Talk:Arveson spectrum

A rapid check shows that $\hat{x}(n)$ does not exactly satisfy the indicated equation but rather with an inverse $$U_z \hat{x}(n) = (z)^{-1} \hat{x}(n)$$

Even if there is no ambiguity, it would also be better if we make it clear that $\hat{x}(n)$ is still a function on $T$ (while usually in Fourier transform, the transform is a function on the "Fourier space" which I admit is just a word that doesn't explain anything)