MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was reading the proof of Isodiametric inequality in the book by Evans and Gariepy. As a remark they said that the set, say A, is not necessarily contained in a ball of diameter diamA. I thought every set has to be contained in a ball of radius the diam of the set. Can some one explain this to me?

share|cite|improve this question
I know it's not directly relevant to your question, but why not state the inequality too, for the sake of background? – Scott Morrison Oct 22 '09 at 4:01
Consider the three vertices of an equilateral triangle with side length a. The diameter of the set is a, but the smallest ball which can contain all three points has radius \sqrt{3}a/3 and hence diameter 2\sqrt{3}a/3 > a – j.c. Oct 22 '09 at 4:06
For all sets A the subset of R_n, the n-dimensional Lebesgue measure is less or equal to alpha(n)(diamA/2)^n , where alpha(n) is (pi^n/2 )/gamma fn(n/2 +1) Probably it is messy but that the inequality. – user962 Oct 22 '09 at 4:10
Looks like the informal statement in words of that inequality is: Any subset A of R^n has volume bounded by above by the ball with diameter diam(A). – j.c. Oct 22 '09 at 4:13

I don't know what the isodiametric inequality is, but I think I can answer your question. It's true that any set A is contained in a ball of radius diam(A). In fact, by the definition of diameter, any ball of radius diam(A) centered at a point of A must contain A. But A need not be contained in a ball of diameter diam(A).

For example, take A to be an equilateral triangle in the plane, with side-length 1. The diameter of this set is 1, but it is not contained in any ball of diameter 1. If it were, one of the sides of the ball would have to pass through the center of the ball (so that the ball contains to two endpoints), but then the other vertex is not in the ball.

share|cite|improve this answer
Is the simplex in some sense the worst case. In other words, is it true that for any A in R^n we have that (diameter of smallest ball containing A)/diam(A) is at most what it is for the simplex? – Kevin P. Costello Oct 22 '09 at 20:59
I think it's pretty easy to see that the worst case is some simplex. It's almost definitely the regular simplex, but I haven't thought of a slick way to prove that. Whatever the worst case is, though, there is a hard bound diam(A) >= sqrt(2)*rad(A) for a subset A of a real inner product space (where rad(A), the "radius" of A, is the radius of the smallest ball that contains A). – Darsh Ranjan Oct 23 '09 at 20:50

Regarding the question of the worst case:

Jung's Theorem states that every set of diameter $d$ in $n$-dimensional Euclidean space is contained in a ball of radius $\le d \sqrt{\frac{n}{2n+2}}$, and that this is best possible. Equality is attained by the regular $n$-dimensional simplex.

share|cite|improve this answer

Consider an equilateral triangle of edge length 1.

  1. What is its diameter?

  2. What is the diameter of the smallest circle which contains it?

share|cite|improve this answer
Thank you very much!! I got it. – user962 Oct 22 '09 at 4:16

As people have said, obviously in general you don't have A contained in the ball of diameter diamA, more clearly r=diam A/2. What you can have immediately is A contained in the ball of of radius diam A, i.e. r= diamA.

In the proof of Evans - Gariepy, they use Steiner symmetry lemma, meaning that for any set A Lebesgue measurable, you take any line d. You try to make the set to be symmetric w.r.t. the line d but keep the same volume (this can be make by using Fubini's theorem). It's not so hard to show that, by doing that procedure, the volume remains the same while diamA decreases (philosophically it's correct since make somethings more round, more nice will decrease the diameter).

After using Steiner symmetry lemma for the n axis e_1,...,e_n, you will have the new object, name A' that is symmetric w.r.t. e_1,...,e_n and |A|=|A'| while diamA' \le diamA. Notice that A' need not to be the sphere, it can have some diamond shape for example. But, the nice thing is A' \subset B(0,diamA'/2).

So, we get the result.

share|cite|improve this answer

For a really nice survey of the isodiametric problem for polygons, and a lot of cool examples and calculations see a recent paper of Mike Mossinghoff "Isodiametric Problems for Polygons"

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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