# A Generalization of the Concavity of Rényi Entropy Power

Laigang Guo, Chun-Ming Yuan, Xiao-Shan Gao
KLMM, Academy of Mathematics and Systems Science
Chinese Academy of Sciences, Beijing 100190, China
University of Chinese Academy of Sciences, Beijing 100049, China

###### Abstract

Recently, Savaré-Toscani proved that the Rényi entropy power of general
probability densities solving the $p$-nonlinear heat equation in
$\mathbb{R}^{n}$ is always a concave function of time, which extends Costa’s
concavity inequality for Shannon’s entropy power to Rényi entropies. In this
paper, we give a generalization of Savaré-Toscani’s result by giving a class
of sufficient conditions of the parameters under which the concavity of the
Rényi entropy power is still valid. These conditions are quite general and
include the parameter range given by Savaré-Toscani as special cases. Also,
the conditions are obtained with a systematical approach.

Keywords. Rényi entropy, entropy power inequality, nonlinear heat equation.

##  1 Introduction

The $p$-th Rényi entropy [1, 2] of a probability density function
$f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ is defined as

$\displaystyle H_{p}(f):=\dfrac{1}{1-p}{\rm
log}\int_{\mathbb{R}^{n}}f^{p}(x)dx,$ (1)

for $0<p<+\infty$, $p\neq 1$. The $p$-th Rényi entropy power is given by

$\displaystyle N_{p}(f):={\rm exp}(\frac{\mu}{n}H_{p}(f)),$ (2)

where $\mu$ is a real valued parameter. The Rényi entropy for $p=1$ is defined
as the limit of $H_{p}(f)$ as $p\rightarrow 1$. It follows from definition (1)
that

$\displaystyle H_{1}(f)=\underset{p\rightarrow 1}{\rm
lim}H_{p}(f)=-\int_{\mathbb{R}^{n}}f(x){\rm log}f(x)dx,$

which is Shannon’s entropy. In this case, the proposed Rényi entropy power of
index $p=1,\mu=2$, given by (2), coincides with Shannon’s entropy power

$\displaystyle N_{1}(f):={\rm exp}(\dfrac{2}{n}H_{1}(f)).$ (3)

Shannon’s entropy power inequality (EPI) is one of the most important
information inequalities [3], which has many proofs, generalizations, and
applications [4, 5, 6, 7, 8, 9, 10, 11, 12]. In particular, Costa presented a
stronger version of the EPI [13].

Let $X_{t}\triangleq X+N_{n}(0,tI)$ be the $n$-dimensional random vector
introduced by Costa [13, 19, 20] and $u(x_{t})$ the probability density of
$X_{t}$, which solves the heat equation in the whole space $\mathbb{R}^{n}$,

$\displaystyle\dfrac{\partial}{\partial t}u(x_{t})=\Delta u(x_{t}).$ (4)

Costa’s differential entropy is defined to be

$H(u(x_{t}))=-\int_{{\mathbb{R}}^{n}}u(x_{t})\log
u(x_{t}){\hbox{\rm{d}}}x_{t}.$ (5)

Costa [13] proved that the Shannon entropy power $N(u)=\frac{1}{2\pi
e}e^{(2/n)H(u)}$ is a concave function in $t$, that is
$({\hbox{\rm{d}}}/{\hbox{\rm{d}}}t)N(u)\geq 0$ and
$({\hbox{\rm{d}}}^{2}/{\hbox{\rm{d}}}^{2}t)N(u)\leq 0$. Several new proofs and
generalizations for Costa’s EPI were given [14, 15, 16].

Savaré-Toscani [21] proved that the concavity of entropy power is a property
which is not restricted to the Shannon entropy power (3) in connection with
the heat equation (4), but it holds for the $p$-th Rényi entropy power (2).
They put it in connection with the solution to the nonlinear heat equation

$\displaystyle\dfrac{\partial}{\partial t}u(x_{t})=\Delta u(x_{t})^{p}$ (6)

posed in the whole space $\mathbb{R}^{n}$ and $p\in{\mathbb{R}}_{>0}$.

In this paper, we give a generalization for concavity of $p$-th Rényi entropy
power (CREP). Precisely, we give a propositional logic formula $\Phi(n,p,\mu)$
such that if $n\in{\mathbb{N}},p,\mu\in{\mathbb{R}}$ satisfy this formula,
then the CREP holds. The condition $\Phi(n,p,\mu)$ extends the parameter’s
range of the CREP given by Savaré-Toscani [21] and contains much more cases.

The formula $\Phi$ is obtained using a systematic procedure which can be
considered as a parametric version of that given in [19, 20, 17], where
parameters $n,p,\mu$ exist in the formulas. The procedure reduces the proof of
the CREP to check the semi-positiveness of a quadratic form whose coefficients
are polynomials in the parameters $n,p,\mu$. In principle, a necessary and
sufficient condition for the parameters to satisfy this property can be
computed with the quantifier elimination [22]. In this paper, the problem is
special and an explicit proof is given.

The rest of this paper is organized as follows. In Section 2, we give the
proof procedure and to prove concavity of entropy powers in the parametric
case. In Section 3, we present the generalized version of CREP using the proof
procedure. In Section 4, conclusions are presented.

##  2 Proof Procedure

In this section, we give a procedure to prove the CREP. To make the paper
concise, we only give those steps that are needed in this paper.

###  2.1 Notations

Let $x_{t}=[x_{1,t},x_{2,t},\ldots,x_{n,t}]$ be a set of variables depending
on $t$ and

$\displaystyle{\hbox{\rm{d}}}^{(i)}x_{t}={\hbox{\rm{d}}}x_{1,t}{\hbox{\rm{d}}}x_{2,t}\ldots{\hbox{\rm{d}}}x_{i-1,t}{\hbox{\rm{d}}}x_{i+1,t}\ldots{\hbox{\rm{d}}}x_{n,t},i=1,2\ldots,n.$

Let $[n]_{0}=\\{0,1,\ldots,n\\}$ and $[n]=\\{1,\ldots,n\\}$. To simplify the
notations, we use $u$ to denote $u(x_{t})$ in the rest of the paper. Denote

$\mathcal{P}_{n}=\\{\frac{\partial^{h}u}{\partial^{h_{1}}x_{1,t}\cdots\partial^{h_{n}}x_{n,t}}:h=\sum_{i=1}^{n}h_{i},h_{i}\in{\mathbb{N}}\\}$

to be the set of all derivatives of $u$ with respect to the differential
operators $\frac{\partial}{\partial x_{i,t}},i=1,\ldots,n$ and

${\mathcal{R}}={\mathbb{R}}[n,p,\mu][\mathcal{P}_{n}]$

to be the set of polynomials in $\mathcal{P}_{n}$ with coefficients in
${\mathbb{R}}[n,p,\mu]$, where $n,p,\mu$ are parameters. For
$v\in\mathcal{P}_{h,n}$, let ${\rm{ord}}(v)$ be the order of $v$. For a
monomial $\prod_{i=1}^{r}v_{i}^{d_{i}}$ with $v_{i}\in{\mathcal{P}}_{n}$, its
degree, order, and total order are defined to be $\sum_{i=1}^{r}d_{i}$,
$\max_{i=1}^{r}{\rm{ord}}(v_{i})$, and
$\sum_{i=1}^{r}d_{i}\cdot{\rm{ord}}(v_{i})$, respectively.

A polynomial in ${\mathcal{R}}$ is called a $k$th-order differentially
homogenous polynomial or simply a $k$th-order differential form, if all its
monomials have degree $k$ and total order $k$. Let ${\mathcal{M}}_{k,n}$ be
the set of all monomials which have degree $k$ and total order $k$. Then the
set of $k$th-order differential forms is an ${\mathbb{R}}$-linear vector space
generated by ${\mathcal{M}}_{k,n}$, which is denoted as
${\hbox{\rm{Span}}}_{\mathbb{R}}({\mathcal{M}}_{k,n})$. We will use Gaussian
elimination in ${\hbox{\rm{Span}}}_{\mathbb{R}}({\mathcal{M}}_{k,n})$ by
treating the monomials as variables. We always use the lexicographic order for
the monomials defined in [19, 20].

###  2.2 The proof procedure

In this section, we give the procedure to prove the CREP. The property
$\frac{{\hbox{\rm{d}}}}{{\hbox{\rm{d}}}t}N_{p}(u)\geq 0$ can be easily proved
[21]. We focus on proving
$\frac{{\hbox{\rm{d}}}^{2}}{{\hbox{\rm{d}}}^{2}t}N_{p}(u)\leq 0$. The
procedure consists of four steps.

In step 1, we reduce the proof of CREP into the proof of an integral
inequality, as shown by the following lemma whose proof is given in section
2.4.

######  Lemma 2.1.

Proof of $\frac{{\hbox{\rm{d}}}^{2}}{{\hbox{\rm{d}}}^{2}t}N_{p}(u)\leq 0$ can
be reduced to show

$\begin{array}[]{ll}\displaystyle{\int_{{\mathbb{R}}^{n}}u^{3p-6}E_{2,n}{\hbox{\rm{d}}}x_{t}}\geq
0\end{array}$ (7)

under the condition $p\geq 1-\frac{\mu}{n}$, where
$E_{2,n}=\sum_{a=1}^{n}\sum_{b=1}^{n}E_{2,n,a,b}$ is a $4$th-order
differential form in ${\mathbb{R}}[n,p,\mu][{{\mathcal{P}}}_{2,n}]$ and

${{\mathcal{P}}}_{2,n}=\\{\frac{\partial^{h}u}{\partial^{h_{1}}x_{a,t}\partial^{h_{2}}x_{b,t}}:h\in[3]_{0};a,b\in[n]\\}.$
(8)

Then the problem $\frac{{\hbox{\rm{d}}}^{2}}{{\hbox{\rm{d}}}^{2}t}N_{p}(u)\leq
0$ can be transformed to $\int u^{3p-6}E_{2,n}{\hbox{\rm{d}}}x_{t}\geq 0$.
Thus, Lemma 2.1 is proved.

In step 2, we compute the constraints which are relations satisfied by the
probability density $u$ of $X_{t}$. Since $E_{2,n}$ in (7) is a $4$th-order
differential form, we need only the constraints which are $4$th-order
differential forms. A $4$th-order differential form $R$ is called an
equational or inequality constraint if

${\int_{{\mathbb{R}}^{n}}u^{3p-6}R{\hbox{\rm{d}}}x_{t}}=0\hbox{ or
}{\int_{{\mathbb{R}}^{n}}u^{3p-6}R{\hbox{\rm{d}}}x_{t}}\geq 0.$ (9)

The method to compute the constraints is given in section 2.3. Suppose that
the equational and inequality constraints are respectively

${\mathcal{C}}_{E}=\\{R_{i},\,|\,i=1,\ldots,N_{1}\\}\hbox{ and
}{\mathcal{C}}_{I}=\\{I_{i},\,|\,i=1,\ldots,N_{2}\\}.$ (10)

In step 3, we find a propositional formula $\Phi(n,p,\mu)$ such that when
$n,p,\mu\in{\mathbb{R}}$ satisfy $\Phi$,

$\displaystyle\exists
c_{j},e_{i}\in{\mathbb{R}},s.t.\,\,E_{2,n}-\sum_{i=1}^{N_{1}}e_{i}R_{i}-\sum_{j=1}^{N_{2}}c_{j}I_{j}=S\geq
0\hbox{ and }c_{j}\geq 0,j=1,\ldots,n_{2}.$ (11)

Details of this step and the formula $\Phi(n,p,\mu)$ are given in section 3.

To summarize the proof procedure, we have

######  Theorem 2.2.

The CREP is true if $\Phi(n,p,\mu)$ is valid.

###### Proof.

By Lemma 2.1, we have the following proof:

$\displaystyle\int_{\mathbb{R}}u^{3p-6}{{E}_{2,n}}{\hbox{\rm{d}}}x_{t}$ (12)
$\displaystyle\overset{\eqref{eq-S1}}{=}$
$\displaystyle\int_{\mathbb{R}}u^{3p-6}({\sum_{i=1}^{N_{1}}e_{i}R_{i}+\sum_{j=1}^{N_{2}}c_{j}I_{j}+S}){\hbox{\rm{d}}}x_{t}$
$\displaystyle\overset{S1}{=}$
$\displaystyle\int_{\mathbb{R}}u^{3p-6}(\sum_{j=1}^{N_{2}}c_{j}I_{j}+S){\hbox{\rm{d}}}x_{t}$
$\displaystyle\overset{S2}{\geq}$
$\displaystyle\int_{\mathbb{R}}u^{3p-6}{S}{\hbox{\rm{d}}}x_{t}\overset{S3}{\geq}0.$

Equality S1 is true, because $R_{i}$ is a equational constraint. Inequality S2
is true, because $I_{j}$ is an inequality constraint. Inequality S3 is true,
because $S\geq 0$ under the condition $\Phi(n,p,\mu)$. ∎

###  2.3 The equational constraints

In this section, we show how to find the second order equational constraints.
A second order equational constraint is a $4$th-order differential form in
${\mathbb{R}}[n,p,\mu][{\mathcal{P}}_{2,n}]$ such that
$\int_{{\mathbb{R}}^{n}}\ u^{3p-6}{R}\ {\hbox{\rm{d}}}x_{t}=0$.

We introduce the following notations

${\mathcal{V}}_{a,b}=\\{\frac{\partial^{h}u}{\partial^{h_{1}}x_{a,t}\partial^{h_{2}}x_{b,t}}:h=h_{1}+h_{2}\in[3]_{0}\\},$
(13)

where $a,b$ are variables taking values in $[n]$. Then
${\mathcal{P}}_{2,n}=\cup_{a=1}^{n}\cup_{b=1}^{n}{\mathcal{V}}_{a,b}$.

We need the following property.

######  Property 2.3.

Let $a,r,m_{i},k_{i}\in{\mathbb{N}}_{>0}$ and $u^{(m_{i})}$ an $m_{i}$th-order
derivative of $u$. If $u(x_{t})$ is a smooth, strictly positive and rapidly
decaying probability density, then

$\displaystyle{\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty}u^{3p-2}\left[\prod\limits_{i=1}^{r}\frac{[u^{(m_{i})}]^{k_{i}}}{u^{k_{i}}}\right]\Bigg{|}_{x_{a,t}=-\infty}^{\infty}{\hbox{\rm{d}}}^{(a)}}x_{t}=0,$
(14)

with $\sum_{i=1}^{r}k_{i}m_{i}=4,\ \sum_{i=1}^{r}k_{i}=4$.

