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Bokuan Li
2026-09-24 11:56:58 -04:00
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\begin{definition}[Unital Banach Algebra] \begin{definition}[Unital Banach Algebra]
\label{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. 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 $\norm{\cdot}_{1}: A \to [0, \infty)$ such that $\norm{1}_1 = 1$, and $A$ is always assumed to be equipped with this norm.
\end{definition} \end{definition}
\begin{proof}[Proof, {{\cite[Proposition I.1.3]{Takesaki1}}}. ] \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 $\norm{\cdot}_A$. 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 $\norm{\cdot}_A$.
@@ -82,7 +82,7 @@
The algebra $\tilde A$ is the \textbf{unitisation} of $A$. The algebra $\tilde A$ is the \textbf{unitisation} of $A$.
\end{definition} \end{definition}
\begin{proof} \begin{proof}
(U): For each $\lambda + x \in \tilde A$, let $\tilde \phi(\lambda + x) = \lamdba + \phi(x)$. (U): For each $\lambda + x \in \tilde A$, let $\tilde \phi(\lambda + x) = \lambda + \phi(x)$.
\end{proof} \end{proof}