2. An improved Hardy’s inequality with shifts

We begin this section by introducing shifting backward and forward difference operators acting on a sequence $A=\{A_n\}$ of real or complex numbers. Let us choose $m, n\in \mathbb{N}$ and $r$ , $s$ be two real numbers such that $rs\gt 0$ . Then the $m$ th shifting backward $T_m^{(r, s)}$ and forward difference operators $T_m^{*(r, s)}$ acting on a sequence $\{A_n\}$ are defined as follows:

\begin{align*} (T_m^{(r, s)}A)_{n} &= \left \{ \begin{array}{lll} -sA_{n} & \quad \mbox{if}\,\,n=1, 2, \ldots, m,\\[5pt] rA_{n-m}-sA_{n} & \quad \mbox{if}\,\,n\gt m, \end{array}\right . \end{align*}

and for $n\geq m$

\begin{align*} (T_m^{*(r, s)}A)_{n}&=rA_{n+m}-sA_{n}. \end{align*}

Using the above difference operators, we introduce the $m$ th shift discrete Dirichlet’s Laplacian operator $(\!-\Delta _{m}^{(r, s)})$ acting on $A=\{A_n\}$ as below:

\begin{align*} (\!-\Delta _{m}^{(r, s)}A)_n=(T_m^{*(r, s)}T_m^{(r, s)}A)_n=(r^{2}+s^{2})A_{n}-rsA_{n-m}-rsA_{n+m}. \end{align*}

Note that the $1$ st shift discrete Dirichlet’s Laplacian operator $(\!-\Delta _{1}^{(r, s)})$ with $r=s=1$ coincides with the well-known discrete Dirichlet’s Laplacian operator $(\!-\Delta )$ and has been considered earlier in [Reference Gerhat, Krejčiřík and Štampach5, Reference Krejčiřík, Laptev and Štampach16] (see also [Reference Das and Manna3]) in connection with the study of Hardy and Rellich inequalities. Now for a strictly positive sequence $\mu =\{\mu _n\}$ of real numbers, we define a weight sequence $w_{n}(r, s, \mu )$ as

\begin{align*} w_{n}(r, s, \mu )&=\frac{(\!-\Delta _{m}^{(r, s)}\mu )_{n}}{\mu _n}=r^{2}+s^{2}-rs\Big (\frac{\mu _{n+m}}{\mu _{n}}+\frac{\mu _{n-m}}{\mu _{n}}\Big ), \,\,\,m\in \mathbb{N}. \end{align*}

Suppose further that $\lambda =\{\lambda _n\}$ is a strictly positive sequence of real numbers. Then we introduce a more accurate and general form of the weight sequence $w_{n}(r, s, \mu )$ as follows:

(2.1) \begin{align}{} w_{n}(\lambda, r, s, \mu )&=\frac{1}{\lambda _{n}}\Big (r^{2}-rs\frac{\mu _{n-m}}{\mu _{n}}\Big ) +\frac{1}{\lambda _{n+m}}\Big (s^2-rs\frac{\mu _{n+m}}{\mu _{n}}\Big ), \,\,\,m\in \mathbb{N}. \end{align}

With this definition, we have the following result, which gives a general improved discrete Hardy’s inequality with shifts. Let us begin with the following statement.

Theorem 2.1. Let $A=\{A_n\}$ be any sequence of complex numbers such that $A\in C_c(\mathbb{N}_0)$ with $A_{n}=0$ for $0\leq n\leq m-1$ and $r, s\in \mathbb{R}$ be such that $rs\gt 0$ . Then we have

(2.2) \begin{align}{} &\sum _{n=m}^{\infty }\frac{|rA_{n}-sA_{n-m}|^{2}}{\lambda _{n}}\geq \sum _{n=m}^{\infty }w_{n}(\lambda, r, s, \mu )|A_{n}|^{2}, \end{align}

where the sequence $w_{n}(\lambda, r, s, \mu )$ is defined in ( 2.1 ).

Proof. We first calculate the following sum and proceed with an approach initiated by [Reference Krejčiřík and Štampach15]. Let us assume that $A_{n}=0$ , $0\leq n\leq m-1$ and $\frac{\mid A_{0}\mid ^{2}}{\lambda _{m}\mu _{0}}$ is zero. Then

\begin{align*} &\sum _{n=m}^{\infty }w_{n}(\lambda, r, s, \mu )|A_{n}|^{2}\\[5pt] &=\sum _{n=m}^{\infty }\Big (\frac{r^{2}}{\lambda _{n}}+\frac{s^{2}}{\lambda _{n+m}}\Big )|A_{n}|^{2}-\sum _{n=m}^{\infty } \Big (\frac{rs}{\lambda _{n+m}}\frac{\mu _{n+m}}{\mu _{n}}+\frac{rs}{\lambda _{n}}\frac{\mu _{n-m}}{\mu _{n}}\Big )|A_{n}|^{2}\\[5pt] &=\sum _{n=m}^{\infty }\frac{r^{2}}{\lambda _{n}}|A_{n}|^{2}+\sum _{n=2m}^{\infty }\frac{s^{2}}{\lambda _{n}}|A_{n-m}|^{2} -rs\sum _{n=2m}^{\infty }\frac{\mu _{n}}{\lambda _{n}\mu _{n-m}}|A_{n-m}|^{2}-rs\sum _{n=m}^{\infty }\frac{\mu _{n-m}}{\lambda _{n}\mu _{n}}|A_{n}|^{2}\\[5pt] &=\sum _{n=m}^{\infty }\Big (\frac{r^{2}}{\lambda _{n}}|A_{n}|^{2}+\frac{s^{2}}{\lambda _{n}}|A_{n-m}|^{2}\Big )-rs\sum _{n=m}^{\infty } \Big (\frac{\mu _{n}}{\lambda _{n}\mu _{n-m}}|A_{n-m}|^{2}+\frac{\mu _{n-m}}{\lambda _{n}\mu _{n}}|A_{n}|^{2}\Big ). \end{align*}

