Can every symmetric function be factorized through symmetric polynomials? A symmetric function is a function $f:\mathbb R^n\to \mathbb R$ such that $f(x_1,\ldots,x_n)=f(\sigma(x_1,\ldots,x_n))$ for every permutation $\sigma\in S_n.$
The most commonly encountered symmetric functions are polynomial functions.
Question 1: Is it true that, given a symmetric function $f:\mathbb R^n\to \mathbb R,$ there exist symmetric polynomials $p_1,\ldots,p_m:\mathbb R^n\to \mathbb R$ and functions $h:\mathbb R^m\to \mathbb R$ and $g_1,\ldots,g_n:\mathbb R\to \mathbb R$ such that $f$ can be factorized as $$f(x_1,\ldots,x_n)=h(p_1(g_1(x_1),\ldots, g_1(x_n)),p_2(g_2(x_1),\ldots, g_2(x_n)),\ldots,p_m(g_m(x_1),\ldots, g_m(x_n)))?$$
Question 2: If $f$ has some regularity, e.g., continuous, $C^\infty$ or analytic, can we find the above $g_i$ and $h$ to be also regular?
To illustrate the idea behind the question I furnish an example.
If $f:\mathbb R^2\to \mathbb R$ with $$f(x,y)=(x^2+y^2)\ln(\frac{xy}{x+y})+\sin(x)+\sin(y),$$ we have $m=4$, $p_1(x,y)=x^2+y^2,$ $p_2(x,y)=xy$, $p_3(x,y)=x+y$, $p_4(x,y)=x+y$, $g_1,g_2,g_3=\operatorname{id}_{\mathbb R}$, $g_4(x)=\sin(x)$ and $h(x,y,z,t)=x\ln(\frac{y}{t})+t$.
Notice that Question 1 seems to be true thanks to an argument similar to that in the answer of Qi Zhu to Can every symmetric function be written as some function of a sum?.
Question 3: does this argument works here?
 A: $\newcommand\R{\mathbb R}$For any $(x_1,\dots,x_n)\in\R^n$,
$$P(x):=\prod _{i=1}^{n}(x-x_{i})=\sum _{k=0}^{n}(-1)^{k}e_k(x_1,\dots,x_n)x^{n-k},$$
where the $e_k$'s are the elementary symmetric polynomials in $x_1,\dots,x_n$. Therefore and because "the roots of a polynomial are determined by its coefficients", there is a function $g\colon\R^m\to[\R^n]$ such that
$$[x_1,\dots,x_n]=g(e_0(x_1,\dots,x_n),\dots,e_n(x_1,\dots,x_n))$$
for all $(x_1,\dots,x_n)\in\R^n$, where $m:=n+1$,
$$[x_1,\dots,x_n]:=\{(x_{\pi(1)},\dots,x_{\pi(n)})\colon\pi\in S_n\},$$
$$[\R^n]:=\{[x_1,\dots,x_n]\colon(x_1,\dots,x_n)\in\R^n\},$$
and $S_n$ is the set of all permutations of the set $\{1,\dots,n\}$.
On the other hand, if a function $f\colon\R^n\to\R$ is symmetric, then there clearly exists a function $u\colon[\R^n]\to\R$ such that
$$f(x_1,\dots,x_n)=u([x_1,\dots,x_n])$$
for all $(x_1,\dots,x_n)\in\R^n$.
Thus,
$$f(x_1,\dots,x_n)=h(e_0(x_1,\dots,x_n),\dots,e_n(x_1,\dots,x_n)) \tag{1}\label{1}$$
for all $(x_1,\dots,x_n)\in\R^n$, where $h:=u\circ g$. This answers your Question 1. $\quad\Box$

As an illustration of the previous consideration, here is the particular case of identity \eqref{1} for the symmetric function $f\colon\R^2\to\R$ given by the formula $f(x,y):=x\sin y+y\sin x$:
$$x\sin y+y\sin x \\
=\Big(-\frac{p(x,y)}2+\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\
\times\sin\Big(-\frac{p(x,y)}2-\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\ 
+\Big(-\frac{p(x,y)}2-\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\
\times\sin\Big(-\frac{p(x,y)}2+\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\ 
$$
for all $(x,y)\in\R^2$, where $p(x,y):=x+y$ and $q(x,y):=xy$ are elementary symmetric polynomials in $x,y$.

It is now an easy exercise to show that $h$ is continuous if $f$ is continuous. Hints: (i) note that the polynomial $P$ is monic and use a bound on the roots of a monic polynomial in terms of the bounds on its coefficients; (ii) using compactness and the continuity of the value of a polynomial with respect to the coefficients and the argument of the polynomial, show that the map $g$ is continuous; (iii) the map $f\mapsto u$ is clearly continuous in any reasonable topology; (iv) so, $h=u\circ g$ is continuous.
A: Q1 and Q2 both have positive answers. I believe these are established results in the literature. Have a look at the papers of Glaeser and Schwarz that I referenced in this answer.