When $p\geq 2$, Property 2.3 follows from [23]. While $0<p<2,p\neq 1$, we make
the assumption that $u(x_{t})$ also satisfies Property 2.3.

Using Property 2.3, we can compute the second order equational constraints
using the method given in [19, 20]:

${\mathcal{C}}_{2,n}=\\{R_{i,a,b}\,:\,i=1,\ldots,28\\}\subset{\mathbb{R}}[n,p,\mu][{\mathcal{V}}_{a,b}],$
(15)

where $R_{i,a,b}$ can be found in the Appendix. Note that $a,b$ are variables
taking values in $[n]$.

We use an example to show how to obtain these constraints. Starting from a
monomial $u\frac{\partial^{2}u}{\partial^{2}x_{a,t}}(\frac{\partial
u}{\partial x_{a,t}})^{2}$ with degree 4 and total order 4, by integral by
parts, we have

$\begin{array}[]{ll}\int
u^{3p-6}u\frac{\partial^{2}u}{\partial^{2}x_{a,t}}(\frac{\partial u}{\partial
x_{a,t}})^{2}{\hbox{\rm{d}}}x_{t}\\\
=\int_{-\infty}^{\infty}\ldots\int_{-\infty}^{\infty}[u^{3p-5}\frac{\partial
u}{\partial x_{a,t}}(\frac{\partial u}{\partial
x_{a,t}})^{2}]\Big{|}_{x_{a,t}=-\infty}^{\infty}{\hbox{\rm{d}}}^{(a)}x_{t}\\\
-\int\frac{\partial u}{\partial x_{a,t}}[\frac{\partial}{\partial
x_{a,t}}(u^{3p-5}(\frac{\partial u}{\partial
x_{a,t}})^{2})]{\hbox{\rm{d}}}x_{t}\\\
\overset{\eqref{lem4}}{=}-\int\frac{\partial u}{\partial
x_{a,t}}[\frac{\partial}{\partial x_{a,t}}(u^{3p-5}(\frac{\partial u}{\partial
x_{a,t}})^{2})]{\hbox{\rm{d}}}x_{t}.\end{array}$ (16)

Then,

$\displaystyle\int
u^{3p-6}u\frac{\partial^{2}u}{\partial^{2}x_{a,t}}(\frac{\partial u}{\partial
x_{a,t}})^{2}+\frac{\partial u}{\partial x_{a,t}}[\frac{\partial}{\partial
x_{a,t}}(u^{3p-5}(\frac{\partial u}{\partial
x_{a,t}})^{2})]{\hbox{\rm{d}}}x_{t}$ (17) $\displaystyle=\int
u^{3p-6}[3p(\frac{\partial u}{\partial
x_{a,t}})^{4}+3u\frac{\partial^{2}u}{\partial^{2}x_{a,t}}(\frac{\partial
u}{\partial x_{a,t}})^{2}-5(\frac{\partial u}{\partial
x_{a,t}})^{4}]{\hbox{\rm{d}}}x_{t}=0.$

We obtain a 2th-order constraint: $R_{1,a,b}=3p(\frac{\partial u}{\partial
x_{a,t}})^{4}+3u\frac{\partial^{2}u}{\partial^{2}x_{a,t}}(\frac{\partial
u}{\partial x_{a,t}})^{2}-5(\frac{\partial u}{\partial x_{a,t}})^{4}$. The
other 27 constraints in ${\mathcal{C}}_{2,n}$ are obtained in the same way.

###  2.4 Proof of Lemma 2.1

We first prove several lemmas.

######  Lemma 2.4.

$\begin{array}[]{ll}\frac{{\hbox{\rm{d}}}H_{p}(u)}{{\hbox{\rm{d}}}t}=\frac{p}{1-p}\frac{\int
u^{p-1}\Delta u^{p}{\hbox{\rm{d}}}x_{t}}{\int u^{p}{\hbox{\rm{d}}}x_{t}},\\\
\end{array}$ (18)
$\frac{{\hbox{\rm{d}}}^{2}H_{p}(u)}{{\hbox{\rm{d}}}^{2}t}=\frac{p}{1-p}\frac{\frac{\partial}{\partial
t}(\int u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}-p(\int u^{p-1}\frac{\partial u}{\partial
t}{\hbox{\rm{d}}}x_{t})^{2}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}.$ (19)

###### Proof.

By the definition of $p$-Rényi entropy (1), we have

$\displaystyle\frac{{\hbox{\rm{d}}}H_{p}(u)}{{\hbox{\rm{d}}}t}$
$\displaystyle=\frac{p}{1-p}\frac{\int u^{p-1}\frac{\partial u}{\partial
t}{\hbox{\rm{d}}}x_{t}}{\int
u^{p}{\hbox{\rm{d}}}x_{t}}=\frac{p}{1-p}\frac{\int u^{p-1}\Delta
u^{p}{\hbox{\rm{d}}}x}{\int u^{p}{\hbox{\rm{d}}}x},$
$\displaystyle\frac{{\hbox{\rm{d}}}^{2}H_{p}(u)}{{\hbox{\rm{d}}}^{2}t}$
$\displaystyle=\frac{p}{1-p}\frac{\frac{\partial}{\partial t}(\int
u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}-\int u^{p-1}\frac{\partial u}{\partial
t}{\hbox{\rm{d}}}x_{t}\frac{\partial}{\partial t}(\int
u^{p}{\hbox{\rm{d}}}x_{t})}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle\\!\\!\\!\\!=\frac{p}{1-p}\frac{\frac{\partial}{\partial t}(\int
u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}-\int u^{p-1}\frac{\partial u}{\partial
t}{\hbox{\rm{d}}}x_{t}\int pu^{p-1}\frac{\partial u}{\partial
t}{\hbox{\rm{d}}}x_{t}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle\\!\\!\\!\\!=\frac{p}{1-p}\frac{\frac{\partial}{\partial t}(\int
u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}-p(\int u^{p-1}\frac{\partial u}{\partial
t}{\hbox{\rm{d}}}x_{t})^{2}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}.$

∎

######  Lemma 2.5.

We have

$\displaystyle\int u^{p-1}\Delta u^{p}{\hbox{\rm{d}}}x_{t}=\int\Delta
u^{p-1}u^{p}{\hbox{\rm{d}}}x_{t}.$ (20)

###### Proof.

Integrating by parts[21], we have

$\int u^{p-1}\Delta u^{p}{\hbox{\rm{d}}}x_{t}=-\int\nabla u^{p-1}\nabla
u^{p}{\hbox{\rm{d}}}x_{t}=\int\Delta u^{p-1}u^{p}{\hbox{\rm{d}}}x_{t}.$

∎

######  Lemma 2.6.

By Cauchy-Schwarz inequality we have

$\displaystyle(\int\Delta u^{p-1}u^{p}{\hbox{\rm{d}}}x_{t})^{2}\leq\int
u^{p}{\hbox{\rm{d}}}x_{t}\int(\Delta u^{p-1})^{2}u^{p}{\hbox{\rm{d}}}x_{t}.$
(21)

Then, we obtain

$\begin{array}[]{ll}\frac{{\hbox{\rm{d}}}^{2}}{{\hbox{\rm{d}}}^{2}t}N_{p}(u)&\\!\\!\\!=\frac{\mu}{n}\frac{{\hbox{\rm{d}}}^{2}H_{p}(u)}{{\hbox{\rm{d}}}^{2}t}{\rm
e}^{\frac{\mu}{n}H_{p}(u)}+\left(\frac{\mu}{n}\frac{{\hbox{\rm{d}}}H_{p}(u)}{{\hbox{\rm{d}}}t}\right)^{2}{\rm
e}^{\frac{\mu}{n}H_{p}(u)}\\\\[5.69046pt] &\\!\\!\\!=\frac{\mu}{n}{\rm
e}^{\frac{\mu}{n}H_{p}(u)}I_{2,n},\end{array}$ (22)

where
$I_{2,n}=\frac{{\hbox{\rm{d}}}^{2}H_{p}(u)}{{\hbox{\rm{d}}}^{2}t}+\dfrac{\mu}{n}\left(\frac{{\hbox{\rm{d}}}H_{p}(u)}{{\hbox{\rm{d}}}t}\right)^{2}$.
So, by (18),(19),we have

$\displaystyle I_{2,n}$
$\displaystyle=\frac{p}{1-p}\frac{\frac{\partial}{\partial t}(\int
u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}-p(\int u^{p-1}\frac{\partial u}{\partial
t}{\hbox{\rm{d}}}x_{t})^{2}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$ (23)
$\displaystyle+\frac{\mu}{n}(\frac{p}{1-p}\frac{\int u^{p-1}\Delta
u^{p}{\hbox{\rm{d}}}x_{t}}{\int u^{p}{\hbox{\rm{d}}}x_{t}})^{2}$
$\displaystyle=\frac{\mu p^{2}}{n(1-p)^{2}}\frac{(\int u^{p-1}\Delta
u^{p}{\hbox{\rm{d}}}x_{t})^{2}}{(\int
u^{p}{\hbox{\rm{d}}}x_{t})^{2}}+\frac{p}{1-p}\frac{\frac{\partial}{\partial
t}(\int u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle-\frac{p^{2}}{1-p}\frac{(\int u^{p-1}\Delta
u^{p}{\hbox{\rm{d}}}x_{t})^{2}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle=(\frac{\mu p^{2}}{n(1-p)^{2}}-\frac{p^{2}}{1-p})\frac{(\int
u^{p-1}\Delta u^{p}{\hbox{\rm{d}}}x_{t})^{2}}{(\int
u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle+\frac{p}{1-p}\frac{\frac{\partial}{\partial t}(\int
u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle\overset{\eqref{T1}}{=}\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}\frac{(\int\Delta
u^{p-1}u^{p}{\hbox{\rm{d}}}x_{t})^{2}}{(\int
u^{p}{\hbox{\rm{d}}}x_{t})^{2}}+\frac{p}{1-p}\frac{\frac{\partial}{\partial
t}(\int u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle\overset{(i)}{\leq}\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}\frac{\int
u^{p}{\hbox{\rm{d}}}x_{t}\int(\Delta
u^{p-1})^{2}u^{p}{\hbox{\rm{d}}}x_{t}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle+\frac{p}{1-p}\frac{\frac{\partial}{\partial t}(\int
u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\int
u^{p}{\hbox{\rm{d}}}x_{t}}{(\int u^{p}{\hbox{\rm{d}}}x_{t})^{2}}$
$\displaystyle=\frac{1}{\int
u^{p}{\hbox{\rm{d}}}x_{t}}\left(\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}\int(\Delta
u^{p-1})^{2}u^{p}{\hbox{\rm{d}}}x_{t}+\frac{p}{1-p}\frac{\partial}{\partial
t}(\int u^{p-1}\frac{\partial u}{\partial t}{\hbox{\rm{d}}}x_{t})\right)$
$\displaystyle=\frac{1}{\int
u^{p}{\hbox{\rm{d}}}x_{t}}\int\left(\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}(\Delta
u^{p-1})^{2}u^{p}+\frac{p}{1-p}\frac{\partial}{\partial
t}(u^{p-1}\frac{\partial u}{\partial t})\right){\hbox{\rm{d}}}x_{t}$
$\displaystyle=\frac{1}{\int u^{p}{\hbox{\rm{d}}}x_{t}}\int
F_{2,n}{\hbox{\rm{d}}}x_{t},$

where $F_{2,n}=\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}(\Delta
u^{p-1})^{2}u^{p}+\frac{p}{1-p}\frac{\partial}{\partial
t}(u^{p-1}\frac{\partial u}{\partial t})$.

Remark: In (23), the step $(i)$ is according to (21), and
$\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}\geq 0$ should be satisfied, which is
true under condition $p\geq 1-\frac{\mu}{n}$. When $\mu:=2+n(p-1)$,
$\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}\geq 0$ yields $p\geq 1-\frac{1}{n}$.
Savaré-Toscani [21] also used the inequality (21), but ignore the
nonnegativity of the coefficient $\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}$, thus
the parameter’s range $p>1-\frac{2}{n}$ in [21] should be corrected to $p\geq
1-\frac{1}{n}$.

Further, we can get

$\displaystyle F_{2,n}$
$\displaystyle=\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}u^{p}\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}[(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p-1})(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p-1})]$
(24) $\displaystyle+\frac{p}{1-p}\frac{\partial}{\partial
t}[u^{p-1}\sum\limits_{a=1}^{n}(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p})]$
$\displaystyle=\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}u^{p}\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}[(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p-1})(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p-1})]$
$\displaystyle+\frac{p}{1-p}\sum\limits_{a=1}^{n}[(p-1)u^{p-2}(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p})\frac{\partial
u}{\partial
t}+pu^{p-1}\frac{\partial^{2}}{\partial^{2}x_{a,t}}(u^{p-1}\frac{\partial
u}{\partial t})]$
$\displaystyle=\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}u^{p}\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}[(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p-1})(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p-1})]$
$\displaystyle+\frac{p}{1-p}\sum\limits_{a=1}^{n}\left[(p-1)u^{p-2}(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p})\sum\limits_{b=1}^{n}(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p})\right.$
$\displaystyle\left.+pu^{p-1}\frac{\partial^{2}}{\partial^{2}x_{a,t}}(u^{p-1}\sum\limits_{b=1}^{n}(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p}))\right]$
$\displaystyle=\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}u^{p}\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}[(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p-1})(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p-1})]$
$\displaystyle+\frac{p}{1-p}\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}\left[(p-1)u^{p-2}(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p})(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p})\right.$
$\displaystyle\left.+pu^{p-1}\frac{\partial^{2}}{\partial^{2}x_{a,t}}\left(u^{p-1}(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p})\right)\right]$
$\displaystyle=\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}\mathcal{T}_{a,b},$

where

$\displaystyle\mathcal{T}_{a,b}$
$\displaystyle=\frac{(\mu-n(1-p))p^{2}}{n(1-p)^{2}}u^{p}[(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p-1})(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p-1})]$
(25)
$\displaystyle+\frac{p}{1-p}\left[(p-1)u^{p-2}(\frac{\partial^{2}}{\partial^{2}x_{a,t}}u^{p})(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p})\right.$
$\displaystyle\left.+pu^{p-1}\frac{\partial^{2}}{\partial^{2}x_{a,t}}\left(u^{p-1}(\frac{\partial^{2}}{\partial^{2}x_{b,t}}u^{p})\right)\right].$