Further assume that $\mu _{n}=0$ for $0\leq n\leq m-1$ and term $\sqrt{\frac{\mu _{m}}{\lambda _{m}\mu _{0}}}A_{0}$ is also zero. Since $rs\gt 0$ , so the following computation of differences gives

\begin{align*} &\sum _{n=m}^{\infty }\frac{|rA_{n}-sA_{n-m}|^{2}}{\lambda _{n}}-\sum _{n=m}^{\infty }w_{n}(\lambda, r, s, \mu )|A_{n}|^{2}\\[5pt] =&\sum _{n=m}^{\infty }\Big (\frac{r^{2}}{\lambda _{n}}|A_{n}|^{2}+\frac{s^{2}}{\lambda _{n}}|A_{n-m}|^{2}-\frac{rs}{\lambda _{n}}2\mathbb{R}(A_{n}\bar A_{n-m})\Big )\\[5pt] &-\sum _{n=m}^{\infty }\Big (\frac{r^{2}}{\lambda _{n}}|A_{n}|^{2}+\frac{s^{2}}{\lambda _{n}}|A_{n-m}|^{2}-rs\frac{\mu _{n}}{\lambda _{n}\mu _{n-m}}|A_{n-m}|^{2} -rs\frac{\mu _{n-m}}{\lambda _{n}\mu _{n}}|A_{n}|^{2}\Big )\\[5pt] =&rs\sum _{n=m}^{\infty }\Big (\frac{\mu _{n}}{\lambda _{n}\mu _{n-m}}|A_{n-m}|^{2} +\frac{\mu _{n-m}}{\lambda _{n}\mu _{n}}|A_{n}|^{2}\Big )-\frac{rs}{\lambda _{n}}\sum _{n=m}^{\infty }2\mathbb{R}(A_{n}\bar A_{n-m})\\[5pt] =&rs\sum _{n=2m}^{\infty }\frac{1}{\lambda _{n}}\mid \big (\frac{\mu _{n-m}}{\mu _{n}}\big )^{1/2}A_{n}-\big (\frac{\mu _{n}}{\mu _{n-m}}\big )^{1/2}A_{n-m}\mid ^{2}\geq 0. \end{align*}

This proves the desired inequality:

\begin{align*} &\sum _{n=m}^{\infty }\frac{|rA_{n}-sA_{n-m}|^{2}}{\lambda _{n}}\geq \sum _{n=m}^{\infty }w_{n}(\lambda, r, s, \mu )|A_{n}|^{2}. \end{align*}

Hence the theorem.

Observe that Theorem2.1 establishes several important results and some of them are given in the following corollaries.

Corollary 2.1. Suppose that $\lambda _{n}=1$ and $\mu _{n}=n^{\alpha }$ , $\alpha \in (0,1)$ for each $n\in \mathbb{N}$ . Then $w_{n}(\lambda, r, s, \mu )$ reduces to $w_{n}(r, s, \alpha )$ (say), where

\begin{align*} w_{n}(r, s, \alpha )&=r^{2}+s^{2}-rs\Big (1-\frac{m}{n}\Big )^{\alpha }-rs\Big (1+\frac{m}{n}\Big )^{\alpha }. \end{align*}

Expanding $w_{n}(r, s, \alpha )$ , one gets

\begin{align*} w_{n}(r, s, \alpha ) &= \left \{ \begin{array}{lll} r^{2}+s^{2}-2^{\alpha }rs & \quad \mbox{if}\,\,n=m,\\[5pt] (r-s)^{2}+2rs\alpha \sum _{k=1}^{\infty }\Big (\frac{1}{(2k!)}\frac{m^{2k}}{n^{2k}}\big (\prod _{i=1}^{2k-1}(i-\alpha )\big )\Big ) & \quad \mbox{if}\,\,n\geq m+1,\,\, n,m\in \mathbb{N}. \end{array}\right . \end{align*}

Hence we have $w_{n}(r, s, \alpha )\gt rs\alpha (1-\alpha )\frac{m^{2}}{n^{2}}$ . Therefore we obtain a different look of improved discrete Hardy’s inequality from ( 2.2 ) as below:

(2.3) \begin{align}{} &\sum _{n=m}^{\infty }|rA_{n}-sA_{n-m}|^{2}\geq \sum _{n=m}^{\infty }w_n(r, s, \alpha )_{n}|A_{n}|^{2}\gt \sum _{n=m}^{\infty } rs\alpha (1-\alpha )\frac{m^{2}}{n^{2}}|A_{n}|^{2}. \end{align}

Note that in the case of $m=1$ , $r=s$ and $\alpha =\frac{1}{2}$ , inequality (2.3) is nothing but an improved discrete Hardy’s inequality (1.3) established by Keller, Pinchover and Pogorzelski [Reference Keller, Pinchover and Pogorzelski13].

Corollary 2.2. Let us choose $\lambda _{n}=1$ for each $n\in \mathbb{N}$ , $m=1$ and $r=s=1$ in inequality ( 2.2 ). Then we have the following inequality:

(2.4) \begin{align}{} &\sum _{n=1}^{\infty }|A_{n}-A_{n-1}|^{2}\geq \sum _{n=1}^{\infty }w_{n}(\mu )|A_{n}|^{2}, \end{align}

where $w_{n}(\mu )=2-\frac{\mu _{n+1}}{\mu _{n}}+\frac{\mu _{n-1}}{\mu _{n}}$ .

The above inequality (2.4) was established by Krejčiřík et al. [Reference Krejčiřík, Laptev and Štampach16], and an elementary proof of (2.4) for $\mu _{n}=\sqrt{n}$ was presented by Krejčiřík and Štampach [Reference Krejčiřík and Štampach15].

Corollary 2.3. If one chooses $\lambda _{n}=\frac{1}{n^{\alpha }}$ , $\alpha \in \mathbb{R}$ and $\mu _{n}=n^{\beta }$ , $\beta \in \mathbb{R}$ , then the corresponding weight sequence $w_{n}(\lambda, r, s, \mu )$ reduces to $w_{n}(\alpha, r, s, \beta )$ , where

\begin{align*} w_{n}(\alpha, r, s, \beta )&=n^{\alpha }\Big (r^{2}+s^{2}\Big (1+\frac{m}{n}\Big )^{\alpha }-rs\Big (1-\frac{m}{n}\Big )^{\beta }-rs\Big (1+\frac{m}{n}\Big )^{\alpha +\beta }\Big ), \end{align*}

and inequality ( 2.2 ) becomes a power type improved Hardy’s inequality with shifts as below:

(2.5) \begin{align}{} &\sum _{n=m}^{\infty }n^{\alpha }|ru_{n}-su_{n-m}|^{2}\geq \sum _{n=m}^{\infty }w(\alpha, r, s, \beta )_{n}|u_{n}|^{2}. \end{align}

Note that the above inequality (2.5) strengthen the inequality of Gupta [Reference Gupta6], who obtained this inequality (2.5) for the case when $m=1$ and $r=s$ .

