Problem Statement
For $x_1,\ldots,x_n\in [-1,1]$ let\[l_k(x)=\frac{\prod_{i\neq k}(x-x_i)}{\prod_{i\neq k}(x_k-x_i)},\]which are such that $l_k(x_k)=1$ and $l_k(x_i)=0$ for $i\neq k$.
Let $x_1,x_2,\ldots\in [-1,1]$ be an infinite sequence, and let\[L_n(x) = \sum_{1\leq k\leq n}\lvert l_k(x)\rvert,\]where each $l_k(x)$ is defined above with respect to $x_1,\ldots,x_n$.
Must there exist $x\in (-1,1)$ such that\[L_n(x) >\frac{2}{\pi}\log n-O(1)\]for infinitely many $n$?
Is it true that\[\limsup_{n\to \infty}\frac{L_n(x)}{\log n}\geq \frac{2}{\pi}\]for almost all $x\in (-1,1)$?
Let $x_1,x_2,\ldots\in [-1,1]$ be an infinite sequence, and let\[L_n(x) = \sum_{1\leq k\leq n}\lvert l_k(x)\rvert,\]where each $l_k(x)$ is defined above with respect to $x_1,\ldots,x_n$.
Must there exist $x\in (-1,1)$ such that\[L_n(x) >\frac{2}{\pi}\log n-O(1)\]for infinitely many $n$?
Is it true that\[\limsup_{n\to \infty}\frac{L_n(x)}{\log n}\geq \frac{2}{\pi}\]for almost all $x\in (-1,1)$?
Categories:
Analysis Polynomials
Progress
A result of Bernstein [Be31] implies that the set of $x\in(-1,1)$ for which\[\limsup_{n\to \infty}\frac{L_n(x)}{\log n}\geq \frac{2}{\pi}\]is everywhere dense.Erdős [Er61c] proved that, for any fixed $x_1,\ldots,x_n\in [-1,1]$,\[\max_{x\in [-1,1]}\sum_{1\leq k\leq n}\lvert l_k(x)\rvert>\frac{2}{\pi}\log n-O(1).\]See also [1129] for more on $L_n(x)$.
Source: erdosproblems.com/1132 | Last verified: January 19, 2026