Jensen inequality as a functional inequality

Remarks

The idea of the proof is Cauchy's forward and backward induction. By induction on $k$, the inequality holds for $m = 2^k$ points with equal weights $1/m$. For the step, split the points into two halves of $2^{k-1}$ points and apply midpoint convexity to the two averages. For $n < 2^k$ points with mean $\bar x = \frac{1}{n} \sum x_i$, add $2^k - n$ copies of $\bar x$; the case $m = 2^k$ gives $f(\bar x) \le \frac{1}{2^k}\big(\sum f(x_i) + (2^k - n) f(\bar x)\big)$, which rearranges to $f(\bar x) \le \frac1n \sum f(x_i)$. For rational weights $q_i = k_i / N$, repeat each point $x_i$ exactly $k_i$ times and use the equal-weight case with $N$ points. The step that rearranges uses that $f(\bar x)$ is a real number.

Example: with $f(x) = x^2$, $q_1 = 1/3$ and $q_2 = 2/3$, the statement gives $\big(\frac{x_1 + 2x_2}{3}\big)^2 \le \frac{x_1^2 + 2x_2^2}{3}$. The statement is the form of Jensen's inequality that holds without any regularity, as a consequence of the functional inequality of midpoint convexity alone. For real weights it can fail: a discontinuous additive function is midpoint convex, but it is not convex, so the inequality fails for some real weights.

Going deeper

For a student

The result is the first step in the regularity theory of midpoint convex functions. Passing from rational to real weights needs some continuity, and results such as the Bernstein–Doetsch theorem and Sierpiński's theorem supply it under weak hypotheses. The same forward and backward induction proves the inequality of arithmetic and geometric means.