3. Improvement of some general Hardy inequalities

This section is devoted to the study of improvement of general Hardy’s inequality (1.6) in some particular cases. In fact, we study the improvement of inequalities (1.7) and (1.8) and prove that these sharp inequalities are not optimal; that is, they admit improvement. Indeed, we prove the following two successive theorems.

Theorem 3.1. Let $A=\{A_n\}$ be any sequence of complex numbers such that $A\in C_c(\mathbb{N}_0)$ with $A_{0}=0$ . Then we have

(3.1) \begin{align} &\sum _{n=1}^{\infty }\frac{|A_{n}-A_{n-1}|^{2}}{n^{2}}\geq \sum _{n=1}^{\infty }\eta _{n}^{(1)}|A_{n}|^{2} \gt 9\sum _{n=1}^{\infty }\frac{|A_{n}|^{2}}{(2n+1)^{2}(n+1)^{2}}, \end{align}

where the weight sequence $\eta _{n}^{(1)}$ is defined as below:

\begin{align*} \eta _{n}^{(1)}&=\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}-\frac{1}{n^{2}}\Big (\frac{(n-1)(2n-1)}{(n+1)(2n+1)}\Big )^{\frac{1}{2}}-\frac{1}{(n+1)^{2}} \Big (\frac{(n+2)(2n+3)}{n(2n+1)}\Big )^{\frac{1}{2}}. \end{align*}

Theorem 3.2. Suppose that $A=\{A_n\}\in C_c(\mathbb{N}_0)$ be such that $A_{0}=0$ . Then

(3.2) \begin{align} &\sum _{n=1}^{\infty }\frac{|A_{n}-A_{n-1}|^{2}}{n^{3}}\geq \sum _{n=1}^{\infty }\eta _{n}^{(2)}|A_{n}|^{2} \gt 4\sum _{n=1}^{\infty }\frac{|A_{n}|^{2}}{n(n+1)^{4}}, \end{align}

holds, where the improved weight $\eta _{n}^{(2)}$ is read as follows:

\begin{align*} \eta _{n}^{(2)}&=\frac{1}{n^{3}}+\frac{1}{(n+1)^{3}}-\frac{1}{n^{3}}\big (\frac{n-1}{n+1}\big )-\frac{1}{(n+1)^{3}}\big (\frac{n+2}{n}\big ). \end{align*}

We first begin with the proof of Theorem3.1. Before proceed to prove it, we need to establish some results given in the form of lemma. Let us start with the following lemma.

Lemma 3.3. Suppose that $n\in \mathbb{N}$ . Then

\begin{align*} &4n^{4}+6n^{3}-\frac{n^{2}}{2}+5n+1\gt (2n^{2}+3n+1)\sqrt{4n^{4}-5n^{2}+1} \,\,\,\mbox{holds}. \end{align*}

Proof. Let us denote $T_{1}=4n^{4}+6n^{3}-\frac{n^{2}}{2}+5n+1$ and $T_{2}=(2n^{2}+3n+1)\sqrt{4n^{4}-5n^{2}+1}$ . Then the difference of the squares of $T_1$ and $T_2$ gives

\begin{align*} &T_{1}^{2}-T_{2}^{2}=\frac{n}{4}(280n^{4}+501n^{3}+100n^{2}+64n+16)\gt 0, \end{align*}

which implies that $\big (T_{1}+T_{2}\big )\big (T_{1}-T_{2}\big )\gt 0$ . Since $(T_{1}+T_{2})\gt 0$ always holds, so we conclude that $(T_{1}-T_{2})\gt 0$ , which leads to the desired inequality.

We have another lemma to prove.

Lemma 3.4. Suppose that $n\in \mathbb{N}$ . Then

\begin{align*} &4n^{4}+10n^{3}+\frac{11n^{2}}{2}+n\gt (2n^{2}+n)\sqrt{4n^{4}+16n^{3}+19n^{2}+6n} \,\,\,\mbox{holds}. \end{align*}

Proof. Here we denote $S_{1}=4n^{4}+10n^{3}+\frac{11n^{2}}{2}+n$ and $S_{2}=(2n^{2}+n)\sqrt{4n^{4}+16n^{3}+19n^{2}+6n}$ . Then the following computation gives

\begin{align*} &S_{1}^{2}-S_{2}^{2}=\frac{n^{2}}{4}(8n^{3}+29n^{2}+20n+4)\gt 0, \end{align*}

which means $\big (S_{1}+S_{2}\big )\big (S_{1}-S_{2}\big )\gt 0$ . Since $(S_{1}+S_{2})\gt 0$ , so we get $(S_{1}-S_{2})\gt 0$ . This proves the desired inequality.

The next lemma is quite important and is mainly concerned with the improvement of inequality (1.7) via the weight sequence $\eta _{n}^{(1)}$ . Indeed, we prove the following point-wise result.

Proof. We denote $\tilde{f}(n)=\eta _{n}^{(1)}-\frac{9}{(2n+1)^{2}(n+1)^{2}}$ . Then it is enough to show that $\tilde{f}(n)\gt 0$ holds for each $n\in \mathbb{N}$ . We have

\begin{align*} &\tilde{f}(n)\\[5pt] &=\eta _{n}^{(1)}-\frac{9}{(2n+1)^{2}(n+1)^{2}}\\[5pt] &=\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}-\frac{1}{n^{2}}\Big (\frac{(n-1)(2n-1)}{(n+1)(2n+1)}\Big )^{\frac{1}{2}}-\frac{1}{(n+1)^{2}} \Big (\frac{(n+2)(2n+3)}{n(2n+1)}\Big )^{\frac{1}{2}}-\frac{9}{(2n+1)^{2}(n+1)^{2}}\\[5pt] &=\frac{H(n)}{n^{2}(n+1)^{2}(2n+1)^{2}}, \end{align*}

where we denote

\begin{align*} H(n)=&(n+1)^{2}(2n+1)^{2}+n^{2}(2n+1)^{2}-(n+1)(2n+1)\sqrt{(n-1)(2n-1)(n+1)(2n+1)}\\[5pt] &-n(2n+1)\sqrt{n(n+2)(2n+3)(2n+1)}-9n^{2}. \end{align*}

