%gmChap01.tex

\graphicspath{{./figures/}}

\chapter{The Heat Equation\label{Chap1}}

The heat equation is the prototype  of all the PDEs used in image
analysis. There are strong reasons for that and it is the aim of
this chapter to explain some of them.  Some more will be given in
Chapter \ref{Chap17}. Our first section  is dedicated to a simple
example of linear smoothing illustrating the relation between
linear smoothing and the Laplacian. In the next section,
 we prove the existence and uniqueness of its
solutions, which incidentally establishes the equivalence between
the convolution with a Gaussian and the heat equation.
\section{Linear smoothing and the Laplacian}

Consider a continuous and bounded function $u_0$ defined on
$\mathbb R^2$. If we wish to smooth $u_0$, then the simplest way
to do so without favoring a particular direction is to replace
$u_0(\fxx)$ with the average of the values of $u_0$ in a disk
$D(\fxx, h)$ of radius $h$ centered at $\fxx$. This means that we
replace $u_0(\fxx)$ with
\begin{equation}\label{Mh}
M_hu_0(\fxx)=\frac1{\pi h^2}\int_{D(\fxx, h)}u_0(\fxy)\,
\rmd\fxy=\frac1{\pi h^2}\int_{D(0, h)}u_0(\fxx+\fxy)\,\rmd\fxy.
\end{equation}

Although the operator $M_h$ is  quite simple, it exhibits
important characteristics of a general linear isotropic smoothing
operator. For example, it is localizable: As $h$ becomes small,
$M_h$ becomes more localized, that is, $M_hu_0(\fxx)$ depends only
on the values of $u_0(\fxx)$ in a small neighborhood of $\fxx$.
 Smoothing an image by averaging over a small symmetric area is
illustrated in Figure \ref{smoothing}.

\begin{figure}[!t]
\setkeys{Gin}{width=10cm}
\centerline{
\includegraphics{fig01.01.1.eps}
} \caption[Local averaging algorithm]{\label{smoothing} Local
averaging algorithm.   Left to right: original image; result of
replacing the grey level at each pixel by  the average of the grey
levels over the neighboring pixels. The shape of the neighborhood
is shown by the black spot displayed in the upper right-hand
corner.}
\end{figure}

Our objective is to point out the relation between the action of
$M_h$ and the action of the Laplacian, or the heat equation.  To
do so, we assume enough regularity for $u_0$, namely that it is
$C^2$. We shall actually prove in Theorem \ref{theorem2.2} that
under that condition
\begin{equation}\label{Mhx}
M_hu_0(\fxx)=u_0(\fxx)+\frac{h^2}8\Delta
u_0(\fxx)+h^2\varepsilon(\fxx,h),
\end{equation}
where $\varepsilon(\fxx,h)$ tends to 0  when $h\to 0.$   As we
have seen in the introduction, \eqref{Mhx} provides the
theoretical basis for deblurring an image by subtracting a small
amount of its Laplacian.  It also suggests that $M_h$ acts as  one
step forward in the heat equation starting with initial condition
$u_0$,
\begin{equation}\label{heat}
\frac{\partial u}{\partial t}(t,\fxx)=\frac{1}8\Delta
u(t,\fxx),\quad u(0,\fxx)=u_0(\fxx).
\end{equation}
This statement is made more precise in Exercise \ref{Mhetheat}.
  Equation {\eqref{Mhx} actually suggests that if we
let $n\to+\infty$ and at the same time require that $nh^2\to t$,
then
\begin{equation}\label{convergenceMhitere}
(M_h^nu_0)(\fxx)\to u(t,\fxx)
\end{equation}
where $u(t,x)$ is a solution of \eqref{heat}.

