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Matching C1 pieces across a hyperplane have no jump derivative

Sources

  • Juha Kinnunen, Sobolev Spaces, Chapter 1 §1.1, Example 1.7, printed pp. 3–4, proves the weak derivative identity for a continuous piecewise affine function with a matching value at its single break point by splitting the one-dimensional integral and applying integration by parts and the fundamental theorem of calculus. This is a one-dimensional model only.
  • John K. Hunter, Notes on Partial Differential Equations, Chapter 3 §3.2, Example 3.3, printed p. 48, computes the one-dimensional test pairing for the continuous positive-part function and its step-function weak derivative. That calculation is the one-dimensional slice model used here; it does not state the higher-dimensional result.
  • Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Chapter 8 §8.2, Examples (i) and the following sentence, printed pp. 202–203, states as exercises that ∣x∣ lies in W1,p for every 1≤p≤∞ and that a continuous piecewise-C1 function on a closed interval lies in W1,p for all such p. The source gives no proof of those exercises and treats only one dimension. The multidimensional claim below is proved by coordinate slices.

Statement

Assume the Axiom of Countable Choice. Let n≥2, let Q=(−1,1)n, and put K+=[−1,1]n−1×[0,1],K−=[−1,1]n−1×[−1,0]. For K∈{R,C}, let f±:K±→K be C1 up to the boundary: each is continuous on its closed half-box and each first partial derivative on the interior extends continuously to that half-box. Suppose f+(y,0)=f−(y,0)(y∈(−1,1)n−1). Define f~ on [−1,1]n by f~(x)=f+(x) when xn≥0 and f~(x)=f−(x) when xn<0, and let f=f~∣Q; matching traces give continuity across the interface inside Q. For j=1,…,n, define gj(x)={∂jf+(x),xn>0,∂jf−(x),xn<0,0,xn=0. using the continuous boundary extensions in the first two cases. Then for every 1≤p≤∞, f∈W1,p(Q;K),Djf=[gj](1≤j≤n). Thus the weak first derivatives agree almost everywhere with the classical derivatives on the two open half-boxes; their values on the interface are irrelevant.

Facts & Assumptions

Given: The Axiom of Countable Choice, n≥2, the two closed half-boxes, the functions f± and their matching traces, and a test function φ∈Cc∞(Q;C).

[F1]

The only choice assumption declared here is the Axiom of Countable Choice, which says that every countable family of nonempty sets has a choice function (The Axiom of Countable Choice (ACω)).

[F2]

The weak derivative identity for a first coordinate derivative is ∫Qu ∂jφ dx=−∫Qv φ dx for every test function (Weak derivative of a locally integrable function, Test function space d of an open set). Test functions have compact support in Q and extend by zero to smooth compactly supported functions on Rn; their boundary values on ∂Q vanish. The pairing is complex bilinear, without conjugation (Test function space d of an open set).

[F3]

Membership in W1,p requires an Lp class for the function and for each weak first derivative; the zero multi-index is the function itself (Integer-order Sobolev spaces and their norms, Ck maps and multi-index derivative notation in Euclidean space). A weak derivative value class is unique almost everywhere under Countable Choice (Uniqueness of a weak derivative as an almost-everywhere class).

[F5]

A continuous map has Borel preimages of Borel sets, and the Borel sigma algebra on a subspace is the trace of the ambient Borel sigma algebra (A continuous map has Borel preimages of Borel sets, The Borel sigma-algebra of a subspace is the trace of the ambient Borel sigma-algebra). The half-boxes are closed and hence Borel (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, The Borel sigma-algebra of a topological space). Consequently finite piecewise gluing on the two open half-boxes and the interface is Borel. Continuous test functions and their derivatives are Borel (Continuous functions on Euclidean spaces are Borel measurable); Borel functions on Rn are Lebesgue measurable under Countable Choice (Borel measurable and Lebesgue measurable functions on Rn, Assuming countable choice, every Borel subset of Rn is Lebesgue measurable). Sums and products of real and complex measurable functions remain measurable by the componentwise arithmetic rules (Arithmetic and lattice operations preserve measurability whenever they are defined, Complex Lp classes and Euclidean test-function conventions).

[F6]

Under Countable Choice, Q has measure 2n and every box with a degenerate side, including the interface inside a bounded box, is null (A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included). Lebesgue measure on each Euclidean factor is sigma-finite (Lebesgue measure is sigma-finite, and every metrically bounded subset of Rn has finite outer measure).