Simplifying $H(n)$ , one gets

\begin{align*} H(n)=&8n^{4}+16n^{3}+5n^{2}+6n+1-(2n^{2}+3n+1)\sqrt{4n^{4}-5n^{2}+1}\\[5pt] &-(2n^{2}+n)\sqrt{4n^{4}+16n^{3}+19n^{2}+6n}\\[5pt] =&\Big (4n^{4}+6n^{3}-\frac{n^{2}}{2}+5n+1-(2n^{2}+3n+1)\sqrt{4n^{4}-5n^{2}+1}\Big )\\[5pt] &+\Big (4n^{4}+10n^{3}+\frac{11n^{2}}{2}+n-(2n^{2}+n)\sqrt{4n^{4}+16n^{3}+19n^{2}+6n}\Big )\\[5pt] &=I_{1}+I_{2}, \end{align*}

where

\begin{align*} I_{1}&=4n^{4}+6n^{3}-\frac{n^{2}}{2}+5n+1-(2n^{2}+3n+1)\sqrt{4n^{4}-5n^{2}+1},\\[5pt] I_{2}&=4n^{4}+10n^{3}+\frac{11n^{2}}{2}+n-(2n^{2}+n)\sqrt{4n^{4}+16n^{3}+19n^{2}+6n}. \end{align*}

Using Lemma 3.3 and Lemma 3.4, we get $I_{1}\gt 0$ and $I_{2}\gt 0$ . Therefore, $H(n)\gt 0$ for all $n\in \mathbb{N}$ . This shows that $\tilde{f}(n)\gt 0$ for all $n\in \mathbb{N}$ . This completes proof of the lemma.

Proof. (of Theorem 3.1):

The L.H.S. part of the inequality is easily derived by choosing $r=s$ , $m=1$ , $\lambda _{n}=n^{2}$ and $\mu _{n}=\big (\frac{n(n+1)(2n+1)}{6}\big )^{\frac{1}{2}}$ in inequality (2.2). The R.H.S. part of the inequality is an immediate consequence of Lemma 3.5. This finishes proof of the theorem.

We now proceed to prove Theorem3.2. The following result is required to prove the theorem.

Proof. Denote $\tilde{\tilde{f}}(n)=\eta _{n}^{(2)}-\frac{4}{n(n+1)^{4}}$ . Then it is sufficient to prove that $\tilde{\tilde{f}}(n)\gt 0$ for each $n\in \mathbb{N}$ . We write

\begin{align*} \tilde{\tilde{f}}(n)&=\eta _{n}^{(2)}-\frac{4}{n(n+1)^{4}}\\[5pt] &=\frac{1}{n^{3}}+\frac{1}{(n+1)^{3}}-\frac{1}{n^{3}}\big (\frac{n-1}{n+1}\big )-\frac{1}{(n+1)^{3}}\big (\frac{n+2}{n}\big )-\frac{4}{n(n+1)^{4}}\\[5pt] &=\frac{2}{n^{3}(n+1)}-\frac{2}{n(n+1)^{3}}-\frac{4}{n(n+1)^{4}}. \end{align*}

Simplifying the terms, one gets

\begin{align*} \tilde{\tilde{f}}(n)&=\frac{6n+2}{n^{3}(n+1)^{4}}\gt 0. \end{align*}

This proves the lemma.

Proof. (of Theorem 3.2):

Let us choose $r=s$ , $m=1$ , $\lambda _{n}=n^{3}$ and $\mu _{n}=\frac{n(n+1)}{2}$ in inequality (2.2). Then we immediately establish the L.H.S. part of the inequality. On the other hand, the R.H.S. follows easily from Lemma 3.6. This completes proof of the theorem.

4. Improvement of the variant Hardy inequalities

In this section, an improvement study is being considered for the variant inequalities (1.12)–(1.15). We begin with the study of improvement for inequality (1.12), and subsequently, we consider the other inequalities. We prove that each of these inequalities (1.12)–(1.15) admits an improvement for a corresponding weight sequence. Let us state our first result in this section.

Theorem 4.1. Suppose that $A=\{A_n\}$ be any sequence of complex numbers such that $A\in C_c(\mathbb{N}_0)$ with $A_{0}=A_{1}=0$ . Then

(4.1) \begin{align} &\sum _{n=2}^{\infty }(n-1)^{2}|A_{n}-A_{n-1}|^{2}\geq \sum _{n=2}^{\infty }\beta _{n}|A_{n}|^{2}\gt \sum _{n=2}^{\infty }\frac{1}{4}|A_{n}|^{2}\,\,\, \mbox{holds} \end{align}

where the improved weight $\beta _n$ for $n\geq 2$ is defined as below:

\begin{align*} \beta _{n}&=n^{2}\Big [1+\Big (1-\frac{1}{n}\Big )^{2}-\Big (1+\frac{1}{n}\Big )^{-\frac{1}{2}}-\Big (1-\frac{1}{n}\Big )^{\frac{3}{2}}\Big ]. \end{align*}

Before proving the above theorem, we need to establish the following lemma, statement of which is presented below.

Proof. Note that $\beta _{2}=\frac{5\sqrt{6}-2\sqrt{3}-8}{\sqrt{6}}\gt \frac{1}{4}$ and $\beta _{3}=\frac{25\sqrt{3}-12\sqrt{2}-27}{2\sqrt{3}}\gt \frac{1}{4}$ . Hence $\beta _{n}\gt \frac{1}{4}$ for $n=2, 3$ . Let us now choose $n\geq 4$ . We choose a function $Q(x)$ as below:

(4.2) \begin{align} Q(x)&=1+(1-x)^{2}-(1-x)^{\frac{3}{2}}-(1+x)^{-\frac{1}{2}}-\frac{x^{2}}{4} \end{align}

where $x\in [0,\frac{1}{4}]$ . Now successive differentiation of $Q(x)$ gives the following:

\begin{align*} Q'(x)&=\frac{3x}{2}-2+\frac{3}{2}(1-x)^{\frac{1}{2}}+\frac{1}{2}(1+x)^{\frac{-3}{2}},\\[5pt] Q''(x)&=\frac{3}{2}-\frac{3}{4}(1-x)^{-\frac{1}{2}}-\frac{3}{4}(1+x)^{-\frac{5}{2}},\\[5pt] Q'''(x)&=-\frac{3}{8}(1-x)^{-\frac{3}{2}}+\frac{15}{8}(1+x)^{-\frac{7}{2}},\\[5pt] Q''''(x)&=-\frac{9}{16}(1-x)^{-\frac{5}{2}}-\frac{105}{16}(1+x)^{-\frac{9}{2}}. \end{align*}

It is easy to observe that for any $x\in [0, \frac{1}{4}]$ , we have $Q''''(x)\lt 0$ . This shows that $Q'''(x)$ is decreasing in $x\in [0,\frac{1}{4}]$ . Since $Q'''(1/4)\gt 0$ , so we conclude that $Q'''(x)\gt 0$ for $x\in [0, \frac{1}{4}]$ , and hence $Q''(x)$ is increasing in $[0, \frac{1}{4}]$ . Again since $Q''(0)=0$ , so $Q''(x)\gt 0$ for $x\in (0,\frac{1}{4}]$ .This shows that $Q'(x)$ is increasing in $[0, \frac{1}{4}]$ . Further, we see that $Q'(0)=0$ and $Q(0)=0$ . Therefore, we get $Q(x)\geq 0$ for $x\in [0,\frac{1}{4}]$ . Hence $Q(x)\gt 0$ in $x\in (0,\frac{1}{4}]$ .

