PROVED
This has been solved in the affirmative.
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\[\lambda(x)=\sum_k \lvert l_k(x)\rvert.\]Is it true that, for any fixed $-1\leq a< b\leq 1$,\[\max_{x\in [a,b]}\lambda(x)> \left(\frac{2}{\pi}-o(1)\right)\log n?\]
A question originally asked by Erdős and Turán
[ErTu61]. Bernstein
[Be31] proved this for $a=-1$ and $b=1$, and Erdős
[Er61c] improved this to\[\max_{x\in [-1,1]}\lambda(x)> \frac{2}{\pi}\log n-O(1).\]This is best possible, since taking the $x_i$ as the roots of the $n$th Chebyshev polynomial yields\[\max_{x\in [-1,1]}\lambda(x)<\frac{2}{\pi}\log n+O(1).\]Erdős and Szabados
[ErSz78] proved that\[\max_{x\in [a,b]}\lambda(x)\gg \log n.\]This was resolved by Tao
[Ta26b] who proved that in fact for any fixed $-1\leq a< b\leq 1$\[\max_{x\in [a,b]}\lambda(x)\geq \frac{2}{\pi}\log n-O(1).\]See also
[1129] and
[1132].
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This page was last edited 01 April 2026. View history
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Additional thanks to: Wouter van Doorn and Terence Tao
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #1153, https://www.erdosproblems.com/1153, accessed 2026-04-11
