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Erdős #1151 · analysis / polynomials

Given a1,…,an∈[−1,1]a_1,\ldots,a_n\in [-1,1] let

Lnf(x)=∑1≤i≤nf(ai)ℓi(x)\mathcal{L}^nf(x) = \sum_{1\leq i\leq n}f(a_i)\ell_i(x)

be the unique polynomial of degree n−1n-1 which agrees with ff on aia_i for 1≤i≤n1\leq i\leq n (that is, the Lagrange inte…

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Problem statement

Given a1,…,an∈[−1,1]a_1,\ldots,a_n\in [-1,1] let

Lnf(x)=∑1≤i≤nf(ai)ℓi(x)\mathcal{L}^nf(x) = \sum_{1\leq i\leq n}f(a_i)\ell_i(x)

be the unique polynomial of degree n−1n-1 which agrees with ff on aia_i for 1≤i≤n1\leq i\leq n (that is, the Lagrange interpolation polynomial).

Let aia_i be the set of Chebyshev nodes. Prove that, for any closed A⊆[−1,1]A\subseteq [-1,1], there exists a continuous function ff such that AA is the set of limit points of Lnf(x)\mathcal{L}^nf(x).

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Background

This is as the problem is given in [Va99], but I am unclear exactly what is intended here - is this meant for fixed, arbitrary, x∈[−1,1]x\in [-1,1]?

Erdős [Er41] proved that, if x=cos⁡(πp/q)x=\cos(\pi p/q) for some odd integers p,q≥1p,q\geq 1, then there is a continuous function ff such that

lim⁡n→∞Lnf(x)=∞,\lim_{n\to \infty}\mathcal{L}^nf(x)=\infty,

where the limit is taken over the sequence of Chebyshev nodes as n→∞n\to \infty. In [Er43] he claims (without proof) that for any closed set AA there exists a continuous ff such that the limit points of Lnf(x)\mathcal{L}^nf(x) is AA (for specific xx of this shape).

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The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #1151 link

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