This heuristics justifies the need for a thorough  analysis of the
heat equation. The next chapter will prove that
(\ref{convergenceMhitere}) is true under fairly general
conditions. In the next section, we shall prove that the heat
equation has a unique solution for a given continuous initial
condition $u_0$, and that this solution at time $t$ is equal to
the convolution $G_t*u_0$, where $G_t$ is the Gaussian (Figure
\ref{gaussian}). The effect on level lines of smoothing with the
Gaussian is shown in Figure \ref{heatlevellines}.

\begin{figure}[!t]
\setkeys{Gin}{width=7cm} \centerline{
\includegraphics{fig01.02.1.eps}
}
\caption[The Gaussian]{\label{gaussian}The Gaussian in two dimensions.}
\end{figure}





%%%%%%%%%%%%%%%%%%%%%%%


\section{Existence and uniqueness of solutions of the heat equation}


\begin{definition}
We say that a function $g$ defined on $\RR^N$ is {\rm rapidly decreasing}, or {\rm has fast decay}, if for each
multi-index $\beta$ there is a constant $C$ such that
\[
|\fxx|^\beta| g(\fxx)|\leq C.
\] We say that $g$ belongs to the
Schwartz class $\mathcal S$ if $g\in C^\infty(\RR^N)$ and if $\partial^\alpha g$ has fast decay  for each multi-index
$\alpha$.
\end{definition}

%\vspace{-8pt}

\begin{proposition}\label{proposition1.4}
If $g\in\mathcal S$, then $g\in L^1(\RR^N)$, that is,
$\int_{\RR^N}|g(\fxx)|d\fxx<+\infty$. For each pair of
multi-indices $\alpha,\beta$, the function
$\fxx^\beta\partial^\alpha g$ also belongs to $\mathcal S$, and
$\partial^\alpha g$ is uniformly continuous on $\RR^N$.
\end{proposition}
\begin{proof}
The second statement follows from the Leibnitz rule for differentiating a product.  By the  definition of $\mathcal S$,
there is a constant $C$ such that $|\fxx|^{N+2}|g(\fxx)|\leq C$. Thus there is another $C$ such that $|g(\fxx)|\leq
{C}/({1+|\fxx|^{N+2}})$; since ${C}/({1+|\fxx|^{N+2}})\in L^1(\RR^N)$, $g\in L^1(\RR^N)$. Finally, note that
$|\partial^\alpha g(\fxx)|\to0$ as $|\fxx|\to\infty$. But any continuous function on $\RR^N$ that tends to zero at
infinity is uniformly continuous.\hfill\end{proof}

\begin{proposition}[The Gaussian and the heat
equation]\label{gaussianheat} For all $t>0$, the  function
$\fxx\mapsto G_t(\fxx)=(1/{(4\pi t)^{N/2}}){\rm e}^{-|\fxx|^2
/4t}$ belongs to $\mathcal S$ and satisfies the heat equation
\[
\frac{\partial G_t}{\partial t}-\Delta G_t=0.
\]
\end{proposition}
\begin{proof}
It is sufficient to prove the first  statement for the function
$g(\fxx)={\rm e}^{-|\fxx|^2}$. An induction argument shows that
$\partial^\alpha g(\fxx)=P_\alpha (\fxx){\rm e}^{-|\fxx|^2}$,
where $P_\alpha (\fxx)$ is a polynomial of degree $|\alpha|$ in
the variables $x_1, x_2,\ldots,x_N$. The fact that, for every
$k\in\mathbb N$, $x^k{\rm e}^{-x^2}\to0$ as $|x|\to+\infty$
finishes the proof. Differentiation shows that $G_t$ satisfies the
heat equation.\hfill\end{proof}

\begin{exercise}
Check that $G_t$ is solution of the heat equation. \finex
\end{exercise}
Linear image filtering is mainly done by convolving an image $u$
with a positive integrable kernel $g$. This means that the
smoothed image is given by the function $g*u$ defined as
\[
g*u(\fxx)=\int_{\RR^N} g(\fxx-\fxy)u(\fxy)\,\rmd\fxy=\int_{\RR^N}
g(\fxy)u(\fxx-\fxy)\,\rmd\fxy.
\]
\begin{exercise}
  Prove that the convolution, when it makes sense, is translation invariant. This means that
$g*u(\fxx-\fxz)=g_{\fxz}*u(\fxx)$, where $g_{\fxz}(\fxx)=g(\fxx-\fxz)$.\finex
\end{exercise}
\begin{exercise}
Check that $G_t*G_s=G_{t+s}$. \finex
\end{exercise}

