%gmChap11.tex. 

\graphicspath{{./figures/chapter11/}}

\chapter[Asymptotic Behavior of Contrast-Invariant Operators in $\mathbb R^N$]{Asymptotic Behavior of Contrast-Invariant Isotropic Operators, Dimension N}

We are going to generalize to $N$ dimensions the asymptotic results of Chapter 10. Our aim is to show that the action of any isometric inf-sup filter, when properly scaled, is a motion of the $N$-dimensional image that is controlled by its principal curvatures. Once we have established the main asymptotic result, we will, as in Chapter 10, analyze several multidimensional filters. In particular, we will relate the median filter to the mean curvature of the level surface. The action of median filtering in three dimensions is illustrated in Figures 11.1 and 11.2.

\section{Asymptotic behavior theorem in $\RR^N$}

The main theorem, Theorem 11.1, is a straightforward generalization of Theorems 10.1 and 10.2; the proof is basically the same, once the notation is established. For $\fxx\in\RR^N$, we write $\fxx=(x,y_2,\ldots,y_N)=(x,\fxy)$, so $\fxy\in\RR^{N-1}$. We let $\fxs=(s_2,\ldots,s_N)$ and define
\begin{equation}\label{eq121}
H(\fxs)=T[x+s_2y_2^2+\cdots+s_Ny_N^2](0),
\end{equation}
where the operator $T$ is defined by
\[
Tu(\fxx)=\inf_{B\in\mathcal B} \sup_{\fxy\in\fxx+ B}u(\fxy).
\]
Throughout the chapter, the set of structuring elements $\mathcal B$ will be bounded and isotropic. Thus, we assume that the elements of $\mathcal B$ are contained in the ball $D(0,M)$ and that, for any linear isometry $R$, $RB\in\mathcal B$ if and only if $B\in\mathcal B$. Arguments similar to those given in section 10.1 show that $H$ is a nondecreasing function is each variable $s_i$, and that it satisfies the Lipschitz condition 
\begin{equation}\label{eq122}
|H(\fxs_1)-H(\fxs_2)|\leq M^2|\fxs_1-\fxs_2|.
\end{equation}

Let $u:\RR^N\to\RR$ be $C^3$ and assume that $Du(\fxx)\not=0$. Then we denote the vector whose terms are the $N-1$ principal curvatures of the level surface $\{\fxy\mid u(\fxy)=u(\fxx)\}$ that passes through $\fxx$
by ${\bm \kappa}(u)(\fxx)={\bm \kappa}(u)=(\kappa_2,\ldots,\kappa_N)$. The terms $\kappa_i(u)(\fxx)|Du(\fxx)|$ are then the eigenvalues of the restriction of $D^2u(\fxx)$ to $Du(\fxx)^\perp$.  (We refer to section 4.6 for information about principal curvature.)

As in Chapter 10, we will state and prove the main results, given in Theorem 11.1, for operators written in their inf-sup form. We leave it as exercise to show that the relation $T^-u=-T(-u)$ implies the result for sup-inf operators.

\begin{theorem}Let $\mathcal B$ be a family of structuring elements belonging to $\RR^N$ and assume that $\mathcal B$ is bounded and isotropic. Define the operator $T$ and its scaled versions $T_h$, $0<h\leq1$, by 
\[
Tu(\fxx)=\inf_{B\in\mathcal B} \sup_{\fxy\in\fxx+ B}u(\fxy)\quad\quad\text{and}\quad\quad T_hu(\fxx)=\inf_{B\in h\mathcal B} \sup_{\fxy\in\fxx+ B}u(\fxy),
\]
respectively, and write $H(\fxs)=T[x+s_2y_2^2+\cdots+s_Ny_N^2](0)$. Then for every $C^3$ function $u:\RR^N\to\RR$, we have the following results:
\begin{enumerate}[$(i)$]\setlength\itemsep{2pt}
\item It is always true that 
\[
T_hu(\fxx)=u(\fxx)+hH(0)|Du(\fxx)|+O(\fxx, h^2);
\]
\item If $H(0)=0$, then on every compact set $K$ contained in $\{\fxx\mid Du(\fxx)\not=0\}$
\[
T_hu(\fxx)=u(\fxx)+hH\Bigg(h\frac1{2}{\bm \kappa}(u)(\fxx)\Bigg)|Du(\fxx)|+O(\fxx, h^3)
\]
where $|O(\fxx, h^3)|\leq C_Kh^3$ for some constant $C_K$ that depends only on $u$ and $K$;
\item If $H(0)=0$, then on every compact set $K\subset\RR^N$,
\[
|T_hu(\fxx)-u(\fxx)|\leq \frac{h^2M^2}{2}\|D^2u(\fxx)\|+O(\fxx,h^3),
\]
where $|O(\fxx, h^3)|\leq C_Kh^3$ for some constant $C_K$ that depends only on $u$ and $K$.
\end{enumerate}
\end{theorem}

