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\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 47, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}


\begin{document}
\title[\hfilneg EJDE-2006/47\hfil Viscosity solutions]
{Viscosity solutions of the Cauchy problem for second-order
nonlinear partial differential equations in Hilbert spaces}
\author[T. V. Bang \& T. D. Van \hfil EJDE-2006/47\hfilneg]
{Tran Van Bang,  Tran Duc Van}

\address{Tran Van Bang\hfill\break
Department of Mathematics, Hanoi pedagogical University number 2, Vietnam}
\email{bangtv75@yahoo.com}

\address{Tran Duc Van\hfill\break
Hanoi Institute of Mathematics\\
P.O. Box 631, BoHo, Hanoi, Vietnam}
\email{tdvan@math.ac.vn}


\date{}
\thanks{Submitted November 2, 2005. Published April 11, 2006.}
\subjclass[2000]{35D99}
\keywords{Hamilton-Jacobi equations; viscosity solutions}

\begin{abstract}
 In this paper we prove the existence and uniqueness of viscosity
 solutions of the Cauchy problem for the second order nonlinear
 partial differential equations in Hilbert spaces.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{definition}{Definition}[section]
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}



\section{Introduction}

The theory of scalar partial differential equations in infinite
dimensional Hilbert spaces has been developing very rapidly in
recent years. The object of its study is first and second
order PDE's of the form
\begin{equation}
G(x,u(x),Du(x),D^2u(x))=0\quad\text{in }\Omega,\label{A1}
\end{equation}
where $\Omega$ is a subset of Hilbert space $H, u$ is a real valued
and $Du(x)$ and $D^2u(x)$ correspond respectively to the first and
second order Fr\'echet derivatives of $u$. Identifying $H$ with
its dual, $Du(x)$ corresponds to an element of $H$ and $D^2u(x)$
to an element of $S(H)$, the space of bounded, self-adjoint
operators equipped with the operator norm. Therefore,
$$
G:W\subset H\times\mathbb{R}\times H\times S(H)\mapsto\mathbb{R}
$$
is appropriate. If the set $W$ is open in $H\times\mathbb{R}\times H\times S(H)$
and $G$ is locally bounded we call equation (\ref{A1}) {\it bounded}.
It may however happen that $W$ is just dense in
$H\times\mathbb{R}\times H\times S(H )$ and $G$ is not locally bounded.
In such case we refer to (\ref{A1}) as to being {\it unbounded}.

The unbounded equations are of importance since they appear as
dynamic programming equations associated with problems of optimal
control and differential games. Roughly speaking, if one controls
an infinite dimensional system governed by an ODE in a Hilbert
space, one has to deal with a first order stationary or time
dependent PDE, while controlling a system for which the state
equation is a stochastic PDE gives rise to a second order
stationary or time dependent PDE. ``Unboundedness'' arises when
the state equation of a system involves unbounded operators or, in
the stochastic case, for instance, so called ``white noise''.

More precisely, we will study the Cauchy problem for the fully
nonlinear PDE's having the form
\begin{equation}
\begin{gathered}
u_t(t,x)+\langle Ax,Du(t,x)\rangle +F(t,x,Du(t,x),D^2u(t,x))=0\\
(t,x)\in (0,T)\times H\\
u(0,x)=g(x)\quad \text{for }x\in H,
\end{gathered}\label{CP}
\end{equation}
where $H$ is a real, separable Hilbert space endowed with the
inner product $\langle.,.\rangle$ and the norm $|.|$ and
$A:\mathscr{D}(A)\subset H\to H$ is closed linear operator that generates
an analytic $C_0$-semigroup $e^{-tA}$ on $H$. Moreover, we assume
that $A$ is positive definite and self-adjoint and has compact
resolvent $R(\mu,A)$.

There is an increasing interest in and a growing literatures on
Hamilton-Jacobi equations in infinite dimensions. These equations
were first studied by Barbu and Da Prato \cite{b1}, setting the
problem in classes of convex functions and using semigroup and
perturbation methods. Much progress has been made recently due to
the introduction of notion of viscosity solutions. We refer the
reader to \cite{c3,g2,i1,t1,t2,t3} for the first order equations.
As regards the second order, ``bounded'' equations have been
investigate in \cite[Parts I and III]{l2}, and ``unbounded'' in
\cite[Part II]{l2}, \cite{g1,g3,i2,s1,v1}. Except for
\cite{g1,g2,g3,v1} the unboundedness in the studied equations was
always coming from the term $\langle Ax,Du(x)\rangle$. This paper
is concerned with equations that exhibit ``bad behavior'' in the
$F$ also in $Du$ and $D^2u$ as the same as in \cite{g1,g2,g3,v1}.
We notice that, \cite{g1} studied the stationary version of
\eqref{CP}, \cite{g3} studied \eqref{CP} with
$F(t,x,Du(t,x),D^2u(t,x))$ has form
$-\mathop{\rm Trace}(QD^2u(t,x))+G(t,x,Du(t,x))$ and used a
different test functions.

The plan of the paper is the following. In section 2 we give
some preliminaries. In section 3 we present the definition of
viscosity  solutions and prove a general uniqueness and existence
results for \eqref{CP}.

\section{Preliminaries}

For any Hilbert spaces $X,Y$ and $E$, we denote
\begin{gather*}
UC(X)=\{ u:X\to \mathbb{R}; u\text{ is uniformly continuous}\},\\
BUC(X)=\{ u\in UC(X); u\text{ is bounded}\},\\
UC_x([0,T]\times X)=\{ u\in C([0,T]\times X); u(t,.)\in UC(X)
\text{ uniformly in } t\in [0,T]\},\\
BUC_x([0,T]\times X)=\{ u\in UC_x([0,T]\times X); u \text{ is
bounded}\}.
\end{gather*}
Let $u:(0,T)\times E\to \mathbb{R}$. If $(\hat t,\hat{x})\in (0,T)\times E$
and $(a,p,Z)\in \mathbb{R}\times E\times S(E)$ we say that $(a,p,Z)\in
P^{2,+}u(\hat t,\hat{x})$ provided that (see \cite{c2})
\begin{equation*}
\begin{split}
u(t,x)&\leqslant u(\hat t,\hat{x})+a(t-\hat t)+\langle p,x-\hat{x}\rangle
+\frac 1 2\langle Z(x-\hat{x}),x-\hat{x}\rangle\\
&\quad +o(|x-\hat{x} |^2+|t-\hat t|)\quad\text{as } (t,x)\to (\hat
t,\hat{x}).
\end{split}
\end{equation*}
The closure of $P^{2,+},\bar P^{2,+}$, is defined as follows:
\begin{equation*}
\begin{split}
\bar P^{2,+}u(t,x)
=\big\{& (a,p,Z)\in \mathbb{R}\times E\times S(E):
\exists (t_n,x_n,a_n,p_n,Z_n)\text{ in}\\
& (0,T)\times E\times\mathbb{R}\times E\times S(E):
(a_n,p_n,Z_n)\in P^{2,+}u(t_n,x_n)\\
&\text{ and }(t_n,x_n,u(t_n,x_n),a_n,p_n,Z_n) \to (t,x,u(t,x),a,p,Z)\big\}.
\end{split}
\end{equation*}

We are interested in the situation where $E=E_1\times E_2$ is the
product of two spaces and $u(t,x_1,x_2)=u_1(t,x_1)+u_2(t,x_2)$.
Proposition below is a straightforward corollary from
\cite[Theorem 8.3]{c1}. In this paper, identify operator in any space is denoted
by a same symbol $I$.

