From c66c3bc73d5d627ec7051e9ada6316c98ae072e0 Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Wed, 29 May 2024 14:04:59 +0200
Subject: weiter

---
 analytic2.tex | 10 +++++-----
 1 file changed, 5 insertions(+), 5 deletions(-)

(limited to 'analytic2.tex')

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
-- 
cgit v1.2.3-70-g09d2