LECTURE NOTES ON DVORETZKY S THEOREM

Transcription

1 LECTURE NOTES ON DVORETZKY S THEOREM STEVEN HEILMAN Abstract. We present the first half of the paper [S]. In particular, the results below, unless otherwise state, shoul be attribute to G. Schechtman. Below, P an E enote probability an expectation, respectively. We try to make explicit the measure space in question by aing an appropriate subscript to P, E whenever necessary.. Introuction Our goal is to prove the following theorem Theorem 3 from [S], which proves Dvoretzky s Theorem Thm..7 with the best known ε epenence: Theorem.3: c > such that n N, ε >, every n-imensional norme linear space X amits a subspace Y X such that Y, l k + ε, an k > cε log/ε log n. Let us sketch the proof. If M := S n x µx is large, then we can use the proof an conclusion of Milman s Dvoretzky theorem Thm..5 to get a sphere of large imension. If M is small, then via our main Lemma Lemma., the assumptions of the Alon-Milman theorem Thm.. are satisfie. That is, we can fin a cube of large imension, which itself has a sphere of large imension. Thus, in either case, we have our esire result. Note that the new ingreient in this proof is the application of the Alon-Milman Theorem, which is enable by the crucial Lemma.. In the last class, Talagran s proof [T] of this theorem was presente by Evan Chou an Lukas Koehler. In Section, we prove Theorem.3 in more etail. Section 3 gives some preliminaries. Unfortunately, we will use a result Theorem.5 which was not yet proven in class. In the following lecture, Sean Li will prove this result, which is known as Milman s extension of Dvoretzky s theorem, with best known constants. Before we begin, we nee to cite some relevant theorems an efinitions. Remark.. Below, g, g,... will always esignate stanar normal real value ranom variables on a probability space Ω. Eg i =, Eg i =, P g i < t = t e x / x/ π. Also, r j t := sign sin j t, t [, ] enote the Raemacher functions. Moreover, c will be a constant that is allowe to change from line to line. Remark.. In the proofs of Lemma. an Theorem.3, we are going to use the triangle inequality in the form a+b + a b a + b + a b = b. Let α an β be inepenent ranom variables, with symmetric istributions, so P α < t = P α > t for all t >. Integrating separately for β < an β > an taking expecte values shows that E αa + βb E βb. That is, throwing out vectors Date: December 7th,.

2 STEVEN HEILMAN only makes the expectation smaller. One can also see this result as a consequence of the Contraction Principle. Remark.3. In the proof of Theorem.3, we nee: l m contains a subspace of c imension at least k = log/ε log m which is +ε-isomorphic to Eucliean space. This was almost proven in class actually it was assigne as an exercise. Recall: we consier an embeing T : l k l m efine by T x := { x, x i } m, {x i } m is an ε-net of Sk of size m = 3 k. ε That is, we flatten the sphere into an approximating polytope. Finally, such an ε-net exists by volume consierations which we prove in class, an taking logs shows log m = k log3/ε, as esire. Definition.4. Let X be a norme space. Define EX by { } n EX := sup E Ω g i ue i : n N, u: l n X, u =. X Theorem.5. Milman s Dvoretzky Theorem a function cε > such that, for all k cεex, l k +ε X. Actually, one may take k cε EX. That is, a linear invertible T : l k Y X with T T + ε. Remark.6. Sean Li will prove this theorem inepenently of our work. We will therefore use this result without further comment Theorem.7. Classical Dvoretzky Let X be a norme space of imension n. There exists a function cε > such that, for all k cε log n, l k +ε X. For an expression for cε, see Theorem.3. Now, recall three results from class. Lemma.8. Johnson-Linenstrauss Lemma Let H be a Hilbert space. ε,, cε > such that, n N, if x,..., x n H, then k cε log n an y,..., y n l k such that i, j x i x j H y i y j + ε x i x j H. Proof: Rough Sketch Using the probabilistic metho on the orthogonal group, fin a ranom projection that oes what we want. Lemma.9. Dvoretzky-Rogers Lemma Let be a norm on R N with unit ball K. Let E K be the John ellipsoi i.e. the ellipsoi of maximal volume in K. Then x,..., x N R N which are orthonormal with respect to, E, an such that x i N/N i+, i =,..., N. Proof: Rough Sketch Inuctively fin new vectors of maximal norm, an pay attention to the volume of the ellipsois. Theorem.. Alon-Milman, Talagran With the assumptions of Lemma., for any < ε <, a subspace of {span{e i } n } of imension k cn cε/ log L that is + ε-isomorphic to l k. Here c > is a universal constant. Remark.. In class, we prove k cn c log+ε log Mn see. Using the assumptions an conclusion of Lemma., we have M n L, an then using log + ε ε gives Theorem..

