
\section{From curve motion to image motion and viscosity solutions.}
%%Curves evolution and viscosity solutions.}
\begin{th1} \label{viscoimpliquegrayson}
If $u$ is a viscosity solution of 
$$ \frac{\partial u}{\partial t} = |Du| F(curv(u)) $$
and if $\fxx(t)$ is a $C^\infty$ (in time and space) level line of $u$, then this level line also is
a classical solution of 
$$ \frac{\partial \fxx}{\partial t} =  F(Curv(\fxx)) \fxn $$
\end{th1}

\begin{lem}
\label{functiondistancecourbe}
Let $u(t,\fxx)$ be a continuous function whose isolevelset $C(t) = \partial\{\fxx, u(\fxx)\geq\lambda\} = \partial \{\fxx, u(\fxx)<\lambda\}$, is, for some level $\lambda$ and some interval of time around $t_0$, a finite union of disjoint closed $C^\infty$ Jordan curves. Then there exists $\epsilon>0$, a $C^\infty$ function $\varphi(t,\fxx)$ and a continuous non decreasing contrast change $g$, such that 
\begin{itemize}
\item $\varphi \geq g(u)$, for all $[t_0 -\epsilon, t_0 + \epsilon]$ and $\fxx \in \real^2$,
\item $\varphi(t_0,\fxx_0) = u(t_0,\fxx_0)$,
\item $C(t)$ is a isolevelset of both $g(u)$ and $\varphi$ for $t \in [t_0 -\epsilon, t_0 + \epsilon]$.
\end{itemize}
\end{lem}

\paragraph{Proof of lemma \ref{functiondistancecourbe}}
We consider for $\fxx \in \real^2$, $\psi(t,\fxx) = d((t,\fxx), C(t))$, where $d$ denotes the signed Hausdorff distance, i.e. 
\begin{itemize}
\item $d((t,\fxx), C(t)) > 0$ if $\fxx$ is surrounded by $C(t)$
\item $d((t,\fxx), C(t)) < 0$ otherwise
\end{itemize}
Then, there exists $\epsilon$ such that for all $t\in [t_0-\epsilon, t_0+\epsilon]$ and for all $\fxx$ such that $ |d((t,\fxx),C(t)) | < \epsilon$, $\psi(t,\fxx)$ is a $C^\infty$ function.
We can extend $\psi$ into a $C^\infty$ function on $\real^+ \times \real^2$, $\varphi$.
By Lemma \ref{lemmedecomparaison}, we can find a continuous and nondecreasing function $g$ such that
 $\varphi \geq g(u)$ and $\varphi = g(u)$ on $C(t)$ for all $t \in [t_0 -\epsilon, t_0 + \epsilon]$.
(Notice that Lemma \ref{lemmedecomparaison} is stated for functions defined on $\real^N$, but still holds for functions defined in a closed subset of $\real^N$, without any change.)
\endproof

\paragraph{Proof of theorem \ref{viscoimpliquegrayson}}
We apply lemma \ref{functiondistancecourbe} and the fact that $u$ is a viscosity solution.
We deduce that
$$\frac{\partial \varphi}{\partial t} \leq F(curv(\varphi))|D\varphi|$$
Let us consider the normal flow $\fxx(t)$ associated with $\varphi(t,\fxx)$, satisfying therefore the relation (\ref{flotnormalcommeODE}),
$$\frac {\partial \fxx}{\partial t}= -(\frac{\partial \varphi}{\partial
t}\frac{D\varphi}{|D\varphi|^2})(t,\fxx(t)).
$$
Since the level lines of $g(u)$ and $\varphi$ coincide for $t \in [t_0 -\epsilon, t_0 + \epsilon]$,
we have $curv(\varphi) = curv(u) = Curv(\fxx)$. Thus,
$$\frac{\partial \fxx}{\partial t}.D\varphi \leq F(Curv(\fxx)) |D\varphi|,$$
which yields, 
$$\frac{\partial \fxx}{\partial t}.\fxn \leq F(Curv(\fxx)). $$
We get the converse inequality by similar arguments.




