Bourgain s Theorem. Computational and Metric Geometry. Instructor: Yury Makarychev. d(s 1, s 2 ).

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1 Bourgain s Theorem Computationa and Metric Geometry Instructor: Yury Makarychev 1 Notation Given a metric space (X, d) and S X, the distance from x X to S equas d(x, S) = inf d(x, s). s S The distance between two sets S 1, S 2 X equas d(s 1, S 2 ) = inf s 1 S 1,s 2 S 2 d(s 1, s 2 ). Exercise 1. Show that distances between sets do not necessariy satisfy the triange inequaity. That is, it is possibe that d(s 1, S 2 ) + d(s 2, S 3 ) > d(s 1, S 3 ) for some sets S 1, S 2 and S 3. Exercise 2. Prove that d(x, y) d(s, x) d(s, y) and thus d(x, y) d(s, x) d(s, y). Proof. Fix ε >. Let y S be such that d(y, y) d(s, y) + ε (if S is a finite set, there is y S s.t. d(y, y ) = d(s, y)). Then d(x, S) d(x, y ) d(x, y) + d(y, y ) d(x, y) + d(s, y) + ε. We proved that d(x, S) d(x, y) + d(s, y) + ε for every ε >. Therefore, d(x, S) d(x, y) + d(s, y). Definition 1.1. Let (X, d) be a metric space, x X and r >. The (cosed) ba of radius r around x is B r (x ) = Ba r (x ) = {x : d(x, x ) r}. 1

2 2 Warm-up Consider two normed spaces (U, U ) and (V, V ). Let f be a inear operator between U and V. What is the Lipschitz norm of f? It is equa f(x) f(y) V sup x,y U x y U x y by inearity of f f(x y) V = sup x,y U x y U x y f(z) V = sup. z U z U z f(z) The expression sup V z U z z U is caed the operator norm of f. The above computation shows that the Lipschitz norm of a inear operator equas its operator norm. At the previous ecture, we proved that q p and L p [, s] L q [, s] when p < q and a, b R. These embeddings define incusion maps i 1 : p q and i 2 : L q [, s] L p [, s] defined by i 1 (a) = a for every a p i 2 (f) = f for every f L q [, s]. Note that even though maps i 1 and i 2 do not do much they just map every eement to itsef they are not ow distortion maps! Exercise 3. Compute the Lipschitz norm and distortion of map i 1 : p q. Soution. Consider a p. Note that a j a p and a j q = a j p a j q p a j p a q p p. Therefore, a q q = a j q a j p a q p p = ( aj p ) a q p p = a p p a q p p = a q p. We get that a q a p. That is, i 1 Lip 1. On the other hand, e 1 p = e 1 q = 1, where e 1 = (1,,... ). We get, i 1 Lip = 1. Now et n 1. Consider a = ( 1,..., 1 }{{},,... ) p. We have, a p = n 1/p and n coordinates a q = n 1/q. Thus i 1 1 Lip a p / a q = n 1/p /n 1/q = n 1/p 1/q. Since n 1/p 1/q as n, the norm f 1 Lip is unbounded. Answer: i 1 Lip = 1, i 1 has infinite distortion. We wi need the foowing inequaity. Theorem 2.1 (Lyapunov s inequaity). Let 1 p < q =. For every random variabe α with finite q-th moment, we have α p α q. Proof. The statement is obvious for q = since α < α amost surey. Let us assume that q <. Let f(x) = x q/p for x. Note that f(x) is a convex function. Let β = α p (β is a random variabe). We have α q q = E [ α q ] = E [ β q/p] = E [f( β )] We concude that α q α p as required. by Jensen s Inequaity f(e [ β ]) = (E [ α p ]) q/p. 2

3 Exercise 4. Compute the Lipschitz norm and distortion of map i 2 : L q [, s] L p [, s]. Proof. First, consider f(x) = 1, a constant function defined on [, s]. We have f Lp = s 1/p and f Lq = s 1/q. Therefore, i 2 Lip i 2(f) p f q = s 1/p 1/q. Now consider f L q [, s]. Let ξ be a random variabe uniformy distributed on [, s]. Note that for every function h on [, s], we have s h(x)dx = s 1 h(ys)dy = se [h(ξ)]. Therefore, f p L p = se [ f(ξ) p ] = s f(ξ) p p, and f Lp = s 1/p f(ξ) p (here, f(ξ) is a random variabe). Simiary, f Lq = s 1/q f(ξ) q. By Lyapunov s inequaity for α = f(ξ), f Lp = s 1/p f(ξ) p s 1/p f(ξ) q = s 1/p 1/q ( s 1/q f(ξ) q ) = s 1/p 1/q f Lq. We get that f Lip s 1/p 1/q and therefore f Lip = s 1/p 1/q Let ε (, 1). Consider f ε (x) = x 1 ε q. We compute its p and q norms and get that f ε p (qs q p q q p ) 1/p < as ε, f ε q as ε. Therefore, i 2 has infinite distortion. Answer: i 2 Lip = s 1/p 1/q, i 2 has infinite distortion. 3 Bourgain s Theorem Definition 3.1. Let X be a finite metric space and p 1. Suppose that Z is a random subset of X (chosen according to some probabiity distribution). For every u X, define random variabe ξ u = d(u, Z) = min z Z d(u, z). Consider the map f from X to the space of random variabes L p (Ω, µ) that sends u to ξ u (where Ω is the probabiity space and µ is the probabiity measure on Ω). We say that f is a Fréchet embedding. Lemma 3.2. Every Fréchet embedding f is non-expanding. That is, f Lip 1. Proof. Consider a Fréchet embedding that sends u to ξ u = d(u, Z). For every u, v X, we have ξ u ξ v p = (E [ d(u, Z) d(v, Z) p by Exercise 2 1/p ]) (E [ d(u, v) p ]) 1/p = d(u, v). 3

