Added the inverse function theorem.

src/op/banach/definitions.tex

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+\section{Banach Algebras}
+\label{section:banach-algebras}
+
+\begin{definition}[Banach Algebra]
+\label{definition:banach-algebra}
+    Let $A$ be an associative algebra over $\complex$ and $\norm{\cdot}_A: A \to [0, \infty)$ be a norm, then $A$ is a \textbf{Banach algebra} if:
+    \begin{enumerate}
+        \item $A$ is complete with respect to $\norm{\cdot}_A$. 
+        \item For any $x, y \in A$, $\norm{xy}_A \le \norm{x}_A\norm{y}_A$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Unital Banach Algebra]
+\label{definition:unital-banach-algebra}
+    Let $A$ be a Banach algebra, then $A$ is \textbf{unital} if there exists $1 \in A$ such that for any $x \in A$, $x1 = 1x = x$. In which case, $1$ is the unique \textbf{multiplicative identity} of $A$. 
+\end{definition}
+
+

src/op/banach/index.tex

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+\chapter{$C*$-Algebras}
+\label{chap:banach-algebras}
+
+\input{./definitions.tex}
+\input{./invertible.tex}

src/op/banach/invertible.tex

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+\section{Invertible Elements}
+\label{section:invertible-elements}
+
+
+\begin{definition}[Invertible]
+\label{definition:banach-algebra-invertible}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $x$ is \textbf{invertible} if there exists $x^{-1} \in A$ such that $xx^{-1} = x^{-1}x = 1$. The set $G(A)$ denotes the collection of all invertible elements in $A$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:neumann-series}
+    Let $A$ be a unital banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
+    \[
+    x^{-1} = \sum_{n = 0}^\infty (1 - x)^n
+    \]
+\end{lemma}
+\begin{proof}
+    Since $\norm{1 - x}_A < 1$, the series converges absolutely. Let $y = \sum_{n = 0}^\infty (1 - x)^n$, then
+    \[
+    (1 - x) \sum_{n = 0}^\infty (1 - x)^n = \sum_{n = 0}^\infty (1 - x)^n - 1 = \sum_{n = 0}^\infty (1 - x)^n (1 - x)
+    \]
+
+    so $(1 - x)y = y - 1 = y(1 - x)$, and $xy = yx = 1$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:banach-algebra-inverse}
+    Let $A$ be a unital Banach algebra, then:
+    \begin{enumerate}
+        \item $G(A)$ is open. 
+        \item For any $x \in G(A)$ and $y \in B_A(0, \normn{x^{-1}}_A^{-1})$, 
+        \[
+        (x - y)^{-1} = x^{-1}\sum_{n = 0}^\infty (yx^{-1})^n
+        \]
+
+        \item The map $G(A) \to G(A)$ defined by $x \mapsto x^{-1}$ is $C^\infty$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): For any $x \in G(A)$ and $y \in B(0, \normn{x^{-1}}_A^{-1})$, $(x - y) = (1 - yx^{-1})x$. By \autoref{lemma:neumann-series}, 
+    \[
+    (1 - yx^{-1})^{-1} = \sum_{n = 0}^\infty (yx^{-1})^n
+    \]
+
+    so
+    \[
+    (x - y)^{-1} = x^{-1}\sum_{n = 0}^\infty (yx^{-1})^n
+    \]
+
+    (3): Since the inversion map is locally a power series, it is $C^\infty$ by \autoref{theorem:termwise-differentiation}. 
+\end{proof}
+

src/op/index.tex

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+\part{Operator Algebras}
+\label{part:operator-algebras}
+
+\input{./banach/index.tex}
+\input{./notation.tex}

src/op/notation.tex

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+\chapter{Notations}
+\label{chap:op-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    $1$ & Identity element of a unital algebra. & \autoref{definition:unital-banach-algebra} \\
+\end{tabular}