In the following, our purpose is to define an image evolution by handling separately all level lines of the initial image. The level lines evolution is defined by e.g. the intrinsic heat equation, see chapter \ref{chaptermainequation} or any curve scale space ensuring the existence of a smooth curve evolution. We need to define a kind of image $u$ for which all iso level sets $\{\fxx, u(\fxx)=\lambda\}$ are uniquely described as finite unions of Jordan curves.



\begin{Defi} \label{definitionshapelocalinclusionprinciple}
We say that a curve scale-space satisfies a uniform local comparison principle, if there is a 
function $h(r,\epsilon)>0$ satisfying the following condition:

Let $c_1$ and $c_2$ any two Jordan curves which can be locally parameterized as graphs $y=f_1(x)$, $y=f_2(x)$, $-r \leq x \leq r$ and such that 
$$ r \geq f_1(x) \geq f_2(x)+\epsilon x^2 \geq -r $$
for every $-r \leq x \leq r$. Then the evolution of $c_1$ and $c_2$ by the curve scale-space $T_t$ defines locally two graphs for $t \leq h(r,\epsilon)$ and $-r/2 \leq x \leq r/2$:
$y=f_1(t,x)$, $y=f_2(t,x)$ such that 
$$f_1(t,x) \geq f_2(t,x).$$
\end{Defi}


\begin{remark}
The Grayson evolution and the affine shortening (respectively $G(s)=s$ and $G(s)=s^\frac 1 3$, satisfy this property. See \cite{Grayson:1987:HESa}, lemma ??, page ??, \cite{Angenent:1998:AHF} lemma ??, page ??
\end{remark}

\begin{Defi} {\bf Curve evolution equation and Curve Scale Space}.
\label{evolutiongrayson}

We consider a curve evolution equation
\fb& \frac{\partial \fxx}{\partial t} = G(Curv(\fxx)) \fxn, \label{curvescalespace}\fe&
and its associated function evolution equation
\fb& \frac{\partial u}{\partial t} = |Du|G(curv(u)) = F(D^2u,Du)\fe&
where $G$ is a non-decreasing continuous function and $F$ is admissible (Definition \ref{extensionsemicontinue}).
(Note that the admissibility implies that $r(G(1/r)$ is bounded for $r$ around $0$).

 We say that (\ref{curvescalespace}) is a smoothing scale-space, if we can assert that 
\begin{enumerate}
\item Every closed Jordan curve with finite length $\fxx_0(s)$ evolves by (\ref{curvescalespace}) into a $C^2$ closed Jordan curve $\fxx(t,s)$ such that $\fxx(0,s)= \fxx_0(s)$.
\item The curve map $\fxx_0(.) \rightarrow \fxx(t,.)$ is continuous for the Hausdorff distance,
\item satisfies the global shape inclusion principle
\item and the uniform shape local inclusion principle.
%%(Definition \ref{definitionshapelocalinclusionprinciple})
%%For every $r >0$, there exists $h(r)>0$ such that if $\fxc_1$ and $\fxc_2$ are two Jordan curves which do not cross inside some disk $D(\fxx, r)$, then for all $h\leq h(r)$, $\fxc_1(h)$ and $\fxc_2(h)$ do not cross inside $D(\fxx,r/2)$. 
%%In addition, $h(+\infty)=+\infty$, so that we also have a global shape inclusion principle.
\end{enumerate}
\end{Defi}

