\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 74, pp. 1--14.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/74\hfil A Holmgren type theorem] {A Holmgren type theorem for partial differential equations whose coefficients are Gevrey functions} \author[M. Kawagishi \hfil EJDE-2010/74\hfilneg] {Masaki Kawagishi} \address{Masaki Kawagishi \newline Department of General Education College of Science and Technology, Nihon University\\ 1-8-14 Kanda-Surugadai, Chiyoda-ku, Tokyo 101-8308, Japan} \email{masaki@suruga.ge.cst.nihon-u.ac.jp} \dedicatory{Dedicated to Prof. Takesi Yamanaka on his 77th birthday} \thanks{Submitted March 28, 2010. Published May 21, 2010.} \subjclass[2000]{35A05, 35A10, 35G10} \keywords{Banach scale; Gevery function; Holmgren type; Uniqueness theorem} \begin{abstract} In this article, we consider a uniqueness theorem of Holmgren type for $p$-th order Kovalevskaja linear partial differential equations whose coefficients are Gevrey functions. We prove that the only $C^p$-solution to the zero initial-valued problem is the identically zero function. To prove this result we use the uniqueness theorem for higher-order ordinary differential equations in Banach scales. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} In this article we consider the linear partial differential equation \begin{equation} \label{1.1} \frac{\partial^{p} v(t,x)}{\partial t^{p}}= \sum_{\alpha \in \mathbb{Z}_{+}^{n},\, \lambda |\alpha|+j\leq p,\, j\leq p-1} a_{\alpha, j}(t,x)\frac{\partial^{j+|\alpha|}v(t,x)} {\partial t^{j}\partial x^{\alpha}}, \end{equation} where $p$ is an integer $\geq 1$, $\lambda$ is a real constant $>1$ and $a_{\alpha, j}(t, x), v(t, x)$ are $\mathbb{C}$-valued functions of $(t,x)\in \mathbb{R}\times \mathbb{R}^{n}$. We denote by $\mathbb{Z}_{+}^{n}$ the set of $n$-dimensional multi-indices. We write $|\alpha|=\alpha_{1}+\dots +\alpha_{n}$ for $\alpha=(\alpha_{1},\dots,\alpha_{n}) \in \mathbb{Z}_{+}^{n}$. Each $a_{\alpha, j}(t,x)$ is assumed to be continuous in $t$ and Gevrey function of order $\lambda$ in $x$. The purpose of this paper is to prove that the only $C^{p}$-solution of \eqref{1.1} which satisfies the following condition \begin{equation} \label{1.2} v(0,x)=\frac{\partial v(0,x)}{\partial t}=\dots =\frac{\partial^{p-1}v(0,x)}{\partial t^{p-1}}=0 \end{equation} is $v(t,x)\equiv 0$. An exact statement of the above result will be given in \S{4} as Theorem \ref{thm3.1}. To prove the result mentioned above, we use the result given in \cite{kawagishi} on the uniqueness of the solution of a non-linear ODE in a Banach scale. The outline of the proof of the Holmgren type uniqueness theorem in this paper is as follows: First we construct a Banach scale consisting of the duals of some normed spaces of Gevrey functions, and define the adjoint equation to the equation \eqref{1.1} on the above dual Banach scale. Then it is shown, by the uniqueness, that the only solution of $0$-initial value problem of the adjoint equation in a Banach scale is identically $0$. On the other hand the given solution $v(t,x)$ of the problem \eqref{1.1}--\eqref{1.2} gives rise to a solution $L(t)$ of the $0$-initial value problem for the adjoint equation. From these facts we conclude that $v(t,x)\equiv 0$. As a preparation for performing our plan as mentioned above, we shall prepare in \S{2} some important properties of Gevrey functions. Especially, a problem of approximation in Gevrey class is important for our purpose. We shall state that problem in Lemma \ref{lem2.4} and Theorem \ref{thm2.2}, and at the end of \S{2} we construct a dual Banach scale. \S{3} is devoted to showing the uniqueness of the solution of initial value problem in a Banach scale. In \S{4} we shall state and prove the main result in the paper, i.e., a Holmgren type uniqueness theorem for the initial value problem \eqref{1.1}--\eqref{1.2}. Here we review the definition of Banach scale. Let $J$ be an interval of real numbers. A family $\{E_{\sigma}\}_{\sigma\in J}$ of Banach spaces $E_{\sigma}$ is called a Banach scale, if $\delta<\sigma$ $(\sigma, \delta\in J)$, then $E_{\sigma}\subset E_{\delta}$ and $\|u\|_{\delta}\leq \|u\|_{\sigma}\ (u\in E_{\sigma})$. Finally we define the notation for partial differential operators. If $f(x)$ is a function of $x=(x_{1},\dots,x_{n})\in \mathbb{R}^{n}$, we write $\partial_{i}f(x)=\partial f(x)/\partial x_{i}$ and $\partial^{\alpha}f(x)= \partial_{1}^{\alpha_{1}}\dots \partial_{n}^{\alpha_{n}}f(x) =\partial^{|\alpha|}f(x)/\partial x_{1}^{\alpha_{1}}\dots \partial x_{n}^{\alpha_{n}}$ for $\alpha=(\alpha_{1},\dots,\alpha_{n})\in \mathbb{Z}_{+}^{n}$. \section{Preparation: Gevrey functions} Throughout this paper, $\lambda$ denotes a fixed real constant $>1$ and $\Omega$ a fixed open set $\subset \mathbb{R}^{n}$. We write $\alpha !