It is known that, if a function $f$ from a planar domain $D$ to a Banach space $A$ is weakly analytic [i.e. $l(f)$ is analytic for every $l$ in $A^*$], then $f$ is strongly analytic [i.e. $\lim_{h \to 0} h^{-1}[f(z+h)-f(z)]$ exists in norm for every $z$ in $D$].
Now the question is, if above $f$ is assumed to be weakly continuous [i.e.$l(f)$ is continuous for every $l$ in $A^*$], then is it true that $f$ will be strongly continuous.[i.e. $\lim_{h \to 0} [f(z+h)-f(z)] = 0$ in norm for every $z$ in $D$.]
Best Answer
Regarding the clarified question with finite dimensional domain. Let $X$ be an infinite dimensional separable and reflexive Banach space. Its unit ball $B$ is weakly compact and metrizable. It is also convex.
So by Hahn–Mazurkiewicz theorem there exists a continuous function $f\colon [0,1]\to B$ that is onto $B$. This function is not norm continuous as $B$ is not norm compact.