Probabilistic Methods in Asymptotic Geometric Analysis.

Probabilistic Methods in Asymptotic Geometric Analysis. C. Hugo Jiménez PUC-RIO, Brazil September 21st, 2016. Colmea. RJ

1 Origin 2 Normed Spaces 3 Distribution of volume of high dimensional convex bodies 4 Concentration of measure 5 Geometry of Log-Concave Functions

Asymptotic Geometric Analysis. Origin Asymptotic Geometric Analysis has its origin in the interaction of Convex Geometry and Functional Analysis.

Convexity x YES x y y YES x x y y YES NO!

Convex bodies A convex body is a subset K R n which is convex, compact and has non-empty interior.

Classical Convex Geometry The main interest lies in the study of the geometry of convex bodies and related geometric inequalities in Euclidean Space of fixed dimension.

The Minkowski sum The minkowski sum of two sets is defined as A + B = {x + y : x A, y B}

The Minkowski sum The minkowski sum of two sets is defined as A + B = {x + y : x A, y B} = x A(x + B)

The Minkowski sum The minkowski sum of two sets is defined as A + B = {x + y : x A, y B} = x A(x + B) = {x R n : A (x B) }.

The Minkowski sum A B A + B

Brunn-Minkowski inequality Brunn-Minkowski inequality (1887) For any convex bodies K, L K + L 1 n K 1 n + L 1 n. with equality if and only if K and L are homothetic.

Brunn-Minkowski inequality Brunn-Minkowski inequality (1887) For any convex bodies K, L K + L 1 n K 1 n + L 1 n. with equality if and only if K and L are homothetic. Equivalent statement: Brunn-Minkowski inequality (1887) For any 0 λ 1 and any convex bodies K, L, λk + (1 λ)l K λ L 1 λ.

Functional Analysis Classical Functional Analysis is (often) devoted to the study of Infinite dimensional Spaces.

Functional Analysis Classical Functional Analysis is (often) devoted to the study of Infinite dimensional Spaces. However, a latter approach paid more attention to finite dimensional structures within these spaces. In the first few decades of its development, this approach was called local theory of normed spaces, which stood for investigating infinite dimensional Banach spaces via their finite dimensional features, for example subspaces or quotients.

1 Origin 2 Normed Spaces 3 Distribution of volume of high dimensional convex bodies 4 Concentration of measure 5 Geometry of Log-Concave Functions

Connection of Convex Geometry with normed spaces Given a convex body 0 int(k) we define its Minkowski Functional denoted by K as x K = min{λ > 0 : x λk} x R n

Connection of Convex Geometry with normed spaces Given a convex body 0 int(k) we define its Minkowski Functional denoted by K as x K = min{λ > 0 : x λk} x R n x

Connection of Convex Geometry with normed spaces Given a convex body 0 int(k) we define its Minkowski Functional denoted by K as x K = min{λ > 0 : x λk} x R n x If K = K then K defines a norm whose unit ball is precisely K.On the other hand, the unit ball of any normed space in R n is a centrally symmetric convex body

Some examples ( R n, x 1 = ) n x i i=1 ( n R n, x 2 = i=1 x 2 i ) 1/2 ( ) R n, x = max x i i

Banach-Mazur distance Definition X, Y Banach Spaces. The Banach-Mazur distance between them is defined as d BM (X, Y ) = inf{ T T 1 : T : X Y is an isomorphism}

Banach-Mazur distance Definition K = K, L = L R n convex o-symmetric bodies. infimum: over all T GL n. d BM (K, L) = inf{λ > 0 : K T (L) λk}, K L

Banach-Mazur distance Definition K = K, L = L R n convex o-symmetric bodies. infimum: over all T GL n. d BM (K, L) = inf{λ > 0 : K T (L) λk}, λk K T(L)

Banach-Mazur distance Definition K = K, L = L R n convex o-symmetric bodies. infimum: over all T GL n. d BM (K, L) = inf{λ > 0 : K T (L) λk}, Question: What is the maximum? When is it attained?

Banach-Mazur distance Definition K = K, L = L R n convex o-symmetric bodies. infimum: over all T GL n. d BM (K, L) = inf{λ > 0 : K T (L) λk}, Question: What is the maximum? When is it attained? (John)(1948) showed d BM (K, B n 2 ) n. As a consequence d BM (K, L) n

Banach-Mazur distance How good is the bound d BM (K, L) n?

