\magnification\magstep1
\documentstyle{amsppt}
\def\version{V2.5}

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%%%%%%%%%%   DEFINITIONS   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\define\A{{\Cal A}}
\define\ab{[a,b]}
\define\An{{{\Cal A}(\Rn)}}
\define\BB{{\Cal B}}
\define\BF{{\Cal B}\left(\F\right)}
\define\BFF{{\Cal B}\left(\F_F\right)}
\define\BFloc{{\Cal B}_{\text{loc}}\left(\F\right)}
\define\C{{\Cal C}}
\define\Cn{{\C(\Rn)}}
\define\dE{{\partial E}}
\define\dist{{\text {dist}}}
\define\Emr{{E^-_\varrho}}
\define\Epr{{E^+_\varrho}}
\define\eps{\epsilon}
\define\enI{\frac{\eps}{n} \text{Id}}
\define\F{{\Cal F}}
\define\fea{{f_\eps(a)}}
\define\FF{{\Cal F}_F}
\define\FFm{{\Cal F}_F^>}
\define\G{{\Cal G}}
\define\Ga{\Gamma}
\define\grad{\nabla}
\define\HH{{\Cal H}}
\define\iE{{\text{int}(E)}}
\define\la{\lambda}
\define\M{{\Cal M}}
\define\MEFHt{{\Cal M}\left(E,\F_F,t_0\right)}
\define\MEFt{{\Cal M}\left(E,\F,t_0\right)}
\define\misE{{\M^\star\left(E,\F\right)}}
\define\MiEFHt{{\M_\star\left(E,\F_F,t_0\right)}}
\define\MsEFHt{{\M^\star\left(E,\F_F,t_0\right)}}
\define\ndim{n}
\define\oE{{\overline E}}
\define\oi{[0,+\infty[}
\define\Om{\Omega}
\define\op{\overline p}
\define\ot{\overline t}
\define\ox{\overline x}
\define\oX{\overline X}
\define\PP{{\Cal P}}
\define\p{\phi}
\define\Pn{{\PP}(\Rn)}
\define\R{{\bold R}}
\define\rga{\rightarrow}
\define\ro{\varrho}
\define\Rn{{\bold R}^n}
\define\SSn{{\bold S}^{n-1}}

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\topmatter
\title
Comparison results between minimal barriers and 
viscosity solutions for geometric evolutions
\endtitle
\rightheadtext{Comparison results between minimal barriers ...}

\author
G. Bellettini,
M. Novaga
\endauthor

\address
Dipartimento di Matematica Applicata ``U. Dini'',
Universit\`a di Pisa
\endaddress
\email
bellettini\@sns.it
\endemail

\address
Scuola Normale Superiore di Pisa
\endaddress
\email
novaga\@cibs.sns.it
\endemail

\endtopmatter

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\document

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\head
{1}. Introduction      
\endhead
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

In [{12}] De Giorgi introduced a notion of weak solution,
called minimal barrier, for a wide class of evolution problems.
In the particular case of geometric flows of subsets of $\Rn$,
the concept of minimal barrier can  be described as follows (see
Section {2.1} for precise definitions). First we choose
a nonempty family $\F$ of maps which take some time interval into the set
$\Pn$ of all subsets of $\Rn$: for instance $\F$ can be the family
of all smooth local evolutions with respect to a given geometric
law. Then we define the class ${\Cal B}(\F)$ of all maps $\p:
\oi \rga \Pn$ which are barriers for $\F$ in $\oi$
with respect to the inclusion of sets, that is,
if $f : \ab \subseteq \oi \rga \Pn$ belongs to $\F$ and $f(a) \subseteq \p(a)$,
then it must hold $f(b) \subseteq \p(b)$. Finally, we define
the minimal barrier $\M(E,\F)(t)$ with origin the set 
$E \subseteq \Rn$,  
with respect to $\F$, at time $t \in \oi$ as 
%
$$
\M(E,\F)(t) := \bigcap \{\p(t) : \p : \oi \rga \Pn, 
\p \in {\Cal B}(\F), \p(0) \supseteq E\}.
\tag{1.1}
$$
%
We stress the dependence on $\F$ of the minimal barrier
(see Example {2.1}) and also the fact that 
the minimal barrier is unique and globally defined, for 
an arbitrary initial set $E$.
Therefore, given any initial function $u_0: \Rn \rga \R$, 
({1.1})
yields a unique global
evolution function $\M_{u_0,\F}(t,x)$ (assuming $u_0$ as initial datum), 
defined as the
function which, for any $\lambda \in \R$, has
$\M(\{u_0 < \lambda\},\F)(t)$ 
as $\lambda$-sublevel set at time $t \in \oi$.

The aim of this paper is to compare the minimal barrier with the viscosity
solution of geometric fully nonlinear parabolic problems of the form
%
$$
\frac{\partial u}{\partial t} + F(t,x,\grad u,\grad^2 u) =0.
\tag{1.2}
$$
%
The definition of viscosity solution has been introduced 
by Crandall and P.-L. Lions [{11}] 
(we refer to [{10}] for a bibliography on this argument).
It has been exploited by Evans-Spruck [{13}] in the 
case of motion by mean curvature and by Chen-Giga-Goto [{9}],
Giga-Goto-Ishii-Sato [{16}] in the case of
geometric evolutions of the form ({1.2}).
We recall that, in order to define the viscosity evolution 
$V(E)(t)$ of a bounded open set $E \subseteq \Rn$ for problem ({1.2}),
first we find the unique continuous viscosity solution of ({1.2})
(with a suitable initial datum) and then we recover $V(E)(t)$ by setting 
$V(E)(t) := \{x \in \Rn : v(t,x) < 0\}$. 

A comparison result for sets $E$ with compact boundary
in case of driven motion by mean curvature (whose corresponding 
function $F$ is given by $F(t,x,p,X) = 
- \text{tr}((\text{Id} - p\otimes p/\vert p\vert^2) X) 
+ g(t,x) \vert p\vert$, $g$ being the driving force)
has been proved in  [{7}],
and shows that the two weak definitions are essentially equivalent.
The proofs of [{7}] 
rely on a paper by Ilmanen
[{20}], where 
viscosity solutions are compared, in the case
of motion by mean curvature, with the so called
set theoretic subsolutions.

The results of [{20},{7}] are based
on Ilmanen's interposition lemma and on Huisken's estimates
[{18}]
of the existence time for the evolution of a 
smooth compact hypersurface in dependence
on the $L^\infty$ norm of its second fundamental form, 
without requiring bounds on further derivatives of the curvatures.
The above results of Ilmanen and Huisken apply basically 
to the case of motion by mean curvature;
it seems
difficult to recover the time estimates
of [{18}] for a general evolution
law of the form ({1.2}) 
(some generalizations of Huisken's results can be found in 
[{2},{3}]).
This is the main reason for which we follow, in this paper, 
a completely different approach to the  problem, 
which allows us to compare minimal barriers with  viscosity
solutions for a general $F$.

A further remark on the definition of minimal barrier is the following:
denoting by $\FF$ the family of all local smooth geometric
supersolutions of ({1.2}) (see Definition
{2.5}), 
to ensure that 
$\M(E,\FF)$ is well defined 
we do not need to assume that $F$, 
if considered as a function on symmetric matrices, is decreasing
(degenerate ellipticity condition);
it turns out [{6}] that when
$E$ is open we have
%
$$
\M(E,\FF) = \M(E,\F_{F^+}),
\tag{1.3}
$$
%
where $F^+$ is defined as the smallest function which is 
degenerate elliptic and greater than or equal to $F$,
i.e., 
%
$$
F^+(t,x,p,X) 
:= \sup \{F(t,x,p,Y) : Y \geq X\}.
\tag{1.4}
$$
%
Such  a result is obtained in the present paper by 
passing through the viscosity theory (Corollary
{6.2}) and allows to remove 
the degenerate ellipticity assumption from the hypotheses
of all results of Sections {3} and {5}, provided that
also $F^+$ 
satisfies the assumptions listed in [{16}]. 

Finally we observe that $\M(E,\FF)$ and $\M_{u_0,\FF}$ verify 
by definition the comparison principle  and it is 
immediate to check that, if $\partial E$ is smooth,
$\M(E,\FF)$ coincides with the classical evolution of $E$,
as long as the latter exists, provided that 
the classical evolutions
are barriers (which is the case, for instance, 
for uniformly elliptic smooth functions $F$).

Let us briefly summarize the content and the main results
of the present paper. In Section {2}
we introduce some notation and the notion
of minimal barrier and regularized minimal barriers with respect
to a family $\F$ (Definitions {2.2}, {2.3}, 
{2.4}). In Proposition {2.2} we show that 
the minimal barrier extend the smooth evolutions
whenever the latter exist.
We conclude Section {2} 
with two examples of minimal barriers
obtained with particular choices of $\F$: Example {2.1}
concerns motion by mean curvature whenever $\F$ consists
of smooth convex evolutions; 
in Example {2.2} we consider
the case of
inverse mean curvature flow. 
Sections {3}-{5} are
concerned  with geometric evolutions of 
the form ({1.2}) where $F$ satisfies some of the assumptions
made by Giga-Goto-Ishii-Sato in [{16}].
In Section {4} we prove some auxiliary
results on barriers used throughout the paper. 
The comparison result between barriers and viscosity solutions
is divided into two parts. In Section {3} we prove
that the sublevel sets of a viscosity subsolution of ({1.2})
 are barriers (Theorem
{3.2}) and in Section {5} we prove that a function
whose sublevel sets are barriers is a viscosity subsolution of
({1.2}) (Theorem
{5.1}).
In Theorems {3.2} and
{5.1}, 
in order to simplify the proofs,
we distinguish the case in which $F$ does not depend explicitly on
$x$ with the general case; if $F$ is not degenerate elliptic we extend
the results to the function $F^+$.
In Corollary {6.1} we summarize the comparison results
whenever there exists a unique uniformly continuous viscosity
solution $v$ of ({1.2}) having a given initial datum. 
More precisely, if
$E \subseteq \Rn$ is a bounded set,
for any $t \in \oi$ we have
%
$$
\aligned
& \M_*(E,\FF)(t) = \{x \in \Rn : v(t,x) < 0\}, 
\cr
& \M^*(E,\FF)(t)= \{x \in \Rn : v(t,x) \leq 0\},
\endaligned
\tag{1.5}
$$
%
where $\M_*(E,\FF)$ and $\M^*(E,\FF)$ are the lower and upper 
regularized minimal barriers (see Definition {2.3}).
In particular 
%
$$
\M^*(E,\FF)(t) \setminus \M_*(E,\FF)(t) = \{x \in \Rn : 
v(t,x) = 0\}.
\tag{1.6}
$$
%
Equality ({1.6}) is connected with the so called
fattening phenomenon (see ({2.5}) and Remark {6.1}).

In case of nonuniqueness of viscosity solutions, 
we show in Corollary {6.3}
that $\M_{u_0,\FF}$
coincides with the maximal viscosity
subsolution, see also Example {6.1}. 
If $F$ is not necessarily a  
degenerate elliptic function, and if $F^+$ verifies
the assumptions of Corollary {6.1}, 
then ({1.5})
holds when $v$ is the viscosity solution of ({1.2})
with $F^+$ in place of $F$ (Corollary {6.2}).
In Remark {6.6}
we extend our results to the case in which $F$ has superlinear
growth and $E$ is unbounded, where 
the notion of viscosity evolution is the one
introduced by Ishii-Souganidis in [{22}].

\smallskip
{\bf Acknowledgements.} 
We wish to thank Ennio De Giorgi for many useful
suggestions and advices.
We are also grateful to Luigi Ambrosio,
Gerhardt Huisken, Tom Ilmanen and Alessandra Lunardi for interesting
discussions.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\head
{2}. Notation and main definitions      
\endhead
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\noindent 
In the following we let $I := [t_0, +\infty[$, for a fixed
$t_0 \in \R$; in Sections {3}-{6} we will
take $t_0 =0$.
We denote by $\Pn$ the family of all subsets of $\Rn$, 
$n \geq 1$.
Given  a set $C \subseteq \Rn$, we denote by $\text{int}(C)$, 
$\overline C$ and $\partial C$ the interior part, the closure
and the boundary of $C$, respectively; $\chi^{}_C$ is the characteristic
function of $C$, i.e., $\chi_C(x) = 1$ if $x \in C$, $\chi_C(x) = 
0$ if $x \notin C$.
If $C\neq\Rn$ and $C\neq\emptyset$, we set 
%
$$
d_C(x) := \text{dist}(x,C) - \text{dist}(x,\Rn \setminus C),
$$
%
and for any $\ro >0$ 
%
$$
\align
& C^-_\ro := 
\{x \in \Rn : \dist\left(x,\Rn \setminus C\right) > \ro\},
\tag{2.1}
\\
& C^+_\ro := \{x \in \Rn : \dist(x,C) < \ro\}.
\tag{2.2}
\endalign
$$
%
Given a map $\p : J \rga \Pn$,
where $J\subseteq  \R$ is a convex set,
if $\p (t)\neq\Rn$ and $\p (t)\neq\emptyset$ for any $t\in J$
we let $d_\p : J \times \Rn  \rga \R$
be the function defined by
%
$$
d_\p(t,x) := \text{dist}(x,\p(t)) - \text{dist}(x, \Rn\setminus \p(t))= 
d_{\p(t)}(x).
$$
%
If $\p_1,\p_2 : J \rga \Pn$, by 
$\p_1 \subseteq \p_2$ (resp. $\p_1 = \p_2$) we mean $\p_1(t) 
\subseteq \p_2(t)$
(resp. $\p_1(t) = \p_2(t)$) for any $t \in J$.

\noindent 
Given a function $v : J \times \Rn \rga \R$ we denote by $v_*$ 
(resp. $v^*$) the lower (resp. upper) semicontinuous envelope
of $v$.

\noindent 
For $x \in \Rn$ and $R>0$ we set $B_R(x) := \{y \in \Rn :\vert y- x\vert
< R\}$ and $\SSn := \{x \in \Rn : \vert x\vert =1\}$.

\noindent 
If $c_1, c_2 \in \R$, we let $c_1 \wedge c_2 = \min(c_1,c_2)$
and  $c_1 \vee c_2 = \max(c_1,c_2)$. 

\noindent 
We denote by $\text{Sym}(n)$ the space of all symmetric real
$(n\times n)$-matrices, endowed with the norm
$\vert X\vert^2 = \displaystyle \sum_{i,j=1}^n X_{ij}^2$, 
where $X = (X_{ij})$.

\noindent Given $p \in \Rn \setminus \{0\}$,
we set $P_p := \text{Id} - p \otimes p/\vert p\vert^2$.

\noindent 
Finally we define 
%
$$
J_0 := I \times \Rn \times (\Rn
\setminus \{0\}) 
\times \text{Sym}(n), \qquad 
J_1 := I \times (\Rn
\setminus \{0\}) 
\times \text{Sym}(n), \qquad 
$$
%
\proclaim{Remark {2.1}}
All results of this paper still hold when $I$
in $J_0$ and $J_1$ is replaced by $[t_0,t_0 + T[~$, for some $T > 0$.
\endproclaim
%
\subhead 
{2.1}. Definitions of minimal barriers
\endsubhead
The following two definitions are a particular case of the definition
proposed in [{12}].

\proclaim{Definition {2.1}}
Let $\F$ be a family of functions
with the following property: for any $f \in \F$
there exist $a,b \in \R$, $a < b$, such that $f : \ab \rga \Pn$.
A function $\p$ is a barrier with respect to $\F$
if and only if
there exists a convex set
$L \subseteq I$ such that
$\p : L \rga \Pn$ 
and the following property holds:
if $f : \ab \subseteq L  \rga \Pn$ belongs
to $\F$ and $f(a)\subseteq \p(a)$ then $f(b) \subseteq \p(b)$. 
We denote by $\BF$ the family of all barriers $\p$ 
such that $L = I$ (that is, barriers on the whole of $I$).
\endproclaim

\proclaim{Definition {2.2}}
Let $E \subseteq \Rn$ be a given set.
The minimal barrier
$\MEFt: I \rga \Pn$ (with origin in $E$ at time $t_0$) 
with respect to the family $\F$ at any time $t \in I$ 
is defined by
%
$$
\MEFt(t) := \bigcap \left\{ \p(t) : \  \ \p : I \rga \Pn, \ 
\p \in \BF, \ \p(t_0) \supseteq E
\right\}.
\tag{2.3}
$$
%
\endproclaim

\noindent 
Let us observe that  
$\MEFt \in {\Cal B}(\F)$ (uniqueness of the minimal barrier), 
$\M(E,\F,t_0)(t_0) = E$, 
and that $E_1 \subseteq E_2$ implies
$\M(E_1,\F,t_0) \subseteq \M(E_2,\F,t_0)$ (comparison property).

For simplicity of notation, we drop the dependence on 
$t_0$ of the minimal barrier, thus we write $\M(E,\F)$
in place of $\M(E,\F,t_0)$.

