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How can I formalize and prove the following intuition?:

Picture a very skinny rectangle, one with base length 1 and sides length $\epsilon$. Or imagine a very flattened ellipse. The interiors of these objects are "skinny" (or we might say $\epsilon$-skinny) in the sense that each point in the interior is very close (or within $\epsilon$) to a point on the boundary. This notion seems easy to formalize.

Now think of the sides of the rectangle, or the boundary of the ellipse, as the image of a Jordan curve $f : S^1 \rightarrow \mathbb{R}^2$ . It's intuitively "obvious" (though not at all obvious) that what makes the bounded region skinny (in the sense of the above paragraph) is the fact that: For every $\theta \in S^1$ there exists $\theta ' \in S^1$ such that $f (\theta )$ is close to $f (\theta ' )$ but $\theta$ is not very close to $\theta '$.

I have been thinking about the right way to formalize these ideas and I'm a bit stuck. I'd like to formulate them in the $C^0$ setting, that is to say without assuming tangent vectors or any calculus-related things. Do you have any thoughts? Thanks!

Also: This is motivated by the completely well-defined question here:why is an annulus close to it's boundary when it's boundary curves are close?

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For a notion of 'not very close'-ness, I think the concept you're looking for here is the topological notion of connectedness. Something like "there is a $\delta$-ball $B(x,\delta)=f(S)$ around each $f(\theta)$ such that $S$ is disconnected" might be useful. – Eric Stucky May 25 '13 at 11:34
@EricStucky Note that this does not work near the short sides of the rectangle. – Abel May 25 '13 at 12:02
Ah, that is true. At least there are others who seem to have good insight for us :) – Eric Stucky May 25 '13 at 12:27

Part I Non-parametric discussion of thinness.

each point in the interior is very close to a point on the boundary.

This says precisely that the inradius is small. The inradius (for inscribed radius) is the maximal radius of an open disk enclosed by the curve. For example, the rectangle of dimensions $a,b$ has inradius $\frac12 \min(a,b)$.

It remains to answer: small compared to what? There are two very reasonable options:

  • outradius: the minimal radius of a closed disk containing the curve.
  • diameter: the maximal distance between any pair of points on the curve.

The former is appealing because of its symmetry with inradius. The latter is easier to find in practice. It does not really matter which one you choose, because they are comparable for any planar set: $$ \sqrt{3} \operatorname{outradius} \le \operatorname{diameter} \le 2\operatorname{outradius} \tag1$$

Part II Relation with bi-Lipschitz parameterization

Now think of the sides of the rectangle, or the boundary of the ellipse, as the image of a Jordan curve $f:S^1→\mathbb R^2$.

This changes the game a bit, because a very nice curve (e.g., a circle) has parametrizations which horribly distort pairwise distances. Apparently, you want a description of skinniness in terms of parametrization $f$. Consider the bi-Lipschitz condition: $$ L^{-1}\le \frac{|f(\theta)-f(\theta')|}{|\theta-\theta'|} \le L \tag2 $$ I prefer to use the chordal metric for $|\theta-\theta'|$: that is, the Euclidean distance between the two points $\theta,\theta'$. With this choice, (2) implies that the diameter of $f(S^1)$ is at most $2 L$ (obviously) and its inradius is at least $L^{-1}$ (not as obviously). Therefore, $L$ controls the diameter/inradius ratio: if $L$ is not very large, then the image is not too skinny.

The converse does not work: a non-skinny curve does not in general allow a bi-Lipschitz parameterization. For example, a tiniest cusp added to a disk of radius $1$ precludes the possibility of a bi-Lipschitz parameterization. To get around this, you may want to relax the bi-Lipschitz requirement by allowing an additive constant: $$ L^{-1}|\theta-\theta'|-M\le {|f(\theta)-f(\theta')|} \le L|\theta-\theta'|+M \tag3 $$ but this is just a vague idea on my part; I don't know if it leads to a characterization of non-skinniness in terms of parametrization.

Part III Tentative parametric definition of thinness: a simple closed curve $\Gamma$ is $\epsilon$-thin if there exist two homeomorphisms $f:S^1\to \Gamma$ and $g:S^1\to\Gamma$ such that

  1. $f$ and $g$ traverse $\Gamma$ in opposite directions (clockwise and counterclockwise)
  2. $\max |f-g| \le \epsilon$

Indeed, suppose that 1 and 2 hold, and $a$ is a point in the interior of $\Gamma$. Then the argument I gave elsewhere applies: the curves $f$ and $g$ are not homotopic in $\mathbb R^2\setminus \{0\}$, since their winding numbers are different. In particular, the straight-line homotopy between $f$ and $g$ must pass through $a$. It follows that there exist $\theta\in S^1$ such that $f(\theta)-a$ and $g(\theta)-a$ are vectors pointing in opposite directions. Since $$|f(\theta)-a|+|g(\theta)-a| = |f(\theta)-g(\theta)|\le \epsilon$$ it follows that the inradius of $\Gamma$ is at most $\epsilon/2$.

An equivalent, and possibly more attractive form of the above definition: for any homeomorphism $f:S^1\to \Gamma$ there exists a sense-reversing homeomorphism $\psi:S^1\to S^1$ such that $\max|f-f\circ \psi|\le \epsilon$.

It is easy to see that both ellipses and rectangles satisfy the above definition of thinness. But unfortunately, there are curves of small inradius that do not satisfy it, such as this one:


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Thanks! I will have to mull this over. This was motivated by a question about why the annular region between two $\epsilon$-close Jordan curves is contained in the $\epsilon$ neighborhood of either curve. I think I will post that as a separate question. – Rimbaud May 25 '13 at 11:50
@Rimbaud If you have to mull this over, you should not mark the answer as accepted. Accepting means you found that the answer solved your problem, which I'm sure it hasn't. By the way, long ellipses/rectangles do not satisfy the condition with $\theta,\theta'$ that you gave in the post, at least not with arclength parametrization. Consider what happens when $f(\theta)$ is at an end of the major axis of the ellipse. – ˈjuː.zɚ79365 May 25 '13 at 12:00
You are right on both counts. Unaccepted :) Maybe one needs some upper bound on the length of a segment whose points violate the condition. – Rimbaud May 25 '13 at 12:12
@user79365 Your last condition relaxing the bi-Lipschitz requirement is the important condition in the definition of a quasi-isometry, c.f. . What you describe is a quasi-isometric embedding of $S^1$ with the chordal metric into the plane. – Neal May 25 '13 at 13:15
@Rimbaud I added Part III, which attempts to define thinness via parametrization, but falls short due to the last example (cross-shaped domain). I guess the domain in this example should be considered skinny? – ˈjuː.zɚ79365 May 26 '13 at 0:34

How about something like this:

Definition: A Jordan curve $f\colon S^1\to\mathbb{R}^2$ is called $\epsilon$-boundary-skinny if there exists a homeomorphism $k\colon S^1 \to S^1$ such that $\|(f\circ k)(\theta)-(f\circ k)(-\theta)\|<2\epsilon$ for all $\theta\in S^1$.

Note: Here $-\theta$ is to be understood in the sense of $S^1\cong\mathbb{R}/2\pi\mathbb{Z}$, not $S^1\cong \{z\in\mathbb{C}:|z|=1\}$.

The result you are looking for would then be something like: 'Any $\epsilon$-boundary-skinny Jordan curve is $\epsilon$-skinny.'

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Thanks for your feedback, Abel. For some reason my phone is not rendering tex right now but Im sure I will benefit from your response when i get back to my desktop :) – Rimbaud May 25 '13 at 11:59

How about: a region is "skinny" if the ratio of its area to its perimeter is small.

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How can I talk meaningfully about a "perimeter" without a differentiable curve? – Rimbaud May 25 '13 at 11:06
I think I would at least need to assume I have a "rectifiable" curve. Which I'd prefer not to do. – Rimbaud May 25 '13 at 11:14
Also it's not really skinniness that I need to define (sorry, I should have made that more clear :). I'm happy with the definition of $\epsilon$ skinniness as "each interior point is within $\epsilon$ of the boundary." It's the condition sketched in the second paragraph that needs to be formalized. Hopefully a proof that that condition implies skinniness will follow. – Rimbaud May 25 '13 at 11:18

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