Let us now put $x=\frac{1}{n}$ for $0\lt x\leq \frac{1}{4}$ , and inserting it in $Q(x)$ , we get

\begin{align*} Q\Big (\frac{1}{n}\Big )&=1+(1-\frac{1}{n})^{2}-(1+\frac{1}{n})^{-\frac{1}{2}}-(1-\frac{1}{n})^{\frac{3}{2}}-\frac{1}{4n^{2}}\gt 0\,\,\mbox{for}\,\, n\geq 4. \end{align*}

Since $\beta _{n}-\frac{1}{4}=n^{2}Q\Big (\frac{1}{n}\Big )$ , so we get $\beta _{n}\gt \frac{1}{4}$ for $n\geq 4$ . Hence for each $n\geq 2$ , we have $\beta _{n}\gt \frac{1}{4}$ . This proves the lemma.

Proof. (of Theorem 4.1):

The L.H.S. of the above inequality is easily followed from inequality (2.2) by choosing $r=s$ , $m=1$ , $\lambda _{n}=\frac{1}{(n-1)^{2}}$ and $\mu _{n}=\frac{1}{\sqrt n}$ . To establish the R.H.S part, it is sufficient to prove that $\beta _{n}\gt \frac{1}{4}$ holds point-wise for $n\geq 2$ , $n\in \mathbb{N}$ and which is enclosed in Lemma 4.2. This completes proof of the theorem.

In the next theorem, we show that inequality (1.13) gets an improvement for a suitable weight sequence $\beta _{n}^{(1)}$ for $n\geq 2$ . In fact, we have the following improved inequality.

Theorem 4.3. Let $A=\{A_n\}$ be any sequence of complex numbers such that $A\in C_c(\mathbb{N}_0)$ with $A_{0}=A_{1}=0$ . Then

(4.3) \begin{align} &\sum _{n=2}^{\infty }(n-1)n^{2}|A_{n}-A_{n-1}|^{2}\geq \sum _{n=2}^{\infty }\beta _{n}^{(1)}|A_{n}|^{2}\gt \sum _{n=2}^{\infty }n|A_{n}|^{2} \end{align}

holds, where the improved weight is defined as below:

\begin{align*} \beta _{n}^{(1)}&=n^{2}(n-1)+n(n+1)^{2}-n^{2}(n-1)\Big (\frac{n-1}{n+1}\Big )^{-\frac{1}{2}}-n(n+1)^{2}\Big (\frac{n+2}{n}\Big )^{-\frac{1}{2}}. \end{align*}

We establish this theorem through some important inequalities stated in the following lemmas.

Lemma 4.4. Suppose that $n\in \mathbb{N}$ . Then we have

\begin{align*} &(a)\,\,(n^{3}+n^{2}+\frac{n}{2})\sqrt{n+2}\gt n\sqrt{n}(n+1)^{2},\\[5pt] &(b)\,\,n^{3}-\frac{n}{2}\gt n^{2}\sqrt{n^{2}-1}. \end{align*}

Proof. $(a)$ Let us denote $L_{1}=(n^{3}+n^{2}+\frac{n}{2})\sqrt{n+2}$ and $L_{2}=n\sqrt{n}(n+1)^{2}$ . Then difference of squares of $L_{1}$ and $L_{2}$ gives

\begin{align*} &L_{1}^{2}-L_{2}^{2}=\frac{n^{2}}{4}\Big (4n^{2}+5n+2\Big )\gt 0. \end{align*}

This implies that $\big (L_{1}+L_{2}\big )\big (L_{1}-L_{2}\big )\gt 0$ . Since $(L_{1}+L_{2})\gt 0$ always holds for any $n\in \mathbb{N}$ , so we conclude that $(L_{1}-L_{2})\gt 0$ . This proves the desired inequality in $(a)$ .

(b) Here we first denote $J_{1}=n^{3}-\frac{n}{2}$ and $J_{2}=n^{2}\sqrt{n^{2}-1}$ . Then direct computation gives

\begin{align*} &J_{1}^{2}-J_{2}^{2}=\frac{n^{2}}{4}\gt 0, \end{align*}

which means $\big (J_{1}+J_{2}\big )\big (J_{1}-J_{2}\big )\gt 0$ . Since for each $n\in \mathbb{N}$ , $(J_{1}+J_{2})\gt 0$ , so we have $(J_{1}-J_{2})\gt 0$ . Hence we have the desired inequality in $(b)$ . This also completes proof of the lemma.

Proof. We denote $G(n)=\beta _{n}^{(1)}-n$ . It is then enough to prove that $G(n)\gt 0$ for each $n\geq 2$ . Note that $G(n)$ can be simplified into the following form:

\begin{align*} G(n)&=n^{2}(n-1)+n(n+1)^{2}-n^{2}(n-1)\Big (\frac{n-1}{n+1}\Big )^{-\frac{1}{2}}-n(n+1)^{2}\Big (\frac{n+2}{n}\Big )^{-\frac{1}{2}}-n=\frac{D(n)}{\sqrt{n+2}}, \end{align*}

where $D(n)=(2n^{3}+n^{2})\sqrt{n+2}-n^{2}\sqrt{n^{2}-1}\sqrt{n+2}-n^{\frac{3}{2}}(n+1)^{2}$ . Again $D(n)$ can be further simplified into the following $D(n)=\Big ((n^{3}+n^{2}+\frac{n}{2})\sqrt{n+2}-n\sqrt{n}(n+1)^{2}\Big )+\sqrt{n+2}\Big (n^{3}-\frac{n}{2}-n^{2}\sqrt{n^{2}-1}\Big )=k_{1}+\sqrt{n+2} k_{2}$ (say), where $k_{1}=(n^{3}+n^{2}+\frac{n}{2})\sqrt{n+2}-n\sqrt{n}(n+1)^{2}$ and $k_{2}=n^{3}-\frac{n}{2}-n^{2}\sqrt{n^{2}-1}$ . Using Lemma 4.4, we get $k_{1}\gt 0$ and $k_{2}\gt 0$ . Therefore, $D(n)\gt 0$ for all $n\in \mathbb{N}$ . This shows that $G(n)\gt 0$ for all $n\geq 2$ . This completes proof of the lemma.

