Gaussian width of intersection of cube and ball in high-dimensional euclidean space

Question

Let $d$ be a large positive integer and fix $r \ge 0$. Set $S := B_2^n \cap [-r,r]^d$, where $B_2^d$ is the euclidean unit-ball in $\mathbb R^d$. Finally, let $\omega(S)$ be the Gaussian width of $S$, defined by

$$ \omega(S) := \mathbb E \sup_{x \in S} x^\top z, $$

where the expectation is over $z \sim N(0,I_d)$.

Question. What is a good upper-bound for $\omega(T)$, valid for large $d$ ?

Note. Using Proposition 1 of this manuscript with $T = [-1/\sqrt{d},1/\sqrt{d}]^d$ (the convex hull of $m=2^d$ points in $B_2^d$), and $s=1/(r\sqrt{d})$, I'm able to obtain the following upper-bound $$ \omega(S) = s\cdot\omega(s B_2^d \cap T) \lesssim r\sqrt{d\log(em)} \land \sqrt{d} = rd \land \sqrt{d}. $$

Unfortunately, the above bound is not very good for my purposes.

Answer 1

The answer is, up to a constant factor $\omega(S) = \Theta(\min(\sqrt{d}, rd))$.

To see the upper bound, we can use the fact that if $S \subset Q$, then $\omega(S) \leq \omega(Q)$, therefore $$ \omega(B_2^n \cap r B_\infty^n) \leq \min(\omega(B_2^n), r \omega(B_\infty^n)) = \min( \mathbb{E}_{z \sim \mathcal{N}(0, I)} \|z\|_2, r\, \mathbb{E}_{z \sim \mathcal{N}(0, I)} \|z\|_1) = \mathcal{O}(\min(\sqrt{d}, rd)). $$

For the lower bound, note that for any given vector $z$, we can chose $\tilde{z} \in S$ to be $\tilde{z}_i := \min(1/\sqrt{d}, r) \mathrm{sgn}(z_i)$. With this choice, we have for any fixed $z$ $$\sup_{x \in S} x^T z \geq \tilde{z}^T z = \min(1/\sqrt{d}, r) \sum_i |z_i| = \min(1/\sqrt{d}, r) \|z\|_1,$$ and therefore $\omega(S) \geq \min(1/\sqrt{d}, r) \mathbb{E}_{z\sim\mathcal{N}(0, I)} \|z\|_1 = \Omega(\min(\sqrt{d}, rd)$.

Note that this is not inconsistent with the upper bound given by Proposition 1 in the cited paper --- the hypercube $[-r, r]^d$ has $2^d$ vertices, not $2d$.