Let \(U\) be a unitary operator on \(H\) a Hilbert space, let \(M \da \ts{x\in H \st Ux = x}\), let \(P\) be the orthogonal projection onto \(M\), and define \[ S_N \da {1\over N} \sum_{n=0}^{N-1} U^n .\] Show that for all \(x\in H\), \[ \norm{ S_N x - Px}_H \converges{N\to \infty } \to 0 .\]
Solution
1. \(M = \ker(U - I)\) is a closed subspace, and \(M^\perp = \overline{\operatorname{ran}(U - I)}\).
\(M\) is the kernel of the bounded operator \(U - I\), hence a closed subspace. For a bounded operator \(T\), \((\operatorname{ran} T^*)^\perp = \ker T\), so \((\ker T)^\perp = \overline{\operatorname{ran} T^*}\). Take \(T = U - I\). Then \(T^* = U^{-1} - I = -U^{-1}(U - I)\), so \(\operatorname{ran}T^* = U^{-1}\operatorname{ran}(U - I)\). Since \(U^{-1}(Uy - y) = Uz - z\) with \(z = U^{-1}y\), and \(Uy - y = U^{-1}(Uw - w)\) with \(w = Uy\), \(U^{-1}\operatorname{ran}(U - I) = \operatorname{ran}(U - I)\).
2. \(S_N x = x\) for \(x \in M\) and every \(N\).
\(U^n x = x\) for all \(n\).
3. \(\|S_N\| \le 1\), and \(\|S_N x\| \to 0\) for \(x \in \operatorname{ran}(U - I)\).
\(\|U^n\| = 1\), so \(\|S_N\| \le \frac1N\sum_{n<N}\|U^n\| = 1\). For \(x = Uy - y\), the sum telescopes: \(S_N x = \frac{1}{N}\sum_{n=0}^{N-1}(U^{n+1}y - U^ny) = \frac{1}{N}(U^N y - y)\), so \(\|S_N x\| \le \frac{2\|y\|}{N}\).
4. \(\|S_N x\| \to 0\) for \(x \in M^\perp\).
Given \(\eps > 0\), step 1 gives \(z \in \operatorname{ran}(U - I)\) with \(\|x - z\| < \eps/2\). By step 3, \(\|S_N x\| \le \|S_N (x - z)\| + \|S_N z\| < \eps/2 + \|S_N z\| < \eps\) for \(N\) large.
5. Q.E.D.
Write \(x = Px + (x - Px)\) with \(Px \in M\) and \(x - Px \in M^\perp\). By step 2, \(S_N x - Px = S_N(x - Px)\), which tends to \(0\) by step 4.