I’ve have provided a cleaner proof of Lemma 3 than the one presented in my previous post. I have also included a stronger result, which shows that a midpoint convex function is either continuous everywhere or discontinuous everywhere.

In what follows, we assume that is an open convex set of some normed space. We can actually replace normed space by metric vector space, which does not change the proof.

**Lemma 1** * If is midpoint convex and discontinuous at , then . *

*Proof:* Replacing by , we may assume that . Replacing by , we may assume that . Set , and let be a sequence converging to so that . I claim that . Otherwise, by midpoint convexity,

But , which contradicts the definiion of . Discarding finitely many terms, we may assume that the sequence is contained in . By midpoint convexity,

Taking the liminf of both sides, we obtain

Suppose we have shown that . Discarding finitely many terms, we may assume that the sequence is contained in . By midpoint convexity,

Taking the liminf of both sides, we obtain

By induction, we obtain for all .

**Theorem 2Â ***Let be a midpoint convex function defined on a convex open set . If is discontinuous at a point , then is unbounded on every open convex subset of . Hence, is discontinuous everywhere.*

*Proof:* Replacing by the midpoint convex function , we may assume that . By the same argument, we may assume that is symmetric. By Lemma 1, there exists a sequence and . Let be an arbitrary point. Since is open, there exists a neighborhood of which contains almost all terms of the sequence . Relabeling, if necessary, we may assume that it contains the entire sequence. Observe that .

Taking the liminf of both sides, we see that . Since was arbitrary, we conclude that is unbounded in every neighborhood of its domain.

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