The following regularizations have been introduced in [{7}]
for driven motion by mean curvature, and will be useful in the sequel.

\proclaim{Definition {2.3}}
Let $E \subseteq \Rn$.
If $t \in I$ 
we set 
%
$$
\aligned
& \displaystyle \M_*(E,\F)(t) := 
\bigcup_{\ro >0} \M(E^-_\ro,\F)(t),
\\
& \displaystyle \M^*(E,\F)(t) := 
\bigcap_{\ro>0} \M(E^+_\ro,\F)(t).
\endaligned
\tag{2.4}
$$
%               
\endproclaim

\noindent
Following [{5}], we say that the set $E$ develops
fattening (with respect to $\F$) at time $t \in I$ if 
%
$$
{\Cal H}^n\Big(\M^*(E,\F)(t)
\setminus \M_*(E,\F)(t)\Big) > 0,
\tag{2.5}
$$
%
where ${\Cal H}^n$ denotes the $n$-dimensional Hausdorff
measure (note that one could define the $m$-dimensional fattening
by replacing ${\Cal H}^n$ with ${\Cal H}^m$, $0 \leq m \leq n$).

Once the evolution of an arbitrary set is uniquely defined, we can define
the unique evolution of an arbitrary initial function $u_0$.
%
\proclaim{Definition {2.4}}
Let $u_0 : \Rn \rga \R$ be a given function. 
The two functions $\M_{u_0,\F}, \overline \M_{u_0,\F} : I \times \Rn \rga
\R \cup \{\pm \infty\}$ are defined by
%
$$
\aligned
& \M_{u_0,\F}(t,x) := 
\inf \{\lambda \in \R : {\Cal M}(\{u_0 < \lambda\},\F)(t) \ni
x \},
\\
& \overline {\Cal M}_{u_0,\F}(t,x) := 
\inf \{\lambda \in \R : {\Cal M}_*(\{u_0 < \lambda\},\F)(t) \ni
x \}.
\endaligned
\tag{2.6}
$$
%
\endproclaim

If $\F$ consists of functions $f : \ab \subseteq I \rga \Pn$ 
such that $f(t)$ 
is compact for any $t \in \ab$, if
$\M(A,\F)(t)$ is open for any open set $A \subseteq \Rn$, and if
$u_0 : \Rn \rga \R$ is upper semicontinuous, 
then there holds
%
$$
\{x \in \Rn : \M_{u_0,\F}(t,x) < \lambda\} = \M(\{u_0 < \lambda\}, \F)(t),
\qquad t\in I.
\tag{2.7}
$$
%
Hence under these assumptions 
 $\M_{u_0,\F}(t,\cdot)$ is upper semicontinuous; if we drop the 
upper semicontinuity assumption on $u_0$, we have
$\overline\M_{u_0,\F} = \M_{u_0^*,\F}$.

\noindent General properties of minimal barriers will appear in 
a forthcoming paper [{6}].

\noindent The definitions of the minimal barriers for geometric evolutions
described by a function $F$
are a particular case of the previous definitions,
by choosing a suitable family $\FF$, and read as follows.

Let $F : J_0 \rga \R$ be an arbitrary function.

\proclaim{Definition {2.5}} 
Let $a,b \in \R$, $a< b$, $\ab \subseteq I$ and let $f : \ab \rga \Pn$.
We write $f \in \FF$ if and only if the following conditions
hold: $f(t)$ is compact for any $t \in \ab$, 
there exists an open set
$A \subseteq \Rn$ such that
$d_f \in \C^\infty(\ab \times A)$, $\partial f(t) \subseteq A$ for any
$t \in \ab$, and 
%
$$
\frac{\partial d_f}{\partial t}(t,x) + F(t,x,\grad d_f(t,x), \grad^2 d_f(t,x)) 
\geq 0 
\qquad t \in ~]a,b[, \  x \in \partial f(t).
\tag{2.8}
$$
%
We write $f \in \FFm$ if and only if $f \in \FF$ and the strict
inequality holds in ({2.8}).
\endproclaim
%
\noindent Obviously ${\Cal B}(\FF) \subseteq {\Cal B}(\FFm)$, hence
$\M(E,\FF) \supseteq \M(E,\FFm)$.

\proclaim{Remark {2.2}}
Definition {2.5} (and consequently
the definition of minimal barriers)
can be adapted to geometric flows on a riemannian
manifold $(V,g)$ by 
substituting $\Pn$ with the family of all subsets of
$V$ ordered by the inclusion, the euclidean distance with 
the geodesic distance on $(V,g)$, and the operators
$\grad, \grad^2$ with the corresponding intrinsic operators.
\endproclaim
%

We recall that 
$F$ is {\it geometric} 
[{9}, (1.2)] if
%
$$
F(t,x,\lambda p, \lambda X + \sigma p \otimes p) 
  =\lambda F(t,x,p,X),
$$
%
for any $\lambda > 0$, $\sigma \in \R$, $(t,x,p,X) \in J_0$.

If we define $\widetilde F(t,x,p,X) = 
\vert p\vert F(t,x,\frac{p}{\vert p\vert}, \frac{P_p X P_p}{\vert p\vert})$
for $(t,x,p,X) \in J_0$,
then $\widetilde F$ is geometric and 
$\widetilde F(t,x,\grad d_f,\grad^2 d_f)
= F(t,x,\grad d_f,\grad^2 d_f)$ for $f \in \FF$, so
$\FF = \F_{\widetilde F}$.
Hence in what follows, without loss of generality, 
we can always assume that the function $F$ is geometric.

A concept similar to the minimal barrier (without the regularization
defined in ({2.4}))
in the case of motion
by mean curvature (i.e., $F(t,x,p,X) = -\text{tr} ( P_p X P_p ) $) 
for compact sets was introduced by Ilmanen
in [{20}] and was called set-theoretic subsolution; in that case
$\FF$ is essentially the family of all local smooth evolutions
by mean curvature,  and ({2.8}) is considered 
with the equality instead of
the inequality.

Notice that to define $\M(E,\FF)$ we need only that $\FF$ is 
nonempty, which is guaranteed under very mild assumptions on $F$.

\proclaim{Proposition {2.1}}
Assume that 
there exists a function $F_1 : J_0\rga\R$ 
which is bounded on compact subsets of $J_0$ and $F_1 \leq F$.
Then the family $\F_{F_1}$ (hence $\FF$)
is nonempty.
\endproclaim
\demo{Proof}
Let $R >0$ and $0 < \eps < 1$ be such that
%
$$
\frac{1}{\eps}\!>\!\sup \{\vert F_1(t,x,p,X)\vert : [t_0,t_0+R/2], 
\vert x\vert \in [R/2,R], \vert p\vert=1,
0 \leq  \vert X\vert \!\leq\! 2\sqrt{n-1}/R\}.
$$
%
Let $R(t) := -(t-t_0)/\eps  + R_0$ and $d(t,x) := 
\vert x\vert - R(t)$. When $t \in [t_0,t_0+\eps R/2]$
then $R(t) \in [R/2,R]$, and therefore
$\displaystyle 
\sup_{t \in [t_0,t_0+\eps R/2], \vert x\vert = R(t)} 
\vert \grad^2 d
\vert \leq 2\sqrt{n-1}/R$. We then have
%
$$
\frac{\partial d}{\partial t}(t,x) = \frac{1}{\eps} 
> - F_1(t,x,\grad d(t,x), \grad^2 d(t,x))
\qquad t \in ~]t_0,t_0+\eps R/2[~, \  
\vert x\vert = R(t).
$$
%
It follows that the map $t \in [t_0,t_0+\eps R/2] \rga B_{R(t)}$ belongs to 
$\F_{F_1}^>$.
\qed\enddemo

If $F$ is of class $\C^\infty$, if it does not depend explicitly
on $x$ and 
is geometric and uniformly elliptic 
then, as proved in [{15}], any compact boundary of class $\C^\infty$
has a unique smooth evolution  for small
times. Hence we have the following proposition, which shows 
in particular that the minimal barrier coincides with the smooth
evolution whenever the latter exists (see ({2.11})).

\proclaim{Proposition {2.2}}
Assume
that  $F : J_1 \rga \R$
does not depend on $x$, is geometric, 
uniformly elliptic and of class $\C^\infty$.
We write 
$f \in \F_F^=$ 
if and only if $f \in \FF$ and equality holds in ({2.8}).
Then for any $E \subseteq \Rn$ we have
%
$$
\M(E,\F_F^=) = \M(E,\FF)
\tag{2.9}
$$
%
and if $E$ is open we have also
%
$$
\M(E,\FFm) = \M(E,\FF).
\tag{2.10}
$$
%
Moreover for any $f : \ab \subseteq I \rga \Pn$, $f \in \F_F^=$, we have
%
$$
\M(f(a),\FF,a)(t) = f(t), \qquad t \in \ab.
\tag{2.11}
$$
%
\endproclaim
\demo{Proof}
To prove ({2.9}) it is enough to show 
$\M(E,\F_F^=) \supseteq \M(E,\FF)$.
Hence we are reduced to show that $\M(E,\FF^=) \in \BFF$.
Let $f : \ab \subseteq I \rga \Pn$, $f \in \FF$,
$f(a) \subseteq \M(E,\F_F^=)(a)$. 
We have to prove that
$f(b) \subseteq \M(E,\F_F^=)(b)$. The set
$\partial f(s)$ is of class $\C^\infty$ and compact
for any $s \in \ab$, therefore the $L^\infty$
norm of the 
second fundamental form
of $\partial f(s)$ (and of $\grad^3_{ijk} d_f$, if necessary)
is uniformly bounded with respect to $s \in \ab$.  Hence there is
$\tau >0$, independent of $s$, so that 
the evolution of $\partial f(s)$ by ({2.8}), with the 
equality,
is of class $\C^\infty$ in $[s,s+\tau]$ for any $s \in [a,b]$. Write 
$\ab = \displaystyle \bigcup_{i=1}^m [t_i,t_{i+1}]$
where $a = t_1 < \dots <t_{m+1} =b$ e $t_{i+1} - t_i \leq \tau$.
Let us denote by $f^i(t)$ the geometrical evolution of
$f(t_i)$ by means of ({2.8}) with the equality. 
Then, using the comparison principle 
between smooth evolutions, we have
$f(t_{i+1}) \subseteq f^i(t_{i+1})$. 
Reasoning by induction on 
$i$, we have $f^i(t_i) \subseteq \M(E,\F_F^=)(t_i)$ for any
$i=1,\dots,m+1$, hence $f(t_{i+1}) \subseteq 
\M(E,\F_F^=)(t_{i+1})$ for any $i=1,\dots,m$. 
For $i=m$ we conclude the proof of ({2.9}).

Let $E$ be open.
To prove ({2.10})
we need to show 
that $\M(E,\FFm) \in \BFF$. Let $f : \ab \subseteq I\rga \Pn$,
$f \in \FF$, $f(a) \subseteq \M(E,\FFm)(a) =: A$.
We have to show that $f(b) \subseteq \M(E,\FFm)(b)$.
As $A$ is open (see ({4.24})) and since
$f(a)$ is compact,
we have $\text{dist}(f(a), \Rn \setminus A) > 0$.
For any $t \in \ab$ we can find a bounded tubular neighbourhood
of $\partial f(t)$, of thickness $c(t)$, each point of which
has a unique projection on $\partial f(t)$; we set
$c := \inf \{c(t), t\in \ab\}$, which is strictly positive.
Let $L$ be the Lipschitz constant of $F(t,\grad d_f(t,x), \grad^2 d_f(t,x))$
and $M$ be the supremum of $\vert \grad^2 d_f(t,x) \vert^2$ when
$t \in \ab$ and $x$ belongs to the $c(t)$-tubular neighbourhood
of $\partial f(t)$.
Pick a $\C^\infty$ function $\ro : \ab \rga
~]0,+\infty[$ so that $\ro(a) < \min(c,\text{dist}(f(a),
\Rn \setminus A))$ and $\dot \ro + 2 M L \ro < 0$.
The map $g : \ab \rga \Pn$, $g(t) := 
\overline {f^+_{\ro(t)}(t)} = \{x \in \Rn : 
\text{dist}(x,f(t)) \leq \ro(t)\}$ is of class $\C^\infty$, and each
point $y \in \partial g(t)$ is of the form $y = x + \ro(t) 
\grad d_f(t,x)$ for a unique $x \in \partial f(t)$. We observe that 
$g \in \FFm$. Indeed for any $y \in \partial g(t)$,
$y = x + \ro(t) \grad d_f(t,x)$, $x \in \partial f(t)$,
we have $\grad^2 d_g(t,y) = \grad^2 d_f(t,x) (\text{Id} - \ro(t) \grad^2 
d_f(t,x))^{-1}$, so that
%
$$
\vert \grad^2 d_g(t,y) - \grad^2 d_f(t,x)\vert
\leq 2 M \ro(t).
$$
%
Therefore, recalling that $f \in \FF$, 
for any $t\in ~]a,b[~$ we have 
%
$$
\aligned
& - \frac{\partial d_g}{\partial t}(t,y) = 
- \frac{\partial d_f}{\partial t}(t,x)+ \dot \ro(t) 
\leq F(t,\grad d_f(t,x),\grad^2 d_f(t,x)) + \dot \ro(t) 
\\
& =  F(t,\grad d_g(t,y),\grad^2 d_f(t,x)) + \dot \ro(t)  
\\
& \leq F(t,\grad d_g(t,y),\grad^2 d_g(t,y)) +  2 L M \ro(t) + \dot \ro(t) 
\\
& < F(t,\grad d_g(t,y),\grad^2 d_g(t,y)),
\endaligned
\tag{2.12}
$$
%
so that $g \in \FFm$. Hence $f(b) \subseteq g(b) \subseteq
\M(E,\FFm)(b)$.

Let us prove ({2.11}). 
It is enough to show that for any $f : \ab \rga \Pn$,
$f \in \F_F^=$, we have $\M(f(a),\F_F^=,a)(t) \subseteq f(t)$
for any $t \in \ab$; this follows by the comparison
principle between smooth evolutions, since $f$ is a barrier on $\ab$.
\qed\enddemo 

\proclaim{Remark {2.3}}
To prove
({2.10}) for open sets $E$ we only need that $F$ is locally
Lipschitz in the $X$-variable. As we shall see 
in ({6.5}), equality ({2.10}) holds true
under weaker assumptions on $F$ (which may also depend on $x$).
\endproclaim
 
\example{Example {2.1}}
Let 
$F(p,X) = - \text{tr}(P_p X P_p)$ (i.e., motion by mean curvature) and
%
$$
\align
& \C_F := \{f : \ab \rga \Pn, f \in \FF, f(t) \text{ is convex for any }
t \in \ab\},
\\
& {\Cal D}_F := \{f : \ab \rga \Pn, f \in \F_F^=, f(a) \text{ is convex}\}.
\endalign
$$
%
Then for any $E \subseteq \Rn$ we have
%
$$
\align
& \M_*(E,\C_F) = \M_*(E,{\Cal D}_F) = \M_*(E,\F_G),
\\
& \M^*(E,\C_F) = \M^*(E,{\Cal D}_F) = \M^*(E,\F_G),
\endalign
$$
%
where 
%
$$
G(p,X) := 
\cases
F(p,X) & \text{if } X \geq 0,
\\
0 & \text{elsewhere.}
\endcases
$$
%
Note that if $n=2$ then $G = F \wedge 0$.

\demo{Proof} Let $E \subseteq \Rn$.
Using [{18}] (see also [{14}]) we know that a smooth convex
set flowing by mean curvature remains convex, hence we have that
$\M(E,\C_F) \supseteq \M(E, {\Cal D}_F)$. Reasoning as in the 
proof of ({2.9}) we also have $\M(E,\C_F) = \M(E, {\Cal D}_F)$.
Furthermore, as ${\Cal C}_F = {\Cal C}_G \subseteq
\F_G$, we have $\M(E, \C_F) \subseteq \M(E, \F_G)$.

To complete the proof it is enough to show
%
$$
\M(A,\C_F) \supseteq \M(A,\F_G),
\tag{2.13}
$$
%
for any open set $A \subseteq \Rn$.
We will prove that, given $g : \ab\subseteq I\rga \Pn$, $g \in \F_G^>$,
we have
%
$$
g(t) \subseteq \M(g(a), \C_F, a)(t), \qquad t \in \ab,
\tag{2.14}
$$
%
which implies $\M(E, \C_F) \supseteq \M(E,\F_G^>)$
for any $E \subseteq \Rn$, which in turn, thanks to
({2.10}), implies ({2.13}).

