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1 Convex Geometry Introduction Martin Henk TU Berlin Winter semester 2014/15 webpage

2 CONTENTS i Contents Preface ii WS 2014/15 0 Some basic and convex facts 1 1 Support and separate 7 2 Radon, Helly, Caratheodory and (a few) relatives 17 3 The multifaceted world of polytopes 25 4 The space of compact sets / convex bodies 39 5 A glimpse on mixed volumes 43 6 Ellipsoids and the others 63 Index 79

3 ii CONTENTS Preface The material presented here is stolen from different excellent sources: First of all: A manuscript of Ulrich Betke on convexity which is partially based on lecture notes given by Peter McMullen. The inspiring books by Alexander Barvinok, A course in Convexity Günter Ewald, Combinatorial Convexity and Algebraic Geometry Peter M. Gruber, Convex and Discrete Geometry Peter M. Gruber and Cerrit G. Lekkerkerker, Geometry of Numbers Jiri Matousek, Discrete Geometry Rolf Schneider, Convex Geometry: The Brunn-Minkowski Theory Günter M. Ziegler, Lectures on polytopes and some original papers!! and they are part of lecture notes on Discrete and Convex Geometry jointly written with Maria Hernandez Cifre but not finished yet.

4 Some basic and convex facts 1 0 Some basic and convex facts 0.1 Notation. R n = { x = (x 1,..., x n ) : x i R } denotes the n-dimensional Euclidean space equipped with the Euclidean inner product x, y = n x i y i, x, y R n, and the Euclidean norm x = x, x. 0.2 Definition [Linear, affine, positive and convex combination]. Let m N and let x i R n, λ i R, 1 i m. i) m λ i x i is called a linear combination of x 1,..., x m. ii) If m λ i = 1 then m λ i x i is called an affine combination of x 1,..., x m. iii) If λ i 0 then m λ i x i is called a positive combination of x 1,..., x m. iv) If λ i 0 and m λ i = 1 then m λ i x i is called a convex combination of x 1,..., x m. v) Let X R n. x R n is called linearly (affinely, positively, convexly) dependent of X, if x is a linear (affine, positive, convex) combination of finitely many points of X, i.e., there exist x 1,..., x m X, m N, such that x is a linear (affine, positive, convex) combination of the points x 1,..., x m. 0.3 Definition [Linearly and affinely independent points]. x 1,..., x m R n are called linearly (affinely) dependent, if one of the x i is linearly (affinely) dependent of {x 1,..., x m }\{x i }. Otherwise x 1,..., x m are called linearly (affinely) independent. 0.4 Proposition. Let x 1,..., x m R n. i) x 1,..., x m are affinely dependent if and only if ( x 1 ) ( 1,..., xm1 ) R n+1 are linearly dependent. ii) x 1,..., x m are affinely dependent if and only if there exist µ i R, 1 i m, with (µ 1,..., µ m ) (0,..., 0), m µ i = 0 and m µ i x i = 0. iii) If m n + 1 then x 1,..., x m are linearly dependent. iv) If m n + 2 then x 1,..., x m are affinely dependent. Proof. By definition we have that x 1,..., x m are affinely dependent if and only if there exists an x i, say, and scalars λ j, 1 j i m, such that x i = j i λ j x j and j i λ j = 1. This can be reformulated as ( ) 0 = 0 ( xi 1 ) + ( ) xj λ j 1 j i which is equivalent to the linear dependency of the vectors ( x j 1 ). For ii) we just observe that the equation above is just a reformulation of what is to show if we

5 2 Some basic and convex facts set µ i = 1 and µ j = λ j, j i. Fianlly, iv) follows from i) and iii), which is trivial. 0.5 Definition [Linear subspace, affine subspace, cone and convex set]. X R n is called i) linear subspace (set) if it contains all x R n which are linearly dependent of X, ii) affine subspace (set) if it contains all x R n which are affinely dependent of X, iii) (convex) cone if it contains all x R n which are positively dependent of X, iv) convex set if it contains all x R n which are convexly dependent of X. 0.6 Notation. C n = {K R n : K convex} denotes the set of all convex sets in R n. The empty set is regarded as a convex, linear and affine set. 0.7 Theorem. K R n is convex if and only if λx + (1 λ) y K, for all x, y K and 0 λ 1. Proof. Of course, if K is convex and x, y K then for any λ [0, 1] the point λ x + (1 λ) y is convexly dependent of K and hence it must belong to K. Conversely, let v R n be convexly dependent of K. Then there exist x 1,..., x m K and scalars λ 1,..., λ m 0 with λ i = 1 and v = λ i x i. We show by induction on m that v K. The case m = 1 is trivial, and so let m 2 and λ m < 1. Then ( m m 1 ) λ i v = λ i x i = (1 λ m ) x i +λ m x m = (1 λ m ) x + λ m x m. 1 λ m }{{} Since x m 1 m 1 λ i 1 = λ i = 1 λ m = 1, 1 λ m 1 λ m 1 λ m we get by our inductive argumention x K and thus, by our assumption v K. 0.8 Example. The closed n-dimensional ball B n (a, ρ) = { x R n : x a ρ } with centre a and radius ρ > 0 is convex. The boundary of B n (a, ρ), i.e., { x R n : x a = ρ } is non-convex. In the case a = 0 and ρ = 1 the ball B n (0, 1) is abbreviated by B n and is called n-dimensional unit ball. Its boundary is denoted by S n 1.

6 Some basic and convex facts Corollary. Let K i C n, i I. Then i I K i C n. Proof. Let x, y i I K i and let λ [0, 1]. Since K i is a convex set for all i I, we have λ x + (1 λ) y K i for all i I and hence λ x + (1 λ) y i I K i. By Theorem 0.7 we obatin the convexity of i I K i Definition [Linear, affine, positive and convex hull, dimension]. Let X R n. i) The linear hull lin X of X is defined by lin X = L R n, L linear, X L ii) The affine hull aff X of X is defined by aff X = A R n, A affine, X A iii) The positive (conic) hull pos X of X is defined by pos X = L. A. C R n, C convex cone, X C iv) The convex hull conv X of X is defined by conv X = K R n, K convex, X K v) The dimension dim X of X is the dimension of its affine hull, i.e., dim aff X Theorem. Let X R n. Then { m } m conv X = λ i x i : m N, x i X, λ i 0, λ i = 1. Proof. Let M(X) bet the set on the right hand side of the equation above. For the inclusion conv X M(X) we note that by definition conv X is contained in any convex set containing X, and so it suffices to show that M(X) is convex. So let x, y M(X) with x = m 1 ν i x i and y = m 2 j=1 µ j y j, x i, y j X, ν i, µ j 0, and m 1 ν i = m 2 j=1 µ j = 1. Hence for λ [0, 1] we find C. K. m 1 m 2 λx + (1 λ)y = λν i x i + (1 λ)µ j y j, j=1

