\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{cite}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 113, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/113\hfil Noncontinuous solutions]
{Noncontinuous solutions to degenerate parabolic inequalities}

\author[K. A. Topolski \hfil EJDE-2015/113\hfilneg]
{Krzysztof A. Topolski}

\address{Krzysztof A. Topolski \newline
Institute of Mathematics, University of Gda\`nsk,
Wit Stwosz 57, 80-952 Gda\`nsk, Poland}
\email{matkt@mat.ug.edu.pl}

\thanks{Submitted October 2, 2014. Published April 28, 2015.}
\makeatletter
\@namedef{subjclassname@2010}{\textup{2010} Mathematics Subject Classification}
\makeatother
\subjclass[2010]{35D30, 35K51, 35R45}
\keywords{Parabolic equations; Cauchy problem; generalized solution}

\begin{abstract}
 We consider the initial value problem for degenerate parabolic
 equations.  We prove theorems on differential inequalities and
 comparison theorems in unbounded domain. As a solution of differential
 inequality we consider upper absolutely (lower absolutely) continuous
 in $t$ function (we admit discontinuity in time variable). In the last
 section we compare our notion of subsolutions  to the notion of viscosity
 subsolutions smooth in space variable. By giving a counterexample we show that
 upper absolutcontinuity plays  crucial role in the equivalence of the two notions.

\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}


The  aim of this paper is to investigate generalized inequalities
and  comparison problems for  degenerate
parabolic equations  with nonlinear comparison function.
 As it is well known they found numerous applications in differential
problems. The basic examples are estimates of solutions, estimates of the 
domain of existence of solutions,
uniqueness and error estimates for approximate solutions. Upper and lower 
functions are important in existence
results. In the paper we admit non-continuous in time solutions
for differential inequalities (semiabsolutely continuous in time
variable) and noncontinuous comparison functions.

In our opinion this paper is probably the first where
semiabsolutely in time variable solutions of partial differential
inequalities are considered. In the case of ordinary differential
inequalities non-continuous solutions where considered in
\cite{LPu} where some class of piecewise continuous  solutions
(lower and upper functions) were defined and in \cite{Pu} where
some special functions  of bounded variation were investigated.
Both cases are covered by our definition.


First order partial differential inequalities were first treated in \cite{H,N}.
Second order  inequalities of parabolic type were first treated in \cite{N2,N3,We}
where the first step involves
a strict inequality, and the result for weak inequality is then obtained by 
introduction of a suitable perturbation.
For the first time the classical theory of parabolic differential inequalities 
was widely described  in \cite{Sz0,W}. Comparison theorems for viscosity 
solutions  of  first order were first investigated in \cite{CL}. 
Uniqueness results were only obtainable at the time.
The best general reference for second order viscosity solutions is \cite{CIL}.

Our paper is divided into three main parts. In Section 2  we
present  a definition and properties of semiabsolutely continuous
functions of one variable. In Section 3, using this definition we
consider noncontinuous solutions of parabolic inequalities in
unbounded domain. We prove comparison theorems for generalized
subsolutions (supersolutions) and  solutions. In the last section we
compare our  notions of subsolutions to the notions of viscosity subsolutions 
smooth in space variable. We give sufficient conditions under which these  
two types of solutions are equivalent. We show that these  conditions are 
optimal and  that upper absolutcontinuity plays  crucial role in the equivalence.

\section{Semiabsolutely continuous functions}


The genesis of the notion of semiabsolutely continuous functions
dates to \cite{R} (see also \cite{CML} and \cite{P} where the term
absolute upper (lower) semicontinuous was used). In this paper we
base on the definition which is the most convenient in our
investigation. We follow some of the notation proposed in \cite{CML}.
In our notation the word semiabsolutly continiuous means
that a function is upper absolutely continuous or lower absolutely continuous.


\begin{definition} \label{def2.1} \rm
Let $a,b\in \mathbb{R}$, $a<b$,  $z:[a,b]\to \mathbb{R}$.
We write $z\in UAC([a,b],\mathbb{R})$  (upper absolutely continuous) if
$z'(t)$ exists a.e. in $[a,b]$ is integrable and for all $ s,t\in [a,b]$,
\begin{equation} \label{upper}
s\le t \Rightarrow z(t)-z(s)\le \int_{s}^{t}z'(\tau){d\tau}.
\end{equation}
Similarly, we write  $z\in LAC([a,b],\mathbb{R})$ (lower absolutely continuous ) if
$z'(t)$ exists a.e. in $[a,b]$ is integrable and for all $ s,t\in [a,b]$,
\begin{equation} \label{lower}
s\le t \Rightarrow z(t)-z(s)\ge \int_{s}^{t}z'(\tau){d\tau}.
\end{equation}
\end{definition}

It is clear that
 $AC([a,b],\mathbb{R})=UAC([a,b],\mathbb{R})\cap LAC([a,b],\mathbb{R})$, where $AC([a,b],\mathbb{R})$ 
is the set of all absolutely continuous
scalar function in $[a,b]$. Moreover, $z\in UAC([a,b],\mathbb{R})$ if
and only if $-z\in LAC([a,b],\mathbb{R})$.


\begin{remark} \label{left} \rm
If $z\in UAC([a,b],\mathbb{R})$ ($z\in LAC([a,b],\mathbb{R})$), then $z$ is
left-hand side lower (upper) semicontinuous and right-hand side
upper (lower) semicontinuous.
\end{remark}



\begin{remark}\label{monot1} \rm
Notice that if  $z\in UAC([a,b],\mathbb{R})$ then $z(t)-\int_{a}^{t}z'(\tau){d\tau} $ 
is nonincreasing. From the property of   monotone
functions and by the continuity and a.e. differentiability of
$\int_{a}^{t}z'(\tau){d\tau}$ we see that,
upper absolutely continuous function has at most
countably many points of discontinuity and one-sided limits in
every point of $[a,b]$.
\end{remark}



\begin{proposition} \label{prop2.4}
  $z\in UAC([a,b],\mathbb{R})$ if and only if
 there exists an integrable function $l:[a,b]\to \mathbb{R}$ such that
for all $ s,t\in [a,b]$,
\begin{equation} \label{upperint}
s\le t \Rightarrow z(t)-z(s)\le
\int_{s}^{t}l(\tau){d\tau}.
 \end{equation}
\end{proposition}

\begin{proof} 
The proof  (only  ``$\Leftarrow$'' is not obvious) follows from  \cite[Theorem 1]{P}.
\end{proof}

\begin{corollary}\label{monot}
A function $z:[a,b]\to \mathbb{R}$ is non-increasing if and only if
$z\in UAC([a,b],\mathbb{R})$  and $z'\le 0$
a.e. in $[a,b]$.
\end{corollary}


