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A new discrete Hilbert-type inequality involving partial sums

Abstract

In this paper, we derive a new discrete Hilbert-type inequality involving partial sums. Moreover, we show that the constant on the right-hand side of this inequality is the best possible. As an application, we consider some particular settings.

Introduction

The Hilbert inequality [5] asserts that

$$ \sum_{n=1}^{\infty }\sum _{m=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{ \pi }{\sin (\pi /p)} \Biggl( \sum_{m=1}^{\infty }a_{m}^{p} \Biggr)^{ \frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }b_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(1)

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.

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].

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

$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{s}}< B(1-p \alpha _{2}, p\alpha _{2}+s-1) \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}a_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n ^{pq\alpha _{2}-1}b_{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(2)

where the constant \(B(1-p\alpha _{2}, p\alpha _{2}+s-1)\) is the best possible.

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.

Preliminaries and lemma

Recall that the Gamma function \(\varGamma (\theta )\) and the Beta function \(B ( \mu ,\nu ) \) are defined, respectively, by

$$\begin{aligned}& \varGamma (\theta )= \int _{0}^{\infty }t^{\theta -1}e^{-t}\,dt, \quad \theta >0, \\& B ( \mu ,\nu ) = \int _{0}^{\infty }\frac{t^{\mu -1}}{ ( t+1 ) ^{\mu +\nu }}\,dt, \quad \mu ,\nu >0, \end{aligned}$$

and they satisfy the following relation

$$ B ( \mu ,\nu ) =\frac{\varGamma (\mu )\varGamma (\nu )}{\varGamma (\mu +\nu )}. $$

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

$$\begin{aligned} \sum_{m=1}^{\infty }e^{-tm}a_{m} =&\lim_{m\to \infty }A_{m}e^{-t(m+1)}+ \sum _{m=1}^{\infty }A_{m}\bigl(e^{-tm}-e^{-t(m+1)} \bigr) \\ =& \biggl(1-\frac{1}{e^{t}} \biggr)\sum_{m=1}^{\infty }e^{-tm}A_{m} \\ \leq &t\sum_{m=1}^{\infty }e^{-tm}A_{m}. \end{aligned}$$

The lemma is proved. □

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

$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< C \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr) ^{\frac{1}{q}}, $$
(5)

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).

Proof

Using (3), the left-hand side of inequality (5) can be expressed in the following form:

$$\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }} =&\frac{1}{\varGamma (\lambda )}\sum _{m=1}^{\infty }\sum_{n=1} ^{\infty }a_{m}b_{n} \biggl( \int _{0}^{\infty }t^{\lambda -1}e^{-(m+n)t}\,dt \biggr) \\ =&\frac{1}{\varGamma (\lambda )} \int _{0}^{\infty }t^{\lambda -1} \Biggl(\sum _{m=1}^{\infty }e^{-tm}a_{m} \Biggr) \Biggl(\sum_{n=1}^{\infty }e^{-tn}b _{n} \Biggr)\,dt. \end{aligned}$$
(6)

Now, by applying inequality (4) and equality (3) to the previous equality, we have

$$\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }} \leq &\frac{1}{\varGamma (\lambda )} \int _{0}^{\infty }t^{ \lambda +1} \Biggl(\sum _{m=1}^{\infty }e^{-tm}A_{m} \Biggr) \Biggl(\sum_{n=1} ^{\infty }e^{-tn}B_{n} \Biggr)\,dt \\ =&\frac{1}{\varGamma (\lambda )}\sum_{m=1}^{\infty }\sum _{n=1}^{\infty }A_{m}B_{n} \biggl( \int _{0}^{\infty }t^{\lambda +1}e^{-(m+n)t}\,dt \biggr) \\ =&\frac{\varGamma (\lambda +2)}{\varGamma (\lambda )}\sum_{m=1}^{\infty } \sum _{n=1}^{\infty }\frac{A_{m}B_{n}}{(m+n)^{\lambda +2}}. \end{aligned}$$
(7)

