Properties of a $B^\ast$-algebra Defining a complex Banach algebra as a $B^\ast$-algebra when it is equipped with an application $\ast:B\to B$ such that for any $x,y\in B$ $$(x+y)^\ast=x^\ast+y^\ast,\quad(xy)^\ast=x^\ast y^\ast,\quad(\alpha x)^\ast=\bar{\alpha}x^\ast,\quad(x^\ast)^\ast=x,\quad\|xx^\ast\|=\|x\|^2$$I read and would like to find a proof that


*

*a commutative $B^\ast$-algebra is regular, i.e. $\|x^2\|=\|x\|^2$ [edit: I hope I'm right in proving it by noticing that, because of commutativity and since $\|y\|^2=\|yy^\ast\|$, it follows from $\|x\|^4=\|xx^\ast\|^2=\|(xx^\ast)(xx^\ast)^\ast\|=\|x^2(x^2)^\ast\|=\|x^2\|^2$]

*for any $B^\ast$-algebra and for any non-trivial continuous multiplicative linear functional $f_M:B\to\mathbb{C}$ defined on it we have $\forall x\in B\quad \overline{f_M(x)}=f_M(x^\ast)$.


Could anybody help me with a proof or a link to one? $\infty$ thanks!!!
 A: B*-algebra is just an old-fashioned name for an abstract C*-algebra. Can you use the Gelfand–Naimark theorem? 
If so, then the first clause follows directly from it as in this case your algebra is isometrically *-isomorphic to the space of continuous functions vanishing at infinity $C_0(X)$ defined on some locally compact Hausdorff space $X$. You can use it also to derive the second clause. Indeed, all non-zero multiplicative linear functionals on $C_0(X)$ are just point evaluations for which this is clear.
A: *

*Suppose that $xx^{\star}=x^{\star}x$ for some $x \in B$. Then $\|x^{2}\|=\|x\|^{2}$ holds because
$$
   \|x^{2}\|^{2}=\|x^{2}(x^{2})^{\star}\|=\|(xx^{\star})(xx^{\star})\|=\|(xx^{\star})(xx^{\star})^{\star}\|=\|xx^{\star}\|^{2}=\|x\|^{4}.
$$

*The second part of showing that $\overline{f(x)}=f(x^{\star})$ for all $x$ is equivalent to showing that $f(x)$ is real if $x=x^{\star}$ because, in this case, $x+x^{\star}$ and $i(x-x^{\star})$ are real and, hence,
$$
         f(x)+f(x^{\star})=\overline{f(x)}+\overline{f(x^{\star})} \\
         if(x)-if(x^{\star})=-i\overline{f(x)}+i\overline{f(x^{\star})}
$$
Adding the first to $-i$ times the second then yields $f(x)=\overline{f(x^{\star})}$. I'll carry out this part only in the case where $B$ has a unit $u$, because then one may use the exponential
$$
                e^{itx} = u+\frac{(itx)}{1!}+\frac{(itx)^{2}}{2!}+\frac{(itx)^{3}}{3!}+\cdots.
$$
This series converges for all complex $t$, but is needed here only for real $t$. The usual exponential properties hold for this series, including $e^{itx}e^{it'x}=e^{i(t+t')x}$ for all $t,\;t'\in\mathbb{R}$. And $e^{itx}e^{-itx}=u$ is the unit. Adjoint is isometric and, therefore, $(e^{itx})^{\star}=e^{-itx^{\star}}$. Finally, because $f$ is continuous and because $f(u)=1$ must hold,
$$
        f(e^{itx})=f(u)+\frac{(itf(x))}{1!}+\frac{(itf(x))^{2}}{2!}+\frac{(itf(x))^{3}}{3!}+\cdots= e^{itf(x)}.
$$
Therefore, if $x=x^{\star}$,
$$
\begin{align}
     |e^{itf(x)}|^{2}=|f(e^{itx})|^{2}
                 & \le \|f\|^{2}\|e^{itx}\|^{2} \\
                 & =\|f\|^{2}\|e^{itx}(e^{itx})^{\star}\| \\
                 & =\|f\|^{2}\|e^{itx}e^{-itx}\| \\
                 & =\|f\|^{2}\|u\|.
\end{align}
$$
Because the above is uniformly bounded for all $t\in\mathbb{R}$, then it must be that $f(x)$ is real whenever $x=x^{\star}$, which finishes the proof.