\begin{proposition} \label{prop2.1}
Let $u_i, i=1,2$ be upper semicontinuous on $(0,T)\times\mathbb{R}^N$ and 
$\varphi:(0,T)\times \mathbb{R}^{2N}\to\mathbb{R}$ be once continuously 
differentiable in $t$ and twice continuously differentiable in $x$. 
Suppose that
$$
u_1(t,x_1)+u_2(t,x_2)-\varphi(t,x_1,x_2)
$$
has a local maximum at $(\hat t,\hat{x})=(\hat t,\hat{x}_1,\hat{x}_2)
\in (0,T)\times\mathbb{R}^{2N}$ and that
$$
D^2\varphi(\hat t,\hat{x})=D=D_1-D_2,
$$
where $D_1,D_2\in S(\mathbb{R}^N)$ and $D_1,D_2\geq 0$. Assume, moreover,
that

\begin{equation}
\begin{aligned}
\exists r>0:\forall M>0,\exists c>0 \text{ such that for }i=1,2\\
b_i\leqslant c\text{ if } (b_i,q_i,Z_i)\in P^{2,+}u_i(t,x_i),\ |x_i-\hat{x}_i|
+|t-\hat t|<r\\
\text{and }|u_i(t,x_i)|+|q_i|+\| Z_i\|\leqslant M.
\end{aligned}\label{B1}
\end{equation}
Then, for every $\alpha >0$ there are $Z_1,Z_2\in S(\mathbb{R}^N)$ such that
\begin{itemize}
\item[(i)] $(b_i,D_{x_i}\varphi (\hat t,\hat{x}),Z_i)\in
\bar{P}^{2,+}u_i(\hat t,\hat{x}_i)$,  for $i=1,2$;
\item[(ii)]  $-(\|D_1\|+\|D_2\|)(1+\frac 2 \alpha)I\leqslant
\begin{pmatrix}
Z_1&0\\
0&Z_2
\end{pmatrix}
\leqslant D+\alpha(D_1+D_2)$;

\item[(iii)] $b_1+b_2=\varphi_t(\hat t,\hat{x})$.
\end{itemize}
\end{proposition}

The norm of the symmetric matrix used above is
$$
\|\phi\|=\sup\big\{|\lambda |:\lambda \text{ is an eigenvalue of
}\phi\big\} =\sup\big\{|\langle\phi\xi,\xi\rangle |:|\xi|\leqslant
1\big\}.
$$

\begin{remark} \rm
The condition (\ref{B1}) will be satisfied if $u_i$ are the viscosity
subsolutions (see \cite{c2}) of
$$
(u_i)_t(t,x_i)+F(t,x_i,u_i(t,x_i),Du_i(t,x_i),D^2u_i(t,x_i))\leqslant 0
\text{ in }(0,T)\times\mathbb{R}^N
$$
with $F$ bounded on bounded sets.
\end{remark}

We say that a function $\rho :[0,+\infty)\to [0,+\infty)$ is a {\it
modulus} if $\rho $ is continuous, nondecreasing, subadditive, and
$\rho (0)=0$. Subadditivity in particular implies that for all
$\varepsilon >0$, there exists $C_\varepsilon >0$ such that
$$
\rho (r)\leqslant\varepsilon +C_\varepsilon r,\quad\text{for every } r\geq 0.
$$
Moreover, a function $\omega:[0,+\infty)\times [0,+\infty)\to [0,+\infty)$ is
a {\it local modulus} if $\omega$ is continuous, nondecreasing in
both variables, subadditive in the first variable, and
$\omega(0,r)=0$, for every $ r\geq0$.

We assume the following hypothesis:
\begin{itemize}
\item (A1):$\mathscr{D}(A)\subset H\to H$ is a self-adjoint
operator, there exists $a>0$ such that $\langle Ax,x\rangle\geq a|x|^2$
for all $x\in \mathscr{D}(A)$, and $A^{-1}$ is compact.
\end{itemize}
\begin{remark} \rm
Hypothesis (A1) implies in particular that $-A$ is the infinitesimal
generator of an analytic semigroup with compact resolvent
satisfying $\| e^{-tA}\|\leqslant e^{-at}$ for all $t\geq 0$ and that
there is an orthonormal basis of $H$ made of eigenvectors of $A$
such that the corresponding sequence of eigenvalues diverges to
$+\infty$ as $n\to \infty$. It also follows that
\end{remark}

We also assume the {\it Interpolation inequality:}
Let $\gamma \in (0,1], \alpha \in (0,\gamma )$. For every $\sigma >0$,
there exists $C_\sigma >0$ such that
\begin{equation}
|A^\alpha z|\leqslant  \sigma |A^\gamma z|+C_\sigma |z|,\quad
\forall z\in \mathscr{D}(A^\gamma ).\label{B3}
\end{equation}
Let $ H_1\subset  H_2\subset \cdots$ be finite dimensional
subspaces of $ H$ generated by eigenvectors of $A$ such that
$\overline{\cup_{N=1}^\infty  H_N}= H$. Given $N\in \mathbb{N}$, denote by
$P_N$ the orthogonal projection in $ H$ onto $ H_N$, let
$Q_N=I-P_N$ and let $ H_N^\bot =Q_N H$. We then have an orthogonal
decomposition $ H= H_N\times  H_N^\bot $ and we will denote by
$x_N$ an element of $ H_N$ and by $x_N^\bot $ an element of $
H_N^\bot $. For $x\in  H$, we will write $x=(P_Nx,Q_Nx)$. We make
the following assumptions about $F$.

\begin{itemize}

\item[(F1)] There exists $\beta \in (0,1)$ such that the
function $F:[0,T]\times\mathscr{D}(A^{\frac \beta 2}) \times 
\mathscr{D}(A^{\frac
\beta 2})\times S(H)\to \mathbb{R}$ is continuous (in the topology of
$[0,T]\times\mathscr{D}(A^{\frac \beta 2}) \times \mathscr{D}(A^{\frac \beta
2})\times S(H)$);

\item[(F2)] $F(t,x,p,S_1)\leqslant  F(t,x,p,S_2), \forall
t\in (0,T),\forall x,p\in \mathscr{D}(A^{\frac \beta 2})$, and all 
$S_1\geq S_2$;

\item[(F3)] There exists a modulus $\rho $ such that
\begin{align*}
&| F(t,x,p,S_1)- F(t,x,q,S_2)|\\
&\leqslant \rho\big ((1+ |A^{\frac \beta 2}x| ) |A^{\frac\beta 2}(p-q)|
+(1+ |A^{\frac \beta 2}x| ^2)\| S_1-S_2\|\big),
\end{align*}
for all $t\in (0,T)$, all $x,p,q\in \mathscr{D}(A^{\frac \beta 2}) $ and
all $S_1,S_2\in S(H)$;

\item[(F4)] There exist $0<\eta<1-\beta$ and a modulus $\omega$ such that,
for all $\varepsilon>0$, all $N\geq 1$, all $t\in (0,T)$, all
$x,y\in \mathscr{D}(A^{\frac \beta 2}) $ and $X,Y\in S(H_N)$ such that
\begin{equation}
\begin{pmatrix}
X&0\\
0&-Y
\end{pmatrix}
\leqslant\frac 2 \varepsilon
\begin{pmatrix}
P_NA^{-\eta }P_N&-P_NA^{-\eta }P_N\\
-P_NA^{-\eta }P_N&P_NA^{-\eta }P_N
\end{pmatrix}
\label{B4}
\end{equation}
we have
\begin{equation*}
\begin{split}
&F\Big( t,x,\frac {A^{-\eta }(x-y)}\varepsilon ,X\Big)
-F\Big( t,y,\frac {A^{-\eta }(x-y)}\varepsilon ,Y\Big)\\
&\geq -\omega\Big(|A^{\frac \beta 2}(x-y)| 
\big(1+\frac {|A^{\frac \beta 2}(x-y)|}\varepsilon \big)\Big);
\end{split}
\end{equation*}

\item[(F5)] For every $R<+\infty$, $|\lambda |\leqslant R$,
$t\in (0,T)$, $x,p\in \mathscr{D}(A^{\frac \beta 2}) $, we have
\begin{equation*}
\sup \big\{|F(t,x,p ,S+\lambda Q_N)-F(t,x,p ,S)|:
\| S\|\leqslant R, S=P_NSP_N\big\}\to 0
\end{equation*}
as $N\to\infty$.
\end{itemize}

\begin{remark} \rm
By the properties of moduli, condition (F3)
guarantees that there exists a constant $C$ such that for all
$t\in (0,T)$, all $x,p\in \mathscr{D}(A^{\frac \beta 2})$, all
$S\in S(H)$,
\begin{equation}
| F(t,x,p,S)|\leqslant C\Big(1+(1+ |A^{\frac \beta 2}x| )|A^{\frac \beta 2}p|
+(1+ |A^{\frac \beta 2}x|^2)\| S\|\Big)
+| F(t,x,0,0)|. \label{B5}
\end{equation}
\end{remark}

