%gmChap07.tex

\graphicspath{{./figures/}}

\begin{slide}

\section{From set operators to contrast invariant function
operators}

\begin{definition}\label{definitionuppersemicontinuousoperator}
We say that a set operator $\setT$ is upper semicontinuous if for
every strictly decreasing sequence of compact sets $X_n$, we have
$$ \label{vii.2} \setT(\bigcap_n X_n) = \bigcap_n \setT(X_n).
$$ (Strictly decreasing : closure of $X_{n+1}$ contained in
$X_n^\circ$).
\end{definition}

\end{slide}
\begin{slide}
\begin{theorem} \label{3.5bis}
Let $\setT $:   be a monotone upper semicontinuous set operator on
compact periodic sets. Then the function operator defined on
continuous periodic functions by superposition principle,

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

\begin{slide}
\section{Conversely   : the Evans--Spruck extension}




\begin{theorem} [Evans--Spruck]
Let $T$ be a contrast-invariant monotone operator defined on
continuous $C$-periodic functions. Then there is a unique set
operator $\mathcal T$ defined on $C$-periodic compact sets
 satisfying the
commutation relation
\begin{equation}
\mathcal T\levelset_{\lambda}u=\levelset_{\lambda}Tu.
\end{equation}
In addition, $\mathcal T$ is upper semicontinuous on compact sets.
\end{theorem}


\end{slide}
\begin{lemma}Let $u$  and $v$ be  continuous periodic functions bounded above by zero
 and assume that $\levelset_0u=\levelset_0v\,(\ne\emptyset)$. Then there
 is a continuous contrast change $h$ such that $h(0)=0$ and $u\geq h(v)$.
\end{lemma}

\end{slide}
\begin{slide}






\end{slide}


\begin{slide}


\begin{proof} (Evans-Spruck) It is sufficient to prove that
 $\levelset_{0}u=\levelset_{0}v$ implies $\levelset_{0}Tu=\levelset_{0}Tv$ when
  $u$ and $v$ are   bounded from above by zero.

The proof follows directly from the preceding lemma : Let $h$ be a
continuous contrast change such that $h(0)=0$ and $u\geq h(v)$.
Since $T$ is monotone and contrast invariant, $Tu\leq0$ and
$Tu\geq Th(v)=h(Tv)$. Using the fact that $h(0)=0$, we see that
$Tv(\fxx)=0$ implies that $Tu(\fxx)=0$. By interchanging the roles
of $u$ and $v$ we see that $Tu(\fxx)=0$ implies that $Tv(\fxx)=0$
and conclude that $\levelset_0 Tu=\levelset_0 Tv$.
\end{proof}


\end{slide}


\begin{slide}
Let us prove that $\mathcal T$ is upper semicontinuous on compact
sets.

\begin{lemma}\label{Urysohn} (Urysohn). Let $X_n$
be a strictly decreasing sequence of periodic compact sets, namely
$X_{n+1} \subset X_n^\circ.$ Then
 there exists a continuous periodic function $u$ such that
$\levelset_{1-\frac 1n} u= X_n$ and $\levelset_1u=\cap_{n}X_n.$
\end{lemma}
 By assumption, $$\levelset_1(Tu)=\mathcal
 T(\levelset_1u)=\mathcal T(\cap_n\levelset_{1-\frac 1n}u)=\mathcal
 T(\cap_n X_n).$$  On the other hand,
 $\levelset_1(Tu)=\cap_n\levelset_{1-\frac 1n}Tu=\cap_n\mathcal
 T(\levelset_{1-\frac 1n}u)=\cap_n\mathcal  T(X_n).$
 Combining both results, we get
 $$\mathcal
 T(\cap_n X_n)=\cap_n\mathcal  T(X_n).$$

\end{slide}


\begin{figure}[!t]
\setkeys{Gin}{width=3cm} \centerline{
\includegraphics{fig07.01.1.eps}
\includegraphics{fig07.01.2.eps}
\includegraphics{fig07.01.3.eps}
\includegraphics{fig07.01.4.eps}
} \caption[Evans-Spruck extensions and the chessboard dilemma]
{Evans-Spruck extension and the chessboard dilemma. Left: a
chessboard image. Next: result with the finite difference scheme
for the curvature motion, applied up to a fixed scale. The
creation of a new gray level proves that the scheme is not fully
contrast invariant.
   The next two images are obtained by
the Evans--Spruck extension of the curvature motion, first applied
to the white set considered as closed, second to the black set
considered as closed.  The median filters $Med$ and  $Med^-$ will
yield exactly the same dilemma.}
\end{figure}



\begin{slide}
\section{Contrast-Invariant Monotone Operators as Sup-Inf
Operators}

Two representation theorems: If $T$ is a monotone contrast
invariant translation invariant function operator, then it has a
``sup-inf'' representation of the form
\[
Tu(\fxx)=\sup_{B\in\mathcal B} \inf_{\fxy\in B}u(\fxx+\fxy),
\]
where $\mathcal B$ is a family of subsets of $\mathcal P(\RR^N)$, the set of all subsets of $\RR^N$. This theorem is a nonlinear version of the Riesz theorem that states that a continuous linear translation-invariant operator can be represented as a convolution
\[
Tu(\fxx)=\int_{\RR^N}u(\fxx-\fxy)k(\fxy)\,\rmd\fxy.
\]
In this case, the kernel $k$ is called the impulse response.  In
the same way, $\mathcal B$ is an impulse response for the
nonlinear operator.

