weakly continuous vs weakly sequentially continuous operator

Question

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 $X$).

Answer 1

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