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?

3 Answers 3


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.

  • $\begingroup$ Nice! Do you have any thoughts on sufficient conditions? If not, I'd still happily accept the answer, since this was the main point of my question. $\endgroup$
    – user155731
    Apr 6, 2020 at 17:29
  • $\begingroup$ I'm afraid I don't have anything intelligent to say about sufficient conditions ... I will think about it. $\endgroup$
    – Nik Weaver
    Apr 6, 2020 at 17:43
  • $\begingroup$ I guess for each distinct $s$ and $t$ in the range you want to look at the value ${\rm inf}\{|x_1-y_1|+\cdots+|x_r-y_r|: f(x_1) = s$, $f(y_r) = t$, and $f(y_i) = f(x_{i+1})\}$. If it's ever zero there's no hope, but if not this value describes a metric on the range which makes $f$ a contraction. It shouldn't be hard to use this to get a metric on $\mathbb{R}^m$ that makes $f$ a contraction. $\endgroup$
    – Nik Weaver
    Apr 6, 2020 at 19:38
  • $\begingroup$ Yes, this makes sense. Thanks (to both you and Pietro Majer) for the answers. I'm accepting this one just because it was first and you were kind enough to think about the sufficiency, but Pietro Majer's is also worth accepting! $\endgroup$
    – user155731
    Apr 6, 2020 at 20:11
  • $\begingroup$ You are welcome! $\endgroup$
    – Nik Weaver
    Apr 6, 2020 at 21:48

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 $d(0,1)=0$.

  • $\begingroup$ Thank you, this is also a very straightforward example. $\endgroup$
    – user155731
    Apr 6, 2020 at 20:10
  • 2
    $\begingroup$ So the idea of your example is that a Lipschitz function maps measure-zero sets into measure-zero sets, and whatever metric we choose on the target, the Hausdorff measure of an open set $(0,1)$ has to be positive. Perhaps this can be enhanced somehow to get a sufficient condition? $\endgroup$
    – erz
    Apr 7, 2020 at 2:58
  • $\begingroup$ Exact, that was what I had vaguely in mind, though I could not find the right words :) $\endgroup$ Apr 7, 2020 at 8:38
  • $\begingroup$ @erz Indeed your idea of a sufficient condition in terms of the Hausdorff measure should be the right one. I tried to expand it a bit. $\endgroup$ Apr 9, 2020 at 23:08

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.

  • $\begingroup$ This goes far beyond my little comment. Very interesting! $\endgroup$
    – Nik Weaver
    Apr 10, 2020 at 2:24
  • 1
    $\begingroup$ Could you please elaborate on the extension? As for topological equivalence, here is a proof if $K$ is "nice". Assume $d(y_n,y)\to 0$ and $\varepsilon>0$. Since $f$ is uniformly continuous on $K$, there is $\delta$ such that $|x-z|<\delta\Rightarrow |f(x)-f(z)|<\varepsilon$. For large $n$ $d(y_n,y)<\delta$, and so there are connected $S_n\subset K$ and $x_n,z_n\in S_n$ such that $f(x_n)=y,~ f(z_n)=y_n$ and $|y_n-z_n|\le H^1(S_n)<\delta$, from where $|y_n-y|\to0$. Assume $|y_n-y|\to0$ but $d(y_n,y)\not\to 0$. WLOG $d(y_n,y)>a>0$. Choose $z_n\in f^{-1}(y_n)$. From compactness WLOG $z_n\to z$. $\endgroup$
    – erz
    Apr 12, 2020 at 3:32
  • 1
    $\begingroup$ If $K$ is nice so that we can join $z_n$ and $z$ with short curves, then we get a contradiction with $d(y_n,y)\not\to 0$. So $K$ should be such that the length metric is equivalent with an Euclidean one. If you indeed can extend $f$ to a closed neighborhood of $K$ that would be enough. $\endgroup$
    – erz
    Apr 12, 2020 at 3:35
  • $\begingroup$ I'd like to add that this condition is necessary to existence of metric on $f(K)$ such that $f$ is Lipschitz, and also the coordinate projections on $f(K)$ are Lipschitz. Perhaps it is also sufficient. What I am saying is that the condition depends on the Euclidean metric on the image, and is not purely topological, as one would like to. $\endgroup$
    – erz
    Apr 12, 2020 at 4:03
  • $\begingroup$ @erz very good. For the issue of the extension, one can make it locally Lipschitz, or even smooth, on the complement of $K$: isn't this sufficient? For any $H^1$-null set $S$ in the domain, $f(S\cap K)$ is $H^1$-null by hypothesis, and since $S\setminus K$ is covered by countably many balls on which $f$ is Lipschitz, the set $f(S\setminus K)$ is $H^1$-null as well. $\endgroup$ Apr 12, 2020 at 20:04

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