\begin{proof}The proof follows exactly the proofs of Theorems 10.1 and 10.2; we simply have to relate the notation used for the $N$-dimensional case to that used in the two-dimensional case. We begin by assuming that $Du(\fxx)\not=0$. We then establish the local coordinate system at $\fxx$ defined by ${\bm i}_1=Du(\fxx)/|Du(\fxx)$ and ${\bm i}_2,\ldots,{\bm i}_N$, where ${\bm i}_2,\ldots,{\bm i}_N$ are the eigenvectors of the restriction of $D^2u(\fxx)$ to the hyperplane $Du(\fxx)^\perp$. Then in a neighborhood of $\fxx$ we can expand $u$ as follows:
\begin{equation}\label{eq123}
u(\fxx+\fxy)=u(\fxx)+px+ax^2+b_2y_2^2+\cdots+b_Ny_N^2+(\fxc\cdot\fxy)x+R(\fxx,\fxy),
\end{equation}
where $\fxy=x{\bm i}_1+y_2{\bm i}_2+\cdots+y_N{\bm i}_N$, $p=|Du(\fxx)|>0$, and for $j=2,\ldots,N$, 
\begin{equation}
\begin{aligned}
a&=\frac{1}{2}\frac{\partial^2 u}{\partial x^2}(\fxx)=\frac{1}{2}D^2u(\fxx)({\bm i}_1,{\bm i}_1),\\
b_j&=\frac{1}{2}\frac{\partial^2 u}{\partial y_j^2}(\fxx)=\frac{1}{2}D^2u(\fxx)({\bm i}_j,{\bm i}_j),\\
c_j&=\frac{\partial^2u}{\partial x\partial y_j}(\fxx)=D^2u(\fxx)({\bm i}_1,{\bm i}_j).
\end{aligned}
\end{equation}
We can also write $b_j$ as
\begin{equation}\label{eq125}
b_j=\frac1{2}|Du(\fxx)|\kappa_j(u)(\fxx).
\end{equation}

For the proof of $(i)$, we write $u(\fxx+\fxy)=u(\fxx)+px+O(\fxx,|\fxy|^2)$ and just follow the steps of the proof of Theorem 11.1. The proof of $(ii)$ and $(iii)$ follows, step by step, the proof of Theorem 10.2. We need only make the following identifications: $cxy\leftrightarrow (\fxc\cdot\fxy)x$, $by^2\leftrightarrow b_2y_2^2+\cdots+b_Ny_N^2$, and ${\rm curv}(u)\leftrightarrow\kappa(u)$. 
\end{proof}

\begin{figure}[!t]
\setkeys{Gin}{width=10cm}
\centerline{
\includegraphics{fig11.01.1.eps}
}
\caption[Three-dimensional median filter]{Three-dimensional median filter. The original three-dimensional image (not shown) is of 20 slices of a vertebra. Three successive slices are displayed in the left column. The next column shows their level lines (multiples of 20). 
The third column shows these three slices after one iteration
of the median filter based on the three-dimensional ball of radius two. The resulting level lines are shown in the last column.
}
\end{figure}

\section{Asymptotic behavior of median filters in $\RR^N$}

The median filters we consider will be defined in terms of a continuous weight function $k:\RR^N\to[0,+\infty)$ that is radial, $k(\fxx)=k(|\fxx|)$, and that is normalized, $\int_{\RR^N}k(\fxx)\,\rmd\fxx=1$. Recall that, by definition,
\[
|B|_k=\int_B k(\fxx)\,\rmd\fxx
\]
and that the corresponding median operator is defined by
\begin{equation}\label{eq126}
\fcnM_ku(\fxx)=\sup_{|B|_k\geq1/2}\inf_{\fxy\in\fxx+B}u(\fxy).
\end{equation}
We also assume that $k$ is nonseparable, which is the case if $\{\fxx\mid k(\fxx)>0\}$ is connected. Then by Proposition 9.7, $\fcnM_ku=\fcnM_k^-u$ if $u$ is continuous. This means that we can represent $\fcnM_k$ as an inf-sup operator using the same structuring elements, namely, $\{B\mid |B|_k\geq1/2\}$. In fact, we need consider only these $B$ for which $|B|_k=1/2$. 

