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A distribution with a vanishing derivative is a polynomial

Statement

Assume Countable Choice. Let $n \in N_{0}$ and let $T$ be a distribution on the connected open set $R$ . Test functions are complex-valued and distributions are complex-linear; here, real distribution means that $⟨ T, ψ ⟩ \in R$ for every real-valued test function $ψ$ . If $T$ is a real distribution and its $(n + 1)$ st distributional derivative is zero, then there is a real algebraic polynomial $p$ of degree at most $n$ whose regular distribution equals $T$ .

Statement references

Dyatlov, Lecture Notes for 18.155, distributions with vanishing derivative

Hörmander, The Analysis of Linear Partial Differential Operators I, Chapter 2

Facts & Assumptions

Given: Countable Choice, $n \in N_{0}$ , and a real distribution $T \in D^{'} (R)$ with $D^{n + 1} T = 0$ .

[F1]

A distribution on a connected open set whose first derivative vanishes is a constant regular distribution, under Countable Choice (A distribution with zero derivatives on a connected open set is constant).

[F2]

The distributional derivative acts by signed differentiation of test functions; derivatives preserve realness because derivatives of real-valued tests are real-valued (Distributional derivative).

[F3]

A smooth regular distribution differentiates to the regular distribution of its classical derivative, under Countable Choice (Distributional differentiation is continuous and commutes).

[F4]

The formal derivative of $x^{n} / n!$ is $x^{n - 1} / (n - 1)!$ for $n \geq 1$ , and its $n$ th derivative is $1$ (The formal derivative of a polynomial).

[A1]

Countable Choice means every sequence of nonempty sets has a function choosing one element from each (The Axiom of Countable Choice ( ${A C}_{ω}$ )); it is assumed by the supplied regular-distribution derivative theorem.

[F5]

A compact subset of an open set admits a nonnegative smooth cutoff equal to one near it; for ${0} \subset R$ this gives a real test function with positive finite integral (positivity follows from the positive measure of an interval), which can be normalized to have integral one (Test function cutoffs and euclidean localization, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

Proof

1.1givenF1F5A1base

Base case $n = 0$ : the hypothesis is $D T = 0$ , so [F1] gives a constant regular distribution with some value $c \in C$ . Choose a real test function $η$ with integral one by [F5]. Since $c = ⟨ T, η ⟩$ and $T$ is real, $c \in R$ . Thus $T$ is represented by a real polynomial of degree at most zero.

1.2givenF1F2F5A1IH

Inductive step: let $n \geq 1$ and assume the result for $n - 1$ . The distribution $S = D^{n} T$ satisfies $D S = D^{n + 1} T = 0$ and is real by [F2]. By [F1], $S$ is the constant distribution with some value $c \in C$ . Choose a real test function $η$ with integral one by [F5]. Since $c = ⟨ S, η ⟩$ and $S$ is real, $c \in R$ .

2.1step 1.2F3F4A1IH

The smooth polynomial $q (x) = c x^{n} / n!$ has $D^{n} q = c$ as a regular distribution by [F3] and [F4]. Thus $U = T - q$ satisfies $D^{n} U = 0$ . The induction hypothesis applied with index $n - 1$ gives a polynomial $r$ of degree at most $n - 1$ whose regular distribution is $U$ .

3.1step 2.1algebradischarge-induction∎

The polynomial $p = q + r$ has degree at most $n$ and its regular distribution is $T$ , since regular distributions and distributional derivatives are linear. This completes the induction.

Depends on

Used by

Polynomial convolutions force activation polynomiality almost everywhere Lemma

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Sources

L. Hörmander, The Analysis of Linear Partial Differential Operators I, Chapter 2 (standard reference, not scraped)
Semyon Dyatlov, Lecture Notes for 18.155, distributions and derivatives (standard reference, not scraped)