TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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A distribution with zero derivatives on a connected open set is constant

Statement

Assume Countable Choice for the Lebesgue regular-distribution convention. If $Ω \subseteq R^{n}$ is nonempty, open and connected and $\partial_{j} u = 0$ for every $1 \leq j \leq n$ , then there is a unique $c \in C$ such that $u (ψ) = c \int ψ$ for every test. Conversely every constant regular distribution has zero first derivatives. On a disconnected open set constants may differ on different connected components.

Facts & Assumptions

[F1]

Smooth local convolutions satisfy $\partial_{j} (u * ρ_{ε}) = (\partial_{j} u) * ρ_{ε}$ (Convolution with a test function is smooth).

[F2]

Unit-mass mollifications converge locally in distribution pairings under Countable Choice (Mollifier approximation in distributions).

[F3]

A compact ball admits a nonnegative smooth compact cutoff, so a nonzero such bump inside any open ball can be normalized by its positive finite integral (Test function cutoffs and euclidean localization).

[F4]

The mean-value inequality for $C = R^{2}$ implies that a smooth function with zero gradient is constant on a ball (The mean value inequality: if $f : [a, b] \to R^{m}$ is continuous and differentiable on $(a, b)$ with $∥ f^{'} ∥_{2} \leq M$ , then $∥ f (b) - f (a) ∥_{2} \leq M (b - a)$ ).

[F5]

Distributions equal on an open cover are equal globally (Distributions form a sheaf).

[F6]

Classical derivatives of smooth regular distributions agree with distribution derivatives under Countable Choice (Distributional differentiation is continuous and commutes). We assume The Axiom of Countable Choice ( ${A C}_{ω}$ ) for F2 and this regular interpretation.

Proof

Given: the vanishing first derivatives and the stated choice assumption.

1.1

Take any open ball $B$ with compact closure in $Ω$ . For all sufficiently small $ε > 0$ , F1 gives a smooth mollification on a neighborhood of $\overline{B}$ with every first derivative zero. Along the segment between any two points of $B$ , the chain rule gives derivative zero; F4 with bound zero makes their values equal. Denote this value by $c_{ε}$ .

givenF1F2F4

2.1

Choose $η \in D (B)$ with $\int η = 1$ using F3. Then F2 gives $c_{ε} = \int (u * ρ_{ε}) η \to u (η) = c_{B}$ . For every $ψ \in D (B)$ , F2 therefore gives $u (ψ) = \lim c_{ε} \int ψ = c_{B} \int ψ$ . A unit-integral test also shows uniqueness of this constant on $B$ .

step 1.1F2F3

3.1

If two such balls overlap, their intersection contains an open ball and hence a unit-integral test by F3. Evaluation on that test proves that their constants agree. Consequently there is a well-defined locally constant function $c (x)$ on $Ω$ , whose value is the unique constant of any sufficiently small ball about $x$ . For a fixed $x_{0}$ , the set ${x : c (x) = c (x_{0})}$ and its complement are open. It is nonempty, so connectedness forces the complement empty. F5 applied to the ball cover yields $u = c (x_{0})$ as a regular distribution. A unit-integral test anywhere proves uniqueness.

step 2.1givenF3F5

4.1

F6 proves the converse, since the classical first derivatives of a constant vanish. In any open subset of Euclidean space each connected component is open: a ball about a point is connected and belongs to its component. Applying the preceding argument in each component gives independent constants. Conversely such a componentwise constant function is locally constant, hence smooth, and F6 gives zero derivatives. On the empty domain the sole distribution is zero but its representing constant is not unique; this explains the nonempty hypothesis. Countable Choice is inherited only from the integral mollification and regular-derivative clauses, not from the overlap argument. $□$

step 3.1F6

Depends on

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Sources

Semyon Dyatlov, Lecture notes for 18.155 (2022) (standard reference, not scraped)