Let $(M,d)$ and $(N,\rho)$ be metric spaces. A function $f: M \to N$ is Lipschitz if there exists some constant $\kappa \geq 0$ so that $\rho(f(x),f(y))$ is smaller than $\kappa d(x,y)$ for all points $x,y \in M$.

If $f$ and $g$ are two Lipschitz maps from $M$ to $N$, a Lipschitz homotopy between them is a Lipschitz map $H: M \times [0,1] \to N$ for which $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for each $x \in M$. Here we assume that the metric $\Delta$ on $M \times [0,1]$ is given by

$$\Delta\left((x,t),(y,s)\right) = \max\left\{d(x,y),|s-t|\right\}.$$

Here's the question which has been driving me nuts:

Is there an example of metric spaces $M, N$ and Lipschitz functions $f,g:M \to N$ which are homotopic, but not Lipschitz-homotopic?

If there is any justice in the world, the answer is an easy "yes". And in this case, I ask

Does the answer continue to be "yes" if $M$ and $N$ are assumed compact?

  • $\begingroup$ My answer below shows the weak spot of the question. Let's ask now about homotopies such that only $H_t: M\rightarrow N$ maps are assumed to be Lipschitz while $H$ does not have to. Alexandre Eremenko's example will still work (but will be harder to prove). What about the case of $M=N=I$ ? $\endgroup$ – Włodzimierz Holsztyński Jun 30 '13 at 20:49
  • $\begingroup$ If $I$ is the unit interval, then of course $tf(x)+(1-t)g(x)$ will be a Lipschitz homotopy. $\endgroup$ – Alexandre Eremenko Jul 1 '13 at 6:48
  • $\begingroup$ @ Alexandre Eremenko, I meant a custom made metric which still induces Euclidean topology. Then in general your homotopy will not be Lipschitz even in my weaker sense. $\endgroup$ – Włodzimierz Holsztyński Jul 1 '13 at 9:31

There is no "justice" in the world of arbitrary metric spaces:-) Let your first space $M$ be the ordinary circle with its ordinary metric, and the second space $N$ be homeomorphic to a cylinder with two border circles. Let the metric be usual on the border circles, but very rough (bumpy) on the middle part of the cylinder. You can certainly make it so rough that intermediate curves between the border circles will simply have infinite length... And of course you can make this example compact.

  • $\begingroup$ Thanks, Alexandre! This is a nice and simple example that I wish I'd thought of. $\endgroup$ – Vidit Nanda Jun 30 '13 at 16:12
  • $\begingroup$ Alexandre, do you mean an $S^1$ of infinite length? $\endgroup$ – Fernando Muro Jun 30 '13 at 16:47
  • 1
    $\begingroup$ @FernandoMuro Something like a Koch snowflake. $\endgroup$ – Baby Dragon Jul 1 '13 at 4:47
  • $\begingroup$ Hard to imagine a non-rectifiable curve? Look at the pictures of Koch's snowflakes or Julia sets of $z^2+c$ for small $c$ on the Internet. $\endgroup$ – Alexandre Eremenko Jul 1 '13 at 7:37
  • $\begingroup$ Right, bunches of fractals provide examples upon examples. However, below I have provided a very regular example. $\mathbb R$ with a metric which has the same Euclidean isometries, and similarities, ... (and each of its non-degenerated intervals is non-rectifiable). $\endgroup$ – Włodzimierz Holsztyński Jul 1 '13 at 9:50

There are many situations where you can get Lipshitz homotopies, at least with restrictions on the topology of the spaces in question. See the paper "Quantitative algebraic topology and Lipschitz homotopy" by Ferry and Weinberger, available here.

They discuss a number of results and conjectures. One relevant one is the following, which is Theorem 1 in their paper. In it, simplicial complexes are given the path metric obtained by giving each simplex the standard Euclidean metric:

Theorem : Let $Y$ be a finite simplicial complex with finite homotopy groups in dimensions less than or equal to $d$. Then there exists $C(d)$ so that for all simplicial complexes $X$ of dimension less than or equal to $d$, if $f,g : X \rightarrow Y$ are homotopic $L$-Lipschitz maps with $L \geq 1$, then there is a $C(d)L$-Lipschitz homotopy $F$ from $f$ to $g$.