\noindent 
For any $x \in \partial g(a)$, let $C_x \subseteq g(a)$
be a convex set with smooth boundary such that $\partial C_x 
\cap \partial g(a) = \{x\}$, $\grad d_{C_x}(x) = 
\grad d_{g(a)}(x)$, $\grad^2 d_{C_x}(x)= \grad^2 d_{g(a)}(x)$
and $\sup_{y \in \partial C_x} \vert \grad^2 d_{C_x}(y) \vert
\leq 2 \vert \grad^2 d_{C_x}(x)\vert$, which implies 
$\vert \grad^2 d_{C_x}\vert\leq K$, for a positive constant
$K$ independent of $x \in \partial g(a)$. 
By [{18}] we can find $\tau >0$, independent
of $x\in \partial f(a)$, such that 
there exists a smooth mean curvature 
evolution $f_x : [a,a+\tau] \rga \Pn$,
$f_x(a) = C_x$ and $f_x \in \C_F \cap \F_F^=$. 
Note that 
%
$$
\frac{\partial d_g}{\partial t}(a,x) > \frac{\partial d_{C_x}}{\partial t}(a,x).
\tag{2.15}
$$
%
Using ({2.15}) and 
 an argument similar to the one in Lemma {5.1} 
(see [{6}]) 
we have $\partial g(t) \subseteq 
\displaystyle \bigcup_{x \in 
\partial g(a)} f_x(t)$ for any $t \in [a,a+\tau]$, that implies,
as $g(t)$ is compact for any $t \in \ab$, $g(t) \subseteq
\M(g(a), \C_F,a)(t)$ for any $t \in [a, a+\tau]$.
Now ({2.14}) follows by an induction argument and 
the compactness of $\ab$.
\qed\enddemo
\endexample

\example{Example {2.2}}
Let us define the family ${\Cal G}$ as follows.
 A function $f : \ab \rga \Pn$ belongs to 
${\Cal G}$ if and only if $f(t)$ is compact for any
$t \in \ab$, 
there exists an open set
$A \subseteq \Rn$ such that
$d_f \in \C^\infty(\ab \times A)$, $\partial f(t) \subseteq A$ for any
$t \in \ab$, and 
%
$$
\Delta d_f > 0, \qquad
\frac{\partial d_f}{\partial  t} 
+ \frac{1}{\Delta d_f} \geq 0 \qquad t \in ~]a,b[~, x \in \partial f(t).
$$
%
Then the associated minimal barrier $\M(E,{\Cal G},t_0)$
provides a definition of weak evolution
of  any convex set $E \subseteq \Rn$
 by the inverse mean curvature (see [{19}]).
\endexample

We conclude this section by recalling
that in [{12}] 
a suitable choice of $\F$ is suggested to obtain
motion by mean curvature
of manifolds of arbitrary codimension, see
Remark {6.4} and [{1}].


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\head
{3}. Level sets of 
subsolutions are barriers
\endhead
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let us begin 
the comparison between 
the  minimal barriers and the viscosity evolution.
>From now on we take $I = [0,+\infty[$ (i.e., $t_0 =0$) and 
all barriers we consider are barriers on $\oi$. 
Moreover we use
the word subsolution 
to mean viscosity subsolution (and similarly for
solution and supersolution).
The function $F$ is always geometric, and
we denote by $F_*$ (resp. $F^*$) the lower (resp. upper)
semicontinuous envelope of $F$.

We list here some assumptions we use in the sequel. 
We follow the notation of [{16}, pp. 462-463]; 
we omit those properties in [{16}]
which are not useful in our context.

\item{(F1)} $F: J_0 \rga \R$ is continuous;

\item{(F2)} $F$ is degenerate elliptic, i.e.,
%
$$
F(t,x,p,X) \geq F(t,x,p,Y)
$$
%
for any $(t,x,p,X) \in J_0$, $Y \in \text{Sym}(n)$, $Y \geq X$;

\item{(F3)} $-\infty < F_*(t,x,0,0) = F^*(t,x,0,0) < +\infty$ 
for all $t \in \oi$, $x \in \Rn$;

\item{(F4)} for every $R >0$, $\sup\{\vert F(t,x,p,X)\vert :
\vert p\vert, \vert X\vert \leq R, (t,x,p,X) \in J_0\} < +\infty$.
One can check that if $F$ is geometric and satisfies
(F4), then $F_*(t,x,0,0) \leq 0$, $F^*(t,x,0,0) \geq 0$ for any
$t \in \oi$, $x \in \Rn$.

\item{(F6)} For every $R > \ro > 0$ there is a constant $c= 
c_{R,\ro}$ such that 
%
$$
\vert F(t,x,p,X) - F(t,x,q,Y)\vert \leq c (\vert p-q\vert
+\vert X-Y\vert)
$$
%
for all $t \in \oi$, $x \in \Rn$,
$\ro \leq \vert p\vert, \vert q\vert \leq R$,
$\vert X\vert, \vert Y\vert \leq R$;

\item{(F6')} for every $R > \ro > 0$ there is a constant 
$c = c_{R,\ro}$ 
such that
%
$$
\vert F(t,x,p,X)-F(t,x,q,X)\vert \leq c \vert p-q\vert
$$
%
for any $t \in \oi$, 
$x \in \Rn$, $\ro \leq \vert p\vert,\vert q\vert \leq R$,
$\vert X\vert \leq R$;

\item{(F7)} there are $\ro_0>0$ and a modulus $\sigma_1$ such that
%
$$
\align
& F^*(t,x,p,X) - F^*(t,x,0,0) \leq \sigma_1(\vert p\vert + \vert X\vert),
\\
& F_*(t,x,p,X) - F_*(t,x,0,0) \geq -\sigma_1(\vert p\vert + \vert X\vert),
\endalign
$$
%
provided $t \in \oi$, $x \in \Rn$,
$\vert p\vert, \vert X\vert \leq \ro_0$;

\item{(F8)} there is a modulus $\sigma_2$ such that 
%
$$
\vert F(t,x,p,X)-F(t,y,p,X)\vert \leq \vert x-y\vert \vert p\vert
\sigma_2(1 + \vert x-y\vert)
$$
%
for $y \in \Rn$, $(t,x,p,X) \in J_0$;

\item{(F8')}  for any $R \geq 0$ 
there is a modulus $\sigma_R$ such that 
%
$$
\vert F(t,x,p,X)-F(t,y,p,X)\vert \leq \vert x-y\vert \vert p\vert
\sigma_R(1 + \vert x-y\vert)
$$
%
for $y \in \Rn$, $(t,x,p,X) \in J_0$, $\vert X\vert \leq R$;

\item{(F9)} there is a modulus $\sigma_2$ 
such that
$F_*(t,x,0,0)-F^*(t,y,0,0) \geq -\sigma_2(\vert x-y\vert)$
for any $t \in \oi$, $x,y \in \Rn$;

\item{(F10)} suppose that 
$\displaystyle -\mu \left(\matrix \text{Id} & 0 \\
0 & \text{Id} \endmatrix \right)
 \leq  \left(\matrix X & 0 \\
0 & Y \endmatrix \right) \leq
 \nu  \left(\matrix \text{Id} & -\text{Id} \\
-\text{Id} & \text{Id} \endmatrix \right)$
with $\mu,\nu \geq 0$. Let $R \geq 2 \nu \vee \mu$ and let 
$\ro >0$; then 
%
$$
F_*(t,x,p,X) - F^*(t,y,p,-Y) \geq - \vert x-y\vert \vert p\vert
\overline \sigma(1 + \vert x-y\vert
+ \nu \vert x-y\vert^2)
$$
%
for $(t,x) \in \oi \times \Rn$, 
$\ro \leq \vert p\vert \leq R$, with some modulus
$\overline \sigma=\overline \sigma_{R,\ro}$ independent of
$t,x,y,X,Y,\mu,\nu$.

\proclaim{Remark {3.1}}
One can check that, if $F$ is geometric,
then condition (F6)
(resp. (F6'), (F8), (F10))
is equivalent to the analogous condition in [{16}].
Moreover (F4) implies conditions $(6.3_\pm)$ of [{9}]
and (F10) implies (F2) (see [{16}, proof
of Theorem 2.1, case 2]) and (F8').
Furthermore, it is proved in [{16}, Proposition 4.3]
that
(F3), (F8) imply (F9) and (F2), (F6), (F8) imply (F10).
\endproclaim
%

Let $A \subseteq \Rn$.
We recall [{9},{10}]
that a function $u : \oi \times A \rga \R$ is called
a viscosity sub-(super) solution of 
%
$$
\frac{\partial u}{\partial t} + F(t,x,\grad u, \grad^2 u) =0
\tag{3.1}
$$
%
in $]0,+\infty[ \times A$ if $u^* < +\infty$ (resp. $u_* > -\infty$)
in $\oi \times A$ and 
%
$$
\frac{\partial \psi}{\partial t}(\ot,\ox) + 
F_*(\ot,\ox,\grad \psi(\ot,\ox), \grad^2 \psi(\ot,\ox)) \leq 0
\tag{3.2}
$$
%
%
$$
\left(\text{resp. }
\frac{\partial \psi}{\partial t}(\ot,\ox) + 
F^*(\ot,\ox,\grad \psi(\ot,\ox), \grad^2 \psi(\ot,\ox)) 
\geq 0\right)
\tag{3.3}
$$
%
for any function $\psi \in \C^\infty(]0,+\infty[ \times A)$ such that
$u^* - \psi$ (resp. $u_* - \psi$) has a maximum (resp. minimum)
at $(\ot,\ox) \in ~]0,+\infty[ \times A$ (one achieves an equivalent
definition of viscosity sub- and supersolution by taking $\C^2$ test
functions).

\noindent
Finally, we define
%
$$
F_c(t,x,p,X) := - F(t,x,-p,-X)
$$
%
for any $(t,x,p,X) \in J_0$.
Note that if $F$ is degenerate elliptic then $F_c$ is degenerate
elliptic.

The following theorem is proved in [{16}, Theorem 4.9].
%
\proclaim{Theorem {3.1}}
Assume that $F : J_0 \rga \R$ is geometric and satisfies 
either (F1)-(F4), (F8), or (F1), (F3), (F4), (F9), (F10).
Let $v_0: \Rn \rga \R$ be a continuous function which is constant
outside a bounded subset of $\Rn$. Then there exists a unique
continuous viscosity solution (constant outside a bounded subset
of $\Rn$) of ({3.1}) with $v(0,x) = v_0(x)$. 
\endproclaim

Given a bounded open set $E \subseteq \Rn$
 we define the viscosity
evolutions $V(E)(t)$, $\Gamma(t)$ of $E$, $\partial E$ respectively
(the so called level-set flow) as
%
$$
V(E)(t) := \{x \in \Rn : v(t,x) < 0\}, 
\qquad
\Gamma(t) := \{x \in \Rn : v(t,x) = 0\}, 
\tag{3.4}
$$
%
where $v$ is as in Theorem {3.1} with $v_0(x) 
:= (-1) \vee d_E(x) \wedge 1$. It is proved in 
[{16}] that, if $u$ denotes the solution
of ({3.1}) with $u(0,\cdot) = u_0(\cdot)$,
where 
$u_0 : \Rn \rga \R$ is 
an admissible initial function
such that $\{u_0 \leq 0 \} = \{v_0 \leq 0\}$ and
$\{u_0 = 0 \} = \{ v_0 =0\}$, then 
$\{u(t,\cdot) \leq 0 \} = \{v(t,\cdot) \leq 0\}$ for any $t \in \oi$.
Applying the same argument to $-u,-v$, which are solutions
of ({3.1}) with $F$ replaced by $F_c$, we also 
have
$\{u(t,\cdot) \geq 0 \} = \{v(t,\cdot) \geq 0\}$ for any 
$t \in \oi$. We then conclude
that $u_0$ and $v_0$ give raise to the same
level set flow.

When $F : J_1 \rga \R$ does not depend on $x \in \Rn$ all
previous definitions are consequently modified in the obvious way.

The following result can be proved reasoning 
as in [{1}, Lemma 3.11].

\proclaim{Lemma {3.1}}
Assume that $F : J_0 \rga \R$ 
is geometric and satisfies (F2) and (F4).
Let $\Omega \subseteq \Rn$ be an open set and let
$u : \oi \times \Omega \rga~ ]-\!\infty~,0]$
(resp. $u : \oi \times \Omega \rga~ [0, +\infty[~$)
 be an upper (resp. lower) semicontinuous
function satisfying the following properties:
\item{(i)} for every $(t,x) \in ~ ]0,+\infty[ \times \Omega$ 
with $u(t,x) =0$, there
is a sequence $\{(t_m,x_m)\}$ of points 
of $~]0,+\infty[ \times \Omega$ converging to $(t,x)$ 
such that $u(t_m,x_m) =0$ and $t_m < t$;
\item{(ii)} $u$ is a viscosity sub (resp. super) solution  of 
({3.1})
in the set $\{(t,x) \in ~]0,+\infty[ \times \Omega : u(t,x) < 0\}$
$($ 
resp. in the set $\{(t,x) \in ~]0,+\infty[ \times \Omega : u(t,x) > 0\}$ $)$;
\item{(iii)} $\vert u(t,x) - u(t,y) \vert \leq \vert x-y\vert$ for 
any $t \in ~]0,+\infty[~$, $x,y \in \Omega$. 

\noindent 
Then $u$ is a viscosity subsolution 
(resp. supersolution)
of ({3.1})
in $~]0,+\infty[ \times \Omega$.
\endproclaim
%

The main result of this section reads as follows.

%
\proclaim{Theorem {3.2}}
The following two statements hold.

\noindent 
\item{A)} Assume that $F: J_1 \rga \R$
does not depend on $x$,
is geometric and satisfies (F1)-(F4), (F6), (F7). Let $u$
and $v$ be, respectively, 
 a viscosity sub- and supersolution of %
$$
\frac{\partial u}{\partial t} + F(t,\grad u, \grad^2 u) =0
\tag{3.5}
$$
%
in $]0,+\infty[ \times \Rn$.
Then for any $\lambda \in \R$ we have
%
$$
\align
& \{x \in \Rn: u^*(\cdot,x) < \lambda\} \in \BFF,
\tag{3.6}
\\
& \{x \in \Rn: u^*(\cdot,x) \leq \lambda\} \in \BFF.
\tag{3.7}
\\
& \{x \in \Rn: v_*(\cdot,x) > \lambda\} \in {\Cal B}(\F_{F_c}),
\tag{3.8}
\\
& \{x \in \Rn: v_*(\cdot,x) \geq \lambda\} \in {\Cal B}(\F_{F_c}).
\tag{3.9}
\endalign
$$
%
Let $w\!$ be the unique uniformly continuous
viscosity solution of ({3.5}) in $]0,\!+\infty[\!\times\!\Rn$ 
with $w(0,x) = u_0(x)$ a given continuous function which is constant 
outside a bounded subset of $\Rn$ (see Theorem {3.1}). 
Then for any $\lambda\in\R$ 
we have ({3.6}), ({3.7}) with $u^*$ replaced
by $w$ and ({3.8}), ({3.9}) with $v_*$ replaced
by $w$. If additionally $F = F_c$ then for any $\lambda
\in \R$ we have also
%
$$
\{x \in \Rn : w(\cdot,x) = \lambda\} \in \BFF.
\tag{3.10}
$$
%
\item{B)} Assume that $F: J_0 \rga \R$
is geometric and satisfies 
(F1), (F3), (F4), (F6'), (F7), (F9), (F10).
Then, if we substitute ({3.5}) with ({3.1}) and
$F(t,p,X)$ with $F(t,x,p,X)$ all assertions of statement A)
hold.
\endproclaim

\demo{Proof} 
Statement A). 

To prove ({3.6}) it is enough
to consider the case $\lambda =0$. 
Let $f : \ab \subseteq \oi 
\rga \Pn$, $f \in \FF$, and $f(a) \subseteq \{x \in \Rn :
u^*(a,x) < 0\}$; 
we have to show that $f(b) \subseteq \{x \in \Rn : u^*(b,x) < 0\}$. 
Reasoning as in
[{1}, Corollary 3.9, step 7] and using Lemma {3.1}
one can check that the function $\delta := d_f \vee 0$ 
is a continuous supersolution of ({3.5})
in $~ ]a,b[ \times \Rn$ (see also [{4}, Theorem 3.1]). Then 
$\delta$ 
is a continuous supersolution of ({3.5})
in $~ ]a,b] \times \Rn$ (see, for instance, 
[{9}, Lemma 5.7]).
Moreover, since subsolutions are preserved by the composition
with a continuous nondecreasing function (see [{9}, Theorem 5.2]),
we have that 
$u^* \wedge 0$ is an upper semicontinuous 
subsolution of ({3.5}) in 
$~ ]a,b] \times \Rn$. 
As 
$f(a) \subseteq \{x \in \Rn : u^*(a,x) <0\}$ and $f(a)$ is compact, 
there is $\eps > 0$ such that
$\delta(a,\cdot) - \eps \geq (u^*\wedge 0)(a,\cdot)$ on $\Rn$.
We can
apply the viscosity 
comparison principle in [{16}, Theorem 4.1]
to $u^* \wedge 0$ and $\delta -\eps$, and we obtain
%
$$
\delta(t,x) \geq (u^*\wedge 0)(t,x) +\eps, 
\qquad (t,x) \in  \ab \times \Rn.
\tag{3.11}
$$
%
This implies 
%
$$
f(t) \subseteq 
\{x \in \Rn : u^*(t,x) \leq -\eps\} \subseteq \{x \in \Rn:u^*(t,x) < 0\},
\qquad t \in \ab,
$$
%
and ({3.6}) is proved. The relation in
({3.7}) follows 
from ({3.6}) by observing that 
$\{x \in \Rn :
u^*(\cdot,x) \leq 0\} = \displaystyle
 \bigcap_{\eps > 0} \{x \in \Rn : u^*(\cdot,x) < \eps\}$.

