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\newtheorem{thm}{Exercise}
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\title{Functional analysis problem sheet 7}
\author{Max Klinger and Kilian Klebes}
\begin{document}
\maketitle
\setcounter{thm}{17}
\thm{
Let $A:\CC^n \rightarrow \CC^n, \; n > 1$ be selfadjoint, with an Eigenvalue $\lambda$ which sports an algebraic multiplicity of 2 or more. Let $e_1, e_2\in \CC^n$ be two corresponding eigenvectors. Then A doesn't have a cyclic vector.
}
\proof{
Assume it had a cyclic vector v. Then by the spectral theorem, we have for all $f \in C\left(\spec(A)\right):$
\begin{align*}
f(A) e_1 = f(\lambda) e_1 && f(A) e_2 = f(\lambda) e_2
\end{align*}
or using the notation of Lemma 1.47:
$
Ae_1 = \lambda e_1 \Leftrightarrow UAe_1 = \lambda U e_1 \Leftrightarrow UAe_1 U^* = UAU^* Ue_1 U^* = M_{id} \underbrace{Ue_1U^*}_{\phi} = \lambda \phi(\lambda) = \lambda U e_1 U^*, \; \mu_v$-almost surely
and analogously for $e_2$.
Thus using $U:L^2(\spec(A), \mu_v)\rightarrow \CC^n$ we get that
$$\overline{\{A^n v: n \in \NN\}} = \{A^n v: n \in \NN\}= \text{Im}(U) \subset \{v\in \CC^n| \scalprod{v}{e_1} = \scalprod{v}{e_2}\} \neq \CC^n$$
and thus v cannot be cyclic. Contradiction. So v cannot have been a cyclic vector in the first place.
}
\thm{
Let $\HH$ be a Hilbert space and $A:\HH\rightarrow \HH$ with $\exists K \subset \RR$ compact, $\mu:K\Rightarrow \RR$ Borel and $U:\HH\rightarrow L^2(K)$ unitary with $UAU^*:L^2(K,d\mu(x)) \rightarrow L^2(K,d\mu(x)), \psi\mapsto \left(x\mapsto x\psi(x)\right)$. Then A is bounded and self-adjoint.
}
\proof{
\begin{description}
\item[Self-adjointness]
Write $\psi\mapsto\left(x\mapsto x\psi(x)\right)$ as $M$. First note $M^* = M$ since $x \in K$ and $\psi \in L^2(K)$ thus giving $(M\psi)(x) \in \RR\; \forall x \in K$. Now use the unitary equivalence of A to $M_\psi$ to rewrite it as such:
$$UAU^* = M \Longleftrightarrow A = U^{-1} M U^{*-1} \stackrel{U^* =U^{-1}}{\Longleftrightarrow} A= U^*MU
$$
Thus giving $A^* = (U^*MU)^* = U^* M^* (U^*)^* = U^* M U = A$, i.e. A is self-adjoint.
\item[Boundedness] $U$ and $U^*$ as unitary operators are isometric and thus bounded. Now consider $$\norm{M} = \!\!\sup_{\psi \in L^2(K)} \frac{\norm{M(\psi)}}{\norm{\psi}} = \!\sup_{\psi \in L^2(K)} \frac{\sup_{x\in K} x\psi(x)}{\sup_{x\in K}\psi(x)} \leq \sup_{\psi \in L^2(K)} \sup_{x\in K} \frac{x\psi(x)}{\psi(x)} = \sup_{x\in K} x$$ which is clearly bounded (by Heine-Borel) since K is a compact subset of $\RR$. Of course this only works because of Corollary 33 in section 3.3.