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

$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{A_{m}B_{n}}{(m+n)^{ \lambda +2}}< B(1-p\alpha _{2}, p\alpha _{2}+\lambda +1) \Biggl(\sum _{m=1} ^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr)^{\frac{1}{q}}, $$

so by (7) we get

$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< C \Biggl(\sum _{m=1}^{\infty }m^{pq\alpha _{1}-1}A_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{pq\alpha _{2}-1}B_{n}^{q} \Biggr) ^{\frac{1}{q}}. $$

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

$$ \tilde{A}_{m}=\sum_{k=1}^{m} \tilde{a}_{k}=\sum_{k=1}^{m}m^{-q\alpha _{1}-1-\frac{\varepsilon }{p}} \leq \int _{0}^{m}x^{-q\alpha _{1}-1-\frac{ \varepsilon }{p}}\,dx= \frac{m^{-q\alpha _{1}-\frac{\varepsilon }{p}}}{-q \alpha _{1}-\frac{\varepsilon }{p}}, $$

and similarly

$$ \tilde{B}_{n}=\sum_{k=1}^{n} \tilde{b}_{k}\leq \frac{n^{-p\alpha _{2}-\frac{ \varepsilon }{q}}}{-p\alpha _{2}-\frac{\varepsilon }{q}}. $$

Inserting the above sequences in (5), the right-hand side of (5) becomes

$$\begin{aligned} &K \Biggl(\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1} \tilde{A}_{m}^{p} \Biggr) ^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty }n^{pq\alpha _{2}-1}\tilde{B} _{n}^{q} \Biggr)^{\frac{1}{q}} \\ & \quad \leq K \Biggl(\sum_{m=1}^{\infty }m^{pq\alpha _{1}-1} \frac{m^{-pq\alpha _{1}-\varepsilon }}{(-q\alpha _{1}-\frac{\varepsilon }{p})^{p}} \Biggr) ^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty }n^{pq\alpha _{2}-1}\frac{n ^{-pq\alpha _{2}-\varepsilon }}{(-p\alpha _{2}-\frac{\varepsilon }{q})^{q}} \Biggr) ^{\frac{1}{q}} \\ & \quad =\frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \Biggl(\sum_{m=1}^{\infty }m^{-1-\varepsilon } \Biggr) ^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{-1-\varepsilon } \Biggr) ^{\frac{1}{q}} \\ & \quad =\frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \Biggl(1+\sum_{m=2}^{\infty }m^{-1-\varepsilon } \Biggr) \\ & \quad \leq \frac{K}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})} \biggl(1+ \int _{1}^{\infty }x^{-1-\varepsilon }\,dx \biggr) \\ & \quad =\frac{K(1+\varepsilon )}{\varepsilon (-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{\varepsilon }{q})}. \end{aligned}$$
(8)

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

$$\begin{aligned} \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{\tilde{a}_{m}\tilde{b} _{n}}{(m+n)^{\lambda }} &=\sum _{m=1}^{\infty }\sum_{n=1}^{\infty } \frac{m ^{-q\alpha _{1}-1-\frac{\varepsilon }{p}}n^{-p\alpha _{2}-1-\frac{ \varepsilon }{q}}}{(m+n)^{\lambda }} \\ &\geq \int _{1}^{\infty } \int _{1}^{\infty }\frac{x^{-q\alpha _{1}-1-\frac{ \varepsilon }{p}}y^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(x+y)^{ \lambda }}\,dx \,dy \\ &= \int _{1}^{\infty }x^{-1-\varepsilon } \int _{1/x}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du \,dx \\ &= \int _{1}^{\infty }x^{-1-\varepsilon } \biggl( \int _{0}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du- \int _{0}^{1/x}\frac{u^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{ \lambda }}\,du \biggr)\,dx \\ &\geq \int _{1}^{\infty }x^{-1-\varepsilon } \biggl( \int _{0}^{\infty }\frac{u ^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}}{(1+u)^{\lambda }}\,du- \int _{0}^{1/x}u^{-p\alpha _{2}-1-\frac{\varepsilon }{q}}\,du \biggr)\,dx \\ &=\frac{1}{\varepsilon }\cdot B \biggl(-p\alpha _{2}- \frac{\varepsilon }{q},-q\alpha _{1}+\frac{\varepsilon }{q} \biggr)- \frac{1}{ (p \alpha _{2}+\frac{\varepsilon }{q} ) (p\alpha _{2}-\frac{ \varepsilon }{p} )}. \end{aligned}$$
(9)