Assertions ({3.8}), ({3.9}) follow from
({3.6}), ({3.7}) by recalling that 
$-u^*$ is a supersolution of ({3.5}) with
$F$ replaced by $F_c$, and 
({3.10}) follows from
({3.7}) and ({3.9}).

Statement B).

Following the proof and the 
notation of statement A) and using
the viscosity comparison principle in
[{16}, Theorem 4.2] and 
the fact that a supersolution in $]a,b[ \times \Rn$ 
is a supersolution in $]a,b] \times \Rn$
(see [{9}, Lemma 5.7]), 
in order to show that $f(b) \subseteq \{x \in \Rn : u^*(b,x) < 0\}$
it is enough to prove that the function $\chi(t,x) :=
1- \chi^{}_{f(t)}(x)$ is a supersolution of ({3.1})
in $]a,b[ \times \Rn$.

Let $(\ot,\ox) \in ~]a,b[ \times  \Rn$
and let $\psi$ be a smooth
function such that $(\chi - \psi)$ has a minimum at 
$(\ot,\ox)$. 
Assume first that $\ox \in \text{int}(f(\ot))$.
We can suppose that 
$\chi(\ot,\ox) = \psi(\ot,\ox)=0$.
As $\chi$ is twice differentiable at $(\ot,\ox)$ we have
$\grad \psi(\ot,\ox) =0$, 
$\grad^2 \psi(\ot,\ox) \leq 0$. 
Moreover there exists two 
sequences $\{(t^{(1)}_m,x_m)\}$,
$\{(t^{(2)}_m,x_m)\}$ 
converging to $(\ot,\ox)$, with $t^{(1)}_m < \ot < t^{(2)}_m$ for any
$m \in {\bold N}$, 
such that $\chi(t_m^{(i)},x_m) =0 \geq \psi(t^{(i)}_m,x_m)$,
$i= 1,2$. 
Therefore $\frac{\partial \psi}{\partial t}(\ot,\ox) =0$, and we conclude
$\frac{\partial \psi}{\partial t}(\ot,\ox) + F^*(\ot,\ox, \grad \psi(\ot,\ox),
\grad^2 \psi(\ot,\ox))  = F^*(\ot,\ox,0, \grad^2 \psi(\ot,\ox))
\geq F^*(\ot,\ox,0,0) =0$.

The case $\ox \in \Rn \setminus f(\ot)$ is similar.
It remains to consider the case $\ox \in \partial f(\ot)$.
Pick $\eps >0$ and $\tau >0$ small
enough 
so that each point
of $B_{\eps/2}(\ox)$ 
has a unique smooth projection on $\partial f(t)$
belonging to $B_\eps(\ox)$ 
for any $t \in [\ot-\tau, \ot + \tau] \subseteq ~]a,b[$. 
Define
%
$$
F_\eps(t,x,p,X) := \cases
\displaystyle
\sup_{x \in \overline{B_\eps(\ox)}} F^*(t,x,p,X) & 
\text{if } x \in \overline{B_\eps(\ox)},
\\
F^*(t,x,p,X) & \text{elsewhere}.
\endcases
$$
%
Note that $f \in \F_{F_\eps}$, $F_\eps$ 
is geometric, upper semicontinuous and satisfies (F2) and (F4).
We claim that the function
$\delta:= d_f \vee 0$ is a supersolution of 
$\frac {\partial u} {\partial t} + 
F_\eps(t,x,\grad u,\grad^2 u) =0 $ in $]\ot - \tau, \ot +\tau[ \times
B_{\eps/2}(\ox)$.
To prove the claim we follow
[{1}, Corollary 3.9, step 7]. Thanks to
Lemma {3.1} it is sufficient to prove that 
$\delta$ is a supersolution in $\{\delta >0\}$. 
Let $\xi$ be a smooth function and
$(t_0,x_0) \in ~]\ot - \tau, \ot +\tau[ \times
B_{\eps/2}(\ox)$ be a minimizer of $(\delta -\xi)$, with
$\delta(t_0,x_0) > 0$.
Choose $y_0 \in \partial f(t_0) \cap
B_{\eps}(\ox)$ satisfying $\delta(t_0,x_0) = \vert x_0 - y_0\vert$
and set $\zeta(t,y) := \xi(t,y+x_0 - y_0)$. Then $(t_0,y_0)$
is a minimizer of $(\delta - \zeta)$ by the triangular
property of $\delta$. Let $V := H_\sigma(\delta)$,
where $\sigma > 0$ is such that $d_f$ is smooth on
$\{x \in \Rn : \text{dist}(x,\partial f(t)) < \sigma\}$,
$t \in ~]\ot - \tau, \ot + \tau[$,
 $H_\sigma(r) := r \wedge \sigma/2$, $r \geq 0$. As
$y_0 \in \partial f(t_0)$,
$(t_0,y_0)$ is also a minimizer of $(V - \zeta)$. Reasoning 
as in [{1}, Thorem 3.8 and Corollary 3.9], we have that
$V$ is a supersolution of 
$\frac {\partial u} {\partial t} + 
F_\eps(t,x,\grad u,\grad^2 u) =0 $ in $(]\ot - \tau, \ot +\tau[ \times
B_{\eps/2}(\ox)) \cap \{V >0\}$; by Lemma {3.1}
it is a supersolution 
in $]\ot - \tau, \ot +\tau[ \times
B_{\eps/2}(\ox)$. Therefore
%
$$
\align
& 0 \leq \frac{\partial \zeta}{\partial t}(t_0,y_0) 
+ F_\eps(t_0,y_0,\grad \zeta(t_0,y_0), \grad^2
\zeta(t_0,y_0)) 
\\
& = 
\frac{\partial \xi}{\partial t}(t_0,x_0) 
+ F_\eps(t_0,y_0,\grad \xi(t_0,x_0), \grad^2
\xi(t_0,x_0)) 
\\
& = \frac{\partial \xi}{\partial t}(t_0,x_0) 
+ F_\eps(t_0,x_0,\grad \xi(t_0,x_0), \grad^2
\xi(t_0,x_0)),
\endalign
$$
%
and this proves the claim.

Using the stability properties of viscosity supersolutions
it then follows that 
also the function $\chi$ is a supersolution of 
$\frac {\partial u} {\partial t} + 
F_\eps(t,x,\grad u,\grad^2 u) =0 $ on $]\ot - \tau, \ot +\tau[ \times
B_{\eps/2}(\ox)$ (see for instance
[{1}, Lemma 4.3]).
Therefore
%
$$
\frac{\partial \psi}{\partial t}(\ot,\ox) + 
F_\eps(\ot,\ox, \grad \psi(\ot,\ox),
\grad^2 \psi(\ot,\ox)) \geq 0.
$$
%
Letting $\eps \rga 0$ we get
({3.3}).
\qed\enddemo

We recall that $F^+ : J_0 \rga \R$ is defined 
by 
%
$$
F^+(t,x,p,X) := \sup\{F(t,x,p,Y) : Y \geq X\}, \qquad
(t,x,p,X) \in J_0.
$$
%
Notice that $F^+$ is the smallest degenerate elliptic
function greater than or equal to $F$; moreover,
if $F$ is geometric (resp. lower semicontinuous)
then $F^+$ is geometric (resp. lower semicontinuous).

\proclaim{Remark {3.2}}
The theses ({3.6}), ({3.7}) still hold
if we assume that $F^+$, in place of $F$, satisfies the assumptions
in Theorem {3.2}, statements A), B) $($recall
that $F^+ \geq F$, hence ${\Cal B}(\F_{F^+}) \subseteq {\Cal B}(\F_F))$.
\endproclaim

\proclaim{Remark {3.3}}
In Theorem {3.2},
if $F$ does not depend explicitly on $(t,x)$, then $u_0$ 
can be taken uniformly continuous (see 
[{22}, Theorem 2.2] and also
[{1}, Theorem 2.4]).
\endproclaim

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\head
{4}. 
Some useful results on barriers
\endhead
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

All results of this section will be used to prove Theorem
{5.1}, which is the converse of Theorem
{3.2}.

The next lemma shows that
we can construct arbitrarily small elements of $\FFm$
with assigned normal and curvatures in a
suitable neighbourhood of a 
given point. Note that 
we will not assume that $F$ is degenerate elliptic.

\proclaim{Lemma {4.1}}
Assume that $F : (\Rn \setminus \{0\}) \times \text{Sym}(n) \rga \R$ 
does not depend on $(t,x)$, is geometric and
lower semicontinuous.
Let $L \subseteq \Rn$ be a closed set with smooth boundary. Let
$\overline x \in \partial L$ and $\alpha \in \R$ be such that
%
$$ 
\alpha + F\big(\grad d_L(\overline x), 
\grad^2 d_L(\overline x)\big) > 0.
\tag{4.1}
$$
%
Then for any $R >0$ there exist $\tau >0$, 
$f : [a,a+\tau] \rga \Pn$ and $\sigma > 0$ such that 
%
$$
\align
& f(a) \subseteq L, \quad
\partial f(a) \cap B_\sigma(\overline x) = 
\partial L \cap B_\sigma(\overline x), 
\quad
\alpha = \frac{\partial d_f}{\partial t}(a,\overline x),
\tag{4.2}
\\ 
& f \in \FFm, \qquad
f(t) \subseteq B_R(\overline x), \qquad t \in [a,a+\tau].
\tag{4.3}
\endalign
$$
%
\endproclaim
\demo{Proof}
As $F$ is lower semicontinuous, it is 
the pointwise supremum of a family of continuous functions,
and since $F_1 \leq F_2 \Rightarrow \F_{F_1}^> \subseteq \F_{F_2}^>$,
we can assume that $F$ is continuous. 
Fix $R >0$ and set
$(\overline p, \overline X) := \big(\grad d_L(\overline x), 
\grad^2 d_L(\overline x)\big)$.

{\smc case 1}. Assume that $F$ is degenerate elliptic. 
Choose any smooth compact set, that we denote by $f(a)$,
such that $f(a) \subseteq L \cap
B_\sigma(\overline x)$ and 
$\partial f(a) \cap B_\sigma(\overline x) = 
\partial L \cap B_\sigma(\overline x)$,  
for a suitable
choice of $0 < \sigma < R$.
We claim that there exists a function $\widetilde F : (\{\Rn \setminus \{0\}) 
\times \text{Sym}(n) \rga \R$ with the following  properties:
$\widetilde F$ is geometric, of class $\C^\infty$, uniformly elliptic
and 
%
$$
\aligned
& \alpha + \widetilde F(\overline  p,\overline X)= 0,
\\
& \widetilde F\big(\grad d_{f(a)}(x),\grad^2 d_{f(a)} (x)\big) 
< F\big(\grad d_{f(a)}(x),\grad^2 d_{f(a)} (x)\big), 
\qquad x \in \partial f(a).
\endaligned
\tag{4.4}
$$
%
Let us prove the claim.
Fix $0 < \eps < (\alpha+F(\overline p, \overline X))/2$; 
approximating $F$ by convolution and using
the compactness of $\partial f(a)$,
we can find a degenerate elliptic function
$G_\eps : (\Rn \setminus \{0\}) \times \text{Sym}(n) \rga
\R$ of class $\C^\infty$ such that 
%
$$
\big\vert 
G_\eps\big(\grad d_{f(a)},\grad^2 d_{f(a)}\big) 
- F\big(\grad d_{f(a)},\grad^2 d_{f(a)}\big) \big\vert <
\eps/2 \qquad \text{on } \partial f(a).
\tag{4.5}
$$
%
Set 
$F_\eps (p,X) :=  \vert p\vert G_\eps\Big(\frac{p}{\vert p\vert}, 
\frac{P_p X P_p}{\vert p\vert}\Big)$ for  $(p,X) \in 
(\Rn \setminus \{0\}) \times \text{Sym}(n)$; then
$F_\eps$ is geometric, of class $\C^\infty$ and
degenerate elliptic. In addition
$G_\eps\big(\grad d_{f(a)}, \grad^2 d_{f(a)}\big) =
F_\eps\big(\grad d_{f(a)}, \grad^2 d_{f(a)}\big)$ on $\partial
f(a)$, hence
({4.5}) holds with $G_\eps$ replaced by $F_\eps$. 
Let $\eta >0$ be such that 
%
$$
\eta \vert \Delta d_{f(a)} (x) \vert < \eps/2, \qquad
x \in \partial f(a).
\tag{4.6}
$$
%
Set $c := -\alpha  - 
F_\eps (\overline p, \overline X) + \eta \Delta d_{f(a)} (\overline x)$.
Then $c < -\eps$, since by ({4.5}), ({4.6}) and 
({4.1})
%
$$
-c 
= \alpha + F_\eps(\overline p, \overline X) - \eta \Delta d_{f(a)}(\overline x)
> \alpha + F(\overline p, \overline X) - \eps
> \eps.
$$
%
Define 
$\widetilde F : (\Rn \setminus \{0\}) \times \text{Sym}(n)
\rga \R$ as 
%
$$
\widetilde F(p,X) := F_\eps(p,X) - \eta \text{tr}(P_p X P_p) + c.
$$
%
Then $\widetilde F$ is geometric, of class $\C^\infty$,   
uniformly elliptic, $\alpha + \widetilde F(\overline p,
\overline X) =0$; in addition, for any $x \in \partial f(a)$,
we have
%
$$
\align
& \widetilde F\big(\grad d_{f(a)}(x),\grad^2 d_{f(a)} (x)\big) 
< F\big(\grad d_{f(a)}(x),\grad^2 d_{f(a)} (x)\big) + \eps/2
- \eta \Delta d_{f(a)} (x) + c 
\\
& < 
F\big(\grad d_{f(a)}(x),\grad^2 d_{f(a)} (x)\big) + \eps + c
< F\big(\grad d_{f(a)}(x),\grad^2 d_{f(a)} (x)\big),
\endalign
$$
%
and the claim is proved.
 
Denote by $\Om$ an open set containing $\partial f(a)$ and such that
$d_{f(a)} \in \C^\infty(\Om)$.
Let $u$ be 
the unique smooth solution [{15}, Theorem 2.1] of 
%
$$
\cases
\displaystyle \frac{\partial u}{\partial t}
+  \widetilde F(\grad u,\grad^2 u(\text{Id} - u \grad^2 u)^{-1}) =0 ,
\\
u(a,x) = d_{f(a)}(x)
\endcases
\tag{4.7}
$$
%
on $[a,a+\tau] \times \Omega$, $\tau > 0$
sufficiently small.
Let $\partial f(t):= \{x : u(x,t) = 0\}$
and $f(t)$ be the closure of the bounded connected component
of $\Rn \setminus \partial f(t)$. 
In [{15}, Lemma 2.3] it is also proven that $\vert
\grad u \vert = 1$, hence $f \in {\Cal F}_{\widetilde F}^=$.
Notice that $\frac{\partial d_f}{\partial t}(a,\overline x) = \alpha$ and 
$f(t) \subseteq B_R(\overline x)$ for $t \in [a,a+\tau]$, 
provided that $\tau$ is small enough. 
It remains to show that $f \in \FFm$. 
As $d_f$ is of class $\C^\infty$ in a neighbourhood of $\partial 
f(\cdot)$ and the inequality in ({4.4}) holds,
possibly reducing $\tau$ we can assume 
%
$$
\widetilde F\big(\grad d_f(t,x),\grad^2 d_f(t,x)\big)
< F\big(\grad d_f(t,x),\grad^2 d_f(t,x)\big),
\qquad t \in [a,a+\tau], \ x \in \partial f(t).
$$
%
Therefore, by ({4.4}), on $\partial f(t)$, $t \in [a,a+\tau]$,
we have
%
$$
0 = \frac{\partial d_f}{\partial t} + \widetilde 
F\big(\grad d_f,\grad^2 d_f\big)
< \frac{\partial d_f}{\partial t} + F\big(\grad d_f,\grad^2 d_f\big),
$$
%
so that $f\in \FFm$. The proof of ({4.2}) and 
({4.3}) is complete.

{\smc case 2}. Assume that $F$ is not necessarily degenerate elliptic.
The proof is divided into three steps.

