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TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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On a convex open set, a uniform bound Df(z)v2Mv2\|Df(z)v\|_2\le M\|v\|_2 implies f(y)f(x)2Myx2\|f(y)-f(x)\|_2\le M\|y-x\|_2

Statement

Let URmU\subseteq\mathbb R^m be convex and open, and let f:URnf:U\to\mathbb R^n be totally differentiable at every point. If M0M\ge0 satisfies Df(z)v2Mv2\|Df(z)v\|_2\le M\|v\|_2 for every zUz\in U and vRmv\in\mathbb R^m, then

f(y)f(x)2Myx2(x,yU).\|f(y)-f(x)\|_2\le M\|y-x\|_2\qquad(x,y\in U).

Facts & Assumptions

Given: The stated convex open domain, total differentiability, and uniform derivative bound.

[L1]

A convex subset contains every line segment between two of its points (A convex subset of Rm\mathbb{R}^m contains every line segment between two of its points).

[L2]

The chain rule for total derivatives is D(gf)(a)=Dg(f(a))Df(a)D(g\circ f)(a)=Dg(f(a))\circ Df(a) (The chain rule for total derivatives: D(gf)(a)=Dg(f(a))Df(a)D(g\circ f)(a)=Dg(f(a))\circ Df(a)).

[L4]

Total differentiability implies continuity at the point of total differentiability (Total differentiability gives a local O(h2)O(\|h\|_2) increment bound and therefore continuity).

Proof

technique · direct
1.1

If x=yx=y the conclusion is immediate. Otherwise put γ(t)=x+t(yx)\gamma(t)=x+t(y-x) for 0t10\le t\le1; [L1] keeps γ([0,1])\gamma([0,1]) in UU.

L1L2L3
2.1

The chain rule gives (fγ)(t)=Df(γ(t))(yx)(f\circ\gamma)'(t)=Df(\gamma(t))(y-x) for 0<t<10<t<1, whose norm is at most Myx2M\|y-x\|_2 by hypothesis.

step 1.1L2algebra
3.1

By [L4] the curve fγf\circ\gamma is continuous at the endpoints, so [L3] applied on [0,1][0,1] yields f(y)f(x)2Myx2\|f(y)-f(x)\|_2\le M\|y-x\|_2.

step 1.1step 2.1L3L4

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 115 results over 22 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.

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