Lectures on
Convex Sets
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PST˙ws
Lectures on
Convex Sets Valeriu Soltan George Mason University, USA
World Scientific NEW JERSEY
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LONDON
9508_9789814656689_tp.indd 2
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SINGAPORE
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BEIJING
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SHANGHAI
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HONG KONG
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TA I P E I
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CHENNAI
17/2/15 10:28 am
Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE
Library of Congress Cataloging-in-Publication Data Soltan, V. P. (Valerii Petrovich) Lectures on convex sets / Valeriu Soltan, George Mason University, USA. pages cm ISBN 978-9814656689 -- ISBN 978-9814656696 1. Convex sets. 2. Convex geometry. I. Title. QA640.S645 2015 516'.08--dc23 2015002770 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.
Copyright © 2015 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.
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Preface
As any other well-established mathematical discipline, convex geometry is divided into various subfields, like generalized convexity, finite-dimensional Banach spaces, asymptotic theory of convex bodies, combinatorial convexity, mixed volumes, etc. Among all these subfields, algebraic and topological properties of convex sets in the Euclidean space Rn have a place apart: they serve the whole convex geometry and are widely used in various mathematical disciplines. The role of these properties outside convex geometry is especially visible in convex analysis, optimization, operations research, and their applications. The first systematic treatment of algebraic and topological properties of convex sets in Rn was given in the seminal monograph of Rockafellar [184], published in 1970. Since then, the list of chapters on convex sets used in convex analysis became almost canonical (see, e. g., the books of Florenzano and Le Van [87], G¨ uler [107], Hiriart-Urruty and Lemar´echal [114], Stoer and Witzgall [210], and some others). The topics usually include general properties of convex sets and convex hulls, cones and conic hulls, polyhedral sets, the extreme structure and separation properties of convex sets. Various monographs on convex geometry provide another source for the study of the above properties of convex sets, predominantly for the case of convex bodies (see, for instance, Gardner [93], Gruber [102], Schneider [191], and many others). Despite the presence of a large number of monographs on convex geometry, there are quite few books on this topic, suitable for the undergraduate study. Examples could be the books of Lay [147] and Webster [222] on convex sets, most recent of them published in 1994. The present book provides a systematic treatment of the above topics, which concern arbitrary convex sets in Rn , possibly non-closed or unv
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bounded. An essential part of the text is adapted from various research articles or is written ad hoc by the author for the purpose of completeness. Comments on further related topics and results, as well as some bibliographic notes, are given at the end of each chapter. A selection of exercises (with solutions given at the end of the book) provides additional results to the main text. This book grew up out of various courses on geometry and convexity, taught by the author at George Mason University. The audience usually consisted of students interested in all fields of mathematics and operations research. The text can be used for a one-semester undergraduate or an entry-level graduate course on convex sets, or as a supplementary book for any course on convex geometry or convex analysis. Also, the book may be viewed as a source for independent study of the subject, suitable to non-geometers. Prerequisites include proof-based undergraduate courses on linear algebra, analysis, and elementary topology. The text is divided into ten chapters. The introductory Chapter 0 gives a brief account of necessary prerequisites. It is assumed that the reader starts the book from Chapter 1 and uses the first one as a reference material. Chapter 1 deals with the affine structure of the Euclidean space Rn . It describes various properties of planes, affine spans, and affine transformations. Assuming that students have little knowledge of n-dimensional geometry, this chapter develops the topic as a natural extension of linear algebra and sets up the methods of proof for the entire book. The next three chapters are devoted to basic constructions concerning algebraic and topological properties of convex sets (see Chapter 2) and corresponding properties of convex and conic hulls (Chapters 3 and 4). Chapters 5–8 deal with general properties of convex sets: recession and normal cones, support and separation, the extreme and exposed structures. The concluding Chapter 9 demonstrates how the results from previous chapters can be applied to the study of polyhedral sets, which are important in many applications. Further material on polyhedral sets and polytopes can be found in the books of Gr¨ unbaum [104], Brøndsted [41], and Ziegler [228]. For comments, information, and corrections to this book, please contact the author at
[email protected]. V.S.
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Notation
|a| sgn (a) R [a, b] [a, b) (a, b) [a, ∞) (−∞, a) X ∪Y X ∩Y X \Y ∅ card X f :U →V f −1 g ◦f f (X) f −1 (Y ) rng f Rn X +Y X ⊕Y span X null f x·y X⊥
absolute value of a scalar a sign value of a scalar a set of real numbers (scalars) closed interval in R semi-open interval in R open interval in R closed halfline in R open halfline in R union of sets X and Y intersection of sets X and Y set difference of sets X and Y empty set cardinality of a finite set X mapping from a set U into a set V inverse of a mapping f composition of mappings f and g f -image of a set X inverse f -image of a set Y range of a mapping f n-dimensional coordinate space (Minkowski) sum of sets X and Y direct sum of sets X and Y span of a set X null space of a linear transformation f dot product of vectors x and y orthogonal complement of a set X
vii
1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 4 4 5 5 5
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kxk kx − yk Bρ (c) Uρ (c) Sρ (c) Bρ (X) int X cl X δ(X, Y ) sub L dim L hx, yi [x, yi (x, yi [x, y] [x, y) (x, y) aff X sub X dim X rint X rbd X conv X convr X ap C cones X rec X lin X pK (x) NK (z) nor K bar K FK (X) ext K extr K GK (X) exp K expr K
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norm of a vector x distance between points x and y closed ball with center c and radius ρ open ball with center c and radius ρ sphere with center c and radius ρ ρ-neighborhood of a set X interior of a set X closure of a set X inf -distance between sets X and Y characteristic subspace of a plane L dimension of a plane L line through points x and y closed halfline through y with endpoint x open halfline through y with endpoint x closed segment with endpoints x and y semi-open segment with endpoints x and y open segment with endpoints x and y affine span of a set X characteristic subspace of a set X dimension of a set X relative interior of a set X relative boundary of a set X convex hull of a set X union of convex hulls of r-pointed subsets of X apex set of a cone C conic hull with apex s of a set X recession cone of a set X linearity space of a set X metric projection of point x on convex set K normal set of K at point z ∈ K normal cone of a convex set K barrier cone of a convex set K extreme face of K generated by a set X ⊂ K set of extreme points of K set of r-extreme points of K exposed face of K generated by a set X ⊂ K set of exposed points of K set of r-exposed points of K
6 6 7 7 7 7 7 8 9 12 13 25 25 25 26 26 26 26 42 55 78 104 117 122 153 161 187 197 208 209 211 212 260 267 278 298 299 312
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Contents
Preface
v
Notation
vii
0.
Prerequisites
1.
The affine structure of Rn
11
1.1 1.2 1.3
11 37 57
2.
3.
71
2.1 2.2 2.3
71 78 93
Algebraic properties of convex sets . . . . . . . . . . . . . Relative interior of convex sets . . . . . . . . . . . . . . . Closure and relative boundary of convex sets . . . . . . .
Convex hulls
117
Algebraic properties of convex hulls . . . . . . . . . . . . . 117 Topological properties of convex hulls . . . . . . . . . . . . 128
Convex cones and conic hulls 4.1 4.2
5.
Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Affine spans . . . . . . . . . . . . . . . . . . . . . . . . . . Affine transformations . . . . . . . . . . . . . . . . . . . .
Convex sets
3.1 3.2 4.
1
145
Convex cones . . . . . . . . . . . . . . . . . . . . . . . . . 145 Conic hulls . . . . . . . . . . . . . . . . . . . . . . . . . . 161
Recession and normal directions 5.1 5.2
183
Recession cones . . . . . . . . . . . . . . . . . . . . . . . . 183 Linearity spaces . . . . . . . . . . . . . . . . . . . . . . . . 197 ix
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5.3 6.
9.
287
Exposed faces . . . . . . . . . . . . . . . . . . . . . . . . . 287 Exposed representations . . . . . . . . . . . . . . . . . . . 299
Polyhedra 9.1 9.2
255
Extreme faces . . . . . . . . . . . . . . . . . . . . . . . . . 255 Extreme representations . . . . . . . . . . . . . . . . . . . 267
The exposed structure of convex sets 8.1 8.2
227
Bounds, supports, and asymptotes . . . . . . . . . . . . . 227 Containing and supporting halfspaces . . . . . . . . . . . . 235 Separation by hyperplanes and slabs . . . . . . . . . . . . 241
The extreme structure of convex sets 7.1 7.2
8.
Normal and barrier cones . . . . . . . . . . . . . . . . . . 206
Support and separation properties 6.1 6.2 6.3
7.
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Intersections of halfspaces . . . . . . . . . . . . . . . . . . 319 Properties of polyhedra . . . . . . . . . . . . . . . . . . . 328
Solutions to Exercises
345
Bibliography
387
Author Index
399
Subject Index
403
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Chapter 0
Prerequisites
Basic Set Theory and Notation In what follows, R stands for the set of real numbers, called scalars. Scalars a and b, with a < b, determine the closed interval [a, b], two semi-open intervals [a, b) and (a, b], and open interval (a, b) of the number line R. Similarly, [a, ∞), (a, ∞), (−∞, a], and (−∞, a) are the halflines of R determined by a. The absolute value and sign value of a scalar a are denoted |a| and sgn (a), respectively. The symbol marks the end of a proof or of a statement without proof. Given a universal set U , the union, intersection, and set difference of sets X and Y from U are denoted X ∪ Y , X ∩ Y , and X \ Y , respectively. The symbol ∅ stands for the empty set. In a standard way, x ∈ X means that an element x of U belongs to X, and Y ⊂ X shows that Y is a subset of X (possibly, X = Y ). In what follows, a subset Y of X is called proper if ∅ 6= Y 6= X. A set X ⊂ U is often described by its indexed elements, like X = {xα }, or by a given condition P (x) on U , as X = {x ∈ U : P (x)}. We distinguish finite, denumerable, countable (finite or denumerable), and uncountable sets. The number of elements in a finite set X is denoted card X. If F = {Xα : α ∈ A} is an indexed family of subsets of a universal set U , then De Morgan’s Laws state that U \ ∪ Xα = ∩ (U \ Xα ) and U \ ∩ Xα = ∪ (U \ Xα ). α
α
α
α
The family F is called disjoint if ∩ Xα = ∅, and is called pairwise disjoint α provided Xβ ∩ Xγ = ∅ for any distinct indices β, γ ∈ A. We say that F is nested if Xβ ⊂ Xγ or Xγ ⊂ Xβ whenever β, γ ∈ A. To simplify the arguments, we considered only nonempty families of sets. Assuming the axiom of choice, Zorn’s lemma implies that a family F of 1
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subsets of a given set U contains a maximal element provided that every totally ordered subfamily of F has an upper bound in F. A mapping f from a set U into a set V (notation f : U → V ) assigns to every element x ∈ U a unique element f (x) ∈ V . The mapping f : U → V is onto if every element v ∈ V is an f -image of an element u ∈ U ; it is oneto-one if f (x) 6= f (y) whenever x 6= y; finally, f is invertible (equivalently, is a bijection) if it is both onto and one-to-one. A bijection f : U → V has the inverse mapping f −1 : V → U , defined by the identities f −1 (f (x)) = x and f (f −1 (y)) = y for all x ∈ U and y ∈ V. For a mapping f : U → V (not necessarily invertible), the image of a set X ⊂ U and the inverse image of a set Y ⊂ V are defined by and f −1 (Y ) = {x : f (x) ∈ Y }.
f (X) = {f (x) : x ∈ X}
Obviously, f (∅) = ∅ and f −1 (∅) = ∅. Furthermore, X ⊂ f −1 (f (X)),
f −1 (Y ) = f −1 (Y ∩ rng f ),
f (f −1 (Y )) = Y ∩ rng f.
The set f (U ), denoted rng f , is called the range of f . For a family {Xα } of subsets of U , one has f (∪ Xα ) = ∪ f (Xα ) α
α
and f (∩ Xα ) ⊂ ∩ f (Xα ), α
α
with equality if f is one-to-one. Similarly, f −1 (∪ Yα ) = ∪ f −1 (Yα ) α
α
and f −1 (∩ Yα ) = ∩ f −1 (Yα ) α
α
for a family {Yα } of subsets of V . If X1 , X2 ⊂ U and Y1 , Y2 ⊂ V , then f (X1 ) \ f (X2 ) ⊂ f (X1 \ X2 )
and f −1 (Y1 ) \ f −1 (Y2 ) = f −1 (Y1 \ Y2 ),
with f (X1 ) \ f (X2 ) = f (X1 \ X2 ) provided f is one-to-one. The composition g ◦ f of mappings f : U → V and g : V → W is defined by (g ◦ f )(x) = g(f (x)). The Vector Space Rn Throughout this book we deal with elements and subsets of the n-dimensional vector space Rn , n > 1, consisting of all n-tuples x = (x1 , . . . , xn ), where x1 , . . . , xn are scalars, named the coordinates of x. The elements of Rn are called vectors, or points. We make no difference between these two notions, and their use should support the reader’s intuition. To distinguish similarly looking elements, 0 will stand for the number zero, and o for the origin (zero vector) of Rn .
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Given vectors x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) and a scalar λ, the sum x + y and the product λx are defined, respectively, as x + y = (x1 + y1 , . . . , xn + yn )
and λx = (λx1 , . . . , λxn ).
These operations can be expanded to the case of nonempty sets X and Y in Rn : X + Y = {x + y : x ∈ X, y ∈ Y }
and λX = {λx : x ∈ X}.
The set X + Y is often called the Minkowski sum of sets X and Y , and λX is the scalar multiple of X. Put X + Y = ∅ if at least one of the sets X and Y is empty; also, let λ ∅ = ∅ for all scalars λ. For sets X, Y, Z ⊂ Rn and scalars λ, µ, one has X + Y = Y + X,
(X + Y ) + Z = X + (Y + Z),
(X ∪ Y ) + Z = (X + Z) ∪ (Y + Z), (X ∩ Y ) + Z ⊂ (X + Z) ∩ (Y + Z), λ(X + Y ) = λX + λY,
(λ + µ)X ⊂ λX + µX.
If X consists of a single vector x, we use x + Y instead of {x} + Y and say that x + Y is a translate of Y . We write −x and −X for (−1)x and (−1)X, respectively. X +Y
@ @ *H Y X x + yHH @ H @H HH Y Y H * x HH y H q H o Fig. 0.1
QQ Q λX Q Q Q * Q Q X Q q o
The sum X + Y and a scalar multiple λX.
A nonempty set S ⊂ Rn is called a subspace if it is closed under vector addition and scalar multiplication: x+y ∈ S and λx ∈ S whenever x, y ∈ S and λ ∈ R. The zero subspace {o} and the whole space Rn are trivial subspaces. Subspaces S and T of Rn are called independent provided S ∩ T = {o} (possibly, S = {o} or T = {o}); equivalently, every vector x in S + T is uniquely expressible as x = y + z, where y ∈ S and z ∈ T . The
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sum of independent subspaces S and T is called direct and is denoted S ⊕T . A similar notation, X ⊕ Y , is used for the sum of nonempty subsets X and Y of independent subspaces S and T , respectively. If a subspace P of Rn is the sum of independent subspaces S and T , then we say that S and T are complementary in P . In particular, S and T are called complementary if S ⊕ T = Rn . The intersection of all subspaces containing a given set X ⊂ Rn is called the span of X (clearly, the span of ∅ is {o}). The span of a nonempty set X ⊂ Rn can be described in terms of linear combinations: span X = {λ1 x1 + · · · + λk xk : k > 1, λ1 , . . . , λk ∈ R, x1 , . . . , xk ∈ X}. For nonempty sets X and Y in Rn , one has span (X + Y ) ⊂ span (X ∪ Y ) = span X + span Y. Furthermore, span (X + Y ) = span X + span Y if o ∈ X ∩ Y . A set {x1 , . . . , xr } of vectors in Rn is called linearly dependent if there are scalars λ1 , . . . , λr , not all zero, such that λ1 x1 +· · ·+λr xr = o; otherwise {x1 , . . . , xr } is linearly independent. Each linearly independent subset of Rn contains at most n vectors. A basis for a subspace S ⊂ Rn is a linearly independent set whose span is S (the set ∅ is linearly independent and forms a basis for {o}). The vectors e1 = (1, 0, . . . , 0), e2 = (0, 1, . . . , 0), . . . , en = (0, 0, . . . , 1) form the standard basis for Rn . The (unique) number of vectors in a basis for a subspace S is called the dimension of S and is denoted dim S. A subspace S ⊂ Rn of dimension n − 1 is called a hypersubspace. For subspaces S and T of Rn , one has dim (S + T ) = dim S + dim T − dim (S ∩ T ). Furthermore, if a basis b1 , . . . , br for S ∩ T is completed into bases b 1 , . . . , b r , c1 , . . . , cp
and b1 , . . . , br , e1 , . . . , eq
for S and T , respectively, then the combined list b1 , . . . , br , c1 , . . . , cp , e1 , . . . , eq is a basis for S + T . A mapping f : Rn → Rm is called a linear transformation provided f (x + y) = f (x) + f (y)
and f (λx) = λf (x)
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5
whenever x, y ∈ Rn and λ ∈ R. A linear transformation f : Rn → R is called a linear functional. Any linear transformation f preserves linear combinations of finitely many vectors, and f (X + Y ) = f (X) + f (Y )
and f (λX) = λf (X)
for any sets X, Y ⊂ Rn and a scalar λ. The subspaces null f = {x ∈ Rn : f (x) = o}
and
rng f = {f (x) : x ∈ Rn }
are called, respectively, the null space and the range of f . One has dim (null f ) + dim (rng f ) = n. Furthermore, the following statements hold. 1. If {c1 , . . . , cr } is a linearly independent set in Rn and f is one-toone, then the set {f (c1 ), . . . , f (cr )} is linearly independent. 2. If {a1 , . . . , ar } is a linearly independent set in Rm and vectors c1 , . . . , cr ∈ Rn satisfy the conditions f (ci ) = ai , 1 6 i 6 r, then the set {c1 , . . . , cr } is linearly independent. For a linearly independent set {c1 , . . . , cr } in Rn and vectors a1 , . . . , ar in Rm , there is a linear transformation f : Rn → Rm such that f (ci ) = ai for all 1 6 i 6 r. If, additionally, r = n, then f is uniquely determined. Furthermore, if n = m = r and {a1 , . . . , an } is a basis for Rn , then f is invertible. If S and T are complementary subspaces of Rn and x = y + z is a (unique) expression of a vector x ∈ Rn as the sum of vectors y ∈ S and z ∈ T , then the linear transformation f : Rn → Rn defined by f (x) = y is called the linear projection on S along T . The Euclidean Structure of Rn The dot product x·y of vectors x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) is defined by x·y = x1 y1 + · · · + xn yn . Vectors x and y are called orthogonal provided x·y = 0 (possibly, x = o or y = o). Nonempty sets X and Y in Rn are orthogonal if x·y = 0 whenever x ∈ X and y ∈ Y . The orthogonal complement of a nonempty set X ⊂ Rn is defined by X ⊥ = {y ∈ Rn : x·y = 0 for all x ∈ X}. The set X ⊥ is a subspace of Rn ; moreover, (X ⊥ )⊥ = span X
and
span X ⊕ X ⊥ = Rn .
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If S and T are subspaces such that S ⊂ T , then the subspace S ⊥ ∩ T is the orthogonal complement of S within T . If S ⊂ Rn is a subspace, then S ⊕ S ⊥ = Rn , which gives dim S + dim S ⊥ = n. For subspaces S1 and S2 , one has (S1 + S2 )⊥ = S1⊥ ∩ S2⊥
and
(S1 ∩ S2 )⊥ = S1⊥ + S2⊥ .
