Let $f:[a,b]\rightarrow[c,d]$ be a continuous and piecewise $C^{\infty}$ function sending $a$ to $c$ and $b$ to $d$ and let $g:[c,d]\rightarrow \mathbb{R}$ be a real continuous and piecewise $C^{\infty}$ function. Is the composition $g\circ f$ a continuous and piecewise $C^{\infty}$ function?
The difficulty is to prove that the preimage via $f$ of a partition of $[c,d]$ can "generate" a partition of $[a,b]$. The matter consists to prove that the set of zeroes of a function $h\in C^{\infty}([a,b],\mathbb{R})$ is a finite set of isolated points united a finite number of closed intervals: $h^{-1}(0)=\{x_1,x_2,…,x_m \} \cup I_1 \cup…\cup I_s $ where $I_j$ is a closed interval in $[a,b]$.
Thanks for any useful hint/answer.
NOTE:
A function $f:[a,b]\rightarrow\mathbb{R}$ is called piecewise $C^{l}$ if there is a partition $a=a_0 < a_1 < …< a_n =b$ of the interval $[a,b]$ such that $f|_{[a_k,a_{k+1}]}\in C^l({[a_k,a_{k+1}]},\mathbb{R})$ for every $0\le k \le n-1$.
Best Answer
The property you want of the set of zeros of a smooth function does not hold in general: If you define $f\colon \mathbb R\to \mathbb R$ $$ f(x) = \left\{\begin{array}{cc} \sin(1/x)\exp(-\frac{1}{x}) & \text{ if } x>0\\ 0, & \text{ if } x\leq 0, \end{array} \right. $$ then using the fact that if $p(x)$ is a polynomial function of $x$ then $$ \lim_{x\to 0} p(1/x)\exp(-1/x) =0, $$ you can show that $f(x)$ is a smooth function on all of $\mathbb R$. Clearly $f^{-1}(0)$ contains no intervals of positive length, but $0$ is it is standard that $f$ is infinitely differentiable on all of $\mathbb R$.