On a strengthened version of Hardy’s inequality

Abstract

On the basis of analytic method and skill, a strengthened version of well-known Hardy’s inequality is obtained. In comparison with some similar results given in recent decades, it successfully reduced the related coefficient.

MSC:26D15.

1 Introduction

There are many applications with well-known Hardy’s inequality in analytics, which refers to the following: let $a_k>0$ ($k=1,2,\dots$), $p>1$, then

$${\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}{a}_n^p>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p.$$

(1.1)

In recent decades, there have also been many results due to the extension and refinement of this inequality (cf. [1–5]), especially the monograph [6], which summarized part of the research done before 2005. In research on the coefficient of (1.1), the following conclusion in the case $p=2$ was drawn in [6]:

$$4\sum _{n=1}^{\mathrm{\infty }}(1-\frac{1}{3\sqrt{n}+5})a_n^2>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^2.$$

(1.2)

In [7], by using the method of weight-coefficient, the following inequality is proved with $p\in [\frac{7}{6},2]$:

$${\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}(1-\frac{15}{196}\frac{1}{n^{1-1/p}+3,436})a_n^p>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p.$$

(1.3)

In the following section, let $p>1$ and

$$Z_p=\{\begin{array}{cc}p-1-\frac{{(p-1)}^2}{p}{2}^{\frac{1}{p}},\hfill & 1<p\le 2,\hfill \\ 1-{(\frac{p-1}{p})}^{p-1}{2}^{\frac{p-1}{p}},\hfill & p>2.\hfill \end{array}$$

We shall strengthen Hardy’s inequality to

$${\left(\frac{p-1}{p}\right)}^p\sum _{n=1}^{\mathrm{\infty }}(1-\frac{Z_p}{2{(n-\frac{1}{2})}^{1-\frac{1}{p}}})a_n^p>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p.$$

2 Relevant lemmas

Some characters of a convex function will be cited in this section.

Definition 2.1 $L\subseteq \mathbb{R}$ is an interval, and $f:I\to \mathbb{R}$ is continuous. If

$$f\left(\frac{x+y}{2}\right)\le (\ge )\frac{f(x)+f(y)}{2}$$

holds for all $x,y\in I$, then f is called a convex (concave) function.

The sufficient and necessary condition for a second-order differential function f to be convex (concave) function is that $f^{″}(x)\ge (\le )0$ always holds for any $x\in I$. The famous Hadamard inequality is as follows. Let f be a convex (concave) function on $[a,b]$, then the equality

$$f\left(\frac{b+a}{2}\right)\le (\ge )\frac{1}{b-a}{\int }_a^{b}f(x)dx\le (\ge )\frac{f(b)+f(a)}{2}$$

(2.1)

holds if and only if f is a linear function.

Lemma 2.1 Let $p>1$, and $Z_p$ is defined in the first section, then $0<Z_p<\frac{1}{2}$.

Proof The proof includes two parts.

Part 1: When $1<p\le 2$, if we can obtain $f(p):=2^{-\frac{1}{p}}-1+\frac{1}{p}>0$, then we can get $Z_p>0$ obviously. Since

$$f^{\prime }(p)=\frac{1}{p^2}{2}^{-\frac{1}{p}}ln2-\frac{1}{p^2}=\frac{1}{p^2}{2}^{-\frac{1}{p}}(ln2-2^{\frac{1}{p}})<0,$$

$f(p)$ is monotone decreasing on $(1,2]$, then $f(p)\ge f(2)=\frac{\sqrt{2}}{2}-\frac{1}{2}>0$.