=\alpha_{1}!\dots \alpha_{n}!$ for $\alpha=(\alpha_{1},\dots,\alpha_{n})\in \mathbb{Z}_{+}^{n}$. A $C^{\infty}$-function $f: \Omega \to \mathbb{C}$ is said to be in the Gevrey class of order $\lambda$ if there exists a positive constant $\sigma $ such that \[ \sup_{\alpha \in \mathbb{Z}_{+}^{n} ,\,x\in \Omega} {|\partial^{\alpha}f(x)| \frac{\sigma^{|\alpha|}}{(\alpha !)^{\lambda}}}<\infty. \] It is easily seen that a $C^{\infty}$-function $f: \Omega \to \mathbb{C}$ is in the Gevrey class of order $\lambda$ if and only if there exists a positive constant $\rho$ such that \begin{equation}\label{2.1} \sup_{\alpha \in \mathbb{Z}_{+}^{n},\, x\in \Omega} {|\partial^{\alpha}f(x)|\frac{\rho^{|\alpha|}}{(\alpha !)^{\lambda}}} (1+|\alpha|)^{2n}<\infty. \end{equation} We denote by $\mathcal{G}_{\lambda, \rho}(\Omega)$ the set of all $C^{\infty}$-functions $f: \Omega \to \mathbb{C}$ which satisfy the condition \eqref{2.1} and define the norm $|f|_{\lambda, \rho}$ of $f\in \mathcal{G}_{\lambda, \rho}(\Omega)$ by $$ |f|_{\lambda, \rho}= \sup_{\alpha \in \mathbb{Z}_{+}^{n},\,x\in \Omega} {|\partial^{\alpha}f(x)| \frac{\rho^{|\alpha|}}{(\alpha !)^{\lambda}}}(1+|\alpha|)^{2n}. $$ Then $\mathcal{G}_{\lambda, \rho}(\Omega)$ is a Banach space with the norm $|\cdot|_{\lambda, \rho}$ and the family $\{\mathcal{G}_{\lambda, \rho}(\Omega) \}_{\rho >0}$ forms a Banach scale. In the following two theorems some important properties of elements of the Gevrey class $\mathcal{G}_{\lambda, \rho}(\Omega)$ are stated. These results were given in \cite{kawagishi_yamanaka} and \cite{yamanaka}. However, it may not be considered quite suitable for our present situation without some modification. So we shall state and prove these results as Theorem \ref{thmofprod} and \ref{thmofdiff} \begin{theorem} \label{thmofprod} If $f,g\in \mathcal{G}_{\lambda, \rho}(\Omega)$, then the product $fg$ is again in $\mathcal{G}_{\lambda, \rho}(\Omega)$ and \begin{equation}\label{2.2} |fg|_{\lambda, \rho}\leq 2^{3n}|f|_{\lambda, \rho}|g|_{\lambda, \rho}. \end{equation} \end{theorem} \begin{proof} For any $\alpha\in \mathbb{Z}_{+}^{n}$, we have \begin{align*} |\partial^{\alpha}(fg)(x)| &\leq \sum_{\beta\in \mathbb{Z}_{+}^{n},\, \beta\leq \alpha} {}_\alpha \text{C}_{\beta}|\partial^{\beta}f(x)| |\partial^{\alpha -\beta}g(x)| \\ &\leq |f|_{\lambda, \rho}|g|_{\lambda, \rho} \frac{(\alpha !)^{\lambda}}{\rho^{|\alpha|}} \frac{1}{(2+|\alpha|)^{2n}} \sum_{\beta\leq \alpha} ({}_\alpha \text{C}_{\beta})^{1-\lambda} \big\{\frac{1}{1+|\beta|}+ \frac{1}{1+|\alpha -\beta|}\big\}^{2n} \\ &\leq |f|_{\lambda, \rho}|g|_{\lambda, \rho} \frac{(\alpha !)^{\lambda}}{\rho^{|\alpha|}} \frac{2^{2n}}{(1+|\alpha|)^{2n}} \sum_{\beta\leq \alpha} \big\{\frac{1}{1+|\beta|}\big\}^{2n} \\ &\leq |f|_{\lambda, \rho}|g|_{\lambda, \rho} \frac{(\alpha !)^{\lambda}}{\rho^{|\alpha|}} \frac{2^{2n}}{(1+|\alpha|)^{2n}} \Big(\sum_{k=0}^{\infty} \frac{1}{(1+k)^{2}}\Big)^{n} \\ &\leq |f|_{\lambda, \rho}|g|_{\lambda, \rho} \frac{(\alpha !)^{\lambda}}{\rho^{|\alpha|}} \frac{2^{3n}}{(1+|\alpha|)^{2n}}, \end{align*} which shows the theorem. \end{proof} In what follows we restrict the range of the scale parameter $\rho$ for the sake of simplicity of calculation. We restrict $\rho$ to the range $m^{-1}\leq \rho \leq em^{-1}$, where $m$ is a positive integer. \begin{theorem} \label{thmofdiff} If $f\in \mathcal{G}_{\lambda, \rho}(\Omega)$, $m^{-1}\leq \sigma<\rho \leq em^{-1}$, then, for $\alpha\in \mathbb{Z}_{+}^{n}$, $\partial^{\alpha}f \in \mathcal{G}_{\lambda, \sigma}(\Omega)$ and \begin{equation} \label{2.4} |\partial^{\alpha}f|_{\lambda, \sigma}\leq (\lambda |\alpha|)^{\lambda |\alpha|} \frac{|f|_{\lambda, \rho}}{(\rho -\sigma)^{\lambda |\alpha|}}. \end{equation} \end{theorem} \begin{proof} Note first that, if $a,b\in \mathbb{Z}_{+}$ and $0<\sigma<\rho \leq em^{-1}$, then \begin{equation} \label{ineq1} \begin{aligned} \left( \frac{\sigma}{\rho} \right)^{a+b} \left( \frac{(a+b)!}{b!} \right)^{\lambda} &\leq \sup_{s\in \mathbb{R},\, s\geq 0} {\left(\frac{\sigma}{\rho} \right)^{s}s^{\lambda a}}\\ &\leq\left( \frac{\lambda \rho a}{e} \right)^{\lambda a} \frac{1}{(\rho -\sigma)^{\lambda a}}\\ &\leq \frac{1}{m^{\lambda a}} \frac{(\lambda a)^{\lambda a}}{(\rho -\sigma)^{\lambda a}}. \end{aligned} \end{equation} If $\beta \in \mathbb{Z}_{+}^{n}$, we have, using \eqref{ineq1} and the condition $m^{-1}\leq \sigma$, \begin{align*} |\partial^{\beta}(\partial^{\alpha}f)(x)| &\leq |f|_{\lambda, \rho} \frac{((\alpha +\beta)!)^{\lambda}}{\rho ^{|\alpha +\beta|}} \frac{1}{(1+|\alpha +\beta|)^{2n}} \\ &\leq |f|_{\lambda, \rho}\frac{(\beta !)