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Radial basis and kernel interpolation

1 · Prerequisites

2 · Summary

This page defines strict positive-definite interpolation, strict conditional-positive-definite polynomial augmentation, site fill and separation, and an exact augmented RBF-FD procedure. For arbitrary fixed d≥1, ν>0, and a>0, it derives the normalized Matérn spectral kernel and proves its weighted-Fourier native-space identity, the restriction-quotient equivalence with Hν+d/2(Ω), and the projection and power-function formulas. On bounded Lipschitz domains with a stated uniform interior cone condition, stable polynomial norming stencils give a complete scattered-zeros estimate and the hν interpolation and power-function rate. The fill-distance rate needs no mesh-ratio condition. A separate compact-frequency-window argument proves the q2ν Gram eigenvalue lower bound for separated sites with 0<q≤1 and the exact-arithmetic conditioning consequence for 0<h≤1 under bounded mesh ratio. The RBF-FD item proves polynomial reproduction and a Taylor truncation estimate retaining the actual weight norm; unisolvence alone is not claimed to bound that norm. Choice assumptions inherited from the weighted-L2 and Hilbert-projection suppliers are stated locally. No Gaussian flat-limit stability, RBF-QR algorithm, PDE stability, or floating-point guarantee for the augmented indefinite solve is asserted.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Radial kernels, polynomial augmentation, and site geometry

Definition

Fix an integer d≥1. An admitted domain Ω⊂Rd is a nonempty bounded Lipschitz domain (its boundary is locally the graph of a Lipschitz function, with Ω on one side) with fixed r0>0 and 0<θ<π/2 such that for every x∈Ω‾ there is an orthogonal map Qx for which x+Qx{z∈Rd:0<∣z∣<r0, z1>∣z∣cos⁡θ}⊂Ω. An orthogonal map is a real linear map with QxTQx=I, equivalently one preserving the Euclidean inner product. The map Qx may depend on x; no simultaneous selection of the maps is part of this definition.

A real radial kernel on Rd has the form K(x,y)=ϕ(∣x−y∣) for a function ϕ:[0,∞)→R. For a finite ordered list of distinct sites X=(x1,…,xN), its Gram matrix is KX=(K(xi,xj))i,j=1N. The kernel is strictly positive definite if cTKXc>0 for every such X and every nonzero c∈RN; the nonstrict positive-semidefinite condition replaces > by ≥. Only strict positivity is used for the ordinary interpolation solve below.

For an integer m≥1, let Πm−1(Rd) be the real polynomials of total degree at most m−1, with an ordered basis p1,…,pJ, where J=dim⁡Πm−1(Rd). The evaluation matrix on X is PX=(pj(xi))i,j. The sites are m-unisolvent for this polynomial space when PX has linearly independent columns. A kernel is strictly conditionally positive definite of order m when, for every finite distinct X and this evaluation matrix, cTKXc>0 for every nonzero c∈RN satisfying PXTc=0. Thus strict conditional positivity is a strict inequality on the moment-null subspace; it is stronger than the corresponding nonstrict inequality.

For data y∈RN, the strictly positive definite interpolation system is KXa=y. For an order-m strictly conditionally positive definite kernel and m-unisolvent sites, the polynomially augmented interpolation system is [KXPXPXT0][ab]=[y0]. Whenever the relevant system has a unique solution, its kernel-plus-polynomial function x↦∑iaiK(x,xi)+∑jbjpj(x) is called the interpolant. Invertibility for the stated strict hypotheses is supplied by Strict positive and conditionally positive augmented interpolation systems are invertible ↗.

For a finite nonempty site set X⊂Ω, the fill distance is hX=sup⁡x∈Ωmin⁡xi∈X∣x−xi∣. If N≥2, the separation radius and mesh ratio are qX=12min⁡i≠j∣xi−xj∣ and ρX=hX/qX, respectively. These are exact geometric quantities, independent of any floating-point solver.

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Strict positive and conditionally positive augmented interpolation systems are invertible

Statement

Let X=(x1,…,xN) be a finite ordered list of distinct sites in Rd. If K is strictly positive definite, then KX is invertible and KXa=y has a unique solution for every y∈RN. If K is strictly conditionally positive definite of order m and X is m-unisolvent for Πm−1(Rd), then the augmented matrix in Radial kernels, polynomial augmentation, and site geometry is invertible and its system has a unique solution for every right-hand side.

Facts & Assumptions

Given: A finite distinct site list X, a real radial kernel, and either strict positive definiteness or strict conditional positive definiteness of order m together with m-unisolvent sites.

[F1]

Strict positive definiteness means cTKXc>0 for every nonzero c on distinct sites. (Radial kernels, polynomial augmentation, and site geometry)

[F2]

Strict conditional positive definiteness means cTKXc>0 for each nonzero c with PXTc=0; m-unisolvence means PX has linearly independent columns. (Radial kernels, polynomial augmentation, and site geometry)

[F3]

For a linear map with finite-dimensional domain, dimension equals nullity plus rank. (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T)

[F4]

The real numbers form a field. (The reals form a field)

[F5]

A real square matrix is invertible when it has a two-sided matrix inverse. (Invertible matrices and the general linear group GL⁡n(F))

Proof

Given: The hypotheses above, with KX∈RN×N and, in the augmented case, PX∈RN×J.

1.1F1algebra

In the strictly positive definite case, if KXa=0, then aTKXa=0, so [F1] forces a=0. Thus the linear map a↦KXa has zero kernel.

1.2F2algebra

In the augmented case, put M=[KXPXPXT0] and suppose M[ab]=0. The lower block gives PXTa=0, and multiplying the upper block by aT gives aTKXa+aTPXb=0. Since aTPX=(PXTa)T=0, strict conditional positivity [F2] forces a=0.

2.1step 1.1F3F4F5algebrachoose

By [F3], the map a↦KXa has rank N, so it is onto RN as well as one-to-one. For the standard basis e1,…,eN, choose vj with KXvj=ej and let B have columns vj. Then KXB=IN; for every x, injectivity of KX applied to KX(BKXx−x)=0 gives BKXx=x. Thus B is a two-sided inverse by [F5], and every system KXa=y has a unique solution.

3.1step 1.2F2F3F4F5algebrachoose∎

With a=0, the upper block reduces to PXb=0. The columns of PX are linearly independent by [F2], so b=0. The augmented square matrix therefore has zero kernel, and [F3] gives rank N+J and surjectivity. For the standard basis of RN+J, choose the unique preimage of each basis vector and use those preimages as columns of a matrix B. Then MB=IN+J; injectivity gives BM=IN+J by applying M to BMx−x. Hence M is invertible by [F5], and its system has a unique solution for every right-hand side.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

The normalized Matérn kernel in the 2π Fourier convention

Definition

Fix the dimension d≥1 of an admitted domain, a smoothness parameter ν>0, and an inverse length a>0. Assume Countable Choice for the Lebesgue-integration suppliers used to justify this formula; this definition does not assume full Choice. Put s=ν+d/2 and A=Γ(s)(4π)d/2a2νΓ(ν),w(ξ)=A(a2+4π2∣ξ∣2)−s. All integrals are Lebesgue integrals with respect to λd (Lebesgue measurable sets, the family L(Rn), and the restricted set function λn), and Γ is the Euler integral of The real Gamma function by Euler's integral. With the negative-sign Fourier transform convention U^(ξ)=∫U(x)e−2πix⋅ξ dx, define the normalized stationary kernel by k(z)=∫Rdw(ξ)e2πiz⋅ξ dξ,K(x,y)=k(x−y). The corresponding length parameter in the usual Matérn notation is ℓ=2ν/a. For b>0, define the modified Bessel function in its integral normalization by Kν(b)=12(b2)ν∫0∞u−ν−1e−u−b2/(4u) du. The classical Bessel expression for k(z) when z≠0, its value k(0)=1, absolute convergence, continuity, and radial symmetry are established by The Matérn spectral integral, normalization, and radial formula ↗. This definition makes no finite-precision stability claim.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Fourier Bessel-potential and restriction-quotient Sobolev spaces