Define the scaled weight function $k_h$, $0<h\leq1$, by $k_h(\fxx)=h^{-N}k(\fxx/h)$. Then a change of variable shows that $|B|_k=1/2$ if and only if $|hB|_{k_h}=1/2$, and this implies that $(\fcnM_k)_h=\fcnM_{k_h}$ (see Exercise 10.5). Since we consider only one weight function at a time, there should be no confusion if we write $\fcnM_h$ for the scaled operator.

We analyzed the asymptotic behavior of a median filter in $\RR^2$ whose weight function was the characteristic function of the unit disk in Chapter 10. This can be generalized to $\RR^N$ by taking $k$ to be the normalized characteristic function of the unit ball, but we not do that here. We will go in a different direction by taking smooth weight functions. On the other hand, our analysis will not be as general as possible because this would be needlessly complicated. The $k$ we consider will be smooth ($C^\infty$) and have compact support. This means that the structuring elements are bounded and the results of Theorem 11.1 apply. On the other hand, $k$ that do not have compact support are important; consider, for example, the Gaussian.

As in the case of $\RR^2$, it is that action of $\fcnM_k$ on a ``test function'' that controls the first term of the expansion of $(\fcnM_k)_h$, and as before, we deal with this in a lemma. 

\begin{lemma}Let $k$ be a nonnegative radial function belonging to the Schwartz class $\mathcal S$. Assume that $\int_{\RR^N} k(\fxx)\,\rmd\fxx=1$ and that the support of $k$ is connected. Then the function $H(h\fxb)=\fcnM_k[x+h(b_2y_2^2+\cdots+b_Ny_N^2)](0)$ can be expressed as
\[
H(h\fxb)=hc_k\Big(\sum_{j=2}^Nb_j\Big)+O(h^2),
\]
where 
\[
c_k=\frac{\int_{\RR^{N-1}}y_2^2k(\fxy)\,\rmd\fxy}{\int_{\RR^{N-1}}k(\fxy)\,\rmd\fxy},
\]
$\fxy=(y_2,\ldots,y_N)$, and $\fxb=(b_2,\ldots,b_N)$.
\end{lemma}

\begin{proof}Before beginning the proof, note that we have not assumed that $k$ has compact support, so the result applies to the Gaussian, for example.

We will use the abbreviation $\fxb(\fxy,\fxy)=b_2y_2^2+\cdots+b_Ny_N^2$, since $\fxb$ is, in fact, a diagonal matrix. Our proof is based on an analysis of the function $f(\lambda, h)=|\levelset_\lambda(x+h\fxb(\fxy,\fxy))|_k$. Since $\levelset_\lambda(x+h\fxb(\fxy,\fxy))=\{(x,\fxy)\mid x+h\fxb(\fxy,\fxy)\geq\lambda\}$, we can express $f$ as an integral,
\[
f(\lambda,h)=\int_{\RR^{N-1}}\int_{\lambda-h\fxb(\fxy,\fxy)}^\infty k(x,\fxy)\,\rmd x\,\rmd\fxy.
\]
It follows from the assumption that $k$ is in the Schwartz class 
that $f:\RR^2\to\RR$ is bounded and $C^\infty$. Also, for every $h\in\RR$, $\lim_{\lambda\to-\infty}f(\lambda,h)=1$ and $\lim_{\lambda\to+\infty}f(\lambda,h)=0$. Thus, for every $h\in\RR$, there is at least one $\lambda$ such that $f(\lambda,h)=1/2$. In fact, there is only one such $\lambda$;  this is a consequence of the assumption that the $k$ is continuous and that its support is connected, which implies that it is nonseparable (see Exercise 9.2). To see that $\lambda$ is unique, assume that there are $\lambda<\lambda'$ such that $f(\lambda,h)=1/2$ and $f(\lambda',h)=1/2$. Then the two sets $\{(x,\fxy)\mid x+h\fxb(\fxy,\fxy)\geq\lambda'\}$ and $\{(x,\fxy)\mid x+h\fxb(\fxy,\fxy)\leq\lambda\}$ both have $k$-measure $1/2$, but their intersection is empty. This contradicts the fact that $k$ is nonseparable. This means that the relation $f(\lambda,h)=1/2$ defines implicitly a well-defined function $h\mapsto\lambda(h)$. 

