Problem 1129

Erdős Problem #1129: declared status 'proved'. Formalized: no. 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 $i\neq k$. Describe which choice of $x_i$ minimise$$\Lambda(x_1,\ldots,x_n)=\max_{x\in [-1,1]} \sum_k \lvert l_k(x)\rvert.$$ Current best: Erd\H{o}s [Er61c] improved this to$$\Lambda(x_1,\ldots,x_n)> \frac{2}{\pi}\log n-O(1).$$This is best possible, since taking the $x_i$ as the roots of the $n$th Chebyshev polynomial yields$$\Lambda(x_1,\ldots,x_n)< \frac{2}{\pi}\log n+O(1).$$Erd\H{o}s thought that the minimising choice is characterised by the property that the sums$$\max_{x\in [x_i,x_{i+1}]}\sum_k \lvert l_k(x)\rvert$$are all equal for $0\leq i\leq n$ (where $x_0=-1$ and $x_{n+1}=1$). (Such a choice is called canonical.) The minimising canonical choice is known only for $n\leq 4$. Prize: no. Tags: analysis, polynomials.