Recognizing Lipschitz functions up to change of target metric

Question

Let $K$ be a compact subset of $\mathbb{R}^n$ (for simplicity, I am happy to take $K=\overline{B(0,1)}$ for now if it is easier).

Let $f:K \rightarrow \mathbb{R}^m$ be a continuous function.

Is there a new metric $d$ on $\mathbb{R}^m$ (compatible with the topology) such that $$ f: K \rightarrow (\mathbb{R}^m,d) $$ is Lipschitz?

(Here $K$ is still equipped with the usual Euclidean metric, and only the metric on the target space $\mathbb{R}^m$ has changed.)

If not,

• What are obstructions?
• Are there sufficient conditions?

Answer 1

It can't be done. Here's a counterexample with $n = 2$ and $m = 1$.

For each $k \in \mathbb{N}$, define a function $f_k: [0,1] \to [0,1]$ by linearly interpolating between the values $f_k(\frac{i}{k}) = \frac{i}{k}$ ($0 \leq i \leq k$) and $f_k(\frac{i}{k} + \frac{1}{k^2}) = \frac{i+1}{k}$ ($0 \leq i \leq k -1$). They look like staircases.

I claim there is no metric on the target $[0,1]$ that makes all of the $f_k$ contractions. If $f_k$ is a contraction this forces the distance from $\frac{i}{k}$ to $\frac{i+1}{k}$ to be at most $\frac{1}{k^2}$, and thus the distance from $0$ to $1$ to be at most $\frac{1}{k}$. If this were true for all $k$ then the distance from $0$ to $1$ would have to be zero and the metric would not be compatible with the topology.

Now define $f(t,\frac{1}{k}) = f_k(t)$ for all $k$ and $f(t,0) = t$. Then $f$ is a continuous function defined on a compact subset of $[0,1]^2$, so by Tietze it extends to a continuous function from the unit square to $[0,1]$. If there were a change of metric on the target $[0,1]$ that made $f$ Lipschitz then by scaling there would be a change of metric that made $f$ a contraction, and this would make each $f_k$ a contraction, contradicting the claim.

Answer 2

For instance, no equivalent distance on $\mathbb{R}^1$ can make Lipschitz the Cantor function $f:[0,1]\to\mathbb{R}^1$. Suppose by contradiction $d$ is a distance that makes $f$ $L$-Lip.

Let $F_n$ be the closed sets in the usual construction of the cantor set $C=\bigcap_{n\ge\mathbb{N}}F_n$, that is $F_0=[0,1]$ and $F_{n+1}=\frac{1}{3}F_n \cup (\frac{1}{3}F_n+ \frac{2}{3})$.

If $0=x_1<\dots<x_{2^n}=1$ are the endpoints of the components of $F_n$, then $x_{2i}-x_{2i-1}=\Big(\frac{2}{3}\Big)^n$ and $f(x_{2i})=f(x_{2i+1})$ because $f$ is constant on the connected components of $F_n^c$. But then for all $n$ $$d(1,0)=d(f(0),f(1))\le L\sum_{i=1}^{2^n}|x_i-x_{i-1}|=L \Big(\frac{2}{3}\Big)^n$$ so $d(0,1)=0$.

Answer 3

Here are some suggestions trying to expand erz's and Nik Weaver's comments. I do not have time to work out the details (although stuck at home arrests like many of us), but I'd be glad if nevertheless they turn out to be useful.

Recall that a set $S\subset\mathbb{R}^m$ is $\mathcal{H}^1$- null iff for all $\epsilon>0$ it has a countable partition $\{S_j\}_{j\in\mathbb{N}}$ such that $\sum_{j\in\mathbb{N}}\rm{diam}(S_j)\le\epsilon$.

Say the that a map $f:A\subset\mathbb{R}^m\to\mathbb{R}^n$ is $\mathcal{H}^1$- Absolutely Continuous iff it is continuous and for any $\mathcal{H}^1$-null set $S\subset A$, the image $f(S)$ is $\mathcal{H}^1$-null in $\mathbb{R}^n$ .

[edit 4.24.20]: as user erz pointed out to me, this definition is too weak (the function in Nik Weaver's counterexample satisfies it). A more suitable definition is maybe: iff it is continuous and for any $\epsilon>0$ there exists $\delta>0$ such that for any $S\subset A$, $\mathcal{H}^1(S)<\delta$ implies that $\mathcal{H}^1 (f(S))<\epsilon$ .

The conjecture should be:

Any $\mathcal{H}^1$-AC map defined on a compact subset $K$ is Lipschitz up to a choice of an equivalent distance on the target space.

We may always assume $n=m$ (via inclusion $K\subset \mathbb{R}^n\subset\mathbb{R}^m$, or $\mathbb{R}^m\subset\mathbb{R}^n$ if needed). Then

• It does not seem problematic to extend an $\mathcal{H}^1$-absolutely continuous map $f$ defined on a compact $K$ to a $\mathcal{H}^1$-absolutely continuous map $f:\mathbb{R}^n\to\mathbb{R}^n$ that is also a surjective and proper map (it should sufficient to make it locally lipschitz on the complement of $K$, and equal to the identity outside a large ball.

• For $x,y$ in $\mathbb{R}^n$ define

$$d(x,y):=\inf\{\mathcal{H}^1(S): f(S)\,{ \rm connected},\, \{x,y\}\subset f(S)\}$$

• Then $d:\mathbb{R}^n\times \mathbb{R}^n\to[0,\infty)$ is clearly symmetric, and satisfies the triangular inequality.

• To show $d(x,y)=0$ implies $x=y$, is where the hypotheses enter. The idea should be that $d(x,y)=0$ implies the existence of a $\mathcal{H}$-null subset $S$ such that $f(S)$ is connected and $\{x,y\}\subset f(S)$, which forces $x=y$ since $f(S)$ is also $\mathcal{H}$-null. To this end one should start from a minimizing bounded sequence of compact subsets $S_j$ with $\mathcal{H}^1(S_j)\to0$, and $\{x,y\}\subset f(S_j)$. Compact subsets of a given compact are a compact in the Hausdorff distance, the Hausdorff measure is lower semicontinuous, connected sets are a closed set, so I think it could be done. In fact, it seems OK that the infimum in the definition of this distance be always attained, by the same argument.

• By definition of $d$, taking as $S$ the segment $[u,v]$ one has $d(f(u),f(v))\le \mathcal{H}^1([u,v])= \|u-v\|$.

• It remains to show that $d$ is topologically equivalent to the Euclidean distance, which seems true, although I see it less clearly at the moment.