Recall that $\fcnM_k$ was originally defined in terms of the superposition formula
\[
\fcnM_ku(\fxx)=\sup\{\lambda\mid\fxx\in\setM_k\levelset_\lambda u\}.
\]
This translates for our case into the relation
\[
\fcnM_k[x+h\fxb(\fxy,\fxy)](0)=\sup\{\lambda\mid0\in\setM_k\levelset_\lambda [x+h\fxb(\fxy,\fxy)]\}=\lambda(h)
\]
because $0\in\setM_k\levelset_\lambda [x+h\fxb(\fxy,\fxy)]$ if and only if $|\levelset_\lambda[x+h\fxb(\fxy,\fxy)]|_k\geq1/2$.

We are interested in the behavior of $h\mapsto\lambda(h)$ near the origin. The first thing to note is that $\lambda(0)=0$. To see this, write
\[
f(\lambda(0),0)=\int_{\RR^{N-1}}\int_{\lambda(0)}^\infty k(x,\fxy)\,\rmd x\,\rmd\fxy=\frac1{2}.
\]
Since $k$ is radial, the value $\lambda=0$ solves the equation $\int_{\RR^{N-1}}\int_{\lambda}^\infty k(x,\fxy)\,\rmd x\,\rmd\fxy=1/2$. We have just shown that this equation has a unique solution, so $\lambda(0)=0$.

\begin{figure}[!t]
\setkeys{Gin}{width=10cm}
\centerline{
\includegraphics{fig11.02.1.eps}
}
\caption[Median filtering of a three-dimensional image]{Median filtering of a three-dimensional image. The first image is a representation of the horizontal slices of a three-dimensional level surface of the three-dimensional image of a vertebra. Right to left, top to bottom: 1, 2, 5, 10, 20, 30, 60, 100 iterations of a three-dimensional median filter based on a ball with radius three. This scheme is a possible implementation of the mean curvature motion, originally proposed as such by Bence, Merriman, and Osher.}
\end{figure}

Now consider the first partial derivatives of $f$:
\begin{align}
\frac{\partial f}{\partial\lambda}(\lambda,h)&=-\int_{\RR^{N-1}}k\Big(\big((\lambda-h\fxb(\fxy,\fxy))^2+\fxy\cdot\fxy\big)^{1/2}\Big)\,\rmd\fxy.\label{eq11a}\\
\frac{\partial f}{\partial h}(\lambda,h)&=\int_{\RR^{N-1}}\fxb(\fxy,\fxy)k\Big(\big((\lambda-h\fxb(\fxy,\fxy))^2+\fxy\cdot\fxy\big)^{1/2}\Big)\,\rmd\fxy.\label{eq11b}
\end{align}
These functions are $C^\infty$ because $k$ is in the Schwartz class; also, $(\partial f/\partial\lambda)(0,0)\not=0$. Then by the implicit function theorem, we know that the function $h\mapsto\lambda(h)$ that satisfies $f(\lambda(h),h)=1/2$ is also $C^\infty$ and that 
\[
\lambda'(h)\frac{\partial f}{\partial\lambda}(\lambda(h),h)+\frac{\partial f}{\partial h}(\lambda(h),h)=0.
\]
Thus, for small $h$,
\[
\lambda'(h)=-\frac{\frac{\partial f}{\partial h}(\lambda(h),h)}{\frac{\partial f}{\partial\lambda}(\lambda(h),h)},
\]
and, using equations \eqref{eq11a} and\eqref{eq11b}, we see that
\[
\lambda'(0)=\frac{\int_{\RR^{N-1}}\fxb(\fxy,\fxy)k\big((\fxy\cdot\fxy)^{1/2}\big)\,\rmd\fxy}{\int_{\RR^{N-1}}k\big((\fxy\cdot\fxy)^{1/2}\big)\,\rmd\fxy}.
\]
Now expand $\lambda$ for small $h$:
\[
\lambda(h)=\lambda(0)+\lambda'(0)h+O(h^2).
\]
Since  $\int_{\RR^{N-1}}\fxb(\fxy,\fxy)k\big((\fxy\cdot\fxy)^{1/2}\big)\,\rmd\fxy=\big(\sum_{j=2}^{N-1}b_j\big)\int_{\RR^{N-1}}y_2^2 k\big((\fxy\cdot\fxy)^{1/2}\big)\,\rmd\fxy$, $H(h\fxb)=\lambda(h)$, and $\lambda(0)=0$, this proves the lemma.
\end{proof}

