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.

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