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

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Chebyshev interpolation error is controlled by best polynomial error

Statement

For either Chebyshev node family, every $n \geq 0$ , and every $f \in C [- 1, 1]$ , $∥ f - I_{n}^{σ} f ∥_{\infty} \leq (1 + Λ_{n}^{σ}) E_{n} (f) .$ Here $E_{n}$ is the best degree- $n$ uniform error and $Λ_{n}^{σ}$ is the operator norm defined in Best uniform polynomial error and Chebyshev interpolation operators.

Facts & Assumptions

Given: $f \in C [- 1, 1]$ , $n \geq 0$ , and $σ \in {R, L}$ .

[F1]

Best uniform polynomial error and Chebyshev interpolation operators defines $E_{n}$ , $I_{n}^{σ}$ , and $Λ_{n}^{σ}$ .

[F2]

Chebyshev cardinal interpolation reproduces polynomials and its norm is the cardinal sum: $I_{n}^{σ} p = p$ for every $p \in P_{n}$ , and $∥ I_{n}^{σ} ∥ = Λ_{n}^{σ}$ .

Proof

1.1F1F2givenalgebra

Fix $p \in P_{n}$ and put $h = f - p$ . By [F2], $I_{n}^{σ} p = p$ , so $f - I_{n}^{σ} f = h - I_{n}^{σ} h$ . If $∥ h ∥_{\infty} = 0$ , then $h = 0$ and the asserted bound follows; otherwise $g = h / ∥ h ∥_{\infty}$ lies in $C [- 1, 1]$ with $∥ g ∥_{\infty} = 1$ , so the definition in [F1] gives $∥ I_{n}^{σ} h ∥_{\infty} = ∥ h ∥_{\infty} ∥ I_{n}^{σ} g ∥_{\infty} \leq Λ_{n}^{σ} ∥ h ∥_{\infty}$ . Pointwise, $∣ h - I_{n}^{σ} h ∣ \leq ∣ h ∣ + ∣ I_{n}^{σ} h ∣$ , so taking suprema gives $∥ f - I_{n}^{σ} f ∥_{\infty} \leq (1 + Λ_{n}^{σ}) ∥ f - p ∥_{\infty}$ .

2.1givenF1step 1.1algebra∎

Step 1.1 holds for every $p \in P_{n}$ . The unit ball in [F1] contains the zero function and each norm is nonnegative, so $Λ_{n}^{σ} \geq 0$ and $1 + Λ_{n}^{σ} > 0$ . Taking the infimum of the right-hand side over the nonempty set $P_{n}$ therefore gives $(1 + Λ_{n}^{σ}) E_{n} (f)$ , proving the claim.

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Sources

Lloyd N. Trefethen, Approximation Theory and Approximation Practice, Ch. 5 (standard reference, not scraped)
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Polynomial interpolation (standard reference, not scraped)