3 LECTURE NOTES ON DVORETZKY S THEOREM 3. Proof of Main Theorem With the preliminary results of the Appenix Section 3 an the following crucial Lemma., the proof of Theorem.3 will follow quickly. In the following Lemma, we use the notation of [T] in particular, the constants M an M n. Lemma.. Main Lemma, [S] Let be a norm on R N containing the unit Eucliean ball so that. Let {e i } n be an orthonormal sequence in RN with respect to satisfying e i /4 for all i, an n nm = EΩ g i e i L log n small rounness Recall M := S N a µa, with µ normalize Haar measure. Then, for all isjoint subsets σ,..., σ n {,..., n} with σ j = n for all j, a further subset J {,..., n } of carinality at least n/, an there are {x j } j J with x j = i σ j λ i e i, x j =, an M n := E t [,] r j tx j L large cube-ness j J Remark.. To summarize, we can rearrange an a together our n orthonormal vectors to get n/ vectors with small expecte length. Proof: Ieas: Gaussian concentration, the probabilistic metho, an symmetry. Corollary 3.6 applie to m = n an using e i /4 shows that E g i e i i σ j log n log n Define T : R N, R, by T {λ i } i σj = i σ j λ i e i. Using the orthonormality of the e i an that, the map T is -Lipschitz. So, we can apply 3 an Lemma 3.8 to get P i σj g i e i log n P g i e i E g i e i i σ j i σ j > log n e,π log n. So, for n 4 5,π, this probability is /, for every j. Let A j enote the event { i σ j g i e i > log n }. Let A enote the following event: there exists at least one subset J {,..., n } with J n / an A j occurs for all j J. Then P A /, from Proposition 3.9. Now, exten the probability space Ω to

4 4 STEVEN HEILMAN inclue inepenent Raemacher functions, apply our assumptions, an observe L n log n E g g i e i, by j= i σ j n = E r E g r j g i e i, by symmetry of the g i j= i σ j n E r E g r j g i e i j= i σ j A n E g E r r j g i e i j= i σ j A. Here we use the efinition of conitional expectation, an that P A /. In conclusion, by the efinition of A, if we choose ω A, then J {,..., n } with J n/ as above, so that A j hols for all j J. That is, x j := i σ j g i ωe i satisfies x j > log n, an by the inequality above an Remark., there exists ω A such that E r r j x j L log n. j J So, taking x j := x j / x j completes the Lemma, since then E r r j x j log n L, log n which is, as esire. j J Theorem.3. Dvoretzky, Best Constants, [S] c, > such that for all n N an all < ε < <, every n-imensional norme space amits a subspace Y with Y, l k cε + ε, an k > log/ε log n. Equivalently, every symmetric convex boy in R n amits a k-imensional section that contains an Eucliean ball an is containe in + ε times that ball, where cε k > log/ε log n. Proof: We may assume by taking an appropriate linear transformation that X = R n, an S n is the ellipsoi of maximal volume in K := B X,. By Dvoretzky-Rogers Lemma.9, {e,..., e n/ } R n orthonormal, with e i /4, i =,..., n/. Since S n K,. Define E := E n g ie i. Applying Theorem.5 recall the efinition of EX an E, Y X with Y, l k + ε, an k > cε E. So, we have two cases: Case large ball: ε E ε log/ε log n. Case large cube: ε E ε < log/ε log n. cε For Case, we have k > log/ε log n by assumption, so the theorem is proven. For Case, we apply Remark., use the efinition of E, an rewrite the assumption for this case to get small rounness: n/ E Ω g i e i E ε log ε log n 4

