Erdős Problem #671

The problem

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

$$\mathcal{L}^nf(x) \to f(x)?$$

Statement quoted from erdosproblems.com/671, maintained by Thomas Bloom. Original source: [Er82e][Er97f][Va99,2.40]. Current status there: open.

AI contributions

As recorded in Terence Tao's wiki AI contributions to Erdős problems (data as of 30 Jun 2026). 🟢 full, 🟡 partial, ⚪ unverified, 🔴 incorrect.

Caveats

The wiki's classifications are provisional and are not a benchmark. Many Erdős problems have never had a thorough literature review, so a result that looks new may later turn out to be known; some problems were also stated ambiguously or incorrectly. See the problem page for the full discussion.

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