Selfadjoint bounded isometric operators and orthogonal projections

Question

Let $H$ a Hilbert space, $ T = T^{\star} \in B(H)$ with $ || T || = 1$. Prove that for any $v\in H$ $ \lim_{n \to \infty} T^nv= Pv$, where $P\in B(H)$ is the orthogonal projection onto $\text{ker}(Id-T) $.

Answer 1

This is false. If $T=\pmatrix{1&0\cr 0&-1}$, then $$ T^n\pmatrix{x\cr y}=\pmatrix{x\cr (-1)^ny}, $$ which does not converge if $y\neq0$.

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EDIT: From now on let us assume that $T≥0$.

It is well known that $σ(T)$ consists of a sequence of eigenvalues converging to zero in $[0, 1]$, including 0 (which may or may not be an eigenvalue). Moreover the spectral radius of $T$ coincides with its norm, so 1 is necessarily in the spectrum of $T$.

Denoting by $H_λ$ the eigenspace for each $λ∈σ(T)$, we have that $$ H=\bigoplus_{λ∈σ(T)}H_λ = H_1\oplus K, $$ where $$ K=\bigoplus_{λ∈σ(T)\setminus \{1\}}H_λ. $$

Letting $S=T|_K$, observe that the spectrum of $S$ coincides with $σ(T)\setminus \{1\}$ and, since 1 is necessarily an isolated point of $σ(T)$, we see that $σ(S)⊆[0,a]$, for some $a<1$. In particular $$ \|S\|\leq a<1. $$

Observing that $T=I_{H_1}\oplus S$, we have that $$ T^n=I_{H_1}\oplus S^n \to I_{H_1}\oplus 0, $$ because $\|S^n\|≤\|S\|^n\to 0$. Finally notice that $I_{H_1}\oplus 0$ is the orthogonal projection onto $H_1$, as desired.

This in fact proves a little more than required, namely that $T$ converges to $I_{H_1}$ in norm.