Definition

Assume the Axiom of Choice when using the well-definedness results below; it is inherited conservatively from the published L2 completeness supplier. Let d≥1 and s>d/2. Write Ls2(Rd) for the equivalence classes modulo Lebesgue-almost-everywhere equality of measurable complex functions g satisfying ∥g∥Ls22=∫Rd(1+∣ξ∣2)s∣g(ξ)∣2 dξ<∞. For such g, the inverse Fourier representative is the function Ug(x)=∫Rdg(ξ)e2πix⋅ξ dξ. The whole-space Bessel-potential space Hs(Rd) consists of these representatives, with norm ∥Ug∥Hs(Rd)=∥g∥Ls2. For the admitted domain Ω of Radial kernels, polynomial augmentation, and site geometry, define Hs(Ω) to be the restrictions U∣Ω of whole-space members, with the quotient norm ∥u∥Hs(Ω)=inf⁡{∥U∥Hs(Rd):U∣Ω=u}. The whole-space inner product is ⟨Ug,Uh⟩Hs=∫(1+∣ξ∣2)sg(ξ)h(ξ)‾ dξ, linear in the first variable as in Real and complex inner-product spaces and their induced length. The inverse Fourier map is injective, whole-space representatives are continuous, the whole-space model is Hilbert, and Schwartz functions are dense in that norm; these facts are supplied by The weighted Fourier model is an injective Hilbert representation ↗ and Schwartz functions are dense in the Fourier Bessel-potential norm ↗. The restriction quotient is a norm by bounded point evaluation; its Hilbert realization as the restricted native space is proved in The restricted Matérn native space is the Sobolev quotient space ↗. No extension operator or identification with an intrinsic fractional Sobolev norm on Ω is part of this definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Uniform cone domains admit stable polynomial norming stencils

Statement

Let Ω⊂Rd be an admitted domain of Radial kernels, polynomial augmentation, and site geometry, let X⊂Ω be finite and nonempty, and put h=hX. Fix an integer m≥1 and let J=dim⁡Πm−1(Rd). There are constants R,B,h0>0, depending only on d,m,r0,θ, such that if 0<h≤h0, then for every x∈Ω‾ there are distinct nodes y1,…,yJ∈X and real weights λ1,…,λJ with ∣yj−x∣≤Rh,∑j=1J∣λj∣≤B,p(x)=∑j=1Jλjp(yj)(p∈Πm−1(Rd)). No lower bound on site separation is assumed.

Facts & Assumptions

Given: An admitted domain with uniform cone constants r0,θ, a finite nonempty site set X, fill distance h, an integer m≥1, and a point x∈Ω‾.

[F1]

The cone constants give, at each x, an orthogonal map Qx that sends the radius-r0 cone into Ω−x; the fill distance is sup⁡z∈Ωmin⁡y∈X∣z−y∣. (Radial kernels, polynomial augmentation, and site geometry)

[F2]

A nonzero univariate polynomial over an integral domain has at most its degree many distinct roots. (A nonzero polynomial of degree n over an integral domain has at most n distinct roots)

[F3]

For a linear map with finite-dimensional domain, dimension is nullity plus rank. (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T)

[F4]

The real numbers form a totally ordered field. In an interval I=(a,b), the m numbers a+km+1(b−a), 1≤k≤m, are distinct and lie in I. (The reals form a totally ordered field)

[A1]

For one fixed x, choose its cone orientation and finitely many nearest sites; these are finite choices, and the argument below never selects a family of stencils indexed by all x.

Proof

Given: The data above and the finite-dimensional polynomial space Πm−1(Rd) with its monomial basis.

1.1F2F3F4givenchoose

The evaluations at points of the open unit cone C={z:0<∣z∣<1, z1>∣z∣cos⁡θ} span the dual of Πm−1(Rd). Indeed, if fewer than J independent evaluations have been chosen, the map from Πm−1 to their finite value list has nonzero kernel by [F3]; choose a nonzero polynomial p in that kernel. Suppose, for contradiction, that p vanishes throughout C. Since C is nonempty and open (it contains e1/2), it contains a box I1×⋯×Id of nonempty open intervals. Write p as a polynomial in its first coordinate, of degree at most m−1. For every fixed tuple of the other coordinates in their intervals, this univariate polynomial vanishes at the m distinct points a1+km+1(b1−a1), 1≤k≤m, where I1=(a1,b1). These points lie in I1 by [F4], so the root bound [F2] forces every coefficient to vanish at that tuple. Applying the same argument inductively to each coefficient polynomial on I2×⋯×Id shows every coefficient is identically zero, contrary to the choice of p. Thus p is nonzero at some point of C, and evaluation there increases the independent list. After the finite number J of steps, choose z1,…,zJ∈C with invertible evaluation matrix.

2.1step 1.1algebrachoose

For the chosen nodes, the weights representing evaluation at zero are the row p(0)TE−1, where E is their evaluation matrix and p is the monomial vector. Choose 0<ε<1; if J>1, also require it to be smaller than one third of the minimum distance between distinct reference nodes. On the bounded perturbation neighborhood, every monomial entry changes by at most a constant times ε by telescoping its finite product; the determinant, being a finite signed sum of products of entries, therefore changes by at most Cε. Choose ε smaller if necessary so Cε<∣det⁡E∣/2. Then ∣det⁡E′∣≥∣det⁡E∣/2 for every perturbed matrix E′. Its cofactors are bounded there because the entries are bounded, so the adjugate formula E′−1=adj⁡(E′)/det⁡E′ gives a uniform bound B on the sum of absolute evaluation-at-zero weights.

3.1F1A1step 2.1givenchoose

Choose L>1/ε and set h0≤r0/L. For the fixed x, use [F1] to choose Qx. Each target x+LhQxzj lies in Ω when h≤h0, so the fill-distance definition supplies a site yj∈X within distance h. The normalized points zj′=(Lh)−1Qx−1(yj−x) then satisfy ∣zj′−zj∣≤1/L<ε; hence they are distinct and their evaluation weights have sum at most B.

4.1step 3.1algebra∎

For any p∈Πm−1(Rd), the affine pullback q(z)=p(x+LhQxz) has degree at most m−1. The weights from step 3.1 represent evaluation at zero for every such polynomial, so p(x)=q(0)=∑jλjq(zj′)=∑jλjp(yj). Also ∣yj−x∣≤Lh∣zj∣+h≤(L+1)h. Taking R=L+1 proves the bounds with constants independent of X,h,x; no separation condition entered the construction.

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The Matérn spectral integral, normalization, and radial formula

Statement

Assume Countable Choice, and fix d≥1, ν>0, a>0, s=ν+d/2, and A,w,k as in The normalized Matérn kernel in the 2π Fourier convention. Then w>0, w∈L1(Rd), and ∫Rdw(ξ) dξ=1. The integral defining k(z) is absolutely convergent for every z; k is real, even, radial, and continuous, and k(0)=1. For every z∈Rd, k(z)=a2νΓ(ν)∫0∞tν−1exp⁡ ⁣(−a2t−∣z∣24t)dt. For z≠0 this is equivalently k(z)=21−νΓ(ν)(a∣z∣)νKν(a∣z∣), where Kν has the integral normalization in The normalized Matérn kernel in the 2π Fourier convention. Under the stated Fourier convention, the d-dimensional Gaussian identity is, for t>0, ∫Rde−4π2t∣ξ∣2e2πiz⋅ξ dξ=(4πt)−d/2e−∣z∣2/(4t).

Facts & Assumptions

Given: Countable Choice, the normalized Matérn parameters and density in The normalized Matérn kernel in the 2π Fourier convention, and the real Euler integral for Γ.

[F1]

For q>0, Γ(q)=∫0∞tq−1e−t dt; it converges and is positive. (The real Gamma function by Euler's integral, Euler's Gamma integral converges exactly for positive real parameters)

[F2]

∫−∞∞e−x2dx=π. (The Gaussian integral ∫−∞∞e−x2 dx=π)

[F3]

Tonelli/Fubini holds for nonnegative or absolutely integrable functions on completed Euclidean product measures under Countable Choice, and λm×λn agrees with λm+n on Borel sets. (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n})

[F4]

Polar coordinates express a nonnegative radial Lebesgue integral as its radial integral against rd−1dr and finite sphere measure. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

[F6]

Pointwise convergence under an integrable majorant permits passage of limits through Lebesgue integrals. (Dominated convergence)

[F7]

For b>0, Kν(b)=12(b/2)ν∫0∞u−ν−1e−u−b2/(4u)du. (The normalized Matérn kernel in the 2π Fourier convention)

[F8]

Under Countable Choice, complex C1 factors obey Lebesgue integration by parts on finite intervals; on R the boundary-free identity holds if both derivative products are in L1 and the product tends to zero at both ends. (Complex integration by parts on intervals and decaying lines)

[A1]

Countable Choice is the precise choice principle used by [F3] and [F4]; no full Axiom of Choice is used in this spectral calculation. (The Axiom of Countable Choice (ACω))

Proof

technique · direct

Given: The hypotheses of the statement. Write λd for Lebesgue measure and use the negative-sign 2π Fourier convention from The normalized Matérn kernel in the 2π Fourier convention.

