Erdős Problem #671

declared status 'open'. Formalized: no. Given $a_{i}^n\in [-1,1]$ for all $1\leq i\leq n<\infty$ we define $p_{i}^n$ as the unique polynomial of degree $n-1$ such that $p_{i}^n(a_{i}^n)=1$ and $p_{i}^n(a_{i'}^n)=0$ if $1\leq i'\leq n$ with $i\neq i'$. We similarly define$$\mathcal{L}^nf(x) = \sum_{1\leq i\leq n}f(a_i^n)p_i^n(x),$$the unique polynomial of degree $n-1$ which agrees with $f$ on $a_i^n$ for $1\leq i\leq n$ (that is, the sequence of Lagrange interpolation polynomials). Is there such a sequence of $a_i^n$ such that for every continuous $f:[-1,1]\to \mathbb{R}$ there exists some $x\in [-1,1]$ where$$\limsup_{n\to \infty} \sum_{1\leq i\leq n}\lvert p_{i}^n(x)\rvert=\infty$$and yet$$\mathcal{L}^nf(x) \to f(x)?$$Is there such a sequence such that$$\limsup_{n\to \infty} \sum_{1\leq i\leq n}\lvert p_{i}^n(x)\rvert=\infty$$for every $x\in [-1,1]$ and yet for every continuous $f:[-1,1]\to \mathbb{R}$ there exists $x\in [-1,1]$ with$$\mathcal{L}^nf(x) \to f(x)?$$ Current best: Bernstein [Be31] proved that for any choice of $a_i^n$ there exists $x_0\in [-1,1]$ such that$$\limsup_{n\to \infty} \sum_{1\leq i\leq n}\lvert p_{i}^n(x)\rvert=\infty.$$Erd\H{o}s and V\'{e}rtesi [ErVe80] proved that for any choice of $a_i^n$ there exists a continuous $f:[-1,1]\to \mathbb{R}$ such that$$\limsup_{n\to \infty} \lvert \mathcal{L}^nf(x)\rvert=\infty$$for almost all $x\in [-1,1]$. Prize: $250. Tags: analysis.

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