For convenience, introduce the notation
$u_{i,j}:=\frac{\partial^{i+j}u}{\partial^{i}x_{a,t}\partial^{j}x_{b,t}}$. By
simple computation, we have
$\mathcal{T}_{a,b}=-\frac{u^{3p-6}p^{2}}{(p-1)n}T_{a,b}$, and

$\displaystyle T_{a,b}$
$\displaystyle=4np^{4}u_{0,1}^{2}u_{1,0}^{2}+2np^{3}uu_{0,1}^{2}u_{2,0}+8np^{3}uu_{0,1}u_{1,0}u_{1,1}$
(26)
$\displaystyle+4np^{3}uu_{0,2}u_{1,0}^{2}-15np^{3}u_{0,1}^{2}u_{1,0}^{2}-\mu
p^{3}u_{0,1}^{2}u_{1,0}^{2}+2np^{2}u^{2}u_{0,1}u_{2,1}$
$\displaystyle+2np^{2}u^{2}u_{0,2}u_{2,0}+4np^{2}u^{2}u_{1,0}u_{1,2}+2np^{2}u^{2}u_{1,1}^{2}-3np^{2}uu_{0,1}^{2}u_{2,0}$
$\displaystyle-20np^{2}uu_{0,1}u_{1,0}u_{1,1}-8np^{2}uu_{0,2}u_{1,0}^{2}-\mu
p^{2}uu_{0,1}^{2}u_{2,0}-\mu p^{2}uu_{0,2}u_{1,0}^{2}$
$\displaystyle+16np^{2}u_{0,1}^{2}u_{1,0}^{2}+5\mu
p^{2}u_{0,1}^{2}u_{1,0}^{2}+npu^{3}u_{2,2}-2npu^{2}u_{0,1}u_{2,1}$
$\displaystyle-
npu^{2}u_{0,2}u_{2,0}-4npu^{2}u_{1,0}u_{1,2}-2npu^{2}u_{1,1}^{2}-\mu
pu^{2}u_{0,2}u_{2,0}$ $\displaystyle-
npuu_{0,1}^{2}u_{2,0}+12npuu_{0,1}u_{1,0}u_{1,1}+2npuu_{0,2}u_{1,0}^{2}+3\mu
puu_{0,1}^{2}u_{2,0}$ $\displaystyle+3\mu
puu_{0,2}u_{1,0}^{2}-npu_{0,1}^{2}u_{1,0}^{2}-8\mu
pu_{0,1}^{2}u_{1,0}^{2}-nu^{2}u_{0,2}u_{2,0}$ $\displaystyle+\mu
u^{2}u_{0,2}u_{2,0}+2nuu_{0,1}^{2}u_{2,0}+2nuu_{0,2}u_{1,0}^{2}-2\mu
uu_{0,1}^{2}u_{2,0}$ $\displaystyle-2\mu
uu_{0,2}u_{1,0}^{2}-4nu_{0,1}^{2}u_{1,0}^{2}+4\mu u_{0,1}^{2}u_{1,0}^{2},$

which is a 4th-order differential form.

From (22), (23), (24), and (25), we have

$\frac{{\hbox{\rm{d}}}^{2}}{{\hbox{\rm{d}}}^{2}t}N_{p}(u)\leq-\frac{p^{2}\mu}{n^{2}}e^{\frac{\mu}{n}H_{p}(u)}\frac{1}{\int
u^{p}{\hbox{\rm{d}}}x_{t}}\int u^{3p-6}E_{2,n}{\hbox{\rm{d}}}x_{t}$ (27)

where $E_{2,n}=\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}\frac{T_{a,b}}{p-1}$
and $T_{a,b}$ is defined in (26). Then the problem
$\frac{{\hbox{\rm{d}}}^{2}}{{\hbox{\rm{d}}}^{2}t}N_{p}(u)\leq 0$ can be
transformed to $\int u^{3p-6}E_{2,n}{\hbox{\rm{d}}}x_{t}\geq 0$. Thus, Lemma
2.1 is proved.

##  3 A generalized version of CREP

In this section, we prove a generalized CREP using the procedure given in
section 2.

######  Theorem 3.1.

Let $u(x_{t})$ be a probability density in ${\mathbb{R}}^{n}$ solving (6) and
satisfying (14). Then we give a formula $\Phi(n,p,\mu)$ such that the $p$-th
Rényi entropy power defined in (2) satisfies

$\displaystyle\frac{{\hbox{\rm{d}}}^{2}}{{\hbox{\rm{d}}}^{2}t}N_{p}(x_{t})\leq
0,$ (28)

under the condition $\Phi(n,p,\mu)$, that is $N_{p}(x_{t})$ is concave under
$\Phi(n,p,\mu)$.

The proof of the above theorem consists three steps which will be given in the
following subsections.

###  3.1 Reduce to a finite problem

We first give an inequality constraint. Denote that
$|\nabla^{2}f|^{2}=\sum_{i,j}(\frac{\partial^{2}f}{\partial x_{i}\partial
x_{j}})^{2}$. Then, based on the trace inequality
$|\nabla^{2}f|^{2}\geq\frac{1}{n}(\Delta f)^{2}$, we give a nonnegative
constraint:

$\displaystyle
I_{1}=\frac{u^{p}}{u^{3p-6}}\left[|\nabla^{2}u^{p-1}|^{2}-\frac{1}{n}(\Delta
u^{p-1})^{2}\right]=\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}I_{1,a,b}\geq 0$
(29)

where $I_{1,a,b}=u^{6-2p}\left[(\frac{\partial^{2}u^{p-1}}{\partial
x_{a,t}\partial
x_{b,t}})^{2}-\frac{1}{n}\frac{\partial^{2}u^{p-1}}{\partial^{2}x_{a,t}}\frac{\partial^{2}u^{p-1}}{\partial^{2}x_{b,t}}\right]$.

From (27) and (29), in order for (28) to be true, it suffices to solve
Problem I. Find a formula $\Phi(n,p,\mu)$ such that

$\begin{array}[]{l}{E}_{2,n}\geq\widetilde{E}_{2,n}=E_{2,n}+c_{1}I_{1}=\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}(\frac{1}{p-1}T_{a,b}+c_{1}I_{1,a,b})\geq
0\end{array}$

under the conditions $c_{1}\leq 0$, $p\geq 1-\frac{\mu}{n}$,
$R_{i,a,b}=0,i=1,\ldots,28$ given in (15).

Since
$\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}T_{a,b}=\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}T_{b,a}$
and $I_{1,a,b}=R^{(I)}_{1,b,a}$, we have

$\displaystyle\widetilde{E}_{2,n}$
$\displaystyle=\frac{1}{2}\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}[\frac{1}{p-1}(T_{a,b}+T_{b,a})+c_{1}(I_{1,a,b}+I_{1,b,a})]=\frac{1}{2}\sum\limits_{a=1}^{n}\sum\limits_{b=1}^{n}L_{a,b},$
(30)

where $L_{a,b}=\frac{1}{p-1}(T_{a,b}+T_{b,a})+c_{1}(I_{1,a,b}+I_{1,b,a})$.

###  3.2 Simplify the problem with the constraints

From (30), to solve Problem I, it suffices to solve
Problem II. Find a formula $\Phi(n,p,\mu)$ such that $L_{a,b}\geq 0$ under the
conditions $c_{1}\leq 0$, $p\geq 1-\frac{\mu}{n}$, and
$R_{i,a,b}=0,i=1,\ldots,{28}$.

In this section, we simplify $L_{a,b}$ in Problem II with the constraints.
Note that the subscripts $a$ and $b$ are fixed and will be treated as symbols.

Our goal is to reduce $L_{a,b}$ into a quadratic form in certain new
variables. The new variables are all the monomials in
${\mathbb{R}}[{\mathcal{V}}_{a,b}]$ with degree 2 and total order 2, where
${\mathcal{V}}_{a,b}$ is defined in (13).

$\displaystyle m_{1}=uu_{0,2},\ m_{2}=uu_{1,1},\ m_{3}=uu_{2,0}$
$\displaystyle m_{4}=u_{0,1}^{2},\ m_{5}=u_{1,0}u_{0,1},\ m_{6}=u_{1,0}^{2}.$

We will simplify the constraints in (15) as follows. A quadratic monomial in
$m_{i}$ is called a quadratic monomial. Write monomials in
${\mathcal{C}}_{2,n}=\\{R_{i},i=1,\ldots,N_{1}\\}$ as quadratic monomials if
possible. Doing Gaussian elimination to ${\mathcal{C}}_{2,n}$ by treating the
monomials as variables and according to a monomial order such that a quadratic
monomial is less than a non-quadratic monomial, we obtain

$\widetilde{{\mathcal{C}}}_{2,n}={{\mathcal{C}}}_{2,n,1}\cup{{\mathcal{C}}}_{2,n,2},$

where ${{{\mathcal{C}}}}_{2,n,1}$ is the set of quadratic forms in $m_{i}$,
${{{\mathcal{C}}}}_{2,n,2}$ is the set of non-quadratic forms, and
${\hbox{\rm{Span}}}_{\mathbb{R}}({\mathcal{C}}_{2,n})={\hbox{\rm{Span}}}_{\mathbb{R}}(\widetilde{{\mathcal{C}}}_{2,n})$.
We obtain ${{\mathcal{C}}}_{2,n,1}=\\{\widehat{R}_{i},i=1,\ldots,9\\}$ and
${{\mathcal{C}}}_{2,n,2}=\\{\widetilde{R}_{i},i=1,\ldots,13\\}$, where

$\begin{array}[]{ll}\widehat{R}_{1}=2m_{1}m_{5}+\frac{2(3p-5)}{3}m_{4}m_{5},\widehat{R}_{2}=m_{2}m_{6}+\frac{3p-5}{3}m_{5}m_{6},\\\
\widehat{R}_{3}=-6m_{3}m_{5}+2(5-3p)m_{5}m_{6},\widehat{R}_{4}=(3p-5)m_{4}^{2}+3m_{1}m_{4},\\\
\widehat{R}_{5}=(3p-5)m_{6}^{2}+3m_{3}m_{6},\widehat{R}_{6}=(3p-5)m_{4}m_{5}+3m_{2}m_{4},\\\
\widehat{R}_{7}=(3p-5)m_{5}^{2}+2m_{2}m_{5}+m_{3}m_{4},\\\
\widehat{R}_{8}=m_{1}m_{3}-m_{2}^{2}+\frac{9p-12}{2}m_{3}m_{4}+\frac{9p^{2}-27p+20}{2}m_{5}^{2},\\\
\widehat{R}_{9}=m_{1}m_{6}-m_{3}m_{4}.\\\
\widetilde{R}_{1}=u^{3}u_{0,4}+(3-3p)m_{1}^{2}+(9p^{3}-36p^{2}+47p-20)m_{4}^{2},\\\
\widetilde{R}_{2}=u^{3}u_{1,3}+(3-3p)m_{1}m_{2}+(9p^{3}-36p^{2}+47p-20)m_{4}m_{5},\\\
\widetilde{R}_{3}=u^{3}u_{3,1}+(3-3p)m_{2}m_{3}+(-9p^{2}+21p-12)m_{3}m_{5},\\\
\widetilde{R}_{4}=u^{3}u_{4,0}+(3-3p)m_{3}^{2}+(9p^{3}-36p^{2}+47p-20)m_{6}^{2},\\\
\widetilde{R}_{5}=u^{2}u_{0,1}u_{0,3}+m_{1}^{2}+\frac{-9p^{2}+27p-20}{3}m_{4}^{2},\\\
\widetilde{R}_{6}=u^{2}u_{0,1}u_{1,2}+m_{1}m_{2}+\frac{-9p^{2}+27p-20}{3}m_{4}m_{5},\\\
\widetilde{R}_{7}=u^{2}u_{0,1}u_{3,0}+m_{2}m_{3}+\frac{-9p^{2}+27p-20}{3}m_{5}m_{6},\\\
\widetilde{R}_{8}=u^{2}u_{1,0}u_{0,3}+m_{1}m_{2}+\frac{-9p^{2}+27p-20}{3}m_{4}m_{5},\\\
\widetilde{R}_{9}=u^{2}u_{1,0}u_{2,1}+m_{2}m_{3}+\frac{-9p^{2}+27p-20}{3}m_{5}m_{6},\\\
\widetilde{R}_{10}=u^{2}u_{1,0}u_{3,0}+m_{3}^{2}+\frac{-9p^{2}+27p-20}{3}m_{6}^{2},\\\
\widetilde{R}_{11}=u^{3}u_{2,2}+(3-3p)m_{2}^{2}+\frac{9p^{2}-21p+12}{2}m_{3}m_{4}+\frac{27p^{3}-108p^{2}+141p-60}{2}m_{5}^{2},\\\
\widetilde{R}_{12}=u^{2}u_{0,1}u_{2,1}+m_{2}^{2}+\frac{4-3p}{2}m_{3}m_{4}+\frac{-9p^{2}+27p-20}{2}m_{5}^{2},\\\
\widetilde{R}_{13}=u^{2}u_{1,0}u_{1,2}+m_{2}^{2}+\frac{4-3p}{2}m_{3}m_{4}+\frac{-9p^{2}+27p-20}{2}m_{5}^{2}.\end{array}$

We now simplify $L_{a,b}$ using ${\mathcal{C}}_{2,n,1}$ and
${\mathcal{C}}_{2,n,2}$. Eliminating the non-quadratic monomials in $L_{a,b}$
using ${\mathcal{C}}_{2,n,2}$, and doing further reduction by
${\mathcal{C}}_{2,n,1}$, we have

$\displaystyle\widehat{L}_{a,b}=L_{a,b}-2(p^{3}c_{1}+4np^{2}-4p^{2}c_{1}-6np+5pc_{1}-2c_{1})\widehat{R}_{7}$
(31)
$\displaystyle-\frac{2}{n}(2n^{2}p-p^{2}c_{1}+n^{2}-n\mu+2pc_{1}-c_{1})\widehat{R}_{8}$
$\displaystyle-\frac{1}{n}(6n^{2}p^{2}-2p^{3}c_{1}-5n^{2}p-2np\mu+8p^{2}c_{1}-4n^{2}+4n\mu-10pc_{1}+4c_{1})\widehat{R}_{9}$
$\displaystyle-\frac{2np}{p-1}\widetilde{R}_{11}-6np\widetilde{R}_{12}-6np\widetilde{R}_{13}$
$\displaystyle=(2np+2n-2\mu)m_{2}^{2}+(5np-5np^{2}+5p\mu+4n-4\mu)m_{3}m_{4}$
$\displaystyle+(18np^{2}-7np^{3}+7p^{2}\mu-3np-19p\mu-12n+12\mu)m_{5}^{2}$
$\displaystyle+\frac{c_{1}}{n}[{(2n-2)(p^{2}-2p+1)}m_{2}^{2}+(4n-2np+5p-4)(p^{2}-2p+1)m_{3}m_{4}$
$\displaystyle+(14np-4np^{2}+7p^{2}-12n-19p+12)(p^{2}-2p+1)m_{5}^{2}]$

In order for $\widehat{L}_{a,b}\geq 0$ to be true, we need to eliminate the
monomial $m_{3}m_{4}$ from $\widehat{L}_{a,b}$, which can be done with
$\widehat{R}_{7}$ as follows.