\begin{Defi} \label{definitionregulargrayson}
We define a class $\cal T$ of subsets of $\real^2$ which we call ``regular'' sets. A regular set is a closed subset of $\real^2$ whose boundary is a locally finite union of disjoint closed $C^2$ Jordan curves.
If $\fxc$ is a Jordan curve we denote by $\fInt(\fxc)$ the unique closed bounded subset of $\real^2$ whose boundary is $\fxc$. If $X$ belongs to $\cal T$ has no unbounded connected component, we can write
\fb& \label{ensembleregulier} X= \cup_{i=1}^k \left( \fInt(\fxc_i) \setminus \cup_{j=1}^{l_i} \interior{\fInt(\fxc_{i,j})} \right), \fe&
where the index $i$ runs over all connected components of $X$, the index $j \in \{1, ..., l_i\}$ runs over all bounded connected components of the complementary set to the $i$-th connected component (its holes). If $X$ has an unbounded connected component, then $X^c$ has no unbounded connected component and satisfies (\ref{ensembleregulier}).

We associate with $\cal T$ the class $\cal F$ of continuous functions $u$ such that $\levelset_\lambda u$ belongs to $\cal T$ for every $\lambda$ except a finite number of them which we call critical levels. The functions of $\cal F$ will be called ``regular''.
\end{Defi}

\begin{Exa}It is easily seen that a periodic Morse function, that is, a $C^2$ function such that $det D^2u(\fxx) \neq 0$ for all $\fxx$, is an instance of regular function. 
\end{Exa}

%%\begin{Exa}
%%As examples of functions of $\cal F$, we can mention periodic Morse functions, that is $C^2$ periodic functions such that $D^2u(\fxx)$ is non degenerate for all $\fxx$. Notice that every continuous and periodic function can be approximated uniformly by a periodic Morse function.
%%\end{Exa}

\begin{Defi} \label{evolutiongraysonensembles} (Curve evolution as a set operator $\setT_t$). \newline
We consider a smoothing curve scale-space $\fxc \rightarrow \fxc(t)$, as defined in Definition (\ref{evolutiongrayson}). For every set in $\cal T$, if $X$ satisfies: 
$$X = \cup_{i=1}^k \left( \fInt(\fxc_i) \setminus \cup_{j=1}^{l_i} \interior{\fInt(\fxc_{i,j})} \right),$$
we set
$$\setT_t(X) = \cup_{i=1}^k \left( \fInt(\fxc_i(t)) \setminus \cup_{j=1}^{l_i} \interior{\fInt(\fxc_{i,j}(t))} \right),$$
where $\fxc_i(t)$ and $\fxc_{i,j}(t)$ denote the curves evolved at time $t$ by the curve scale-space.

If $X \in \cal T$ has an unbounded component, we set $\setT_t(X) = (\setT_t(X^c))^c$. 

We also set $\setT(\emptyset) = \emptyset$ and therefore $\setT(\real^2) = \real^2$.
\end{Defi}


\begin{lem} \label{monotoniedegraysonensemble}
$\setT_t$ is monotone on $\cal T$ and transforms a set of $\cal T$ into a set pf $\cal T$.
\end{lem}

\paragraph{Proof}
If $X \subset Y$, then each external Jordan curve of a connected component of $Y$ is enclosed by an exterior Jordan curve of a connected component of $X$ and each internal Jordan curve of $X$ is enclosed by an internal Jordan curve of $Y$. Since the evolution by a smoothing scale-space preserves the inclusion of curves, we easily deduce the announced statement.
\endproof


\begin{Defi} \label{evolutiongraysonimages}
Let $u \in \cal F$ be a regular function. We call level line evolution of $u$ the unique monotone functional extension $T$ to $\cal F$ of the set evolution $\setT_t$ given by Proposition \ref{setoperatortoimageoperatorfaible}. We then have \newline
(i) $T(u)$ is regular.
\newline
\noindent (ii) If $u$ is $K-$Lipschitz, then $T(u)$ is equal almost everywhere to a $K-$Lipschitz function. \newline
\noindent (iii) $\setT(\levelset_\lambda u) = \levelset_\lambda(T(u)),$ a.e and for almost every $\lambda \in \real$ such that $\lambda$ is not critical for $u$. \newline
\noindent (iv) As a consequence, $T_t$ is recursive: $T_{t+t'} = T_{t} T_{t'}$ for all $t,t' \geq 0$.
\end{Defi}