4 Remark 3.3. If X is infinite, then the random variabe ξ u = d(u, Z) does not necessariy beong to L p (Ω, µ) (its p-norm might be infinite). However, we can define ξ u as ξ u = d(u, Z) d(x, Z), where x is some point in X. Then the proof of Lemma 3.2 shows that ξ u p d(u, x ) < and the map f : u ξ u is non-expanding. Theorem 3.4 (Bourgain s Theorem). Every metric space X on n points embeds into L p (X, µ) with distortion O(og n) (for every p 1). That is, c p (X) = O(og n). Proof. Let = og 2 n + 1. Construct a random set Z as foows. Choose s uniformy at random from {1,..., }. Initiay, et Z =. Add every point of X to Z with probabiity 1/2 s, independenty. Now et f be the Fréchet embedding that maps u X to random variabe ξ u = d(z, u). By Lemma 3.2, f is non-expanding. We are going to prove that for every u and v, f(u) f(v) p c d(u, v), for some absoute constant c. Note that it is sufficient to prove this statement for p = 1, since by Lyapunov s inequaity f(u) f(v) p f(u) f(v) 1. Consider two points u and v. Let = d(u, v)/2. Write, [ ] f(u) f(v) 1 = E [ d(u, Z) d(v, Z) ] = E dt [d(u,z),d(v,z)) [d(v,z),d(u,z)) (by Fubini s theorem) = Pr (d(u, Z) t < d(v, Z) or d(v, Z) t < d(u, Z)) dt Pr (d(u, Z) t < d(v, Z) or d(v, Z) t < d(u, Z)) dt. We now prove that Pr (d(u, Z) t < d(v, Z) or d(v, Z) t < d(u, Z)) Ω(1) if t (, ). That wi impy that f(u) f(v) 1 Ω(1) = Ω(1) d(u, v). We fix t (, ). Consider bas B t (u) and B t (v). Note that they are disjoint since 2t < 2 = d(u, v). Assume without oss of generaity that B t (u) B t (v). Denote m = B t (u). Let s = og 2 m + 1. Then m < 2 s 2m. Let E u be the event that d(u, Z) > t and E v be the event that d(v, Z) t. We have, Pr (d(u, Z) < t < d(v, Z) or d(v, Z) < t < d(u, Z)) Pr (E u and E v ). Note that the event E v occurs if and ony if there is a point in Z at distance at most t from v; that is, when B t (v) Z. The event E u occurs if and ony if B t (u) Z =. 4

5 Consider the event s = s. It happens with probabiity 1/. Conditioned on this event, events E u and E v are independent (since B t (u) and B t (v) are disjoint) and Pr(E u s = s ) = Pr (w / Z s = s ) = ( 1 1 ) ( = 1 1 ) m 1 2 s 2 s e. w B t(u) w B t(u) Pr(E v s = s ) = 1 Pr (w / Z s = s ) = 1 ( 1 1 ) ( ) m 2 s 2 s We get w B t(v) 1 1 e 1/2. w B t(v) Pr (d(u, Z) < t < d(v, Z) or d(v, Z) < t < d(u, Z)) Pr (E u and E v ) 1 Pr (E u s = s ) Pr (E v s = s ) Ω ( ) 1. Exercise 5. The set Z might be equa to in our proof, then random variabes ξ u = d(u, Z) are not we defined. Show how to fix this probem. Proof. There are many ways to fix this probem. For instance, we can add an extra point x to the metric space X, and define d(u, x ) = 2 diam(x), where diam(x) = max u,v X d(u, v). Then construct the set Z as before, except that aways add x to Z. Thus we ensure that Z. In other words, we can define ξ u as before if Z, and ξ u = 2 diam(x) if Z =. The rest of the proof goes through without any other changes. The proof of Bourgain s theorem provides an efficient randomized procedure for generating set Z. As presented here, this procedure gives an embedding ony in L p (Ω, µ) and not in N p. We aready know that if a set of n points embeds in L p (Ω, µ) with distortion D then it embeds in (n 2) p with distortion D. However, in fact, we need ony N = O((og n) 2 ) dimensions: for every vaue of s {1,..., } we make Θ(og n) sampes of the set Z. Then the tota number of sampes equas Θ((og n) 2 ). Using the Chernoff bound, it is easy to show that the distortion of the obtained embedding is O(og n) w.h.p. Fact 3.5 (Matoušek). Let D n,p be the smaest number D such that every metric space on n points embeds in p with distortion at most D n,p. Then ( ) og n D n,p = Θ. p 5