Any real-valued linear functional ϕ(x) on Rn is expressible in the form ϕ(x) = x·c, where c ∈ Rn is a suitable vector. The norm, or length, of a vector x = (x1 , . . . , xn ) is defined by q √ kxk = x·x = x21 + · · · + x2n . Basic properties of the norm: 1. kxk > 0, with kxk = 0 if and only if x = o. 2. kλxk = |λ| kxk. 3. |x·y| 6 kxk kyk, with |x·y| = kxk kyk if and only if x and y are linearly dependent (Cauchy-Schwarz inequality). 4. kx + yk 6 kxk + kyk, with kx + yk = kxk + kyk if and only if one of x and y is a positive scalar multiple of the other. A unit vector is a vector of unit norm. A set X ⊂ Rn of pairwise orthogonal unit vectors is called orthonormal. For a pair of points x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ), the norm p kx − yk = (x1 − y1 )2 + · · · + (xn − yn )2 is called the distance between x and y. Clearly, 5. kx − yk > 0, with kx − yk = 0 if and only if x = y. 6. kλx − λyk = |λ| kx − yk. 7. kx − yk 6 kx − zk + kz − yk whenever x, y, z ∈ Rn . The linear projection p of Rn on a subspace S along its orthogonal complement S ⊥ is called the orthogonal projection on S. For any vectors x ∈ Rn and y ∈ S, one has kx − yk2 = kx − p(x)k2 + kp(x) − yk2 . Consequently, p(x) is the unique nearest to x point in S and x − p(x) is orthogonal to S. If, additionally, dim S = n − 1, then S = {x ∈ Rn : x·c = 0}, where c is a certain nonzero vector, and the orthogonal projection p on S can be written as x·c c for all x ∈ Rn . p(x) = x − kck2
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7
Elementary Topology in Rn Given a point c ∈ Rn and a scalar ρ > 0, the sets Bρ (c) = {x ∈ Rn : kx − ck 6 ρ}, Uρ (c) = {x ∈ Rn : kx − ck < ρ}, Sρ (c) = {x ∈ Rn : kx − ck = ρ} are called, respectively, the closed ball, open ball, and sphere with center c and radius ρ. The sets B = B1 (o) and U = U1 (o) are called, respectively, the closed and open unit balls of Rn . Remark. To simplify certain geometric arguments, we consider neighborhoods of sets and points in terms of closed balls. Because of the inclusions Uδ (c) ⊂ Bγ (c) ⊂ Uρ (c) whenever 0 < δ < γ < ρ, our approach is topologically equivalent to the standard one. The ρ-neighborhood of a set X ⊂ Rn , denoted Bρ (X), is defined as Bρ (X) = ρB + X = ∪ (Bρ (x) : x ∈ X). $
'
X &
Bρ (X) %
A point x ∈ Rn is called interior for a set X ⊂ Rn provided there is a scalar ρ > 0 such that Bρ (x) ⊂ X. A set Y ⊂ Rn is said to be open if every point y ∈ Y is interior for Y . The union of interior points of a given set X, denoted int X, is called the interior of X (put int ∅ = ∅). It is easy to see that int X is the largest open set contained in X; furthermore, int X ⊂ X,
int (int X) = int X,
int X ⊂ int Y if X ⊂ Y.
n
For sets X1 , . . . , Xr in R , one has int (X1 ∩ · · · ∩ Xr ) = int X1 ∩ · · · ∩ int Xr . The union of a family of open sets and the intersection of finitely many open sets are open sets. Moreover, int X + Y ⊂ int (X + Y ) for sets X and Y in Rn . If a set X ⊂ Rn is covered with a family F of open sets, then F contains a countable subfamily of sets whose union also covers X (Lindel¨ of’s theorem).
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We say that an infinite sequence of points x1 , x2 , . . . converges to a point x (and write lim xi = x, or xi → x) if lim kxi − xk = 0. With i→∞
i→∞
(i)
x = (x1 , . . . , xn ) and xi = (x1 , . . . , xn(i) ), i > 1, one has lim xi = x if and only if
i→∞
(i)
lim xj = xj for all 1 6 j 6 n.
i→∞
Clearly, kxi k → kxk provided xi → x. A point x ∈ Rn is called a closure point of a nonempty set X ⊂ Rn if every ball Bρ (x), ρ > 0, meets X. All closure points of X form the closure of X, denoted cl X, which can be expressed as cl X = ∩ (ρB + X : ρ > 0), where B is the closed unit ball (let cl ∅ = ∅). Given sets X and Y in Rn , X ⊂ cl X,
cl (cl X) = cl X,
cl X ⊂ cl Y if X ⊂ Y.
For sets X1 , . . . , Xr in Rn , one has cl (X1 ∪ · · · ∪ Xr ) = cl X1 ∪ · · · ∪ cl Xr . A set X ⊂ Rn is called closed if cl X = X. Equivalently, X is closed if and only if its complement Rn \ X is open. The intersection of a family of closed sets and the union of finitely many closed sets are closed sets. A subset Y of a set X is called dense in X provided X ⊂ cl Y . Every set X in Rn has a countable dense subset. The boundary of a set X ⊂ Rn , denoted bd X, is the set of all points x ∈ Rn such that every ball Bρ (x), ρ > 0, meets both X and Rn \ X. Clearly, X is closed if and only if bd X ⊂ X. Furthermore, bd X 6= ∅ if and only if X is a proper subset of Rn (that is, ∅ 6= X 6= Rn ). A set X ⊂ Rn is called bounded if it lies in a ball of positive radius. A nonempty set X ⊂ Rn is bounded if and only if every infinite sequence of points from X contains a convergent subsequence. The diameter of a bounded set X is defined by diam X = sup{ku − vk : u, v ∈ X}. A set X ⊂ Rn is called compact if every family of open sets in Rn whose union covers X contains a finite subfamily that also covers X. Criteria for the compactness of a nonempty set X in Rn : 1. X is bounded and closed. 2. Every infinite sequence of points from X contains a subsequence converging to a point in X (the Heine-Borel theorem).
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If X ⊂ Rn is a nonempty bounded set and ρ > 0, then there is a finite set {x1 , . . . , xr } ⊂ X satisfying the condition X ⊂ Bρ (x1 ) ∪ · · · ∪ Bρ (xr ). The inf -distance between nonempty sets X and Y in Rn is defined by δ(X, Y ) = inf {kx − yk : x ∈ X, y ∈ Y }. We say that X and Y are strongly disjoint provided δ(X, Y ) > 0. A mapping f : Rn → Rm is called continuous at a point x0 ∈ Rn if for any ε > 0 there is a δ = δ(ε) > 0 such that kf (x) − f (x0 )k 6 ε whenever kx − x0 k 6 δ. The mapping f is said to be continuous on Rn if it is continuous at every point of Rn ; the latter occurs if and only if any of the following equivalent conditions is satisfied: 1. f (cl X) ⊂ cl f (X) for every set X ⊂ Rn . 2. cl f −1 (Y ) ⊂ f −1 (cl Y ) for every set Y ⊂ Rm . 3. The inverse image f −1 (Y ) of every closed set Y ⊂ Rm is a closed set. 4. f −1 (int Y ) ⊂ int f −1 (Y ) for every set Y ⊂ Rm . 5. The inverse image f −1 (Y ) of every open set Y ⊂ Rm is an open set. If a continuous mapping f : Rn → Rm is one-to-one, or if a set X ⊂ Rn is bounded, then f (cl X) = cl f (X). In particular, f (X) is compact if X is compact. A continuous real-valued function f on Rn attains its maximum and minimum values on every nonempty compact subset of Rn . A countable union of closed sets is called an Fσ -set. A set X ⊂ Rn is an Fσ -set if and only if it is a countable union of a nested family of compact sets. Countable unions and finite intersections of Fσ -sets are Fσ -sets. A countable intersection of open sets is called a Gδ -set. A nonempty set X ⊂ Rn is called connected if it cannot be partitioned into two nonempty subsets such that each of them does not meet the closure of the other. Chapter 0 Exercise 0.1. Let f : Rn → Rm be a linear transformation and S be a subspace of Rn complementary to null f . Show that the mapping h : S → rng f defined by h(x) = f (x) is an invertible linear transformation and that the inverse image f −1 (x) of every vector x ∈ rng f equals h−1 (x) + null f .
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Exercise 0.2. Let f : Rn → Rm be a linear transformation and L and M be subspace in Rn and Rm , respectively. Show that dim f (L) = dim L − dim (L ∩ null f ), dim f −1 (M ) = dim (M ∩ rng f ) + dim (null f ). Exercise 0.3. Show that a linear transformation f : Rn → Rn is a linear projection if and only if f 2 = f . Exercise 0.4. Given a vector c ∈ Rn , show that the linear functional ϕ(x) = x·c is continuous on Rn . Exercise 0.5. Show that kx − yk − kz − yk 6 kx − zk for any points x, y, z in Rn . Conclude from here that for a given point c ∈ Rn , the distance function δc (x) = kx − ck is continuous on Rn . Exercise 0.6. Show that every linear transformation f : Rn → Rm is continuous on Rn . Exercise 0.7. Let {b1 , . . . , br } ⊂ Rn be a linearly independent set. Show (i) (i) that an infinite sequence xi = λ1 b1 + · · · + λr br , i > 1, converges to (i) the vector x = λ1 b1 + · · · + λr br if and only if limi→∞ λj = λj for all 1 6 j 6 r. Exercise 0.8. Let X and Y be nonempty set in Rn such that at least one of them is bounded. Show the existence of points x ∈ cl X and y ∈ cl Y with the property kx − yk = δ(X, Y ). Exercise 0.9. Given nonempty sets X and Y in Rn , show the equivalence of the following conditions: (1) δ(X, Y ) > 0, (2) o ∈ / cl (X − Y ), (3) there is a scalar ρ > 0 such that Bρ (X) ∩ Bρ (Y ) = ∅. Exercise 0.10. Show that cl X + cl Y ⊂ cl (X + Y ) for sets X and Y in Rn . Furthermore, cl X + cl Y = cl (X + Y ) provided any of the following conditions is satisfied: (1) at least one of X and Y is bounded, (2) X and Y lie in independent subspaces. Exercise 0.11. Let X and Y be Fσ -sets in Rn . Show that the sum X + Y is an Fσ -set. Also, for a linear transformation f : Rn → Rm , the set f (X) is an Fσ -set.
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Chapter 1
The Affine Structure of Rn
1.1
Planes
Definition and Basic Properties Definition 1.1. A nonempty set L ⊂ Rn is called a plane provided it is a translate of a subspace: L = a + S (= {a + x : x ∈ S}), where a is a point in Rn and S is a subspace of Rn . The empty set is defined to be a plane. A plane L ⊂ Rn is called proper if ∅ 6= L 6= Rn .
Fig. 1.1
a 6
L=a+S
ro
S
A subspace S and a plane L.
For example, planes in R3 are the empty set, singletons, lines, usual planes, and R3 itself. Synonyms for “plane”, used by various authors, are “affine set”, “affine subspace”, “affine variety”, and “flat.” Ways to describe a nonempty plane L = a + S derive from those for the subspace S. Thus, if S is the span of vectors b1 , . . . , br , then L can be expressed parametrically: L = a + span {b1 , . . . , br } = {a + λ1 b1 + · · · + λr br : λ1 , . . . , λr ∈ R}. 11
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Similarly, if S is the solution set of a system of homogenous linear equations ai1 x1 + · · · + ain xn = 0,
1 6 i 6 m,
then L is the solution set of a suitable system ai1 x1 + · · · + ain xn = bi ,
1 6 i 6 m.
One more way to describe planes in terms of affine spans of points is studied in Section 1.2. Theorem 1.2. Any nonempty plane L ⊂ Rn is a translate of a unique subspace S of Rn , given by S = L − L (= {x − y : x, y ∈ L}).
(1.1)
n
For a vector c ∈ R , the equality L = c + S holds if and only if c ∈ L. Furthermore, L = X + S and S = L − X for a nonempty subset X of L. Consequently, L is a subspace if and only if o ∈ L. Proof. By the definition, L can be expressed as L = a + S, where a ∈ Rn and S is a subspace of Rn . To prove (1.1), choose points x and y in L. Then x = a + x0 and y = a + y0 for certain points x0 , y0 ∈ S. This argument gives x − y = (a + x0 ) − (a + y0 ) = x0 − y0 ∈ S. Hence L − L = {x − y : x, y ∈ L} ⊂ S. On the other hand, every point x ∈ S can be written as x = (a + x) − (a + o) ∈ L − L, which implies the opposite inclusion S ⊂ L − L. The equality S = L − L shows the desired uniqueness of S. If L = c + S for a point c ∈ Rn , then c = c + o ∈ c + S = L. Conversely, if c ∈ L, then c = a + c0 , where c0 ∈ S. Since c0 + S = S, we have c + S = (a + c0 ) + S = a + (c0 + S) = a + S = L. Now, assume that X is a nonempty subset of L. By the above proved, X + S = ∪ (x + S : x ∈ X) = ∪ (L : x ∈ X) = L. Similarly, L − X = ∪ (L − x : x ∈ X) = ∪ (S : x ∈ X) = S. In view of Theorem 1.2, we introduce the following definition. Definition 1.3. For a nonempty plane L ⊂ Rn , the subspace S = L − L is called the characteristic subspace of L and is denoted sub L.
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Theorem 1.4. The following statements hold. (1) If F = {Lα } is a family of planes in Rn , then the intersection ∩ Lα α is a plane. (2) If L1 , . . . , Lr are planes in Rn and µ1 , . . . , µr are scalars, then the sum µ1 L1 + · · · + µr Lr is a plane. Proof. (1) Suppose that ∩ Lα 6= ∅ and choose a point a ∈ ∩ Lα . Accordα α ing to Theorem 1.2, every plane Lα ∈ F can be written as Lα = a + Sα , where Sα is a subspace. Since ∩ Sα is a subspace, the equality α
∩ Lα = ∩ (a + Sα ) = a + ∩ Sα α
α
α
shows that ∩ Lα is a plane as a translate of ∩ Sα . α
α
(2) If at least one of the planes L1 , . . . , Lr is empty, then the whole sum µ1 L1 + · · · + µr Lr is empty. Suppose that all L1 , . . . , Lr are nonempty. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, 1 6 i 6 r. Then µ1 L1 + · · · + µr Lr = (µ1 a1 + · · · + µr ar ) + (µ1 S1 + · · · + µr Sr ). Hence the sum µ1 L1 + · · · + µr Lr is a plane as a translate of the subspace µ1 S1 + · · · + µr Sr . Additional results on algebraic operations with planes are given in Corollary 1.89 and Exercise 1.2. Dimension of Planes Theorem 1.2 gives a base to the definition below. Definition 1.5. If L ⊂ Rn is a nonempty plane and S is its characteristic subspace, then the dimension of L, denoted dim L, is defined by dim L = dim S. If L = ∅, then we let dim L = −1. Planes of dimension 0 and 1 are called points and lines, respectively. Theorem 1.6. The following statements take place. (1) If L ⊂ Rn is a plane, then for a point c ∈ Rn and a scalar µ 6= 0, the plane c + µL is a translate of L and has the same dimension as L. (2) If planes L1 and L2 in Rn satisfy the inclusion L1 ⊂ L2 , then dim L1 6 dim L2 , with dim L1 = dim L2 if and only if L1 = L2 .
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Proof. Since both statements are obvious for the case of empty planes, we assume that all planes involved are nonempty. (1) Let L = a + S, where a ∈ Rn and S is a subspace. From the equalities c + µL = c + µ(a + S) = c + µa + S = c + (µ − 1)a + L if follows that c + µL is a translate of L and of S. Therefore, dim (c + µL) = dim S = dim L. (2) Choose a point b ∈ L1 . By Theorem 1.2, L1 = b+S1 and L2 = b+S2 for certain subspaces S1 and S2 . Theorem 1.2 shows that S1 = L1 − b ⊂ L2 − b = S2 . From linear algebra we know that dim S1 6 dim S2 , with dim S1 = dim S2 if and only if S1 = S2 . Hence dim L1 = dim S1 6 dim S2 = dim L2 , and dim L1 = dim L2 if and only if L1 = b + S1 = b + S2 = L2 . Theorem 1.7. For nonempty planes L1 and L2 in Rn , one has dim (L1 + L2 ) = dim L1 + dim L2 − dim (sub L1 ∩ sub L2 ). If, additionally, L1 ∩ L2 6= ∅, then dim (L1 + L2 ) = dim L1 + dim L2 − dim (L1 ∩ L2 ). Proof. Let Si = sub Li , i = 1, 2. By Theorem 1.2, we can write Li = ci +Si for suitable points ci ∈ Li , i = 1, 2. Then L1 + L2 = c1 + c2 + (S1 + S2 ). This argument and the equality dim (S1 + S2 ) = dim S1 + dim S2 − dim (S1 ∩ S2 ) give dim (L1 + L2 ) = dim L1 + dim L2 − dim (S1 ∩ S2 ). Suppose that L1 ∩ L2 6= ∅ and choose a point c ∈ L1 ∩ L2 . By Theorem 1.2, Li = c + Si , i = 1, 2. Consequently, L1 + L2 = 2c + (S1 + S2 )
and L1 ∩ L2 = c + S1 ∩ S2 .
Therefore, dim (L1 + L2 ) = dim (S1 + S2 ) = dim L1 + dim L2 − dim (L1 ∩ L2 ). Theorem 1.8. If planes L1 and L2 lie within a nonempty plane L ⊂ Rn such that L1 + L2 is a translate of L, then L1 ∩ L2 6= ∅.
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Proof. Translating all three planes L, L1 , and L2 on the same suitable vector, we may assume that L is a subspace. Then L1 + L2 ⊂ L, which gives L1 + L2 = L due to the assumption on L1 + L2 . Clearly, both planes L1 and L2 are nonempty. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, i = 1, 2. Since ai ∈ Li ⊂ L, i = 1, 2 (see Theorem 1.2), one has a1 + a2 ∈ L. Therefore, S1 + S2 = (L1 + L2 ) − (a1 + a2 ) = L − (a1 + a2 ) = L. With r = dim (S1 ∩ S2 ), p = dim S1 − r, and q = dim S2 − r, a dimension argument yields p + q + r = dim S1 + dim S2 − dim (S1 ∩ S2 ) = dim (S1 + S2 ) = dim L. Choose a basis b1 , . . . , br for S1 ∩ S2 and complete it into bases b 1 , . . . , b r , c1 , . . . , cp
and b1 , . . . , br , e1 , . . . , eq
for S1 and S2 . The combined list b1 , . . . , br , c1 , . . . , cp , e1 , . . . , eq is a basis for L (see page 4). Therefore, we can express a1 and a2 as linear combinations a1 = ξ1 b1 + · · · + ξr br + ξr+1 c1 + · · · + ξp+r cp + ξp+r+1 e1 + · · · + ξn eq , a2 = η1 b1 + · · · + ηr br + ηr+1 c1 + · · · + ηp+r cp + ηp+r+1 e1 + · · · + ηn eq . Let u0 = ηr+1 c1 + · · · + ηp+r cp + ξp+r+1 e1 + · · · + ξn eq , u1 = −(ξ1 b1 + · · · + ξr br ) + (ηr+1 − ξr+1 )c1 + · · · + (ηp+r − ξp+r )cp , u2 = −(η1 b1 + · · · + ηr br ) + (ξp+r+1 − ηp+r+1 )e1 + · · · + (ξn − ηn )eq . Obviously, u1 ∈ S1 , u2 ∈ S2 , and u0 = a1 + u1 = a2 + u2 . Furthermore, u1 + S1 = S1 and u2 + S2 = S2 . Hence Li = ai + Si = ai + (ui + Si ) = (ai + ui ) + Si = u0 + Si , i = 1, 2. Then u0 ∈ L1 ∩ L2 according to Theorem 1.2. Independent and Complementary Planes We recall that subspaces S and T are independent provided S ∩ T = {o}. If, additionally, S + T = Rn , then S and T are complementary. Definition 1.9. Nonempty planes L1 and L2 in Rn are called independent (respectively, complementary) if their characteristic subspaces are independent (respectively, complementary).
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S2 rc S1
ro
L1 = a1 + S1
L2 = a2 + S2
Fig. 1.2
Complementary planes L1 and L2 .