^{\lambda}}{\sigma ^{|\beta|}} \frac{1}{(1+|\beta|)^{2n}} \frac{1}{\sigma ^{|\alpha|}} \left( \frac{\sigma}{\rho} \right)^{|\alpha +\beta|} \left( \frac{(\alpha +\beta)!}{\beta !} \right)^{\lambda} \\ &\leq |f|_{\lambda, \rho}\frac{(\beta !)^{\lambda}}{\sigma ^{|\beta|}} \frac{1}{(1+|\beta|)^{2n}}(\lambda |\alpha|)^{\lambda |\alpha|} \frac{1}{(\rho -\sigma)^{\lambda |\alpha|}}, \end{align*} which shows the theorem. \end{proof} Let us state the following important fact which is known as that the Gevrey space $\mathcal{G}_{\lambda, \rho}(\Omega)$ is sufficiently rich. \begin{theorem} \label{thm2.1} Let $\lambda >1$C$\rho$ a positive constant and $\Omega$ an open set in $\mathbb{R}^{n}$. Let a point $x_{0}\in \Omega$ and a neighborhood $V\subset \Omega$ of $x_{0}$ be given. Then there exists an element $\varphi\in\mathcal{G}_{\lambda, \rho}(\Omega)$ such that $$ \mathop{\rm supp}\varphi\subset V,\quad \varphi(x_{0})>0, \quad \varphi(x)\geq 0\quad (\forall x\in \Omega), $$ where $\mathop{\rm supp}\varphi$ means the support of $\varphi$. \end{theorem} \begin{proof} Here we give a sketch of the proof. Let $q$ be an integer $\geq 2$. Define a $C^{\infty}$-function $f_{q} : \mathbb{R} \to \mathbb{R}$ by $$ f_{q}(t)=\begin{cases} \exp{(-1/t^{q})} &(t> 0), \\ 0 &(t\leq 0). \end{cases} $$ We can show that, if $1+q^{-1}<\lambda$, then there exists a positive constant $c$ such that the function $$ f_{c, q}(t)=f_{q}(ct) $$ belongs to $\mathcal{G}_{\lambda, \rho}(\mathbb{R})$. Write $x_{0}=(x_{01},\dots,x_{0n})$. Take $r>0$ such that $$ [x_{01}-r,x_{01}+r]\times \dots \times [x_{0n}-r,x_{0n}+r]\subset V. $$ For $x=(x_{1},\dots,x_{n})\in \Omega$, define $$ \varphi(x)=\prod_{i=1}^{n}f_{c, q}(x_{i}-x_{0i}+r)f_{c, q}(x_{0i}+r-x_{i}). $$ Then $\varphi\in \mathcal{G}_{\lambda, \rho}(\Omega)$ and satisfies all requirements in the theorem. \end{proof} Lemma \ref{lem2.1} and Lemma \ref{lem2.2} stated below will play important roles when we discuss the problem concerning the approximation of functions on Gevrey spaces\ (Lemma \ref{lem2.4}). However, the proofs of two lemmas can be performed by a standard method similar to the case of $C^{\infty}$-functions as can be seen in Treves \cite{Treves}. So we omit the proofs. \begin{lemma} \label{lem2.1} Let $\varphi\in \mathcal{G}_{\lambda, \rho}(\mathbb{R}^{n})$ be a function such that $$ \mathop{\rm supp}\varphi\subset \{x\in \mathbb{R}^{n} : \|x\|\leq 1\},\quad \varphi(0)>0,\quad \varphi(x)\geq 0\ (\forall x\in \mathbb{R}^{n}). $$ For $\varepsilon >0$, define $$ \varphi_{\varepsilon}(x) =\varepsilon ^{-n}a\varphi(\varepsilon^{-1} x) \quad (x\in \mathbb{R}^{n}), $$ where $$ a=\Big(\int_{\|x\|\leq 1}\varphi(x)dx\Big)^{-1}. $$ Then $\varphi_{\varepsilon}$ has the following properties: \begin{itemize} \item[(i)] $\varphi_{\varepsilon}\in \mathcal{G}_{\lambda, \varepsilon\rho}(\mathbb{R}^{n})$, $|\varphi_{\varepsilon}|_{\lambda, \varepsilon\rho}\leq \varepsilon^{-n}a|\varphi|_{\lambda, \rho}$, \item[(ii)] $\mathop{\rm supp}\varphi_{\varepsilon}\subset \{x\in \mathbb{R}^{n} : \|x\|\leq \varepsilon\}$, $\int_{\mathbb{R}^{n}}\varphi_{\varepsilon}(x)dx=1$. \end{itemize} \end{lemma} \begin{lemma} \label{lem2.2} Let $\varphi_{\varepsilon}\in \mathcal{G}_{\lambda, \varepsilon\rho}(\mathbb{R}^{n})$ be the function defined in Lemma \ref{lem2.1}. Let $f: \mathbb{R}^{n}\to \mathbb{C}$ be a continuous function such that $\mathop{\rm supp}f$ is compact. Then the convolution $$ \varphi_{\varepsilon}\ast f(x)= \int_{\mathbb{R}^{n}}\varphi_{\varepsilon}(y)f(x-y)dy\ \Big(= \int_{\mathbb{R}^{n}}\varphi_{\varepsilon}(x-y)f(y)dy\Big) $$ of $\varphi_{\varepsilon}$ and $f$ satisfies the following properties: \begin{itemize} \item[(i)] $\varphi_{\varepsilon}\ast f \in \mathcal{G}_{\lambda, \varepsilon\rho}(\mathbb{R}^{n})$, $|\varphi_{\varepsilon}\ast f|_{\lambda, \varepsilon\rho} \leq \left\{\int_{\mathbb{R}^{n}}|f(y)|dy\right \} |\varphi_{\varepsilon}|_{\lambda, \varepsilon\rho}$. \item[(ii)] $\mathop{\rm supp}\varphi_{\varepsilon}\ast f \subset\{x\in \mathbb{R}^{n} : d(x,\mathop{\rm supp}f)\leq \varepsilon\}$, \end{itemize} where the letter $d$ denotes distance. \end{lemma} \begin{lemma} \label{lem2.3} Let $f\in \mathcal{G}_{\lambda, \rho}(\mathbb{R}^{n})$ and $y\in \mathbb{R}^{n}$ be given. Define the function $f_{y}$ by $f_{y}(x)=f(x-y)$. If $0<\sigma<\rho$, then $f_{y}\in \mathcal{G}_{\lambda, \sigma}(\mathbb{R}^{n})$ and the map $$ \mathbb{R}^{n}\ni y \mapsto f_{y}\in \mathcal{G}_{\lambda, \sigma}(\mathbb{R}^{n}) $$ is continuous. \end{lemma} \begin{proof} Fix $y_{0},y\in \mathbb{R}^{n}$ and $\alpha\in\mathbb{Z}_{+}^{n}$ arbitrarily. Define the function $k(\theta)$ by $$ k(\theta)=\partial^{\alpha}f(x-y_{0}+\theta (y_{0}-y)) \quad (0\leq \theta \leq 1). $$ Write $y=(y_{1},\dots,y_{n}),y_{0}=(y_{01},\dots,y_{0n})$. Then we obtain \begin{align*} \partial^{\alpha}f_{y}(x)-\partial^{\alpha}f_{y_{0}}(x) &= \partial^{\alpha}f(x-y)-\partial^{\alpha}f(x-y_{0})=k(1)-(0)\\ &=\int_{0}^{1}k'(\theta)d\theta \\ &=\int_{0}^{1}\Big\{\sum_{j=1}^{n} \partial_{j}(\partial^{\alpha}f)(x-y_{0}+\theta(y_{0}-y)) (y_{j}-y_{0j})\Big\}d\theta. \end{align*} On the other hand, by Theorem \ref{thmofdiff}, we know that $\partial_{j}f\in \mathcal{G}_{\lambda, \sigma}(\mathbb{R}^{n})$. Hence, from the above equalities, it follows that \begin{align*} |\partial^{\alpha}f_{y}(x)-\partial^{\alpha}f_{y_{0}}(x)| &\leq \sum_{j=1}^{n}\int_{0}^{1} |\partial^{\alpha}(\partial_{j}f)(x-y_{0}+\theta(y_{0}-y))| |y_{j}-y_{0j}|d\theta \\ &\leq \sum_{j=1}^{n}|\partial_{j}f|_{\lambda, \sigma}|y_{j}-y_{0j}| \frac{(\alpha !)^{\lambda}}{\sigma^{|\alpha|}} \frac{1}{(1+|\alpha|)^{2n}} \\ &\leq \Big\{\sum_{j=1}^{n}|\partial_{j}f|_{\lambda, \sigma}\Big\} \|y-y_{0}\|\frac{(\alpha !)^{\lambda}}{\sigma^{|\alpha|}} \frac{1}{(1+|\alpha|)^{2n}}, \end{align*} and we obtain the inequality $$ |f_{y}-f_{y_{0}}|_{\lambda, \sigma}\leq \Big\{\sum_{j=1}^{n}|\partial_{j}f|_{\lambda, \sigma}\Big\} \|y-y_{0}\|, $$ which shows the continuity of the map $y\mapsto f_{y}\in \mathcal{G}_{\lambda,\sigma}(\mathbb{R}^{n})$. \end{proof} Let us define the subset $\mathcal{G}_{\lambda,\rho}^{c}(\Omega)$ of $\mathcal{G}_{\lambda, \rho}(\Omega)$ by $$ \mathcal{G}_{\lambda, \rho}^{c}(\Omega)= \{f\in \mathcal{G}_{\lambda, \rho}(\Omega) : \mathop{\rm supp}f\ \text{is compact}\}, $$ and give $\mathcal{G}_{\lambda, \rho}^{c}(\Omega)$ the norm $|\cdot|_{\lambda, \rho}$ defined on $\mathcal{G}_{\lambda, \rho}(\Omega)$. \par Lemma \ref{lem2.1}, Lemma \ref{lem2.2} and Lemma \ref{lem2.3} yield the next lemma. \begin{lemma} \label{lem2.4} If $0<\delta<\sigma<\rho$, then $\mathcal{G}_{\lambda, \rho}^{c}(\Omega)$ is a dense subspace of $\mathcal{G}_{\lambda, \sigma}^{c}(\Omega)$ with respect to the norm of $\mathcal{G}_{\lambda, \delta}^{c}(\Omega)$. In other words, if $f\in \mathcal{G}_{\lambda, \sigma}^{c}(\Omega)$ and $\varepsilon>0$, then there exists an element $g\in \mathcal{G}_{\lambda, \rho}^{c}(\Omega)$ such that $$ |f-g|_{\lambda, \delta}<\varepsilon. $$ \end{lemma} \begin{proof} Let $f\in\mathcal{G}_{\lambda, \sigma}^{c}(\Omega)$ and $\varepsilon >0$ be given. If $0<\delta<\sigma$, then, by Lemma \ref{lem2.3}, there exists $r>0$ such that $$ \|y\|0$, define a subspace $G_{\lambda, \sigma}^{c}(\Omega)$ of $\mathcal{G}_{\lambda, \sigma}^{c}(\Omega)$ by $$ G_{\lambda, \sigma}^{c}(\Omega)= \bigcup_{\sigma<\rho}\mathcal{G}_{\lambda, \rho}^{c}(\Omega). $$ In $G_{\lambda, \sigma}^{c}(\Omega)$, we adopt the same norm $|\cdot|_{\lambda, \sigma}$ as in $\mathcal{G}_{\lambda, \sigma}(\Omega)$. If $\sigma<\delta$, then $$ G_{\lambda, \delta}^{c}(\Omega)\subset G_{\lambda, \sigma}^{c}(\Omega) $$ and $G_{\lambda, \delta}^{c}(\Omega)$ is a dense subspace of $G_{\lambda, \sigma}^{c}(\Omega)$. \end{theorem} \begin{proof} The relation $G_{\lambda, \delta}^{c}(\Omega)\subset G_{\lambda, \sigma}^{c}(\Omega)$ is clear. In order to see that $G_{\lambda, \delta}^{c}(\Omega)$ is dense in $G_{\lambda, \sigma}^{c}(\Omega)$, take $f\in G_{\lambda, \sigma}^{c}(\Omega)$ and $\varepsilon>0$. Then $f\in \mathcal{G}_{\lambda, \sigma_{1}}^{c}(\Omega)$ for some number $\sigma_{1}$ such that $\sigma<\sigma_{1}<\delta$. Take two numbers $\sigma_{2}$ and $\delta_{1}$ such that $\sigma<\sigma_{2}<\sigma_{1}<\delta<\delta_{1}$. Then, by Lemma \ref{lem2.4}, there exists $g\in \mathcal{G}_{\lambda, \delta_{1}}^{c}(\Omega)$ such that $|f-g|_{\lambda, \sigma_{2}}<\varepsilon$. On the other hand, $g$ belongs to $G_{\lambda, \delta}^{c}(\Omega)$ and satisfies the inequality $$ |f-g|_{\lambda, \sigma}\leq |f-g|_{\lambda, \sigma_{2}}<\varepsilon. $$ Hence, this shows that $G_{\lambda, \delta}^{c}(\Omega)$ is a dense subspace of $G_{\lambda, \sigma}^{c}(\Omega)$. \end{proof} Concerning the product of two functions and differential operators on a function, the family $\{G_{\lambda, \sigma}^{c}(\Omega)\}_{\sigma>0}$ defined in Theorem \ref{thm2.2} has the same properties as those in the Gevrey class $\{\mathcal{G}_{\lambda, \sigma}(\Omega)\}_{\sigma>0}$. In other words, the implication $$ f,g\in G_{\lambda, \sigma}^{c}(\Omega) \Rightarrow fg \in G_{\lambda, \sigma}^{c}(\Omega) $$ holds and, if $0<\sigma<\rho,\alpha\in\mathbb{Z}_{+}^{n}$, then the implication $$ f\in G_{\lambda, \rho}^{c}(\Omega) \Rightarrow \partial^{\alpha}f\in G_{\lambda, \sigma}^{c}(\Omega) $$ holds. Now, it is easy, by Theorem \ref{thm2.2}, to construct