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 $i\neq k$.
What is the minimal value of\[I(x_1,\ldots,x_n)=\int_{-1}^1 \sum_k \lvert l_k(x)\rvert^2\mathrm{d}x?\]In particular, is it true that\[\min I =2-(1+o(1))\frac{1}{n}?\]
What is the minimal value of\[I(x_1,\ldots,x_n)=\int_{-1}^1 \sum_k \lvert l_k(x)\rvert^2\mathrm{d}x?\]In particular, is it true that\[\min I =2-(1+o(1))\frac{1}{n}?\]
Categories:
Analysis Polynomials
Progress
Erdős first conjectured this minimum was achieved by taking the $x_i$ to be the roots of the integral of the Legendre polynomial, since Fejer [Fe32] had earlier shown these to be minimisers of\[\max_{x\in [-1,1]}\sum_k \lvert l_k(x)\rvert^2.\]This was disproved by Szabados [Sz66] for every $n>3$.Erdős, Szabados, Varma, and Vértesi [ESVV94] proved that\[2-O\left(\frac{(\log n)^2}{n}\right)\leq \min I\leq 2-\frac{2}{2n-1}\]where the upper bound is witnessed by the roots of the integral of the Legendre polynomial as above.
Source: erdosproblems.com/1131 | Last verified: January 19, 2026