Show that $\gamma:[a,b]\rightarrow \mathbb{C}$ has bounded variation iff exists $h:[0,1]\rightarrow [a,b]$ with $\gamma \circ h$ Lipschitz. 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
 A: 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.
A: Your attempt works if $h$ is nondecreasing. I think my argument here shows that you can use $h$ to construct a nondecreasing function $g : [0,1] \to [a,b]$ such that $\gamma \circ g$ is also Lipschitz, after which you can apply your argument. (Please comment if you find errors in my argument.)

WLOG we may assume there exists $0 \le z_0 < z_1 \le 1$ such that $h(z_0)=a$ and $h(z_1)=b$. (If not, replace $h$ with $\tilde{h}(x)=h(1-x)$.)
Let $g:[z_0,1] \to [a,b]$ be defined by $g(x) = \sup_{t \in [z_0, x]} h(t)$. Show that $g$ is continuous, nondecreasing, and surjective onto $[a,b]$.
$\gamma$ is uniformly continuous on $[a,b]$, so there exists $\delta$ such that $|u-u'| \le \delta$ implies $|\gamma(u)-\gamma(u')| \le \epsilon$.
I claim that for $z_0 \le x < y \le 1$, there exist $x',y'$ satisfying $x \le x' \le y' \le y$ such that
$h(x') = g(x)$ and $|g(y)-h(y')| \le \delta$.

Proof of claim: If $g(y) > g(x)$, choose $y' \le y$ such that $|g(y) - h(y')| \le \min\{\delta, \frac{g(y)-g(x)}{2}\}$, by definition of $g(y)$. We must have $y' \ge x$ because $h(y') \ge g(y) - \frac{g(y)-g(x)}{2} =  \frac{g(y) + g(x)}{2} \ge g(x) = \sup_{[z_0, x]} h(t)$. Next, because $h(x) \le g(x) \le h(y')$, the intermediate value theorem implies the existence of some $x'$ between $x$ and $y'$ such that $h(x') = g(x)$.

By uniform continuity of $\gamma$, we have $$|\gamma(g(x)) - \gamma(g(y))| \le \epsilon + |\gamma(h(x'))-\gamma(h(y'))| \le \epsilon + k|x'-y'| \le \epsilon + k|x-y|.$$
Since $\epsilon$ is arbitrary, we see that $\gamma \circ g$ is also Lipschitz.
Finally, if we define the linear map $\tilde{g} : [0,1] \to [z_0, 1]$ by $\tilde{g}(t) = z_0 + t(1-z_0)$, you can note that $(g \circ \tilde{g}) : [0,1] \to [a,b]$ is your continuous nondecreasing surjection such that $\gamma \circ (g \circ \tilde{g})$ is Lipschitz.
