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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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A locally constant step map on the disconnected open set R{0}\mathbb{R}\setminus\{0\} has zero total derivative but is not globally Lipschitz

Statement refuted

A uniform total-derivative bound on every open domain implies a global Lipschitz bound on that domain.

Facts & Assumptions

Given: U=R{0}U=\mathbb R\setminus\{0\} and f:URf:U\to\mathbb R defined by f(x)=0f(x)=0 for x<0x<0 and f(x)=1f(x)=1 for x>0x>0.

[L1]

In the total-derivative definition, the normalized remainder tends to zero as hh tends to zero (The total (Fréchet) derivative Df(a)Df(a) as the linear first-order approximation with o(h2)o(\|h\|_2) remainder).

[L2]

A map is Lipschitz with constant LL when f(x)f(y)Lxy|f(x)-f(y)|\le L|x-y| for every pair of points in its domain (Lipschitz map, α\alpha-Hölder map for rational 0<α10 < \alpha \le 1, and contraction).

Counterexample

technique · direct
1.1

Each xUx\in U has a small interval contained in its own component of UU, on which ff is constant; hence Df(x)=0Df(x)=0 by [L1].

L1L2
2.1

For any L0L\ge0, take t=1/(2(L+1))t=1/(2(L+1)). The points t,tU-t,t\in U satisfy f(t)f(t)=1>2Lt=Lt(t)|f(t)-f(-t)|=1>2Lt=L|t-(-t)|, so [L2] fails for that LL.

step 1.1L2algebra
3.1

The segment from t-t to tt contains 0U0\notin U, so UU is not convex; this is exactly the omitted hypothesis of the mean-value inequality.

step 1.1step 2.1

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