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TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-24 (gpt-6-sol)
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Whitney extension for finite-order Euclidean jets

Statement

Let d,N≥1, let A be a closed subset of Rd, r a nonnegative integer, and, for each a∈A, let Pa be a polynomial of degree at most r with values in RN. Suppose that on each compact part of A, for every multi-index ∣α∣≤r, DαPa(b)−DαPb(b)=o(∣a−b∣r−∣α∣)(a,b∈A, ∣a−b∣→0) uniformly in a,b. Then there is a Cr map H:Rd→RN whose order-r Taylor polynomial at each a∈A is Pa. The construction needs no choice axiom.

Facts & Assumptions

Given: The closed set, integer order, and compatible polynomial jets of the statement.

[F1]

Ordinary finite-order Taylor estimates apply to the polynomial jets and smooth cutoff functions (Multivariable Taylor formula with a Lagrange remainder along a line segment).

Proof

technique · explicit dyadic Whitney-cube construction
1.1

The cases A=∅ and A=Rd are immediate: use zero in the first case, and in the second define H(a)=Pa(a); the compatibility condition is precisely the Taylor criterion for its derivatives DαH(a)=DαPa(a). Assume A is nonempty and proper. Subdivide the standard integer-translated dyadic grid into closed cubes. Retain every dyadic cube Q⊂Rd∖A satisfying 4d ℓ(Q)≤dist⁡(Q,A) which is maximal under dyadic parent inclusion. Every z∉A belongs to the closure of one of these cubes: arbitrarily small cubes about z satisfy the inequality, while sufficiently large ancestors fail it. Their interiors are disjoint. A retained cube has 4d ℓ(Q)≤dist⁡(Q,A)≤10d ℓ(Q): the upper bound follows because its parent fails the retention test and every point of the parent lies within 2d ℓ(Q) of Q. Consequently cubes whose fixed small enlargements meet have comparable side lengths, and only a dimension-dependent bounded number of those enlargements meet any one point. All cubes are obtained from a countable, explicitly ordered grid.

givenconstructalgebra
2.1

Fix one nonnegative C∞ bump on the unit cube, equal to one on that cube and supported in its concentric 9/8 enlargement. Rescale it to each retained cube to get ϕQ. Set ψQ=ϕQ/∑RϕR on Rd∖A. The denominator is at least one because the cubes cover the complement, and the sum is locally finite by step 1.1. Thus ∑QψQ=1 and ∣DαψQ∣≤Cαℓ(Q)−∣α∣; the constants are locally uniform because intersecting enlarged cubes have comparable sizes and bounded overlap. For each cube choose the nearest point aQ∈A to its centre, breaking ties by successive minimum coordinates on the compact nearest-point set. This specifies aQ without arbitrary selections. Define H(z)=∑QψQ(z)PaQ(z) off A, and H(a)=Pa(a) on A.

step 1.1constructalgebra
3.1

Fix a∈A and let z→a through the complement. Write t=dist⁡(z,A) and choose its lexicographically first nearest point az. If ψQ(z)≠0, step 1.1 gives ℓ(Q)≍t and ∣aQ−az∣=O(t). Taylor expansion of the polynomial PaQ−Paz about az, combined with the assumed compatibility of every derivative order, gives Dβ(PaQ−Paz)(z)=o(tr−∣β∣)(∣β∣≤r), locally uniformly as z→a. Differentiating the partition sum and using ∑QψQ=1 and the derivative bound in step 2.1 yields DαH(z)−DαPaz(z)=∑Q∑β≤α(αβ)Dα−βψQ(z)Dβ(PaQ−Paz)(z)=o(tr−∣α∣) for ∣α∣≤r. There are only boundedly many terms at z, and each has the displayed order.

F1step 1.1step 2.1givenalgebra
4.1

Compatibility again gives DαPaz(z)−DαPa(z)=o(∣z−a∣r−∣α∣): ∣az−a∣≤2∣z−a∣, and the polynomial Taylor sum converts every jet discrepancy into that bound. Since t≤∣z−a∣, step 3.1 implies DαH(z)−DαPa(z)=o(∣z−a∣r−∣α∣) as z→a through the complement. The same estimate at points of A is exactly the given compatibility. Starting with ∣α∣=0, these estimates prove by induction that each derivative of order at most r extends continuously across A with value DαPa(a): for ∣α∣<r subtract the linear part of Pa and divide by ∣z−a∣ to verify differentiability, and for ∣α∣=r use the zero-order continuity estimate. Thus H∈Cr and has the required jets. Every selection in the construction was by an ordered grid or a compact-set coordinate minimum.

step 2.1step 3.1givenalgebra∎

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