Proposition 36.12.1.label Let $A$ be a non-unital $C^{*}$-algebra and $\td A$ be its unitisation, then there exists a unique norm $\norm{\cdot}_{\td A}: \td A \to [0, \infty)$ such that:
- (1)
For each $x \in A$, $\norm{x}_{\td A}= \norm{x}_{A}$.
- (2)
$(\td A, \norm{\cdot}_{\td A})$ is a unital $C^{*}$-algebra.
Proof, [Theorem 15.1, Zhu93]. For each $x \in A$ and $\lambda \in \complex$, let
be the operator seminorm corresponding to $\td A$ acting on $A$. If $\norm{xy + \lambda y}_{A} = 0$ for all $y \in A$, then $xy = -\lambda y$ for all $y \in A$. Given that $A$ is non-unital, $\lambda = 0$. Since $A$ is a $C^{*}$-algebra, $\norm{x}_{A}^{2} = \norm{xx^*}_{A} = 0$, and $x = 0$ as well. Thus $\norm{\cdot}_{\td A}$ is indeed a norm on $\td A$.
(1): Let $x \in A \setminus \bracs{0}$, then since $A$ is a Banach algebra, $\norm{x}_{\td A}\le \norm{x}_{A}$. On the other hand, as $A$ is a $C^{*}$-algebra,
(2): As $\norm{\cdot}_{\td A}$ is the operator norm corresponding to $\td A$ acting on $A$, $(\td A, \norm{\cdot}_{\td A})$ is a Banach algebra. Moreover, for any $x \in A$ and $\lambda \in \complex$,
so $(\td A, \norm{\cdot}_{\td A})$ is a $C^{*}$-algebra.
Finally, Corollary 36.4.5 implies that there can be at most one norm on $\td A$ making it a unital $C^{*}$-algebra, so the constructed norm is unique.$\square$
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