Lecture 19. sup y 1,..., yn B d n

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1 STAT 06A: Polyomials of adom Variables Lecture date: Nov Lecture 19 Grothedieck s Iequality Scribe: Be Hough The scribes are based o a guest lecture by ya O Doell. I this lecture we prove Grothedieck s iequality, which states that for all real matrices there is a uiversal costat K < (idepedet of ) such that x 1,...,x [ 1,1] y 1,...,y [ 1,1] A i,j x i y j 1 K i,j=1 x 1,..., x B d y 1,..., y B d A i,j x i y j, (1) i,j=1 where B d = { z d : z 1 } is the ball of radius 1 i d. The optimal costat K for which equatio (1) holds is called the Grothedieck costat, K G. We observe that the rema i (1) are obtaied whe x i, y j { 1, 1}, ad x i, y j = 1, so it is equivalet to replace [ 1, 1] with { 1, 1} ad B d with S d = { z d : z = 1 } i (1). Moreover, by takig the vectors x i, y j to all lie alog a commo lie i d, we see that the reverse iequality to equatio (1) holds with K = 1. 1 Why do we care? Oe motivatio for studyig Grothedieck s iequality is that it provides a method for rapidly approximatig the cut orm of a matrix. Defiitio 1 The cut orm of a matrix A is defied to be A c = max A i,j = max I [] x J [] i I i,y j {0,1} A i,j x i y j i,j=1. () Here [] deotes {1,,..., }. j J Let us deote the left had side of (1) by A 1. A easy computatio verifies that Exercise A c A 1 4 A c. The right had side of (1) ca be computed i polyomial time usig semi-defiite programmig methods. O the other had, computig the cut orm of a matrix is NP-hard. 19-1

2 Grothedieck s costat is kow ot to be too big, so this approach provides a reasoable method for estimatig the cut orm. Grothedieck s iequality also has importat cosequeces i Baach spaces. Ideed, it implies the followig: Corollary 3 Every bouded liear operator T from a L 1 space X to a Hilbert space is absolutely summable. That is, if x 1, x, X have the property that i x i coverges regardless of re-arragemet, the i T x i <. Kow bouds o K G Whe Grothedieck origially proved the iequality i 1956, he showed that 1.57 π K G sih ( ) π.3. iesz ( 74) improved the upper boud slightly to.6, ad Krivie ( 79) showed that K G π , (3) sih ad cojectured that this value is correct. A.M. Davie ( 84, upublished) improved the lower boud. Idepedetly, Jim eeds (c 91, upublished) improved the lower boud to approximately We shall discuss eeds proof below. 3 Proof of lower boud for K G Observe that x i, y j B d i,j=1 A i,j x i y j = f:[] B d g:[] B d f(i), i=1 A i,j g(j). (4) j=1 This quatity is clearly icreasig i d, we are iterested i uderstadig by how much it ca icrease as d icreases from 1 to. Scalig the matrix A (which has o ifluece o the Grothedieck s iequality), the above may be expressed as f,g:[] B d [] f, Ag dµ (5) where µ is the uiform measure o []. For ay value of, this quatity equals f, Ag dγ = f,g: B d f: B d Ag(x) dγ(x). (6) 19-

