A property of unital commutative Banach *-algebra

Question

Problem: Let $\mathcal{B}$ be a unital and commutative Banach $*$-algebra, and $\|x x^* \| = \|x\|^2, \forall x \in \mathcal{B}$. Prove that $\|x^2 \| = \|x \|^2, \forall x \in \mathcal{B}$.

Attempt: This seems like a simple algebra problem I can't seem to get the result. I can easily show $\|x^2 \| \le \|x \|^2$, but the other direction is giving me trouble. Can anyone give me some hints on how to proceed?

Answer 1

First show that if $x=x^*$, then $\|x^2\|=\|x\|^2$ (quite easy). Note that if $(xx^*)^*=xx^*$ for all $x\in\mathcal B$. Then show that for any $x\in\mathcal B$, $\|x\|=\|x^*\|$ (also quite easy). Then combine these results.

Answer 2

$\|x^{\ast}\|^{2}=\|x^{\ast}(x^{\ast})^{\ast}\|=\|x^{\ast}x\|=\|xx^{\ast}\|=\|x\|^{2}$, so $\|x^{\ast}\|=\|x\|$.

Now \begin{align*} \|x\|^{4}&=\|xx^{\ast}\|^{2}\\ &=\|(xx^{\ast})(xx^{\ast})^{\ast}\|\\ &=\|x^{2}(x^{\ast})^{2}\|\\ &\leq\|x^{2}\|\|(x^{\ast})^{2}\|\\ &=\|x^{2}\|\|(x^{2})^{\ast}\|\\ &=\|x^{2}\|\|x^{2}\|\\ &=\|x^{2}\|^{2}. \end{align*}