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# Lyapunov-type inequalities for 2M th order equations under clamped-free boundary conditions

Journal of Inequalities and Applications20122012:242

https://doi.org/10.1186/1029-242X-2012-242

• Accepted: 9 October 2012
• Published:

## Abstract

This paper generalizes the well-known Lyapunov-type inequalities for second-order linear differential equations to certain 2M th order linear differential equations

${\left(-1\right)}^{M}{u}^{\left(2M\right)}\left(x\right)-r\left(x\right)u\left(x\right)=0\phantom{\rule{1em}{0ex}}\left(-s\le x\le s\right)$

under clamped-free boundary conditions. The usage of the best constant of some kind of a Sobolev inequality helps clarify the process for obtaining the result.

## Keywords

• Green Function
• Classical Solution
• Linear Differential Equation
• Sobolev Inequality
• Distributional Sense

## 1 Introduction

Let us consider the second-order linear differential equation
$\left\{\begin{array}{c}{u}^{″}\left(x\right)+r\left(x\right)u\left(x\right)=0\phantom{\rule{1em}{0ex}}\left(-s\le x\le s\right),\hfill \\ u\left(±s\right)=0,\hfill \end{array}$
(1)
where . It is well known that the Lyapunov inequality
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx>\frac{2}{s}$
(2)
gives a necessary condition for the existence of non-trivial classical solutions of (1), where ${r}^{+}\left(x\right)=\left(r\left(x\right)+|r\left(x\right)|\right)/2$. There are various extensions and applications for the above result; see, for example, surveys of Brown and Hinton [1] for relations to other fields and Tiryaki [2] for recent developments. Extensions to higher-order equations
$\left\{\begin{array}{c}{u}^{\left(n\right)}\left(x\right)+r\left(x\right)u\left(x\right)=0\phantom{\rule{1em}{0ex}}\left(-s\le x\le s\right),\hfill \\ \text{Boundary Conditions},\hfill \end{array}$
(3)

will be one important aspect. The first result for the high-order equation (3) is due to Levin [3], which states without proof:

Theorem A Let $n=2M$, and a non-trivial solution of (3) satisfies the clamped boundary condition, ${u}^{\left(i\right)}\left(±s\right)=0$ ($i=0,1,\dots ,M-1$). Then it holds that
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx>\frac{{2}^{2M-1}\left(2M-1\right){\left\{\left(M-1\right)!\right\}}^{2}}{{s}^{2M-1}}.$
Later, Das and Vatsala [4] gave the proof and extended the result by constructing the Green function. Other interesting developments for higher-order equations are seen in [59]. For example, as shown in Yang [8], Lyapunov-type inequalities can be obtained under the following conditions:
Here we note for the condition (c), very recently Çakmak [11], He and Tang [12], He and Zang [13] and [14] improved and extended the results of [5] and [8]. This paper considers the necessary condition for the existence of a non-trivial solution of the 2M th order linear differential equation
${\left(-1\right)}^{M}{u}^{\left(2M\right)}\left(x\right)-r\left(x\right)u\left(x\right)=0\phantom{\rule{1em}{0ex}}\left(-s\le x\le s\right)$
(4)

under yet another boundary condition:

Clamped-free boundary condition
${u}^{\left(i\right)}\left(-s\right)=0,\phantom{\rule{2em}{0ex}}{u}^{\left(M+i\right)}\left(s\right)=0\phantom{\rule{1em}{0ex}}\left(i=0,\dots ,M-1\right).$

The main result is as follows.

Theorem 1 Suppose a non-trivial solution u of (4) exists under the clamped-free boundary condition, then it holds
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx>\frac{{\left\{\left(M-1\right)!\right\}}^{2}\left(2M-1\right)}{{\left(2s\right)}^{2M-1}}.$
(5)

Moreover, the estimate is sharp in the sense that there exists a function $r\left(x\right)$, and for this $r\left(x\right)$, the solution u of (4) exits such that the right-hand side is arbitrarily close to the left-hand side.