7 4 Some basic and convex facts where the coefficients of the above combination satisfy λν i, (1 λ)µ j 0 and m1 λν i + m 2 j=1 (1 λ)µ j = λ + (1 λ) = 1. Hence λ x + (1 λ) y M(X) which shows that M(X) is convex (cf. Theorem 0.7), and so conv X M(X). In order to verify the reverse inclusion M(X) conv X, we observe that each x M(X) is convexly dependent of X, and hence x is contained in any convex set containing X, i.e., x K C n,x K K = conv X Remark. i) conv {x, y} = { λ x + (1 λ) y : λ [0, 1] }. ii) lin X = { m λ ix i : λ i R, x i X, m N }. iii) aff X = { m λ ix i : λ i R, m λ i = 1, x i X, m N }. iv) pos X = { m λ ix i : λ i R, λ i 0, x i X, m N } Definition [(Relative) interior point and (relative) boundary point]. Let X R n. i) x X is called an interior point of X if there exists a ρ > 0 such that B n (x, ρ) X. The set of all interior points of X is called the interior of X and is denoted by int X. ii) x R n is called boundary point of X if for all ρ > 0, B n (x, ρ) X and B n (x, ρ) (R n \X). The set of all boundary points of X is called the boundary of X and is denoted by bd X. iii) Let A = aff X. x X is called a relative interior point of X if there exists a ρ > 0 such that B n (x, ρ) A X. The set of all relative interior points is called the relative interior of X and is denoted by relint X. iv) Let A = aff X. x A is called a relative boundary point of X if for all ρ > 0, B n (x, ρ) X and B n (x, ρ) (A\X). The set of all relative boundary points of X is called relative boundary of X and is denoted by relbd X Remark. Let X R n be closed. Then X = relint X relbd X Theorem. Let K C n, x relint K and y K. Then (1 λ)x + λy relint K for all λ [0, 1). Proof. Let A = aff K, x relint K, and for λ [0, 1) let z λ = (1 λ) x + λ y. Since x = z 0 relint K there exists a ρ > 0 such that B n (x, ρ) A K. By the theorem on intersecting lines it follows immediately that B n ( zλ, (1 λ)ρ ) A K, and hence z λ relint K. Or more explicitly: For a B n ( zλ, (1 λ)ρ ) A we have (1 λ)ρ z λ a = (1 λ)x + λy a = (1 λ)x (a λy).

8 Some basic and convex facts 5 Since λ < 1 we may divide both sides by 1 λ and get x (a λ y)/(1 λ) ρ. Thus (a λ y)/(1 λ) B n (x, ρ) A K and by the convexity of K we finally find a = (1 λ) a λy 1 λ + λy K Corollary. Let K C n be closed. Let x relint K and y aff K \ K. Then the segment conv {x, y} intersects relbd K in precisely one point. Proof. K conv {x, y} is a convex, compact 1-dimensional set. Hence K conv {x, y} = conv {x, y} for some y K. Obviously, y / relint K and so y relbd K. By Theorem 0.15, y is the only point of conv {x, y} lying on relbd K Definition [Polytope and simplex]. Let X R n of finite cardinality, i.e., #X <. i) conv X is called a (convex) polytope. ii) A polytope P R n of dimension k is called a k-polytope. iii) If X is affinely independent and dim X = k then conv X is called a k- simplex Notation. P n = {P R n : P polytope} denotes the set of all polytopes in R n Lemma. Let T = conv {x 1,..., x k+1 } R n be a k-simplex, and let λ i > 0, 1 i k + 1, with λ i = 1. Then λ i x i relint T. Proof. See Exercise?? Corollary. Let K C n, K. Then relint K. Proof. Let k = dim K 0. Then there exist x 1,..., x k+1 K affinely independent such that aff K = aff {x 1,..., x k+1 }. Let T k = conv {x 1,..., x k+1 } K. From Lemma 0.19 we get relint T k, and hence relint K Theorem. Let P = conv {x 1,..., x m } P n. A point x R n belongs to relint P if and only if x admits a representation as x = m λ ix i with λ i > 0, 1 i m, and m λ i = 1. Proof. Let x relint P and let y = m (1/m) x i P. Since x relint P there exists a z P such that x = λ z + (1 λ) y with λ [0, 1). Let z = m µ ix i, with µ i 0, 1 i m, and µ i = 1. Then x = m ( λµi + (1 λ)/m ) x i, where all the scalars λµ i + (1 λ)/m are positive and sum up to 1. Next we assume that x has a representation as x = m λ ix i with λ i > 0, 1 i m, and λ i = 1. Let k = dim P and without loss of generality let

9 6 Some basic and convex facts x 1,..., x k+1 be affinely independent. Setting λ = k+1 λ i, Lemma 0.19 shows that y = k+1 λ i λ x i relint conv {x 1,..., x k+1 } relint P. If λ = 1 then k + 1 = m and hence x = y relint P. If λ < 1 let z = 1/(1 λ) m i=k+1 λ i x i P and with Theorem 0.15 we also find in this case x = λy + (1 λ)z relint P Notation. i) For two sets X, Y R n the vectorial addition X + Y = {x + y : x X, y Y } is called the Minkowski 1 sum of X and Y. If X is just a singleton, i.e., X = {x}, then we write x + Y instead of {x} + Y. ii) For λ R and X R n we denote by λ X the set For instance, B n (a, ρ) = a + ρ B n. λ X = {λ x : x X}. 1 Hermann Minkowski,

10 Support and separate 7 1 Support and separate 1.1 Notation. Let a R n, a 0, and α R. The closed halfspaces H + (a, α), H (a, α) R n are given by H + (a, α) = { x R n : a, x α }, H (a, α) = { x R n : a, x α }. The hyperplane H(a, α) is defined by H(a, α) = { x R n : a, x = α }. 1.2 Definition [Supporting hyperplane]. Let X R n. A hyperplane H(a, α) R n is called supporting hyperplane of X if: i) H(a, α) X and ii) X H (a, α). a is called outer normal vector of X and if, in addition, a = 1 then it is called outer unit normal vector of X. 1.3 Proposition. Let X R n and let H(a, α) be a supporting hyperplane of X. Then H(a, α) conv X = conv ( H(a, α) X ). Proof. Let x H(a, α) conv X. Since x conv X there exist x i X and λ i > 0, i = 1,..., m, with m λ i = 1 and x = m λ ix i. Since x H(a, α) we have a, x = α and from X H (a, α) we get a, x i α, 1 i m. Hence m m α = a, x = λ i a, x i αλ i = α. Thus a, x i = α, 1 i m, and so x i H(a, α) X which implies x conv ( H(a, α) X ). The reverse inclusion is trivial since conv ( H(a, α) X ) conv ( H(a, α) ) conv X = H(a, α) conv X. 1.4 Remark. Let X R n be compact and a R n \{0}. Then there exists a supporting hyperplane of X with outer normal vector a. 1.5 Definition [Nearest point map (or metric projection)]. Let K C n be closed. The map Φ K : R n K, where for x R n the point Φ K (x) K is given by x Φ K (x) = min { x y : y K } is called the nearest point map (metric projection) with respect to K. 1.6 Remark. The nearest point map is well-defined: Notice that since K is closed, for all x R n there exist y x K such that x y x = min { x y : y K }, and y x is uniquely determined. In fact, if there exists y K, y y x, with x y = x y x then we may assume that x y x and x y are linearly independent. Hence x y x + y 2 = 1 2 (x y x) + 1 (x y) 2 < 1 2 x y x x y = x y vx. Since K is convex, (y x + y)/2 K which contradicts the minimality of y x.