Proof. (of Theorem 4.3):

Let us put $r=s$ , $m=1$ , $\lambda _{n}=\frac{1}{(n-1)n^{2}}$ and $\mu _{n}=\big (\frac{n(n+1)}{2}\big )^{-\frac{1}{2}}$ in inequality (2.2). Then we instantly get the L.H.S. part of the inequality. The R.H.S. part is an immediate consequence of Lemma 4.5.

Our next result demonstrates that inequality (1.14) achieves an improvement. In fact, we have the following result.

Theorem 4.6. Suppose $A=\{A_n\}\in C_c(\mathbb{N}_0)$ such that $A_{0}=A_{1}=0$ . Then

(4.4) \begin{align} &\sum _{n=2}^{\infty }(2n-1)^{2}n^{2}|A_{n}-A_{n-1}|^{2}\geq \sum _{n=2}^{\infty }\beta _{n}^{(2)}|A_{n}|^{2}\gt 9\sum _{n=2}^{\infty }n^{2}|A_{n}|^{2}\,\,\mbox{holds} \end{align}

where the improved variant Hardy weight $\beta _{n}^{(2)}$ , $n\geq 2$ is defined as below:

\begin{align*} \beta _{n}^{(2)}=&n^{2}(2n-1)^{2}+(2n+1)^{2}(n+1)^{2}-n^{2}(2n-1)^{2}\Big (\frac{(n-1)(2n-1)}{(n+1)(2n+1)}\Big )^{-\frac{1}{2}}\\[5pt] &-(2n+1)^{2}(n+1)^{2}\Big (\frac{(n+2)(2n+3)}{n(2n+1)}\Big )^{-\frac{1}{2}}. \end{align*}

It is required to establish a lemma before proving Theorem4.6.

Lemma 4.7. Suppose that $n\in \mathbb{N}$ such that $n\geq 2$ . Then we have

\begin{align*} &(a)\,\,(4n^{4}+2n^{3}-\frac{n^{2}}{2}+n+1)\sqrt{2n^{2}-3n+1}\gt n^{2}(2n-1)^{2}\sqrt{2n^{2}+3n+1},\\[5pt] &(b)\,\,(4n^{4}+6n^{3}+\frac{11n^{2}}{2}+5n)\sqrt{2n^{2}+7n+6}\gt (n+1)^{2}(2n+1)^{2}\sqrt{2n^{2}+n}. \end{align*}

Proof. $(a)$ We denote $R_{1}=(4n^{4}+2n^{3}-\frac{n^{2}}{2}+n+1)\sqrt{2n^{2}-3n+1}$ and $R_{2}=n^{2}(2n-1)^{2}\sqrt{2n^{2}+3n+1}$ . Squaring $R_{1}$ and $R_{2}$ and then subtracting them, we get

\begin{align*} &R_{1}^{2}-R_{2}^{2}=\frac{(2n-1)}{4}\Big (8n^{4}(n^{2}-4)+12(n^{5}-n^{3})+(n^{5}-n^{4})+(8n^{2}-4n-4)\Big )\gt 0\,\,for\,\,n\geq 2. \end{align*}

which is positive, and this implies that $\big (R_{1}+R_{2}\big )\big (R_{1}-R_{2}\big )\gt 0$ . Hence we have $(R_{1}-R_{2})\gt 0$ since $(R_{1}+R_{2})\gt 0$ for each $n\in \mathbb{N}$ . This proves the inequality.

$(b)$ Similarly, we denote $P_{1}=(4n^{4}+6n^{3}+\frac{11n^{2}}{2}+5n)\sqrt{2n^{2}+7n+6}$ and $P_{2}=(n+1)^{2}(2n+1)^{2}\sqrt{2n^{2}+n}$ . The following computation gives

\begin{align*} &P_{1}^{2}-P_{2}^{2}=\frac{n}{4}\Big (368n^{6}+1602n^{5}+2787n^{4}+2690n^{3}+1676n^{2}+544n-4\Big )\gt 0, \end{align*}

which finally gives $P_{1}\gt P_{2}$ as $(P_{1}+P_{2})\gt 0$ for any $n\in \mathbb{N}$ . This proves the desired inequality and hence the lemma.

Proof. Let us denote $M(n)=\beta _{n}^{(2)}-9n^{2}$ . Then it is enough to show that $M(n)\gt 0$ holds for each $n\geq 2$ . A direct computation gives

\begin{align*} M(n)&=n^{2}(2n-1)^{2}+(2n+1)^{2}(n+1)^{2}-n^{2}(2n-1)^{2}\Big (\frac{(n-1)(2n-1)}{(n+1)(2n+1)}\Big )^{-\frac{1}{2}}\\[5pt] &-(2n+1)^{2}(n+1)^{2}\Big (\frac{(n+2)(2n+3)}{n(2n+1)}\Big )^{-\frac{1}{2}}-9n^{2}\\[5pt] &=\frac{R(n)}{\sqrt{(n+2)(n-1)(2n-1)(2n+3)}}, \end{align*}

where simplifying $R(n)$ , we get

\begin{align*} R(n)=&\sqrt{(n+2)(n-1)(2n-1)(2n+3)}(8n^{4}+8n^{3}+5n^{2}+6n+1)\\[5pt] &-n^{2}(2n-1)^{2}\sqrt{(n+2)(n+1)(2n+1)(2n+3)}\\[5pt] &-(n+1)^{2}(2n+1)^{2}\sqrt{n(n-1)(2n+1)(2n-1)}\\[5pt] =&C_{1}\sqrt{(n+2)(2n+3)}+C_{2}\sqrt{(n-1)(2n-1)}, \end{align*}

where $C_1$ and $C_2$ are defined as below:

\begin{align*} C_{1}&=(4n^{4}+2n^{3}-\frac{n^{2}}{2}+n+1)\sqrt{2n^{2}-3n+1}-n^{2}(2n-1)^{2}\sqrt{2n^{2}+3n+1},\\[5pt] C_{2}&=(4n^{4}+6n^{3}+\frac{11n^{2}}{2}+5n)\sqrt{2n^{2}+7n+6}-(n+1)^{2}(2n+1)^{2}\sqrt{2n^{2}+n}. \end{align*}

Using Lemma 4.7, we observed that $C_{1}\gt 0$ and $C_{2}\gt 0$ . Therefore, $R(n)\gt 0$ for all $n\geq 2$ . This proves that $M(n)\gt 0$ for all $n\geq 2$ . This proves the lemma.

