character on commutative $C^*$-algebras involutive? Let $A$ be a commutative $C^*$-algebra, $0\neq f:A\to\mathbb{C}$ a linear, multiplicative map (a so called character).  Is $f(a^*)=f(a)^*$ for all $a\in A$? 
I think so, since f can be seen as a point-evaluation map ev$_x$ on $C_0(X)$ for a loccaly compact Hausdorff space X, where $x\in X$. And this point-evaluation maps preserve *. So f must be involutive. Is it correct?
 A: $A$ doesn't even have to be commutative, here's a short outline to how to prove this: Let $A$ be any unital $C^*$-algebra. Consider any character $f:A \to \mathbb{C}$. It's sufficient to prove that $f(x) \in \mathbb{R}$ for all self-adjoint $x \in A$. You can prove that the spectrum of $x$ is real. Because $f(x) \in \sigma(x)$, we must have $f(x) \in \mathbb{R}$. Now if $A$ is non-unital, you can simply extend $f:A \to \mathbb{C}$ to $\bar{f}:A_1 \to \mathbb{C}$ where $A_1$ is its unitisation. It then follows by the previous result. 
As a side note, here's a way to prove $\sigma(x)$ is real when $x$ is self-adjoint without using characters. First we prove the spectra of a unitary element $u$ is contained in $S^1$. Note that $\lVert u \rVert^2 = \lVert u^* u \rVert =1$ so $r(u) \leq 1$ where $r$ denotes the spectral radius. We also have that if $\lambda \in \sigma(u)$, $\lambda^{-1} \in \sigma(u^*)$. So $\lvert \lambda \rvert \leq 1$ and $\lvert \lambda \rvert \geq 1$, from which we conclude that $\lvert \lambda \rvert=1$. Now if $x$ is self-adjoint, by the holomorphic calculus, $e^{ix}$ is unitary. Then by the spectral mapping theorem, $e^{i\sigma(x)} = \sigma(e^{ix}) \subset S^1$. So $\sigma(x) \subset \mathbb{R}$.
As for a reference, I'm pretty sure this argument (or a very similar one) is made in Murphy's $C^*$-algebras and Operator Theory.  
A: Here is another proof. As mentioned by Demophilus, one can assume that $A$ is unital. Furthermore, it suffices to show that $f(a) \in \mathbb{R}$ if $a = a^{\ast}$.
Now, we know that $\|f\| = 1$, so if $t\in \mathbb{R}$, then
$$
|f(a+it)|^2 \leq \|a+it\|^2 = \|(a-it)(a+it)\| = \|a^2+t^2\| \leq \|a^2\|+t^2
$$
So if $f(a) = \alpha + i\beta$, then
$$
|\alpha|^2 + (\beta+t)^2 \leq \|a^2\| + t^2
$$
Hence,
$$
|\alpha|^2 + 2t\beta \leq \|a^2\| \quad\forall t\in \mathbb{R}
$$
Let $t\to \infty$ to conclude that $\beta = 0$, so $f(a) \in \mathbb{R}$.