Theorem 1.10. Nonempty planes L1 and L2 in Rn are independent if and only if dim (L1 + L2 ) = dim L1 + dim L2 . Furthermore, L1 and L2 are complementary if and only if dim (L1 + L2 ) = dim L1 + dim L2 = n. Proof. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, i = 1, 2. Then L1 + L2 = (a1 + a2 ) + (S1 + S2 ), which yields dim (L1 + L2 ) = dim (S1 + S2 ). From linear algebra we know that S1 and S2 are independent if and only if dim (S1 + S2 ) = dim S1 + dim S2 . Furthermore, S1 and S2 are complementary if and only if dim S1 + dim S2 = n. Now, both statements of the theorem follows from the definitions and the equalities dim Li = dim Si , i = 1, 2. Theorem 1.11. For nonempty planes L1 and L2 in Rn , the following conditions are equivalent. (1) L1 and L2 are complementary. (2) L1 + L2 = Rn and L1 ∩ L2 is a singleton. (3) There are complementary subspaces S1 and S2 and a point c ∈ Rn such that L1 = c + S1 and L2 = c + S2 . (4) There is a point c ∈ L1 ∩L2 such that every point x ∈ Rn is uniquely expressible as x = x1 + x2 − c, where x1 ∈ L1 and x2 ∈ L2 . Proof. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, i = 1, 2. (1) ⇒ (2). If L1 and L2 are complementary, then the subspaces S1 and S2 are complementary. Therefore, L1 + L2 = (a1 + a2 ) + (S1 + S2 ) = (a1 + a2 ) + Rn = Rn .
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By Theorem 1.8, L1 ∩ L2 6= ∅ and dim (L1 ∩ L2 ) = dim L1 + dim L2 − n = dim S1 + dim S2 − n = 0. The latter equality shows that L1 ∩ L2 is a singleton. (2) ⇒ (3). Denote by c the unique point of L1 ∩ L2 . Then L1 = c + S1 and L2 = c + S2 (see Theorem 1.2). Furthermore, S1 + S2 = (L1 + L2 ) − 2c = Rn − 2c = Rn , {c} = L1 ∩ L2 = (c + S1 ) ∩ (c + S2 ) = c + S1 ∩ S2 . Hence S1 ∩ S2 = {o}, and S1 and S2 are complementary. (3) ⇒ (4). Clearly, the point c satisfying condition (3) belongs to L1 ∩ L2 . Since the subspaces S1 and S2 are complementary, for every point x ∈ Rn , the point x − c is uniquely expressible as the sum z1 + z2 , where z1 ∈ S1 and z2 ∈ S2 . Consequently, the representation x = x1 + x2 − c, where x1 = z1 + c and x2 = z2 + c, is unique. (4) ⇒ (1). Let S1 = L1 − c and S2 = L2 − c. Condition shows that every vector u ∈ Rn is uniquely expressible as u = u1 + u2 , where u1 ∈ S1 and u2 ∈ S2 . Therefore, the subspaces S1 and S2 are complementary, and Definition 1.9 shows that the planes L1 and L2 are complementary. Parallel Planes Definition 1.12. Nonempty planes L1 and L2 in Rn are called parallel if one of them contains a translate of the other.
L1 L2 Fig. 1.3
L1 L2
Parallel planes L1 and L2 .
Remark. Parallelism of planes defined above is not an equivalence relation on the family of all planes in Rn . For example, the coordinate x- and y-axes of R3 are not parallel, while both are parallel to the coordinate xy-plane. Obviously, any singleton is parallel to each nonempty plane.
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Theorem 1.13. Suppose that nonempty planes L1 and L2 in Rn satisfy the inequality dim L1 6 dim L2 . If S1 and S2 are the characteristic subspaces of L1 and L2 , respectively, then L1 and L2 are parallel if and only if S1 ⊂ S2 . Consequently, if dim L1 = dim L2 , then L1 and L2 are parallel if and only if they are translates of each other. Proof. Let Li = ai + Si for a certain point ai ∈ Rn , i = 1, 2. Suppose first that L1 and L2 are parallel. From Theorem 1.6 and the assumption dim L1 6 dim L2 it follows that namely L2 contains a translate of L1 . Choose a vector c ∈ Rn satisfying the condition c + L1 ⊂ L2 . Put b = c + a1 . Then b = c + a1 ∈ c + a1 + S1 = c + L1 ⊂ L2 . Thus L2 = b + S2 according to Theorem 1.2. Therefore, S1 = L1 − a1 = (c + L1 ) − (c + a1 ) ⊂ L2 − b = S2 . Conversely, if S1 ⊂ S2 , then (a2 − a1 ) + L1 = (a2 − a1 ) + (a1 + S1 ) = a2 + S1 ⊂ a2 + S2 = L2 , implying that L1 and L2 are parallel. XXX L1 XX X
Fig. 1.4
L01 = L1 + sub L2
L2
L02 = L2 + sub L1
Parallel planes L01 and L02 through planes L1 and L2 .
Theorem 1.14. If L1 and L2 are nonempty planes in Rn , then L01 = L1 + sub L2
and
L02 = sub L1 + L2
are parallel planes containing L1 and L2 , respectively, and having the same dimension dim (L1 + L2 ). Furthermore, L01 and L02 are disjoint if and only if L1 and L2 are disjoint. Proof. Choose points ai ∈ Li , i = 1, 2. According to Theorem 1.2, one has Li = ai + Si , where Si = sub Li , i = 1, 2. Hence both sets L01 = L1 + S2 = a1 + (S1 + S2 ), L02 = S1 + L2 = a2 + (S1 + S2 )
(1.2)
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are parallel planes as translates of the same subspace S1 +S2 . Consequently, they have the same dimension dim (S1 + S2 ) = dim (L1 + L2 ). Clearly, L1 ⊂ L1 + S2 = L01
and L2 ⊂ S1 + L2 = L02 .
If L01 and L02 are disjoint, then so are L1 and L2 . Suppose that L01 ∩L02 6= ∅ and choose a point c ∈ L01 ∩ L02 . From Theorem 1.2 and the above argument it follows that L01 = c + S1 + S2 = L02 . Comparing the latter equalities with (1.2), we conclude from the same theorem that a1 − c ∈ L01 − L01 = S1 + S2 , a2 − c ∈ L02 − L02 = S1 + S2 . Therefore, a1 + a2 − 2c ∈ S1 + S2 . Furthermore, Li − c ⊂ Li + Sj − c = L0i − c = S1 + S2 ,
i 6= j,
and (L1 − c) + (L2 − c) = (a1 + S1 − c) + (a2 + S2 − c) = (a1 + a2 − 2c) + (S1 + S2 ) = S1 + S2 . Finally, Theorem 1.8 shows that (L1 − c) ∩ (L2 − c) 6= ∅. Consequently, L1 ∩ L2 = (L1 − c) ∩ (L2 − c) + c 6= ∅. In elementary geometry, two lines in the plane (a line and a plane, or a pair of planes in the 3-space) are defined to be parallel provided they do not meet. Theorem 1.15 below relates Definition 1.12 with this approach. Theorem 1.15. Let L1 and L2 be nonempty planes in Rn such that dim L1 6 dim L2 . Then L1 and L2 are parallel if and only if the following two conditions are satisfied. (1) The union L1 ∪ L2 lies within a plane of dimension dim L2 + 1. (2) Either L1 ∩ L2 = ∅ or L1 ⊂ L2 . Proof. Let Li = ai + Si , where ai ∈ Rn and Si is a subspace, i = 1, 2. Suppose first that L1 and L2 are parallel. From Theorem 1.6 and the assumption dim L1 6 dim L2 it follows that L2 contains a translate of L1 . Therefore, S1 ⊂ S2 by Theorem 1.13. Choose a point c ∈ Rn satisfying the inclusion c + L1 ⊂ L2 , and put b = c + a1 . Then b = c + a1 ∈ c + a1 + S1 = c + L1 ⊂ L2 . Thus L2 = b + S2 according to Theorem 1.2. Let S = span {c ∪ S2 } and L = b + S. Clearly, S1 − c ⊂ S2 − c ⊂ S and dim L = dim S 6 dim S2 + 1 = dim L2 + 1.
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This argument yields the inclusions L1 = a1 + S1 = b + (S1 − c) ⊂ b + S = L, L2 = b + S2 ⊂ b + S = L. If L1 ∩ L2 6= ∅ and a is a point in L1 ∩ L2 , then Theorem 1.2 gives L1 = a + S1 ⊂ a + S2 = L2 . Conversely, let L1 and L2 satisfy conditions (1) and (2) of the theorem. Since the case L1 ⊂ L2 implies the parallelism of L1 and L2 , we may assume that L1 ∩ L2 = ∅. Choose a plane L ⊂ Rn of dimension dim L2 + 1 or less, which contains L1 ∪L2 . Denote by S the characteristic subspace of L. Then S1 ∪ S2 ⊂ S according to Theorem 1.13. Suppose that L1 and L2 are not parallel. Then S1 6⊂ S2 by the same Theorem 1.13. Put m = dim S2 . Choose a vector b0 ∈ S1 \ S2 and a basis b1 , . . . , bm for S2 . Then b0 , b1 , . . . , bm is a basis for S because of the condition dim S 6 m + 1. Since both a1 and a2 belong to L, their difference a1 − a2 is in L − L = S. Hence a1 − a2 can be expressed as a linear combination a1 − a2 = λ0 b0 + λ1 b1 + · · · + λm bm . Put c = λ1 b1 + · · · + λm bm . Then c ∈ S2 and a1 − λ0 b0 = a2 + c. Furthermore, a1 − λ0 b0 ∈ a1 + S1 = L1
and a2 + c ∈ a2 + S2 = L2 ,
contrary to the assumption L1 ∩ L2 = ∅. Hence L1 and L2 are parallel. Hyperplanes Definition 1.16. A hyperplane in Rn is a plane of dimension n − 1. Theorem 1.17. A set H ⊂ Rn is a hyperplane if and only if there is a nonzero vector c ∈ Rn and a scalar γ such that H = {x ∈ Rn : x·c = γ}.
(1.3)
Furthermore, this representation of H is unique up to a common nonzero scalar multiple of c and γ. Proof. A set H ⊂ Rn is a hyperplane if and only if H = a + S, where a ∈ Rn and S is a subspace of dimension n − 1. From linear algebra we know (see page 6) that S can be written as S = {u ∈ Rn : u·c = 0}, c 6= o,
(1.4)
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and this expression is unique up to a nonzero scalar multiple of c. Put γ = a·c. Then H is a hyperplane if and only if H = {a + u ∈ Rn : u·c = 0} = {x ∈ Rn : (x − a)·c = 0} = {x ∈ Rn : x·c = γ}. Assume now that H has another representation H = {x ∈ Rn : x·c0 = γ 0 } for a nonzero vector c0 ∈ Rn and a scalar γ 0 . Then a·c0 = γ 0 because of a ∈ H. From S = H − a, we obtain S = {x − a ∈ Rn : x·c0 = γ 0 } = {u ∈ Rn : (u + a)·c0 = γ 0 } = {u ∈ Rn : u·c0 = 0}. A comparison of the latter equality with (1.4) yields that c0 = λc for a certain scalar λ 6= 0. Finally, γ 0 = a·c0 = a·(λc) = λγ. Corollary 1.18. If a hyperplane H ⊂ Rn is given by (1.3), then a set H 0 ⊂ Rn is a translate of H if and only if H 0 can be expressed as H 0 = {x ∈ Rn : x·c = γ 0 }.
(1.5)
Consequently, a set S ⊂ Rn is the characteristic subspace of H if and only if S = {x ∈ Rn : x·c = 0}.
(1.6)
Proof. A set H 0 is a translate of H if and only H 0 = b + H for a certain vector b ∈ Rn . With γ 0 = γ + b·c, we have H 0 = {b + x ∈ Rn : x·c = γ} = {u ∈ Rn : (u − b)·c = γ} = {u ∈ Rn : u·c = γ 0 }. The second statement of the corollary is obvious. Q Q Q Q S QQ
r
Q
oQ Q
Fig. 1.5
Q Q Q
Q H Q Q Q
Q
Normal vectors to a hyperplane H.
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If a hyperplane H = a + S is given by (1.3), then (1.6) shows that c is orthogonal to the subspace S. In this regard, all vectors of the form λc, λ 6= 0, are called normal to H (see example on page 211). For the next theorem, we recall that S ⊥ denotes the orthogonal complement of a subspace S ⊂ Rn (see page 5). Theorem 1.19. For a hyperplane H ⊂ Rn and a proper plane L ⊂ Rn , the following statements hold. (1) H and L are parallel if and only if either H ∩ L = ∅ or L ⊂ H. (2) H and L are parallel if and only if H + L is a translate of H. (3) H and L are parallel if and only if there is a unique hyperplane H 0 ⊂ Rn containing L and parallel to H. (4) If H is given by (1.3), then H and L are parallel if and only if c belongs to the subspace (sub L)⊥ . (5) H and L are not parallel if and only if ∅ 6= H ∩ L 6= L. (6) H and L are not parallel if and only if H + L = Rn . (7) H and L are not parallel if and only if H ∩L is a plane of dimension dim L − 1. Proof. In what follows, let H = a + S and L = b + T , where a and b are points, and S and T are subspaces. (1) Assume first that H and L are parallel. Then T ⊂ S according to Theorem 1.13. If H ∩ L 6= ∅ and c is a point in H ∩ L, then L = c + T ⊂ c + S = H (see Theorem 1.2). Conversely, suppose that H ∩ L = ∅ or L ⊂ H. Since the case L ⊂ H implies the parallelism of L and H, we may assume that H ∩L = ∅. Assume for a moment that H and L are not parallel. Then T 6⊂ S by Theorem 1.13. Choose a vector b1 ∈ T \ S and a basis b2 , . . . , bn for S. Then b1 , . . . , bn is a basis for Rn . Hence b − a can be expressed as a linear combination b − a = λ1 b1 + · · · + λn bn . Put c = λ2 b2 + · · · + λn bn . Then c ∈ S and b − λ1 b1 = a + c. Furthermore, b − λ1 b1 ∈ b + T = L and a + c ∈ a + S = H, contrary to the assumption H ∩ L = ∅. Hence H and L are parallel. (2) By Theorem 1.13, H and L are parallel if and only if T ⊂ S, which is equivalent to the equality T + S = S. Consequently, H and L are parallel if and only if H + L = (a + b) + (S + T ) = (a + b) + S = b + (a + S) = b + H.
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(3) Assume that H and L are parallel. Since dim L 6 dim H, Theorem 1.6 implies that namely H contains a translate of L. If c + L ⊂ H for a certain point c ∈ Rn , then the hyperplane G = H − c contains L and is parallel to H. For the uniqueness of G, choose a hyperplane G0 containing L and parallel to H. According to Theorem 1.13, G0 is a translate of G, and Theorem 1.2 shows that G − b and G0 − b are identical subspaces. Consequently, G = G0 . Conversely, suppose that H 0 is a unique hyperplane containing L and parallel to H. Then H 0 is a translate of H (see Theorem 1.13). Expressing H 0 as H 0 = u+H for a certain point u ∈ Rn , we obtain L−u ⊂ H 0 −u = H. Hence L is parallel to H. (4) By Theorem 1.13, H and L are parallel if and only if T ⊂ S, which is equivalent to S ⊥ ⊂ T ⊥ . Since S ⊥ = span {c} and T = sub L, the inclusion S ⊥ ⊂ T ⊥ is equivalent to c ∈ (sub L)⊥ . Statement (5) is the contrapositive of (1). (6) According to Theorem 1.13, H and L are not parallel if and only if T 6⊂ S. Since dim S = n − 1, the latter is equivalent to S + T = Rn , or, to H + L = (a + b) + (S + T ) = Rn . (7) Assume first that H and L are not parallel. Then H + L = Rn by statement (6) above, and Theorem 1.8 gives dim (H ∩ L) = dim H + dim L − n = dim L − 1. Conversely, if H ∩ L is a plane of dimension dim L − 1, then ∅ 6= H ∩ L 6= L, and statement (5) above shows that H and L are not parallel. Theorem 1.20. Every proper plane L ⊂ Rn of dimension m can be expressed as the intersection of n − m hyperplanes. Furthermore, if L is the intersection of a certain family H of hyperplanes, then H contains a subfamily of n − m members whose intersection is L.
pp p p p pp p
Fig. 1.6
pppp
ppp p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p p
A line as the intersection of two planes in R3 .
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Proof. Let L = a + S, where a ∈ Rn and S is a subspace of dimension m. From linear algebra we know that S can be expressed as the intersection of certain (n − 1)-dimensional subspaces S1 , . . . , Sn−m ⊂ Rn . The sets Fi = a + Si , 1 6 i 6 n − m, are hyperplanes, and L = a + S = a + S1 ∩ · · · ∩ Sn−m = (a + S1 ) ∩ · · · ∩ (a + Sn−m ) = F1 ∩ · · · ∩ Fn−m . Suppose that L is the intersection of a family H = {Hα } of hyperplanes. Choose a point c ∈ L. According to Theorem 1.2, all sets L − c and Hα − c are subspaces. Clearly, L − c = ∩ Hα − c = ∩ (Hα − c), α
dim (L − c) = m,
α
dim (Hα − c) = n − 1 for all Hα ∈ H.
From linear algebra we know that the family {Hα − c : Hα ∈ H} contains n − m members, say H1 − c, . . . , Hn−m − c, whose intersection is L − c. Consequently, L = c + (L − c) = c + (H1 − c) ∩ · · · ∩ (Hn−m − c) = H1 ∩ · · · ∩ Hn−m . The following result complements Theorem 1.20. Theorem 1.21. The intersection of r, 1 6 r 6 n, hyperplanes in Rn of the form Hi = {x ∈ Rn : x·ci = γi }, 1 6 i 6 r, is an (n − r)-dimensional plane if and only if the set {c1 , . . . , cr } is linearly independent. Proof. The intersection H1 ∩ · · · ∩ Hr is the solution set of the system of r linear equations in x1 , . . . , xn : x·ci = x1 c1i + · · · + xn cni = γi ,
1 6 i 6 r.
This solution set is a translate of an (n − r)-dimensional subspace if and only if the r × n matrix (cij ) has rank r, which occurs if and only if the set {c1 , . . . , cr } is linearly independent. Corollary 1.22. Every plane L in Rn is a closed set.
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Proof. The statement is trivial if L = ∅ or L = Rn . Assume that L is a proper plane. Suppose first that L is a hyperplane. Then L can be expressed by (1.3). Because the linear functional ϕ(x) = x·c is continuous on Rn (see Exercise 0.4), the hyperplane L is closed as the inverse image ϕ−1 (γ) of the singleton {γ}. If dim L 6 n − 2, then L is the intersection of n − m hyperplanes, say H1 , . . . , Hn−m (see Theorem 1.20). Since each hyperplane Hi , 1 6 i 6 n − m, is a closed set, their intersection also is closed. Lines, Halflines, and Segments According to Definitions 1.1 and 1.5, a line in Rn is a translate of a onedimensional subspace. In what follows, we use another description of lines. (1 − λ)x + λy r x
λ60 Fig. 1.7
06λ61
r y
λ>1
The line through distinct points x and y.
Definition 1.23. The line through distinct points x and y in Rn , denoted hx, yi, is defined by hx, yi = {(1 − λ)x + λy : λ ∈ R}.
(1.7)
Remark. For distinct points x, y ∈ Rn , the mapping λ 7→ (1 − λ)x + λy is a bijection from R on the line hx, yi. Rewriting (1.7) as hx, yi = x + {λ(y − x) : λ ∈ R}, we see that hx, yi is a translate of the 1-dimensional subspace span {y −x}. Definition 1.24. For distinct points x and y in Rn , the closed halflines [x, yi and hx, y], and the open halflines (x, yi and hx, y) are defined, respectively, as [x, yi = {(1 − λ)x + λy : λ > 0},
(x, yi = {(1 − λ)x + λy : λ > 0},
hx, y] = {(1 − λ)x + λy : λ 6 1},
hx, y) = {(1 − λ)x + λy : λ < 1}.
