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

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A totally differentiable map with zero derivative on a convex open set is constant

Statement

Let $U\subseteq\mathbb R^m$ be convex and open. If $f:U\to\mathbb R^n$ is totally differentiable at every point and $Df(z)=0$ for every $z\in U$ , then $f$ is constant on $U$ .

Facts & Assumptions

Given: A convex open $U$ and a totally differentiable map with zero total derivative at every point.

[L1]

The total-derivative mean-value inequality implies $\|f(y)-f(x)\|_2\le M\|y-x\|_2$ under a uniform derivative bound $M$ (On a convex open set, a uniform bound $\|Df(z)v\|_2\le M\|v\|_2$ implies $\|f(y)-f(x)\|_2\le M\|y-x\|_2$ ).

Proof

technique · direct

1.1

The zero derivative hypothesis satisfies the bound in [L1] with $M=0$ for arbitrary $x,y\in U$ .

2.1

Hence $\|f(y)-f(x)\|_2\le0$ , so norm separation gives $f(y)=f(x)$ .

step 1.1algebra

3.1

Since $x,y$ were arbitrary, the map is constant; if $U$ is empty this conclusion is vacuous.

step 1.1step 2.1∎

Depends on

On a convex open set, a uniform bound $\|Df(z)v\|_2\le M\|v\|_2$ implies $\|f(y)-f(x)\|_2\le M\|y-x\|_2$

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 31 results over 10 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

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