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.

1 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.

2 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. □

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

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

(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.

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.