Given $(X,d)$ metric space we define the length of a curve as follows :
$$l(\gamma, [a,b])=\sup\limits\limits_{P \in \mathbb{P}([a,b])}l(\gamma,P)$$
($\mathbb{P}[a,b]$ is the set of all possible partition of $[a,b]$)
Where for every partition $P \in \mathbb{P}([a,b])$, $l(\gamma, P)=\sum\limits_{k=1}^n d(\gamma(x_k),\gamma(x_{k+1}))$
I would prove that this last one is continuous as operator from $C^0([a,b],X)$ to $\mathbb{R}$.
Trying with the definition didn't see how to end : Let $\gamma_{1},\gamma_{2}$ be such that given $P = (x_{i})_{i=1,\cdots,n+1}$ with $d_{\infty}(\gamma_{1},\gamma_{2}) < \delta$ I'd like to state that $$|l(\gamma_{1},P)-l(\gamma_{2},P)| \leq \sum\limits_{k=1}^{n} |d(\gamma_{1}(x_{k}),\gamma_{1}(x_{k+1}))-d(\gamma_{2}(x_{k}),\gamma_{2}(x_{k+1}))| < \epsilon$$ but I got stuck since triangle's inequality didn't see to bring me somewhere.
Any help, solution or hint would be appreciated.
I don't think the map is continuous. Consider $X=\mathbb R^2$ and compare two paths, $\gamma_1$ and $\gamma_{n}$ both going from the origin $(0,0)$ to $(0,1)$.
Make $\gamma_1$ the straight line with length $\ell(\gamma_1) = 1$. Take $\gamma_{n}$ to be a right-angled saw tooth with $n$ teeth, spaced $1/n$ apart and of side $\frac{1}{n\sqrt 2}$. This can be formalised is you like, but by construction, $$\ell(\gamma_n) = n\cdot \frac{\sqrt 2 }{n} = \sqrt 2.$$ But $\lVert \gamma_1 - \gamma_n\rVert_\infty = 1/(2n)$ and for $n$ large enough this can be made as small as we like but the lengths of the two curves remain stubbornly apart.