Definition 1.25. For distinct points x and y in Rn , the closed segment [x, y], the semi-open segments [x, y) and (x, y], and the open segment
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(x, y) are defined, respectively, as [x, y] = {(1 − λ)x + λy : 0 6 λ 6 1}, [x, y) = {(1 − λ)x + λy : 0 6 λ < 1}, (x, y] = {(1 − λ)x + λy : 0 < λ 6 1}, (x, y) = {(1 − λ)x + λy : 0 < λ < 1}. For every point x ∈ Rn , let [x, x] = {x} and [x, x) = (x, x] = (x, x) = ∅. Clearly, [x, yi = hy, x], (x, yi = hy, x), [x, y] = [y, x], [x, y) = (y, x]. Lemma 1.26. For distinct points x and y in Rn , the following statements hold. (1) If u and v are distinct points in hx, yi, then hu, vi = hx, yi. (2) If u ∈ (x, yi, then [x, ui = [x, yi. (3) If v ∈ [x, y] and w ∈ hx, yi, then one of the inclusions v ∈ [x, w] and v ∈ [y, w] holds. Proof. (1) Since u, v ∈ hx, yi, we can write u = (1 − λ)x + λy
and v = (1 − µ)x + µy
(1.8)
for certain scalars λ and µ. Clearly, λ 6= µ because of u 6= v. To prove the inclusion hu, vi ⊂ hx, yi, choose a point w ∈ hu, vi. Then w = (1 − ξ)u + ξv for a scalar ξ. With ν = λ − λξ + µξ, one has w = (1 − ξ)((1 − λ)x + λy) + ξ((1 − µ)x + µy) = (1 − λ + λξ − µξ)x + (λ − λξ + µξ)y = (1 − ν)x + νy ∈ hx, yi. Conversely, let z ∈ hx, yi. Then z = (1 − η)x + ηy for a scalar η. Solving (1.8) for x and y, x=
λ µ u− v µ−λ µ−λ
and y =
1−λ 1−µ v− u, µ−λ µ−λ
and letting α = (η − λ)/(µ − λ), we obtain 1−λ µ λ 1−µ z = (1 − η) u− v +η v− u µ−λ µ−λ µ−λ µ−λ η − λ η−λ = 1− u+ v = (1 − α)u + αv ∈ hu, vi. µ−λ µ−λ
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(2) Let u = (1 − λ)x + λy, where λ > 0. To show the inclusion [x, ui ⊂ [x, yi, choose a point v ∈ [x, ui. Then v = (1 − η)x + ηu for a scalar η > 0. Because λη > 0 and v = (1 − η)x + η((1 − λ)x + λy) = (1 − λη)x + ληy, it follows that v ∈ [x, yi. Conversely, let z ∈ [x, yi. Then z = (1−µ)x+µy for a certain scalar µ > 0. Since y = (1 − λ−1 )x + λ−1 u and µλ−1 > 0, one has z = (1 − µ)x + µ((1 − λ−1 )x + λ−1 u) = (1 − µλ−1 )x + µλ−1 u ∈ [x, ui. (3) Let v = (1−λ)x+λy and w = (1−µ)x+µy for scalars λ ∈ [0, 1] and µ ∈ R. Assume first that λ 6 µ. Leaving aside the case λ = µ = 0 (which corresponds to v = w = x), we suppose that µ > 0. Since 0 6 λ/µ 6 1, the equality λ λ λ λ x+ (1 − µ)x + µy = 1 − x+ w v = 1− µ µ µ µ gives v ∈ [x, w]. Suppose now that λ > µ. Excluding the case λ = µ = 1 (which corresponds to v = w = y), we suppose that µ < 1. Then 0 6 (λ−µ)/(1−µ) 6 1, and the equality λ−µ λ − µ 1 − λ λ−µ y = 1− y v= (1 − µ)x + µy + w+ 1−µ 1−µ 1−µ 1−µ yields the inclusion v ∈ [y, w]. Lemma 1.27. For any points x, y, and z in Rn and points u ∈ [x, y] and v ∈ [x, z], the segments [u, z] and [v, y] have a point in common. Similarly, if q ∈ [u, z], then q ∈ [w, y] for a certain point w ∈ [x, z]. z r P P
v w x r r r PP r pP PP r r qP u r y
Proof. Let u = (1 − λ)x + λy and v = (1 − µ)x + µz, where λ, µ ∈ [0, 1]. Excluding the case λ = µ = 1 (which corresponds to u = y and v = z), we
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assume that λµ < 1. Put p=
(1 − λ)(1 − µ) λ(1 − µ) (1 − λ)µ x+ y+ z, 1 − λµ 1 − λµ 1 − λµ 1−µ 1−λ ξ= , η= . 1 − λµ 1 − λµ
Obviously, ξ, η ∈ [0, 1], and the equalities 1−µ 1−µ z+ (1 − λ)x + λy = (1 − ξ)z + ξu, p= 1− 1 − λµ 1 − λµ 1−λ 1−λ p= 1− y+ (1 − µ)x + µz = (1 − η)y + ηv 1 − λµ 1 − λµ imply the inclusion p ∈ [u, z] ∩ [v, y]. Similarly, given a point q ∈ [u, z], we can write q = (1 − γ)z + γu for a certain scalar γ ∈ [0, 1]. Leaving aside the case λ = γ = 1 (corresponding to q = u = y), we assume that λγ < 1. Let ζ = 1 − λγ,
ν=
1−γ , 1 − λγ
w = (1 − ν)x + νz.
Then ζ, ν ∈ [0, 1] and w ∈ [x, z]. Finally, the equalities q = (1 − γ)z + γ u = (1 − γ)z + γ((1 − λ)x + λ y) (1 − λ)γ 1−γ = λγ y + (1 − λγ) x+ z 1 − λγ 1 − λγ = (1 − ζ)y + ζ((1 − ν)x + ν z) = (1 − ζ)y + ζw show that q ∈ [w, y]. Halfspaces and Slabs Definition 1.28. If a hyperplane H ⊂ Rn is given by (1.3), then the sets V1 = {x ∈ Rn : x·c 6 γ}
and V2 = {x ∈ Rn : x·c > γ}
(1.9)
are called the opposite closed halfspaces determined by H, and the sets W1 = {x ∈ Rn : x·c < γ}
and W2 = {x ∈ Rn : x·c > γ}
(1.10)
are the opposite open halfspaces determined by H. The vectors λc, λ > 0, are called normal to both V1 and W1 , and the vectors λc, λ < 0, are normal to both V2 and W2 .
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Topological properties of halfspaces are described in the next theorem. Theorem 1.29. If H ⊂ Rn is a hyperplane expressed by (1.3), then the halfspaces V1 and V2 given by (1.9) are closed sets, and the halfspaces W1 and W2 given by (1.10) are open sets. Furthermore, Vi = cl Wi ,
Wi = int Vi ,
H = bd Vi = bd Wi ,
i = 1, 2.
(1.11)
Proof. The linear functional ϕ(x) = x · c is continuous on Rn (see Exercise 0.4). Therefore, the halfspaces (1.9) are closed sets as ϕ−1 -images of the closed halflines (−∞, γ] and [γ, ∞) of R. Similarly, the halfspaces (1.10) are open sets as ϕ−1 -images of the respective open halflines of R. For the equalities (1.11), it suffices to consider the case i = 1. From the inclusion W1 ⊂ V1 it follows that W1 = int W1 ⊂ int V1 . Hence bd V1 = V1 \ int V1 ⊂ V1 \ W1 = H. For the opposite inclusion, choose a point u ∈ H and a scalar ρ > 0. Let ρ ρ c and v2 = u + c. v1 = u − kck kck Since kv1 − uk = kv2 − uk = ρ, both points v1 and v2 belong to the closed ball Bρ (u). Furthermore, from the inequalities ρ v1 ·c = (u − c)·c = γ − ρkck < γ kck and v2 ·c = (u +
ρ c)·c = γ + ρkck > γ kck
it follows that v1 ∈ V1 and v2 ∈ W2 . Hence Bρ (u) meets both sets V1 and Rn \ V1 , implying that u ∈ bd V1 . Consequently, bd V1 = H
and
int V1 = V1 \ bd V1 = V1 \ H = W1 .
Similarly, the inclusion W1 ⊂ V1 gives cl W1 ⊂ V1 . Therefore, bd W1 = cl W1 \ W1 ⊂ V1 \ W1 = H. To show the opposite inclusion, choose a point u ∈ H. By the above argument, every point u ∈ H belongs to bd W1 . Therefore, bd W1 = H
and V1 = W1 ∪ H = W1 ∪ bd W1 = cl W1 .
Remark. There are obvious relations between closed and open halfspaces: (a) a closed halfspace {x ∈ Rn : x·c 6 γ} is the intersection of denumerably many open halfspaces {x ∈ Rn : x·c < γ + 1/k}, k > 1,
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(b) an open halfspace {x ∈ Rn : x·c < γ} is the union of denumerably many closed halfspaces {x ∈ Rn : x·c 6 γ − 1/k}, k > 1. The next result complements Theorem 1.19. Theorem 1.30. If H ⊂ Rn is a hyperplane and L ⊂ Rn is a proper plane, then the following conditions are equivalent. (1) H and L are parallel. (2) Either L ⊂ H, or L lies in an open halfspace determined by H. Proof. Let H = a + S and L = b + T , where a and b are points, and S and T are subspaces. Assume that H is given by (1.3). Then Corollary 1.18 shows that S is expressed by (1.6). (1) ⇒ (2). If H and L are parallel, then, according to Theorem 1.19, either H ∩ L = ∅ or L ⊂ H. Theorem 1.13 yields the inclusion T ⊂ S. Let H ∩ L = ∅. Then b ∈ / H, since otherwise L = b + T ⊂ b + S = H (see Theorem 1.2). Put β = b·c. Clearly, β 6= γ because of b ∈ / H. Suppose that β < γ (the case β > γ is similar). We state that L lies in the open halfspace W = {x ∈ Rn : x · c < γ}. Indeed, for a point x ∈ L, we can write x = b + x0 , where x0 ∈ T , which gives x·c = b·c + x0 ·c = β + 0 < γ. (2) ⇒ (1). Since the case L ⊂ H is obvious, we may suppose that L lies in an open halfspace determined by H. Then H ∩L = ∅, and Theorem 1.19 shows that H and L are parallel. The theorem below plays an important role in separation properties of convex sets. Theorem 1.31. For a hyperplane H ⊂ Rn and distinct points y, z ∈ Rn , the following statements hold. (1) If both y and z belong to H, then the line hy, zi lies in H. (2) If y ∈ H and z belongs to an open halfspace determined by H, then the open halfline (y, zi lies in this halfspace. (3) If both y and z belong to an open halfspace determined by H, then at least one of the halflines [y, zi, hy, z] lies in this halfspace. (4) If y and z belong, respectively, to distinct open halfspaces determined by H, then hy, zi meets H at a single point, which belongs to the open segment (y, z).
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Proof. Suppose that H is given by (1.3). Any point u ∈ hy, zi can be written as u = (1 − λ)y + λz, and this expression implies u·c = ((1 − λ)y + λz)·c = (1 − λ)y·c + λ z·c.
(1.12)
(1) If y, z ∈ H and u ∈ hy, zi, then y ·c = z ·c = γ, and (1.12) gives u·c = γ. Hence hy, zi ⊂ H. (2) Assume that y ∈ H and z belongs to the halfspace W = {x ∈ Rn : x·c < γ}. Then y·c = γ and z·c < γ. If u ∈ (y, zi, then λ > 0, and (1.12) gives u·c = y·c + λ(z·c − y·c) < y·c = γ. Therefore, (y, zi ⊂ W . (3) Suppose that y and z belong to the same open halfspace W determined by H. As above, we assume that W = {x ∈ Rn : x · c < γ}. If z·c 6 y·c and u ∈ [y, zi, then λ > 0, and (1.12) gives u·c = y·c + λ(z·c − y·c) 6 y·c < γ. Hence [y, zi ⊂ W . If y·c 6 z·c and u ∈ hy, z], then λ 6 1, and (1.12) gives u·c = y·c + λ(z·c − y·c) 6 y·c + (z·c − y·c) = z·c < γ. Hence hy, z] ⊂ W . (4) Suppose that y and z belong, respectively, to distinct open halfspaces determined by H. Without loss of generality, we assume that y·c < γ < z·c. Put µ=
γ − y·c . z·c − y·c
(1.13)
Since 0 < µ < 1, the point u = (1 − µ)y + µz belongs to (y, z). Combining (1.12) and (1.13), we obtain u·c = (1 − µ)y·c + µz·c γ − y·c γ − y·c = 1− y·c + z·c = γ. z·c − y·c z·c − y·c Hence u ∈ H. By statement (2) above, the open halflines (u, yi and (u, zi lie, respectively, in distinct open halfspaces determined by H. So, u is the unique point from hy, zi which belongs to H. Corollary 1.32. Let y and z be distinct points in a closed halfspace V determined by a hyperplane H ⊂ Rn . If (y, z) ∩ H 6= ∅, then [y, z] ⊂ H.
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Proof. From Theorem 1.31 it follows that [y, z] ⊂ V . Assume for a moment that at least one of the points y and z belongs to V \ H. According to Theorem 1.29, the set W = V \ H is an open halfspace determined by H. Theorem 1.31 shows that the open segment (y, z) lies in W , contrary to the assumption (y, z) ∩ H 6= ∅. Hence both points y and z belong to H, and the same Theorem 1.31 implies the inclusion [y, z] ⊂ H. Definition 1.33. Let H and H 0 be distinct parallel hyperplanes in Rn , given, respectively, by (1.3) and (1.5), where γ < γ 0 . The set M = {x ∈ Rn : γ 6 x·c 6 γ 0 }
(1.14)
is called the closed slab between H and H 0 . Similarly, the set N = {x ∈ Rn : γ < x·c < γ 0 }
(1.15)
is called the open slab between H and H 0 . Furthermore, the sets {x ∈ Rn : x·c 6 γ}
and {x ∈ Rn : γ 0 6 x·c}
are the opposite closed halfspaces determined by M , and the sets {x ∈ Rn : x·c < γ}
and {x ∈ Rn : γ 0 < x·c}
are the opposite open halfspaces determined by M . Theorem 1.29 implies that if M and N are slabs defined by (1.14) and (1.15), respectively, then M = cl N,
N = int M,
and
bd M = bd N = H ∪ H 0 .
Theorem 1.34. For a closed slab M ⊂ Rn and distinct points y, z ∈ Rn , the following statements take place. (1) If both y and z belong to M , then the segment [y, z] lies in M . (2) If y ∈ int M and z ∈ Rn \ M , then [y, z] meets bd M at exactly one point. (3) If y and z belong, respectively, to opposite open halfspaces determined by M , then [y, z] meets bd M at exactly two points. (4) If the halfline [y, zi lies in M (respectively, in int M ), then the whole line hy, zi lies in M (respectively, in int M ). Proof. Assume that M is given by (1.14). Then M can be expressed as the intersection of closed halfspaces {x ∈ Rn : x·c > γ}
and {x ∈ Rn : x·c 6 γ 0 }.
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Statements (1)–(3) follow from Theorem 1.31. Hence it remains to prove statement (4). For this, choose a point u ∈ [y, zi. Then u = (1 − λ)y + λz, where λ > 0. The inclusion [y, zi ⊂ M implies that the inequalities γ 6 u·c = (1 − λ)y·c + λ z·c = y·c + λ(z·c − y·c) 6 γ 0
(1.16)
hold for all λ > 0. Clearly, this is possible only if y·c = z·c. Let β = y · c. Then γ 6 β 6 γ 0 . If v is a point in hy, zi, then v = (1 − µ)y + µz for a certain scalar µ. So, v·c = (1 − µ)y·c + µz·c = β, implying that the entire line hy, zi lies in M . Similarly, if [y, zi ⊂ int M , then both inequalities in (1.16) are strict, which yields γ < β < γ 0 . Consequently, hy, zi ⊂ int M . Halfplanes and Plane Slabs As we will see below, given a plane L ⊂ Rn of positive dimension m, its subplanes of dimension m−1 have properties similar to those of hyperplanes in Rn . Theorem 1.35. For a plane L ⊂ Rn of positive dimension m and a proper subplane M of L, the following statements hold. (1) dim M = m − 1 if and only if there is a hyperplane H ⊂ Rn such that M = H ∩ L. (2) dim M = m − 1 if and only if there is a vector c in Rn \ (sub L)⊥ such that M = {x ∈ L : x·c = γ} for a suitable scalar γ. (3) If dim M = m − 1 and c is a vector in (Rn \ (sub L)⊥ ) ∩ (sub M )⊥ , then M = {x ∈ L : x·c = γ} for a suitable scalar γ. Proof. The “if” parts in statements (1) and (2) follow from Theorems 1.17 and 1.19, while the “only if” parts in these statements follow from statement (3) of the theorem. So, it remains to prove statement (3). (3) Let dim M = m − 1 and c be a vector in (Rn \ (sub L)⊥ ) ∩ (sub M )⊥ . Choose a point u ∈ M . Then L = u+S and M = u+T , where S and T are the characteristic subspaces of L and M , respectively (see Theorem 1.2). Clearly, T ⊂ S, dim S = m, dim T = m − 1, and c ∈ (Rn \ S ⊥ ) ∩ T ⊥ . Consider the hypersubspace P = {x ∈ Rn : x·c = 0}. Then T ⊂ {c}⊥ = P , implying the inclusion T ⊂ P ∩ S. Furthermore, S 6⊂ P because c ∈ / S⊥.
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Theorem 1.19 shows that dim (P ∩ S) = m − 1. Hence T = P ∩ S. Since the set H = u + P is a hyperplane, and since M = u + T = u + P ∩ S = (u + P ) ∩ (u + S) = H ∩ L, we obtain a desire expression of M . Finally, with γ = u·c, one has M = H ∩ L = {x ∈ L : x·c = γ}.
V1 D1
Fig. 1.8
H p p p p p p p p p p p p p p pp pp pp pp p p p ppp
V2 D2
L
Closed halfplanes D1 and D2 of the plane L.
Definition 1.36. Let L ⊂ Rn be a plane of positive dimension (possibly, L = Rn ). A closed halfplane of L is the intersection of L and a closed halfspace V of Rn satisfying the condition ∅ 6= L ∩ V 6= L. Similarly, an open halfplane of L is the intersection of L and an open halfspace W of Rn such that ∅ 6= L ∩ W 6= L. Combining Theorems 1.19 and 1.30, we obtain two corollaries. Corollary 1.37. Let V (respectively, W ) be a closed (respectively, open) halfspace determined by a hyperplane H ⊂ Rn . For a proper plane L ⊂ Rn of positive dimension, the following conditions are equivalent. (1) (2) (3) (4)
∅ 6= L ∩ H 6= L. ∅ 6= L ∩ V 6= L (respectively, ∅ 6= L ∩ W 6= L). L meets both open halfspaces of Rn determined by H. No normal vector of V (respectively, of W ) belongs to the characteristic subspace (sub L)⊥ .
Corollary 1.38. Let L ⊂ Rn be a plane of positive dimension. A subset of L is a closed halfplane of L if and only if it can be expressed as D1 = {x ∈ L : x·c 6 γ}
or
D2 = {x ∈ L : x·c > γ},
(1.17)
⊥
where c ∈ / (sub L) and γ ∈ R. Similarly, a subset of L is an open halfplane of L if and only if it can be expressed as E1 = {x ∈ L : x·c < γ}
or
E2 = {x ∈ L : x·c > γ},
(1.18)
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where c ∈ / (sub L)⊥ and γ ∈ R. Consequently, if Vi and Wi are halfspaces given by (1.9) and (1.10), respectively, then Di = Vi ∩ L
and
Ei = Wi ∩ L = int Vi ∩ L,
i = 1, 2.
