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\begin{document}


\begin{center}
  \Large{\textbf{Notes for 18.117\\Elliptic operators}}
\end{center}

\vspace{4ex}
\section{Differential operators on $\RR^n$}
\label{sec:1}

Let $U$ be an open subset of $\RR^n$ and let $D_k$ be the
differential operator,
%
\begin{displaymath}
  \frac{1}{\sqrt{-1}} \,\, \frac{\partial}{\partial x_k} \, .
\end{displaymath}
%
For every multi-index, $\alpha =\alpha_1 , \ldots , \alpha_n$, we
define
%
\begin{displaymath}
  D^{\alpha} = D^{\alpha_1}_1 \cdots D^{\alpha_n}_n \, .
\end{displaymath}
%
A differential operator of order $r$:
%
\begin{displaymath}
  P : \Cinf (U) \to \Cinf (U)\, ,
\end{displaymath}
%
is an operator of the form
%
\begin{displaymath}
  Pu = \sum_{|\alpha | \leq r} a_{\alpha} D^{\alpha}u \, , 
          \quad a_{\alpha} \in \Cinf (U) \, .
\end{displaymath}
%
Here $|\alpha | = \alpha_1 + \cdots \alpha_n$.

The \emph{symbol} of $P$ is roughly speaking its ``$r$\st{th} order
part''.  More explicitly it is the function on $U \times \RR^n$
defined by
%
\begin{displaymath}
  (x,\xi) \to \sum_{|\alpha |=r} a_{\alpha}(x) \xi^{\alpha}
      = : p (x,\xi) \, .
\end{displaymath}
%
The following property of symbols will be used to define the
notion of ``symbol'' for differential operators on manifolds.
Let $f: U \to \RR$ be a $\Cinf$ function.

\begin{theorem}
  \label{th:1.1}
The operator
%
\begin{displaymath}
  u \in \Cinf (U) \to e^{-itf} P e^{itf} u
\end{displaymath}
%
is a sum 
%
\begin{equation}
\label{eq:1.1}
\sum^r_{i=0} t^{r-i} P_i u
\end{equation}
%
$P_i$ being a differential operator of order~$i$ which doesn't
depend on~$t$.  Moreover, $P_0$ is multiplication by the function
%
\begin{displaymath}
  p_0 (x) = : p(x,\xi)
\end{displaymath}
%
with $\xi_i = \frac{\partial f}{\partial x_i}$, $i=1,\ldots n$.
\end{theorem}


\begin{proof}
  It suffices to check this for the operators~$D^{\alpha}$.
  Consider first $D_k$:
%
  \begin{displaymath}
    e^{-itf} D_k e^{itf}u = D_k u 
         + t \frac{\partial f}{\partial x_k} \, .
  \end{displaymath}
%
Next consider $D^{\alpha}$
%
\begin{eqnarray*}
  e^{-itf} D^{\alpha} e^{itf} u 
     &=& e^{-itf} (D^{\alpha_1}_1 \cdots
           D^{\alpha_n}_n)e^{itf}u\\
%
     &=& (e^{-itf} D_1 e^{itf})^{\alpha_1} \cdots
           (e^{-itf} D_n e^{itf})^{\alpha_n}u
\end{eqnarray*}
%
which is by the above
%
\begin{displaymath}
  \big( D_1 + t \frac{\partial f}{\partial x_1}\big)^{\alpha_1}
      \cdots
  \big( D_n +t \frac{\partial f}{\partial x_n} \big)^{\alpha_n}
\end{displaymath}
%
and is clearly of the form (\ref{eq:1.1}).  Moreover the $t^r$
term of this operator is just multiplication by
%
\begin{equation}
  \label{eq:1.2}
  \big( \frac{\partial}{\partial x_1}f\big)^{\alpha_1}\cdots
     \big( \frac{\partial f}{\partial x_n}\big)^{\alpha_n}\, .
\end{equation}

\end{proof}


\begin{corollary}
  \label{cor:1.2}
  If $P$ and $Q$ are differential operators and $p(x,\xi)$ and
  $q(x,\xi)$ their symbols, the symbol of $PQ$ is $p(x,\xi) \, q(x,\xi )$.

\end{corollary}



\begin{proof}
Suppose $P$ is of the order $r$ and $Q$ of the order $s$.  Then
%
\begin{eqnarray*}
  e^{-itf}PQ e^{itf} u 
     &=& \big( e^{-itf}Pe^{itf}\big)
            \big( e^{-itf}Qe^{itf}\big)u\\[1ex]
%
     &=& \big( p(x, \, df) t^r + \cdots \big)
             \big( q(x,\, df) t^{s}+\cdots \big) u\\[1ex]
%
     &=& \big( p(x,\, df)q(x,\, df)t^{r+s}+\cdots \big) u \, .
\end{eqnarray*}

\end{proof}


Given a differential operator
%
\begin{displaymath}
  P=\sum_{|\alpha |\leq r} a_{\alpha}D^{\alpha}
\end{displaymath}
%
we define its \emph{transpose} to be the operator
%
\begin{displaymath}
  u \in \Cinf (U) \to \sum_{|\alpha |\leq r}
      D^{\alpha}\overline{a}_{\alpha}u =: P^t u \, .
\end{displaymath}
%

\begin{theorem}
  \label{th:1.3}

For $u,v \in \Cinf_0 (U)$
%
\begin{displaymath}
  \langle P u,v \rangle  =: \int Pu \overline{v} \, dx
      = \langle u,P^t v \rangle \, .
\end{displaymath}

\end{theorem}



\begin{proof}
By integration by parts
%
\begin{eqnarray*}
  \langle D_k u,v \rangle 
      &=&\int D_k u \overline{v} \, dx =
         \frac{1}{\sqrt{-1}} \int \frac{\partial}{\partial x_k}
         u \overline{v} \, dk\\
%
      &=& -\frac{1}{\sqrt{-1}}\int u\frac{\partial}{\partial x_k}
          \overline{v} \, dx =\int u \overline{D_kv}\, dx\\
%
      &=& \langle u,\, D_kv \rangle \, .    
\end{eqnarray*}
%
Thus
%
\begin{eqnarray*}
  \langle D^{\alpha} u,v \rangle 
       &=& \langle u,D^{\alpha}v \rangle\\
%
\noalign{\hbox{and}}\\
%
   \langle a_{\alpha} D^{\alpha} u,v  \rangle
       &=& \langle D^{\alpha}u,\overline{a}_{\alpha}v\rangle
          = \langle u,D^{\alpha}\overline{a}_{\alpha}v\rangle , .
\end{eqnarray*}
   

\end{proof}



\subsection*{Exercises.}

If $p (x,\xi)$ is the symbol of $P$, $\overline{p} (x,\xi)$ is
the symbol of $P^t$.

\subsection*{Ellipticity.}

$P$ is elliptic if $p(x,\xi) \neq 0$ for all $x \in U$ and $\xi
\in \RR^n-0$.


\section{Differential operators on manifolds.}
\label{sec:2}

Let $U$ and $V$ be open subsets of $\RR^n$ and $\varphi :U \to V$
a diffeomorphism.

\noindent{\textbf{Claim.}}~~If $P$ is a differential operator of
order $m$ on $U$ the operator
%
\begin{displaymath}
  u \in \Cinf (V) \to (\varphi^{-1})^* P \varphi^* u
\end{displaymath}
%
is a differential operator of order $m$ on $V$.

\begin{proof}
$(\varphi^{-1})^* D^{\alpha}\varphi^* =
\big( (\varphi^{-1})^* D_1 \varphi^* \big)^{\alpha_1}
  \cdots \big( (\varphi^{-1})^* D_n \varphi^* \big)^{\alpha_n}$ so
  it suffices to check this for $D_k$  and for $D_k$ this follows
  from the chain rule
%
  \begin{displaymath}
    D_k \varphi^* f = \sum 
        \frac{\partial \varphi_i}{\partial x_k}
        \varphi^* D_i f \, .
  \end{displaymath}

\end{proof}


This invariance under coordinate changes means we can define
differential operators on manifolds.



\begin{definition}
  \label{def:2.1}

Let $X=X^n$ be a real $\Cinf$ manifold.  An operator, $P:\Cinf
(X) \to \Cinf (X)$, is an $m$\st{th} order differential operator
if, for every coordinate patch, $(U,x_1 ,\ldots ,x_n)$ the
restriction map
%
\begin{displaymath}
  u \in \Cinf (X) \to Pu \upharpoonleft U
\end{displaymath}
%
is given by an $m$\st{th} order differential operator,
i.e.,~restricted to $U$,
%
\begin{displaymath}
  Pu = \sum_{|\alpha |\leq m} a_{\alpha}D^{\alpha} u\, , 
  \quad a_{\alpha} \in \Cinf (U) \, .
\end{displaymath}

\end{definition}


\noindent{\textbf{Remark.}}~~Note that this is a non-vacuous definition.  More explicitly let
$(U,x_1,\ldots ,x_n)$ and $(U',x'_1 , \ldots , x'_n)$ be
coordinate patches.  Then the map
%
\begin{displaymath}
  u \to Pu \upharpoonleft U \cap U'
\end{displaymath}
%
is a differential operator of order $m$ in the $x$-coordinates if
and only if it's a differential operator in the $x'$-coordinates.





