I’ve written a couple posts now on continuity as a sufficient condition for midpoint convexity to imply that a function , where is an interval, is convex. I now want to weaken the hypothesis of continuity to that of lower semicontinuity.

Recall that an extended real-valued function defined on a topological space is said to be *lower semicontinuous (l.s.c.)* if for every , the level set is closed in . If is a second-countable topological space, then a function is lower semicontinuous if and only if

Lemma.If is a compact interval and is a l.s.c. function, then attains its infimum at a point .

*Proof. * Set , and let be a sequence in such that . Since is compact, the sequence has a convergent subsequence with limit . By l.s.c.,

,

which implies that .

Proposition.A lower-semicontinuous function is convex if and only if is midpoint convex.

*Proof. *Necessity is obvious, so we concern ourselves for sufficiency. Let be a compact interval. By the preceding lemma, attains its infimum at a point . We will show that is continuous on and on .

We have that and therefore by midpoint convexity, . By definition of , we have the inequality

Suppose that we have shown that

for some . Then by midpoint convexity, we have that

By definition of , we have that

Suppose we have shown that, for some even , we have the inequality

Then by midpoint convexity,

which implies that ; and

Define the set of “dyadic points” in by . By induction, we see that and imply that . Hence,

exists by the monotone convergence theorem. Observe that defines a monotone function on and . If we can show that and is continuous, then it follows from previous results (see my earlier posts on midpoint convexity and continuity) that is convex. Choose sequences and in such that

, , and

Set . Since , we have that

,

which implies that and therefore is continuous, being monotone. To see that on , observe that by lower semicontinuity, . For the reverse inequality, let be a sequence in such that , and define so that . Then by midpoint convexity,

Letting and applying the continuity of , we conclude that

Applying the same argument to on the interval , we arrive at the desired conclusion.