\paragraph{Proof}

\noindent (i) By Exercise \ref{exerciceutile}, we know that $T_t(u)$ is obtained from $u$ by a sup-inf operator, so that the Lipschitz constant of $u$ is preserved. \newline
\noindent (ii) Let $\lambda_0 \notin \{\lambda_1, ..., \lambda_l\}$ critical levels of $u$, that is levels for which $\levelset_\lambda u$ is not in $\cal T$. As $\lambda$ tends to $\lambda_0$, $\levelset_\lambda u$ tends to $\levelset_{\lambda_0} u$ locally for the Hausdorff distance. (Each Jordan curve of the boundary of $\levelset_\lambda u$ tends to a corresponding curve of the boundary of $\levelset_{\lambda_0}u$).
By Definition \ref{evolutiongrayson}.2,  we then have
$$\setT_t(\levelset_\lambda u) \to \setT_t(\levelset_{\lambda_0} u)$$
locally for the Hausdorff distance.
Since we can choose an increasing sequence $\lambda_k  \to \lambda_0$ of non critical values for $u$, we then have
by (\ref{evolutiongraysonimages}.ii) $\setT_t(\levelset_{\lambda_k} u) = \levelset_{\lambda_k}(T_t u)$, a.e. By passing to the limit and using (\ref{II.1}.ii) we obtain
$$\setT_t(\levelset_{\lambda_0} u = \lim_{\lambda_k \to \lambda_0} \levelset_{\lambda_k}(T_t u) = \bigcap_{\lambda_k \to \lambda_0} \levelset_{\lambda_k}(T_t u) = \levelset_{\lambda_0} (T_t u)  \hbox{ a.e.}$$
(iii) Obvious consequence of (ii) and the pyramidality of the curve scale-space. 
\endproof

%% fonctionsradialessuffisent

\begin{lem}
\label{principedecomparaisonpourgrayson}
Let $u$ and $v \in \cal F$ be two regular functions such that $u(\fxy) > v(\fxy)$ for every $\fxy \in D(\fxx,r)\setminus \{\fxx\}$. Then there is some $h_0>0$ such that for all $h<h_0$, $(T_h u)(\fxx) \geq (T_h v)(\fxx)$. In other terms the level line evolution $T_t$ satisfies the local comparison principle. 
\end{lem}

\paragraph{Proof}
The set operator $\setT_t$ satisfies by definition the shape local inclusion principle (Definition
\ref{evolutiongrayson}.3). By, comparing the level lines of $u$ and $v$ in the disk $D(\fxx,r)$, it is easily checked that the function operator $T_t$ inherits a local comparison principle.
\endproof


\begin{lem} \label{regularitepourfonctionsradiales}
(i) Let $G$ be a function associated with a smoothing curve scale-space (Definition \ref{evolutiongrayson}). Then, for every $C^2$ increasing radial function $Q$ we have for $t \to 0$ and when $DQ(\fxx) \neq 0$
$$ \frac{T_t(Q) (\fxx) - Q(\fxx)}{t} \to |DQ(\fxx)| G(curv(Q)(\fxx)),$$
(ii) The same result is true for any $C^\infty \cap \cal F$ function $f$. \newline
(iii) Let $f$ be a $C^\infty \cap \cal F$ function such that $f(\fxx)=0$ for all $\fxx \in D(\fxx_0,r)$. Then
$$ \lim_{t \to 0} |\frac{T_t(f) (\fxx_0) - f(\fxx_0)}{t}| \to 0.$$
\end{lem}

\begin{remark}
We choose the above functions in $\cal F$, so that their evolution by the curve scale-space is imediately described by the evolution of their level lines.
\end{remark}