11 8 Support and separate 1.7 Theorem. Let K C n be closed and let x R n \ K. Let a = x Φ K (x) and α = a, Φ K (x). Then H(a, α) is a supporting hyperplane of K with outer normal vector a. Proof. By the choice of a and α it Φ K (x) K H(a, α) and hence it remains to show K H (a, α), i.e., that a, y α for all y K. Suppose the opposite and let y K with a, y > α. For λ [0, 1] we consider the distance of x to z(λ) = (1 λ)φ K (x) + λy K h(λ) = x z(λ) 2 = x Φ K (x) + λ(φ K (x) y) 2 = x Φ K (x) 2 + 2λ x Φ K (x), Φ K (x) y + λ 2 Φ K (x) y 2 = x Φ K (x) 2 + 2λ a, Φ K (x) y + λ 2 Φ K (x) y 2. Since a, Φ K (x) y < 0 there exists a positive λ (0, 1] with h(λ ) < h(0). This, however, contradicts the fact that h(0) is the minimal (squared) distance between x and K. 1.8 Corollary. Let K C n, K R n, be closed. Then K = H (a, α), H(a,α) supporting hyperplane of K i.e., K is the intersection of all its supporting halfspaces. Proof. Clearly K H(a,α) H (a, α). In order to prove the reverse inclusion we take x K, and let H(a, α) be the supporting hyperplane defined in Theorem 1.7. Then x H + (a, α) but x H(a, α), i.e, x H (a, α) and hence x is not contained in the intersection on the right hand side. 1.9 Corollary. Let X R n such that conv X is closed and conv X R n. Then conv X = H (a, α), X H (a,α) i.e., conv X is the intersection of all halfspaces containing X. Proof. Since each halfspace containing X also contains conv X, we certainly have that conv X is contained in the intersection of halfspaces above. On the other hand, if x conv X, by Corollary 1.8 there exists a supporting hyperplane H(a, α) of conv X with x H (a, α). Since X conv X H (a, α), x is also not contained in the intersection of the right hand side Lemma [Busemann-Feller Lemma]. 2,3 Let K C n be closed. Then Φ K (x) Φ K (y) x y for all x, y R n, i.e., the nearest point map does not increase distances. In particular, it is a continuous map. 2 Herbert Busemann, William Feller,

12 Support and separate 9 Proof. We suppose that Φ K (x) Φ K (y) and let a = Φ K (x) Φ K (y), α x = a, Φ K (x) and α y = a, Φ K (y). It suffices to show that x H + (a, α x ) and y H + ( a, α y ), because then a, x α x and a, y α y, which implies a, x y α x + α y = a, Φ K (x) Φ K (y) = Φ K (x) Φ K (y) 2. By the Cauchy-Schwarz inequality we conclude x y Φ K (x) Φ K (y). So we assume the contrary and without loss of generality let x / H + (a, α x ), i.e., a, x < α x. Then the ray R(x) = { Φ K (x) + λ ( x Φ K (x) ) : λ 0 } has to intersect the hyperplane H( a, α y ) in a point x, say, and by the Pythagorean theorem we obtain x Φ K (y) < x Φ K (x). On the other hand, by Exercise?? we have Φ K (z) = Φ K (x) for all z R(x), and so we get the contradiction x Φ K (x) = x Φ K (x) > x Φ K (y) Theorem. Let K C n be compact and let ρ > 0 such that K int (ρb n ). The nearest point map on ρ S n 1, i.e., Φ K : ρ S n 1 bd K is surjective. Proof. Let x bd K. Then Φ K (x) = x and for i N let x i int (ρb n ) such that x i K and lim i x i = x. By Lemma 1.10 we have x Φ K (x i ) = Φ K (x) Φ K (x i ) x x i. and so lim i Φ K (x i ) = x as well. By Exercise??, the intersection point z i of the ray R(x i ) = { Φ K (x i ) + λ ( x i Φ K (x i ) ) : λ 0 } with ρ S n 1 verifies Φ K (z i ) = Φ K (x i ), and thus lim i Φ K (z i ) = x. By the compactness of ρ S n 1 we may asssume (after restricting to a convergent subsequence) that (z i ) is convergent, and so let lim i z i = z ρ S n 1. Finally, the continuity of the nearest map point gives Φ K (z) = lim Φ K (z i ) = x. i 1.12 Corollary. Let K C n be closed and let x relbd K. Then there exists a supporting hyperplane H(a, α) of K with x H(a, α). Proof. Without loss of generality let dim K = n. Let x bd K and let γ > 0 with x int (γb n ). The convex set K = K (γb n ) is compact and x bd K. Let ρ > γ be such that K int (ρb n ). By Theorem 1.11 we can find z ρs n 1 with x = Φ K (z). Then Theorem 1.7 ensures that the hyperplane H(a, α), with a = z x and α = a, x, supports K at x. Finally we have to check that H(a, α) is also a supporting hyperplane of K at x. By definition we have K H (a, α) and suppose that there exists y K with a, y > α. Since a, x = α it holds a, (1 λ)x + λy > α for any λ (0, 1]. Let λ > 0 be sufficiently small such that y = (1 λ)x + λy γb n. Since y K we have y K with a, y > α, a contradiction.

13 10 Support and separate H (a, α) H + (a, α) K 1 K 2 H(a, α) Figure 1: A strictly separating hyperplane of two compact convex sets 1.13 Theorem [Separation theorem]. Let K 1, K 2 C n with K 1 K 2 =. Then there exists a separating hyperplane H(a, α) of K 1 and K 2, i.e., K 1 H + (a, α) and K 2 H (a, α). If K 1 is closed and K 2 is compact, then there exists even a strictly separating hyperplane H(a, α) of K 1 and K 2, i.e., K 1 int H + (a, α) and K 2 int H (a, α). Proof. If both, K 1 and K 2 are compact the statement is certainly true by standard compactness arguments (see also Exercise??). If K 1 is closed and K 2 is compact we take the intersection K 1 = K 1 ρb n for ρ > 0 sufficiently large such that the distance between K 2 and K 1 equals the distance between K 2 and K 1. Hence we are back in the case of two compact sets, and we finally have just to observe that a strictly separating hyperplane of K 2 and K 1 is also one for K 2 and K 1 (see also the proof of Corollary 1.12). Next we consider the case of arbitrary disjoint convex sets K 1 and K 2. Let x 1 relint K 1 and x 2 relint K 2, and for i N let [ ( ( Kj i = cl 1 1 ) ) ] (K j x j ) + x j (ib n ), for j = 1, 2. i Clearly Kj i Ki+1 j K j, for j = 1, 2 and any i N, and for every x relint K j there exists an index i x such that x Kj i for all i i x. Moreover, K1 i and Ki 2 are compact convex sets with K1 i Ki 2 = for any i N. By Exercise?? we know that there exists a separating hyperplane H(a i, α i ) of K1 i and Ki 2 with a i = 1, and thus a i, x α i for all x K i 1 and a i, x α i for all x K i 2. Since a i, x 1 α i a i, x 2 for i large and a i = 1, the sequence (α i ) i N is bounded and hence {( a i ) } α i : i N R n+1 is also a bounded sequence. Without loss of generality we can assume that this sequence is convergent and we write a = lim i a i and α = lim i α i. In order to prove that the hyperplane H(a, α) separates K 1 and K 2, let x K 1. If x relint K 1 then there exists an index i x such that a i, x α i for all i i x, which implies that a, x α. For x relbd K 1, we approach x by the points x λ = (1 λ)x + λx 1 relint K 1, λ (0, 1]. By the previous discussion we have a, x λ α for all λ (0, 1] and