Proof. (of Theorem 4.6):

By choosing $r=s$ , $m=1$ , $\lambda _{n}=\frac{1}{(2n-1)^{2}n^{2}}$ and $\mu _{n}=\big (\frac{n(n+1)(2n+1)}{6}\big )^{-\frac{1}{2}}$ in inequality (2.2), we can easily establish the L.H.S. of the desired inequality. Since by Lemma 4.8, we have $\beta _{n}^{(2)}\gt 9n^{2}$ point-wise, so R.H.S. of the inequality easily follows.

Now we prove a last result in this section. It is shown that the improvement of inequality (1.15) is possible. We have the following result.

Theorem 4.9. Let $A=\{A_n\}\in C_c(\mathbb{N}_0)$ be such that $A_{0}=A_{1}=0$ . Then

(4.5) \begin{align} &\sum _{n=2}^{\infty }(n-1)n^{4}|A_{n}-A_{n-1}|^{2}\geq \sum _{n=2}^{\infty }\beta _{n}^{(3)}|A_{n}|^{2} \gt 4\sum _{n=2}^{\infty }n^{3}|A_{n}|^{2}\,\mbox{holds} \end{align}

where for $n\geq 2$

\begin{align*} \beta _{n}^{(3)}&=n^{4}(n-1)+n(n+1)^{4}-n^{4}(n-1)\Big (\frac{n-1}{n+1}\Big )^{-1}-n(n+1)^{4}\Big (\frac{n+2}{n}\Big )^{-1}. \end{align*}

Proof. Suppose that $r=s$ , $m=1$ , $\lambda _{n}=\frac{1}{(n-1)n^{4}}$ and $\mu _{n}=\big (\frac{n(n+1)}{2}\big )^{-1}$ , and inserting them in inequality (2.2), one immediately obtains the L.H.S. part of the desired inequality. To establish the R.H.S. part of the same, it is sufficient to prove that $V(n)=\beta _{n}^{(3)}-4n^{3}\gt 0$ for each $n\geq 2$ . The term $V(n)$ can be simplified into the following form:

\begin{align*} V(n)&=n^{4}(n-1)+n(n+1)^{4}-n^{4}(n-1)\Big (\frac{n-1}{n+1}\Big )^{-1}-n(n+1)^{4}\Big (\frac{n+2}{n}\Big )^{-1}-4n^{3}\\[5pt] &=\frac{2n(2n^{2}+4n+1)}{n+2}\gt 0 \end{align*}

for each $n\geq 2$ . Hence the proof.

5. An improved multivariate discrete $p$ -Hardy’s inequality

In the following subsection, we prove that inequality (1.17) admits an improvement. For the sake of clarity, let us first recall inequality (1.4) as below:

\begin{align*} \sum _{n=1}^{\infty }|A_n-A_{n-1}|^{p}&\geq \sum _{n=1}^{\infty } w_n^{FKP}(p)|A_n|^{p}\gt \sum _{n=1}^{\infty } \frac{|A_n|^{p}}{C_pn^{p}}. \end{align*}

Now onwards, $w_n^{FKP}(p)$ and $w_n(p)$ will share the same expression. For brevity, we will use the latter.

5.1. Improved Hardy’s inequality for two variables

Let us begin with the following theorem.

Theorem 5.1. Let $\{a_{mn}\}$ be any sequence of complex numbers and inequality ( 1.4 ) of one dimension holds. Then

(5.1) \begin{align}{} &\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }\frac{1}{(C_p)^{2}m^{p}n^{p}}|A_{mn}|^{p}\lt \sum _{m=1}^{\infty } \sum _{n=1}^{\infty }w_{mn}(p)|A_{mn}|^{p} \leq \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }|a_{mn}|^{p}, \end{align}

where the sequence $w_{mn}(p)$ is defined as below:

\begin{align*} w_{mn}(p)&=w_{m}(p)w_{n}(p)=\prod _{\theta =m, n}\Big \{\big (1-(1-\frac{1}{\theta })^{\frac{p-1}{p}}\big )^{p-1}-\big ((1+\frac{1}{\theta })^{\frac{p-1}{p}}-1\big )^{p-1}\Big \}\gt \frac{1}{(C_{p})^{2}m^{p}n^{p}}. \end{align*}

Proof. Since $w_{mn}(p)\gt \frac{1}{(C_{p})^{2}m^{p}n^{p}}$ holds point-wise for each $m, n\in \mathbb{N}$ so L.H.S. of inequality (5.1) is established. To prove the R.H.S. of inequality (5.1), we begin with

\begin{align*} &\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }w_{mn}(p)|A_{mn}|^{p}\\[5pt] &=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }w_{m}(p)w_{n}(p)\Big |\sum _{i=1}^{m}\sum _{j=1}^{n}a_{ij}\Big |^{p}\\[5pt] &=\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }w_{m}(p)w_{n}(p)\Big |\sum _{j=1}^{n}B_{j}\Big |^{p}\,\,\,\,\,\,where\,\,\, B_{j}=\sum _{i=1}^{m}a_{ij}\\[5pt] &=\sum _{m=1}^{\infty }w_{m}(p)\sum _{n=1}^{\infty }w_{n}(p)\Big |\sum _{j=1}^{n}B_{j}\Big |^{p}\leq \sum _{m=1}^{\infty } w_{m}(p)\sum _{n=1}^{\infty }|B_{n}|^{p}, \end{align*}

where the last inequality easily follows from $\sum _{n=1}^{\infty }w_{n}(p)\Big |\sum _{j=1}^{n}B_{j}\Big |^{p}\leq \sum _{n=1}^{\infty }|B_{n}|^{p}$ .

Therefore, we have

\begin{align*} &\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }w_{mn}(p)|A_{mn}|^{p}\leq \sum _{m=1}^{\infty }w_{m}(p)\sum _{n=1}^{\infty } \Big |\sum _{i=1}^{m}a_{in}\Big |^{p}, \end{align*}

and hence

\begin{align*} &\sum _{m=1}^{\infty }\sum _{n=1}^{\infty }w_{mn}(p)|A_{mn}|^{p}\leq \sum _{n=1}^{\infty }\sum _{m=1}^{\infty }w_{m}(p) \Big |\sum _{i=1}^{m}a_{in}\Big |^{p}\leq \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }|a_{mn}|^{p}. \end{align*}

This proves the desired inequality and completes the proof of the theorem.

