Originally Posted by

**Mauritzvdworm** This has to be something stupid which I'm just not seeing. Let $\displaystyle R_{\theta}(g)(z)=g(e^{2\pi i \theta}z)$ with $\displaystyle g\in C(\mathbb{T})$ and $\displaystyle z\in\mathbb{T}$

Further, let $\displaystyle M_f$ be the multiplication operator on $\displaystyle B(L^2(\mathbb{T}))$ with $\displaystyle f\in C(\mathbb{T})$

$\displaystyle R_{\theta}$ can be extended to a unitary operator on $\displaystyle B(L^2(\mathbb{T}))$

show that the following commutation relation holds

$\displaystyle R_{\theta}M_{id}=e^{2\pi i \theta}M_{id}R_{\theta}$

Where $\displaystyle id$ denotes the identity function

This is what I have:

$\displaystyle (R_{\theta}M_{id})(g)(z)=R_{\theta}(M_{id}(g)(z))= R_{\theta}(id(z)g(z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)$

$\displaystyle M_{id}R_{\theta}(g)(z)=M_{id}(R_{\theta}(g)(z))=M_ {id}(g(e^{2\pi i \theta}z))=id(e^{2\pi i \theta}z)g(e^{2\pi i \theta}z)$

but here the commutation relation does not hold. Am I using the definitions wrongly?