\section{Viscosity solutions}

The definition of a viscosity solution proposed here has its
predecessors in  \cite{g1,g2,s1}.


\begin{definition} \label{def3.1} \rm
A function $\psi : H\to \mathbb{R}$ is a  {\it test function} for the
equation in \eqref{CP} if
$$\psi(t,x)=\varphi(t,x)+\delta(t)(1+|x|^2),$$
where
\begin{enumerate}
\item $\delta\in C^1((0,T))$ and $\delta>0$ in $(0,T)$;

\item $\varphi \in C^{1,2}((0,T)\times H)$ and is weakly sequentially 
lower semicontinuous;

\item $D\varphi(t,.) \in UC( H, H)\cap UC(\mathscr{D}(A^{\frac 1 2-k}),
\mathscr{D}(A^{1/2}))$, for some $k =k(\varphi )>0$ and for all $t\in (0,T)$;

\item $D^2\varphi(t,.) \in BUC( H,S(H))$, for all $t\in (0,T)$.
\end{enumerate}
\end{definition}

\begin{definition} \rm
A weakly sequentially upper (lower) semicontinuous function 
$u: (0,T)\times H\to \mathbb{R}$ is a {\it viscosity subsolution} 
(respectively: {\it viscosity supersolution}) of the equation in \eqref{CP} 
if for every test function $\psi$, whenever $u-\psi$ has a local maximum  
(respectively: $u+\psi$ has a local minimum) at $(t,x)$ then 
$x\in \mathscr{D}(A^{1/2})$ and
$$
\psi_t(t,x)+\langle A^{1/2}x,A^{1/2}D\psi(t,x)\rangle
+ F(t,x,D\psi(t,x),D^2\psi(t,x))\leqslant 0
$$
(resp.
$$
-\psi_t(t,x)+\langle A^{1/2}x,-A^{1/2}D\psi(t,x)\rangle
+ F(t,x,-D\psi(t,x),-D^2\psi(t,x))\geq 0).
$$
A function $u$ is a {\it viscosity solution} of the equation in
\eqref{CP} if it is both a viscosity subsolution and a viscosity
supersolution.
\end{definition}

The main result of this paper is theorem below.

\begin{theorem} \label{thm3.1}
Let the Hypothesis (A1) and (F1)-(F5) hold.

\noindent{\bf Comparison:} Let $u,v:(0,T)\times H\to \mathbb{R}$ be
respectively a viscosity subsolution and a  viscosity
supersolution of the equation in \eqref{CP}. Assume that there
exists a constant $C$ such that
\begin{equation}
u(t,x),-v(t,x),|g(x)|\leqslant C(1+|x|)\label{C1}
\end{equation}
and
\begin{equation}
\begin{gathered}
(i)\quad \underset{t\downarrow 0}\lim (u(t,x)-g(x))^+=0\\
(ii)\quad \underset{t\downarrow 0}\lim (v(t,x)-g(x))^-=0
\end{gathered}\label{C2}
\end{equation}
uniformly on the bounded subsets in $H$. Then we have that
$u\leqslant v$ in $(0,T)\times H$.

\noindent {\bf Existence:} Let $g\in BUC(H)$ and
\begin{equation}
F_R=\sup\{ |F(t,x,p,X)|:(t,x)\in [0,T]\times \mathscr{D}(A^{1/2}),
|p|,\| X\|\leqslant R\}<+\infty.\label{C3}
\end{equation}
Then \eqref{CP} has a unique solution $u\in BUC_x([0,T]\times
H)\cap BUC_x([\tau,T]\times \mathscr{D}(A^{-\frac\eta 2} ))$ for $\tau>0$,
satisfying $\lim_{t\downarrow 0} u(t,x)=g(x)$ in $H$. Moreover,
there is a modulus $m$ such that
\begin{equation*}
|u(t,x)-u(s,e^{-(t-s)A}x)|\leqslant m(t-s)
\end{equation*}
for $0\leqslant s\leqslant t\leqslant T$ and $x\in H$.
\end{theorem}


Before we can attempt to prove the above theorem we would like
to begin with some facts about viscosity solutions of parabolic partial 
differential equations in finite dimensional spaces.
Those facts will be needed in the proofs of Theorem \ref{thm3.1}.
For the definition of viscosity solutions in this case, we refer to
\cite{c1}.

\begin{proposition}[{\cite[Proposition 3.4]{s1}}] \label{prop3.2}
Let an upper semicontinuous function $u$ and a lower semicontinuous
function $v$ on $(0,T)\times\mathbb{R}^N$ be respectively a viscosity
subsolution and a viscosity supersolution of
\begin{equation}
u_t(t,x)+F(t,x,Du(t,x),D^2u(t,x))=0\quad
\text{for }t\in (0,T),x\in \mathbb{R}^N,
\label{C4}
\end{equation}
where $F:([0,T]\times\mathbb{R}^N\times\mathbb{R}^N\times S(\mathbb{R}^N)\to 
\mathbb{R}$ is continuous
and satisfies the following three conditions:
\begin{itemize}
\item[(i)] $F(t,x,p,S_1)\leqslant  F(t,x,p,S_2)$,
for all $t\in (0,T)$, all $x,p\in \mathbb{R}^N$, all $S_1\geq S_2$;

\item[(ii)] There exists $\mu\in C^2(\mathbb{R}^N)$, radial, nondecreasing,
nonnegative, $\mu\to\infty$  as $\| x\|\to\infty$,
$D\mu, D^2\mu$ are bounded and
\begin{equation} \label{C5}
F(t,x,p+\alpha D\mu(x),X+\alpha D^2\mu(x))\geq F(t,x,p,X)
-\sigma(\alpha,|p|+\|X\|)\forall x,p,X,\forall\alpha\geq 0,
\end{equation}
where $\sigma$ is a local modulus;

\item[(iii)]  There exists a modulus
$\omega$: so that for all $t\in (0,T)$, all $x,y\in \mathbb{R}^N$,
all $X,Y\in S(\mathbb{R}^N)$ such that
\[
-c_1I \leqslant  \begin{pmatrix}
X&0\\
0&Y
\end{pmatrix}\leqslant
c_2 \begin{pmatrix}
I &-I \\
-I &I
\end{pmatrix},
\]
with the constants $c_1,c_2\geq 0$,  we have
\[
F(t,x,c_3(x-y),X)- F(t,y,c_3(x-y),-Y)\geq
-\omega (|x-y| (1+ (c_1+c_2+|c_3|)| x-y|)).
\]
\end{itemize}
Let $g\in BUC(\mathbb{R}^N)$. Then
\begin{itemize}
\item[(i)]
$u(t,x)-v(t,x)\leqslant\sup_{z\in\mathbb{R}^N}(u(0,z)-v(0,z))^+$
for all $t\in [0,T]$ and $x\in\mathbb{R}^N$.

\item[(ii)] If
$u(0,x)\leqslant g(x)\leqslant v(0,x)$
and $u,-v\leqslant M$, then there is a modulus of continuity $m$,
depending only on $M, \omega$ and a modulus of continuity of $g$,
such that
$$ u(t,x)-v(t,y)\leqslant m(|x-y|)
$$
for all $t\in [0,T)$ and $x,y\in\mathbb{R}^N$. Moreover, if $u=v$, then
$u\in C([0,T]\times\mathbb{R}^N)$.