Meanwhile, $Z_p<\frac{1}{2}$ is equivalent to

$$p-\frac{3}{2}<\frac{{(p-1)}^2}{p}\cdot 2^{\frac{1}{p}},-\frac{3}{2}<(p-2+\frac{1}{p})\cdot 2^{\frac{1}{p}}-p.$$

The above two inequalities obviously hold with $1<p<\frac{3}{2}$. If $\frac{3}{2}<p<2$, then $p-2+\frac{1}{p}$ is increasing about p, $2^{\frac{1}{p}}$ and −p are decreasing in relation to p. Then

Part 2: When $p>2$, then it should be proved that ${(\frac{p-1}{p})}^{p-1}{2}^{\frac{p-1}{p}}<1$ and ${(\frac{p-1}{p})}^{p-1}{2}^{\frac{p-1}{p}}>\frac{1}{2}$ are respectively equivalent to $2<{(1+\frac{1}{p-1})}^p$ and ${(1+\frac{1}{p-1})}^p<2^{2+\frac{1}{p-1}}$. Since ${(1+\frac{1}{p-1})}^p$ is strictly decreasing for $p\in (2,+\mathrm{\infty })$, then it is proved that

$${(1+\frac{1}{p-1})}^p>\underset{p\to \mathrm{\infty }}{lim}{(1+\frac{1}{p-1})}^p=e>2$$

and

$${(1+\frac{1}{p-1})}^p<{(1+\frac{1}{2-1})}^2=4<2^{2+\frac{1}{p-1}}.$$

The proof of Lemma 2.1 is completed. □

Lemma 2.2

• (i) Let $x\ge 1$, $p>2$, then

  $${(1-Z_p{x}^{\frac{1}{p}-1})}^{\frac{2-p}{p-1}}(1-\frac{p-2}{p-1}{Z}_p{x}^{\frac{1}{p}-1})>1.$$

  (2.2)

• (ii) Let $p>1$, then $f:x\in [1,+\mathrm{\infty })\to x^{\frac{1}{p}-2}-Z_p{x}^{\frac{2}{p}-3}$ is a convex function.

Proof (i) The proposition is equivalent to

$$1-\frac{p-2}{p-1}{Z}_p{x}^{\frac{1}{p}-1}>{(1-Z_p{x}^{\frac{1}{p}-1})}^{\frac{p-2}{p-1}}.$$

Bernoulli’s inequality refers to the following: ${(1+t)}^{\alpha }\le 1+\alpha t$ ($t>-1$, $0<\alpha <1$) holds if $t=0$. If $t=-Z_p{x}^{-1+\frac{1}{p}}$, $\alpha =\frac{p-2}{p-1}$, then formula (2.2) holds.

• (ii)
  $$\begin{array}{rcl}{f}^{″}(x)& =& (\frac{1}{p}-2)(\frac{1}{p}-3)x^{\frac{1}{p}-4}-Z_p(\frac{2}{p}-3)(\frac{2}{p}-4)x^{\frac{2}{p}-5}\\ =& \frac{(2p-1)x^{\frac{2}{p}-5}}{p^2}[(3p-1)x^{1-\frac{1}{p}}-2Z_p(3p-2)]\\ \ge & \frac{(2p-1)x^{\frac{2}{p}-5}}{p^2}[(3p-1)-2Z_p(3p-2)].\end{array}$$

According to Lemma 2.1,

$$f^{″}(x)\ge \frac{(2p-1)x^{\frac{2}{p}-5}}{p^2}[(3p-1)-(3p-2)]>0,$$

so, f is a convex function. The proof of Lemma 2.2 is completed. □

Lemma 2.3 Let n be a positive natural number, $p>1$.

• (i) If $1<p\le 2$, then

  $$\sum _{k=1}^n\frac{1}{{(k-\frac{1}{2})}^{\frac{1}{p}}}\le \frac{p}{p-1}(n^{1-\frac{1}{p}}-\frac{Z_p}{p-1}).$$

  (2.3)

• (ii) If $p>2$, then

  $$\sum _{k=1}^n\frac{1}{{(k-\frac{1}{2})}^{\frac{1}{p}}}\le \frac{p}{p-1}{n}^{1-\frac{2}{p}}{(n^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}.$$

  (2.4)