1.1F1F4A1algebra

Since 2s>d, polar coordinates [F4] and comparison with rd−1−2s for r≥1 show ∫Rd(a2+4π2∣ξ∣2)−sdξ<∞; the integrand is bounded near 0. Since A>0 by [F1] and the positive parameters, w>0 and w∈L1.

1.2F2F5F6F8A1algebra

For one dimension let It(z)=∫Re−4π2tξ2e2πizξdξ. Differentiation under the integral is justified by [F6], since 2π∣ξ∣e−4π2tξ2 is integrable. Apply [F8] with u(ξ)=e−4π2tξ2 and v(ξ)=e2πizξ: both are C1, u′v and uv′ are integrable Gaussian multiples, and uv→0 at both ends. Thus −8π2t∫ξuv=−2πizIt(z), which gives It′(z)=−zIt(z)/(2t). The initial value from [F2] and scaling is It(0)=(4πt)−1/2. Set J(z)=It(z)ez2/(4t); the differential equation gives J′(z)=0. The real and imaginary parts of J are constant by the mean value theorem [F5], so J(z)=J(0)=It(0) and It(z)=(4πt)−1/2e−z2/(4t).

2.1F1F2F3A1step 1.1algebra

For B>0, [F1] and the substitution u=Bt give B−s=Γ(s)−1∫0∞ts−1e−Btdt. Apply this with B=a2+4π2∣ξ∣2 and use Tonelli [F3]. The product Gaussian integral in [F2] is (4πt)−d/2 by scaling and the Borel product identity in [F3]. A second substitution u=a2t gives ∫w=AΓ(s)(4π)−d/2∫0∞tν−1e−a2tdt=AΓ(ν)Γ(s)(4π)d/2a2ν=1.

2.2F3step 1.2

The integrand in d dimensions is a product of the one-dimensional integrands in step 1.2. Absolute integrability and [F3] therefore give the stated d-dimensional Gaussian Fourier identity.

3.1F1F3A1step 1.1step 2.1step 2.2algebra

Absolute Fubini is valid in the gamma representation of w because its absolute double integral is the finite normalization integral in steps 1.1 and 2.1. Substitute that representation into k(z) and apply step 2.2; the result is k(z)=AΓ(s)(4π)−d/2∫0∞ts−d/2−1e−a2t−∣z∣2/(4t)dt=a2νΓ(ν)∫0∞tν−1e−a2t−∣z∣2/(4t)dt. The same absolute bound proves convergence for every z.

4.1F6step 2.1step 3.1algebra

From w∈L1, dominated convergence [F6] makes k continuous. The integral formula in step 3.1 is real and depends on z only through ∣z∣, so k is real, even, and radial. Step 2.1 gives k(0)=1.

5.1step 3.1F7algebra∎

In step 3.1 substitute u=a2t. For b=a∣z∣>0, then substitute v=b2/(4u) in the resulting integral; [F7] gives ∫0∞uν−1e−u−b2/(4u)du=21−νbνKν(b). Thus the displayed Bessel expression follows for z≠0.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Exact augmented RBF-FD weights for a local differential functional

Definition

The exact augmented RBF finite-difference weight procedure has the following inputs: a dimension d≥1; an integer polynomial degree r≥0; a finite ordered stencil Y=(y1,…,yN)⊂Rd of distinct sites that is (r+1)-unisolvent for Πr(Rd); a real kernel K strictly conditionally positive definite of order r+1; an ordered basis p1,…,pJ of Πr(Rd); a center x∈Rd; a homogeneous constant-coefficient differential functional L=∑∣α∣=tcαDα with 0≤t≤r; and a neighborhood of x on which every section z↦K(z,yi) has continuous derivatives through order t. Here α∈N0d, ∣α∣=∑jαj, and Dα is the corresponding partial derivative. All arithmetic in this procedure is exact real arithmetic.

The input is supplied with total, correct oracles for admissibility and exact evaluation. The admissibility oracle returns whether all the stated input conditions hold, including the polynomial-basis condition, global strict conditional positive definiteness, and continuous section differentiability on the supplied neighborhood. On an affirmative answer, the evaluation oracles return the exact real values K(yi,yj), pj(yi), (LzK(z,yi))∣z=x, and (Lpj)(x) for the finitely many requested indices. Oracle calls are primitive operations in this model; neither global kernel admissibility nor differentiation is claimed to be decidable by real arithmetic alone.

Query the admissibility oracle. If it returns false, return inadmissible-input. Otherwise use the evaluation oracles to form Pij=pj(yi), bi=(LzK(z,yi))∣z=x, ℓj=(Lpj)(x), and M=[KYPPT0],g=[bell]. Solve Mz=g by exact Gaussian elimination with deterministic nonzero-pivot row selection: set A=M, c=g, and n=N+J. For k=1,…,n, choose the least i≥k with Aik≠0; if none exists, return singular-system. Swap rows i and k in both A and c. For each j=k+1,…,n in increasing order, set μ=Ajk/Akk, replace row j by row j−μ rowk in A and c, and set Ajk=0. After all pivots, set zn=cn/Ann; then, for k=n−1,n−2,…,1 in decreasing order, set zk=(ck−∑j=k+1nAkjzj)/Akk. Return success with the first N coordinates w of z and the remaining J polynomial multipliers. The stated strict hypotheses imply that singular-system is unreachable, as proved by Strict positive and conditionally positive augmented interpolation systems are invertible.

The returned w is an exact mathematical weight vector. The nonsingularity lemma Strict positive and conditionally positive augmented interpolation systems are invertible makes singular-system unreachable for every input that passes the stated strict hypotheses. This definition makes no floating-point stability or solver-error claim.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

The weighted Fourier model is an injective Hilbert representation

Statement

Assume the Axiom of Choice as a conservative cover of the published L2 completeness dependency, which declares it in its transitive dependencies although its statement requires only Countable Choice. For every s>d/2, each g∈Ls2(Rd) lies in L1(Rd), its inverse Fourier representative Ug is bounded and continuous, the map g↦Ug is injective, and the whole-space Fourier Bessel-potential space of Fourier Bessel-potential and restriction-quotient Sobolev spaces is a complex Hilbert space. In particular, for a constant Cd,s<∞, ∣Ug(x)∣≤Cd,s∥g∥Ls2(x∈Rd).

Facts & Assumptions

Given: The Axiom of Choice, an integer d≥1, a real s>d/2, and g∈Ls2(Rd).

[F1]

The weighted Fourier norm is ∥g∥Ls22=∫(1+∣ξ∣2)s∣g(ξ)∣2dξ, and Ug(x)=∫g(ξ)e2πix⋅ξdξ. (Fourier Bessel-potential and restriction-quotient Sobolev spaces)

[F2]

Cauchy--Schwarz holds in L2(λd). (Cauchy-Schwarz inequality for L2)

[F3]

The integral ∫Rd(1+∣ξ∣2)−sdξ is finite when s>d/2, by polar coordinates. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

[F4]

The Matérn calculation supplies the d-dimensional Gaussian Fourier transform and its normalized Gaussian Gϵ(y)=ϵ−d/2e−π∣y∣2/ϵ of integral one. (The Matérn spectral integral, normalization, and radial formula)

[F5]

Fubini/Tonelli applies to the absolute integrals on completed product measure spaces. (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability)

[F7]

Increasing nonnegative integrals converge to the integral of their pointwise limit, and dominated convergence applies under an integrable majorant. (Monotone convergence for the integral, Dominated convergence)

[F8]

L2 with the integral pairing is a Hilbert space under Countable Choice. Its declared transitive dependencies include the Axiom of Choice. (L2 with the integral pairing is a Hilbert space, The Axiom of Countable Choice (ACω), The Axiom of Choice)

[F9]

A nonnegative measurable function has integral zero exactly when it vanishes almost everywhere. (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere)

[A1]

Full AC is carried conservatively to match the legacy transitive dependency of the published L2 Hilbert-space item; the Fourier injectivity argument itself is choice-free. (The Axiom of Choice)

Proof technique: direct.

Proof

technique · direct

Given: The hypotheses in the Statement and the Fourier convention in [F1].

1.1F1F2F3F7

Write Ws(ξ)=(1+∣ξ∣2)s/2. By [F2] and [F3], ∫∣g(ξ)∣dξ≤(∫Ws(ξ)2∣g(ξ)∣2dξ)1/2(∫Ws(ξ)−2dξ)1/2<∞. Hence Ug is bounded by ∥g∥1 and continuous by [F7] applied to the pointwise-continuous exponential integrand dominated by ∣g∣. The same estimate gives the displayed evaluation bound with Cd,s=∥Ws−1∥2.