\begin{theorem}Let $k$ be a nonnegative radial function belonging to the Schwartz class $\mathcal S$. Assume that $\int_{\RR^N} k(\fxy)\,\rmd\fxy=1$ and that the support of $k$ is compact and connected. Then for every $C^3$ function $u:\RR^N\to\RR$, we have the following results:
\begin{enumerate}[$(i)$]\setlength\itemsep{2pt}  
\item On every compact set $K\subset\{\fxx\mid Du(\fxx)\not=0\}$,
\[
\fcnM_hu(\fxx)=u(\fxx)+h^2\frac1{2}c_k\Big(\sum_{i=2}^N\kappa_i(u)(\fxx)\Big)|Du(\fxx)|+O(\fxx,h^3),
\]
where $|O(\fxx,h^3)|\leq C_Kh^3$ for some constant that depends only on $u$ and $K$.
\item On every compact set $K\subset\RR^N$,
\[
|\fcnM_hu(\fxx)-u(\fxx)|\leq \frac{h^2M^2}{2}\|D^2u(\fxx)\|+O(\fxx,h^3),
\]
where $|O(\fxx, h^3)|\leq C_Kh^3$ for some constant $C_K$ that depends only on $u$ and $K$.
\end{enumerate}
\end{theorem}
\begin{proof}Recall that $\fcnM_h=(\fcnM_k)_h=\fcnM_{k_h}$. Since $k$ is not separable and since we are restricting the domain of $\fcnM$ to continuous functions, we can represent it as an inf-sup operator (Proposition 9.7). Furthermore, the set of structuring elements may be taken to be $\mathcal B=\{B\mid |B|_k=1/2, B\subset \text{support of }k\}$. Since the support of $k$ is compact, $\mathcal B$ is bounded. The set $\mathcal B$ is invariant under linear isometries of $\RR^N$ because $k$ is radial. Thus, Theorem 11.1 is applicable. We know from Lemma 11.2 that $H(0)=0$, so we can read $(ii)$ directly from Theorem 11.1$(iii)$.

By Lemma 11.2,  
\[
H(h{\bm\kappa}(u))=hc_k\Big(\sum_{i=2}^N\kappa_i(u)|Du|\Big)+O(h^2).
\]
From this and Theorem 11.1$(ii)$, we have
\[
\fcnM_hu(\fxx)=u(\fxx)+h^2\frac1{2}c_k\Big(\sum_{i=2}^N\kappa_i(u)(\fxx)\Big)|Du(\fxx)|+O(\fxx,h^3), 
\]
and we know that the estimate is uniform on any compact set $K\subset\{\fxx\mid Du(\fxx)\not=0\}$.
\end{proof}

As mentioned above, this is by no means the most general result. The following exercise generalizes the result in one direction.

\begin{exercise}Prove Lemma 11.2 and Theorem 11.3 assuming that $k$ has compact support and is $C^2$, radial, and nonseparable. \finex
\end{exercise}

\section{Applications in three dimensions}

This section contains several applications of Theorem 11.1 in three dimensions. A level surface of a $C^3$ function in three dimensions has two principal curvatures, and this provides an extra degree of freedom for constructing contrast-invariant operators base on curvature motion. We develop the applications in three exercises. For each case, we will assume that the principal curvatures $\kappa_1$ and $\kappa_2$ are ordered such that $\kappa_1\leq\kappa_2$. In each example, the set of structuring element $\mathcal B$ is constructed from a single set $B$ in $\RR^2$ by rotating $B$ is all possible ways, that is, $\mathcal B=\{RB\mid B\in\RR^2, R\in SO(3)\}$. For each example we write
\[
SI_hu(\fxx)=\sup_{B\in \mathcal B}\inf_{\fxy\in\fxx+hB}u(\fxy)
\]
and
\[
IS_hu(\fxx)=\inf_{B\in \mathcal B}\sup_{\fxy\in\fxx+hB}u(\fxy),
\]
where $0<h\leq1$.

