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/duality/polar.tex

@@ -22,15 +22,108 @@
 \end{definition}
 
 \begin{proposition}
-\label{proposition:polar-properties}
+\label{proposition:polar-gymnastics}
     Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, then:
     \begin{enumerate}
-        \item $\emptyset^\circ = \emptyset^\square = F$ and $F^\circ = F^\square = \bracs{0}$. 
-        \item For any $A, B \subset E$ and $\lambda \ne 0$, if $\lambda A \subset B$, then $B^\circ \subset \lambda^{-1}A^\circ$. 
+        \item $\emptyset^\circ = F$ and $E^\circ = \bracs{0}$. 
+        \item For any $\alpha \in K \setminus \bracs{0}$ and $A \subset E$, $(\alpha A)^\circ = \alpha^{-1} \cdot A^\circ$.
+        \item For any $\seqi{A} \subset E$, $\paren{\bigcup_{i \in I}A_i}^\circ = \bigcap_{i \in I}A_i^\circ$. 
+        \item For any $A \subset B \subset E$, $A^\circ \supset B^\circ$. 
+        \item For any saturated ideal $\sigma \subset \mathfrak{B}(E, \sigma(E, F))$, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology on $F$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): For any $\lambda \in K \setminus \bracs{0}$ and $A \subset E$, 
+    \begin{align*}
+        (\lambda A)^\circ &= \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in \alpha A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1 \forall x \in \alpha A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{x, \alpha y}{\lambda} \le 1 \forall x \in A} \\
+        &= \bracs{y \in F|\text{Re}\dpn{\alpha x,  y}{\lambda} \le 1 \forall x \in A} \\
+        &= \bracs{\alpha^{-1} y \in F|\text{Re}\dpn{x,  y}{\lambda} \le 1 \forall x \in A} \\
+        &= \alpha^{-1} \cdot A^\circ
+    \end{align*}
+
+    (5): Let $S \in \sigma$, then
+    \[
+    (\aconv(S))^\circ = \bracs{y \in F|\ |\dpn{x, y}{\lambda}| \le 1 \forall x \in A} \subset S^\circ
+    \]
+
+    Since $\sigma$ is saturated, $\bracs{S^\circ|S \in \sigma}$ is a fundamental system of neighbourhoods at $0$ for the $\sigma$-uniform topology.     
+\end{proof}
+
+\begin{proposition}[{{\cite[IV.1.4]{SchaeferWolff}}}]
+\label{proposition:polar-properties}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
+    \begin{enumerate}
+        \item $A^\circ$ is a $\sigma(F, E)$-closed convex subset of $F$ containing $0$. 
+        \item If $A$ is circled, then so is $A^\circ$. 
+        \item If $A$ is a subspace of $E$, then
+        \[
+        A^\circ = A^\perp = \bracs{y \in F| \dpn{x, y}{E} = 0 \forall x \in A}
+        \]
+
+        and $A^\circ$ is a subspace of $F$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $x \in E$, 
+    \[
+    \bracs{x}^\circ = \bracs{y \in F|\text{Re}\dpn{x, y}{\lambda} \le 1}
+    \]
+
+    is the sublevel set of a continuous $\real$-linear functional, so it is $\sigma(F, E)$-closed and convex. Since $A^\circ = \bigcap_{x \in A}\bracs{x}^\circ$, $A$ is also $\sigma(F, E)$-closed and convex. 
+
+    (2): If $A$ is circled, then by \autoref{proposition:polar-gymnastics},
+    \[
+    A^\circ = \bigcap_{\substack{\alpha \in K \\ |\alpha| \ge 1}}\alpha A^\circ
+    \]
+
+    For any $y \in A^\circ$ and $\alpha \in K \setminus \bracs{0}$ with $|\alpha| \le 1$. Since $y \in \alpha^{-1}A^\circ$, $\alpha y \in A^\circ$, so $A^\circ$ is circled. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:equicontinuous-polar}
+    Let $E$ be a TVS over $K \in \RC$, $\dpn{E, E^*}{E}$ be the canonical duality, and $A \subset E^*$, then the following are equivalent
+    \begin{enumerate}
+        \item $A$ is equicontinuous.
+        \item $A^\circ \in \cn_E(0)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:equicontinuous-linear}, $A$ is equicontinuous if and only if
+    \[
+    \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1)) \in \cn_E(0)
+    \]
+
+    (1) $\Rightarrow$ (2): $A^\circ \supset \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1))$.
+
+    (2) $\Rightarrow$ (1): Since $A^\circ \in \cn_E(0)$, there exists $V \in \cn_E(0)$ circled with $V \subset A^\circ$, so $V \subset \bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1))$, and $\bigcap_{\phi \in A}\phi^{-1}(B_K(0, 1)) \in \cn_E(0)$. 
+\end{proof}
+
+
+\begin{theorem}[Bipolar Theorem]
+\label{theorem:bipolar}
+    Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$. For each $A \subset F$,
+    \[
+    A^{\circ\circ} = \ol{\conv}(A \cup \bracs{0})
+    \]
+
+    with respect to the $\sigma(E, F)$-topology. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[IV.1.5]{SchaeferWolff}}}. ]
+    By \autoref{proposition:polar-properties}, $A^{\circ \circ}$ is a $\sigma(E, F)$-closed, convex set that contains $0$. Since $A^{\circ \circ} \supset A$, it is sufficient to show that $A^{\circ\circ} \subset \ol{\conv}(A \cup \bracs{0})$. 
+
+    Let $x_0 \in E \setminus \ol{\conv}(A \cup \bracs{0})$, then by the \hypreref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi: E \to \real$ such that:
+    \begin{enumerate}
+        \item $\phi$ is $\sigma(E, F)$-continuous.
+        \item $\phi(\ol{\conv}(A \cup \bracs{0})) \subset (-\infty, 1)$ and $\phi(x_0) > 1$. 
     \end{enumerate}
 
+    If $K = \real$, let $\Phi = \phi$. If $K = \complex$, then by \autoref{proposition:polarisation-linear}, the mapping $\Phi(x) = \phi(x) - i\phi(ix)$ is a $\sigma(E; F)$-continuous $K$-linear map on $E$ such that $\text{Re}(\Phi) = \phi$. By \autoref{lemma:duality-dual}, there exists $y \in F$ such that $\dpn{x, y}{\lambda} = \Phi(x)$ for all $x \in E$. In which case, for any $x \in E$,
+    \[
+    \text{Re}\dpn{x, y}{\lambda} = \text{Re}\Phi(x) = \phi(x)
+    \]
 
-\end{proposition}
-
-
+    Therefore $y \in A^{\circ}$, but $\text{Re}\dpn{x_0, y}{\lambda} > 1$, so $x_0 \not\in A^{\circ\circ}$.
+\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}