14 Support and separate 11 so also a, x 0 = a, x = α. Analogously we obtain a, x α for all x K Definition [Support function, breadth]. Let K C n, K. The function h(k, ) : R n R given by h(k, u) = sup { u, x : x K } is called support function of K. For u S n 1 the breadth of K in the direction u is defined by h(k, u) + h(k, u) Proposition. Let K C n be non-empty and compact. Then K = { x R n : u, x h(k, u) }. u S n 1 Proof. By Corollary (1.8) it suffices to observe that the intersection is just taken over all supporting hyperplanes of K. In fact, given a supporting hyperplane H(a, α) of K we may assume a S n 1 and since K H (a, α) but K H(a, α), we have α = max x K a, x = h(k, a) Definition [Convex function]. Let K C n. A function f : K R is called convex if f ( λx + (1 λ)y ) λf(x) + (1 λ)f(y), for all x, y K, λ (0, 1). f is called strictly convex when the above inequality holds as a strict inequality if x y. If f is convex then f is called concave Proposition. Let f : K R be a function differentiable on an open convex set K R n. i) f is convex if and only if f(x) f(y) + f(y), x y, for all x, y K. (1.17.1) ii) Let f be twice differentiable. Then f is convex if and only if its Hessian is positive semi-definite for all points in K. Proof. i) We suppose first that f is convex. Then for any x, y K and any λ (0, 1) it holds f ( y + λ(x y) ) = f ( (1 λ)y + λx ) (1 λ)f(y) + λf(x), which implies f ( y + λ(x y) ) f(y) λ f(y), x y f(x) f(y) f(y), x y. The left hand side approaches zero when λ 0, which proves (1.17.1). Conversely, if (1.17.1) holds then interchanging the roles of x and y and adding both inequalities we get f(x) f(y), x y 0, for all x, y K. (1.17.2)

15 12 Support and separate Let x, y K. We define the function g : [0, 1] R given by g(λ) = f ( y + λ(x y) ), which is differentiable. Then for 0 λ 0 < λ 1 1 we get g (λ 1 ) g (λ 0 ) = f ( y + λ 1 (x y) ), x y f ( y + λ 0 (x y) ), x y 0 by using (1.17.2). Thus g is an increasing function which implies that g is a convex function, and so g(λ) λ g(1) + (1 λ) g(0) = λ f(x) + (1 λ) f(y). ii) For x, y K we consider the same function g x,y (λ) as in the previous proof. Then f is convex if and only if all g x,y are convex for all x, y K, which is equivalent to g x,y 0. Hence, f is convex if and only if for all x, y K and λ [0, 1] (x y) [(Hess f)(x + λ(x y))] (x y) 0. This is certainly fulfilled if the Hessian is positive semi-definite. Otherwise, for a given z in the open set K and a given vector v R n we can find x, y K and λ [0, 1] and µ R >0 such that v = µ(x y), z = x + λ(x y) and so v (Hess f)(z)v = µ 2 (x y) [(Hess f)(x + λ(x y))] (x y) 0, which shows that the Hessian is positive definite for every z K Theorem [Jensen s inequality]. 4 Let K C n and let f : K R be convex. For all x 1,..., x m K and 0 λ 1,..., λ m with m λ i = 1 it holds ( m ) f λ i x i m λ i f(x i ). Proof. Induction on m Remark. Let f : K R be defined on a convex set K. The set epi f = { ( ) x x n+1 R n+1 : x K, x n+1 f(x)} is called the epigraph of f. Then f is convex if and only if its epigraph epi f is convex Theorem*. Let K C n be open and let f : K R be convex. Then f is continuous Theorem. Let K C n be bounded and let K. i) h(k, ) is a convex function. ii) h(k, ) is positively homogeneous of degree 1, i.e., h(k, λu) = λh(k, u) for all λ 0 and u R n. iii) If h : R n R is a function satisfying i) and ii) then there exists a convex closed and bounded set K C n such that h(k, u) = h(u) for all u R n. 4 Johan Jensen,

16 Support and separate 13 Proof. Let u, v R n and λ [0, 1]. Then h ( K, λu + (1 λ)v ) = sup { λ u, x + (1 λ) v, x : x K } sup { λ u, x : x K } + sup { (1 λ) v, x : x K } = λh(k, u) + (1 λ)h(k, v), which shows i), and the last step also gives ii). For iii) and the given function h : R n R we consider K = v R n { x R n : x, v h(v) }, which is a closed convex(cf. Corollary 0.9) and bounded(consider v = e i ) set. If it is non-empty then clearly h(k, u) h(u) for all u R n, and so it suffices to prove that for every u R n there exist an a K with u, a = h(u) and thus h(k, u) = h(u). Since h satisfies i) and ii), its epigraph epi h is a closed (cf. Theorem 1.20) convex cone in R n+1. Let u R n. Since ( u, h(u) ) bd epi h, we can find by Corollary 1.12 a supporting hyperplane H ( (a, t), α ) of epi h through ( u, h(u) ) such that epi h H ( (a, t), α ). Since epi h is a cone containing 0, H ( (a, t), α ) must pass through 0, hence α = 0. If t = 0 then (b, h(b)) / H ( (a, t), 0 ) for b, a > 0. If t > 0, then for some x n+1 > h(u) the point (u, x n+1 ) would not be contained in H ( (a, t), 0 ). Hence t < 0 and so we may assume t = 1. Then we can write epi h H ( (a, 1), 0 ), and since (v, h(v)) epi h we have a, v h(v) for all v R n, which shows a K. Moreover, by construction we have a, u = h(u) and so h(k, u) = a, u = h(u) Definition [Polar set]. Let X R n. is called the polar set of X Proposition. X = { y R n : x, y 1 for all x X } i) X is a convex and closed set and 0 X. ii) If X 1 X 2 then X 2 X 1. iii) Let M be a regular n n matrix. Then (MX) = M X. iv) Let X i R n, i I. Then ( i I X ) i = i I X i. v) X (X ). vi) Let X R n. Then X = X if and only if X = B n.

17 14 Support and separate Proof. i) X is an intersection of closed and convex sets and hence X is closed and convex(cf. Corollary 0.9). Obviously, 0 X. ii) Let y X2. Then x, y 1 for all x X 2. In particular, x, y 1 for all x X 1 which proves that y X1. iii) Let y (MX). By definition, Mx, y 1 for all x X, i.e., x, M y 1 for all x X. This condition is equivalent to M y X, which says that y M X. iv) Just notice that ( ) { X i = y R n : x, y 1 for all x } X i i I i I = { y R n } : x, y 1 for all x X i = Xi. i I i I v) Let x X. By definition of polar body y, x 1 for all y X, i.e., x (X ). vi) We suppose first that X = B n. Then B n = { } y R n : x, y 1 for all x B n. If y Bn, it is clear that y, x 1 for all x B n, which shows B n B n. If y B n with y > 1 then y/ y B n. Since y/ y, y = y > 1 we get y B n, a contradiction. Conversely, let x X = X. By the definition of the polar body we know x, x 1, which shows X B n. Now part ii) implies X = X B n = B n, and thus X = B n Proposition. i) Let P = conv {x 1,..., x m } R n. Then P = { y R n : x i, y 1, 1 i m }. ii) Let P = { x R n : a i, x 1, 1 i m } with a i R n. Then P = conv {0, a 1,..., a m }. Proof. i) P { y R n : x i, y 1, 1 i m } holds by the definition of polar body. So let y R n with x i, y 1, 1 i m. For any x P there exist λ 1,..., λ m 0 with m λ i = 1 such that x = m λ ix i. Then x, y = m λ i x i, y m λ i = 1, i.e., y P. ii) Clearly P conv {0, a 1,..., a m }. So suppose there exists y P with y conv {0, a 1,..., a m }. Theorem 1.13 ensures the existence of a strictly separating hyperplane H(a, α) with a, x < α for all x conv {0, a 1,..., a m } and a, y > α. Since α > 0 we have, in particular, a/α, a i < 1 for 1 i m, which shows that a/α P. Since a/α, y > 1 we conclude y P, a contradiction Lemma. Let K C n be closed with 0 K. Then (K ) = K.

