A glimpse into convex geometry. A glimpse into convex geometry

A glimpse into convex geometry 5 \ þ ÏŒÆ

Two basis reference: 1. Keith Ball, An elementary introduction to modern convex geometry 2. Chuanming Zong, What is known about unit cubes Convex geometry lies at the intersection of geometry, linear algebra, functional analysis, probability, etc.

Definition A subset K R n is called convex if tx + (1 t)y K, ( x, y K and 0 t 1). Convex body: compact convex subset with nonempty interior. It is called symmetric, if x K implies x K.

Basis facts about convex bodies Supporting hyperplane: each boundary point of a convex body admits a supporting hyperplane (not necessarily unique).

Basis facts about convex bodies Separation of disjoint convex bodies by hyperplane.

Hahn-Banach theorem Above two statements sound intuitively true, but an instructive proof invokes the famous Hahn-Banach theorem. Theorem If g : V R is a sublinear function, and f : U R is a linear functional on a subspace U of V, with f g, then there exists a linear extension f of f such that f = f on U; f g on V.

A function g : V R is sublinear if Positive homogeneity: f (λx) = λf (x), λ > 0; Subadditivity: f (x + y) f (x) + f (y). A convex body K containing origin defines a sublinear function (Minkowski functional): g(x) := inf{λ > 0 : x λk}. A symmetric convex body defines a norm on the underlying vector space, i.e., it is sublinear (triangle inequality) and λx = λ x.

Proof of existence of supporting hyperplane Pick x 0 K, then g(x 0 ) = 1 by definition. Let f : Rx 0 R be f (λx 0 ) = λg(x 0 ), λ R. It is clearly linear and dominated by g, thus there is a linear extension f, and f g. x K = g(x) 1 = f (x) 1. Thus, K { f 1}, i.e., f = 1 is a supporting hyperplane at x 0.

John s lemma Theorem Every n-dimensional convex body K contains a unique ellipsoid E of largest volume, and E K ne.

Distance between convex bodies How to say two convex bodies are alike? Cube and sphere seem to be two extremal examples: for symmetric convex bodies, a cube is made of fewest possible faces, a ball can be regarded as made of infinite many faces. We need to introduce a quantitative measurement for the distance between convex bodies. A general principle for introducing distance for convex bodies is that a convex body and its images under affine transformation are all regarded as one same object. In other words, the distance should be affine invariant.

Affine transformation: y = A x + x 0, where A is an invertible n n matrix. It is a combination of a linear transformation and a translation. It preserves linear structure and convexity. Distance: d(k, L) = inf{λ : K L λ K, } where K, L are linear images of K and L. For example: d(ball, cube) = n, and d(k, L) = 1 means K and L are identical up to affine transformation.

Even though a cube looks less and less like a ball as dimension grows, we will see that there do exist slice of cube which is almost round. More precisely, The cube in R n has almost spherical sections whose dimension k is roughly log n. In fact, this statement is true for all convex bodies. This is the content of Dvoretzky s Theorem which we are aiming for. A slice of n-cube has at most n-pair of faces. If the slice dimension is low, this sounds more plausible. So the significance is that k log n.

Brunn-Minkowski inequality Original form of Brunn: Theorem (Brunn) Let K be a convex body in R n, let u be a unit vector in R n, and for each r, let H r be the hyperplane x u = r. Then the function is concave on its support. r vol(k H r ) 1 n 1

1-d illustration

A novel view by Minkowski: the slice A r := K H r viewed as convex sets in R n, then for r < s < t with s = (1 λ)r + λt, A s includes the Minkowski sum (1 λ)a r + λa t := {(1 λ)x + λy : x A r, y A t }.

Brunn s inequality says: vol(a s ) 1 n 1 (1 λ)vol(ar ) 1 n 1 + λvol(at ) 1 n 1. Minkowski s formulation only involves two sets, which gets rid of the cross-section of convex body. Theorem (Brunn-Minkowski) If A and B are nonempty compact subsets of R n, then vol((1 λ)a + λb) 1 n (1 λ)vol(a) 1 n + λvol(b) 1 n.

