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Choose a maximal subset $U\subset S^{n-1}$ whose distinct points have Euclidean distance greater than one. The volumetric bound for Euclidean metric nets makes the construction finite: the open Euclidean balls of radius $1/2$ about the points of $U$ are disjoint and all lie in $(3/2)B^n$, soMaximality means that $U$ is a metric net of radius one on the unit sphere. Thus every $\theta \in S^{n-1}$ has some $u\in U$ with $\Vert u-\theta \Vert \le 1$. Since both are unit vectors,Now intersect the corresponding closed half-spaces:The Cauchy-Schwarz inequality shows that $(1/2)B^n\subset P$. Conversely, write any nonzero $x\in P$ as $x=t\theta$ and choose $u$ as above. Then $t/2\le t(u\cdot \theta )=u\cdot x\le 1/2$, so $t\le 1$. Hence $P\subset B^n$, which also proves boundedness. This convex polytope has at most $|U|\le 3^n$ facets, since redundant inequalities can only reduce their number. It meets the required inclusions with $r=2$.

For a symmetric matrix $H$ and an $\eta$-metric net $U$ of the unit sphere, with $0<\eta <1/2$, approximate a maximizing unit vector $x$ by $u\in U$. Expanding $x^{\top }Hx-u^{\top }Hu=(x-u{)}^{\top }Hx+u^{\top }H(x-u)$ bounds its absolute value by $2\eta \Vert H{\Vert }_{\mathrm{op}}$. Rearrangement proves the bound. The volumetric bound for Euclidean metric nets limits the number of needed directions, allowing a union bound to control a random matrix.

For independent Bernoulli distribution upper-triangular entries of a symmetric zero-diagonal adjacency matrix of a graph, put $W=A-\mathbb{E}A$. For a unit vector $x$, $x^{\top }Wx=2{\sum }_{i<j}{x}_i{x}_j{W}_{ij}$ has sub-Gaussian variance proxy at most ${\sum }_{i<j}{x}_i^2{x}_j^2\le 1/2$, by the Hoeffding lemma. Hence $\mathbb{P}(|x^{\top }Wx|>u)\le 2e^{-u^2}$. The volumetric bound for Euclidean metric nets gives a $1/4$-metric net of the unit sphere with at most $9^n$ points. The quadratic form net bound gives $\Vert W{\Vert }_{\mathrm{op}}\le 2{\max }_x|x^{\top }Wx|$ on that net. The union bound gives $\mathbb{P}(\Vert W{\Vert }_{\mathrm{op}}>2\sqrt{n\log 9+\log 2+s})\le e^{-s}$; integrating this tail proves the displayed expectation bound.