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Let $\gamma_n$ be the standard Gaussian measure on $\mathbb R^n$. It is well-known (e.g see Proposition 1) that for a given Gaussian volume content, half-spaces $H=\{x \in \mathbb R^n | a^Tx \le b\}$ have the smallest "boundary" content (aka surface areas). More formally

[Gaussian Isoperimetry] If $A \subseteq \mathbb R^n$ is Borel and $H$ is a half-space with $\gamma_n(A) = \gamma_n(H)$, then $$ \gamma_n(A^\epsilon) \ge \gamma_n(H^\epsilon),\;\forall \epsilon > 0. $$

where

  • $A^\epsilon := \{x \in \mathbb R^n | d(x,A) \le \epsilon\}$ is the $\epsilon$-blowup of $A$, and

  • $d(x,A):=\inf_{x' \in \mathbb R^n}\|x'-x\|_2$ is the distance of $x$ from $A$.

Question

In the above result, if the norm $p$ is changed from $\ell_2$ to some general $\ell_p$ with $p \in [1, \infty]$, the minimal boundary shapes $H$ change from half-space to what ?

N.B.: I'm particularly interested in the cases $p = 1$ and $p=\infty$.

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  • $\begingroup$ Is this answered by equivalence of norms? Probably I miss something... $\endgroup$
    – Dirk
    Mar 20, 2019 at 18:18
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    $\begingroup$ The minimizers are restricted classes of halfspaces. For $p \geq 2$, the blowup by an $\ell_p$ ball is no smaller than the blowup by an $\ell_2$ ball of the same radius, and this is tight for axis-aligned halfspaces. For $p < 2$, the blowup is no smaller than the blowup by an $\ell_2$ ball of radius $n^{1/2 - 1/p}\varepsilon$, and this is tight for halfpsaces with $a = \mathbf{1}$, the all-ones vector. $\endgroup$
    – Alf
    Mar 20, 2019 at 18:32
  • $\begingroup$ @Dirk, Jon. Thanks for the input. Yes "equivalence of norms" can help get rough bounds (upper and lower) on the boundary measure of an $\ell_p$ blowup (and the dimension pops up in a perhaps sub-optimal manner). I am aiming for something more direct and exact. In particular, i'm really interested in what the shapes $H$ are. $\endgroup$
    – dohmatob
    Mar 20, 2019 at 18:47
  • $\begingroup$ @Jon Are we saying the extremal shapes $H$ for $\ell_p$ distances are also half-spaces ? $\endgroup$
    – dohmatob
    Mar 20, 2019 at 18:49
  • $\begingroup$ Yes, but not all halfspaces. Because the $\ell_p$ distances are not rotation-invariant for $p \neq 2$ not all halfspaces have the same blowup. $\endgroup$
    – Alf
    Mar 20, 2019 at 20:10

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