\item[(iii)] If
$\sup_{t\in (0,T),x\in\mathbb{R}^N} |F(t,x,0,0)|=K<+\infty$,
then there exists a unique solution $u\in BUC_x([0,T]\times\mathbb{R}^N)$
of \eqref{C4} such that $u(0,x)=g(x)$ and $\|u\|_\infty$ only depends
on $\| g\|_\infty$ and $K$.
\end{itemize}
\end{proposition}

The Proposition below is needed in the proof of existence.

\begin{proposition}[{\cite[Lemma 2.8]{s1}}] \label{prop3.3}
If $F:(0,T)\times H\times H\times S(H)\to \mathbb{R}$ is uniformly
continuous on bounded sets, and satisfies (F2) and (F5) then for
every $(t,x,p,X)\in (0,T)\times H\times H\times S(H)$,
$$
F(t,P_Nx,P_Np,P_NXP_N)\to F(t,x,p,X)\quad\text{as }N\to \infty.
$$
\end{proposition}


\subsection*{Proof of Theorem \ref{thm3.1}: Comparison}
Given $\mu>0$, define
$$
u_\mu(t,x)=u(t,x)-\frac \mu {T-t},\quad
v_\mu(t,x )=v(t,x)+\frac \mu {T-t}.
$$
Then $u_\mu$ and $v_\mu$ satisfy respectively
$$
(u_\mu)_t(t,x)+\langle Ax,Du_\mu(t,x)\rangle+F(t,x,Du_\mu(t,x),D^2u_\mu(t,x))
\leqslant -\frac \mu{(T-t)^2}
$$
and
$$
(v_\mu)_t(t,x)+\langle Ax,Dv_\mu(t,x)\rangle+F(t,x,Dv_\mu(t,x),D^2v_\mu(t,x))
\geq \frac \mu{(T-t)^2}
$$
For $\epsilon ,\delta,\gamma >0, 0<t_\delta<T$ we consider the function
\begin{equation*}
\begin{split}
\Phi (t,s,x,y):=&u_\mu(t,x)-v_\mu(s,y)-\frac {|A^{-\frac \eta 2}(x-y)|^2}{2\varepsilon }\\
&- \delta e^{K_\mu t}(1+|x|^2)- \delta e^{K_\mu s}(1+|y|^2)-\frac {(t-s)^2}{2\gamma}.
\end{split}
\end{equation*}
The constant $K_\mu$ will be chosen later.

Since the function $\Phi$ is weakly sequentially upper
semicontinuous in $(0,T)\times(0,T)\times H\times  H$ and
(\ref{C1}), $\Phi$ has a global maximum over
$[t_\delta,T)\times[t_\delta,T)\times H\times H$ at some points
$(\bar{t},\bar{s},\bar{x},\bar{y})$, where $\bar{t},\bar{s}<T$ and 
$\bar{x},\bar{y}$ bounded
independently of $\varepsilon$ for a fixed $\delta$. We can assume
this maximum to be strict and (see \cite{g2,i1})
\begin{gather}
\limsup_{\delta  \to 0} \limsup_{\varepsilon  \to 0} \limsup_{\gamma  \to 0}
\delta(|\bar{x}|^2+|\bar{y}|^2 )=0 \label{C6}
\\
\limsup_{\varepsilon  \to 0} \limsup_{\gamma  \to 0}
\frac {|A^{-\frac \eta 2}(\bar{x}-\bar{y})|^2}{2\varepsilon} =0
\quad \text{for fixed }\delta >0. \label{C7} \\
\limsup_{\gamma  \to 0}  \frac {(\bar{t}-\bar{s})^2}{2\gamma}=0
\quad \text{for fixed }\varepsilon,\delta.
\label{C8}
\end{gather}
If $u\not\leqslant v$ it then follows from  (\ref{C7}), (\ref{C8}) and
(\ref{C2}) that for small $\mu$ and $\delta$, and $t_\delta$
sufficiently close $0$ we have $\bar{t},\bar{s}>t_\delta$ if $\gamma$ and
$\varepsilon$ sufficiently small.

We will now use a rather standard technique of reduction to finite
dimensional spaces to produce appropriate test functions.

We now fix $N\in\mathbb{N}$. Then obviously
\begin{equation*}
|A^{-\frac \eta 2}(x-y)|^2=\langle P_NA^{-\eta}P_N(x-y),x-y\rangle
+|A^{-\frac \eta 2}Q_N(x-y)|^2,
\end{equation*}
and we have
\begin{equation*}
\begin{split}
|A^{-\frac \eta 2}Q_N(x-y)|^2\leqslant& 2\langle Q_NA^{-\eta}Q_N(\bar{x}-\bar{y}),
x-y\rangle
-\langle Q_NA^{-\eta}Q_N(\bar{x}-\bar{y}),\bar{x}-\bar{y}\rangle\\
&+2|A^{-\frac \eta 2}Q_N(x-\bar{x})|^2+2|A^{-\frac \eta 2}Q_N(y-\bar{y})|^2
\end{split}
\end{equation*}
with equality if and only if  $x=\bar{x}$, $y=\bar{y}$. Therefore, if we
define
\begin{equation*}
\begin{split}
u_1(t,x)=&u_\mu(t,x)-\frac {\langle x,Q_NA^{-\eta}Q_N(\bar{x}-\bar{y})\rangle}\varepsilon\\
& +\frac {\langle Q_NA^{-\eta}Q_N(\bar{x}-\bar{y}),\bar{x}-\bar{y}\rangle}{2\varepsilon }
-\frac {|A^{-\frac {\eta}2}Q_N(x-\bar{x})|^2}\varepsilon - \delta e^{K_\mu t}(1+|x|^2)
\end{split}
\end{equation*}
and
\[
v_1(s,y)=v_\mu(s,y)-\frac {\langle y,Q_NA^{-\eta}Q_N(\bar{x}-\bar{y})
\rangle}\varepsilon
+\frac {|A^{-\frac {\eta}2}Q_N(y-\bar{y})|^2}\varepsilon + \delta e^{K_\mu s}(1+|y|^2),
\]
it follows that the function
\begin{equation}
\tilde{\Phi}(t,s,x,y):=u_1(t,x)-v_1(s,y)
-\frac {\langle P_NA^{-\eta}P_N(x-y),x-y\rangle}{2\varepsilon}
-\frac {(t-s)^2}{2\gamma}\label{C9}
\end{equation}
always satisfies $\tilde{\Phi}\leqslant\Phi$ and attains a strict global
maximum  over $[t_\delta,T)\times [t_\delta,T)\times H\times H$ at
$(\bar{t},\bar{s},\bar{x},\bar{y})$. Moreover,
$$
\tilde{\Phi}(\bar{t},\bar{s},\bar{x},\bar{y})=\Phi(\bar{t},\bar{s},\bar{x},\bar{y}).
$$
We now define, for $x_N,y_N\in  H_N$, the functions
$$
\tilde{u}_1(t,x_N):=\sup_{x_N^\bot \in  H_N^\bot} u_1(t,x_N,x_N^\bot),\quad
\tilde{v}_1(s,y_N):=\inf {y_N^\bot\in  H_N^\bot} v_1(s,y_N,y_N^\bot).
$$
Since the assumptions about $u,-v$ and the weakly sequentially
continuity of inner product,  we obtain that $\tilde{u}_1$ and $-\tilde{v}_1$
are upper semicontinuous on $(0,T)\times H_N$ (see \cite{c2}).
Moreover, by definition of $u_1, v_1$ and by the form of $\tilde{\Phi}$,
it follows that
\begin{equation}
\tilde{u}_1(\bar{t},P_N\bar{x})=u_1(\bar{t},\bar{x}),\quad\quad \tilde{v}_1(\bar{s},
P_N\bar{y})=v_1(\bar{s},\bar{y}). \label{C10}
\end{equation}
Defining now the map $\Phi_N: (0,T)\times (0,T)\times H_N\times  H_N\to 
\mathbb{R}$
as
\begin{equation*}
\begin{split}
&\Phi_N(t,s,x_N,y_N)\\
&:=\tilde{u}_1(t,x_N)-\tilde{v}_1(s,y_N)
-\frac {\langle P_NA^{-\eta}P_N(x_N-y_N),x_N-y_N)\rangle}{2 \varepsilon}
-\frac {(t-s)^2}{2\gamma}\\
&=\sup_{x_N^\bot,y_N^\bot\in  H_N^\bot} \tilde{\Phi}\big(t,s,(x_N,x_N^\bot),
(y_N,y_N^\bot)\big).
\end{split}
\end{equation*}
It is not difficult to check that $\Phi_N$ attains a strict global
maximum  at 
$(\bar{t},\bar{s},\bar{x}_N,\bar{y}_N)=(\bar{t},\bar{s},P_N\bar{x},P_N\bar{y})$. By
the finite dimensional results (see \cite{c1}) for every $n\in\mathbb{N}$, there
exist points  $t^n,s^n\in (0,T);x_N^n,y_N^n\in  H_N$ such that
\begin{gather}
t^n\to\bar{t},\quad s^n\to\bar{s};\quad x_N^n\to \bar{x}_N,\quad y_N^n\to \bar{y}_N,
\quad\text{as }n\to \infty.\label{C11}
\\
\tilde{u}_1(t^n,x_N^n)\to \tilde{u}_1(\bar{t},\bar{x}_N),\quad \tilde{v}_1(s^n,y_N^n)\to
\tilde{v}_1(\bar{s},\bar{y}_N) \quad \text{as }n\to \infty.\label{C12}
\end{gather}
and there exist functions $\varphi _n,\psi _n\in C^{1,2}((0,T)\times H_N)$
with uniformly continuous derivatives such that $\tilde{u}_1-\varphi _n$
and $-\tilde{v}_1+\psi_n$ have unique, strict, global maxima
at $(t^n,x_N^n)$ and $(s^n,y_N^n)$ respectively, and
\begin{equation}
\begin{gathered}
(\varphi_n)_t(t^n,x_N^n)\;\to\; \frac {\bar{t}-\bar{s}}\gamma,\\
D\varphi _n(t^n,x_N^n)\;\to\; \frac 1 \varepsilon P_NA^{-\eta}P_N(\bar{x}_N-\bar{y}_N),\\
(\psi_n)_t(s^n,y_N^n)\;\to\; \frac {\bar{t}-\bar{s}}\gamma,\\
D\psi _n(s^n,y_N^n)\;\to\; \frac 1 \varepsilon P_NA^{-\eta}P_N(\bar{x}_N-\bar{y}_N),\\
D^2\varphi _n(t^n,x_N^n)\;\to\; X_N,\\
D^2\psi _n(s^n,y_N^n)\;\to\; Y_N
\end{gathered}
\label{C13}
\end{equation}
where $X_N,Y_N$ satisfy (\ref{B4}).