 Linear filtering with the Gaussian at several
scales is illustrated in Figure \ref{blurs}. The next result establishes properties of the convolution that we need for
our treatment of the heat equation.

\begin{figure}[!t]
\setkeys{Gin}{width=11cm} \centerline{
\includegraphics{fig01.04.1.eps}
} \caption[Convolution with Gaussian kernels  (heat
equation)]{\label{blurs}Convolution with Gaussian kernels (heat
equation).  Displayed from top-left to bottom-right are the
original image and the results of convolutions with Gaussians of
increasing variance. A grey level representation of the
convolution kernel is put on the right of each convolved image to
give an idea of the size of the involved neighborhood.}
\end{figure}

\begin{proposition}\label{proposition1.6}
Assume that $u\in \mathcal F$ and that $g\in L^1(\RR^N)$. Then the
function $g*u$ belongs to $\mathcal F$ and  satisfies the
inequality
\begin{equation}\label{g*u1}
\|g*u\|_{\mathcal F}\leq\|g\|_{L^1(\RR^N)}\|u\|_{\mathcal F}.
\end{equation}
\end{proposition}
\begin{proof}
\[
|g*u(\fxx)| \leq\int_{\RR^N}|g(\fxx-\fxy)||u(\fxy)|\, \rmd\fxy\leq\|u\|_{\mathcal F}\int_{\RR^N}|
g(\fxx-\fxy)|\,\rmd\fxy=\|u\|_{\mathcal F}\|g\|_{L^1(\RR^N)}.
\]
 \hfill\end{proof}

%\vspace{-12pt}
\begin{exercise}
Verify that $g*u$ indeed is continuous and  tends to $u(\infty)$ at infinity : this a direct application of Lebesgue
Theorem. \finex
\end{exercise}
  We are now going to focus on kernels that, like the Gaussian, belong to $\mathcal S$.

\begin{proposition}\label{proposition1.8}
If $u\in \mathcal F$ and $g\in\mathcal S$, then $g*u\in
C^{\infty}(\RR^N)\cap \mathcal F$ and
\begin{equation}\label{1.19}
\partial^{\alpha}(g*u)=(\partial^{\alpha}g)*u
\end{equation}
for every multi-index $\alpha$.
\end{proposition}

\begin{proof}
Since $g\in\mathcal S$, $g$ is in  $ L^1(\RR^N)$, and so is $\partial^{\alpha}g$ for any multi-index $\alpha$
(Proposition \ref{proposition1.4}). Thus by Proposition \ref{proposition1.6}, $(\partial^{\alpha}g)*u$ belongs to
$\mathcal F$. To prove \eqref{1.19}, it is sufficient to prove it for $\alpha=(1,0,\ldots,0)$. Indeed, we know that
$\partial^{\alpha}g$ is in $\mathcal S$ if $g$ is in $S$, so the general case follows from the case
$\alpha=(1,0,\ldots,0)$ by induction. Letting ${\bm e}_1=(1,0,\ldots,0)$ and using Taylor's formula with Lagrange's
form for the remainder, we can write

