Another question about $C^\ast$ algebra

If $A$ is a Banach algebra then let $\Omega (A)$ denote the character space of $A$. Apparently there exist abelian Banach algebras such that $\Omega (A) = \varnothing$. Also apparently, if $A$ is a (commutative) $C^\ast$ algebra then $\Omega (A) \neq \varnothing$. I am trying to prove this but got stuck.

Let $A \neq 0$ be a commutative $C^\ast$ algebra. Then there exists $a \in A$ with $\|a\|\neq 0$. By the Hahn Banach theorem there exists a continuous linear functional $\tau : A \to \mathbb C$ such that $|\tau (a)|=\|a\|$ and $\|\tau\|\le 1$. That's how it's stated on Wikipedia but of course $|\tau (a)|=\|a\|$ and $\|\tau\|\le 1$ implies $\|\tau\|=1$. Hence $\Omega(A) \neq \varnothing$.

To me this seems to be a correct proof. But in Murphy's book there is something quite different. Could someone help me and enlighten me about what is wrong with my proof above? I am including the text given in the book: Note that your Hahn-Banach functional $\tau$ will in general not be multiplicative, hence not a character.
• Thank you. If it does not follow from Hahn-Banach then why does $r(a) = \|a\|$ imply the existence of a character? – Student Jul 3 '14 at 11:53
• Note that for any complex Banach algebra and any $x \in A$ we have $\sigma(x) = \{\gamma(x) \mid \gamma \in \Omega(A)\}$, now use that your $a$ has real spectrum (as it is hermitian) and hence $r(x) = \max\sigma(x)$ or $r(x) = -\min\sigma(x)$. – martini Jul 3 '14 at 12:05
• To see the statement about the spectrum, let $\lambda \in \sigma(x)$. Then $x-\lambda$ is not invertible in $\hat A$, hence there is an maximal ideal $I$ of $\hat A$ containing $x-\lambda$. Now restrict the morphism $\hat A \to \hat A/I \cong \mathbb C$ to $A$. – martini Jul 3 '14 at 12:19
• Ok. I'm not sure I understand, can you tell me if this is the argument you have in mind: If $r(a) = \max \sigma (a) > 0$ then $\{\gamma (x) \mid \gamma \in \Omega (A)\}\neq \varnothing$ and therefore $\Omega (A) \neq \varnothing$? – Student Jul 4 '14 at 9:33
• Exactly.${}{}{}$ – martini Jul 4 '14 at 10:23