Characters on a unital commutative C star algebra preserve star

First $1^{\ast }=1$: taking adjoints of $1b=b=b1$ and using surjectivity of the involution shows $1^{\ast }$ is a two-sided identity, hence equals $1$ by uniqueness. For a self-adjoint $s\in A$ and real $t$ the element $s+it1$ satisfies $(s+it1{)}^{\ast }(s+it1)=s^2+t^{2}1$: indeed $(s+it1{)}^{\ast }=s-it1$ by [F3] and conjugate-linearity of the involution, and multiplying out in the commutative algebra gives $s^2+its-its+t^{2}1=s^2+t^{2}1$.

$\chi$ is complex-linear with $\chi (1)=1$ and $\mathrm{\mid }\chi (x)\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }$ for all $x$; in particular $\chi (s+it1)=\chi (s)+it$.

For $a\in A$ set $s=(a+a^{\ast })\mathrm{/}2$ and $t=(a-a^{\ast })\mathrm{/}(2i)$. Conjugate-linearity and involutivity give $s^{\ast }=(a^{\ast }+a)\mathrm{/}2=s$ and $t^{\ast }=-(a^{\ast }-a)\mathrm{/}(2i)=t$, while $s+it=a$ and $s-it=a^{\ast }$. Thus both parts are self-adjoint by [F3].

For a self-adjoint $s$ and real $t$: $\mathrm{\mid }\chi (s)+it{\mathrm{\mid }}^2=\mathrm{\mid }\chi (s+it1){\mathrm{\mid }}^2\le \mathrm{\parallel }s+it1{\mathrm{\parallel }}^2=\mathrm{\parallel }(s+it1{)}^{\ast }(s+it1)\mathrm{\parallel }=\mathrm{\parallel }{s}^2+t^{2}1\mathrm{\parallel }\le \mathrm{\parallel }{s}^2\mathrm{\parallel }+t^2\le \mathrm{\parallel }s{\mathrm{\parallel }}^2+t^2$, using [step 1.1], [step 1.2], the C*-identity, the triangle inequality and submultiplicativity.

Writing $\chi (s)=u+iv$ with $u,v$ real, the inequality of [step 2.1] reads $u^2+(v+t{)}^2\le \mathrm{\parallel }s{\mathrm{\parallel }}^2+t^2$, that is, $u^2+v^2+2vt\le \mathrm{\parallel }s{\mathrm{\parallel }}^2$ for every real $t$. If $v>0$ then $t\to +\mathrm{\infty }$ makes the left side tend to $+\mathrm{\infty }$; if $v<0$ then $t\to -\mathrm{\infty }$ does the same; both contradict the uniform upper bound. Hence $v=0$ and $\chi (s)\in \mathbb{R}$ for every self-adjoint $s$.

For arbitrary $a=s+it$ as in [step 1.3]: $\chi (a^{\ast })=\chi (s-it)=\chi (s)-i\chi (t)=\overline{\chi (s)+i\chi (t)}=\overline{\chi (a)}$ by [step 3.1] and linearity.