\subsection*{The symbol of a differential operator}

\begin{theorem}
  \label{the:2.2}

Let $f : X \to \RR$ be  $\Cinf$ function.  Then the operator
%
\begin{displaymath}
  u \in \Cinf (X) \to e^{-itf}P e^{-itf}u
\end{displaymath}
%
can be written as a sum
%
\begin{displaymath}
  \sum^m_{i=0} t^{m-i} P_i
\end{displaymath}
%
$P_i$ being a differential operator of order $i$ which doesn't
depend on $t$.

\end{theorem}



\begin{proof}
We have to check that for every coordinate patch $(U,x_1 ,\ldots
,x_n)$ the operator
%
\begin{displaymath}
  u \in \Cinf (X) \to e^{-itf}P e^{itf}\upharpoonleft U
\end{displaymath}
%
has this property.  This, however, follows from Theorem~\ref{th:1.1}.

\end{proof}

In particular, the operator, $P_0$, is a zero\st{th} order
operator, i.e.,~multiplication by a $\Cinf$ function, $p_0$.

\begin{theorem}
  \label{th:2.3}
There exists $\Cinf$ function
%
\begin{displaymath}
  \sigma (P) : T^* X \to \CC
\end{displaymath}
%
not depending on $f$ such that
%
\begin{equation}
  \label{eq:2.1}
  p_0 (x) = \sigma (P) (x,\xi)
\end{equation}
%
with $\xi = df_x$.

\end{theorem}


\begin{proof}
It's clear that the function, $\sigma (P)$, is uniquely
determined at the points, $\xi \in T^*_x$ by the
property~(\ref{eq:2.1}), so it suffices to prove the local
existence of such a function on a neighborhood of~$x$.  Let
$(U,x_1,\ldots ,x_n)$ be a coordinate patch centered at $x$ and
let $\xi_1 ,\ldots ,\xi_n$ be the cotangent coordinates on $T^*U$
defined by
%
\begin{displaymath}
  \xi \to \xi_1 \, dx_1 + \cdots + \xi_n \, dk_n \,.
\end{displaymath}
%
Then if
%
\begin{displaymath}
  P=\sum a_{\alpha}D^{\alpha}
\end{displaymath}
%
on $U$ the function, $\sigma (P)$, is given in these coordinates
by $p(x,\xi) = \sum a_{\alpha}(x) \xi^{\alpha}$.  (See (\ref{eq:1.2}).)


\end{proof}

\subsection*{Composition and transposes}

\begin{list}{}{}
\item %%1
If $P$ and $Q$ are differential operators of degree $r$ and $s$,
$PQ$ is a differential operator of degree~$r+s$, and $\sigma (PQ)
= \sigma (P) \sigma (Q)$.

\item %%2
Let $\F_X$ be the sigma field of Borel subsets of $X$.  A
\emph{measure, $dx$, on }$X$ is a measure on this sigma field.  A
measure, $dx$, is \emph{smooth} if for every coordinate patch
%
\begin{displaymath}
  (U, x_1 , \ldots , x_n) \, .
\end{displaymath}
%
The restriction of $dx$ to $U$ is of the form
%
\begin{equation}
  \label{eq:2.2}
  \varphi \, dx_1 \ldots dx_n
\end{equation}
%
$\varphi$ being a non-negative $\Cinf$ function and $dx_1 \ldots
dx_n$ being Lebesgue measure on $U$.  $dx$ is
\emph{non-vanishing} if the $\varphi$ in (\ref{eq:2.2}) is
strictly positive.

\end{list}

Assume $dx$ is such a measure.  Given $u$ and $v \in \Cinf_0 (X)$
one defines the $L^2$ inner product
%
\begin{displaymath}
  \langle u,v \rangle
\end{displaymath}
%
of $u$ and $v$ to be the integral
%
\begin{displaymath}
    \langle u,v \rangle = \int u \overline{v} \, dx \, .
\end{displaymath}

\begin{theorem}
  \label{th:2.4}

If $P : \Cinf (X) \to \Cinf (X)$ is an $m$\st{th} order
differential operator there is a unique $m$\st{th} order
differential operator, $P^t$, having the property
%
\begin{displaymath}
  \langle Pu,v \rangle = \langle u,P^t v \rangle
\end{displaymath}
%
for all $u,v \in \Cinf_0 (X)$.

\end{theorem}

\begin{proof}

Let's assume that the support of $u$ is contained in a coordinate
patch, $(U,x_1,\ldots ,x_n)$.  Suppose that on $U$
%
\begin{eqnarray*}
  P &=& \sum a_{\alpha} D^{\alpha}\\
%
\noalign{\hbox{and}}\\
%
   dx &=& \varphi dx_1 \ldots dx_n \, .
\end{eqnarray*}
%
Then
%
\begin{eqnarray*}
  \langle Pu,v \rangle 
      &=& \sum_{\alpha} \int a_{\alpha} D^{\alpha}u
               \overline{v} \varphi dx_1 \ldots dx_n\\
%
      &=& \sum_{\alpha} \int a_{\alpha} \varphi D^{\alpha}
              u \overline{v} dx_1 \ldots dx_n\\
%
      &=& \sum \int u \overline{D^{\alpha}
             \overline{a}_{\alpha} \varphi v}
               dx_1 \ldots dx_n\\
%
      &=& \sum \int u \overline{\frac{1}{\varphi} D^{\alpha}
               \bar{a}_{\alpha} \varphi v} \varphi dx_1 \ldots dx_n\\
%
      &=& \langle u,P^t v\rangle\\
%
\noalign{\hbox{where}}\\
%
  P^tv &=& \frac{1}{\varphi} \sum D^{\alpha}
      \overline{a}_{\alpha} \varphi v \, .
\end{eqnarray*}

This proves the local existence \emph{and local uniqueness} of
$P^t$ (and hence the \emph{global} existence of $P^t$!).

\end{proof}

\subsection*{Exercise.} $\sigma (P^t)(x,\xi) =
 \overline{\sigma (P) (x,\xi)}$.

\subsection*{Ellipticity.}

$P$ is elliptic if $\sigma (P) (x,\xi) \neq 0$ for all $x \in X$
and $\xi \in T^*_x-0$.

\vspace{2ex}

The main goal of these notes will be to prove:

\begin{theorem}[Fredholm theorem for elliptic operators.]
  \label{th:2.5}

If $X$ is compact and
%
\begin{displaymath}
  P : \Cinf (X) \to \Cinf (X)
\end{displaymath}
%
is an elliptic differential operator, the kernel of $P$ is finite
dimensional and $u \in \Cinf (X)$ is in the range of $P$ if and
only if
%
\begin{displaymath}
  \langle u,v \rangle =0
\end{displaymath}
%
for all $v$ in the kernel of $P^t$.

\begin{remark*}

Since $P^t$ is also elliptic its kernel is finite dimensional.

