Show that if $c : \mathbb R → \mathbb R^n$ is a parametrisation by arc-length of a closed curve, then c is periodic.
My attempt:- I know the definition of periodic curve as A parametrised curve $c : \mathbb R → \mathbb R^n$ is called periodic with period $L$ if for all $t ∈ \mathbb R$ we have $c(t + L) = c(t), L > 0,$ and there is no $0 < L' < L$ such that $c(t + L' ) = c(t)$ for all $t ∈ \mathbb R$ as well. A curve is called closed if it has a periodic regular parametrisation.
I know here Period will be arc length($S$) of the curve. Let $c:[0,S]\to \mathbb R^n$ represent the closed curve. $c(s+S)=c(\int_0^s||\dot{c}(t)||dt+ S)$. I don't know How to proceed further. Please help me.
Suppose you have an arc-length parametrization $c(s)$ of a regular closed curve. Being closed, according to your definition, means that there is a regular parametrization $\gamma:\mathbb{R}\to \mathbb{R}^n$ which is periodic, $\gamma(t+L)=\gamma(t)$ for any $t$ ($L$ is the principal period, as in your definition). Choose $t_0$ such that $\gamma(t_0)=c(0)$. By regularity, $\|\gamma'(t)\|>0$ and then $$ s(t)=\int\limits_{t_0}^t\|\gamma'(r)\|\,dr $$ is a strictly increasing function, with $s(t_0+L)=S$ (total length of the curve) and $s(t)=at+f(t)$, where $a=S/L$ and $f$ is periodic with period $L$ (this is a general fact: the integral of a periodic function is linear + periodic, you can easily prove this yourself). Therefore, for any $s=s(t)$, $$ c(s+S)=\gamma(t+L)=\gamma(t)=c(s), $$ thus proving the claim.