Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02
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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Integer-order Sobolev extension from a half-space

Statement

Assume the Axiom of Choice and n≥1. In the proof write en for the last canonical basis vector en−1. Let H=Rn−1×(0,∞)⊆Rn be the upper half-space, written x=(x′,t) with x′∈Rn−1 and t>0. For every k∈N0, every 1≤p≤∞ and every K∈{R,C} there is a bounded linear extension operator Ek,p:Wk,p(H;K)⟶Wk,p(Rn;K),(Ek,pu)∣H=u almost everywhere on H. For k≥1 one may take, with the unique coefficients (a1,…,ak)∈Rk satisfying ∑j=1kaj(−j)m=1 for m=0,1,…,k−1, Ek,pu(x′,t)=u(x′,t)  (t>0),Ek,pu(x′,t)=∑j=1kaju(x′,−jt)  (t<0), and for k=0 one may take the even reflection E0,pu(x′,t)=u(x′,∣t∣). In particular H is a Wk,p-extension domain in the sense of Sobolev extension domains and extension operators for every k and every p, and no density assertion is made or needed.

Facts & Assumptions

Given: the Axiom of Choice; the half-space H=Rn−1×(0,∞); k∈N0; 1≤p≤∞; K∈{R,C}; a class u∈Wk,p(H;K); and a test function φ∈Cc∞(Rn).

[F1]

Sobolev classes on an open set: u∈Wk,p(H;K) means that for every multi-index α with ∣α∣≤k there is a class Dαu∈Lp(H;K) satisfying the weak identity ∫Hu Dαψ=(−1)∣α∣∫H(Dαu)ψ for all ψ∈Cc∞(H;K), with D0u=u; the norm is the ℓp sum over ∣α∣≤k, and the maximum of essential bounds for p=∞ (Integer-order Sobolev spaces and their norms).

[F2]

Weak differentiation is local and linear: for open V⊆Rn, v=Dαu weakly on Rn implies v∣V=Dα(u∣V) weakly on V, and the sum of two weakly differentiable classes is weakly differentiable with the sum of the derivatives (Linearity, locality, and commutation of weak derivatives).

[F3]

Linear changes of variables: for the invertible linear map Tj(x′,t)=(x′,−jt), whose determinant has absolute value j, and every nonnegative measurable f, ∫Rnf dλn=j∫Rnf∘Tj dλn; in particular ∫{t<0}∣w(x′,−jt)∣p dx′dt=j−1∫H∣w(x′,τ)∣p dx′dτ for every measurable w (A linear map T of Rn sends Lebesgue measurable sets to Lebesgue measurable sets, with λn(T[E])=∣det⁡T∣ λn(E) when T is invertible and T[E] Lebesgue null when it is not).

[F4]

Traces of normal sections. Let u∈Wk,p(H;K) with k≥1 and 1≤p≤∞. For every multi-index γ with ∣γ∣≤k−1 and almost every x′∈Rn−1 the section s↦Dγu(x′,s) has a representative that is absolutely continuous on every compact interval [0,T], extends continuously to s=0, and satisfies ddsDγu(x′,s)=Dγ+enu(x′,s) for almost every s, where en is the last unit vector; at p=∞ one first restricts to a finite exponent on compact sets, since Wloc1,∞⊆Wloc1,q. The trace tr⁡γ(x′):=lim⁡s→0+Dγu(x′,s) exists for almost every x′, is measurable, and satisfies the estimate ∣tr⁡γ(x′)∣≤2T∫0T∣Dγu(x′,s)∣ ds+∫0T∣Dγ+enu(x′,s)∣ ds for every 0<T<∞, whose right-hand side is a.e. finite and integrable in x′ over compact sets (The ACL characterisation of W1,p, One-dimensional W1,p functions have unique absolutely continuous representatives, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, The Euclidean Lebesgue measure is the completion of the product of the factor Lebesgue measures, Holder's inequality for integrals, including the endpoint cases).

[F5]

Half-space integration by parts with traces. Let u∈Wk,p(H;K), let α=(β,m) with ∣α∣≤k, and let φ∈Cc∞(Rn). Then ∫Hu Dαφ=(−1)∣α∣∫H(Dαu) φ+∑r=0m−1(−1)r+1∫Rn−1tr⁡ren(x′) ∂tm−1−rDβφ(x′,0) dx′. For the reflected function f(x′,t)=u(x′,−jt) on the lower half-space, whose α-derivative is Dαf(x′,t)=(−j)mDαu(x′,−jt) and whose traces of the normal derivatives at t=0− are (−j)rtr⁡ren(x′), and the same formula holds with the boundary signs (−1)r in place of (−1)r+1. In both formulas the boundary term keeps the tangential derivatives on the test function; they are not also applied to the trace. This follows by integrating in the normal variable first, then moving the tangential derivatives in the interior term onto u. The one-dimensional integrations are justified by the absolutely continuous representatives of [F4] and assembled with Fubini's theorem (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, A box in Rn with parameters ai≤bi is Lebesgue measurable of measure ∏i<n(bi−ai), whichever of its faces are included).