a Banach scale consisting of dual Banach spaces. If $(X,\|\cdot\|_{X})$ is a normed space, then the dual space $X^{\ast}$ of $X$ is defined as usual. Let $Y$ be a linear subspace of $X$ and $(Y,\|\cdot\|_{Y})$ a normed space such that $\|y\|_{X}\leq \|y\|_{Y}$ for $y\in Y$. Then the map $i : Y\ni y \mapsto y \in X$ is continuous, and the adjoint operator $i^{\ast} : X^{\ast} \to Y^{\ast}$ of $i$ satisfies the inequality \begin{equation} \label{2.ast} \|i^{\ast}(u)\|_{Y}\leq \|u\|_{X}\ (u\in X^{\ast}). \end{equation} Here we want to identify $X^{\ast}$ with $i^{\ast}(X^{\ast})$. This is possible if $Y$ is dense in $X$, since $i^{\ast}$ is injective in that case. Then we have, by \eqref{2.ast}, $$ \|u\|_{Y}\leq \|u\|_{X}\ (u\in X^{\ast}). $$ We have the following theorem. \begin{theorem} \label{thm2.3} Fix a number $\rho>0$ arbitrarily. For $\sigma$ such that $0<\sigma<\rho$, put $$ \mathcal{D}_{\lambda, \sigma}(\Omega)= \{G_{\lambda, \rho -\sigma}^{c}(\Omega)\}^{\ast}, $$ and denote by $\|\cdot\|_{\lambda, \sigma}$ the norm on $\mathcal{D}_{\lambda, \sigma}(\Omega)$. Then the family $\{\mathcal{D}_{\lambda, \sigma}(\Omega)\}_{0<\sigma<\rho}$ forms a Banach scale. \end{theorem} The proof of the above theorem is obvious from the arguments preceding the theorem. \section{Uniqueness of the solution of the initial value problem in a Banach scale} In this section, we shall prove the uniqueness of the solution of the initial value problem in a Banach scale, and we use this result in showing our main theorem of this paper. Let $\{E_{\sigma}\}_{\sigma_{0}<\sigma\leq\delta_{0}}$ be a scale of Banach spaces, where $0\leq \sigma_{0}<\delta_{0}<\infty$. Let $I$ be an interval which contains $0$ as an inner point. Let $F$ be a map of the form $$ F: I\times \bigcup_{\sigma_{0}<\sigma\leq\delta_{0}} \underbrace{E_{\sigma}\times \dots \times E_{\sigma}} _{p-\text{times}} \to \bigcup_{\sigma_{0}<\sigma\leq\delta_{0}}E_{\sigma} $$ such that if $\sigma_{0}<\delta <\sigma \leq\delta_{0}$, then $F(I\times E_{\sigma}\times \dots \times E_{\sigma})\subset E_{\delta}$ and the map $$ F : I\times E_{\sigma}\times \dots \times E_{\sigma}\to E_{\delta} $$ is continuous. Further we assume that there exists a positive constant $C$ such that if $\sigma_{0}<\delta <\sigma \leq\delta_{0}$, then the inequality \begin{equation} \label{1.3} \|F(t,u)-F(t,v)\|_{\delta}\leq C\sum_{j=0}^{p-1} \frac{\|u_{j}-v_{j}\|_{\sigma}}{(\sigma -\delta)^{p-j}} \quad \quad (\forall t\in I) \end{equation} holds for $u=(u_{0},\dots,u_{p-1}),v=(v_{0},\dots,v_{p-1}) \in E_{\sigma}\times \dots \times E_{\sigma}$. For such a map $F$ we can consider an initial value problem of the form \begin{equation} \label{1.4} u^{(p)}(t)=F(t,u(t),u'(t),\dots,u^{(p-1)}(t)), \end{equation} \begin{equation} \label{1.5} u(0)=b_{0},u'(0)=b_{1},\dots,u^{(p-1)}(0)=b_{p-1}, \end{equation} where $$ b_{0},\dots,b_{p-1}\in E_{\delta_{0}}. $$ Concerning the above mentioned problem, the author proved, in \cite{kawagishi}, the existence and the uniqueness of the solution. However, in this paper, we only use the uniqueness of the solution. Here we state and prove a uniqueness result which is simpler than the uniqueness part of the result given in \cite{kawagishi}. \begin{theorem} \label{thm3'} Let $T>0$ such that $[-T,T]\subset I$. Let $u, v$ be $C^{p}$- maps from the interval $[-T,T]$ to $E_{\delta_{0}}$. If $u$ and $v$ are the solutions of the problem \eqref{1.4}-\eqref{1.5}, then for $t\in [-T,T]$ $$ u(t)=v(t). $$ \end{theorem} \begin{proof} The problem \eqref{1.4}-\eqref{1.5} is rewritten as a system of first order equations of the form \begin{equation} \label{1.6} \begin{cases} u_{1}'(t)&=u_{2}(t) \\ &\vdots \\ u_{p-1}'(t)&=u_{p}(t) \\ u_{p}'(t)&=F(t,u_{1}(t),\dots,u_{p}(t)) \end{cases} \end{equation} \begin{equation} \label{1.7} u_{1}(0)=b_{0},\dots,u_{p}(0)=b_{p-1} \end{equation} in the unknown functions $u_{1}(t),\dots,u_{p}(t)$. Further the problem \eqref{1.6}-\eqref{1.7} is equivalent to the integral equation of the form \begin{equation} \label{1.8} \begin{bmatrix} u_{1}(t) \\ \vdots \\ u_{p}(t) \end{bmatrix} = \begin{bmatrix} b_{0}\\ \vdots \\ b_{p-1} \end{bmatrix} +\int_{0}^{t} \begin{bmatrix} u_{2}(\tau) \\ \vdots \\ u_{p}(\tau)\\ F(\tau,u_{1}(\tau),\dots,u_{p}(\tau)) \end{bmatrix} d\tau. \end{equation} From \eqref{1.8} it follows that each $u_{j}(t)$ satisfies \begin{equation} \label{1.9} \begin{aligned} u_{j}(t)&=b_{j-1}+t b_{j}+\dots +\frac{t^{p-j}}{(p-j)!}b_{p-1}\\ &\quad + \int_{0}^{t}\frac{(t-\tau)^{p-j}}{(p-j)!