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By D. Kannan

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8), but perhaps the reader deems this example artificial. In this section we will illustrate this gap again using ellipsoids in Hilbert space. It is hard to argue that ellipsoids are unnatural or unimportant. Given a sequence (ai )i≥1 , ai > 0, we consider the set E = t∈ 2 ; i≥1 t2i ≤1 . 1. We have 1 L a2i 1/2 ≤ E sup Xt ≤ a2i t∈E i≥1 1/2 . 20) i≥1 Proof. We observe that, by the Cauchy-Schwarz inequality we have Y := sup Xt = sup t∈T t∈E i≥1 ti g i ≤ a2i gi2 1/2 . 21) i≥1 Taking ti = a2i gi / j≥1 a2j gj2 , we see that actually Y = ( i≥1 a2i gi2 )1/2 and thus EY 2 = i≥1 a2i .

8) 36 2 Gaussian Processes and Related Structures Consider an integer s ≥ 2 and the process (Xi )i≤Ns so the index set is T = {2, 3, . . , Ns }. The distance d associated to the process satisfies for p=q 2 1 ≤ d(p, q) ≤ . 9) we have d(p, Tn ) ≥ 2−n/2 /L, and thus en (T ) ≥ −n/2 /L. Thus 2 s−1 . 10) 2n/2 en (T ) ≥ L n On the other hand, for n ≤ s let us define Tn = {2, 3, . . , Nn , Ns }. Consider integers p ∈ T and m ≤ s − 1 such that Nm < p ≤ Nm+1 . 9) and since p, Ns ≥ Nm . Hence we have L2n/2 2−m/2 ≤ L .

Part (a) is obvious. 5. 4. We define 2k/α ∆(Ak (t)) γα,n (T, d) = inf sup t∈T k≥n where the infimum is over all admissible sequences (Ak ). We consider the functionals Fn (A) = sup γα,n (G, d) where the supremum is over G ⊂ A and G finite. 2 with β = 1 , θ(n + 1) = 2n/α−1 , τ = 1, and r = 4. 31), consider m = Nn+1 and consider points (t ) ≤m of T , with d(t , t ) ≥ a if = . Consider sets H ⊂ B(t , a/4) and c < min ≤m Fn+1 (H ). For ≤ m, consider finite sets G ⊂ H with γα,n+1 (G , d) > c, and G = ≤m G .

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