Consider finally the map
$\Phi_N^n: (0,T)\times (0,T)\times H\times  H\to \mathbb{R}$ defined as
\begin{equation}
\Phi_N^n(t,s,x,y):=u_1(t,x)-v_1(s,y)-\varphi _n(t,P_Nx)+\psi_n(s,P_Ny). \label{C14}
\end{equation}
This map has the variables split and, by the definition of $u_1$
and $v_1$, attains its global maximum (which we can assume to be
strict) at some point $(\hat{t}^n,\hat{s}^n,\hat{x}^n,\hat{y}^n)$. This point
depends also on $N$ but we will drop this dependence since $N$ is
now fixed. Repeating now the arguments of \cite[page 409]{g1}
(see also \cite{v1}) it is not difficult to show that
\begin{gather}
u_1(\hat{t}^n,\hat{x}^n)\to u_1(\bar{t},\bar{x}),\quad v_1(\hat{s}^n,\hat{y}^n)
\to v_1(\bar{s},\bar{y})
\label{C15}
\\
\hat{t}^n=t^n,\quad \hat{s}^n=s^n;\quad \hat{x}_N^n=x_N^n,\quad \hat{y}_N^n=y_N^n,
\quad (\hat{x}^n,\hat{y}^n)\to (\bar{x},\bar{y})
\label{C16}
\end{gather}
as $n\to \infty$. Moreover $\bar{x},\bar{y}\in\mathscr{D}(A^{1/2})$ and
\begin{equation}
A^{1/2}\hat{x}^n\rightharpoonup A^{1/2}\bar{x},\quad A^{1/2}\hat{y}^n\rightharpoonup
A^{1/2}\bar{x}\quad \text{as }n\to \infty.\label{C17}
\end{equation}
We define
\begin{equation*}
\psi(t,x)=\frac{\langle x,Q_NA^{-\eta}Q_N(\bar{x}-\bar{y})\rangle}\varepsilon
+\frac{|A^{-\frac \eta 2}Q_N(x-\bar{x})|^2}\varepsilon
+\varphi_n(t,P_Nx)+\delta e^{K_\mu t}(1+|x|^2).
\end{equation*}
Then $\psi$ satisfies the conditions of a test function
(Definition \ref{def3.1}) and it follows from (\ref{C14}) and the definitions of
$u_1,v_1$ that $u_\mu-\psi$ has a maximum  at $(\hat{t}^n,\hat{x}^n)$.
Thus we have
\begin{equation}
\begin{split}
&\psi_t(\hat{t}^n,\hat{x}^n)+\langle A^{1/2}\hat{x}^n,A^{1/2}D\psi(\hat{t}^n,\hat{x}^n)
\rangle\\
&+F(\hat{t}^n,\hat{x}^n,D\psi(\hat{t}^n,\hat{x}^n),D^2\psi(\hat{t}^n,\hat{x}^n))
\leqslant-\frac \mu{(T-\hat{t}^n)^2}.
\end{split}\label{C18}
\end{equation}