\begin{equation}\label{proof18}
\begin{split}
g*u(\fxx+h{\bm e}_1)-g*u(\fxx)=&\int_{\RR^N}(g(\fxx+h{\bm e}_1-\fxy)-g(\fxx-\fxy))u(\fxy)\,\rmd\fxy\\
=&\int_{\RR^N}(g(\fxy+h{\bm e}_1)-g(\fxy))u(\fxx-\fxy)\,\rmd\fxy\\
=&\,\,h\!\!\int_{\RR^N}\frac{\partial g}{\partial x_1}(\fxy)u(\fxx-\fxy)\,\rmd\fxy\\&+\frac{h^2}{2}\int_{\RR^N}\frac{\partial^2 g}{\partial x_1^2}(\fxy+\theta(\fxy) h{\bm e}_1)u(\fxx-\fxy)\,\rmd\fxy,
\end{split}
\end{equation}
where $0\leq\theta(\fxy)\leq 1$. To complete the proof, we wish to have a bound on the last integral that is
independent of $\fxx\in C$. This last integral is of the form $f*u$, where $f$ is defined by $f(\fxy)=({\partial^2
g}/{\partial x_1^2})(\fxy+\theta(\fxy) h{\bm e}_1)$. Since $g\in\mathcal S$, ${\partial^2 g}/{\partial
x_1^2}\in\mathcal S$, and from this it is a simple computation to show that $f$ is bounded and decays fast  at
infinity. Having done this, Proposition \ref{proposition1.6} applies, and we deduce that $g*u$ is differentiable in
$x_1$ and that ${\partial( g*u)}/{\partial x_1}=({\partial g}/{\partial x_1})*u$.\hfill\end{proof}
%\vspace{-12pt}
\begin{proposition} \label{dirac}
Assume that $g$ decreases rapidly at infinity,  that $g(\fxx)\geq
0$ for all $\fxx\in\RR^N$, and that $\int_{\RR^N}g(\fxx)\,\rmd\fxx
= 1$ and set, for $t>0$, $g_t(\fxx)=(1/ {t^N})g(\fxx/ t)$. Then:
 If $u_0\in \mathcal F$, $g_t*u_0$ converges to $u_0$
uniformly as $t \to 0$. In addition, we have a maximum principle~:
\begin{equation}
\label{meanvalue}
\inf_{\fxx\in \RR^N}u_0(\fxx)\leq g_t*
u_0(\fxx)\leq\sup_{\fxx\in \RR^N}u_0(\fxx).
\end{equation}
 \end{proposition}

\begin{proof}
  Note first that $g_t$ is normalized so that
\begin{equation}\label{norm}
\int_{\RR^N}g_t(\fxy)\,\rmd\fxy=1.
\end{equation}
A change of variable $\fxx\to t\fxx$ and an application of Lebesgue's theorem shows that, for any $\eta>0$,
\begin{equation}\label{concentration}
\int_{|\fxy|\geq\eta}g_t(\fxy)\,\rmd\fxy\to0\;\text{ as }\;t\to0.
\end{equation}
Using \eqref{norm}, we have
\begin{equation}\label{diff}
g_t*u_0(\fxx)-u_0(\fxx)=\int_{\RR^N}g_t(\fxy)(u_0(\fxx-\fxy)-u_0(\fxx))\,\rmd\fxy.
\end{equation}
As already mentioned,  $u_0\in \mathcal F$ is uniformly
continuous. Thus, for any $\varepsilon>0$, there is an
$\eta=\eta(\varepsilon)>0$ such that
$|u_0(\fxx-\fxy)-u_0(\fxx)|\leq\varepsilon$ when $|\fxy|\leq\eta$.
Using this inequality, we have
\begin{align*}
|g_t*u_0(\fxx)-u_0(\fxx)|\leq&\int_{|\fxy|<\eta}g_t(\fxy)|u_0(\fxx-\fxy)-u_0(\fxx)|\,\rmd\fxy\\&+\int_{|\fxy|\geq\eta}g_t(\fxy)|u_0(\fxx-\fxy)-u_0(\fxx)|\,\rmd\fxy\\
\leq&\varepsilon\int_{|\fxy|<\eta}g_t(\fxy)\,\rmd\fxy+2\|u\|_{L^{\infty}(C)}\int_{|\fxy|\geq\eta}g_t(\fxy)\,\rmd\fxy.
\end{align*}
By \eqref{norm}  and \eqref{concentration}, we conclude that $g_t*u$ tends to $u_0$ uniformly in $\fxx$ as $t\to0$.
Relation \eqref{meanvalue} is an immediate consequence of the assumption that $g_t(\fxx)\geq0$ and equation
\eqref{norm}.
 \hfill\end{proof}

\begin{lemma}\label{convergenceuniformeversinfini}
Let $u_0\in  \mathcal F$ and  $u(t,\fxx)=(G_t\ast  u_0)(\fxx).$
Then for  every $t_0>0$, $u(t,\fxx)\to u_0(\infty)$ uniformly for
$t\leq t_0$ as $\fxx\to  \infty$.
\end{lemma}

