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Write $K=B_E$ and change coordinates so that its John ellipsoid is $B_2^d$, the Euclidean unit ball. Then the standard symmetric-body inequalities areWe give the contact argument underlying the construction. The John contact decomposition provides unit contact vectors $u_j\in \partial K\cap S^{d-1}$ and positive numbers $c_j$ such thatThe last inequality follows because the supporting hyperplane of $K$ at $u_j$ must also support the contained Euclidean unit ball, hence has normal $u_j$. For completeness, the decomposition follows from maximal volume as follows. If $I$ were not in the convex cone generated by $u\otimes u$ at contact points, finite-dimensional separation would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract $\delta I$ for small $\delta >0$, retaining positive trace and obtaining strict negative values. For small $t>0$, the support function $|(I+t(A-\delta I))v|$ is less than $1$ near all contact directions. On the compact complement of those neighbourhoods, $h_K(v)-1$ has a positive minimum, so the perturbed ellipsoid remains inside $K$ there too. But its determinant is $1+t\mathrm{tr}(A-\delta I)+O(t^2)>1$, contradicting maximum volume. The cone is closed: its generating contact matrices form a compact set of trace $1$, so bounded traces bound their nonnegative coefficient sums. Finite-dimensional convexity then supplies finitely many contacts in the decomposition. Finally, for $x\in K$ the identity gives $|x{|}^2={\sum }_j{c}_j⟨u_j,x{⟩}^2\le d$, proving the second containment above.

Construct an orthonormal basis recursively. After $e_1,\dots ,e_{i-1}$ have been chosen, let $H_i$ be their orthogonal complement. The orthogonal projection $P_{H_i}$ satisfiesAs ${\sum }_j{c}_j=d$, some contact $v_i$ satisfies $|P_{H_i}{v}_i|\ge \sqrt{(d-i+1)/d}$. Choose $e_i=P_{H_i}{v}_i/|P_{H_i}{v}_i|$. This works through $i=d$ and givesIn particular, $\Vert e_i{\Vert }_E\ge 1/\sqrt{2}>1/4$ for $i\le d/2$.

Keep the contact vectors $v_i$ used for this basis. They lie in $\mathrm{span}(e_1,\dots ,e_i)$. Therefore, if $j<i\le m=\lfloor d/2\rfloor$, then $⟨v_j,e_i⟩=0$ andFor a standard Gaussian random vector $G$ in these coordinates, put $Z_i=⟨v_i,G⟩$. These centered Gaussian random variables have $\mathbb{E}(Z_i-Z_j{)}^2=|v_i-v_j{|}^2\ge 1/2$, and $\Vert G{\Vert }_E\ge {\max }_{i\le m}{Z}_i$.

Here is a proof of the Sudakov-Fernique inequality needed for this finite family. Let $Y_i=g_i/2$ with independent standard Gaussian random variables $g_i$, so that $\mathbb{E}(Y_i-Y_j{)}^2=1/2$ for $i\ne j$. Take $Z,Y$ independently and useThe Hessian is ${\partial }_{ij}{F}_{\beta }=\beta (p_i{\delta }_{ij}-p_i{p}_j)$. Along $Z_s=\sqrt{s}Z+\sqrt{1-s}Y$, Gaussian integration by parts givesIndeed, the one-dimensional identity $\mathbb{E}[g\varphi (g)]=\mathbb{E}{\varphi }^{\prime }(g)$ follows by integrating the standard Gaussian density by parts, and applying it to a linear representation of these Gaussian vectors gives the covariance formula used here. Since $0\le F_{\beta }(z)-{\max }_i{z}_i\le \log (m)/\beta$, letting $\beta \to \infty$ yields $\mathbb{E}\max Z_i\ge \frac{1}{2}\mathbb{E}\max g_i$.

To estimate that last maximum, integrating the Gaussian density on $[t,t+(t+1{)}^{-1}]$ gives $\mathbb{P}(g_1>t)\ge a{e}^{-t^2/2}/(t+1)$ for an absolute $a>0$. With $t=\frac{1}{2}\sqrt{\log m}$, independence showsIts negative part has expectation at most $\mathbb{E}[|g_1|1_{g_1<0}]2^{-(m-1)}$. Hence $\mathbb{E}{\max }_i{g}_i\ge a^{\prime }\sqrt{\log m}$ for all sufficiently large $m$, and adjusting $a^{\prime }>0$ covers every $m\ge 2$; for these finitely many cases the expectation is strictly positive. For $d\ge 4$, $\log \lfloor d/2\rfloor \ge \frac{1}{3}\log d$. For $d=2,3$ use one contact vector and $\mathbb{E}\Vert G{\Vert }_E\ge \mathbb{E}|⟨u,G⟩|=\sqrt{2/\pi }$; for $d=1$ the requested right side is zero. Thusfor an absolute $c>0$, independent of both dimension and the norm.