The result is obtained using Takemura [[15], Theorem 1], which computes the best constant of some kind of a Sobolev inequality. In Section 4, we give a concise proof for an ${L}^{p}$ extension of Theorem 1 of [15].

## 2 Proof of Theorem 1

Now, let us introduce the following ${L}^{p}$-type Sobolev inequality:
${\left(\underset{-s\le x\le s}{sup}|{u}^{\left(m\right)}\left(x\right)|\right)}^{p}\le C{\int }_{-s}^{s}{|{u}^{\left(M\right)}\left(x\right)|}^{p}\phantom{\rule{0.2em}{0ex}}dx,$
(6)
where u belongs to
$W\left(M,p\right):=\left\{u|{u}^{\left(M\right)}\in {L}^{p}\left(-s,s\right),{u}^{\left(i\right)}\left(-s\right)=0\phantom{\rule{0.25em}{0ex}}\left(i=0,\dots ,M-1\right)\right\},$

$1, m runs over the range $0\le m\le M-1$, and ${u}^{\left(i\right)}$ is the i th derivative of u in a distributional sense. We denote by ${C}_{CF}\left(M,m,p\right)$ the best constant of the above Sobolev inequality (6). Here, we note that in [15], Takemura obtained the best constant for $p=2$, $m=0$ by constructing the Green function of the clamped-free boundary value problem. Although, for the proof of Theorem 1, we simply need the value ${C}_{CF}\left(M,0,2\right)$, we would like to compute ${C}_{CF}\left(M,m,p\right)$ for general p and m since the proof presented in Section 4 does not depend on special values of p and m and quite simplifies the proof of Theorem 1 of [15]. Now, we have the following propositions.

Proposition 1 The best constant of (6) is
${C}_{CF}\left(M,m,p\right)=\frac{1}{{\left\{\left(M-m-1\right)!\right\}}^{p}}{\left(\frac{\left(p-1\right){\left(2s\right)}^{\frac{p\left(M-m\right)-1}{p-1}}}{p\left(M-m\right)-1}\right)}^{p-1},$
(7)
and it is attained by
${u}_{\ast }\left(x\right)={\int }_{-s}^{x}\frac{{\left(x-t\right)}^{M-1}}{\left(M-1\right)!}\cdot {\left\{\frac{{\left(s-t\right)}^{M-1-m}}{\left(M-1-m\right)!}\right\}}^{p-1}\phantom{\rule{0.2em}{0ex}}dt.$
(8)
Proposition 2 Suppose a ${C}^{2M}\left[-s,s\right]$ solution of (4) with the clamped-free boundary condition exists, then it holds that
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx>\frac{1}{{C}_{CF}\left(M,0,2\right)}.$
(9)

Moreover, the estimate is sharp.

Proof of Theorem 1 Clearly, Theorem 1 is obtained from Propositions 1 and 2. □

Thus, all we have to do is to show Propositions 1 and 2. Before proceeding with the proof of these propositions, we would like to show a corollary obtained from Proposition 1.

Corollary 1 Suppose a non-trivial solution u of the non-linear equation
${\left(-1\right)}^{M}u\left(x\right){u}^{\left(2M\right)}\left(x\right)-r\left(x\right){\left({u}^{\left(m\right)}\left(x\right)\right)}^{2}=0$
(10)
exists under the clamped-free boundary condition, where m satisfies ($1\le m\le M-1$), then it holds
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx>\frac{{\left\{\left(M-1-m\right)!\right\}}^{2}\left(2\left(M-m\right)-1\right)}{{\left(2s\right)}^{2\left(M-m\right)-1}}.$
(11)

The following are the examples of Theorem 1 and Corollary 1.