Proof (i) If $n=1$, inequality (2.3) is proved easily. Assume that when $n=m\ge 1$, the following equality holds for $n=m+1$:

$$\begin{array}{rcl}\sum _{k=1}^{m+1}\frac{1}{{(k-\frac{1}{2})}^{\frac{1}{p}}}& \le & \frac{p}{p-1}(m^{1-\frac{1}{p}}-\frac{Z_p}{p-1})+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}\\ =& \frac{p}{p-1}[{(m+1)}^{1-\frac{1}{p}}-\frac{Z_p}{p-1}]-{\int }_m^{m+1}{x}^{-\frac{1}{p}}dx+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}.\end{array}$$

Because $x^{-\frac{1}{p}}$ is a convex function on $[m,m+1]$, then according to Hadamard’s inequality of a convex function, formula (2.3) also holds if $n=m+1$.

• (ii) If $n=1$, inequality (2.4) is proved easily. Assume that when $n=m\ge 1$, the inequality holds. For $n=m+1$,

  $$\begin{array}{rcl}\sum _{k=1}^{m+1}\frac{1}{{(k-\frac{1}{2})}^{\frac{1}{p}}}& \le & \frac{p}{p-1}{m}^{1-\frac{2}{p}}{(m^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}\\ =& \frac{p}{p-1}{(m+1)}^{1-\frac{2}{p}}{({(m+1)}^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}\\ -{\int }_m^{m+1}{\left[\frac{p}{p-1}{x}^{1-\frac{2}{p}}{(x^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}\right]}^{\prime }dx+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}\\ =& \frac{p}{p-1}{(m+1)}^{1-\frac{2}{p}}{({(m+1)}^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}\\ -\frac{p}{p-1}{\int }_m^{m+1}[(1-\frac{2}{p})x^{-\frac{2}{p}}{(x^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}\\ +\frac{1}{p-1}(1-\frac{1}{p})x^{1-\frac{2}{p}}{(x^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}-1}\cdot x^{-\frac{1}{p}}]dx\\ =& \frac{p}{p-1}{(m+1)}^{1-\frac{2}{p}}{({(m+1)}^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}\\ -\frac{p}{p-1}{\int }_m^{m+1}{x}^{-\frac{2}{p}}{(x^{1-\frac{1}{p}}-Z_p)}^{\frac{2-p}{p-1}}\\ \times [(1-\frac{1}{p})x^{1-\frac{1}{p}}-(1-\frac{2}{p})Z_p]dx\\ =& \frac{p}{p-1}{(m+1)}^{1-\frac{2}{p}}{({(m+1)}^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}\\ -{\int }_m^{m+1}{x}^{-\frac{1}{p}}{(1-Z_p{x}^{-1+\frac{1}{p}})}^{\frac{2-p}{p-1}}[1-\frac{p-2}{p-1}{Z}_p{x}^{-1+\frac{1}{p}}]dx.\end{array}$$

By inequality (2.2) and the fact that $x^{-\frac{1}{p}}$ is a convex function on $[m,m+1]$, we have

$$\begin{array}{rcl}\sum _{k=1}^{m+1}\frac{1}{{(k-\frac{1}{2})}^{\frac{1}{p}}}& <& \frac{p}{p-1}{(m+1)}^{1-\frac{2}{p}}{({(m+1)}^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}-{\int }_m^{m+1}{x}^{-\frac{1}{p}}dx\\ \le & \frac{p}{p-1}{(m+1)}^{1-\frac{2}{p}}{({(m+1)}^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}+\frac{1}{{(m+\frac{1}{2})}^{\frac{1}{p}}}-{(m+\frac{1}{2})}^{-\frac{1}{p}}\\ =& \frac{p}{p-1}{(m+1)}^{1-\frac{2}{p}}{({(m+1)}^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}.\end{array}$$