18 Support and separate 15 Proof. In view of Proposition 1.23, part v), it suffices to show (K ) K. We suppose there exists y (K ) with y K. Then by Theorem 1.13 we can find a strictly separating hyperplane H(a, α) such that a, y > α and a, x < α for all x K. Since α > 0 we have a/α, x < 1 for all x K and so a/α K. From a/α, y > 1 we get the contradiction y (K ).

19 16 Support and separate

20 Radon, Helly, Caratheodory and (a few) relatives 17 2 Radon, Helly, Caratheodory and (a few) relatives 2.1 Theorem [Radon]. 5 Let X R n. If #X n + 2 then there exist X 1, X 2 X with X 1 X 2 = and conv X 1 conv X 2. Proof. Since #X n + 2, X contains n + 2 affinely dependent points x 1,..., x n+2 X. Thus there exist λ i R, i = 1,..., n + 2, not all zero, with n+2 λ i = 0 and n+2 λ i x i = 0. Without loss of generality let λ 1,..., λ k > 0 and λ k+1,..., λ n+2 0. Then k λ i x i = n+2 i=k+1 ( λ i) x i, and with λ = k λ i = n+2 i=k+1 ( λ i) we may write k λ i λ x i = n+2 ( λ ) i x i conv {x 1,..., x k } conv {x λ }{{} k+1,..., x n+2 } }{{} X 1 i=k+1 X Theorem [Helly]. 6 Let K 1,..., K m C n, m n + 1, such that for each (n + 1)-index set I {1,..., m} we have i I K i. Then all sets K i have a point in common, i.e., m j=1 K i. Proof. We use induction on m. The case m = n + 1 is certainly true by assumption, and for the proof of the induction step m 1 m we set L j = m,i j K i, for j = 1,..., m. By induction hypothesis L j, and let x j L j, j = 1,..., m. Hence x j K i for all i j. Since m n + 2, we may apply Radon s Theorem 2.1 to {x 1,..., x m } and hence, without loss of generality, we may suppose that there exists x conv {x 1,..., x k } conv {x k+1,..., x m }, for a suitable index k. Now conv {x 1,..., x k } K i for i > k and conv {x k+1,..., x m } K i for i k, and so x m K i. 2.3 Remark. i) Without any further restrictions/assumptions Helly s theorem is not true for infinitely many convex sets K i. For instance, let K i = (0, 1 i ], i N. ii) Helly s theorem, however, can be e generalised to any collection of compact (bounded and closed) convex sets due to the finite intersection property of compact sets: a family o compact sets has nonempty intersection proviede each of its finite subfamilies has. 2.4 Corollary. Let C C n be compact. Then there exists t R n with 5 Johann Karl August Radon, Eduard Helly, C t + n C.

21 18 Radon, Helly, Caratheodory and (a few) relatives Proof. For c C we set L c = {t R n : c t + nc}. Clearly, L c = c nc, and so it is a convex and compact set. We are loking for a t c C L c. According to Helly s Theorem 2.2 (cf. also Remark 2.3 ii)) it suffcies to show that for c 1,..., c n+1 C n+1 L ci. For it we consider t = n+1 c i. Then for j = 1,..., n + 1, n+1 n+1 c j t = c i = n 1 n c i nc, i.e., t L ci, 1 i n + 1.,i j,i j 2.5 Definition [Centerpoint]. For a finite point set X R n a point c R n is called centerpoint if every closed halfspace containing c containes at least #X points of X. 1 n Theorem. Every finite set X R n has a centerpoint. Proof. With M = {U X : #U > (n/(n + 1))#X }, we first notice that c is a centerpoint of X if and only if c conv U. U M For if, c is not a centerpoint if and only if there exists a halfspace H containing c such that #(H X) < 1/(n+1)X. Hence c is not contained in conv U with U = (int H + X) M. On the other hand, if c / conv U for some U M, then Theorem 1.13 gives a halfspace H containing c and H X = X \ U. Hence we have to show that the intersection U M conv U, consisting of finitely many compact convex sets, is nonempty. Due to the cardinality of the sets U M the intersection of each n + 1 of theses sets is nonempty, and so the intersection of their convex hulls is non-empty. Observe, that for U 1,..., U m M n m + 1 #(U 1 U m ) > #X. n + 1 Hence, Helly s Theorem 2.2 yields U M conv U. 2.7 Theorem [Carathéodory]. 7 Let X R n. Then { n+1 } n+1 conv X = λ i x i : λ i 0, λ i = 1, x i X, i = 1,..., n Constantin Carathéodory,

22 Radon, Helly, Caratheodory and (a few) relatives 19 Proof. Let x conv X. By Theorem 0.11 there exist a minimal m N and x 1,..., x m X such that x = m λ i x i for certain λ i > 0, i = 1,..., m, with n λ i = 1. We have to show m n + 1. So we assume that m n + 2. Then {x 1,..., x m } are affinely dependent and there exist µ 1,..., µ m R not all of them zero with m µ i = 0 and m µ i x i = 0. Hence we can write x = m (λ i αµ i ) x i for any α R. Let the index k be chosen such that { } λ k λi = min : µ i > 0. µ k 1 i m µ i Then λ i (λ k /µ k )µ i 0 for i = 1,..., m, m,i k (λ i (λ k /µ k )µ i ) = 1 and x = m,i k ( λ i λ ) k µ i x i, µ k which contradicts the minimality of m. The reverse inclusion is certainly true by Theorem Remark. Let X R n. Then conv X = { dim X+1 λ i x i : λ i 0, dim X+1 λ i = 1, x i X As a direct consequence of Carathéodory s Theorem 2.7 we get 2.9 Corollary. A polytope is the union of simplices Corollary. The convex hull of a compact set is compact. Proof. Let X be a compact set and we suppose that X B n (0, ρ). Then conv X B n (0, ρ), which shows conv X is bounded. In order to see that conv X is closed let x = lim i x i, with x i conv X, i N. By Carathéodory s Theorem 2.7 we can write x i = n+1 j=1 λ ij x ij, with x ij X and λ ij 0, n+1 j=1 λ ij = 1. Since 0 λ ij 1 and X is bounded we may assume that for each j {1,..., n + 1} the limits lim i λ ij = λ j and lim i x ij = x j exist. By the closeness of X we have x j X, and moreover, λ j 0 and n+1 j=1 λ j = 1. So we know n+1 n+1 x = lim x i = lim λ ij x ij = λ j x j, i i which proves that x conv X. j=1 j=1 } Theorem [(Weak) Fractional Helly theorem]. Let K 1,..., K m C n, m n + 1, and let α (0, 1] such that for at least α ( m n+1) of the (n + 1)-index sets I {1,..., m} we have i I K i. Then there exists a point in common α of at least n+1 m sets K i.

