Is the volume of $L^2$ norm ball larger than $L^∞$ norm ball in 784 dimensional space with same epsilon (unit) value?

Question

While I was reading a blog post, I came across with an explanation saying that "ϵ you need to consider for $ℓ^2$ norm perturbations is larger than what you need for $ℓ^∞$ perturbations, because the volume of the $ℓ^2$ ball is proportional to $\sqrt{n}$ times the volume of the $ℓ^∞$ ball, where n is the input dimension."

I could not understand the explanation well (the volume of the $ℓ^2$ ball is proportional to $\sqrt{n}$ times the volume of the $ℓ^∞$ ball).

I wonder how the volume of $ℓ^2$ norm ball is larger than $ℓ^∞$ ball in 784-dimensional space.

I was thinking that the volume of $ℓ^∞$ norm ball should be bigger, so I am a bit confused. Can somebody help and explain please?

Here is the link for the post: https://adversarial-ml-tutorial.org/adversarial_examples/

Answer 1

The volume of an $\ell_2^n$ ball of radius $r$ is $r^n\cdot\frac{\pi^{n/2}}{\Gamma(n/2+1)}$. The volume of the $\ell_{\infty}^n$ unit-ball, i.e., the cube $[-1,1]^n$, is $2^n$. A straight-forward application of Stirling's formula shows that $$\lim_{n\to\infty}\left(\frac{\pi^{n/2}}{\Gamma(n/2+1)}\right)^{1/n}\sqrt{n}=\sqrt{2e\pi}$$ Therefore, what you wrote is incorrect. A correct statement is this: (the $n$'th root of the volume of the $\ell_2^n$ unit ball) times $\sqrt{n}$ is proportional to the $n$'th root of the volume of the unit cube.

There is a concept in $n$-dimensional space called "volume ratio", defined as $\left(\frac{|B_1|}{|B_2|}\right)^{1/n}$. One takes the $n'th$ root as because if you multiply an $n$-dimensional set that has volume by $\alpha>0$, its volume scales by $\alpha^n$. So in this terminology, the volume ratio of the cube to the ball is roughly $\sqrt{n}$, but one should keep in mind that the ratio is not between the volumes themselves, but between their respective $n$'th roots.