$\widehat{L}_{a,b}+p_{7}\widehat{R}_{7}=A_{1}m_{2}^{2}+A_{2}m_{2}m_{5}+A_{3}m_{5}^{2}$
(32)

where

$\displaystyle
p_{7}=(2np^{3}c_{1}+5n^{2}p^{2}-8np^{2}c_{1}-5p^{3}c_{1}-5n^{2}p-5np\mu+10npc_{1}$
$\displaystyle\quad+14p^{2}c_{1}-4n^{2}+4n\mu-4nc_{1}-13pc_{1}+4c_{1})/n.$
$\displaystyle
A_{1}=-2c_{1}p^{2}/n+4c_{1}p/n+2np+2c_{1}+2c_{1}p^{2}-4c_{1}p-2c_{1}/n-2\mu+2n,$
$\displaystyle
A_{2}=4c_{1}p^{3}-16c_{1}p^{2}-10c_{1}p^{3}/n-10p\mu+20c_{1}p+28c_{1}p^{2}/n-26c_{1}p/n$
$\displaystyle\quad+10np^{2}-10np+8\mu-8c_{1}+8c_{1}/n-8n,$ $\displaystyle
A_{3}=-8\mu+8n+26c_{1}p^{2}-24c_{1}p-8c_{1}/n-12c_{1}p^{3}+2c_{1}p^{4}+8c_{1}-52c_{1}p^{2}/n$
$\displaystyle\quad+34c_{1}p/n+34c_{1}p^{3}/n-8c_{1}p^{4}/n+18p\mu+8np^{3}-22np^{2}-8p^{2}\mu+10np.$

###  3.3 Compute $\Phi(n,p,\mu)$

From (32), in order to solve Problem II, it suffices to solve

Problem III: Find a formula $\Phi(n,p,\mu)$ such that

$\Phi(n,p,\mu)=\exists c_{1}\forall m_{1},m_{2},m_{3}(c_{1}\leq 0\wedge p\geq
1-\frac{\mu}{n}\wedge A_{1}m_{2}^{2}+A_{2}m_{2}m_{5}+A_{3}m_{5}^{2}\geq 0).$
(33)

In principle, Problem III can be solved with the quantifier elimination [22].
In this paper, the problem is special and an explicit proof is given.

By the knowledge of linear algebra, we have
$A_{1}m_{2}^{2}+A_{2}m_{2}m_{5}+A_{3}m_{5}^{2}\geq 0$ is equivalent to
$\Delta_{1}=A_{1}=\frac{2}{n}s_{1}\geq 0$,
$\Delta_{2}=A_{3}=\frac{2}{n}s_{2}\geq 0$,
$\Delta_{3}=A_{1}A_{3}-\frac{1}{4}A_{2}^{2}=\frac{p}{n^{2}}s_{3}\geq 0$, where

$\displaystyle s_{1}=(p-1)^{2}(n-1)c_{1}+n^{2}(p+1)-n\mu,$ $\displaystyle
s_{2}=(p-1)^{2}(n(p-2)^{2}-4p^{2}+9p-4)c_{1}$
$\displaystyle+n^{2}(4p^{3}-11p^{2}+5p+4)-(4p^{2}-9p+4)n\mu,$ $\displaystyle
s_{3}=(4-9p)n^{2}(\mu-\mu_{3})(\mu-\mu_{4}),$

where $\mu_{3}$ and $\mu_{4}$ are defined in (37) and $p\neq\frac{4}{9}$ is
assumed in $\mu_{4}$. We thus proved

######  Lemma 3.2.

We can eliminate $m_{1},m_{2},m_{3}$ in (33):

$\Phi(n,p,\mu)=\exists c_{1}(s_{1}\geq 0\wedge s_{2}\geq 0\wedge s_{3}\geq
0\wedge c_{1}\leq 0\wedge p\geq 1-\mu/n).$ (34)

We will give an explicit formula for $\Phi$ in (34). First, introduce the
following parameters.

$\begin{array}[]{lll}n_{1}=\frac{9-\sqrt{17}}{8},&n_{2}=\frac{9+\sqrt{17}}{8},&n_{3}=(\sqrt{17}+1)/2,\\\
\theta_{1}=-\frac{2n}{(p-1)^{2}},&\theta_{2}=\frac{2n^{2}p(9p-13)}{(p-1)^{2}(4n+9p-4)},&\theta_{3}=(\sqrt{17}-9)n,\\\
\theta_{4}=\frac{n^{2}p-n^{2}-n\mu}{(p-1)^{2}},&\theta_{5}=\frac{n^{2}(9p^{2}-13p-4)-n(9p-4)\mu}{(p-1)^{2}(4n+9p-4)},&\theta_{6}=\frac{-4n(\sqrt{17}-1)\mu+8n^{2}}{\sqrt{17}+1},\\\
\theta_{7}=n(1-p),&\theta_{8}=5n/9,&\theta_{9}=-\frac{162n}{25},\\\
\theta_{10}=\frac{64n}{\sqrt{17}-9},&\theta_{11}=\frac{8(9\sqrt{17}+23)n^{2}}{-32n-49+9\sqrt{17}},&\theta_{12}=-\frac{16(11\sqrt{17}+47)n\mu+152n^{2}}{26\sqrt{17}n+73\sqrt{17}+118n+305},\\\
\theta_{13}=-\frac{4n(\mu\sqrt{17}+2n+\mu)}{\sqrt{17}-1},&\theta_{14}=-\frac{8n(22\mu\sqrt{17}-19n-94\mu)}{26\sqrt{17}n+73\sqrt{17}-118n-305},&\theta_{15}=-\frac{9}{5}n^{2}-\frac{81}{25}\mu
n,\\\ \theta_{16}=(\sqrt{17}-1)n/8,&&\\\ \phi_{1}\triangleq p\geq
1-\frac{1}{n},&\phi_{2}\triangleq\mu=2+n(p-1),&\phi_{3}\triangleq
p>1-\frac{1}{n}.\\\ \end{array}$ (35)

$p>n_{2}$ | $\theta_{4}>\theta_{1}\wedge\theta_{5}\leq 0$ | * | $\phi_{1}\wedge\phi_{2}$
---|---|---|---
$p=n_{2}$ | $\theta_{6}>\theta_{3}\wedge\theta_{12}\leq 0$ | * | $\phi_{1}\wedge\phi_{2}$
$\frac{13}{9}<p<n_{2}$ | $\theta_{4}>\theta_{1}\wedge\theta_{5}\leq 0$ | * | $\phi_{1}\wedge\phi_{2}$
$n_{1}<p\leq\frac{13}{9}$, $p\neq 1$ | $\phi_{3}\wedge\theta_{4}>\theta_{1}\wedge\theta_{5}\leq\theta_{2}$ | $\phi_{3}\wedge\theta_{7}\leq\mu\wedge\theta_{4}>\theta_{2}$ | $\phi_{1}\wedge\phi_{2}$
$p=n_{1}$ | $n<n_{3}\wedge\theta_{13}>\theta_{10}\wedge\theta_{14}\leq\theta_{11}$ | $n<n_{3}\wedge\theta_{13}>\theta_{11}\wedge\mu\geq\theta_{16}$ | $\phi_{1}\wedge\phi_{2}$
$\frac{4}{9}<p<n_{1}$ | $\phi_{3}\wedge\theta_{4}>\theta_{1}\wedge\theta_{5}\leq\theta_{2}$ | $\phi_{3}\wedge\theta_{7}\leq\mu\wedge\theta_{4}>\theta_{2}$ | $\phi_{1}\wedge\phi_{2}$
$p=\frac{4}{9}$ | $n=1\wedge\theta_{15}>\theta_{9}\wedge\mu\geq\theta_{8}$ | * | $n=1\wedge-p\leq\mu-1\leq p$
$0<p<\frac{4}{9}$ | $n=1\wedge\theta_{7}\leq\mu\wedge\theta_{4}>\theta_{1}$ | $n=1\wedge\theta_{5}<\theta_{1}\wedge\mu\geq\theta_{7}$ | $n=1\wedge-p\leq\mu-1\leq p$

Table 1: The description for $\Phi$ in (34)

With these notations, we introduce the conditions for defining $\Phi$ in Table
1, where $*$ means $\varnothing$. Define $T(i,j)$ to be the formula in the
$i$-th row and the $j$-th column in Table 1. Then we denote

$\mathbb{T}(i,j)\triangleq T(i,1)\wedge T(i,j)\hbox{ for
}i=1,\ldots,8,j=2,\ldots,4.$ (36)

For examples,
$\mathbb{T}(1,2)=(p>n_{2}\wedge\theta_{4}>\theta_{1}\wedge\theta_{5}\leq 0)$,
which means that if $p,n,\mu$ satisfy $\mathbb{T}(1,2)$ then there exists a
$c_{1}\leq 0$ such that (33) is true and the CREP is valid.
$\mathbb{T}(1,3)=\varnothing$, which means that there exist no values for
$p,n,\mu$ such that (33) and the CREP are true.

We now give the main result of the paper, which implies Theorem 3.1. The proof
for the theorem can be found in section 3.5. Also, from the proof we can see
that for the cases $\mathbb{T}(i,4),i=1,\ldots,8$, the corresponding value for
$c_{1}$ is $c_{1}=\theta_{1}$.

######  Theorem 3.3.

The sufficient and necessary condition for Problem III, that is, (33) to be
true, is

$\Phi(n,p,\mu)=\vee_{i=1}^{8}\vee_{j=2}^{4}\mathbb{T}(i,j),$

where $\mathbb{T}(i,j)$ is defined in (36) and $\vee$ means disjunction.

###  3.4 Compare with existing results

We will show that our result includes the result proved in [21], and
essentially more results.

In [21], CREP was proved under the conditions $\mu=2+n(p-1)$ and $p\geq
1-\frac{1}{n}$. Obviously, the result proved in [21] corresponds to
$\mathbb{T}(i,4),i=1,\ldots,8$ in Table 1.

We can also prove the result in [21] directly as follows. Set $\mu=2+n(p-1)$
and $c_{1}=\theta_{1}\leq 0$ in (31), we obtain $\widehat{L}_{a,b}=0$. Also,
the condition $p\geq 1-\frac{\mu}{n}$ implies $p\geq 1-\frac{1}{n}$. So, when
$\mu=2+n(p-1)$ and $p\geq 1-\frac{1}{n}$, the CREP is proved based on our
proof procedure.

We can use the SDP code in [19][APPENDIX B] to verify the result in Table 1.
For instance, for $\mu=2,p=\frac{11}{5},n=2$, the condition $p\geq
1-\frac{\mu}{n}$ is satisfied naturally. With the SDP code in [19], we obtain
$\widehat{L}_{a,b}+\frac{172}{25}\widehat{R}_{7}=(2\sqrt{2}m_{2}+\frac{344}{100\sqrt{2}}m_{5})^{2}+\frac{22}{625}m_{5}^{2}\geq
0$ with $c_{1}=-\frac{5}{9}$. Thus, the CREP is proved when
$\mu=2,p=\frac{11}{5},n=2$. This case
$[\mu=2,p=\frac{11}{5},n=2,c_{1}=-\frac{5}{9}]$ is included in
$\mathbb{T}(1,1)$ in Table 1. Note that $\mu=2+n(p-1)$ is not satisfied for
these parameters.

###  3.5 Proof of Theorem 3.3

Introduce more parameters.

$\displaystyle\mu_{1}$ $\displaystyle=((p-1)^{2}(n-1)c_{1}+n^{2}(p+1))/n,$
(37) $\displaystyle\mu_{2}$
$\displaystyle=((p-1)^{2}(n(p-2)^{2}-4p^{2}+9p-4)c_{1}+n^{2}(4p^{3}-11p^{2}+5p+4))/(n(4p^{2}-9p+4))$
$\displaystyle\mu_{3}$
$\displaystyle=(n^{2}p-p^{2}c_{1}-n^{2}+2pc_{1}-c_{1})/n,$
$\displaystyle\mu_{4}$
$\displaystyle=(9n^{2}p^{2}-4np^{2}c_{1}-9p^{3}c_{1}-13n^{2}p+8npc_{1}+22p^{2}c_{1}-4n^{2}-4nc_{1}-17pc_{1}+4c_{1})/(n(9p-4)),$
$\displaystyle\mu_{5}$
$\displaystyle=-(nc_{1}\sqrt{17}-c_{1}\sqrt{17}+136n^{2}+17nc_{1}-17c_{1})/(4n(\sqrt{17}-17)),$
$\displaystyle\mu_{6}$
$\displaystyle=-(c_{1}\sqrt{17}-8n^{2}+c_{1})/(4n(\sqrt{17}-1)),$
$\displaystyle\mu_{7}$
$\displaystyle=-(26nc_{1}\sqrt{17}+73c_{1}\sqrt{17}+152n^{2}+118nc_{1}+305c_{1})/(16n(11\sqrt{17}+47)),$
$\displaystyle\mu_{8}$
$\displaystyle=-(nc_{1}\sqrt{17}-c_{1}\sqrt{17}-136n^{2}-17nc_{1}+17c_{1})/(4n(\sqrt{17}+17)),$
$\displaystyle\mu_{9}$
$\displaystyle=-(c_{1}\sqrt{17}+8n^{2}-c_{1})/(4n(\sqrt{17}+1)),$
$\displaystyle\mu_{10}$
$\displaystyle=-(26nc_{1}\sqrt{17}+73c_{1}\sqrt{17}-152n^{2}-118nc_{1}-305c_{1})/(16n(11\sqrt{17}-47)),$
$\displaystyle\mu_{11}$ $\displaystyle=(117n^{2}+25nc_{1}-25c_{1})/(81n),$
$\displaystyle\mu_{12}$
$\displaystyle=(7218n^{2}+1225nc_{1}-400c_{1})/(1296n),$
$\displaystyle\mu_{13}$ $\displaystyle=-(5(9n^{2}+5c_{1}))/(81n),$
$\displaystyle\eta_{1}$
$\displaystyle=\frac{2n^{2}}{p-1},\eta_{2}=-\frac{16n^{2}}{\sqrt{17}-1},\eta_{3}=-\frac{18}{5}n^{2}.$