Banach-Mazur distance How good is the bound d BM (K, L) n? For reference d BM (B n 2, Bn 1 ) = d BM(B n 2, Bn ) = n

Banach-Mazur distance How good is the bound d BM (K, L) n? For reference d BM (B n 2, Bn 1 ) = d BM(B n 2, Bn ) = n (Gluskin) There exists a universal constant c > 0 such that n we can always find K, L R n symmetric convex bodies with d BM (K, L) cn.

Idea He constructed n-dimensional convex bodies for which d BM (K, L) cn in the following way. Let K = conv{ k i } n i=1 and L = conv{ l i} n i=1 where every k i S n 1 is taken independently and uniformly with respect to the Lebesgue measure.

Hyperplane Conjecture A convex body K R n is called isotropic if it has volume K = 1, it is centered (i.e. its barycenter is at the origin) and there exists a constant L K > 0 such that x, y 2 dx = L K y 2 2, for all y R n. Equivalently, K K x, θ 2 dx = L K, for all θ S n 1. The constant L K is called isotropic constant of K.

Hyperplane Conjecture Given K we consider a random vector X uniformly distributed in K and, for every θ S n 1 the real random variable x, θ with density f θ (t) = K θ + tθ

Hyperplane Conjecture Given K we consider a random vector X uniformly distributed in K and, for every θ S n 1 the real random variable x, θ with density f θ (t) = K θ + tθ K is isotropic if all x, θ are centered and have the same variance.

Hyperplane Conjecture The hyperplane conjecture asks whether there exists a universal constant C > 0 such that L K C for every isotropic convex body K.

Hyperplane Conjecture The hyperplane conjecture asks whether there exists a universal constant C > 0 such that L K C for every isotropic convex body K. The previous conjecture is equivalent with the following. There exists an absolute constant c > 0 with the following property: for every n 1 and every centered convex body K of volume 1 in R n there exists θ S n 1 such that K θ c.

Hyperplane conjecture This question is rather hard and the only successful approach uses random polytopes.

Hyperplane conjecture This question is rather hard and the only successful approach uses random polytopes. Klartag and Kozma proved that if N > n and if G 1,..., G N are independent standard Gaussian random vectors in R n, then the isotropic constant of the random polytopes K N := conv{±g 1,..., ±G N } and C N := conv{g 1,..., G N } is bounded by an absolute constant C > 0 with probability greater than 1 Ce cn

1 Origin 2 Normed Spaces 3 Distribution of volume of high dimensional convex bodies 4 Concentration of measure 5 Geometry of Log-Concave Functions

distribution of volume of high dimensional convex bodies In very high dimension some counter intuitive phenomena occurs. σ The Lebesgue measure normalized on the ball B, σ(a) = vol(a B) vol(b)

distribution of volume of high dimensional convex bodies In very high dimension some counter intuitive phenomena occurs. σ The Lebesgue measure normalized on the ball B, σ(a) = vol(a B) vol(b) Where are the points of B typically located? (with respect to σ) σ(tb) = t n

The Euclidean Ball To compute the volume of Euclidean unit ball B2 n coordinates. we use polar

The Euclidean Ball To compute the volume of Euclidean unit ball B2 n we use polar coordinates. We can write the integral of a function on R n as f = nω n f (rθ)r n 1 dσ(θ)dr, R n r=0 S n 1 where in the previous integral we are normalizing by pulling the surface area on the sphere nω n so that σ = σ n 1 is the rotation-invariant measure on S n 1 of total mass 1.

The Euclidean Ball To find ω n we integrate the function x exp( 1 2 n x 2 ) 1 with both sides of the equality. f = R n R n n e xi 2 /2 dx = 1 n ( ) e xi 2 /2 dx i = ( 2π) n 1 and Hence nω n 0 S n 1 e r 2 /2 r n 1 dσdr = ω n 2 n/2 Γ ω n = πn/2 Γ ( n 2 + 1). ( n 2 + 1 ).

The Euclidean Ball ω n = πn/2 Γ ( n 2 + 1)

The Euclidean Ball From Stirling s formula ( n ) Γ 2 + 1 ω n = ( ) n 2πe So that ω n is roughly n, πn/2 Γ ( n 2 + 1) ( 2πe n/2 n ) (n+1)/2 2

The Euclidean Ball From Stirling s formula ( n ) Γ 2 + 1 ( So that ω n is roughly volume 1 has radius about which is pretty big. ω n = πn/2 Γ ( n 2 + 1) ) (n+1)/2 ( 2πe n/2 n 2 ) n 2πe n,or equivalently, the Euclidean ball of n 2πe,

The Euclidean Ball Let s go back to the question: how is the mass distributed?