1.2F4F7algebra

The kernels Gϵ are nonnegative and have integral one by [F4]. For each fixed δ>0, scaling y=ϵz and [F7] show ∫∣y∣≥δGϵ(y)dy=∫∣z∣≥δ/ϵe−π∣z∣2dz⟶0(ϵ↓0).

2.1F4F5step 1.1

Suppose Ug(x)=0 for every x. For ϵ>0 and η∈Rd, the double integral of ∣g(ξ)∣e−πϵ∣x∣2 is finite by step 1.1 and [F4]. Thus [F5] and [F4] give 0=∫Ug(x)e−πϵ∣x∣2e−2πix⋅ηdx=∫g(ξ)ϵ−d/2e−π∣ξ−η∣2/ϵdξ=(g∗Gϵ)(η).

2.2F5F6step 1.2algebra

Given τ>0, translation continuity [F6] supplies δ>0 such that ∥g(⋅−y)−g∥1<τ when ∣y∣<δ. Tonelli and the triangle inequality then give ∥g∗Gϵ−g∥1≤∫Gϵ(y)∥g(⋅−y)−g∥1dy≤τ+2∥g∥1∫∣y∣≥δGϵ(y)dy. Step 1.2 makes the last term tend to zero; as τ was arbitrary, g∗Gϵ→g in L1.

3.1F9step 2.1step 2.2

Step 2.1 says g∗Gϵ=0 for every ϵ, whereas step 2.2 says these convolutions converge to g in L1. Hence ∥g∥1=0, so ∫∣g∣=0 and [F9] gives g=0 almost everywhere. This proves injectivity.

4.1F1F8A1step 1.1step 3.1algebra∎

Multiplication g↦Wsg is a linear isometry of the weighted Ls2 space onto ordinary complex L2(λd), with inverse h↦Ws−1h. By [F8] the latter is Hilbert under the declared choice assumption. Step 3.1 makes the inverse Fourier image isometric and one-to-one, so the whole-space Hs is Hilbert; step 1.1 makes its restriction quotient norm positive on nonzero functions.

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Separated Matérn sites have a positive Gram lower bound

Statement

Assume Countable Choice. Fix d≥1, ν>0, a>0, s=ν+d/2, and the normalized Matérn kernel with spectral density w(ξ)=A(a2+4π2∣ξ∣2)−s from The normalized Matérn kernel in the 2π Fourier convention. For a list of N≥2 distinct sites X=(x1,…,xN)⊂Rd, put q=12min⁡i≠j∣xi−xj∣. There is c=c(d,ν,a)>0 such that if 0<q≤1, then λmin⁡(KX)≥cq2ν,λmax⁡(KX)≤N,κ2(KX)≤c−1Nq−2ν. The eigenvalues and condition number are those of the exact real symmetric positive-definite matrix. For N=1, KX=[1] and κ2(KX)=1.

Facts & Assumptions

Given: Countable Choice, fixed Matérn parameters, N≥2 distinct sites, and 0<q≤1.

[F1]

The spectral density is positive, w(ξ)=A(a2+4π2∣ξ∣2)−s, and k(z)=∫w(ξ)e2πiz⋅ξdξ. (The normalized Matérn kernel in the 2π Fourier convention, The Matérn spectral integral, normalization, and radial formula)

[F2]

The Matérn kernel is real and even, with k(0)=1, so its Gram matrix is real symmetric and every diagonal entry is one. (The Matérn spectral integral, normalization, and radial formula)

[F5]

A smooth bump can be chosen equal to one on the half-radius ball and supported in the unit ball. (A smooth bump between concentric Euclidean balls, Euclidean balls have positive finite Lebesgue measure)

[F6]

The induced matrix 2-norm of a symmetric matrix is its largest absolute eigenvalue. (A symmetric matrix operator norm is its largest absolute Rayleigh quotient, The matrix norm induced by a published vector p-norm)

[F7]

The nonnegative integral is monotone and the modulus of an integral is bounded by the integral of the modulus. (Monotonicity and nonnegative homogeneity of the nonnegative integral, The modulus of an integral is bounded by the integral of the modulus)

[F8]

Smooth compactly supported functions are Schwartz, and the negative-sign, 2π-normalized Fourier transform maps Schwartz functions to Schwartz functions. (Schwartz space and its seminorms, Fourier transform acts continuously on Schwartz space)

[F9]

Euler's formula gives ∣eit∣=1 for real t and eit‾=e−it. (exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0)

[F11]

The positive p-series ∑n≥1n−p converges for p>1. (The p-series for a real exponent p converges exactly when p is greater than one)

[A1]

Countable Choice is inherited from the spectral normalization, ball and Lebesgue measure, and Schwartz Fourier-transform suppliers; no choice of sites or window functions is made from an arbitrary family. (The Axiom of Countable Choice (ACω))

Proof technique: localized spectral window, packing, and Rayleigh quotients.

Proof

technique · localized spectral window, packing, and Rayleigh quotients

Given: The hypotheses above.

1.1F3F5F7F9A1algebrachoose

By [F5] choose a smooth function ρ:Rd→[0,1] equal to one on B‾(0,1/2) and supported in B(0,1). Set η(ξ)=(ρ(ξ)+ρ(−ξ))/2. Then η is smooth, even, nonnegative, at most one, still equals one on the half-radius ball, and is supported in the unit ball. Define ψ(z)=∫η(ξ)e2πiz⋅ξdξ. Reflection invariance of Lebesgue measure [F3] and Euler's formula [F9] make ψ real and even, and monotonicity [F7] gives ψ(0)=∫η≥λ(B(0,1/2))>0.

2.1F7F8F9step 1.1algebra

Since η is smooth and compactly supported, it is Schwartz by [F8], so η^ is Schwartz and ψ(z)=η^(−z) is Schwartz as well. For ∣z∣≥1, choose a coordinate j with ∣zj∣≥∣z∣/d; the Schwartz seminorm sup⁡ξ∣ξjMη^(ξ)∣<∞ gives ∣ψ(z)∣≤CM∣z∣−M. For ∣z∣≤1, [F7] and ∣e2πiz⋅ξ∣=1 from [F9] give ∣ψ(z)∣≤∫η. Combining the two bounds yields ∣ψ(z)∣≤CM′(1+∣z∣)−M.

3.1F3F11A1step 2.1givenchoose

Fix i. The open balls B(xj,q) are pairwise disjoint because center distances are at least 2q. If n≤∣xi−xj∣/q<n+1, then B(xj,q)⊂B(xi,(n+2)q). Translation invariance and dilation change of variables in [F3] give λ(B(x,r))=rdλ(B(0,1)); this unit-ball measure is finite and positive. Finite additivity on the disjoint balls therefore bounds the number of such j by (n+2)d. Hence, for any integer M>d+1 and T≥1, [step 2.1] yields the uniform row-sum estimate ∑j≠i∣ψ ⁣(T(xi−xj)/q)∣≤CMT−M∑n≥1(n+2)dn−M. For n≥1, (n+2)d≤3dnd, so this series is bounded by 3d∑n≥1n−(M−d), which converges by [F11] because M−d>1. Choose one fixed T large enough that this upper bound is at most ψ(0)/2.

4.1F3step 3.1algebra

For any c∈CN, the quadratic form of Hij=ψ(T(xi−xj)/q) satisfies c∗Hc≥(ψ(0)−max⁡i∑j≠i∣Hij∣)∥c∥22≥ψ(0)2∥c∥22, because each off-diagonal pair is bounded using 2∣ci∣∣cj∣≤∣ci∣2+∣cj∣2. By the change of variables ζ=(q/T)ξ, ∫∣ξ∣≤T/q∣∑jcje−2πixj⋅ξ∣2dξ≥∫η(qξ/T)∣∑jcje−2πixj⋅ξ∣2dξ=(T/q)dc∗Hc. The first inequality uses 0≤η≤1 and support in the unit ball.

5.1F1F4F10step 4.1algebra

On ∣ξ∣≤T/q and q≤1, set B=(a2+4π2T2)q−2. Then a2+4π2∣ξ∣2≤a2+4π2T2/q2≤B. Monotonicity and the power laws [F10] give w(ξ)≥AB−s=A(a2+4π2T2)−sq2s. Using the Fourier quadratic-form identity from [F1] and step 4.1 gives c∗KXc≥c0q2s−d∥c∥22=c0q2ν∥c∥22 for a constant c0>0 independent of X,q,N. Thus [F4] yields λmin⁡(KX)≥c0q2ν.

