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})}. $$
(10)
Considering Theorem 3, equipped with parameters \(\lambda =1\), \(\alpha _{1}=-\frac{1}{q^{2}}\), \(\alpha _{2}=-\frac{1}{p^{2}}\), we obtain the following result.
Corollary 4
Let
\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), and let
\(a_{m}, b_{n}>0\)
with
\(a_{m}, b_{n}\in \ell ^{1}\). If
\(\sum_{m=1}^{\infty }m^{-p}A_{m}^{p}< \infty \)
and
\(\sum_{n=1}^{\infty }n^{-q}B_{n}^{q}<\infty \), then
where the constant
\(\frac{\pi }{pq\sin (\pi /p)}\)
is the best possible.
$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{ \pi }{pq\sin (\pi /p)} \Biggl( \sum_{m=1}^{\infty } \biggl(\frac{A_{m}}{m} \biggr) ^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty } \biggl(\frac{B _{n}}{n} \biggr)^{q} \Biggr)^{\frac{1}{q}}. $$
(11)
Remark 5
It should be noticed here that if sequences \(a_{m}, b_{n}\in \ell ^{1}\) such that \(\sum_{m=1}^{\infty }a_{m}^{p}<\infty \) and \(\sum_{n=1}^{\infty }b_{n}^{q}<\infty \), inequality (11) provides refinement of the Hilbert inequality. Indeed, by Hardy’s inequality, the series \(\sum_{m=1}^{\infty } (\frac{A_{m}}{m} ) ^{p}\) and \(\sum_{n=1}^{\infty } (\frac{B_{n}}{n} )^{q}\) are converge. So, inequality (11) holds. The Hilbert inequality becomes after applying Hardy’s inequality on the right-hand side of inequality (11).
Letting \(\alpha _{1}=\alpha _{2}=\frac{-\lambda }{pq}\) in Theorem 3, we can obtain the following Hilbert-type inequality.
Corollary 6
Let
\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), and let
\(0<\lambda \leq \min \{p,q \}\), \(a_{m}, b_{n}>0\)
with
\(a_{m}, b_{n}\in \ell ^{1}\). If
\(\sum_{m=1} ^{\infty }m^{-\lambda -1}A_{m}^{p}<\infty \)
and
\(\sum_{n=1}^{\infty }n ^{-\lambda -1}B_{n}^{q}<\infty \), then
where the constant
\(\frac{\lambda ^{2}}{pq} B (\frac{\lambda }{q}, \frac{\lambda }{p} )\)
is the best possible.
$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< \frac{\lambda ^{2}}{pq} B \biggl(\frac{\lambda }{q}, \frac{ \lambda }{p} \biggr) \Biggl(\sum _{m=1}^{\infty }m^{-\lambda -1}A_{m} ^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{-\lambda -1}B _{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(12)
4 Conclusion
In the present study, we have established a discrete Hilbert-type inequality involving partial sums. Moreover, we have proved that the constant on the right-hand side of this inequality is the best possible. As an application, we considered some particular settings.
Competing interests
The authors declare that they have no competing interests.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.