For some context, I am referring to a proof of the Mean Ergodic Theorem from Ergodic Theory and Dynamical Systems by Yves Coudene.
Suppose $H$ is a Hilbert space, and let $U \colon H \to H$ be a linear map with
$|| Uf || \leq ||f||$. Set
$$S_n(f) = \sum_{k=0}^{n-1} U^k f \qquad Inv = \{ f \in H : Uf = f\}.$$
Then
$$ \frac{1}{n} S_n(f) \to Pf $$
where $P$ is the orthogonal projection onto $Inv$.
We want to prove that $|| \tfrac{1}{n}S_n (f) || \to 0$ if $f \in Inv^\perp$. The statement I am having difficulty understanding is
We have the equality
$$ || \tfrac{1}{n}S_n (f) ||^2 = \langle f, \tfrac{1}{n}{S_n}^*\tfrac{1}{n}S_n(f)\rangle. $$
We must therefore verify, for every $f \in Inv^\perp$, that the sequence
$\tfrac{1}{n}{S_n}^*\tfrac{1}{n}S_n(f)$ converges weakly to $0$, or equivalently
that the accumulation points of this sequence are all $0$.
Now I do not understand the equivalence here. The left to right implication is clear.
Best Answer
Here's the proof with details for future reference.
Let $H$ be a Hilbert space, $U : H \rightarrow H$ a linear map satisfying the property that for all $f \in H$ we have $\|Uf\| \leq \|f\|$. Such linear maps are called contracting (maybe a bad name -- generally this is reserved for linear maps with $\|Uf\| < \|f\|$). Denote the Birkhoff average of $f$ with respect to $U$ by
$$P_n(f) := \frac{1}{n}\sum_{k=0}^{n-1} U^k(f), \ \ n \geq 1$$
Note this is a linear operator $P_n : H \rightarrow H$, and if $U$ is contracting then so is the Birkhoff average. Let $I : H \rightarrow H$ be the identity map. Define
$$\text{Inv} := \{f \in H \ | \ Uf = f\} = \{f \in H \ | \ (U-I)f = 0\} = \ker(U-I).$$
Since this is a kernel, it is a closed subspace, hence we can write
$$H = \text{Inv} \oplus \text{Inv}^\perp.$$
It is not too hard to see that if $f \in \text{Inv}$, then $Uf \in \text{Inv}$, so we have that $U$ restricts to maps on $\text{Inv}$ and $\text{Inv}^\perp$. We also have that the same holds for $P_n$.
Let $P : H \rightarrow \text{Inv}$ be the projection map. That is, if $f \in H$, then we can uniquely write it as $f = f_0 + f_1$ with $f_0 \in \text{Inv}$ and $f_1 \in \text{Inv}^\perp$, so $P(f) = f_1$.
Before proving the theorem, we need the following lemma.
Lemma: Let $U : H \rightarrow H$ be a contracting linear map. For every $g \in H$, $Ug = g$ if and only if $U^* g = g$.
Proof: By replacing $U$ with $U^*$ and noting that $(U^*)^* = U$, it suffices to show that $Ug = g$ implies $U^*g = g$. Assume $Ug = g$. Then
$$ \|g - U^*g\|^2 = \langle g - U^*g, g - U^*g \rangle = \|g\|^2 + \|U^*g\|^2 - 2 \langle g, U^*g \rangle \leq 2 \|g\|^2 - 2 \langle Ug, g \rangle = 2\|g\|^2 - 2 \|g\|^2 = 0.$$
Thus $\|g - U^*g\| = 0$ and hence $g = U^*g$. $\blacksquare$
With this, we can now prove the theorem.
Theorem (Mean Ergodic Theorem): If $U : H \rightarrow H$ is contracting, then $P_n f \rightarrow Pf$ in norm.
