Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A totally differentiable map with zero derivative on a convex open set is constant

Statement

Let U⊆Rm be convex and open. If f:U→Rn is totally differentiable at every point and Df(z)=0 for every z∈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≤M∥y−x∥2 under a uniform derivative bound M (On a convex open set, a uniform bound ∥Df(z)v∥2≤M∥v∥2 implies ∥f(y)−f(x)∥2≤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∈U.

L1
2.1

Hence ∥f(y)−f(x)∥2≤0, 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

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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