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Bootstrapping a smooth Poisson problem

Example

Assume the Axiom of Choice (inherited from the embedding theorem used below) together with Countable Choice. Let Ω⊆Rn be open, let f∈C∞(Ω), and let u∈H1(Ω) be a local weak solution of −Δu=f on Ω in the sense of Local weak solutions of a divergence-form operator; for instance, when Ω is a nonempty bounded open set and also f∈L2(Ω), u may be the zero-trace weak Dirichlet solution, whose existence and uniqueness under the Axiom of Choice are Existence and uniqueness for the weak Dirichlet Poisson problem. Iterating Interior Hk+2 elliptic regularity with the constant coefficients of the Laplacian gives u∈Hlocm(Ω) for every m, hence a representative of class C∞ by Higher-order Sobolev embedding, and for that representative the equation −Δu=f holds pointwise on Ω. No boundary data and no boundary regularity are used: the smoothness of f alone permits the induction to continue at every order.

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; an open set Ω⊆Rn; f∈C∞(Ω); and a local weak solution u∈H1(Ω) of −Δu=f on Ω.

[F1]

A class u∈H1(Ω) is a local weak solution of −Δu=f when a(u,v)=∫Ωfv‾ dx for every v∈Cc∞(Ω), with a the form of the divergence-form operator; by the closure definition this is equivalent to the same identity for every bounded Ω2⋐Ω and every v∈H01(Ω2). (Local weak solutions of a divergence-form operator)

[F2]

Assume Countable Choice. The Laplacian L=−Δ has constant coefficients aij=δij, b=c=0, hence aij∈Wlock+1,∞(Ω) and b,c∈Wlock,∞(Ω) for every k≥0 with zeroth-order principal bound M0=1 and all positive-order derivative bounds zero; it is uniformly elliptic with θ=1. (Uniformly elliptic divergence-form operators and their sesquilinear forms)

[F3]

Assume Countable Choice. Let k≥0, let L have aij∈Wlock+1,∞(Ω) and bi,c∈Wlock,∞(Ω) with all coefficient derivatives through the indicated orders bounded by constants Mℓ, let f∈Hlock(Ω), and let u∈H1(Ω) be a local weak solution of Lu=f. Then u∈Hlock+2(Ω). (Interior Hk+2 elliptic regularity)

[F4]

Assume the Axiom of Choice. For n≥2, let Ω0 be a bounded extension domain in Rn, let k≥1, 1≤p<∞ with kp>n. If m≥0 and 0<α<1 satisfy m+α<k−n/p, then every u∈Wk,p(Ω0) has a representative in Cm,α(Ω0‾) with norm bounded by C∥u∥Wk,p(Ω0). In particular, p=2 and an integer k>n/2 give a continuous representative on Ω0. (Higher-order Sobolev embedding)

[F5]

Every open ball in Rn, n≥2, is a bounded extension domain: it is a bounded C∞ domain in the graph sense, so Bounded C^k domains admit integer-order Sobolev extension supplies an extension operator. For n=1, on each bounded open interval I⋐Ω every class in W1,2(I) has a unique continuous, locally absolutely continuous representative by One-dimensional W1,p functions have unique absolutely continuous representatives. (Sobolev extension domains and extension operators)

[F6]

Assume Countable Choice. If u∈Hloc2(Ω) is a local weak solution of Lu=f with aij∈W1,∞(Ω), bounded first coefficient derivatives, and f∈Lloc2(Ω), then the equation holds pointwise almost everywhere, with Di(aijDju) understood through the a.e. defined product (Diaij)Dju+aijDiDju; for the constant coefficients of the Laplacian this is the a.e. identity −∑iDiDiu=f. (Interior H2 regularity for divergence-form equations)

Verification

1.1F1F2F3algebragiven

Local smoothness at every order. Fix k≥0 and a bounded open Ω′′⋐Ω. By [F2] the Laplacian satisfies the coefficient hypotheses of [F3] at order k, and f∈C∞(Ω) gives f∈Hlock(Ω); since u is a local weak solution, [F3] gives u∈Hlock+2(Ω), hence u∈Hk+2(Ω′′). As k was arbitrary, u∈Hlocm(Ω) for every m≥1.

2.1F4F5step 1.1algebra

A smooth representative. If n≥2, fix a ball B⋐Ω and an integer m≥0. Choose a slightly larger ball B′⋐Ω and an integer s>m+n/2. Step 1.1 gives u∈Hs(B′), and [F5] makes B′ a bounded extension domain; [F4] then gives a Cm,α representative on B′‾ for some α>0. For different m these representatives agree everywhere on overlaps: they are continuous and represent the same almost-everywhere class. Thus they define a C∞ representative on B. If n=1, fix a bounded open interval I⋐Ω. Step 1.1 gives Dju∈H1(I) for every j≥0. By [F5] each has a unique continuous, locally absolutely continuous representative gj. The weak derivative of gj is Dj+1u, so the fundamental theorem gives gj(y)−gj(x)=∫xygj+1(t) dt for x<y in I; continuity of gj+1 implies gj∈C1(I) and gj′=gj+1. Iterating, g0 is C∞(I). These local representatives agree on overlaps, yielding a smooth representative on all components of Ω.

3.1F6step 2.1algebragiven∎

The equation holds pointwise. For the C∞ representative of step 2.1, u∈Hloc2(Ω) and f∈Lloc2(Ω), so by [F6] the equation −Δu=f holds pointwise almost everywhere, the Laplacian's expression being the a.e. function −∑iDiDiu. Both sides, −∑iDiDiu by step 2.1 and f by hypothesis, are continuous on Ω, and two continuous functions that agree almost everywhere on an open set agree at every point. Thus the smoothed representative solves the classical equation pointwise.

Source notes

Hunter's Corollary 4.29 and Laugesen's Theorems 5.8-5.9 (printed pp. 114 and 111-112, read in full) iterate the interior estimate to obtain Hlocm for every m and then a smooth representative; the same two-step pattern is used above. The example separates the two inputs: the coefficient regime of the Laplacian never obstructs the induction, and the smoothness of f is exactly what allows the data order k to increase; the Axiom of Choice is carried only by the Sobolev embedding used for the representative.

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