\begin{proposition} \label{AC}
Suppose that $z\in UAC([a,b],\mathbb{R})$ has a left-hand side local
maximum  in $\hat t\in (a,b]$ and $A$ is a full measure subset of
$[a,b]$. Then there exists a sequence $t_{m}\to \hat t^{-}$ such
that $t_{m}\in A$ for every $m$,  and  $z'(t_{m})\ge 0$.
\end{proposition}

\begin{proof}
 There exists $\delta_{1}>0$ such that $z(t)\le z(\hat t)$
in $[\hat t-\delta_{1},\hat t]$. The conclusion follows from the fact
that the following sentence is false: there exists
$0<\delta<\delta_{1}$ such that the set $\{t\in [\hat t-\delta,\hat
t]:z'(t)<0\}$ has Lebesgue measure $\delta$. Of course, it
follows from Corollary~\ref{monot} that $z$ is non-increasing in
$[\hat t-\delta,\hat t]$ for such $\delta$. This implies that $z$
is constant and $z'(t)=0$ in $[\hat t-\delta,\hat t]$, a contradiction.
\end{proof}

 We write $c^{+}=\max\{c,0\}$ and
$c^{-}=\max\{-c,0\}$ for  $c\in \mathbb{R}$. For a given scalar function
$z$ we define functions  $z^{+},\ z^{-}$ in an obvious way.

\begin{proposition} \label{plus}
If $z\in UAC([a,b],\mathbb{R})$, then $z^{+}\in UAC([a,b],\mathbb{R})$ and
$(z^{+})'=(\operatorname{sgn} z^+) z'$ a.e.
\end{proposition}

\begin{proof} Since  $z$ satisfies \eqref{upper},  for $t\ge s$, 
$z^{+}(t)-z^{+}(s)\le (z(t)-z(s))^{+}\le
\int_{s}^{t}{l^{+}(\tau){d\tau}}$ and $z^{+}\in UAC([a,b])$. 
Consider the  set $B\subseteq (a,b)$ of all $t$ such that $z'(t)$ and 
$(z^{+})'(t)$ exist.
 Of course, $B$ has  Lebesgue measure $b-a$. It is
not difficult to show that for $t\in B$ we have $(z^{+})'(t)=z'(t)$ if $z(t)>0$, 
and $(z^{+})'(t)=0$ if $z(t)<0$. Moreover, the set 
$\{t\in B: z(t)=0, z'(t)\neq 0\}$ contains only isolated points, hence 
it has at most countable many elements. On the other hand, if $t\in B$ 
is such that
$z(t)=0$,  $z'(t)=0$, then 
\[
(z^{+})'(t)=\lim_{h\to 0}{\frac{z^{+}(t+h)}{h}}
= \lim_{h\to 0}{\frac{(\operatorname{sgn} z^{+}(t+h))z(t+h)}{h}}=0.
\]
\end{proof}

Similarly, if $z\in LAC([a,b])$, then $z^{-}\in UAC([a,b])$ and
$(z^{-})' ={\rm -(sgn z^-)}z'$ a.e.

\section{Comparison theorems}

Define  $E_{T}=[0,T]\times\mathbb{R}^{n}$, $T > 0$,\ $E_{0}=\{0\}\times\mathbb{R}^{n}$, 
$\Theta_{T}=E_{T}\setminus E_{0}$. Let $S[n]$ be the set of all symmetric
$n\times n$ real matrices. For $X,\ Y\in S[n],\ X\le Y$ means that $Y-X$ is a  
positive semidefinite matrix. For $X\in S[n]$, $|X|$ is any matrix norm of
$X$ and for $p\in\mathbb{R}^n,\ |p|$ is any vector norm of $p$. 
Suppose that $g:\Theta_{T}\times \mathbb{R}\times \mathbb{R}^{n} \times S[n]\to \mathbb{R}$ is monotone 
in matrix variable i.e. if $X\le Y$, then 
$g(t,x,z,p,X)\le  g(t,x,z,p,Y)$ and $\psi:E_{0}\to\mathbb{R}$.
Consider problem
\begin{gather} \label{zag2}
D_{t}v  =  g(t,x,v,Dv,D^{2}v) \quad\text{in } \Theta_{T},\\
 \label{pocz2}
v  =  \psi \quad \text{in } E_{0}
\end{gather}
(we write  $Dv=D_{x}v$, $D^2v=D^2_{x}v$ and $D_{t}v=\frac{\partial}{\partial t}v$).

Notice  that our formulation includes  as a particular case the first order
equation ($g$ does not depend on a matrix argument).

\begin{definition}  \label{subsol} \rm
We say that  $v:E_{T}\to \mathbb{R}$  is a subsolution (supersolution, solution) of
\eqref{zag2} if
\begin{enumerate}
\item[(i)] for every $x\in \mathbb{R}^{n}$, $c\in (0,T)$,
$v(\cdot, x)\in UAC([c,T],\mathbb{R})$ $(LAC([c,T],\mathbb{R})$,  $AC([c,T],\mathbb{R}))$,

\item[(ii)] there  exist $Dv, D^{2}v$ in $\Theta_{T}$,

 \item[(iii)]  for every $x\in \mathbb{R}^{n},\ Dv(\cdot,x), D^{2}v(\cdot,x)$ 
are left-hand side continuous in $(0,T]$,

\item[(iv)] for every $x\in \mathbb{R}^{n}$ $v$ satisfies 
\begin{equation} \label{zag2sub}
 D_{t}v  \le g(t,x,v,Dv,D^{2}v)
\quad\text{a.e in } t\in (0,T]\quad (``\ge ",\ `` =").
\end{equation}
\end{enumerate}
(In case of  first order equations ($g$ does not depend on $X$) we assume
that (ii) and (iii) are satisfied only for $Dv$.)
\end{definition}

We write $ v\in \operatorname{Sub}(g,\psi)\ (\operatorname{Sup}(g,\psi), 
\ \operatorname{Sol}(g,\psi))$ if $v$ is a subsolution (supersolution, solution) of
\eqref{zag2}, and in
addition $v\le \psi$  ($``\ge", ``=" $) in $E_{0}$. Of course,
$\operatorname{Sol}(g,\psi)=\operatorname{Sub}(g,\psi)\cap
 \operatorname{Sup}(g,\psi)$.

We say  that  $ v$ satisfies \eqref{zag2sub} (\eqref{zag2sub}  with
a reversed inequality) in a generalized sense if it is a subsolution
(supersolution) of \eqref{zag2}.

