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

Statement

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

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

Facts & Assumptions

Given: A midpoint-convex $f:I\to\mathbb R$ 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 $2j$ reduces to the induction hypothesis; an odd numerator $2j+1$ 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 $0\le k\le2^n$ .

step 1.1step 2.1discharge-induction∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 43 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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/2^nk/2n