\paragraph{Proof of Lemma \ref{regularitepourfonctionsradiales}}
(i) This is an easy computation which we leave as exercise. \newline
(ii) Consider a $C^\infty$ function $f$, such that $Df(\fxx_0) \neq 0$. It is easily seen that we
can find a radial function $v_{\pm \epsilon}(\fxx)$ such that 
$$v_{\pm \epsilon}(\fxx_0) = f(\fxx_0), \qquad Dv_{\pm \epsilon}(\fxx_0) = Df(\fxx_0),\qquad curv (v_{\pm \epsilon})(\fxx_0) = curv(f(\fxx_0) \pm \epsilon)$$
and, in addition, 
$$v_{-\epsilon}(\fxx) \leq f(\fxx) \leq v_{\epsilon}(\fxx)$$
in a neighborhood $D(\fxx_0,r)$.
By lowering $r$ if necessary we can assume by the implicit function theorem that for every
$\lambda \in [v(\fxx_0)-\eta,v(\fxx_0)+\eta]$ and $\eta$ small enough, the level lines 
$c_1$, $c_2$ with level $\lambda$ of $v_\epsilon$ and $f$ respectively satisfy the assumptions
of Definition (\ref{definitionshapelocalinclusionprinciple}). (The same applies to the level lines of $f$ and $v_{-\epsilon}$.

Thus, for $t \leq h(r,\epsilon)$ the inclusion of level lines of $v_{\pm \epsilon}$ and $f$ is preserved for all levels close to $\lambda_0 = f(\fxx_0)$. We deduce that in a neighborhood of $\fxx_0$, say $D(\fxx_0,r/4)$ we have
$$ T_h(v_{-\epsilon})(\fxx_0) \leq T_h(f)(\fxx_0) \leq  T_h(v_{\epsilon})(\fxx_0).$$
Substracting $f(\fxx_0) = v_{\pm \epsilon}(\fxx_0)$, dividing by $h$ and letting $h \to 0$ we deduce from (i) that
$$G( (curv(f))(\fxx_0) - \epsilon ) \leq \liminf_{h\to 0} \frac{T_h(f)(\fxx_0) -f(\fxx_0)}{h} 
$$
$$ \leq \limsup_{h\to 0} \frac{T_h(f)(\fxx_0) -f(\fxx_0)}{h} \leq G( (curv(f))(\fxx_0) + \epsilon ).$$
We conclude by letting $\epsilon \to 0$ and using the continuity of $G$. \newline
(iii) No level line of $f$ meets $D(\fxx_0,r)$. Since the evolution by the curve scale-space is continuous for the Hausdorff distance, we deduce that for $t\leq t_0(r)$, no level line of $f(t)=T_t(f)$ meets $D(\fxx_0,r/2)$. Thus $f(t)(\fxx_0) = 0$ for $t \leq t_0(r)$.
\endproof



\begin{th1}
We consider a smoothing curve scale-space $\frac{\partial\fxc}{\partial t} = G(Curv(\fxc)) \fxn$, as defined in Definition (\ref{evolutiongrayson}) and its unique extension $T_t$ to the set of regular function $\cal F$.
Let $u_0 \in \cal F$. Then $u(t,\fxx) = T_t(u_0)(\fxx)$ is viscosity solution
of
$$\frac{\partial u}{\partial t} = |Du| G(curv(u)).$$
\end{th1}