} F(\tau,u_{1}(\tau),\dots,u_{p}(\tau))d\tau. \end{aligned} \end{equation} If $u(t),v(t)$ are $C^{p}$-solutions of the problem \eqref{1.4}-\eqref{1.5} , then $(u_{1}(t),\dots,u_{p}(t))=(u(t),u'(t),\dots,u^{(p-1)}(t))$ and $(v_{1}(t),\dots,v_{p}(t)) =(v(t),v'(t),\dots,v^{(p-1)}(t))$ are solutions of the integral equation \eqref{1.8}. $u_{j}(t),v_{j}(t)$ are continuous maps from $[-T,T]$ to the Banach space $E_{\delta_{0}}$. To prove the theorem, it is sufficient to prove $u_{j}(t)=v_{j}(t)$. If $\sigma_{0}<\nu<\mu\leq \delta_{0}$, then, by \eqref{1.3} and by \eqref{1.9}, we have, for $t\in [-T,T]$, \begin{equation} \label{1.10} \begin{aligned} \|u_{j}(t)-v_{j}(t)\|_{\nu} &\leq \int_{0}^{|t|}\frac{(|t|-\tau)^{p-j}}{(p-j)!} \|F(\tau,u_{1}(\tau).\ \dots,u_{p}(\tau)) \\ &\quad -F(\tau,v_{1}(\tau).\ \dots,v_{p}(\tau))\|_{\nu}d\tau \\ &\leq C\sum_{i=1}^{p}\frac{1}{(\mu -\nu)^{p+1-j}} \int_{0}^{|t|}\frac{(|t|-\tau)^{p-j}}{(p-j)!} \|u_{i}(\tau)-v_{i}(\tau)\|_{\mu}d\tau. \end{aligned} \end{equation} For the moment we assume that $0\leq t\leq T$ and put $$ M= \max_{0\leq t \leq T}{\sum_{j=1}^{p} \|u_{j}(t)-v_{j}(t)\|_{\delta_{0}}}. $$ Let $\mu=\delta_{0},\nu=\delta_{0}-\xi\ (0<\xi<\delta_{0}-\sigma_{0})$. We have, by \eqref{1.10}, \begin{equation} \begin{aligned} \|u_{j}(t)-v_{j}(t)\|_{\delta_{o}-\xi} &\leq C\sum_{i=1}^{p}\frac{1}{\xi ^{p+1-i}} \int_{0}^{t}\frac{(t-\tau)^{p-j}}{(p-j)!} \|u_{i}(\tau)-v_{i}(\tau)\|_{\delta_{0}}d\tau \\ &\leq MC\sum_{l=1}^{p}\frac{1}{\xi^{j-l}}\frac{1}{\xi^{p+1-j}} \frac{t^{p+1-j}}{(p+1-j)!} \\ &\leq \label{1.11} % MC\sum_{l=1}^{p}\frac{1}{\xi^{j-l}} \sum_{k=1}^{p}\frac{1}{\xi^{k}} \frac{t^{k}}{k!}. \end{aligned} \end{equation} Next let $\mu=\delta_{0}-\xi, \nu=\delta_{0}-2\xi\ (0<2\xi<\delta_{0}-\sigma_{0})$, we have, by \eqref{1.10} and \eqref{1.11}, \begin{align*} \|u_{j}(t)-v_{j}(t)\|_{\delta_{0}-2\xi} &\leq MC^{2}\sum_{i=1}^{p}\frac{1}{\xi^{p+1-i}} \int_{0}^{t}\frac{(t-\tau)^{p-j}}{(p-j)!} \Big\{ \sum_{l=1}^{p}\frac{1}{\xi^{i-l}}\sum_{k=1}^{p}\frac{1}{\xi^{k}} \frac{\tau^{k}}{k!} \big\}d\tau \\ &= MC^{2}\sum_{i=1}^{p}\frac{1}{\xi^{p+1-i}} \Big( \sum_{l=1}^{p}\frac{1}{\xi^{i-l}} \Big) \sum_{k=1}^{p}\frac{1}{\xi^{k}}\int_{0}^{t}\frac{(t-\tau)^{p-j}}{( p-j)!} \frac{\tau^{k}}{k!}d\tau \\ &= MpC^{2}\sum_{l=1}^{p}\frac{1}{\xi^{p+1-l}}\sum_{k=1}^{p} \frac{1}{\xi^{k}}\frac{t^{p+1+k-j}}{(p+1+k-j)!} \\ &= MpC^{2}\sum_{l=1}^{p}\frac{1}{\xi^{j-l}}\sum_{k=1}^{p} \frac{1}{\xi^{p+1+k-j}}\frac{t^{p+1+k-j}}{(p+1+k-j)!} \\ &\leq MpC^{2}\sum_{l=1}^{p}\frac{1}{\xi^{j-l}}\sum_{k=2}^{2p} \frac{1}{\xi^{k}}\frac{t^{k}}{k!}. \end{align*} Repeating this process, we can show that, if $n$ is a natural number and $\xi$ is a positive number such that $n\xi<\delta_{0}-\sigma_{0}$, then the inequality \begin{equation} \label{1.12} \|u_{j}(t)-v_{j}(t)\|_{\delta_{0}-n\xi} \leq Mp^{n-1}C^{n}\sum_{l=1}^{p}\frac{1}{\xi^{j-l}} \sum_{k=n}^{np}\frac{1}{\xi^{k}}\frac{t^{k}}{k!} \end{equation} holds for $0\leq t\leq T$. Writing $\mu=\delta_{0}-n\xi$, \eqref{1.12} is rewritten as \begin{equation} \label{1.12+1} \|u_{j}(t)-v_{j}(t)\|_{\mu} \leq Mp^{n-1}C^{n}\sum_{l=1}^{p}\frac{n^{j-l}}{(\delta_{0}-\mu)^{j-l}} \sum_{k=n}^{np}\frac{n^{k}}{(\delta_{0}-\mu)^{k}}\frac{t^{k}}{k!}. \end{equation} In \eqref{1.12+1}, $\mu$ can be any number such that $\sigma_{0}<\mu<\delta_{0}$. Moreover, since $e^{n}>n^{k}/k!$, we see that the inequality $$ \|u_{j}(t)-v_{j}(t)\|_{\mu} \leq \frac{M}{p} \Big( \frac{peCt}{\delta_{0}-\mu} \Big)^{n} \Big( \sum_{l=1}^{p}\frac{n^{j-l}}{(\delta_{0}-\mu)^{j-l}} \Big) \sum_{k=0}^{n(p-1)} \big( \frac{t}{\delta_{0}-\mu} \big)^{k} $$ holds. Let $L=\min{\{1/peC,1\}}$. For each $t\in [0,T]$ such that $t1$. Write $\rho_{0}=m^{-1}(1+e)$, where $m$ is a positive integer. In the differential equation \eqref{1.1}, assume that each coefficient $a_{\alpha, j}(t,x)$ satisfies the condition that the function $x\mapsto a_{\alpha, j}(t,x)$ belongs to $\mathcal{G}_{\lambda, \rho_{0}}(\Omega)$ for each fixed $t\in I$ and the map $I\ni t\mapsto a_{\alpha, j}(t,\cdot)\in \mathcal{G}_{\lambda, \rho_{0}}(\Omega)$ is continuous. Then the only $C^{p}$-solution $v(t,x)$ in the domain $(t,x)\in I\times\Omega$ of the initial value problem \eqref{1.1}--\eqref{1.2} is $v(t,x)\equiv 0$. \end{theorem} \begin{proof} We first introduce an initial value problem on the dual Banach scale which is called `adjoint' to the problem \eqref{1.1}-\eqref{1.2}. Let $\Psi\subset \mathbb{R}^n$ be an open set such that its closure $\overline{\Psi}$ is compact and is contained in $\Omega$. For $\delta$ such that $m^{-1}<\delta\leq m^{-1}e$, put $\mathcal{D}_{\lambda, \delta}(\Psi)= \{G_{\lambda, \rho_{0}-\delta}^{c}(\Psi)\}^{\ast}$. Then, by Theorem \ref{thm2.3}, $\{\mathcal{D}_{\lambda, \delta}(\Psi)\} _{m^{-1}<\delta\leq m^{-1}e}$ forms a Banach scale, and we use this scale throughout the rest of this paper. Using the {\it