Let $F: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function that is coercive, that is, $\lim\_{||x|| \to \infty} F(x) = +\infty$.


Given any rectifable curve $c: [0, 1] \to \mathbb R^n$, define the $F$-arc length of the curve, $A(F, c)$ as $\sup \sum\_{i=1}^n |F(c(x\_i) - F(c(x\_{i-1}))|$, where the sup is taken over all points $x\_i \in [0, 1]$ such that $0 = x\_0 < .. < x\_n = 1$.


Given any such $F$ and two points $a, b$ in $\mathbb R^n$, does there always exist a “geodesic” curve $c\_0$ such that $A(F, c\_0) = \inf A(F, c)$? Where the infimum is taken over all rectifiable curves such that $c(0) = a$ and $c(1) = b$.


If not, what extra conditions on $F$ need to be imposed for this to hold true?



Question on existence of “geodesic” curves
I assume you want a *continuous* minimiser $c:[0,1]\to\mathbb{R}^n$ (otherwise $c\_0:=a\chi\_{[0,1/2)}+b\chi\_{[1/2,1]}$ is a trivial solution with $A(F,c\_0)=|F(b)-F(a)|$).


In general the infimum is not attained by a continuous curve. Consider e.g. $n=2$ and let $F$ be the distance function from the [topologist sine curve](https://en.wikipedia.org/wiki/Topologist%27s_sine_curve) $\Gamma$. Let $a:=(0,0)\in\Gamma$ and $b:=(1/\pi,0)\in\Gamma$. For any $\epsilon>0$ you can join $a$ an $b$ with a continuous curve 
$c\_\epsilon$ with $A(F,c\_\epsilon)\le\epsilon$: a straight horizontal segment from $a$ to an $\epsilon$-close point on the graph of $\sin(1/x)$ on $\mathbb{R}\_+$ (e.g. $(1/N\pi,0)$ with $N>1/\pi\epsilon$) , juxtaposed to the arc on $\Gamma$ connecting this point to $b$ . Since $F$ is $1$-Lipschitz, $A(F,c\_\epsilon)\le\epsilon$. On the other hand, any continuous $c:[0,1]\to\mathbb{R}^2$ joining $a$ and $b$ can't be included in $\Gamma$, which forces $A(F,c)>0$.


