If a linear operator in a Hilbert space preserves weak convergence, it must be continuous

Question

Let $H$ be a Hilbert space and $T:H \to H$ be a linear operator. Suppose that $T$ maps weakly convergent sequences to weakly convergent sequences. Is it true that $T$ must be bounded (i.e. continuous)? The other direction of this assertion is true (and easily proved). I'm not able to prove this direction.

I've tried using the closed graph theorem. If I can show that $T(x_{n})$ weakly goes to $T(x)$ then CGT will give continuity, but I'm not able to show this. Is this the correct approach. If it isn't, what is?

Answer 1

If $f$ is weak-to-weak continuous, then the assertion is valid:

Assume that $x_{n}\rightarrow x$ and $Tx_{n}\rightarrow y$, we are to show that $y=Tx$.

Now $T$ is weak-to-weak continuous, so for any $u$, $\left<Tx_{n},u\right>\rightarrow\left<Tx,u\right>$, but we have also that $\left<Tx_{n},u\right>\rightarrow\left<y,u\right>$, so $\left<y,u\right>=\left<Tx,u\right>$ is valid for every $u$, this shows that $y=Tx$.

Answer 2

The answer is yes, weakly convergent operators are bounded.

Assume that it is not bounded, that means exists $h_n$ such that $\|Th_n\| > n$ for all $n \in \mathbb{N}$, you can assume that $h_n \in \overline{B(0;1)}$ (why?)

Now since $\overline{B(0;1)}$ is weakly compact we have that exists a converging subsequent $h_{n_k} \to h$ (weakly), now since $T$ is weakly convergent, we have $Th_{n_k} \to Th$ (weakly), but also not that

$$\|Th_{n_k}\| = {n_k} \to \infty$$