3 where γ deotes the -dimesioal Gaussia distributio, ad A is ow a appropriate liear operator o fuctios i Gaussia space. Coversely, as, ay liear operator o Gaussia space ca be approximated arbitrarily well by discrete operators, hece it suffices to cosider by how much (6) ca icrease as d icreases from 1 to. I particular, to obtai a good lower boud for K G, we shall exhibit a operator A for which (6) is large as d, ad small for d = 1. To defie this operator A, we first recall that ay g : B d has a Hermite expasio. That is: g(x) = S N ĝ(s)h S (x) (7) where ĝ(s) B d ad H S (x) = i=1 h S i (x i ) where h j are the 1-dimesioal Hermite polyomials. It is easy to show that the H S s are orthoormal: { 1 if S = T H S (x)h T (x) = (8) 0 if S T Moreover, H 0 = 1 ad if e i deotes the i th stadard basis vector i, the H ei = x i. Let us refer to S = S i as the level of S, ad defie P 1 to be the projectio to level 1 operator. That is, P 1 g(x) = i=1 ĝ(e i)x i. Defie A = P 1 λ 1, where 1 deotes the idetity operator, ad we shall choose λ later. Our first claim gives a lower boud for (6) for large d (ote we take = d here for coveiece). Propositio 4 g: B Ag 1 λ O ( ) 1. (9) Proof: To prove this lower boud, it suffices to exhibit a fuctio g for which the claimed lower boud holds. Set g(x) = ad compute Now we have that x x (P 1 g)(x) = ĝ(e i ) = = x i ĝ(e i ). (10) i=1 y i g(y)dγ(y) (11) y y dγ(y), (1) y y y i ad a straightforward computatio shows that { y i y j y dγ(y) = 0 if a := 1 x dγ(x) if i j i = j. (13) 19-3

4 Therefore, (P 1 g)( x) = a x ad we compute Ag dγ( x) = (P 1 g)( x) λg(x) dγ( x) (14) = a x λ x dγ( x) (15) x = (a λ ) x dγ( x) (16) x = a x λ (17) = 1 ( ) x λ (18) Theorem 5 The quatity is maximized by a fuctio g of the followig form: 1. g (x) depeds oly o x 1. g: { 1,1} Ag dγ( x) (19). g (x) is a odd fuctio of x 1, ad there exists a costat a so that g( x) = 1 for x 1 > a ad g( x) = 1 for 0 x 1 a. emark 6 Gratig Theorem 5, oe ca solve for the optimal costats a ad λ ad derive a lower boud for K G. The resultig expressio is ugly, so we wo t repeat the computatio here. Proof: If g : { 1, 1}, the P 1 g is liear so (P 1 g)(x) = i=1 b ix i for some costat b i. otatig space appropriately, we may assume that b = (b, 0,..., 0) for some 0 b 1. This rotatio will ot ifluece the itegral Ag dγ( x), sice γ is rotatioally symetric. Defie α : [ 1, 1] by α(z) = E[g P 1 g = z], ad observe that 1 α(z) 1 for all z, ad P[g = 1 P 1 g = z] = α(z) (0) P[g = 1 P 1 g = z] = 1 1 α(z). (1) Defie σ = ĝ(e 1 ) = x 1 g( x)dγ( x) = zα(z)ϕ(z)dz () 19-4

5 where ϕ(z) is the desity of a 1-dimesioal Gaussia radom variable. We compute: Ag dγ( x) = (P 1 g)( x) λg( x) dγ( x) (3) = [( α(z)) σz λ + (1 1 α(z)) σz + λ ]ϕ(z)dz (4) 1 = ( σz λ + σz + λ )ϕ(z)dz + α(z)ψ(z)ϕ(z) (5) where ψ(z) = 1 ( σz λ σz + λ ). To prove our theorem, we assume that σ is fixed, ad cosider the followig liear program: LP σ : Maximize: subject to the costraits: Cost. + ψ(z)ϕ(z)α(z)dz (6) 1 α(z) 1 for all z 0 ad σ = zϕ(z)α(z)dz. (7) Sice ϕ(z) is eve ad ψ(z) is odd, we see that there will be a optimizig α which is odd (replace α(z) with ( 1 ψ(z) (α(z) α( z)) if ecessary). Now, sice z is strictly icreasig o [0, ] we see that the optimal α will satisfy α(z) = 1 for z > a ad α(z) = 1 for 0 z a. (If α(z) is ot of this form, the we ca perturb α slightly to icrease the value of ψ(z)ϕ(z)α(z)dz while still satisfyig the costraits). It ow follows from (0) ad (1) that g must have the same form. 19-5