\begin{proof}
By assumption,
\begin{equation}\label{limiteinfini}
\forall \varepsilon>0,\;\exists R, \; |\fxx|\geq R\Rightarrow
|u_0(\fxx)-u_0(\infty)|< \varepsilon.
\end{equation}
Because of the fast decay of the gaussian at infinity (or using  Lebesgue's theorem, as in the former proof), we have
\begin{equation}\label{gaussienneconcentre}
\forall\varepsilon>0,\;\exists r(\varepsilon),\; r\geq r(\varepsilon) \Rightarrow\int_{|\fxy|\geq
r}G_{t_0}(\fxy)d\fxy<\varepsilon.
\end{equation}
By using $\int G_t(\fxy)d\fxy=1,$ we  deduce that
\begin{equation}\label{coupeendeux}|u(t,\fxx)-u(\infty)| \leq \!\!
\int_{|\fxy|\leq r}\!\!\!\!\!\!G_t(\fxy)|u_0(\fxx-\fxy)-u_0(\infty)|d\fxy+\!\int_{|\fxy|\geq
r}\!\!\!\!\!\!G_t(\fxy)|u_0(\fxx-\fxy)-u_0(\infty)|d\fxy.
\end{equation}
Using \eqref{gaussienneconcentre}, the second term in \eqref{coupeendeux} is bound from above for $r\geq
r(\varepsilon)$ and $t\leq t_0$ by
$$(2\sup|u_0|)\int_{|\fxy|\geq r}G_{t_0}(\fxy)
\leq (2\sup|u_0|)\varepsilon.$$ Fix therefore $r\geq
r(\varepsilon)$. Then using $\int G_t=1$, the first term in
\eqref{coupeendeux} is bound by $\varepsilon$ by
\eqref{limiteinfini} for $|\fxx|\geq R+r$.

\end{proof}
\begin{lemma}\label{deriveetemporellegaussienne}
Let $u_0\in \mathcal F$ and $G_t$ the  gaussian. Then $$({\partial G_t}/{\partial t})*u_0={\partial
(G_t*u_0)}/{\partial t}.$$
\end{lemma}

\begin{proof}
Proposition \ref{proposition1.8} does not apply directly, since it
applies to the spatial partial derivatives of $G_t$ but not to the
derivative with respect to $t$.  Observe, however, that a slight
modification of the proof of this proposition does the job:
Replace $g$ with $G_t$ and $\fxx_1$ with $t$. Then the crux of the
matter is to notice that, given an interval $0<t_0<t_1$, there is
a rapidly decreasing function $f$ such that $ |({\partial^2
G_t}/{\partial t^2})(t+\theta(t)h, \fxy)|\leq f(\fxy) $ uniformly
for $t\in[t_0,t_1]$, where $f$ depends  on $t_0$ and $t_1$ but not
on $h$. Then Proposition \ref{proposition1.6} applies, and the
last integral in equation \eqref{proof18} is uniformly bounded.
\end{proof}

All of the tools are in place to state and prove the main theorem
of this chapter.
\begin{theorem}[Existence and uniqueness of solutions of the heat equation]
\label{existenceuniquenessheat} Assume that $u_0\in \mathcal F$
and define for $t>0$ and $\fxx\in \RR^N$,
$u(t,\fxx)=(G_t*u_0)(\fxx)$, $u(t,\infty)=u_0(\infty)$ and
$u(0,\fxx)=u_0(\fxx)$. Then
 \begin{description}
\item[(i)] $u$ is $C^\infty$ and bounded on
$(0,+\infty)\times\RR^N$;

\item[(ii)] $\fxx\to u(t,\fxx)$ belongs to $\mathcal F$ for every
$t\geq 0$;

\item[(iii)] for any $t_0\geq 0$,  $u(t,\fxx)$ tends uniformly for
$t\leq t_0$ to $u(\infty)$ as $\fxx\to \infty$;

\item[(iv)] $u(t,\fxx)$ tends uniformly to $u_0(\fxx)$ as $t\to
0$;

