Erdős Problem #1130

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 $i\neq k$. 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 $\Upsilon(x_1,\ldots,x_n)$. Current best: Erd\H{o}s [Er47] could prove$$\Upsilon(x_1,\ldots,x_n)< \sqrt{n}.$$Erd\H{o}s thought that the maximising 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 $x_{n+1}=1$), which would be the same (conjectured) characterisation as [1129]. Prize: no. Tags: analysis, polynomials.

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