Let $U:H\to H$ be a unitary transformation of a Hilbert space $H$. We know that for each $x\in H$ there is a unique positive Borel measure $\rho_x$ on $S^1$ such that $\hat \rho_x(n) = \langle x, U^nx\rangle$ for all $n\in \mathbb Z$. We will refer to this measure as the spectral measure associated to $x$.
I am trying to see why the following is true:
Theorem. If $x, y\in H$ are such that the associated spectral measures are mutually singular then $x$ and $y$ are orthogonal.
This is mentioned in Chapter 11 of Handbook of Dynamical Systems. The proof is mentioned as an immediate corollary of the following:
Fact. Let $\rho$ be a positive Borel measure on $S^1$. Let $M:L^2(S^1, \rho)\to L^2(S^1, \rho)$ be defined as $M\varphi = \chi\varphi$, where $\chi:S^1\to S^1$ is the identity map (note that $M$ is unitary). If $K$ is an $M$-invariant subspace of $L^2(S^1, \rho)$, then there is a Borel set $E \subseteq S^1$ such that $K = \{\varphi\in L^2(S^1, \rho):\ \varphi \text{ vanishes on } E^c\}$.
I am unable to see how to apply the above to our question.
Let $\rho_x$ and $\rho_y$ be the spectral measures associated to $x$ and $y$ and assume that $\rho_x$ and $\rho_y$ are mutually singular. If we could somehow show that $\rho_x + \rho_y$ is the spectral measures associated to some vector in $H$ then we could in fact apply the above fact (because $L^2(S^1, \rho_x)$ and $L^2(S^1, \rho_y)$ are orthogonal subspaces of $L^2(S^1, \rho_x + \rho_y)$). So I was trying to show this but that evades me.
I didn't really read your question but doesn't it follow from the functional calculus?
You have the Gelfand map $\Phi:C(\sigma) \to \mathcal{L}(H)$ which extends to a map (a $*$-homomorphism) on 'bounded Borel functions' $B(\sigma) \to \mathcal{L}(H)$ (call it $\Phi$ again). Here $\sigma$ is the spectrum of $U$. We have the spectral measures $E_{x,x}$ and $E_{y,y}$ which are induced by the requirement $$ (\Phi (f)x|x) = \int f dE_{x,x},\ (\Phi(f)y|y) = \int fdE_{y,y} $$ for all $f \in C(\sigma)$. Now observe that the same formulas holds for bounded Borel functions on $\sigma$. Let $f$ be the indicator function of $S:=\operatorname{supp}(E_{x,x})$ and $g$ be the indicator function of $T:=\operatorname{supp}(E_{y,y})$. Observe then \begin{equation} \begin{split} (\Phi(1-f)x|x) = \int_{S^c} dE_{x,x} = 0. \end{split} \end{equation} The operator $\Phi(1-f)$ is a self-adjoint projection so that the equation above shows that $\Phi(1-f)x=0$. That is, $\Phi(f)x=x$. Similarly, $\Phi(g)y=y$. We now see, $$ (x|y)= (\Phi(f)x|\Phi(g)y) = (\Phi(gf)x|y). $$ Now, if $S\cap T = \varnothing$, then $fg$ is the indicator function of the empty set.
What does it have to do with the fact that $U$ is unitary? I don't know. I'll delete this post soon if it's not helpful.