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Floater–Hormann barycentric denominator lower bound

Statement

Let $n \geq 1$ , $0 \leq d < n$ , $x_{0} < \dots < x_{n}$ , and $h = \max_{j} (x_{j + 1} - x_{j})$ . Let $D_{d} (x) = \sum_{j = 0}^{n} w_{j} / (x - x_{j})$ be the Floater–Hormann denominator from Floater–Hormann barycentric interpolant. For every open gap $x_{α} < x < x_{α + 1}$ , $∣ D_{d} (x) ∣ \geq {\begin{cases} \frac{1}{d! h^{d + 1}}, & d \geq 1, \\ \frac{1}{h (1 + β)}, & d = 0, \end{cases}$ where $β$ is the local ratio of Local mesh ratio for degree-zero Floater–Hormann interpolation.

Facts & Assumptions

Given: Ordered distinct real nodes, $n \geq 1$ , $0 \leq d < n$ , an open node gap, and the associated Floater–Hormann denominator.

[F1]

Floater–Hormann weights equal the local-polynomial blend weights: $D_{d} (x) = \sum_{i} λ_{i} (x)$ and $\sum_{i} μ_{i} (x) = (- 1)^{n - d} (\prod_{j} (x - x_{j})) D_{d} (x)$ away from nodes.

[F2]

Floater–Hormann blend partial sums are positive in each node gap: each nonempty block sum over $I_{1}, I_{2}, I_{3}$ is positive; for $d = 0$ , the first two terms of $I_{3}$ and the last two terms of $I_{1}$ form positive pairs.

[F3]

Local mesh ratio for degree-zero Floater–Hormann interpolation: for $d = 0$ , $β$ is the maximum of the interior minima of the two adjacent-gap ratios, and is zero when $n \leq 2$ .

Proof

1.1givenF1F2choosealgebra

Suppose $d \geq 1$ . The middle block $I_{2}$ contains at least one index: $j = \max (0, α - d + 1)$ satisfies $j \leq α$ , $j + d \geq α + 1$ , and $j \leq n - d$ . Thus its window contains both endpoints of the gap and $μ_{j} (x) > 0$ . By [F2], $s (x) = s_{1} (x) + s_{2} (x) + s_{3} (x) \geq s_{2} (x) \geq μ_{j} (x)$ , with empty block sums interpreted as zero. From [F1], $∣ D_{d} (x) ∣ = \frac{s (x)}{\prod_{k = 0}^{n} ∣ x - x_{k} ∣} \geq \frac{μ_{j} (x)}{\prod_{k = 0}^{n} ∣ x - x_{k} ∣} = ∣ λ_{j} (x) ∣ = \frac{1}{\prod_{k = j}^{j + d} ∣ x - x_{k} ∣} .$ This reduces the denominator estimate to one window product.

1.2givenF1F2algebra

Now let $d = 0$ . On the first gap, $I_{1} = {0}$ and the other nonempty block sums are positive by [F2], hence $s \geq μ_{0}$ and $∣ D_{0} (x) ∣ \geq ∣ λ_{0} (x) ∣ = 1 / (x - x_{0}) \geq 1 / h$ . On the last gap it gives $I_{3} = {n}$ and the other nonempty block sums are positive, hence $s \geq μ_{n}$ and $∣ D_{0} (x) ∣ \geq 1 / (x_{n} - x) \geq 1 / h$ . These bounds imply the claimed inequality because $β \geq 0$ .

2.1givenstep 1.1algebra

Put $u = α - j + 1$ and $v = j + d - α$ , so $u, v \geq 1$ and $u + v = d + 1$ . For $j \leq k \leq α$ , $∣ x - x_{k} ∣ \leq x_{α + 1} - x_{k} \leq (α + 1 - k) h$ ; for $α + 1 \leq k \leq j + d$ , $∣ x - x_{k} ∣ \leq x_{k} - x_{α} \leq (k - α) h$ . Multiplying these $d + 1$ bounds gives $\prod_{k = j}^{j + d} ∣ x - x_{k} ∣ \leq u! v! h^{d + 1} \leq d! h^{d + 1} .$ The last inequality follows from $(\binom{d + 1}{u}) = (d + 1)! / (u! v!) \geq d + 1$ . Substitution in step 1.1 proves the $d \geq 1$ bound.

3.1givenF1F2F3step 1.2algebra∎

If $1 \leq α \leq n - 2$ , both $I_{1}$ and $I_{3}$ are nonempty. The first two terms from $I_{3}$ and first two read backward from $I_{1}$ form positive pairs. With $Δ = x_{α + 1} - x_{α}$ , $δ_{R} = x_{α + 2} - x_{α + 1}$ and $δ_{L} = x_{α} - x_{α - 1}$ , [F1]–[F2] yield $∣ D_{0} (x) ∣ \geq \max {\frac{1}{x_{α + 1} - x} - \frac{1}{x_{α + 2} - x}, \frac{1}{x - x_{α}} - \frac{1}{x - x_{α - 1}}} .$ Writing $t = x_{α + 1} - x \leq Δ$ in the first difference gives $δ_{R} / [t (t + δ_{R})] \geq 1 / [h (1 + Δ / δ_{R})]$ . Writing $u = x - x_{α} \leq Δ$ in the second gives $δ_{L} / [u (u + δ_{L})] \geq 1 / [h (1 + Δ / δ_{L})]$ . Hence $∣ D_{0} (x) ∣ \geq \frac{1}{h (1 + \min {Δ / δ_{L}, Δ / δ_{R}})} \geq \frac{1}{h (1 + β)}$ by [F3]. For $n \leq 2$ there is no such interior index, so steps 1.2 cover all gaps.

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7 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

Michael S. Floater and Kai Hormann, Barycentric rational interpolation with no poles and high rates of approximation, proofs of Theorems 2–3 (standard reference, not scraped)