{\smc step 1}. For any $\eps >0$ there exist a smooth compact 
set, that we denote by $f_\eps(a)$, 
and $\sigma = \sigma(\eps) >0$ such that
%
$$
f_\eps(a) \subseteq L, \qquad 
\partial f_\eps(a) \cap B_\sigma(\overline x) = 
\partial  L\cap B_\sigma(\overline x),
\tag{4.8}
$$
%
%
$$
\grad^2 d_{f_\eps(a)}(x) > \overline X -\frac{\eps}{n} \text{Id} \qquad
\text{for any } x \in \partial f_\eps(a) \text{ with } 
\grad d_{f_\eps(a)}(x) = \overline p;
\tag{4.9}
$$
%
moreover we can find a  constant $k < 0$ independent of $\eps$ 
such that $(1 + k) \text{Id} < \overline X$ and
%
$$
\grad^2 d_{f_\eps(a)} (x) > k \text{Id}, \qquad
\eps \in ~]0,1], \ x \in \partial f_\eps(a).
\tag{4.10}
$$
%
Let $\eps >0$; 
up to a rotation and a translation, we can assume that there exist
a neighbourhood $U' = U_\eps'$ of $0$ in $\R^{n-1}$, a smooth function 
$l : U' \rga \R$ such that $\overline x = (0,l(0))$ and 
$\grad l(0) = 0$, and a neighbourhood $U = U_\eps \subseteq
B_R(\ox)$ of $\overline x$ 
in $\Rn$ such that
%
$$
\align
& U \cap \partial L = \{(x', l(x')) : x' \in U'\},
\qquad
U \cap L \subseteq \{(x', y) : y \geq l(x'), x' \in U'\},
\\
& \grad^2 l(x') > \grad^2 l(0) - \enI, \qquad x' \in U'.
\endalign
$$
%
Given $\ro > 0$, we choose  a function
$g = g_\ro : \oi \rga \oi$ with the following properties:
$g \in \C^\infty(\oi)$, $g$ is convex, $g = 0$ on $[0,\ro]$,
$g(s) = s^4/\ro$ for $s \in [2 \ro, +\infty[$.
We define $h = h_\ro : U' \rga \R$ as
$h(x') := l(x') + g(\vert x'\vert)$.
Notice that for $x' \in U'$ we have
%
$$
\grad^2 h(x') = \grad^2 l(x') 
+ \Big[
\frac{1}{\vert x'\vert} g'(\vert x'\vert) \Big(\text{Id} - 
\frac{x' \otimes x'}{\vert x'\vert^2}\Big) +
g''\big(\vert x'\vert\big)
\frac{x' \otimes x'}{\vert x'\vert^2} \Big] \geq \grad^2 l(x').
$$
%
Define $H = H_\ro := \{(x',y) : y \geq h(x'), x' \in U'\}$. 
Let us observe that, at each point $(x_0', h(x_0'))$ with 
$x_0' \in U'$ and $\grad h(x_0') =0$, the second fundamental
form of $\partial H$ is $\grad^2 h(x_0')$; therefore
%
$$
\grad^2 d_H(x_0', h(x_0')) > \oX - \enI.
$$
%
To have ({4.8}), ({4.9}) it is then enough
to define $\fea \subseteq U$ as a $\C^\infty$ regularization 
of $H \cap \overline{B_\mu(\ox)}$, where $\mu = \mu(\eps)$
and $\ro = \ro(\eps)$ are suitable positive numbers 
sufficiently small.

\noindent Eventually, property ({4.10}) holds by construction.

{\smc step 2}. Let $\eps >0$ and $\partial \fea$ be as in {\smc step 1}.
Then there exists $\delta = \delta(\eps, \overline p)$ 
such that 
%
$$
x \in \partial \fea, \ \grad d_{\fea}(x) = p, \
\vert p-\op\vert < \delta \Rightarrow
\grad^2 d_{\fea}(x) > \oX - \enI.
\tag{4.11}
$$
%
Indeed, assume by contradiction that there exists a sequence
$\{p_m\}\subset \SSn$ such that $p_m = \grad d_{\fea}(x_m)$
for $x_m \in \partial \fea$, 
$\lim_{m \rga +\infty} p_m = \op$ and $\grad^2 d_{\fea}(x_m)
\leq \oX - \enI$. Passing to a (not relabelled)
subsequence,
we have
%
$$
\lim_{m \rga +\infty} x_m = \widetilde x \in \partial f_\eps(a), \quad
\grad d_{\fea}(\widetilde x) = \op.
$$
%
By ({4.9}) we have
%
$$
\oX - \enI \geq \lim_{m \rga +\infty} \grad^2 d_{\fea}(x_m) = 
\grad^2 d_{\fea}(\widetilde x) > \oX - \enI,
$$
%
a contradiction. 
\smallskip

To continue the proof of the lemma, we introduce 
some notation: $Y$ will be always an element
of $\text{Sym}(n)$, we set
%
$$
\beta := \min_{q \in \SSn} F(q, k\text{Id}),
\qquad
c := \text{tr}\big(\oX - k \text{Id}\big) > 0,
\tag{4.12}
$$
%
and define
$m_0, m_1 : \oi \rga \oi$ as
%
$$
\aligned
& m_0(s) := \max\{\vert F(q,k \text{Id}) - F(q, k \text{Id} + Y)\vert
: q \in \SSn, Y \geq 0, \text{tr}(Y) \leq s
\},
\\
& m_1(s) := \max\Big\{\vert F(q,\oX) - F(q, \oX + Y) \vert
: q \in \SSn, \vert Y \vert \leq s\Big\}.
\endaligned
\tag{4.13}
$$
%
Then 
$m_0(0) =
m_1(0) =0$, and $m_0$, $m_1$ are continuous and nondecreasing. 

Moreover, recalling
({4.1}), we choose
$0 < \eps < \min(1,c)$ in such a way that 
%
$$
F\Big(p, \oX - \enI\Big) + \alpha > 2 m_1(\sqrt{2} \eps) + m_0(c)
- m_0(c-\eps), \qquad \vert p-\op\vert \leq \eps.
\tag{4.14}
$$
%

{\smc step 3}. Let $\eps$ be chosen as in ({4.14}) and let
$f_\eps(a)$ be the corresponding set given by {\smc step 1}. Then there
exists
a geometric,
uniformly elliptic function $\widetilde F :(\Rn \setminus \{0\})
\times \text{Sym}(n) \rga \R$ of class $\C^\infty$  such that 
%
$$
\aligned
& \alpha + \widetilde F(\overline  p,\overline X)= 0,
\\
& \widetilde F\big(\grad d_{\fea}(x),\grad^2 d_{\fea} (x)\big) 
< F\big(\grad d_{\fea}(x),\grad^2 d_{\fea} (x)\big), 
\qquad x \in \partial \fea.
\endaligned
\tag{4.15}
$$
%
Let us define the function $m_2:\oi \rga \oi$ as $m_2 = m_0$ 
on $[0,c-\eps]$ and, if $s \geq c-\eps$,
%
$$
\align
m_2(s) := 
& \max \Big(
m_0(s), m_0(c-\eps) + 
\max\Big\{\Big\vert F\Big(q, \oX -\enI\Big) - 
F\Big(q, \oX - \enI + Y\Big)\Big\vert : 
\\
& q \in \SSn, Y \geq 0, 
\text{tr}(Y) \leq s-c+\eps\Big\}\Big).
\endalign
$$
%

Choose $\beta' < \min (\beta, m_2(c) - \alpha)$, 
let $0 < \delta = \delta(\eps) \leq \eps$ be given by {\smc step 2}
and pick 
a function $g = g_\delta \in \C^\infty(\SSn)$ such that
%
$$
g(\op) = \max_{q \in \SSn} g(q) := m_2(c) - \alpha, \qquad
g(p) := \beta'  \qquad \text{for } \vert p-\op\vert\geq\delta.
\tag{4.16}
$$
%
Let us prove that for any $x\in\partial \fea$ we have
%
$$
g\big(\grad d_{\fea} (x)\big)
- m_2\big(\text{tr}(\grad^2d_{\fea} (x)-k\text{Id})\big) <
F\big(\grad d_{\fea} (x),\grad^2d_{\fea} (x)\big).
\tag{4.17}
$$
%
Let $x\in\partial \fea$ and set 
$(p,X):=(\grad d_{\fea} (x),\grad^2d_{\fea} (x))$.

If $\vert p-\op\vert\geq\delta$ then, recalling ({4.16}) 
and the definition of $\beta,\beta'$, 
we have
%
$$
g(p)=\beta'<\beta \leq F(p,k\text{Id}).
\tag{4.18}
$$
%
Moreover
%
$$
\aligned
& m_2(\text{tr}(X-k\text{Id}))\geq m_0(\text{tr}(X-k\text{Id}))
\\
& \geq \max\big\{F(p,k\text{Id}+Y) -F(p,k\text{Id}) : Y\geq 0, 
\text{tr}(Y)\leq\text{tr}(X-k\text{Id})\big\}.
\endaligned
\tag{4.19}
$$
%
Taking $Y=X-k\text{Id}$, by ({4.18}) and
({4.19}) we have 
%
$$
g(p)-m_2(\text{tr}(X-k\text{Id}))<F(p,k\text{Id})+F(p,X)
-F(p,k\text{Id})=F(p,X)
$$
%
and ({4.17}) is proved.

Assume now that $\vert p-\op\vert <\delta \leq \eps$. Then by 
({4.16}) we have
%
$$
g(p) \leq g(\op)=m_2(c)-\alpha.
\tag{4.20}
$$
%
Moreover, by ({4.11}) we also have $X\geq\oX -\enI$, 
hence $\text{tr}(X - k \text{Id}) \geq c-\eps$.
Therefore, recalling the definition of $m_2$, we have
%
$$
m_2(\text{tr}(X-k\text{Id}))\geq m_0(c-\eps)+
F\Big(p,\oX -\enI\Big)-F(p,X).
\tag{4.21}
$$
%
Then, 
by ({4.20}), ({4.21}) and ({4.14}) we find
%
$$
\aligned
& g(p) - m_2\big(\text{tr}(X-k\text{Id})\big) 
\leq m_2(c)-\alpha - m_0(c-\eps) - 
F\Big(p,\oX -\enI\Big) + F(p,X)  
\\
& < F(p,X) + m_2(c) - 2 m_1(\sqrt{2}\eps) - m_0(c).
\endaligned
$$
%
To prove ({4.17}) it is enough to 
show that
%
$$
m_2(c) - 2 m_1(\sqrt{2}\eps) - m_0(c) \leq 0.
\tag{4.22}
$$
%
If $m_2(c) = m_0(c)$ then ({4.22}) is 
obvious; otherwise
%
$$
\aligned
& m_2(c)-m_0(c)= - m_0(c) + m_0(c-\eps) 
\\
& + \max\Big\{\Big\vert F\Big(q,\oX-\enI\Big)-
F\Big(q,\oX-\enI+Y\Big)\Big\vert : 
q\in\SSn, Y\geq 0, \text{tr}(Y)\leq\eps\Big\} 
\\
& \leq \max\Big\{\Big\vert F\Big(q,\oX-\enI\Big)-
F\Big(q,\oX-\enI+Y\Big)\Big\vert : 
q\in\SSn, Y\geq 0, \text{tr}(Y)\leq\eps\Big\} 
\\
& \leq \max\Big\{\Big\vert F\Big(q,\oX-\enI\Big)-
F\Big(q,\oX\Big)\Big\vert : 
q\in\SSn\Big\} 
\\
& + \max\Big\{\Big\vert F\Big(q,\oX\Big)-
F\Big(q,\oX-\enI+Y\Big)\Big\vert : 
q\in\SSn, Y\geq 0, \text{tr}(Y)\leq\eps\Big\} 
\\
& \leq m_1(\eps/\sqrt{n}) + m_1(\sqrt{2}\eps) 
\leq 2 m_1(\sqrt{2}\eps).
\endaligned
$$
%
Therefore ({4.17}) is proved.

By ({4.17}) and the continuity of $F$, 
there exists $\eps'>0$ such that
%
$$
F(\grad d_\fea(x),\grad^2d_\fea(x)+Y)-g(\grad d_\fea(x))
+m_2(\text{tr}\big(\grad^2d_\fea(x)-k\text{Id})\big)>\eps',
$$
%
for any $x\in\partial\fea$, $Y\geq 0$, tr$(Y)\leq\eps'$.

Choose now an odd function $m \in \C^\infty(\R)$ such that 
$m \geq  m_2$ on $[c+\eps', +\infty[~$, 
$\vert m-m_2\vert <\eps'$ on $[0, c+\eps']$, 
$m(c) = m_2(c)$ and 
$m'(x) \geq \lambda$ for any $x \in \R$ and for
a suitable constant $\lambda > 0$.
Eventually, we set
%
$$
\align
& G(p,X) := g(p) - m(\text{tr}(X- k\text{Id})), \qquad (p,X) \in 
\SSn \times \text{Sym}(n),
\\
& \widetilde F(p,X) := \vert p\vert G\Big(\frac{p}{\vert p\vert}, 
\frac{P_p X P_p}{\vert p\vert}\Big), \qquad (p,X) \in 
(\Rn \setminus \{0\}) \times \text{Sym}(n).
\endalign
$$
%
Then one can check that $\widetilde F$ 
is smooth,  geometric, uniformly elliptic and
$\alpha + \widetilde F(\op,\oX) = 0$. 
The inequality in ({4.15}) follows from 
({4.17}) and the definition of 
$\widetilde F$. The proof of {\smc step 3} is concluded.

Now the thesis follows reasoning as in {\smc case 1}
replacing $f(a)$ with $\fea$ (see 
({4.7}) and below).
\qed\enddemo

\proclaim{Remark {4.1}}
Following [{24}, Theorem 8.5.4], one can show that
the number $\tau$ in the statement of Lemma {4.1} 
can be chosen depending
in a continuous way on the initial datum $u(a,x)$ in ({4.7})
in a $\C^{2+\alpha}$ neighbourhood of $d_{f(a)}$, 
for any $\alpha \in ~]0,1[$.
\endproclaim

\proclaim{Remark {4.2}}
>From Lemma {4.1} one can check that the following
holds. Let $\alpha, \alpha_m \in \R$ with $\alpha_m \rga \alpha$,
$\ox, x_m \in \Rn$ with $x_m \rga \ox$, and $\partial L,
\partial L_m$ be a family of smooth closed hypersurfaces
such that 
$\ox \in \partial L$, $x_m \in \partial L_m$,
$\partial L_m \rga \partial L$ locally in 
$\C^\infty$, and 
%
$$
\alpha + F(\grad d_L (\ox), \grad^2 d_L (\ox)) > 0,
\qquad
\alpha_m + F(\grad d_{L_m}(x_m), \grad^2 d_{L_m}(x_m)) > 0
$$
%
for any $m\in {\bold N}$.
Then we can find corresponding $\tau, \tau_m >0$, $f,f_m \in \FFm$
given by Lemma {4.1}, such that $\partial f_m(a)
\rga \partial f(a)$ in $\C^\infty$, hence $\tau_m \rga
\tau$ by Remark {4.1}.
\endproclaim

Let $F : J_0 \rga
\R$ be a given function. 
Following 
[{25}, Section 3] and slightly changing
the notation, given a map $\p : \oi \rga \Pn$,  we set
%
$$
\phi_-(t) 
:= \bigcup_{\eps > 0} \text{int}\Big(
\bigcap_{s \in [t-\eps,t+\eps] \cap \oi} \phi(s)\Big), 
\qquad t \in \oi.
\tag{4.23}
$$
%
Given $\ro > 0$,
we also let $\psi_\ro^{\pm} : \oi \rga \Pn$ be the map
defined by $\psi_\ro^{\pm}(t) := (\psi(t))_\ro^{\pm}$.


