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

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Finite Jensen inequality for a convex function and nonnegative weights summing to one

Statement

Let $f : I \to R$ be convex. If $N \geq 1$ , $x_{1}, \dots, x_{N} \in I$ , and $λ_{1}, \dots, λ_{N} \geq 0$ satisfy $\sum_{i = 1}^{N} λ_{i} = 1$ , then

$f (\sum_{i = 1}^{N} λ_{i} x_{i}) \leq \sum_{i = 1}^{N} λ_{i} f (x_{i}) .$

Facts & Assumptions

Given: A convex $f : I \to R$ , a positive finite family $x_{1}, \dots, x_{N} \in I$ , and nonnegative weights summing to $1$ .

[L1]

A function is convex when the convex-combination inequality holds for every weight in $[0, 1]$ (Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval).

Proof

technique · induction

1.1

For $N = 1$ , the sole weight is $1$ , so the two sides are both $f (x_{1})$ .

baseL1

2.1

Assume the assertion for $N - 1$ terms. If $λ_{N} = 1$ , all earlier nonnegative weights vanish and the assertion is immediate; otherwise put $s = 1 - λ_{N} > 0$ and normalize the earlier weights as $μ_{i} = λ_{i} / s$ .

ihstep 1.1algebra

3.1

The induction hypothesis bounds $f (\sum_{i < N} μ_{i} x_{i})$ by $\sum_{i < N} μ_{i} f (x_{i})$ ; applying [L1] to this point and $x_{N}$ with weights $s, λ_{N}$ gives the asserted $N$ -term inequality.

step 1.1step 2.1discharge-induction∎

Depends on

Used by

Finite Jensen for x↦ x² gives that the square of a weighted mean is at most the weighted mean of the squares Example

Dependency tree · two levels

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Sources

S. Boyd and L. Vandenberghe, Convex Optimization, §3.1 (standard reference, not scraped)