Commit 3c1c3f51 authored by Jannes Bantje's avatar Jannes Bantje

added an important proof

parent 1d5330c8
......@@ -286,7 +286,7 @@ Another important tool, which will be useful in other places as well, are \Index
Let $\phi \in C^\infty_c(\mathbb{R}^n)$ be a function with $\phi \ge 0$, $\int \phi(x) \intmathd x = 1$ and $\phi(-x)=\phi(x)$.
For $\varepsilon>0$ we let $\phi_\varepsilon(x) = \frac{1}{\varepsilon^n} \phi \enbrace*{\frac{x}{\varepsilon}}$.
\begin{propositiondef}
\begin{propositiondef}\label{prop:friedrichs}
The \Index{Friedrichs' mollifier} is the operator $F_\varepsilon \colon \Schwartz \to \Schwartz$, $u \mapsto \phi_\varepsilon * u$.
It has the following properties:
\begin{enumerate}[(i)]
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......@@ -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 $\norm*{F_t} \le 1$,
\item for each $u \in L^2(K,E)$, $F_t u$ is smooth with compact support,
\item for each $u \in L^2(K,E)$, $F_tu \grenzw{t \to 0} u$ 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.
\end{lemma}
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