\proclaim{Proposition {4.1}}
If $F : J_1 \rga \R$
does not depend on $x$,  
then
%
$$
\p \in \BFF \ \Rightarrow \ \text{int}(\p) \in \BFF.
\tag{4.24}
$$
%
If $F : J_0 
\rga \R$ is lower semicontinuous, then
%
$$
\p \in {\Cal B}(\FFm) \ \Rightarrow \ \text{int}(\p) \in {\Cal B}(\FFm), ~
\ \p_- \in {\Cal B}(\FFm).
\tag{4.25}
$$

\endproclaim

\demo{Proof}
Assertion ({4.24}) follows from the spatial
translation invariance of the family $\FF$ and from
the definition of barrier.
Assume that $F$ depends on $x$ 
and is lower semicontinuous,
and let $\p \in {\Cal B}(\FFm)$.
Let $f : \ab \rga \Pn$, $f \in \FFm$, $f(a) \subseteq \text{int}(\p(a))$.
As $f$ is smooth, 
$f(a)$ is compact and $F$ is lower semicontinuous, we can 
pick $\ro >0$ small enough
such that the map $\overline{f^+_\ro}$ belongs to $\FFm$ and
$\overline{f^+_\ro(a)} \subseteq \p(a)$.
Then $\overline{f^+_\ro(b)} \subseteq \p(b)$, which implies
$f(b) \subseteq \text{int}(\p(b))$, hence $\text{int}(\p) 
\in {\Cal B}(\FFm)$.
\noindent
Assume now that $f(a) \subseteq \p_-(a)$. 
As $f$ is smooth, $f(a)$ is compact and $\p_-(a)$ is open,
there exists $\ro > 0$ such that $g := \overline{f^+_\ro}$ 
belongs to $\FFm$ and $g(a) 
\subseteq \p_-(a)$. 
By the definition of $\p_-$, there is $\eps > 0$ 
such that $g(a) \subseteq \p(a+\tau)$ for
any $\tau \in [-\eps,\eps]$. Let $\tau \in [-\eps,\eps]$
and define $h(t) := g(t-\tau)$ for $t \in [a+\tau, b+\tau]$.
As $F$ is lower semicontinuous, possibly reducing $\eps$, 
we have $h \in \FFm$; moreover $h(a+\tau) \subseteq \p(a+\tau)$,
hence $g(b) = h(b+\tau) \subseteq \p(b+\tau)$. Hence 
$\displaystyle
g(b) \subseteq \bigcap_{s \in [b-\eps,b+\eps]\cap \oi} \p(s)$,
therefore $\displaystyle 
f(b) \subseteq \text{int}
\Big( \bigcap_{s \in [b-\eps,b+\eps] \cap \oi} \p(s) \Big)
\subseteq \p_-(b)$.
\qed\enddemo


\proclaim{Lemma {4.2}}
Given $\p : \oi \rga \Pn$ and $\ro > 0$ we have
$(\p_-)_\ro^- = (\p_\ro^-)_-$. In particular
$(\p_-)_\ro^- = (\p_{--})_\ro^- = 
((\p_-)_\ro^-)_-$. 
\endproclaim
%
\demo{Proof}
Let $\ro > 0$ and $t \in \oi$.
Given $\eps >0$ set $I(t,\eps) := [t-\eps, t+\eps] \cap \oi$. 
 Let us prove that 
%
$$
L_\ro := 
\Big[ \bigcup_{\eps > 0} \text{int}\Big( \bigcap_{s \in I(t,\eps)} 
\p(s) \Big) \Big]^-_\ro 
 = \bigcup_{\eps > 0} \Big[ \bigcap_{s \in I(t,\eps)}
\p(s) \Big]^-_\ro =: R_\ro.
\tag{4.26}
$$
%
It is enough to show ({4.26}) when, instead of $\eps > 0$,
we take unions over a sequence $\{\eps_m\}_{m\in {\bold N}}$ 
of positive numbers
converging to zero as $m \rga +\infty$. Let us define
%
$$
\Om_m := \text{int}\Big( \bigcap_{s \in I(t,\eps_m)} \p(s) \Big).
$$
%
We can assume that $\Om_m \neq \Rn$ for any
$m \in {\bold N}$, otherwise the result it trivial.
Let $x \in L_\ro$. Then
%
$$
\text{dist}\Big(x, \bigcap_m~ (
\Rn \setminus \Om_m)
\Big) > \ro.
\tag{4.27}
$$
%
To prove that $x \in R_\ro$ we need to show that there exists $m_1 \in 
{\bold N}$ such that 
%
$$
\text{dist}\Big(x, \Rn \setminus 
\bigcap_{s \in I(t,\eps_{m_1})} \p(s) 
\Big) = \text{dist}(x, \Rn \setminus \Om_{m_1}) > \ro.
$$
%
Assume by contradiction that  
$\text{dist}\left(x, \Rn \setminus \Om_m
\right) \leq \ro$ for any $m \in {\bold N}$. Let $y_m 
\in \Rn \setminus \Om_m$ be such that $\vert y_m - x\vert \leq 
\ro$. Possibly passing to a subsequence 
(still denoted by $\{\eps_m\}$)
we have $\lim_{m \rga +\infty} y_m = y$ with 
$\vert x-y\vert \leq \ro$; moreover
$y \in \bigcap_m (\Rn\setminus \Om_m)$, 
since $y_m \in \bigcap_{k=1}^m (\Rn\setminus \Om_k)$.
We then have a contradiction with ({4.27}).
We have proved that $L_\ro \subseteq R_\ro$. The opposite inclusion
follows from the fact that $\displaystyle 
\Big[\bigcap_{s \in I(t,\eps)}\p(s) \Big]^-_\ro 
= \Big[\text{int}\Big( \bigcap_{s \in I(t,\eps)} 
\p(s) \Big) \Big]^-_\ro 
\subseteq
\Big[ \bigcup_{\eps > 0} \text{int}\Big( \bigcap_{s \in I(t,\eps)} 
\p(s) \Big) \Big]^-_\ro$ for any $\eps >0$.

Let us show now that for any $\eps > 0$
%
$$
l^\eps_\ro := \Big[ \bigcap_{s \in I(t,\eps)
} \p(s) \Big]^-_\ro = \text{int}\Big(\bigcap_{s \in I(t,\eps)} 
\p(s)^-_\ro\Big) =: r_\ro^\eps.
$$
%
Let $x \in r_\ro^\eps$. Then there is $c >0$ so that
$\text{dist}\big(x, \bigcup_{s \in I(t,\eps)} 
\Rn \setminus \p(s)_\ro^-\big)= c$.
Hence for any $s \in I(t,\eps)$
we have 
$\text{dist}(x, \Rn \setminus \p(s)_\ro^-)\geq c$,
which implies 
$\text{dist}(x, \Rn \setminus \p(s))\geq c + \ro ~$. Therefore 
%
$$
\text{dist}\Big(x, \bigcup_{s \in I(t,\eps)} 
\big(\Rn \setminus \p(s)\big)\Big)\geq c + \ro > \ro,
$$
%
hence
$x \in l_\ro^\eps$. We have proved that $r_\ro^\eps \subseteq l_\ro^\eps$.
The opposite inclusion follows from the fact that 
$\p(s)_\ro^- \supseteq l^\eps_\ro$ for any $s \in I(t,\eps)$.

>From ({4.26}) and ({4.27}) we then have
%
$$
(\phi_-)^-_\ro(t) = L_\ro = R_\ro = \bigcup_{\eps >0} l_\ro^\eps = 
\bigcup_{\eps >0} r_\ro^\eps = (\phi_\ro^-(t))_-.
\tag{4.28}
$$
%
The last assertion of the lemma is a consequence of 
({4.28}) and the equality $\p_{--} = \p_-$.
\qed\enddemo

We conclude this section with the following proposition.
\proclaim{Proposition {4.2}}
Let $F : J_0 \rga \R$ be a function such that for any $R > 0$
%
$$
C_R := \sup \{\vert F(t,x,p,X)\vert : t \in \oi, 
x \in \Rn, \vert p\vert =1, \vert X \vert \leq R \} < +\infty.
\tag{4.29}
$$
%
Let $\p \in {\Cal B}(\FFm)$ and $(\ot,\ox) \in ~]0,+\infty[ \times \Rn$
be such that $\ox \in \Rn \setminus \p(\ot)$. Then there exists a sequence
$\{(t_m,x_m)\}$ of points of $]0,+\infty[ \times \Rn$ with $x_m \in
\Rn \setminus \p(t_m)$ and $t_m < \ot$ such that $(t_m,x_m) \rga
(\ot,\ox)$ as $m \rga +\infty$.
\endproclaim

\demo{Proof} 
Let $h : \oi \rga ~]0,+\infty[$
be any strictly increasing $\C^\infty$ function such that $h(R)
> C_R$ for any $R\geq 0$. For any $\ro >0$
define $H(\ro) := \displaystyle 
\int_0^\ro \frac{1}{h(\sqrt{n-1}/r)}~ dr$. 
Then $H : \oi \rga \oi$ is strictly increasing, surjective, 
$H(0) =0$, $H \in \C^0([0,+\infty[) \cap \C^\infty(]0,+\infty[)$.
Let $\ro_F^{} := H^{-1}$. Given $0 \leq a < b$, $\eps >0$, $x \in \Rn$,
one can check that the function $g : \ab \rga \Pn$ defined
by $g(t) := \{y \in\Rn: \vert y-x\vert \leq \ro_F^{}(\eps+ b-t)\}$
belongs to $\FFm$.
\noindent
Let now $\ox \in \Rn \setminus \p(\ot)$. To prove the proposition
it is enough to show that there exists a sequence 
$\{t_m\}$ converging to $\ot$ with $t_m <\ot$, such that 
$\overline{B_{2\ro_F^{}(\ot-t_m)}(\ox)} \cap (\Rn \setminus \p(t_m)) \neq
\emptyset$. 
Assume by contradiction that for $t_m \uparrow \ot$
we have
$\overline{B_{2\ro_F^{}(\ot-t_m)}(\ox)} \subseteq \p(t_m)$.
Let $t^* > \ot - t_m$ be such that $\ro_F^{}(t^*) = 
2 \ro_F^{}(\ot - t_m)$. The map $t \in [t_m, \ot~] 
\rga \overline{B_{\ro_F^{}(t^* + t_m - t)}(\ox)}$ belongs to $\FFm$. 
Hence, as $\p \in {\Cal B}(\FFm)$
and $\overline{B_{\ro_F^{}(t^*)} (\ox)} 
= \overline{B_{2\ro_F^{}(\ot - t_m)} (\ox)}
\subseteq \p(t_m)$,  
 we have
$\ox \in \overline{B_{\ro_F^{}(t^* + t_m-\ot)}(\ox)} \subseteq \p(\ot)$,
a contradiction.
\qed\enddemo

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\head
{5}. A function whose Level sets are barriers
is a subsolution
\endhead
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Our aim is now to prove a converse of Theorem {3.2} 
(Theorem {5.1}). To do this
we need several preliminary results.

\proclaim{Definition {5.1}} 
Let $f : \ab \subseteq \oi \rga \Pn$.
We say that $f$ is a smooth compact flow
if and only if 
$f(t)$ is compact for any $t \in \ab$ and 
there exists an open set
$A \subseteq \Rn$ such that
$d_f \in \C^\infty(\ab\times A)$ and  $\partial f(t) \subseteq A$ for any
$t \in \ab$.
\endproclaim
%

\proclaim{Lemma {5.1}}
Let $f,g : \ab\rga \Pn$ be two smooth compact flows,
$x \in \Rn$ and $\ro >0$.
Assume that
%
$$
\aligned
& x = \partial f(a) \cap \partial g(a) \cap \overline{B_\ro(x)},
\\
& (g(a) \setminus \{x\}) \cap \overline{B_\ro(x)} 
\subseteq \text{int}(f(a)) \cap \overline{B_\ro(x)},
\\
& \frac{\partial d_f}{\partial t}(a,x) <
\frac{\partial d_g}{\partial t}(a,x).
\endaligned
$$
%
Then there exists $0 < \tau \leq b-a$ such that
%
$$
g(t) \cap \overline{B_\ro(x)} \subseteq \text{int}(f(t)) \cap 
\overline{B_\ro(x)}
\qquad t\in~ ]a,a+\tau].
\tag{5.1}
$$
%
Moreover, $\tau$ depends in a continuous way on 
small perturbations of $f$ and $g$ in the 
$\C^2$ norm.
\endproclaim

\demo{Proof} 
Let $c := \frac{1}{2}[\frac{\partial d_g}{\partial t}(a,x)
-\frac{\partial d_f}{\partial t}(a,x)]$ and 
$\eta(t):=
\text{dist}(\partial 
g(t)\cap\overline{B_\ro(x)}, \partial f(t)\cap\overline{B_\ro(x)})$,
for $t \in \ab$.
Since $f$, $g$ are smooth compact flows, using the hypotheses
we can find $0<\sigma<\ro$ and $\tau>0$ such that, 
for $t \in [a,a+\tau]$,
$$
\frac{\partial d_g}{\partial t}(t,y)-
\frac{\partial d_f}{\partial t}(t,z)> c \qquad
y \in \partial g(t) \cap \overline{B_\sigma(x)}, \ 
z \in \partial f(t) \cap \overline{B_\sigma(x)}, 
\tag{5.2}
$$
%
and 
$\eta(t)=|y-z| \Rightarrow y,z\in B_\sigma(x)$. 
Reasoning as in [{7}, Lemma 4.2] one can check that
for any $t \in [a,a+\tau]$ we have
$\liminf_{s \rga 0^+} \frac{\eta(t+s) - \eta(t)}{s} \geq c$,
which in turn implies 
$\eta(t)\geq c(t-a)$ for any $t\in[a,a+\tau]$ and 
({5.1}) follows. 

The continuity of $\tau$ follows by construction.
\qed\enddemo

The following proposition plays a crucial r\^ole in the 
proof of Theorem {5.1} and 
is based on Lemma {4.1}; note that we will not assume that 
$F$ is degenerate elliptic.
%
\proclaim{Proposition {5.1}}
Assume that $F: J_0 \rga R$ is geometric and
lower semicontinuous.
Let $\p \in {\Cal B}(\FFm)$ and let $f : \ab 
\rga \Pn$ be a smooth compact flow.
Assume that there exist $\theta \in ~]a,b[$ and $x \in \Rn$ such that
%
$$
\aligned
& \partial f(\theta) \cap \partial \p(\theta) = \{x\},
\\
& f(\theta) \setminus \{x\} \subseteq \text{int}(\p(\theta)),
\\
& f(t) \subseteq \text{int}(\p(t)), \qquad  t \in
\ab \setminus \{\theta\}.
\endaligned
\tag{5.3}
$$ 
% 
Then
%
$$
\frac{\partial d_f}{\partial t}(\theta,x)
+ F\big(\theta, x, \grad d_f(\theta,x), \grad^2 d_f(\theta,x)\big)
\leq 0.
\tag{5.4}
$$
%
\endproclaim
%
\demo{Proof} 
{\smc case 1}. 
Suppose that $F$ does not depend
on $(t,x)$.

\noindent Assume by contradiction that
%
$$
\frac{\partial d_f}{\partial t} + F\big(\grad d_f,\grad^2 d_f\big)
= 2 c >0 \qquad \text{at } (\theta,x).
$$
%
As $f$ is a smooth compact
flow, there exists $\theta_1>0$ such that, for every $t\in 
[\theta-\theta_1, \theta+ \theta_1] = : I(\theta)$, 
each point $y \in \partial f(t)$ has a unique smooth projection
$\pi(t,y)$ on $\partial f(\theta)$. 
>From now on we restrict to the interval $I(\theta)$.
Set $x(t) := \pi^{-1}(t,x)$, 
%
$$
p(t):= \grad d_f(t,x(t)), \qquad
X(t) := \grad^2 d_f(t,x(t)), \qquad
\alpha := c - F(p(\theta), X(\theta)).
$$
%
We can assume that 
%
$$
\frac{\partial d_f}{\partial t}(t,x(t)) > \alpha
\tag{5.5}
$$
%
and, as $F$ is lower semicontinuous,
possibly taking a smaller $\theta_1$, we can also assume
%
$$
\alpha + F(p(t),X(t)) > 0, \qquad t \in I(\theta).
\tag{5.6}
$$
%
Choose a function $\ro: \partial f(\theta) \rga [0,+\infty[$ of class
$\C^\infty$ verifying the following properties:
\itemitem{(i)} $\ro(y) =0$ if and only if $y = x$;
\itemitem{(ii)} $\grad \ro(x)=0$, $\grad^2 \ro(x) = 0$;
\itemitem{(iii)} the map $t \in I(\theta) \rga
f_1(t)$
is a smooth compact flow, where $\partial f_1(t) 
:= \{z \in \Rn : z = y - \ro(\pi(t,y)) \grad d_f(t,y), y 
\in \partial f(t)\}$.

