
\begin{slide}
\section{From set operators to contrast invariant function
operators. \label{setoptoimop}}
\def\setT{\mathcal T}


Can we extend a set monotone operator into a contrast invariant
and monotone operator defined on an adequate set of functions ? We
start from a set operator $\setT $, defined on a subset $\cal L$
of ${\cal P}(\real^N)$ and with values  in $\cal L$.  We assume
that $\setT $ is a {\it monotone} operator, that is, $X \subset Y
\Rightarrow \setT (X) \subset \setT (Y)$.   We define $\cal F$  as
the set of all real functions defined on $\real^N$ whose level
sets $\levelset_\lambda u$ belong to $\cal L$. As a consequence,
$\cal F$ contains the characteristic functions of elements of
$\cal L$.

\end{slide}

\begin{slide}
 A natural definition for the function operator $T$
associated with $\setT$ is given  by the threshold superposition
principle,
$$
Tu(\fxx) = \sup\{ \lambda, \fxx \in \setT(\levelset_\lambda u)\}
$$
 We have now to check whether $T$ is contrast invariant.

\begin{proposition} \label{setoperatortoimageoperatorfaible}
Let $\setT$, ${\cal T} \rightarrow {\cal T}$ be a monotone
operator satisfying
$$ T(\emptyset) = \emptyset,  \setT(\real^N)= \real^N $$
Then the operator $T$ defined on $\cal F$ by superposition
principle, \fb&  \label{superpositionprinciple2}
 Tu(\fxx) = \sup \{\lambda, \fxx \in \setT(\levelset_\lambda u) \}
 \fe&
 satisfies,  for almost every $\lambda \in \real$,
\fb& \levelset_\lambda (Tu) = \setT(\levelset_\lambda(u)) \hbox{
almost everywhere in }  \fxx \fe& and for any nondecreasing
continuous contrast change  $g$, \fb& g(Tu) = T(g(u)). \fe&
\end{proposition}
\end{slide}
\begin{slide}
\paragraph{Proof}

By Corollary \ref{II.1.bis}, we have for almost every $\lambda$
\fb& \levelset_\lambda (Tu) = \setT(\levelset_\lambda u) \hbox{
almost everywhere} \label{ae1} \fe& that is for all $\lambda$ in a
subset $\Lambda$ of $\real$ such that $\meas(\real \setminus
\Lambda) = 0$. We notice that $u \leq v$ if and only if
$\levelset_\lambda u \subset \levelset_\lambda v$ for all
$\lambda$ in a dense countable subset of $\real$.  We deduce
immediately that $T$ is monotone, ie $u \geq v \; \Rightarrow Tu
\geq Tv$.
\end{slide}

\begin{slide}
 Let us now  show that $T$ commutes with contrast changes.
Assume first that $g$ is strictly increasing and set
$g(+\infty)=\lim_{s\to +\infty}g(s)$ and $g(-\infty)=\lim_{s\to
-\infty}g(s).$
\newline For $\lambda>g(+\infty),$ we have
$\levelset_\lambda g(u)=\emptyset$ and therefore
$\setT(\levelset_\lambda g(u))=\emptyset.$
\newline
For $\lambda < g(-\infty)$, we  have $\levelset_\lambda
g(u)=\real^N$ and therefore $\setT(\levelset_\lambda
g(u))=\real^N.$
\newline As a consequence, using \ref{superpositionprinciple2} we can restrict the range of $\lambda$ in the definition
of $T(g(u(\fxx)))$ :
$$T(g(u(\fxx)))=\sup \{\lambda, \; g(-\infty)\leq \lambda\leq g(+\infty), \; \fxx\in\setT(\levelset_\lambda g(u))\}$$
$$=\sup\{g(\mu),\;  \fxx\in \setT(\levelset_{g(\mu)}g(u))\}$$
$$=\sup\{g(\mu),\; \fxx\in \setT(\levelset_\mu u)\}= g(Tu(\fxx)).$$
Let us now check that $T$ commutes with general  nondecreasing
contrast changes $g$. We can find strictly increasing continuous
functions $g_n$ and $h_n$ such that $g_n(s)\to g(s)$, $h_n(s)\to
g(s)$ for all $s$ and $g_n\leq g\leq h_n$.  Thus, by using the
just proven commutation of $T$ with increasing contrast changes,
we have
$$T(g(u))\geq T(g_n(u))=g_n(Tu)\to g(Tu)\; \hbox {  and  }$$
$$T(g(u))\leq T(h_n(u))=h_n(Tu)\to g(Tu), $$
which yields  $T(g(u))=g(Tu).$ \end{slide}
\begin{slide}