The next corollary, which follows from Theorems 1.19 and 1.29, describes topological properties of halfplanes. Corollary 1.39. Let L ⊂ Rn be a plane of positive dimension m. A set E ⊂ L is an open halfplane of L if and only if there is a subplane M ⊂ L of dimension m − 1 such that E is one of the two components of L \ M . A set D ⊂ L is a closed halfplane of L if and only if there is a subplane M ⊂ L of dimension m − 1 such that D = M ∪ E, where E is an open halfplane of L determined by M . Remark. Any closed halfplane of a proper plane L ⊂ Rn is a closed set as the intersection L and a suitable closed halfspace (see Corollary 1.22 and Theorem 1.29). On the contrary, no open halfplane E of L is an open set if dim L 6 n − 1. As we will see later (cf. Definition 2.13), open halfplanes are relatively open sets. One more corollary follows from Theorems 1.19 and 1.30. Corollary 1.40. Let L ⊂ Rn be a plane of positive dimension m and M be a subplane of L of dimension m − 1. If N is a proper subplane of L, then M and N are parallel if and only if either N ⊂ M , or N lies in an open halfplane of L determined by M . A set of useful properties of halfplanes follows from Theorem 1.31 and Corollary 1.32. Corollary 1.41. Let L ⊂ Rn be a plane of positive dimension m and M be a subplane of L of dimension m − 1. For distinct points y, z ∈ L, the following statements hold. (1) If both y and z belong to M , then the line hy, zi lies in M . (2) If y ∈ M and z belongs to an open halfplane of L determined by M , then the open halfline (y, zi lies in this halfplane. (3) If both y and z belong to an open halfplane of L determined by M , then at least one of the halflines [y, zi, hy, z] lies in this halfplane. (4) If y and z belong, respectively, to distinct open halfplanes of L determined by M , then hy, zi meets M at a single point, which belongs to the open segment (y, z).
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(5) If both y and z belong to a closed halfplane of L determined by M such that (y, z) ∩ M 6= ∅, then [y, z] ⊂ M . An algebraic description of halfplanes is given by the following theorem. Theorem 1.42. Let L ⊂ Rn be a plane of positive dimension m. A set X ⊂ Rn is a closed (respectively, open) halfplane of L if and only if X can be expressed as X = h + M , where M is a subplane of L of dimension m − 1 and h is a closed (respectively, open) halfline with endpoint o, which lies in sub L and is not parallel to M . Proof. Assume first that X is a closed halfplane of L. By Corollary 1.38, X can be expressed as X = {x ∈ L : x·c 6 γ},
c∈ / (sub L)⊥ ,
γ ∈ R.
Let H = {x ∈ Rn : x·c = γ}
and M = {x ∈ L : x·c = γ}.
According to Theorem 1.17, H is a hyperplane. Clearly, M = H ∩ L, and Theorem 1.19 implies that M is a plane of dimension m−1. Choose a point a ∈ M and let X 0 = X − a,
S = L − a,
T = M − a.
Theorem 1.2 shows that S and T are subspaces of dimension m and m − 1, respectively. One can write T = {x ∈ S : x·c = 0}
and X 0 = {x ∈ S : x·c 6 0}.
If b1 , . . . , bm−1 is a basis for T and bm ∈ X 0 \ T , then b1 , . . . , bm is a basis for S. Clearly, bi ·c = 0 for all 1 6 i 6 m − 1, and bm ·c < 0. Since every vector x ∈ S can be written as x = ξ1 b1 + · · · + ξm bm , one has X 0 = {x ∈ S : x·c 6 0} = {ξ1 b1 + · · · + ξm bm : (ξm bm )·c 6 0} = {ξ1 b1 + · · · + ξm bm : ξm > 0}. Let h = {λbm : λ > 0} be the closed halfline generated by bm . Obviously, h ⊂ S = sub L such that h is not parallel to M . From the argument above it follows that a point x ∈ S belongs to X 0 if and only if x = y + z, where y ∈ h and z ∈ S. Consequently, X 0 = h + M , and X = a + h + T = h + M. Conversely, if X = h + M , then the above argument shows that X is a closed halfplane of L determined by the subplane M . The case of open halfplanes is similar.
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Various descriptions of halfplanes and their properties are given in Theorem 1.71, Lemma 9.17, and Exercises 9.3 and 9.4. Definition 1.43. Let L ⊂ Rn be a plane of positive dimension. A closed slab of L is the intersection of L and a closed slab M of Rn such that ∅ 6= L ∩ M 6= L. Similarly, an open slab of L is the intersection of L and an open slab N of Rn such that ∅ 6= L ∩ N 6= L. M H H0 p p p p p p p p p p p p p p p pp p p p p p p p p p p p p p p pp pp pp F p pp p pp p p p p p p p ppp ppp
Fig. 1.9
L
A closed slab F of the plane L.
Definition 1.33 and Corollary 1.38 imply the following description of plane slabs. Corollary 1.44. Let L ⊂ Rn be a plane of positive dimension. F ⊂ L is a closed slab of L if and only if it can be expressed as F = {x ∈ L : γ 6 x·c 6 γ 0 }, c ∈ / (sub L)⊥ , γ < γ 0 . A set G ⊂ L is an open slab of L if and only if it can be expressed G = {x ∈ L : γ < x·c < γ 0 }, c ∈ / (sub L)⊥ , γ < γ 0 .
1.2
A set (1.19) as (1.20)
Affine Spans
Affine Combinations of Points and Planes Definition 1.45. An affine combination of points x1 , . . . , xr in Rn is a linear combination λ1 x1 + · · · + λr xr , where λ1 + · · · + λr = 1. For example, the affine combinations of distinct points x and y in Rn fulfill the line hx, yi. The next two theorems show that affine combinations provide a suitable characteristic property of planes. Theorem 1.46. A nonempty set X ⊂ Rn is a plane if and only if it contains all points (1 − λ)x + λy whenever x, y ∈ X and λ ∈ R.
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Proof. Suppose first that X is a plane and express it as X = a + S for a certain a ∈ Rn and a subspace S. Choose points x, y ∈ X and a scalar λ. Then x = a + x0 and y = a + y0 , where x0 , y0 ∈ S. Consequently, (1 − λ)x + λy = a + ((1 − λ)x0 + λy0 ) ∈ a + S = X. Conversely, assume that (1 − λ)x + λy ∈ X whenever x, y ∈ X and λ ∈ R. Choose a point c ∈ X and let T = X − c. Clearly, o = c − c ∈ T . We state that (1 − λ)x + λy ∈ T for all x, y ∈ T and λ ∈ R. Indeed, expressing x and y as x = x0 − c and y = y 0 − c, with x0 , y 0 ∈ X, we obtain (1 − λ)x + λy = ((1 − λ)x0 + λy 0 ) − c ∈ X − c = T. Based on this argument, for points x, y ∈ T and a scalar λ, one has λx = (1 − λ)o + λx ∈ T,
x + y = 2((1 − 21 )x + 12 y) ∈ T.
Hence T is a subspace, which shows that X = c + T is a plane. Remark. We can reformulate Theorem 1.46, saying that a set X ⊂ Rn with two or more points is a plane if and only if with every pair of distinct points x, y ∈ X it contains the whole line hx, yi. The next result expands Theorem 1.46 to the case of affine combinations of finitely many points. Theorem 1.47. A nonempty set X ⊂ Rn is a plane if and only if it contains all affine combinations of points from X. Proof. If X contains all affine combinations of its points, then λ1 x1 + λ2 x2 ∈ X whenever x1 , x2 ∈ X and λ1 + λ2 = 1. By Theorem 1.46, X is a plane. Conversely, let X be a plane. By induction on k, we are going to prove that X contains all affine combinations of k points from X. The case k = 1 is obvious, and the case k = 2 is proved in Theorem 1.46. Assuming that the statement holds for all positive integers k 6 r − 1 (r > 2), choose an affine combination y = λ1 x1 + · · · + λr xr of points x1 , . . . , xr ∈ X. One of the scalars λ1 , . . . , λr is not 1, since otherwise λ1 + · · · + λr = r > 2. Let, for example, λ1 6= 1. By the induction hypothesis, the affine combination λr λ2 x2 + · · · + xr z= 1 − λ1 1 − λ1 belongs to X. Finally, the equality y = λ1 x1 + (1 − λ1 )z and Theorem 1.46 imply the inclusion y ∈ X.
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Theorem 1.48. A nonempty set X ⊂ Rn is a plane if and only if any of the following conditions holds. (1) µ1 X + · · · + µr X = (µ1 + · · · + µr )X for every choice of scalars µ1 , . . . , µr , where µ1 + · · · + µr 6= 0. (2) µ1 X + · · · + µr X = (µ1 + · · · + µr − 1)c + X for every choice of c ∈ X and scalars µ1 , . . . , µr , where µ1 + · · · + µr 6= 0. Proof. (1) Suppose X is a plane. Since the inclusion (µ1 + · · · + µr )X ⊂ µ1 X + · · · + µr X holds for every set X ⊂ Rn and every choice of scalars µ1 , . . . , µr , it remains to prove that µ1 X + · · · + µr X ⊂ (µ1 + · · · + µr )X provided X is a plane and µ1 + · · · + µr 6= 0. For this, put µ = µ1 + · · · + µr
and λi = µi /µ, 1 6 i 6 r.
Then λ1 + · · · + λr = 1. According to Theorem 1.47, λ1 X + · · · + λr X = {λ1 x1 + · · · + λr xr : x1 , . . . , xr ∈ X} ⊂ X. Multiplying both sides of this inclusion by µ, one has (µ1 + · · · + µr )X ⊂ µ1 X + · · · + µr X. Conversely, let X satisfy condition (1) of the exercise. Choose points x, y ∈ X and a scalar λ. Then (1 − λ)x + λy ∈ (1 − λ)X + λX = X, and Theorem 1.46 implies that X is a plane. (2) If X is a plane, then, according to Theorem 1.2, the set S = X − c is a subspace. This argument gives µ1 X + · · · + µr X = (µ1 + · · · + µr )c + (µ1 + · · · + µr )S = (µ1 + · · · + µr )c + S = (µ1 + · · · + µr − 1)c + X. Conversely, let X satisfy condition (2) of the exercise. Choose points x, y ∈ X and a scalar λ. Then (1 − λ)x + λy ∈ (1 − λ)X + λX = (1 − λ + λ − 1)c + X = X, and Theorem 1.46 shows that X is a plane.
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Affine Spans Definition 1.49. For a given set X ⊂ Rn , the intersection of all planes containing X is called the affine span of X and denoted aff X. Theorem 1.4 implies that the affine span of any set X ⊂ Rn exists and is the smallest plane containing X. Furthermore, aff ∅ = ∅. Example. The affine span of a point x ∈ Rn is the singleton {x}, and the affine span of two distinct points x, y ∈ Rn is the line hx, yi. Example. If L ⊂ Rn is a plane of positive dimension, then the affine span of a halfplane or a slab of L coincides with L (see Corollary 1.80). Some elementary properties of affine spans are described below. Theorem 1.50. For sets X and Y in Rn , the following statements hold. (1) (2) (3) (4) (5) (6)
X ⊂ aff X, with X = aff X if and only if X is a plane. aff (aff X) = aff X. aff X ⊂ span X = aff ({o} ∪ X). aff X = span X if and only if o ∈ aff X. aff X ⊂ aff Y if X ⊂ Y . aff X = aff Y if X ⊂ Y ⊂ aff X.
Furthermore, if {Xα } is a family of sets in Rn , then the statements below are true. (7) aff (∩ Xα ) ⊂ ∩ aff Xα . α
α
(8) ∪ aff Xα ⊂ aff (∪ Xα ). α
α
(9) ∪ aff Xα = aff (∪ Xα ) if the family {Xα } is nested. α
α
(10) aff (∪ aff Xα ) = aff (∪ Xα ). α
α
Proof. Let P(X) denote the family of all planes containing a given set X ⊂ Rn . The proofs of statements (1)–(10) derive from the following simple arguments. (a) (b) (c) (d)
aff X is the smallest element in P(X). P(X) is exactly the family of all planes containing aff X. P({o} ∪ X) is exactly the family of all subspaces containing X. If X ⊂ Y , then P(Y ) ⊂ P(X).
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The next theorem gives an important description of affine spans in terms of affine combinations of points. Theorem 1.51. The affine span of a nonempty set X ⊂ Rn is the collection of all affine combinations of points from X: aff X = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 + · · · + λk = 1}. Proof. Since aff X is a plane containing X, Theorem 1.47 yields that the set L = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 + · · · + λk = 1} lies in aff X. We state that L also is a plane. For this, choosing points x, y ∈ L and expressing them as affine combinations x = γ1 x 1 + · · · + γp x p
and y = µ1 y1 + · · · + µq yq
of certain points x1 , . . . , xp , y1 , . . . , yq from X, we see that (1 − λ)x + λy also is an affine combination of these points for every choice of λ ∈ R: (1 − λ)x + λy = (1 − λ)γ1 x1 + · · · + (1 − λ)γp xp + λµ1 y1 + · · · + λµq yq . Hence (1 − λ)x + λy ∈ L, and Theorem 1.46 shows that L is a plane. Since X ⊂ L (every point x ∈ X can be written as 1x), the inclusions X ⊂ L ⊂ aff X and Theorem 1.50 give aff X = L. Corollary 1.52. For points x1 , . . . , xr ∈ Rn , one has aff {x1 , . . . , xr } = {λ1 x1 + · · · + λr xr : λ1 + · · · + λr = 1}. Proof. By Theorem 1.51, a point x ∈ Rn belongs to aff {x1 , . . . , xr } if and only if it can be written as an affine combination x = λi1 xi1 + · · · + λit xit of certain points xi1 , . . . , xit from {x1 , . . . , xr }. If any of these points, say xip and xiq , coincide, then we replace λip xip + λiq xiq with (λip + λiq )xip . Since such a replacement can be performed at most finitely many times, one may assume that all xi1 , . . . , xit are pairwise distinct. Finally, for every index i ∈ {1, . . . , r} \ {i1 , . . . , it }, we add 0xi to the right-hand side of x = λi1 xi1 + · · · + λit xit to express x as an affine combination of all points x1 , . . . , xr . Theorem 1.51 allows to establish the following important algebraic property of affine spans. Theorem 1.53. If X1 , . . . , Xr are sets in Rn and µ1 , . . . , µr are scalars, then aff (µ1 X1 + · · · + µr Xr ) = µ1 aff X1 + · · · + µr aff Xr .
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Proof. Excluding the trivial case when at least one of the sets X1 , . . . , Xr is empty, we assume that all these sets are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 (if r = 1, then we can write µ1 X1 = µ1 X1 + µ2 o). Since the set µ1 aff X1 + µ2 aff X2 is a plane (see Theorem 1.4), the obvious inclusion µ1 X1 + µ2 X2 ⊂ µ1 aff X1 + µ2 aff X2 and Theorem 1.50 give aff (µ1 X1 + µ2 X2 ) ⊂ µ1 aff X1 + µ2 aff X2 . For the opposite inclusion, choose an x ∈ µ1 aff X1 + µ2 aff X2 . Then x = µ1 x1 + µ1 x2 for certain points x1 ∈ aff X1 and x2 ∈ aff X2 . By Theorem 1.51, x1 and x2 can be written as affine combinations x1 = λ1 u1 + · · · + λp up
and x2 = γ1 v1 + · · · + γq vq ,
where u1 , . . . , up ∈ X1 and v1 , . . . , vq ∈ X2 . Because µ1 ui + µ2 vj ∈ µ1 X1 + µ2 X2
for all
16i6p
and
1 6 j 6 q,
the equalities x = µ1 x1 + µ2 x2 =
p P q P i=1 j=1
λi γj (µ1 ui + µ2 vj ),
p P q P
λ i γj = 1
i=1 j=1
show that x is an affine combination of points from µ1 X1 +µ2 X2 . Therefore, Theorem 1.51 gives x ∈ aff (µ1 X1 + µ2 X2 ). A sharper version of Theorem 1.53 for the case of large sums is considered in Exercise 1.7. Definition 1.54. For a nonempty set X ⊂ Rn , the subspace sub X = aff X − aff X is called the characteristic subspace of X. Corollary 1.55. For a nonempty set X ⊂ Rn and a point c ∈ aff X, the characteristic subspace of X can be expressed in the following ways. (1) sub X = aff X − aff X = aff (X − X) = span (X − X). (2) sub X = aff X − c = aff (X − c) = span (X − c). (3) sub X = {λ1 x1 + · · · + λk xk : x1 , . . . , xk ∈ X, λ1 + · · · + λk = 0}. Proof. Statements (1) and (2) immediately follow from a combination of Theorems 1.50 and 1.53. (3) Choose a point z ∈ X. If x ∈ sub X, then x ∈ aff X − z, and Theorem 1.51 shows that x can be expressed as x = λ1 x1 + · · · + λm xm + (−1)z, x1 , . . . , xm ∈ X, λ1 + · · · + λm = 1.
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Therefore, x belongs to the set S(X) = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 + · · · + λk = 0}. Conversely, if x ∈ S(X), then x = µ1 x1 + · · · + µk xk for suitable x1 , . . . , xk ∈ X and scalars µ1 , . . . , µk whose sum is 0. Then x + z ∈ aff X, implying that x = (x + z) − z ∈ aff X − z = sub X. Affine Span and Union of Sets For the sake of simplicity, we consider the affine spans of two sets in Rn . Theorem 1.56. For nonempty set X and Y in Rn , the following statements hold. (1) aff (X ∪ Y ) = (aff X + sub Y ) ∪ (sub X + aff Y ) ∪ L(X, Y ), where L(X, Y ) = {λx + µy : x ∈ aff X, y ∈ aff Y, λ + µ = 1}. (2) If aff X ∩ aff Y 6= ∅ and c ∈ aff X ∩ aff Y , then aff (X ∪ Y ) = L(X, Y ), aff (X ∪ Y ) = aff (X + Y ) − c = aff (X + Y ) − X ∩ Y.
(1.21) (1.22)
(3) If aff X ∩ aff Y = ∅, c ∈ aff X, e ∈ aff Y , and l is the line through c and e, then aff (X ∪ Y ) = aff X + aff Y + l − c − e.
(1.23)
Proof. (1) By Theorem 1.51, a point z ∈ Rn belongs to aff (X ∪ Y ) if and only if it can be expressed as an affine combination z = λ1 x1 + · · · + λp xp + µ1 y1 + · · · + µq yq , where x1 . . . , xp ∈ X and y1 , . . . , yq ∈ Y . If one of the sums λ = λ1 + · · · + λp
and µ = µ1 + · · · + µq
equals 0, then x belongs to the respective set aff Y +sub X or aff X +sub Y . Suppose that both λ and µ are distinct from 0. Then x0 = λ−1 (λ1 x1 + · · · + λp xp ) ∈ aff X, y 0 = µ−1 (µ1 y1 + · · · + µq yq ) ∈ aff Y, implying the inclusion z = λx0 + µy 0 ∈ L(X, Y ).
