Edit: The reference I am reading is Yves Coudène's Ergodic Theory and Dynamical Systems, chapter 1, proof of theorem 1.1 on page 5.
Let $H$ be a Hilbert space and $U:H\to H$ a bounded linear operator with $\|U\|\leq 1$. Define $\mathrm{Inv} = \{f\in H\mid Uf = f\}$. I am currently trying to understand a proof that $\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}U^k f = Pf$ where $P:H\to \mathrm{Inv}$ is an orthogonal projection.
Denote $S_nf := \sum_{k=0}^{n-1}U^kf$. If $f\in \mathrm{Inv}$ the claim is clear. The proof I am reading says that since
$$\big\|\tfrac{1}{n}S_n(f)\big\|^2 = \left<f, \tfrac1nS_n^*\tfrac1n S_nf\right>$$
it suffices to show that for every $f\in \mathrm{Inv}^\perp$, the sequence $f_n := \frac{1}{n}S_n^*\frac{1}{n}S_n(f)$ converges weakly to $0$, or that all accumulation points of the sequence $f_n$ are $0$. Up to this point I am fully with the proof, since if what is said about the limit points of $f_n$ is true then $\left<f, f_n\right>\to 0, n\to \infty$ by e.g. Cauchy-Schwarz.
(Questions:) But then the mystery begins with the following claim made in the proof:
Because they [(the accumulation points)] are in $\mathrm{Inv}^\perp$, it suffices to prove that they are invariant under $U$ or $U^*$.
1.) I don't really follow why the accumulation points of $f_n$ are necessarily in $\mathrm{Inv}^\perp$ given everything that has been said so far. But this is potentially only a minor issue if the weak convergence can be concluded without really considering where the limit points are. With this in mind, the following (bold marked part later in this post) confuses me even more:
It is quite simple to conclude that for every $h\in H:(I – U^*)\frac{1}{n}S_n^*h = \frac{1}{n}(I – U^*)h$. The author uses this fact to choose $h = \frac{1}{n}S_nf$ and writes
$$\big\|(I – U*)\tfrac1n S_n^*\tfrac1n S_nf\big\|\leq \tfrac1n \|I – U^{*n}\|\big\|\tfrac1n S_nf\big\|\leq \tfrac2n\|f\|\to 0,n\to\infty$$
(Confusing part:) and states
Convergence in norm implies weak convergence.
2.) I don't understand why this shows what we want, that $f_n$ tends weakly to zero. The norm inequality shows precisely that $\lim_{n\to\infty}(I – U^*)\left(f_n\right) = 0$. My current understanding is that this would be great if we would know that $\frac{1}{n}S_n^*\frac{1}{n}S_nf$ converges to something, as then we could say that the limit of $\frac{1}{n}S_n^*\frac{1}{n}S_nf$ is invariant under $U^*$ and consequently under $U$.
How can the weak convergence be deduced?
Best Answer
The contraction $\,U\in\mathscr L(H)\,$ under consideration determines an apt orthogonal decomposition of $H$ for the proof to take off.
$$\mathrm{Inv}:=\ker(U-Id)\,=\,\{f\in H\mid Uf=f\}\,$$ is the closed subspace of $\,U$-invariant vectors, or fixed points of $\,U.$
Two properties are needed in the proof: By means of $\|U^*\|\leqslant 1\,$ and $\,Ug=g\,$ one has $$\|U^*g-g \|^2 \;=\; \|U^*g\|^2 +\|g\|^2 -\langle U^*g, g\rangle -\langle g, U^*g\rangle \;\leqslant\; 2\|g\|^2 -\langle g, Ug\rangle -\langle Ug, g\rangle\;=\; 0\,,$$ hence $\,U^*$ has the same set of invariant vectors as $\,U.$
Secondly, it follows from $$g\in\mathrm{Inv}^\perp\;\iff\; \langle f, U^*g\rangle = \langle Uf, g\rangle = \langle f, g\rangle = 0 \quad\forall\, f\in\mathrm{Inv}\;\iff\; U^*g\in\mathrm{Inv}^\perp\,,$$ that the orthogonal complement $\mathrm{Inv}^\perp$ is invariant under $\,U^*$.
Note that both statements and argumentations are symmetric in $U$ and $U^*$, in particular $\mathrm{Inv}^\perp$ is also invariant under $\,U$.
This implies $(f_n)\subset\mathrm{Inv}^\perp$ if $f=f_1\in\mathrm{Inv}^\perp$. Since $\mathrm{Inv}^\perp\subset\mathrm{Inv}\oplus\mathrm{Inv}^\perp =H\,$ is also weakly closed (the non-zero linear functionals arising from $\mathrm{Inv}$ all evaluate to zero on $\mathrm{Inv}^\perp$), all the accumulation points lie in $\mathrm{Inv}^\perp$. And if an accumulation point is proven to be invariant under $U^*$, i.e., it also belongs to $\mathrm{Inv}$, then it must be zero because $\mathrm{Inv}\cap\mathrm{Inv}^\perp =\,\{0\}.$
Your 2.) refers to the last step in the proof which yields that the sequence $\big((I - U^*)f_n\big)$ converges in norm to zero, hence it does so weakly. This does$\,$ n o t$\,$ mean that $(f_n)$ is weakly convergent, but it does imply that its accumulation points are $U^*$-invariant (after picking a subsequence converging to an accumulation point).