Convex Geometry - analysis.math.uni-kiel.deanalysis.math.uni-kiel.de/schuett/ConvexGeometry.pdf4 CONTENTS 0.1 Convex sets As a norm on Rn we consider the Euclidean norm x = n i=1 |x

Convex Geometry

Carsten Schutt

November 25, 2006

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Contents

0.1 Convex sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40.2 Separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90.3 Extreme points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150.4 Blaschke selection principle . . . . . . . . . . . . . . . . . . . . . . . 180.5 Polytopes and polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . 230.6 Volume and Surface Area of a Convex Body . . . . . . . . . . . . . . 310.7 Steiner and Schwarz Symmetrizations . . . . . . . . . . . . . . . . . . 380.8 A Characterization of the Ellipsoid . . . . . . . . . . . . . . . . . . . 470.9 Theorem of F. John . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500.10 Blaschke-Santalo inequality . . . . . . . . . . . . . . . . . . . . . . . 570.11 Packing of the Euclidean Sphere . . . . . . . . . . . . . . . . . . . . . 580.12 Approximation of the Euclidean Ball . . . . . . . . . . . . . . . . . . 62

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4 CONTENTS

0.1 Convex sets

As a norm on Rn we consider the Euclidean norm

‖x‖ =

√√√√ n∑i=1

|xi|2

The ball with center x and radius r is

B(x, r) = y|‖x − y‖ ≤ r

The boundary of a set A is ∂A = A\A. A set K in a real vector space is said to be

convex if we have for all x, y ∈ K that

tx + (1 − t)y| t ∈ [0, 1] ⊆ K

Examples of convex sets are the n-dimensional unit balls Bnp of lnp .

Bnp =

x ∈ Rn

∣∣∣∣∣‖x‖p = (n∑

i=1

|xi|p)1p ≤ 1

Let x1, . . . , xn be elements of a real vector space and t1, . . . tn real numbers from theinterval [0, 1] with

∑ni=1 ti = 1. Then

n∑i=1

tixi

a convex combination of the vectors x1, . . . , xn.

Lemma 1 Let C be a convex set in Rn, k ∈ N, and x1, . . . , xk ∈ C. Then for allt ∈ Rn with 0 ≤ ti ≤ 1 and

∑ki=1 ti = 1 we have

k∑i=1

tixi ∈ C

Proof. We use induction over the number of vectors x1, . . . , xk. The case k = 2follows from the definition of the convexity of C.

k+1∑i=1

tixi = (k∑

j=1

tj)

(k∑

i=1

ti∑kj=1 tj

xi

)+ tk+1xk+1

By induction hypotheses we have

k∑i=1

ti∑kj=1 tj

xi ∈ C

0.1. CONVEX SETS 5

Lemma 2 An arbitrary intersection of convex sets is again convex.

A hyperplane in Rn is a set x| < x, ξ >= t. There are two halfspaces associatedwith a hyperplane

H+ = x| < x, ξ >≤ t H− = x| < x, ξ >≥ t

Halfspaces are assumed to be closed sets.A supporting hyperplane H of a convex set C is a hyperplane such that one of

its halfspaces contains C, C ⊆ H+, and such that C is not contained in a halfspacethat is properly contained in H+.

Lemma 3 Let A and B be convex sets in Rn. Then

A + B = x + y|x ∈ A, y ∈ B

is a convex set. If A and B are compact, then A + B is also compact.

Proof. Let x, y ∈ A + B. Then there are x1, y1 ∈ A and x2, y2 ∈ B with

x = x1 + x2 y = y1 + y2

Then

tx + (1 − t)y = t(x1 + x2) + (1 − t)(y1 + y2) = (tx1 + (1 − t)y1) + (tx2 + (1 − t)y2)

Lemma 4 The closure and the interior of a convex set K in Rn are convex.

Proof. Let x, y ∈ K. Then there are sequences xi ∈ K, i ∈ N, and yi ∈ K, i ∈ N,with limi→∞ xi = x and limi→∞ yi = y. By the convexity of K for all i ∈ N and allt with 0 < t < 1

txi + (1 − t)yi ∈ K

Thereforetx + (1 − t)y = lim

i→∞(txi + (1 − t)yi) ∈ K

Now we show that the interior of a convex set is convex. Let x and y be interiorpoints of K. Then there is ε > 0 such that

B(x, ε) ⊆ K B(y, ε) ⊆ K

We have to show that [x, y] ⊆K, i.e. the points tx+(1− t)y|0 ≤ t ≤ 1 are interior

points. Since K is convex we have for all z, w ∈ B(0, ε)

t(z + x) + (1 − t)(w + y) = tz + (1 − t)w + (tx + (1 − t)y) ∈ K

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Thus

(tx+(1−t)y)+tB(0, ε)+(1−t)B(0, ε) = tz+(1−t)w+(tx+(1−t)y)|z, w ∈ B(0, ε) ⊆ K

It followsB(0, ε) + (tx + (1 − t)y) ⊆ K

Lemma 5 For all x, y ∈ Rn we have(i) | < x, y > | ≤ ‖x‖‖y‖(ii) ‖x + y‖ ≤ ‖x‖ + ‖y‖(iii) |‖x‖ − ‖y‖| ≤ ‖x − y‖(iv) If there is ε > 0 such that for all t with 0 < t < ε we have ‖x‖ ≤ ‖x + ty‖, then0 ≤< x, y >.

Proof.‖x‖2 ≤ ‖x + ty‖2 = ‖x‖2 + 2t < x, y > +t2‖y‖2

0 ≤< x, y > +1

2t‖y‖2

Taking the limit of t towards 00 ≤< x, y >

The convex hull CH(A) of a set A is the smallest convex set containing A, i.e.the intersection of all convex sets containing A.

Lemma 6 Let A be a subset of Rn. Then

CH(A) =

k∑

i=1

tixi

∣∣∣∣∣ k ∈ N, xi ∈ A, ti ≥ 0, i = 1, . . . , k,k∑

i=1

ti = 1

Proof. We put

A =

k∑

i=1

tixi

∣∣∣∣∣ k ∈ N, xi ∈ A, ti ≥ 0, i = 1, . . . , k,k∑

i=1

ti = 1

Since CH(A) is a convex set that contains A we get by Lemma 1

CH(A) ⊇ A

We show now that A is convex. Let x, y ∈ A. Then there are xi ∈ A, yj ∈ A suchthat

x =k∑

i=1

tixi y =m∑

j=1

sjyj

0.1. CONVEX SETS 7

Moreover, let r ∈ R, 0 ≤ r ≤ 1. Then

k∑i=1

rti +m∑

j=1

(1 − r)sj = r + (1 − r) = 1

Consequently,

rx + (1 − r)y ∈

k∑i=1

tixi

∣∣∣∣∣ k ∈ N, xi ∈ A, ti ≥ 0, i = 1, . . . , k,k∑

i=1

ti = 1

Lemma 7 (Caratheodory) Let A be a subset of Rn. Then

CH(A) =

n+1∑i=1

tixi

∣∣∣∣∣ xi ∈ A, ti ≥ 0, i = 1, . . . , n + 1,n+1∑i=1

ti = 1

The important point in the lemma of Caratheodory is that we need at most n+1to write a point of the convex hull as a convex combination.

Proof. By Lemma 6 for every x ∈ CH(A) there are k ∈ N, x1, . . . , xk ∈ A andt1, . . . , tk ≥ 0 with

x =k∑

i=1

tixi andk∑

i=1

ti = 1

We consider now that this representation has the smallest possible k. (We assumethat all ti are strictly bigger than 0.) We show that k ≤ n + 1. Suppose thatk ≥ n + 2. Then, since x1, . . . , xn+2 are linearly dependent there are real numberss1, . . . , sn+2 such that

(1)k∑

i=1

sixi = 0k∑

i=1

si = 0

and not all of the coefficients si are 0. We prove this. Since x1, . . . , xk−1 are linearlyindependent there are real numbers r1, . . . , rk−1 such that not all of the coefficientsare 0 and

k−1∑i=1

rixi = 0

If∑k−1

i=1 ri = 0 we are done. Therefore we may suppose that∑k−1

i=1 ri = 0. Moreover,we may assume that r1 = 0. (If not, we rearrange the vectors.) Again, sincex2, . . . , xk are linearly dependent there are coefficients t2, . . . , tks such that not all ofthe ti are 0 and

k∑i=2

tixi = 0

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If∑k

i=2 ti = 0 we are done. If not, we may assume that

n+1∑i=1

ri =n+2∑i=2

ti

Now we choose as our coefficients

s1 = r1, s2 = r2 + t2, . . . , sk−1 = rk−1 + tk−1, sk = tk

Clearly, these coefficients sum up to 0 and not all of them equal 0 since r1 = 0. Thuswe have shown (1).

Now we consider

x =k∑

i=1

(ti + λsi)xi

We can choose λ > 0 in such a way that all the coefficients ti − λsi are nonnegativeand at least one of them equals 0. Therefore k was not chosen to be minimal.

As the volume of a convex, compact C set we take the Lebesgue measure ofC which is in this case the same as the Riemann integral over the characteristicfunction of C.

Example 1voln(Bn

p ) =

Example 2 [StR] Let Cnp be the Schatten class of dimension n. Then the volume

of the unit ball equals

2−n

(n∏

k=1

kπ

k2

Γ(k2

+ 1)

)2 ∫Ωp

∏1≤j<k≤n

(s2k − s2

j)ds1 · · · dsn

where

Ωp =

s

∣∣∣∣∣n∑

j=1

spj < 1

The center of gravity or barycenter of a convex body is

cg(K) =1

voln(K)

∫K

xdx

0.2. SEPARATION 9

0.2 Separation

Let A be a closed nonempty set in Rn and x ∈ Rn. We call a point a0 ∈ A nearestpoint to x or best approximation of x if

‖x − a0‖ = inf‖x − a‖|a ∈ A

If x ∈ A then the nearest point is x itself.

Lemma 8 Let A be a closed, convex, nonempty set in Rn and x ∈ Rn.(i) There is a nearest point a0 of x in A.(ii) For all a ∈ A we have

< x − a0, a − a0 >≤ 0

(Geometrically this means that A is contained in the halfspace y| < x− a0, y >≤<x − a0, a0 >.)(iii) The nearest point a0 is unique.

Proof. (i) If x ∈ A then we choose a0 = x. Suppose now that x /∈ A. Then thereis a sequence an ∈ A, n ∈ N, such that

limn→∞

‖x − an‖ = inf‖x − a‖|a ∈ A

Almost all an are contained in the ball B(x, 2ρ) where

ρ = inf‖x − a‖|a ∈ A

SinceA ∩ B(x, 2ρ)

is a compact set there is a convergent subsequence ani, i ∈ N. We have

limi→∞

ani= a0

limi→∞

‖x − ani‖ = inf‖x − a‖|a ∈ A

This implies‖x − a0‖ = inf‖x − a‖|a ∈ A

(ii) Since A is convex we have for all t with 0 < t < 1 and a ∈ A that (1−t)a0+ta ∈ A.Since a0 is nearest point to x

‖x − a0‖2 ≤ ‖x − ((1 − t)a0 + ta)‖2 = ‖(x − a0) + t(a0 − a)‖2

By Lemma 50 ≤< x − a0, a0 − a >

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(iii) Suppose there is another point a1 such that

‖x − a1‖ = inf‖x − a‖|a ∈ A

By (ii) we get

< x − a0, a1 − a0 >≤ 0

< x − a1, a0 − a1 >≤ 0

We add the two inequalities

0 ≥< x − a0, a1 − a0 > + < x − a1, a0 − a1 >= ‖a0 − a1‖2

Let A be a nonempty, closed, convex set. The metric projection pA : Rn → Amaps a point x onto its nearest point in A.

Lemma 9 Let A be a nonempty, closed, convex set in Rn. The metric projectionpA is a 1-Lipschitz map, i.e. for all x, y ∈ Rn

‖pA(x) − pA(y)‖ ≤ ‖x − y‖

Proof. pA is well defined since by Lemma 8 for every x ∈ Rn there is a unique pointof minimal distance.

Let x, y ∈ Rn. By Lemma 8 for all a ∈ A

< x − pA(x), a − pA(x) >≤ 0

< y − pA(y), a − pA(y) >≤ 0

In particular

< x − pA(x), pA(y) − pA(x) >≤ 0

< y − pA(y), pA(x) − pA(y) >≤ 0

Thus

< (x − pA(x)) − (y − pA(y)), pA(x) − pA(y) >≥ 0

With this

‖x − y‖2 = ‖(x − pA(x)) − (y − pA(y)) + (pA(x) − pA(y))‖2

= ‖(x − pA(x)) − (y − pA(y))‖2

+2 < (x − pA(x)) − (y − pA(y)), pA(x) − pA(y) > +‖pA(x) − pA(y)‖2

≥ ‖pA(x) − pA(y)‖2

0.2. SEPARATION 11

Proposition 1 Let A and B be nonempty, disjoint, convex sets in Rn. Moreover, letA be closed and B compact. Then A and B can be strictly separated by a hyperplane,i.e. there exists ξ ∈ Rn such that

supa∈A

< ξ, a >< infb∈B

< ξ, b >

Proof. There are two points a0 ∈ A and b0 ∈ B such that

‖a0 − b0‖ = inf‖a − b‖|a ∈ A, b ∈ B

We show this. Clearly, there are two sequences an ∈ A and bn ∈ B, n ∈ N, suchthat

limn→∞

‖an − bn‖ = inf‖a − b‖|a ∈ A, b ∈ B

By compactness of B there is a subsequence bni, i ∈ N, that converges

limi→∞

bni= b0

We have

‖ani− bni

‖ − ‖bni− b0‖ ≤ ‖ani

− b0‖ ≤ ‖ani− bni

‖ + ‖bni− b0‖

It follows thatlimi→∞

‖ani− b0‖ = inf‖a − b‖|a ∈ A, b ∈ B

Almost all of ani, i ∈ N, are elements of the ball B(b0, 2ρ) with

ρ = inf‖a − b‖|a ∈ A, b ∈ B

thus the sequence is contained in a compact set and has a convergent subsequence.The limit is denoted by a0.

Now we choose ξ = a0 − b0. Since a0 is nearest point for b0 and vice versa wehave by Lemma 8 for all a ∈ A and b ∈ B

< a0 − b0, a − b0 >≤ 0

< b0 − a0, b − a0 >≤ 0

It follows

< a0 − b0, b > ≥< a0 − b0, a0 >= ‖a0‖2− < b0, a0 >

= 12‖a0‖2 − ‖b0‖2 + ‖a0 − b0‖2

> 12‖a0‖2 − ‖b0‖2

> 12‖a0‖2 − ‖b0‖2 − ‖a0 − b0‖2

=< a0, b0 > −‖b0‖2 =< a0 − b0, b0 >≥< a0 − b0, a >

Therefore

infb∈B

< a0 − b0, b >> 12‖a0‖2 − ‖b0‖2 > sup

a∈A< a0 − b0, a >

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Corollary 1 Let A be a nonempty, closed, convex set in Rn and x ∈ Rn with x /∈ A.Then there is a hyperplane that separates A and x strictly.

Corollary 2 Let A be a closed convex set in Rn. Then A is the intersection of allclosed halfspaces H+ containing A, i.e.

A =⋂

A⊆H+

H+

Proof. Let x ∈ Rn with x /∈ A. Then there is a hyperplane separating x and Astrictly.

Lemma 10 Let K be a nonempty, closed, convex set in Rn and let x ∈ ∂K. Thenthere is y /∈ K such that p(y) = x.

Proof. There is a sequence xi ∈ Rn \ K, i ∈ N, with

limi→∞

xi = x

We put

yi = p(xi) +xi − p(xi)

‖xi − p(xi)‖We have

‖yi − p(xi)‖ =

∥∥∥∥ xi − p(xi)

‖xi − p(xi)‖

∥∥∥∥ = 1

andp(yi) = p(xi)

We verify that p(yi) = p(xi). Since p(xi) is the nearest point to xi and p(yi) to yi

we have by Lemma ?? that for all ξ, η ∈ K

< xi − p(xi), ξ − p(xi) >≤ 0

< yi − p(yi), η − p(yi) >≤ 0

Choosing ξ = p(yi) and η = p(xi) we get

< xi − p(xi), p(yi) − p(xi) >≤ 0

and

0 ≥< yi − p(yi), p(xi) − p(yi) >

=< p(xi) − p(yi) +xi − p(xi)

‖xi − p(xi)‖, p(xi) − p(yi) >

= ‖p(xi) − p(yi)‖2 +

⟨xi − p(xi)

‖xi − p(xi)‖, p(xi) − p(yi)

⟩

0.2. SEPARATION 13

Thus we get

0 ≥⟨

xi − p(xi)

‖xi − p(xi)‖, p(yi) − p(xi)

⟩≥ ‖p(xi) − p(yi)‖2

which implies p(xi) = p(yi). Therefore

p(x) = p(

limi→∞

xi

)= lim

i→∞p(xi) = lim

i→∞p(yi)

The sequence yi, i ∈ N, is bounded since

‖yi‖ =

∥∥∥∥p(xi) +xi − p(xi)

‖xi − p(xi)‖

∥∥∥∥ ≤ 1 + ‖p(xi)‖

and xi, i ∈ N, is a convergent sequence. Therefore the sequence yi, i ∈ N, has aconvergent subsequence yij , j ∈ N,

y = limj→∞

yij

We have

p(y) = limj→∞

p(yij) = limj→∞

p(xij) = p( limj→∞

yij) = p(x) = x

Moreover we have x = y. Indeed,

‖x − y‖ = limj→∞

‖xij − yij‖ = limj→∞

∥∥∥∥xij − p(xij) +xij − p(xij)

‖xij − p(xij)‖

∥∥∥∥Since

limj→∞

xij − p(xij) = 0

we get ‖x − y‖ = 1.

Lemma 11 Let K be a nonempty, convex set in Rn that does not contain the origin0. Then there is a hyperplane H with

0 ∈ H− and K ⊆ H+

Proof. Suppose 0 /∈ K. By Lemma ?? K is convex. Now we can apply Corollary1.

Now suppose that 0 ∈ K. By Lemma ?? there is x ∈ Rn \ K with p(x) = 0. Asa hyperplane we choose now H = ξ| < x, ξ >= 0. Clearly, 0 ∈ H and therefore0 ∈ H− and 0 ∈ H+. Since 0 is the nearest point to x we have by Lemma 8 for allξ ∈ K

< ξ, x >≤ 0

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Proposition 2 Let K and C be nonempty, convex, disjoint sets in Rn. Then thereis a hyperplane separating K and C.

If one of the sets is lower dimensional then it might happen that the set iscontained in the separating hyperplane.

Proof. The setC − K = x − y|x ∈ C, y ∈ K

is convex and 0 /∈ C−K. We verify this. Let x1−y1, x2−y2 ∈ C−K and 0 < t < 1.Then

t(x1 − y1) + (1 − t)(x2 − y2) = (tx1 + (1 − t)x2) − (ty1 + (1 − t)y2) ∈ C − K

We have 0 /∈ C −K because C and K are disjoint. Therefore, there is a hyperplaneseparating 0 from C − K, i.e. there is a ξ ∈ Rn, ξ = 0, such that for all x ∈ C − K

0 ≤< ξ, x >

Thus there is ξ such that for all y ∈ K and z ∈ C

0 ≤< ξ, y − z >

and consequently< ξ, z >≤< ξ, y >

A hyperplane H is called a support hyperplane to the convex set K if H∩K = ∅and K ⊆ H+ or K ⊆ H−.

Proposition 3 Let K be a nonempty, convex set in Rn and x ∈ ∂K. Then thereis a supporting hyperplane of K that contains x.

Proof. Suppose that K has no interior point. Then every point of K is a boundarypoint and K is contained in a hyperplane. We verify this.

Suppose that there is no hyperplane that contains K. We may assume that0 ∈ K. We choose a point x1 ∈ K with 0 = x1. Now we choose a point x2 ∈ Kthat is not contained in the subspace spanned by x1. Such a point exists becauseotherwise K would be contained in this subset. By induction we get that thereare n linearly independent points x1, . . . , xn that are contained in K. Therefore[0, x1, . . . , xn] ⊆ K and [0, x1, . . . , xn] contains an interior point.

Suppose now that K has an interior point and let x be a boundary point of K.

Then we consider the disjoint, convex setsK and x. By Proposition 2 there is a

separating hyperplane H, i.e.

H+ ⊇K and x ∈ H−

It follows H+ ⊇ K and x ∈ H.

0.3. EXTREME POINTS 15

0.3 Extreme points

A point x of a convex set C in Rn is not an extreme point of C if there are y, z ∈ C,y = z, such that

x =y + z

2

In fact this is equivalent to the following. A point x of a convex set C in Rn is notan extreme point of C if there are y, z ∈ C, y = z, and t with 0 < t < 1 such that

x = ty + (1 − t)z

We show this. We may suppose that 12≤ t ≤ 1. We choose the points y and 2x− y.

Then we havex = 1

2y + 1

2(2x − y)

and 2x − y ∈ C because

2x − y = 2ty + 2(1 − t)z − y = (2t − 1)y + (2 − 2t)z

where 2t − 1 ≥ 0 and (2t − 1) + (2 − 2t) = 1.These considerations can be extended further. Suppose x ∈ C, x1, . . . , xk ∈ C

and t1, . . . , tk > 0 with∑k

i=1 ti = 1 and

x =k∑

i=1

tixi

Then x is not an extreme point. We choose y = x1 and

z =k∑

i=2

ti∑ki=2 ti

xi

Then

x = t1y +

(k∑

i=2

ti

)z

Therefore, x is not an extreme point. The following theorem is a special case of theKrein-Milman theorem.

Theorem 0.3.1 A convex, compact subset of Rn equals the closed, convex hull ofits extreme points.

This result can be improved. In fact, it can be shown that in Rn a convex,compact set equals the convex hull of its extreme points. For infinite dimensionalvector spaces one has to take the closed convex hull.

Lemma 12 Let C be a convex subset of Rn and H a supporting hyperplane of C.The the extreme points of C∩H are exactly the extreme points of C that are containedin H.

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Proof. Clearly, all extreme points of C that are contained in H are also extremepoints of C ∩ H.

Let x be an extreme point of C ∩H and not an extreme point of C. Then thereare y and z with y, z ∈ C and x = y+z

2. We show that y, z ∈ H.

Let H be given by < ·, ξ >= t and suppose that C is a subset of v| < v, ξ >≤ t.Then

t =< x, ξ >=1

2(< y, ξ > + < z, ξ >) ≤ t

This means that we have equality in the above equation which means y, z ∈ H.

Lemma 13 Let C be a nonempty, compact, convex set in Rn. Then C has anextreme point.

Proof. If C consists of one point only, this point is also extreme point of C. Thuswe may assume that C consists of more than just one point.

Since C is compact, there is z ∈ C such that the Euclidean norm is maximal,i.e.

‖z‖ = maxx∈C

‖x‖

We claim that z is an extreme point. We show this. Suppose that z is not anextreme point of C. Then there are x, y ∈ C and t, 0 < t < 1, with

z = tx + (1 − t)y

We have ‖x‖ = ‖y‖ = ‖z‖ because if ‖x‖ < ‖z‖ or ‖y‖ < ‖z‖ we get

‖z‖ = ‖tx + (1 − t)y‖ ≤ t‖x‖ + (1 − t)‖y‖ < ‖z‖

Therefore, we get

‖z‖2 = ‖tx + (1 − t)y‖2

= t2‖x‖2 + 2t(1 − t) < x, y > +(1 − t)2‖y‖2

≤ t2‖x‖2 + 2t(1 − t)‖x‖‖y‖ + (1 − t)2‖y‖2

= (t‖x‖ + (1 − t)‖y‖)2 = ‖z‖2

Thus we have2t(1 − t) < x, y >= 2t(1 − t)‖x‖‖y‖

Since 0 < t < 1 we get< x, y >= ‖x‖‖y‖

This implies that there is s with x = sy. Since ‖x‖ = ‖y‖ we have x = y or x = −y.Since < x, y >≥ 0 we conclude x = y and thus z is an extreme point.

Lemma 14 Let C be a compact, convex set in Rn. Then the set of extreme points ofC is not empty and every support hyperplane H of C contains at least one extremepoint of C.

0.3. EXTREME POINTS 17

Proof. Let H be a support hyperplane of C. C ∩ H is a compact, convex set. ByLemma 13 C ∩ H has an extreme point. By Lemma 12 this is an extreme point ofC.

Proof of Theorem 0.3.1 We denote the set of extreme points of C by

Ext(C)

Since C is compact and closed we have

CH(Ext(C)) ⊆ C

Now we show the opposite inclusion, namely

CH(Ext(C)) ⊇ C

Suppose that this inclusion does not hold, then there is z ∈ C with z /∈ CH(Ext(C)).Since CH(Ext(C)) is a closed convex set by Corollary 1 there is a hyperplane H =x| < ξ, x >= t that separates CH(Ext(C)) and z strictly, i.e.

maxx∈CH(Ext(C))

< ξ, x ><< ξ, z >

Consider now the hyperplane

L = x| < ξ, x >= maxy∈C

< ξ, y >

This hyperplane is a support hyperplane of C and it contains by Lemma ?? anextreme point v of C. In particular,

v ∈ L ∩ CH(Ext(C))

It follows< ξ, v ><< ξ, z >

On the other hand, since z ∈ C

< ξ, z >≤ maxx∈C

< ξ, x >=< ξ, v ><< ξ, z >

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0.4 Blaschke selection principle

A convex body in Rn is a convex, compact set with an interior point. The supportfunction hC : Rn → R of a convex body C is defined by

hC(ξ) = supx∈C

< ξ, x >

The Hausdorff distance between two convex, compact sets K and C is

dH(C, K) = max infρ|C ⊆ K + Bn2 (0, ρ), infρ|K ⊆ C + Bn

2 (0, ρ)

Lemma 15 Let C and K be two convex, compact sets in Rn. Then we have(i)

dH(C, K) = max

supx∈C

infy∈K

‖x − y‖, supy∈K

infx∈C

‖x − y‖

(ii)dH(C, K) = sup

‖ξ‖=1

|hC(ξ) − hK(ξ)|

Proof. (i) Suppose we have C ⊆ K + Bn2 (0, ρ). Then, for every x ∈ C there are

y ∈ K and z ∈ Bn2 (0, ρ) with x = y + z. This implies ‖x− y‖ = ‖z‖ ≤ ρ. Therefore,

for every x ∈ C we have infy∈K ‖x − y‖ ≤ ρ and subsequently

supx∈C

infy∈K

‖x − y‖ ≤ ρ

and thereforesupx∈C

infy∈K

‖x − y‖ ≤ infρ|C ⊆ K + Bn2 (0, ρ)

It follows that

dH(C, K) ≥ max

supx∈C

infy∈K

‖x − y‖, supy∈K

infx∈C

‖x − y‖

(ii)C ⊆ K + Bn

2 (0, ρ) and K ⊆ C + Bn2 (0, ρ)

is equivalent to: For all ξ, ‖ξ‖ = 1, we have

hC(ξ) ≤ hK(ξ) + ρ and hK(ξ) ≤ hC(ξ) + ρ

We prove the equivalence.

hC(ξ) = supx∈C

< ξ, x >≤ supx∈K+Bn

2 (0,ρ)

< ξ, x >= supy ∈ K

z ∈ Bn2 (0, ρ)

< ξ, y + z >

= supy∈K

< ξ, y > + supz∈Bn

2 (0,ρ)

< ξ, z >= hK(ξ) + ρ

0.4. BLASCHKE SELECTION PRINCIPLE 19

On the other hand, if C K + Bn2 (0, ρ) then there is x ∈ C but x /∈ K + Bn

2 (0, ρ).By the Theorem of Hahn-Banach there is a hyperplane H = y| < ξ, y >= t thatseparates x and K + Bn

2 (0, ρ), i.e.

supy∈K+Bn

2 (0,ρ)

< ξ, y ><< ξ, x > .

It follows that hK(ξ) + ρ < hC(ξ).

The symmetric difference distance is

dS(C, K) = voln((C \ K) ∪ (K \ C))

Lemma 16 (i) dH is a metric on the space K of all compact, convex subsets of Rn.(ii) dS is a metric on the space K of all compact, convex subsets of Rn.

Proof. For compact sets C and K we have

dH(C, K) = max‖ξ‖2=1

|hC(ξ) − hK(ξ)|

From this it follows immediately that dH(C, C) = 0, dH(C, K) ≥ 0, and dH(C, K) ≤dH(C, M) + dH(M, K). We have that dH(C, K) > 0 if C = K since two compact,convex sets are equal provided that their support functions are equal. We verifythis. Suppose that C = K. Then, w.l.o.g there is x ∈ C with x /∈ K. Then byCorollary 1 there is ξ ∈ Rn with

supy∈K

< ξ, y ><< ξ, x >

This implies

hK(ξ) = supy∈K

< ξ, y ><< ξ, x >≤ hC(ξ)

Lemma 17 Let Ki, i ∈ N, be a sequence of compact, nonempty, convex sets in Rn

such that we have for all i ∈ N

Ki+1 ⊆ Ki

Then∞⋂i=1

Ki

is a convex, compact, nonempty set and the sequence Ki, i ∈ N, converges to it withrespect to the Hausdorff metric.

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Proof. We have to show that for all ε > 0 there is l ∈ N such that we have for allj ≥ l

dH

( ∞⋂i=1

Ki, Kj

)< ε

This means

max

sup

x∈⋂∞

i=1 Ki

infy∈Kj

‖x − y‖, supy∈Kj

infx∈

⋂∞i=1 Ki

‖x − y‖

= supy∈Kj

infx∈

⋂∞i=1 Ki

‖x − y‖ < ε

Suppose this is not true. Then there is a subsequence Kjm , m ∈ N and elementsxjm ∈ Kjm such that

infx∈∩∞

i=1Ki

‖xjm − x‖ ≥ ε

Then there is a subsequence of xjm , m ∈ N, that converges to an element z. Wehave

z ∈∞⋂i=1

Ki and infx∈∩∞

i=1Ki

‖z − x‖ ≥ ε

This cannot be.

Lemma 18 The space K of all nonempty, compact, convex subsets of Rn with themetric dH is complete.

Proof. Let Ki, i ∈ N, be a Cauchy sequence. We put

Cj = CH

( ∞⋃i=j

Ki

)

The sequence Cj, j ∈ N, is a decreasing sequence of compact, convex sets. The setsCj are obviously convex and closed. We show that they are bounded. Since thesequence Ki, i ∈ N, is a Cauchy sequence we have that there is i0 such that for alli ≥ i0

dH(Ki0 , Ki) ≤ 1

This means that for all i ≥ i0

Ki ⊆ Ki0 + Bn2 (0, 1)

Since Ki0 + Bn2 (0, 1) is a bounded set, the set

⋃i≥i0

Ki

is also bounded.

0.4. BLASCHKE SELECTION PRINCIPLE 21

By Lemma 17 the sequence Cj, j ∈ N, converges in the metric space. We claimthat the sequence Ki, i ∈ N, converges to the same limit. It suffices to show thatfor every ε > 0 there is j0 so that we have for all j ≥ j0

dH(Cj, Kj) < ε

Since Ki, i ∈ N, is a Cauchy sequence there is i0 so that we have for all l, m ≥ i0

dH(Kl, Km) < ε

orKl ⊆ Km + Bn

2 (0, ε)

Therefore we get for all j ≥ i0

Kj ⊆ Cj = CH

( ∞⋃i=j

Ki

)⊆ Kj + Bn

2 (0, ε) = Kj + Bn2 (0, ε)

ord(Kj, Cj) ≤ ε

Lemma 19 (Blaschke selection principle) Let Ki, i ∈ N, be a sequence of nonempty,compact, convex sets in a bounded subset of Rn. Then there is a subsequence Kij ,j ∈ N, that converges with respect to the Hausdorff distance to a nonempty, compact,convex set.

Moreover, if all sets Ki, i ∈ N, are contained in a compact set C, then the limitis contained in C.

Proof. We show first that for any ε > 0 and any sequence Ki, i ∈ N, in a boundedset there is a subsequence Kij , j ∈ N, such that we have for all j, j′ ∈ N

dH(Kij , Kij′ ) < ε

By assumption the sequence Ki, i ∈ N, is contained in a set [a, b]n. We may assumethat a = 0 and b = 1. We choose k ∈ N such that

√n2−k < ε and we put

Ql =n∏

i=1

[li − 1

2k,

li2k

]

where 1 ≤ li ≤ 2k, l = (l1, . . . , ln) and 1 ≤ i ≤ n. Then we have

[0, 1]n =⋃

l

Ql

We considerIi = l|Ql ∩ Ki = ∅

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There are 2kn different vectors . Therefore there are at most 22kndifferent sets Ii.

Thus there are only finitely many different sets Ii and therefore there is a subsequenceij, j ∈ N, such that we have for all j, j′ ∈ N

Iij = Iij′

Let x ∈ Kij . Then there is Ql with x ∈ Ql. Since Iij = Iij′ there is y ∈ Ql ∩ Kij′ .

Therefore ‖x − y‖2 ≤√

n2k < ε, or x ∈ Bn

2 (y, ε). Thus we get

Kij ⊆ Kij′ + Bn2 (0, ε)

We concludedH(Kij , Kij′ ) ≤ ε

Now we select from the sequence Ki, i ∈ N, a subsequence that is a Cauchy sequence.From the sequence Ki, i ∈ N, we choose a subsequence K1,j such that we havedH(K1,j, K1,j′) < 1

2for all j ∈ N. After having chosen the subsequence K(k,j), j ∈ N,

we choose the sequence Kk+1,j, j ∈ N, as a subsequence of Kk,j, j ∈ N, such thatwe have for all j ∈ N

dH(Kk+1,j, Kk+1,j′) <1

2k+1

This implies that the sequence Kj,j, j ∈ N, is a Cauchy sequence.

0.5. POLYTOPES AND POLYHEDRA 23

0.5 Polytopes and polyhedra

A polytope P is the convex hull of of finitely many points in Rn

P = [x1, . . . , xk]

We have thatExt(P ) ⊆ x1, . . . , xk

We verify this. P is the convex hull of the points x1, . . . , xk. By Lemma ?? theconvex hull consists of all convex combinations

∑ki=1 tixi of the points x1, . . . , xk.

All points that are convex combinations∑k

i=1 tixi where at least 2 coefficients ti arestrictly larger than 0 are not extreme points. That leaves all those points whereexactly one coefficient is strictly larger than 0 and thus 1.

The extreme points of a polytope are also called vertices.A face of a convex body is the intersection of a support hyperplane with the

boundary of K. The dimension of a face is the smallest dimension of all affinespaces containing that face.

A simplex in Rn is the convex hull of n+1 points. A regular simplex is a simplexfor which all the distances between two vertices are equal.

A polyhedral set is the intersection of finitely many closed halfspaces in Rn.

Lemma 20 Each face of a polytope P is again a polytope and its vertices are exactlythe vertices of P that are contained in the hyperplane defining the face.

Proof. Let P = [x1, . . . , xk] and let H be a support hyperplane of P . Let usnote that P ∩ H is a convex set. Then Ext(P ) ⊆ x1, . . . , xk. By choosing a newnumbering we my assume that x1, . . . , xm, m ≤ k, are the extreme points of P . Bythe Theorem of Krein-Milman we have P = [x1, . . . , xm]. By Lemma ?? the extremepoints of P ∩ H are the extreme points of P contained in H. By the Theorem ofKrein-Milman P ∩ H is the convex hull of the extreme points of P contained in H.

Corollary 3 Let P be a convex polytope in Rn. Then P has only finitely manyfaces.

Lemma 21 Let P be a polyhedral set in Rn that has an interior point

P =N⋂

i=1

x| < ξi, x >≤ ti

Suppose that none of the halfspaces can be dropped, i.e. the intersection of thehalfspaces would be different from P . Then the sets

Fi = P ∩ x| < ξi, x >= ti

24 CONTENTS

have the dimension n − 1 and

∂P =N⋃

i=1

Fi

Proof. We show first that

P=

N⋂i=1

x| < ξi, x >< ti

Clearly, it is an open set contained in P and thus

P⊇

N⋂i=1

x| < ξi, x >< ti

On the other hand, let x be such that there is i with < x, ξi >= ti. Then in everyneighborhood of x there is a point not in P . The point x + εξi is not an element ofP :

< x + εξi, ξi >= ti + ε

Thus we have shown that

P=

N⋂i=1

x| < ξi, x >< ti

and, consequently,

∂P = P\P= P ∩ (

P )c = P ∩

N⋃i=1

x| < ξi, x >= ti =N⋃

i=1

Fi

We show now that the faces Fi have dimension equal to n − 1. Since none of thehalfspaces can be omitted there are xi /∈ P , 1 ≤ i ≤ N , such that

< xi, ξj >≤ tj 1 ≤ j = i ≤ N

< xi, ξi >> ti 1 ≤ i ≤ N

Let x0 be an interior point of P . Then there is ε with Bn2 (x0, ε) ⊆ P . By continuity

there is a point yi ∈ ∂P with < ξi, yi >= ti. Indeed, we consider all points sx0 +(1 − s)xi with 0 ≤ s ≤ 1. By convexity we have

< sx0 + (1 − s)xi, ξj >≤ tj

so that yi ∈ P . By convexity

CH(xi, Bn2 (x0, ε)) ⊆

⋂j =i

x| < ξj, x >< tj

0.5. POLYTOPES AND POLYHEDRA 25

It follows that there is δ > 0

Bn2 (yi, δ) ⊆

⋂j =i

x| < ξj, x >< tj

ThusBn

2 (yi, δ) ∩ x| < ξi, x >= ti ⊆ P

Therefore the boundary of P contains a n − 1-dimensional Euclidean ball and thecorresponding face is n − 1-dimensional.

Lemma 22 A set in Rn is a polytope if and only if it is a bounded polyhedral set.

Proof. We may assume that the set has an interior point. Otherwise, we restrictourselves to the smallest affine subspace containing the set.

Let P be a poytope. According to the lemma following the Theorem of Krein-Milman each face of P is the closed, convex hull of extreme points of P . There areonly finitely many extreme points and thus there are only finitely many faces. LetFi, i = 1, . . . , N , be all the faces of P and let x| < x, ξi >= ti, i = 1, . . . , N , bethe hyperplanes giving the faces

Fi = P ∩ x| < x, ξi >= ti

The normals ξi are oriented in such a way that

P ⊆ x| < x, ξi >≤ ti

Clearly, we have

P ⊆N⋂

i=1

x| < x, ξi >≤ ti

We show that the sets are equal. Suppose that they are not equal. Then there isx0 /∈ P with

x0 ∈N⋂

i=1

x| < x, ξi >≤ ti

Moreover, there is an interior point x1 of P . Therefore there is y0 ∈ [x0, x1] withy0 ∈ ∂P , i.e.

y0 = λx0 + (1 − λ)x1 0 ≤ λ ≤ 1

Since x0 /∈ P we have y0 = x0 and since x1 is an interior point of P we have y0 = x1.Thus

y0 = λx0 + (1 − λ)x1 0 < λ < 1

By the Theorem of Hahn-Banach there is a supporting hyperplane of P at y0. There-fore y0 is an element of a face of P . Therefore, there is i0 such that

ti0 =< y0, ξi0 >= λ < x0, ξi0 > +(1 − λ) < x1, ξi0 >

26 CONTENTS

Since x1 is an interior point of P we have < x1, ξi0 < ti0 . Moreover, 0 < λ < 1.Thus

ti0 =< y0, ξi0 >< ti0

We show now that a bounded polyhedral set is a polytope. For this we show thatan extreme point e of

⋂Ni=1x| < x, ξi >≤ ti satisfies

e =⋂

<e,ξi>=ti

x| < x, ξi >= ti

Since e is an extreme point it is an element of the boundary. There are sets I andJ such that

I = i| < e, ξi >= ti J = i| < e, ξi >< tiSuppose that

e =⋂

<e,ξi>=ti

x| < x, ξi >= ti

Then there are at least two different elements e and f in the set⋂

<e,ξi>=tix| <

x, ξi >= ti. Both are also elements of

N⋂i=1

x| < x, ξi >= ti

For all λ ∈ R we have

(1 − λ)e + λf ∈⋂

<e,ξi>=ti

x| < x, ξi >= ti

Indeed, for all i ∈ I

< (1 − λ)e + λf, ξi >= (1 − λ) < e, ξ > +λ < f, ξi >= ti

Moreover, there is ε > 0 such that for all i ∈ J

< (1 − ε)e + εf, ξi >= (1 − ε) < e, ξi > +ε < f, ξi >< ti

< (1 + ε)e − εf, ξi >= (1 + ε) < e, ξi > −ε < f, ξi >< ti

Thus we have

e =(1 − ε)e + εf

2+

(1 + ε)e − εf

2

and e is not an extreme point.Therefore, every extreme point is the intersection of some of the hyperplanes

defining the halfspaces. Since there are only N there are at most 2N extreme points.By the Theorem of Minkowski/Krein-Milman a compact convex set is the closedconvex hull of its extreme points. The convex hull of finitely many points is apolytope (which is a closed set).