\paragraph{Proof}
%% viscositeformesquadratiquessuffisent

In order to prove that $u(t,\fxx) = T_t(u_0)(\fxx)$ is a viscosity
solution of (\ref{eqgen}), we only to check that $u$
is a viscosity subsolution of (\ref{eqgen}), the supersolution property
being shown in the same way. 
Let
$(t_0, \fxx_0)$ in $[0, \infty]\times\real^N$ be a strict local maximum point of
$u-\phi$ where $\phi(t,\fxx) = f(\fxx)+g(t)$ where $f$ is regular, $C^\infty(\real^2)$ and equal to a quadratic form in a neighborhood of $\fxx_0$, and $g$ is
$C^\infty(\real)$. By Exercise \ref{viscositeformesquadratiquessuffisent}, we are allowed to do this restriction on test functions.
By Lemma \ref{lemmaBarles}, it is enough to show
that when $Df(\fxx_0) \neq 0$
\fb&	\frac{\partial g}{\partial t}(t_0, \fxx_0) - |Df(\fxx_0)| G(curv(f)(\fxx_0)) \leq 0.
\fe&
and when $Df(\fxx_0)=0$ and $D^2f(\fxx_0)=0$,
\fb&	\frac{\partial g}{\partial t}(t_0, \fxx_0)  \leq 0.
\fe&
By Lemma \ref{localauglobal} we can assume (w.l.o.g.) that the inequality between $u$ and $\phi$
is global, i.e.  $u\leq \phi$ on $[0, \infty]\times \real^N$. 
 Let $h > 0$. We set $u(t) = u(t, .)$ for all functions $u$ on $\real^N*[0, \infty[$.  We have $u(t) \leq
\phi(t)$, then $u(t_0-h, \fxx) \leq \phi(t_0-h, \fxx)$ for all $h<t_0$
and $\fxx$. Using the monotonicity of $T_h$ (Definition \ref{evolutiongraysonimages}), we obtain
$$
T_h(u(t_0-h) )(\fxx_0) \leq
T_h(\phi(t_0-h))(\fxx_0)
$$ 
Now, by the commutation of $T_t$ with the addition of constants and since $\phi(t,\fxx)  =
f(\fxx) + g(t)$ we have
$$
 T_h(u(t_0-h) )(\fxx_0) \leq
T_h(\phi(t_0-h))(\fxx_0) \leq T_h(f)(\fxx_0)
+ g(t_0-h).
$$ 
We deduce that 
$$
 u(t_0,\fxx_0) -
(T_h(f)(\fxx_0) + g(t_0-h)) \leq 0
$$ 
Since  $u(t_0,\fxx_0) = \phi(t_0,\fxx_0)$ and then
 
$$
g(t_0) - g(t_0-h) \leq T_h(f)(\fxx_0) - f(\fxx_0)
$$
Then,
$$ 
(g(t_0) - g(t_0-h))/h \leq
(T_h(f)(\fxx_0) - f(\fxx_0))/h
$$

If $Df(\fxx_0) \neq 0$ then letting
$h$ go to $0$, using (i), we recover
$$
\frac{\partial g}{\partial t}(t_0-h) - |Df(\fxx_0)| G(curv(f)(\fxx_0)) \leq 0
$$  
By letting $h$ tends to zero, we conclude.

If $Df(\fxx_0) = 0$ and $D^2f(\fxx_0) = 0$ then by the assumption on $f$ we know that 
$f$ is zero on a neighborhood of $\fxx_0$, and by Lemma \ref{regularitepourfonctionsradiales}.(iii) we deduce that for $h$ small enough,
$|T_h(f)(\fxx_0) - f(\fxx_0)| =0.$
\endproof



\section{Motion by curvature and Total Variation}

\begin{lem} \label{decroissanceduperimetre}
Let $\fxx(t,s)$ satistying the curve equation evolution
$$ \frac{\partial \fxx}{\partial t} =  F(Curv(\fxx)) \fxn $$. Assume that for some $(t,s) \in [t_1, t_2]\times \real$, $\fxx$ is $C^\infty$. As we have seen, this holds for e.g. $F(x)=x$.