dual scale}, we can define the adjoint equation to the problem \eqref{1.1}. For $\delta,\sigma$ such that $m^{-1}<\delta<\sigma\leq m^{-1}e$, we define a map $$ F: I\times \underbrace{\mathcal{D}_{\lambda, \sigma}(\Psi)\times \dots \mathcal{D}_{\lambda, \sigma}(\Psi)}_{p-\text{times}} \to \mathcal{D}_{\lambda, \delta}(\Psi) $$ by \begin{equation} \label{3.1+1} F(t,L_{0},\dots,L_{p-1}) =\sum_{\alpha \in \mathbb{Z}_{+}^{n},\, \lambda |\alpha|+j\leq p,\,j\leq p-1} (-1)^{|\alpha|}A_{\alpha, j}^{\ast}(t)(L_{j}), \end{equation} where $A_{\alpha, j}^{\ast}(t)$ is the adjoint of the linear map $A_{\alpha, j}(t)$ which is defined by $$ A_{\alpha, j}(t) : G_{\lambda, \rho_{0}-\delta}^{c}(\Psi) \to G_{\lambda, \rho_{0}-\sigma}^{c}(\Psi),\ A_{\alpha, j}(t)(\varphi)(x)= \partial^{\alpha}(a_{\alpha, j}(t,x)\varphi(x)). $$ We have to verify that $F$ is well defined. If $m^{-1}<\delta<\sigma\leq m^{-1}e$ and $\varphi\in G_{\lambda, \rho_{0}-\delta}^c(\Psi)$, then, by Theorem \ref{thmofprod}, the product $a_{\alpha, j}(t,\cdot)\varphi(\cdot)$ is in $G_{\lambda, \rho_{0}-\delta}^{c}(\Psi)$. Then, by Theorem \ref{thmofdiff}, $\partial^{\alpha}(a_{\alpha, j}(t,\cdot)\varphi(\cdot))$ is in $G_{\lambda, \rho_{0}-\sigma}^{c}(\Psi)$. Hence the map $A_{\alpha, j}(t)$ is well defined. The continuity of $A_{\alpha, j}(t)$ as a linear map from $G_{\lambda, \rho_{0}-\delta}^{c}(\Psi)$ to $G_{\lambda, \rho_{0}-\sigma}^{c}(\Psi)$ follows immediately from Theorem \ref{thmofprod} and Theorem \ref{thmofdiff}. This shows that if $\delta<\sigma$, then the adjoint map $$ A_{\alpha, j}^{\ast}(t): \mathcal{D}_{\lambda, \sigma}(\Psi)\to \mathcal{D}_{\lambda, \delta}(\Psi) $$ of $A_{\alpha, j}(t)$ is well defined, and the map $F$ is well defined, too. Here, by the above arguments, we can define an ODE on the Banach scale $\{\mathcal{D}_{\lambda, \sigma}(\Psi)\}_{m^{-1}<\sigma\leq m^{-1}e}$, by \begin{equation} \label{3.2} L^{(p)}(t)=F(t,L(t),L'(t),\dots,L^{(p-1)}(t)), \end{equation} which is the `adjoint' equation to the given equation \eqref{1.1}. Now we show the following fact as the first step of the proof. \noindent {\bf Assertion ($\star$)}: {\it The only $C^{p}$ -solution of the equation \eqref{3.2} such that} \begin{equation} \label{3.3} L(0)=L'(0)=\dots =L^{(p-1)}(0)=0 \end{equation} {\it is} $L(t)\equiv 0$. To prove the above result, we use Theorem \ref{thm3'}. We first show that the map $F$ is continuous. Fix the numbers $\delta,\sigma$ such that $m^{-1}<\delta<\sigma\leq m^{-1}e$. We show the continuity of the adjoint operator $$ A_{\alpha, j}^{\ast} : I\times \mathcal{D}_{\lambda, \sigma}(\Psi) \to \mathcal{D}_{\lambda, \delta}(\Psi). $$ For $t\in I$ and $L,L_{0}\in \mathcal{D}_{\lambda, \sigma}(\Psi)$, the inequality \begin{equation} \label{3.5} \begin{aligned} &\|A_{\alpha, j}^{\ast}(t)(L-L_{0})\|_{\lambda, \delta} \\ & \leq \|L-L_{0}\|_{\lambda, \sigma} \sup{\{|A_{\alpha, j}(t)\varphi|_{\lambda, \rho_{0}-\sigma} : \varphi\in G_{\lambda, \rho_{0}-\delta}^{c}(\Psi),\ |\varphi|_{\lambda, \rho_{0}-\delta}\leq 1\}} \end{aligned} \end{equation} holds. If $|\varphi|_{\lambda, \rho_{0}-\delta}\leq 1$, then, by \eqref{2.2} and \eqref{2.4} in \S{2}, we obtain \begin{equation} \label{3.6} \begin{aligned} |A_{\alpha, j}(t)\varphi|_{\lambda, \rho_{0}-\sigma} &=|\partial^{\alpha}(a_{\alpha, j}(t,\cdot)\varphi(\cdot))| _{\lambda, \rho_{0}-\sigma} \\ &\leq (\lambda |\alpha|)^{\lambda |\alpha|} \frac{|a_{\alpha, j}(t,\cdot)\varphi(\cdot)| _{\lambda, \rho_{0}-\delta}} {(\rho_{0}-\delta-\rho_{0}+\sigma)^{\lambda |\alpha|}} \\ &\leq p^{p} \frac{2^{3n}|a_{\alpha, j}(t,\cdot)|_{\lambda, \rho_{0}-\delta}} {(\sigma-\delta)^{\lambda |\alpha|}}. \end{aligned} \end{equation} Put $$ K=\sup{\{|a_{\alpha, j}(t,\cdot)|_{\lambda, \rho_{0}}: \lambda |\alpha|+j\leq p,j\leq p-1,t\in I\}}. $$ Since the map $t\mapsto a_{\alpha, j}(t,\cdot) \in \mathcal{G}_{\lambda, \rho_{0}}(\Omega)$ is continuous, we know that $K$ is finite. It follows, from \eqref{3.6}, that $$ |A_{\alpha, j}(t)\varphi|_{\lambda, \rho_{0}-\sigma} \leq p^{p}2^{3n}K\frac{1}{(\sigma-\delta)^{\lambda|\alpha|}} $$ holds. By the last inequality and \eqref{3.5}, we obtain \begin{equation} \label{3.7} \|A_{\alpha, j}^{\ast}(t)(L-L_{0})\|_{\lambda, \delta}\leq \|L-L_{0}\|_{\lambda, \sigma} \frac{2^{3n}p^{p}K}{(\sigma-\delta)^{\lambda |\alpha|}}. \end{equation} Next, we look at the inequality \begin{equation} \label{3.8} \begin{aligned} &\|(A_{\alpha, j}^{\ast}(t)-A_{\alpha, j}^{\ast}(t_{0})) L_{0}\|_{\lambda, \delta} \\ &\leq \|L_{0}\|_{\lambda, \sigma} \sup{\{|A_{\alpha, j}(t)\varphi-A_{\alpha, j}(t_{0})\varphi |_{\lambda, \rho_{0}-\sigma} : \varphi\in G_{\lambda, \rho_{0}-\delta}^{c}(\Psi), |\varphi|_{\lambda, \rho_{0}-\delta}\leq 