Prékopa-Leindler inequality Brunn-Minkowski inequality can be derived by an inequality involving three functions: Theorem (Prékopa-Leindler) If f, g and h are nonnegative measurable functions on R n, λ (0, 1) and for all x and y, h((1 λ)x + λy) f (x) 1 λ g(y) λ, then h ( f ) 1 λ ( g) λ.

Take f, g, h to be the characteristic functions for A, B, and (1 λ)a + λb respectively, then Prekopa-Leindler implies the multiplicative form of Brunn-Minkowski: vol((1 λ)a + λb) vol(a) 1 λ vol(b) λ.

Theorem (Isoperimetric inequality) Among bodies of a given volume, Euclidean balls have lease surface area. The Minkowski sum A ɛ := A + ɛb can be regarded as ɛ neighborhood of A Surface area of a compact set A: vol( A) = lim ɛ 0 vol(a+ɛb n) vol(a) ɛ.

Then Brunn-Minkowski implies vol(a + ɛb n ) (vol(a) 1 n + ɛvol(bn ) 1 n ) n vol(a) + nɛvol(a) n 1 n vol(bn ) 1 n. Take a compact set A with same volume of B n, we get vol( A) nvol(b n ) = vol(s n 1 ). In other words, if B is a Euclidean ball of same volume with A, then vol(a ɛ ) vol(b ɛ ), ɛ > 0.

Concentration of measure on S n This formulation makes it possible to talk about isoperimetric inequality in a measured metric space (X, d, µ). P. Lévy proved isoperimetric inequality on sphere. As one might guess, if B is a spherical cap of same measure as A, then vol(a ɛ ) vol(b ɛ ), ɛ > 0.

On S n 1, the measure is the surface area normalized to be 1, the distance is the Euclidean distance inherited from R n. Take a subset A S n 1 with vol(a) = 1 2, then B is a half sphere and vol(a ɛ ) vol(b ɛ ) 1 e nɛ2 /2.

An estimate of the ɛ cap: ɛ cap S n 1 = spherical cone B n. (1 ɛ 2 ) n 2 e nɛ 2 /2.

A more striking fact occurs when taking a function f : S n 1 R which is 1-Lipschitz: i.e., f (θ) f (φ) θ φ. Then for the median M of f, we have vol({f M}) and vol({f M}) have measure at least half. At a point x distance within ɛ from {f M}({f M}), we have f (x) M + ɛ, (f (x) M ɛ) thus vol( f M > ɛ) 2e nɛ2 /2. Even though f could vary by 2 from definition, but f is almost constant!!!

Dvoretzky s Theorem Theorem There is a positive number c such that, for every ɛ > 0 and every natural number n, every symmetric convex body of dimension n has a slice of dimension k > cɛ 2 log(1 + ɛ 1 ) log n which is within distance 1 + ɛ of the k-dimensional Euclidean ball.

Strategy of the proof The distance is measured up to affine transformations, so assume the maximal ellipsoid of K is B 1, i.e., B 1 K nb 1. (John s Lemma) The symmetric convex body defines a norm on R n. K becomes the unit sphere under this norm. Take a k-dimensional slice, and consider the S k 1 in this slice. We will show that we can choose a slice, such that the restriction of the norm on this copy of S k 1 is almost constant. Consequently, that slice of K is close to S k 1.

: S n 1 R is 1-Lipschitz. Let be the Euclidean norm, since B 1 K, then x x. Thus x y x y x y. Due the concentration of measure discussed above, such function is close to a constant on a large portion, say N S n 1, and this constant is the median of the function, but it can be shown the average M = x dθ S n 1 has the same effect. Find a k-dimensional subspace H, such that H S n 1 lies (almost) in N.

