OPEN
This is open, and cannot be resolved with a finite computation.
For $n\geq 1$ fix some sequence of $n$ distinct numbers $x_{1n},\ldots,x_{nn}\in [-1,1]$. Let $\epsilon=\epsilon(n)\to 0$.
Does there always exist a continuous function $f:[-1,1]\to \mathbb{R}$ such that if $p_n$ is a sequence of polynomials, with degrees $\deg p_n<(1+\epsilon(n))n$, such that $p_n(x_{kn})=f(x_{kn})$ for all $1\leq k\leq n$, then $p_n(x)\not\to f(x)$ for almost all $x\in [-1,1]$?
Erdős, Kroó, and Szabados
[EKS89] proved that, if $\epsilon>0$ is fixed and does not $\to 0$, then there exist sequences $x_{ij}$ such that, for any continuous function $f$, there exists a sequence of polynomials $p_n$, with degrees $\deg p_n<(1+\epsilon)n$, such that $p_n(x_{kn})=f(x_{kn})$ for all $1\leq k\leq n$, and $p_n(x)\to f(x)$ uniformly for all $x\in [-1,1]$.
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This page was last edited 23 January 2026.
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