Application of continuous functional calculas

Question

I'm going through the topic C* algebra and facing few questions . It would be great if you people could help me to clear the doubts. Q2. Let $x$ and $y$ be two positive elements in a C* algebra such that ( Then $x$ and $y$ are normal so we can apply continuous functional calculus) $xy^5=y^5x$. Does it imply $xy= yx$. Any hint will be appreciated. I have tried to approach which may be redundant: $xy^5=y^5x$ $=> x(Pol y) = (Pol y)x$. Now how to apply continuous functional calculus to get $xf(y)=f(y)x$

Answer 1

The approach you outline works. Write $y_0=y^5$. We want to show that if $xy_0=y_0x$, then $xy_0^{1/5}=y_0^{1/5}x$. To this end, assume $x\neq0$, fix $\varepsilon>0$, and choose a polynomial $p$ such that $|p(\lambda)-\lambda^{1/5}|<\varepsilon/(2\|x\|)$ for all $\lambda$ in the spectrum of $y_0$. Then we have \begin{align*}\|xy_0^{1/5}-y_0^{1/5}x\|&=\|xy_0^{1/5}-xp(y_0)+p(y_0)x-y_0^{1/5}x\|\\ &\leq\|xy_0^{1/5}-xp(y_0)\|+\|p(y_0)x-y_0^{1/5}x\|\\ &\leq2\|x\|\|y_0^{1/5}-p(y_0)\|\\ &<\varepsilon. \end{align*} Since $\varepsilon$ was arbitrary, it follows that $xy_0^{1/5}=y_0^{1/5}x$.