6.1F2F4F6step 5.1algebra∎

By [F4], choose an orthonormal eigenbasis and write KX=QΛQT. If Q=(qij), then tr⁡(KX)=∑i(KX)ii=∑i,jqij2λj=∑jλj=N, since each column of Q has unit norm and [F2] gives diagonal entries k(0)=1. By step 5.1 all eigenvalues are positive, so λmax⁡(KX)≤N. In this eigenbasis the induced 2-norms of KX and KX−1 are λmax⁡ and 1/λmin⁡, respectively, by [F6]. The definition κ2(KX)=∥KX∥2∥KX−1∥2 therefore gives κ2(KX)≤c−1Nq−2ν. The N=1 case follows directly from k(0)=1 and the same definition.

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Augmented RBF-FD weights reproduce polynomials and obey a local Taylor bound

Statement

Let p≥0 be an integer, let Y=(y1,…,yN) be distinct and (p+1)-unisolvent for Πp(Rd), and let K be a real strictly conditionally positive-definite kernel of order p+1. Fix a center x, an ordered basis p1,…,pJ of Πp, and a homogeneous constant-coefficient differential functional L of order t≤p for which all section derivatives in Exact augmented RBF-FD weights for a local differential functional exist. Let (w,v) be the exact output of that procedure and put rY=max⁡i∣yi−x∣. Then the procedure returns success, and for every q∈Πp, ∑i=1Nwiq(yi)=Lq(x). If f∈Cp+1(U) on an open convex set U containing the compact convex hull H=conv⁡{x,y1,…,yN}, then ∣∑iwif(yi)−Lf(x)∣≤Cd,p(∑i∣wi∣)rYp+1max⁡∣α∣=p+1sup⁡z∈H∣Dαf(z)∣. Consequently, when rY>0, a separately proved bound ∑i∣wi∣≤CwrY−t yields the order Cd,pCwrYp+1−t truncation bound. Unisolvence alone is not asserted to provide this weight-norm bound.

Facts & Assumptions

Given: The sites, kernel, polynomial basis, center, functional, and function hypotheses in the Statement.

[F1]

The exact algorithm forms the augmented matrix with blocks KY,P,PT,0, right side (b,ℓ), and applies deterministic Gaussian elimination with row pivoting; the lower block of any solution is PTw=ℓ. (Exact augmented RBF-FD weights for a local differential functional)

[F2]

Under strict conditional positive definiteness of order p+1 and (p+1)-unisolvence, the augmented matrix is invertible for every right-hand side. (Strict positive and conditionally positive augmented interpolation systems are invertible, Radial kernels, polynomial augmentation, and site geometry)

[F3]

Taylor's theorem gives a degree-p polynomial Tpf at x whose value remainder at y is bounded by Cd,p∣y−x∣p+1 times the supremum of all order-(p+1) partial derivatives on the segment. (Multivariable Taylor formula with a Lagrange remainder along a line segment)

Proof

technique · exact elimination, moment equations, and Taylor cancellation

Given: The hypotheses above. All linear algebra and arithmetic in the algorithm are exact.

1.1F1F2inductionalgebra

Let M be the augmented matrix and g its right side in [F1]. It is square and invertible by [F2]. Initially the leading 0×0 pivot block is invertible. Inductively suppose the first k−1 pivots have been nonzero. After their row swaps and elimination, the row-equivalent matrix has block form [UB0S] with U invertible and S the remaining Schur complement. The block matrix is invertible, so S is invertible: if Sz=0, then (−U−1Bz,z) would be a nonzero vector in its kernel. If the first column of S were zero, its first coordinate vector would be a nonzero kernel vector. Thus the finite set of remaining entries in column k contains a nonzero entry, and the algorithm's least-index rule finds a nonzero pivot. Row swaps and adding multiples of pivot rows are invertible elementary row operations, so they preserve the solution set. After N+J pivots the matrix is upper triangular with nonzero diagonal, and descending substitution returns the unique solution. Therefore the singular-system status is unreachable under the hypotheses.

2.1F1step 1.1algebra

Writing the returned vector as (w,v), its lower block equation is PTw=ℓ. For each basis polynomial pj, this says ∑iwipj(yi)=Lpj(x). Every q∈Πp is a linear combination q=∑jajpj, so multiplying these identities by aj and summing, using linearity of L, gives ∑iwiq(yi)=Lq(x).

3.1F3step 2.1algebra

Let Tpf be the Taylor polynomial of f at x of total degree at most p. Since t≤p, its derivatives at x through order t agree with those of f, so LTpf(x)=Lf(x). Applying step 2.1 to Tpf gives ∑iwiTpf(yi)=Lf(x).

4.1F3step 3.1algebra∎

Subtract the identity in step 3.1 from the sampled sum for f. For every i, [F3] bounds ∣f(yi)−Tpf(yi)∣ by Cd,p∣yi−x∣p+1 times the derivative supremum on H, since each segment lies in the convex hull H. The triangle inequality and ∣yi−x∣≤rY now give the stated truncation estimate. When rY>0, the additional weight bound yields the final conditional rate.

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Schwartz functions are dense in the Fourier Bessel-potential norm

Statement

Assume the Axiom of Choice inherited from the declared L2 dependency. For every s>d/2, every Schwartz function f belongs to Hs(Rd), and the inverse Fourier transforms of Cc∞(Rd) are Schwartz functions and are dense in Hs(Rd) of Fourier Bessel-potential and restriction-quotient Sobolev spaces. The Fourier transform uses the negative-sign 2π convention.

Facts & Assumptions

Given: The Axiom of Choice, s>d/2, and g∈Ls2(Rd).

[F1]

Hs is defined by the weighted Fourier norm and inverse Fourier representative. (Fourier Bessel-potential and restriction-quotient Sobolev spaces)

[F2]

Smooth compactly supported functions are dense in L2(O) for every open O⊂Rd. (Smooth compactly supported functions of an open set are dense in L2)

[F3]

Truncating an integrable nonnegative function to increasing balls makes the integrals increase to its full integral. (Monotone convergence for the integral)

[F4]

The inverse Fourier map is injective and the weighted representation is a Hilbert space under the declared choice assumption. (The weighted Fourier model is an injective Hilbert representation)

[F5]

A Schwartz function and all its derivatives decay faster than every polynomial, with seminorms sup⁡x∣xβDαf(x)∣ finite. (Schwartz space and its seminorms)

[F6]

The Matérn calculation gives the d-dimensional Gaussian Fourier transform and the complex exponential derivative; Tonelli/Fubini, polar integration, monotone convergence and dominated convergence hold under their stated hypotheses. (The Matérn spectral integral, normalization, and radial formula, The complex exponential is entire and its complex derivative is itself, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, Monotone convergence for the integral, Dominated convergence)

[F7]

Under Countable Choice, complex C1 factors obey Lebesgue integration by parts on finite intervals and on R when the derivative products are integrable and the product tends to zero at both ends. (Complex integration by parts on intervals and decaying lines)

[A1]

Full AC conservatively covers the inherited L2 Hilbert-space dependency. It also supplies the Countable Choice required by the smooth-density supplier [F2], the spectral Gaussian and integration suppliers [F6], and integration by parts [F7]. (The Axiom of Choice, The Axiom of Countable Choice (ACω))

Proof technique: direct.

Proof

technique · direct

Given: The hypotheses above and the Fourier convention in [F1].

1.1F1F3choose

Put gR=g1B(0,R). By [F3], ∥g−gR∥Ls22=∫∣ξ∣>R(1+∣ξ∣2)s∣g(ξ)∣2dξ⟶0(R→∞). Choose R so this norm is less than a prescribed ε/2.

1.2F5F6F7algebra

Let f be any Schwartz function and define f^(ξ)=∫f(x)e−2πix⋅ξdx. Polynomial multiples of derivatives of f have arbitrarily rapid decay by [F5], so they and their derivatives are integrable, including on each coordinate line with the other coordinates fixed. Differentiation in ξ is justified by dominated convergence [F6] using these integrable moments. Apply [F7] on each coordinate line to such a function and its Fourier exponential: the derivative products are integrable and the boundary product tends to zero. Absolute Fubini [F6] then combines the coordinate identities and shows that every ξγDξαf^(ξ) is a finite linear combination of Fourier transforms of derivatives of polynomial multiples of f. Each such function is in L1 by [F5], so its Fourier transform is bounded; hence f^ is Schwartz and in L1. In particular ∫(1+∣ξ∣2)s∣f^(ξ)∣2dξ<∞, by polar integration [F6].

2.1F1F2step 1.1construct

On the bounded ball O=B(0,R+1), the weight (1+∣ξ∣2)s is bounded above by a finite constant CR. The restriction of gR to O lies in L2(O), so [F2] gives ϕ∈Cc∞(O) with ∥gR−ϕ∥L2(O)<ε/(2CR). Extending ϕ by zero is still smooth and compactly supported on Rd because its support is compactly contained in O, and ∥gR−ϕ∥Ls22≤CR∥gR−ϕ∥L2(O)2<ε2/4.

