Let $f:\mathbb{R}^n\to \mathbb{R}^{n-1}$ and $K\subseteq \mathbb{R}^n$ be a set of positive Lebesgue measure. What kind of regularity do we have to impose on $f$ (e.g., $C^1$, Lipschitz) to conclude that $f$ cannot be one-to-one on $K$?

Continuity is (in general) not enough, as demonstrated here.

On the other hand, a nonvanishing Jacobian on a subset of $K$ of positive measure allows us to construct a contradiction by the coarea formula.

But what if we cannot assume anything about the Jacobian? Is, e.g., Lipschitz continuous sufficient to construct a contradiction? Or do there exist Lipschitz continuous examples of one-to-one mappings?

Edit: This seems to have a connection to singularity theory. Unfortunately, things like Sard's Theorem also don't help as it only tells me something about singular values but I would need some information about the possible size of singular points of a one-to-one mapping.

  • $\begingroup$ I don't have a proof off the top of my head, and I gotta go now, but I would think that you can't do it with Lipschitz maps (remember Lip maps have derivatives a.e., and a Sard's theorem holds, so from a geometric measure theory point of you, they are not too far from being $C^1$). My comment is useless, I am aware. $\endgroup$ – Silvia Ghinassi Nov 16 '15 at 15:38
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    $\begingroup$ Lipschitz should be sufficient, as the coarea formula is true for Lipschitz maps (see e. g. here) $\endgroup$ – martini Nov 16 '15 at 16:24
  • $\begingroup$ The coarea formula holds, but if I don't know that the Jacobian does not vanish a.e. I was not able to construct a contradiction using it. Any ideas? $\endgroup$ – KoliG Nov 16 '15 at 16:32
  • $\begingroup$ I don't think i'm wrong by saying that a lipschitz function which jacobian vanish almost everywhere is constant. It is true in 1D because lipschitz functions are absolutely continuous. This adapt easily in any convex subset of $\mathbf R^n$ . $\endgroup$ – Renart Feb 25 '16 at 14:23
  • $\begingroup$ Another thing : if a map is bi-lipschitz (ie $a|x-y|<|f(x)-f(y)|<b|x-y|$ for some $0<a<b$) then it preserves the Hausdorff dimension. So for bi-lipschitz maps you can even strenghten the assumptions on $K$ : If the Hausdorff dimension of $K$ is strictly greater than $n-1$ there is no bi-lipschitz functions from $K$ to $\mathbb R^{n-1}$ (a bi-lipschitz function is automatically bijective). $\endgroup$ – Renart Feb 25 '16 at 14:27

Lipschitz is not sufficient (so, since Lipschitz maps are $C^1$ on the complement of a small set, $C^1$ is not sufficient too). I will build an example of an injective map for $n=2$, which clearly gives an example for any $n\ge 2$.

Let $C$ be the perfect set obtained in this way: start from $C_0:=[0,1]$, then remove a centered open interval with length $\frac{1}{2}$ obtaining $C_2:=\left[0,\frac{1}{4}\right]\cup\left[\frac{3}{4},1\right]$, then, from the remaining two intervals, remove two centered open intervals whose length is $\frac{1}{8}$, etc. It is analogous to the usual construction of the Cantor set, but the proportions are $\left(\frac{1}{4},\frac{1}{2},\frac{1}{4}\right)$. So for example at the second step we have a union of $4$ intervals: $$ C_2:=\left[0,\frac{1}{16}\right]\cup\left[\frac{3}{16},\frac{1}{4}\right]\cup\left[\frac{3}{4},\frac{13}{16}\right]\cup\left[\frac{15}{16},1\right]. $$ $C$ is defined as the intersection of the compact sets obtained at each step, i.e. $C:=\cap_i C_i$.

It is easy to see that the orthogonal projection $\pi:(C\times C)\setminus (V\times V)\to L$ is injective, where $L$ is the line $\{(x,y):y=2x\}$ and $V$ is the countable set of the endpoints of the several intervals appearing in the construction of $C$ (i.e. $V:=\{0,1,\frac{1}{4},\frac{3}{4},\dots\}$). But $|C\times C|=0$, so we need to modify this construction a little. It suffices to prove the following lemma: then we take $f:=\pi\circ(g\times g):A\times A\to L$ and we identify $L$ with $\mathbb{R}$.

Lemma. There exists an injective Lipschitz map $g:A\to C$, where $A\subset\mathbb{R}$ is a compact set having positive Lebesgue measure.

Proof. Start from $[0,2]$ and repeat the same construction as above, by creating holes having the same sizes as in $C$ (but not the same proportions!). So at the first step $A_0:=[0,2]$ becomes $A_1:=\left[0,\frac{3}{4}\right]\cup\left[\frac{5}{4},2\right]$, then at the second step $$ A_2:=\left[0,\frac{5}{16}\right]\cup\left[\frac{7}{16},\frac{3}{4}\right]\cup\left[\frac{5}{4},\frac{25}{16}\right]\cup\left[\frac{27}{16},2\right], $$ and so on. Define $A:=\cap_i A_i$. Clearly $|A|=1$. We now build the required map $g$. Take $$h:=1_{\mathbb{R}\setminus A},\quad g(x):=\int_0^x h(t)\,dt.$$ $g:\mathbb{R}\to\mathbb{R}$ is $1$-Lipschitz and strictly increasing (since $A$ has empty interior). Moreover we claim that $g(A)=C$. Since $g$ is bijective, it suffices to prove that $g(A_i)=C_i$.

Let us check it for example when $i=0,1$ (the general case is analogous). In fact $g(0)=0$ and $g(2)=1$, so $g(A_0)=C_0$. Moreover $g\left(\frac{5}{4}\right)-g\left(\frac{3}{4}\right)=\frac{1}{2}$ and $g\left(\frac{3}{4}\right)-g(0)=g(1)-g\left(\frac{5}{4}\right)$, so $g\left(\frac{3}{4}\right)=\frac{1}{4}$ and $g\left(\frac{5}{4}\right)=\frac{3}{4}$, thus $g(A_1)=C_1$. The general case really boils down simply to the fact that $A$ and $C$ have holes with the same sizes by construction. $\blacksquare$

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Kaufman [1] proved that there exists a surjection $f\in\mathcal{C}^1((0,1)^{n+1},(0,1)^n)$ such that $\operatorname{rank}(Df)\leq 1$ for all $x\in (0,1)^{n+1}$. Therefore I suppose you need less then $\mathcal{C}^1$, and probably less than Lipschtiz, but this is not trivial since Whitney is not a.e.

Moreover, there is a related result for Sobolev functions by Wildrick and Zürcher [2]. Their Theorem 1.5 states that for any $\epsilon > 0$, there is a compact Ahlfors $2$-regular metric space $X$ which supports a $(2 + \epsilon)$-Poincaré inequality, and is such that for any $1 \le p < 2 +\epsilon$, there is a continuous surjection $f : X \to Y$ onto any length-compact metric space $Y$ that is constant off a set of finite measure and has an upper gradient in the space $L^p(X)$.


[1] R. Kaufman. A singular map of a cube onto a square. J. Differential Geom. 14 (1979), no. 4, 593–594 (1981)
[2] K. Wildrick, T. Zürcher. Space filling with metric measure spaces. Math. Z. 270 (2012), no. 1-2, 103–131.

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  • $\begingroup$ obviously bilipschitz condition is related to invertibility and not to regularity. $\endgroup$ – Diesirae92 Mar 1 '16 at 1:01
  • $\begingroup$ Of course. Sorry for spelling it wrong $\endgroup$ – Diesirae92 Mar 1 '16 at 9:05

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