# Function whose range is a bounded subset of an infinite-dimensional Banach space is weakly continuous?

Let $X_1$ and $X_2$ be infinite-dimensional Banach spaces, and consider any function $f : X_1\to X_2$ such that $\mathrm{im}(f)$ is bounded. It seems to me that $f$ should be weakly continuous by the open set definition of continuity, since $\mathrm{im}(f)$ contains no nonempty open sets in the weak topology (as nonempty weakly open sets in infinite-dimensional Banach spaces are unbounded), so the preimage of any weakly open set in $\mathrm{im}(f)$ (which is to say, the preimage of the empty set in $X_2$, which is the empty set in $X_1$) is weakly open in $X_1$. Is this logic correct? Can we conclude that any function between infinite-dimensional Banach spaces with bounded image is weakly continuous? Thanks for your time.

If $O$ is weakly open in $X_2$, $O \nsubseteq \operatorname{Im}(f)$ does not imply that $f^{-1}[O] = \emptyset$. For that you need $\operatorname{Im}(f) \cap O = \emptyset$. But I don't think that's necessarily true.