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\title{Functional analysis problem sheet 12}
\author{Max Klinger and Kilian Klebes}
\begin{document}
\maketitle
\setcounter{thm}{28}
\begin{thm}
Let A be a symmetric operator on a (dense domain of a) Hilbert space $\HH$. Then the essential-selfadjointness of A implies A has exactly one self-adjoint extension.
\end{thm}
\proof{
Let A be as defined and B a self-adjoint extension of A. We already know that B is closed (thm 3.9 i)), i.e. $\overline{B} = B$ and thus by (thm 3.9 ii)) $B =\overline{B} = B^{**}$. Hence we get $A \subset B \Rightarrow A \subset B^{**}$. Since Remark 3.8 ii) implies $B=\footnote{B selfadjoint}B^{*} \subset\footnote{Remark 3.8 ii)} A^{*} = (A^{**})^{*} =\footnote{ $A^{*} = A^{**}$ was stated in our definition in class } A^{**}$, i.e. $B=A^{**}$. Since $A^{**}$ is unique and B was arbitrary it follows that the self-adjoint extension of A is unique.
}
\begin{thm}
On the Hilbert space $L^2(\RR)$ define the Operator $D:H^1(\RR) \rightarrow L^2(\RR), \psi \mapsto i \psi^{'}$ where $\psi^{'}$ is the weak derivative of $\psi$. D is selfadjoint.
\end{thm}
\begin{proof}
Let $\psi \in L^2(\RR)$ and $\phi \in H^1(\RR)$. Through integration by parts of the weak derivative one clearly sees that D is symmetric, i.e. $T \subseteq T^{*}$. All that one needs to show is that $\dom(T^{*}) = H^1(\RR)$. But I'm to tired.
\end{proof}
\end{document}