Condition (iii) in Definition \ref{subsol} has a technical meaning. 
We need it in the proof of the  maximum
principle (see Theorem \ref{subsol}). It can  be relaxed (Remark \ref{class}).
 What is interesting is that it will be
necessary also in the last section where viscosity solutions are considered 
(see Theorem \ref{genvis}).
We give a simple  example of the problem such that (iii) in Definition \ref{subsol} 
is not satisfied.
Consider $D_{t}u+Du=0,\ u(0,x)=\psi(x)$ in $[0,2]\times \mathbb{R}$ where 
$\psi'$ exists is bounded but $\psi'$ is not right hand
continuous at $x=0$.
It is easy to verify that $u(t,x)=\psi(x-t)$ satisfies all condition of  
Definition \ref{subsol} except for (iii)
at point $(1,1)$.

The notion of subsolutions  (supersolutions, solutions) given in
Definition~\ref{subsol} extends the  definition of classical
subsolutions  (supersolutions, solutions) (Remark \ref{class}). 
Moreover, in the case of first order equations it
covers the definition of $CC-$ solutions considered by
Cinquni-Cibrario (see \cite{CC1,CC2}  for
existence results). In \cite{BM} close but not the same extension
of $CC-$ solutions is considered and some comparison results are
proved under the assumption that solutions exist.

The reason why we introduce  Definition~\ref{subsol} is the theorem
on differential inequalities (see Proposition~\ref{16zwycz7}  and
Corollary~\ref{difineq}).
 In the definition we  require as little as is needed in  the proof.
 Our notion of subsolutions (supersolutions, solutions) is placed between 
classical definition and more generalized definitions a.e.
 ``almost everywhere'' in the case of first order equations and viscosity
subsolutions (supersolutions, solutions) in the case of  first and
second order equations. It is, however, still close to the classical
meaning and we need relatively simple assumptions on $g$ to obtain
the theorem on differential inequalities and comparison results.
We cannot say the same when we consider ``almost everywhere''
solutions (first order equations) where convexity in $p$ is
required and uniqueness is proved  under additional ``entropy
condition'' (see \cite{K}). In the case of viscosity solutions
even more complicated assertions  are needed to obtain comparison
theorems (see \cite{CIL,CL}).

Let $C(E_{T},\mathbb{R})$, $LSC(E_{T},\mathbb{R})$, $USC(E_{T},\mathbb{R})$, be  sets of scalar functions
which are resp. continuous, lower semicontinuous  and upper semicontinuous 
in $E_{T}$.
Let $C_{b}(E_{T},\mathbb{R})$, $LSC_{b}(E_{T},\mathbb{R})$, $USC_{b}(E_{T},\mathbb{R})$
be sets of such functions which are in addition  resp.  bounded, bounded 
from below, and bounded from above.

We write $\omega \in \mathcal{M}$ if $\omega:[0,\infty)\to \mathbb{R}\cup \{\infty\}$ and
$\lim_{r\to 0^+}\omega(r)=\omega(0)=0$.


\begin{theorem} \label{16zwycz0plus}
Suppose that
\begin{enumerate}
\item[(i)] $v\in USC_{b}(E_{T},\mathbb{R})$ is a subsolution of \eqref{zag2}, 
\item[(i)] for every  $R>0$
there exists $\omega_{R}\in \mathcal{M} $ such that $\ g(t,x,z,p,X)\le
\omega_{R}(|p|+|X|)$ for  $\ z\in [0,R]$.
\end{enumerate}
Then 
\begin{equation}
\sup_{E_{T}} v^+ = \sup_{E_{0}} v^+.
 \end{equation}
\end{theorem}

\begin{proof} 
It is sufficient to prove that  $\sup_{E_{T}} v \le \sup_{E_{0}} v^+$. 
Define $w(t,x)=v(t,x)-\eta t$  for $\eta>0$. We will show
\begin{equation} \label{zz}
\sup_{E_{T}}{w}\le \sup_{E_{0}}{w^+}=
\sup_{E_{0}}{v^+}.
\end{equation}
 The proof will be completed  by letting
$\eta\to 0$.

Set $M=\sup_{E_{T}}{w}$. We only need  to consider the case $M>0$ for
some $\eta$.
For every $M/2>\delta>0$ there exists 
$(\bar t_{\delta},\bar x_{\delta})\in E_{T}$ such that
\[ 
M\ge w(\bar t_{\delta},\bar x_{\delta})>M-\delta>\frac{M}{2}.
\]
Define
\[
\Phi_{\delta}(t,x)=w(t,x)+2\delta\xi_{\delta}(x),
\]
where $\xi_{\delta}\in C^{2}_{0}(\mathbb{R}^{n}),\ \xi_{\delta}(\bar
x_\delta)=1,\ 0\le\xi_{\delta}\le 1,\  |D \xi_{\delta}|,\
|D^{2}\xi_{\delta}|\le 2$.


Since $\Phi_{\delta}(t,x)=w(t,x)\le M$ when $x\not\in \operatorname{supp}\xi_{\delta}$
and 
\begin{equation} \label{suprem}
\Phi_{\delta}(\bar t_{\delta}, \bar
x_\delta)=w(\bar t_{\delta}, \bar x_\delta)
+2\delta\xi_{\delta}(\bar x_{\delta})> M-\delta+2\delta=M+\delta> M
\end{equation}
$\Phi_{\delta}$ attains its
supremum in some $(t_{\delta},x_{\delta})\in E_{T}$, where
$x_{\delta}\in \operatorname{supp}\xi_{\delta}$ .

It follows from \eqref{suprem} that
\begin{equation} \label{suprem2}
w(t_{\delta}, x_\delta)+2\delta\ge \Phi_{\delta}(t_{\delta}, x_\delta)\ge\Phi_{\delta}(\bar t_{\delta}, \bar x_\delta)>
\frac{M}{2}+2\delta
\end{equation}
and consequently
\begin{equation} \label{suprem3}
v(t_{\delta}, x_\delta)\ge w(t_{\delta}, x_\delta)>\frac{M}{2}.
\end{equation}
Define $A_{\delta}$ as the full measure set of all $t\in [0,T]$
such that $D_{t}v(t,x_{\delta})$ exists and \eqref{zag2sub} is
satisfied at point $(t,x_{\delta})$. Let us fix a sequence
$\delta_{m}\in (0,\frac{M}{2}),\ m\in \mathbb{N}$ approaching  zero.
 We consider two cases (taking a subsequence if necessary).

(i) If $t_{\delta_{m}}=0$ for $m\in \mathbb{N}$ then
$\sup_{E_{T}}{\Phi_{\delta_{m}}}\le \sup_{E_{0}}{\Phi_{\delta_{m}}},\ m\in \mathbb{N}$,
which implies that \eqref{zz} holds true.

