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

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For a convex function and $x<y<z$ , the three secant slopes satisfy $s(x,y)\le s(x,z)\le s(y,z)$

Statement

Let $f:I\to\mathbb R$ be convex on an interval and, for distinct $a,b\in I$ , write $s(a,b):=(f(b)-f(a))/(b-a)$ . If $x<y<z$ lie in $I$ , then

$s(x,y)\le s(x,z)\le s(y,z).$

Facts & Assumptions

Given: A convex $f:I\to\mathbb R$ and $x<y<z$ in $I$ .

[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 · direct

1.1

Put $\lambda=(z-y)/(z-x)\in(0,1)$ ; then $y=\lambda x+(1-\lambda)z$ , so convexity gives $f(y)\le\lambda f(x)+(1-\lambda)f(z)$ .

2.1

Multiplying this inequality by the positive number $z-x$ and rearranging gives $(z-y)(f(y)-f(x))\le(y-x)(f(z)-f(y))$ .

step 1.1algebra

3.1

Dividing step 2.1 successively by the positive products $(y-x)(z-x)$ and $(z-y)(z-x)$ gives $s(x,y)\le s(x,z)\le s(y,z)$ .

step 1.1step 2.1∎

Depends on

Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 6 results over 5 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)

For a convex function and x<y<zx<y<zx<y<z, the three secant slopes satisfy s(x,y)≤s(x,z)≤s(y,z)s(x,y)\le s(x,z)\le s(y,z)s(x,y)≤s(x,z)≤s(y,z)