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

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The best uniform polynomial approximant is unique

Statement

For every $f \in C ([a, b], R)$ with $a < b$ and every $n \geq 0$ , there is exactly one $p_{*} \in P_{n} ([a, b])$ with $∥ f - p_{*} ∥_{\infty, [a, b]} = E_{n} (f; [a, b])$ .

Facts & Assumptions

Given: $I = [a, b]$ with $a < b$ , $n \geq 0$ , and $f \in C (I, R)$ .

[F1]

Best uniform polynomial error on a compact interval: $E_{n} (f; I)$ is the infimum of the uniform errors over $P_{n} (I)$ .

[F2]

Best uniform polynomial approximants exist: at least one polynomial in $P_{n} (I)$ attains $E_{n} (f; I)$ .

[F3]

A nonzero best polynomial error must alternate at n+2 extrema: if the best error is positive, a best polynomial has $n + 2$ ordered alternating extremal points.

[F4]

Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials: every real polynomial of degree at most $n$ has a coefficient-list representative.

[F5]

A nonzero real polynomial of degree $n$ has no more than $n$ distinct real roots: a nonzero real polynomial of degree at most $n$ has at most $n$ distinct real roots.

Proof

1.1givenF1F2algebra

Let $p, q \in P_{n} (I)$ both attain $E : = E_{n} (f; I)$ . If $E = 0$ , then $∥ f - p ∥_{\infty, I} = ∥ f - q ∥_{\infty, I} = 0$ , so $p = f = q$ at every point of $I$ .

1.2F1F3givenchoosealgebra

Suppose $E > 0$ and set $r = (p + q) / 2 \in P_{n} (I)$ . The triangle inequality gives $∥ f - r ∥_{\infty, I} \leq E$ , while [F1] gives $∥ f - r ∥_{\infty, I} \geq E$ ; hence $r$ is also best. By [F3], choose ordered alternating extrema $x_{0} < \dots < x_{n + 1}$ for $f - r$ .

2.1givenF2F3step 1.2algebra

At each $x_{i}$ , $f (x_{i}) - r (x_{i})$ equals either $E$ or $- E$ . It is the average of $f (x_{i}) - p (x_{i})$ and $f (x_{i}) - q (x_{i})$ , each lying in $[- E, E]$ . An average of two numbers in this interval equals $E$ only if both are $E$ , and equals $- E$ only if both are $- E$ . Thus $p (x_{i}) = q (x_{i})$ for all $n + 2$ nodes.

3.1F2F4F5step 2.1algebra∎

Choose coefficient-list representatives $P, Q$ of $p, q$ using [F4]. The polynomial $P - Q$ has degree at most $n$ and vanishes at the $n + 2$ distinct points from step 2.1. If it were nonzero, [F5] would bound its distinct real roots by $n$ , a contradiction. Thus $P - Q = 0$ , so their restrictions satisfy $p = q$ . Existence is [F2], hence the best approximant is unique.

Depends on

Used by

The degree-one minimax approximation to x squared Example

Dependency tree · two levels

17 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)