# $C^1$ Parameter Transformation. Show function and it's inverse are continuously differentiable.

Let $\gamma:[a,b]\rightarrow \mathbb{R}^N$ be a $C^1$-curve with length $L$.

Show that $$f:[a,b]\rightarrow[0,L], \ \ \ t\mapsto \int_a^t \| \gamma'(s) \|ds$$ is a $C^1$ parameter transformation.

By definition, we define a $C^1$ parameter transformation as a bijective map $g:[a,b]\mapsto[c,d]$, with $a,b,c,d\in \mathbb{R}$, such that $g$ and $g^{-1}$ (inverse of $g$), are continuously differentiable.

Trivially, the derivate of $f$ exists, and the norms on $\mathbb{R}$ are continuous. That is, $\|\gamma'(s)\|$ is continuous, and thus $f$ is continuously differentiable.

This is where I am having trouble. I am not sure how to show the inverse of $f$ is continuously differentiable. This is probably really trivial, but I just don't see where to proceed, and would appreciate any help. Thanks in advance!

Consider any differentiable bijection $g:[a,b]\to[c,d]$. By differentiating the identity $g^{-1} \circ g(t) = t$ and using the product rule, we see that
$\frac{d}{dt}(g^{-1})(t) = \frac{1}{g'(g^{-1}(t))}.$
If $g$ is continuously differentiable, then it follows from this expression that $g^{-1}$ is continuously differentiable. This can also be interpreted as the single-variable special case of the inverse function theorem.
Now take $g = f$ for your problem.
Nothing stops from $\gamma$ being a cosntant on a sub-interval in which case f is not injective.
• My apologies, I forgot to mention $\gamma$ is regular. I figured out the question, but appreciate the answer. – The math god Feb 12 '18 at 8:09