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(2) First, we will prove the equalities (1.22). For this, we observe that o ∈ (aff X − c) ∪ (aff Y − c). Therefore, Theorem 1.50 and the properties of spans (see page 4) give aff ((X − c) ∪ (Y − c)) = span ((X − c) ∪ (Y − c)) = span (X − c) + span (Y − c). Using Theorem 1.53 twice, we obtain aff (X ∪ Y ) = aff ((X − c) ∪ (Y − c) + c) = aff ((X − c) ∪ (Y − c)) + c = span (X − c) + span (Y − c) + c = aff (X − c) + aff (Y − c) + c = aff X + aff Y − c = aff (X + Y ) − c. The second part of (1.22), aff (X ∪ Y ) − c = aff (X + Y ) − X ∩ Y, follows from Theorem 1.2. Now, we will prove the equality (1.21). Theorem 1.51 shows that the set L(X, Y ) lies in aff (aff X ∪ aff Y ). Since aff (aff X ∪ aff Y ) = aff (X ∪ Y ) (see Theorem 1.50), one has L(X, Y ) ⊂ aff (X ∪ Y ). For the opposite inclusion, choose points z ∈ aff (X ∪ Y ) and c ∈ X ∩ Y . By the above argument, z − c ∈ aff (X ∪ Y ) − c = span (X − c) + span (Y − c). Therefore, z − c can be written as z − c = u + v, where u ∈ span (X − c) and v ∈ span (Y − c). Because 2u ∈ span (X − c) and 2v ∈ span (Y − c), it follows that 2u + c ∈ span (X − c) + c = aff X, 2v + c ∈ span (Y − c) + c = aff Y. Finally, z = 21 (2u + c) + 12 (2v + c) ∈ L(X, Y ). (3) Since every point x ∈ l can be written as x = (1 − λ)c + λe, Theorem 1.51 shows that l ⊂ aff (aff X ∪ aff Y ) = aff (X ∪ Y ). Hence X ∪Y ⊂ X ∪Y ∪l ⊂ aff (X ∪Y ), and Theorem 1.50 gives aff (X ∪Y ) = aff (X ∪ Y ∪ l). Since e ∈ aff Y ∩ l
and c ∈ aff X ∩ aff (Y ∪ l),
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a combination of Theorem 1.50 and statement (2) above gives aff (X ∪ Y ) = aff (X ∪ Y ∪ l) = aff (X ∪ aff (Y ∪ l)) = aff X + aff (Y ∪ l) − c = aff X + aff Y + l − c − e. Corollary 1.57. For nonempty set X and Y in Rn , the following statements hold. (1) If aff X ∩ aff Y = 6 ∅, then dim (X ∪ Y ) = dim (X + Y ). (2) If aff X ∩ aff Y = ∅, then dim (X ∪ Y ) = dim (X + Y ) + 1. Proof. (1) In terms of statement (2) of Theorem 1.56, one has dim (X ∪ Y ) = dim (aff (X ∪ Y )) = dim (aff (X + Y ) − c) = dim (aff (X + Y )) = dim (X + Y ). (2) Similarly, based on statement (2) of Theorem 1.56, we obtain first dim (X + Y ) = dim (aff (X + Y )) 6 dim (aff X + aff Y + l) = dim (X ∪ Y ) 6 dim (aff (X + Y )) + 1
(1.24)
= dim (X + Y ) + 1. Next, we state that l 6⊂ L ≡ aff X + aff Y − c − e. Indeed, assuming the contrary, one has {c, e} ⊂ L. The set L is a subspace as the sum of subspaces aff X − c and aff Y − e. Therefore, c − e ∈ L, and we can write c − e = α(x − c) + β(y − e),
where
x ∈ aff X, y ∈ aff Y, α, β ∈ R.
Thus c − α(x − c) = e + β(y − e). The left-hand side of the latter equality belongs to aff X, and the right-hand side belongs to aff Y , contrary to the assumption aff X ∩ aff Y = ∅. Summing up, l 6⊂ L. Consequently, dim (X + Y ) = dim (aff X + aff Y ) = dim L < dim (aff X + aff Y + l). Combined with (1.24), this gives dim (X ∪ Y ) = dim (X + Y ) + 1. Affinely Independent Sets Definition 1.58. A set of points x1 , . . . , xr in Rn , with r > 2, is called affinely dependent if there are scalars ν1 , . . . , νr , not all zero, such that ν1 x1 + · · · + νr xr = o
and ν1 + · · · + νr = 0.
(1.25)
The set {x1 , . . . , xr } is called affinely independent if it is not affinely dependent. The empty set ∅ and every singleton {x} are assumed to be affinely independent.
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Occasionally, we will write “affinely independent points x1 , . . . , xr ” instead of “affinely independent set {x1 , . . . , xr }.” The next result refines Theorem 1.51. Theorem 1.59. The affine span of a nonempty set X ⊂ Rn is the collection of all affine combinations of affinely independent points from X. Proof. By Theorem 1.51, a point x ∈ Rn belongs to aff X if and only if is expressible as an affine combination x = λ1 x1 + · · · + λk xk of certain points x1 , . . . , xk ∈ X. We are going to show that if {x1 , . . . , xk } is affinely dependent, then x can be written as an affine combination of k − 1 of fewer points from {x1 , . . . , xk }. So, let {x1 , . . . , xk } be affinely dependent. Then there are scalars ν1 , . . . , νk , not all zero, such that ν1 x1 + · · · + νk xk = o
and ν1 + · · · + νk = 0.
Without loss of generality, we assume that ν1 6= 0. Then νk ν2 x1 = − x2 − · · · − xk . ν1 ν1 Hence x can be written as an affine combination νk ν2 x2 + · · · + λk − λ1 xk , x = λ1 x1 + · · · + λk xk = λ2 − λ1 ν1 ν1 which involves only points x2 , . . . , xk . Consecutively repeating this argument, we obtain a desired expression of x. Theorem 1.60. For points x1 , . . . , xr ∈ Rn , the following conditions are equivalent. (1) The set {x1 , . . . , xr } is affinely independent. (2) None of x1 , . . . , xr is an affine combination of the others. (3) For every index i = 1, . . . , r, the set {x1 − xi , . . . , xi−1 − xi , xi+1 − xi , . . . , xr − xi }
(1.26)
is linearly independent (it is empty if r = 1). (4) There is an index i ∈ {1, . . . , r} such that the set (1.26) is linearly independent. (5) The plane aff {x1 , . . . , xr } has dimension r − 1. (6) Every point x ∈ aff {x1 , . . . , xr } is uniquely expressible as an affine combination of x1 , . . . , xr .
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Proof. Since the case r = 1 is obvious, we assume that r > 2. (1) ⇒ (2). Assume for a moment the existence of a point xi , i ∈ {1, . . . , r}, which is an affine combination of x1 , . . . , xi−1 , xi+1 , . . . , xr : xi = λ1 x1 + · · · + λi−1 xi−1 + λi+1 xi+1 · · · + λr xr , λ1 + · · · + λi−1 + λi+1 + · · · + λr = 1. Rewriting these equalities as λ1 x1 + · · · + λi−1 xi−1 + (−1)xi + λi+1 xi+1 · · · + λr xr = 0, λ1 + · · · + λi−1 + (−1) + λi+1 + · · · + λr = 0, we conclude that the set {x1 , . . . , xr } is affinely dependent, in contradiction with (1). (2) ⇒ (3). Suppose the existence of an index i ∈ {1, . . . , r} such that the set (1.26) is linearly dependent. Then one of these vectors, say x1 − xi , is a linear combination of the others: x1 − xi = λ2 (x2 − xi ) + · · · + λi−1 (xi−1 − xi ) + λi+1 (xi+1 − xi ) + · · · + λr (xr − xi ). With λi = 1 − λ2 − · · · − λi−1 − λi+1 − · · · − λr , the equality x1 = λ2 x2 + · · · + λr xr shows that x1 is an affine combination of x2 , . . . , xr , contradicting (2). Since (3) implies (4), we show next that (4) ⇒ (5). For this, consider the (r − 1)-dimensional subspace S = span {x1 − xi , . . . , xi−1 − xi , xi+1 − xi , . . . , xr − xi }.
(1.27)
A combination of Theorems 1.50 and 1.53 gives S = aff {x1 − xi , . . . , xi−1 − xi , o, xi+1 − xi , . . . , xr − xi } = aff {x1 , . . . , xr } − xi . Hence the plane aff {x1 , . . . , xr } = xi + S has dimension r − 1. (5) ⇒ (6). Choose a point x ∈ aff {x1 , . . . , xr }. By Corollary 1.52, x can be written as an affine combination x = λ1 x1 + · · · + λr xr . To show its uniqueness, choose an affine combination x = µ1 x1 + · · · + µr xr . Consequently, the vector y = x − xi can be written as y = λ1 (x1 − xi ) + · · · + λi−1 (xi−1 − xi ) + λi+1 (xi+1 − xi ) + · · · + λr (xr − xi ), y = µ1 (x1 − xi ) + · · · + µi−1 (xi−1 − xi ) + µi+1 (xi+1 − xi ) + · · · + µr (xr − xi ).
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The subspace (1.27) is (r −1)-dimensional as a translate of aff {x1 , . . . , xr }. Hence the set (1.26) is linearly independent, which implies that the above expressions for y are identical: λj = µj for all j 6= i. Finally, P P λi = 1 − λj = 1 − µj = µi . j6=i
j6=i
(6) ⇒ (1). Assume for a moment that the set {x1 , . . . , xr } is affinely dependent. Then there are scalars ν1 , . . . , νr , not all zero, such that (1.25) holds. Choose a point x ∈ aff {x1 , . . . , xr }. By Corollary 1.52, x can be expressed as an affine combination x = λ1 x1 + · · · + λr xr . Then x = (λ1 + ν1 )x1 + · · · + (λr + νr )xr is another expression of x as an affine combination of x1 , . . . , xr , contrary to (6). Since every m-dimensional subspace of Rn contains at most m linearly independent vectors, Theorem 1.60 implies the following corollary. Corollary 1.61. An affinely independent set X ⊂ Rn which lies within a plane of dimension m consists of m + 1 or fewer points. Theorem 1.62. A set X = {x1 , . . . , xr } ⊂ Rn is affinely independent if and only if it does not contain disjoint subsets whose affine spans meet. Proof. We will prove the contrapositive statement: the set X is affinely dependent if and only if it it contains disjoint subsets whose affine spans meet. Let X be affinely dependent. This means the existence of scalars ν1 , . . . , νr , not all zero, such that ν1 x1 + · · · + νr xr = o,
ν1 + · · · + νr = 0.
Put I = {i : νi > 0} and J = {i : νi < 0}. Then both sets I and J are nonempty and partition the set {1, . . . , r}. Furthermore, P P νi xi = (−νj )xj . (1.28) i∈I
P
Let ν = i∈I νi . Clearly, ν > 0 and of (1.28) by ν, we obtain
i∈J
P
i∈J (−νj )
= ν. Dividing both parts
P (−νj ) P νi xi = xj . ν i∈J i∈I ν
(1.29)
As easily seen, left-hand part of (1.29) is an affine combination of points from the set Y = {xi : i ∈ I}; similarly, right-hand side of (1.29) is an affine
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combination of points from the set Z = {xj : j ∈ J}. Hence Y ∩ Z = ∅ and aff Y ∩ aff Z 6= ∅. Conversely, let X = Y ∪Z be a partition of X such that aff Y ∩aff Z 6= ∅. Choose a point x ∈ aff Y ∩ aff Z. By Theorem 1.51, x can be expressed as affine combinations, x = λ1 y1 + · · · + λk yk = µ1 z1 + · · · + µm zm , of certain points y1 , . . . , yk ∈ Y and z1 , . . . , zm ∈ Z. Definition 1.58 and the equalities λ1 y1 + · · · + λk yk − µ1 z1 − · · · − µm zm = o, λ1 + · · · + λk − µ1 − · · · − µm = 0 show that the set {y1 , . . . , yk , z1 , . . . , zm } is affinely dependent. Hence X is affinely dependent. Corollary 1.63. Let a set {x1 , . . . , xr } ⊂ Rn be affinely independent and xr+1 be a point in Rn . Then the set {x1 , . . . , xr+1 } is affinely independent if and only if xr+1 does not belong to the plane aff {x1 , . . . , xr }. Affine Bases Definition 1.64. We say that points c1 , . . . , cr ∈ Rn form an affine basis for a nonempty plane L ⊂ Rn provided the set {c1 , . . . , cr } is affinely independent and L = aff {c1 , . . . , cr }. The empty set ∅ is assumed to be the affine basis for the empty plane. r
c3
c3 − c1 r c 1
o r Fig. 1.10
r c 2
- c −c 2 1
Affine basis c1 , c2 , c3 and basis c2 − c1 , c3 − c1 for R2 .
Theorem 1.65. Every plane L ⊂ Rn has an affine basis. Furthermore, if dim L = m, then each affine basis for L consists of m + 1 points.
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Proof. Since the case L = ∅ is obvious, we assume that L is nonempty. Let m = dim L. The case m = 0 is clear (L is a singleton, {c}, and c is its affine basis). Let m > 1. Choose a point a ∈ L and consider the m-dimensional subspace S = L − a. If b1 , . . . , bm is a basis for S, then, according to Theorem 1.50, S = span {b1 , . . . , bm } = aff {o, b1 , . . . , bm }. Therefore, L = a + S = a + aff {o, b1 , . . . , bm } = aff {a, a + b1 , . . . , a + bm } (see Theorem 1.53). Finally, Theorem 1.60 shows that a, a + b1 , . . . , a + bm is an affine basis for L. The same theorem implies that each affine basis for L consists of m + 1 points. Theorem 1.60 implies one more corollary. Corollary 1.66. For a nonempty plane L ⊂ Rn of dimension m and points c1 , . . . , cm+1 ∈ L, the following conditions are equivalent. (1) (2) (3) (4)
The points c1 , . . . , cm+1 form an affine basis for L. The set {c1 , . . . , cm+1 } is affinely independent. L = aff {c1 , . . . , cm+1 }. For every index i = 1, . . . , m + 1, the vectors c1 − ci , . . . , ci−1 − ci , ci+1 − ci , . . . , cm+1 − ci
(1.30)
form a basis for the subspace sub L (this set of vectors is empty if m = 1). (5) There is an index i ∈ {1, . . . , m + 1} such that the vectors (1.30) form a basis for the subspace sub L. (6) Every point x ∈ L is uniquely expressible as an affine combination of c1 , . . . , cm+1 . Corollary 1.67. If L ⊂ Rn is a plane of dimension m, −1 6 m 6 n − 1, and c is a point in Rn \ L, then aff (c ∪ L) is a plane of dimension m + 1. Proof. Since the case L = ∅ is obvious, we let L 6= ∅. By Theorem 1.65, L has an affine basis c1 , . . . , cm+1 , and Corollary 1.63 shows that the set {c, c1 , . . . , cm+1 } is affinely independent. Theorem 1.50 gives aff (c ∪ L) = aff (c ∪ aff {c1 , . . . , cm+1 }) = aff {c, c1 , . . . , cm+1 }. Finally, Theorem 1.60 implies that dim (aff (c ∪ L)) = m + 1.
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The next result refines Theorem 1.65 by putting a restriction on the choice of an affine basis. Theorem 1.68. If Y is an affinely independent subset of a set X ⊂ Rn , then there is an affine basis Z for aff X such that Y ⊂ Z ⊂ X. Proof. The case X = ∅ is obvious; so, we assume that X 6= ∅. Choose a point c ∈ Y if Y 6= ∅ or any c ∈ X if Y = ∅. Consider the set Y 0 = (Y − c) \ {o}. Clearly, Y 0 ⊂ X − c. By Theorem 1.60, the set Y 0 is linearly independent (if card Y 6 1, then Y 0 = ∅ is linearly independent by the definition). From linear algebra we know that Y 0 can be expanded to a basis U for the subspace span (X − c). Furthermore, span (X − c) = aff (X − c) because of o ∈ X − c (see Theorem 1.50). Finally, let Z = {c} ∪ (c + U ). Then Y ⊂ Z ⊂ X and, according to Corollary 1.66, Z is an affine basis for the plane c + span (X − c) = c + aff (X − c) = aff X. Corollary 1.69. If L is a plane in Rn , then every affinely independent subset of L can be completed into an affine basis for L. ppp
pp ppp
p
p ppp ppp ppppp prpp c ppp ppp 1 ppp pp ppppp ppp p p ppp ppp
ppp
ppp
r x = 12 c1 − 21 c2 + 1c3
ppp ppp pp ppp p p p p p p p p ppp rp pp pcp p2p p p p p p p p p p p p p p p p p p p p p ppppprpppp pcp p3p p p p p p p ppp ppp ppp p ppp Fig. 1.11
Affine coordinates of x relative to c1 , c2 , c3 .
Theorem 1.60 gives a base to the following definition. Definition 1.70. Let {c1 , . . . , cr } ⊂ Rn be an affinely independent set and x be a point in aff {c1 , . . . , cr }. If x is (uniquely) expressed as an affine combination x = λ1 c1 + · · · + λr cr ,
λ1 + · · · + λr = 1,
then λ1 , . . . , λr are called the affine coordinates of x relative to c1 , . . . , cr .
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The next theorem shows that affine coordinates are useful in descriptions of halfplanes (see Definition 1.36). Theorem 1.71. Let L ⊂ Rn be a plane of positive dimension m. A set D ⊂ Rn is a closed halfplane of L if and only if there is an affine basis c1 , . . . , cm+1 for L such that D = {λ1 c1 + · · · + λm+1 cm+1 : λ1 + · · · + λm+1 = 1, λm+1 > 0}. (1.31) Similarly, a set E ⊂ Rn is an open halfplane of L if and only if there is an affine basis c1 , . . . , cm+1 for L such that E = {λ1 c1 + · · · + λm+1 cm+1 : λ1 + · · · + λm+1 = 1, λm+1 > 0}. (1.32)
rc1 r c2
D = {λ1 c1 + λ2 c2 + λ3 c3 :
r c3
λ1 + λ2 + λ3 = 1, λ3 > 0}
Fig. 1.12
A closed halfplane of R2 .
Proof. Assume first that D is a closed halfplane of L. By Corollary 1.38, D can be expressed as D = {x ∈ L : x·c 6 γ},
c∈ / (sub L)⊥ ,
γ ∈ R.
n
Let H = {x ∈ R : x·c = γ} and M = {x ∈ L : x·c = γ}. According to Theorem 1.17, H is a hyperplane. Clearly, M = H ∩ L, and Theorem 1.19 shows that M is a plane of dimension m − 1. Choose a point a ∈ M and let D0 = D − a, S = L − a, T = M − a. Theorem 1.2 shows that S and T are subspaces of dimension m and m − 1, respectively. One can write T = {x ∈ L0 : x·c = 0}
and D0 = {x ∈ L0 : x·c 6 0}.
If b1 , . . . , bm−1 is a basis for T and bm ∈ D0 \ T , then b1 , . . . , bm is a basis for S. Clearly, bi ·c = 0 for all 1 6 i 6 m − 1, and bm ·c < 0. Since every vector x ∈ S can be written as x = ξ1 b1 + · · · + ξm bm , one has D0 = {x ∈ S : x·c 6 0} = {ξ1 b1 + · · · + ξm bm : (ξm bm )·c 6 0} = {ξ1 b1 + · · · + ξm bm : ξm > 0}. Furthermore, with c1 = a, ci+1 = a + bi , and λ1 = 1 − ξ1 − · · · − ξm , λi+1 = ξi , 1 6 i 6 m,
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the halfplane D can be described as D = a + D0 = a + {ξ1 b1 + · · · + ξm bm : ξm > 0} = {λ1 a + ξ1 (a + b1 ) + · · · + ξm (a + bm ) : ξm > 0} = {λ1 c1 + · · · + λm+1 cm+1 : λ1 + · · · + λm+1 = 1, λm+1 > 0}. Repeating the above argument in the converse order, we obtain that every set of the form (1.31) is a closed halfplane of L. The case of an open halfplane is similar. Affine Bases and Operations with Planes Corollary 1.72. Let L1 = a1 + S1 and L2 = a2 + S2 be nonempty planes in Rn , expressed as translates of subspaces S1 and S2 , respectively. If a basis b1 , . . . , br for S1 ∩ S2 is completed into bases b1 , . . . , br , c1 , . . . , cp
and
b1 , . . . , br , e1 , . . . , eq
(1.33)
for S1 and S2 , respectively, then {(a1 + a2 ), b01 , . . . , b0r , c01 , . . . , c0p , e01 , . . . , e0q }
(1.34)
is an affine basis for L1 + L2 , where b0i = (a1 + a2 ) + bi , c0i = (a1 + a2 ) + ci , e0i = (a1 + a2 ) + ei for all suitable indices i. Proof. From linear algebra we know (see page 4), that the combined list b1 , . . . , br , c1 , . . . , cp , e1 , . . . , eq is a basis for the subspace S1 + S2 . Corollary 1.66 implies that the set (1.34) is an affine basis for the plane L1 + L2 = (a1 + a2 ) + (S1 + S2 ). Corollary 1.73. Let L1 = a1 + S1 and L2 = a2 + S2 be nonempty planes in Rn , expressed as translates of subspaces S1 and S2 , respectively. If a basis b1 , . . . , br for S1 ∩ S2 is completed into bases b1 , . . . , br , c1 , . . . , cp
and
b1 , . . . , br , e1 , . . . , eq
(1.35)
for S1 and S2 , respectively, then the following statements take place. (1) If L1 and L2 are disjoint, then {a1 , a2 , b01 , . . . , b0r , c01 , . . . , c0p , e01 , . . . , e0q } is an affine basis for aff (L1 ∪ L2 ), where b0i = a1 + bi , c0i = a1 + ci , e0i = a2 + ei for all suitable indices i.
