# 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.

• See this. – David Mitra Aug 9 '14 at 18:33
• Edited title to be search friendly. – David Mitra Aug 9 '14 at 18:59

## 1 Answer

Hint:

1. 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.

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

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

• Thank you Nate! I saw the proof of the step 2 this morning in the setting of Hilbert space. Do you suggest that it's not true in a general Banach space? – So Wai Chung Aug 9 '14 at 18:38
• Hi David, I don't see why we can set Xn goes to 0 – So Wai Chung Aug 9 '14 at 18:50
• @DavidMitra: You are quite right, and in fact that makes the former step 2 unnecessary. – Nate Eldredge Aug 9 '14 at 18:52
• @SoWaiChung: It's no longer relevant to this question, but in fact it is not true in general that a bounded sequence in a Banach space has a weakly convergent subsequence. – Nate Eldredge Aug 9 '14 at 18:53
• @SoWaiChung Start with all $x_n$ of norm $1$. Choose $x_{n_k}$ with $\Vert Tx_{n_k}\Vert\ge k^2$. Divide by $k$... – David Mitra Aug 9 '14 at 19:00