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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 $p \in \{1,2,\infty\}$.

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    $\begingroup$ 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
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    $\begingroup$ It's the first time I see "blow-up" used in this sense. $\endgroup$
    – YCor
    May 28, 2019 at 6:20

1 Answer 1

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

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