\item[(v)]  $u(t,\fxx)$ satisfies the heat equation with initial
value $u_0$;
\begin{equation}\label{heatsol}
\frac{\partial u}{\partial t}=\Delta u\quad\text{and}\quad u(0,\fxx) =u_0(\fxx);
\end{equation}

\item[(vi)] More specifically,
\begin{equation}\label{unibound}
\sup_{\fxx\in\RR^N,\;t\geq 0}|u(t,\fxx)|\leq \|u_0\|_{\mathcal F}.
\end{equation}
\end{description}

 Conversely, given  $u_0\in \mathcal F$,
$u(t,\fxx)=(G_t*u_0)(\fxx)$ is the only $C^2$ bounded solution $u$
of \eqref{heatsol} that satisfies properties (ii)-(v).
\end{theorem}

\begin{proof}  Let us prove properties (i)-(vi).
 For each $t>0$, $G_t\in\mathcal S$,
so by Proposition
\ref{proposition1.8} and Lemma \ref{deriveetemporellegaussienne},
\begin{equation}\label{solvesheat}
\frac{\partial u}{\partial t}-\Delta u=
u_0*\bigg(\frac{\partial G_t}{\partial t}-\Delta G_t\bigg).
\end{equation}
Proposition \ref{proposition1.8} also tells us that $u(t,\cdot)\in
C^{\infty}(\RR^N)\cap \mathcal F$ for each $t>0$. The right-hand
side of \eqref{solvesheat} is zero by Proposition
\ref{gaussianheat}, and the fact that
$|u(t,\fxx)-u_0(\fxx)|\to0\;\text{uniformly as }t\to0$ follows
from Proposition \ref{dirac}. The inequality \eqref{unibound} is a
direct application of Proposition \ref{proposition1.6}. Relation
(iii) comes from  Lemma \ref{convergenceuniformeversinfini}.

\medskip\noindent {\bf Uniqueness proof}.
 If
both $v$ and $w$ are solutions of the heat equation with the same
initial condition $u_0\in \mathcal F$, then $u=v-w$ is in
$\mathcal F$ and satisfies \eqref{heatsol} with the initial
condition $u_0(\fxx)=0$ for all $\fxx\in\RR^N$.    Also, by the
assumptions of (ii), $u$ is  bounded on $[0,+\infty)\times\RR^N$
and is $C^2$ on $(0,+\infty)\times\RR^N$. We wish to show that
$u(t,\fxx)=0$ for all $t>0$ and all $\fxx\in\RR^N$. Assume that
this is not the case. Then there is some point $(t,\fxx)$ where
$u(t,\fxx)\not=0$. Assume that $u(t,\fxx)>0$, by changing $u$ to
$-u$ if necessary.

\begin{figure}[!t]
\setkeys{Gin}{width=10cm}
\centerline{
\includegraphics{fig01.05.1.eps}}
\caption[Level lines and the heat equation]{\label{heatlevellines}Level lines and the heat equation. Top, left to right: original 410$\times$270 grey level image; level lines of original image for levels at multiples of 12. Bottom, left to right: original image smoothed by the heat equation (convolution with the Gaussian). The standard deviation of the Gaussian is 4, which means that its spatial range is comparable to a disk of radius 4.  The image gets blurred by the convolution, which averages grey level values and removes all sharp edges.  This can be appreciated on the right, where we have displayed all level lines for levels at multiples of 12. Note how some level lines on the boundaries of the image have split into parallel level lines that have drifted away from each other. The image has become smooth, but it is losing its structure.}
\end{figure}


