By D. Kannan

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**Example text**

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 .