Connecting two points inside the Koch snowflake without getting too close to the boundary

Question

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain. We say $\Omega$ is a uniform domain with constant $c \geq 1$ if for any $x,y \in \Omega$ there is a rectifiable curve $\gamma : [0, l_\gamma] \to \Omega$, parameterized by arc length, such that $\gamma(0) = x$, $\gamma(l_\gamma) = y$, and \begin{align} l_\gamma &\leq c \lVert x - y\rVert, \label{1}\tag{i}\\ \operatorname{min}(t, l_\gamma - t) &\leq c \operatorname{dist}(\gamma(t),\partial\Omega)\label{2}\tag{ii} \end{align} for all $t \in [0,l_\gamma]$.

A typical example of a uniform domain which is not a Lipschitz domain is the interior of the Koch snowflake. Let me call the Koch snowflake $S$ and its interior $\mathrm{int}(S)$. So far, I couldn't find an explicit proof for why $\mathrm{int}(S)$ is a uniform domain as it is often stated as "obvious" fact.

I tried to prove it myself. Here are my ideas:

1. I need to pick two points and connect them via a curve. The length of the curve should be comparable to the distance between the two points and it should not get too close to the boundary. The curve can probably be defined piecewise affine.

2. When picking $x \in \mathrm{int}(S)$, $x$ must lie in exactly one triangle of the construction process of $S$, except when $x$ lies on a common boundary of two triangles.

3. If $x$ and $y$ lie in the same triangle of the construction process, call it $D$, I have an idea on how to construct the curve. First, construct a new triangle $\triangle xyz$, where $z$ is the barycenter of $D$ and then connect $x$ and $y$ with straight lines via the barycenter of $\triangle xyz$.

4. If $x$ and $y$ lie in different triangles of the construction process, I don't really know how to proceed.

Questions. How do you constructively prove that the Koch snowflake is a uniform domain? Is there another proof idea which might be easier?

Any hint is appreciated, thank you!

Answer 1

TL;DR

Very roughly: each approximation of $S$ is a uniform domain. If $x$ and $y$ are in the von Koch snowflake, then there exists some approximation of $S$ which contains both $x$ and $y$, so there is a path in this approximation which does the job. Then, since increasing approximations only "expand" the space, the distance from a point in some approximation to the boundary can only increase.

A more explicit construction (with exercises!)^[1]

For each $j \in \mathbb{N}$, let $S_j$ denote the $j$-th approximation of the von Koch snowflake, with respect to the "standard" construction. That is,

• $S_0$ is an equilateral triangle of side length $1$;
• $S_1$ is the union of $S_1$ and three equilateral triangles of side length $\frac{1}{3}$, where these new triangles are "glued" to the middle of each side of $S_1$;
• $S_2$ is the union of $S_2$ and twelve equilateral triangles of side length $\frac{1}{9}$, where these new triangles are "glued" to the middle of each side of $S_2$;

and so on. The process is illustrated in the following image from Wikipedia:

The von Koch snowflake is, then, $S = \bigcup_{j=0}^{\infty} S_j$.

First (as noted in the question), if $x$ and $y$ are in the same triangle in some approximation of the von Koch snowflake, let $\gamma$ be the straight line path between them.

Otherwise, let $\ast$ denote the point at the barycenter of $S_1$ (the original equilateral triangle). It is sufficient to show that if $x\in S$, then there a construction which gives a unique path $\gamma$ from $x$ to $\ast$ which satisfies the required conditions: if $x,y\in S$, take $\gamma_1$ to be the path from $x$ to $\ast$, take $\gamma_2$ to be the path from $\ast$ to $y$, then take $\gamma$ to be the concatenation of those paths (minus any parts of the path which are traced in both directions).

Exercise 1: Verify the above claim.

For $x \in S$, let $n$ be the smallest natural number such that $x \in S_n$. Construct a path $\gamma$ from $\ast$ to $x$ as follows:

1. The point $x$ is contained in one of the triangles which makes up $S_n \setminus S_{n-1}$; let $\gamma_{n+1}$ be the path from $x$ to the barycenter of that triangle (note that $\ell_{\gamma_{n+1}} < \frac{1}{3^n\sqrt{3}}$).

2. The triangle containing $x$ is "glued" to one of the triangles making up $S_{n} \setminus S_{n-1}$; let $\gamma_{n}$ be the path from the endpoint of $\gamma_{n}$ to the barycenter of this triangle in $S_{n-1}$ (note that $\ell_{\gamma_{n}} = \frac{1}{2\cdot 3^n \sqrt{3}} + \frac{1}{2\cdot 3^{n-1} \sqrt{3}}$).

3. Continue this construction iteratively, so that $\gamma_j$ joins the barycenter of a triangle in $S_j$ to the barycenter of the attached triangle in $S_{j-1}$ (note that $\ell_{\gamma_j} = \frac{1}{2\cdot 3^j \sqrt{3}} + \frac{1}{2\cdot 3^{j-1} \sqrt{3}}$). Note that $\gamma_1$ will connect the barycenter of a triangle in $S_1$ to the point $\ast$.

Take $\gamma$ to be the concatenation of these paths, in the "obvious" manner. Then \begin{align} \ell_{\gamma} = \sum_{j=1}^{n+1} \ell_{\gamma_j} < \frac{1}{3^n\sqrt{3}} + \sum_{j=1}^n \left( \frac{1}{2\cdot 3^j \sqrt{3}} + \frac{1}{2\cdot 3^{j-1} \sqrt{3}} \right). \end{align}

Exercise 2: Compute this sum, and show that there is a constant $c$ such that $\ell_{\gamma} < c \cdot \operatorname{dist}(x,\ast)$.

Exercise 3: Show that $\operatorname{dist}(\gamma(t), \partial(S))$ is an increasing function of $t$. Use this to find a constant $c$ such that $\min(t,\ell_\gamma-t) = t$ is bounded by $c\cdot \operatorname{dist}(\gamma(t), \partial(S))$.

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[1] The exercises exist because I am too lazy to work through the details. Exercise 1 should be fairly straight forward; Exercise 2 is mostly a tedious computation involving geometric series; Exercise 3 is somewhat less trivial, but basically follows from the observation that a particle moving from the barycenter of a triangle to the midpoint of a side of that triangle is moving away from the other two sides.