Example 1 The following example corresponds to the case $M=1$ and $r\left(x\right)=-6/\left(-11{s}^{2}+2sx+{x}^{2}\right)$ of (4) with the clamped-free boundary condition
$\left\{\begin{array}{c}-{u}^{″}\left(x\right)+\frac{6}{-11{s}^{2}+2sx+{x}^{2}}u\left(x\right)=0,\hfill \\ u\left(-s\right)={u}^{\prime }\left(s\right)=0.\hfill \end{array}$
It is easy to see that $u\left(x\right)=-\left(s+x\right)\left(11{s}^{2}-2sx-{x}^{2}\right)$ is the solution of the above equation. Moreover, it holds that
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx={\int }_{-s}^{s}-\frac{6}{-11{s}^{2}+2sx+{x}^{2}}\phantom{\rule{0.2em}{0ex}}dx=\frac{\sqrt{3}log\left(2+\sqrt{3}\right)}{2s}>\frac{1}{{C}_{CF}\left(1,0,2\right)}=\frac{1}{2s}.$
Example 2 The following example corresponds to the case $M=2$, $m=1$ and $r\left(x\right)=\left(3\left(17{s}^{2}-6sx+{x}^{2}\right)\right)/\left(2{\left(7{s}^{2}-4sx+{x}^{2}\right)}^{2}\right)$ of (10) with the clamped-free boundary condition
$\left\{\begin{array}{c}u\left(x\right){u}^{\left(4\right)}\left(x\right)-\frac{3\left(17{s}^{2}-6sx+{x}^{2}\right)}{2{\left(7{s}^{2}-4sx+{x}^{2}\right)}^{2}}{\left({u}^{\prime }\left(x\right)\right)}^{2}=0,\hfill \\ u\left(-s\right)={u}^{\prime }\left(-s\right)={u}^{″}\left(s\right)={u}^{‴}\left(s\right)=0.\hfill \end{array}$
It is easy to see that $u\left(x\right)={\left(s+x\right)}^{2}\left(17{s}^{2}-6sx+{x}^{2}\right)$ is the solution of the above equation. Moreover, it holds that
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx={\int }_{-s}^{s}\frac{3\left(17{s}^{2}-6sx+{x}^{2}\right)}{2{\left(7{s}^{2}-4sx+{x}^{2}\right)}^{2}}\phantom{\rule{0.2em}{0ex}}dx=\frac{3+2\sqrt{3}\pi }{12s}>\frac{1}{{C}_{CF}\left(2,1,2\right)}=\frac{1}{2s}.$

## 3 Proof of Proposition 2

Assuming Proposition 1, we first prove Proposition 2.

