CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02

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A locally constant step map on the disconnected open set $R ∖ {0}$ 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: $U = R ∖ {0}$ and $f : U \to R$ defined by $f (x) = 0$ for $x < 0$ and $f (x) = 1$ for $x > 0$ .

[L1]

In the total-derivative definition, the normalized remainder tends to zero as $h$ tends to zero (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[L2]

A map is Lipschitz with constant $L$ when $∣ f (x) - f (y) ∣ \leq L ∣ x - y ∣$ for every pair of points in its domain (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

Counterexample

technique · direct

1.1

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

L1L2

2.1

For any $L \geq 0$ , take $t = 1 / (2 (L + 1))$ . The points $- t, t \in U$ satisfy $∣ f (t) - f (- t) ∣ = 1 > 2 L t = L ∣ t - (- t) ∣$ , so [L2] fails for that $L$ .

step 1.1L2algebra

3.1

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

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis I, §8.4 (standard reference, not scraped)

A locally constant step map on the disconnected open set R∖{0}R∖{0} has zero total derivative but is not globally Lipschitz