OPEN
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 $1\leq i\leq n$.

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

Bernstein [Be31] proved that for any choice of $a_i^n$ there exists $x_0\in [-1,1]$ such that $\limsup_{n\to \infty} \sum_{1\leq i\leq n}\lvert p_{i}^n(x)\rvert=\infty.$ Erdős and Vértesi [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 $x\in [-1,1]$.