We now like to pass to the limit as $n\to \infty$ in (\ref{C18})
keeping $\varepsilon ,\delta ,N$ fixed.
Since $A^{-\frac {1-\beta} 2}$ and $A^{-\frac \eta 2}$ are compact
we conclude that, as $n\to \infty$,
\begin{equation*}
A^{\frac {1-\eta} 2}\hat{x}^n=A^{-\frac \eta 2}(A^{1/2}\hat{x}^n)
\to A^{\frac{1- \eta} 2}\bar{x},\quad A^{\frac \beta 2}(\hat{x}^n)
= A^{-\frac {1-\beta} 2}(A^{1/2}\hat{x}^n)\to A^{\frac \beta 2}(\bar{x})
\end{equation*}
which together with the weakly semicontinuity of the norm implies
$$
\liminf_{n\to \infty}\langle A^{1/2}\hat{x}^n,A^{1/2}D\psi(\hat{t}^n,\hat{x}^n)
\rangle\geq \Big \langle A^{\frac {1-\eta}2}\bar{x},
\frac {A^{\frac {1-\eta}2}(\bar{x}-\bar{y})}\varepsilon \Big\rangle+2\delta
e^{K_\mu \bar{t}}|A^{1/2}\bar{x}|^2.
$$
On the other hand, using (\ref{C16}), (\ref{C11}) and (\ref{C13})
we have that, as $n\to \infty$,
\begin{gather*}
\psi_t(\hat{t}^n,\hat{x}^n)\;\to\;\frac {\bar{t}-\bar{s}}\gamma
+\delta K_\mu e^{K_\mu \bar{t}}(1+|\bar{x}|^2),\\
D\psi(\hat{t}^n,\hat{x}^n)\;\to\; \frac 1 \varepsilon A^{-\eta }(\bar{x}-\bar{y})
+2\delta e^{K_\mu \bar{t}}\bar{x},\\
D^2\psi(\hat{t}^n,\hat{x}^n)\;\to\; X_N+\frac {2A^{-\eta }Q_N}\varepsilon
+2\delta e^{K_\mu \bar{t}}I
\leqslant  X_N+\frac {2\|A^{-\eta }\| Q_N}\varepsilon +2\delta e^{K_\mu \bar{t}}I,
\end{gather*}
Therefore, using above results, (F1) and (F2), letting $n\to \infty$
in (\ref{C18}) yields
\begin{equation}
\begin{split}
&\frac {\bar{t}-\bar{s}}\gamma+\delta K_\mu e^{K_\mu \bar{t}}(1+|\bar{x}|^2)
+\frac 1 \varepsilon \big \langle A^{\frac {1-\eta}2}\bar{x},
A^{\frac {1-\eta}2}(\bar{x}-\bar{y})\big\rangle +2\delta e^{K_\mu\bar{t}}|A^{1/2}
\bar{x}|^2\\
&+F\Big(\bar{t},\bar{x},\frac 1 \varepsilon A^{-\eta }(\bar{x}-\bar{y}) +2\delta
e^{K_\mu\bar{t}} \bar{x},
X_N+\frac 2 \varepsilon \| A^{-\eta }\| Q_N+2\delta e^{K_\mu\bar{t}} I\Big)\\
&\leqslant -\frac \mu{(T-\bar{t})^2}.
\end{split}\label{C19}
\end{equation}
We now eliminate terms with $\delta $ and $N$. Using (F3) we
have
\begin{equation*}
\begin{split}
&F\Big(\bar{t},\bar{x},\frac 1 \varepsilon A^{-\eta }(\bar{x}-\bar{y}),X_N
+\frac 2 \varepsilon \| A^{-\eta }\| Q_N\Big)-\rho \big( d\delta e^{K_\mu\bar{t}}
(1+|\bar{x}|_\beta ^2)\big)\\
&\leqslant F\Big(\bar{t},\bar{x},\frac 1 \varepsilon A^{-\eta }(\bar{x}-\bar{y})
+2\delta e^{K_\mu\bar{t}} \bar{x},X_N+\frac 2 \varepsilon \| A^{-\eta }\| Q_N+2\delta 
e^{K_\mu\bar{t}} I\Big)
\end{split}
\end{equation*}
for some constant $d>0$. Now, given $\tau >0$, let $K_\tau $ be
such that
$$
\rho (s)\leqslant \tau +K_\tau s.
$$
Applying (\ref{B3}) with $\alpha =\frac \beta 2$ and $\gamma =\frac 1 2$,
we obtain that
$$
\rho \big( d\delta e^{K_\mu\bar{t}}(1+|\bar{x}|_\beta ^2)\big)
\leqslant \tau+2\delta e^{K_\mu\bar{t}}|A^{1/2}\bar{x}|^2
+\delta C_\tau e^{K_\mu\bar{t}}(1+|\bar{x}|^2)
$$
for some constant $C_\tau >0$ independent of $\delta $ and
$\varepsilon $. Therefore, using these results in (\ref{C19}),
(F5) and choosing  $K_\mu=C_\tau$, we obtain
\begin{equation}
\begin{split}
&\frac {\bar{t}-\bar{s}}\gamma +\frac 1 \varepsilon
\big \langle A^{\frac {1-\eta}2}\bar{x},A^{\frac {1-\eta}2}(\bar{x}-\bar{y})\big\rangle
+F\big(\bar{t},\bar{x},\frac 1 \varepsilon A^{-\eta }(\bar{x}-\bar{y}),X_N)\\
&\leqslant \tau +\omega_1(N;\varepsilon ,\delta,\gamma )-\frac \mu {(T-\bar{t})^2}.
\end{split}\label{C21}
\end{equation}
where $\lim_{N\to \infty}\omega_1(N;\varepsilon ,\delta,\gamma )=0$
if $\varepsilon ,\delta,\gamma $ are fixed.
Similarly, we obtain
\begin{equation}
\begin{split}
&\frac {\bar{t}-\bar{s}}\gamma +\frac 1 \varepsilon
\big \langle A^{\frac {1-\eta}2}\bar{y},A^{\frac {1-\eta}2}(\bar{x}-\bar{y})
\big\rangle
+F\big(\bar{s},\bar{y},\frac 1 \varepsilon A^{-\eta }(\bar{x}-\bar{y}),Y_N)\\
&\geq -\tau-\omega_1(N;\varepsilon ,\delta )+\frac \mu {(T-\bar{s})^2}.
\end{split}\label{C22}
\end{equation}
We now subtract (\ref{C21}) from (\ref{C22}), using ($F4$), and
then let $N\to \infty$. We then conclude that
\begin{equation}
\begin{split}
\frac \mu {(T-\bar{t})^2}+\frac \mu {(T-\bar{s})^2}
&\leqslant 2\tau +\omega\Big(|A^{\frac \beta 2}(\bar{x}-\bar{y})|
\big(1+\frac 1 \varepsilon |A^{\frac \beta 2}(\bar{x}-\bar{y})| \big)\Big)\\
&\quad -\frac 1 \varepsilon |A^{\frac {1-\eta} 2}(\bar{x}-\bar{y})|^2
\end{split}
\end{equation}
Set $r=|A^{\frac {1-\eta} 2}(\bar{x}-\bar{y})|$. Using the interpolation
inequality (\ref{B3}), the fact that $|A^{\frac \beta
2}(\bar{x}-\bar{y})|\leqslant c|A^{\frac {1-\eta} 2}(\bar{x}-\bar{y})|$ for some
 $c>0$
and the property of the moduli, we have that, for all
$\alpha,\sigma>0$, there exist $C_\sigma, K_\alpha>0$ such that
$$\frac \mu {(T-\bar{t})^2}+\frac \mu {(T-\bar{s})^2}\leqslant 2\tau  + \alpha 
+cK_\alpha\Big(\sigma\frac {r^2}\varepsilon+C_\sigma\frac {|A^{-\frac \eta 2}
(\bar{x}-\bar{y})|}\varepsilon r+r \Big)- \frac {r^2} \varepsilon.$$
For $\alpha$ fixed, we choose $\sigma$ such that
$cK_\alpha\sigma<1$. Then, in the right -hand side of the previous
inequality, we have a polynomial of order $2$ in  $\frac r
{\sqrt{\varepsilon}} $ which is bounded from above and we get
$$
\frac \mu {(T-\bar{t})^2}+\frac \mu {(T-\bar{s})^2}\leqslant 2\tau  + \alpha 
+\frac {K_\alpha^2c^2\big(\sqrt{\varepsilon} +C_\sigma
\frac {|A^{-\frac \eta 2}(\bar{x}-\bar{y})|}{\sqrt {\varepsilon}}\big)^2} 
{4(1-K_\alpha c\sigma)}.
$$
By sending $\gamma\to 0,\varepsilon\to 0, \delta\to 0$ and using
(\ref{C7}), we obtain a contradiction, which proves that we must
have $u\leqslant v$.

\subsection*{Existence}
To produce a solution of \eqref{CP} we consider approximation
\begin{equation}
\begin{gathered}
(u_N)_t(t,x)+\langle Ax,Du_N(t,x)\rangle
+F(t,x,Du_N(t,x),D^2u_N(t,x))=0,\\
(t,x)\in (0,T)\times H_N,\\
u_N(0,x)=g(x),\quad x\in H_N.
\end{gathered}\label{D3}
\end{equation}
Note that (\ref{D3}) satisfies the assumptions in (\ref{C5})
with constants and moduli independent of $N$. By Proposition \ref{prop3.2}
(iii), there is a unique solution $u_N\in BUC_x([0,T]\times H_N)$
of (\ref{D3}) such that $\| u_N\|_\infty\leqslant M$ for some $M$ which
depends only on $\| g\|_\infty, F_0$ in (\ref{C3}). Moreover, since
$A$ is positive definite, Proposition \ref{prop3.2} (ii) provides a modulus
of continuity $m_1$ such that
\begin{equation}
|u_N(t,x)-u_N(t,y)|\leqslant m_1(|x-y|),\quad\forall t\in (0,T),
\forall x,y\in H_N.\label{T1}
\end{equation}
We now show that for each $\tau>0$ there is a modulus $m_\tau$ such that
\begin{equation}
|u_N(t,x)-u_N(t,y)|\leqslant m_\tau(|A^{-\frac \eta 2}(x-y)|)
\quad\text{for }\tau\leqslant t\leqslant T.\label{T2}
\end{equation}
Given $\mu>0$, set
$$
u_0(t,x)=u_N(t,x)-\frac \mu{T-t}.
$$
Let $w$ be the modulus of continuity in (F4). For every
$\varepsilon>0$ let $K_\varepsilon$ be such that $w(r)\leqslant
\varepsilon/2+K_\varepsilon r$. For $L>M+1$, we set
$$
\psi_L(r)=2L2^{1-\frac 1 {2L}}r^{\frac 1 {2L}}.
$$
The function $\psi_L\in C^2(0,\infty)$ is increasing and concave,
$\psi'_L(r)\geq 1$ for $0<r\leqslant 2, \psi_L(0)=0, \psi_L(1)>2(M+1)$,
and
\begin{equation}
\psi_L(r)>L(\psi'_L(r)r+r)\quad\text{for } 0\leqslant r\leqslant 2.\label{D4}
\end{equation}
We will show that for every $\varepsilon>0$ there exists
$L=L_\varepsilon$ such that
\begin{equation}
u_0(t,x)-u_0(t,y)\leqslant \big(\psi_L(|A^{-\frac \eta 2}(x-y)|)+\varepsilon\big)(1+t)
\label{D5}
\end{equation}
for all $t\in (0,T); x,y\in H_N$.
Indeed, we denote by
$$
\Delta:=\{(x,y)\in H_N\times H_N:\ |A^{-\eta/ 2}(x-y)|<1\}.
$$
It is clear, from the properties of $\psi_L$, that for $(x,y)\not
\in \Delta$, (\ref{D5}) always satisfied independently of $L$.
Assume now by contradiction that (\ref{D5}) is false. Then, given
any $L>M+1$, let
$$
\psi(t,x,y)=\psi_L\big((|A^{-\frac \eta 2}(x-y)|+\varepsilon)\big)(1+t)
$$
we have that
$$
\sup_{t\in (0,T),(x,y)\in \Delta} \Big(u_0(t,x)-u_0(t,y)-\psi(t,x,y)\Big)>0
$$
(if not we are done). Then, for small $\delta>0$,
$$
\sup_{t\in (0,T),(x,y)\in \Delta}
\Big(u_0(t,x)-u_0(t,y)-\psi(t,x,y)-\delta |x|^2-\delta |y|^2\Big)>0
$$
and is attained at a point $(\bar{t},\bar{x},\bar{y})$  with 
$(\bar{x},\bar{y})\in
\Delta,\bar{x}\not =\bar{y}$. It follows from the initial condition and the
definition of $u_0,\psi_L$ that $0<\bar{t}<T$.