23 20 Radon, Helly, Caratheodory and (a few) relatives Proof. For I {1,..., m} let K I = i I K i. First, we argue that we may assume that K i are compact convex sets. To this end let x I K I for any (n+1) index set I with K I and set K i = conv {x I : i I} K i. Then for each (n + 1)-index set J we have j J K j if and only if j J K j. Hence let K i, and thus K I, be compact convex sets. The compactness of a subset D R n, in particular, implies that is has a unique minimal point with respect to the lexicographic ordering < 8 lex. Next we claim: Let K I, #I n, and let y I K I be the unique lexicographical minumim of K I. Then there exists an n-index subset J I such that y I is the lexiographic minimum of K J as well. (Claim) For the proof of the claim we observe that C = {x R n : x < lex y I } is a convex set with C K I =. Application of Helly s theorem 2.2 to the sets C, K i, i I, shows that there exists n + 1 of these sets having an empty intersection. By definition of y I, C be oen of them. Hence, there must exist already an n-index set J I such that C K J = which means that y I K J is the minimum of K J as well. To finish the proof we fix for any of the α ( m n+1) index sets I of cardinality n + 1 an n-index set J I I having the same lexicographic minimum as K I. The number of distinct n tuples is ( m n), hence one of them, J, say, appears as J I for at least α ( ( m n+1) / m ) n = α m n n+1 distinct sets I which are of the type J {i} for some i {1,..., m}. Hence the lexicographic minimum of K J is contained in at least n + α m n n + 1 > α m n + 1 sets K i Theorem* [(strong) Fractional Helly theorem]. Let K 1,..., K m C n, m n + 1, and let α (0, 1] such that for at least α ( m n+1) of the (n + 1)-index sets I {1,..., m} we have i I K i. Then there exists a point in common of at least (1 (1 α) 1/(n+1) ) m sets K i Remark. The (strong and sharp) fractional Helly theorem, which is due to Kalai, gives that (1 (1 α) 1/(n+1) ) m sets have a point in common. Obviously, for α = 1 we get again the classcial Helly Theorem Theorem [Colorful Carathéodory theorem]. Let X 1,..., X n+1 R n such that 0 conv X i, 1 i n + 1. There exist x i X i, 1 i n + 1, such that 0 conv {x 1,..., x n+1 }. Proof. It suffices to prove the statement for finite sets X i, because otherwise we may replace X i be any finite subset X i containing 0 in its convex hull. For x i X i, 1 i n+1, we call {x 1,..., x n+1 } a rainbow set and conv {x 1,..., x n+1 } a rainbow simplex. Suppose there is no rainbow simplex containing 0, and let S = {s 1,..., s n+1 }, s i X i, be a rainbow set such that conv S has minimal 8 x = lex y iff x = y and x < lex y if there exists an i {0,..., n 1} such that x j = y j, 1 j i, and x i+1 < y i+1.

24 Radon, Helly, Caratheodory and (a few) relatives 21 distance to 0 attained at the point s S, say. Let H = {x R n : s, x = s, s } be the hyperplane perpendicular to s and containing s. Then S H + = {x R n : s, x s, s } which does not contain 0. Since conv S H = conv (S H) (cf. Proposition 1.3) we know by Carathéodory s Theorem 2.7 that there exists an n-point set T S H with s conv T. Without loss of generality let s 1 / T. If X 1 H + then 0 / conv X 1, and so we may assume that there exists a z X 1 with s, z < s, s. The new rainbow set S = S \ {s 1 } {z} contains the segment conv {s, z}, which, by the choice of z, contains a point closer to 0 than s a contradiction Theorem [Tverberg]. 9 Let X R n and let k N 1. If #X (k 1)(n+1)+1, k N, then there exist k subsets X 1,..., X k X with X i X j =, i j, but conv X 1 conv X 2 conv X k. Proof. First we show that the following conic version of the theorem is equivalent to Tverberg s result: Let X R n+1, 0 / conv X, and let k N 1. If #X (k 1)(n + 1) + 1, k N, then there exist k subsets X 1,..., X k X with X i X j =, i j, but pos X 1 pos X 2 pos X k {0}. (2.15.1) To see that it implies the theorem we canonically embed our given set X R n into R n+1 by adding a 1 to all points of X as last coordinate. This embedded set X is contained in the hyperplane H = {x R n+1 : x n+1 = 1} and hence 0 / conv X. So, according to (2.15.1), there exist X 1,..., X k X with X i X j =, i j, and a point ỹ 0 with ỹ pos X 1 pos X 2 pos X k. In particular, pos {ỹ} pos X 1 pos X 2 pos X k and so we may assume that ỹ = (y, 1). But then y conv X 1 conv X 2 conv X k, where X i is obtained from X i by removing the last coordinate from all points. The other implication is left as an exercise. For the proof of we set N = (k 1)(n + 1) and we consider the following k linear maps φ j : R n+1 R N : The first k 1 maps just writes the coordinates of a vector x R n+1 at the positions (j 1)(n + 1) + 1,..., j(n + 1) and all other coordinates are zero, or blockwise φ j (x) = (0 0 0 x 0 0), j = 1,..., k 1. The k-th map φ k : R n+1 R N just has x in each block, i.e., φ k (x) = ( x x x). We observe that for x 1,..., x k R n+1 holds k φ j (x j ) = 0 if and only if x 1 = x 2 = = x k. (2.15.2) j=1 9 Helge Arnulf Tverberg, 1935

25 22 Radon, Helly, Caratheodory and (a few) relatives Of course, it suffices to prove for a set X = {x 1,..., x N+1 } R n+1 of cardinality N + 1 with the property 0 / conv X. Let M = φ 1 (X) φ 2 (X) φ k (X) R N and we set M i = {φ 1 (x i ), φ 2 (x i ),..., φ k (x i )}, 1 i N + 1. In view of (2.15.2) we have 0 conv M i, 1 i N + 1, and hence we may apply the colorful Carathéodory Theorem 2.14 and find a rainbow set S M, #S = N + 1 with 0 conv S. Let f(i) {1,..., k} be the index of the point of M i in S, i.e., S = {φ f(1) (x 1 ), φ f(2) (x 2 ),..., φ f(n+1) (x N+1 )}. Since 0 conv S there exist nonnegative numbers α i, 1 i N + 1, such that N+1 0 = α i φ f(i) (x i ) and N+1 α i = 1. (2.15.3) Next for 1 j k let I j = {i : f(i) = j, 1 i N + 1} and let X j = {x i : i I j }. If I j = we set X j = {0}. By the linearity of the maps φ we can rewrite (2.15.3) as N+1 0 = α i φ f(i) (x i ) = k j=1 i I j α i φ j (x i ) = k j=1 φ j α i x i. i Ij Setting u j = i I j α i x i, 1 j k, we have u j pos X j and k j=1 φ j(u j ) = 0. By (2.15.2) we obtain u 1 = u 2 = = u k and so we have found a point in common of all pos X j. In remains to show that u j. By (2.15.3) we can find a representation of u j = i I j α i x i with some α i > 0. Since 0 / conv X we conclude u j Theorem*. Let X R n and let #X = m n + 1. Then there exists a point y R n ( contained in at least γ m n n+1) X-simplices, i.e., simplices of the form conv S, S X, #S = n+1. Here γ n is a positive constant depending only on the dimension, and X-simplices conv S 1, conv S 2 are considered different if S 1 S 2. Proof. We may assume that m is large, e.g., m 2n(n + 1), otherwise we can choose γ n so small that y is contained in a single simplex. Let k = m/(n + 1). Then m (k 1)(n + 1) + 1 and by Tverberg s Theorem 2.15 there exists k pairwise disjoint subsets X i X having a point y k conv X i. For a subset J = {j 0,..., j n } {1,..., k} and with respect to the sets X ji, 0 i n, we may apply the colorful Carathéodory theorem and find an X-simplex S J containing y and having one vertex from each X ji. For different sets J we get different X-simplices S J and so the number of X-