We first treat the three inequalities $s_{1}\geq 0$, $s_{2}\geq 0$, $s_{3}\geq
0$. Firstly, $s_{1}\geq 0$ is equivalent to $\mu\leq\mu_{1}$. Secondly, since
the roots of $4p^{2}-9p+4=0$ are $n_{1}$ and $n_{2}$, we have $s_{2}\geq
0\Leftrightarrow\mu\leq\mu_{2}$ if $p<n_{1}$ or $p>n_{2}$; and $s_{2}\geq
0\Leftrightarrow\mu\geq\mu_{2}$ if $n_{1}<p<n_{2}$. In order to analyse
$s_{3}\geq 0$, we first compute

$\mu_{3}-\mu_{4}=\frac{4((p-1)^{2}c_{1}+2n)}{9p-4}.$ (38)

Therefore, $s_{3}\geq 0$ can be divided into four cases: $s_{3}\geq
0\Leftrightarrow\mu_{4}\leq\mu\leq\mu_{3}$ if $p>\frac{4}{9}$ and
$\theta_{1}<c_{1}$; $s_{3}\geq 0\Leftrightarrow\mu_{3}\leq\mu\leq\mu_{4}$ if
$p>\frac{4}{9}$ and $c_{1}<\theta_{1}$; $s_{3}\geq
0\Leftrightarrow\mu\geq\mu_{3}\ {\rm or}\ \mu\leq\mu_{4}$ if $p<\frac{4}{9}$
and $c_{1}<\theta_{1}$; $s_{3}\geq 0\Leftrightarrow\mu\geq\mu_{4}\ {\rm or}\
\mu\leq\mu_{3}$ if $p<\frac{4}{9}$ and $\theta_{1}<c_{1}$. Finally, $p\geq
1-\frac{\mu}{n}$ is equivalent to $\mu\geq\theta_{7}$.

Based on the above analysis and (34), $\Phi(n,p,\mu)$ can be divided into six
cases:

$\begin{array}[]{l}\Phi(n,p,\mu)\Leftrightarrow\max(\mu_{4},\theta_{7})\leq\mu\leq\min(\mu_{1},\mu_{2},\mu_{3}),\hbox{
if }(p\in(\frac{4}{9},n_{1})\hbox{ or }p>n_{2})\hbox{ and
}\theta_{1}<c_{1}\leq 0;\\\
\Phi(n,p,\mu)\Leftrightarrow\max(\mu_{2},\mu_{4},\theta_{7})\leq\mu\leq\min(\mu_{1},\mu_{3}),\hbox{
if }p\in(n_{1},n_{2})\hbox{ or }\theta_{1}<c_{1}\leq 0;\\\
\Phi(n,p,\mu)\Leftrightarrow\max(\mu_{3},\theta_{7})\leq\mu\leq\min(\mu_{1},\mu_{2},\mu_{4}),\hbox{
if }(p\in(\frac{4}{9},n_{1})\hbox{ or }p>n_{2})\hbox{ or }c_{1}<\theta_{1};\\\
\Phi(n,p,\mu)\Leftrightarrow\max(\mu_{2},\mu_{3},\theta_{7})\leq\mu\leq\min(\mu_{1},\mu_{4}),\hbox{
if }p\in(n_{1},n_{2})\hbox{ or }c_{1}<\theta_{1};\\\
\Phi(n,p,\mu)\Leftrightarrow\theta_{7}\leq\mu\leq\min(\mu_{1},\mu_{2},\mu_{4})\hbox{
or }\max(\mu_{3},\theta_{7})\leq\mu\leq\min(\mu_{1},\mu_{2}),\hbox{ if
}p<\frac{4}{9}\hbox{ or }c_{1}<\theta_{1};\\\
\Phi(n,p,\mu)\Leftrightarrow\theta_{7}\leq\mu\leq\min(\mu_{1},\mu_{2},\mu_{3})\hbox{
or }\max(\mu_{4},\theta_{7})\leq\mu\leq\min(\mu_{1},\mu_{2}),\hbox{ if
}p<\frac{4}{9}\hbox{ or }\theta_{1}<c_{1}\leq 0.\end{array}$ (39)

The special cases $p=\frac{4}{9},n_{1},n_{2}$ and $c=\theta_{1}$ need to be
considered separately. Also, we omit $\exists c_{1}$ in the above
formulations.

In what below, we will give detailed analysis of the above six cases which
leads to the results in Table 1. We first have the formulas:

$\displaystyle\mu_{1}-\mu_{3}$ $\displaystyle=$
$\displaystyle(p-1)^{2}c_{1}+2n,$ (40) $\displaystyle\mu_{1}-\mu_{4}$
$\displaystyle=$ $\displaystyle\frac{9p((p-1)^{2}c_{1}+2n)}{9p-4},$ (41)
$\displaystyle\mu_{2}-\mu_{3}$ $\displaystyle=$
$\displaystyle\frac{2(p-2)^{2}((1/2)(p-1)^{2}c_{1}+n)}{(4p^{2}-9p+4)},$ (42)
$\displaystyle\mu_{2}-\mu_{4}$ $\displaystyle=$
$\displaystyle\frac{p(3p-4)^{2}((p-1)^{2}c_{1}+2n)}{(4p^{2}-9p+4)(9p-4)},$
(43) $\displaystyle\mu_{4}-\theta_{7}$ $\displaystyle=$
$\displaystyle\frac{-(p-1)^{2}(4n+9p-4)c_{1}+2n^{2}p(9p-13)}{n(9p-4)},$ (44)
$\displaystyle\mu_{3}-\theta_{7}$ $\displaystyle=$
$\displaystyle\frac{(p-1)(2n^{2}-pc_{1}+c_{1})}{n},$ (45)
$\displaystyle\theta_{2}-\theta_{1}$ $\displaystyle=$
$\displaystyle\frac{18n(np-n+1)(p-\frac{4}{9})}{(p-1)^{2}(4n+9p-4)},$ (46)
$\displaystyle\theta_{1}-\eta_{1}$ $\displaystyle=$
$\displaystyle-\frac{2n(np-n+1)}{(p-1)^{2}},$ (47)
$\displaystyle\eta_{1}-\theta_{2}$ $\displaystyle=$
$\displaystyle\frac{8n^{2}(np-n+1)}{(p-1)^{2}(4n+9p-4)}.$ (48)

We also have the following formulas which will be used to eliminate $c_{1}$ in
the proof.

$\begin{array}[]{ll}\mu\leq\mu_{3}\Leftrightarrow
c_{1}\leq\theta_{4},&\mu\geq\mu_{4}\Leftrightarrow c_{1}\geq\theta_{5},\hbox{
if }p>\frac{4}{9},\\\ \mu\geq\mu_{4}\Leftrightarrow c_{1}\leq\theta_{5},\hbox{
if }p<\frac{4}{9},&\mu\leq\mu_{4}\Leftrightarrow c_{1}\leq\theta_{5},\hbox{ if
}p>\frac{4}{9},\\\ \mu\leq\mu_{4}\Leftrightarrow c_{1}\geq\theta_{5},\hbox{ if
}p<\frac{4}{9},&\mu\leq\mu_{6}\Leftrightarrow c_{1}\leq\theta_{6},\\\
\mu\geq\mu_{7}\Leftrightarrow
c_{1}\geq\theta_{12},&\mu\leq\mu_{9}\Leftrightarrow c_{1}\leq\theta_{13},\\\
\mu\geq\mu_{10}\Leftrightarrow
c_{1}\geq\theta_{14},&\mu\leq\mu_{13}\Leftrightarrow c_{1}\leq\theta_{15}.\\\
\end{array}$ (49)

We divide the proof into several cases, first according to the values of
$c_{1}$ and then according to the values of $n$.

Case 1: $\theta_{1}<c_{1}\leq 0$. From (40), we have $\mu_{1}>\mu_{3}$ in this
case and from (39), $\Phi(n,p,\mu)$ simplifies to three cases:

$\begin{array}[]{l}\Phi(n,p,\mu)\Leftrightarrow\max(\mu_{4},\theta_{7})\leq\mu\leq\min(\mu_{2},\mu_{3}),\hbox{
if }p\in(\frac{4}{9},n_{1})\hbox{ or }p>n_{2};\\\
\Phi(n,p,\mu)\Leftrightarrow\max(\mu_{2},\mu_{4},\theta_{7})\leq\mu\leq\mu_{3},\hbox{
if }p\in(n_{1},n_{2});\\\
\Phi(n,p,\mu)\Leftrightarrow\theta_{7}\leq\mu\leq\min(\mu_{2},\mu_{3})\hbox{
or }\max(\mu_{4},\theta_{7})\leq\mu\leq\min(\mu_{1},\mu_{2}),\hbox{ if
}p<\frac{4}{9}.\end{array}$ (50)

According to the vales of $p$, we consider seven cases below.

Case 1.1: $\theta_{1}<c_{1}\leq 0$ and $p>n_{2}$. In this case, from (42) and
(44), we have $\mu_{2}\geq\mu_{3}$ and $\mu_{4}>\theta_{7}$. From (50), we
have $\Phi(n,p,\mu)\Leftrightarrow\mu_{4}\leq\mu\leq\mu_{3}$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p>n_{2}\wedge\theta_{1}<c_{1}\leq 0\wedge\mu_{4}\leq\mu\leq\mu_{3})$. By
(49), $\mu_{4}\leq\mu\leq\mu_{3}$ is equivalent to $\theta_{5}\leq
c_{1}\leq\theta_{4}$. $\exists c_{1}(\theta_{5}\leq
c_{1}\leq\theta_{4}\wedge\theta_{1}<c_{1}\leq 0$) is equivalent to
($\theta_{4}>\theta_{1}\wedge\theta_{5}\leq 0$). Therefore, in this case
$\Phi(n,p,\mu)\Leftrightarrow(p>n_{2}\wedge\theta_{4}>\theta_{1}\wedge\theta_{5}\leq
0)$, and $\mathbb{T}(1,2)$ is proved.

Case 1.2: $\theta_{1}<c_{1}\leq 0$ and $p=n_{2}$. When $p=n_{2}$, we have
$\theta_{1}=\theta_{3},\
s_{2}=-\frac{1}{1024}(7\sqrt{17}-33)(c_{1}\sqrt{17}+64n+9c_{1})n$. Then
$s_{2}\geq 0\Leftrightarrow c_{1}\geq\theta_{3}$. Because
$\theta_{1}<c_{1}\leq 0$ and $p=n_{2}>\frac{4}{9}$, we have $s_{3}\geq
0\Leftrightarrow\mu_{4}\leq\mu\leq\mu_{3}$. By (44), we have
$\mu_{4}>\theta_{7}$. When $p=n_{2}$, we have $\mu_{3}=\mu_{6}$ and
$\mu_{4}=\mu_{7}$. Thus $\Phi(n,p,\mu)\Leftrightarrow(\theta_{3}<c_{1}\leq
0,\mu_{7}\leq\mu\leq\mu_{6})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p=n_{2}\wedge\theta_{3}<c_{1}\leq 0\wedge\mu_{7}\leq\mu\leq\mu_{6})$. By
(49), $\mu_{7}\leq\mu\leq\mu_{6}$ is equivalent to $\theta_{12}\leq
c_{1}\leq\theta_{6}$. $\exists c_{1}(\theta_{12}\leq
c_{1}\leq\theta_{6}\wedge\theta_{3}<c_{1}\leq 0)$ is equivalent to
$\theta_{6}>\theta_{3}$ and $\theta_{12}\leq 0$. Therefore, in this case
$\Phi(n,p,\mu)\Leftrightarrow(p=n_{2}\wedge\theta_{6}>\theta_{3}\wedge\theta_{12}\leq
0)$, and $\mathbb{T}(2,2)$ is proved.

Case 1.3: $\theta_{1}<c_{1}\leq 0$ and $p\in(n_{1},n_{2})$, $p\neq 1$. Due to
(44), this case is divided into two sub-cases.

Case 1.3.1: $\theta_{1}<c_{1}\leq 0$ and $p\in(\frac{13}{9},n_{2})$. By (43)
and (44), we have $\mu_{4}>\mu_{2}$ and $\mu_{4}>\theta_{7}$. From (50), we
have $\Phi(n,p,\mu)\Leftrightarrow\mu_{4}\leq\mu\leq\mu_{3}$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p\in(\frac{13}{9},n_{2})\wedge\theta_{1}<c_{1}\leq
0\wedge\mu_{4}\leq\mu\leq\mu_{3})$. Similar to Case 1.1, we have
$\Phi(n,p,\mu)\Leftrightarrow(p\in(\frac{13}{9},n_{2})\wedge\theta_{4}>\theta_{1}\wedge\theta_{5}\leq
0)$, $\mathbb{T}(3,2)$ is proved.

Case 1.3.2: $\theta_{1}<c_{1}\leq 0$ and $p\in(n_{1},\frac{13}{9}]$, $p\neq
1$. By (43), (45), (44) and (46), we have $\mu_{4}\geq\mu_{2}$,
($\mu_{3}\geq\theta_{7}\Leftrightarrow c_{1}\leq\eta_{1}$),
($\mu_{4}\geq\theta_{7}\Leftrightarrow c_{1}\leq\theta_{2}$) and
($\theta_{2}>\theta_{1}\Leftrightarrow\phi_{3}$). Hence
$\Phi(n,p,\mu)\Leftrightarrow\max(\mu_{4},\theta_{7})\leq\mu\leq\mu_{3}$. This
case is further divided into two sub-cases.

