Convex Geometry. Carsten Schütt

Transcription

1 Convex Geometry Carsten Schütt November 25, 2006

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3 Contents 0.1 Convex sets Separation Extreme points Blaschke selection principle Polytopes and polyhedra Volume and Surface Area of a Convex Body Steiner and Schwarz Symmetrizations A Characterization of the Ellipsoid Theorem of F. John Blaschke-Santaló inequality Packing of the Euclidean Sphere Approximation of the Euclidean Ball

4 4 CONTENTS 0.1 Convex sets As a norm on R n we consider the Euclidean norm x = n x i 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 Bp n of lp n. { } n Bp n = x R n x p =( x i p ) 1 p 1 Let x 1,...,x n be elements of a real vector space and t 1,...t n real numbers from the interval [0, 1] with n t i = 1. Then n t i x i a convex combination of the vectors x 1,...,x n. Lemma 1 Let C be a convex set in R n, k N, and x 1,...,x k C. Then for all t R n with 0 t i 1 and k t i =1we have k t i x i C Proof. We use induction over the number of vectors x 1,...,x k. The case k =2 follows from the definition of the convexity of C. ( k+1 k k ) t i t i x i =( t j ) k j=1 t x i + t k+1 x k+1 j j=1 By induction hypotheses we have k t i k j=1 t j x i C

5 0.1. CONVEX SETS 5 Lemma 2 An arbitrary intersection of convex sets is again convex. A hyperplane in R n is a set {x <x,ξ>= t}. There are two halfspaces associated with 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 halfspace that is properly contained in H +. Lemma 3 Let A and B be convex sets in R n. 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 x 1,y 1 A and x 2,y 2 B with x = x 1 + x 2 y = y 1 + y 2 Then tx +(1 t)y = t(x 1 + x 2 )+(1 t)(y 1 + y 2 )=(tx 1 +(1 t)y 1 )+(tx 2 +(1 t)y 2 ) Lemma 4 The closure and the interior of a convex set K in R n are convex. Proof. Let x, y K. Then there are sequences x i K, i N, and y i K, i N, with lim i x i = x and lim i y i = y. By the convexity of K for all i N and all t with 0 <t<1 tx i +(1 t)y i K Therefore tx +(1 t)y = lim (tx i +(1 t)y i ) K i Now we show that the interior of a convex set is convex. Let x and y be interior points of K. Then there is ɛ>0 such that B(x, ɛ) K B(y, ɛ) 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,ɛ) t(z + x)+(1 t)(w + y) =tz +(1 t)w +(tx +(1 t)y) K

6 6 CONTENTS Thus (tx+(1 t)y)+tb(0,ɛ)+(1 t)b(0,ɛ)={tz+(1 t)w+(tx+(1 t)y) z,w B(0,ɛ)} K It follows B(0,ɛ)+(tx +(1 t)y) K Lemma 5 For all x, y R n we have (i) <x,y> x y (ii) x + y x + y (iii) x y x y (iv) If there is ɛ>0 such that for all t with 0 <t<ɛwe have x x + ty, then 0 < x,y>. Proof. x 2 x + ty 2 = x 2 +2t<x,y>+t 2 y 2 0 < x,y>+ 1 2 t y 2 Taking the limit of t towards 0 0 < 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 R n. Then { k CH(A) = t i x i k N,x i A, t i 0,,...,k, } k t i =1 Proof. We put { k } k à = t i x i k N,x i A, t i 0,,...,k, t i =1 Since CH(A) is a convex set that contains A we get by Lemma 1 CH(A) à We show now that à is convex. Let x, y Ã. Then there are x i A, y j A such that k m x = t i x i y = s j y j j=1