26 Radon, Helly, Caratheodory and (a few) relatives 23 simplices containing y is at least ( ) ( ) ( ) ( ) m k m/(n + 1) n+1 m n m n+1 n = n + 1 n + 1 (n + 1)! 1 m (m (n + 1))... (m n(n + 1)) = (n + 1) n+1 (n + 1)! ( ) 1 1 m (n + 1) n+1 2 n, n + 1 in view of our assumption m 2n(n + 1).

27 24 Radon, Helly, Caratheodory and (a few) relatives

28 The multifaceted world of polytopes 25 3 The multifaceted world of polytopes 3.1 Definition [Polyhedron]. The intersection of finitely many closed halfspaces is called a polyhedron. 3.2 Theorem [Minkowski, Weyl]. 10,11 i) A bounded polyhedron is a polytope. ii) A polytope is a bounded polyhedron. Proof. i) Let P = {x R n : a i, x α i, 1 i m} be bounded. We proceed by induction on n. The case n = 1 is obvious. Hence, we may assume n 2, and let F i = P H(a i, α i ), 1 i m. F i is a bounded polyhedron in an affine space of dimension at most n 1 and by our inductive argument there exist finite sets V i such that F i = conv V i, 1 i m. It suffices to show that P = conv (V 1 V 2 V m ). The inclusion follows from V i P and the convexity of P. For the reverse inclusion let x P and let l be a line passing through x. The intersection l P is a non-empty compact convex set of dimension at most 1. Hence we can find y, z P such that l P = conv {y, z}. Since both, y and z, have to lie in the boundary of P we can find k and j with y P H(a k, α k ) = F k = conv V k and z P H(a j, α j ) = F j = conv V j. So we have x conv {y, z} conv (V k V j ) conv (V 1 V 2 V m ). For the second statement ii) we apply polarity to i). Let P = conv {v 1,..., v m }, and here we may assume that dim P = n and 0 int P. By Propositions 1.23 ii) and 1.24 i) we find that P is a bounded polyhedron. Applying i) to P we can find points w 1,..., w l R n such that P = conv {w 1,..., w l }. Next we consider (P ), which can be written as (cf. Proposition 1.24 i)) (P ) = {x R n : w i, x 1, 1 i l}. By Lemma 1.25 we know P = (P ) and we are done. 3.3 Notation [V-Polytope, H-Polytope]. A polytope given as the convex hull of finitely many points is called a V-polytope. If it is given as the bounded intersection of finitely many closed halfspaces, then it is called an H-polytope. 3.4 Corollary. Let P P n. i) Let A R m n and t R m. Then AP + t is a polytope. ii) Let U R n be an affine subspace. Then P U is a polytope. 10 Hermann Minkowski, Hermann Klaus Hugo Weyl,

29 26 The multifaceted world of polytopes Proof. For i) we may assume that P is a V-Polytope, i.e., P = conv {v 1,..., v m } for some points v i R n. Hence AP + t = t + conv {A v 1,..., A v m } = conv {A v 1 + t,..., A v m + t} is again a polytope. For ii) we regard P as an H-polytope, i.e., P = m H (a i, α i ) for suitable hyperplanes H(a i, α i ). Each affine subspace U R n can be written as intersection of hyperplanes and thus as intersection of halfspaces. Hence let U = k H (u i, µ i ), 1 i k, and so P U = m H (a i, α i ) k H (u i, µ i ). Since P is bounded, P U is a bounded polyhedron, i.e., a polytope by Theorem Definition [Faces]. Let K C n be closed and let H be a supporting hyperplane of K. If j = dim(k H), then K H is called a j-face of K. Moreover, K itself is regarded as a (dim K)-face and the empty set as ( 1)- face of K. 3.6 Notation [Vertices, exposed points, edges, facets]. A 0-face of K C n, K closed, is called an exposed point or in the case of a polytope a vertex, a 1-face of a polytope is called edge and a (dim K 1)-faceof K is called facet of K. K itself and the empty set are called improper faces, whereas the remaining faces are called proper faces of K. The set of all vertices of a polytope P is denoted by vert P. 3.7 Remark. i) Let K C n be closed. Every (relative) boundary point of K lies in a suitable j-face, 0 j dim K 1 (cf. Corollary 1.12). ii) Let K C n, dim K = n. Let F be a facet of K and H a supporting hyperplane of K with F = K H. Then H = aff F. iii) A point v K, K C n, is called an extreme point of K, if it can not be written as a convex combination of two other points in K. 3.8 Proposition. Every face of a polytope is a polytope, and a polytope has only finitely many faces. Proof. Let P = conv {v 1,..., v m } and let F = P H(a, α) be a face of F. By Proposition 1.3 we have P H(a, α) = conv (H(a, α) {v 1,..., v m }) which shows that F is a polytope. In particular, we can write each face F of P as conv V F for a suitable subset V F {v 1,..., v m }. Since different faces have different affine hulls and since the affine hull of a j-face, j {0,..., n 1}, is uniquely determined by j + 1 affinely independent points we may bound the number of all proper faces by n 1 ( m i=0 i+1). 3.9 Definition [f-vector]. For P P n let f i (P ) be the number of i-faces of P, 1 i dim P. Furthermore, let f i (P ) = 0 for dim P + 1 i n. The vector f(p ) with entries f i (P ), 1 i n, is called the f-vector of P.

30 The multifaceted world of polytopes Remark. i) Let T n be an n-dimensional simplex. Then f i (T n ) = ( n+1 i+1), i.e., any (i+1) subset of the vertices are the vertices of an i-face. ii) For any n-polytope P P n we have n i= 1 f i(p ) 2 n+1 with equality if and only if P is an n-simplex Lemma. Let P P n. i) v vert P can not be written as a convex combination of two other points of P, i.e., v / conv ( P \{v} ). ii) If P = conv W then vert P W. iii) P = conv (vert P ). Proof. For i) let H(a, α) be a supporting plane of v. Then we have a, v = α and a, x < α for all x P \ {v}. Hence we cannot write v = λ x 1 + (1 λ) x 2 with x i P \ {v}, λ [0, 1]. Statement ii) is a direct consequence of i). For iii) let W R n be a minimal (w.r.t. its cardinality) set with P = conv W. By ii) we know already vert P W, and so for w W it remains to show that w is a vertex of P. By the minimality of W we have w / conv (W \ {w}). By Theorem 1.13 there exists a strictly separation hyperplane H(a, α) of w and conv (W \ {w}), and let a, w > α and a, w < α for all w conv (W \ {w}). With α = a, w this implies firstly P = conv W H (a, α ) and secondly (cf. Proposition (1.3)) Hence w vert P. P H(a, α ) = conv (W H(a, α )) = {w} Lemma. Let P P n be an n-polytope with 0 int P. For a proper face F of P let F = {y P : x, y = 1 for all x F }. Then i) F is a face of P. ii) F = (F ). iii) If G is a face of P and F G, then G F. iv) dim F = n 1 dim F.

