Banach Algebra: $r_\sigma(x)=\|x\| \iff \|x^2\|=\|x\|^2$

Question

I am trying to see that if we have a complex Banach algebra with unity, we will have that $$r_\sigma(x)=\|x\| \iff \|x^2\|=\|x\|^2.$$

I was able to do the first implication: Since we know that $\|x^2||\leq \|x||^2$ always and we have that $r_\sigma(x^2)=r_\sigma(x)^2=\|x\|^2$ so we will have that $\|x^2\|\geq r_\sigma(x^2)=\|x\|^2$.

Now for the other one, I haven't figured out how to do it yet I just can't seem to see why we have $r_\sigma(x)\geq \|x\|$ if $\|x^2\|=\|x\|^2$, so any tips are appreciated. Thanks.

Answer 1

Hint: Since $\|x^2\|=\|x\|^2,$ therefore $\|x^{2^n}\|=\|x\|^{2^n}\\$ for all $n \in \mathbb{N}.$

Edit: To see this, consider $\|x\|^4=(\|x\|^2)^2=\|x^2\|^2=\|x^4\|.$ Now repeated application/induction proves it for all $n \in \mathbb{N}.$

Answer 2

$||x^2||=||x||^2 \implies ||x^2||^{\frac{1}{2}} =||x|| \implies ||x^{2^n}||^{\frac{1}{2^n}} =||x||\\$ $r(x)=\lim_n||x^{2^n}||^{\frac{1}{2^n}} =||x||$.