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Let $\mathbb{R}^n$ be the Euclidean space and $A \subseteq \mathbb{R}^n$ be a sufficiently regular set, e.g., one that has smooth boundary or is convex. We define the $\epsilon$-neighbor of $A$ in the $\ell_2$ sense as \begin{align} A^{\epsilon} = \{ y \in \mathbb{R}^n \colon \text{there exists}~ x \in A ~\text{such that}~ \| x - y \|_{2} \leq \epsilon \}. \end{align} We denote $\gamma$ as the standard $n$-dimensional Gaussian measure $N(0, I_n)$, then the Gaussian surface measure is defined as \begin{align} \tau (A) = \lim_{\epsilon \rightarrow 0} \frac{ \gamma( A^{\epsilon} \setminus A) }{ \epsilon}. \end{align} By results of Keith Ball (the reverse isoperimetric problem for Gaussian measure https://link.springer.com/article/10.1007/BF02573986), there is a universal upper bound of the Gaussian surface area of any convex set.

My questions are: 1) when is Keith Ball's upper bound tight? 2) what is a tight upper bound of the Gaussian surface area of the cone of positive semidefinite matrices?

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    $\begingroup$ What do you mean by the "positive semidefinite cone"? The cone of positive semidefinite matrices? An orthant? For any cone $A$ with apex at the origin, the Gaussian perimeter is proportional to the $n-2$-dimensional measure of $A \cap \{x : |x|=1\}$, the constant of proportionality being $(2\pi)^{-n/2} \int_0^\infty r^{n-2} \exp(-r^2/2) dr = 2^{-3/2} \pi^{-n/2}\Gamma((n-1)/2)$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 24, 2017 at 8:38
  • $\begingroup$ @MateuszKwaśnicki Sorry for the confusion. I meant the cone of positive semidefinite matrices. I will revise my statement in the question. Could you detail your claim in an answer? Thanks a lot! $\endgroup$
    – Minkov
    Commented Aug 24, 2017 at 11:28
  • $\begingroup$ If $A$ is a "cone" (a positively homogeneous set) with Lipschitz boundary, then $\partial A$ is also a "cone". Denote $F=\partial A\cap \partial B(0,1)$. The $n-1$-dimensional Hausdorff measure $\sigma_{n-1}$ on $\partial A$ can be written as $\sigma_{n-1}(dx)=r^{n-2}\sigma_{n-2}(dz)dr$, where $x=rz$, $r>0$, $z\in F$ and $\sigma_{n-2}$ is the $n-2$-dimensional Hausdorff measure on $F$. Now apply this to the equivalent definition of the Gaussian perimeter: $\tau(A)=\int_{\partial A}\gamma(x)\sigma_{n-1}(dx)$, where $\gamma(x)$ is the density of the Gaussian measure. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 24, 2017 at 11:49
  • $\begingroup$ @MateuszKwaśnicki For the cone of positive semidefinite matrices, is it possible to get a closed form or order estimation for the $n-2$-dimensional measure of $A\cap \{x: |x| = 1\}$? $\endgroup$
    – Minkov
    Commented Aug 24, 2017 at 23:04
  • $\begingroup$ I do not know the answer, although I guess it follows from Weyl's integration formula, see mathoverflow.net/a/95256/108637. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 25, 2017 at 8:40

2 Answers 2

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Regarding the tightness of Ball's bound, which is $O(n^{1/4})$ for a convex set in dimension $n$, this was proven to be sharp (up to a constant factor) by Nazarov, in

Nazarov, Fedor, On the maximal perimeter of a convex set in $\mathbb{R}^n$ with respect to a Gaussian measure, Milman, V. D. (ed.) et al., Geometric aspects of functional analysis. Proceedings of the Israel seminar (GAFA) 2001--2002. Berlin: Springer (ISBN 3-540-00485-8/pbk). Lect. Notes Math. 1807, 169-187 (2003). ZBL1036.52014.

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The Gaussian surface area of the positive semidefinite cone $\mathbb{S}_d^{+} = \{ M \in \mathbb{R}^{d\times d} \colon M \succeq 0 \}$ is bounded by a constant in $[0,1]$ uniformly, according to ''Learning Geometric Concepts via Gaussian Surface Area'' by Adam R. Klivans, Ryan O'Donnell and Rocco A. Servedio.

More specifically, as shown in Definition 2 of this paper, the Gaussian surface area of any convex set $A\in \mathbb{R} ^{d\times d} $ can be written as \begin{align} \Gamma(A) = \int_{\partial A} \varphi(x) d \sigma (x),~~~~[1] \end{align} where $\varphi$ is the Gaussian measure on $\mathbb{R}^{d\times d}$ and $\sigma$ is the Euclidean surface area. By Nazarov's inequality, for any convex set $A$ containing the origin, it holds that \begin{align} \int_{\partial A} \frac{\phi (x)}{1 + h(x) } d \sigma (x) \leq 1 - \gamma (A) \leq 1, ~~~~[2] \end{align} where $\gamma $ is the Gaussian measure and function $h$ is defined as the distance from the origin to the tangent hyperplane of $A$ which contains $h\in \partial A$. In more details, for any $h \in \partial A$, let $n_y$ be the unit normal vector to $\partial A$ at $y$. Then we define $h$ by $h(y)= \| y\|_2 \cdot \cos ( \langle y, n_y \rangle )$.

Now we go back to the PSD cone. The boundary of $\mathbb{S}_d^{+}$ is defined as \begin{align} \partial \mathbb{S}_d^{+} = \{ M\colon M \succeq 0, \textrm{rank}(M) < d\}. \end{align} For any $M \in \partial \mathbb{S}_d^{+}$, the tangent plane at $M$ is given by \begin{align} \mathcal{T}(M) = \left \{ X M + MX^{\top} \colon X \in \mathbb{R}^{d\times d} \right \}, \end{align} which implies that $M\in \mathcal{T} (M)$. Thus we have $h(M) = 0$. Combining $[1]$ and $[2]$, we conclude that the Gaussian surface area of the PSD cone is no larger than $1 - \gamma (S_d^{+})$.

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