(ii) Suppose now that
$t_{\delta_{m}}>0,\ m\in \mathbb{N}$. Note that $D_{t}w=D_{t}\Phi_{\delta_{m}}$ (in
the set of existence). It follows from the fact that
$\Phi_{\delta_{m}}(\cdot,x_{\delta_{m}})$ has a local maximum in
$t_{\delta_{m}}$ (left-hand side if $t_{\delta_{m}}=T$) that for
every $m$ there exists $t_{k,m} \in A_{\delta_{m}}$ such that
$D_{t}w(t_{k,m},x_{\delta_{m}})\ge0$ and $t_{k,m}\to
t_{\delta_{m}}^{-}$ when $k\to \infty $ (see
Proposition~\ref{AC}). Moreover, by the left-hand side continuity of
$v$ in $t$ and by \eqref{suprem3} we can assume that $v(t_{k,m},x_{\delta_{m}})>0$
and $Dv(t_{k,m},x_{\delta_{m}}),\ D^2v(t_{k,m},x_{\delta_{m}})$ exist.


Since \eqref{zag2} is satisfied in $(t_{k,m},x_{\delta_{m}})$ and
$D_{t}v(t_{k,m},x_{\delta_{m}})=D_{t}w(t_{k,m},x_{\delta_{m}})+
\eta \ge \eta $ we obtain 
\begin{equation} \label{equat}
\eta \le
g(t_{k,m},x_{\delta_{m}},v(t_{k,m},x_{\delta_{m}}),Dv(t_{k,m},
x_{\delta_{m}}),D^{2}v(t_{k,m},x_{\delta_{m}}))
\end{equation}
 where
\begin{equation} \label{der1} Dv(t_{k,m},x_{\delta_{m}})
=Dv(t_{k,m},x_{\delta_{m}})-Dv(t_{\delta_{m}},x_{\delta_{m}})-2\delta_{m}
D\xi_{\delta}(x_{\delta})
\end{equation}
($D\Phi(t_{\delta_{m}},x_{\delta_{m}})=Dv(t_{\delta_{m}},x_{\delta_{m}})
+2\delta_{m} D\xi_{\delta_{m}}(x_{\delta_{m}})= 0$) and
\begin{equation} \label{der2}
D^2 v(t_{k,m},x_{\delta_{m}})\le  D^2
v(t_{k,m},x_{\delta_{m}})-D^2 v(t_{\delta_{m}},x_{\delta_{m}})
-2\delta_{m} D^{2}\xi_{\delta_{m}}(x_{\delta_{m}})=B_{k,m}
\end{equation}
($D^{2}\Phi(t_{\delta_{m}},x_{\delta_{m}})=D^2
w(t_{\delta_{m}},x_{\delta_{m}})+2\delta
D^{2}\xi_{\delta}(x_{\delta_{m}})\le 0$).

We will show that the above estimation leads to a contradiction.
It follows from $(ii)$ that for every $\varepsilon>0$ and $R>0$  there exists
$\rho>0$ such that $\ g(t,x,z,p,X)\le \varepsilon\ $ if $\ |p|,\ |X|<\rho$
and $z\in [0,R]$. Let $\rho>0 $ be such that $g(t,x,z,p,X)\le
\eta/2$ for $|p|,|X|\le \rho$,
$0 \le z\le  R= \sup_{E_{T}}{|v^+|}$.
Fix $m$ such that $\delta_{m}< \frac{\rho}{8}$. This gives
$2\delta_{m}|D^{2}\xi_{\delta_{m}}(x_{\delta_{m}})|\le
\frac{\rho}{2}$ and 
$2\delta_{m}|D\xi_{\delta_{m}}(x_{\delta_{m}})|\le \rho/2$.

On the other hand, (see Definition \ref{subsol} (iii)) there exists $k$ such that
\begin{gather*}
|Dv(t_{k,m},x_{\delta_{m}})-Dv(t_{\delta_{m}},x_{\delta_{m}})|<
\frac{\rho}{2}, \\
 |D^{2} v(t_{k,m},x_{\delta_{m}})-D^{2}
v(t_{\delta_{m}},x_{\delta_{m}})|\le \frac{\rho}{2}.
\end{gather*}
 Applying
\eqref{der1}, \eqref{der2} we have $|Dv(t_{k,m},x_{\delta_{m}})|, \
|B_{k,m}|< \rho $. Finally, in view of \eqref{equat} and
\eqref{der2} we obtain
\[
\eta \le g(t_{k,m},x_{\delta_{m}},v(t_{k,m},x_{\delta_{m}}),Dv(t_{k,m},
x_{\delta_{m}}), B_{k,m})
\le  \frac{\eta}{2}
\]
 a contradiction.
\end{proof}

\begin{remark}\label{uwag} \rm
The statement of Theorem \ref{subsol} holds  if we assume that
(ii)-(iv) in Definition \ref{subsol} hold only in the set 
$ v^{-1}((0,\infty))\cap \Theta_{T}$.
Note that if $v$ satisfies (i) of Definition \ref{subsol} then
 $v(t,x)>0,\ t>0$ implies that $v(s,x)>0,\ s\in (t-\varepsilon,t]$ for some  
$\varepsilon >0$ (see   Remark \ref{left}).
\end{remark}

\begin{remark}\label{class} \rm
The statement of Theorem \ref{subsol} holds  if in place of Definition 
\ref{subsol} (iii) we assume that
 $Dv(\cdot,x), D^{2}v(\cdot,x)$ are left-hand side continuous
in every point $(t,x)\in (0,T]\times \mathbb{R}^{n}$ such that $D_{t}v(t,x)$ does not exist.
\end{remark}


\begin{proposition}  \label{16zwycz2}
Suppose that
\begin{enumerate}
\item[(i)] $v\in USC_{b}(E_{T},\mathbb{R})$ is a subsolution of \eqref{zag2},
\item[(ii)] there exists integrable  function $h: [0,T]\to \mathbb{R}_{+}$  and for
every $R>0$ there exists $\omega_{R}\in \mathcal{M} $ such that
   $\ g(t,x,z,p,X)\le h(t)+\omega_{R}(|p|+|X|)\ $
if $\ z\in [0,R]$.
\end{enumerate}
Then
\[
  \sup_{E_{t}} v^{+} \le \sup_{E_{0}} v^{+}  + \int_{0}^{t}h(s){ds} \quad
\text{for } t\in [0,T].
\]
\end{proposition}

\begin{proof}
Set $\bar v(t,x)= v(t,x)-\int_{0}^{t}{h(s)}{ds}$ and
\[
\bar g(t,x,z,p,X)=g(t,x,z+\int_{0}^{t}{h(s)}{ds}, p, X)-h(t).
\]
For $z\in [0,R]$ we have
\[
\bar g(t,x,z,p,X)=g(t,x,z+ \int_{0}^{t}{h(s)}{ds}, p, X)-h(t)\le \omega_{R+R_{1}}(|p|+|X|)
\]
where $ R_{1}=\int_{0}^{T}{h(s)}{ds}$.