Conversely, if a Lipschitz homotopy always exists, then the homotopy groups of $Y$ are finite in dimensions less than or equal to $d$.

  • 1
    $\begingroup$ Thanks and +1, Andy. This paper by Ferry and Weinberger looks great. $\endgroup$ – Vidit Nanda Jul 1 '13 at 15:01

Let me give the answer in a clear and clean way.

Define $$ \forall_{\,x\ y\in\mathbb R}\ d(x\ y)\ :=\ |x-y|^\frac 12 $$

Thus $\ d(x\ x)=0\ $ and $\ d(x\ y) = d(y\ x)\ $ and

$$ (d(x\ y) + d(y\ z))^2 \ge |x-y|+|y-z| \ge |x-z| = d^2(x\ z) $$

thus $\ d(x\ y)+d(y\ z)\ge d(x\ z);\ $ function $\ d\ $ is a metric in $\mathbb R.\ $ And so is the induced metric in $\ \mathbb J:=[0;1].$

Thus let $\ M:=\{p\}\ $ be a 1-point metric space, and $\ N:=\mathbb J\ $ with metric $\ d\,|\,\mathbb J.$ Next, let the functions $\ f\ g: M\rightarrow N\ $ be given by $\ f(p):= 0\ $ and $\ g(p) := 1.\ $ These two functions are (topologically) homotopic.

Of course my spaces $\ M\ N\ $ are compact. But there is no Lipschytz homotopy as defined in the Question above.


let the Lipschytz constant $\ \kappa\ge 0\ $ be arbitrary. Then $\ \kappa\le n\ $ for a certain natural $\ n.\ $ Consider a homotopy $\ H:M\times[0;1]\rightarrow N\ $ between $\ f\ $ and $\ g.\ $ Thus $\ H(p,\ 0) = 0\ $ and $\ H(p,\ 1) = 1.$

Thus there exists an integer $\ k\ $ such that $\ 1\le k\le n^2\ $ and

$$ \left|H(p,\ \frac{k-1}{n^2}) - H(p,\ \frac k{n^2})\right|\ \ge\ \frac 1{n^2} $$


$$ d\left(H\left(p,\ \frac{k-1}{n^2}\right), \ H\left(p,\ \frac k{n^2}\right)\right)\ \ge\ \frac 1n \ \ge\ \kappa\cdot\left|\frac k{n^2}-\frac{k-1}{n^2}\right| $$


  • $\begingroup$ Thank you for your continued attention to this old question; however, I think the map $H(p,t) = t$ is $1$-Lipschitz (in fact, an isometry!) from $\{p\} \times [0,1]$ to $[0,1]$. $\endgroup$ – Vidit Nanda Oct 18 '16 at 23:33
  • $\begingroup$ @ViditNanda, $H$ is not an isometry, not even Lipschitz. Indeed, the space of arguments, $\ \{p\}\times \mathbb I,\ $ is isometric to the ordinary Euclidean unit interval, while the space of values is interval $\ \mathbb J\ $ which has infinite length. These are metrically two DRASTICALLY different spaces. (If there is still an interest in this topic, I could redefine it and to extend it somewhat). $\endgroup$ – Włodzimierz Holsztyński Oct 19 '16 at 2:57
  • 1
    $\begingroup$ Sorry, I had misunderstood the distance on J. Thanks for the nice example. $\endgroup$ – Vidit Nanda Oct 19 '16 at 3:16

I agree with Eremenko that there is no "justice" in the world of arbitrary metric spaces:-) However, some justice is provided by the following beautiful result of Lang and Schlichenmaier, Theorem 1.5 in [1]. From this result one can conclude some results about Lipschitz homotopy of maps.

Theorem. Suppose that $X$ and $Y$ are metric spaces such that the Nagata dimension of $X$ satisfies $\dim_N X\leq n$ and $Y$ is Lipschitz $(n-1)$-connected. Then there is a constant $C\geq 1$ such that for any closed set $Z\subset X$ and any $L$-Lipschitz map $f:Z\to Y$ there is a $CL$-Lipschitz extension $F:X\to Y$.

