Spectral radius in Banach algebra is commutative

Question

I want to show that for a Banach algebra $A$ and elements $x,y \in A$, we have $$ r_A(xy) = r_A(yx), $$ where $r_A$ is a spectral radius. This is how I am trying to do that: $$ r_a(xy) = \lim_{n \rightarrow \infty} \| (xy)^n \| ^{\frac1n}=\lim_{n\rightarrow \infty} \|x(yx)^{n-1} y \|^{\frac1n}. $$ And this is where I stuck, since I only know that $\|xy\| \leqslant \|x\| \|y\|$. Could you please suggest any ideas on how to proceed with the proof?

Answer 1

Note that $$\|x(yx)^{n-1}y\|^{1/n}\leq \|x\|^{1/n}\|(yx)^{n-1}\|^{1/n}\|y\|^{1/n}=\|x\|^{1/n}\|y\|^{1/n}(\|(yx)^{n-1}\|^{1/(n-1)})^{(n-1)/n}.$$ Now think about what happens as $n\to\infty$.