\chapter{Analytic Vectors}

\info{Dies ist nur ein Relikt meines Studiums analytischer Vektoren. Wird wieder entfernt, falls nicht benötigt.}

\begin{definition}{Analytic Vector for an Operator}{analytic-vector-operator}
  Let $A : D(A) \to \hilb{H}$ be an unbounded linear operator in a complex Hilbert space $\hilb{H}$.
  A vector $x \in \hilb{H}$ is said to be an \emph{analytic vector} for $A$,
  if $x$ lies in the domain of the power $A^n$ for all $n \in \NN$, and the power series
  \begin{equation*}\tag{power-series-analytic-vector}
    \sum_{n=0}^{\infty} \frac{A^n x}{n!} \, z^n
  \end{equation*}
  has a nonzero radius of convergence.
  If the power series converges for all $z \in \CC$,
  we say that $x$ is an \emph{entire analytic vector} for $A$.
\end{definition}
Note that, if $x$ is analytic for $A$, then the power series
\begin{equation*}
  \sum_{n=0}^{\infty} \frac{\norm{A^n x}}{n!} \, z^n
\end{equation*}
converges for all complex $z$ with $\abs{z} < t$,
that is, in the open disc with radius $t$ centered in the origin of the complex plane.
This is a well-known consequence of the convergence behavior of power series.

\begin{definition}{Analyticity of Vector-Valued Functions}{}
  Let $G \subset \CC$ be open and let $\hilb{H}$ be a Hilbert space.
  A function $f : G \to \hilb{H}$ is called
  \begin{itemize}
    \item \emph{strongly analytic} at $a \in G$, if the limit
      \begin{equation*}
        \lim_{z \to a} \frac{f(z) - f(a)}{z-a}
      \end{equation*}
      exists in norm.
    \item \emph{weakly analytic} in $a \in G$, if for each $w \in \CC$ the scalar-valued function
      \begin{equation*}
        G \longrightarrow \CC, \quad z \longmapsto \innerp{w}{f(z)}
      \end{equation*}
      is analytic in $a$.
  \end{itemize}
\end{definition}

\begin{lemma}{Equivalence of Weak and Strong Analyticity}{}
  Let $G \subset \CC$ be open.
  Then a Banach space-valued function is strongly analytic on $G$ if and only if it is weakly analytic on $G$.
\end{lemma}
\begin{myproof}
  Let $X$ be a Banach space and suppose that the function $f : G \to X$ is weakly analytic.
  By definition, for each $g \in X'$ the scalar valued function $g \circ f : G \to \CC$ is analytic on $G$.
  Consider a point $a \in G$.
  Since $G$ is open, there exists a circular contour $\gamma$ around $a$ such that $\gamma$ and its interior lie wholly inside of $G$.
  By Cauchy’s Integral Formula we have
  \begin{equation*}
    g(f(z)) = \frac{1}{2 \pi i} \int_{\gamma} \frac{g(f(w))}{w-z} \, dw
  \end{equation*}
  for any $z$ in the interior of $\gamma$.
  Writing
  \begin{equation*}
    Q(z) = \frac{f(z) - f(a)}{z - a}
  \end{equation*}
  for the difference quotient, we get
  \begin{equation*}
    g(Q(z)) = \frac{1}{2 \pi i} \int_{\gamma} \frac{g(f(w))}{(w-z)(w-a)} \, dw
  \end{equation*}
  and
  \begin{equation*}
    g \parens*{\frac{Q(z) - Q(z')}{z - z'}} = \frac{1}{2 \pi i} \int_{\gamma} \frac{g(f(w))}{(w-z)(w-z')(w-a)} \, dw
  \end{equation*}
  for all $z,z'$ in the interior of $\gamma$.
  The family of vectors $f(w) \in X$, indexed by complex numbers $w$ on the contour $\gamma$, can be viewed as a family of bounded linear functionals $C(f(w)) : X' \to \CC$
  via the canonical embedding $C : X \to X''$ of $X$ into its bidual. For every fixed $g \in X'$ the set of values $C(f(w))(g) = g(f(w))$ is bounded, because the function $g \circ f$ is continuous and the contour is compact.
  In other words, the family of functionals $C(f(w))$, $w \in \gamma$, is pointwise bounded.
  The Uniform Boundedness Theorem implies that there exists a constant $M > 0$ such that $\abs{g(f(w))} \le M \norm{g}$ for all $w$ on $\gamma$ and all $g \in X'$.
  \begin{equation*}
    \abs*{g \parens*{\frac{Q(z) - Q(z')}{z - z'}}} \le \frac{M}{2 \pi} \norm{g} \int_{\gamma} \frac{dw}{\abs{w-z}\abs{w-z'}\abs{w-a}}
  \end{equation*}
  If we restrict $z,z'$ to a neighborhood $N$ of $a$ that stays away from $\gamma$, then the integral on the right hand side is
  bounded by a constant independent of $z$ and $z'$.
  Absorbing all constants into $M' > 0$ we obtain
  \begin{equation*}
    \abs{g(Q(z) - Q(z'))} \le M' \norm{g} \abs{z-z'} \quad \forall z,z' \in N.
  \end{equation*}
  \begin{equation*}
    \norm{Q(z) - Q(z')} = \sup_{\substack{g \in X'\\ \norm{g} \le 1}} \abs{g(Q(z) - Q(z'))} \le M' \norm{z - z'}.
  \end{equation*}
  Hence, the limit of $Q(z)$ for $z \to a$ exists by completeness of $X$.
\end{myproof}

