LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Midpoint convexity gives the convexity inequality at every dyadic weight $k / 2^{n}$

Statement

Let $f : I \to R$ be midpoint convex. For every $n \in N$ , every $k \in {0, \dots, 2^{n}}$ , and all $x, y \in I$ ,

$f (\frac{k}{2^{n}} x + (1 - \frac{k}{2^{n}}) y) \leq \frac{k}{2^{n}} f (x) + (1 - \frac{k}{2^{n}}) f (y) .$

Facts & Assumptions

Given: A midpoint-convex $f : I \to R$ and $x, y \in I$ .

[L1]

Midpoint convexity is the convexity inequality at weight $1 / 2$ (Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval).

Proof

technique · induction

1.1

For $n = 0$ , the only weights are $0$ and $1$ , and the asserted inequalities are equalities.

baseL1

2.1

Assume the assertion at $n$ . At $n + 1$ , an even numerator $2 j$ reduces to the induction hypothesis; an odd numerator $2 j + 1$ is the midpoint of the adjacent weights $j / 2^{n}$ and $(j + 1) / 2^{n}$ , so [L1] followed by the induction hypothesis proves the assertion.

ihstep 1.1algebra

3.1

The base and successor steps establish the assertion for every natural $n$ and every $0 \leq k \leq 2^{n}$ .

step 1.1step 2.1discharge-induction∎

Depends on

Used by

A continuous midpoint-convex function on an interval is convex Theorem

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

R. Gardner, Convex Functions, Notes 6.6 (standard reference, not scraped)

Midpoint convexity gives the convexity inequality at every dyadic weight k/2nk/2n