\end{remark*}



\end{theorem}


\section{Smoothing operators}
\label{sec:3}

Let $X$ be an $n$-dimensional manifold equipped with a smooth
non-vanishing measure, $dx$.  Given $K \in \Cinf (X \times X)$,
one can define an operator
%
\begin{displaymath}
  T_K : \Cinf (X) \to \Cinf (X)
\end{displaymath}
%
by setting
%
\begin{equation}
  \label{eq:3.1}
  T_K f (x) = \int K (x,y) f (y) \, dy \, .
\end{equation}
%
Operators of this type are called \emph{smoothing} operators.  The
definition~(\ref{eq:3.1}) involves the choice of the measure,
$dx$, however, it's easy to see that the notion of ``smoothing
operator'' doesn't depend on this choice.  Any other smooth
measure will be of the form, $\varphi (x) \, dx$, where $\varphi$
is an everywhere-positive $\Cinf$ function, and if we replace
$dy$ by $\varphi (y) \, dy$ in (\ref{eq:3.1}) we get the
smoothing operator, $T_{K_1}$, where $K_1 (x,y) = K(x,y) \,
\varphi (y)$.

A couple of elementary remarks about smoothing operators:

\begin{list}{}{}
\item 1.~~Let $L (x,y) = \overline{K(y,x)}$.  Then $T_L$ is the
\emph{transpose} of $T_K$.  For $f$ and $g$ in $\Cinf_0 (X)$,
%
\begin{eqnarray*}
  \langle T_K f,g \rangle &=& \int\overline{g}(x)
  \left( \int K (x,y) f(y) \, dy \right) \, dx \\
%
     &=& \int f(y) \overline{(T_L g) (y)} \, dy =
       \langle f,T_L g \rangle \, .
\end{eqnarray*}


\item 2.~~If $X$ is compact, the composition of two smoothing
  operators is a smoothing operator.  Explicitly:
%
  \begin{displaymath}
    T_{K_1} T_{K_2} = T_{K_3}
  \end{displaymath}
%
where
%
\begin{displaymath}
  K_3 (x,y) = \int K_1 (x,z) K_2 (z,y) \, dz \, .
\end{displaymath}

\end{list}


We will now give a rough outline of how our proof of
Theorem~\ref{th:2.5} will go.  Let $I : \Cinf (X) \to \Cinf (X)$
be the identity operator.  We will prove in the next few sections
the following two results.  

\begin{theorem}
  \label{th:3.1}
The elliptic operator, $P$ is right-invertible modulo smoothing
operators, i.e.,~there exists an operator, $Q : \Cinf (X) \to
\Cinf (X)$ and a smoothing operator, $T_K$, such that
%
\begin{equation}
  \label{eq:3.2}
  PQ = I-T_K
\end{equation}

\end{theorem}

and

\begin{theorem}
 \label{th:3.2}

The Fredholm theorem is true for the operator, $I-T_K$, i.e.,~the
kernel of this operator is finite dimensional, and $f \in \Cinf
(X)$ is in the image of this operator if and only if it is
orthogonal to kernel of the operator, $I-T_L$, where $L(x,y) =
\overline{K (y,x)}$.
\end{theorem}


\noindent{\textbf{Remark.}}~~In particular since $T_K$ is the
transpose of $T_L$, the kernel of $I-T_L$ is finite dimensional.

The proof of Theorem~\ref{th:3.2} is very easy, and in fact we'll
leave it as a series of exercises.  (See \S \ref{sec:8}.)  The
proof of Theorem~\ref{th:3.1}, however, is a lot harder and will
involve the theory of pseudodifferential operators on the
$n$-torus, $T^n$.

We will conclude this section by showing how to deduce
Theorem~\ref{th:2.5} from ~Theorems ~\ref{th:3.1} and
\ref{th:3.2}.  Let $V$ be the kernel of $I-T_L$.  By
Theorem~\ref{th:3.2}, $V$ is a finite dimensional space, so every
element, $f$, of $\Cinf (X)$ can be written uniquely as a sum
%
\begin{equation}
  \label{eq:3.3}
  f=g+h
\end{equation}
%
where $g$ is in $V$ and $h$ is orthogonal to $V$.  Indeed, if
$f_1 ,\ldots ,f_m$ is an orthonormal basis of $V$ with respect to
the $L^2$ norm
%
\begin{displaymath}
  g= \sum \langle f,f_i \rangle f_i
\end{displaymath}
%
and $h=f-g$.  Now let $U$ be the orthocomplement of $V \cap
\Image P$ in $V$.

\begin{proposition}
  \label{prop:3.3}
Every $f \in \Cinf (M)$ can be written uniquely as a sum
%
\begin{equation}
  \label{eq:3.4}
  f=f_1 + f_2
\end{equation}
%
where $f_1 \in U$, $f_2 \in \Image P$ and $f_1$ is orthogonal to
$f_2$.
\end{proposition}


\begin{proof}
By Theorem~\ref{th:3.1}
%
\begin{equation}
  \label{eq:3.5}
  \Image P \supset \Image (I-T_K) \, .
\end{equation}
%
Let $g$ and $h$ be the ``$g$'' and ``$h$'' in (\ref{eq:3.3}).
Then since $h$ is orthogonal to $V$, it is in $\Image (I-T_K)$ by
Theorem~\ref{th:3.2} and hence in $\Image P$ by (\ref{eq:3.5}).
Now let $g = f_1 + g_2$ where $f_1$ is in $U$ and $g_2$ is in the
orthocomplement of $U$ in $V$ (i.e.,~in $V \cap \Image P$).  Then
%
\begin{displaymath}
  f=f_1 + f_2
\end{displaymath}
%
where $f_2 = g_2 +h$
is in $\Image P$.  Since $f_1$ is orthogonal to $g_2$ and $h$ it
is orthogonal to $f_2$.

\end{proof}


Next we'll show that
%
\begin{equation}
  \label{eq:3.6}
  U=\Ker P^t \, .
\end{equation}
%
Indeed $f \in U \Leftrightarrow f \perp \Image P \Leftrightarrow
\langle f,Pu \rangle =0$ for all $u \Leftrightarrow \langle P^t
f,u \rangle =0$ for all $u \leftrightarrow P^t f=0$.

This proves that all the assertions of Theorem~{\ref{th:3.2} are true
except for the finite dimensionality of $\Ker P$.  However,
(\ref{eq:3.6}) tells us that $\Ker P^t$ is finite dimensional and
so, with $P$ and $P^t$ interchanged, $\Ker P$ is finite dimensional.


\section{Fourier analysis on the $n$-torus}
\label{sec:4}

In these notes the ``$n$-torus'' will be, by definition, the
manifold:  $T^n = \RR^n /2\pi \ZZ^n$.  A $\Cinf$ function, $f$,
on $T^n$ can be viewed as a $\Cinf$ function on $\RR^n$ which is
\emph{periodic} of period~$2\pi$:  For all $k \in \ZZ^n$
%
\begin{equation}
  \label{eq:4.1}
  f (x+ 2\pi k) = f(x) \, .
\end{equation}
%
Basic examples of such functions are the functions
%
\begin{displaymath}
  e^{ikx}\, , \quad k \in \ZZ^n \, , \quad
      kx=k_1x_1 + \cdots k_nx_n \, .
\end{displaymath}
%
Let $\P = \Cinf (T^n) = \Cinf$ functions on $\RR^n$ satisfying
(\ref{eq:4.1}), and let $Q \subseteq \RR^n$ be the open cube
%
\begin{displaymath}
  0< x_i < 2 \pi \, . \quad i=1,\ldots ,n \, .
\end{displaymath}
%
Given $f \in \P$ we'll define
%
\begin{displaymath}
  \int_{T^n} f\, dx = \left( \frac{1}{2\pi}\right)^n
               \int_Q f \, dx
\end{displaymath}
%
and given $f,g \in \P$ we'll define their $L^2$ inner product by
%
\begin{displaymath}
  \langle f,g \rangle = \int_{T^n} f \overline{g} \, dx \, .
\end{displaymath}
%
I'll leave you to check that
%
\begin{displaymath}
  \langle e^{ikx}\, , \, e^{i\ell x} \rangle
\end{displaymath}
%
is zero if $k \neq \ell$ and $1$ if $k=\ell$.  Given $f \in \P$
we'll define the $k$\st{th} \emph{Fourier coefficient} of $f$ to
be the $L^2$ inner product 
%
\begin{displaymath}
  c_k = c_k (f) = \langle f,e^{ikx}\rangle =
      \int_{T^n} f e^{-ikx} \, dx \, .
\end{displaymath}
%
The \emph{Fourier series} of $f$ is the formal sum
%
\begin{equation}
  \label{eq:4.2}
  \sum c_k e^{ikx} \, , \quad k \in \ZZ^n \, .
\end{equation}


In this section I'll review (very quickly) standard facts about
Fourier series.

