If a compact operator satisfies $T^nx\to0$ weakly for all $x$, then $\|T^n\|\to0$

Question

Let $H$ be a real Hilbert space, $T:H\to H$ be a compact operator. Suppose that for every $x\in H$, sequence $(T^n x)_{n\in \mathbb{N}}$ converges weakly to $0$.

How to prove that $ \lim_{n\to\infty}\|T^n\|=0 $ ?

Though I don't know weather it is fundamental or not, I found that...

By the uniform boundedness principle, we can find that for every $x\in H$, sequence $(T^n x)_{n\in \mathbb{N}}$ is bounded. Since $T$ is compact, sequence $(T^n x)$ have a subsequence strongly converging to 0.

Answer 1

1. First observation: by the uniform boundedness principle, the sequence $(\lVert T^n\rVert)_{n\geqslant 1}$ is bounded.
2. Let $x\in H$. Using compactness of $T$ and the fact that $T^nx\to 0$ weakly, we can find $n_k\uparrow \infty$ such that $\lVert T^{n_k}x\rVert\to 0$. Since $$\sup_{n_k\leqslant n\lt n_{k+1}}\lVert T^nx\rVert\leqslant \sup_{n_k\leqslant n\lt n_{k+1}}\lVert T^{n-n_k}x\rVert\cdot \lVert T^{n_k}x\rVert\leqslant \sup_j\lVert T^j\rVert\cdot \lVert T^{n_k}x\rVert,$$ we obtain that $\lVert T^nx\rVert\to 0$ for each $x$.
3. Suppose that $(\lVert T^n\rVert)$ does not converge to $0$. Then for some sequences $m_k\uparrow\infty$ and $(x_k)$ of elements of $H$ of norm $1$ such that $\inf_k\lVert T^{m_k}x_k\rVert$ is positive. Extract $(x_{k'})$ such that $(Tx_{k'})$ converges to some $y$. Then $$\lVert T^{m_{k'}}x_{k'}\rVert\leqslant \lVert T^{m_{k'}-1}y\rVert+\lVert T^{m_{k'}-1}(Tx_{k'}-y)\rVert\leqslant \lVert T^{m_{k'}-1}y\rVert+\sup_j\lVert T^j\rVert\cdot \lVert Tx_{k'}-y\rVert.$$