I want to show the following theorem:
Let U be an unitary operator and denote $L_0^2(X) := (\mathbb{C}\cdot 1)^\perp$. The following are equivalent:
- $\langle U^n f,g \rangle \to \langle f,1\rangle\cdot\langle 1,g\rangle$ as $n\to\infty$ for every $f,g\in L^2(X)$
- $\langle U^n f,g \rangle \to 0$ as $n\to\infty$ for every $f,g\in L_0^2(X)$
Here $U^1=U$ and $U^n=U^{n-1}\circ U$ for $n\geq 1$
1 $\Rightarrow$ 2 is easy and follows from the definition of $L_0^2(X)$. I'm interested in the the second implication. I tried $L^2(X)=L_0^2(X)\oplus \mathbb{C}\cdot 1$ but that does not seem to be enough.
Edit: Constant functions are $U$-invariant
If the constants are invariant under $U$ then they are also invariant under $U^*$. Now suppose 2. holds and note that for $f\in L_0$ you have:
$$\langle U^nf, 1\rangle = \langle f, (U^*)^n1\rangle = \langle f, 1\rangle =0$$ and then for $f = \tilde f +\lambda1$, $g=\tilde g+\mu1$ with $\tilde f, \tilde g\in L_0$:
$$\langle U^nf,g\rangle = \langle U^n \tilde f, \tilde g\rangle + \langle\lambda U^n1, \tilde g\rangle + \langle U^n\tilde f, \mu 1\rangle +\langle\lambda1,\mu1\rangle$$ Here the first term goes to $0$ as $n\to\infty$ and the second and third terms are $0$ for all $n$ since the constants are invariant under $U$. So:
$$\langle U^nf,g\rangle \to \langle\lambda 1,\mu1\rangle = \langle\lambda 1,1\rangle\langle1,\mu1\rangle = \langle \tilde f+\lambda1,1\rangle\langle1,\tilde g+\mu1\rangle$$ where it is assumed that $\|1\|^2=1$, ie that you are looking at a probability measure. In the last equation we have used that elements of $L_0$ are perpendicular to $1$. The last term is however just equal to $\langle f,1\rangle\langle 1, g\rangle$.