It's clear that $f \in \P \Rightarrow D^{\alpha}f \in \P$ for all
multi-indices, $\alpha$.


\begin{proposition}
  \label{prop:4.1}
If $g = D^{\alpha } f$
%
\begin{displaymath}
  c_k (g) = k^{\alpha} c_k (f) \, .
\end{displaymath}
%
\end{proposition}


\begin{proof}
  \begin{displaymath}
    \int_{T^n} D^{\alpha}fe^{-ikx}\, dx =
        \int_{T^n} f \overline{D^{\alpha}e^{ikx}}\, dx \, .
  \end{displaymath}
%
Now check
%
\begin{displaymath}
  D^{\alpha} e^{ikx} = k^{\alpha} e^{ikx}\, .
\end{displaymath}

\end{proof}


\begin{corollary}
  \label{cor:4.2}

For every integer $r>0$ there exists a constant $C_r$ such that
%
\begin{equation}
  \label{eq:4.3}
  | c_k (f) | \leq C_r (1+|k|^2)^{-r/2}\, .
\end{equation}

\end{corollary}

\begin{proof}
Clearly
%
\begin{displaymath}
  | c_k (f) | \leq \frac{1}{(2\pi)^n} 
      \int_Q |f| \, dx =C_0 \, .
\end{displaymath}
%
Moreover, by the result above, with $g = D^{\alpha}f$
%
\begin{displaymath}
  | k^{\alpha} c_k (f) | = |c_k (g)| \leq C_{\alpha}
\end{displaymath}
%
and from this it's easy to deduce an estimate of the form~(\ref{eq:4.3}).

\end{proof}


\begin{proposition}
  \label{prop:4.3}
The Fourier series (\ref{eq:4.2}) converges and this sum is a
$\Cinf $ function.

\end{proposition}


To prove this we'll need

\begin{lemma}
  \label{lem:4.4}
If $m>n$ the sum
%
\begin{equation}
  \label{eq:4.4}
  \sum \left( \frac{1}{1+ |k|^2}\right)^{m/2}\, , \quad
      k \in \ZZ^n\, ,
\end{equation}
%
converges.
\end{lemma}

\begin{proof}
By the ``integral test'' it suffices to show that the integral
%
\begin{displaymath}
  \int_{\RR^n} \left( \frac{1}{1+|x|^2}\right)^{m/2}\, dx
\end{displaymath}
%
converges.  However in polar coordinates this integral is equal
to
%
\begin{displaymath}
  \gamma_{n-1}\int^{\infty}_0
     \left(\frac{1}{1+|r|^2}\right)^{m/2}
        r^{n-1}\, dr
\end{displaymath}
%
($\gamma_{n-1}$ being the volume of the unit $n-1$ sphere) and
this converges if $m>n$.

\end{proof}


Combining this lemma with the estimate (\ref{eq:4.3}) one sees
that (\ref{eq:4.2}) converges absolutely, i.e.,
%
\begin{displaymath}
  \sum |c_k (f) |
\end{displaymath}
%
converges, and hence (\ref{eq:4.2}) converges uniformly to a
continuous limit.  Moreover if we differentiate (\ref{eq:4.2})
term by term we get
%
\begin{displaymath}
  D^{\alpha}\sum c_k e^{ikx} = \sum k^{\alpha} c_k e^{ikx}
\end{displaymath}
%
and by the estimate (\ref{eq:4.3}) this converges absolutely and
uniformly.  Thus the sum~(\ref{eq:4.2}) exists, and so do its
derivatives of all orders.

Let's now prove the fundamental theorem in this subject, the
identity
%
\begin{equation}
  \label{eq:4.5}
  \sum c_k (f) e^{ikx} =f(x) \, .
\end{equation}


\begin{proof}
Let $\A \subseteq \P$ be the algebra of trigonometric
polynomials:
%
\begin{displaymath}
  f \in \A \Leftrightarrow f(x) = \sum_{|k|\leq m}
      a_k e^{ikx}
\end{displaymath}
%
for some $m$.
\end{proof}


\noindent{\textbf{Claim.}}~~This is an algebra of continuous
functions on $T^n$ having the Stone--Weierstrass properties

\begin{list}{}{}
\item 1)~~Reality:  If $f \in \A$, $\overline{f} \in \A$.

\item 2)~~$1 \in \A$.

\item 3)~~If $x$ and $y$ are points on $T^n$ with $x \neq y$,
  there exists an $f \in \A$ with $f(x) \neq f(y)$.
\end{list}


\begin{proof}
Item 2 is obvious and item~1 follows from the fact that
$\overline{e^{ikx}} = e^{-ikx}$.  Finally to verify item~3 we
note that the finite set, $\{ e^{ix_1}, \ldots , e^{ix_n} \}$,
already separates points.  Indeed, the map
%
\begin{displaymath}
  T^n \to (S^1)^n
\end{displaymath}
%
mapping $x$ to $e^{ix_1}, \ldots ,e^{ix_n}$ is bijective.

Therefore by the Stone--Weierstrass theorem $\A$ is dense in $C^0
(T^n)$.  Now let $f \in \P$ and let $g$ be the Fourier
series~(\ref{eq:4.2}).  Is $f$ equal to $g$?  Let $h=f-g$.  Then
%
\begin{eqnarray*}
  \langle h,e^{ikx}\rangle &=& \langle f,e^{ikx}\rangle
      - \langle g,e^{ikx} \rangle \\
%
      &=&  c_k (f) - c_k (f) =0
\end{eqnarray*}
%
so $\langle h,e^{ikx} \rangle =0$ for all $e^{ikx}$, hence
$\langle h,\varphi \rangle =0$ for all $\varphi \in \A$.
Therefore since $\A$ is dense in $\P$, $\langle h,\varphi \rangle
=0$ for all $\varphi \in \P$.  In particular, $\langle h,h
\rangle =0$, so $h=0$ .

\end{proof}


I'll conclude this review of the Fourier analysis on the
$n$-torus by making a few comments about the $L^2$ theory.

The space, $\A$, is dense in the space of continuous functions on
$T^n$ and this space is dense in the space of $L^2$ functions on
$T^n$.  Hence if $h \in L^2 (T^n)$ and $\langle h,e^{ikx} \rangle
=0$ for all $k$ the same argument as that I sketched above shows
that $h=0$.  Thus
%
\begin{displaymath}
  \{ e^{ikx} \, , \, k \in \ZZ^n \}
\end{displaymath}
%
is an orthonormal basis of $L^2 (T^n)$.  In particular, for every
$f \in L^2 (T^n)$ let
%
\begin{displaymath}
  c_k (f) = \langle f,e^{ikx}\rangle \, .
\end{displaymath}
%
Then the Fourier series of $f$
%
\begin{displaymath}
  \sum c_k (f) e^{ikx}
\end{displaymath}
%
converges in the $L^2$ sense to $f$ and one has the Plancherel
formula
%
\begin{displaymath}
  \langle f,f \rangle = \sum |c_k (f) |^2 \, , \quad
      k \in \ZZ^n \, .
\end{displaymath}


\section{Pseudodifferential operators on $T^n$}
\label{sec:5}

In this section we will prove Theorem~\ref{th:2.5} for elliptic
operators on $T^n$.  Here's a road map to help you navigate this
section.  \S\ref{sec:5.1} is a succinct summary of the material in
\S4.  Sections~\ref{sec:5.2}, \ref{sec:5.3} and \ref{sec:5.4} are a brief
account of the theory of pseudodifferential operators on $T^n$
and the symbolic calculus that's involved in this theory.  In
\S\ref{sec:5.5} and \ref{sec:5.6} we prove that an elliptic
operator on $T^n$ is right invertible modulo smoothing operators
(and that its inverse is a pseudodifferential operator).  Finally,
in \S\ref{sec:5.7}, we prove that pseudodifferential operators
have a property called ``pseudolocality'' which makes them behave
in some ways like differential operators (and which will enable
us to extend the results of this section from $T^n$ to arbitrary
compact manifolds).

