De-Nesting Absolute Value Function into Linear Combination of Absolute Value Functions Context: In formulating problems for secondary school mathematics teachers (and students) about absolute value functions, which we define as functions $\mathbb{R} \rightarrow \mathbb{R}$ that send $x \mapsto a|x-h|+k$ for fixed parameters $a, h, k \in \mathbb{R}$, I was able to rewrite the nested absolute value function
$$f(x) = \Big||x|-1\Big|$$
as a linear combination of absolute value functions,
$$g(x) = |x+1| + |x-1| - (|x|+1)$$
(You can view the graphs of $f$ and $g$ here; although not delved into in this post, my colleagues enjoyed finding similar relationships even when there is a quadratic $x$ term, for example, in the graphs/functions depicted here.)
My question is twofold (although the follow-up question depends on the first answer):

1. Is it true that every nested absolute value function (NAVF) or linear combination of NAVFs can be written as a linear combination of AVFs?
2a. If not, what is a counterexample, and what criteria must be satisfied for de-nesting to be possible?
2b. If so, is there an algorithm for de-nesting, i.e., rewriting an arbitrary NAVF as a linear combination of AVFs?

Pointers to related literature/references would be welcome, even if they do not explicitly answer the questions above. Please edit the questions, title, or tags if you believe it will improve clarity.
 A: $\newcommand{\al}{\alpha}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\om}{\omega}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\renewcommand{\P}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} 
\newcommand{\tf}{\widetilde{f}}$ 
Lemma 3 in Explicit additive decomposition of norms (which is Lemma 1.2 in the arXiv version of that note) states the following: 

Suppose that $f\colon\R\to\R$ is a convex function such that for some real $k$ there exist finite limits 
  \begin{equation*}
 d_+:=d_{f,k;+}:=\lim_{u\to\infty}[f(u)-ku]\quad\text{and}\quad d_-:=d_{f,k;-}:=\lim_{u\to-\infty}[f(u)+ku]. 
\end{equation*}
  Then for all $x\in\R$ 
  \begin{equation*}
 f(x)=\frac{d_++d_-}2+\frac12\,\int_\R|x-t|\,d f'(t). \tag{1}
\end{equation*}

As is clear from the short proof, this lemma holds for any absolutely continuous function $f$ with (possibly infinite) limits $\lim_{x\to\pm\infty}f'(x)$. So, it is easy to see by induction on the nesting depth that the lemma holds for any nested absolute value function. 
Added details: 
For a finite nesting depth, the function $f$ is piecewise-affine: 
\begin{multline*}
 f(x)=(a_1+b_1 x)\ii{x\le t_1}+(a_2+b_2 x)\ii{t_1<x\le t_2}+\dots \\ 
 +(a_n+b_n x)\ii{t_{n-1}<x\le t_n}
+(a_{n+1}+b_{n+1} x)\ii{x>t_n}
\end{multline*}
for some natural $n$, some real "switch points" $t_1<\dots<t_n$, some real $a_i$ and $b_i$'s, and all real $x$, where $\ii\cdot$ denotes the the indicator. 
So, the integral in the lemma reduces to a sum: 
\begin{equation*}
 f(x)=\frac{d_++d_-}2+\frac12\,\sum_1^n|x-t_i|\,\De f'(t_i) \tag{2}
\end{equation*}
for all real $x$, 
where $\De f'(t):=f'(t+)-f'(t-)$, the "jump" of $f'$ at $t$, so that $\De f'(t_i):=b_{i+1}-b_i$ for all $i=1,\dots,n$. 
For the example at the link provided in the comment by the OP, formula (2) is illustrated in this Mathematica notebook and its pdf image.  
