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?
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$\begingroup$ I know it's not directly relevant to your question, but why not state the inequality too, for the sake of background? $\endgroup$– Kim MorrisonCommented Oct 22, 2009 at 4:01
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7$\begingroup$ 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 $\endgroup$– j.c.Commented Oct 22, 2009 at 4:06
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$\begingroup$ 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. $\endgroup$– user962Commented Oct 22, 2009 at 4:10
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1$\begingroup$ 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). $\endgroup$– j.c.Commented Oct 22, 2009 at 4:13
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5 Answers
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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.
answered Oct 22, 2009 at 4:08
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2$\begingroup$ 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? $\endgroup$ Commented Oct 22, 2009 at 20:59
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$\begingroup$ 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). $\endgroup$ Commented Oct 23, 2009 at 20:50
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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.
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Consider an equilateral triangle of edge length 1.
What is its diameter?
What is the diameter of the smallest circle which contains it?
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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.
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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 http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.92.2301 "Isodiametric Problems for Polygons"
answered Aug 31, 2010 at 13:54