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Erdős #1131 · analysis / polynomials

For x1,…,xn∈[−1,1]x_1,\ldots,x_n\in [-1,1] let

lk(x)=∏i≠k(x−xi)∏i≠k(xk−xi),l_k(x)=\frac{\prod_{i\neq k}(x-x_i)}{\prod_{i\neq k}(x_k-x_i)},

which are such that lk(xk)=1l_k(x_k)=1 and lk(xi)=0l_k(x_i)=0 for i≠ki\neq k.

What is the minimal value of

I(x1,…,x…I(x_1,\ldots,x…
Source reports OPENAs of 2026-09-08Source & evidence ↓
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Problem statement

For x1,…,xn∈[−1,1]x_1,\ldots,x_n\in [-1,1] let

lk(x)=∏i≠k(x−xi)∏i≠k(xk−xi),l_k(x)=\frac{\prod_{i\neq k}(x-x_i)}{\prod_{i\neq k}(x_k-x_i)},

which are such that lk(xk)=1l_k(x_k)=1 and lk(xi)=0l_k(x_i)=0 for i≠ki\neq k.

What is the minimal value of

I(x1,…,xn)=∫−11∑k∣lk(x)∣2dx?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))1n?\min I =2-(1+o(1))\frac{1}{n}?

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Background

Erdős first conjectured this minimum was achieved by taking the xix_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∈[−1,1]∑k∣lk(x)∣2.\max_{x\in [-1,1]}\sum_k \lvert l_k(x)\rvert^2.

This was disproved by Szabados [Sz66] for every n>3n>3.

Erdős, Szabados, Varma, and Vértesi [ESVV94] proved that

2−O((log⁡n)2n)≤min⁡I≤2−22n−12-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.

Progress & source assessment

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 #1131 link

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