3

The volume of a $d-$ball of radius $R$ is

$$V_d(R) = C_d R^d$$

with $C_d$ a constant depending on $d$.

This means that

$$\frac{V_d(R-\delta)}{V_d(R)}= \left(1 - \frac{\delta }{R} \right)^d$$

where we choose $\delta \in (0,R)$.

For every choice of $\delta$ we can always find a value of $d$ such that

$$\left(1 - \frac{\delta }{R} \right)^d < \epsilon$$

Since $V_d(R-\delta)$ represents what is left once we have removed from the ball a shell of thickness $\delta$, i.e. what is left once we have "peeled the hyperapple", this means that after we peel the hyperapple, we are left with nothing (if the number of dimensions is large enough).

To say it in another (sloppy) way, for large $d$ all the volume of the $d-$ball is concentrated near the surface.

My question is: how general is this property? Is it valid for any hypervolume we can consider?

valerio
  • 881
  • You need some requirement like convexity, because otherwise you can make artificial counterexamples like a sphere with very long very thin spikes pointing out of it, which will have most of its volume in the 'sphere' part, hence close to the center. But for cubes etc I think it is true. – Vincent Jul 11 '17 at 08:27
  • @Vincent That makes sense. Maybe I should add the constraint that the hypersurface must be convex, because I admit that I imagined it that way when I asked the question. Even in this case, though, it would be interesting to see if it is possible to find counterexamples. – valerio Jul 11 '17 at 09:16
  • As stated, the question is a little hard to address: The phenomenon you note is an asymptotic behavior for a family of objects, namely balls of dimension $d$. Any putative generalization requires not just one object, but a family of objects, one for each dimension, such as cubes. – Andrew D. Hwang Jul 11 '17 at 14:01
  • @AndrewD.Hwang You are right. I will try to explain better what I am interested in. – valerio Jul 11 '17 at 14:16

0 Answers0