Spectral theorem, how to show spectral measure has no atoms This is from Peter Walters' An Introduction to Ergodic Theory. He quotes the "Spectral Theorem for Unitary Operators" which I understand:
If $U$ is a unitary operator on a complex Hilbert space $H$, then for each $f \in H$ there is a unique finite Borel measure $\mu_f$ on the circle $K=\{|z|=1\}\subseteq \mathbb{C}$ such that
$$
\langle U^n f, f\rangle = \int_K \lambda^n \mu_f(d\lambda),\qquad \forall n \in \mathbb{Z}.
$$
But he also remarks that if $T$ is measure preserving with measure preserving inverse then $U_Tf:= f \circ T$ is a unitary operator on $L^2$.
Further (and this is the part I don't understand) if $1$ is the only eigenvalue of $U_T$ with constants the only eigenvectors and $\langle f,1\rangle = 0$, then $\mu_f$ has no atoms.
I don't see how to show this. Clearly I can take $n=0$ and the hypotheses tell me $\mu_f(K) = 0$ but I don't see much else that can be said from this. Can I construct a non-constant eigenvector if I knew $\mu_f\{x\}>0$ for some $x$?
 A: It may helps you:

Theorem.
Let $U$ be a (linear) isometry of a separable Hilbert space $\mathcal{H}$, and $H_{e}(U)$ be the closure of the subspace spanned by the eigenvectors of $U$.
Then for $f \in \mathcal{H}$, $$f \perp H_{e}(U) \Rightarrow \mu_{f} \mbox{ has no atoms.}$$

Proof.
Let $f \perp H_{e}(U)$ and $W = e^{-2\pi i x}U$. Then by von Neumann ergodic theorem, let $$g = \lim \limits _{n \rightarrow \infty} {1 \over n} \sum \limits _{k = 0} ^{n-1} W^{k}f.$$ Then $Wg = g$ so that $g$ is an eigenvector of $U$ with eigenvalue $e^{2\pi i x}$ if $g \neq 0$.
Now we have $$\left\langle g, f \right\rangle  = \lim \limits _{n \rightarrow \infty} {1 \over n} \sum \limits _{k = 0} ^{n-1} e^{-2\pi i x k } \left\langle U^{k}f, f \right\rangle = \lim \limits _{n \rightarrow \infty} \int_{0} ^{1} {1 \over n} \sum \limits _{k = 0} ^{n-1} e^{-2\pi i (x-y) k } \mu_{f}(dy) = \mu_{f}(x).$$
Hence if $\mu_{f}(x) > 0$, then $g \in H_{e}(U)$ and $f \not\perp g$, which contradicts to the assumption $f \perp H_{e}(U)$. This completes the proof.


