# weakly continuous vs weakly sequentially continuous operator

Im reading some papers where the condition of weak sequential continuity is crucial, and this question come to my mind:

Why it's more interesting to use this condition instead of the classical weak continuity?

Context:

• $$T$$ between a Banach space X and itself.
• $$T$$ is weakly sequentially continuous if for any $$(x_n)_{n\in \mathbb N}$$, $$x_n \stackrel{w}{\rightharpoonup}x$$ in $$X$$ $$\Rightarrow$$ $$T(x_n) \stackrel{w}{\rightharpoonup } T( x )$$ in $$X$$).
• All of these conditions boil down to norm continuity of $T$. But of course some might be easier to verify in concrete situations.
– Ruy
Sep 21, 2020 at 20:25
• Can you give more explication, I will appreciate it! Sep 22, 2020 at 9:47

Theorem. Let $$X$$ and $$Y$$ be normed vector spaces and let $$T:X\to Y$$ be a linear map. Then the following are equivalent:

1. $$T$$ is continuous,
2. $$T$$ is weakly continuous,
3. $$T$$ is sequentially weakly continuous,
4. For every sequence $$\{x_n\}_{n\in {\bf N}}$$ in $$X$$ which is weakly convergent to $$0$$, one can find a weakly converging subsequence $$\{T(x_{n_k})\}_{k\in {\bf N}}$$.

Proof. The implications (1) $$\Rightarrow$$ (2) $$\Rightarrow$$ (3) $$\Rightarrow$$ (4) are trivial, so let us prove that (4) $$\Rightarrow$$ (1). Arguing by contradiction assume that $$T$$ is not continuous, hence unbounded on the unit ball of $$X$$. So we can find a sequence $$\{y_n\}_{n\in {\bf N}}$$ in that ball such that $$\Vert T(y_n)\Vert >n^2$$, for every $$n$$. Setting $$x_n=y_n/n$$, we have that $$x_n\to 0$$ in norm, so also weakly, while $$\Vert T(x_n)\Vert = {1\over n} \Vert T(y_n)\Vert > {1\over n}n^2 = n,$$ so no subsequence of $$\{T(x_n)\}_{n\in {\bf N}}$$ is weakly convergent because weakly convergent sequences are bounded (by a well known Corollary of the uniform boundedness principle). This contradicts (4), so the proof is complete. QED

• Unfortunately, $T$ is not supposed to be linear, but thanks for your answer anyway.. Sep 22, 2020 at 13:40
• Oh! I am sorry. But since the majority of facts about functions between Banach spaces involve linear maps, it is perhaps wise to emphasize it when you are considering non-linear maps!
– Ruy
Sep 22, 2020 at 13:44