Given $a_i^n\in [-1,1]$ for all $1\le i\le n<\infty$ we define $p_i^n$ as the unique polynomial of degree $n-1$ such that $p_i^n(a_i^n)=1$ and $p_i^n(a_{i^{\prime }}^n)=0$ if $1\le i^{\prime }\le n$ with $i\ne i^{\prime }$. We similarly define$${\mathcal{L}}^{n}f(x)=\sum _{1\le i\le n}f(a_i^n)p_i^n(x),$$the unique polynomial of degree $n-1$ which agrees with $f$ on $a_i^n$ for $1\le i\le n$ (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$$\underset{n\to \infty }{\mathrm{limsup}}\sum _{1\le i\le n}|p_i^n(x)|=\infty$$and yet$${\mathcal{L}}^{n}f(x)\to f(x)?$$Is there such a sequence such that$$\underset{n\to \infty }{\mathrm{limsup}}\sum _{1\le i\le n}|p_i^n(x)|=\infty$$for every $x\in [-1,1]$ and yet for every continuous $f:[-1,1]\to \mathbb{R}$ there exists $x\in [-1,1]$ with$${\mathcal{L}}^{n}f(x)\to f(x)?$$