Is every path a reparametrization of a path with non zero velocity?

Question

Let $\alpha:[0,1] \to \mathbb{R}^n$ be a non-constant smooth path, and suppose $\alpha'(0)=0$. Do there exist a smooth path $\beta:[a,b]\to \mathbb{R}^n$ and a smooth increasing function $h:[0,1] \to [a,b]$,

such that $\alpha=\beta \circ h,h(0)=0$ and $\beta'(0) \neq 0$. Note $h$ must satisfy $h'(0)=0$.

Example: $n=1,\alpha(t)=t^2$, take $h(t)=\alpha(t)$, $\beta=\text{Id}$.

It's easy to shows that if the answer is positive for $n=1$, then it is positive for every dimension $n$. So we are reduced to the one-dimensional case.

Answer 1

Not necessarily. For a $C^\infty$ counter-example take $\alpha(0)=(0,0)$ and: $$ \alpha(t) = \left( e^{-1/t}\cos(1/t), e^{-1/t}\sin(1/t) \right), \;\; 0<t\leq 1 $$ It spirals around the origin infinitely many times as $t\rightarrow 0$. Whatever the reparametrization, the MVT implies that the direction $\beta'(t)/|\beta'(t)|$ must rotate infinitely many times as $t\rightarrow 0$ which forces $\beta'(0)=(0,0)$ if smooth.

If $\alpha$ is analytic then the claim is partially true. You must have $$\alpha(t) = (at^n, bt^m) + {\rm (l.o.t)}$$ with say, $n\leq m$ and $a,b$ non-zero. Setting $s=h(t)=t^n$ and $$\beta(s)=(as, bs^{m/n}) + {\rm (l.o.t)}$$ you get a $C^1$ path for $\beta$. The smoothness of $\beta$ depends upon ratios of powers (also further away in the series). For example, with $n=2,m=3$ $\beta$ is $C^1$ but not $C^2$.

Answer 2

Let $n=1$. We have for all $x\in(0,1)$ that $\alpha'(x)\neq 0$ (because $\alpha$ is a path), and so $\alpha$ is strictly monotonous. Hence choosing $\beta=\text{Id}$ and $h=\alpha$ suffice.
$\ $

Keep $n=1$. If instead we let $\alpha$ be a smooth, nowhere constant function on $(0,1)$ which is not a path, then there exists $x_0\in(0,1)$ such that $\alpha'(x_0)=0$.

Assume that $\alpha$ equals the reparametrization $h$ of $\beta$, where $\beta$ has non-zero velocity.
That is $\alpha(x)=\beta(h(x))$ and $\beta'(x)\neq 0$.
Hence $h'(x_0)=0$ since $0=\alpha'(x_0) = h'(x_0) \beta '(h(x_0))$.
Also $h''(x_0)=0$ because otherwise $x_0$ would be an extramum contradicting the monotonicity of $h$.
Differentiating once more we get: $$\alpha''(x_0) = h''(x_0) \beta '(h(x_0))+h'(x_0)^2 \beta ''(h(x_0)) \\ \alpha''(x_0) = 0\hspace{6.61cm}$$

So a reparametrization exists only if $\alpha''(x) = 0$ whenever $\alpha'(x)=0$.

If we continue differentiating we see that if in addition $\alpha'''(x)=0$ then $\alpha''''(x)=0$ must hold, etc... Hence one of the following must hold

1. For all $x$ the first non-zero derivative must be an odd order derivative.

2. For those $x$ where 1 doesn't hold, all derivatives of $\alpha$ must be equal to $0$.

A reparametrization may exist only if this condition is satisfied by $\alpha$.