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Let $C$ be a convex subset of a l.s.c space $E$. If $f$ is a continuous linear form on $E$, then $\overline C\subseteq f^{-1}(\overline{f(C)})$, indeed $\overline C$ is the intersection of all closed half spaces that contains $C$.

This remains true if $f$ is continuous on $\operatorname{span} C$ so my question is if, for a linear form that is discontinuous on $\operatorname{span} C$, we have $\overline C\subseteq f^{-1}(\overline{f(C)})$.

If I recall correctly, being continuous is equivalent to being bounded, therefore we get that $\overline{f(C)}$ is one of $]-\infty,+\infty[$, $]-\infty, a]$ or $[a,+\infty[$ assume, toward contradiction and WLOG the last case as well as $x\in\overline C$ is such that $f(x) < a$.

I am not sure on how to finish the proof, or if it is even true, any ideas would be most welcome.

P. Quinton
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  • If $E$ is normed and $C$ has interior points then $f(C)=\mathbb R$, and the inclusion is trivially true., – daw Nov 08 '23 at 19:40
  • @daw Do you have any proof or reference ? Also do you think this can be used to generalize ? – P. Quinton Nov 08 '23 at 22:16
  • see https://math.stackexchange.com/questions/671043/unbounded-linear-functional-maps-every-open-ball-to-mathbbr – daw Nov 09 '23 at 06:30

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