4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge May 10 at 5:40
  • 1
    $\begingroup$ 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. $\endgroup$ – Mateusz Kwaśnicki May 10 at 10:42
  • 1
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge May 10 at 15:54
  • 3
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge May 10 at 16:06
  • 2
    $\begingroup$ 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. $\endgroup$ – Mateusz Kwaśnicki May 10 at 17:26
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.