The probabilistic Waring problem for finite simple groups has positive solution

You need to know: Group, identity element $1$ , finite group, abelian and non-abelian groups, simple group, free group, notation $|S|$ for the number of elements in finite set S.

Background: Let $F_k$ be the free group with k generators $x_1, \dots x_k$ . Any element $w\in F_k$ (which we call word) can be written as $w=\prod\limits_{j=1}^r x_{i_j}^{\epsilon_j}$ for some $1 \leq i_1 < \dots < i_r \leq k$ and $\epsilon_j = \pm 1$ , $j=1,\dots,r$ . For every group G, w induces a word map from $w:G^k \to G$ given by $w(g_1, \dots, g_k)=\prod\limits_{j=1}^r g_{i_j}^{\epsilon_j}$ . If G is a finite group and $g\in G$ , let $N_G^w(g)$ be the number of $(g_1, \dots, g_k)\in G^k$ such that $w(g_1, \dots, g_k)=g$ . Two words $w_1, w_2$ are said to be disjoint if they are words in disjoint sets of variables. For a function $f: G\to{\mathbb R}$ , we write $||f||_{L^1} = \sum\limits_{g \in G}|f(g)|$ .

The Theorem: On 14th August 2018, Michael Larsen, Aner Shalev, and Pham Tiep
submitted to the Annals of Mathematics a paper in which they proved the following result. Let $w_1, w_2 \neq 1$ be disjoint words and let $w = w_1w_2$ . Then $\lim\limits_{|G|\to\infty}\left\|\frac{N_G^w(g)}{|G|^k}-\frac{1}{|G|}\right\|_{L^1}=0$ , where G ranges over all finite non-abelian simple groups.

Short context: In an earlier work, Larsen, Shalev, and Tiep proved that, for every word w as in the Theorem, and for every sufficiently large finite non-abelian simple group G, we have $N_G^w(g)>0$ for every $g\in G$ . In the language of probability theory, this means that if we select $g_1, \dots, g_k$ from G at random, and compute $w(g_1, \dots, g_k)$ , we can get any element $g\in G$ in this way with non-zero probability. In 2013, Shalev further asked whether $w(g_1, \dots, g_k)$ has almost uniform distribution in G with respect to the $L^1$ norm. This question became known as the probabilistic Waring problem for finite simple groups. The Theorem answers this question affirmatively.