It follows from inequalities (8) and (9) that

$$ B \biggl(-p\alpha _{2}-\frac{\varepsilon }{q},-q\alpha _{1}+\frac{\varepsilon }{q} \biggr)-\frac{\varepsilon }{ (p\alpha _{2}+\frac{\varepsilon }{q} ) (p\alpha _{2}-\frac{\varepsilon }{p} )}\leq \frac{K(1+ \varepsilon )}{(-q\alpha _{1}-\frac{\varepsilon }{p})(-p\alpha _{2}-\frac{ \varepsilon }{q})}. $$
(10)

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. □

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

$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{m+n}< \frac{ \pi }{pq\sin (\pi /p)} \Biggl( \sum_{m=1}^{\infty } \biggl(\frac{A_{m}}{m} \biggr) ^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum _{n=1}^{\infty } \biggl(\frac{B _{n}}{n} \biggr)^{q} \Biggr)^{\frac{1}{q}}. $$
(11)

where the constant \(\frac{\pi }{pq\sin (\pi /p)}\) is the best possible.

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

$$ \sum_{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{a_{m}b_{n}}{(m+n)^{ \lambda }}< \frac{\lambda ^{2}}{pq} B \biggl(\frac{\lambda }{q}, \frac{ \lambda }{p} \biggr) \Biggl(\sum _{m=1}^{\infty }m^{-\lambda -1}A_{m} ^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum_{n=1}^{\infty }n^{-\lambda -1}B _{n}^{q} \Biggr)^{\frac{1}{q}}, $$
(12)

where the constant \(\frac{\lambda ^{2}}{pq} B (\frac{\lambda }{q}, \frac{\lambda }{p} )\) is the best possible.

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.

References

  1. 1.

    Adiyasuren, V., Batbold, Ts., Krnić, M.: On several new Hilbert-type inequalities involving means operators. Acta Math. Sin. Engl. Ser. 29, 1493–1514 (2013)

    MathSciNet  Article  Google Scholar 

  2. 2.

    Adiyasuren, V., Batbold, Ts., Krnić, M.: Multiple Hilbert-type inequalities involving some differential operators. Banach J. Math. Anal. 10(2), 320–337 (2016)

    MathSciNet  Article  Google Scholar 

  3. 3.

    Azar, L.E.: Two new forms of Hilbert-type integral inequality. Math. Inequal. Appl. 17(3), 937–946 (2014)

    MathSciNet  MATH  Google Scholar 

  4. 4.

    Batbold, Ts., Krnić, M., Pečarić, J., Vuković, P.: Further Development of Hilbert-Type Inequalities. Element, Zagreb (2017)

    Google Scholar 

  5. 5.

    Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities, 2nd edn. Cambridge University Press, Cambridge (1967)

    Google Scholar 

  6. 6.

    Krnić, M., Pečarić, J.: Extension of Hilbert’s inequality. J. Math. Anal. Appl. 324(1), 150–160 (2006)

    MathSciNet  Article  Google Scholar 

  7. 7.

    Krnić, M., Pečarić, J., Perić, I., Vuković, P.: Recent Advances in Hilbert-Type Inequalities. Element, Zagreb (2012)

    Google Scholar 

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Funding

The authors would like to thank the National University of Mongolia for supporting this research.

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Adiyasuren, V., Batbold, T. & Azar, L.E. A new discrete Hilbert-type inequality involving partial sums. J Inequal Appl 2019, 127 (2019). https://doi.org/10.1186/s13660-019-2087-6

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MSC

  • 26D15

Keywords

  • Hilbert-type inequality
  • Gamma function
  • The best possible constant
  • Conjugate exponents