\begin{exercise}Let $B$ be a segment of length 2 centered at the origin. Our aim is to show that 
\begin{align*}
IS_hu&=u+h^2\frac1{2}\kappa_1^+(u)|Du|+O(h^3),\\
SI_hu&=u+h^2\frac1{2}\kappa_2^-(u)|Du|+O(h^3),
\end{align*}
and that
\[
IS_h(SI_h)u=u+h^2\frac1{2}({\rm sign}(\kappa_1(u))+{\rm sign}(\kappa_2(u)))\min(|\kappa_1(u)|,|\kappa_2(u)|)+O(h^3).
\]
\begin{enumerate}[(i)]\setlength\itemsep{2pt}
\item The first step is to compute $H(h\fxb)$. One way to do this is to write $x=r\sin\phi$, $y_2=r\cos\phi\cos\theta$, $y_3=r\cos\phi\cos\theta$, and use an argument similar to that given in the proof of Theorem 10.5 to show that, for a fixed $\theta$ and small $h$, the ``inf-sup'' of 
\[
r\sin\phi+hb_2r^2\cos^2\phi\cos^2\theta+hb_3r^2\cos^2\phi\cos^2\theta
\]
always occurs at $\phi=0$. Then $H(h\fxb)=hH(\fxb)$ and
\[
H(\fxb)=\inf_{B\in\mathcal B}\sup_{\fxy\in B}\,(b_2y_2^2+b_3y_3^2)=\inf_{\theta}\sup_{0\leq r\leq1}r^2(b_2\cos^2\theta+b_3\sin^2\theta).
\]
Deduce that $b_2<0$ or $b_3<0$ implies $H(\fxb)=0$ and that $0\leq b_2\leq b_3$ implies $H(\fxb)=b_2$.
\item Since $H(0)=0$, deduce from Theorem 11.1 that 
\begin{equation}\label{eq1213}
IS_hu(\fxx)=u(\fxx)+h^2\frac1{2}\kappa_1^+(u)(\fxx)|Du(\fxx)|+O(\fxx,h^3).
\end{equation}
\end{enumerate}
\end{exercise}

\begin{exercise}Let $B$ be the two symmetric points $(1,0,0)$ and $(-1,0,0)$. Use the techniques of Exercise 11.2 to show that
\begin{align*}
IS_hu&=u+h^2\frac1{2}\min\{\kappa_1(u),\kappa_2(u)\}|Du|+O(h^3);\\
SI_hu&=u+h^2\frac1{2}\max\{\kappa_1(u),\kappa_2(u)\}|Du|+O(h^3);\\
IS_h(SI_h)u&=u+h^2\frac1{2}(\kappa_1(u)+\kappa_2(u))|Du|+O(h^3).\\
\end{align*}
The last formula shows that the operator $IS_h(SI_h)$ involves the mean curvature of $u$ at $\fxx$. \finex
\end{exercise}

\begin{exercise}Let $B$ consist of two orthogonal segments of length two centered at the origin. 
\begin{enumerate}[(i)]\setlength\itemsep{2pt}
\item Show that 
\begin{align*}
IS_hu&=u+h^2\frac1{2}\Big(\frac{\kappa_1(u)+\kappa_2(u)}{2}\Big)^+|Du|+O(h^3);\\
SI_hu&=u+h^2\frac1{2}\Big(\frac{\kappa_1(u)+\kappa_2(u)}{2}\Big)^-|Du|+O(h^3).
\end{align*}
\item Show that you can get the mean curvature by simply taking $B$ to be the four endpoints of the orthogonal segments. Check that another possibility for obtaining the mean curvature is to alternate these operators or to add them. \finex
\end{enumerate}
\end{exercise}

\section{Comments and references} 
The references for this chapter are essentially the same as those for Chapter 10. The main theorem on the asymptotic behavior of morphological filters was first stated and proved in \cite{Guichard:1995:PDE} and \cite{Guichard:1996:PDE}. The examples developed in Exercises 11.2, 11.3, and 11.4 have not been published elsewhere. The consistency of Gaussian smoothing followed by thresholding and mean-curvature motion was proved in increasing mathematical sophistication and generality by Merriman, Bence, and Osher \cite{Merriman:1992:DGM}, Mascarenhas \cite{Mascarenhas:1992:DGM}, Barles and Georgelin \cite{Barles:1995:SPC}, and Evans \cite{Evans:1993:CAM}. Our presentation is slightly more general than the ones cited because we allow any nonnegative weight function in the Schwartz class. The most general result was given by Ishii in \cite{Ishii:1994:GBM}.