It follows easily that $\bar v$ is a subsolution of \eqref{zag2} with
 $g$ repleaded by $\bar g$.
 Thus we can apply Theorem \ref{16zwycz0plus} to $\bar v$ and $\bar g$
in the set $E_{t}=\{(s,x)\in E_{T}: s\le t\}$. This gives:
 \[
\sup_{E_{t}} v^+-\int_{0}^{t}{h(s)}{ds} \le \sup_{(\tau,x)\in E_{t}}
\{(v(\tau,x)-\int_{0}^{\tau }{h(s)}{ds})^+\}= \sup_{E_{0}}v^+
\]
which completes the proof.
\end{proof}

Analogically to Remark \ref{uwag} we can formulate

\begin{remark}\label{uwag1} \rm
The statement of Proposition \ref{16zwycz2} holds true if we assume that
 (ii)-(iv) in Definition \ref{subsol} hold only in the set
$Z=\{(t,x)\in \Theta_{T}: v^{+}(t,x)>\int_{0}^{t}{h(s)}{ds}\}$.
\end{remark}


\begin{remark}\label{zero} \rm
 It follows from Proposition~\ref{16zwycz2} (ii) that $g(t,x,z,0,0)\le h(t)$ 
in $\Theta_{T}\times \mathbb{R}_{+}$. Moreover,
if we assume this  and the following: for every $R>0$  there
exists $\omega_{R}\in \mathcal{M}$
 such that $g(t,x,z,p,X)-g(t,x,z,0,0)\le \omega_{R}(|p|+|X|)$
 for $\ z\in [0,R]$, then (ii) is satisfied.
\end{remark}


\begin{definition} \label{om} \rm
For $M\in \mathbb{R}_{+}$ and $\sigma:[0,T]\times \mathbb{R}_+ \mapsto \mathbb{R}_+$ we
write $\sigma\in O_{M}$, if
\begin{enumerate}
\item[(i)] $\sigma=\sigma(t,z)$ is nondecreasing in  $z$,
\item[(ii)] $\sigma(t,z(t))$ is integrable for every nondecreasing
$z:[0,T]\to \mathbb{R}$,
\item[(iii)] there exists $\mu(t)=\mu_{\sigma}(t,M)\in
AC([0,T],\mathbb{R}_{+})$ such that $\mu(t)$
 is a solution of the problem
\begin{equation}  \label{sig}
z'(t)=\sigma(t,z(t)),\quad \text{a.e.  in }[0,T],\quad
 z(0)=M
\end{equation}
 and
\[
\mu(t)= \max \{ z\in
AC([0,T],\mathbb{R}): z'(t)\le\sigma(t,z(t))\text{ a.e in }[0,T],\; z(0)=M\}.
\]
\end{enumerate}
\end{definition}


A typical example of $\sigma\in O_{M}$ (for every $M\ge 0$) is
 $\sigma(t,z)=l(t)z+m(t)$ 
where $l:[0,T]\to \mathbb{R}_{+}$ integrable.
It is not difficult to show  that we can also take $\sigma$ which  is
nondecreasing in both variables and sublinear or if it satisfies the well known
Carath\'eodory conditions (see \cite {HR} for even more general conditions).
Note that $\sigma$ is not supposed to be continuous.

 For $v:E_{T}\to \mathbb{R}$ bounded and $0\le t\le T$  we define 
$\|v\|_{t}= \sup_{E_{t}}{|v|}$.

\begin{proposition} \label{16zwyczpl}
Suppose that
\begin{enumerate}
\item[(i)] $v\in \operatorname{Sub}(g,\psi)\cap USC_{b}(E_{T},\mathbb{R}), M^{+}
=\sup_{E_{0}} \psi^{+}$,
\item[(ii)] there exists  $\sigma\in O_{M^{+}}$ and for every  $R>0$
there exists $\omega_{R}\in\mathcal{M}$ such that $\ g(t,x,z,p,X)\le
\sigma(t,z)+\omega_{R}(|p|+|X|) $ if $ z\in [0,R]$.
\end{enumerate}
Then
\[
 \|v^{+}\|_{t} \le \mu_{\sigma}(t,M^{+}) \quad\text{for } t\in [0,T].
\]
\end{proposition}

\begin{proof}
Define $\bar g(t,x,z,p,X)=g(t,x,v^{+}(t,x),p,X)$. Notice that $\bar g$ 
satisfies assumption (ii) of Proposition \ref{16zwycz2} with
$h(t)=\sigma(t,\|v^{+}\|_{t})\ge 0$ independently of $R$ ($\bar g$ does not 
depend on $z$). Indeed,
\begin{align*}
\bar g(t,x,z,p,X)
&=g(t,x,v^{+}(t,x),p,X)
 \le\sigma(t,v^{+}(t,x))+\omega_{\tilde R}(|p|+|X|)\\
&\le \sigma(t,\|v^{+}\|_{t})+\omega_{\tilde R}(|p|+|X|)
 =h(t)+\omega_{\tilde R}(|p|+|X|)
\end{align*}
where $\tilde R=\|v^{+}\|_{T}$. On the other hand  $v^{+}$ satisfies
\[
D_{t}v\le \bar g(t,x,v,D_{x}v,D_{x}^{2}v)
\]
in the set $Z=\{(t,x)\in \Theta_{T}: v^{+}(t,x)>\int_{0}^{t}{h(s)}{ds}\}$. 
Indeed since $v^{+}=v$, $D_{t}v^{+}=D_{t}v$,
$Dv^{+}=Dv$, $D^{2}v^{+}=D^{2}v$ in $Z$,
we obtain
\begin{align*}
D_{t}v^{+}
&= D_{t}v\le g(t,x,v,Dv,D^{2}v)
=g(t,x,v^{+},Dv^{+},D^{2}v^{+})\\
&=\bar g(t,x,v^{+},Dv^{+},D^{2}v^{+})
\end{align*}
in $Z$ a.e in $t$.


