@@ -311,16 +311,34 @@ This example already suggests, that non-selfadjointness is in some way connected

The same phenomenon shows up in the following statement:

\begin{lemma}

Let $D$ be a symmetric differential operator and $u$ a compactly supported element of $L^2(M,E)$.

Then $u$ belongs to the minimal domain of $D$ if and only if it belongs to the maximal domain of $D$.

Let $D$ be a symmetric differential operator and $u\in L^2(M,E)$\emph{compactly supported}.

Then $u$ belongs to the minimal domain of $D$ if and only if it belongs to the maximal domain.

\end{lemma}

\begin{proof}

See \cite[Lem.~10.2.5]{higson2000analytic}.\todo{mention the use of Friedrich's Mollifiers?}

\end{proof}

In particular we see, that every symmetric differential operator on a closed manifold is essentially selfadjoint.

More generally every compactly supported, symmetric differential operator on an open manifold is essentially selfadjoint.

\begin{proof}

Let $K$ be a compact subset of $M$.

We need \emph{Friedrichs' mollifiers} to proceed:

A local variant is discussed in \cref{prop:friedrichs}, which can be globalised via the usual arguments to operators $F_t \colon L^2(K,E)\to L^2(M,E)$ such that

\begin{enumerate}[(a)]

\item$\norm*{F_t}\le1$,

\item for each $u \in L^2(K,E)$, $F_t u$ is smooth with compact support,

\item for each $u \in L^2(K,E)$, $F_tu \grenzw{t \to0} u$ and

\item the commutator $\benbrace*{D,F_t}$ extends to a bounded operator $L^2(K,E)\to L^2(M,E)$, whose norm is independent of $t$.

\end{enumerate}

Now assume, that $u$ is in the maximal domain of $D$ and supported in $K$.

We have

\[

D F_t u = F_t D u +\benbrace*{D,F_t} u

\]

By \cref{rmk:max_domain_distribution} and (a) we know, that the first summand is uniformly bounded ; the second one is uniformly bounded by (d).

Considering $t=\frac{1}{n}$ we get a sequence of elements $u_n$ in the domain of $D$ converging to $u$ such that $\norm*{D u_n}$ is uniformly bounded.

Hence \cref{lem:charac_minimal_domain} implies, that $u$ is in the minimal domain of $D$.

\end{proof}

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

If $D$ is a symmetric elliptic operator, then its adjoint is elliptic as well.