5.2. Improved Hardy’s inequality for multivariables

We now give an improved version of inequality (1.19) in the following theorem.

Theorem 5.2. Let $\{a_{m_{1}m_{2}\cdots m_{r}}\}$ be a $r$ -fold sequence of complex numbers. Then

\begin{align*} &\sum _{m_{1}=1}^{\infty }\sum _{m_{1}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\frac{1}{(C_{p})^{r}\Big (m_{1}m_{2}\ldots m_{r}\Big )^{p}}\mid A_{m_{1}m_{2}\ldots m_{r}}\mid ^{p}\\[5pt] &\lt \sum _{m_{1}=1}^{\infty }\sum _{m_{1}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }w_{m_{1}m_{2}\ldots m_{r}}(p)\mid A_{m_{1}m_{2}\ldots m_{r}}\mid ^{p}\leq \sum _{m_{1}=1}^{\infty }\sum _{m_{1}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\mid a_{m_{1}m_{2}\ldots m_{r}}\mid ^{p}, \end{align*}

where the weight sequence $\{w_{m_{1}m_{2}\ldots m_{r}}(p)\}$ is defined as below:

\begin{align*} w_{m_{1}m_{2}\ldots m_{r}}(p)&=\prod _{\theta =m_{1}, m_{2}, \ldots, m_{r}}w_{\theta }(p)\gt \frac{1}{(C_{p})^{r}(m_{1}m_{2}\ldots m_{r})^{p}}. \end{align*}

Proof. The proof of this theorem will be established by an induction process. Since Theorem5.2 is true for $r=1$ and $r=2$ , which gives inequalities (1.4) and (5.1), respectively, so we assume that Theorem5.2 is true for $r$ -fold series. We now prove that the above theorem is true for $(r+1)$ -fold series. Before moving further, we first define

(5.2) \begin{align}{} &P_{i_{2}\ldots i_{r}i_{r+1}}=\sum _{i_{1}=1}^{m_{1}}a_{i_{1}i_{2}\ldots i_{r}i_{r+1}} \end{align}

and

(5.3) \begin{align}{} &Q_{m_{2}m_{3}\ldots m_{r}m_{r+1}}=\sum _{i_{2}=1}^{m_{2}}\ldots \sum _{i_{r}=1}^{m_{r}}\sum _{i_{r+1}=1}^{m_{r+1}}P_{i_{2}\ldots i_{r}i_{r+1}}. \end{align}

Using (5.2), we have

\begin{align*} &\sum _{m_{1}=1}^{\infty }\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }w_{m_{1}m_{2}\ldots m_{r+1}}(p)|A_{m_{1}m_{2}\ldots m_{r}m_{r+1}}|^{p}\\[5pt] =&\sum _{m_{1}=1}^{\infty }\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }\Big (\prod _{\theta =m_{1}, m_{2}, \ldots, m_{r+1}}w_{\theta }(p)\Big )\Big |\sum _{i_{1}=1}^{m_{1}}\sum _{i_{2}=1}^{m_{2}}\ldots \sum _{i_{r}=1}^{m_{r}} \sum _{i_{r+1}=1}^{m_{r+1}}a_{i_{1}i_{2}\ldots i_{r}i_{r+1}} \Big |^{p}\\[5pt] =&\sum _{m_{1}=1}^{\infty }\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }\Big (\prod _{\theta =m_{1}, m_{2}, \ldots, m_{r+1}}w_{\theta }(p)\Big )\Big |\sum _{i_{2}=1}^{m_{2}}\ldots \sum _{i_{r}=1}^{m_{r}}\sum _{i_{r+1}=1}^{m_{r+1}}P_{i_{2}\ldots i_{r}i_{r+1}}\Big |^{p}. \end{align*}

Again by using (5.3), one gets

\begin{align*} &\sum _{m_{1}=1}^{\infty }\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }w_{m_{1}m_{2}\ldots m_{r+1}}(p)|A_{m_{1}m_{2}\ldots m_{r}m_{r+1}}|^{p}\\[5pt] =&\sum _{m_{1}=1}^{\infty }w_{m_{1}}(p)\Big (\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty } \Big (\prod _{\theta =m_{2}, \ldots, m_{r+1}}w_{\theta }(p)\Big )|Q_{m_{2}m_{3}\cdots m_{r}m_{r+1}}|^{p}\Big )\\[5pt] \leq &\sum _{m_{1}=1}^{\infty }w_{m_{1}}(p)\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }\mid P_{m_{2}m_{3}\ldots m_{r}m_{r+1}}\mid ^{p}\, (\mbox{by}\,(1.19))\\[5pt] =&\sum _{m_{1}=1}^{\infty }w_{m_{1}}(p)\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty } \mid \sum _{i_{1}=1}^{m_{1}}a_{i_{1}m_{2}\cdots m_{r}m_{r+1}}\mid ^{p}(\mbox{using}\,(5.2)). \end{align*}

Thus Fubini’s theorem implies that

\begin{align*} &\sum _{m_{1}=1}^{\infty }\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }w_{m_{1}m_{2}\ldots m_{r+1}}(p)|A_{m_{1}m_{2}\ldots m_{r}m_{r+1}}|^{p}\\[5pt] \leq &\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }\Big (\sum _{m_{1}=1}^{\infty }w_{m_{1}}(p) \mid \sum _{i_{1}=1}^{m_{1}}a_{i_{1}m_{2}\ldots m_{r}m_{r+1}}\mid ^{p}\Big ) \end{align*}

\begin{align*} \leq &\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }\sum _{m_{1}=1}^{\infty }\mid a_{m_{1}m_{2}\ldots m_{r}m_{r+1}}\mid ^{p}(\mbox{by 1-fold improved Hardy's inequality})\\[5pt] =&\sum _{m_{1}=1}^{\infty }\sum _{m_{2}=1}^{\infty }\ldots \sum _{m_{r}=1}^{\infty }\sum _{m_{r+1}=1}^{\infty }\mid a_{m_{1}m_{2}\ldots m_{r}m_{r+1}}\mid ^{p}. \end{align*}

This proves that Theorem5.2 is true for $(r+1)$ -fold series. Hence, by principal of induction, we conclude that the result is true for all $r\in \mathbb{N}$ . This finishes proof of the theorem.