It is well known that convex implies connected, and it is clear that if $X$ is a locally convex topological vector space and $\emptyset \neq A\subseteq X$ is open then $A$ contains a nonempty open convex set.

Question. Does there exist a topological vector space $X$ and a nonempty open connected set $A\subseteq X$ such that $A$ does not contain any nonempty open convex subset?

  • $\begingroup$ Isn't $X=L^p([0,1])$, with $0<p<1$ (with $A$ being the "unit ball") an example? I'm guessing any $X$ not locally convex should provide an example. Or at least any not locally convex $F$-space. $\endgroup$ – tomasz May 15 at 15:52
  • $\begingroup$ Yes, $L^p([0,1])$ indeed works, since its unit ball is path-connected (we do not need to triangle inequality to hold but only the homogeneity of $\|\cdot\|_p$). Hence the claim works for all non-locally convex tvs for which the topology is induced by a quasi-norm. I don't know what happens with not-locally F-space, but it would be interesting to know. $\endgroup$ – Paolo Leonetti May 15 at 17:44
  • $\begingroup$ @PaulFrost: My guess is that for topological vector spaces over the reals or the complex numbers, it shouldn't be possible. But my knowledge of non-normed vector spaces is cursory at best, so my guess could be very naive. $\endgroup$ – tomasz May 16 at 9:12

Yes. For instance, consider the space $X=L^p([0,1])$ for some $p\in (0,1)$, and take $A:=\{x\in X\mid \lVert x\rVert_p<1 \}$.

Then $A$ is open and path-connected (for every $a\in A$, $t\mapsto ta$ yields a path from $0$ to $a$), but it does not contain any nonempty convex open subset (the only convex open sets in $X$ are $\emptyset$ and $X$).


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