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Download Geometry of isotropic convex bodies by Silouanos Brazitikos PDF

By Silouanos Brazitikos

The learn of high-dimensional convex our bodies from a geometrical and analytic perspective, with an emphasis at the dependence of varied parameters at the size stands on the intersection of classical convex geometry and the neighborhood idea of Banach areas. it's also heavily associated with many different fields, reminiscent of chance concept, partial differential equations, Riemannian geometry, harmonic research and combinatorics. it truly is now understood that the convexity assumption forces many of the quantity of a high-dimensional convex physique to be centred in a few canonical manner and the most query is whether or not, below a few typical normalization, the reply to many primary questions could be autonomous of the size. the purpose of this publication is to introduce a couple of recognized questions concerning the distribution of quantity in high-dimensional convex our bodies, that are precisely of this nature: between them are the cutting challenge, the skinny shell conjecture and the Kannan-Lovász-Simonovits conjecture. This e-book offers a self-contained and recent account of the development that has been made within the final fifteen years

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Hm : R → [0, ∞) are measurable functions, we set ∗ K(h1 , . . , hm ) = where ∗ m Rn m c hjj (θj ) | θj ∈ R , x = sup j=1 θj cj uj dx, j=1 denotes the outer integral. Then, inf K(h1 , . . , hm ) | R hj = 1 , j = 1, . . , m = √ F. 6. BRASCAMP-LIEB INEQUALITY AND ITS REVERSE FORM 23 Although the Brascamp-Lieb inequality and its reverse form do not play a central role in this book, we sketch Barthe’s argument which is very elegant, short and related in spirit with arguments that appear in subsequent chapters.

Zn in Rn such that n − i + 1 1/2 zi zi 2 = 1 n for all 1 i n. y, T y = T, y ⊗ y Proof. We define the zi ’s inductively; z1 can be any contact point of K and B2n . Assume that z1 , . . , zk have been chosen for some k < n. We set Fk = span{z1 , . . , zk }. 9 we find a contact point yk+1 with PFk⊥ yk+1 It follows that PFk yk+1 2 2 = yk+1 , PFk⊥ yk+1 PFk yk+1 2 k/n. n−k . 5. CLASSICAL POSITIONS OF CONVEX BODIES 21 We define zk+1 = PFk⊥ yk+1 / PFk⊥ yk+1 2 . Then, 1 = zk+1 2 | yk+1 , zk+1 | = PFk⊥ yk+1 zk+1 2 n−k n 1/2 .

2 Using the above we see that N (K, tB2n ) N (K, tB2n ) N (K, t4 K ◦ ) 2 N K, t2 K ◦ 2 N (K, 2tB2n )N B2n , 8t K ◦ . We write 1 t (2t)2 log N (K, 2tB2n ) + 64(t/8)2 log N B2n , K ◦ 4 8 1 2 n 2 (2t) log N (K, 2tB2 ) + 64A , 4 and taking sup over all t > 0 we arrive at 3B 2 256A2 . We can now prove Sudakov inequality: for every t > 0 we have t2 log N (K, tB2n ) t2 log N (K, tB2n ) 100A2 cnw2 (K), where c > 0 is an absolute constant. 1 holds true for not necessarily symmetric convex bodies as well.

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