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diff --git a/analytic2.tex b/analytic2.tex
index e0dc68f..13b6700 100644
--- a/analytic2.tex
+++ b/analytic2.tex
@@ -9,7 +9,7 @@
\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 convergece.
+ 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}
@@ -66,13 +66,13 @@ This is a well-known consequence of the convergence behavior of power series.
\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 continous and the contour is compact.
+ 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 restict $z,z'$ to a neighbourhood $N$ of $a$ that stays away from $\gamma$, then the integral on the right hand side is
+ 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*}
@@ -121,9 +121,9 @@ This is a well-known consequence of the convergence behavior of 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. Hernce $x$ is an analytic vector for the operator $A$.
+ is convergent for $t \le \lambda$ by majorization. Hence $x$ is an analytic vector for the operator $A$.
- Coversely, suppose that $x$ is analytic for the generator $A$ of $\sigma$.
+ 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