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Floater–Hormann C^(d+2) parity error bounds

Statement

Let n≥1, 0≤d<n, and x0<⋯<xn be real nodes in [a,b]=[x0,xn]. Put h=max⁡0≤j<n(xj+1−xj), m=n−d, and ϵ=1 if m is even and ϵ=0 if m is odd. Let f∈Cd+2([a,b]) and let rd be the exact Floater–Hormann interpolant. For a continuous q on [a,b], write ∥q∥∞=sup⁡x∈[a,b]∣q(x)∣. If d≥1, then ∥f−rd∥∞≤hd+1((b−a)∥f(d+2)∥∞d+2+ϵ∥f(d+1)∥∞d+1). If d=0, let β be the local mesh ratio in Local mesh ratio for degree-zero Floater–Hormann interpolation. Then ∥f−r0∥∞≤h(1+β)((b−a)∥f′′∥∞2+ϵ∥f′∥∞). No quasi-uniformity hypothesis is needed when d≥1. If d=0, a bound β≤ρ follows from a global mesh-ratio bound ρ by A global mesh ratio bounds the local degree-zero ratio.

Facts & Assumptions

Given: Ordered distinct nodes x0,…,xn, an integer 0≤d<n, f∈Cd+2([x0,xn]), and the exact Floater–Hormann interpolant.

[F1]

Floater–Hormann weights equal the local-polynomial blend weights and Newton remainder for a Floater–Hormann window polynomial: off the nodes, f(x)−rd(x)=(∑i=0n−d(−1)if[xi,…,xi+d,x])/(∑iλi(x)).

[F2]

Adjacent-window divided-difference recurrence: adjacent terms satisfy Gi−Gi+1=−(xi+d+1−xi)f[xi,…,xi+d+1,x].

[F3]

Rolle bound for a divided difference: for f∈Cq, an order-q divided difference has magnitude at most ∥f(q)∥∞/q!.

[F4]

Floater–Hormann barycentric denominator lower bound: on each open gap the denominator has magnitude at least 1/(d!hd+1) for d≥1, and at least 1/[h(1+β)] for d=0.

[F5]

Floater–Hormann interpolants have no real poles: the quotient has no real poles and interpolates the data at every node.

[F6]

Higher derivatives and the classes Ck and C∞: Cd+2 means derivatives through order d+2 exist and are continuous on the closed interval, with one-sided endpoint derivatives.

[F7]

The natural numbers N (von Neumann): N contains 0 and is closed under successor, so d+1,d+2∈N.

[F8]

Extreme value theorem: a continuous real function on a nonempty compact subset of R attains a greatest and a least value: a continuous real function on a nonempty compact interval has a finite attained maximum and minimum.

Proof

1.1givenF1F6F8algebra

Fix a point x in an open node gap. By [F6], the stated smoothness hypothesis supplies the derivatives used below. By [F8], their sup norms are finite. By [F1], the error equals N(x)/D(x), where N(x)=∑i=0m(−1)iGi,Gi=f[xi,…,xi+d,x],D(x)=∑iλi(x). This identity follows by writing f−rd=∑iλi(f−pi)/D and substituting the window remainder; λi∏k=ii+d(x−xk)=(−1)i.

2.1givenF2F3F7F8step 1.1algebra

Pair the terms of N as (G0−G1)+(G2−G3)+⋯. If m is odd, these pairs exhaust the sum. If m is even, the final term Gm remains. By [F2] each pair is −(xi+d+1−xi)f[xi,…,xi+d+1,x]. The order-(d+2) Rolle bound [F3] bounds its divided difference by ∥f(d+2)∥∞/(d+2)!. Each interval length xi+d+1−xi is the sum of d+1 adjacent mesh gaps, and each gap occurs in at most d+1 windows. Hence the sum of lengths over the even indices in the pairing is at most (d+1)(b−a).

3.1givenF3F7F8step 2.1algebra

If m is odd, step 2.1 gives ∣N∣≤(d+1)(b−a)∥f(d+2)∥∞/(d+2)!. If m is even, it gives the same bound plus ∣Gm∣≤∥f(d+1)∥∞/(d+1)! by [F3] and [F7]. Thus both cases are ∣N∣≤(d+1)(b−a)∥f(d+2)∥∞(d+2)!+ϵ∥f(d+1)∥∞(d+1)!. This bounds the numerator in either parity case.

4.1givenF1F4step 3.1algebra

If d≥1, divide the numerator bound in step 3.1 by ∣D∣≥1/(d!hd+1) from [F4]. Simplifying factorials gives ∣f(x)−rd(x)∣≤hd+1((b−a)∥f(d+2)∥∞d+2+ϵ∥f(d+1)∥∞d+1). This is the required open-gap bound for d≥1.

4.2givenF1F4step 3.1algebra

If d=0, the same numerator bound in step 3.1, with ∣D∣≥1/[h(1+β)] from [F4], gives ∣f(x)−r0(x)∣≤h(1+β)((b−a)∥f′′∥∞2+ϵ∥f′∥∞). This is the required open-gap bound for d=0.

5.1givenF5F8step 4.1step 4.2∎

At every node the error is zero by [F5]. The no-real-pole property makes rd continuous on the compact interval, so the pointwise estimates on the open gaps extend to the supremum over all [a,b]; finiteness of that supremum follows from [F8]. The two cases prove the stated bounds.

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