I am thinking about something I just read:
The volume of the n-ball is given by $V_n(r) = \frac{\pi^{n/2}}{\Gamma (\frac n 2 + 1)}r^n$ and its surface area is $S_n(r) = \frac{\pi^{n/2}}{\Gamma (\frac n 2 + 1)}r^{n-1}n$. So far so good. Now there is this statement:
Because the volume is dependent exponentially from n, the majority of points within the ball are very close to the surface. In fact, for $n \rightarrow \infty$, the point sets 'surface' and 'volume' become identical.
Is there any way to explain this statement more precisely? My point is: Taking the fraction $\lim_{n \rightarrow \infty} \frac V S = \lim_{n \rightarrow \infty} \frac r n = 0$. Thus I would assume that the surface becomes larger than the volume.
Yes, this is what happens in high dimensions. But one does not need the formulas with $\Gamma$-function to see this. Consider a ball $B_1$ of radius $1$, and inside of it draw a concentric ball $B_r$ of radius $r<1$. Then the fraction of volume of $B_1$ contained in $B_r$ is simply $r^n$. And this is how the function $r^n$ looks for $n=1,2,4,8,\dots,512$:
As you can see, in 64 dimensions there is practically nothing in $B_{0.9}$, in terms of volume. In 128 dimensions there is almost nothing in $B_{0.95}$.
I don't like this statement at all. In whatever sense we take $n\to\infty$, the Euclidean norm ball ought to have some convexity, which its sphere does not. (I understand the idea of imagining, e.g., the $L^1$ ball as something nonconvex, but here we consider balls for an inner product metric).