From midpoint convexity to full convexity

mathematics analysis

We will prove that midpoint convexity implies full convexity under mild conditions.

First, let’s define these terms. We say a function $f:(0,1) \to \mathbb R$ is $\lambda$-convex if

\[f(\lambda x + (1-\lambda)y) \le \lambda f(x) + (1-\lambda) f(y), \quad \forall x, y \in (0,1).\]

If $f$ satisfies this condition for $\lambda = 1/2$, it is called midpoint convex. If $f$ is $\lambda$-convex for any $\lambda \in (0,1)$, then it is fully convex.

Proposition. If $f:(0,1) \to \mathbb R$ satisfies midpoint convexity and local upper boundedness, then $f$ is fully convex. $\Box$

The proof can be completed by establishing the following smaller results in order:

  • The function $f$ is locally bounded;
  • It is $\lambda$-convex for any dyadic number $\lambda$;
  • $f$ is locally Lipschitz;
  • Finally, $f$ is fully convex.

In the above, the set of dyadic numbers (or dyadic rationals), often denoted by $\mathbb{D}$, is the set of rational numbers of the form:

\(\mathbb{D} = \cup_{n=0}^\infty \mathbb{D}_n,\) where \(\mathbb{D}_n = \left\{ \frac{k}{2^n} \;\middle|\; k = 0, \ldots, 2^n \right\}.\)

The proof is detailed in this PDF.