strongly continuous mapping implies bounded mapping Hi does anyone know how to show the result that if we have a relexive Banach space $X$ and a mapping $A: X \rightarrow X^{*}$ (not necessarily linear), which is strongly continuous, which means  $$u_{n} \rightharpoonup u~~~\text{in }X\implies A(u_{n}) \rightarrow A(u)~~\text{ in }X^{*}$$ Then does it follow that $A$ is also bounded, in the sense that $A$ takes bounded sets to bounded sets?
Thanks for any help.

Thanks for responses in the comments. Is this then okay? 
Proof:
Assume there is some bounded set $B \subset X$, where $A(B)$ is unbounded in $X^{*}$. Then choose a sequence $\{ A(u_{n}) \}_{n \in \mathbb{N}} \subset A(B)$ such that $\| A(u_{n}) \| > n$. Note then that $\{ u_{n}\}_{n} \subset B$ is bounded in $X$, so by Kakutani's Theorem and Emerlein-Smulian Theorem it follows that there exists a subsequence $\{ u_{n_{k}} \}_{k}$ such that $u_{n_{k}} \rightharpoonup u$ in $X$. It follows then from the strong continuity assumption that $A(u_{n_{k}}) \rightarrow A(u)$, the sequence is therefore bounded. But we also have $\| A(u_{n_{k}}) \| \geq k$, which contradicts $\{ A(u_{n_{k}}) \}_{k}$ being bounded.
$\square$