Case 1.3.2.1: If $c_{1}\leq\theta_{2}$, then $\mu_{4}\geq\mu_{5}$, and
$\Phi(n,p,\mu)\Leftrightarrow\mu_{4}\leq\mu\leq\mu_{3}$. So we need
$\theta_{1}<\theta_{2}$, which yields $\phi_{3}$. From (50), we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{1}<c_{1}\leq\theta_{2},\phi_{3},\mu_{4}\leq\mu\leq\mu_{3})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p\in(n_{1},\frac{13}{9})\wedge p\neq
1\wedge\phi_{3}\wedge\theta_{1}<c_{1}\leq\theta_{2}\wedge\mu_{4}\leq\mu\leq\mu_{3})$.
Like Case 1.1, we have
$\Phi(n,p,\mu)\Leftrightarrow(p\in(n_{1},\frac{13}{9})\wedge p\neq
1\wedge\phi_{3}\wedge\theta_{4}>\theta_{1}\wedge\theta_{5}\leq\theta_{2})$,
and $\mathbb{T}(4,2)$ is proved

Case 1.3.2.2: If $c_{1}\geq\theta_{2}$, then $\mu_{4}\leq\theta_{7}$, and
$\Phi(n,p,\mu)\Leftrightarrow\theta_{7}\leq\mu\leq\mu_{3}$. So we need
$\theta_{7}\leq\mu_{3}$, which yields $c_{1}\leq\eta_{1}$. By (47), we know
$\eta_{1}>\theta_{1}$ results $\phi_{3}$, which yields
$\theta_{1}<\theta_{2}<\eta_{1}$. From (50), we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{2}<c_{1}\leq\min(0,\eta_{1}),\phi_{3},\theta_{7}\leq\mu\leq\mu_{3})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p\in(n_{1},\frac{13}{9})\wedge p\neq
1\wedge\phi_{3}\wedge\theta_{2}<c_{1}\leq\min(0,\eta_{1})\wedge\theta_{7}\leq\mu\leq\mu_{3})$.
$\theta_{7}\leq\mu\leq\mu_{3}$ is equivalent to $\theta_{7}\leq\mu$ and
$c_{1}\leq\theta_{4}$. $\exists
c_{1}(c_{1}\leq\theta_{4}\wedge\theta_{2}<c_{1}\leq\min(0,\eta_{1})$) is
equivalent to $\theta_{4}>\theta_{2}$. Therefore, in this case
$\Phi(n,p,\mu)\Leftrightarrow(p\in(n_{1},\frac{13}{9})\wedge p\neq
1\wedge\phi_{3}\wedge\theta_{7}\leq\mu\wedge\theta_{4}>\theta_{2})$, and
$\mathbb{T}(4,3)$ is proved.

Case 1.4: $\theta_{1}<c_{1}\leq 0$ and $p=n_{1}$. When $p=n_{1}$, we have
$\theta_{1}=\theta_{10},\ \theta_{2}=\theta_{11},\ \eta_{1}=\eta_{2},\
s_{2}=-\frac{1}{1024}(33+7\sqrt{17})(c_{1}\sqrt{17}-64n-9c_{1})n$. Then
$s_{2}\geq 0\Leftrightarrow c_{1}\geq\theta_{10}$. Because
$\theta_{1}<c_{1}\leq 0$ and $p=n_{1}>\frac{4}{9}$, we have $s_{3}\geq
0\Leftrightarrow\mu_{4}\leq\mu\leq\mu_{3}$. By (44), we have
$\mu_{4}\geq\theta_{7}\Leftrightarrow c_{1}\leq\theta_{2}$. This case is
divided into two sub-cases.

Case 1.4.1: Similar to Case 1.3.2.1,
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{1}<c_{1}\leq\theta_{2}\wedge\phi_{3}\wedge\mu_{4}\leq\mu\leq\mu_{3})$.
When $p=n_{1}$, we have $\mu_{3}=\mu_{9},\ \mu_{4}=\mu_{10},\
\phi_{3}\Leftrightarrow n<n_{3}$. From (50), we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{10}<c_{1}\leq\theta_{11},n<n_{3},\mu_{10}\leq\mu\leq\mu_{9})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p=n_{1}\wedge\theta_{10}<c_{1}\leq\theta_{11}\wedge
n<n_{3}\wedge\mu_{10}\leq\mu\leq\mu_{9})$. $\mu_{10}\leq\mu\leq\mu_{9}$ is
equivalent to $\theta_{14}\leq c_{1}\leq\theta_{13}$. $\exists
c_{1}(\theta_{14}\leq
c_{1}\leq\theta_{13}\wedge\theta_{10}<c_{1}\leq\theta_{11})$ is equivalent to
$\theta_{13}>\theta_{10}$ and $\theta_{14}\leq\theta_{11}$. Therefore, in this
case $\Phi(n,p,\mu)\Leftrightarrow(p=n_{1}\wedge
n<n_{3}\wedge\theta_{13}>\theta_{10}\wedge\theta_{14}\leq\theta_{11})$, and
$\mathbb{T}(5,2)$ is proved.

Case 1.4.2: Similar to Case 1.3.2.2,
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{2}<c_{1}\leq\min(0,\eta_{1})\wedge\phi_{3}\wedge\mu_{5}\leq\mu\leq\mu_{3})$.
When $p=n_{1}$, we have $\theta_{7}=\theta_{16}$. From (50), we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{11}<c_{1}\leq\eta_{2},n<n_{3},\theta_{16}\leq\mu\leq\mu_{9})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p=n_{1}\wedge\theta_{11}<c_{1}\leq\eta_{2}\wedge
n<n_{3}\wedge\theta_{16}\leq\mu\leq\mu_{9})$. $\theta_{16}\leq\mu\leq\mu_{9}$
is equivalent to $c_{1}\leq\theta_{13}$ and $\mu\geq\theta_{16}$. $\exists
c_{1}(c_{1}\leq\theta_{13}\wedge\theta_{11}<c_{1}\leq\eta_{2})$ is equivalent
to $\theta_{13}>\theta_{11}$. Therefore, in this case
$\Phi(n,p,\mu)\Leftrightarrow(p=n_{1}\wedge
n<n_{3}\wedge\theta_{13}>\theta_{11}\wedge\mu\geq\theta_{16})$, and
$\mathbb{T}(5,3)$ is proved.

Case 1.5: $\theta_{1}<c_{1}\leq 0$ and $p\in(\frac{4}{9},n_{1})$. By (42) and
(44), we have $\mu_{2}>\mu_{3}$ and ($\mu_{4}\geq\theta_{7}\Leftrightarrow
c_{1}\leq\theta_{2}$). Hence
$\Phi(n,p,\mu)\Leftrightarrow\max(\mu_{4},\theta_{7})\leq\mu\leq\mu_{3}$. This
case is divided into two sub-cases.

Case 1.5.1: Similar to Case 1.3.2.1, we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{1}<c_{1}\leq\theta_{2}\wedge\phi_{3}\wedge\mu_{4}\leq\mu\leq\mu_{3})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p\in(\frac{4}{9},n_{1})\wedge\phi_{3}\wedge\theta_{1}<c_{1}\leq\theta_{2}\wedge\mu_{4}\leq\mu\leq\mu_{3})$.
Like Case 1.3.2.1, we have
$\Phi(n,p,\mu)\Leftrightarrow(p\in(\frac{4}{9},n_{1})\wedge\phi_{3}\wedge\theta_{4}>\theta_{1}\wedge\theta_{5}\leq\theta_{2})$,
and $\mathbb{T}(6,2)$ is proved

Case 1.5.2: Similar to Case 1.3.2.2, we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{2}<c_{1}\leq\eta_{1}\wedge\phi_{3}\wedge\mu_{5}\leq\mu\leq\mu_{3})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p\in(\frac{4}{9},n_{1})\wedge\phi_{3}\wedge\theta_{2}<c_{1}\leq\eta_{1}\wedge\theta_{7}\leq\mu\leq\mu_{3})$.
Similar to Case 1.3.2.2, we have
$\Phi(n,p,\mu)\Leftrightarrow(p\in(\frac{4}{9},n_{1})\wedge\phi_{3}\wedge\theta_{7}\leq\mu\wedge\theta_{4}>\theta_{2})$,
and $\mathbb{T}(6,3)$ is proved.

Case 1.6: $\theta_{1}<c_{1}\leq 0$ and $p=\frac{4}{9}$. When $p=\frac{4}{9}$,
we have $\theta_{1}=\theta_{9},\ \eta_{1}=\eta_{3},\ \theta_{7}=\theta_{8},\
s_{3}=-\frac{4n}{6561}(162n+25c_{1})(45n^{2}+81nu+25c_{1})$. Then $s_{3}\geq
0\Leftrightarrow\mu\leq\mu_{13}$ if $c_{1}\geq\theta_{9}$. By (45), we know
$\mu_{3}\geq\theta_{7}\Leftrightarrow c_{1}\leq\eta_{1}$. And
$\theta_{9}<\eta_{3}\Leftrightarrow n=1$. From (50), we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{9}<c_{1}\leq\eta_{3},n=1,\theta_{8}\leq\mu\leq\mu_{13})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p=\frac{4}{9}\wedge\theta_{9}<c_{1}\leq\eta_{3}\wedge
n=1\wedge\theta_{8}\leq\mu\leq\mu_{13})$. $\theta_{8}\leq\mu\leq\mu_{13}$ is
equivalent to $c_{1}\leq\theta_{15}$ and $\mu\geq\theta_{8}$. $\exists
c_{1}(c_{1}\leq\theta_{15}\wedge\theta_{9}<c_{1}\leq\eta_{3})$ is equivalent
to $\theta_{15}>\theta_{9}$. Therefore, in this case
$\Phi(n,p,\mu)\Leftrightarrow(p=\frac{4}{9}\wedge
n=1\wedge\theta_{15}>\theta_{9}\wedge\mu\geq\theta_{8})$, and
$\mathbb{T}(7,2)$ is proved

Case 1.7: $\theta_{1}<c_{1}\leq 0$ and $0<p<\frac{4}{9}$. This case is divided
into two sub-cases.

Case 1.7.1: If we select $\theta_{7}\leq\mu\leq\min(\mu_{2},\mu_{3})$, by
(42), we have $\mu_{2}>\mu_{3}$. Thus
$\Phi(n,p,\mu)\Leftrightarrow\theta_{7}\leq\mu\leq\mu_{3}$. So we need
$\theta_{7}\leq\mu_{3}$, which yields $c_{1}\leq\eta_{1}$. By (47), we know
$\eta_{1}>\theta_{1}$ results $\phi_{3}$, which yields $n=1$ with
$0<p<\frac{4}{9}$. From (50), we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{1}<c_{1}\leq\eta_{1},n=1,\theta_{7}\leq\mu\leq\mu_{3})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p\in(0,\frac{4}{9})\wedge
n=1\wedge\theta_{1}<c_{1}\leq\eta_{1}\wedge\theta_{7}\leq\mu\leq\mu_{3})$.
Similar to Case 1.3.2.2, we have
$\Phi(n,p,\mu)\Leftrightarrow(p\in(0,\frac{4}{9})\wedge
n=1\wedge\theta_{7}\leq\mu\wedge\theta_{4}>\theta_{1})$, and $\mathbb{T}(8,2)$
is proved.

Case 1.7.2: If we select
$\max(\mu_{4},\theta_{7})\leq\mu\leq\min(\mu_{1},\mu_{2})$, by (41), we have
$\mu_{1}<\mu_{4}$, which yields contradiction.

Case 2: $c_{1}<\theta_{1}$. From (40), we have $\mu_{1}<\mu_{3}$ in this case
and from (35), $\Phi(n,p,\mu)$ simplifies to one case:

$\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(\theta_{7}\leq\mu\leq\min(\mu_{1},\mu_{2},\mu_{4}),\hbox{ if
}0<p<\frac{4}{9}\hbox{ and }c_{1}<\theta_{1}).$

Since $p$ satisfies $0<p<\frac{4}{9}$, we need only consider the following
cases.

Case 2.1: $c_{1}<\theta_{1}$ and $0<p<\frac{4}{9}$. By (41), (43) and (44), we
have $\mu_{1}>\mu_{4}$, $\mu_{2}>\mu_{4}$ and
($\mu_{4}\geq\theta_{7}\Leftrightarrow c_{1}\geq\theta_{2}$). Then we need
$\theta_{2}<\theta_{1}$, which yields $\phi_{3}$ by (46). Because $\phi_{3}$
means $n=1$ with $0<p<\frac{4}{9}$, we have
$\Phi(n,p,\mu)\Leftrightarrow(\theta_{2}\leq
c_{1}<\theta_{1},n=1,\theta_{7}\leq\mu\leq\mu_{4})$.

We now eliminate $c_{1}$ from $\Phi(n,p,\mu)\Leftrightarrow\exists
c_{1}(p\in(0,\frac{4}{9})\wedge n=1\wedge\theta_{2}\leq
c_{1}<\theta_{1}\wedge\theta_{7}\leq\mu\leq\mu_{4})$.
$\theta_{7}\leq\mu\leq\mu_{4}$ is equivalent to $c_{1}\geq\theta_{5}$ and
$\mu\geq\theta_{7}$. $\exists c_{1}(c_{1}\geq\theta_{5}\wedge\theta_{2}\leq
c_{1}<\theta_{1})$ is equivalent to $\theta_{5}<\theta_{1}$. Therefore, in
this case $\Phi(n,p,\mu)\Leftrightarrow(p\in(0,\frac{4}{9})\wedge
n=1\wedge\theta_{5}<\theta_{1}\wedge\mu\geq\theta_{7})$, and $\mathbb{T}(8,3)$
is proved

Case 2.2: $c_{1}<\theta_{1}$ and $p=n_{2}$. In Case 1.2, we know
$\theta_{1}=\theta_{3}$ with $p=n_{2}$, and $s_{2}\geq 0\Leftrightarrow
c_{1}\geq\theta_{3}$, which yields contradiction.

Case 2.3: $c_{1}<\theta_{1}$ and $p=n_{1}$. In Case 1.4, we know
$\theta_{1}=\theta_{10}$ with $p=n_{1}$, and $s_{2}\geq 0\Leftrightarrow
c_{1}\geq\theta_{10}$, which yields contradiction.

Case 2.4: $c_{1}<\theta_{1}$ and $p=\frac{4}{9}$. We have
$\theta_{1}=\theta_{9},\ \mu_{2}=\mu_{12},\ \mu_{3}=\mu_{13}$ based on
$p=\frac{4}{9}$. Then we have ($s_{2}\geq 0\Leftrightarrow\mu\leq\mu_{12}$)
and ($s_{3}\geq 0\Leftrightarrow\mu\geq\mu_{13}$ if $c_{1}\leq\theta_{9}$). So
we need $\mu_{12}\geq\mu_{13}$. By (42), we have $\mu_{12}<\mu_{13}$, which
yields contradiction.

