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A glimpse into convex geometry

A glimpse into convex geometry

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A glimpse into convex geometry

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.

A glimpse into convex geometry

Definition

A subset K ⊂ Rn 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.

A glimpse into convex geometry

Definition

A subset K ⊂ Rn 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.

A glimpse into convex geometry

Definition

A subset K ⊂ Rn 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.

A glimpse into convex geometry

Basis facts about convex bodies

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

A glimpse into convex geometry

Basis facts about convex bodies

Separation of disjoint convex bodies by hyperplane.

A glimpse into convex geometry

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

• n U;

˜ f ≤ g

• n V .

A glimpse into convex geometry

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||.

A glimpse into convex geometry

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||.

A glimpse into convex geometry

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||.

A glimpse into convex geometry

Proof of existence of supporting hyperplane

Pick x0 ∈ ∂K, then g(x0) = 1 by definition. Let f : Rx0 → R be f (λx0) = λg(x0), 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 x0.

A glimpse into convex geometry

John’s lemma

Theorem

Every n-dimensional convex body K contains a unique ellipsoid E

• f largest volume, and

E ⊂ K ⊂ nE.

A glimpse into convex geometry

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.

A glimpse into convex geometry

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.

A glimpse into convex geometry

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.

A glimpse into convex geometry

Affine transformation: − → y = A · − → x + − → x0, 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.

A glimpse into convex geometry

Affine transformation: − → y = A · − → x + − → x0, 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.

A glimpse into convex geometry

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 Rn 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

• f 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.

A glimpse into convex geometry

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 Rn 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

• f 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.

A glimpse into convex geometry

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 Rn 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

• f 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.

A glimpse into convex geometry

Brunn-Minkowski inequality

Original form of Brunn:

Theorem (Brunn)

Let K be a convex body in Rn, let u be a unit vector in Rn, and for each r, let Hr be the hyperplane x · u = r. Then the function r → vol(K ∩ Hr)

1 n−1

is concave on its support.

A glimpse into convex geometry

1-d illustration

A glimpse into convex geometry

A novel view by Minkowski: the slice Ar := K ∩ Hr viewed as convex sets in Rn, then for r < s < t with s = (1 − λ)r + λt, As includes the Minkowski sum (1 − λ)Ar + λAt := {(1 − λ)x + λy : x ∈ Ar, y ∈ At}.

A glimpse into convex geometry

Brunn’s inequality says: vol(As)

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 Rn, then vol((1 − λ)A + λB)

1 n ≥ (1 − λ)vol(A) 1 n + λvol(B) 1 n .

A glimpse into convex geometry

Brunn’s inequality says: vol(As)

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 Rn, then vol((1 − λ)A + λB)

1 n ≥ (1 − λ)vol(A) 1 n + λvol(B) 1 n .

A glimpse into convex geometry

Pr´ ekopa-Leindler inequality

Brunn-Minkowski inequality can be derived by an inequality involving three functions:

Theorem (Pr´ ekopa-Leindler)

If f , g and h are nonnegative measurable functions on Rn, λ ∈ (0, 1) and for all x and y, h((1 − λ)x + λy) ≥ f (x)1−λg(y)λ, then

• h ≥ (
• f )1−λ(
• g)λ.

A glimpse into convex geometry

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−λvol(B)λ.

A glimpse into convex geometry

Theorem (Isoperimetric inequality)

Among bodies of a given volume, Euclidean balls have lease surface area. The Minkowski sum Aǫ := A + ǫB can be regarded as ǫ neighborhood of A Surface area of a compact set A: vol(∂A) = limǫ→0

vol(A+ǫBn)−vol(A) ǫ

.

A glimpse into convex geometry

Theorem (Isoperimetric inequality)

Among bodies of a given volume, Euclidean balls have lease surface area. The Minkowski sum Aǫ := A + ǫB can be regarded as ǫ neighborhood of A Surface area of a compact set A: vol(∂A) = limǫ→0

vol(A+ǫBn)−vol(A) ǫ

.

A glimpse into convex geometry

Then Brunn-Minkowski implies vol(A + ǫBn) ≥ (vol(A)

1 n + ǫvol(Bn) 1 n )n

≥ vol(A) + nǫvol(A)

n−1 n vol(Bn) 1 n .

Take a compact set A with same volume of Bn, we get vol(∂A) ≥ nvol(Bn) = vol(Sn−1). In other words, if B is a Euclidean ball of same volume with A, then vol(Aǫ) ≥ vol(Bǫ), ∀ǫ > 0.

A glimpse into convex geometry

Then Brunn-Minkowski implies vol(A + ǫBn) ≥ (vol(A)

1 n + ǫvol(Bn) 1 n )n

≥ vol(A) + nǫvol(A)

n−1 n vol(Bn) 1 n .

