Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Derivatives of piecewise smooth functions include jump deltas

Example

Assume Countable Choice. Let AR be locally finite, and let fLloc1(R) be C1 on each component of RA. Assume finite one-sided limits f(a±) at every aA, and assume that the classical derivative g=f off A, assigned arbitrary finite values on A, belongs to Lloc1. Then Duf=ug+aA(f(a+)f(a))δa. The sum is locally finite. These hypotheses hold, in particular, when f is C1 up to each side of every break point.

Facts & Assumptions

[F1]

Locally integrable functions have regular functionals, and under Countable Choice the embedding theorem makes them distributions; distribution derivatives are signed test transposes and Dirac masses evaluate tests (Regular distribution from a locally integrable function, Locally integrable functions embed in distributions, Distributional derivative, Dirac delta and its derivatives).

[F2]

Complex integration by parts on closed intervals holds under Countable Choice (Complex integration by parts on intervals and decaying lines).

[F3]

Dominated convergence passes limits through integrable complex functions (Dominated convergence).

[F4]

Compactwise finite-order bounds characterize distributions (Local finite order characterization of distributions).

[F5]

Assume The Axiom of Countable Choice (ACω) for the Lebesgue integration interfaces.

Proof

Given: A,f,g and the stated assumptions.

1.1

By F1, uf and ug are distributions. Fix a test φ and a closed interval [b,c] containing its support in its interior, with endpoints outside A. Local finiteness and compactness imply A[b,c] is finite: take a finite subcover of neighborhoods each meeting finitely many points. List these break points in increasing order. On each intervening open interval (s,t) apply F2 to f,φ on [s+ε,tε] for sufficiently small positive ε. This gives [given, F1, F2, F5] s+εtεfφ=s+εtεgφ+f(s+ε)φ(s+ε)f(tε)φ(tε).

2.1

Let ε decrease to zero, for example through the reciprocal integers once the truncated interval is nonempty. F3 applies to the two integrals, with majorants fφ and gφ, integrable on [b,c] by the local integrability assumptions. The boundary values tend to f(s+)φ(s) and f(t)φ(t) by the finite one-sided limits; at b,c the test vanishes. Sum over the finitely many intervals. At each break point a, the left interval contributes f(a)φ(a) and the right contributes f(a+)φ(a). The result is exactly the asserted formula when paired with φ.

step 1.1givenF1F3
3.1

On any fixed compact test support K, the delta sum is finite and bounded in modulus by (aAKf(a+)f(a))p0(φ). It therefore defines a distribution by F4, so the test equality proves the distribution identity. If there are no break points in K the sum is zero, and a zero jump contributes no delta. If f is C1 up to both sides, f and g are bounded on each of the finitely many compact pieces meeting a compact interval, hence locally integrable, verifying the stated sufficient case. Merely being C1 on the open pieces does not supply local integrability of g at the breaks.

step 2.1givenF1F4

Depends on

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Sources