Some notation which will be useful below:  for $a \in \RR^n$ let
%
\begin{displaymath}
  \langle a \rangle = (|a|^2 +1)^{\frac{1}{2}} \, .
\end{displaymath}
%
Thus
%
\begin{displaymath}
  |a| \leq \langle a \rangle
\end{displaymath}
%
and for $|a| \geq 1$
%
\begin{displaymath}
  \langle a \rangle \leq 2 |a| \, .
\end{displaymath}
\vspace{2ex}

\subsection{The Fourier inversion formula}
\label{sec:5.1}

Given $f \in \Cinf (T^n)$, let $c_k (f) = \langle f,e^{ikx}
\rangle$.  Then:

\begin{list}{}{}
\item 1)~~$c_k (D^{\alpha}f)=k^{\alpha} c_k (f)$.

\item 2)~~$| c_k (f) | \leq C_r \langle k \rangle^{-r}$ for all $
 r$.

\item 3)~~$\sum c_k (f) e^{ikx}=f$.

\end{list}


Let $S$ be the space of functions,
%
\begin{displaymath}
  g: \ZZ^n \to \CC
\end{displaymath}
%
satisfying 
%
\begin{displaymath}
  |g(k)| \leq C_r \langle k \rangle^{-r}
\end{displaymath}
%
for all $r$.  Then the map
%
\begin{displaymath}
  F: \Cinf (T^n) \to S \, , \quad Ff(k) = c_k (f)
\end{displaymath}
%
is bijective and its inverse is the map,
%
\begin{displaymath}
  g \in S \to \sum g(k) e^{ikx}\, .
\end{displaymath}



\renewcommand{\theequation}{\arabic{section}.\arabic{subsection}.%
  \arabic{equation}}


\subsection{Symbols}
\label{sec:5.2}

A function $a: T^n \times \RR^n \to \CC$ is in $\Ss^m$ if, for all
multi-indices, $\alpha $ and $\beta$, 
%
\begin{equation}
  \label{eq:5.2.1}
  | D^{\alpha}_x D^{\beta}_{\xi} a| \leq C_{\alpha , \beta}
      \langle \xi \rangle^{m-|\beta |}\, .
\end{equation}

\noindent{\textbf{Examples}}

\begin{list}{}{}
\item 1)~~$a (x,\xi) = \sum_{|\alpha| \leq m} a_{\alpha}(x)
  \xi^{\alpha}$, $a_{\alpha} \in \Cinf (T^n)$.

\item 2)~~$\langle \xi \rangle^m$.

\item 3)~~$a \in \Ss^{\ell}$ and $b \in \Ss^m \Rightarrow ab \in
  S^{\ell +m}$.

\item 4)~~$a \in \Ss^m \Rightarrow D^{\alpha}_x D^{\beta}_{\xi} a
  \in \Ss^{m-|\beta|}$.
\end{list}

\subsection*{The asymptotic summation theorem}

Given $b_i \in \Ss^{m-i}$, $i=0,1,\ldots,$ there exists a $b \in
\Ss^m$ such that
%
\begin{equation}
  \label{eq:5.2.2}
  b-\sum_{j<i}b_j  \in \Ss^{m-i}.
\end{equation}


\begin{proof}
\vspace{2ex}

\noindent{\emph{Step~1.}}~~Let $\ell =m+\epsilon$, $\epsilon
>0$.  Then 
%
\begin{displaymath}
  |b_i (x,\xi)| < C_i \langle \xi \rangle^{m-i}=
  \frac{C_i \langle \xi \rangle^{\ell
      -i}}{\langle\xi\rangle^{\epsilon}}\, .
\end{displaymath}
%
Thus, for some $\lambda_i$,
%
\begin{displaymath}
  | b_i (x,\xi)| < \frac{1}{2^i} \langle \xi \rangle^{\ell-i}
\end{displaymath}
%
for $|\xi | \ >\lambda_i$.  We can assume that $\lambda_i \to +
\infty$ as $i \to +\infty$.  Let $\rho \in \Cinf (\RR)$ be
bounded between $0$ and $1$ and satisfy $\rho (t) =0$ for $t<1$
and $\rho (t)=1$ for $t>2$.  Let
%
\begin{equation}
  \label{eq:5.2.3}
  b= \sum \rho \left( \frac{|\xi|}{\lambda_i}\right)
      b_i (x,\xi) \, .
\end{equation}
%
Then $b$ is in $\Cinf (T^n \times \RR^n)$ since, on any compact
subset, only a finite number of summands are non-zero.  Moreover,
$b-\sum_{j<i} b_j$ is equal to:
%
\begin{displaymath}
  \sum_{j<i} \left( \rho \left(\frac{|\xi|}{\lambda_j}
           \right) -1 \right) b_j + b_i + \sum_{j>i}
           \rho \left( \frac{|\xi|}{\lambda_j}\right) b_j \, .
\end{displaymath}
%
The first summand is compactly supported, the second summand is
in $\Ss^{m-1}$ and the third summand is bounded from  above by
%
\begin{displaymath}
  \sum_{k>i} \frac{1}{2^k} \langle \xi \rangle^{\ell -k}
\end{displaymath}
%
which is less than $\langle \xi \rangle^{\ell-(i+1)}$ and hence,
for $\epsilon <1$, less than $\langle \xi \rangle^{m-i}$.

\vspace{2ex}

\noindent{\emph{Step 2.}}~~For $|\alpha | + |\beta| \leq N$
choose $\lambda_i$ so that
%
\begin{displaymath}
  |D^{\alpha}_x D^{\beta}_{\xi}b_i (x,\xi)| \leq
      \frac{1}{2^i}\langle \xi \rangle^{\ell-i-|\beta |}
\end{displaymath}
%
for $\lambda_i< |\xi|$.  Then the same argument as above implies
that
%
\begin{equation}
  \label{eq:5.2.4}
  D^{\alpha}_x D^{\beta}_{\xi} (b-\sum_{j,i}b_j)
     \leq C_N \langle \xi \rangle^{m-i-|\beta|}
\end{equation}
%
for $|\alpha | + |\beta| \leq N$.

  \vspace{2ex}

\noindent{\emph{Step 3.}}~~The sequence of $\lambda_i$'s in
step~2 depends on $N$.  To indicate this dependence let's denote
this sequence by $\lambda_{i,N}$, $i=0,1,\ldots$.  We can, by
induction, assume that for all $i$, $\lambda_{i,N} \leq
\lambda_{i,N+1}$.  Now apply the Cantor diagonal process to this
collection of sequences, i.e.,~let $\lambda_i = \lambda_{i,i}$ .
Then $b$ has the property~(\ref{eq:5.2.4}) for all $N$.

We will denote the fact that $b$ has the
property~(\ref{eq:5.2.2}) by writing
%
\begin{equation}
  \label{eq:5.2.5}
  b \sim \sum b_i\, .
\end{equation}

The symbol, $b$, is not unique, however, if $b \sim \sum b_i$ and
$b' \sim \sum b_i$, $b-b'$ is in the intersection, $\bigcap
\Ss^{\ell}$, $-\infty < \ell < \infty$.

\end{proof}



\subsection{Pseudodifferential operators}
\label{sec:5.3}


Given $a \in \Ss^m$ let
%
\begin{displaymath}
  T^0_a : S \to \Cinf (T^n)
\end{displaymath}
%
be the operator
%
\begin{displaymath}
  T^0_a g = \sum a(x,k) g (k) e^{ikx}\, .
\end{displaymath}
%
Since
%
\begin{displaymath}
  |D^{\alpha} a(x,k) e^{ikx}| \leq C_{\alpha}
     \langle k \rangle^{m+\langle \alpha \rangle}
\end{displaymath}
%
and
%
\begin{displaymath}
  |g(k) | \leq C_{\alpha} \langle k \rangle^{-(m+n+|\alpha |+1)}
\end{displaymath}
%
this operator is well-defined, i.e.,~the right hand side is in
$\Cinf (T^n)$.  Composing $T^0_a$ with $F$ we get an operator
%
\begin{displaymath}
  T_a : \Cinf (T^n) \to \Cinf (T^n)\, .
\end{displaymath}
%
We call $T_a$ the pseudodifferential operator with symbol~$a$.