(1.36)
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(2) If L1 ∩ L2 6= ∅ and a is a point in L1 ∩ L2 , then {a, b01 , . . . , b0r , c01 , . . . , c0p , e01 , . . . , e0q }
(1.37)
is an affine basis for aff (L1 ∪ L2 ), where b0i = a + bi , c0i = a + ci , e0i = a + ei for all suitable indices i. Proof. (1) According to Corollary 1.66, the sets X1 = {a1 , b01 , . . . , b0r , c01 , . . . , c0p }
and X2 = {a2 , b01 , . . . , b0r , e01 , . . . , e0q }
are affine bases for L1 and L2 , respectively. In particular, L1 = aff X1 and L2 = aff X2 , and Theorem 1.50 gives aff (L1 ∪ L2 ) = aff (aff X1 ∪ aff X2 ) ⊂ aff (aff (X1 ∪ X2 )) = aff (X1 ∪ X2 ) ⊂ aff (L1 ∪ L2 ). Hence aff (L1 ∪ L2 ) = aff (X1 ∪ X2 ). According to Corollary 1.57, dim (L1 ∪ L2 ) = dim (L1 + L2 ) + 1 = p + q + r + 1. Since card (X1 ∪ X2 ) = p + q + r + 2, Corollary 1.66 implies that X1 ∪ X2 is an affine basis for aff (L1 ∪ L2 ). (2) By Theorem 1.2, L1 = a + S1 and L2 = a + S2 . Repeating the above argument, with a1 = a2 = a and p + q + r instead of p + q + r + 1, we obtain the desired conclusion. Affine bases provide a suitable description of complementary planes, as shown below. Theorem 1.74. Nonempty planes L1 and L2 in Rn are complementary if and only if there is an affine basis c1 , . . . , cn+1 for Rn such that L1 = aff {c1 , . . . , cm+1 }
and
L2 = aff {cm+1 , . . . , cn+1 },
(1.38)
where m = dim L1 . Proof. Assume first that the planes L1 and L2 are complementary. By Theorem 1.11, L1 + L2 = Rn and the set L1 ∩ L2 is a singleton. Let cm+1 denote the unique point from L1 ∩ L2 . Corollary 1.69 shows that {cm+1 } can be completed into affine bases c1 , . . . , cm+1 and cm+1 , . . . , cn+1 for L1 and L2 , respectively. Then both equalities (1.38) hold, and a combination of Theorem 1.56 and Corollary 1.73 implies that c1 , . . . , cn+1 is an affine basis for aff (L1 ∪ L2 ) = L1 + L2 − cm+1 = Rn − cm+1 = Rn .
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Conversely, suppose the existence of an affine basis c1 , . . . , cn+1 for Rn such that both equalities (1.38) hold. Combining Theorems 1.50 and 1.56, we obtain L1 + L2 = aff (L1 ∪ L2 ) + cm+1 = aff (aff {c1 , . . . , cm+1 } ∪ aff {cm+1 , . . . , cn+1 }) + cm+1 = aff {c1 , . . . , cn+1 } + cm+1 = Rn . Since dim L1 = m and dim L2 = n − m (see Theorem 1.60), one has dim (L1 + L2 ) = n = dim L1 + dim L2 . By Theorem 1.10, the planes L1 and L2 are complementary. Dimension of Sets Definition 1.75. The dimension of a set X ⊂ Rn , denoted dim X, is defined by dim X = dim (aff X). In particular, dim ∅ = −1. Theorems 1.59, 1.60, and 1.68 imply one more corollary. Corollary 1.76. The dimension of a set X ⊂ Rn equals the maximum number of affinely independent points in X minus 1. If m = dim X, then the following statements hold. (1) X contains an affinely independent subset Y of m + 1 points such that aff Y = aff X. (2) Every affinely independent set of m + 1 points from X is an affine basis for aff X. (3) aff X is the set of all affine combinations of m + 1 or fewer affinely independent points from X. Corollary 1.77. If L ⊂ Rn is a plane of dimension m and X is subset of L, then aff X = L if and only if dim X = m. Proof. If aff X = L, then dim X = m according to the definition. Conversely, let dim X = m. The inclusion X ⊂ L shows that aff X ⊂ L (see Theorem 1.50). Since the planes aff X and L have the same dimension, Theorem 1.6 yields the equality aff X = L. Corollary 1.78. If X ⊂ Rn is a set of dimension m, −1 6 m 6 n − 1, and c is a point in Rn \ aff X, then dim ({c} ∪ X) = m + 1.
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Proof. Let L = aff X. Then dim L = m, and, according to Theorem 1.50, aff ({c} ∪ X) = aff ({c} ∪ aff X) = aff ({c} ∪ L). Because dim (aff ({c} ∪ L)) = m + 1 (see Corollary 1.67), we conclude that dim ({c} ∪ X) = m + 1. Theorem 1.79. If L ⊂ Rn is a nonempty m-dimensional plane and Bρ (c) ⊂ Rn is the closed ball of radius ρ > 0 centered at a point c ∈ L, then aff (Bρ (c) ∩ L) = L
and
dim (Bρ (c) ∩ L) = m.
Proof. Since the case m = 0 is trivial, we assume that m > 1. Consider the subspace S = L − c and the ball Bρ (o) = Bρ (c) − c. Because dim S = m, one can choose in Bρ (o) ∩ S a basis b1 , . . . , bm for S. Put ci = bi + c, 1 6 i 6 m. By Corollary 1.66, the set {c, c1 , . . . , cm } is an affine basis for L. Furthermore, ci ∈ Bρ (c) for all 1 6 i 6 m because of kci − ck = kbi k 6 ρ. From the inclusions {c, c1 , . . . , cm } ⊂ Bρ (c) ∩ L ⊂ L = aff {c, c1 , . . . , cm } and Theorem 1.50 it follows that aff (Bρ (c) ∩ L) = L. Finally, Definition 1.75 shows that dim (Bρ (c) ∩ L) = dim L = m. Remark. A section of a ball Bρ (c) by an m-dimensional plane L ⊂ Rn through c is often called an m-dimensional ball (see Exercise 1.3 for a characterization of planes in terms of m-balls). Corollary 1.80. If L ⊂ Rn is a plane of positive dimension m and F is a halfplane of L (respectively, G is a slab of L), then aff F = aff G = L. Consequently, dim F = dim G = m. Proof. Suppose that F is a closed halfplane of L (the case of an open halfplane is similar). By the definition, F = L ∩ V , where V is a closed halfspace of Rn such that ∅ 6= L ∩ V 6= L. According to Corollary 1.37, the set L ∩ int V is nonempty. Choose a point c ∈ L ∩ int V . Since the set int V is open, there is a scalar ρ > 0 such that the ball Bρ (c) lies in V . Therefore, Bρ (c) ∩ L ⊂ V ∩ L = F . A combination of Theorems 1.50 and 1.79 shows that L = aff (Bρ (c) ∩ L) ⊂ aff F ⊂ L. Hence aff F = L, which gives dim F = m. The proof for G is similar.
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Affine Transformations
Definition and Basic Properties Definition 1.81. A mapping f : Rn → Rm is called an affine transformation provided it can be expressed as f (x) = a + g(x),
x ∈ Rn ,
where a is a point in Rm and g : Rn → Rm is a linear transformation. An affine functional on Rn is an affine transformation f : Rn → R. Theorem 1.82. The following statements take place. (1) Any affine transformation f : Rn → Rm is uniquely expressible as f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. (2) If an affine transformation f : Rn → Rn is invertible, then its inverse also is an affine transformation. (3) If fi : Rn → Rm , i = 1, 2, are affine transformations, then every linear combination µ1 f1 + µ2 f2 is an affine transformation. (4) If f1 : Rn → Rm and f2 : Rm → Rk are affine transformations, then their composition f2 ◦ f1 : Rn → Rk is an affine transformation. Proof. (1) Let f (x) = a1 + g1 (x) = a2 + g2 (x), where a1 , a2 ∈ Rm and both g1 and g2 are linear transformations. Then a1 = f (o) = a2 , and g1 (x) = f (x) − a1 = f (x) − a2 = g2 (x)
for all
x ∈ Rn .
(2) Let f (x) = a + g(x), where a ∈ Rn and g : Rn → Rn is a linear transformation. Clearly, g is invertible, and from linear algebra we know that g −1 is a linear transformation. Consider the affine transformation h(x) = −g −1 (a) + g −1 (x). The identities f (h(x)) = a + g(−g −1 (a) + g −1 (x)) = x, h(f (x)) = −g −1 (a) + g −1 (a + g(x)) = x show that h = f −1 . (3) If fi (x) = ai + gi (x), where ai ∈ Rn and gi : Rn → Rm is a linear transformation, i = 1, 2, then (µ1 f1 + µ2 f2 )(x) = (µ1 a1 + µ2 a2 ) + (µ1 g1 + µ2 g2 )(x). Since µ1 g1 + µ2 g2 is a linear transformation, µ1 f1 + µ2 f2 is an affine one.
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(4) Let f1 and f2 be expressed as f1 (x) = a1 + g1 (x)
and f2 (x) = a2 + g2 (x),
where a1 ∈ Rm , a2 ∈ Rk and g1 and g2 are respective linear transformations. With a = a2 + g2 (a1 ), we have (f2 ◦ f1 )(x) = f2 (f1 (x)) = a2 + g2 (a1 + g1 (x)) = a2 + g2 (a1 ) + g2 (g1 (x)) = a + (g2 ◦ g1 )(x). Since g2 ◦ g1 is a linear transformation, f2 ◦ f1 is an affine one. Theorem 1.83. For an affine transformation f : Rn → Rm , the following statements hold. (1) If {c1 , . . . , cr } is an affinely independent set in Rn and f is oneto-one, then the set {f (c1 ), . . . , f (cr )} is affinely independent. (2) If {a1 , . . . , ar } is an affinely independent set in Rm and points c1 , . . . , cr ∈ Rn satisfy the conditions f (ci ) = ai , 1 6 i 6 r, then the set {c1 , . . . , cr } is affinely independent. Proof. Let f (x) = a + g(x), where a ∈ Rm , and g : Rn → Rm is a linear transformation. Since both statements (1) and (2) are obvious when r = 1, we may assume that r > 2. (1) By Theorem 1.60, the set {c2 − c1 , . . . , cr − c1 } is linearly independent. Because g is one-to-one, the set {g(c2 − c1 ), . . . , g(cr − c1 )} also is linearly independent (see page 5). The equalities f (ci ) − f (c1 ) = g(ci ) − g(c1 ) = g(ci − c1 ),
2 6 i 6 r,
and Theorem 1.60 show that {f (c1 ), . . . , f (cr )} is affinely independent. (2) As above, the set {a2 − a1 , . . . , ar − a1 } is linearly independent, and the equalities ci − c1 = g(ai − a1 ),
2 6 i 6 r,
show that the set {c2 −c1 , . . . , cr −c1 } also is linearly independent. Finally, Theorem 1.60 implies that {c1 , . . . , cr } is affinely independent. Theorem 1.84. If f : Rn → Rm is an affine transformation and X is a set in Rn , then dim X > dim f (X), with dim X = dim f (X) provided f is one-to-one.
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Proof. Let k = dim f (X). Since the case X = ∅ is obvious, we assume that k > 0. Choose in f (X) an affinely independent set {a1 , . . . , ak+1 } (see Corollary 1.76). If c1 , . . . , ck+1 are points in X satisfying the conditions f (ci ) = ai , 1 6 i 6 k + 1, then Theorem 1.83 shows that the set {c1 , . . . , ck+1 } is affinely independent. By the same corollary, dim X > k = dim f (X). Assume now that f is one-to-one, and let m = dim X. As above, X contains m + 1 affinely independent points c1 , . . . , cm+1 . By Theorem 1.83, the set {f (c1 ), . . . , f (cm+1 )} is affinely independent. Corollary 1.76 implies that dim f (X) > m = dim X, and the above argument gives dim f (X) = dim X. Theorem 1.85. For an affinely independent set {c1 , . . . , cr } ⊂ Rn and points a1 , . . . , ar ∈ Rm , there is an affine transformation f : Rn → Rm such that f (ci ) = ai , 1 6 i 6 r. If, additionally, r = n + 1, then f is uniquely determined. Furthermore, if r = n+1 = m+1 and {a1 , . . . , an+1 } is an affine basis for Rn , then f is invertible. Proof. Since the case r = 1 is obvious (let, for instance, f (x) = a1 + o(x), where o : Rn → Rm is the zero linear transformation), we may assume that r > 2. By Theorem 1.60, the set {c2 − c1 , . . . , cr − c1 } is linearly independent. Choose a linear transformation g : Rn → Rm (see page 5) such that g(ci − c1 ) = ai − a1 ,
2 6 i 6 r.
Put f (x) = a1 − g(c1 ) + g(x). Clearly, f is an affine transformation, and f (ci ) = a1 + g(ci − c1 ) = a1 + (ai − a1 ) = ai ,
1 6 i 6 r.
The transformation g is uniquely determined if r = n+1. Consequently, f is uniquely determined if r = n + 1. Finally, if r = n + 1 = m + 1 and {a1 , . . . , an+1 } is an affine basis for Rn , then the set {a2 −a1 , . . . , an+1 −a1 } is a basis for Rn (see Corollary 1.66). In this case, g is invertible, and so if f. Affine Transformations and Affine Combinations Theorem 1.86. For a mapping f : Rn → Rm , the following conditions are equivalent. (1) f is an affine transformation.
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(2) f ((1 − λ)x + λy) = (1 − λ)f (x) + λf (y) whenever x, y ∈ Rn and λ ∈ R. (3) f (λx + µy) = (1 − λ − µ)f (o) + λf (x) + µf (y) whenever x, y ∈ Rn and λ, µ ∈ R. Proof. (1) ⇒ (3). Let f be an affine transformation, expressed as f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. Then f (o) = a. For points x, y ∈ Rn and scalars λ, µ, one has f (λx + µy) = a + g(λx + µy) = (1 − λ − µ)a + λ(a + g(x)) + µ(a + g(y)) = (1 − λ − µ)f (o) + λf (x) + µf (y). The statement (3) ⇒ (2) is obvious; so, it remains to show that (2) implies (1). Assuming (2), we state that the mapping g(x) = f (x) − f (o) is a linear transformation. Indeed, for every x ∈ Rn and λ ∈ R, g(λx) = f ((1 − λ)o + λx) − f (o) = (1 − λ)f (o) + λf (x) − f (o) = λ(f (x) − f (o)) = λg(x). Using the above property for λ = 2, we obtain g(x + y) = 2g( 21 x + 21 y) = 2f ((1 − 12 )x + 12 y) − 2f (o) = 2((1 − 21 )f (x) + 12 f (y)) − 2f (o) = (f (x) − f (o)) + (f (y) − f (o)) = g(x) + g(y). Hence g(x) is a linear transformation, and f (x) = f (o) + g(x) is an affine one. Theorem 1.87. A mapping f : Rn → Rm is an affine transformation if and only if it preserves affine combinations of points: f (λ1 x1 + · · · + λk xk ) = λ1 f (x1 ) + · · · + λk f (xk ) for all k > 1, x1 , . . . , xk ∈ Rn , and λ1 + · · · + λk = 1.
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Proof. If a mapping f : Rn → Rm preserves affine combinations of any two points, then f is an affine transformation (see Theorem 1.86). Conversely, let f : Rn → Rm be an affine transformation. By induction on k, we will show that f preserves affine combinations of any k points from Rn . The case k = 1 is trivial, and the case k = 2 is proved in Theorem 1.86. Assuming that the induction hypothesis holds for all positive integers k 6 r − 1 (r > 2), choose a affine combination λ1 x1 + · · · + λr xr ∈ Rn . At least one of λ1 , . . . , λr is distinct from 1, since otherwise λ1 + · · · + λr = r > 2. Let, for example, λ1 6= 1. Consider the affine combination λr λ2 z= x2 + · · · + xr . 1 − λ1 1 − λ1 By the induction hypothesis, λ2 λr f (z) = f (x2 ) + · · · + f (xr ). 1 − λ1 1 − λ1 Hence f (λ1 x1 + · · · + λr xr ) = f (λ1 x1 + (1 − λ1 )z) = λ1 f (x1 ) + (1 − λ1 )f (z) = λ1 f (x1 ) + · · · + λr f (xr ). Theorem 1.88. For an affine transformation f : Rn → Rm and sets X ⊂ Rn and Y ⊂ Rm , one has aff f (X) = f (aff X), aff f
−1
(Y ) = f
−1
(aff (Y ∩ rng f )).
(1.39) (1.40)
Proof. Excluding the trivial cases X = ∅ and Y ∩ rng f = ∅, we assume that both sets X and Y ∩ rng f are nonempty. Consequently, f −1 (Y ) 6= ∅. For the equality (1.39), choose a point x ∈ aff f (X). By Theorem 1.51, x can be written as an affine combination x = λ1 x1 + · · · + λp xp ,
where
x1 , . . . , xp ∈ f (X).
Let z1 , . . . , zp be points in X such that f (zi ) = xi for all 1 6 i 6 p. Put z = λ1 z1 + · · · + λp zp . Then z ∈ aff X by the same theorem, and x = λ1 x1 + · · · + λp xp = λ1 f (z1 ) + · · · + λp f (zp ) = f (z) according to Theorem 1.87. Hence x = f (z) ∈ f (aff X), which proves the inclusion aff f (X) ⊂ f (aff X). Conversely, let x ∈ aff X. Similarly to the above, x can be expressed as an affine combination x = µ1 x1 + · · · + µq xq ,
where
x1 , . . . , xq ∈ X.
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Then f (x) = µ1 f (x1 ) + · · · + µq f (xq ) according to Theorem 1.87, and Theorem 1.51 gives f (x) ∈ aff f (X). Hence f (aff X) ⊂ aff f (X). In particular, rng f = f (Rn ) = f (aff Rn ) = aff f (Rn ) = aff (rng f ), which shows that rng f is a plane. It remains to prove (1.40). Letting X = f −1 (Y ) in (1.39), one has f (aff f −1 (Y )) = aff f (f −1 (Y )) = aff (Y ∩ rng f ). Hence aff f −1 (Y ) ⊂ f −1 (f (aff f −1 (Y ))) = f −1 (aff (Y ∩ rng f )). For the opposite inclusion, let x ∈ f −1 (aff (Y ∩ rng f )). Then f (x) belongs to aff (Y ∩ rng f ), and Theorem 1.51 implies that f (x) can be written as an affine combination f (x) = γ1 x1 + · · · + γr xr ,
where
x1 , . . . , xr ∈ Y ∩ rng f.
−1
Choose points z1 , . . . , zr ∈ f (Y ) such that f (zi ) = xi , 1 6 i 6 r, and put z = γ1 z1 + · · · + γr zr . Then z ∈ aff f −1 (Y ) and f (z) = γ1 f (z1 ) + · · · + γr f (zr ) = γ1 x1 + · · · + γr xr = f (x). Let ui = zi + (x − z), 1 6 i 6 r. Since ui is an affine combination, one has f (ui ) = f (zi ) + f (x) − f (z) = f (zi ) = xi . Hence ui ∈ f
−1
(xi ) ⊂ f −1 (Y ), 1 6 i 6 r. From the equalities x = ui − zi + z,
1 6 i 6 r,
it follows that x can be written as an affine combination of u1 , . . . , ur : x = (γ1 + · · · + γr )x = γ1 (u1 − z1 + z) + · · · + γr (ur − zr + z) = (γ1 u1 + · · · + γr ur ) − (γ1 z1 + · · · + γr zr ) + (γ1 + · · · + γr )z = γ1 u1 + · · · + γr ur . Therefore, x ∈ aff f −1 (Y ). Summing up, f −1 (aff (Y ∩ rng f )) ⊂ aff f −1 (Y ). Remark. The equality (1.40) implies the inclusion aff f −1 (Y ) ⊂ f −1 (aff Y ), which may be proper. Indeed, let f : R2 → R2 be the orthogonal projection on the x-axis of R2 and Y = {(0, 1), (0, −1)}. Then aff f −1 (Y ) = f −1 (Y ) = ∅, while aff Y is the y-axis and f −1 (aff Y ) = aff Y .