5 LECTURE NOTES ON DVORETZKY S THEOREM 5 We therefore apply Lemma. with L = ε log/ε, giving large cube-ness: M n/ L. logε By Alon-Milman Theorem., we can take m cn/ / log ε an Y X of imension m with Y, l m + ε. Then by Remark.3, Y contains a subspace Z of imension k where Z Y X an cε k c log/ε log m c log ε cε logε / log ε log n cε log log n, ε for ε small, where Z is +ε -isomorphic to l k an +ε +3ε for < ε <. The final statement of the theorem follows since an n-imensional ellipsoi has a spherical section of imension at least n/6 we prove this in class. Remark.4. Concerning the computation of ε near the en of the proof, we have c ε log ε logε / log ε log n cε log log n ε logε / log ε C log ε log ε C logε / log ε ε ε / log C ε ε /C+/ log ε ε /C / log ε For small C = c/ c >, the last inequality is only true for ε small. Acknowlegements: Thanks to Evan Chou an Assaf Naor for fixing an error in the proof of Lemma., an thanks to Sean Li for helpful iscussions. 3. Appenix: Preliminary Things Proposition 3.. Let g be a stanar normal real value gaussian ranom variable. We have the following estimates on the istribution function of g: a / λ e λ / x, for λ >. λ e x b / λ e λ / λ e x x, for λ >. c / λ e x x / λ e λ, as λ.

6 6 STEVEN HEILMAN Proof: Let λ >, an observe λ e x / = = e λ / = e λ / e y+λ / y, c.o.v. e y / e λy/ e λ / y = λ e λ /, e λy y, e y / [ λ e λy ] y= y= which proves a. Now, let λ > an observe, using λ > [λ ] λ 3 e λ / λ = e λ / + λ e λ / λ e λ / + 3λ 4 e λ / = 3λ 4 e λ / Therefore, the Funamental Theorem of Calculus gives λ e x / x λ 3x 4 e x / x, by monotonicity = λ λ 3 e λ / λ e λ /, by ifferentiating the integral,, since λ >, so /λ λ 3 > Actually, from the above sequence of inequalities ening at, we see that we have prove c as well using a too.. Proposition 3.. Let {g,..., g n } be n stanar real value gaussians. Then P max g i < c {, c <, c > log n,...,n, c >. Proof: P max g i < λ = P g < λ n, by efinition of max,...,n n λ = e x x /, by efinition of the g i π = λ So, setting λ = c log n we get λ n e x / x π n π λ e λ /, by Proposition 3.b.

7 LECTURE NOTES ON DVORETZKY S THEOREM 7 n P max g i < λ,...,n c π log n e log n c / n = c π n c /+. log n n Then, using the power series expansion of log recall, log x = x + x / + x 3 /3 + for x < shows that for c n = on log c n /n n = n log c n /n c n + Oc n /n. So, for c, log P max,...,n g i < c log n. An for c >, log P max,...,n g i < c log n, using Proposition 3.a. Proposition 3.3. Let x,..., x n be unit norm vectors in a norme space X. Let ε i be inepenent, P ε i = = P ε i = = /. Then for all a,..., a n R, n P εi ε i a i x i < max a i /. i n Proof: We may assume symmetry an rearrangement that a = max i n a i. Suppose a x + n i= ε ia i x i < a. Then n a n x ε i a i x i = a x + a x a x ε i a i x i i= i= n a x a x + ε i a i x i i= > a, using x i = an our assumption. Let D be the event { n ε ia i x i < a }, an let C be the event efine by { n ε ia i x i > a }. For {ε i } n D, we have shown: there exists a unique {ε i }n = {ε, ε,..., ε n } associate to {ε i } n with {ε i }n C. So P C P D, i.e. n n P ε i a i x i > a P ε i a i x i < a. Thus n P ε i a i x i max a i i n n n = P ε i a i x i < a + P ε i a i x i > a n P ε i a i x i < a, by efinition of a Remark 3.4. Letting x = x, a = a =, an a i = otherwise, we see that the Proposition is sharp.

8 8 STEVEN HEILMAN Proposition 3.5. Let x,..., x n be vectors in a norme space with x i /4 for all i. Let g,..., g n be stanar normal gaussian ranom variables on a probability space Ω. Then n log n P Ω g i x i < /3. Proof: n log n P Ω g i x i < n log n log n P Ω g i x i < < max g i x i i n log n + P Ω max g i x i i n n P Ω,εi ε i g i x i x i x i < max g i x i i n + P Ω max g i log n, since i n 5 x i 4 an g i is symmetric / + o /3, from Proposition 3.3 an Proposition 3.. In the penultimate equality, we first conition on g i, then integrate in Ω. Corollary 3.6. With the assumptions of Proposition 3.5, E Ω m g ix i log m/3 Definition 3.7. Let Y, Z be two Banach spaces finite or infinite imensional. Suppose there exists a linear isomorphism T : Y Z. Define the Banach-Mazur istance Y, Z as Y, Z = inf{ T T : T : Y Z, a linear isomorphism}. Note that ilating T has no effect on T T. Lemma 3.8. Gaussian Concentration Let F : R n, R be a Lipschitz function with constant σ. Then P F g,..., g n EF g,..., g n > C e C /π σ. Proof: Iea: Gaussian integration by parts. By a mollifying argument, we may assume F C. Let H = h,..., h n be equal in istribution to G = g,..., g n. Define G θ := G sin θ + H cos θ, θ, π/. Recall that the gaussian istribution is invariant uner rotation tg i + th i = t+t / g i = g i. So, G θ = G, θ G θ = G cos θ H sin θ = H, an G θ, θ G θ are inepenent, since they are jointly normal with zero covariance. That is G θ, θ G θ = G, H. Let E G, E H, E enote expectation with respect to G,