This gives (see Proposition \ref{16zwycz2} and Remark \ref{uwag1})
\[
\|v^{+}\|_{t}\le \|v^{+}\|_{0} 
+\int_{0}^{t}{\sigma(\tau,\|v^{+}\|_{\tau})}{d\tau}.
\]
Similarly,  for $t\ge s$ we have
\[
\|v^{+}\|_{t}\le \|v^{+}\|_{s} 
+\int_{s}^{t}{\sigma(\tau,\|v^{+}\|_{\tau})}{d\tau}.
\]
This implies that $ \alpha(t)=\|v^{+}\|_{t}$ is in $UAC([0,T])$ and
$\alpha'(t)\le \sigma(\tau,\alpha(t))$ a.e in $[0,T]$. Since $\alpha$
is nondecreasing, it belongs to  $AC([0,T])$. This completes the
proof in view of Definition~\ref{om} (iii).
\end{proof}

\begin{proposition}  \label{16zwyczmin}
Suppose that
\begin{enumerate}
\item[(i)] $\ v\in \operatorname{Sup}(g,\psi)\cap LSC_{b}(E_{T},\mathbb{R})$,
$M^{-}=\sup_{E_{0}} \psi^{-}$,
\item[(ii)] there exists  $\sigma\in O_{M^{-}}$ and for every  $R>0$
there exists $\omega_{R}\in\mathcal{M}$  such that $\ g(t,x,z,p,X)\ge
-\sigma(t,-z)-\omega_{R}(|p|+|X|)\ $ if $ z\in [-R,0]$.
\end{enumerate}
Then
\[
 \|v^{-}\|_{t} \le \mu_{\sigma}(t,M^{-}) \quad \text{for } t\in [0,T].
\]
\end{proposition}

\begin{proof} 
Notice that $v^{-}=(-v)^{+}$, $\psi^{-}=(-\psi)^{+}$, 
$-v \in \operatorname{Sub}(\tilde g,-\psi)$
where 
\[
\tilde g(t,x,z,p,X)=-g(t,x,-z,-p,-X)
\]
 and $\tilde g$ satisfies
all assumptions  of Proposition~\ref{16zwyczpl}.
\end{proof}

\begin{corollary} 
Let assumptions of Propositions \ref{16zwyczmin} hold, 
and $\mu_{\sigma}(\cdot,0)\equiv 0$.
Then $\psi\ge 0$ implies $v\ge 0$.
\end{corollary}

\begin{proposition}  \label{16zwyczsol}
Suppose that
\begin{enumerate}
\item[(i)] $v\in \operatorname{Sol}(g,\psi)\cap C_{b}(E_{T},\mathbb{R})$,
$M=\sup_{E_{0}} |\psi|$,
\item[(ii)] there exists  $\sigma\in O_{M}$ and for every $R>0$ there
exists $\omega_{R}\in\mathcal{M}$ such that
\[
(\operatorname{sgn}z)g(t,x,z,p,X)\le
\sigma(t,|z|)+\omega_{R}(|p|+|X|)\quad{\rm for}\ z\in [-R,R].
\]
\end{enumerate}
Then
\[
\|v\|_{t} \le \mu_{\sigma}(t,M) \quad\text{for } t\in [0,T].
\]
\end{proposition}

\begin{proof} 
Since $|v|= \max{\{v^{+},v^{-}\}},\ M=\max{\{M^{+},M^{-}\}}$ and 
$\mu(t,M^{-}),\mu(t,M^{+})\le \mu(t,M)$ the conclusion  follows from 
Propositions \ref{16zwyczpl} and \ref{16zwyczmin}.
\end{proof}


\begin{proposition}  \label{16zwycz7}
Suppose that
\begin{enumerate}
\item[(i)] $v\in \operatorname{Sub}(g,\psi)\cap USC_{b}(E_{T},\mathbb{R})$,
$\bar v\in \operatorname{Sup}(\bar g,\bar\psi)$,  $v-\bar v\in USC_{b}(E_{T},\mathbb{R})$
$M^{+}=\sup_{E_{0}} (\psi-\bar \psi)^{+}$,

\item[(ii)] there exists  $\sigma\in O_{M^{+}}$ and for every  $R>0$
there exists $\omega_{R}\in\mathcal{M}$ such that
\[
g(t,x,z,p,X)-\bar g(t,x,\bar z,\bar p,\bar X)\le \sigma(t, z-\bar z)+
\omega_{R}(|p-\bar p|+|X-\bar X|)
\]
for $z-\bar z\in [0,R]$.
\end{enumerate}
Then
\[
 \|(v-\bar v)^{+}\|_{t} \le  \mu_{\sigma}(t,M^{+})
\quad\text{for } t\in [0,T].
\]
\end{proposition}

\begin{proof} 
It is easy to check that  $v-\bar v \in
\operatorname{Sub}(G,\psi-\bar\psi)$ where 
\begin{equation} \label{gef}
G(t,x,z,p, X)= g(t,x,z+\bar v,p+D \bar v, X+D^{2}\bar v)-\bar g(t,x,\bar v,D\bar
v,D^{2}\bar v).
\end{equation}
The conclusion follows from Proposition \ref{16zwyczpl}.
\end{proof}

In a similar way Proposition \ref{16zwyczsol} yields the following result.

\begin{proposition}  \label{16zwycz8}
Suppose that
\begin{enumerate}
\item[(i)] $v\in \operatorname{Sol}(g,\psi)$, 
$\bar v\in \operatorname{Sol}(\bar g,\bar \psi)$,  
$v-\bar v\in C_{b}(E_{T},\mathbb{R})$   and 
$M=\sup_{E_{0}} |\psi-\bar \psi|$,

\item[(ii)] there exists  $\sigma\in O_{M}$ and for every $R>0$ there
exists $\omega_{R}\in\mathcal{M}$ such that
\[
\operatorname{sgn}(z-\bar z) [g(t,x,z,p,X)-\bar g(t,x,\bar z,\bar p,\bar X)]\le
\sigma(t,|z-\bar z|)+\omega_{R}(|p-\bar p|+|X-\bar X|)
\]
for $|z-\bar z|\le R$.
\end{enumerate}
Then
\[
 \|v- \bar v\|_{t} \le  \mu_{\sigma}(t,M)
\quad\text{for } t\in [0,T].
\]
\end{proposition}

\begin{corollary}\label{difineq}
If $\mu_{\sigma}(\cdot,0)\equiv 0,\ g=\bar g,\ \psi=\bar \psi$, then
Proposition~\ref {16zwycz7} implies  the theorem on differential
inequalities and Proposition~\ref{16zwycz8} implies the theorem on
the uniqueness for problem \eqref{zag2} \eqref{pocz2}.
\end{corollary}

\section{Viscosity solutions}

\begin{definition} \label{visc} \rm
We say that $u:E_{T}\to \mathbb{R}$ is a viscosity subsolution (resp.
supersolution) of \eqref{zag2} if $u\in USC(E_{T},\mathbb{R})$ (resp. 
$u\in LSC(E_{T},\mathbb{R})$)
 and for each $\phi \in C^{1,2}(\Theta_{T})$ if $u-\phi$ attains a local maximum (resp.
local minimum) at $(\tilde t,\bar x)\in\Theta_{T}$, then
\[
  D_t\phi(\tilde t,\tilde x)\le g(\tilde t,\tilde x,u(\tilde t,\tilde x),
D\phi(\tilde t,\tilde x),D^{2}\phi(\tilde t,\tilde x))
  \quad (\text{resp.} \ge )
\]
 We say that $u$ is a viscosity solution of \eqref{zag2} if it is both
viscosity subsolution  and supersolution.
\end{definition}