The Euclidean Ball Let s go back to the question: how is the mass distributed? Let s estimate the (n 1)-dimensional volume of the slice through the center of the ball of volume 1. The ball has radius r = ωn 1/n. The slice is an (n 1)-dimensional ball of this radius, so its volume is ω n 1 r n 1 = ω n 1 ( 1 ω n ) (n 1)/n e. when n is large The slice at a distance x from the center has volume e (1 x 2 r 2 ) (n 1)/2 Since r is roughly n/(2πe), we get that e (1 2πex 2 ) (n 1)/2 e exp( πex 2 ). n

The Euclidean ball Therefore, if we project the mass distribution of the ball of volume 1 onto a single direction, we get a distribution that is approximately Gaussian with variance 1/(2πe).

The Euclidean ball Therefore, if we project the mass distribution of the ball of volume 1 onto a single direction, we get a distribution that is approximately Gaussian with variance 1/(2πe). Let s note that the variance doesn t depend on n, whereas the radius of the ball of volume 1 grows like n/(2πe). So almost all the volume stays within a slab of fixed width.

For the n-dimensional cube Let K R n a convex body with 0 int(k). If we denote by r(θ) the radius of K in the direction θ then the volume of K is r(θ) nω n s n 1 dsdσ = ω n r(θ) S n dσ(θ). n 1 S n 1 0 For the cube K = [ 1, 1] n that has volume 2 n the equality above says that Sn 1 r(θ) n = 2n ω n ( ) n 2n. πe

Volume distribution The volume on high dimensional convex bodies concentrates in places our low dimensional intuition considers small!

Central Limit Theorem [Klartag] Uniform distribution on high dimensional convex bodies has marginals that are approximately gaussian. Indeed, for an isotropic convex body K R n and a X uniformly distributed in K with some mild additional assumptions (k cn κ ) we have that there exists E G n,k with σ n,k (E) 1 e c n such that for any E E sup A E Prob{Proj E (X ) A} 1 (2π) k/2 e x 2 dx A 1 n κ, where the sup runs over all A E measurables sets and c, κ don t depend on the dimension.

Volumetric shape of Convex bodies A valid question seems to be: How do convex bodies in high dimension look like? V. Milman s picture of high dimensional convex body

1 Origin 2 Normed Spaces 3 Distribution of volume of high dimensional convex bodies 4 Concentration of measure 5 Geometry of Log-Concave Functions

Isoperimetric Inequality [Lévy] A S n+1 R n+2

Isoperimetric Inequality [Lévy] A S n+1 R n+2 B spherical cap with µ n+1 (B) = µ n+1 (A)

Isoperimetric Inequality [Lévy] A S n+1 R n+2 B spherical cap with µ n+1 (B) = µ n+1 (A) µ n+1 (B ε ) µ n+1 (A ε )

Isoperimetric Inequality [Lévy] A S n+1 R n+2 B spherical cap with µ n+1 (B) = µ n+1 (A) µ n+1 (B ε ) µ n+1 (A ε ) µ n+1 (A) 1/2 = µ n+1 (A ε ) 1 π 8 e ε2 n/2

General Problem (Ω, d, µ) metric probability space. For A Ω and ɛ > 0, the ɛ-expansion of A is defined by A d ɛ = {x Ω : d(x, a) ɛ for some a A}.

General Problem (Ω, d, µ) metric probability space. For A Ω and ɛ > 0, the ɛ-expansion of A is defined by α concentration function A d ɛ = {x Ω : d(x, a) ɛ for some a A}. α (Ω,d,µ) (ε) = sup{1 µ(a d ε ) : A Ω Borel, µ(a) 1/2}

Uniformly convex spaces (X, ) a normed space

Uniformly convex spaces (X, ) a normed space δ modulus of convexity { δ(ε) = inf 1 x + y 2 } : x, y B, x y ε

Uniformly convex spaces (X, ) a normed space δ modulus of convexity { δ(ε) = inf 1 x + y 2 } : x, y S, x y = ε

Uniformly convex spaces (X, ) a normed space δ modulus of convexity { δ(ε) = inf 1 x + y 2 } : x, y S, x y = ε (X, ) is uniformly convex if ε > 0, δ(ε) > 0

Result by Arias de Reyna, Ball, and Villa X n-dimensional uniformly convex normed space

Result by Arias de Reyna, Ball, and Villa X n-dimensional uniformly convex normed space B closed unit ball of X

Result by Arias de Reyna, Ball, and Villa X n-dimensional uniformly convex normed space B closed unit ball of X the Lebesgue probability on B

