When is a curve the shortest curve in its own neighborhood?

Question

Consider a parametric curve with bounded curvature, $f: [0,1] \rightarrow \mathbb{R}^n$ and the set of rectifiable curves within distance $\epsilon$ of $f$ under the max norm.

Intuitively, the inf of the arc length of curves in that set should converge to the arc length of $f$ as $\epsilon$ goes to 0.

Is that indeed true, and is there a (non trivially) looser condition we can put on $f$ where this remains true?

Answer 1

Yes.

Suppose $X$ is a metric space, $f:[0,1]\to X$ is rectifiable, and $\epsilon>0$. There exists $\delta>0$ such that if $g:[0,1]\to X$ satisfies $d(f(t), g(t))<\delta$ for all $t$ then $L(g)>L(f)-\epsilon$.

Proof: By definition there exist $0=t_0<\dots<t_n=1$ with $$\sum d(f(t_j),f(t_{j-1})|>L(f)-\epsilon/2.$$ If $d(f(t),g(t))<\frac\epsilon{4n}$ for every $t$ then the triangle inequality shows that $$L(g)\ge\sum d(g(t_j)-g(t_{j-1}))>\sum d(f(t_j),f(t_{j-1}))-\epsilon/2.$$

Regarding looser conditions: I don't think so. If $f$ is not rectifiable then $L(g)>L(f)-\epsilon$ says $L(g)>\infty$, silly. Of course in that case the $\epsilon$ should not be there. A revised version without rectifiability:

Suppose $X$ is a metric space, $f:[0,1]\to X$ is continuous, and $\alpha<L(f)$. There exists $\delta>0$ so that if $d(f(t),g(t))<\delta$ for every $t$ then $L(g)>\alpha$.

(Here the point is that in general $L(\gamma)\in[0,\infty]$.)

The proof is more or less the same.

I suppose this might count as a similar result under weaker hypotheses; if $f$ is not rectifiable it says you can make $L(g)$ arbitrarily large by taking $g$ close enough to $f$.

A one-line version:

$L$ is a semicontinuous map (I never remember which is which, sorry) from $C([0,1])$ to $[0,\infty]$.