Weak continuity for non-linear maps It is known that a linear map $T:H\to H$ on a Hilbert space $H$ is weak-weak continuous if and only if it is strong-strong continuous.
My question is : what if we don't consider a linear map, is it still true ?
Or is there at least one implication still true ?
 A: Both implications are not true in the non-linear setting.
1) Define $T(x) = \|x\| x_0$, where $x_0\ne0$ is given. Then $T$ is strong-strong but not weak-weak continuous: If there is a sequence $(x_n)$ that converges weakly but not strongly to $x$ then $\|x_n\|\not\to \|x\|$ and $T(x_n)\not\to T(x)$.
2) This is going to be more complicated. Take $H=l^2$ and denote by $(e_n)$ the standard sequence of unit vectors, $e_n\rightharpoonup0$.
Define a mapping $f:\mathbb R\to l^2$ by
$$
f(s) = \begin{cases} e_1 & \text{ if } s\ge1,\\
 (n+1-s^{-1})e_n + (s^{-1}-n)e_{n+1}& \text{ if } \frac1{n+1}\le s < \frac1n, n\in \mathbb N\\
0 & \text{ for } s=0.
\end{cases}
$$
Then $f$ maps converging sequences to weakly converging sequences: If $(s_n)$ converges to $s\ne0$, then $f(s_n)$ converges strongly to $f(s)$.
If $(s_n)$ converges to $0$, then $f(s_n)$ converges weakly to $0$ in $l^2$.
Now define $T(x):= f(\langle x, e_1\rangle)$. Then $T$ maps weakly convergent sequences to weakly convergent sequences. However $T(n^{-1}e_1)=f(n^{-1})=e_n$ does not converge strongly. Hence $T$ is not strong-strong continuous.
