The minimum Gaussian width set of a fixed area The Gaussian width of a set $S\subseteq \mathbb{R}^d$ is defined as $\mathbb{E} \sup_{x\in S}|\langle x, g\rangle|$ where $g\sim \mathcal{N}(0,I_d).$
I am interested in the subset $S$ of the sphere $\mathbb{S}^{d-1}$, of spherical measure $m$ for which the Gaussian width is minimized.  It seems plausible that the minimum should be achieved for a spherical cap, but are there standard results or techniques which prove this? Alternatively are there (perhaps slightly weaker) inequalities which lower bound the Gaussian width of a set in terms of its measure?
 A: Not sure if this helps, but I believe it is possible to show that spherical caps minimize the Gaussian width up to a factor of 6, i.e. for any $\alpha\in [0,1]$ and any $A\subseteq S^{n-1}$ such that the $\sigma(A)=\alpha$, any spherical cap $B$ satisfying $\sigma(A)=\alpha$ satisfies
$$ \mathbb{E}\left[\sup_{x\in B}\vert \langle x,g\rangle\vert\right] \leq 6\cdot \mathbb{E}\left[\sup_{x\in A}\vert \langle x,g\rangle\vert\right].$$
I have no idea how good this bound is. Note that for small $\alpha$, it is clearly better to take the measure of two antipodal spherical caps of size $\alpha/2$, so one could in principle get a tighter bound if desired.
For convenience of notation following Vershynin's book, for any subset $A$, let $w(A)=\mathbb{E}\left[\sup_{x\in A}\langle x,g\rangle\right]$ and $\gamma(A)=\mathbb{E}\left[\sup_{x\in A}\vert \langle x,g\rangle\vert\right]$, so that we wish to show $\gamma(B)\leq 6\cdot \gamma(A)$ where $B$ is a spherical cap as above and $A$ has the same measure. First, we claim that spherical caps minimize $w(\cdot)$ for any prescribed spherical measure $\alpha$. Note that $w(\cdot)$ is exactly proportional to the same expectation replacing $g$ by a uniform $z\sim S^{n-1}$ by rotational invariance of Gaussians, so it first suffices to show that spherical caps minimize the expectation under this replacement (note that $w(A)\geq 0$ for all $A$ by monotonicity). To see this, we claim that for $z$ uniform on $S^{n-1}$, the random variable $\sup_{x\in B}\langle x,z\rangle$ is stochastically dominated by $\sup_{x\in A}\langle x,z\rangle$. Indeed, for any $t\in [-1,1]$,
\begin{align*}
\Pr(\sup_{x\in B}\langle x,z\rangle\geq t)&= \Pr(\sup_{x\in B}\cos(\angle(x,z))\geq t)\\
&=\Pr(\cos(\inf_{x\in B}\angle (x,z))\geq t)\\
&=\Pr(\inf_{x\in B} \angle (x,z)\leq \arccos(t))\\
&=\Pr(z\in B_{\arccos(t)})\\
&\leq \Pr(z\in A_{\arccos(t)})\\
&=\Pr(\sup_{x\in A}\langle x,z\rangle\geq t).
\end{align*}
Here, we just straightforwardly rewrite inner products in terms of angles, and then apply the standard isoperimetric inequality on the sphere, which is usually stated in terms of geodesic (angular) distance. By the reduction to uniform $z$, this implies that $w(B)\leq w(A)$.
To translate this to $\gamma(\cdot)$, by Exercise 7.6.9 of Vershynin, we note that for any subset $A$ of $S^{n-1}$, it holds that
\begin{equation*}
\frac{1}{3}(w(A)+1)\leq \gamma(A)\leq 2(w(A)+1)
\end{equation*}
This implies that
\begin{equation*}
\gamma(B)\leq 2(w(B)+1)\leq 2(w(A)+1)\leq 6\gamma(A),
\end{equation*}
as claimed.
