Weakly continuous function implies some sort of triviality So, I asked a similar question at math stackexchange and was directed here. Please let me know if this question is better suited elsewhere.
Let $U$ and $V$ be (infinite-dimensional) Banach spaces. Assume we have a sequence $(u_n)$ in $U$ converging weakly to $u_0$ and a nonlinear function $F:U \to V$, where we can assume for instance that $\|F(u)\|_V \leq C \|u\|_U$, it is not really important for the question.
Moving on to the question, in $L^p(0,1)$, weak continuity of a function $\psi: \mathbb{R} \to \mathbb{R}$ implies that that $\psi$ is affine (see 2.10 in these lecture notes )
I am wondering if there are similar results for other Banach (or Hilbert spaces), that weak continuity implies some sort of triviality? I suspect there is no general theorem, but I am particularly concerned with spaces such as $L^2([a,b],V)$ where $V$ is a reflexive, separable Banach space (perhaps even finite-dimensional), or $C^1([a,b],V)$, $W^{k,p}([a,b],V)$. In the first case my guess is that it does carry over and weakly continuous functions are affine, in the other cases, I am not so sure.
 A: From the question and from the OP's related question on Mathematics StackExchange, I infer that the OP is in general interested in the weak continuity of nonlinear mappings. So here are two general facts the seem to be relevant:
Let $X$, $Y$ be Banach spaces (over the same field). Assume that $X$ is infinite-dimensional and that $Y$ is non-zero. Then:
(a) There exists a norm-continuous nonlinear mapping $F: X \to Y$ that is not weakly continuous.
(b) There exists a non-linear mapping $G: X \to Y$ that is weakly continuous.
Proof. (a) Choose a non-zero vector $y \in Y$ and set $F(x) = \|x\| \cdot y$ for each $x \in X$. Obviously, $F$ is norm-continuous. But if $F$ were weakly continuous, then the preimage of the set $\{y\}$ under $F$ - which is the unit sphere in $X$ - would be weakly closed. However, the unit sphere an an infinite-dimensional Banach space is never weakly closed.
(b) Let $x' \in X'$ be a non-zero bounded linear functional on $X$ and fix a non-zero vector $y \in Y$. We set $G(x) = \langle x', x \rangle^2 \cdot y$ for each $x \in X$. Then $G$ is not linear, but it is weakly continuous since it is the composition of the weakly continuous mappings
$$
  X \overset{x'}{\longrightarrow} \mathbb{F} \; \overset{s \mapsto s^2}{\longrightarrow} \; \mathbb{F} \; \overset{t \mapsto t \cdot y}{\longrightarrow} \; Y
$$
(where $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$ denotes the underlying scalar field of $X$ and $Y$). $\square$
Remarks.
(1) The mapping $F$ from (a) is not even sequentially weakly continuous. This follows from the fact that, in an infinite-dimensional Banach space $X$, there always exists a sequence $(x_n)$ in the unit sphere that converges weakly to $0$.
(2) It is maybe worthwhile to recall that, for a linear mapping $T$ between two Banach spaces $X$ and $Y$, the following are equivalent:
(i) $T$ is norm continuous (i.e., bounded).
(ii) $T$ is weakly continuous.
(iii) $T$ is sequentially weakly continuous.
(Assertion (i) implies (ii) due to the existence of the dual operator $T'$, (ii) obviously implies (iii), and (iii) implies (i) due to the closed graph theorem.)
Disclaimer. Actually, I think that this would be a better fit for Mathematics StackExchange, but since the OP was directed to MathOverflow from there, it thought would be a bit unfair to send them back without an answer.