\vspace{2ex}

Note that
%
\begin{displaymath}
  T_a e^{ikx} = a(x,k) e^{ikx} \, .
\end{displaymath}
%
Also note that if
%
\begin{eqnarray}
  \label{eq:5.3.1}
  P &=& \sum_{|\alpha|\leq m} a_{\alpha} (x) D^{\alpha}\\
%
\noalign{\nonumber{\hbox{and}}}\\
%
  \label{eq:5.3.2}
   p(x,\xi) &=& \sum_{|\alpha | \leq m}
      a_{\alpha} (x) \xi^{\alpha}\, .
\end{eqnarray}
%
Then
%
\begin{displaymath}
  P=T_p \, .
\end{displaymath}


\subsection{The composition formula}
\label{sec:5.4}


Let $P$ be the differential operator (\ref{eq:5.3.1}).  If $a$
is in $\Ss^r$ we will show that $PT_a$ is a pseudodifferential
operator of order $m+r$.  In fact we will show that
%
%
\begin{eqnarray}
  \label{eq:5.4.1}
   PT_a &=& T_{p \circ a}\\
%
\noalign{\nonumber{\hbox{where}}}\\
%
\label{eq:5.4.2}
  p \circ a (x,\xi) &=& \sum_{|\beta| \leq m}
     \frac{1}{\beta !} \partial^{\beta}_{\xi} p(x,\xi)
       D^{\beta}_x a (x,\xi)
\end{eqnarray}
%
and $p(x,\xi)$ is the function~(\ref{eq:5.3.2}).

\begin{proof}
By definition
%
\begin{eqnarray*}
  PT_a e^{ikx} &=& P a (x,k) e^{ikx}\\
%
     &=& e^{ikx}(e^{-ikx} Pe^{ikx}) a(x,k)\, .
\end{eqnarray*}
%
Thus $PT_a$ is the pseudodifferential operator with symbol
%
\begin{equation}
  \label{eq:5.4.3}
  e^{-ix\xi} P e^{ix\xi} a (x, \xi )\, .
\end{equation}
%
However, by (\ref{eq:5.3.1}):
%
\begin{eqnarray*}
  e^{-ix\xi} P e^{ix\xi} u(x) 
     &=& \sum a_{\alpha}(x) e^{-ix\xi} D^{\alpha}e^{ix\xi}u(x)\\
%
     &=& \sum a_{\alpha}(x) (D+\xi)^{\alpha} u(x)\\
%
     &=& p (x,D+\xi) u(x) \, .
\end{eqnarray*}
%
Moreover,
%
\begin{displaymath}
  p(x,\eta +\xi) = \sum \frac{1}{\beta !}\,
  \frac{\partial}{\partial \xi^{\beta}} p(x,\xi) \eta^{\beta}\, ,
\end{displaymath}
so
\begin{displaymath}
p(x,D+\xi ) u(x) = \sum \frac{1}{\beta !} \, 
\frac{\partial}{\partial \xi^{\beta}} p(x,\xi)D^{\beta}u(x) 
\end{displaymath}
%
and if we plug in $a(x,\xi)$ for $u(x)$ we get, by
(\ref{eq:5.4.3}), the formula~(\ref{eq:5.4.2}) for the symbol of
$PT_a$.

\end{proof}


\subsection{The inversion formula}
\label{sec:5.5}

Suppose now that the operator~(\ref{eq:5.3.1}) is elliptic.  We
will prove below the following inversion theorem.

\begin{theorem}
  \label{th:5.1}
There exists an $a \in \Ss^{-m}$ and an $r \in \bigcap S^{\ell}$,
$-\infty <\ell < \infty$, such that
%
\begin{displaymath}
  PT_a = I-T_r \, .
\end{displaymath}

\end{theorem}


\begin{proof}
Let
%
\begin{displaymath}
  p_m (x,\xi) = \sum_{|\alpha|=m} a_{\alpha}(x) \xi^{\alpha}\, .
\end{displaymath}
%
By ellipticity $p_m (x,\xi) \neq 0$ for $\xi \not\in 0$.  Let
$\rho \in \Cinf (\RR)$ be a function satisfying $\rho (t)=0$ for
$t<1$ and $\rho (t) =1$ for $t>2$.  Then the function
%
\begin{equation}
  \label{eq:5.5.1}
  a_0 (x,\xi) = \rho (|\xi|) \frac{1}{p_m (x,\xi)}
\end{equation}
%
is well-defined and belongs to $S^{-m}$.  To prove the theorem we
must prove that there exist symbols $a \in \Ss^{-m}$ and $r \in
\bigcap \Ss^{\ell}$, $-\infty < \ell <\infty$, such that
%
\begin{displaymath}
  p \circ q = 1-r \, .
\end{displaymath}
%
We will deduce this from the following two lemmas.


\begin{lemma}
  \label{lem:5.2}
If $b \in \Ss^i$ then
%
\begin{displaymath}
  b-p \circ a_0 b
\end{displaymath}
%
is in $\Ss^{i-1}$.
\end{lemma}



\begin{proof}

Let $q=p-p_m$.  Then $q \in \Ss^{m-1}$ so $q \circ a_0 b$ is in
$\Ss^{i-1}$ and by (\ref{eq:5.4.2})
%
\begin{eqnarray*}
  p\circ a_0b &=& p_m \circ a_0 b + q \circ a_0 b\\
     &=& p_m a_0 b + \cdots =b+\cdots
\end{eqnarray*}
%
where the dots are terms of order~$i-1$.

\end{proof}


\begin{lemma}
  \label{lem:5.3}

There exists a sequence of symbols $a_i \in \Ss^{-m-i}$,
$i=0,1,\ldots$, and a sequence of symbols $r_i \in \Ss^{-i}$,
$i=0,\ldots,$ such that $a_0$ is the symbol~(\ref{eq:5.5.1}),
$r_0 =1$ and
%
\begin{displaymath}
  p \circ a_i = r_i - r_{i+1}
\end{displaymath}
%
for all $i$.


\end{lemma}
 
\begin{proof}
Given $a_0, \ldots , a_{i-1}$ and $r_0 , \ldots r_i$, let
$a_i=r_ia_0$ and $r_{i+1}=r_i-p \circ a_i$.  By
Lemma~\ref{lem:5.2}, $r_{i+1} \in \Ss^{-i-1}$.

\end{proof}


Now let $a \in \Ss^{-m}$ be the ``asymptotic sum'' of the $a_i$'s
%
\begin{displaymath}
  a \sim \sum a_i \, .
\end{displaymath}
%
Then
%
\begin{displaymath}
     p \circ a \sim \sum p \circ a_i = \sum^{\infty}_{i=0}
        r_i - r_{i+1} = r_0 =1 \, , 
\end{displaymath}
%
so $1-p \circ a \sim 0$, i.e.,~$r=1-p \circ q$ is in $\bigcap
\Ss^{\ell}$, $-\infty < \ell < \infty$.

\end{proof}

\subsection{Smoothing properties of $\Psi DO$'s}
\label{sec:5.6}


Let $a \in \Ss^{\ell}$, $\ell <-m-n$.  We will prove in this
section that the sum
%
\begin{equation}
  \label{eq:5.6.1}
  K_a (x,y) = \sum a(x,k) e^{ik(x-y)}
\end{equation}
%
is in $C^m (T^n \times T^n)$ and that $T_a$ is the integral
operator associated with $K_a$, i.e.,
%
\begin{displaymath}
  T_a u(x) = \int K_a (x,y) u (y) \, dy \, .
\end{displaymath}

\begin{proof}

For $|\alpha| + |\beta| \leq m$
%
\begin{displaymath}
  D^{\alpha}_x D^{\beta}_y a(x,k) e^{ik (x-y)}
\end{displaymath}
%
is bounded by $\langle k \rangle^{\ell +|\alpha|+|\beta|}$ and
hence by $\langle k \rangle^{\ell +m}$.  But $\ell +m < -n$, so
the sum
%
\begin{displaymath}
\sum D^{\alpha}_x D^{\beta}_y a (x,k) e^{ik(x-y)}
\end{displaymath}
%
converges absolutely.  Now notice that
%
\begin{displaymath}
  \int K_a (x,y) e^{iky}\, dy = a (x,k)e^{ikx}= T_{\alpha}
  e^{ikx}\, .
\end{displaymath}
%
Hence $T_a$ is the integral operators defined by $K_a$.  Let 
%
\begin{equation}
  \label{eq:5.6.2}
  \Ss^{-\infty} = \bigcap \Ss^{\ell}\, ,\quad
    -\infty < \ell< \infty \, .
\end{equation}
%
If $a$ is in $\Ss^{-\infty}$, then by (\ref{eq:5.6.1}), $T_a$ is a
smoothing operator.