2.2F1F6step 1.2algebra

For ϵ>0, absolute Fubini [F6] and the Gaussian transform [F6] give ∫f^(ξ)e2πix⋅ξe−πϵ∣ξ∣2dξ=∫f(y)ϵ−d/2e−π∣x−y∣2/ϵdy=(f∗Gϵ)(x), where Gϵ(z)=ϵ−d/2e−π∣z∣2/ϵ. Since f^∈L1, dominated convergence sends the left side to ∫f^(ξ)e2πix⋅ξdξ. The right side tends to f(x): given ε>0, continuity at x gives δ>0 such that ∣f(x−z)−f(x)∣<ε for ∣z∣<δ. Since Gϵ has total mass one, the contribution from this ball is at most ε; boundedness of f and the Gaussian tail estimate from scaling make the contribution on ∣z∣≥δ tend to zero. Thus Fourier inversion holds and f=Uf^∈Hs.

3.1F1F6F7step 2.1algebra

For any multi-indices α,β, differentiation under the integral is justified by the compactly supported integrable moments. For each coordinate, apply [F7] with the smooth compactly supported factor (2πiξ)αϕ(ξ) (or a derivative of it) and the inverse Fourier exponential; both derivative products are integrable and the boundary product is zero outside a bounded interval. Repetition and absolute Fubini [F6] give xβDxαUϕ(x)=(−1)∣β∣(2πi)−∣β∣∫∂ξβ((2πiξ)αϕ(ξ))e2πix⋅ξdξ. The right side is uniformly bounded by the L1 norm of a smooth compactly supported function. Hence every Schwartz seminorm of Uϕ is finite, so Uϕ is Schwartz.

4.1F1F4A1step 1.1step 2.1step 3.1step 1.2step 2.2∎

Steps 1.1 and 2.1 show that Cc∞ frequency functions are dense in the weighted Fourier norm. Step 3.1 shows their inverse transforms are Schwartz. Steps 1.2 and 2.2 show every Schwartz function belongs to Hs, and [F4] identifies the normed Fourier images injectively with the functions in Hs. Thus Schwartz functions are dense in Hs(Rd).

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A Sobolev function vanishing on scattered sites has an h^ν bound

Statement

Assume full AC conservatively for the declared weighted-L2 Hilbert dependency. Let Ω be an admitted domain with its uniform cone constants, fix d≥1, ν>0, s=ν+d/2, and choose an integer m>ν. There exist C,h0>0, depending only on d,s,m and the cone constants, such that for every finite nonempty X⊂Ω of fill distance h=hX≤h0 and every u∈Hs(Ω) with u∣X=0, ∥u∥L∞(Ω)≤Chν∥u∥Hs(Ω). The estimate has no separation or mesh-ratio hypothesis.

Facts & Assumptions

Given: The hypotheses above, a site set X, and u∈Hs(Ω) vanishing on X.

[F1]

Every u∈Hs(Ω) is the restriction of a whole-space inverse Fourier representative Ug with weighted norm; point evaluations are well defined and continuous. (Fourier Bessel-potential and restriction-quotient Sobolev spaces, The weighted Fourier model is an injective Hilbert representation)

[F2]

For every x∈Ω‾ there are distinct nodes yj∈X and weights with ∣yj−x∣≤Rh, ∑j∣λj∣≤B, and p(x)=∑jλjp(yj) for all polynomials of degree below m, once h≤h0. (Uniform cone domains admit stable polynomial norming stencils)

[F3]

Polar coordinates evaluate nonnegative radial integrals; multivariable Taylor's theorem bounds the degree-(m−1) remainder by a dimension- and-order-dependent constant times ∣y−x∣m times the supremum of order-m derivatives along the segment. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, Multivariable Taylor formula with a Lagrange remainder along a line segment)

[F4]

Differentiation under the bounded-frequency inverse Fourier integral is justified by dominated convergence, using the derivative formula for the complex exponential. (Dominated convergence, The complex exponential is entire and its complex derivative is itself)

[F5]

Cauchy--Schwarz holds for the L2 integral pairing. (Cauchy-Schwarz inequality for L2)

[F6]

The reals are an ordered field, so for d≥1 every nonempty open ball in Rd contains infinitely many points along a coordinate line. (The reals form a totally ordered field)

[A1]

Full AC covers the conservative weighted-L2 completeness/injectivity dependency used to define the Fourier representatives. The stencil proof makes only a finite choice for each fixed x, with uniform constants before taking suprema. (The Axiom of Choice, The Axiom of Countable Choice (ACω))

Proof technique: direct Fourier split and polynomial cancellation.

Proof

technique · direct Fourier split and polynomial cancellation

Given: The hypotheses above. Decrease h0 if necessary so h0≤1.

1.1F1F3F5F6A1algebra

Since Ω is a nonempty open domain, it contains a ball B(z,r). For d≥1 this ball contains the infinitely many distinct points z+r2ne1, n≥1, by [F6]. The finite set X omits one such point x, so min⁡y∈X∣x−y∣>0 and therefore hX≥min⁡y∈X∣x−y∣>0. In particular h−1 is defined. Fix any whole-space extension U=Ug of u, and split g=glo+ghi with glo=g1{∣ξ∣≤h−1}. Let V=Uglo and W=Ughi. By Cauchy--Schwarz [F5], polar integration [F3], and s=ν+d/2, ∥W∥∞≤(∫∣ξ∣>h−1(1+∣ξ∣2)−sdξ)1/2∥U∥Hs≤Chν∥U∥Hs. Indeed, for r≥h−1≥1 the radial tail is at most C∫h−1∞r−2ν−1dr=C′h2ν.

2.1F1F3F4F5step 1.1algebra

For each multi-index α with ∣α∣=m, the bounded frequency support makes DαV well defined and continuous by differentiation under an absolutely integrable Fourier integral. Cauchy--Schwarz [F5] and polar coordinates give ∥DαV∥∞≤C(∫∣ξ∣≤h−1∣ξ∣2m(1+∣ξ∣2)−sdξ)1/2∥U∥Hs≤Chν−m∥U∥Hs. For the last estimate, the radial integral on r≤1 is finite and on 1≤r≤h−1 is bounded by C∫1h−1r2m−2ν−1dr≤C′h−2(m−ν), using m>ν.

3.1F2F3step 2.1choosealgebra

Fix x∈Ω and choose the stencil from [F2]. Taylor-expand the globally defined smooth V at x to degree m−1. The segment from x to each yj lies in Rd, where V is smooth, so [F3] bounds each remainder by C∣yj−x∣mmax⁡∣α∣=m∥DαV∥∞. The polynomial part cancels under the stencil's reproduction identities. Therefore ∣V(x)−∑jλjV(yj)∣≤CB(Rh)mhν−m∥U∥Hs≤C′hν∥U∥Hs.

4.1F1F2step 1.1step 3.1algebra∎

Since U(yj)=u(yj)=0, we have V(yj)=−W(yj). Thus U(x)=(V(x)−∑jλjV(yj))+W(x)−∑jλjW(yj). Use step 3.1, step 1.1 and ∑j∣λj∣≤B to obtain ∣u(x)∣=∣U(x)∣≤C′′hν∥U∥Hs. The constants do not depend on x or the chosen extension. Taking the supremum over x∈Ω, then the infimum over all extensions U, gives the claimed quotient-norm estimate.

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Bounded mesh ratio gives an h^{-2s} exact condition-number bound

Statement

Assume Countable Choice. Fix Matérn parameters d≥1, ν>0, a>0, s=ν+d/2, an admitted bounded domain Ω, and a mesh-ratio bound ρ0>0. There is C=C(Ω,d,ν,a,ρ0) such that every finite nonempty distinct site set X⊂Ω with N≥2, 0<qX≤1, 0<hX≤1, and ρX=hX/qX≤ρ0 satisfies κ2(KX)≤ChX−2s. The one-site case has condition number 1. This is an exact-arithmetic condition-number estimate for the Gram matrix; it does not bound the error of a floating-point solver.

Facts & Assumptions

Given: Countable Choice, fixed parameters, a bounded domain, a finite distinct site set with hX≤1, and the stated mesh-ratio bound.

[F1]

For N≥2 and separation radius 0<qX≤1, κ2(KX)≤C0NqX−2ν; for N=1 the condition number is 1. (Separated Matérn sites have a positive Gram lower bound)

[F3]

The mesh ratio and separation radius satisfy ρX=hX/qX, and the condition number is finite for each invertible Gram matrix. (Radial kernels, polynomial augmentation, and site geometry, The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system)

[A1]

Countable Choice is inherited from the Matérn separation estimate [F1] and the Lebesgue measure and ball-packing suppliers used in [F2]. (The Axiom of Countable Choice (ACω))

Proof technique: packing followed by the separation estimate.

