TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Exact polynomial equioscillation certifies best approximation

Statement

Let $n \geq 0$ , let $f \in C ([a, b], R)$ with $a < b$ , and let $p \in P_{n} ([a, b])$ . If $E : = ∥ f - p ∥_{\infty, [a, b]} > 0$ and there are $n + 2$ ordered nodes $x_{0} < \dots < x_{n + 1}$ in $[a, b]$ and a sign $σ \in {- 1, 1}$ such that $f (x_{i}) - p (x_{i}) = σ (- 1)^{i} E (0 \leq i \leq n + 1),$ then no polynomial in $P_{n} ([a, b])$ has smaller uniform error than $p$ . Thus the displayed residual values are an a posteriori optimality certificate.

Facts & Assumptions

Given: $n \geq 0$ , a nondegenerate closed interval $[a, b]$ , a continuous real function $f$ , a polynomial $p$ of degree at most $n$ , its positive error $E = ∥ f - p ∥_{\infty, [a, b]}$ , and the stated ordered alternation nodes in $[a, b]$ .

[F1]

Best uniform polynomial error on a compact interval: the comparison class is $P_{n} ([a, b])$ and its error is the uniform norm on that interval.

[F2]

Alternating polynomial signs force the Haar zero-count contradiction: no nonzero polynomial of degree at most $n$ can have nonzero alternating signs at $n + 2$ ordered nodes.

Proof

1.1givenF1algebra

Suppose $q \in P_{n} ([a, b])$ has $∥ f - q ∥_{\infty, [a, b]} < E$ . At a node with $f (x_{i}) - p (x_{i}) = σ (- 1)^{i} E$ , the strict inequality gives $∣ f (x_{i}) - q (x_{i}) ∣ < E$ , so $q (x_{i}) - p (x_{i}) = (f (x_{i}) - p (x_{i})) - (f (x_{i}) - q (x_{i}))$ has the same nonzero sign as $σ (- 1)^{i}$ .

2.1givenF2step 1.1algebra∎

The polynomial $q - p$ has degree at most $n$ and, by step 1.1, has nonzero alternating signs at $n + 2$ ordered nodes. This contradicts [F2]. Therefore no such $q$ exists, and $p$ is a best approximant.

Depends on

Used by

Finite-budget polynomial Remez exchange with an exact extrema oracle Algorithm
A two-node degree-zero Remez certificate Example
The degree-one minimax approximation to x squared Example

Dependency tree · two levels

11 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

Lloyd N. Trefethen, Approximation Theory and Approximation Practice, Chapter 10 (standard reference, not scraped)
András Kroó and Allan Pinkus, Strong Uniqueness, §4.4 (standard reference, not scraped)