Uniqueness of the involution on a $C^*$-algebra indication please
Let $A$ be a C*-algebra. Suppose that there exists on $A$ another involution $x\rightarrow x^{\#}$ such that $\|xx^{\#}\|=\|x\|^2$, for all $x\in A$. Prove that $x^{\ast}=x^{\#}$, for all $x \in A$.
 A: After all, it can be done with standard $C^*$-facts. Here is a detailed sketch.
Without loss of generality, we may assume that $A$ is unital. I'll denote $x'$ the second involution.
Let $\phi$ be a linear functional on $A$. Then $\phi$ is positive (preserves positive elements) if and only if $\phi$ is bounded and $\|\phi\|=\phi(1)$ (classic fact).
It follows that the positive linear functionals are the same for both involutions.
To prove that the two involutions are equal, it is clear that it suffices to show that they have the same self-adjoint elements.
So let $h=h^*$. Then for every positive functional $\phi$, we have $\phi(h)\in \mathbb{R}$. On the other hand, $\phi(h')=\overline{\phi(h)}$ (why?). So $\phi(h')=\phi(h)$, whence $\phi(i(h'-h))=0$ for every positive functional. 
Now for every self-adjoint element $x$, there exists a pure state (extreme point of the state space which is the set of positive functionals of norm $1$) such that $|\phi(x)|=\|x\|$ (classic fact). 
Applying this to $i(h'-h)$ yields 
$$
0=|\phi(i(h'-h))|=\|i(h'-h)\|\qquad\Rightarrow \qquad h'=h.
$$
By symmetry, we get $h^*=h$ if and only if $h'=h$, which completes the proof.