Take a compact set A with same volume of Bn, we get vol(∂A) ≥ nvol(Bn) = vol(Sn−1). In other words, if B is a Euclidean ball of same volume with A, then vol(Aǫ) ≥ vol(Bǫ), ∀ǫ > 0.

A glimpse into convex geometry

Concentration of measure on Sn

This formulation makes it possible to talk about isoperimetric inequality in a measured metric space (X, d, µ). P. L´ evy proved isoperimetric inequality on sphere. As one might guess, if B is a spherical cap of same measure as A, then vol(Aǫ) ≥ vol(Bǫ), ∀ǫ > 0.

A glimpse into convex geometry

On Sn−1, the measure is the surface area normalized to be 1, the distance is the Euclidean distance inherited from Rn. Take a subset A ⊂ Sn−1 with vol(A) = 1

2, then B is a half sphere and

vol(Aǫ) ≥ vol(Bǫ) ≥ 1 − e−nǫ2/2.

A glimpse into convex geometry

An estimate of the ǫ cap: ǫ−cap

Sn−1 = spherical cone Bn

. (1 − ǫ2)

n 2 ≤ e−nǫ2/2.

A glimpse into convex geometry

A more striking fact occurs when taking a function f : Sn−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 ǫ from {f ≤ M}({f ≥ M}), we have f (x) ≤ M + ǫ, (f (x) ≥ M − ǫ) thus vol(|f − M| > ǫ) ≤ 2e−nǫ2/2. Even though f could vary by 2 from definition, but f is almost constant!!!

A glimpse into convex geometry

A more striking fact occurs when taking a function f : Sn−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 ǫ from {f ≤ M}({f ≥ M}), we have f (x) ≤ M + ǫ, (f (x) ≥ M − ǫ) thus vol(|f − M| > ǫ) ≤ 2e−nǫ2/2. Even though f could vary by 2 from definition, but f is almost constant!!!

A glimpse into convex geometry

Dvoretzky’s Theorem

Theorem

There is a positive number c such that, for every ǫ > 0 and every natural number n, every symmetric convex body of dimension n has a slice of dimension k > cǫ2 log(1 + ǫ−1) log n which is within distance 1 + ǫ of the k-dimensional Euclidean ball.

A glimpse into convex geometry

Strategy of the proof

The distance is measured up to affine transformations, so assume the maximal ellipsoid of K is B1, i.e., B1 ⊂ K ⊂ nB1. (John’s Lemma) The symmetric convex body defines a norm || · || on Rn. ∂K becomes the unit sphere under this norm. Take a k-dimensional slice, and consider the Sk−1 in this slice. We will show that we can choose a slice, such that the restriction of the norm || · || on this copy of Sk−1 is almost constant. Consequently, that slice of K is close to Sk−1.

A glimpse into convex geometry

Strategy of the proof

The distance is measured up to affine transformations, so assume the maximal ellipsoid of K is B1, i.e., B1 ⊂ K ⊂ nB1. (John’s Lemma) The symmetric convex body defines a norm || · || on Rn. ∂K becomes the unit sphere under this norm. Take a k-dimensional slice, and consider the Sk−1 in this slice. We will show that we can choose a slice, such that the restriction of the norm || · || on this copy of Sk−1 is almost constant. Consequently, that slice of K is close to Sk−1.

A glimpse into convex geometry

Strategy of the proof

The distance is measured up to affine transformations, so assume the maximal ellipsoid of K is B1, i.e., B1 ⊂ K ⊂ nB1. (John’s Lemma) The symmetric convex body defines a norm || · || on Rn. ∂K becomes the unit sphere under this norm. Take a k-dimensional slice, and consider the Sk−1 in this slice. We will show that we can choose a slice, such that the restriction of the norm || · || on this copy of Sk−1 is almost constant. Consequently, that slice of K is close to Sk−1.

A glimpse into convex geometry

◮ || · || : Sn−1 → R is 1-Lipschitz. Let | · | be the Euclidean

norm, since B1 ⊂ 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 ⊂ Sn−1, and this constant is the median of the function, but it can be shown the average M =

• Sn−1 ||x||dΘ

has the same effect.

◮ Find a k-dimensional subspace H, such that H ∩ Sn−1 lies

(almost) in N.

A glimpse into convex geometry

◮ || · || : Sn−1 → R is 1-Lipschitz. Let | · | be the Euclidean

norm, since B1 ⊂ 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 ⊂ Sn−1, and this constant is the median of the function, but it can be shown the average M =

• Sn−1 ||x||dΘ

has the same effect.

◮ Find a k-dimensional subspace H, such that H ∩ Sn−1 lies

(almost) in N.

