\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 $(A, \norm{\cdot}_A)$ 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, there exists an equivalent norm $\norm{\cdot}_{1}: A \to [0, \inftu)$ such that $\norm{1}_1 = 1$, and $A$ is always assumed to be equipped with this norm.  
\end{definition}
\begin{proof}[Proof, {{\cite[Proposition I.1.3]{Takesaki1}}}. ]
    For each $x \in A$, let $L_x \in L(A; A)$ be defined by $y \mapsto xy$, and let $\norm{x}_1 = \norm{L_x}_{L(A; A)}$, then $\norm{x}_1 \le \norm{x}_A$ and $\norm{1}_1 = 1$. On the other hand, $\frac{\norm{x}_A}{\norm{1}_A} \le \norm{x}_1$, so $\norm{\cdot}_1$ is equivalent to $\norm{\cdot}_A$.
\end{proof}

\begin{definition}[Homomorphism]
\label{definition:banach-algebra-homomorphism}
    Let $A, B$ be Banach algebras and $\phi: A \to B$, then $\phi$ is a \textbf{homomorphism} if:
    \begin{enumerate}
        \item $\phi \in L(A; B)$. 
        \item For each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. 
    \end{enumerate}
\end{definition}

\begin{definition}[Unital Homomorphism]
\label{definition:banach-algebra-unital-homomorphism}
    Let $A, B$ be unital Banach algebras and $\phi: A \to B$ be a homomorphism, then $\phi$ is a \textbf{unital homomorphism} if $\phi(1) = 1$. 
\end{definition}

\begin{definition}[Unitisation]
\label{definition:unitisation}
    Let $A$ be a Banach algebra over $\complex$, and $\tilde A = \complex \oplus A$ with
    \[
    \iota: A \to \complex \oplus A \quad x \mapsto 0 + x
    \]
    
    For each $\lambda + x, \mu + y \in \tilde A$, define
    \[
    (\lambda + x)(\mu + y) = \lambda \mu + (\lambda x + \mu x + xy)
    \]
    
    and
    \[
    \norm{\lambda + x}_{\tilde A} = |\lambda| \norm{x}_A
    \]

    then
    \begin{enumerate}
        \item $\tilde A$ is a unital associative algebra over $\complex$.
        \item $\iota: A \to \tilde A$ is a homomorphism. 
        \item[(U)] For any pair $(B, \phi)$ satisfying (1) and (2), there exists a unique continuous unital homomorphism $\tilde \phi: \tilde A \to B$ such that $\phi(1) = 1$ and the following diagram commutes:
        \xymatrix{
        \tilde A \ar@{->}[r]^{\tilde \phi } & B \\
        A \ar@{->}[u]^{\iota} \ar@{->}[ru]_{\phi} & 
        }
        \item $\iota(A)$ is a closed two-sided ideal of $\tilde A$. 
    \end{enumerate}

    The algebra $\tilde A$ is the \textbf{unitisation} of $A$. 
\end{definition}
\begin{proof}
    (U): For each $\lambda + x \in \tilde A$, let $\tilde \phi(\lambda + x) = \lamdba + \phi(x)$. 
\end{proof}