7 0.1. CONVEX SETS 7 Moreover, let r R, 0 r 1. Then k m rt i + (1 r)s j = r +(1 r) =1 j=1 Consequently, { k } k rx +(1 r)y t i x i k N,x i A, t i 0,,...,k, t i =1 Lemma 7 (Carathéodory) Let A be a subset of R n. Then { n+1 } n+1 CH(A) = t i x i x i A, t i 0,,...,n+1, t i =1 The important point in the lemma of Carathéodory is that we need at most n+1 to write a point of the convex hull as a convex combination. Proof. By Lemma 6 for every x CH(A) there are k N, x 1,...,x k A and t 1,...,t k 0 with k k x = t i x i and t i =1 We consider now that this representation has the smallest possible k. (We assume that all t i are strictly bigger than 0.) We show that k n + 1. Suppose that k n + 2. Then, since x 1,...,x n+2 are linearly dependent there are real numbers s 1,...,s n+2 such that k k (1) s i x i =0 s i =0 and not all of the coefficients s i are 0. We prove this. Since x 1,...,x k 1 are linearly independent there are real numbers r 1,...,r k 1 such that not all of the coefficients are 0 and k 1 r i x i =0 If k 1 r i = 0 we are done. Therefore we may suppose that k 1 r i 0. Moreover, we may assume that r 1 0. (If not, we rearrange the vectors.) Again, since x 2,...,x k are linearly dependent there are coefficients t 2,...,t ks such that not all of the t i are 0 and k t i x i =0 i=2

8 8 CONTENTS If k i=2 t i = 0 we are done. If not, we may assume that Now we choose as our coefficients n+1 n+2 r i = i=2 t i s 1 = r 1,s 2 = r 2 + t 2,...,s k 1 = r k 1 + t k 1,s k = t k Clearly, these coefficients sum up to 0 and not all of them equal 0 since r 1 0. Thus we have shown (1). Now we consider k x = (t i + λs i )x i We can choose λ>0 in such a way that all the coefficients t i λs i are nonnegative and 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 of C which is in this case the same as the Riemann integral over the characteristic function of C. Example 1 vol n (B n p )= Example 2 [StR] Let Cp n be the Schatten class of dimension n. Then the volume of the unit ball equals ( n ) 2 2 n π k 2 k Γ( k +1) 2 k=1 (s 2 k s 2 j)ds 1 ds n Ω p 1 j<k n where { } n Ω p = s s p j < 1 j=1 The center of gravity or barycenter of a convex body is 1 cg(k) = xdx vol n (K) K

9 0.2. SEPARATION Separation Let A be a closed nonempty set in R n and x R n. We call a point a 0 A nearest point to x or best approximation of x if x a 0 = 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 R n and x R n. (i) There is a nearest point a 0 of x in A. (ii) For all a A we have <x a 0,a a 0 > 0 (Geometrically this means that A is contained in the halfspace {y <x a 0,y > < x a 0,a 0 >}.) (iii) The nearest point a 0 is unique. Proof. (i) If x A then we choose a 0 = x. Suppose now that x/ A. Then there is a sequence a n A, n N, such that lim n x a n = inf{ x a a A} Almost all a n are contained in the ball B(x, 2ρ) where Since ρ = inf{ x a a A} A B(x, 2ρ) is a compact set there is a convergent subsequence a ni, i N. We have This implies lim i a n i = a 0 lim x a n i = inf{ x a a A} i x a 0 = 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)a 0 +ta A. Since a 0 is nearest point to x x a 0 2 x ((1 t)a 0 + ta) 2 = (x a 0 )+t(a 0 a) 2 By Lemma 5 0 < x a 0,a 0 a>

10 10 CONTENTS (iii) Suppose there is another point a 1 such that x a 1 = inf{ x a a A} By (ii) we get <x a 0,a 1 a 0 > 0 <x a 1,a 0 a 1 > 0 We add the two inequalities 0 < x a 0,a 1 a 0 > + <x a 1,a 0 a 1 >= a 0 a 1 2 Let A be a nonempty, closed, convex set. The metric projection p A : R n A maps a point x onto its nearest point in A. Lemma 9 Let A be a nonempty, closed, convex set in R n. The metric projection p A is a 1-Lipschitz map, i.e. for all x, y R n p A (x) p A (y) x y Proof. p A is well defined since by Lemma 8 for every x R n there is a unique point of minimal distance. Let x, y R n. By Lemma 8 for all a A <x p A (x),a p A (x) > 0 <y p A (y),a p A (y) > 0 In particular <x p A (x),p A (y) p A (x) > 0 <y p A (y),p A (x) p A (y) > 0 Thus < (x p A (x)) (y p A (y)),p A (x) p A (y) > 0 With this x y 2 = (x p A (x)) (y p A (y)) + (p A (x) p A (y)) 2 = (x p A (x)) (y p A (y)) 2 +2 < (x p A (x)) (y p A (y)),p A (x) p A (y) > + p A (x) p A (y) 2 p A (x) p A (y) 2