A glimpse into convex geometry

◮ || · || : Sn−1 → R is 1-Lipschitz. Let | · | be the Euclidean

norm, since B1 ⊂ 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 ⊂ Sn−1, and this constant is the median of the function, but it can be shown the average M =

• Sn−1 ||x||dΘ

has the same effect.

◮ Find a k-dimensional subspace H, such that H ∩ Sn−1 lies

(almost) in N.

A glimpse into convex geometry

◮ Fix a slice say Sk−1 := H ∩ Sn−1, if we have suitably many

points on this Sk−1 which also lie in N, we are actually done given that points are well distributed all over Sk−1, namely, the collection of points forms a δ-net.

◮ A collection of points {x1, · · · , xl} is called a δ-net of Sk−1, if

∀y ∈ Sk−1, there exists some xk, such that |y − xk| ≤ 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.

A glimpse into convex geometry

◮ Fix a slice say Sk−1 := H ∩ Sn−1, if we have suitably many

points on this Sk−1 which also lie in N, we are actually done given that points are well distributed all over Sk−1, namely, the collection of points forms a δ-net.

◮ A collection of points {x1, · · · , xl} is called a δ-net of Sk−1, if

∀y ∈ Sk−1, there exists some xk, such that |y − xk| ≤ 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.

A glimpse into convex geometry

◮ Fix a slice say Sk−1 := H ∩ Sn−1, if we have suitably many

points on this Sk−1 which also lie in N, we are actually done given that points are well distributed all over Sk−1, namely, the collection of points forms a δ-net.

◮ A collection of points {x1, · · · , xl} is called a δ-net of Sk−1, if

∀y ∈ Sk−1, there exists some xk, such that |y − xk| ≤ 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.

A glimpse into convex geometry

Lemma

If S ⊂ Sk−1 is a δ-net, such that M(1 − γ) ≤ ||φ|| ≤ M(1 + γ), ∀φ ∈ S, then ∀θ ∈ Sk−1, M 1 − γ − 2δ 1 − δ ≤ ||θ|| ≤ M 1 + γ 1 − δ . By choosing γ and δ small, || · || ∼ M on whole Sk−1.

A glimpse into convex geometry

◮ The chance of each point lies outside of N is at most e−nǫ2/2

i.e., vol({|||θ|| − M| > ǫ}) ≤ e−nǫ2/2.

◮ So if

(2 δ )k−1 · e−nǫ2/2 < 1, we can definitely find a k-dimensional slice Sk−1 on which there exists a δ-net S, of at most ( 2

δ)k−1 members, such that

∀x ∈ S, |||θ|| − M| ≤ ǫ.

◮ let ǫ = Mγ, we get k ∼ nM2 γ2 log(2/δ), and by the lemma,

M 1 − γ − 2δ 1 − δ ≤ ||θ|| ≤ M 1 + γ 1 − δ , ∀θ ∈ Sk−1.

A glimpse into convex geometry

◮ The chance of each point lies outside of N is at most e−nǫ2/2

i.e., vol({|||θ|| − M| > ǫ}) ≤ e−nǫ2/2.

◮ So if

(2 δ )k−1 · e−nǫ2/2 < 1, we can definitely find a k-dimensional slice Sk−1 on which there exists a δ-net S, of at most ( 2

δ)k−1 members, such that

∀x ∈ S, |||θ|| − M| ≤ ǫ.

◮ let ǫ = Mγ, we get k ∼ nM2 γ2 log(2/δ), and by the lemma,

M 1 − γ − 2δ 1 − δ ≤ ||θ|| ≤ M 1 + γ 1 − δ , ∀θ ∈ Sk−1.

A glimpse into convex geometry

◮ The chance of each point lies outside of N is at most e−nǫ2/2

i.e., vol({|||θ|| − M| > ǫ}) ≤ e−nǫ2/2.

◮ So if

(2 δ )k−1 · e−nǫ2/2 < 1, we can definitely find a k-dimensional slice Sk−1 on which there exists a δ-net S, of at most ( 2

δ)k−1 members, such that

∀x ∈ S, |||θ|| − M| ≤ ǫ.

◮ let ǫ = Mγ, we get k ∼ nM2 γ2 log(2/δ), and by the lemma,

M 1 − γ − 2δ 1 − δ ≤ ||θ|| ≤ M 1 + γ 1 − δ , ∀θ ∈ Sk−1.

A glimpse into convex geometry

◮ It remains to show a lower bound of M which is roughly like

• log n

n . This is done by computation

M =

• Sn−1 ||x||dΘ.

◮ We shift the computation to Rn with the Gaussian measure,

namely a probability measure µ with density (2π)− n

2 e− |x|2 2 .

• Rn e− 1

2 (x2 1 +···+x2 n)dx1 · · · dxn = (

∞

−∞

e− x2

2 dx)n = (

√ 2π)n.

A glimpse into convex geometry

◮ It remains to show a lower bound of M which is roughly like

• log n

n . This is done by computation

M =

• Sn−1 ||x||dΘ.

◮ We shift the computation to Rn with the Gaussian measure,

namely a probability measure µ with density (2π)− n

2 e− |x|2 2 .

• Rn e− 1

2 (x2 1 +···+x2 n)dx1 · · · dxn = (

∞

−∞

e− x2

2 dx)n = (

√ 2π)n.

A glimpse into convex geometry

Volume of unit ball: vn = 1

nvol(Sn−1).

vn : = Vol(Bn

1 ) =

• Bn

1

dx1 · · · dxn = 1 rn−1dr π/2 sin ϕn−2

1

dϕ1 · · · 2π dϕn−1 = vol(Sn−1) n .

A glimpse into convex geometry

◮ Using polar coordinates:

• Rn e− 1

2 (x2 1 +···+x2 n)dx1 · · · dxn = (nvn)

∞ e− r2

2 rn−1dr,

which implies vn = π

n 2

Γ( n

2 + 1). ◮ Thus,

M =

• Sn−1 ||x||dΘ =

Γ( n

2)

√ 2Γ( n+1

2 )

• Rn ||x||dµ(x).

A glimpse into convex geometry

◮ Using polar coordinates:

• Rn e− 1

2 (x2 1 +···+x2 n)dx1 · · · dxn = (nvn)

∞ e− r2

2 rn−1dr,

which implies vn = π

n 2

Γ( n

2 + 1). ◮ Thus,

M =

• Sn−1 ||x||dΘ =

Γ( n

2)

√ 2Γ( n+1

2 )

• Rn ||x||dµ(x).

A glimpse into convex geometry

Γ( n

2 )

√ 2Γ( n+1

2 ) is about

1 √n, so it remains to show

• Rn ||x||dµ(x) >
• log n.

◮ Recall the norm || · || is defined by the symmetric convex body

K whose maximal ellipsoid is Bn

1 . ◮ Among all convex bodies K whose maximal ellipsoid is Bn 1 ,

the cube is the one which minimize

• Rn ||x||dµ(x).

◮ So just need to focus on the norm defined by cube, which is

||x|| = max1≤i≤n |xi|.

A glimpse into convex geometry

Γ( n

2 )

√ 2Γ( n+1

2 ) is about

1 √n, so it remains to show

• Rn ||x||dµ(x) >
• log n.

◮ Recall the norm || · || is defined by the symmetric convex body

K whose maximal ellipsoid is Bn

1 . ◮ Among all convex bodies K whose maximal ellipsoid is Bn 1 ,

the cube is the one which minimize

• Rn ||x||dµ(x).

◮ So just need to focus on the norm defined by cube, which is

||x|| = max1≤i≤n |xi|.

A glimpse into convex geometry

◮ Rn ||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 2π R

−R

e− |x|2

2 dx)n = 1

2. So R

−R e− |x|2

2 dx should be about 1 − log 2/n, which implies

R ∼ √log n.

◮ Finally,

• Rn ||x||dµ(x) ≥ R

2 > c

• log n,

and the proof is completed.

A glimpse into convex geometry

◮ Rn ||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 2π R

−R

e− |x|2

2 dx)n = 1

2. So R

−R e− |x|2

2 dx should be about 1 − log 2/n, which implies

R ∼ √log n.

◮ Finally,

• Rn ||x||dµ(x) ≥ R

2 > c

• log n,

and the proof is completed.

A glimpse into convex geometry

◮ Rn ||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 2π R

−R

e− |x|2

2 dx)n = 1

2. So R

−R e− |x|2

2 dx should be about 1 − log 2/n, which implies

R ∼ √log n.

◮ Finally,

• Rn ||x||dµ(x) ≥ R

2 > c

• log n,

and the proof is completed.

A glimpse into convex geometry

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?

A glimpse into convex geometry

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?

A glimpse into convex geometry

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 Hi denote an i-dimensional subspace, then vi(Hi ∩ I n) ≥ 1.

A glimpse into convex geometry

Concerning the upper bound, we have

Theorem (Ball)

vi(Hi ∩ I n) ≤ (n i )

i 2 ,

the upper bound is best possible if i|n.

Theorem (Ball)

vi(Hi ∩ I n) ≤ 2

n−i 2 ,

the upper bound is optimal if i ≥ n

2.

A glimpse into convex geometry

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

A glimpse into convex geometry

Thank you!