# A continuous bi-Lipschitz shrinking of a domain into a compact subset

Let $$\Omega \subset \mathbb{R}^n$$ be a bounded domain. My main problem/question is:

(1) Show there exist a sequence of bi-Lipschitz (i.e injective Lipschitz function with Lipschitz inverse) maps $$F_n : \overline{\Omega} \to F_n(\overline{\Omega})$$ with the image $$F_n(\overline{\Omega})$$ compactly contained in $$\Omega$$, and such that $$\lim_{n\to \infty} F_n(x) = x$$ for all $$x \in \overline{\Omega}$$.

This will likely require the addition of assumptions on (the boundary of) $$\Omega$$.

I have a solution which involves adding the assumption that the boundary of $$\Omega$$ be the zero set of a $$C^2$$ (actually I think $$C^{1,1}$$ will suffice) function $$G$$ and letting $$F_n$$ be the flow (say to time $$1/n$$) of the ODE $$X' = -\nabla G(X)$$. This leads to a few more questions, such as;

(2) Has this been done before? (it must have been, so I guess I'm asking for a source)

(3) Is the existence of $$G$$ implied if I impose $$\Omega$$ to have a $$C^2$$ boundary? (I may ask this in a separate post)

[Edit: It has been pointed out that $$\nabla G$$ must be non-vanishing on $$\partial\Omega$$.]

I have a weaker proof for (1) which adds only the assumption that $$\Omega$$ be star-shaped (say at the origin). I then set $$F_n(x) = (1-\frac{1}{n})x$$.

Any and all other ideas, new ideas, references, modifications/improvements to mine, and improved generality in assumptions is appreciated, thank you all.

• I don't think any assumptions on $\Omega$ should be needed. I think, for instance, that you could choose $\mu_n$ to be supported on a finite set of points in $\Omega$. Partition $\overline{\Omega}$ into a finite number of disjoint Borel sets $A_i$, $i=1,\dots, k$ of diameter less than $1/n$, each containing at least one point $x_i$ of $\Omega$. Then let $\mu_n$ put mass $\mu(A_i)$ at the point $x_i$. I think such $\mu_n$ should converge vaguely to $\mu$. Note that your test functions $\varphi$ are uniformly continuous. – Nate Eldredge May 10 at 5:40
• For the existence of $G$, regularised distance might be a good search term; see, for example, here for construction and basic properties, and here for further developments. – Mateusz Kwaśnicki May 10 at 10:42
• I'm wondering about topological obstructions. Say $\Omega$ is a punctured disk in $\mathbb{R}^2$. Then $\overline{\Omega}$ is a closed disk, and it's contractible, so your map $F_n$ can't just "enlarge the hole". You have to map the whole thing to a contractible blob that doesn't surround the missing point, and I'm having a hard time visualizing how those could converge to the identity. – Nate Eldredge May 10 at 15:54
• In particular, this suggests that your approach with $G$ has a gap. If $\Omega$ is a punctured disk, then $\partial \Omega$ is the zero set of the $C^\infty$ function $G(x) = |x|^2(1-|x|^2)$. But $\nabla G$ vanishes at the origin, and so your flows will just fix 0 instead of mapping it inside $\Omega$. You'd really need a $G$ whose gradient doesn't vanish on $\partial \Omega$, and the implicit function theorem says you can't have that at the puncture. – Nate Eldredge May 10 at 16:06
• If $\Omega$ is $C^{1,1}$, then also the signed distance to the boundary is $C^{1,1}$ (near the boundary), and, appropriately regularised away from the boundary, can serve as the function $G$: the gradient flow does the job in this case. However, my feeling is that $C^{1,1}$ is way too much, the same should be true for $C^{0,\alpha}$ domains for any $\alpha > 0$. If I find time (and if this is what you are looking for), I will try to sketch the argument. – Mateusz Kwaśnicki May 10 at 17:26

Here is a sketch of the argument for $$C^{0,\alpha}$$ domains.

1. If $$f$$ is a $$C^{0,\alpha}$$ function and $$\Omega$$ is the region above the graph of $$f$$ then for every $$r > 0$$ the function $$x \mapsto x + (0, \ldots, 0, r)$$ is a diffeomorphism of $$\mathbb{R}^N$$ which maps $$\overline{\Omega}$$ into $$\Omega$$.

2. If $$\Omega$$ is a bounded $$C^{0,\alpha}$$ domain, then there is a finite collection of balls $$B(x_i, r_i)$$ which cover $$\overline{\Omega}$$, and such that for each $$i$$, either $$x_i$$ is far away from the boundary, or $$x_i$$ lies on the boundary of $$\Omega$$ and $$\Omega$$ near $$x_i$$ looks like a region above a graph of a $$C^{0,\alpha}$$ function. More precisely, we assume that either

(a) $$B(x_i, 2 r_i) \subseteq \Omega$$, or

(b) $$x_i \in \partial \Omega$$, and for some $$C^{0,\alpha}$$ function $$f_i$$ and an isometry $$O_i$$ of $$\mathbb{R}^N$$, we have $$\Omega \cap B(x_i, 2 r_i) = O_i(\Omega_i),$$ with $$\Omega_i = \{x : x_N \ge f_i(x_1, \ldots, x_{N-1}\} \cap B(0, 2 r_i)\} .$$

3. We fix a smooth partition of unity $$\rho_i$$ on $$\overline{\Omega}$$ (extended smoothly to all of $$\mathbb{R}^N$$) in such a way that $$\rho_i$$ is supported in $$B(x_i, r_i)$$.

4. We fix a small $$r > 0$$. For each $$i$$ we define $$\phi_i(x) = x$$ for $$i$$ corresponding to case (a), and we let $$\phi_i$$ to be a local version of the "shift away from the boundary" from point 1 when $$i$$ corresponds to case (b). More precisely, in the latter case we define $$v_i = O((0, \ldots, 0, r)) - O((0, \ldots, 0, 0))$$ to be the vector "normal" to the boundary (in a very vague sense), and we let $$\phi_i(x) = x + \rho_i(x) v_i .$$ Finally, we define $$F$$ to be the composition of all $$\phi_i$$'s.

5. If $$r > 0$$ is small enough, then every $$\phi_i$$ is a diffeomorphism of $$\mathbb{R}^N$$, and hence $$F$$ is a diffeomorphism of $$\mathbb{R}^N$$. Furthermore, by making $$r > 0$$ sufficiently small, we can make $$\sup |F(x) - x|$$ as small as we please. Each $$\phi_i$$ maps $$\Omega$$ into $$\Omega$$ and $$\overline{\Omega}$$ into $$\overline{\Omega}$$, and so $$F$$ also maps $$\Omega$$ into $$\Omega$$ and $$\overline{\Omega}$$ into $$\overline{\Omega}$$. Finally, if $$x \in \partial\Omega$$ and $$i$$ is the first index such that $$\rho_i(x) > 0$$, then $$\phi_i(x)$$ is in $$\Omega$$, and hence it follows that $$F(x)$$ is in $$\Omega$$. Thus, $$F$$ maps $$\overline{\Omega}$$ into $$\Omega$$.

Thus, $$F$$ has all the desired properties.