1\}}, \end{aligned} \end{equation} where $t,t_{0}\in I$. If we use \eqref{2.2}, \eqref{2.4} and \eqref{3.6}, then \begin{align*} |A_{\alpha, j}(t)\varphi-A_{\alpha, j}(t_{0})\varphi | _{\lambda, \rho_{0}-\sigma} &\leq (\lambda |\alpha|)^{\lambda |\alpha|} \frac{|(a_{\alpha, j}(t,\cdot)-a_{\alpha, j}(t_{0},\cdot)) \varphi(\cdot)|_{\lambda, \rho_{0}-\delta}} {(\rho_{0}-\delta-\rho_{0}+\sigma)^{\lambda |\alpha|}} \\ &\leq p^{p}\frac{2^{3n}|a_{\alpha, j}(t,\cdot)-a_{\alpha, j}(t_{0}, \cdot)| _{\lambda, \rho_{0}}} {(\sigma-\delta)^{\lambda |\alpha|}}. \end{align*} The last inequality and \eqref{3.8} imply \begin{equation} \label{3.9} \|(A_{\alpha, j}^{\ast}(t)-A_{\alpha, j}^{\ast}(t_{0}))L_{0}\| _{\lambda, \delta} \leq \|L_{0}\|_{\lambda, \sigma}p^{p} \frac{2^{3n}|a_{\alpha, j}(t,\cdot)-a_{\alpha, j}(t_{0}, \cdot)| _{\lambda, \rho_{0}}}{(\sigma-\delta)^{\lambda |\alpha|}}. \end{equation} From \eqref{3.7} and \eqref{3.9} it follows that \begin{equation} \label{3.10} \begin{aligned} &\|A_{\alpha, j}^{\ast}(t)L-A_{\alpha, j}^{\ast} (t_{0})L_{0}\|_{\lambda, \delta} \\ &\leq \|A_{\alpha, j}^{\ast}(t)(L-L_{0})\|_{\lambda, \delta}+ \|(A_{\alpha, j}^{\ast}(t)-A_{\alpha, j}^{\ast}(t_{0})) L_{0}\|_{\lambda, \delta} \\ &\leq p^{p} \frac{2^{3n}}{(\sigma-\delta)^{\lambda |\alpha|}} \{K\|L-L_{0}\|_{\lambda, \sigma} +|a_{\alpha, j}(t,\cdot) -a_{\alpha, j}(t_{0},\cdot)|_{\lambda, \rho_{0}} \|L_{0}\|_{\lambda, \sigma}\}. \end{aligned} \end{equation} Since the map $t\mapsto a_{\alpha, j}(t,\cdot)\in \mathcal{G}_{\lambda, \rho_{0}}(\Omega)$ is continuous, \eqref{3.10} implies that the map $$ A_{\alpha, j}^{\ast}: I\times \mathcal{D}_{\lambda, \sigma}(\Psi) \to \mathcal{D}_{\lambda, \delta}(\Psi) $$ is continuous. Hence the map $F$ is also continuous. Let us show that the map $F$ satisfies the condition \eqref{1.3} in \S{3}. Put $L_{0}=0$ in \eqref{3.10}. Then, for $t\in I$ and $L\in \mathcal{D}_{\lambda, \sigma}(\Psi)$, the inequality \begin{equation} \label{3.11} \|A_{\alpha, j}^{\ast}(t)L\|_{\lambda, \delta} \leq p^{p} 2^{3n}K \frac{\|L\|_{\lambda, \sigma}}{(\sigma-\delta)^{\lambda |\alpha|}} \end{equation} holds. For $$ \mathcal{L}=(L_{0}, \ \dots,L_{p-1}), \mathcal{M}=(M_{0},\dots,M_{p-1})\in \underbrace{\mathcal{D}_{\lambda, \sigma}(\Psi) \times \dots \times \mathcal{D}_{\lambda, \sigma}(\Psi)}_{p-\text{times}} $$ we have $$ \|F(t,\mathcal{L})-F(t,\mathcal{M})\|_{\lambda, \delta} \leq \sum_{\alpha\in\mathbb{Z}_{+}^{n},\, \lambda |\alpha|+j\leq p ,\,j\leq p-1} \|A_{\alpha, j}^{\ast}(t)(L_{j}-M_{j})\|_{\lambda, \delta}. $$ Hence, by the last inequality and \eqref{3.11}, we obtain \begin{equation} \label{3.12} \|F(t,\mathcal{L})-F(t,\mathcal{M})\|_{\lambda, \delta} \leq p^{p}2^{3n}K\sum_{\alpha, j} \frac{\|L_{j}-M_{j}\|_{\lambda, \sigma}}{(\sigma-\delta) ^{\lambda|\alpha|}}. \end{equation} Here we note that the inequality $$ \frac{1}{(\sigma-\delta)^{\lambda |\alpha|}} =\frac{(\sigma-\delta)^{p-j-\lambda|\alpha|}}{(\sigma- \delta)^{p-j}} \leq \frac{e^{p}}{(\sigma-\delta)^{p-j}} $$ holds and we put $$ C=(ep)^{p}2^{3n}K\sum_{\alpha\in \mathbb{Z}_{+}^{n},\, \lambda|\alpha|\leq p} 1_{\alpha}, $$ where $1_{\alpha}=1$. Then, by \eqref{3.12}, we obtain $$ \|F(t,\mathcal{L})-F(t,\mathcal{M})\|_{\lambda, \delta} \leq C\sum_{j=0}^{p-1} \frac{\|L_{j}-M_{j}\|_{\lambda, \sigma}}{(\sigma-\delta)^{p-j}}, $$ which shows that the map $F$ satisfies the condition \eqref{1.3} in \S{3}. Consequently, we can use the Theorem \ref{thm3'} to the problem \eqref{3.2}-\eqref{3.3}. As a result, we conclude that the Assertion $(\star)$ is true. \noindent {\bf Proof of the identity $v(t,x)\equiv 0$}: Let us construct a solution of the problem \eqref{3.2}-\eqref{3.3} by the given solution $v(t,x)$ of the problem \eqref{1.1}-\eqref{1.2}. We define $$ L_{v}(t)(\varphi)=\int_{\Psi}v(t,x)\varphi(x)dx $$ for $\varphi \in G_{\lambda, \rho_{0}-m^{-1}e}^{c}(\Psi)= G_{\lambda, m^{-1}}^{c}(\Psi)$ and $t\in I$. Then we can see that $L_{v}(t)\in \mathcal{D}_{\lambda, m^{-1}e}(\Psi)$ ($=\{G_{\lambda, m^{-1}}^{c}(\Psi)\}^{\ast}$) and the map $I\ni t\mapsto L_{v}(t)\in \mathcal{D}_{\lambda, m^{-1}e}(\Psi)$ is $C^{p}$-class. The derivative $L_{v}^{(j)}(t)$ of $L_{v}(t)$ has the form $$ L_{v}^{(j)}(t)(\varphi)= \int_{\Psi}\partial_{t}^{j}v(t,x)\varphi(x)dx \quad (j=1, \dots,p-1). $$ These facts can be proved without difficulty by the compactness of $\overline{\Psi}$ and by the uniform continuity of $\partial_{t}^{j}v(t,x)$ on $I\times \Psi$. Now let us show that $L_{v}(t)$ is a solution of the problem \eqref{3.2}-\eqref{3.3}. It is obvious by \eqref{1.2} that $L_{v}(t)$ satisfies the initial condition \eqref{3.3}. Let $\sigma$ be a number such that $m^{-1}<\sigma