We now consider the function $u^{\varepsilon}$ defined by $u^{\varepsilon}(t,\fxx)={\rm e}^{-\varepsilon t}u(t,\fxx)$.
This function tends to zero uniformly in $\fxx$ as $t\to 0$ and as $t\to+\infty$. It also tends uniformly to zero for
each $t\leq t_0$ when $\fxx\to\infty$. These conditions imply that $u^{\varepsilon}$ attains its supremum at some point
$(t_0,\fxx_0)\in(0,+\infty)\times \RR^N$, and this means that $\Delta u^{\varepsilon}(t_0,\fxx_0)={\rm e}^{-\varepsilon
t}\Delta u(t_0,\fxx_0)\leq0$ and $({\partial u^{\varepsilon}}/{\partial t})(t_0,\fxx_0)=0$. Here is the payoff: Using
the fact that $u$ is a solution of the heat equation, we have the following relations:
\begin{align*}
0=\frac{\partial u^{\varepsilon}}{\partial t}(t_0,\fxx_0)&=-\varepsilon
 u^{\varepsilon}(t_0,\fxx_0)+{\rm e}^{-\varepsilon t}\frac{\partial u}{\partial t}(t_0,\fxx_0)\\
&=-\varepsilon u^{\varepsilon}(t_0,\fxx_0)+{\rm e}^{-\varepsilon
t}\Delta u(t_0,\fxx_0)\leq -\varepsilon
u^{\varepsilon}(t_0,\fxx_0)<0.
\end{align*}
This contradiction completes the uniqueness proof.\hfill\end{proof}



\section{Exercises}

\begin{exercise}\label{Mhetheat} The  aim of this exercise is to prove relation
\eqref{Mhx} and its  consequence: A local average is equivalent to one step forward of the heat equation. Theorem
\ref{theorem2.2} yields actually a more general statement.

\medskip
\noindent 1) Expanding $u_0$ around the point $\fxx$ using
Taylor's formula, write
\begin{equation}\label{taylor}
u_0(\fxx+\fxy)=u_0(\fxx)+Du_0(\fxx)\cdot\fxy+\frac1{2}D^2u_0(\fxx)(\fxy,\fxy)+o(|\fxy|^2).
\end{equation}
Expand the various terms using the coordinates $(x,y)$ of $\fxx$.

\medskip
\noindent 2)  Apply $M_h$ to both sides of this expansion and
deduce relation (\ref{Mhx}).

\medskip \noindent 3) Assume $u_0\in\mathcal  F$ and consider the solution $u(t,\fxx)$
of the heat equation (\ref{heat})
 Then, for fixed $t_0>0$ and $\fxx$,
  apply $M_h$ to the  function   $u^{t_0} : \fxx\to u(t_0,\fxx)$ and
  write equation (\ref{Mhx})
  for  $u^{t_0}$. Using that $u(t,\fxx)$ is  a solution of the heat
  equation and its Taylor expansion between $t_0$ and $t_0+h$, deduce
  that
\begin{equation}\label{Mhut}
M_hu(t_0,\fxx)=u(t_0+h^2, \fxx)+h^2\varepsilon(t_0,\fxx,h).
\end{equation}\finex
\end{exercise}

\begin{exercise}\label{exodefSN} Consider
the sphere $S_N=\{\fxz\in \RR^{N+1},\; ||\fxz||=1\}.$  Prove that the mapping $T:\; \RR^N\cup\{\infty\}\to S_N$ defined
in Proposition \ref{defSN} by
$$T(\fxx)=
\left(\frac{2\fxx}{1+\fxx^2},\;\frac{\fxx^2-1}{\fxx^2+1}\right),\;
T(\infty)=(0,1).
$$
is a homeomorphism.
\end{exercise}

%\begin{exercise}
% A natural norm for $\mathcal F\cap C^1$ is
%\begin{equation}
%  \|u\|_{\mathcal F\cap C^1}=
%  \sup_{\fxx\in \RR^N}|u(\fxx)|+|Du(\fxx)|.
%  \end{equation}
%  Prove that $\mathcal F\cap C^1$ is complete, namely that if
%  $u_n\to u$ for the preceding norm, then $u(\fxx)$ tends to a constant
%   and $Du(\fxx)$
%  tends to zero as $|\fxx|$ tends to infinity.
%\end{exercise}

\begin{exercise}Let $u_0$ be a continuous function defined
on $\RR^N$ having the property that there exist a constant
$C>0$ and an integer $k$ such that
\[
|u_0(\fxx)|\leq C(1+|\fxx|^k)
\]
for all $\fxx\in\RR^N$. Show that the function $u$ defined by
$u(t,\fxx)=G_t*u_0(\fxx)$ is well defined and $C^\infty$ on
$(0,\infty)\times\RR^N$ and that it is a classical solution of the
heat equation. Hints: Everything follows from the fact that the
Gaussian and all of its derivatives decay exponentially at
infinity.  \finex
\end{exercise}