It follows easily that if $u\in C^{1,2}(\Theta_{T})$, then
 $u$ is a viscosity subsolution (supersolution, solution) of \eqref{zag2} 
if and only if it is a classical
 subsolution (supersolution, solution). We will extend this result to 
 subsolutions (supersolutions, solutions) given by
 Definition \ref{subsol}

First we present rather known  result which generalize this find in
 \cite{E} (for first order equations).

\begin{lemma}\label{deriv} 
Suppose that $u:E_{T} \to \mathbb{R}$ and $D_{t}u, Du,\ D^2u$  exist at 
$(\bar t,\bar x)\in \Theta_{T}$ and
in some neighborhood of $(\bar t,\bar x)$ we have
\begin{equation} \label{differ}
\begin{aligned}
u(t,x)&=u(\bar t, \bar x)+ D_{t}u(\bar t,\bar x)(t-\bar t)
+\langle Du(\bar t,\bar x),(x-\bar x)\rangle \\
&\quad  +\langle D^{2}u(\bar t,\bar x)(x-\bar x),x-\bar x\rangle +
o(|t-\bar t|+|x-\bar x|^2).
\end{aligned}
\end{equation}
Then there exists $\phi\in C^{1,2}(\Theta_{T})$ such that $u-\phi$ has a
maximum at $(\bar t,\bar x)$ and
 $D_{t}u(\bar t,\bar x)=D_{t}\phi (\bar t,\bar x)$,
$Du(\bar t,\bar x)=D\phi (\bar t,\bar x)$,
$D^2u(\bar t,\bar x)=D^2\phi(\bar t,\bar x) $.
\end{lemma}

\begin{proof}
 Without loss of generality we can assume that
$u:\mathbb{R}^{1+n}\to \mathbb{R}$,  $D_{t}u, Du,\ D^2u$ exist at $(0,0)$ and
$u(0,0)=D_{t}u(0,0)=Du(0,0)=D^2u(0,0)=0$.

Define  $\alpha(t,x)=\frac{|x|^2+t^{2}}{\sqrt{|x|^4+t^2}}$ for
$(t,x)\neq (0,0)$ and $\alpha(0,0)=0$. Set $\bar u=\alpha u$ $y=(t,x)$
and $\bar u=\bar u(y)$. Since $\frac{1}{2}(|x|^2+|t|)\le
\sqrt{|x|^4+t^2}\le |x|^2+|t|$ we easily see that 
$D_{y}\bar u$, $D^2_{y}\bar u$ exist at $0\in \mathbb{R}^{1+n}$ and 
$\bar u(0)=D_{y}\bar u(0)= D^2_{y}\bar u(0)=0$.
 We define
$\rho(y)=\frac{\bar u(y)}{|y|^2}, y \neq 0$
and $\rho(0)=0$. Set
 $\rho_{1}(r)=\max_{ |y|\le r}{|\rho(y)|}$ and
\[
\bar\phi(y)=\int_{|y|}^{2|y|}{d\tau}\int_{\tau}^{2\tau}{\rho_{1}(r)}{dr}.
\]
Since
$\bar \phi(y)\le 2|y|^2\rho_{1}(4|y|)$, we obtain 
$\bar\phi(0)= D\bar\phi(0)=D^2\bar\phi(0)=0$.
Since
$\bar\phi(y)\ge \rho_{1}(|y|)|y|^2\ge \rho(|y|)|y|^2=\bar u(y)$,
  $\bar u-\bar\phi$ attains a maximum point at  $0$.
It is easily seen that $\bar\phi\in C^{1}(\mathbb{R}^{n})$. In order to
have $\bar\phi\in C^{2}(\mathbb{R}^{n})$ we repeat the procedure with
$\bar u$ replaced by $\bar\phi$.

Define $\phi(t,x)=[\alpha(t,x)]^{-1}\bar\phi(t,x)$, 
$(t,x)\neq (0,0)$, $\phi(0,0)=0$. It is not difficult to verify
that  $\phi$ is a desired function.
\end{proof}

\begin{proposition} \label{prop4.3}
Suppose that $u$ satisfies (i), (ii) in Definition~\ref{subsol}
and for every $x\in \mathbb{R}^n$ \eqref{differ} is satisfied for a.e $t\in [0,T]$. 
If $u$ is a viscosity subsolution (supersolution, solution) of
\eqref{zag2} then $u$ is a  subsolution
(supersolution, solution) of \eqref{zag2}.
\end{proposition}

\begin{proof} 
Suppose that $u$ satisfies (i), (ii) in
Definition~\ref{subsol} and for every $x\in \mathbb{R}^n$ \eqref{differ} 
is satisfied for a.e $t\in [0,T]$. Let $(\bar t,\bar x)\in \Theta_{T}$ be such that
$D_{t}u(\bar t,\bar x)$ exists and \eqref{differ} is satisfied. 
Since $Du(\bar t,\bar x)$,
$D^2u(\bar t,\bar x) $ exist by Lemma \ref{deriv} we have 
$\phi\in C^{1,2}(\Theta_{T})$ such that $u-\phi$ has a maximum at 
$(\bar t,\bar x)$ and $D_{t}u(\bar t,\bar x)=D_{t}\phi (\bar t,\bar x)$,
$ Du(\bar t,\bar x)=D\phi (\bar t,\bar x)$, $D^2u(\bar t,\bar x)=D^2\phi(\bar
t,\bar x) $. By Definition \ref{visc} this gives
\[
D_{t}\phi(\bar t,\bar x)   \le  g(\bar t, \bar x,u(\bar t,\bar x),
D\phi(\bar t,\bar x),D^{2}\phi(\bar t,\bar x))
\]
and consequently $u$ satisfies \eqref{zag2sub} in $(\bar t,\bar x)$.
\end{proof}

\begin{theorem}\label{genvis}
Suppose that for  $(\hat t,\hat x,\hat u,\hat p,\hat X) \in
\Theta_{T}\times \mathbb{R}\times \mathbb{R}^{n} \times S[n]$ 
\begin{equation} \label{equisub}
\limsup_{(t,u,p,X)\to (\hat t^{-},\hat u,\hat p,\hat X)} {g(t,\hat
x,u,p,X)}\le g(\hat t,\hat x,\hat u,\hat p,\hat X).
\end{equation}
 Then if $u\in USC(E_{T},\mathbb{R})$ is a  subsolution of \eqref{zag2}, then it is a
viscosity subsolution of \eqref{zag2}.
\end{theorem}

