TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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

Statement

Let $f:I\to\mathbb R$ be convex. If $N\ge1$ , $x_1,\ldots,x_N\in I$ , and $\lambda_1,\ldots,\lambda_N\ge0$ satisfy $\sum_{i=1}^N\lambda_i=1$ , then

$f\left(\sum_{i=1}^N\lambda_i x_i\right)\le\sum_{i=1}^N\lambda_i f(x_i).$

Facts & Assumptions

Given: A convex $f:I\to\mathbb R$ , a positive finite family $x_1,\ldots,x_N\in I$ , 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 $\lambda_N=1$ , all earlier nonnegative weights vanish and the assertion is immediate; otherwise put $s=1-\lambda_N>0$ and normalize the earlier weights as $\mu_i=\lambda_i/s$ .

ihstep 1.1algebra

3.1

The induction hypothesis bounds $f(\sum_{i<N}\mu_i x_i)$ by $\sum_{i<N}\mu_i f(x_i)$ ; applying [L1] to this point and $x_N$ with weights $s,\lambda_N$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 11 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

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