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I'm not quite sure if this is true, and I'm thinking about whether I can weaken the assumptions needed for local convexity in a TVS $V$. My question is this:

Suppose there exists a neighborhood $U$ of $0$ such that for all $W \subseteq U$ such that $W$ is a neighborhood of $0$ we know that there exists $w_1, \dots, w_n \in W$ and $t_1, \dots, t_n \in [0, 1]$ with $\sum t_i = 1$ such that $\sum t_i w_i \notin U$.

Is this condition equivalent to $V$ not being locally convex? If we changed the last $U$ above to a $W$, this would precisely be the negation of having a locally convex neighborhood base around $0$, but this is sort of failing to be locally convex in a "uniform" way. The few non-locally convex TVSs I know seem to satisfy the above condition, but I'm not sure if this holds in general.

Subhasish Mukherjee
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Yes, this is equivalent to non-local-convexity (NLC). Indeed, your condition clearly implies NLC. On the other hand, if your condition fails, every $0$-neighbourhood $U$ contains a $0$-neighbourhood $W$ whose convex hull Conv$(W)$ is contained in $U$. Since Conv$(W)$ is a convex neighbourhood of $0$ contained in $U$, this shows local convexity.

Jochen
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