I've been trying to solve the following exercise from Foundations of Ergodic Theory – Krerley Oliveira and Marcelo Viana:
Let $R_{\theta}\colon S^{1}\to S^{1}$ be an irrational rotation and $m$ be the Lebesgue measure on the circle. Calculate the eigenvalues and the eigenvectors of the Koopman operator $U_{\theta}\colon L^{2}\left(m\right)\to L^{2}\left(m\right).$ Show that the spectrum of $U_{\theta}$ coincides with the unit circle $\left\{z\in \mathbb{C}: |z|=1 \right\}.$
I can show that the set of functions on the form $\varphi_{k}\left(x\right)=e^{2\pi i k x}$ form a Hilbert basis for $L^{2}\left(m\right)$ and are all eigenfunctions (with eigenvalue $\lambda_{k}=e^{2\pi i k \theta}$). As $\theta \in \mathbb{R}\setminus \mathbb{Q}$ this basis is a dense subset in $S^{1}.$
So, because the spectrum is a compact set and $U_{\theta}$ is an isometry, we have the following: $$S^{1}=\overline{\left\{\text{eigenvalues of }U_{\theta}\right\}} \subseteq \text{spec}\left(U_{\theta}\right) \subseteq \mathbb{D},$$
where $\mathbb{D}=\left\{z\in \mathbb{C}: |z|\leq 1\right\}.$
But this doesn't answer neither questions stated above: are those the only eigenvalues and eigenvector? Is that true that $\text{spec}\left(U_{\theta}\right)\subseteq S^{1}?$ I can't see why this are/aren't true… If someone can help I would appreciate a lot!
Thanks!
Best Answer
Note that the collection $\{\varphi_k\}_{k\in\mathbb Z}$ is an orthonormal basis for $L^2(S^1)$. Since eigenvectors of normal operators with distinct eigenvalues are orthogonal, this implies that there are no other eigenvectors/eigenvalues for $U_\theta$.
Since the set $\{e^{2\pi ik\theta}:k\in\mathbb Z\}$ is dense in $S^1$ (as $\theta$ is irrational), you have $S^1\subset\operatorname{spec}(U_\theta)$. Since $U_\theta$ is a unitary, i.e. $U_\theta^*U_\theta=U_\theta U_\theta^*=1$, every element $z$ of the spectrum of $U_\theta$ satisfies $|z|^2=1$, that is, $z\in S^1$. Thus we have $\operatorname{spec}(U_\theta)\subset S^1$, and therefore equality holds.