Case 3: $c_{1}=\theta_{1}$. When $c_{1}=\theta_{1}$, we have
$s_{1}=n(np-n-\mu+2)$, $s_{2}=n(4p^{2}-9p+4)(np-n-\mu+2)$ and
$s_{3}=-n^{2}(9p-4)(np-n-\mu+2)^{2}$. Thus, $s_{1}\geq 0\Leftrightarrow\mu\leq
2+n(p-1)$ and $s_{3}\geq 0\Leftrightarrow(p\leq\frac{4}{9}\ {\rm or}\
\mu=2+n(p-1))$. This case is divided into two sub-cases.

Case 3.1: If $\mu=2+n(p-1)$, then $s_{1}=s_{2}=s_{3}=0$. And $p\geq
1-\frac{\mu}{n}\Leftrightarrow\phi_{1}$. Thus
$\Phi(n,p,\mu)\Leftrightarrow(c_{1}=\theta_{1}\wedge\phi_{1}\wedge\phi_{2})$,
and $\mathbb{T}(i,4),\ i=1,\ldots,6$ are proved.

Case 3.2: If $p\leq\frac{4}{9}$, then $s_{2}\geq 0\Leftrightarrow\mu\leq
2+n(p-1)$. Then we need $2+n(p-1)\geq\mu_{5}$, which yields $\phi_{1}$. And
$\phi_{1}$ implies $n=1$ with $p\leq\frac{4}{9}$. Thus
$\Phi(n,p,\mu)\Leftrightarrow(c_{1}=\theta_{1}\wedge n=1\wedge-p\leq\mu-1\leq
p)$, and $\mathbb{T}(7,4)$, $\mathbb{T}(8,4)$ are proved.

##  4 Conclusion

This paper is an extension of the work [19, 20] to the case where the entropy
power involves parameters. The basic idea is to prove entropy power
inequalities in a systematical way. Precisely, the concavity of Rényi entropy
power is considered, where the probability density $u_{t}$ solve the nonlinear
heat equation with two parameters $p$ and $\mu$. Our procedure reduces the
proof of the CREP to check the semi-positiveness of a quadratic form (33)
whose coefficients are polynomials in the parameters $n,p,\mu$. In principle,
a necessary and sufficient condition on parameters $n,p,\mu$ for this can be
computed with the quantifier elimination [22].

Based on the above method, we give a sufficient condition $\Phi(n,p,\mu)$ for
CREP, which extends the parameter’s range of the CREP given by Savaré-Toscani
[21]. In fact, our results give the necessary and sufficient condition for
CREP under certain conditions. But in the general case, Theorem 2.2 only gives
a sufficient condition for the following reasons: 1\. we use inequalities (21)
in step (23); 2\. there might exist more constraints; 3\. Problem II may not
equivalent to Problem III.

## Acknowledgments

This work is partially supported by NSFC 11688101 and NKRDP 2018YFA0704705,
Beijing Natural Science Foundation (No. Z190004), and China Postdoctoral
Science Foundation (No. 2019TQ0343, 2019M660830).

## Appendix. Constraints in (15)

In this appendix, we give the constraints in (15), where
$u_{h_{1},h_{2}}=\frac{\partial^{h_{1}+h_{2}}u}{\partial^{h_{1}}x_{a,t}\partial^{h_{2}}x_{b,t}}$.

$R_{1,a,b}=3pu_{1,0}^{4}+3uu_{2,0}u_{1,0}^{2}-5u_{1,0}^{4},$

$R_{2,a,b}=3pu_{0,1}^{4}+3uu_{0,2}u_{0,1}^{2}-5u_{0,1}^{4},$

$R_{3,a,b}=3pu^{2}u_{1,0}u_{3,0}+u^{3}u_{4,0}-3u^{2}u_{1,0}u_{3,0},$

$R_{4,a,b}=3pu^{2}u_{0,1}u_{3,0}+u^{3}u_{3,1}-3u^{2}u_{0,1}u_{3,0},$

$R_{5,a,b}=3pu^{2}u_{1,0}u_{2,1}+u^{3}u_{3,1}-3u^{2}u_{1,0}u_{2,1},$

$R_{6,a,b}=3pu^{2}u_{0,1}u_{2,1}+u^{3}u_{2,2}-3u^{2}u_{0,1}u_{2,1},$

$R_{7,a,b}=3pu^{2}u_{1,0}u_{1,2}+u^{3}u_{2,2}-3u^{2}u_{1,0}u_{1,2},$

$R_{8,a,b}=3pu^{2}u_{0,1}u_{1,2}+u^{3}u_{1,3}-3u^{2}u_{0,1}u_{1,2},$

$R_{9,a,b}=3pu^{2}u_{1,0}u_{0,3}+u^{3}u_{1,3}-3u^{2}u_{1,0}u_{0,3},$

$R_{10,a,b}=3pu^{2}u_{0,1}u_{0,3}+u^{3}u_{0,4}-3u^{2}u_{0,1}u_{0,3},$

$R_{11,a,b}=3pu_{0,1}u_{1,0}^{3}+3uu_{1,1}u_{1,0}^{2}-5u_{1,0}^{3}u_{0,1},$

$R_{12,a,b}=3pu_{0,1}^{3}u_{1,0}+3uu_{1,1}u_{0,1}^{2}-5u_{0,1}^{3}u_{1,0},$

$R_{13,a,b}=3pu_{0,1}^{2}u_{1,0}^{2}+2uu_{0,1}u_{1,0}u_{1,1}+uu_{1,0}^{2}u_{0,2}-5u_{1,0}^{2}u_{0,1}^{2},$

$R_{14,a,b}=3puu_{1,0}^{2}u_{2,0}+u^{2}u_{1,0}u_{3,0}+u^{2}u_{2,0}^{2}-4uu_{2,0}u_{1,0}^{2},$

$R_{15,a,b}=3puu_{0,1}u_{1,0}u_{2,0}+u^{2}u_{0,1}u_{3,0}+u^{2}u_{1,1}u_{2,0}-4uu_{2,0}u_{1,0}u_{0,1},$

$R_{16,a,b}=3puu_{1,0}^{2}u_{1,1}+u^{2}u_{1,0}u_{2,1}+u^{2}u_{1,1}u_{2,0}-4uu_{1,1}u_{1,0}^{2},$

$R_{17,a,b}=3puu_{0,1}u_{1,0}u_{2,0}+u^{2}u_{1,0}u_{2,1}+u^{2}u_{1,1}u_{2,0}-4uu_{2,0}u_{1,0}u_{0,1},$

$R_{18,a,b}=3puu_{0,1}^{2}u_{2,0}+u^{2}u_{0,1}u_{2,1}+u^{2}u_{0,2}u_{2,0}-4uu_{0,1}^{2}u_{2,0},$

$R_{19,a,b}=3puu_{0,1}u_{1,0}u_{1,1}+u^{2}u_{0,1}u_{2,1}+u^{2}u_{1,1}^{2}-4uu_{0,1}u_{1,0}u_{1,1},$

$R_{20,a,b}=3puu_{1,0}^{2}u_{0,2}+u^{2}u_{1,0}u_{1,2}+u^{2}u_{0,2}u_{2,0}-4uu_{1,0}^{2}u_{0,2},$

$R_{21,a,b}=3puu_{0,1}u_{1,0}u_{1,1}+u^{2}u_{1,0}u_{1,2}+u^{2}u_{1,1}^{2}-4uu_{0,1}u_{1,0}u_{1,1},$

$R_{22,a,b}=3puu_{0,1}^{2}u_{1,1}+u^{2}u_{0,1}u_{1,2}+u^{2}u_{0,2}u_{1,1}-4uu_{1,1}u_{0,1}^{2},$

$R_{23,a,b}=3puu_{0,1}u_{1,0}u_{0,2}+u^{2}u_{0,1}u_{1,2}+u^{2}u_{0,2}u_{1,1}-4uu_{0,2}u_{1,0}u_{0,1},$

$R_{24,a,b}=3puu_{0,1}u_{1,0}u_{0,2}+u^{2}u_{1,0}u_{0,3}+u^{2}u_{0,2}u_{1,1}-4uu_{0,2}u_{1,0}u_{0,1},$

$R_{25,a,b}=3puu_{0,1}^{2}u_{0,2}+u^{2}u_{0,1}u_{0,3}+u^{2}u_{0,2}^{2}-4uu_{0,2}u_{0,1}^{2},$

$R_{26,a,b}=3pu_{0,1}u_{1,0}^{3}+2uu_{2,0}u_{1,0}u_{0,1}+uu_{1,1}u_{1,0}^{2}-5u_{1,0}^{3}u_{0,1},$

$R_{27,a,b}=3pu_{0,1}^{2}u_{1,0}^{2}+uu_{0,1}^{2}u_{2,0}+2uu_{0,1}u_{1,0}u_{1,1}-5u_{1,0}^{2}u_{0,1}^{2},$

$R_{28,a,b}=3pu_{0,1}^{3}u_{1,0}+uu_{1,1}u_{0,1}^{2}+2uu_{0,2}u_{1,0}u_{0,1}-5u_{0,1}^{3}u_{1,0}.$

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# The Lexicographically Least Binary Rich Word Achieving the Repetition
Threshold

James Currie  and  Narad Rampersad  Department of Mathematics and Statistics
University of Winnipeg
515 Portage Avenue
Winnipeg, Manitoba R3B 2E9 (Canada<EMAIL_ADDRESS>
The work of James Currie is supported by the Natural Sciences and Engineering
Research Council of Canada (NSERC), [funding reference number 2017-03901].

The work of Narad Rampersad is supported by the Natural Sciences and
Engineering Research Council of Canada (NSERC), [funding reference number
2019-04111].

##  1\. Introduction

A major branch of combinatorics on words studies words avoiding various powers
or patterns. A typical question is whether there exists an infinite word over
a certain alphabet avoiding a certain pattern. The earliest known result of
this type is by Thue [16], who proved that there is an infinite word over a
three-letter alphabet containing no factor of the form $h(xx)$ with $h$ a non-
erasing morphism.

We use $\Sigma_{n}$ to denote the $n$-letter alphabet
$\Sigma_{n}=\\{0,1,2,\ldots,n-1\\}$. Let $p$ be an arbitrary finite string.
Several generalizations of Thue’s result have been explored.

  1. (1)

Does there exist some $n$ such that there is an infinite word over
$\Sigma_{n}$ containing no factor of the form $h(p)$ with $h$ a non-erasing
morphism?

  2. (2)

For a fixed $n$, is there an infinite word over $\Sigma_{n}$ containing no
factor of the form $h(p)$ with $h$ a non-erasing morphism?

The first of these problems was shown to be decidable by Bean et. al. [3] and
independently by Zimin [17]. It is unknown whether the second problem is
decidable.

A word of length $\ell$ and period $p$ is called a $k$-power, where
$k=\ell/p$. A reformulation of Thue’s result is that there is an infinite word
over $\Sigma_{3}$ not containing a $2$-power. For integer $n\geq 2$, the
repetitive threshold function is defined by

$\operatorname{RT}(n)=\sup\\{k:\text{ every infinite word over $\Sigma_{n}$
contains a $k$-power}\\}.$

Thus Thue showed that $\operatorname{RT}(3)\leq 2$. Dejean [9] showed that in
fact $\operatorname{RT}(3)=7/4$, and conjectured that

$\operatorname{RT}(n)=\begin{cases}7/4,&\text{ if $n=3$;}\\\ 7/5,&\text{ if
$n=4$;}\\\ n/(n-1),&\text{ if $n\neq 3,4$.}\\\ \end{cases}$

Dejean’s conjecture was finally proved by Rao, and independently by Currie and
Rampersad [14, 8]. Words over an alphabet which realize the repetition
threshold of the alphabet are called threshold words and are extremal objects.
In the case $n=2$, the threhold words are the binary overlap-free words, which
have a large literature. (A good reference is the thesis of Rampersad [13].)
With the solution of Dejean’s conjecture, an indexed family of similar
languages present themselves for study. As an example of such study, for
threshold words on $\Sigma_{n}$ with $n\geq 27$, Currie et. al.[6] have shown
that the number of words grows exponentially with length.

Also branching off from the solution of Dejean’s conjecture is the study of
repetition thresholds for various classes of words. For example, various
authors have found the repetition thresholds for binary rich words, for
balanced sequences, and for circular words [7, 10, 12]. Other types of
repetition thresholds have also been studied, such as undirected repetition
thresholds and Abelian repetition thresholds [5, 15].

When investigating the existence of an infinite word over $\Sigma_{n}$ with
some property, a natural approach is to generate and study long finite words
with the property. Such words are typically generated by backtracking, and are
therefore the lexicographically least words of a given length. Practically
speaking then, solving avoidance problems often involves generating and
parsing prefixes of the lexicographically least infinite word with a given
property. Allouche et. al.[2] characterized the lexicographically least
infinite overlap-free binary word starting with any specified prefix. Currie
[4] characterized the lexicographically least infinite good word, where the
good words are closely related to the period-doubling morphism. However, the
general study of lexicographically least infinite words with avoidance
properties is in its infancy, and more examples are needed.

The current note combines the theme of repetition threshold with that of
lexicographically least words. The 2020 paper of Currie et. al.[7] established
the repetition threshold for binary rich words. Studying such words by
backtracking leads naturally to the question: What is the lexicographically
least infinite binary rich word? We answer this question in this note.

##  2\. Preliminaries

A word over alphabet $\Sigma_{n}$ is a finite or infinite sequence over
$\Sigma_{n}$. We use lower case letters for finite words, and write, e. g.,
word $w=w_{1}w_{2}\cdots w_{m}$, where each $w_{i}\in\Sigma_{n}$. The length
of $w$ is denoted by $|w|=m$. The word of length $0$ is called the empty word,
and is denoted by $\epsilon$. The concatenation of two words
$u=u_{1}u_{2}\cdots u_{s}$ and $v=v_{1}v_{2}\cdots v_{t}$ is given by
$uv=u_{1}u_{2}\cdots u_{s}v_{1}v_{2}\cdots v_{t}$. If $u,v,w,z$ are words and
$w=uzv$, we call word $z$ a factor of $w$, word $u$ a prefix of $w$, and word
$v$ a suffix of $w$. If $w=uv$, we define $u^{-1}w=v$.