\begin{definition}{Analytic Vector for an Unitary Group}{analytic-vector-unitary-group}
  Let $\sigma : \RR \to U(\hilb{H})$ be a strongly continuous one-parameter unitary group on a complex Hilbert space $\hilb{H}$.
  A vector $x \in \hilb{H}$ is said to be an \emph{analytic vector} for $\sigma$, if there exist
  \begin{itemize}
    \item a number $\lambda > 0$, defining a strip $I_{\lambda} = \braces{z : \abs{\Im z} < 1}$, and
    \item a vector-valued function $f : I_{\lambda} \to \hilb{H}$,
  \end{itemize}
  with the properties that
  \begin{itemize}
    \item $f(t) = \sigma_t(x)$ for all $t \in \RR$,
    \item $f$ is weakly analytic on $I_{\lambda}$.
  \end{itemize}
  In this case we write $f(z) = \sigma_z(x)$ for $z \in I_{\lambda}$.
\end{definition}

\begin{proposition}{}{}
  Let $\sigma : \RR \to U(\hilb{H})$ be a strongly continuous one-parameter unitary group on a complex Hilbert space $\hilb{H}$
  and let $A$ be its infinitesimal generator.
  Then a vector $x \in \hilb{H}$ is analytic for $\sigma$ if and only if it is analytic for $A$.
\end{proposition}

\begin{myproof}
  First, suppose that $x$ is an analytic vector for $\sigma$.
  Then, there exist a number $\lambda > 0$ and a function $f : I_{\lambda} \to X$ as in \cref{definition:analytic-vector-unitary-group}.
  In particular, $f$ is (strongly) analytic on the strip $I_{\lambda}$, which contains the disk $\braces{z : \abs{z} \le r}$ when $r \le \lambda$.
  Hence we have Cauchy estimates
  \begin{equation*}
    \norm{f^{(n)}(0)} \le \frac{n!}{r^n} M \quad \forall n \in \NN,
    \quad \text{where} \ M = \sup_{\abs{z} = r} \norm{f(z)}.
  \end{equation*}
  For real $t$ we have $f(t) = \sigma_t(x) = \exp(itA) x$ and
  the mapping $t \mapsto f(t)$ is strongly differentiable with derivatives
  $f^{(n)}(0) = (iA)^n x$.
  This implies that the power series
  \begin{equation*}
    \sum_{n=0}^{\infty} \frac{\norm{A^n x}}{n!} t^n \le M \sum_{n=0}^{\infty} \frac{t^n}{r^n} 
  \end{equation*}
  is convergent for $t \le \lambda$ by majorization. Hence $x$ is an analytic vector for the operator $A$.

  Conversely, suppose that $x$ is analytic for the generator $A$ of $\sigma$.
  Then, by \cref{definition:analytic-vector-operator}, $x$ lies in the domains of all powers $A^n$, $n \in \NN$, and the power series
  \begin{equation*}
    \sum_{n=0}^{\infty} \frac{\norm{A^n x}}{n!} z^n
  \end{equation*}
  has a positive radius of convergence $t>0$.
\end{myproof}

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