Added the successive approximations.

src/fa/norm/index.tex

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+\chapter{Normed Spaces}
+\label{chap:normed-spaces}
+
+\input{./src/fa/norm/normed.tex}

src/fa/norm/normed.tex

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+\section{Normed and Banach Spaces}
+\label{section:normed-banach}
+
+
+\begin{theorem}[Successive Approximation]
+\label{theorem:successive-approximation}
+    Let $E, F$ be normed spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:
+    \begin{enumerate}
+        \item[(a)] $\norm{x}_E \le C\norm{y}_F$.
+        \item[(b)] $\norm{y - Tx}_F \le \gamma \norm{y}_F$.
+    \end{enumerate}
+    then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
+    \begin{enumerate}
+        \item $\sum_{n \in \natp}\norm{x_n}_E < C\norm{y}/(1 - \gamma)$.
+        \item $\sum_{n = 1}^\infty Tx_n = y$.
+    \end{enumerate}
+    In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}/(1 - \gamma)$ and $Tx = y$.
+\end{theorem}
+\begin{proof}
+    Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}} \le \gamma \norm{y_n}_F$.
+
+    For each $n \in \nat$,
+    \[
+    \norm{y_n}_F \le \gamma^{n - 1}\norm{y_1}_F = \gamma^{n - 1}\norm{y}_F
+    \]
+    Since $\norm{x_n}_E \le C\norm{y_n}_F$,
+    \[
+    \sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}}{1 - \gamma}
+    \]
+    In addition,
+    \[
+    \norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}} \le \gamma^n \norm{y}_F
+    \]
+    so $\sum_{n = 1}^\infty Tx_n = y$.
+\end{proof}