# $C^1$ 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