Proof of Proposition 2 Let u be a solution of equation (4). Since u satisfies the clamped-free boundary condition, multiplying (4) by u and integrating it over $\left[-s,s\right]$, we have
$\begin{array}{rcl}{\int }_{-s}^{s}{\left({u}^{\left(M\right)}\left(x\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dx& =& {\int }_{-s}^{s}{\left(-1\right)}^{M}u\left(x\right){u}^{\left(2M\right)}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx={\int }_{-s}^{s}r\left(x\right){\left(u\left(x\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dx\\ \le & {\left(\underset{-s\le x\le s}{sup}|u\left(x\right)|\right)}^{2}{\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ \le & C\left(M,0,2\right){\int }_{-s}^{s}{\left({u}^{\left(M\right)}\left(x\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dx{\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(12)
Here, if ${u}^{\left(M\right)}\equiv 0$, then there exists ($i=0,\dots ,M-1$) such that $u\left(x\right)={\sum }_{i=0}^{M-1}{a}_{i}{x}^{i}$. Since u satisfies the clamped boundary condition at $x=-s$, we have $u\equiv 0$. This contradicts the assumption that u is a non-trivial solution of (4). So, canceling ${\parallel {u}^{\left(M\right)}\parallel }_{2}^{2}$, we obtain
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\ge \frac{1}{{C}_{CF}\left(M,0,2\right)}.$
(13)
Next, we show that the inequality (13) is strict. To see this, we note that in (12), if the equality holds for the first inequality, then u is a constant. But, again from the clamped boundary condition at $x=-s$, we have $u\equiv 0$. Thus, the inequality is strict. Finally, we see (5) is sharp. For this purpose, let us define the functional
$J\left(\varphi \right):=\frac{{\int }_{-s}^{s}{|{\varphi }^{\left(M\right)}|}^{2}\phantom{\rule{0.2em}{0ex}}dx}{{\int }_{-s}^{s}\stackrel{˜}{r}{|\varphi |}^{2}\phantom{\rule{0.2em}{0ex}}dx}\phantom{\rule{1em}{0ex}}\left(\varphi \in W\left(M,2\right),\varphi \not\equiv 0\right),$
where is defined later. By the standard argument of the variational method, J has the minimizer $u\in W\left(M,2\right)$ (see, for example, [[16], Lemma 3]), i.e.,
${\lambda }_{1}:=\underset{\varphi \in W\left(M,2\right),\varphi \not\equiv 0}{min}J\left(\varphi \right)=J\left(u\right).$
Hence, it satisfies the Euler-Lagrange equation (as a classical solution by the regularity argument)
${\left(-1\right)}^{M}{u}^{\left(2M\right)}\left(x\right)={\lambda }_{1}\stackrel{˜}{r}\left(x\right)u\left(x\right)\phantom{\rule{1em}{0ex}}\left(-s\le x\le s\right).$
(14)
Further, it holds that
$\begin{array}{rcl}{\lambda }_{1}& =& \underset{\varphi \in W\left(M,2\right),\varphi \not\equiv 0}{min}\frac{{\int }_{-s}^{s}{|{\varphi }^{\left(M\right)}|}^{2}\phantom{\rule{0.2em}{0ex}}dx}{{\int }_{-s}^{s}\stackrel{˜}{r}{|\varphi |}^{2}\phantom{\rule{0.2em}{0ex}}dx}>\frac{{\int }_{-s}^{s}{|{\varphi }^{\left(M\right)}|}^{2}\phantom{\rule{0.2em}{0ex}}dx}{{\left({sup}_{-s\le x\le s}|\varphi |\right)}^{2}{\int }_{-s}^{s}\stackrel{˜}{r}\phantom{\rule{0.2em}{0ex}}dx}\\ \ge & \frac{1}{{C}_{CF}\left(M,0,2\right){\int }_{-s}^{s}\stackrel{˜}{r}\phantom{\rule{0.2em}{0ex}}dx}.\end{array}$
(15)
Here, let us fix $\stackrel{˜}{r}$ as
$\stackrel{˜}{r}\left(x\right):=\left\{\begin{array}{cc}\frac{x-s}{\delta }+1\hfill & \left(s-\delta
For such $\stackrel{˜}{r}$, let us substitute $\varphi ={u}_{\ast }$ (of Proposition 1) into (15). It is easy to see that ${u}_{\ast }$ takes its maximum at $x=s$, hence by taking δ sufficiently small, we see that the right-hand side of (15) can be arbitrarily close to the left-hand side, i.e., for a small positive ${ϵ}_{1}$, ${\lambda }_{1}$ can be written as
${\lambda }_{1}=\frac{1}{{C}_{CF}\left(M,0,2\right){\int }_{-s}^{s}\stackrel{˜}{r}\phantom{\rule{0.2em}{0ex}}dx}+{ϵ}_{1}.$
(16)
Putting $r={\lambda }_{1}\stackrel{˜}{r}$, we see from (14) a solution u of
${\left(-1\right)}^{M}{u}^{\left(2M\right)}\left(x\right)=r\left(x\right)u\left(x\right)\phantom{\rule{1em}{0ex}}\left(-s\le x\le s\right)$
(17)
exists, and from (16) r satisfies
${\int }_{-s}^{s}r\left(x\right)\phantom{\rule{0.2em}{0ex}}dx=\frac{1}{C\left(M,0,2\right)}+{ϵ}_{1}{\int }_{-s}^{s}\stackrel{˜}{r}\phantom{\rule{0.2em}{0ex}}dx=\frac{1}{C\left(M,0,2\right)}+{ϵ}_{2}.$
(18)

