# Isoperimetry on $[0, 1]^n$ w.r.t $\ell_p$ distance, with $p \in [1,\infty]$

Let $$A$$ be a measurable subset of the metric space $$\mathcal X = ([0, 1]^n,\ell_p)$$ with $$1 \le p \le \infty$$, and define its $$\varepsilon$$-blowup by $$A^\varepsilon:=\{x \in \mathcal X \mid \|x-a\|_p \le \epsilon\text{ for some }a \in A\}$$.

# Question

• If $$\operatorname{vol}(A) > 0$$, what is a good lower bound on $$\operatorname{vol}(A^\epsilon)$$ ?

• Same question with $$\operatorname{vol}(A) \ge 1/2$$.

N.B.: I'm mostly interested in the cases $$p \in \{1,2,\infty\}$$.

• It appears that the euclidean case $p=2$ is solved by (1.1) of www-users.math.umn.edu/~bobko001/papers/… May 28, 2019 at 5:39
• It's the first time I see "blow-up" used in this sense.
– YCor
May 28, 2019 at 6:20

I'll only handle the euclidean case $$p=2$$, as the other cases will follow by equivalence of $$\ell_p$$-norms (it's possible this "delegation procedure" is not very optimal for...).
So, consider the function $$T:\mathbb R^n \rightarrow [0, 1]^n$$ defined by $$T(z_1,\ldots,z_n)=(\Phi(z_1),\ldots,\Phi(z_n))$$ where $$\Phi$$ is the standard Gaussian CDF. It's easy to see that $$u_n=\gamma_n \circ T^{-1}$$. We will take for granted that $$\Phi$$ (and therefore $$T$$) is $$(2\pi)^{-1/2}$$-Lipschitz continuous.
Let $$A$$ be a measurable subset of $$[0,1]^n$$ and let $$B:=T^{-1}A$$. Lipschitzness of $$T$$ implies $$A^\varepsilon \supseteq B^{\varepsilon\sqrt{2\pi}}$$, and so $$u_n(A^\varepsilon) \ge \gamma_n(B^{\varepsilon\sqrt{2\pi}})$$. Now, by Gaussian Isoperimetry, $$u_n(A^\varepsilon) \ge \gamma_n(B^{\varepsilon\sqrt{2\pi}}) \ge \Phi^{-1}(\gamma_n(B)+\varepsilon\sqrt{2\pi}) = \Phi^{-1}(u_n(A)+\varepsilon\sqrt{2\pi})$$