Result by Arias de Reyna, Ball, and Villa X n-dimensional uniformly convex normed space B closed unit ball of X the Lebesgue probability on B A 1/2 = A ε > 1 2e 2nδ(ε)

Result by Arias de Reyna, Ball, and Villa X n-dimensional uniformly convex normed space B closed unit ball of X the Lebesgue probability on B A 1/2 = A ε > 1 2e 2nδ(ε) ϕ(ε) 2e 2nδ(ε)

Proof using B-M by A-B-V Geometric Brunn-Minkowski inequality A + A 2 A 1/2 A 1/2

Proof using B-M by A-B-V Geometric Brunn-Minkowski inequality A + A 2 A 1/2 A 1/2 A B

Proof using B-M by A-B-V Geometric Brunn-Minkowski inequality A + A 2 A 1/2 A 1/2 A B the complementary of A ε : A = {y B : d(y, A) > ε}

Proof using B-M by A-B-V Geometric Brunn-Minkowski inequality A + A 2 A 1/2 A 1/2 A B the complementary of A ε : A = {y B : d(y, A) > ε} A + A (1 δ(ε))b 2

Proof using B-M by A-B-V Geometric Brunn-Minkowski inequality A + A 2 A 1/2 A 1/2 A B the complementary of A ε : A = {y B : d(y, A) > ε} A + A 2 (1 δ(ε))b A A (1 δ(ε)) 2n e 2nδ(ε)

Dvoretzky s Theorem Theorem Let X = R n, be an n-dimensional normed space with unit ball K. Consider M = S x dµ, b = sup n 1 x S n 1 x. Then there exists a subspace E of dimension k c(ε)n ( ) M 2 b such that for all x E or equivalently (1 ε)m x x (1 + ε)m x 1 (1 + ε)m (Bn 2 E) K E which implies d BM (E, l k 2 ) 1+ε 1 ε. 1 (1 ε)m (Bn 2 E)

Dvoretzky s Theorem Milman s picture of high dimensional convex body adapted by R. Vershynin V.

1 Origin 2 Normed Spaces 3 Distribution of volume of high dimensional convex bodies 4 Concentration of measure 5 Geometry of Log-Concave Functions

Log-concave functions f : R n [0, ) is log-concave if f (x) = e u(x) with u : R n (, ] convex.

Log-concave functions f : R n [0, ) is log-concave if f (x) = e u(x) with u : R n (, ] convex. K convex body χ K (x) is log-concave

Log-concave functions f : R n [0, ) is log-concave if f (x) = e u(x) with u : R n (, ] convex. K convex body χ K (x) is log-concave The class of log-concave functions is the smallest class that contains the densities of the marginals of uniform probabilities on convex bodies.

Prékopa-Leindler inequality Brunn-Minkowski inequality (1887) For any 0 λ 1 λk + (1 λ)l K λ L 1 λ.

Prékopa-Leindler inequality Brunn-Minkowski inequality (1887) For any 0 λ 1 Prékopa-Leindler inequality (1971) λk + (1 λ)l K λ L 1 λ. For any three integrable functions f, g, h : R n R + and λ [0, 1] such that for any x, y R n h(λx + (1 λ)y) f λ (x)g 1 λ (y) we have ( h(z)dz ) λ ( f (x)dx g(y)dy) 1 λ.

Prékopa-Leindler inequality If K, L convex bodies, taking f = χ K, g = χ L, h = χ λk+(1 λ)l we obtain λk + (1 λ)l K λ L 1 λ. Multiplicative Brunn-Minkowski inequality

Translation of notation {K convex bodies} {χ K characteristic functions}

Translation of notation {K convex bodies} {χ K characteristic functions} Volume K integral R n f

Translation of notation {K convex bodies} {χ K characteristic functions} Volume K integral R n f ( ) χ K = K R n

Translation of notation {K convex bodies} {χ K characteristic functions} Volume K integral R n f ( ) χ K = K R n K + L sum f g(z) = sup f (x)g(y) Asplund product 2z=x+y

Translation of notation {K convex bodies} {χ K characteristic functions} Volume K integral R n f ( ) χ K = K R n K + L sum f g(z) = sup f (x)g(y) Asplund product 2z=x+y (χ K χ L = χ K+L )

Translation of notation {K convex bodies} {χ K characteristic functions} Volume K integral R n f ( ) χ K = K R n K + L sum f g(z) = sup f (x)g(y) Asplund product 2z=x+y (χ K χ L = χ K+L ) K (x + L) f g(x) = f (z)g(x z)dz convolution R n

Thank you!