The definitions of the Nagata dimension and Lipschitz $k$-connectivity are provided below:

Definition. A metric space $Y$ is Lipschitz $n$-connected for some integer $n\geq 0$ if there is a constant $\gamma\geq 1$ such that for each $k\in \{0,1,\ldots,n\}$, every $L$-Lipschitz map $f:\mathbb{S}^k\to Y$ admits a $\gamma L$-Lipschitz extension $F:\mathbb{B}^{k+1}\to Y$.

Definition. The Nagata dimension $\dim_N X$ of a metric space $X$ is the least integer $n$ with the property that there is $C>0$ such that for any $s>0$, there is a covering $X=\bigcup_{i\in I} A_i$ such that

  • $\operatorname{diam} A_i\leq Cs$ for all $i\in I$;
  • Every ball $\mathbb{B}(x,s)$ intersects at most $n+1$ sets $A_i$.

If no such integer $n$ exists, then $\dim_N X=+\infty$.

The Nagata dimension can be regarded, in some sense, as a quantitative version of the topological dimension.

[1] U. Lang, T. Schlichenmaier, Nagata dimension, quasisymmetric embeddings, and Lipschitz extensions. Int. Math. Res. Not. 2005, no. 58, 3625-3655.


Two examples

(I am not able to select one of them over the other one, sorry):

EXAMPLE 1   Let   $M=N=[-1;1]$   have the same metric, which induces the ordinary topology in   $[-1;1]$,   and such that the length of   $[-1;1]$   is infinite. Let   $f$   be the identity map, and let   $g:=-f$.

EXAMPLE 2   The same, except that   $M:={1}$,   $f(1):=1$,   $g(1):=-1$.

In both cases the image of the length $1$ interval   $\{1\}\times[0;1]$   is   $N$   which has infinite length. This shows that there is no Lipschitz homotopy under the given circumstances.

A distance function in   $\mathbb R$

Every non-degenerated interval in   $\mathbb R$   has infinite length with respect to the distance   $d:\mathbb R^2\rightarrow\mathbb R$   given by:

$$\forall_{x\ y\in\mathbb R}\quad d(x\ y)\ :=\ \sqrt{|x-y|}$$

Distance   $d$   induces the Euclidean topology in   $\mathbb R$. This distance is of course invariant with respect to all translations (in this sense it is   $\mathbb R$-invariant;   and never mind my harmless typo   $\mathbb Z$   in a comment below); its isometry group is simply the same as the group of Euclidean isometries.

  • $\begingroup$ How does $N$ have infinite length? $\endgroup$ – Vidit Nanda Jun 30 '13 at 21:16
  • $\begingroup$ Also, in Example 1, the homotopy $H(x,t)=(2t−1)x$ for $t \in [0,1]$ and $x \in [−1,1]$ appears to have Lipschitz constant $\leq 5$. When estimating $|H(x,t) - H(y,s)|$, we end up with needing an estimate for $\alpha = 2|tx - st|$ which can be easily obtained by adding and subtracting $ty$ from this expression, so $\alpha \leq 2t|x-y| + 2|y||s-t|$. Now note that $|t| < 1 > |y|$. $\endgroup$ – Vidit Nanda Jun 30 '13 at 21:39
  • $\begingroup$ @Vidit, if you believe that any compact parametrized path/curve can have an infinite length, then you must believe that it is possible also for the interval. And if you could disprove my Example 1 (I doubt it :-) then you must have even an easier time disproving Example 2--does, please, concentrate on Example 2 first of all. $\endgroup$ – Włodzimierz Holsztyński Jul 1 '13 at 0:08
  • $\begingroup$ @Vidit, if you still want me to exhibit an explicit metrics in the interval which forces infinite length I'll do it. (Somehow you have accepted a similar thing in the first answer, by Alexandre). $\endgroup$ – Włodzimierz Holsztyński Jul 1 '13 at 0:12
  • $\begingroup$ I might have misunderstood your construction, so let me explain my point of confusion. You would like, in both your examples, for us to assume the existence of a metric on $[-1,1]$ which induces the ordinary topology but results in infinite length. Since the ordinary topology makes $[-1,1]$ compact but infinite length does not, I am not sure what this metric might be. Could you explain? $\endgroup$ – Vidit Nanda Jul 1 '13 at 1:19

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