Thus, the inequality (2.4) also holds if $n=m+1$. □

Lemma 2.4 Let i be any positive natural number and $p>1$, then

$$\sum _{n=i}^{\mathrm{\infty }}(n^{-2+\frac{1}{p}}-Z_p{n}^{-3+\frac{2}{p}})<\frac{p}{p-1}[{(i-\frac{1}{2})}^{\frac{1}{p}-1}-\frac{Z_p}{2}{(i-\frac{1}{2})}^{\frac{2}{p}-2}].$$

Proof Let $i=1,2,\dots$ and

$$f(i)=\frac{p}{p-1}[{(i-\frac{1}{2})}^{\frac{1}{p}-1}-\frac{Z_p}{2}{(i-\frac{1}{2})}^{\frac{2}{p}-2}]-\sum _{n=i}^{\mathrm{\infty }}(n^{\frac{1}{p}-2}-Z_p{n}^{\frac{2}{p}-3}).$$

Then

$$\begin{array}{rcl}f(i)-f(i+1)& =& \frac{p}{p-1}[{(i-\frac{1}{2})}^{\frac{1}{p}-1}-\frac{Z_p}{2}{(i-\frac{1}{2})}^{\frac{2}{p}-2}]-(i^{\frac{1}{p}-2}-Z_p{i}^{\frac{2}{p}-3})\\ -\frac{p}{p-1}[{(i+\frac{1}{2})}^{\frac{1}{p}-1}-\frac{Z_p}{2}{(i+\frac{1}{2})}^{\frac{2}{p}-2}]\\ =& -\frac{p}{p-1}{\int }_{i-\frac{1}{2}}^{i+\frac{1}{2}}{(x^{\frac{1}{p}-1}-\frac{Z_p}{2}{x}^{\frac{2}{p}-2})}^{\prime }dx-(i^{\frac{1}{p}-2}-Z_p{i}^{\frac{2}{p}-3})\\ =& {\int }_{i-\frac{1}{2}}^{i+\frac{1}{2}}(x^{\frac{1}{p}-2}-Z_p{x}^{\frac{2}{p}-3})dx-(i^{\frac{1}{p}-2}-Z_p{i}^{\frac{2}{p}-3}).\end{array}$$

According to the conclusion of Lemma 2.2 and Hadamard’s inequality of a convex function, the sequence ${\{f(i)\}}_{i=1}^{+\mathrm{\infty }}$ is a strictly decreasing sequence. It is also known that ${lim}_{i\to +\mathrm{\infty }}f(i)=0$, then $f(i)>0$ always holds. The proof of Lemma 2.4 is completed. □

3 A new strengthened version of Hardy’s inequality

Theorem 3.1 Let $a_k>0$, $n\ge 1$, $n\in N$, $p>1$, and

$$Z_p=\{\begin{array}{cc}p-1-\frac{{(p-1)}^2}{p}{2}^{\frac{1}{p}},\hfill & 1<p\le 2,\hfill \\ 1-{(\frac{p-1}{p})}^{p-1}{2}^{\frac{p-1}{p}},\hfill & p>2,\hfill \end{array}$$

then

$$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p<{\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}[1-\frac{Z_p}{2{(n-\frac{1}{2})}^{1-\frac{1}{p}}}]a_n^p.$$

(3.1)

Proof Let $r=\frac{p}{p-1}$, then $\frac{1}{r}+\frac{1}{p}=1$. According to Holder’s inequality,

$$\begin{array}{rcl}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p& =& \frac{1}{n^p}{\left(\sum _{k=1}^n{(k-\frac{1}{2})}^{-\frac{p-1}{p^2}}{(k-\frac{1}{2})}^{\frac{p-1}{p^2}}{a}_k\right)}^p\\ \le & \frac{1}{n^p}{\left(\sum _{k=1}^n{(k-\frac{1}{2})}^{-\frac{p-1}{p^2}r}\right)}^{\frac{p}{r}}{\left(\sum _{k=1}^n{(k-\frac{1}{2})}^{\frac{p-1}{p^2}p}{a}_k^p\right)}^{\frac{p}{p}}\\ =& \frac{1}{n^p}{\left(\sum _{k=1}^n{(k-\frac{1}{2})}^{-\frac{1}{p}}\right)}^{p-1}\sum _{k=1}^n{(k-\frac{1}{2})}^{\frac{p-1}{p}}{a}_k^p.\end{array}$$