To use Proposition \ref{prop2.1}, we denote 
$s=|A^{-\frac \eta 2}(\bar{x}-\bar{y})|$ and compute
\begin{gather*}
\psi_t(\bar{t},\bar{x},\bar{y}) =\psi_L(s)+\varepsilon,\\
D_x\psi(\bar{t},\bar{x},\bar{y})  =\psi'_L(s)\frac{A^{-\eta}(\bar{x}-\bar{y})}s 
(1+\bar{t})\\
\begin{aligned}
D^2_{xx}\psi(\bar{t},\bar{x},\bar{y})
&=\psi''_L(s)\frac{A^{-\eta}(\bar{x}-\bar{y})\otimes
 A^{-\eta}(\bar{x}-\bar{y})}{s^2}
(1+\bar{t}) +\psi'_L(s)\frac{P_NA^{-\eta}}s (1+\bar{t})\\
&\quad -\psi'_L(s)\frac{A^{-\eta}(\bar{x}-\bar{y})\otimes
A^{-\eta}(\bar{x}-\bar{y})}{s^3}(1+\bar{t} )\\
&=B_1+B_2+B_3.
\end{aligned}
\end{gather*}
Since $\psi_L$ is nondecreasing and concave, $B_2\geq 0$ and
$B_1,B_3\leqslant 0$. Using this notation we have
\begin{align*}
D^2\psi(\bar{t},\bar{x},\bar{y})&=
\begin{pmatrix}
B_2&-B_2\\-B_2&B_2
\end{pmatrix}
-\begin{pmatrix}
-B_1-B_3&B_1+B_3\\
B_1+B_3&-B_1-B_3
\end{pmatrix}\\
&=D_1-D_2
\end{align*}
where $D_1,D_2\geq 0$.
Proposition \ref{prop2.1} applied with $\varepsilon=1,
u_1(t,x)=u_0(t,x)-\delta|x|^2,\ u_2(t,y)=-u_0(t,y)-\delta|y|^2 $ tells
us that there exist $a,b\in \mathbb{R}$, and matrices $X,Y\in S(\mathbb{R}^N)$
such that
\begin{gather*}
(a,D_x\psi(\bar{t},\bar{x},\bar{y}),X)\in \bar P^{2,+}u_1(\bar{t},\bar{x});\\
(-b,-D_y\psi(\bar{t},\bar{x},\bar{y}),-Y)\in \bar P^{2,-}(-u_2)(\bar{t},\bar{y})
\end{gather*}
where
\begin{gather*}
a+b=\psi_L(s)+\varepsilon,\\
\begin{pmatrix}
X&0\\0&Y
\end{pmatrix}
\leqslant (1+\bar{t})\frac{2\psi'(s)}s
\begin{pmatrix}
P_NA^{-\eta}P_N&-P_NA^{-\eta}P_N\\
-P_NA^{-\eta}P_N&P_NA^{-\eta}P_N
\end{pmatrix}
\end{gather*}
It follows from the properties of $\bar P^{2,+}$ and $\bar P^{2,-}$ that
\begin{gather*}
(a+\frac\mu{(T-\bar{t})^2},D_x\psi(\bar{t},\bar{x},\bar{y})+2\delta \bar{x},X+2\delta I )
\in \bar P^{2,+}u_N(\bar{t},\bar{x});\\
(-b+\frac\mu{(T-\bar{t})^2},-D_y\psi(\bar{t},\bar{x},\bar{y})-2\delta \bar{y},
-Y-2\delta I )
\in \bar P^{2,-}u_N(\bar{t},\bar{y}).
\end{gather*}
By the definition of viscosity solutions in case of finite dimensional,
we obtain
\begin{align*}
&a+\frac\mu{(T-\bar{t})^2}+\frac {\psi'_L(s)} s(1+\bar{t})
\langle A\bar{x},A^{-\eta}(\bar{x}-\bar{y})\rangle+2\delta\langle A^{1/2}\bar{x},
A^{1/2}\bar{x}\rangle\\
&+F(\bar{t},\bar{x},\frac {\psi'_L(s)} s(1+\bar{t})A^{-\eta}(\bar{x}-\bar{y})
+2\delta \bar{x},X+2\delta I )\leqslant 0
\end{align*}
and
\begin{align*}
&-b+\frac\mu{(T-\bar{t})^2}+\frac {\psi'_L(s)}
s(1+\bar{t})\langle A\bar{y},A^{-\eta}(\bar{x}-\bar{y})\rangle-2\delta\langle A^{1/2}\bar{y},A^{1/2}\bar{y}\rangle\\
&+F(\bar{t},\bar{y},\frac {\psi'_L(s)} s(1+\bar{t})A^{-\eta}(\bar{x}-\bar{y})-2\delta
\bar{y},-Y-2\delta I )\geq 0\,.
\end{align*}
Repeating the arguments from the proof of comparison we obtain that
\begin{equation*}
\begin{split}
a+b\leqslant& -\frac {\psi'_L(s)} s(1+\bar{t})|A^{\frac{1-\eta} 2}(\bar{x}-\bar{y})|^2\\
&+w\Big(|A^{\frac \beta 2}(\bar{x}-\bar{y})|\big(1+\frac {\psi'_L(s)} s(1+\bar{t})|A^{\frac\beta 2}(\bar{x}-\bar{y})|\big)\Big)+2w_1(L,\delta)\\
\leqslant& -\frac {\psi'_L(s)} s|A^{\frac{1-\eta} 2}(\bar{x}-\bar{y})|^2+\frac \varepsilon 2\\
&+K_\varepsilon\Big(|A^{\frac \beta 2}(\bar{x}-\bar{y})|\big(1+\frac {\psi'_L(s)} s(1+T)|A^{\frac\beta 2}(\bar{x}-\bar{y})|\big)\Big)+2w_1(L,\delta)
\end{split}
\end{equation*}
where $\limsup_{\delta\to 0} w_1(L,\delta)=0$. Therefore, using the
interpolation inequality (\ref{B3}) with a sufficiently small
$\sigma$, it follows that
\begin{equation*}
\begin{split}
\psi_L(s)+\varepsilon \leqslant& -\frac {\psi'_L(s)} {2s}|A^{\frac{1-\eta} 2}(\bar{x}-\bar{y})|^2+\frac \varepsilon 2\\
&+C_\varepsilon(\psi'_L(s)s+s)+\frac c 2 |A^{\frac {1-\eta} 2}(\bar{x}-\bar{y})|+2w_1(L,\delta)
\end{split}
\end{equation*}
where $C_\varepsilon$ depends only on $K_\varepsilon$ and the
interpolation constants (but not on $L$), and $c$ is such that
$|A^{\frac {1-\eta} 2}x|\geq c|A^{-\frac \eta 2}x|$ for all
$x\in \mathcal{D}(A^{\frac {1-\eta}2})$ \cite{g1}. Thus, we eventually have
$$\psi_L(s)\leqslant C_\varepsilon(\psi'_L(s)s+s)-\frac \varepsilon 2+2w_1(L,\delta),$$
which becomes, choosing $L=C_\varepsilon$ and letting $\delta\to 0$,
$$
\psi_L(s)\leqslant L(\psi'_L(s)s+s)-\frac \varepsilon 2.
$$
This leads to a contradiction in light of (\ref{D4}). Thus we have (\ref{D5}),
 which implies
$$
u_0(t,x)-u_0(t,y)\leqslant \psi_L(|A^{-\frac\eta 2}(x-y)|)(1+T)
+2M|A^{-\frac\eta 2}(x-y)|
$$
for all $x,y\in H_N$ and $t\in [0,T]$. We obtain the required
modulus of $u_N$ by letting $\mu\to 0$.