Hence, (5) is sharp. □

Proof of Corollary 1 Integrating equation (10), we have
$\begin{array}{rcl}{\int }_{-s}^{s}{\left({u}^{\left(M\right)}\left(x\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dx& =& {\int }_{-s}^{s}r\left(x\right){\left({u}^{\left(m\right)}\left(x\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dx\le {\left(\underset{-s\le x\le s}{sup}|{u}^{\left(m\right)}\left(x\right)|\right)}^{2}{\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ \le & C\left(M,m,2\right){\int }_{-s}^{s}{\left({u}^{\left(M\right)}\left(x\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dx{\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(19)
As in the proof of Proposition 2, by canceling ${\parallel {u}^{\left(M\right)}\parallel }_{2}^{2}$, we have
${\int }_{-s}^{s}{r}^{+}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \frac{1}{{C}_{CF}\left(M,m,2\right)}.$

Next, we show that the inequality (11) is strict. To see this, we note that in (19), the equality holds for the first inequality if and only if ${u}^{\left(m\right)}$ is a constant. Hence, from the clamped boundary condition at $x=-s$, we have ${u}^{\left(m\right)}\equiv 0$. So, there exists ($i=0,\dots ,m-1$) such that $u\left(x\right)={\sum }_{i=0}^{m-1}{a}_{i}{x}^{i}$. But, again from the clamped boundary condition at $x=-s$, we have $u\equiv 0$. Thus, inequality (11) is strict. □

## 4 Proof of Proposition 1

We prepare the following lemmas for the proof of Proposition 1.

Lemma 1 Suppose there exists a function ${u}_{\ast }\in W\left(M,p\right)$ which attains the best constant $C\left(M,n,p\right)$ of (6), then it holds that
$\underset{-s\le x\le s}{max}|{u}_{\ast }^{\left(m\right)}\left(x\right)|=|{u}_{\ast }^{\left(m\right)}\left(s\right)|.$
Proof Suppose it holds that
$\underset{-s\le x\le s}{max}|{u}_{\ast }^{\left(m\right)}\left(x\right)|=|{u}_{\ast }^{\left(m\right)}\left(a\right)|,$
where $a\not\equiv s$. Further, let us define
$\stackrel{˜}{u}\left(x\right):=\left\{\begin{array}{cc}0\hfill & \left(-s\le x\le -a\right),\hfill \\ {u}_{\ast }\left(x+a-s\right)\hfill & \left(-a\le x\le s\right).\hfill \end{array}$
Then it holds $\stackrel{˜}{u}\in W\left(M,p\right)$ and
$\underset{-s\le x\le s}{max}|{\stackrel{˜}{u}}^{\left(m\right)}\left(x\right)|=\underset{-s\le x\le s}{max}|{u}_{\ast }^{\left(m\right)}\left(x\right)|=|{u}_{\ast }^{\left(m\right)}\left(a\right)|$
and ${\parallel {\stackrel{˜}{u}}^{\left(M\right)}\parallel }_{{L}^{p}\left(-s,s\right)}<{\parallel {u}_{\ast }^{\left(M\right)}\parallel }_{{L}^{p}\left(-s,s\right)}$. Hence,
$C\left(M,m,p\right)=\frac{{\left({max}_{-s\le x\le s}|{u}_{\ast }^{\left(m\right)}\left(x\right)|\right)}^{p}}{{\parallel {u}_{\ast }^{\left(M\right)}\parallel }_{{L}^{p}\left(-s,s\right)}}<\frac{{\left({max}_{-s\le x\le s}|{\stackrel{˜}{u}}^{\left(m\right)}\left(x\right)|\right)}^{p}}{{\parallel {\stackrel{˜}{u}}^{\left(M\right)}\parallel }_{{L}^{p}\left(-s,s\right)}}.$

This contradicts the assumption that $C\left(M,m,p\right)$ is the best constant of (6). □

Lemma 2 Let
${H}_{m}\left(x\right):=\frac{{\left(-1\right)}^{M-1-m}{\left(x-s\right)}^{M-1-m}}{\left(M-1-m\right)!},$
then for $u\in W\left(M,p\right)$ it holds that
${u}^{\left(m\right)}\left(s\right)={\int }_{-s}^{s}{u}^{\left(M\right)}\left(x\right){H}_{m}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.$
(20)