A morphism from $\Sigma_{n}^{*}$ to $\Sigma_{m}^{*}$ is a function $f$
respecting concatenation; i.e., $f(xy)=f(x)f(y)$ for all
$x,y\in\Sigma_{n}^{*}$. If $f^{-1}(\epsilon)=\\{\epsilon\\}$, we call $f$ non-
erasing.

We use bold-face letters for infinite words, writing
${\boldsymbol{w}}=w_{1}w_{2}w_{3}\cdots$, where each $w_{i}\in\Sigma_{n}$. The
set of finite words over $\Sigma_{n}$ is denoted by $\Sigma_{n}^{*}$, and the
set of infinite words is denoted by $\Sigma_{n}^{\omega}$.

Iteration of a morphism $f$ is written as exponentiation:

$f^{i}(x)=\begin{cases}x,&\text{ if $i=0$;}\\\ f(f^{i-1}(x)),&\text{ if
$i>0$.}\end{cases}$

If $f:\Sigma_{n}^{*}\rightarrow\Sigma_{n}^{*}$ is a morphism such that for
some $a\in\Sigma_{n}$, $f(a)=au$,$u\neq\epsilon$, then $f^{n-1}(a)$ is a
proper prefix of $f^{n}(a)$ for every positive integer $n$. We can then define
${\boldsymbol{w}}=\lim_{n\rightarrow\infty}f^{n}(a)$ to be the infinite word
such that, for each $n$, word $f^{n}(a)$ is a prefix of ${\boldsymbol{w}}$.

Let $w$ be a finite word over $\Sigma_{n}$. Write $w=w_{1}w_{2}\cdots w_{m}$
where each $w_{i}\in\Sigma_{n}$. The reversal of $w$ is the word
$w^{R}=w_{m}w_{m-1}\cdots w_{1}$. We call word $w$ a palindrome if $w=w^{R}$.
Any word $w$ contains at most $|w|$ distinct palindromic factors. If $w$ in
fact contains $|w|$ distinct palindromic factors, we say that $w$ is rich. A
good reference on rich words is the paper of Glen et. al.[11]. One of their
results which we will use is

######  Theorem 1.

[11, Theorem 2.14] For any finite or infinite word w, the following properties
are equivalent:

  * i

$w$ is rich;

  * ii

for any factor $u$ of $w$, if $u$ contains exactly two occurrences of a
palindrome $p$ as a prefix and as a suffix only, then $u$ is itself a
palindrome.

A factor $u$ of $w$ containing exactly two occurrences of a factor $p$ as a
prefix and as a suffix is called a return word of $p$. An infinite word is
defined to be rich if each of its finite factors is rich.

Let $w$ be a finite or infinite word. The critical exponent of $w$ is defined
to be

$\operatorname{ce}(w)=\sup\\{k:\text{ $w$ contains a $k$-power}\\}.$

Let $L$ be a set of infinite words. The repetitive threshold of $L$ is defined
to be

$\operatorname{RT}(L)=\sup\\{k:\text{ every word of $L$ contains a
$k$-power}\\}=\inf\\{\operatorname{ce}(w):w\in L\\}.$

Thus
$\operatorname{RT}(\Sigma_{n}^{*})=\operatorname{RT}(\Sigma_{n}^{\omega})=\operatorname{RT}(n)$.

Baranwal and Shallit [1] showed that there is an infinite binary rich word
with critical exponent $2+\sqrt{2}/2$, and Currie et. al.[7] proved that this
word achieves the repetition threshold for infinite binary rich words. Thus,
if $L$ is the set of binary rich words, $\operatorname{RT}(L)=2+\sqrt{2}/2$.
Let ${\boldsymbol{L}}$ be the set of infinite binary rich words. The set
${\boldsymbol{T}}$ of threshold words is the set of infinite binary rich words
whose critical exponent is the repetition threshold. Thus

${\boldsymbol{T}}=\\{{\boldsymbol{w}}\in{\boldsymbol{L}}:\operatorname{ce}({\boldsymbol{w}})=2+\sqrt{2}/2\\}.$

Define morphisms $f:\Sigma_{3}^{*}\rightarrow\Sigma_{2}^{*}$ and
$g,h:\Sigma_{3}^{*}\rightarrow\Sigma_{3}^{*}$ by

$\displaystyle f(\tt{0})$ $\displaystyle=\tt{0}$ $\displaystyle f(\tt{1})$
$\displaystyle=\tt{01}$ $\displaystyle f(\tt{2})$ $\displaystyle=\tt{011}$
$\displaystyle g(\tt{0})$ $\displaystyle=\tt{011}$ $\displaystyle g(\tt{1})$
$\displaystyle=\tt{0121}$ $\displaystyle g(\tt{2})$
$\displaystyle=\tt{012121}$ $\displaystyle h(\tt{0})$ $\displaystyle=\tt{01}$
$\displaystyle h(\tt{1})$ $\displaystyle=\tt{02}$ $\displaystyle h(\tt{2})$
$\displaystyle=\tt{022}$

Word $f(h^{\omega}(\tt{0}))$ is the word constructed by Baranwal and Shallit
[1]. The word $f(g(h^{\omega}(\tt{0})))$ was shown to be a binary rich word
with the same critical exponent by Currie et. al. [7].

The lexicographic order on $\Sigma_{n}^{*}$ and $\Sigma_{n}^{\omega}$ is
defined as follows:

  * •

We order letters in the natural way: $0<1<2<\cdots<n-1.$ We also insist that
$\epsilon<0$.

  * •

Let the longest common prefix of $u$ and $v$ be $p$. We say that $u<v$ if and
only if the first letter of $p^{-1}u$ is less than the first letter of
$p^{-1}v$, where the first letter of $\epsilon$ is taken to be $\epsilon$.

One checks that morphisms $f$, $g$, and $h$ are order-preserving: Let
$\phi\in\\{f,g,h\\}$. If $u\leq v$ then $\phi(u)\leq\phi(v)$.

######  Theorem 2 (Main Theorem).

Word $\ell=f(01g(h^{\omega}(\tt{0})))$ is the lexicographically least word in
${\boldsymbol{T}}$.

##  3\. Proof of Main Theorem

We say that a word $w\in\Sigma_{2}^{*}$ is good if it is both rich and
$14/5$-free. We say that an infinite word over $\Sigma_{2}$ is good if its
factors are good.

######  Observation 3.

Since an infinite good word ${\boldsymbol{w}}$ must be $3$-free, it can be
written as $pf({\boldsymbol{u}})$, where $p\in\\{\epsilon,1,11\\}$, and
${\boldsymbol{u}}\in\Sigma_{3}^{\omega}$.

We use slight amplifications of the results of Currie et. al. [7]:

######  Lemma 4.

Suppose $f({\boldsymbol{u}})$ is good, where ${\boldsymbol{u}}\in
011\Sigma_{3}^{\omega}$. Then

  1. (1)

Word ${\boldsymbol{u}}=g({\boldsymbol{W}})$ for some word
${\boldsymbol{W}}\in\Sigma_{3}^{\omega}$.

  2. (2)

Word ${\boldsymbol{W}}$ has the form $h({\boldsymbol{U}})$ for some word
${\boldsymbol{U}}\in\Sigma_{3}^{\omega}$.

Lemma 4 follows from the proof of Lemma 9 of Currie et. al. [7].

######  Lemma 5.

Let ${\boldsymbol{u}}\in 0\Sigma_{3}^{\omega}$. Suppose that for some positive
integer $n$, one of $f(g(h^{n}({\boldsymbol{u}})))$ and
$f(h^{n}({\boldsymbol{u}}))$ is good. Then
${\boldsymbol{u}}=h({\boldsymbol{W}})$ for some word
${\boldsymbol{W}}\in\Sigma_{3}^{\omega}$.

Lemma 5 follows from the proofs of Lemmas 10 and 11 of Currie et. al. [7].

######  Theorem 6.

The infinite binary word ${\boldsymbol{v}}=f(g(h^{\omega}(0)))$ is good.

Theorem 6 follows from Theorems 15 and 17 of Currie et. al. [7].

We begin with a preliminary lemma.

######  Lemma 7.

The lexicographically least infinite good word with prefix $001010$ is

${\boldsymbol{v}}=f(g(h^{\omega}(0))).$

###### Proof.

Suppose that ${\boldsymbol{V}}$ is an infinite good word with prefix $001010$,
and ${\boldsymbol{V}}\leq{\boldsymbol{v}}$. By Observation 3, write
${\boldsymbol{V}}=f({\boldsymbol{u}})$, where
${\boldsymbol{u}}\in\Sigma_{3}^{\omega}$. Since ${\boldsymbol{V}}$ has prefix
$001010$, word ${\boldsymbol{u}}$ has prefix $011$. It follows from Lemma 4
that ${\boldsymbol{u}}=g({\boldsymbol{U}})$ for some word
${\boldsymbol{U}}\in\Sigma_{3}^{\omega}$, where ${\boldsymbol{U}}$ has the
form $h({\boldsymbol{W}}_{1})$ for some word
${\boldsymbol{W}}_{1}\in\Sigma_{3}^{\omega}$.

Since $f$ is order-preserving, ${\boldsymbol{u}}\leq g(h^{\omega}(0))$. Since
$g$ is order-preserving, ${\boldsymbol{U}}\leq h^{\omega}(0)$. Since $h$ is
order-preserving, ${\boldsymbol{W}}_{1}\leq h^{\omega}(0)$. In particular,
since the first letter of $h^{\omega}(0)$ is $0$, the first letter of
${\boldsymbol{W}}_{1}$ is $0$. Using Lemma 5, write
${\boldsymbol{W}}_{1}=h({\boldsymbol{W}}_{2})$ for some word
${\boldsymbol{W}}_{2}\in\Sigma_{3}^{\omega}$. Again, since $h$ is order-
preserving, the first letter of ${\boldsymbol{W}}_{2}$ is $0$. By induction,
we find that for each positive integer $n$ we have
${\boldsymbol{W}}_{1}=h^{n-1}({\boldsymbol{W}}_{n})$, for some word
${\boldsymbol{W}}_{n}\in 0\Sigma_{3}^{\omega}$. It follows that $h^{n}(0)$ is
a prefix of ${\boldsymbol{W}}_{1}$ for each $n$, so that
${\boldsymbol{W}}_{1}=h^{\omega}(0)$.

We conclude that the lexicographically least infinite good word with prefix
$001010$ is

${\boldsymbol{v}}=f(g(h^{\omega}(0))).$

∎

The Main Theorem follows from the following three lemmas.

######  Lemma 8.

Let ${\boldsymbol{m}}$ be an infinite good word. Let

${\boldsymbol{\ell}}=f(01g(h^{\omega}(0))).$

Then

${\boldsymbol{\ell}}\leq{\boldsymbol{m}}.$

###### Proof.

The least binary $3$-free word of length 8 is $00100100$. However, $00100100$
cannot be extended on the right to a binary $3$-free word. It follows that
$001001010\leq{\boldsymbol{m}}.$ If ${\boldsymbol{m}}\leq{\boldsymbol{\ell}}$,
then

$f(01g(0))0=001001010\leq{\boldsymbol{m}}\leq{\boldsymbol{\ell}}=f(01g(h^{\omega}(0))),$

forcing ${\boldsymbol{m}}$ to have prefix $001001010$. Then
$(001)^{-1}{\boldsymbol{m}}$ is an infinite good word with prefix $001010$. By
Lemma 7, this forces

$f(g(h^{\omega}(0)))\leq(001)^{-1}{\boldsymbol{m}}$

forcing

${\boldsymbol{\ell}}=001f(g(h^{\omega}(0)))\leq{\boldsymbol{m}}.$

∎

######  Lemma 9.

Word $01f(g(h^{\omega}(0)))$ is recurrent.

###### Proof.

Word $g(h^{\omega}(0))$ is recurrent. However, the only letter preceding a $0$
in $g(h^{\omega}(0))$ is $1$, so that if $p$ is a prefix of
$g(h^{\omega}(0))$, word $1p$ must be a (necessarily recurrent) factor of
$g(h^{\omega}(0))$. Any $u$ factor of $01f(g(h^{\omega}(0)))$ is a factor of
$01f(p)=f(1p)$ for some prefix $p$ of $g(h^{\omega}(0))$. Since $1p$ is
recurrent in $g(h^{\omega}(0))$, $f(1p)$ is recurrent in
$f(g(h^{\omega}(0)))$, and so is $u$. Then $u$ is recurrent in
$01f(g(h^{\omega}(0)))$. we conclude that $01f(g(h^{\omega}(0)))$ is
recurrent. ∎

######  Lemma 10.

The word

${\boldsymbol{\ell}}=f(01g(h^{\omega}(0)))$

is $14/5$-free.

###### Proof.

Currie et. al.[7] proved that $f(g(h^{\omega}(0)))$ is $14/5$-free. By Lemma
9, $01f(g(h^{\omega}(0)))$ has the same factors as $f(g(h^{\omega}(0)))$ and
is also $14/5$-free. Therefore, any $14/5^{+}$ power in
$f(g(h^{\omega}(0)))=001f(g(h^{\omega}(0)))$ must be a prefix. The word
$00100$ is a prefix of $f(01g(h^{\omega}(0)))$, but does not occur in
$f(g(h^{\omega}(0)))$. It follows that any $14/5^{+}$ power which is a prefix
of $f(g(h^{\omega}(0)))$ has period 4 or less. A very short finite check shows
no such $14/5^{+}$ power is a prefix of $f(01g(h^{\omega}(0)))$.

∎

######  Lemma 11.

The word

${\boldsymbol{\ell}}=f(01g(h^{\omega}(0)))$

is rich.

###### Proof.

Currie et. al.[7] proved that $f(g(h^{\omega}(0)))$ is rich. By Lemma 9,
$01f(g(h^{\omega}(0)))$ has the same factors as $f(g(h^{\omega}(0)))$ and is
also rich.

Suppose that $001f(g(h^{\omega}(0)))$ is not rich. It will therefore have a
complete return which is not a palindrome. Since $01f(g(h^{\omega}(0)))$ is
rich, some prefix of $001f(g(h^{\omega}(0)))$ must be a complete return to a
palindrome which is not a palindrome. Let this prefix be $pqp$ where $p$ is a
palindrome and $q$ is not. The palindrome $00100$ is a prefix of
$f(01g(h^{\omega}(0)))$, but does not occur in $f(g(h^{\omega}(0)))$. It
follows that $|p|\leq 4$. The only possibility is seen to be $p=00$. However
the complete return to $00$ is $00100$, which is a palindrome. This is a
contradiction. ∎

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