\begin{exercise}\label{Riesz}
We want to prove the general principle that any linear, translation invariant and continuous operator $T$ is a
convolution, that is $Tu=g*u$ for some kernel $g$. This is one of the fundamental principles of both mechanics and
signal processing, and it has many generalizations that depend on the domain, range, and continuity properties of $T$.
For instance, assume that $T$ is translation invariant (commutes with translations) and is continuous from $L^2(\RR^N)$
into $L^{\infty}(\RR^N) \cap C^0(\RR^N)$. Show that $Tu=g*u$, where the convolution kernel $g$ is in $L^2(\RR^N)$. This
is a direct consequence
 of Riesz theorem, which states that every bounded linear functional on $L^2(\RR^N)$ has the form $f\mapsto \int_{\RR^N}f(\fxx)g(\fxx)\,\rmd\fxx$ for
some $g\in L^2(\RR^N)$. Show that if $u\geq0$ ($u(\fxx)\geq0$ for all $\fxx$) implies $Tu\geq0$, then $g\geq0$. \finex
\end{exercise}


\section{Comments and references}
\paragraph*{The heat equation.}One should not conclude from
Theorem \ref{existenceuniquenessheat} that the solutions of the
heat equation are always unique. The assumption in (ii) that the
solution was bounded is crucial. In fact, without this assumption,
there are solutions $u$ that grow so fast that $gu$ is not in
$L^1(\RR^N)$ for $g\in\mathcal S$ (see, for example, \cite[page
217]{taylor}).  The existence and uniqueness proof of Theorem
\ref{existenceuniquenessheat} is classic and can be found in most
textbooks on partial differential equations, such as Evans
\cite{Evans:1998:PDE}, Taylor \cite{taylor}, or Brezis
\cite{Brezis:1983:AFT}.

\paragraph*{Convolution.}The heat equation---its
solutions and their uniqueness---has been the main topic in this chapter, but to approach this, we have studied several
aspects of the convolution, such as the continuity property  \eqref{g*u1}. We also noted that the convolution commutes
with translation. Conversely, as a general principle, any linear, translation invariant and continuous operator $T$ is
a convolution, that is, $Tu=g*u$ for some kernel $g$.  This is a direct consequence of  a result discovered
independently by F. Riesz and M. Fr\'{e}chet in 1907 (see \cite[page 61]{RN} and exercise \ref{Riesz}). Since we want
smoothing to be translation invariant and continuous in some topology, this means that linear smoothing
operators---which are called filters in the context of signal and image processing---are described by their convolution
kernels. The Gaussian serves as a model for linear filters because it is the only one whose shape is stable under
iteration. Other positive filters change their shape when iterated. This fact will be made precise
 in the next chapter where we show
that a large class of iterated linear filters behaves asymptotically as a convolution with the Gaussian.

\paragraph*{Smoothing and the Laplacian.}
One of the first tools proposed in the  early days of image
processing in the 1960s came, not surprisingly, directly from
signal processing. The idea was to restore an image by averaging
the gray levels locally (see, for example, \cite{Golay:1972:SDL}
and \cite{Helstrom:1967:IRM}). The observation that the difference
between an image and its local average is proportional to the
Laplacian of the image has proved to be one of the most fruitful
contributions to image processing. As noted in the Introduction,
this method for deblurring an image was introduced by Kovasznay
and Joseph in 1955 \cite{Kovasznay:1955}, and it was studied and
optimized by Gabor in 1965 \cite{Gabor:1965} (information taken
from \cite{Lindenbaum:1994:GCI}). (See also \cite{Hummel:1983:GBH}
and \cite{Hummel:1987:DGB}.) Burt and Adelson based their
Laplacian pyramid algorithm on this idea, and this was one of the
results that led to multiresolution analysis and wavelets
\cite{Burt:1983:LPC}.

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