Proof: As an easy observation, note that if $f \in H$, then $f = f_0 + f_1$ as above, and
$$P_n(f) = P_n(f_0) + P_n(f_1) \text{ with } P_n(f_0) = \frac{1}{n} \sum_{k=0}^{n-1} U^k f_0 = \frac{1}{n} \sum_{k=0}^{n-1} f_0 = f_0.$$
So it's clear that $P_n(f_0) = P(f)$. We just need to show that $P_n(f_1) \rightarrow 0$ in norm. Observe that
$$ \|P_n(f)\|^2 = \langle P_n(f), P_n(f) \rangle = \langle f, P_n^* P_n(f) \rangle.$$
Let $Q_n := P_n^* P_n$, so we can rewrite this as
$$ \|P_n(f)\|^2 = \langle f, Q_n(f) \rangle.$$
If $Q_n(f) \rightarrow 0$ in the weak topology (on $\text{Inv}^\perp$), then this means that for all $h \in \text{Inv}^\perp$ we have
$$ \langle h, Q_n(f) \rangle \rightarrow 0.$$
Notice that $Q_n(f) \rightarrow 0$ in the weak topology forces $P_n(f) \rightarrow 0$ in norm. We need the following claim.
Claim: For every $h \in H$, we have
$$ \|(I - U^*)Q_n(h)\| \rightarrow 0 \text{ as } n \rightarrow \infty.$$
Proof: Notice that
$$ (I - U^*) P_n^*(h) = \frac{1}{n} (I - (U^*)^n) h.$$
Hence
$$ \|(I - U^*)Q_n(h)\| \leq \frac{1}{n} \| (I - (U^*)^n)\| \cdot \|P_n(h)\| \leq \frac{2}{n} \|h\| \rightarrow 0 \text{ as } n \rightarrow \infty.$$
This completes the proof. $\blacksquare$
Using Cauchy-Schwarz, we have the norm convergence implies weak convergence. Hence if $Q_n(h)$ converges to $g$ weakly, it converges weakly to a $g$ such that $U^*g = g$, hence it converges to weakly to a $g \in \text{Inv}$ by the above lemma. Thus for any $h \in \text{Inv}^\perp$, if $|\langle Q_n(f_1), h \rangle|$ converges, then it converges to $0$. Observe that for any $h \in \text{Inv}^\perp$ we have by Cauchy-Schwarz and the contracting property that
$$|\langle Q_n(f_1), h \rangle| \leq \|h\| \cdot \|f_1\|.$$
So if we let $x_n := |\langle Q_n(f_1), h \rangle|$, then the sequence $(x_n)$ is bounded, hence $(x_n) \subseteq K$ for some compact set $K$ of $\mathbb{R}$. Now we know that the only accumulation point of $x_n$ is going to be $0$. We need the following observation.
Claim: Suppose $(x_n) \subseteq K$ a compact set and there is only one accumulation point $x$. Then $x_n \rightarrow x$.
Proof: We have that for every open neighborhood $V$ of $x$ there are infinitely many $n$ so that $x_n \in V$. Suppose that for every $N \geq 0$ there is an $k \geq N$ so that $x_k \notin V$. Then we have a subsequence $(x_k) \subseteq K$ so that $x_k \notin V$ for all $k$. Since $K$ is compact, we can find a convergent subsequence of this sequence. Without loss of generality, just assume that $x_k$ converges to $y$. Since $x_k \notin V$ for each $k$, we have $y \neq x$, a contradiction to the fact that there is one limit point. Hence there must be an $N \geq 0$ so that for all $k \geq N$ we have that $x_k \in V$. This works for every open neighborhood $V$ of $x$, so $x_n \rightarrow x$. $\blacksquare$
Thus we have that $\langle Q_n(f_1), h \rangle \rightarrow 0$ for every $h \in \text{Inv}^\perp$, hence $Q_n(f_1) \rightarrow 0$ weakly, hence $P_n(f_1) \rightarrow 0$ in norm. The theorem now follows. $\blacksquare$