We shall now examine stronger conditions on $\setT$, which ensure
commutation of $T$ with all thresholds $\levelset_\lambda$ (and
not only with almost all of them).

\begin{defintion}\label{definitionuppersemicontinuousoperator}
We say that a set operator $\setT$ is upper semicontinuous if for
every nonincreasing sequence of sets $X_n \in \cal T$, we have
\fb& \label{vii.2} \setT(\bigcap_n X_n) = \bigcap_n \setT(X_n).
\fe&
\end{definition}

\begin{theorem} \label{3.5bis}
Let $\setT $: $\cal T$ $\to \cal T$ be a monotone upper
semicontinuous set operator satisfying $ \setT(\emptyset) =
\emptyset$ and  $\setT(\real^N)= \real^N.$ Then the function
operator defined by

\fb& Tu(\fxx) = \sup \{\lambda, \fxx \in \setT (\levelset_\lambda
u) \}. \fe& is contrast invariant and satisfies, for every
$\lambda,$ \fb& \label{FGencore2} \levelset_\lambda (Tu) =
\setT(\levelset_\lambda(u)). \fe&
\end{theorem}
\end{slide}

\paragraph{Proof of Theorem \ref{3.5bis}}
Let us check that (\ref{FGencore2}) holds.
 By Proposition  \ref{II.1}, this is true if and only if the family of sets
$\setT (\levelset_\lambda u)$ satisfies (\ref{II.1}.i) and
(\ref{II.1}.ii).  Now, (\ref{II.1}.i) is true by monotonicity of
$\setT $  and the inclusion  relations between the $X_\lambda u$.
  Relation (\ref{II.1}.ii) follows
immediately from (\ref{vii.2}) and  $\levelset_\lambda u =
\bigcap_{\mu<\lambda}\levelset_\mu u $, which is (\ref{II.1}.ii)
applied to $u$. The monotonicity of $T$ is obvious : one has
$$u\leq v \Leftrightarrow (\forall \lambda, \;\levelset_\lambda
u\subset \levelset_\lambda v).$$ Since $\setT $ is set monotone,
we deduce that $$\forall \lambda, \; \setT (\levelset_\lambda u)
\subset \setT (\levelset_\lambda v) $$ and therefore $\forall
\lambda, \levelset_\lambda (Tu) \subset \levelset_\lambda (Tv)$,
that is $Tu\leq Tv.$  In order to show that $T$ is contrast
invariant, i.e. commutes with all continuous nondecreasing
functions, we can now directly apply  Proposition
\ref{setoperatortoimageoperatorfaible}.
\endproof

\begin{slide}
The upper semicontinuity is necessary to ensure that a monotone
set operator defines a function operator such that the commutation
with thresholds $\levelset_\lambda (Tu) =
\setT(\levelset_\lambda(u)) $ holds for every $\lambda$. Let us
choose for example the following set operator $\setT $,
$$ \setT (X) = X   \; \hbox{ if meas}(X)>a \hbox{ and } \setT (X) = \emptyset \hbox{
otherwise }.
$$
Let  $u$ be the function from $\real$ into $\real$ defined by
$u(x) = -|x|$.  Then
$$T(u)(x) = \sup\{\lambda, x \in \setT (\levelset_\lambda u)\} = \min( -|x|, -a/2) $$
Therefore
$$\levelset_{-a/2} T(u) = [-a/2,a/2]. $$
Now, $\levelset_{-a/2} u = [-a/2,a/2]$ ; its measure is $a$. Thus
$$T(\levelset_{-a/2} u) = \emptyset \neq \levelset_{-a/2} T(u),$$
which means that $T$ is not contrast invariant.

\end{slide}