\begin{proof} 
Suppose that $u$ is a  subsolution of
\eqref{zag2} and $\phi\in C^{1,2}(\Theta_{T},\mathbb{R})$ such that $u-\phi$ has a
local maximum point in $(\tilde t, \tilde x)\in \Theta_{T}$. Let $A$ be
the full measure set of all $t$ such that $D_{t}u(t,\tilde x)$
exists and
\[
D_{t}u(t, \tilde x)  \le  g(t,\tilde x,u(t,\tilde x),
Du(t,\tilde x),D^{2}u(t,\tilde x)).
\]
By  Preposition \ref{AC}
there exists a sequence $t_{m}\to \tilde t^{-}$ such that $t_{m}\in A$ 
for every $m$,  $D_{t}(u-\phi)(t_{m},\tilde x)\ge 0$ and
\[
D_{t}\phi(t_{m}, \tilde x) \le D_{t}u(t_{m}, \tilde x) 
 \le  g(t_{m},\tilde x,u(t_{m},\tilde x),Du(t_{m},\tilde x),D^{2}u(t_{m},\tilde x)).
\]
Notice that by Definition~\ref{subsol}~(i) and Remark~\ref{left}
$u(\cdot,\tilde x)$ is left-hand side continuous. Using this,
Definition~\ref{subsol}~(iii) and \eqref{equisub} we obtain by
letting $t_{m}\to \tilde t$
\[
D_{t}\phi(\tilde t, \tilde x)   
\le  g(\tilde t,\tilde x,u(\tilde t,\tilde x),Du(\tilde t,
\tilde x),D^{2}u(\tilde t,\tilde x)).
\]
Since $Du(\tilde t,\tilde x)=D\phi(\tilde t,\tilde x)$,
$D^{2}u(\tilde t,\tilde x)\le D^{2}\phi(\tilde t,\tilde x) $ we
have
\[
D_{t}\phi(\tilde t, \tilde x)   \le  g(\tilde t,\tilde x,u(\tilde t,\tilde x),D\phi(\tilde t,\tilde x),D^{2}\phi(\tilde t,\tilde x)).
\]
Hence, $u$ is  a viscosity subsolution of \eqref{zag2}.
\end{proof}

\begin{theorem}\label{genvis2}
Suppose that for $(\hat t,\hat x,\hat u,\hat p,\hat X) \in
\Theta_{T}\times \mathbb{R}\times \mathbb{R}^{n} \times S[n]$ 
\begin{equation} \label{equisuper}
\liminf_{(t,u,p,X)\to (\hat t^{-},\hat u,\hat p,\hat X)} {g(t,\hat
x,u,p,X)}\ge g(\hat t,\hat x,\hat u,\hat p,\hat X).
\end{equation}
 Then if $u\in LSC(E_{T},\mathbb{R})$ is a supersolution of \eqref{zag2}, then it is a
viscosity supersolution of \eqref{zag2}.

If \eqref{equisub}, \eqref{equisuper} hold (i.e. $g$ is continuous 
(left-hand side in $t$ ))  and $u\in C(E_{T},\mathbb{R})$ is  a  solution of \eqref{zag2}
then it is a viscosity solution of \eqref{zag2}.
\end{theorem}

It is evident that for Theorem \ref{genvis} and Theorem \ref{genvis2} a remark, 
similar to Remark \ref{class} holds.

\begin{remark} \rm
It is worth mentioning that by virtue of Theorem \ref{genvis} and 
Theorem \ref{genvis2} we can use a subsolution and supersolution
as an upper and lower function in Perron method which is a main tool 
in proving existence results for viscosity solutions.
\end{remark}

In the following we  discuss the optimality of assumptions in Theorem \ref{genvis}.

\begin{example}[the necessity of \eqref{equisub} in $t$] \rm
Let $a\ge 0, b\in R$. Consider the  equation
\begin{equation} \label{intt}
D_{t}u-aD^2u+bDu=g(t)\quad \text{in } E_{T}=[0,2]\times \mathbb{R},
\end{equation}
 where $g(t)=0,\  t\in [0,1)$ and
$g(t)=-1$, $t\in [1,2]$. Set $u(t,x)=0$,   $t\in (0,1]$ and $u(t,x)=1-t$,
$t\in (1,2]$. It is not difficult to verify that $u$ is a
 subsolution of \eqref{intt}. To show that $u$ is not a viscosity
subsolution w set $\tilde t=1,\ \phi\equiv 0$ in Definition \ref{visc}.
(it will change if
we redefine $g$ by setting $g(1)=0$).
\end{example}

\begin{example}[the necessity of  \eqref{equisub} in $u$]\rm
Consider   equation \eqref{intt}  with $g(t)$ replaced by $g(u)=u$,
$u\in (-\infty,e)$ and $g(u)=0$,
$u\in [e,\infty)$. Set $u(t,x)=e^t$,  $t\in [0,1]$ and   $u(t,x)=e$,
$t\in (1,2]$.
 It is not difficult to verify that $u$ is a
 subsolution. To show that $u$ is not a viscosity
subsolution we set $\tilde t=1$, $\phi(t)=e^{t}$ in Definition \ref{visc}
(it will change if we redefine $g$ by setting $g(e)=e$).
\end{example}

\begin{example}[the necessity of  \eqref{equisub} in $p$] \rm
Consider the  equation
\begin{equation} \label{intt2}
D_{t}u=h(Du)x\quad {\rm in}\ E_{T}=[0,2]\times \mathbb{R},
\end{equation}
 where $h(p)=p,\  p\in (-\infty,e)$ and
$g(p)=0$, $p\in [e,\infty)$. Set $u(t,x)=e^tx$,   $t\in [0,1]$ and
$u(t,x)=ex$, $t\in (1,2]$.
It is not difficult to verify that $u$ is a
 subsolution (solution) of \eqref{intt2}. To show that $u$ is not a viscosity
subsolution w set $(\tilde t,\tilde x)=(1,\tilde x)$, $\tilde x>0$, $\phi(t,x)=e^tx$
in Definition \ref{visc}. (it will change if
we redefine $h$ by setting $h(e)=e$).
\end{example}

In a similar way, considering the equation $D_{t}u=\frac{1}{2}h(Du)x^2$ we can show
the necessity of  \eqref{equisub} in $X$.
The last example  shows that
upper absolutcontinuity plays  crucial role in Theorem \ref{genvis}.

\begin{example}[the necessity of (i) in Definition \ref{subsol}]\rm
Let $z:[0,1]\to [0,1]$ be the Cantor function. Of course, $z\notin
UAC([c,1],\mathbb{R})$ for  $c\in (0,1)$ and $z'=0$ a.e in $[0,1]$. We will
show that $z$ is not a viscosity subsolution of $z'\le 0$. Set
$\phi(t)=t$. It follows from the construction of the Cantor
function that there exists $\tilde t\in (0,1)$  such that
$z(\tilde t)-\tilde t=\sup\{z(t)-t:t\in[0,1]\}>0$. Since
$\phi'(\tilde t)=1$ and $\phi\in C^{1}(0,1)$ we have a
contradiction with Definition~\ref{visc} ($g\equiv 0$).
\end{example}


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\end{thebibliography}
\end{document}
