TITLE: Show that $\gamma:[a,b]\rightarrow \mathbb{C}$ has bounded variation iff exists $h:[0,1]\rightarrow [a,b]$ with $\gamma \circ h$ Lipschitz.
QUESTION [6 upvotes]: Let $~\gamma:[a,b]\rightarrow \mathbb{C}~$ be a continuous curve: show that $~\gamma~$ has bounded variation if and only if there exists $~h:[0,1]\rightarrow [a,b]~$ continuous and surjective such that $~\gamma \circ h~$ is a Lipschitz curve.
My only idea is for the proof of the reciprocal proposition: if $$
|\gamma(h(x))-\gamma(h(y))|\leq k|x-y|~\forall x,y\in [0,1],$$
then, given a partition $P=\{a=x_0,x_1,\dots, x_n=b\}$, I have that exists $~y_i~$ with $~h(y_i)=x_i~$ for each $i$ from $1$ to $n$. Then,
$$\sum_{i=1}^n|\gamma(x_i)-\gamma(x_{i-1})|\leq \sum_{i=1}^n|\gamma(h(y_i))-\gamma(h(y_{i-1}))|\leq k\sum_{i=1}^n|y_i-y_{i-1}|$$
But I don't know how to continue, because I think that $h$ is not necessarily monotonic.
Edit: For the direct implication I found a reference that helped me a lot, here goes https://mathoverflow.net/questions/253024/from-bounded-variation-to-1-lipschitz-function

REPLY [3 votes]: The intuition here: $\gamma$ has finite variation if and only if the curve $\gamma$ has finite length. Now the new parametrization is by arc-length, and it should be clear that it is Lipschitz with constant $1$. Works for maps into metric spaces $\gamma\colon [a,b] \to X$, continuous, and of finite variation.