\end{slide}
\begin{slide}

\section{Translation-invariant monotone set operators}

\begin{theorem}[Matheron] Let $\mathcal T$  be a translation-invariant monotone set
operator acting on the set of periodic compact sets $\mathcal L$.
  Then
\begin{equation}\label{eq81}
\mathcal T X=\bigcup_{B\in\mathcal B}\bigcap_{\fxy\in B}(X-\fxy)=\{\fxx\mid\fxx+B\subset X \text{ for some }B\in\mathcal B\},
\end{equation}
where $\mathcal B=\{X\mid 0\in\mathcal TX\}$.  Conversely, if
$\mathcal B$ is any nonempty subset of $\mathcal P(\RR^N)$, then
formula \eqref{eq81} defines a translation-invariant monotone set
operator on $\mathcal P(\RR^N)$.
\end{theorem}
 $\mathcal B$ is called set of
\textit{structuring elements} in the language of mathematical
morphology.


\end{slide}
\begin{slide}
\begin{proof
Now let $\mathcal B=\{X\mid 0\in\mathcal T X\}$.
\begin{align*}
\fxx\in\mathcal T X&\stackrel{\mathrm(1)}{\Longleftrightarrow}0\in\mathcal T X-\fxx\stackrel{\mathrm(2)}{\Longleftrightarrow}0\in\mathcal T(X-\fxx)\stackrel{\mathrm(3)}{\Longleftrightarrow}X-\fxx\in\mathcal B\\
&\stackrel{\mathrm(4)}{\Longleftrightarrow} X-\fxx=B\text{ for some }B\in\mathcal B\stackrel{\mathrm(5)}{\Longleftrightarrow}\fxx+B\subset X\text{ for some }B\in\mathcal B.
\end{align*}
The equivalence (2) follows  from the translation invariance of
$\mathcal T X$; (3) is just the definition of $\mathcal B$; and
(4) is a restatement of (3). The implication from left to right in
(5) is obvious. The implication from right to left in (5) is the
point where the monotonicity of $\mathcal T$ is used: Since
$B\subset X-\fxx$, it follows from the monotonicity of $\mathcal
T$ that $X-\fxx\in\mathcal B$. This proves \eqref{eq81}.



\end{slide}
\begin{slide}


\section{Sup-inf operators}

\begin{theorem}
Let $T$ be a monotone function  contrast invariant operator
  on $\mathcal F$ (continuous periodic functions). Let $\mathcal T$ be the set operator associated
with $T$ by the Evans-Spruck extension $\mathcal T X=\levelset_1
T\bmone_X$. For $\fxx\in\RR^N$ and $X\in\mathcal L$, let $\mathcal
B_{\fxx}=\{X\mid\fxx\in\mathcal T X\}$. Then for $u\in\mathcal F$,
\begin{equation}\label{eq83}
Tu(\fxx)=\sup_{B\in\mathcal B_{\fxx}}\inf_{\fxy\in B}u(y)\; \text{ for a.e. } \fxx\in\RR^N.
\end{equation}
If $T$ is also translation invariant, then
\begin{equation}\label{eq84}
Tu(\fxx)=\sup_{B\in\fxx+\mathcal B_0}\inf_{\fxy\in B}u(y)\; \text{ for a.e. } \fxx\in\RR^N.
\end{equation}
Conversely, equation \eqref{eq83} defines a contrast-invariant
monotone function operator, and equation \eqref{eq84} defines a
contrast-invariant monotone function operator that is also
translation invariant.
\end{theorem}

\end{slide}
\begin{slide}



\begin{proof}

For $u\in\mathcal F$,  let $\tilde T u(\fxx)=\sup_{B\in\mathcal
B_{\fxx}}\inf_{\fxy\in B}u(y)$, where $\mathcal
B_{\fxx}=\{X\mid\fxx\in\mathcal T X\}$.  We know   that
$\levelset_\mu Tu=\mathcal T\levelset_\mu u$.  We are going to
prove that $\tilde T u(\fxx)=Tu(\fxx)$ by showing that
\begin{equation}\label{eq7iff}
\tilde T u(\fxx)\geq\lambda \Longleftrightarrow  T u(\fxx)\geq\lambda
\end{equation}
for $\lambda\in\Lambda$  and $\fxx\in\RR^N\setminus N$.

