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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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Distributional differentiation is continuous and commutes

Statement

Distributional derivatives satisfy αβu=α+βu and are continuous linear maps on D(Ω) for both the weak and strong distribution topologies. These claims hold in ZF. Assuming Countable Choice for the cited Riemann-to-Lebesgue comparison, if fCk(Ω;C) and αk, then αuf=uαf. In particular distributional differentiation extends classical smooth differentiation.

Facts & Assumptions

[F1]

Derivatives are signed transposes of test derivatives (Distributional derivative).

[F2]

Weak seminorms test one function, and strong seminorms test bounded sets (Weak and strong topologies on distributions).

[F3]

Test differentiation is continuous and preserves compact supports with pm(αφ)pm+α(φ) (Test function operations are continuous).

[F5]

Regular functionals pair by the bilinear Lebesgue integral (Regular distribution from a locally integrable function).

[F6]

Compact sets admit smooth compact cutoffs equal to one near them (Test function cutoffs and euclidean localization).

[F7]

One-dimensional integration by parts holds when the factors and their derivatives are continuous on a closed interval (If u,v are differentiable on [a,b] with u,v integrable, then abuv=u(b)v(b)u(a)v(a)abuv).

[F8]

For Riemann-integrable functions on boxes with integrable sections, iterated and multiple Riemann integrals agree (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections).

[F9]

Under Countable Choice, a bounded Borel Riemann-integrable real function on a nondegenerate box has the same Lebesgue integral (Riemann–Lebesgue comparison for distribution test integrands).

[F10]

Countable Choice, needed only for the classical-compatibility clause below, is The Axiom of Countable Choice (ACω).

Proof

Given: a distribution u and multi-indices α,β.

1.1

Evaluating consecutive derivatives on a test gives (1)α+βu(βαφ). By F4 this is (1)α+βu(α+βφ), which is α+βu(φ) by F1. Linearity follows from the same formula.

givenF1F4
2.1

For a single test φ, the weak seminorm of αu equals u(αφ), a weak seminorm of u. For a bounded test set B, the set αB is bounded: any zero-neighborhood has a zero-neighborhood inverse image under the continuous linear test derivative of F3, and absorption of B by that inverse image gives absorption of its image. Hence pB(αu)=pαB(u), a strong seminorm. These equalities prove continuity in both topologies, including for nets, without choice.

step 1.1F1F2F3
3.1

Now assume F10 and fC1(Ω;C), and fix a test φ. Use F6 to choose χ=1 near suppφ with compact support in Ω. Extend g=χf by zero to Rn; it is C1 because it vanishes near the complement of Ω. Choose a nondegenerate box containing the supports of g and φ in its interior. On each coordinate segment F7 gives giφ=(ig)φ, because φ is zero at both endpoints. For complex functions expand into real and imaginary parts and apply the real identity to the four products. All integrands and sections are continuous on compact boxes and are Riemann integrable: uniform continuity makes their oscillation Darboux sums arbitrarily small on sufficiently fine uniform grids. For n>1, F8 integrates the one-coordinate identity over the remaining coordinates; for n=1 this is already the required identity. F9 componentwise converts the two multiple Riemann integrals to Lebesgue integrals. Near the test support g=f and ig=if, so the resulting equality is Ωfiφ=Ω(if)φ.

step 2.1F6F7F8F9F10
4.1

Continuous functions and their continuous derivatives are locally integrable by compact boundedness, and their regular functionals are continuous since their absolute pairings are bounded by Kfp0 on each fixed support (or directly by the global Q0 seminorm on that stage). F5 and step 3.1 therefore give iuf=uif. Iterating this identity through at most k derivatives for fCk, and using step 1.1, proves the classical-compatibility formula. Degree zero is the identity. Zero functions and the empty domain give zero distributions. The additional axiom enters only through the integral comparison in the classical argument; the algebra and dual continuity remain choice-free.

step 3.1step 1.1F1F5

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