Let $(X, \|\cdot\|) $ be a Banach space.
A function $g:X \longrightarrow X$ is said to be sequentially weakly continuous if for every sequence $(x_n)$ in $X$ such that $x_n \rightharpoonup x$, we have $g(x_n) \rightharpoonup g(x)$.
What's an example of a function $g:X \longrightarrow X$ which is strongly continuous (meaning continuous as a map $X \longrightarrow X$ where on both $X$'s we take the topology induced by $\|\cdot\|$) but not sequentially weakly continuous?
Best Answer
As an example let $X=\ell^2(\Bbb N)$, denote with $e_n$ the standard ONB. Then $e_n\to0$ weakly while $\|e_n\|=1$ for all $n$. Now let $$g:\ell^2(\Bbb N)\to\ell^2(\Bbb N), \quad x\mapsto \|x\|\, e_1,$$ this is clearly a norm continuos function, but $g(e_n)=e_1$, which does not converge to $0$ weakly. Hence $g$ is not weak-weak sequentially continuous. You can adapt this example to any space $X$ admiting a sequence that converges weakly but not in norm.
If you want $g$ to be linear then this is impossible, for every linear map $g:X\to X$ that is norm continuous will also be weak-weak continuous, in particular weak-weak sequentially continuous. You can check here for a proof of that.