Proof Integrating by parts, we obtain the result. □

Proof of Proposition 1 From Lemma 2, we see that if the function attains the best constant $C\left(M,m,p\right)$, it belongs to ${W}_{\ast }\left(M,m,p\right)\subset W\left(M,p\right)$:
${W}_{\ast }\left(M,m,p\right)=\left\{u\in W\left(M,p\right)|\underset{-s\le x\le s}{max}|{u}^{\left(m\right)}\left(x\right)|=|{u}^{\left(m\right)}\left(s\right)|\right\}.$
Let $u\in {W}_{\ast }\left(M,m,p\right)$. Then applying Hölder’s inequality to (20), we have
$\underset{-s\le x\le s}{max}|{u}^{\left(m\right)}\left(x\right)|=|{u}^{\left(m\right)}\left(s\right)|\le {\parallel {H}_{m}\parallel }_{{L}^{q}\left(-s,s\right)}{\parallel {u}^{\left(M\right)}\parallel }_{{L}^{p}\left(-s,s\right)},$
(21)
where q satisfies $1/p+1/q=1$. Hence, if there exists the function ${u}_{\ast }\in {W}_{\ast }\left(M,m,p\right)$ which attains the equality of (21), it holds that $C\left(M,m,p\right)={\parallel {H}_{m}\parallel }_{{L}^{q}\left(-s,s\right)}^{p}$. On the contrary, we see that the equality holds for (21) if and only if u satisfies
${u}^{\left(M\right)}\left(x\right)=\left(sgn{H}_{m}\left(x\right)\right){|{H}_{m}\left(x\right)|}^{q-1}.$
(22)
It is easy to see that
${u}_{\ast }\left(x\right)={\int }_{-s}^{x}\frac{{\left(x-t\right)}^{m-1}}{\left(M-1\right)!}\left(sgn{H}_{m}\left(t\right)\right){|{H}_{m}\left(t\right)|}^{q-1}\phantom{\rule{0.2em}{0ex}}dt={\int }_{-s}^{x}\frac{{\left(x-t\right)}^{m-1}}{\left(M-1\right)!}{\left\{\frac{{\left(s-t\right)}^{M-1-m}}{\left(M-1-m\right)!}\right\}}^{\frac{1}{p-1}}\phantom{\rule{0.2em}{0ex}}dt$
satisfies (22) and belongs to ${W}_{\ast }\left(M,m,p\right)$. Thus, we have shown $C\left(M,m,p\right)={\parallel {H}_{m}\parallel }_{{L}^{q}\left(-s,s\right)}^{p}$. Now, we compute ${\parallel {H}_{m}\parallel }_{{L}^{q}\left(-s,s\right)}^{p}$. It is
$\begin{array}{rcl}{\parallel {H}_{m}\parallel }_{{L}^{q}\left(-s,s\right)}^{p}& =& \frac{1}{{\left\{\left(M-1-m\right)!\right\}}^{p}}{\left\{{\int }_{-s}^{s}{\left(s-x\right)}^{q\left(M-1-m\right)}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{p}{q}}\\ =& \frac{1}{{\left\{\left(M-1-m\right)!\right\}}^{p}}{\left\{\frac{\left(p-1\right){\left(2s\right)}^{\frac{p\left(M-m\right)-1}{p-1}}}{p\left(M-m\right)-1}\right\}}^{p-1}.\end{array}$

This completes the proof. □

## Authors’ Affiliations

(1)
Department of Computer Science, National Defense Academy, 1-10-20 Hashirimizu, Yokosuka 239-8686, Japan
(2)
Liberal Arts and Basic Sciences, College of Industrial Technology, Nihon University, 2-11-1 Shinei, Narashino 275-8576, Japan
(3)
Graduate School of Mathematical Sciences, Faculty of Engineering Science, Osaka University, 1-3 Matikaneyamacho, Toyonaka 560-8531, Japan
(4)
Tokyo Metropolitan College of Industrial Technology, 1-10-40 Higashi-ooi, Shinagawa, Tokyo 140-0011, Japan

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