Extending a local distance preserving map $\gamma:[0,L[\rightarrow X$ to $[0,L]$

Question

Let $L$ be a positive real, $(X,d)$ a complete metric space, and $\gamma:[0,L[\rightarrow X$ be a local distance preserving map(where $[0,L[$ is equipped with the Euclidean metric). Must $\gamma$ be extendable to a local distance preserving map from $[0,L]$ to $X$ ? Can at least one extend gamma continuously to $[0,L]$ ?

Thank you,

Answer 1

Here is a sketch of the construction of a counter-exmaple. Consider the map $\gamma : [0, 1) \to \Bbb{C}$ defined by

$$ \gamma(t) = \frac{\exp(2\pi i \cdot 2^{k+1} t) - 1}{2\pi \cdot 2^{k+1}} \qquad \text{for } t \in [1 - 2^{-k}, 1 - 2^{-k-1}). $$

Its image $X = \gamma([0, 1))$ looks like the following figure:

$\hspace{12em}$

Equip $X$ with the intrinsic metric $d$. Then it is possible to check that $\gamma : I \to X$ is a local distance-preserving map. Intuitively, this is because $\gamma$ is distance-preserving on each circular arc.

Now, since $\gamma(1 - 2^{-k}) = 0$ for each $k$, the path $\gamma$ extends to a continuous function $\tilde{\gamma} : [0, 1] \to X$ with $\tilde{\gamma}(1) = 0$. However, this also shows that $\tilde{\gamma}$ cannot be distance-preserving near $1$.

Answer 2

OK. I will answer this part of the question: Can at least one extend gamma continuously to $[0,L]$ ?

Lemma: For every $s,t\in [0,1]$ we have $d(\gamma(s),\gamma(t))\leq |s-t|. $

Proof: Assume WLOG that $s<t$. By applying Lebesuge number lemma on the compact subspace $[s, t] $ and the local distance preserving property of $\gamma$, let $s_0\leq s_1\leq...\leq s_n$ be an arithmetic sequence of real numbers s.t.$s_0=s,s_{n+1}=t$ and $n$ is chosen large enough so that the restriction of $\gamma$ to $[s_i,s_{i+1}]$ is distance preserving. Thus, we have: $$d(\gamma(s),\gamma(t))=d(\gamma(s_0),\gamma(s_{n+1}))\leq\sum_{i=0}^nd(\gamma(s_i),\gamma(s_{i+1}))=\sum_{i=0}^ns_{i+1}-s_i=s_{n+1}-s_0=t-s $$

$\square$

The sequence $\{\gamma(L(1-2^{-k}))\}_{k\geq 1}$ is Cauchy, because by the above lemma and the famous geometric series trick for showing that sequences are cauchy we have:

$$ d(\gamma(L(1-2^{-k})),\gamma(L(1-2^{-(k+1)}))\leq |L(1-2^{-k})-L(1-2^{-(k+1)})|=\frac{L}{2^{k+1}}$$ By completeness of $X$, extend $\gamma$ to $L$ and define $\gamma(L)$ to be the limit of the above Cauchy sequence. For any $t\in [0,L[, k\in \mathbb{Z}^+$, we have by the lemma:

$$ d(\gamma(L(1-2^{-k})),\gamma(t))\leq |L(1-2^{-k})-t|$$

Let $k$ go to infinity, use continuity of the metric $d$,absolute value, subtraction,..., and that $\gamma(L)$ is by definition $\lim_{k\rightarrow \infty}\gamma(L(1-2^{-k})) $:

$$ d(\gamma(L),\gamma(t))\leq |L-t|$$

The last inequality, along with the lemma tells us that the extension of $\gamma$ is actually Lipschitz therefore continuous.

If I get any ideas about proving that $\gamma$ must be a local distance preserving, I will write it down.