In particular $f_1 : I(\theta) \rga \Pn$
satisfies ({5.3}) with $[a,b]$ replaced by
$I(\theta)$,  and 
%
$$
f_1(t) \subseteq f(t), \quad
\partial f_1(t) \cap \partial f(t) = \{x(t)\}, \quad
\grad^2 d_{f_1}(t,x(t))\!=\!X(t), \quad t \in I(\theta).
\tag{5.7}
$$
%
Fix $t \in ~]\theta- \theta_1,\theta[$. 
Recalling also ({5.6}),
we apply Lemma {4.1} 
with $\overline x$ and $L$ replaced
by $x(t)$ and  $f(t)$ in the order.
Hence there exist $\tau_t > 0$, $\sigma_t > 0$, 
$g_t : [t,t+\tau_t] \rga \Pn$, so that
%
$$
\aligned
& g_t \in \FFm,
\qquad
g_t(t)\subseteq f(t) \subseteq \p(t),
\qquad
g_t(t) \cap \overline{B_{\sigma_t}(x)} = f(t) \cap 
\overline{B_{\sigma_t}(x)},
\\
\\
& x(t) \in \partial g_t(t), \
(\alpha, p(t), X(t)) \!=\! \Big(\frac{\partial d_{g_t}}{\partial t}(t,x(t)),
\grad d_{g_t}(t,x(t)), 
\grad^2 d_{g_t}(t,x(t))\Big)
\endaligned
\tag{5.8}
$$
%
(possibly reducing 
$\tau_t$ and $\sigma_t$, 
we can use $x$ instead of $x(t)$ in the first equality
in ({5.8})).
Using the first equality in ({5.8}) and the second relation in
({5.7}) we have
%
$$
(f_1(t)\setminus\{ x(t)\})\cap 
\overline{B_{\sigma_t}(x)}\subseteq\text{int}(g_t(t)).
$$
%
Using Remark {4.2}
we have $\tau_t \rga \tau_{\theta} > 0$ as $t \rga \theta$, so that there
exists $t_1 < \theta$ such that $\tau_t > \theta - t$ for any
$t \in [t_1,\theta]$. Fix 
$t\in~ ]t_1,\theta]$; let us apply Lemma {5.1} to the 
flows $f_1$, $g_t$ (recall that 
$\frac{\partial d_{f_1}}{\partial t}(t,x(t)) > \alpha
= \frac{\partial d_{g_t}}{\partial t}(t,x(t))$ by ({5.5})). 
Then there exists $0 < \tau_t^\prime < \tau_t$ such that
%
$$
f_1(s) \cap \overline{B_{\sigma_t}(x)}
\subseteq\text{int}(g_t(s)), \qquad s \in~ ]t,t+\tau_t^\prime].
\tag{5.9}
$$
%
Using Remark {4.2} we get that $\tau_t^\prime
 \rga \tau^\prime_{\theta} > 0$
as $t \rga \theta$. 
Choose $t_2 \in~ ]t_1, \theta[$ 
such that $\tau^\prime_{t_2} > \theta - t_2$; 
as $g_{t_2}(t_2) \subseteq \p(t_2)$ by ({5.8}) and
$g_{t_2} \in \FFm$, $\p \in {\Cal B}(\FFm)$,
by ({5.9}) 
we have, for $s = \theta$,
%
$$
x \in \text{int}(g_{t_2}(\theta)) 
\subseteq \text{int}(\p(\theta)),
$$
which contradicts $x \in \partial \p(\theta)$.

{\smc case 2}. Suppose that $F$ depends on $(t,x)$.
Assume
by contradiction that 
%
$$
\frac{\partial d_f}{\partial t}(\theta,x) + 
F\big(\theta, x, \grad d_f(\theta,x),
\grad^2 d_f(\theta,x)\big)>0.
$$
%
Let $U \subseteq \oi \times \Rn$ be a compact neighbourhood
of $(\theta,x)$ such that 
%
$$
\frac{\partial d_f}{\partial t}(\theta,x)
+ \min_{(t,y) \in U}
F\big( t,y,\grad d_f(\theta,x),\grad^2 d_f(\theta,x) \big) > 0,
\tag{5.10}
$$
%
and define
%
$$
G(s,z,p,X) := \cases
\displaystyle \min_{(t,y) \in U}
F(t,y, p,X) & \text{if } (s,z) \in U,
\\
F(s,z,p,X) & \text{elsewhere}.
\endcases
$$
%
Notice that $G$ is lower semicontinuous and 
that $\p \in {\Cal B}(\FFm)$ implies  
$\p \in {\Cal B}(\F_G^>)$. 
Applying {\smc case 1} localized in $U$ 
(with $F$ replaced by $G$)
we then get a contradiction with ({5.10}).
\qed\enddemo

%
\proclaim{Proposition {5.2}}
Assume that $F : J_1 \rga \R$ 
is geometric, lower semicontinuous and   
satisfies (F4).
Let $\phi \in {\Cal B}(\FFm)$.
The following statements hold:
\item{(i)} if $F$ satisfies (F2) then
 the function $(t,x) \rga - \chi_{\p(t)}(x)$
is a viscosity subsolution 
of ({3.5}) in $]0,+\infty[ \times \Rn$;
\item{(ii)} if $F^+$ satisfies (F4) then 
the function $(t,x) \rga - \chi_{\p(t)}(x)$
is a viscosity subsolution 
of 
%
$$
\frac{\partial u}{\partial t} + F^+(t,\grad u, \grad^2 u) = 0
\tag{5.11}
$$
%
in $]0,+\infty[ \times \Rn$.
\endproclaim
%
\demo{Proof}
It is enough to show (ii).
Let $T := \sup \{t \geq 0 : 
\phi(t) \neq \emptyset, \phi(t) \neq \Rn\}$. 
To prove the thesis, it is enough to  show that the function 
$d_\phi \wedge 0$ is a subsolution of
({5.11}) in $~ ]0,T[ \times \Rn$. Indeed, using
[{9}, Lemma 5.7] we have that 
$d_\phi \wedge 0$ is a subsolution of
({5.11}) in $~ ]0,T] \times \Rn$;
moreover, using [{1}, Lemma 4.3] we deduce that the function
$(t,x) \rga -\chi_{\p(t)}(x)$ is also 
a subsolution of
({5.11}) in $~ ]0,T] \times \Rn$, hence in 
$]0,+\infty[ \times \Rn$.

\noindent By [{25}, Lemmas 3.1,3.2] we have that 
$(d_\phi \wedge 0)^* = d_{\phi_-} \wedge 0$. 
We let $d := d_{\phi_-} \wedge 0$.
Let $(\ot,\ox) 
\in ~ ]0,T[ \times \Rn$. We have to prove ({3.2}) 
(with $F^+(t,p,X)$ instead of $F(t,x,p,X)$)
for any function $\psi \in \C^\infty(]0,T[ \times \Rn)$ such that
$(d-\psi)$ 
has a strict global maximum at the point $(\ot,\ox)$. 
Set $(\alpha,p,X) := (\frac{\partial \psi}{\partial t}(\ot,\ox),
\grad \psi(\ot,\ox), \grad^2 \psi (\ot,\ox))$.

In view of Proposition {4.2} and Lemma {3.1},
it is enough to consider the case $d(\ot,\ox) < 0$, 
i.e., $\ox \in \p_-(\ot)$. Let
$\ox \in \phi_-(\ot)$;
then 
$\vert p\vert =1$ (we use the fact that $d$ 
is locally semiconvex in $\phi_-(\ot)$,
see [{23}]).
Since $F$ is geometric, 
possibly replacing $X$ with $P_p X P_p$,
we can suppose $X p = 0$.
Let $\overline y \in \partial \phi_-(\ot)$ be such that
$\vert \ox - \overline y\vert = -d(\ot,\ox)$, and 
%
$$
\Psi(t,x) := \psi(t,x + \ox - \overline y) - \psi(\ot,\ox), 
\qquad (t,x)\in ~]0,T[ \times \Rn.
$$
%
Clearly 
$(\alpha,p,X) = (\frac{\partial \Psi}{\partial t}(\ot,\overline y), 
\grad \Psi(\ot,\overline y), 
\grad^2 \Psi(\ot,\overline y))$.
Moreover, using the triangular property of the distance,
we have $\text{dist}(x+ \ox - \overline y, \Rn \setminus 
\phi_-(t)) \leq \text{dist}(x,\Rn \setminus \phi_-(t)) 
+ \vert \ox - \overline y\vert$, hence
%
$$
d(t,x) \leq d(t,\overline y) + d(t,x + \ox - \overline y) + \vert \ox -
\overline y\vert.
$$
%
Therefore 
%
$$
\align
& d(t,x) - \Psi(t,x) - \psi(\ot,\ox)
 = d(t,x) - \psi(t,x + \ox - \overline y)
\leq d(\ot,\overline y) + d(t,x + \ox - \overline y) 
\\
& + \vert \ox - \overline y\vert - \psi(t,x + \ox - \overline y)
< d(\ot,\overline y) + d(\ot,\ox) -\psi(\ot,\ox) 
+ \vert \ox - \overline y\vert
\\
& = d(\ot,\overline y) - \psi(\ot,\ox) = d(\ot,\overline y) - 
\Psi(\ot,\overline y) - \psi(\ot,\ox),
\endalign
$$
%
which implies that $(d-\Psi)$ has a strict global maximum
at the point $(\ot, \overline y)$.

Pick $\tau > 0$ with 
$[\ot - \tau, \ot + \tau] \subseteq ~]0,T[$ 
and a smooth function $\zeta :  
[\ot -\tau, \ot + \tau] \times \Rn \rga \R$ with the following properties:
$\zeta \geq \Psi$,
$\zeta(\ot,\overline y) = \Psi(\ot,\overline y) = 0$,
%
$$
(\alpha,p,X) = \left(\frac{\partial \zeta}{\partial t}(\ot,\overline y), 
\grad \zeta(\ot,\overline y), 
\grad^2 \zeta(\ot,\overline y)\right),
$$
%
$\zeta^2 + \vert\grad \zeta\vert^2 > 0$ (recall that 
$\vert \grad \zeta(\ot,\overline y)\vert = \vert
\grad \psi(\ot,\ox)\vert =1$) and 
$\zeta$ is positive outside a compact subset of $\Rn$.

Let us define $f : [\ot -\tau, \ot + \tau] \rga \Pn$,
$f(t) := \{x \in \Rn : \zeta(t,x) \leq 0\}$. Then $f$ is a smooth
compact flow and 
%
$$
(\alpha,p,X) 
= \left(\frac{\partial d_f}{\partial t}(\ot,\overline y), 
\grad d_f(\ot,\overline y), \grad^2 d_f(\ot,\overline y)\right)
$$
%
(recall that $\vert p\vert =1$ and  $X p =0$).
Then assumptions ({5.3}) (with $\theta$,
$\ab$ and $\phi$ replaced by $\ot$, $[\ot - \tau, \ot + \tau]$ and
$\phi_-$ in the order)
of Proposition {5.1} are fulfilled (recall that 
$\p_-\in {\Cal B}(\FFm)$ by ({4.25})). Then, from
({5.4}) it follows
%
$$
\alpha + F(\ot,p,X) \leq 0
$$
%
and 
therefore $d$ is a subsolution of ({3.5}) 
in $\{(t,x) \in ~]0,T[ \times \Rn : d(t,x) < 0\}$.

Assume now by contradiction that there exists $0 < c < +\infty$
such that
%
$$
\alpha  + F^+(\ot,p,X) = 2 c.
$$
%
Let $Y \in \text{Sym}(n)$  be such that $Y \geq X$
and 
%
$$
F^+(\ot,p,X) \leq 
F(\ot,p,Y) + c.
$$
% 
Define
%
$$
\Phi(t,x) := \psi(t,x) + \frac{1}{2} \langle 
(x-\ox), (Y- X)(x-\ox) \rangle.
$$
%
Then $\grad^2 \Phi(\ot,\ox) = Y$ and $(d - \Phi)$ has
a maximum at $(\ot,\ox)$. Therefore, 
as $d$ is a subsolution of ({3.5}),
at $(\ot,\ox)$ we have
%
$$
0 \geq \frac{\partial \Phi}{\partial t} + F(\ot,\grad \Phi, \grad^2 \Phi) = 
\alpha + F(\ot,p,Y)  \geq \alpha +  
F^+(\ot,p,X)  - c = c
> 0,
$$
%
a contradiction.
\qed\enddemo

We are now in a position to prove the converse of
Theorem {3.2}.

\proclaim{Theorem {5.1}}
Let $u : \oi \times \Rn \rga \R$ be a function
such that $u^* < +\infty$. 
The following statements hold.
\item{A)} Assume that 
$F: J_1 \rga \R$ 
does not depend on $x$,
is 
geometric, lower semicontinuous and satisfies (F4). Suppose that 
for any $\lambda \in \R$ 
%
$$
\{x \in \Rn : u^*(\cdot,x) <\lambda\} \in {\Cal B}(\FFm).
\tag{5.12}
$$
%
\itemitem{(i)} If $F$ satisfies (F2) then 
$u$  is a viscosity subsolution of ({3.5}) in 
$]0,+\infty[ \times \Rn$;
\itemitem{(ii)} if $F^+$ satisfies (F4) then 
$u$  is a viscosity subsolution of ({5.11}) in 
$]0,+\infty[ \times \Rn$.
\item{B)} Assume that $F: J_0 \rga \R$
is geometric,
lower semicontinuous and satisfies (F4). Assume that 
for any $\lambda \in \R$ relation ({5.12})
holds.
\itemitem{(iii)} If $F$ satisfies 
(F2), (F8')
then $u$ is viscosity subsolution
of ({3.1}) in $]0,\!+\infty[ \times\!\Rn\!$;
\itemitem{(iv)} if $F^+$ satisfies (F4), (F8')
then $u$ is viscosity subsolution
of
%
$$
\frac{\partial u}{\partial t} + F^+(t,x,\grad u, \grad^2 u) = 0
\tag{5.13}
$$
%
in $]0,+\infty[ \times \Rn$.
\endproclaim
%
\demo{Proof} Statement A). It is enough to prove (ii).
Let $(\ot,\ox) \in~ ]0,+\infty[ \times \Rn$; we have to prove
({3.2}) (with $F^+(t,p,X)$ instead of $F(t,x,p,X)$)
 for any smooth function $\psi$
such that $(u^*-\psi)$ has a maximum at $(\ot,\ox)$.
Let $\overline \lambda := u^*(\ot,\ox)$; we define the function
$z : \oi \times \Rn \rga \R$ as
%
$$
z(t,x) := \cases
\overline \lambda & \text{if } u^*(t,x) \geq \overline \lambda,
\\
\overline \lambda - 1 & \text{elsewhere}.
\endcases
$$
%
By ({5.12}) and 
Proposition {5.2} (ii), setting
$\phi(t) := \{x \in \Rn : u^*(t,x) < \overline \lambda\}$,
it follows that 
the function $(t,x) \rga -\chi_{\phi(t)}(x)$
is a subsolution of ({5.11}) 
in $]0,+\infty[ \times \Rn$. Therefore
also
$z$ is a subsolution of ({5.11}) 
in $]0,+\infty[ \times \Rn$.
Since $(z-\psi)$ has a maximum at 
$(\ot,\ox)$, ({3.2}) follows.

Statement B). It is enough to prove (iv). Following
the arguments of the proof
of statement A), it is sufficient
to show the following
assertion: given $\p \in {\Cal B}(\FFm)$, the function
%
$$
\chi(t,x) := - \chi^{}_{\phi_-(t)}(x)
$$
%
is a subsolution of ({5.13}) in $]0,+\infty[ \times \Rn$.
For any $0 < \eps <1$ we define $d_\eps(t,x) := (-\eps) \vee
d_{\phi_-}(t,x) \wedge 0$.

We shall prove that 
$d_\eps$ is a subsolution of 
%
$$
\frac{\partial u}{\partial t} + F_\eps(t,x,\grad u,\grad^2 u) =0 
\tag{5.14}
$$
%
in $]0,+\infty[ \times \Rn$, where
%
$$
F_\eps(t,x,p,X) := 
F^+(t,x,p,X) - \eps \sigma_{\vert X\vert}(1 + \eps),
\tag{5.15}
$$
%
and $\sigma_{\vert X\vert}$ is the modulus of 
continuity defined in (F8'). In view of Lemma
{3.1} it is enough to check that $d_\eps$  is a subsolution
in $\{d_\eps < 0\}$. 

Let $(\ot,\ox) \in~ ]0,+\infty[ \times \Rn$. Let $\psi$ be a smooth 
function such that $(d_\eps - \psi)$ has a strict global maximum
at $(\ot,\ox)$ and $d_\eps(\ot,\ox) = \psi(\ot,\ox)$.
Set 
%
$$
(\alpha,p,X) := 
\left(\frac{\partial \psi}{\partial t}(\ot,\ox), 
\grad \psi(\ot,\ox), \grad^2 \psi(\ot,\ox)\right) .
$$
%

{\smc case 1}. $\ox \in \phi_-(\ot)$ and $\text{dist}(\ox, \Rn \setminus
\p_-(\ot)) > \eps$.
Then $d_\eps$ is twice differentiable at $(\ot,\ox)$
with respect to $x$, therefore $p=0$, $X \geq 0$.
Moreover by Lemma {4.2} we have $(\p_-)^-_\eps 
= ((\p_-)^-_\eps)_-$, hence, as $\ox \in  (\p_-(\ot))^-_\eps$,
there exists
a sequence $t_m \uparrow \ot$ such that $d_\eps(t_m,\ox) = -\eps
 \leq \psi(t_m,\ox)$, which yields $\alpha \leq 0$.
Therefore 
%
$$
\alpha + (F_\eps)_*(\ot,\ox,p,X) \leq 
(F^+)_*(\ot,\ox,0,X) \leq (F^+)_*(\ot,\ox,0,0) \leq 0.
$$
%

{\smc case 2}. $\ox \in \p_-(\ot)$ and $\text{dist}(x,\Rn \setminus 
\p(\ot)) \leq \eps$. 
As $d_\eps$ is locally semiconvex in $\phi_-(\ot)$,
we have $\vert p\vert =1$. Let $\overline y \in \partial
\p(\ot)$ be such that $\vert \ox - \overline y\vert = -d_\eps(\ot,\ox)
\leq \eps$. Following the proof of Proposition {5.2} and 
applying
Proposition {5.1} we get
%
$$
\alpha +  F^+(\ot,\overline y,p,X) \leq 0.
$$
%
Therefore, using (F8') and recalling ({5.15}), we have
%
$$
\alpha +  (F_\eps)_*(\ot,\ox,p,X) \leq 
\alpha +  F^+(\ot,\overline y,p,X) \leq 0.
$$
%
We have proved that $d_\eps$ is a subsolution of ({5.14})
in $~]0,+\infty[\times \Rn$.