[F8]

If a measurable function is bounded by C1Q and λn(Q)<∞, then its modulus and each finite positive power have finite integral: the majorant is a simple function with integral Cλn(Q), and the nonnegative integral is monotone (Integrable real and complex functions, and their integrals, Integral over a measurable subset, The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions, Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F9]

On a closed interval, if a continuous function G is differentiable except at finitely many interior points and an integrable extension h agrees with G′ elsewhere, then ∫abh=G(b)−G(a) (Newton–Leibniz remains valid across finitely many exceptional interior points when the primitive is continuous). Bounded functions continuous except at finitely many points are Riemann integrable (A bounded function on [a,b] that is continuous except at finitely many points is Riemann integrable), and bounded Riemann integrable functions have the same Riemann and Lebesgue integrals under Countable Choice (A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral). The product rule holds on each smooth real-valued piece; the complex case is obtained componentwise (Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0).

[F10]

The Euclidean Lebesgue measure on Rn−1×R is the completion of the product of the factor Lebesgue measures under Countable Choice (The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures). Tonelli-Fubini applies to integrable functions for that completed product and gives measurable integrable sections outside factor-null sets (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability).

Proof

technique · direct coordinate slices
1.1F1F4F5given

The two half-boxes are compact, so [F4] gives a finite pointwise bound B for f± and every continuous extension of ∂jf±. Extend f and the gj by zero outside Q. On each closed half-box, the source functions and derivative extensions are continuous; using the Borel trace fact in [F5], their level preimages on each piece are Borel. The piecewise definitions on the strict half-boxes, the interface (where gj=0), and the complement of Q therefore make these extensions Borel. The test functions and their first derivatives are Borel by [F5]. Hence f, gj, and the integrands formed from them and the tests are Lebesgue measurable under the exact assumption [F1].

2.1F1F4F6F7F8step 1.1

For each test φ, set Hj=f ∂jφ+gjφon Q, and extend Hj by zero off Q. The bounds from [F4] and compact support of the test give a finite constant Cj with ∣Hj∣≤Cj1Q, so Hj∈L1(Rn) by [F6, F8]. Put C:=1+B; each representative satisfies ∣f∣,∣gj∣≤C1Q. For finite p, [F7] gives ∣f∣p,∣gj∣p≤Cp1Q, so [F6, F8] proves their Lp membership; for p=∞ the pointwise bound proves essential boundedness. The p=1 case also gives local integrability. The interface is null by [F6], so the chosen values gj=0 there do not change the a.e. classes. All measure claims here use [F1].

3.1F1F2F9F11givenstep 2.1

Fix j=n and y∈(−1,1)n−1, and put Gy(t)=f~(y,t)φ(y,t) on [−1,1]. It is continuous at t=0 because the two traces agree; on each side the product rule [F9] gives Gy′(t)=Hn(y,t). The section Hn(y,⋅) is bounded and continuous away from at most 0, so it is Riemann integrable by [F9]. Apply finite-exception Newton-Leibniz [F9] with exceptional set {0}. Compact support makes Gy(−1)=Gy(1)=0, so the Riemann integral of the section is zero; under Countable Choice its Lebesgue integral is the same by [F9] and [F1]. Whenever Gy,Hn(y,⋅) are complex-valued, split both into real and imaginary parts; componentwise integration in [F11] preserves zero.

4.1F1F6F9F11givenstep 2.1step 3.1

Fix j<n and reorder only the first n−1 coordinates so xj is first and xn remains last. This orthogonal coordinate permutation preserves the integral of Hj by [F11]. For fixed other coordinates z with zn≠0, Gz(t)=f~(x1,…,xj−1,t,xj+1,…,xn)φ(x1,…,t,…,xn) is continuously differentiable on [−1,1] and has derivative Hj along the section by [F9]. Its endpoints vanish, so finite-exception Newton-Leibniz gives zero section integral, first as a Riemann integral and then as a Lebesgue integral. The excluded parameter set zn=0 is a degenerate box in Rn−1 and is null by [F6] under [F1], so the section integral is zero for almost every z; for complex-valued products split into real and imaginary parts as in step 3.1.

5.1F1F2F3F6F10F11step 2.1step 3.1step 4.1∎

Under [F1], [F10] applies Fubini to Hn in Rn−1×R and to each tangential Hj after the permutation in step 4.1 in R×Rn−1. Steps 3.1 and 4.1 give zero section integrals almost everywhere, so ∫Qf ∂jφ dx+∫Qgjφ dx=∫RnHj dx=0(j=1,…,n). For α=0 the weak identity is the identity itself; for α=ej it is exactly the weak-derivative test identity [F2], with gj locally integrable by step 2.1. Uniqueness [F3] identifies this value class as Djf, and the Lp bounds of step 2.1 with the Sobolev definition [F3] give f∈W1,p(Q;K) for every 1≤p≤∞.

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Sources