11 0.2. SEPARATION 11 Proposition 1 Let A and B be nonempty, disjoint, convex sets in R n. Moreover, let A be closed and B compact. Then A and B can be strictly separated by a hyperplane, i.e. there exists ξ R n such that sup <ξ,a><inf <ξ,b> a A b B Proof. There are two points a 0 A and b 0 B such that a 0 b 0 = inf{ a b a A, b B} We show this. Clearly, there are two sequences a n A and b n B, n N, such that lim n a n b n = inf{ a b a A, b B} By compactness of B there is a subsequence b ni, i N, that converges We have lim i b n i = b 0 a ni b ni b ni b 0 a ni b 0 a ni b ni + b ni b 0 It follows that lim a n i b 0 = inf{ a b a A, b B} i Almost all of a ni, i N, are elements of the ball B(b 0, 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 a 0. Now we choose ξ = a 0 b 0. Since a 0 is nearest point for b 0 and vice versa we have by Lemma 8 for all a A and b B It follows Therefore <a 0 b 0,a b 0 > 0 <b 0 a 0,b a 0 > 0 <a 0 b 0,b> < a 0 b 0,a 0 >= a 0 2 <b 0,a 0 > = 1{ a b a 0 b 0 2 } > 1{ a b 0 2 } > 1{ a b 0 2 a 0 b 0 2 } =< a 0,b 0 > b 0 2 =<a 0 b 0,b 0 > < a 0 b 0,a> inf <a 0 b 0,b>> 1{ a b B b 0 2 } > sup <a 0 b 0,a> a A

12 12 CONTENTS Corollary 1 Let A be a nonempty, closed, convex set in R n and x R n with x/ A. Then there is a hyperplane that separates A and x strictly. Corollary 2 Let A be a closed convex set in R n. Then A is the intersection of all closed halfspaces H + containing A, i.e. A = A H + H + Proof. Let x R n with x/ A. Then there is a hyperplane separating x and A strictly. Lemma 10 Let K be a nonempty, closed, convex set in R n and let x K. Then there is y/ K such that p(y) =x. Proof. There is a sequence x i R n \ K, i N, with We put We have and lim x i = x i y i = p(x i )+ x i p(x i ) x i p(x i ) y i p(x i ) = x i p(x i ) x i p(x i ) =1 p(y i )=p(x i ) We verify that p(y i )=p(x i ). Since p(x i ) is the nearest point to x i and p(y i )toy i we have by Lemma?? that for all ξ,η K <x i p(x i ),ξ p(x i ) > 0 <y i p(y i ),η p(y i ) > 0 Choosing ξ = p(y i ) and η = p(x i )weget and <x i p(x i ),p(y i ) p(x i ) > 0 0 < y i p(y i ),p(x i ) p(y i ) > =< p(x i ) p(y i )+ x i p(x i ) x i p(x i ),p(x i) p(y i ) > = p(x i ) p(y i ) 2 xi p(x i ) + x i p(x i ),p(x i) p(y i )

13 0.2. SEPARATION 13 Thus we get 0 xi p(x i ) x i p(x i ),p(y i) p(x i ) p(x i ) p(y i ) 2 which implies p(x i )=p(y i ). Therefore ( ) p(x) =p lim x i = lim p(x i ) = lim p(y i ) i i i The sequence y i, i N, is bounded since y i = p(x i)+ x i p(x i ) x i p(x i ) 1+ p(x i) and x i, i N, is a convergent sequence. Therefore the sequence y i, i N, has a convergent subsequence y ij, j N, y = lim j y ij We have p(y) = lim p(y ij ) = lim p(x ij )=p( lim y ij )=p(x) =x j j j Moreover we have x y. Indeed, x y = lim x ij y ij = lim j j x i j p(x ij )+ x i j p(x ij ) x ij p(x ij ) Since we get x y =1. lim x i j p(x ij )=0 j Lemma 11 Let K be a nonempty, convex set in R n that does not contain the origin 0. 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 Corollary 1. Now suppose that 0 K. By Lemma?? there is x R n \ K with p(x) = 0. As a hyperplane we choose now H = {ξ <x,ξ>=0}. Clearly, 0 H and therefore 0 H and 0 H +. Since 0 is the nearest point to x we have by Lemma 8 for all ξ K <ξ,x> 0

14 14 CONTENTS Proposition 2 Let K and C be nonempty, convex, disjoint sets in R n. Then there is a hyperplane separating K and C. If one of the sets is lower dimensional then it might happen that the set is contained in the separating hyperplane. Proof. The set C K = {x y x C, y K} is convex and 0 / C K. We verify this. Let x 1 y 1,x 2 y 2 C K and 0 <t<1. Then t(x 1 y 1 )+(1 t)(x 2 y 2 )=(tx 1 +(1 t)x 2 ) (ty 1 +(1 t)y 2 ) C K We have 0 / C K because C and K are disjoint. Therefore, there is a hyperplane separating 0 from C K, i.e. there is a ξ R n, 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 R n and x K. Then there is a supporting hyperplane of K that contains x. Proof. Suppose that K has no interior point. Then every point of K is a boundary point and K is contained in a hyperplane. We verify this. Suppose that there is no hyperplane that contains K. We may assume that 0 K. We choose a point x 1 K with 0 x 1. Now we choose a point x 2 K that is not contained in the subspace spanned by x 1. Such a point exists because otherwise K would be contained in this subset. By induction we get that there are n linearly independent points x 1,...,x n that are contained in K. Therefore [0,x 1,...,x n ] K and [0,x 1,...,x n ] 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 sets K and {x}. By Proposition 2 there is a separating hyperplane H, i.e. It follows H + K and x H. H + K and x H

15 0.3. EXTREME POINTS Extreme points A point x of a convex set C in R n 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 R n is not an 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 1 t 1. We choose the points y and 2x y. 2 Then we have x = 1y + 1 (2x y) 2 2 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, x 1,...,x k C and t 1,...,t k > 0 with k t i = 1 and x = k t i x i Then x is not an extreme point. We choose y = x 1 and Then z = k i=2 t i k i=2 t i x i ( k ) x = t 1 y + t i z Therefore, x is not an extreme point. The following theorem is a special case of the Krein-Milman theorem. Theorem A convex, compact subset of R n equals the closed, convex hull of its extreme points. i=2 This result can be improved. In fact, it can be shown that in R n a convex, compact set equals the convex hull of its extreme points. For infinite dimensional vector spaces one has to take the closed convex hull. Lemma 12 Let C be a convex subset of R n and H a supporting hyperplane of C. The the extreme points of C H are exactly the extreme points of C that are contained in H.

16 16 CONTENTS Proof. Clearly, all extreme points of C that are contained in H are also extreme points of C H. Let x be an extreme point of C H and not an extreme point of C. Then there are y and z with y, z C and x = y+z. We show that y, z H. 2 Let H be given by <,ξ >= t and suppose that C is a subset of {v <v,ξ> t}. Then t =< x,ξ>= 1 (<y,ξ>+ <z,ξ>) t 2 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 R n. extreme point. Then C has an Proof. If C consists of one point only, this point is also extreme point of C. Thus we 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 = max x x C We claim that z is an extreme point. We show this. Suppose that z is not an extreme 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 Therefore, we get Thus we have z = tx +(1 t)y t x +(1 t) y < z z 2 = tx +(1 t)y 2 = t 2 x 2 +2t(1 t) <x,y>+(1 t) 2 y 2 t 2 x 2 +2t(1 t) x y +(1 t) 2 y 2 =(t x +(1 t) y ) 2 = z 2 2t(1 t) <x,y>=2t(1 t) x y Since 0 <t<1weget <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 R n. Then the set of extreme points of C is not empty and every support hyperplane H of C contains at least one extreme point of C.

17 0.3. EXTREME POINTS 17 Proof. Let H be a support hyperplane of C. C H is a compact, convex set. By Lemma 13 C H has an extreme point. By Lemma 12 this is an extreme point of C. Proof of Theorem We denote the set of extreme points of C by Since C is compact and closed we have Ext(C) 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. Consider now the hyperplane max < ξ, x ><< ξ, z > x CH(Ext(C)) L = {x <ξ,x>= max y C <ξ,y>} This hyperplane is a support hyperplane of C and it contains by Lemma?? an extreme point v of C. In particular, v L CH(Ext(C)) It follows On the other hand, since z C < ξ, v ><< ξ, z > <ξ,z> max x C <ξ,x>=< ξ, v ><< ξ, z >

18 18 CONTENTS 0.4 Blaschke selection principle A convex body in R n is a convex, compact set with an interior point. The support function h C : R n R ofaconvexbodyc is defined by h C (ξ) = sup <ξ,x> x C The Hausdorff distance between two convex, compact sets K and C is d H (C, K) = max {inf{ρ C K + B n 2 (0,ρ)}, inf{ρ K C + B n 2 (0,ρ)}} Lemma 15 Let C and K be two convex, compact sets in R n. Then we have (i) { d H (C, K) = max inf (ii) sup x C y K } x y, sup inf x y y K x C d H (C, K) = sup h C (ξ) h K (ξ) ξ =1 Proof. (i) Suppose we have C K + B n 2 (0,ρ). Then, for every x C there are y K and z B n 2 (0,ρ) with x = y + z. This implies x y = z ρ. Therefore, for every x C we have inf y K x y ρ and subsequently sup x C inf y K x y ρ and therefore It follows that sup x C inf x y inf{ρ C K + y K Bn 2 (0,ρ)} { d H (C, K) max sup x C inf y K } x y, sup inf x y y K x C (ii) C K + B2 n (0,ρ) and K C + B2 n (0,ρ) is equivalent to: For all ξ, ξ =1,wehave h C (ξ) h K (ξ)+ρ and h K (ξ) h C (ξ)+ρ We prove the equivalence. h C (ξ) = sup <ξ,x> sup <ξ,x>= x C x K+B2 n(0,ρ) = sup y K sup y K z B n 2 (0,ρ) <ξ,y>+ sup <ξ,z>= h K (ξ)+ρ z B2 n(0,ρ) <ξ,y+ z>

19 0.4. BLASCHKE SELECTION PRINCIPLE 19 On the other hand, if C K + B n 2 (0,ρ) then there is x C but x/ K + B n 2 (0,ρ). By the Theorem of Hahn-Banach there is a hyperplane H = {y <ξ,y>= t} that separates x and K + B n 2 (0,ρ), i.e. It follows that h K (ξ)+ρ<h C (ξ). The symmetric difference distance is sup <ξ,y><<ξ,x>. y K+B2 n(0,ρ) d S (C, K) =vol n ((C \ K) (K \ C)) Lemma 16 (i) d H is a metric on the space K of all compact, convex subsets of R n. (ii) d S is a metric on the space K of all compact, convex subsets of R n. Proof. For compact sets C and K we have d H (C, K) = max h C (ξ) h K (ξ) ξ 2 =1 From this it follows immediately that d H (C, C) =0,d H (C, K) 0, and d H (C, K) d H (C, M)+d H (M,K). We have that d H (C, K) > 0ifC K since two compact, convex sets are equal provided that their support functions are equal. We verify this. Suppose that C K. Then, w.l.o.g there is x C with x / K. Then by Corollary 1 there is ξ R n with sup <ξ,y><<ξ,x> y K This implies h K (ξ) = sup <ξ,y><<ξ,x> h C (ξ) y K Lemma 17 Let K i, i N, be a sequence of compact, nonempty, convex sets in R n such that we have for all i N K i+1 K i Then is a convex, compact, nonempty set and the sequence K i, i N, converges to it with respect to the Hausdorff metric. K i

20 20 CONTENTS Proof. We have to show that for all ɛ>0 there is l N such that we have for all j l ( ) d H K i,k j <ɛ This means { max sup x K i y K j inf x y, sup inf y K j x x y K i } = sup inf y K j x x y <ɛ K i Suppose this is not true. Then there is a subsequence K jm, m N and elements x jm K jm such that inf x x K jm x ɛ i Then there is a subsequence of x jm, m N, that converges to an element z. We have z K i and inf z x ɛ x K i This cannot be. Lemma 18 The space K of all nonempty, compact, convex subsets of R n with the metric d H is complete. Proof. Let K i, i N, be a Cauchy sequence. We put ( ) C j = CH K i The sequence C j, j N, is a decreasing sequence of compact, convex sets. The sets C j are obviously convex and closed. We show that they are bounded. Since the sequence K i, i N, is a Cauchy sequence we have that there is i 0 such that for all i i 0 d H (K i0,k i ) 1 i=j This means that for all i i 0 K i K i0 + B2 n (0, 1) Since K i0 + B2 n (0, 1) is a bounded set, the set is also bounded. i i 0 K i

21 0.4. BLASCHKE SELECTION PRINCIPLE 21 By Lemma 17 the sequence C j, j N, converges in the metric space. We claim that the sequence K i, i N, converges to the same limit. It suffices to show that for every ɛ>0 there is j 0 so that we have for all j j 0 d H (C j,k j ) <ɛ Since K i, i N, is a Cauchy sequence there is i 0 so that we have for all l, m i 0 d H (K l,k m ) <ɛ or K l K m + B n 2 (0,ɛ) Therefore we get for all j i 0 or ( ) K j C j = CH K i K j + B2 n (0,ɛ)=K j + B2 n (0,ɛ) i=j d(k j,c j ) ɛ Lemma 19 (Blaschke selection principle) Let K i, i N, be a sequence of nonempty, compact, convex sets in a bounded subset of R n. Then there is a subsequence K ij, j N, that converges with respect to the Hausdorff distance to a nonempty, compact, convex set. Moreover, if all sets K i, i N, are contained in a compact set C, then the limit is contained in C. Proof. We show first that for any ɛ>0 and any sequence K i, i N, in a bounded set there is a subsequence K ij, j N, such that we have for all j, j N d H (K ij,k ij ) <ɛ By assumption the sequence K i, i N, is contained in a set [a, b] n. We may assume that a = 0 and b = 1. We choose k N such that n2 k <ɛand we put Q l = n [ li 1, l ] i 2 k 2 k where 1 l i 2 k, l =(l 1,...,l n ) and 1 i n. Then we have [0, 1] n = l Q l We consider I i = {l Q l K i }

22 22 CONTENTS There are 2 kn different vectors l. Therefore there are at most 2 2kn different sets I i. Thus there are only finitely many different sets I i and therefore there is a subsequence i j, j N, such that we have for all j, j N I ij = I ij Let x K ij. Then there is Q l with x Q l. Since I ij = I ij there is y Q l K ij. Therefore x y 2 n <ɛ,orx B n 2 k 2 (y, ɛ). Thus we get K ij K ij + B n 2 (0,ɛ) We conclude d H (K ij,k ij ) ɛ Now we select from the sequence K i, i N, a subsequence that is a Cauchy sequence. From the sequence K i, i N, we choose a subsequence K 1,j such that we have d H (K 1,j,K 1,j ) < 1 for all j N. After having chosen the subsequence K 2 (k,j), j N, we choose the sequence K k+1,j, j N, as a subsequence of K k,j, j N, such that we have for all j N d H (K k+1,j,k k+1,j ) < 1 2 k+1 This implies that the sequence K j,j, j N, is a Cauchy sequence.

23 0.5. POLYTOPES AND POLYHEDRA Polytopes and polyhedra A polytope P is the convex hull of of finitely many points in R n P =[x 1,...,x k ] We have that Ext(P ) {x 1,...,x k } We verify this. P is the convex hull of the points x 1,...,x k. By Lemma?? the convex hull consists of all convex combinations k t ixi of the points x 1,...,x k. All points that are convex combinations k t ixi where at least 2 coefficients t i are strictly larger than 0 are not extreme points. That leaves all those points where exactly 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 affine spaces containing that face. A simplex in R n is the convex hull of n+1 points. A regular simplex is a simplex for which all the distances between two vertices are equal. A polyhedral set is the intersection of finitely many closed halfspaces in R n. Lemma 20 Each face of a polytope P is again a polytope and its vertices are exactly the vertices of P that are contained in the hyperplane defining the face. Proof. Let P =[x 1,...,x k ] and let H be a support hyperplane of P. Let us note that P H is a convex set. Then Ext(P ) {x 1,...,x k }. By choosing a new numbering we my assume that x 1,...,x m, m k, are the extreme points of P.By the Theorem of Krein-Milman we have P =[x 1,...,x m ]. By Lemma?? the extreme points of P H are the extreme points of P contained in H. By the Theorem of Krein-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 R n. faces. Then P has only finitely many Lemma 21 Let P be a polyhedral set in R n that has an interior point N P = {x <ξ i,x> t i } Suppose that none of the halfspaces can be dropped, i.e. halfspaces would be different from P. Then the sets the intersection of the F i = P {x <ξ i,x>= t i }