\end{proof}



\subsection{Pseudolocality}
\label{sec:5.7}

We will prove in this section that if $f$ and $g$ are $\Cinf$
functions on $T^n$ with non-overlapping supports and $a$ is in
$\Ss^m$, then the operator
%
\begin{equation}
  \label{eq:5.7.1}
  u \in \Cinf (T^n) \to f T_a gu
\end{equation}
%
is a smoothing operator.  (This property of pseudodifferential
operators is called \emph{pseudolocality}.)  We will first prove:

\begin{lemma}
  \label{lem:5.4}
If $a(x,\xi)$ is in $\Ss^m$ and $w \in \RR^n$, the function,
%
\begin{equation}
  \label{eq:5.7.2}
  a_w (x,\xi) = a(x,\xi +w) -a(x,\xi)
\end{equation}
%
is in $S^{m-1}$.

\end{lemma}

\begin{proof}
Recall that $a \in \Ss^m$ if and only if 
%
\begin{displaymath}
  |D^{\alpha}_x D^{\beta}_{\xi} a(x,\xi)|
      \leq C_{\alpha ,\beta} \langle \xi \rangle^{m-|\beta|}\, .
\end{displaymath}
%
From this estimate is is clear that if $a$ is in $\Ss^m$,
$a(x,\xi+w)$ is in $\Ss^m$ and $\frac{\partial a}{\partial \xi_i}
(x,\xi)$ is in $\Ss^{m-1}$, and hence that the integral
%
\begin{displaymath}
  a_w (x,\xi) = \int^1_0 \sum_i \frac{\partial a}{\partial \xi_i}
     (x,\xi +tw) \, dt
\end{displaymath}
%
in $\Ss^{m-1}$.




Now let $\ell$ be a large positive integer and let $a$ be in
$\Ss^m$, $m<-n-\ell$.  Then
%
\begin{displaymath}
  K_a (x,y) = \sum a (x,k) e^{ik(x-y)}
\end{displaymath}
%
is in $C^{\ell} (T^n \times T^n)$, and $T_a$ is the integral
operator defined by $K_a$.  Now notice that  for $w \in \ZZ^n$
%
\begin{equation}
  \label{eq:5.7.3}
  (e^{-i (x-y) w}-1) K_a (x,y) 
      = \sum a_w  (x,k) e^{ik(x-y)}\, , 
\end{equation}
%
so by the lemma the left hand side of (\ref{eq:5.7.3}) is in
$C^{\ell +1} (T^n \times T^n)$.  More generally,
%
\begin{equation}
  \label{eq:5.7.4}
  (e^{-i(x-y)w}-1)^N K_a (x,y)
\end{equation}
%
is in $C^{\ell +N} (T^n \times T^n)$.  In particular, if $x
\neq y$, then for some $1 \leq i \leq n$, $x_i -y_i \not\equiv 0
\mod 2\pi Z$, so if
%
\begin{displaymath}
  w=(0,0,\ldots ,1,0,\ldots ,0) \, ,
\end{displaymath}
%
($a$ ``$1$'' in the i\st{th}-slot), $e^{i(x-y)w}\neq 1$ and, by
(\ref{eq:5.7.4}), $K_a (x,y)$ is $C^{\ell +N}$ in a neighborhood
of $(x,y)$.  Since $N$ can be arbitrarily large we conclude 

\begin{lemma}
  \label{lem:5.5}

$K_a (x,y)$ is a $\Cinf$ function on the complement of the diagonal
in $T^n \times T^n$.
\end{lemma}

Thus if $f$ and $g$ are $\Cinf$ functions with non-overlapping
support,  $fT_a g$ is the smoothing operator, $T_K$, where
%
\begin{equation}
  \label{eq:5.7.5}
  K (x,y) = f(x) K_a (x,y) g (y) \, .
\end{equation}


\end{proof}

We have proved that $T_a$ is pseudolocal if $a \in \Ss^m$, $m <
-n-\ell$, $\ell$ a large positive integer.  To get rid of this
assumption let $\langle D \rangle^N$ be the operator with symbol
$\langle \xi \rangle^N$.  If $N$ is an even positive integer
%
\begin{displaymath}
  \langle D \rangle^N = (\sum D^2_i +I)^{\frac{N}{2}}
\end{displaymath}
%
is a differential operator and hence is a \emph{local} operator:
if $f$ and $g$ have non-overlapping supports, $f \langle D
\rangle^N g$ is identically zero.  Now let $a_N (x,\xi) =
a(x,\xi) \langle \xi \rangle^{-N}$.  Since $a_N \in \Ss^{m-N}$,
$T_{a_N}$ is pseudolocal for $N$~large.  But $T_a = T_{a_N}
\langle D \rangle^N$, so $T_a$ is the composition of an operator
which is pseudolocal with an operator which is local, and
therefore $T_a$ itself is pseudolocal.


\section{Elliptic operators on open subsets of $T^n$}
\label{sec:6}

\numberwithin{equation}{section}
%%%\setcounter{equation}{section}****

Let $U$ be an open subset of $T^n$.  We will denote by $\iota_U :
U \to T^n$ the inclusion map and  by $\iota^*_U : \Cinf (T^n) \to
\Cinf (U)$ the restriction map:  let $V$ be an open subset of
$T^n$ containing $\overline{U}$ and
%
\begin{displaymath}
  P= \sum_{|\alpha |\leq m} a_{\alpha} (x) D^{\alpha} \, , 
     \quad a_{\alpha} (x) \in \Cinf (V)
\end{displaymath}
%
an elliptic $m$\st{th} order differential operator.  Let 
%
\begin{displaymath}
  P^t = \sum_{|\alpha | \leq m} D^{\alpha}
      \overline{a}_{\alpha} (x)
\end{displaymath}
%
be the transpose operator and
%
\begin{displaymath}
  p_m (x,\xi) = \sum_{|\alpha | =m}
     a_{\alpha} (x) \xi^{\alpha}
\end{displaymath}
%
the symbol of $P$.  We will prove below the following localized
version of the inversion formula of \S~\ref{sec:5.5}.


\begin{theorem}
  \label{th:6.1}
There exist symbols, $a \in \Ss^{-m}$ and $r \in \Ss^{-\infty}$
such that
%
\begin{equation}
  \label{eq:6.1}
  P \iota^*_U T_a = \iota^*_U (I-T_r)\, .
\end{equation}

\end{theorem}

\begin{proof}

Let $\gamma \in \Cinf_0 (V)$ be a function which is bounded
between $0$ and $1$ and is identically $1$ in a neighborhood of
$\overline{U}$.  Let
%
\begin{displaymath}
  Q=PP^t \gamma + (1-\gamma)(\sum D^2_i)^n \, .
\end{displaymath}
%
This is a globally defined $2m$\st{th} order differential
operator in $T^n$ with symbol,
%
\begin{equation}
  \label{eq:6.2}
  \gamma (x) | p_m (x,\xi)|^2 + (1-\gamma (x))|\xi |^{2m}
\end{equation}
%
and since (\ref{eq:6.2}) is non-vanishing on $T^n \times (\RR^n
-0)$, this operator is elliptic.  Hence, by Theorem~\ref{th:5.1},
there exist symbols $b \in \Ss^{-2m}$ and $r \in \Ss^{-\infty}$
such that
%
\begin{displaymath}
  QT_b = I-T_r \, .
\end{displaymath}
%
Let $T_a = P^t \gamma T_b$.  Then since $\gamma \equiv 1$ on a
neighborhood of $\overline{U}$,
%
\begin{eqnarray*}
  \iota^*_U (I-T_r) &=& \iota^*_U Q T_b\\
%
  &=& \iota^*_U (PP^t \gamma T_b + (1-\gamma)\sum D^2_i T_b)\\
%
  &=& \iota^*_U PP^t \gamma T_b\\
%
  &=& P \iota^*_U P^t \gamma T_b = P \iota^*_U T_a \, .
\end{eqnarray*}



\section{Elliptic operators on compact manifolds}
\label{sec:7}

Let $X$ be a compact $n$ dimensional manifold and 
%
\begin{displaymath}
  P : \Cinf (X) \to \Cinf (X)
\end{displaymath}
%
an elliptic $m$\st{th} order differential operator.  We will show
in this section how to construct a \emph{parametrix} for $P$:  an
operator
%
\begin{displaymath}
  Q : \Cinf (X) \to \Cinf (X)
\end{displaymath}
%
such that $I-PQ$ is smoothing.

