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.
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...