1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.