If ${\lambda }_n=\frac{1}{n^p}{({\sum }_{k=1}^n{(k-\frac{1}{2})}^{-\frac{1}{p}})}^{p-1}$ and ${\mu }_k={(k-\frac{1}{2})}^{\frac{p-1}{p}}$, then

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p& \le & \sum _{n=1}^{\mathrm{\infty }}{\lambda }_n\cdot \sum _{k=1}^n{\mu }_k{a}_k^p\\ =& \sum _{n=1}^{\mathrm{\infty }}\left(\sum _{k=n}^{\mathrm{\infty }}{\lambda }_k\right){\mu }_n{a}_n^p.\end{array}$$

(3.2)

Thus,

$$\sum _{k=n}^{\mathrm{\infty }}{\lambda }_k=\sum _{k=n}^{\mathrm{\infty }}\frac{1}{k^p}{\left(\sum _{j=1}^k{(j-\frac{1}{2})}^{-\frac{1}{p}}\right)}^{p-1}.$$

If $1<p\le 2$, according to formula (2.3) and Bernoulli’s inequality,

$$\begin{array}{rcl}\sum _{k=n}^{\mathrm{\infty }}{\lambda }_k& =& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{k^p}{(k^{1-\frac{1}{p}}-\frac{Z_p}{p-1})}^{p-1}\\ =& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}{k}^{-2+\frac{1}{p}}{(1-\frac{Z_p}{p-1}{k}^{-1+\frac{1}{p}})}^{p-1}\\ <& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}{k}^{-2+\frac{1}{p}}(1-Z_p{k}^{-1+\frac{1}{p}})\\ =& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}(k^{-2+\frac{1}{p}}-Z_p{k}^{-3+\frac{2}{p}}).\end{array}$$

If $p>2$, by using formula (2.4), we obtain

$$\begin{array}{rcl}\sum _{k=n}^{\mathrm{\infty }}{\lambda }_k& \le & {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}\frac{1}{k^p}{\left(k^{1-\frac{2}{p}}{(k^{1-\frac{1}{p}}-Z_p)}^{\frac{1}{p-1}}\right)}^{p-1}\\ =& {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}(k^{-2+\frac{1}{p}}-Z_p{k}^{-3+\frac{2}{p}}).\end{array}$$

Therefore, for any $p>1$, the following inequality holds:

$$\sum _{k=n}^{\mathrm{\infty }}{\lambda }_k\le {\left(\frac{p}{p-1}\right)}^{p-1}\sum _{k=n}^{\mathrm{\infty }}(k^{-2+\frac{1}{p}}-Z_p{k}^{-3+\frac{2}{p}}).$$

By Lemma 2.4, we get

$$\sum _{n=k}^{\mathrm{\infty }}{\lambda }_k\le {\left(\frac{p}{p-1}\right)}^p({(n-\frac{1}{2})}^{\frac{1}{p}-1}-\frac{Z_p}{2}{(n-\frac{1}{2})}^{\frac{2}{p}-2}).$$

(3.3)

So, from inequalities (3.2) and (3.3), the following result can be obtained:

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^n{a}_k\right)}^p& \le & {\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}({(n-\frac{1}{2})}^{\frac{1}{p}-1}-\frac{Z_p}{2}{(n-\frac{1}{2})}^{\frac{2}{p}-2}){(n-\frac{1}{2})}^{\frac{p-1}{p}}{a}_n^p\\ =& {\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}(1-\frac{Z_p}{2{(n-\frac{1}{2})}^{1-\frac{1}{p}}})a_n^p.\end{array}$$

The proof of Theorem 3.1 is completed. □