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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?

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    $\begingroup$ Here is the proof strategy I would try: Let $M=A(F,t\rightarrow at+(1-t)b)$. Let $S$ be the space of differentiable curves $\gamma$ from $a$ to $b$ with $\max(F(\gamma))<\max(F(a),F(b))+M$, with the Sobolev norm $W^{1,n}$. Then (I think) the closure of $S$ is a compact subset of the rectifiable curves, and $A$ is a continuous function on it, so $A$ has a minimum on that set which is the desired $c_0$. $\endgroup$
    – user44143
    Commented Jul 21, 2019 at 1:37

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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 $\Gamma$. 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 $A(F,c)>0$.

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  • $\begingroup$ What I think is true in your hypotheses is perhaps that there exists a (continuous) minimiser connecting the connected component of the level set of $a$ to the the connected component of the level set of $b$ (in the above example, $a$ and $b$ are already in the same connected component of $F^{-1}(0)$). This should also be true for the analogous problem of connecting fivers of an $F:\mathbb{R}^n\to\mathbb{R}^k$. $\endgroup$
    – Pietro Majer
    Commented Jul 21, 2019 at 16:33

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