This is exercise 6.7.8. from Abbott's Understanding analysis 2nd ed.
As the question title says, this section is about the Weierstrass Approximation Theorem (WAT). Abbott guides us to prove WAT through the following logical sequence: a) prove that any function can be approximated using a polygonal function, b) prove that you can approximate the absolute value function with a polynomial, c) prove that you can approximate any polygonal function with the absolute value function. If we have (a), (b) and (c), then WAT follows.
I have done part a) and part b) (link for the MSE question for b). This question pertains to part c). The statement of the exercise is the following:
a) Fix $a \in [-1,1]$ and sketch $h_{a}(x)=\frac{1}{2}(\lvert x-a\rvert + (x-a))$ over $[-1,1]$. Note that $h_{a}(x)$ is polygonal and satisfies $h_{a}(x)=0$ for all $x\in [-1,a]$.
b) Explain why we know $h_{a}(x)$ can be uniformly approximated with a polynomial on $[-1,1]$.
c) Let $\phi$ be a polygonal function that is linear on each subinterval of the partition $-1=a_{0}<a_{1}<a_{2}<\dots <a_{n}=1$. Show there exists constants $b_{0}, b_{1}, \dots , b_{n-1}$ so that $\phi(x) = \phi(-1) + b_{0}h_{a_{0}}(x) + b_{1}h_{a_{1}}(x) + \dots + b_{n-1}h_{a_{n-1}}(x)$ for all $x\in [-1,1]$.
d) Complete the proof of WAT for the interval $[-1,1]$, and then generalize to an arbitrary interval $[a,b]$.
For completeness, the definition of a polygonal function given in the book is:
Defintion A continuous function $\phi: [a,b] \rightarrow \mathbb{R}$ is polygonal if there is a partition $a=x_{0}<x_{1}<x_{2}<\dots <x_{n}=b$ of $[a,b]$ such that $\phi$ is linear on each subinterval $[x_{i-1}, x_{i}]$ where $i=1,\dots, n$.
My current solutions and questions
a) With Wolfram alpha I reproduce the plot for $a=-1,-\frac{1}{2}, 0, \frac{1}{2}, 1$. As the exercise statement writes, the function equals $0$ before $a$ and then it is $\lvert x \rvert$.
b) I am a bit stuck in this question. I have already proved that I can approximate the absolute value with a polynomial. So I know I can approximate $h_{a}(x)$ on $x \in [a,1]$, however I can't see how a single polynomial could approximate all of $h_{a}(x)$. Potentially, we would need to subtract $h_{-a}(x)$ to $\lvert x \rvert$, but that assumes what we want to prove!
c) We can conclude this because of the way $h_{a_{i}}$ is built. We need to choose $b_{i-1}$ so that $\sum_{j=0}^{i-1}b_{j}$ equals the value of the slope for $[x_{i-1}, x_{i}]$. We can do that recursively. Suppose $c_{i-1}$ is the slope of $\phi$ in $[x_{i-1}, x_{i}$. Start with $b_{0}=c_{0}$, $b_{1} = c_{1}-b_{0}$. In general $b_{i} = c_{i} - \sum_{j=0}^{i}b_{j}$.
d) This is the implication that I stated in the introduction.
Any corrections on my solution and help with (b) are highly appreciated.
EDIT
I realized I might have been overthinking the solution too much. So I will provide one potential answer to (b). I would still appreciate it if someone could verify the answers (including this) I have given.
By the properties of series, $\frac{1}{2} \lvert x \rvert = \frac{1}{2} \sum_{n=0}^{\infty} a_{n}x^{n}$, where $a_{n}$ are the taylor coefficients that approximate the absolute function. Then we can do $\frac{1}{2} \sum_{n=0}^{\infty} a_{n}x^{n} + \frac{1}{2}x$, which would give us $h_{0}(x)$, and would still be a polynomial. Because we can approximate the absolute function in an arbitrary interval $[a,b]$, and we can shift the vertex of the function arbitrarily, then we can approximate $h_{a}(x)$ for any $a\in[-1,1]$.
