Eigenfunction of $T^n$ vs. eigenfunction of $T$ when $T$ is ergodic Let $(X, \mathcal{B}, \mu, T)$ be an ergodic system, and suppose $f \in L^2(X, \mathcal{B}, \mu)$ is an eigenfunction of $T^n$ for some integer $n > 1$. Is it possible to write $f$ as a linear combination (finite or infinite) of eigenfunctions of $T$?
It seems like a fairly easy question (and I believe the answer is yes), but I've been stuck with this for a couple of days. Any thoughts would be appreciated.
Thanks!
 A: You do not need $T$ to be ergodic for this to hold. Measure-preserving is enough. And you get finite sums.
We are interested in the unitary operator
$$
u_T:f\longmapsto f\circ T
$$
on $L^2$ via the so-called Koopman representation.
By the spectral mapping, the spectrum of $u_T^n=u_{T^n}$ is $\sigma(u_T^n)=\{\lambda^n\;;\;\lambda\in \sigma(u_T)\}$. In particular, if $\mu=\lambda^n$ is an eigenvalue in the latter, we have
$$
\{0\}\neq \mbox{Ker}(u_T^n-\mu Id)=\bigoplus_{\lambda^n=\mu}\mbox{Ker}(u_T-\lambda Id). 
$$
So any eigenfunction of $u_T^n$ is a finite sum of eigenfunctions of $u_T$. Not every summand of the rhs decomposition must be nonzero, but at least one must be.
Note: the above really is for the case $\mu\neq 0$ and is due to the fact that $X^n-\mu$ has $n$ distinct roots. Since $u_T$ is unitary, so is $u_T^n$ and their spectra are both contained in the unit circle. So the case $\mu=0$ never occurs. The lemma we used is, for an operator $u$, and a polynomial $p(X)=\prod_{j=1}^np_j(X)$ whose factors are pairwise coprime, the one that says: $$\mbox{Ker}\,p(u)=\bigoplus_{j=1}^n\mbox{Ker}\,p_j(u).$$
