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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 $xf(y)=f(y)x$

Leo
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  • $xy^5=y^5x$; you can cancel out the $x$ here. –  Apr 21 '20 at 10:21
  • Also, yes $xy$ = $yx$; it's just the $x$ and $y$ are the other way around. –  Apr 21 '20 at 10:28
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    Sorry but how could I cancel out x? Please explain. And even , how could cancelling out x gonna help me to get the result. – Leo Apr 21 '20 at 11:04
  • Well $x=x$, so if you divide both sides by $x$, then it'll equal $1$. I'm not so sure how it would help you answer the question, but simplifying things down and understanding the formulae can help you see clearer and find the answer you are looking for. –  Apr 21 '20 at 11:10
  • @JosephHulme: do you even know what a C$^*$-algebra is? – Martin Argerami Apr 21 '20 at 19:29
  • @MartinArgerami no, sorry i didn't realize what i was getting involved in –  Apr 21 '20 at 19:31

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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 $xy_0^{1/5}=y_0^{1/5}x$.

Aweygan
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  • The proof is really nice. Can you suggest anything for this question: x and y is positive and x.y^2 = y^2.x , does it imply e^y.x=x.e^y, with the help continuous functional calculas – Leo Apr 22 '20 at 05:11
  • In this case for any t belongs to the spectrum of y^2 , does the map f(t)=e^sqrt(t)gonna work similar way? – Leo Apr 22 '20 at 05:19
  • Yes, the proof will look very similar. – Aweygan Apr 22 '20 at 05:43