Proof

technique · packing followed by the separation estimate

Given: The hypotheses above.

1.1F2A1givenalgebra

Suppose N≥2 and write q=qX≤1. Choose z0 and R>0 with Ω⊂B(z0,R). The site balls B(xi,q) are pairwise disjoint and lie in B(z0,R+1). Their measures are qdλ(B(0,1)) by [F2], so finite additivity and the positive finite unit-ball measure give Nqdλ(B(0,1))≤(R+1)dλ(B(0,1)), and therefore N≤CΩq−d.

2.1F1step 1.1algebra

Since q≤1, combine [F1] with step 1.1 to get κ2(KX)≤C1q−(d+2ν)=C1q−2s.

3.1F1F3step 2.1algebra∎

The mesh-ratio bound implies q=hX/ρX≥hX/ρ0, so q−2s≤ρ02shX−2s. Step 2.1 gives the desired estimate.

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The Matérn native space has its weighted Fourier norm and strict Gram positivity

Statement

Assume the Axiom of Choice conservatively for the declared L2 Hilbert-space dependency. Fix d≥1, ν>0, a>0, and s=ν+d/2, and let k,w be the normalized Matérn kernel and positive spectral density of The normalized Matérn kernel in the 2π Fourier convention. Let Hk(Rd) consist of the inverse Fourier representatives Ug with ∫∣g(ξ)∣2/w(ξ) dξ<∞, with inner product ⟨Ug,Uh⟩k=∫Rdg(ξ)h(ξ)‾w(ξ)dξ. Then this is a Hilbert space, each kernel section k(⋅−y) reproduces evaluation at y, and its span is dense. There are constants 0<c≤C<∞, depending only on d,ν,a, such that c∥U∥Hs(Rd)≤∥U∥k≤C∥U∥Hs(Rd). For every finite list of distinct sites X=(x1,…,xN), the Matérn Gram matrix KX=(k(xi−xj)) is strictly positive definite.

Facts & Assumptions

Given: The Axiom of Choice, fixed d,ν,a, the definitions of k,w,Hs, and distinct sites X.

[F1]

w(ξ)=A(a2+4π2∣ξ∣2)−s is positive and integrable with integral one; the inverse Fourier representation of Hs is injective. (The normalized Matérn kernel in the 2π Fourier convention, The Matérn spectral integral, normalization, and radial formula, The weighted Fourier model is an injective Hilbert representation)

[F2]

The Hs Fourier norm is ∫(1+∣ξ∣2)s∣g(ξ)∣2dξ, and the inverse Fourier representative is injective. (Fourier Bessel-potential and restriction-quotient Sobolev spaces, The weighted Fourier model is an injective Hilbert representation)

[F3]

A closed subspace of a Hilbert space has an orthogonal complement, and a subspace whose orthogonal complement is zero is dense. (Orthogonal decomposition by a closed subspace)

[F4]

A nonzero univariate polynomial has at most its degree many roots. The real numbers form an ordered field, hence are infinite. (A nonzero polynomial of degree n over an integral domain has at most n distinct roots, The reals form a totally ordered field, The reals form a field)

[F6]

A nonnegative measurable function with integral zero vanishes almost everywhere. (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere)

[A1]

Full AC is carried only to conservatively cover the legacy declared dependency of the published L2 completeness result; the Fourier calculations and finite-site selection use no arbitrary-index choice. (The Axiom of Choice, The Axiom of Countable Choice (ACω))

Proof technique: direct.

Proof

technique · direct

Given: The data above; all Fourier transforms use the negative-sign 2π convention.

1.1F1F2F7A1algebra

Since w=A(a2+4π2∣ξ∣2)−s, the two positive functions (a2+4π2∣ξ∣2)s and (1+∣ξ∣2)s have ratio bounded above and below by positive constants: compare a2+4π2r2 with 1+r2 separately using the minimum and maximum of a2 and 4π2, and use monotonicity of the positive power [F7]. Multiplying by A−1 and taking square roots gives the displayed norm equivalence. Thus the weighted L2 norms defining Hk and Hs are equivalent and have the same representatives by injectivity [F1,F2]; completeness of Hs implies completeness of Hk.

1.2F1F5F6algebra

For a complex coefficient vector c=(cj), the Hermitian quadratic form of the Gram matrix is c∗KXc=∫Rdw(ξ)∣∑j=1Ncje−2πixj⋅ξ∣2dξ. If it is zero, [F6] says w times the squared modulus of the continuous exponential sum vanishes almost everywhere. Since w>0, the sum is zero almost everywhere. A continuous function zero almost everywhere is zero everywhere, because any point of nonzero value would have a neighborhood on which its modulus is bounded below, and such a neighborhood contains a positive-measure box [F5].

2.1F1F5step 1.1algebra

The spectral representative of k(⋅−y) is w(ξ)e−2πiy⋅ξ. Its Hk norm squared is ∫w=1, and for Ug∈Hk, ⟨Ug,k(⋅−y)⟩k=∫g(ξ)e2πiy⋅ξdξ=Ug(y). The integral exists because Cauchy--Schwarz bounds ∫∣g∣ by ∥g∥k(∫w)1/2. Hence point evaluation is continuous and the stated kernel sections reproduce it.

3.1F1F3step 2.1

If U is orthogonal to every section centered at y∈Rd, step 2.1 gives U(y)=0 for every y. Fourier injectivity [F1] then makes U=0 in Hk. By [F3] the span of all sections is dense.

4.1F4F5step 1.2choosealgebra∎

Choose v=(1,t,…,td−1) so that the numbers v⋅xj are pairwise distinct. For each pair i≠j, the polynomial (xi−xj)⋅(1,t,…,td−1) is nonzero because xi−xj≠0; [F4] says its zero set is finite. The ordered field R has 0<1, so the natural multiples 0,1,1+1,… are strictly increasing and R is infinite. Thus a real t exists outside the finite union of root sets. Restrict the vanishing exponential sum to ξ=uv and differentiate at u=0 for orders r=0,…,N−1. This gives ∑jcj(v⋅xj)r=0. For each j, the Lagrange polynomial Lj(t)=∏i≠jt−v⋅xiv⋅xj−v⋅xi has degree at most N−1 and satisfies Lj(v⋅xi)=1 if i=j and 0 otherwise. Taking the same linear combination of the moment equations as the coefficients of Lj yields cj=0. Thus all cj=0, and the real Gram matrix is strictly positive definite.

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The restricted Matérn native space is the Sobolev quotient space

Statement

Assume full AC conservatively for the declared Hilbert projection dependency. Fix the Matérn kernel of The normalized Matérn kernel in the 2π Fourier convention on a nonempty domain Ω of Radial kernels, polynomial augmentation, and site geometry. Let NΩ={U∈Hk(Rd):U∣Ω=0}. The restricted native space is the quotient Hk(Rd)/NΩ with infimal extension norm, and it is the reproducing-kernel Hilbert space of k∣Ω×Ω. Its norm is equivalent to the restriction-quotient norm of Hs(Ω): for constants c,C>0 depending only on d,ν,a, c∥u∥Hs(Ω)≤∥u∥Hk(Ω)≤C∥u∥Hs(Ω). No extension theorem or intrinsic fractional-domain norm is asserted.

Facts & Assumptions

Given: Full AC, fixed Matérn parameters, a nonempty domain Ω, and the quotient definitions above.

[F1]

The whole-space Matérn native space and Hs(Rd) have the same representatives and equivalent norms with constants c0,C0>0. (The Matérn native space has its weighted Fourier norm and strict Gram positivity)

[F2]

Point evaluations are continuous in the whole-space native norm; each section k(⋅−y) reproduces evaluation and the sections centered in Rd span densely. (The Matérn native space has its weighted Fourier norm and strict Gram positivity)

[F3]

A closed subspace of a Hilbert space has a closed orthogonal complement, each vector has a unique orthogonal projection, and the quotient by the subspace is isometric to that complement. (Orthogonal decomposition by a closed subspace, Hilbert projections are linear, self-adjoint and contractive)

[A1]

Full AC is included to match the conservative dependency branch of [F3] and the published L2 completeness supplier; no family of extensions is selected. (The Axiom of Choice, The Axiom of Countable Choice (ACω))

Proof

technique · direct

Given: The hypotheses and notation of the Statement.

1.1F2algebra

The set NΩ is a linear subspace. If Un∈NΩ converges to U in native norm, then for each x∈Ω, [F2] gives ∣Un(x)−U(x)∣≤Cx∥Un−U∥k→0; hence U(x)=0 and U∈NΩ. Thus NΩ is closed.