Fix a slice say S k 1 := H S n 1, if we have suitably many points on this S k 1 which also lie in N, we are actually done given that points are well distributed all over S k 1, namely, the collection of points forms a δ-net. A collection of points {x 1,, x l } is called a δ-net of S k 1, if y S k 1, there exists some x k, such that y x k δ, The most economical way to arrange a δ-net is to take a maximum collection of points so that their δ/2 caps are disjoint (maximal δ-separated). Each cap has volume ( δ 2 )k 1, so there are at most ( 2 δ )k 1 many points in this δ-net.

Lemma If S S k 1 is a δ-net, such that M(1 γ) φ M(1 + γ), φ S, then θ S k 1, M 1 γ 2δ 1 δ θ M 1 + γ 1 δ. By choosing γ and δ small, M on whole S k 1.

The chance of each point lies outside of N is at most e nɛ2 /2 i.e., vol({ θ M > ɛ}) e nɛ2 /2. So if ( 2 δ )k 1 e nɛ2 /2 < 1, we can definitely find a k-dimensional slice S k 1 on which there exists a δ-net S, of at most ( 2 δ )k 1 members, such that x S, θ M ɛ. let ɛ = Mγ, we get k nm 2 M 1 γ 2δ 1 δ γ2 log(2/δ), and by the lemma, θ M 1 + γ 1 δ, θ S k 1.

It remains to show a lower bound of M which is roughly like log n n. This is done by computation M = x dθ. S n 1 We shift the computation to R n with the Gaussian measure, namely a probability measure µ with density (2π) n 2 e x 2 2. 1 2 (x2 1 + +x2 n ) dx 1 dx n = ( e x2 2 dx) n = ( 2π) n. R n e

Volume of unit ball: v n = 1 n vol(s n 1 ). v n : = Vol(B1 n ) = dx 1 dx n = 1 0 r n 1 dr B1 n π/2 0 2π sin ϕ1 n 2 dϕ 1 0 dϕ n 1 = vol(s n 1 ). n

Using polar coordinates: e 1 R n which implies 2 (x2 1 + +x2 n ) dx 1 dx n = (nv n ) v n = π n 2 Γ( n 2 + 1). 0 e r2 2 r n 1 dr, Thus, M = Sn 1 x dθ = Γ( n 2 ) 2Γ( n+1 2 ) x dµ(x). R n

Γ( n 2 ) is about 2Γ( n+1 n 1, so it remains to show 2 ) R n x dµ(x) > log n. Recall the norm is defined by the symmetric convex body K whose maximal ellipsoid is B n 1. Among all convex bodies K whose maximal ellipsoid is B n 1, the cube is the one which minimize R n x dµ(x). So just need to focus on the norm defined by cube, which is x = max 1 i n x i.

R n x dµ(x) is the average of x, the median R is about the same order. The median R : µ( x R) = µ( x R) = 1 2. µ( x R) = 1 2 implies the cube [ R, R]n has measure 1 2, i.e., µ([ R, R] n ) = ( 1 R e x 2 2 dx) n = 1 2π R 2. So R R R log n. Finally, x 2 e 2 dx should be about 1 log 2/n, which implies R n x dµ(x) R 2 > c log n, and the proof is completed.

Specific cube section Droretzky s theorem is a deep general theory on the existence of almost spherical slice of a convex body. we conclude this talk by presenting some questions concerning specific sections of cube. Two extremal questions present themselves very naturally: Problem What is the maximal or minimal area of an i-dimensional cross-section of the unit cube?

The lower bound is referred as Good s conjecture, and was proved by Hensley (1979) for i = n 1 and Vaaler (around the same time) for all i. Theorem Let I n denote a unit cube centered at origin, and H i denote an i-dimensional subspace, then v i (H i I n ) 1.

Concerning the upper bound, we have Theorem (Ball) v i (H i I n ) ( n i ) i 2, the upper bound is best possible if i n. Theorem (Ball) v i (H i I n ) 2 n i 2, the upper bound is optimal if i n 2.

α(n, i) maximal area of i-dimensional cross-section

Thank you!