CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A defective rational minimizer can have fewer than m+n+2 extrema

Statement refuted

Every best real rational approximant of type $(m, n)$ with positive error has at least $m + n + 2$ distinct alternating global extrema.

Facts & Assumptions

Given: $I = [- 1, 1]$ , $f (x) = x^{2}$ , weight $w (x) = 1$ , type $(m, n) = (1, 1)$ , and $r (x) = 1 / 2$ .

[F1]

Weighted rational minimax families and defect: type $(m, n)$ candidates have degree bounds $(m, n)$ ; the defect of a nonzero reduced candidate is $d = \min (m - \deg p, n - \deg q)$ .

[F2]

Weighted rational minimizers satisfy the defect-sensitive alternation criterion: a candidate with positive error is best exactly when its error has $m + n + 2 - d$ ordered alternating global extrema.

Counterexample

1.1F1givenalgebra

The pair $p = 1 / 2, q = 1$ is reduced, has $q > 0$ , coefficient norm $1$ , degrees $(0, 0)$ , and error $f - r = x^{2} - 1 / 2$ . Since $x^{2} \in [0, 1]$ , this residual lies in $[- 1 / 2, 1 / 2]$ and its absolute value reaches $1 / 2$ exactly at $x = - 1, 0, 1$ . The extremal residual signs in order are $+, -, +$ .

2.1F1F2step 1.1algebra∎

Here $d = \min (1 - 0, 1 - 0) = 1$ , so the defect-sensitive count is $1 + 1 + 2 - 1 = 3$ . By [F2], the three alternating extrema make $r = 1 / 2$ a best type- $(1, 1)$ rational approximant. The extremal set is exactly ${- 1, 0, 1}$ , so there are not four distinct global extrema; the refuted $m + n + 2 = 4$ count fails.

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Sources

Lloyd N. Trefethen, Approximation Theory and Approximation Practice, Chapter 24 (standard reference, not scraped)
András Kroó and Allan Pinkus, Strong Uniqueness, §6 (standard reference, not scraped)