\begin{align*}
Tu(\fxx)\geq\lambda&\stackrel{\mathrm(1)}{\Longleftrightarrow}Tu(\fxx)\geq\mu \text{ for all } \mu<\lambda\stackrel{\mathrm(2)}{\Longleftrightarrow}\fxx\in\levelset_{\mu} Tu \text{ for all } \mu<\lambda\\
&\stackrel{\mathrm(3)}{\Longleftrightarrow}\fxx\in\mathcal T \levelset_{\mu} u \text{ for all } \mu<\lambda\stackrel{\mathrm(4)}{\Longleftrightarrow}\levelset_{\mu} u\in{\mathcal B}_{\fxx} \text{ for all } \mu<\lambda\\&\stackrel{\mathrm(5)}{\Longleftrightarrow}\text{There is a } B\in{\mathcal B}_{\fxx}\text{ such that }\inf_{\fxy\in B}u(\fxy)\geq\mu\text{ for all }\mu<\lambda\\
&\stackrel{\mathrm(6)}{\Longleftrightarrow}\sup_{B\in{\mathcal B}_{\fxx}}\inf_{\fxy\in B}u(\fxy)\geq\lambda\stackrel{\mathrm(7)}{\Longleftrightarrow}\tilde T u(\fxx)\geq\lambda.
\end{align*}

If $T$ is also translation  invariant, then  $\mathcal
B_{\fxx}=\fxx+\mathcal B_0$.



\end{slide}
\begin{slide}


To prove the converse, define $T$ on $\{u\mid u:\RR^N\to\RR\}$ by
\[
Tu(\fxx)=\sup_{B\in\mathcal B_{\fxx}}\inf_{\fxy\in B}u(y).
\]
The operator $T$ is well  defined, since we have assumed that
$\mathcal B_{\fxx}$ is not empty. Clearly, $T$ is monotone: If
$u\geq v$, then $\inf_{\fxy\in B}u(y)\geq\inf_{\fxy\in B}v(y)$,
and it follows that $Tu\geq Tv$. If $g$ is any continuous contrast
change, then for any nonempty set $I\subset\RR$, $\inf_{x\in
I}g(x)=g( \inf_{x\in I}x)$ and $\sup_{x\in I}g(x)=g( \sup_{x\in
I}x)$. This implies that $T$ commutes with $g$. If we replace
$\mathcal B_{\fxx}$ with $\fxx+\mathcal B_{0}$, then it is easy to
check that $T$ is also translation invariant.
\end{proof}




\end{slide}
\begin{slide}


\begin{exercise}Use these last results to show that the extrema killer $T_a$ can be represented as a sup-inf function operator with the structuring elements
\[
\mathcal B=\{B\mid B\text{ is compact, connected, meas\,$(B)=a$, and } 0\in B\}. \finex
\]
\end{exercise}

\end{slide}

\begin{slide}
 Sup-inf operators can be
represented as inf-sup operators.
\begin{proposition}
Assume that $T$ can be represented by $Tu(\fxx)=\sup_{B\in\mathcal B}\inf_{\fxy\in\fxx+B}u(\fxy)$. Then $T$ also has an inf-sup representation
\begin{equation}\label{eq86}
Tu(\fxx)=\inf_{B\in\mathcal B'}\sup_{\fxy\in\fxx+B}u(\fxy),
\end{equation}
where $\mathcal B'$ is a new structuring set.
\end{proposition}

\end{slide}
\begin{slide}


\begin{proof}  A map $\phi:\mathcal
B\to\bigcup_{B\in\mathcal B}B$ with the property that $\phi(B)\in
B$ is called a \textit{selection function}. Let $\mathcal
S_{\mathcal B}$ denote the set of all selection functions. We then
have
\[
\sup_{B\in\mathcal B}\inf_{\fxy\in B}u(\fxy)=\inf_{\phi\in\mathcal S_{\mathcal B}}\sup_{B\in\mathcal B}u(\phi(B))=\inf_{\phi\in\mathcal S_{\mathcal B}}\sup_{\fxy\in\phi(\mathcal B)}u(\fxy).
\]
For $\phi\in\mathcal S_{\mathcal B}$ write $\phi(\mathcal B)=\bigcup_{B\in\mathcal B}\phi(B)$ and let $\mathcal D=\{\phi(\mathcal B)\mid\phi\in\mathcal S_{\mathcal B}\}$. Then
\[
\sup_{B\in\mathcal B}\inf_{\fxy\in B}u(\fxy)=\inf_{\phi(\mathcal B)\in\mathcal D}\sup_{\fxy\in\phi(\mathcal B)}u(\fxy).
\]
 \end{proof}

\end{slide}


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