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 $x_0=-1$ and $x_{n+1}=1$ and\[\Upsilon(x_1,\ldots,x_n)=\min_{0\leq i\leq n}\max_{x\in[x_i,x_{i+1}]} \sum_k \lvert l_k(x)\rvert.\]Is it true that\[\Upsilon(x_1,\ldots,x_n)\ll \log n?\]Describe which choice of $x_i$ maximise $\Upsilon(x_1,\ldots,x_n)$.
Let $x_0=-1$ and $x_{n+1}=1$ and\[\Upsilon(x_1,\ldots,x_n)=\min_{0\leq i\leq n}\max_{x\in[x_i,x_{i+1}]} \sum_k \lvert l_k(x)\rvert.\]Is it true that\[\Upsilon(x_1,\ldots,x_n)\ll \log n?\]Describe which choice of $x_i$ maximise $\Upsilon(x_1,\ldots,x_n)$.
Categories:
Analysis Polynomials
Progress
The functions $l_k(x)$ are sometimes called the fundamental functions of Lagrange interpolation.Erdős [Er47] could prove\[\Upsilon(x_1,\ldots,x_n)< \sqrt{n}.\]Erdős thought that the maximising choice is characterised by the property that the sums\[\lambda_i=\max_{x\in [x_i,x_{i+1}]}\sum_k \lvert l_k(x)\rvert\]are all equal for $0\leq i\leq n$ (where $x_0=-1$ and $x_{n+1}=1$), which would be the same characterisation as [1129].
This is true, and was proved by de Boor and Pinkus [dBPi78]. It follows by the bounds discussed in [1129] that\[\Upsilon(x_1,\ldots,x_n)\leq \frac{2}{\pi}\log n+O(1).\]See also [1129].
Source: erdosproblems.com/1130 | Last verified: January 19, 2026