2.1F3A1step 1.1

By [F3], each equivalence class modulo NΩ has a unique representative in NΩ⊥, namely its orthogonal projection onto that complement; its norm is the infimum of the norms of all extensions because adding an element of NΩ changes only the nullspace component. Therefore the restriction quotient is a Hilbert space isometric to NΩ⊥.

3.1F2step 2.1

For y∈Ω, the section ky=k(⋅−y) lies in NΩ⊥: for every N∈NΩ, ⟨N,ky⟩k=N(y)=0 by reproduction [F2]. Its induced quotient element therefore represents the restricted kernel section, and it reproduces evaluation of each restricted function at y.

4.1F2step 1.1step 3.1

If U∈NΩ⊥ is orthogonal to every ky with y∈Ω, then [F2] gives U(y)=0 for all y∈Ω, so U∈NΩ. Thus U∈NΩ∩NΩ⊥={0}. The restricted kernel sections span densely in NΩ⊥, which proves that the quotient is exactly the RKHS of the restricted kernel.

5.1F1step 2.1algebra∎

Every extension U of u satisfies c0∥U∥Hs(Rd)≤∥U∥k≤C0∥U∥Hs(Rd) by [F1]. Taking the infimum over the same set of extensions on both sides gives c0∥u∥Hs(Ω)≤∥u∥Hk(Ω)≤C0∥u∥Hs(Ω).

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Native Matérn interpolation is an orthogonal projection with power function

Statement

Assume full AC conservatively for the declared Hilbert projection dependencies. Let X=(x1,…,xN)⊂Ω be distinct and nonempty, and let Hk(Ω) be the restricted Matérn native RKHS. For every f∈Hk(Ω), the unique kernel interpolant sXf=∑jcjk(⋅−xj) is the orthogonal projection of f onto VX=span⁡{k(⋅−xj):1≤j≤N}. Define PX(x)=∥(I−ΠX)k(⋅−x)∥Hk(Ω), where ΠX is that projection. Then for every x∈Ω, ∣f(x)−sXf(x)∣≤PX(x)∥f∥Hk(Ω),PX(x)2=k(0)−kX(x)∗KX−1kX(x)≥0. where KX=(k(xi−xj)) and kX(x)=(k(xi−x))i=1N.

Facts & Assumptions

Given: Full AC, a finite nonempty distinct site list X⊂Ω, and f∈Hk(Ω).

[F1]

The restricted native space is a Hilbert RKHS; its kernel sections reproduce evaluation and its norm is the quotient native norm. (The restricted Matérn native space is the Sobolev quotient space)

[F2]

The Matérn Gram matrix on distinct sites is strictly positive definite, hence invertible; k(0)=1. (The Matérn native space has its weighted Fourier norm and strict Gram positivity, The Matérn spectral integral, normalization, and radial formula)

[F3]

Finite-dimensional subspaces of a Hilbert space are closed, orthogonal projection is self-adjoint and contractive, and orthogonal residuals satisfy Pythagoras. (Orthogonal decomposition by a closed subspace, Hilbert projections are linear, self-adjoint and contractive)

[A1]

Full AC is carried conservatively for the projection suppliers' declared choice branch. (The Axiom of Choice, The Axiom of Countable Choice (ACω))

Proof technique: Hilbert projection and Cauchy--Schwarz.

Proof

technique · Hilbert projection and Cauchy--Schwarz

Given: The hypotheses above.

1.1F1F2F3A1

The span VX has finite dimension and is therefore closed. Let ΠXf be the orthogonal projection of f onto VX. For each site xi, the residual is orthogonal to k(⋅−xi); by reproduction [F1], (f−ΠXf)(xi)=0. Thus ΠXf interpolates the data. Its coefficients solve KXc=(f(xi))i, which has a unique solution by [F2], so it is the unique kernel interpolant.

2.1F1F3F4step 1.1algebra

Put e=f−sXf=(I−ΠX)f. Self-adjointness of I−ΠX and reproduction give e(x)=⟨e,k(⋅−x)⟩=⟨e,(I−ΠX)k(⋅−x)⟩. Cauchy--Schwarz [F4] yields ∣e(x)∣≤∥e∥PX(x)≤∥f∥PX(x) by contractivity of the orthogonal projection.

3.1F1F2F3step 1.1algebra∎

The projection of k(⋅−x) has the form ∑jvjk(⋅−xj). Reproduction and orthogonality at all sites give KXv=kX(x), so v=KX−1kX(x) by [F2]. Expanding the squared residual norm and using this equation gives PX(x)2=∥k(⋅−x)∥2−∥ΠXk(⋅−x)∥2=k(0)−kX(x)∗KX−1kX(x). It is nonnegative because it is a squared norm.

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Matérn interpolation and its power function converge at rate h^ν

Statement

Assume full AC conservatively for the Hilbert projection and weighted-L2 suppliers. Fix arbitrary d≥1, ν>0, a>0, s=ν+d/2, the normalized Matérn kernel k of The normalized Matérn kernel in the 2π Fourier convention, and an admitted uniform-cone domain Ω. There are constants C,h0>0, depending only on d,ν,a and the domain's cone constants, such that for every finite nonempty distinct site set X⊂Ω with fill distance hX≤h0 and every f∈Hs(Ω), ∥f−sXf∥L∞(Ω)≤ChXν∥f∥Hs(Ω),sup⁡x∈ΩPX(x)≤ChXν. Here sXf and PX are the native interpolant and power function of Native Matérn interpolation is an orthogonal projection with power function. The rate requires no mesh-ratio bound. The constants are uniform over site sets satisfying the fill-distance condition.

Facts & Assumptions

Given: Full AC, the fixed kernel and domain, a finite distinct set X, its fill distance hX≤h0, and f∈Hs(Ω).

[F1]

The restricted native norm and restriction-quotient Sobolev norm satisfy c∥u∥Hs(Ω)≤∥u∥Hk(Ω)≤C∥u∥Hs(Ω). (The restricted Matérn native space is the Sobolev quotient space)

[F2]

A Sobolev function vanishing on X obeys ∥u∥∞≤ChXν∥u∥Hs(Ω) on the uniform-cone domain, without a separation assumption. (A Sobolev function vanishing on scattered sites has an h^ν bound)

[F3]

Native interpolation is an orthogonal projection, its residual vanishes on X, and its pointwise error is bounded by PX(x) times the native norm. (Native Matérn interpolation is an orthogonal projection with power function)

[F4]

The Cauchy-sequence reals have the least-upper-bound property and are a complete ordered field; hence the Archimedean theorem supplies an integer m>ν for the fixed real ν. (The Cauchy-sequence reals have the least-upper-bound property, Every complete ordered field is Archimedean)

[A1]

Full AC is carried conservatively for the inherited Hilbert-space suppliers; the estimate makes no additional choice from an indexed family. (The Axiom of Choice, The Axiom of Countable Choice (ACω))

Proof technique: apply the zeros estimate to the native projection residual.

Proof

technique · apply the zeros estimate to the native projection residual

Given: The hypotheses above.

1.1F1F2F3F4A1algebra

By [F4] choose an integer m>ν and apply the zeros estimate [F2] with this fixed m. Regard f as an element of the same set of continuous restrictions Hs(Ω)=Hk(Ω) using [F1], and let e=f−sXf. By [F3], e∣X=0 and ∥e∥Hk(Ω)≤∥f∥Hk(Ω). Therefore [F1] and [F2] give ∥e∥L∞(Ω)≤C0hXν∥e∥Hs(Ω)≤C0c−1hXν∥e∥Hk(Ω)≤C0c−1C1hXν∥f∥Hs(Ω). This is the first estimate.

2.1F1F2F3step 1.1algebra∎

Fix x∈Ω and set rx=(I−ΠX)k(⋅−x), so PX(x)=∥rx∥. Self-adjointness and reproduction in [F3] give f(x)−sXf(x)=⟨f,rx⟩. The bound in [F3] shows that this residual evaluation functional has norm at most PX(x). If rx=0, it is the zero functional and its norm is PX(x)=0. If rx≠0, the unit vector fx=rx/∥rx∥ satisfies fx(x)−sXfx(x)=⟨fx,rx⟩=∥rx∥=PX(x), so the functional norm is exactly PX(x). Applying step 1.1 to every f of native norm at most one therefore gives PX(x)≤C2hXν, uniformly in x; taking the supremum proves the second estimate. The constant does not depend on X except through the upper bound on hX, and no mesh ratio entered either [F2] or this argument.

5 · Examples, counterexamples and false statements

None yet.

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