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 $X$).

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

*

*$T$ is continuous,

*$T$ is weakly continuous,

*$T$ is sequentially weakly continuous,

*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
