Added power series.

src/dg/derivative/higher.tex

@@ -130,7 +130,7 @@
 
 \begin{theorem}[Power Rule]
 \label{theorem:power-rule}
-    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space, $T \in B^n_\sigma(E; F)$.
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $T \in B^n_\sigma(E; F)$.
 
     For each $1 \le m \le n$, let $\text{Inj}([m]; [n])$ be the set of injective mappings from $[m]$ to $[n]$. For any $\phi \in \text{Inj}([m]; [n])$, denote $\phi^c \in \text{Inj}([n-m]; [n] \setminus \phi([m]))$ as the unique increasing injective map. For each $h \in E^{n-m}$, $k \in E^m$, and $1 \le j \le n$, write
     \[

src/dg/derivative/index.tex

@@ -6,3 +6,4 @@
 \input{./mvt.tex}
 \input{./higher.tex}
 \input{./taylor.tex}
+\input{./power.tex}

src/dg/derivative/power.tex

@@ -0,0 +1,82 @@
+\section{Power Series}
+\label{section:power-series}
+
+\begin{definition}[Power Series]
+\label{definition:power-series}
+    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, $\bracsn{T_n}_0^\infty$ with $T_n \in L^n(E; F)$ for each $n \in \natz$, and $a \in E$, then the \textbf{power series} of $\bracsn{T_n}_0^\infty$ about $a$ is the function
+    \[
+    f(x) = \sum_{n = 0}^\infty T_n(x - a)^{(n)}
+    \]
+
+    defined on points on which the series converges. 
+\end{definition}
+
+\begin{definition}[Radius of Convergence]
+\label{definition:radius-of-convergence}
+    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, and $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$ be a power series about $a \in E$, then $R \in [0, \infty]$ be defined by\footnote{Under the abuse that $1/\infty = 0$ and $1/0 =\infty$.}
+    \[
+    \frac{1}{R} = \limsup_{n \to \infty}\norm{T_n}_{L^n(E; F)}^{1/n}
+    \]
+
+    is the \textbf{radius of convergence of $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$}. For each $0 < r < R$, the series converges uniformly and absolutely on $B_E(a, r)$. 
+\end{definition}
+\begin{proof}
+    For all $x \in B_E(a, r)$, 
+    \[
+        \sum_{n = 0}^\infty \normn{T_n(x - a)^{(n)}}_F \le \sum_{n \in \natz} \norm{T_n}_{L^n(E; F)} \norm{x - a}_E^n \le \sum_{n \in \natz} \norm{T_n}_{L^n(E; F)} r^n
+    \]
+
+    For any $s \in (r, R)$, there exists $N \in \natp$ such that $\norm{T_n}_{L^n(E; F)}^{1/n} \le 1/s$ for all $n \ge N$. In which case, 
+    \[
+    \sum_{n = 0}^\infty \norm{T_n}_{L^n(E; F)} r^n \le \sum_{n = 0}^N \norm{T_n}_{L^n(E; F)}r^n + \sum_{n \ge N}\frac{r^n}{s^n} < \infty
+    \]
+
+    As this estimate holds uniformly on $B_E(a, r)$, the series converges uniformly and absolutely on $B_E(a, r)$. 
+\end{proof}
+
+\begin{remark}
+\label{remark:radius-of-convergence}
+    In \autoref{definition:radius-of-convergence}, the radius of convergence appears to be an arbitrary lower bound on the domain of convergence. However, in the more specialised case of power series from $\complex$ to $\complex$ or in a Banach algebra, $R$ is the largest constant such that the series converges uniformly and absolutely on all $B(0, r)$ where $0 < r < R$. The lack of this "maximum" claim is why the above statement is a definition. 
+\end{remark}
+
+
+\begin{theorem}[Termwise Differentiation]
+\label{theorem:termwise-differentiation}
+    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, $f(x) = \sum_{n = 0}^\infty T_n(x - a)^{(n)}$ a power series about $a \in E$, and $R$ be its radius of convergence, then 
+    \begin{enumerate}
+        \item $f \in C^\infty(B(a, R); F)$.
+        \item For each $x \in B(a, R)$ and $h \in E$, 
+        \[
+        Df(x)(h) = \sum_{n = 0}^\infty \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})
+        \]
+
+        \item The radius of convergence of the above series is at least $R$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (3): For each $n \in \natz$, let
+    \[
+    S_n(x_1, \cdots, x_{n})(h) = \sum_{k = 1}^{n+1}T_{n+1}(((x_1, \cdots, x_n), h)^{\bracs{k}})
+    \]
+
+    then $\norm{S_n}_{L^n(E; L(E; F))} \le (n+1)\norm{T_{n+1}}_{L^{n+1}(E; F)}$. Since $(n+1)^{1/n}$ is convergent and $\{||T_n||_{L^n(E; F)}^{1/n}\}$ is bounded, 
+    \[
+    \limsup_{n \to \infty} \norm{S_n}_{L^n(E; L(E; F))}^{1/n} \le \limsup_{n \to \infty}(n+1)^{1/n}\norm{T_{n+1}}_{L^{n+1}(E; F)}^{1/n} = \frac{1}{R}
+    \]
+
+    so the radius of convergence of the proposed series is at least $R$. 
+    
+    (2): By the \autoref{theorem:power-rule}, for each $N \in \natp$, 
+    \[
+    D\braks{\sum_{n = 0}^N T_n(x - a)^{(n)}}(h) = \sum_{n = 0}^N \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})
+    \]
+
+    By \autoref{definition:radius-of-convergence}, the proposed series converges uniformly on $B(a, r)$ for each $0 < r < R$. Thus by \autoref{theorem:differentiable-uniform-limit}, $f$ is differentiable on $B(a, R)$ with 
+    \[
+    Df(x)(h) = \sum_{n = 0}^\infty S_n(x - a)(h) = \sum_{n = 0}^\infty \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})
+    \]
+
+    (1): By (2), (3) applied inductively to $D^nf$. 
+\end{proof}
+
+

src/fa/lc/spaces-of-linear.tex

@@ -24,7 +24,7 @@
 \label{definition:saturated-ideal}
     Let $E$ be a locally convex space over $K \in \RC$ and $\sigma \subset 2^E$ be an ideal, then $\sigma$ is \textbf{saturated} if:
     \begin{enumerate}
-        \item For each $\lambda \in K$ and $S \in \sigma$, $\lamdba S \in \sigma$. 
+        \item For each $\lambda \in K$ and $S \in \sigma$, $\lambda S \in \sigma$. 
         \item For each $S \in \sigma$, $\ol{\aconv}(S) \in \sigma$. 
     \end{enumerate}