Continuous Midpoint-Convex Function is Convex

Theorem
Let $f$ be a real function which is defined on a real interval $I$.

Let $f$ be midpoint-convex and continuous on $I$.

Then $f$ is convex.

Proof
Let $x, y \in I$ and let $t \in \closedint 0 1$.

Let $\sequence {t_n}_{n \mathop \in \N}$ be a sequence of rational numbers in $\closedint 0 1$ converging to $t$.

It follows that:

and:

From Midpoint-Convex Function is Rational Convex, $f$ is rational convex.

From Continuous Mapping is Sequentially Continuous, $f$ is sequentially continuous.

As $f$ is rational convex:
 * $\forall n \in \N: \map f {t_n x + \paren {1 - t_n} y} \le t_n \map f x + \paren {1 - t_n} \map f y$

From sequential continuity of $f$ and Lower and Upper Bounds for Sequences/Corollary:

which completes the proof.

Also see

 * Continuous Strictly Midpoint-Convex Function is Strictly Convex


 * Continuous Midpoint-Concave Function is Concave
 * Continuous Strictly Midpoint-Concave Function is Strictly Concave