Let $V_i$, $i=1,\ldots ,N$ be a covering of $X$ by coordinate
patches and let $U_i$, $i=1,\ldots ,N$,  $\overline{U}_i \subset
V_i$ be an open covering which refines this covering.  We can,
without loss of generality, assume that $V_i$ is an open subset
of the hypercube
%
\begin{displaymath}
  \{ x \in \RR^n \, \quad 0 < x_i < 2\pi \, 
     \quad i=1,\ldots ,n \}
\end{displaymath}
%
and hence an open subset of $T^n$.  Let
%
\begin{displaymath}
  \{ \rho_i \in \Cinf_0 (U_i)\, , \quad
      i=1,\ldots ,N \}
\end{displaymath}
%
be a partition of unity and let $\gamma_i \in \Cinf_0 (U_i)$ be a
function which is identically one on a neighborhood of the
support of $\rho_i$.  By Theorem~\ref{th:6.1}, there exist
symbols $a_i \in \Ss^{-m}$ and $r_i \in \Ss^{-\infty}$ such that
on $T^n$:
%
\begin{equation}
  \label{eq:7.1}
  P \iota^*_{U_i} T_{a_i} = \iota^*_{U_i} (I-T_{r_i})\, .
\end{equation}
%
Moreover, by pseudolocality $(1-\gamma_i) T_{a_i} \rho_i$ is
smoothing, so
%
\begin{displaymath}
  \gamma_i T_{a_i} \rho_i -\iota^*_{U_i} T_{a_i}\rho_i
\end{displaymath}
%
and
%
\begin{displaymath}
  P \gamma_i T_{a_i} \rho_i - P \iota^*_{U_i}T_{a_i}\rho_i
\end{displaymath}
%
are smoothing.  But by (\ref{eq:7.1})
%
\begin{displaymath}
  P \iota^*_{U_i} T_{a_i} \rho_i - \rho_i I
\end{displaymath}
%
is smoothing.  Hence
%
\begin{equation}
  \label{eq:7.2}
  P \gamma_i T_{a_i} \rho_i - \rho_i I
\end{equation}
%
is smoothing as an operator on $T^n$.  However, $P \gamma_i
T_{a_i} \rho_i$ and $\rho_i I$ are \emph{globally defined} as
operators on $X$ and hence (\ref{eq:7.2}) is a globally defined
smoothing operator.  Now let $Q = \sum \gamma_i T_{a_i}\rho_i$
and note that by (\ref{eq:7.2})
%
\begin{displaymath}
  PQ-I
\end{displaymath}
%
is a smoothing operator.

\end{proof}


This concludes the proof of Theorem~\ref{th:3.1}, and hence,
modulo proving Theorem~\ref{th:3.2}.  This concludes the proof of
our main result:  Theorem~\ref{th:2.5}.  The proof of
Theorem~\ref{th:3.2} will be outlined, as a series of exercises,
in the next section.


\section{The Fredholm theorem for smoothing operators}

\label{sec:8}

Let $X$ be a compact $n$-dimensional manifold equipped with a
smooth non-vanishing measure, $dx$.  Given $K \in \Cinf (X \times
X)$ let
%
\begin{displaymath}
  T_K : \Cinf (X) \to \Cinf (X)
\end{displaymath}
%
be the smoothing operator~(3.1).

\vspace{2ex}

\noindent{\textbf{Exercise 1.}}  Let $V$ be the volume of $X$
(i.e.,~the integral of the constant function, $1$, over $X$).  Show that if
%
\begin{displaymath}
  \max |K (x,y )| <\frac{\epsilon}{V} \, , \quad
     0< \epsilon <1
\end{displaymath}
%
then $I-T_K$ is invertible and its inverse is of the form,
$I-T_L$, $L \in \Cinf (X \times X)$.

\noindent{\emph{Hint 1.}  Let $K_i = K \circ \cdots \circ K$} ($i$
products).  Show that $\sup |K_i (x,y)| < C \epsilon^i$ and
conclude that the series
%
\begin{equation}
  \label{eq:8.1}
  \sum K_i (x,y)
\end{equation}
%
converges uniformly.

\noindent{\emph{Hint 2.}  Let $U$ and $V$ be coordinate patches
  on $X$.  Show that on $U \times V$ 
%
  \begin{displaymath}
    D^{\alpha}_x D^{\beta}_y K_i (x,y)
       = K^{\alpha}\circ K_{i-2} \circ K^{\beta} (x,y)
  \end{displaymath}
%
where $K^{\alpha}(x,z) =D^{\alpha}_x K (x,z)$ and $K^{\beta}(z,y)
= D^{\beta}_y K(z,y)$.  Conclude that not only does $(8.1)$
converge on $U \times V$ but so do its partial derivatives of
\emph{all} orders with respect to $x$ and $y$.

\vspace{2ex}

\noindent{\textbf{Exercise 2.  (finite rank operators.)}}  $T_K$
is a finite rank smoothing operator if $K$ is of the form:
%
\begin{equation}
  \label{eq:8.2}
  K(x,y) = \sum^N_{i=1} f_i (x) g_i (y)\, .
\end{equation}

\alphaparenlist
\begin{enumerate}
\item %%a
Show that if $T_K$ is a finite rank smoothing operator and $T_L$
is any smoothing operator, $T_KT_L$ and $T_LT_K$ are finite rank
smoothing operators.

\item %%b
Show that if $T_K$ is a finite rank smoothing operator, the
operator, $I-T_K$, has finite dimensional kernel and co-kernel.
\end{enumerate}

\noindent{\emph{Hint.}  Show that if $f$ is in the kernel of this
  operator, it is in the linear span of the $f_i$'s and that $f$
  is in the image of this operator if
%
  \begin{displaymath}
    \int f(y) g_i (y) \, dy =0 \, , \quad
    i=1,\ldots, N \, .
  \end{displaymath}

\vspace{2ex}

\noindent{\textbf{Exercise 3.}}  Show that for every $K \in \Cinf
(X \times X)$ and every $\epsilon >0$ there exists a function,
$K_1 \in \Cinf (X \times X)$ of the form (\ref{eq:8.2}) such that
%
\begin{displaymath}
  \sup | K-K_1 | (x,y) <\epsilon \, .
\end{displaymath}


\noindent{\emph{Hint.}  Let $\A$ be the set of all functions of
  the form (\ref{eq:8.2}).  Show that $\A$ is a \emph{subalgebra}
  of $C (X \times X)$ and that this subalgebra separates points.
  Now apply the Stone--Weierstrass theorem to conclude that $\A$
  is dense in $C (X \times X)$.

\noindent{\textbf{Exercise 4.}}  Prove that if $T_K$ is a
smoothing operator the operator
%
\begin{displaymath}
  I-T_K : \Cinf (X) \to \Cinf (X)
\end{displaymath}
%
has finite dimensional kernel and co-kernel.


\noindent{\emph{Hint.}  Show that $K=K_1+K_2$ where $K_1$ is of
  the form (\ref{eq:8.2}) and $K_2$ satisfies the hypotheses of
  exercise~1.  Let $I-T_L$ be the inverse of $I-T_{K_2}$.  Show
  that the operators
%
  \begin{eqnarray*}
    (I-T_K) \circ (I-T_L)\\
%
    (I-T_L) \circ (I-T_K)
  \end{eqnarray*}
%
are both of the form:  identity minus a finite rank smoothing
operator.  Conclude that $I-T_K$ has finite dimensional kernel
and co-kernel.

\vspace{2ex}

\noindent{\textbf{Exercise 5.}}  Prove Theorem~\ref{th:3.2}.





\end{document}


% LocalWords:  Munkres multivariable Spivak analogue Fredholm Plancherel