[F6]

Vandermonde systems. For k≥1 the matrix ((−j)m)0≤m≤k−1, 1≤j≤k is invertible, because its determinant is the Vandermonde product ∏i<i′((−i′)−(−i))≠0 over the distinct nodes −1,−2,…,−k; hence the moment system ∑j=1kaj(−j)m=1, m=0,…,k−1, has exactly one solution. This is finite linear algebra over R and uses no choice.

[F7]

Extension domain and operator: Ω is a Wk,p-extension domain when there is a bounded linear E:Wk,p(Ω;K)→Wk,p(Rn;K) with (Eu)∣Ω=u almost everywhere for every class u (Sobolev extension domains and extension operators).

Choice use. The Axiom of Choice is invoked only through the ACL, one-dimensional absolutely-continuous and Fubini interfaces cited in [F4]–[F5]; the Vandermonde coefficients of [F6] and the reflection formula are explicit.

Proof

technique · direct
1.1F6given

For k≥1 set Jk={1,…,k} and let (aj)j∈Jk be the unique solution of the moment system supplied by [F6]. For k=0 set J0={1} and a1=1, with no moment condition. In either case define Eu:=u on H and Eu(x′,t):=∑j∈Jkaju(x′,−jt)(t<0). For k=0 this is exactly the even reflection u(x′,∣t∣).

1.2F4

Traces exist as in [F4]: for every ∣γ∣≤k−1 the trace tr⁡γ(x′) of Dγu at t=0 exists for almost every x′, is measurable, and obeys the displayed estimate, so each trace is integrable against the compactly supported traces of Dβφ that appear below.

2.1step 1.1

Candidate derivatives. For every multi-index α=(β,m) with ∣α∣≤k define gα:=Dαu on H and gα(x′,t):=∑j∈Jkaj(−j)mDαu(x′,−jt)(t<0). When k=0, this gives g0=Eu on the lower half-space as well.

3.1F1F3step 2.1

Membership and bounds. By [F3] each reflected summand satisfies ∫{t<0}∣Dαu(x′,−jt)∣p=j−1∫H∣Dαu∣p for p<∞, so ∥gα∥Lp(Rn)≤(1+∑j∈Jk∣aj∣jm−1/p)∥Dαu∥Lp(H),∥Eu∥Lp(Rn)≤(1+∑j∈Jk∣aj∣j−1/p)∥u∥Lp(H), with the corresponding essential-supremum bounds (1+∑j∈Jk∣aj∣jm) and (1+∑j∈Jk∣aj∣) when p=∞; in particular Eu and all gα lie in Lp(Rn;K).

3.2F2F5step 1.1step 1.2step 2.1

Interface cancellation. Fix φ∈Cc∞(Rn) and α=(β,m). Apply [F5] on H and to each reflected summand on {t<0}. The tangential derivatives of φ remain in the boundary terms; the tangential weak-derivative identity moves them onto the interior terms, giving ∫RnEu Dαφ=(−1)∣α∣∫Rngαφ+∑r=0m−1(−1)r∫Rn−1tr⁡ren(x′) ∂tm−1−rDβφ(x′,0)[(−1)+∑j∈Jkaj(−j)r]dx′. If k≥1, every index r≤m−1≤k−1 satisfies ∑j∈Jkaj(−j)r=1 by step 1.1, so each bracket vanishes; if k=0, then m=0 and the boundary sum is empty. In either case ∫RnEu Dαφ=(−1)∣α∣∫Rngαφ. The traces in the sum are integrable by step 1.2.

4.1F1F2F7step 1.1step 3.1step 3.2∎

Since φ was an arbitrary test function, step 3.2 exhibits gα as the weak α-derivative of the Lp class Eu for every ∣α∣≤k; with Eu∈Lp(Rn) and gα∈Lp(Rn) from step 3.1 and the norm formula of [F1], this gives Eu∈Wk,p(Rn;K) and ∥Eu∥Wk,p(Rn)≤Ck,p∥u∥Wk,p(H) for the finite constant determined by the coefficients. The map u↦Eu is linear because the reflection formula is linear in u on each half-space, and (Eu)∣H=u holds by construction; hence E=Ek,p is a bounded linear extension operator and H is a Wk,p-extension domain in the sense of [F7]. The case k=0 is the even reflection with no interface terms, and the case n=1 is the same argument with Rn−1 a single point and the traces taken at 0.

Depends on

Used by

Dependency tree · two levels

106 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources