We had a lengthy discussion yesterday, on how to prove that the circumference of a circle is $2\pi r$. By using google, the most commonly found proof starts in the following way.

"Consider the regular n-gon inside the circle touching the circle at its vertices, and the regular n-gon outside the circle touching the circle at its edges. Then the circumference of the inner n-gon is smaller than the circumference of the circle, which is smaller than the circumference of the outer n-gon."

While this apparently is true for n-gons, it is not true for arbitrary geometric shapes inside and outside the circle. So we conjectured it was the convexity of the n-gon that is of importance. Which leads to the question:

Let $C$ be a circle with radius $r$, and $A_n$ be a family of convex sets with $A_n \subset C$ for each $n$. If, for $n \rightarrow \infty$, the volume of these sets converges to $\pi r^2$ (the volume of $C$), will the circumference necessarily converge to $2\pi r$?

  • 1
    $\begingroup$ If the sets $A_n$ need not be closed, the answer is no. Let the left hemisphere of the circle be filled in, but without boundary. Let the right half of the hemisphere be the right half of an $n$-gon. Then, when $n \to \infty$, the area converges to $\pi r^2$, but the area converges to $\pi r$. $\endgroup$ – JavaMan Jan 11 '12 at 18:27
  • $\begingroup$ Note also that the question seems ill-posed. For example, one could probably also concoct sets where the volume converges but the circumference neet not converge. A proper formulation of the question may then ask one to prove that for closed convex sets $A_n \subset C$, we have $$\lim_{n \to \infty} vol(A_n) = \pi r^2 \quad \implies \quad \limsup_{n \to \infty} \text{ }Circ(A_n) = 2 \pi r$$ where $vol(A_n)$ and $Circ(A_n)$ obviously stand for the volume and circumference of $A_n$, respectively. $\endgroup$ – JavaMan Jan 11 '12 at 18:30
  • $\begingroup$ I'm actually not sure what the precise definition of the circumference of an arbitrary set $A \subset \mathbb{R}^2$ is, but I would assume its some function of its boundary, which should always be closed. In this case, in your example above the circumference would converge to $2 \pi r$, unless I'm mistaken? $\endgroup$ – Benno Jan 11 '12 at 18:41
  • $\begingroup$ Probably the right term would be "perimeter" instead of "circumference". In this case this is the one-dimensional Hausdorff measure of the boundary of the set. $\endgroup$ – Dirk Jan 11 '12 at 19:24
  • $\begingroup$ @Benno: You are right. My counterexample yields a perimeter of $2 \pi r$ if by perimeter you mean the $1$-dimensional Lebesgue measure of the boundary of the set $A = \lim_{n \to \infty} A_n$. In my counterexample, I was instead thinking of the left hemisphere as not contributing at all to the perimeter. This made intuitive sense to me at the time, but it depends on how you define perimeter. $\endgroup$ – JavaMan Jan 11 '12 at 22:15


If you don't mind, I'll do the problem in $\mathbb R^d$. Write $B_2^d$ for the unit Euclidean ball (centred at the origin, radius 1). Let $A_n$, as before, be a sequence of convex subsets of $B_2^d$ with $V(A_n)\to V(B_2^d)$, where $V(\cdot)$ denotes volume. Write $$ V_1(K,B_2^d) = V(\underbrace{K,\dotsc,K}_{d-1},B_2^d) $$ for that mixed volume. I will use the following facts about mixed volume:

  1. $V_1(K,K) = V(K)$
  2. $V_1(K,L)$ is increasing in both arguments.
  3. The surface area of a convex body $K$ is $$ S(K) = \lim_{\epsilon\downarrow 0} \frac{V(K+\epsilon B_2^d) - V(K)}{\epsilon} = dV_1(K,B_2^d) $$ (where $K+\epsilon B_2^d$ is the Minkowski sum).
  4. Minkowski's first inequality: $$ V(K)^{(d-1)/d} V(L)^{1/d} \le V_1(K,L) $$

Using (4), then (2) (we assumed $A_n\subset B_2^d$), then (1) yields $$ V(A_n)^{(d-1)/d} V(B_2^d)^{1/d} \le V_1(A_n,B_2^d) \le V_1(B_2^d,B_2^d) = V(B_2^d) $$ Sending $n\to\infty$ yields, by the squeeze law, $$ \lim_{n\to\infty} V_1(A_n,B_2^d) = V(B_2^d) = V_1(B_2^d,B_2^d) $$ Multiplying by $d$ and using (3) twice yields $$ \lim_{n\to\infty} S(A_n) = S(B_2^d) $$ as desired.


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