Let $L(t)$ be the perimeter of $\fxx(t,.)$.  Then $\forall t \in ]t_1,t_2[$, we have
$$ \frac{\partial L}{\partial t}(t) = - \int_0^L(t) Curv(\fxx(t,s))F( Curv(\fxx(t,s))) ds$$
where $s$ is the arc length parametrization.
As consequence the curve equation evolution makes the perimeter of the curves decrease.
\end{lem}

\paragraph{Proof}
We choose to parametrize the curve with $z$ between $0$ and $1$. We thus have
$$L(t) = \int_0^1 |\frac{\partial \fxx}{\partial z}(t,z)| dz$$
Thus
$$\frac{\partial L}{partial t}(t) = \int_0^1 \frac{\partial}{\partial t}|  |\frac{\partial \fxx}{\partial z}(t,z)| dz$$
Denoting by $'$ the partial derivation with respect to $z$:
$$=  \int_0^1 \frac{\fxx'(z,t)}{|fxx'(z,t)|} \frac{\partial}{\partial t}(\fxx') (t,z) dz$$
By integration by parts
$$ = - \int_0^1 Curv(\fxx(t,s)) \fxn \frac{\partial\fxx}{\partial t}(z,t) \fxx'(t,z) dz$$
Using the curve equation evolution, we obtain:
$$\frac{\partial L}{partial t}(t) = - \int_0^1 Curv(\fxx(t,s)) F(Curv(\fxx(t,s)) dz$$
$$= - \int_0^L(t) Curv(\fxx(t,s)) F(Curv(\fxx(t,s))) ds$$
\endproof


\begin{th1} {\bf Total Variation of a Lipschitz function}. \cite{EvansGariepy} Let $u$ be a K-Lipschitz function defined on a square $C$. Then $Du$ is integrable. And the total variation of $u$ satisfies:
$$TV(u)= \int_C |Du(\fxx)| d\fxx \leq K area(C)$$
\end{th1} 


\begin{Defi} {\bf Total Variation by the Coarea formula}
Let $u$ be a Lipschitz and regular (see definition \label{definitionregulargrayson} in the preceding section) function defined on a square $C=[0,1[^2$, then
we define the Total Variation of the image $u$ by:
$$\int_C |Du(\fxx)| d\fxx = \int_{\lambda \in \real}\hbox{Perimeter}(\partial \levelset_\lambda(u))\cap C d\lambda $$
\end{Defi}

This last definition makes sense since the regularity of $u$ insures that the Perimeter of the boundaries of the $\lambda$ level set of $u$  is finite except for a set of $\lambda$ whose measure is zero.



We consider a regular and K-Lipschitz function $u_0$ and its evolution $u(t)$ by a regular curve scale space.  By \ref{evolutiongraysonimages} we know that $u(t)$ is K-Lipschitz and regular, so that we can estimate its total variation.

\begin{th1}
We consider a regular and K-Lipschitz function $u_0$ and its evolution $u(t)$ by a regular curve scale space.  By \ref{evolutiongraysonimages} we know that $u(t)$ is K-Lipschitz and regular, so that we can estimate its total variation.

One has $$\frac{\partial TV(u(t))}{\partial t} \leq 0$$
or in other terms, a regular curve scale space decreases the total variation.
\end{th1}

\paragraph{Proof}
This is an imediat consequence of \ref{decroissanceduperimetre} and the coarea formula.
\endproof.

\begin{Remark}
Let us give a heuristic that illustrates (but not proves) the same result. Consider the total variation as an energy, and compute the gradient of the energy.
(Assume that $u(t,\fxx)$ is $C^\infty$ and $Du \neq 0$). 
Denoting by $G_{\hbox{TV}}(u)$ the gradient of the total variation at function $u$. It is easily proved (integration by parts) that $G(u) = - div(Du/|Du|) = curv(u)$.

$$TV(u(t+dt)) = TV(u(t)) +   \int_C (u(t+dt)-u(t)) (G_{\hbox{TV}}u)(\fxx)) d\fxx$$
$$ \frac{\partial TV(u(t))}{\partial t} = - dt \int_C \frac{\partial u}{\partial t}(t,\fxx) curv(u) d\fxx$$




\end{Remark}