\noindent Reasoning as in [{1}, Lemma 4.3] we then obtain that 
$\eps \chi$ is a subsolution of ({5.14})
in $]0,+\infty[ 
\times \Rn$, hence $\frac{1}{\eps}(\eps \chi) = 
\chi$ is also a subsolution of ({5.14}) in $]0,+\infty[
\times \Rn$. 
Letting $\eps \rga 0$ and using [{9}, Proposition
2.4], we get that $\chi$ is a subsolution of ({5.13})
in  $]0,+\infty[ \times \Rn$. 
\qed\enddemo

We also have a similar statement of Theorem {5.1} 
for supersolutions.
%
\proclaim{Remark {5.2}}
Assume that $F: J_0 \rga \R$ is geometric, upper
semicontinuous and satisfies (F4).
Let $v : \oi \times \Rn \rga \R$ be a function
such that $v_* > -\infty$ and
$\{x \in \Rn : v_*(\cdot,x) >\lambda\} \in {\Cal B}(\F_{F_c}^>)$
for any $\lambda \in \R$.
If $F^-$ satisfies (F4), (F8') then
$v$ is a viscosity supersolution of 
%
$$ 
\frac{\partial v}{\partial t} + F^-(t,x,\grad v, \grad^2 v) = 0
$$
%
in $~]0,+\infty[ \times \Rn$, where
$F^-(t,x,p,X) := \inf\{F(t,x,p,Y) : Y \leq X\} $.
\endproclaim

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\head
{6}. 
Conclusions 
\endhead
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The following result shows the connection between minimal barriers
and the continuous viscosity solution whenever the latter exists and is unique,
see Theorem {3.1}.

\proclaim{Corollary {6.1}}
Assume that 
$F:J_0 \rga \R$ is geometric and 
satisfies (F1), (F3), (F4), (F6'), (F7), (F9), (F10). 
Let $E \subseteq \Rn$ be a bounded set and denote
with $v : \oi \times \Rn \rga \R$ the unique 
uniformly continuous viscosity solution
of ({3.1}) with  $v(0,x) = v_0(x) := (-1) \vee d_E(x) \wedge 1$. 
Then for any $t \in \oi$ we have
%
$$
\align
& \M_*(E,\FFm)(t) = \M_*(E,\FF)(t) = \{x \in \Rn : v(t,x) < 0\}, 
\tag{6.1}
\\
& \M^*(E,\FFm)(t) = \M^*(E,\FF)(t)= \{x \in \Rn : v(t,x) \leq 0\}.
\tag{6.2}
\endalign
$$
%
In particular ({1.6})
holds true and $\M_{v_0,\FF} = v$.
Moreover if $F = F_c$ then
%
$$
\M^*(E,\FF) \setminus \M_*(E,\FF) \in \BFF.
\tag{6.3}
$$
%
\endproclaim
%
\demo{Proof}
It is enough to show that for any bounded open set $A \subseteq \Rn$
%
$$
\M(A,\FFm) = \M(A,\FF) = V(A),
\tag{6.4}
$$
%
where $V(A)$ is defined  in ({3.4}).
By statement B) of Theorem {3.2} 
we have $V(A) \in \BFF$, hence $V(A) \supseteq
\M(A,\FF)$.

\noindent 
Let
%
$$
\chi(t,x) := - \chi_{\M(A,\FFm)(t)}(x),
\qquad (t,x) \in \oi \times \Rn.
$$
%
By statement B) of Theorem {5.1}, 
$\chi$ is a subsolution of ({3.1})
in $]0,+\infty[ \times \Rn$
 (note that 
$\chi(\cdot,x)$ 
is upper semicontinuous by ({4.25}) and [{25}, Lemma 3.1]).
Applying 
the viscosity comparison theorem 
[{16}, Theorem 2.1] we get
$\chi(t,x) \leq v(t,x)$ for any $(t,x) \in \oi \times \Rn$, hence
$V(A) \subseteq \M(A,\FFm)$. 

\noindent We conclude that 
%
$$
\M(A,\FFm) \supseteq V(A) \supseteq \M(A,\FF) \supseteq \M(A,\FFm),
\tag{6.5}
$$
%
and the proof is ({6.1}), ({6.2}), ({1.6}) is 
complete.
Finally, ({6.3})  follows from ({3.10}).
\qed\enddemo

\proclaim{Remark {6.1}}
Equality ({1.6}) 
proved in Corollary {6.1}
shows that definition 
({2.5}) is consistent with the definition of 
fattening given by means of the (unique) viscosity solution, 
see [{13}], [{7}]. Notice that, 
if we adopt definition ({2.5}), fattening
can be defined also when there is non uniqueness
of viscosity solutions, see Example {6.1} below.
\endproclaim

\proclaim{Remark {6.2}}
>From ({6.5}) it follows that,
under the assumptions of Corollary {6.1},
if
$E \subseteq \Rn$ 
is bounded and open then $\M(E,\FF)(t)$ is open for any $t \in \oi$.
\endproclaim

\proclaim{Remark {6.3}}
Corollary {6.1} in the case of driven motion by mean curvature
in codimension one
has been proved in [{7}], where the minimal
barriers are compared with any generalized evolution of sets
satifying the semigroup property, the comparison principle, 
and the extension of smooth evolutions.
\endproclaim

\proclaim{Remark {6.4}}
Corollary {6.1} also applies to the case 
of motion by mean curvature in arbitrary codimension, i.e., when
$F$ has the form
$F(p,X)=\sum_{i=1}^{n-k}\lambda_i$,
where $1\leq k \leq n-1$ is the codimension and 
$\lambda_1\leq\ldots\leq\lambda_{n-1}$ are
the eigenvalues of the matrix $P_p X P_p$ corresponding 
to eigenvectors orthogonal to $p$. In [{1}]
it has been proved that for such a function $F$ it
holds $V(A) \supseteq \M(A,\FF)$ for any bounded open set $A
\subseteq \Rn$. 
\endproclaim

The following results generalize Corollary {6.1}.

\proclaim{Corollary {6.2}}
Assume that $F : J_0 \rga \R$ is geometric, 
lower semicontinuous and satisfies (F4). Assume that  $F^+$  
satisfies (F1), (F3), (F4), (F6'), (F7), (F9), (F10).
Then for any bounded set $E \subseteq \Rn$ and any $t \in \oi$
we have
%
$$
\align
& \M_*(E,\FF)(t) = \M_*(E,\FFm)(t) = \{x \in \Rn : v(t,x) < 0\},
\\
& \M^*(E,\FF)(t) = \M^*(E,\FFm)(t) = \{x \in \Rn : v(t,x) \leq 0\},
\endalign
$$
%
where $v$ is the unique 
uniformly continuous 
viscosity solution of ({5.13}) 
and $v(0,x) = v_0(x) := (-1) \vee
d_E(x) \wedge 1$.
In particular, thanks to Corollary {6.1}, 
we have
$$
\M_*(E,\FF) = \M_*(E,\F_{F^+}), \qquad
\M^*(E,\FF) = \M^*(E,\F_{F^+}). 
$$
%
\endproclaim

\proclaim{Corollary {6.3}}
Assume that
$F:J_0 \rga \R$ is geometric 
and satisfies (F1), (F3), (F4), (F6'), (F7), (F9), (F10). 
Let $u_0 : \Rn\rga \R$ be a given function such that
$u_0^* < +\infty$. 
Define
%
$$
S_{u_0} :=
\{v : v \text{ is a
viscosity subsolution of } ({3.1}) \text{ in } 
]0,+\infty[~ \times \Rn, v^*(0,x) = u_0^*(x)\}.
$$
%
If $u_0$ is upper semicontinuous then
%
$$
\M_{u_0,\FF} = \M_{u_0,\FFm}  
= \sup \{v : v  \in S_{u_0}\}.
\tag{6.6}
$$
%
In the general case we have
%
$$
\overline \M_{u_0,\FF} = 
\overline \M_{u_0,\FFm} 
= \sup \{v : v \in S_{u_0}\}.
\tag{6.7}
$$
%
\endproclaim
\demo{Proof} Let $w_{u_0} := \sup \{v : v  \in S_{u_0}\}$. 
Let $u_0$ be upper semicontinuous. Given any set
$E\subseteq \Rn$ one can verify that $\M(E,\FFm)(0) = E$.
Moreover, given $\lambda \in \R$,
as $\{u_0 < \lambda\}$ is open, by ({4.25}) the set
$\M(\{u_0 < \lambda\},\FFm)(t)$ is open. Then, by 
({4.25}) and ({2.7}) we have
%
$$
\{x \in \Rn : \M_{u_0,\FFm}(\cdot,x) < \lambda \} = \M(\{u_0 < \lambda\},
\FFm)(\cdot) \in {\Cal B}(\FFm).
\tag{6.8}
$$
%
In particular $\{x \in \Rn : \M_{u_0,\FFm}(0,x) < \lambda \} = 
\{u_0 < \lambda\}$, hence $\M_{u_0,\FFm}(0,x) = u_0(x)$ for any
$x \in \Rn$.
Moreover, by ({6.8}) and statement B) of Theorem
{5.1} it follows that $\M_{u_0,\FFm}$ is a 
subsolution of ({3.1}). Hence $\M_{u_0, \FFm} \leq
w_{u_0}$.

Let now $v$ be any subsolution of ({3.1}) such 
that $v^*(0,x) = u_0(x)$. Then, given $\lambda \in \R$, by 
statement B) of
Theorem {3.2} we have
$\{x \in \Rn : v^*(\cdot,x) < \lambda\} \in \BFF$. Therefore
%
$$
\{x \in \Rn : v^*(\cdot,x) < \lambda\} \supseteq 
\M(\{u_0 < \lambda\},\FF)(\cdot) = \{x \in \Rn : \M_{u_0,\FF}(\cdot,x) 
< \lambda\},
$$
%
which implies $v^* \leq \M_{u_0,\FF}$. 
Hence $\M_{u_0,\FF} \geq w_{u_0}$. Since
$\M_{u_0,\FFm} \geq \M_{u_0,\FF}$, ({6.6}) follows.

Let now $u_0$ be arbitrary. It is not difficult to show
[{6}] that 
given any set $E \subseteq \Rn$ we have
%
$$
\M_*(E,\FFm) = \M_*(\text{int}(E),\FFm) = \M(\text{int}(E),\FFm),
$$
%
and that $\M_*(E,\FFm)(t)$ is open for any $t \in \oi$. Therefore,
given $\lambda \in \R$, we have
%
$$
\align
& \M_*(\{u_0 < \lambda\},\FFm) = 
\M_*(\text{int}(\{u_0 < \lambda\}),\FFm) 
\\
& = 
\M_*(\{u_0^* < \lambda\},\FFm) = 
\M(\{u_0 < \lambda\},\FFm).
\endalign
$$
%
Then ({6.7}) follows from ({6.6}).
\qed\enddemo

\proclaim{Remark {6.5}}
A similar assertion of Corollary {6.3} (under the same hypotheses)
holds for supersolutions.
Precisely, if $u_0$ is lower semicontinuous (resp. arbitrary)
such that ${u_0}_* > -\infty$ we have
that, for any $(t,x) \in \oi \times \Rn$, the function
%
$$
\align
& \sup \{\mu : \M(\{u_0 > \mu\},\FF)(t) \ni x\}
\\
& (\text{resp. } 
\sup \{\mu : \M_*(\{u_0 > \mu\},\FF)(t) \ni x\} )
\endalign
$$
%
coincides with the infimum of $u(t,x)$, where $u$ varies over all
viscosity supersolutions of ({3.1}) in $]0,+\infty[~ \times \Rn$
 such that 
$u_*(0,x) = u_0(x)$ $($resp. $u_*(0,x) = {u_0}_*(x))$
and same assertions with $\FF$ replaced by $\FFm$.
\endproclaim

The following remark shows the connections between the 
barriers and the viscosity evolution without growth
conditions on $F$ (see [{22},{17}]) and
for unbounded sets $E$.

\proclaim{Remark {6.6}}
Assume that $F : (\Rn \setminus \{0\}) \times \text{Sym}(n) \rga \R$ 
does not depend
on $(t,x)$, is geometric and satisfies (F1), (F2).
Let $u$ and $v$ be, respectively, a viscosity sub- and supersolution
of
%
$$
\frac{\partial u}{\partial t} + F(\grad u, \grad^2 u) = 0
\tag{6.9}
$$
%
in $]0,+\infty[~ \times \Rn$, in the sense of [{22}, Definition
1.2]. Then, reasoning as in Theorem {3.2} and using
[{22}, Proposition 1.6, Theorem 1.7], one
can check that ({3.6})-({3.9}) 
hold.
Moreover, using also [{22}, Proposition 1.3], it turns out
that Lemma {3.1} is still true and that, given $\p
\in {\Cal B}(\FFm)$, $d_\p \wedge 0$ is a viscosity subsolution
of ({6.9}). Therefore, as [{1}, Lemma 4.3]
still holds, if $u : \oi \times \Rn \rga \R$
is a function such that $u^* < +\infty$ and satisfies
({5.12}) for any $\lambda \in \R$, then u is a 
viscosity subsolution of ({6.9}) in $]0,+\infty[~ \times \Rn$.
Finally, in view of Remark {3.3}, Corollary {6.1}
still holds, even if $E$ is unbounded.
\endproclaim

In particular we have the following result.
 
\proclaim{Corollary {6.4}}
Assume that 
$F: (\Rn \setminus \{0\}) \times \text{Sym}(n) \rga \R$ 
does not depend on $(t,x)$,
is geometric
and satisfies (F1),(F2).
Let $E \subseteq \Rn$ and let
$v : \oi \times \Rn \rga \R$ be the unique 
uniformly continuous viscosity solution
 of ({6.9}) with  $v(0,x) = v_0(x) := d_E(x)$. 
Then for any $t \in \oi$  we have ({6.1}) and 
({6.2}).
In particular $\M^*(E,\FF)(t) \setminus \M_*(E,\FF)(t) = \{x \in \Rn : 
v(t,x) = 0\}$ and $\M_{v_0,\FF} = v$.
\endproclaim

\example{Example {6.1}}
Let $n = 2$, $F(p,X) := - \text{tr}(P_p X P_p)$ (i.e.,
motion by mean curvature) and let
%
$$
v_0(x_1,x_2) := x_2^2 (1 + x_1^2)^2.
$$
%
Then $u_0$ 
is not uniformly continuous and 
 we have nonuniqueness of continuous viscosity
solutions of ({6.9}) with $v(0,x) = v_0(x)$, see
[{21}].
In this case $\M_{v_0,\FF}$ is, by Corollary
{6.3}, the maximal viscosity (sub) solution. One
can check, following [{21}], that there exist $t \in \oi$ and
$x \in \Rn$ such that $\M_{v_0,\FF}(t,x) > -\M_{-v_0,\FF}(t,x)$,
where 
$-\M_{-v_0,\FF}$ represents the minimal viscosity (super) solution.

Note that for any $\lambda > 0$
the set $\{v_0 < \lambda\}$ develops
fattening (with respect to $\FF$).
\endexample


\proclaim{Remark {6.7}}
Let $n=2$ and consider the anisotropic motion by mean curvature
given by 
%
$$
F(p,X) = - \text{tr}(P_p X P_p) \psi(\theta)
(\psi(\theta) + \psi''(\theta)),
$$
%
where $\psi: {\bold S}^1 \rga \R$ is a smooth function and 
$p = (p_1,p_2) = (\cos \theta, \sin \theta)$ 
(see [{8}]). Then, if $\psi + \psi'' \geq 0$ on
${\bold S}^1$ (i.e., convex anisotropy), we have $F^+ = F$.
If the anisotropy is not convex, then there exists $\overline \theta
\in {\bold S}^1$ such that $\psi(\overline \theta) + \psi''(\overline \theta) 
< 0$,
which implies 
$F^+(\overline p, X) = +\infty$
for any $X \in \text{Sym}(2)$, where $\overline p = (\cos \overline \theta,
\sin \overline \theta)$.
\endproclaim

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%   REFERENCES   %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% p2tex: btx file follows...
%
% References written authomatically by BIB
%
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% p2tex: end btx file



\enddocument

\end