The equal-variance form of Slepian's lemma is the following. If $U,V$ are centered vectors with multivariate normal distributions, $\mathbb{E}{U}_i^2=\mathbb{E}{V}_i^2$ for every $i$, and $\mathbb{E}{U}_i{U}_j\ge \mathbb{E}{V}_i{V}_j$ whenever $i\ne j$, thenfor every threshold vector $(u_i)$. In particular $\mathbb{P}({\max }_i{U}_i\le u)\ge \mathbb{P}({\max }_i{V}_i\le u)$. No nonsingularity of the covariance matrices is required.

Here the pointwise variances $2t^{2H}$ and $2t^{2H^{\prime }}$ differ. To make the comparison rigorous, use the increment form, namely the Sudakov-Fernique inequality: if centered multivariate normal distributions satisfy $\mathbb{E}(U_i-U_j{)}^2\le \mathbb{E}(V_i-V_j{)}^2$ for every pair, then $\mathbb{E}{\max }_i{U}_i\le \mathbb{E}{\max }_i{V}_i$. Unlike the preceding form of Slepian's lemma, this statement has no equal-pointwise-variance hypothesis.

For completeness, the increment form follows by Gaussian integration by parts. Make $U,V$ independent, define $Z_r=\sqrt{1-r}U+\sqrt{r}V$ for $0<r<1$, and letThe $p_i$ are softmax function weights, with $p_i\ge 0$ and ${\sum }_i{p}_i=1$. Direct differentiation gives ${\partial }_i{F}_{\tau }=p_i$ and ${\partial }_{ij}{F}_{\tau }=\tau (p_i{1}_{i=j}-p_i{p}_j)$. The Gaussian integration by parts identity therefore yieldsThe last summands are nonnegative because $D_{ii}+D_{jj}-2D_{ij}=\mathbb{E}(V_i-V_j{)}^2-\mathbb{E}(U_i-U_j{)}^2$. Sinceintegration in $r$ and the limit $\tau \to \infty$ prove the Sudakov-Fernique inequality. All derivatives needed here are bounded for a fixed $\tau$. Singular covariance matrices can be handled by adding small independent centered normal noise with the same variance to both vectors, proving the identity for the nonsingular vectors, and letting the noise variance decrease to zero.

Use the printed normalization of fractional Brownian motion, which omits the conventional factor $1/2$. Its increment variance isFor $s,t\in [0,1]$ and $H>H^{\prime }$, $|t-s{|}^{2H}\le |t-s{|}^{2H^{\prime }}$. Enumerate $T$ by increasing finite subsets $T_n$ whose union is $T$, with $0\in T_n$. The Sudakov-Fernique inequality givesBoth Gaussian processes vanish at zero almost surely. Thus these maxima are nonnegative and increase with $n$; monotone convergence theorem givesTheir continuous modifications agree with the original Gaussian processes simultaneously on the countable set $T$, after discarding a single null event. Continuity and density of the rational parameters make each rational supremum equal to the supremum over $[0,1]$ for these modifications.

The expected suprema are finite. Indeed, the canonical distance of $Y_{H^{\prime }}$ is $d(s,t)=\sqrt{2}|t-s{|}^{H^{\prime }}$. A Euclidean grid supplies a closed-ball metric covering number bound $N(\epsilon )\le 1+\lceil (\sqrt{2}/\epsilon {)}^{1/H^{\prime }}\rceil$ for $0<\epsilon \le \sqrt{2}$. The Dudley entropy integral bound for an anchored centered separable Gaussian process isThe integral is finite since its integrand grows at most like a constant times $\sqrt{1+\log (1/\epsilon )}$ near zero. The same argument applies to $X_H$.

For any real path vanishing at zero,A centered Gaussian process has the same probability law as its negative. Combining this symmetry with the increment comparison gives the stronger boundIn particular the requested comparison isThe common normalization factor $\sqrt{2}$ affects both sides equally. The comparison is valid for every $0<H^{\prime }<H<1$, including the equality cases $s=t$ or $|s-t|=1$ in the increment bounds.

For centered multivariate normal distributions represented by vectors $U,V$ with equal corresponding variances and $\mathbb{E}[U_i{U}_j]\ge \mathbb{E}[V_i{V}_j]$ for all distinct indices, $\mathbb{P}(U_i\le u_i\text{ for all }i)\ge \mathbb{P}(V_i\le u_i\text{ for all }i)$ for every threshold vector. In particular the supremum of $U$ is stochastically smaller than the supremum of $V$. Equal corresponding variances are essential to this formulation. The related Sudakov-Fernique inequality compares expected values of maxima using variances of increments and does not require equal pointwise variances.