Let $x_1,x_2,..,x_m$ be equally spaced points in [-1,1] defined as follows: $x_i = -1 + \frac{2(i-1)}{m}$. Let $p(z)$ be a degree $d$ polynomial. Define $\ell_1$ norm of $p$ as $\int_{-1}^{1} p(z)dz$.
What is the minimum $m$ we need (in terms of $d$, $\epsilon$), so that $$\sum_{i \in [m]} \frac{2}{m}|p(x_i)| = (1 \pm \epsilon)\|p\|_1? $$
The 2 in lhs appears because the interval [-1,1] has length 2.
Lemma $3.1$ of this paper by Kane, Karmalkar, and Price considers this question when $x_i$'s are arbitrary points from the partition made by the Chebyshev extrema $\cos(\pi j/m)$. I am trying to find an easier proof for this lemma for the case $q=1$ (and other variants of this lemma for the $q=1$ case) that does not go via a theorem by Nevai (lemma A.1 (page 16) of the Kane-Karmalkar-Price paper). The equally spaced points variant seemed to me an easier and natural question to try first.