$\textbf{Problem}$ An ergodic measure preserving transformation $T$ on $(X,B,\mu)$ is called to have ${discrete \ spectrum}$ if there exists an orthonormal basis for $L^2$ which consists of eigenfunctions of $T$. Show that there is a sequence of integers $n_k$ with $n_k \rightarrow \infty$ such that for any $f \in L^2$, \begin{align*} \Vert T^{n_k}f - f \Vert _{L^2} \rightarrow 0 \end{align*}
$\textbf{Attempt}$ We know that for $\lambda \in \mathbb{C}$ with $\vert \lambda \vert =1$, there is $n_k$ such that $\lambda^{n_k} \rightarrow 1$.
Suppose that $\{f_1,f_2,...,\}$ is an orthonormal basis for $L^2$ which consists of eigenfunctions of $T$.
Then, $T f_i = \lambda_i f_i $ and $\vert \lambda_i \vert =1$. Moreover, for each $i$, there exists $\{i_j\}_{j=1}$ such that $\lambda_i^{i_j} \rightarrow 1$.
(*) Then, by using a diagonal argument, there is a sequence of integers $n_k$ such that for any $f \in L^2$,
\begin{align*}
T^{n_k}f = \lambda ^{n_k} f \rightarrow f
\end{align*}
as $n_k \rightarrow \infty$.
I am not sure whether (*) is correct or not.
Any help is appreciated...
Thank you!
The 'diagonal' argument is correct. To complete the proof you can use the following: $\|T^{n_k}f-f\| \leq \|T^{n_k}g-g\|+\|T^{n_k}f-T^{n_k}g\|+\|f-g\| \leq \|T^{n_k}g-g\|++2\|f-g\|$ becasue $T$ is measure preserving. In view of this it is enough to prove that $T^{n_k}f-f \to 0$ when $f$ is a finite linear combination of eigen functions. (Any $f \in L^{2}$ has an $L^{2}$ expansion of the type $\sum a_j f_j$). Of course, convergence does hold for finite linear combination s of $f_j$'s, so we are done.