Next, we show that there is a modulus $m$ depending only on
$m_1$ and the function $F_R$, such that
\begin{equation}
|u_N(t,x)-u_N(s,e^{-(t-s)A}x)|\leqslant m(t-s)\label{D6}
\end{equation}
for $x\in H_N, 0\leqslant t\leqslant T$. Because of (\ref{T1}) it is enough
to show (\ref{D6}) for $s=0$ since all the estimates can be
reapplied at later time. To do this we begin with $g\in
C^{1,1}(H)$ such that $\|g\|_\infty<\infty$. We denote the Lipschitz
constant of $Dg$ by $L_{Dg}$. We use the fact that
$h(t,x)=g(e^{-tAx})$ solves
\begin{gather*}
h_t(t,x)+\langle Ax,Dh(t,x)\rangle=0\text{ in }(0,T]\times H_N,\\
h(0,x)=g(x) \quad \text{in }H_N
\end{gather*}
which implies
$$
u=h+tF_{\max(L_{Dg},\|Dg\|_\infty)}, \quad
v=h-tF_{\max(L_{Dg},\|Dg\|_\infty)}
$$
are respectively a viscosity supersolution and a viscosity
subsolution of (\ref{D3}). To see this we note that $h$ is
$C^{1,1}$ in $x, \|Dh\|_\infty\leqslant \|Dg\|_\infty$, and $L_{Dh}\leqslant
L_{Dg}$. It then suffices to observe (for a subsolution $u$) the
obvious fact that if for a test function $\varphi,\ u-\varphi$ has
a local maximum at $(t,x)$ then $D^2\varphi(t,x)\geq -L_{Dh}I \geq
-L_{Dg}I $. Hence, comparison gives
\begin{equation}
|u_N(t,x)-g(e^{-tA}x)|\leqslant tF_{\max(L_{Dg},\|Dg\|_\infty)}.\label{D7}
\end{equation}
If $g\in BUC(H)$ we can approximate it by a function $\tilde g\in
C^{1,1}(H)$ such that $\|D\tilde g\|_\infty<\infty$ \cite{l1}. Hence,
if $\tilde{u}_N$ solves (\ref{D3}) with $\tilde g$, then
\begin{equation*}
\begin{split}
|u_N(t,x)-g(e^{-tA}x)|&\leqslant |u_N(t,x)-\tilde{u}_N(t,x)|+|\tilde{u}_N(t,x)-\tilde g(e^{-tA}x)|\\
&\quad +|\tilde g(e^{-tA}x)-g(e^{-tA}x)|\\
&\leqslant 2\|g-\tilde g\|_\infty+tF_{\max(L_{Dg},\|Dg\|_\infty)},
\end{split}
\end{equation*}
where we have used (\ref{D7}) and Proposition \ref{prop3.2} (i).
This gives us (\ref{D6}).

Now set $v_N(t,x)=u_N(t,P_Nx)$. Since $A^{-\frac \eta 2}$ is
compact, (\ref{T2}) and (\ref{D6}) we have the equicontinuity of
$\{v_N\}$ in the weak topology on bounded subsets of
$[\tau,T]\times H$ for $\tau>0$. The Arzela-Ascoli theorem then
provides a subsequence (still denoted by $v_N$) converging
uniformly on bounded subsets of $[\tau,T]\times H$ to a function
$u$ that obviously satisfies the same estimates as $u_N's$
\cite{s1}.  Moreover, (\ref{D6}) imply that
$\lim_{t\downarrow 0} u(t,x)=g(x), x\in H$. It remains to show that
$u$ solves the limiting equation in \eqref{CP}. Let
$\psi(t,x)=\varphi(t,x)+\delta(t)(1+|x|^2)$ is a test function of the
equation in \eqref{CP} and let $u(t,x)-\psi(t,x)$ have a maximum
at $(\hat{t},\hat{x})$ which we may assume to be strict. It follows that
there exists a sequence $\hat{x}_N=P_N\hat{x} \to \hat{x}$ as $N\to \infty$ such
that, for every $x\in H_N$,
$$v_N(t,x)-\psi(t,x)\leqslant v_N(\hat{t},\hat{x}_N)-\psi(\hat{t},\hat{x}).$$
Therefore, since $AP_N=P_NA$,
\begin{equation}
\begin{split}
&\psi_t(\hat{t},\hat{x}_N)+\langle A^{1/2}\hat{x}_N,A^{1/2}D\varphi(\hat{t},\hat{x}_N)\rangle
+\delta(t)|A^{1/2}\hat{x}_N|^2\\
&+F(\hat{t},\hat{x}_N,P_ND\varphi(\hat{t},\hat{x}_N)+2\delta(\hat{t})\hat{x}_N,
P_N(D^2\varphi(\hat{t},\hat{x}_N)+2\delta(\hat{t})I)P_N)\leqslant 0.
\end{split}\label{D8}
\end{equation}
Since $\hat{x}_N\in H_N$ and $\psi$ is a test function we have
\begin{equation}
|A^{1/2}D\varphi(\hat{t},\hat{x}_N)|\leqslant B+C|A^{\frac 1 2-k}\hat{x}_N|\label{D9}
\end{equation}
for some independent constants $B,C$. Also, by (\ref{D9}),
(\ref{B3}), (\ref{B4}) and (\ref{C3}),
\begin{equation*}
\begin{split}
&\Big| F(\hat{t},\hat{x}_N,P_ND\varphi(\hat{t},\hat{x}_N)+\delta(\hat{t})\hat{x}_N,
P_N(D^2\varphi(\hat{t},\hat{x}_N)+\delta(\hat{t})I)P_N)\Big|\\
&\leqslant C_1\Big( 1+|A^{\frac \beta 2}\hat{x}_N|^2+|A^{\frac 1 2-k}\hat{x}_N|^2\Big)\\
&\leqslant C_2+\frac {\delta(\hat{t})} 4 |A^{1/2}\hat{x}_N|^2.
\end{split}
\end{equation*}
Using this, (\ref{D9}) and the interpolation inequality (\ref{B3}), we therefore obtain from (\ref{D8}) that
$|A^{1/2}\hat{x}_N|\leqslant C_3$
for some constant $C_3$ independent of $N$. Thus,
$A^{1/2}\hat{x}_N\rightharpoonup A^{1/2}\hat{x}$ (so $\hat{x}\in \mathscr{D}(A^{1/2}$)) and hence
$$
A^{\frac \beta 2}\hat{x}_N\to A^{\frac \beta 2}\hat{x},
\quad\text{and}\quad A^{1/2}D\varphi(\hat{t},\hat{x}_N)\to A^{1/2}D\varphi(\hat{t},\hat{x}).
$$
These convergence and Proposition \ref{prop3.3} allow us to pass to the
limit in (\ref{D8}) as $N\to \infty$ to conclude that
\begin{equation*}
\begin{split}
&(\psi)_t(\hat{t},\hat{x}) +\langle A^{1/2}\hat{x},A^{1/2}D\varphi(\hat{t},\hat{x})\rangle+\delta(\hat{t})|A^{1/2}\hat{x}|^2\\
&+F(\hat{t},\hat{x},D\varphi(\hat{t},\hat{x})+\delta(\hat{t})\hat{x},D^2\varphi(\hat{t},\hat{x})+\delta(\hat{t})I)\leqslant 0.
\end{split}
\end{equation*}
This proves that $u$ is a viscosity subsolution. Similarly, we
obtain that $u$ is a viscosity supersolution and therefore it is a
viscosity solution of the equation in \eqref{CP}. The comparison
gives us the uniqueness of $u$.

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\end{document}