parametrisation of a curve Is it always possible to choose a continuous parametrisation (only continuous) of a piecewise smooth curve to make it smooth? Assume we have this curve in $\mathbb{R}^{n}$.
 A: Let $f$ be a $C^\infty$ increasing function from [0,1] to [0,1] such that $f(0)=0$, $f(1)=1$ and all derivatives vanish at 0 and at 1. Maybe the simplest example is $f(x)=\int_{-\infty}^x g(t)dt$, here $g(t)$ denotes a smooth bump function supported in [0,1] and $\int_{-\infty}^{\infty}g(t)=1$. Maybe a analytic description of g(t) can be given as following:
$g(x)=\frac{1}{c}e^{-1/(x-x^2)}$ for $x\in (0,1)$, and $g(x)=0$ for $x\leq0$ or $x\geq1$, here $c=\int_{0}^1e^{-1/(x-x^2)}dx$.
For any piecewise linear curve $\gamma=\gamma_1+\cdots+\gamma_n$ where $\gamma_i:[a_{i-1}, a_i]\rightarrow \mathbb{R}^n$ is a smooth curve, we define $\gamma^*:[0,n]\rightarrow\mathbb{R}^n$ is defined as follows:
$\gamma^*(i-1+x)=(\gamma_i\circ f)(a_{i-1}+(a_i-a_{i-1})x)$.
Then $\gamma^*$ is well-defined and continuous by the compatibility of $\gamma_i$ at endpoints, and $\gamma^*$ is of course smooth at all non-integer points due to the smoothness of $\gamma_i$. 
For every  integer $i$ satisfying $0\leq i\leq n$, all the left(and right) derivatives (of order 1,2,$\cdots$) vanish by chain rule and the vanishness of all derivatives of $f$. Therefore, $\gamma^*$ is also smooth at every integer point, hence smooth in the whole interval $[0,n]$.