24 24 CONTENTS have the dimension n 1 and P = N F i Proof. We show first that P = N {x <ξ i, x >< t i } Clearly, it is an open set contained in P and thus P N {x <ξ i, x >< t i } On the other hand, let x be such that there is i with <x,ξ i >= t i. Then in every neighborhood of x there is a point not in P. The point x + ɛξ i is not an element of P : <x+ ɛξ i,ξ i >= t i + ɛ Thus we have shown that and, consequently, P = N {x <ξ i, x >< t i } N N P = P \ P = P ( P ) c = P {x <ξ i,x>= t i } = We show now that the faces F i have dimension equal to n 1. Since none of the halfspaces can be omitted there are x i / P,1 i N, such that <x i,ξ j > t j 1 j i N <x i,ξ i >> t i 1 i N Let x 0 be an interior point of P. Then there is ɛ with B n 2 (x 0,ɛ) P. By continuity there is a point y i P with <ξ i,y i >= t i. Indeed, we consider all points sx 0 + (1 s)x i with 0 s 1. By convexity we have so that y i P. By convexity <sx 0 +(1 s)x i,ξ j > t j CH(x i,b n 2 (x 0,ɛ)) j i{x <ξ j, x >< t j } F i

25 0.5. POLYTOPES AND POLYHEDRA 25 It follows that there is δ>0 B n 2 (y i,δ) j i{x <ξ j, x >< t j } Thus B n 2 (y i,δ) {x <ξ i,x>= t i } P Therefore the boundary of P contains a n 1-dimensional Euclidean ball and the corresponding face is n 1-dimensional. Lemma 22 A set in R n 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 restrict ourselves 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 are only finitely many extreme points and thus there are only finitely many faces. Let F i, i =1,...,N, be all the faces of P and let {x <x,ξ i >= t i }, i =1,...,N,be the hyperplanes giving the faces F i = P {x <x,ξ i >= t i } The normals ξ i are oriented in such a way that Clearly, we have P {x <x,ξ i > t i } P N {x <x,ξ i > t i } We show that the sets are equal. Suppose that they are not equal. Then there is x 0 / P with N x 0 {x <x,ξ i > t i } Moreover, there is an interior point x 1 of P. Therefore there is y 0 [x 0,x 1 ] with y 0 P, i.e. y 0 = λx 0 +(1 λ)x 1 0 λ 1 Since x 0 / P we have y 0 x 0 and since x 1 is an interior point of P we have y 0 x 1. Thus y 0 = λx 0 +(1 λ)x 1 0 <λ<1 By the Theorem of Hahn-Banach there is a supporting hyperplane of P at y 0. Therefore y 0 is an element of a face of P. Therefore, there is i 0 such that t i0 =<y 0,ξ i0 >= λ<x 0,ξ i0 > +(1 λ) <x 1,ξ i0 >

26 26 CONTENTS Since x 1 is an interior point of P we have <x 1,ξ i0 Thus <t i0. Moreover, 0 <λ<1. t i0 =<y 0,ξ i0 >< t i0 We show now that a bounded polyhedral set is a polytope. For this we show that an extreme point e of N {x <x,ξ i > t i } satisfies {e} = {x <x,ξ i >= t i } <e,ξ i >=t i Since e is an extreme point it is an element of the boundary. There are sets I and J such that I = {i <e,ξ i >= t i } J = {i <e,ξ i >< t i } Suppose that {e} {x <x,ξ i >= t i } <e,ξ i >=t i Then there are at least two different elements e and f in the set <e,ξ i >=t i {x < x, ξ i >= t i }. Both are also elements of For all λ R we have Indeed, for all i I N {x <x,ξ i >= t i } (1 λ)e + λf <e,ξ i >=t i {x <x,ξ i >= t i } < (1 λ)e + λf, ξ i >=(1 λ) <e,ξ>+λ <f,ξ i >= t i Moreover, there is ɛ>0 such that for all i J < (1 ɛ)e + ɛf, ξ i >=(1 ɛ) <e,ξ i > +ɛ <f,ξ i >< t i < (1 + ɛ)e ɛf, ξ i >=(1+ɛ) <e,ξ i > ɛ <f,ξ i >< t i Thus we have (1 ɛ)e + ɛf (1 + ɛ)e ɛf e = 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 2 N extreme points. By the Theorem of Minkowski/Krein-Milman a compact convex set is the closed convex hull of its extreme points. The convex hull of finitely many points is a polytope (which is a closed set).