9 LECTURE NOTES ON DVORETZKY S THEOREM 9 H, an both G an H, respectively. For a convex function φ E G φf G E H F H E G E H φf G F H π/ π/ = Eφ θ F G θθ = Eφ π/ = Eφ π π = Eφ π/ π π Eφ π R nf G, H R nf G θ, θ G θ R nf G θ, θ G θ θ, by Jensen s R nf G θ, θ G θ θ, Jensen, Fubini, by equality of joint istributions. θ Let λ R an set φt = e λt. From basic properties of gaussians ag + bg = a + b / g, an R etx γx = R e tx γx = e t / we have E G expλf G E H F H E exp λ π n F Gh i x i = E exp λ π n / F n G h = E G exp λ π F G x i 8 x i expλ π σ /8, since F σ. This is our esire moment estimate, which gives our result via Markov s inequality. Let λ = 4C/π σ > an observe P F G EF H > C = P e λ F G EF H > e λc e λc Ee λ F G EF H e λc λf G EF H Ee, e λc e λ π σ /8 = e 4C /π σ e C /π σ = e C /π σ. using e λ F e λf + e λf Proposition 3.9. Let {A j } k j= be inepenent events on a probability space Ω with P A j / an k even. Define A as the event k { A := Aj k/ = #A j k }. Then P A /. j= Proof: The iea here is to compare the case P A j / to the case P Ãj = /. To o this, we nee to exten the probability space Ω. By extening Ω, let {y j } k j= be inepenent ranom variables that are uniform on [, ]. Define B j := {y j P A j }. By the uniformity of the y j, Bj = yj P A j = y j [,P A j] = Aj.

10 STEVEN HEILMAN So, by inepenence of the y j an A j which implies inepenence of the yj P A j an the A j, Bj = Aj. We therefore conclue that j j P #B j k/ = P #A j k/. Now, efine C j := {y j /}. Since P A j /, C j B j. So, {#C j k/} {#B j k/}. Since P C j = / for all j, an the C j are inepenent, P {#C j k/} /. One can see this, for example, by writing Ω = l C l Cl k k =: l D l, where this isjoint union runs over all multi-inices l with l = l,..., l k, an l i is either the complement operator C li i = Ci c, or the ientity Cli l = C i. Note that P D l = / k. Define l as the number of l i which are ientity operators. Note that D l {#C j k/} if an only if l k/. Now, by symmetry, for every D l = C l Cl k k with l > k/, there exists a unique l efine by D l = C lc C l kc k, which satisfies l < k/. This association l l therefore partitions Ω less sets with l = k/ into two isjoint sets of equal measure: l : l >k/ D l an l : l <k/ D l. Therefore, P #C j k/ = P l : l k/ D l = P l : l =k/d l + P l : l =k/ D l /. Combining,, an, we get P A /, as esire. References [D] R. Durrett, Probability: theory an examples. Thir eition. Duxbury Press, Belmont, CA, 5. [MS] V. Milman, an G. Schechtman, Asymptotic theory of finite-imensional norme spaces. With an appenix by M. Gromov. Lecture Notes in Mathematics,. Springer-Verlag, Berlin, 986. [S] G. Schechtman, Two observations regaring embeing subsets of Eucliean spaces in norme spaces. Av. Math. 6, no., 535. [T] M. Talagran, Embeing of l k an a theorem of Alon an Milman. Geometric aspects of functional analysis Israel, 99994, 8993, Oper. Theory Av. Appl., 77, Birkhäuser, Basel, 995. Suppose a = c, b =, an a, b, c, are inepenent. Then a + b = c +. P a + b < λ = γ Q P a < λ γ, b < γ = γ Q P a < λ γp b < γ, by inepenence = γ Q P c < λ γp < γ, by equality of istributions = γ Q P c < λ γ, < γ, by inepenence = P c + < λ