character space of C*-algebra I have a question about a passage in the book C*-algebras and Operator Theory by Murphy. On page 41, between theorems 2.1.9 and 2.1.10 he proves that the character space of a non-unital, non-zero, abelian C*-algebra $A$ is non-empty. The proof is as follows:
Let $A'$ be the unitization of $A$. $A$ contains a non-zero hermitian element $a$. Since $r(a)=\|a\|$. Then by theorem 2.1.1, it follows that there is a character $\tau$ on $A'$ such that $|\tau(a)|=\|a\|\neq0$. Then the restriction of $\tau$ to $A$ is a non-zero homomorphism from $A$ to $\mathbb{C}$, i.e. a character on $A$.
My question is: why does $r(a)=\|a\|$ imply that there exists a character $\tau$ such that $|\tau(a)|=\|a\|$? I konw $r(a)=\|\hat{a}\|_\infty$, where $\hat{a}$ is the Gelfand transform of $a$. This would imply the existence of a character $\tau_2$ such that $|\tau_2(a)|>r(a)-\varepsilon$ for any $\varepsilon>0$. But why equal to $r(a)$. Is this because $\sigma(a)=\hat{a}(\Omega(A'))$, and $\hat{a}$ is continuous and $\Omega(A')$ is compact since $A'$ is unital? (implying $\sigma(a)\subset\mathbb{C}$ is compact, hence closed)
Is the use of the unitization $A'$ really necessary here, or could the $\tau_2$ i suggesed above suffice?
 A: Yes, that is one way to look at it. Here's another: even before the Gelfand transform, we already know from Lemma 1.2.4 that $\sigma(a)$ is compact. Here it is only proven for unital Banach algebras, but that is simply because Murphy only defines the spectrum of an element in a non-unital algebra on page 13, at the very end of section 1.2. (Think about it: how could we even define a spectrum if there is no unit?) Since $\sigma_{\tilde A}(a)$ is compact and we have $\sigma_A(a) = \sigma_{\tilde A}(a)$ by definition, we see that $\sigma_A(a)$ is also compact.
Now we use theorem 1.3.4. The image of the compact set $\sigma(a)$ under the continuous map $z \mapsto |z|$ is again compact, so the maximum is attained. (The maximum is of course equal to $r(a)$.) Also, the maximum does not arise from the extra $\{0\}$ we adjoined in theorem 1.3.4(2), since we have $r(a) = ||a|| > 0$. Thus, there is some $\tau \in \Omega(A)$ such that $|\tau(a)| = r(a)$ holds. Now we have $|\tau(a)| = r(a) = ||a||$, as desired.
I think this is essentially the same argument as the one you suggested. Most of the machinery that Murphy develops up to the Gelfand transform can again be derived as a consequence of the Gelfand transform.

As for your second question, it seems that the use of the unitisation is not really necessary, as both of our methods show. But then again, whenever you're talking about the spectrum of an element in a non-unital algebra, you're implictly using some information about its unitisation.
