Spectral theorem and strong convergence

Question

Question:Let $H$ be a Hilbert space, and $T$ be a self-adjoint operator. If $||T||\leq 1$ and $(Tx,x)\geq 0$ for all $x\in H$, then $T^n$ strongly convergent.

My idea : By spectral theorem and $||T||\leq 1$, $Tx=\int _{-1}^{1} \lambda dE_\lambda x$. I guess $T^n$ strongly convergent to $0$. But I can't compute $T^nx$.

Answer 1

Note that $T^n$ is a decreasing net of positive elements, by Vigier's theorem(Theorem 4.1.1, C*-algebras and operator theory by Murphy), $T^n$ is strongly convergent.

Answer 2

The spectrum of $T$ is contained on $[0,1]$ because $(Tx,x)\ge 0$ and $\|T\| \le 1$. So, $$ T^n x = \int_{0}^{1}\lambda^n dE(\lambda)x \rightarrow E\{1\}x, $$

which is the projection onto the eigenspace of $T$ with eigenvalue $1$. This strong convergence is obvious from

$$ \|T^n x - E\{1\}x\|^2 \\ = \left\|\int_{0}^{1}(\lambda^n-\chi_{\{1\}}(\lambda))dE(\lambda)x\right\|^2 \\ = \int_{0}^{1}|\lambda^n-\chi_{\{1\}}(\lambda)|^2 d\|E(\lambda)x\|^2. $$