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\title{Functional analysis problem sheet 11}
\author{Max Klinger and Kilian Klebes}
\begin{document}
\maketitle
\setcounter{thm}{28}
\begin{thm}
Let T be a closable, densely defined linear operator on a Hilbert space and let $\overline{T}$ be its closure. $\mathcal{G}(\overline{T}) = \overline{\mathcal{G}(T)}$
\end{thm}
\proof{
Let S be a closed extension of T. Then clearly we have $\overline{\mathcal{G}(T)} \subset \mathcal{G}(S)$ since $\overline{\mathcal{G}(T)}$ is the smallest closed extension of $\mathcal{G}(T)$. BThus if $(0,\psi) \in \overline{\mathcal{G}(T)}$ then $\psi =0$. Now define R with $D(R)= \{\psi| (\psi,\phi) \in \overline{\mathcal{G}(T)}\text{ for some}\}$ by $R\psi =\phi$ where $\phi \in \HH$ is the unique vector so that $(\psi,\phi) \in \overline{\mathcal{G}(T)}$. Then $\overline{R} = \overline{\mathcal{G}(T)}$ so R is a closed extension of T, but $R\subseteq S$ which was any arbitrary closed extension thus this relation holds for all S and therefore $R = \overline{T}$.
}
\begin{thm}
Let T be a densely defined linear operator on a Hilbert space $\HH$. $\rho(T)\neq 0 \Rightarrow T$ is closed.
\end{thm}
\begin{proof}
By the definition of the resolvent set of T: $\rho$ we know that $T-z\1$ is bijective and that its inverse (the resolvent) is bounded (Functional analysis I). By the closed graph theorem we therefore have that the resolvent is also closed. Now assume T is not closed, i.e. $\exists \{\phi_n\}: \phi_n \xrightarrow{n \rightarrow \infty} \phi$ but $T(\phi_n) \not \rightarrow T(\phi)$ since the resolvent is closed, we have $(T-\lambda \1)^{-1}(\phi_n) =\psi_n \rightarrow (T-\lambda \1)^{-1}(\phi) = \psi \stackrel{(T-\lambda \1)^{-1} bij.}{\Longleftrightarrow} $. Therefore $(T-\lambda \1) (\psi_n) \rightarrow (T-\lambda \1) (\psi)$ but $T\psi_n \not \rightarrow T\psi$. This is impossible since the resolvent is bounded and thus T must be closed.
If i am not mistaken, the Hamiltonian of a free particle should provide a counter example. It is densely defined, non closed and if you assume no bounding potential we have that its resolvent is empty.
\end{proof}
\begin{thm}
Let $\NN \ni d\geq 1$ and $s \in \RR, s>0$: $H^{-s}(\RR^d) = H^s(\RR^d)^{*}$
\end{thm}
\begin{proof}
\begin{itemize}
\item [``$\subset$``] Linearity is clear by definition, all we need to show is continuity which is equivalent to boundedness when operating on $H^s(\RR^d)$ which in turn follows from the fact that $H^{-s}(\RR^d) \subset L^2$ and Riesz's Lemma.
\item [``$\supset$``] Linearity is clear again. Let $\phi \in H^s(\RR^d)^{*}$ i.e. $\phi(\psi) \stackrel{\text{Riesz}}{=} \scalprod{\phi^{'}}{\psi} = \int_{\RR^d} \phi^{'} \psi dx< \infty$ $\forall \psi \in H^{s}(\RR^d)$. Since $\psi$ was arbitrary we get that $\phi \in L^2$. Left to show: $D^\alpha\phi \in L^2(\RR^d) \forall \alpha \in \NN^d_0$ where $\abs{\alpha} < s$ Hmm Times up again dammit.
\end{itemize}
\end{proof}
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