31 28 The multifaceted world of polytopes Proof. First we fix some notation. Let vert P = {v 1,..., v m }. Then P = conv (vert P ) (Lemma 3.11) and P = {y R n : v i, y 1, 1 i m} (Proposition 1.24). We may assume dim F = k, F = conv {v 1,..., v l }, l {k + 1,..., m}, v 1,..., v k+1 are affinely independent. Moreover, let H(a, 1) be a supporting hyperplane of F, i.e., F = {x P : a, x = 1} and a, x 1 for all x P. Hence a P and, in particular, a F and a, v i < 1 for l + 1 i m. For i) we observe that we may write F = {y P : v i, y = 1, 1 i l}. Since v i, y 1 for all y P we conclude F = {y P : v 1 + v v l, y = l} and P {y R n : v 1 + v v l, y l}. Hence F is a face of P which shows i). By definition we have (F ) = {x P : y, x = 1 for all y F } and so v i (F ), 1 i l, which implies F (F ). For the reverse inclusion we recall that a F and so (F ) {x P : a, x = 1} = F. iii) is obvious. For iv) let U = {y R n : v i, y = 0, 1 i l} = {y R n : v i, y = 0, 1 i k + 1}. Then we have F = (a + U) P. Since the vectors v 1,..., v k+1 are, in fact, linearly independent, because otherwise 0 aff F contradicting 0 int P, we have dim U = n (k + 1) and so dim F n (k + 1). Now let z U arbitrary and for µ R we consider a + µ z. Then we have { 1, 1 i l, v i, a + µz = v i, a + µ v i, z, i > l. Since v i, a < 1 for i > l we can find and ɛ z > 0 such that a + µ z P, and thus in F for all µ < ɛ z. Since z U was arbitrary we conclude dim F n (k + 1) Theorem. Let P P n be an n-polytope with 0 int P. Then f n 1 i (P ) = f i (P ), 1 i n. Proof. For i { 1, n} there is nothing to prove. By Lemma 3.12 we can associate to every i-face F injevtively a dual n (i + 1)-face F of P. Hence f n i 1 (P ) f i (P ). Applying the same argument to the faces of P yields the assertion Theorem. Let P P n be an n-polytope with facets F 1,..., F m and let H(a i, α i ), 1 i m, be the supporting hyperplanes of F i, 1 i m. Then P = m H (a i, α i ).

32 The multifaceted world of polytopes 29 Proof. We may assume that 0 int P. In view of Theorem 3.13, P has m vertices v 1,..., v m, say, and by Lemma 3.11 iii) we have P = conv {v 1,..., v m }. By Lemma 1.25 and Proposition 1.24 we also have m P = P = {x R n : v i, x 1, 1 i n} = H (v i, 1), and according to Lemma 3.12, {v i } = {x P : v i, x = 1} = P H(v i, 1) is an (n 1)-face of P. Hence, after an appropriate renumbering we must have H(a i, α i ) = H(v i, 1), 1 i m Theorem. Let P P n be an n-polytope. i) The boundary of P is the union of all its facets. ii) A k-face is the intersection of (at least) (n k) facets. iii) An (n 2)-face is contained in exactly two facets. iv) If F, G are faces of P with F G, then F is a face of G. v) A face of P is also a face of a facet of P. Proof. We may assume that 0 int P. According to Theorem 3.14, each boundary point of P is contained in a supporting hyperplane of a facet, which shows i). For ii) let F be a k-face of P, let P = conv {v 1,..., v m } with v i vertex of P and let F = conv {v 1,..., v l }. By Lemma 3.12 iv) we have dim F = n (k+1) and so l n k. Moreover, with Lemma 3.12 ii) we also have F = (F ) = {x P : v i, x = 1, 1 i l} = l {x P : v i, x = 1}. Since v i is a vertex of P, {v i } = {x P : v i, x = 1} is a facet of P, which yields ii). In particular, for an (n 2)-face F we have dim F = 1 and so l = 2, which gives iii). For iv) let H(a, α) be a supporting hyperplane of F, i.e., F = P H(a, α). Then F = G F = G P H(a, α) = G H(a, α), which shows that F is a face of G. Finally, we observe that ii) shows that any face is contained in a facet and thus, by iv), it is a face of that facet Theorem. Let P P n be an n-polytope. i) Let G be a face of P and let F be a face of G. Then F is a face of P. ii) Let F j be a j-face of P and let F k be a k-face of P with F j F k. There exist i-faces F i of P, j < i < k, such that F j F j+1 F k 1 F k.

33 30 The multifaceted world of polytopes Proof. For i) let F and G be proper faces and let 0 F G. Let H(a F, 0) and H(a G, 0) be supporting hyperplanes of F and G, respectively, i.e., F = H(a F, 0) G, G = H(a G, 0) P and G H(a F, 0), P H(a G, 0). For µ 0 we consider the Hyperplane H(a F + µ a G, 0) and observe that { = 0, for x F, a F + µ a G, x = a F, x + µ a G, x < 0, for x G \ F. Since for all x P \ G we have a G, x < 0 there exists a µ > 0 such that for all v (vert P ) \ G we have a F + µ a G, v < 0. Hence for x P which we can write as a convex combination x = v (vert P )\G λ v v + v ((vert P ) G)\F λ v v + v (vert P ) F λ v v we find a F + µ a G, x = + 0, v (vert P )\G v ((vert P ) G)\F λ v a F + µ a G, v λ v a F + µ a G, v + v (vert P ) F λ v a F + µ a G, v with equality if and only if x F. Hence F is a face of P. In order to verify ii) we may assume j k 2 and we first note that by Theorem 3.15 iv) F j is a face of F k. According to Theorem 3.15 v) F j is contained in a facet G, say, of F k, i.e., G is a (k 1) face of F K and hence it is a face of P by i). With F k 1 = G we have shown F j F k 1 F k. In the case j < k 2 we can apply the same reasoning as above to the pair F j, F k 1, and obtain so recursively the desired chain of faces Remark. Let v 0 be a vertex of an n-polytope P and let {v 1,..., v r } be all adjacent vertices of v 0, i.e., conv {v 0, v i } is an edge of P. In other words, {v 1,..., v r } are the neighbours of v 0. Then i) P v 0 + pos {v 1 v 0,..., v r v 0 }. ii) Let c R n with c, v 0 c, v i, 1 i r. Then max{ c, x : x P } = c, v Theorem [Euler-Poincaré formula] Let P P n. Then n ( 1) i f i (P ) = 0. (3.18.1) i= 1 In particular, in the 3-dimensional case, i.e., dim P = 3, it holds f 0 f 1 +f 2 = Leonhard Euler, Henri Poincaré,