Example of an ergodic transformation with some properties in the spectrum of the Koopman operator Let $(X,\mathcal{B},\mu)$ a probability space, for simplicity we can assume $X$ a metric space. Let $T: (X,\mathcal{B},\mu) \to (X,\mathcal{B},\mu) $ be an invertible, measure-preserving transformation. We can now define the (linear) operator $U_T : L^2 (X,\mathcal{B},\mu) \to L^2(X,\mathcal{B},\mu)$ by the following formula:
$$U_T f = f\circ T \text{ for }f \in L^2(X,\mathcal{B},\mu).$$
I would like to know if there is any example of an operator $T$ satisfying the above hypotheses and

*

*$\mathbf{1}$ is a eigenvalue of $U_T$;

*the dimension of the eigenspace of $\mathbf{1}$ is $1$;

*the dimension of the generalized eigenspace of $\mathbf{1}$ is greater than $1$.

Notes:

*

*The operator $U_T$ is known as the Koopman Operator;

*This question arose because I have a doubt about the definition of simple eigenvalue, since it is known that if $1$ is a simple eigenvalue of $U_T$, then $T$ is ergodic. In my head, there is a question whether the fact of being simple is talking about eigenspace or generalized eigenspace. In the literature, I always see it being treated as the dimension of the eigenspace, so there should be an example satisfying the requirements of my question, but I haven't been able to come up with one.

 A: There is no such example.  This is because the Koopman operator $U_T$ is unitary and hence $1-U_T$ is normal.
If $\xi$ is a vector such that  $(1-U_T)^n\xi=0$, for some $n$, one would also have that $(1-U_T)\xi=0$.  Thus the generalized eigenspace coincides with the standard eigenspace.

EDIT:
The proof I had in mind for the fact that the generalized eigenspaces coincide with standard eigenspaces uses
the Spectral Theorem, but here is a more pedestrian argument:
Lemma 1. If  $T$ is a bounded, self-adjoint operator on a Hilbert space $H$, and if $\xi $ is a vector in $H$ such that
$T^2\xi =0$, then $T\xi =0$.
Proof. We have
$$
  \|T\xi \|^2= \langle T\xi ,T\xi \rangle = \langle T^*T\xi ,\xi \rangle =\langle T^2\xi ,\xi \rangle =0.
  $$
QED.
Lemma 2. If  $N$ is a bounded, normal operator on a Hilbert space $H$, and if $\xi $ is a vector in $H$ such that
$N^2\xi =0$, then $N\xi =0$.
Proof. By hypothesis
$$
  0 = (N^*)^2N^2\xi  = N^*N^*NN\xi = N^*NN^*N\xi  = (N^*N)^2\xi ,
  $$
so $N^*N\xi =0$, by Lemma 1.  Consequently,
$$
  \|N\xi \|^2= \langle N\xi ,N\xi \rangle = \langle N^*N\xi ,\xi \rangle =0.
  $$
QED.
Lemma 3. If  $N$ is a bounded, normal operator on a Hilbert space $H$, and if $\xi $ is a vector in $H$ such that
$N^q\xi =0$, for some integer $q>0$,  then $N\xi =0$.
Proof. Choose $p>0$ such that $p+q$ is a power of 2, say $p+q=2^k$.  Then
$$
  N^{2^k}\xi =
  N^pN^q\xi =0.
  $$
Observing that  $N^{2^k}$ is the square of the normal operator $N^{2^{k-1}}$, it follows from Lemma 2 that
$N^{2^{k-1}}\xi =0$, and then induction implies that $N\xi =0$,  as desired.  QED.
Theorem. If  $N$ is a bounded, normal operator on a Hilbert space $H$, and if $\xi $ is a generalized eigenvector of
$N$ for the eigenvalue $\lambda $, then
$\xi $ is an eigenvector of
$N$ for $\lambda $.
Proof. By definition,  there exists some integer $q>0$, such that $(N-\lambda )^q\xi =0$, so $(N-\lambda )\xi =0$, by Lemma 3.  QED
PS:  I wonder if there is a shorter, equally elementary argument for this...
