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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-10-02
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Zero extension of W_0^{1,p} has no boundary derivative

Statement

Assume Countable Choice. For any open Ω⊆Rn, n≥1, any 1≤p≤∞ and any K∈{R,C}, extension by zero E0u(x)=u(x)  (x∈Ω),E0u(x)=0  (x∉Ω), acted on representatives, sends W01,p(Ω;K) linearly into W1,p(Rn;K). For every u∈W01,p(Ω;K) and every coordinate direction i, the class Di(E0u) is the zero extension of the class Diu, and every Lp component norm is preserved: ∥E0u∥Lp(Rn)=∥u∥Lp(Ω),∥Di(E0u)∥Lp(Rn)=∥Diu∥Lp(Ω), so that ∥E0u∥W1,p(Rn)=∥u∥W1,p(Ω).

Facts & Assumptions

Given: Countable Choice; an open set Ω⊆Rn with n≥1; 1≤p≤∞; K∈{R,C}; a class u∈W01,p(Ω;K); and a test function φ∈Cc∞(Rn).

[F1]

Membership u∈W01,p(Ω;K) means that for every δ>0 there is ψ∈Cc∞(Ω;K) with ∥u−ψ∥W1,p(Ω)<δ, and W01,p⊆W1,p; the norm is the Sobolev norm of Integer-order Sobolev spaces and their norms (Zero-boundary Sobolev space as a norm closure).

[F2]

For w∈Lp(Ω;K) and its zero extension E0w: the extension is measurable, ∣E0w∣=∣w∣ on Ω and E0w=0 off Ω, so ∥E0w∥Lp(Rn)=∥w∥Lp(Ω) for 1≤p≤∞ because the integral over a measurable set is the integral of the indicator product; the map E0 is linear on classes; and for v∈Cc∞(Ω;K) the extension E0v lies in Cc∞(Rn) with ∂i(E0v)=E0(∂iv) for every i (Integral over a measurable subset, Complex Lp classes and Euclidean test-function conventions).

[F3]

The defining weak identity on an open set V is ∫Vw Dαψ=(−1)∣α∣∫V(Dαw)ψ for all ψ∈Cc∞(V;K) (Weak derivative of a locally integrable function).

[F4]

A C1 (indeed C∞) function on an open set has each of its classical partial derivatives as its weak derivative there (Classical derivatives agree with weak derivatives).

[F5]

Hölder's inequality: for conjugate exponents p,q and measurable f,g, ∫∣fg∣≤∥f∥p∥g∥q whenever the norms on the right are finite (Holder's inequality for integrals, including the endpoint cases).

[F6]

A class in W1,p(Rn;K) is exactly an Lp class whose coordinate weak derivatives exist as Lp classes, with the norm of Integer-order Sobolev spaces and their norms; weak derivatives are unique up to null sets, and changing representatives does not change the classes (Uniqueness of a weak derivative as an almost-everywhere class, Weak differentiation ignores null-set changes).

Choice use. Countable Choice is used once, in step 1.3, to select a single test-function approximant for each precision 1/j; the rest of the argument is explicit and choice-free.

Proof

technique · direct
1.1F2given

The zero extension E0 is linear on Lp classes, preserves Lp norms, and sends Cc∞(Ω;K) into Cc∞(Rn) with ∂i(E0v)=E0(∂iv); in particular E0vj−E0vk=E0(vj−vk) and ∥E0w∥Lp(Rn)=∥w∥Lp(Ω) for every Lp class w.

1.2F2F3F4given

For v∈Cc∞(Ω;K) the classical partial derivative ∂i(E0v) is the weak i-derivative of E0v on Rn, so for every test φ∈Cc∞(Rn) one has ∫RnE0v ∂iφ=−∫RnE0(∂iv) φ.

1.3F1given

Since u∈W01,p(Ω;K), [F1] with δ=1/j provides for each integer j≥1 some vj∈Cc∞(Ω;K) with ∥u−vj∥W1,p(Ω)<1/j; Countable Choice selects one such sequence (vj)j≥1.

2.1F1step 1.1step 1.3

The linearity and isometry of E0 give E0vj→E0u and E0(∂ivj)→E0(Diu) in Lp(Rn) as j→∞, because vj→u and ∂ivj→Diu in Lp(Ω) by step 1.3 and the definition of the Sobolev norm.

3.1F5step 1.2step 2.1

For the fixed test φ, step 1.2 gives ∫E0vj∂iφ=−∫E0(∂ivj)φ for every j. Hölder's inequality on the compact support of φ turns the Lp convergences of step 2.1 into convergence of both integrals: ∫RnE0vj∂iφ→∫RnE0u ∂iφ and ∫RnE0(∂ivj)φ→∫RnE0(Diu)φ; hence ∫RnE0u ∂iφ=−∫RnE0(Diu) φ.

4.1F6step 1.1step 3.1∎

Since φ was an arbitrary test function, step 3.1 exhibits E0(Diu) as a weak i-derivative of the Lp class E0u on Rn; by [F6] therefore E0u∈W1,p(Rn;K) with Di(E0u)=E0(Diu) almost everywhere, the component norms agree by step 1.1, and u↦E0u is linear because E0 is. Changing representatives on null sets changes no class by [F6], the case k=0 does not arise here, and complex scalars are covered by the same bilinear pairing.

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