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 $A^\varepsilon:=\{x \in \mathcal X \mid \|x-a\|_p \le \epsilon\text{ for some }a \in A\}$.


  • 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 $p \in \{1,2,\infty\}$.

  • 1
    $\begingroup$ It appears that the euclidean case $p=2$ is solved by (1.1) of www-users.math.umn.edu/~bobko001/papers/… $\endgroup$
    – dohmatob
    May 28, 2019 at 5:39
  • 2
    $\begingroup$ It's the first time I see "blow-up" used in this sense. $\endgroup$
    – YCor
    May 28, 2019 at 6:20

1 Answer 1


I managed to piece together a solution to my problem by reading the first page of this paper http://www-users.math.umn.edu/~bobko001/papers/2010_JMS-165_Conc.on.the.cube.pdf.

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, $$ u_n(A^\varepsilon) \ge \gamma_n(B^{\varepsilon\sqrt{2\pi}}) \ge \Phi^{-1}(\gamma_n(B)+\varepsilon\sqrt{2\pi}) = \Phi^{-1}(u_n(A)+\varepsilon\sqrt{2\pi}) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.