\item[Cyclic vector] We see that a necessary condition for f to be a cyclic vector for $UAU^*$ on $L^2[K]$ is $f(x) \neq 0 \mu$ almost everywhere, because if f vanishes then so does $(UAU^*)^n f$ and therefore if f vanishes on a set with measure >0 then so does $(UAU^*)^n f \forall n$ and every linear combination thereof. Thus we might for example pick the constant function $f:K\rightarrow \RR, x\mapsto 1$ and we get a cyclic vector for $(UAU^*)$ ( $(UAU^*)^n f = L^2 \rightarrow L^2, x\mapsto x^n$ and the polynomials where shown to be a basis of $L^2$ in Functional Analysis ). Using the isometricity of U again we hereby obtain that $U^*(f)U$ is a cyclic vector for $A = U^*(M)U$, since it in particular is a bijective mapping.
\end{description}
}
\thm{
Let $\HH$ be a Hilbert space, $\dim\HH= \infty$, $\HH$ separable and $\{\psi_n\}_{n=1}^\infty$ an orthonormal basis of $\HH$. Let $\{a_n\}^\infty_{n=1} \subset l^\infty(\RR), \; a_i\neq a_j, \; \forall i\neq j$, $\;A^{'}:\{\psi_n\}^\infty_{n=1}\rightarrow \{\psi_n\}^\infty_{n=1}, \psi_n \mapsto A\psi_n :=a_n \psi_n$ and $A:\HH\rightarrow \HH$ the linear extension of $A^{'}$. A is bounded and self-adjoint.
}
\proof{
\begin{description}
\item [Boundedness]
By the Riesz representation theorem one obtains that every Hilbert space is canonically isomorphic to its dual. Corollary 8 to the Hahn Banach theorem (Funky Analysis I, Section 4-1) and Corollary 12 from section 5-2 guarantee us the boundedness of the extension of $A^{'}$ when extended to $\HH^* = \HH$, thus A is bounded.
\item[Self-adjointness]
\footnotetext[1]{Linearity and definition of A}
\footnotetext[2]{Linearity of the scalar product}
\footnotetext[3]{Orthonormality of the Basis}
\footnotetext[4]{$a_n \in \RR$ }
Let $\psi= \lambda_n \psi_n; \; \phi = \mu_n \psi_n$ be two arbitraty vectors in $\HH$. Then
\begin{align*}
\scalprod{\phi}{A\psi} &=\footnotemark[1] \scalprod{\sum_{n\in\NN}\mu_n\psi_n}{\sum_{n\in\NN}a_n \lambda_n\phi_n} =\footnotemark[2] \sum_{n\in\NN} \sum_{l\in\NN}\bar \mu_n a_l\lambda_l \scalprod{\psi_n}{\phi_l} =\footnotemark[3] \\
&= \sum_{n\in\NN} \sum_{l\in\NN}\bar \mu_n a_l\lambda_l \delta_{n,l} = \sum_{n\in \NN} \bar \mu_n a_n \lambda_n =\footnotemark[4] \sum_{n\in\NN} \sum_{l\in\NN}\bar\mu_n \bar a_n\lambda_l \delta_{n,l} =\footnotemark[3] \\
&=\sum_{n\in\NN} \sum_{l\in\NN}\bar \mu_n \bar a_n \lambda_l \scalprod{\psi_n}{\phi_l} =\footnotemark[2] \scalprod{\sum_{n\in\NN}a_n \mu_n\psi_n}{\sum_{n\in\NN} \lambda_n\phi_n} =\footnotemark[1] \scalprod{A\phi}{\psi}
\end{align*}
Thus A is selfadjoint.
\item[Cyclic vector]
Consider $K = {\frac{1}{n}: n\in \NN} \cup {\{0\}}$ K is obviously bounded and since it contains the only nontrivial limit (0) K might have, we can conclude by Heine Borel, that K is indeed compact in $\RR$. The only measure that really makes sense is the counting measure, so let $\mu(A) = |{A}|$. $L^2(K) \simeq l^2$ if you shift the whole sequence one to the right and take $\{0\}$ as the zeroth basis vector.
\end{description}
}
\end{document}