Weakly sequentially continuous operators in Hilbert space are norm continuous. Suppose I have a linear operator T from a Hilbert space H to itself, and T maps every weak convergent sequence to a weak convergent sequence. Show that T is continuous.
I feel that this statement will not be true for general Banach spaces. Weak convergence in a Hilbert space means the inner product of Xn and u converges to the inner product of X and u for any u in H. But I don't see how it helps me in proving the claim. And I think closed graph theorem will not help either, since the assumption is about strong convergence.
Any help is appreciated.

Hint:


*

*Try proving the contrapositive.  If $T$ is not continuous, show that it maps some bounded sequence $\{x_n\}$ to a sequence $\{T x_n\}$ with $\|T x_n\| \to \infty$.  Passing to a subsequence and rescaling, we can even assume $\|x_n\| \to 0$, and in particular $x_n \to 0$ weakly.

*Every weakly convergent sequence is bounded.  (This comes from the uniform boundedness principle.)

*Conclude that if $T$ is not continuous, it maps some weakly convergent sequence to a weakly non-convergent sequence.
