Prove strong convergence in Hilbert space

Question

Let $H$ be Hilbert space, $T:H \to H$ be bounded self adjoint operator. Suppose $\|T\|=1$ and for all $x \in H, \langle Tx,x \rangle \geq 0$. Then $\{T^n\}$ strongly convergent, but there exists $T$ not convergent in operator norm.

If $T=\operatorname{Id}$, it meets conditions, so I think $T^n$ strongly convergent $\operatorname{Id}$, but I can't prove.

Answer 1

By the Spectral Theorem, $T^n$ strongly convergent to $id$, so it is proved.

Answer 2

Here is a proof without spectral theorem. Let $x\in H$.

Since $\|T\|\le 1$, the sequence $(\|T^nx\|)$ is monotonically descreasing, hence convergent.

Take $m,n$ such that $m+n$ is even. Then $$ \|T^mx-T^nx\|^2 = \|T^nx\|^2 + \|T^mx\|^2 -2 \|T^{\frac{m+n}2}x\|^2 \to 0 $$ for $m,n\to\infty$. Hence, the sequences $(T^{2n}x)$ and $(T^{2n+1}x)$ are Cauchy sequences, thus converging. Let $y$ be such that $T^{2n}x\to y$, then $T^{2n+1}x\to Ty$, $T^{2n+2}x\to T^2y=y$. So the limit satisfies $T^2y=y$.

It remains to prove $Ty=y$: By assumption $$ 0 \le \langle T(Ty-y),Ty-y\rangle = \langle y-Ty,Ty-y\rangle = -\|Ty-y\|^2, $$ hence $Ty=y$ and $T^nx \to y$.