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Corollary 1.89. If f : Rn → Rm is an affine transformation and L ⊂ Rn and M ⊂ Rm are planes, then both sets f (L) and f −1 (M ) also are planes. In particular, the set rng f is a plane in Rm . Proof. By Theorem 1.88, f (L) = f (aff L) = aff f (L), which shows that f (L) is a plane. The set N = M ∩ rng f is a plane as the intersection of the planes M and rng f (see Theorem 1.4), and the same Theorem 1.88 gives f −1 (M ) = f −1 (N ) = f −1 (aff N ) = aff f −1 (N ) = aff f −1 (M ). Hence f −1 (M ) is a plane. Theorem 1.90. Show that for an affine transformation f : Rn → Rm and nonempty planes L1 and L2 in Rn , the following statements hold. (1) If L1 and L2 are parallel, then the planes f (L1 ) and f (L2 ) are parallel. (2) If L1 and L2 are independent and f is one-to-one, then the planes f (L1 ) and f (L2 ) are independent. (3) If L1 and L2 are complementary, n = m, and f is invertible, then the planes f (L1 ) and f (L2 ) are complementary. Proof. Let f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. (1) Let Li = ci + Si , where ci ∈ Rn and Si is a subspace of Rn , i = 1, 2. Then f (Li ) = a + g(ci ) + g(Si ), i = 1, 2. Without loss of generality, we may suppose that dim L1 6 dim L2 . From Theorem 1.6 it follows that namely L2 contains a translate of L1 , and Theorem 1.13 yields the inclusion S1 ⊂ S2 . Because g(S1 ) ⊂ g(S2 ), the same lemma implies that the planes f (L1 ) and f (L2 ) are parallel. (2) Suppose that L1 and L2 are independent, and f is one-to-one. From Theorem 1.86 it follows that f (L1 + L2 ) = {f (x1 + x2 ) : x1 ∈ L1 , x2 ∈ L2 } = {f (x1 ) + f (x2 ) − a : x1 ∈ L1 , x2 ∈ L2 } = f (L1 ) + f (L2 ) − a, and Theorem 1.6 gives dim (f (L1 ) + f (L2 )) = dim f (L1 + L2 ) = dim(L1 + L2 ). Since f is one-to-one, a combination of Theorems 1.10 and 1.84 gives dim (f (L1 ) + f (L2 )) = dim f (L1 + L2 ) = dim(L1 + L2 ) = dim L1 + dim L2 = dim f (L1 ) + dim f (L2 ).
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Theorem 1.10 shows that the planes f (L1 ) and f (L2 ) are independent. (3) Let L1 and L2 be complementary, n = m, and f be invertible. By statement (2) above, dim (f (L1 ) + f (L2 )) = dim f (L1 ) + dim f (L2 ) = dim L1 + dim L2 = n = m. Consequently, the planes f (L1 ) and f (L2 ) are complementary in Rm , as follows from Theorem 1.10. Homotheties and Projections Homotheties and projections make important classes of affine transformations used in the study of convex sets. Definition 1.91. Given a point a ∈ Rn and a scalar µ 6= 0, the affine transformation f (x) = a + µx,
x ∈ Rn ,
is called a homothety with ratio µ. Furthermore, f is a positive homothety if µ > 0 (respectively, a negative homothety if µ < 0). Among all homotheties, those with µ = 1 are the simplest ones. An affine transformation f (x) = a + x is called the translate on a vector a.
@
@ @
@ -@ a+x
@ x Fig. 1.13
The translate on a vector a.
A homothety f (x) = a + µx, µ 6= 1, can be written as f (x) = c + µ(x − c),
where
c = a/(1 − µ)
is the center of f . Furthermore, a homothety with center is called a contraction if 0 < |µ| < 1 (respectively, an expansion if |µ| > 1). Theorem 1.92. The following statements hold. (1) Every homothety f (x) = a + µx is invertible, and its inverse is a homothety: f −1 (x) = −µ−1 a + µ−1 x.
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(2) The composition of homotheties f (x) = a + µx and h(x) = c + γx again is a homothety: (f ◦ h)(x) = (a + µc) + µγ x. Proof. (1) With g(x) = −µ−1 a + µ−1 x, we have (f ◦ g)(x) = f (−µ−1 a + µ−1 x) = a + µ(−µ−1 a + µ−1 x) = x, (g ◦ f )(x) = g(a + µx) = −µ1 a + µ−1 (a + µx) = x. Hence f is invertible, and f −1 (x) = −µ−1 a + µ−1 x. (2) (f ◦ h)(x) = f (c + γx) = a + µ(c + γx) = (a + µc) + µγx, implying that (f ◦ h) is a homothety. @ @ @ `X @ @ X XX c XXX XXX XX x z c + µ(x − c) Fig. 1.14
A positive homothety with center c.
A characteristic property of homotheties is given in Exercise 1.8. Definition 1.93. Let L and N be complementary planes in Rn , expressed as translates of complementary subspaces S and T , respectively. Let g denote the linear projection on S along T (see page 5). Given a point c ∈ L, the affine transformation f (x) = c + g(x − c) is called the (affine) projection on L along N . If, additionally, T is the orthogonal complement of S, then f is called the orthogonal projection on L.
r x
N
L Fig. 1.15
r
? f (x)
The projection on a plane L along a complementary plane N .
Remark. The projection f in Definition 1.93 does not depend on the choice of c in L. Indeed, let f1 (x) = c1 + g(x − c1 ) for another point c1 ∈ L.
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Then c1 − c ∈ S (see Theorem 1.2). Therefore, g(c1 − c) = c1 − c because g is a linear projection on S. Consequently, c1 − g(c1 ) = c − g(c) due to the linearity of g. Hence f1 (x) = c1 + g(x − c1 ) = (c1 − g(c1 )) + g(x) = (c − g(c)) + g(x) = c + g(x − c) = f (x). Theorem 1.94. Let f be the projection on a plane L ⊂ Rn along a complementary plane N ⊂ Rn , and let c1 , . . . , cn+1 be an affine basis for Rn such that L = aff {c1 , . . . , cm+1 } and N = aff {cm+1 , . . . , cn+1 }, where m = dim L. If a point x ∈ Rn is written as an affine combination x = λ1 c1 + · · · + λn+1 cn+1 , then f (x) = λ1 c1 + · · · + λm cm + (λm+1 + · · · + λn+1 ) cm+1 . Proof. Since both cases m = 0 and m = n are obvious, we may assume that 1 6 m 6 n − 1. Put bi = ci − cm+1 , 1 6 i 6 m,
bi = ci+1 − cm+1 , m + 1 6 i 6 n.
The subspaces S = L − cm+1 and T = N − cm+1 are complementary, and b1 , . . . , bm and bm+1 , . . . , bn are, respectively, their bases (see Corollary 1.66). According to Definition 1.93 and the above remark, one can write f (x) = cm+1 + g(x − cm+1 ), where g is the linear projection on S along T . From the equality x − cm+1 = λ1 b1 + · · · + λm bm + λm+2 bm+1 + · · · + λn+1 bn , it follows that g(x − cm+1 ) = λ1 b1 + · · · + λm bm . Therefore, f (x) = g(x − cm+1 ) + cm+1 = λ1 b1 + · · · + λm bm + cm+1 = λ1 c1 + · · · + λm cm + (1 − λ1 − · · · − λm ) cm+1 = λ1 c1 + · · · + λm cm + (λm+1 + · · · + λn+1 ) cm+1 . Theorem 1.95. An affine transformation f : Rn → Rn is a projection if and only if f 2 = f . Proof. Let f be a projection; that is f (x) = c + g(x − c), where g : Rn → Rn is a linear projection and c ∈ Rn . Since g 2 = g (see Exercise 0.3), one has f 2 (x) = f (f (x)) = f (c + g(x − c)) = a + g(c + g(x − c) − c) = c + g(g(x − c)) = c + g(x − c) = f (x).
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Conversely, let f : Rn → Rn be an affine transformation with the property f 2 = f . Put e = f (o) and g(x) = f (x) − e. Then g(x) is a linear transformation such that g(e) = f (e) − e = f (f (o)) − f (o) = o. Furthermore, g 2 (x) = g(g(x)) = g(f (x) − e) = g(f (x)) = f (f (x)) − e = f (x) − e = g(x). Hence g is a linear projection (see Exercise 0.3), and f (x) = e + g(x) = e + g(x − e) is a projection. Theorem 1.96. Let L ⊂ Rn be a nonempty plane and f be the orthogonal projection of Rn on L. For any points u ∈ Rn and v ∈ L, one has ku − vk2 = ku − f (u)k2 + kf (u) − vk2 . Consequently, f (u) is the nearest to u point in L, and v = f (u) if and only if u − v ∈ (sub L)⊥ . Proof. Let S be the characteristic subspace of L (see Theorem 1.2). Then f is the projection on L along the subspace T = S ⊥ (see remark on page 65). By Theorem 1.11, the set S ∩ T is a singleton, say {c}. Theorem 1.2 shows that L = c + S. If p is the orthogonal projection on S, then p(c) = o, and f (x) = c + p(x − c) = c + p(x) for all x ∈ Rn . Let v = c + w, where w ∈ S. From linear algebra we know that k(u − c) − wk2 = k(u − c) − p(u − c)k2 + kp(u − c) − wk2 . Combining this equality with p(u − c) = p(u), we obtain ku − vk2 = k(u − c) − wk2 = k(u − c) − p(u)k2 + kp(u) − wk2 = ku − f (u)k2 + kf (u) − vk2 . Finally, since both vectors u − p(u) and c are orthogonal to S, their difference, u − f (u), also is orthogonal to S = L − L. Theorem 1.97. The orthogonal projection f of Rn on a hyperplane H = {x ∈ Rn : x·c = γ} ⊂ Rn is given by γ − u·c f (u) = u + c, u ∈ Rn . kck2 Furthermore, f (u) is the nearest to u point in H. Consequently, δ(u, H) = ku − f (u)k = |γ − u·c|/kck.
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Proof. Let S = {x ∈ Rn : x · c = 0} be the characteristic subspace of H (see Corollary 1.18). From linear algebra we know that the orthogonal projection on S is given by u·c p(u) = u − c kck2 (see page 6). Since the point a = (γ/kck2 )c belongs to H, Definition 1.93 shows that the orthogonal projection on H is expressed as γ γ c + p(u − c) f (u) = a + p(u − a) = 2 kck kck2 γ γ γ c + (u − c) − (u − c)·c/kck2 c = 2 2 2 kck kck kck γ − u·c c. =u+ kck2 From Theorem 1.96 it follows that f (u) is the nearest to u point in H. Therefore, γ − u·c |γ − u·c| . δ(u, H) = ku − f (u)k = kck = 2 kck kck Theorem 1.98. If L1 and L2 are nonempty disjoint planes in Rn , then δ(L1 , L2 ) > 0. Proof. Put Li = ci + Si , where Si is the characteristic subspace of Li , i = 1, 2. Let L02 = L2 − c1 and L = S1 + L02 . Then S1 ∩ L02 = ∅, since otherwise L1 ∩ L2 = (c1 + S1 ) ∩ (c1 + L02 ) = c1 + S1 ∩ L02 6= ∅, contrary to the assumption L1 ∩ L2 = ∅. We state that S1 ∩ L = ∅. Indeed, assume for a moment the existence of a point x ∈ S1 ∩ L. Then x = y + z, where y ∈ S1 and z ∈ L02 . Consequently, z = x − y ∈ S1 , which is impossible because of S1 ∩ L02 = ∅. Denote by a the orthogonal projection of o on L. Then a is the nearest to o point in L (see Theorem 1.96). We state that δ(S1 , L) = kak. Indeed, for any points u ∈ S1 and v ∈ L, one has ku − vk = ko − (v − u)k > kak due to v − u ∈ L − S1 = L. Finally, the inclusion L02 ⊂ L implies that δ(L1 , L2 ) = δ(L1 − c1 , L2 − c1 ) = δ(S1 , L02 ) > δ(S1 , L) = kak > 0. Remark. Theorem 1.98 can be sharpened by showing the existence of points u1 ∈ L1 and u2 ∈ L2 with the property ku1 − u2 k = δ(L1 , L2 ).
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Exercises for Chapter 1 Exercise 1.1. Show that a proper plane L ⊂ Rn is not a hyperplane if and only if its complement Rn \ L is connected. Exercise 1.2. Let F = {Lα } be a countable family of planes in Rn . Show that the set L = ∪ Lα is a plane if and only if one of the planes from F α contains all the others. Exercise 1.3. Show that a nonempty set X ⊂ Rn of dimension m is a plane if and only if there is a scalar ρ > 0 satisfying the condition: for every point c ∈ X, the set Bρ (c) ∩ X is an m-dimensional ball (see page 56). If, additionally, X is closed, then the value of ρ may depend on c ∈ X. Exercise 1.4. Let H ⊂ Rn be a hyperplane, a and b be points in Rn \ H, and h be a halfline with endpoint a, which meets H. Show that the halfline h0 = (b − a) + h meets H if and only if a and b belong to the same open halfspace determined by H. Exercise 1.5. Let F = {Fα } be a family of halfplanes of a nonempty plane L ⊂ Rn of positive dimension (each Fα may be closed or open). Show that the following statements hold. (1) If the union U = ∪ Fα is a proper subset of L, then U is a halfplane α of L if and only if F is nested. (2) If the intersection Z = ∩ Fα is nonempty, then Z is a halfplane of α L if and only if F is nested. Exercise 1.6. Let X and Y be nonempty sets in Rn such that a translate of Y lies in aff X. Show that for any point y ∈ Y , aff (X + Y ) = aff X + Y = aff X + y. Exercise 1.7. Let X1 , . . . , Xr ⊂ Rn be nonempty sets in Rn , µ1 , . . . , µr scalars, and ai ∈ Xi , 1 6 i 6 r. Let also r > m = dim (µ1 X1 + · · · + µr Xr ). Show the existence of an index set I ⊂ {1, . . . , r}, card I 6 m, satisfying the condition P P aff (µ1 X1 + · · · + µr Xr ) = µi ai + µi aff Xi . i∈I /
i∈I
Exercise 1.8. Show that a mapping f : Rn → Rn , n > 2, is a homothety if and only if it takes every line in Rn onto a parallel line.
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Exercise 1.9. Show that for points x, y, z ∈ Rn and a scalar λ ∈ [0, 1], k(1 − λ)x + λy − zk 6 (1 − λ)kx − zk + λky − zk. Furthermore, if x 6= y and 0 < λ < 1, then the equality k(1 − λ)x + λy − zk = (1 − λ)kx − zk + λky − zk holds if and only if z ∈ hx, yi such that either x ∈ [z, y] or y ∈ [x, z]. Notes for Chapter 1 The affine structure of Rn . A substantial part of Chapter 1 is scattered through various books on linear algebra and affine geometry and can be viewed as a body of “common knowledge.” Affine spans. Given a nonempty set X ⊂ Rn , let aff r X denote the collection of all affine combinations of r points from X: aff r X = {λ1 x1 + · · · + λi xi : x1 , . . . , xi ∈ X, λ1 + · · · + λr = 1}. In particular, aff 2 X is the union of points and lines through distinct points of X. From Corollary 1.76 it follows that aff m+1 X = aff X for every m-dimensional set X in Rn . It is easy to see that aff r (aff s X) ⊂ aff rs X, and, unlike convex hulls (see Theorem 3.10), this inclusion may be proper. For example, if X consists of four affinely independent points x, y, z, u in R3 , then aff 4 X \ aff 2 (aff 2 X) is the four-point set 1 {−x 2
+ y + z + u, x − y + z + u, x + y − z + u, x + y + z − u}.
Given a set X ⊂ Rn and integers r1 , . . . , rk > 2, Klee [129] studies properties of sets aff r1 (. . . (aff rk X) . . . ). Collineations. One of the fundamental theorems of affine geometry states that an invertible mapping f : Rn → Rn is an affine transformation if and only if it takes every three collinear points x, y, z ∈ Rn onto collinear points, f (x), f (y), f (z) (see, e. g., Veblen and Whitehead [220]). Lenz [149] showed that the statement still holds if “invertible” is replaced with “one-to-one”, while Chubarev and Pinelis [57] succeeded to replace the condition “invertible” with that of “onto.” Frenkel [88, p. 91] (see also Aboubabdullah [2, Article 5]) proved that a mapping f : Rn → Rm , where n > 2 and m > 2, is an affine transformation provided its range is at least 2-dimensional and f (aff {x, z}) = aff {f (x), f (z)} for all x, z ∈ Rn .
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Algebraic Properties of Convex Sets
Definition and Basic Properties Definition 2.1. A nonempty set X ⊂ Rn is called convex provided it contains all points (1 − λ)x + λy whenever x, y ∈ X and λ ∈ [0, 1]. The empty set ∅ is assumed to be convex. r Q Q Q
r r Fig. 2.1
Q
Qr
Convex and nonconvex sets.
We can reformulate Definition 2.1, saying that a nonempty set X in Rn is convex if and only if it contains all segments [x, y] with endpoints x, y ∈ X. The condition λ ∈ [0, 1] in Definition 2.1 can be weakened provided the set X is closed or open (see Exercises 2.2 and 2.3). Example. Every singleton in Rn is a convex set. Convex subsets of a line l ⊂ Rn are the empty set, all singletons, segments and halflines of all kinds (see Definitions 1.24 and 1.25), and the whole line l. Example. Every plane L in Rn is a convex set. Indeed, the case L = ∅ is trivial. If L 6= ∅ and x and y are points in L, then [x, y] ⊂ L according to Theorem 1.46. 71
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Example. Every halfplane of a plane L ⊂ Rn of positive dimension is a convex set. Indeed, if D is a closed halfplane of L, then, by Corollary 1.38, it can be expressed as D = {x ∈ L : x·c 6 γ},
c∈ / (sub L)⊥ ,
γ ∈ R.
Choose points x, y ∈ D and a scalar λ ∈ [0, 1]. Then (1 − λ)x + λy ∈ L because L is convex, and the inequality ((1 − λ)x+ λ y)·c = (1 − λ)x·c + λy·c 6 (1 − λ)γ + λγ = γ shows that (1 − λ)x + λy ∈ D. Similarly, every open halfplane of L is a convex set. Example. Every slab of a plane L ⊂ Rn of positive dimension is a convex set. Let, for instance, F be a closed slab of L. According to Corollary 1.44, F can be described as F = {x ∈ L : γ 6 x·c 6 γ 0 },
c∈ / (sub L)⊥ ,
γ < γ0.
If x, y ∈ F and λ ∈ [0, 1], then (1 − λ)x + λy ∈ L because L is convex, and the inequalities γ 6 ((1 − λ)x+ λ y)·c 6 γ 0 show that (1 − λ)x + λy ∈ F . In a similar way, every open slab of L is a convex set. Example. Every closed ball Bρ (c) ⊂ Rn is a convex set. Indeed, for points x, y ∈ Bρ (c) and a scalar λ ∈ [0, 1], one has k((1 − λ)x + λy) − ck = k(1 − λ)(x − c) + λ(y − c)k 6 (1 − λ)kx − ck + λky − ck 6 (1 − λ)ρ + λρ = ρ. Hence (1 − λ)x + λy ∈ Bρ (c), implying the convexity of Bρ (c). Definition 2.2. A convex combination of points x1 , . . . , xr in Rn is a linear combination λ1 x1 + · · · + λr xr , where λ1 , . . . , λr > 0 and λ1 + · · · + λr = 1. If, additionally, all scalars λ1 , . . . , λr are positive, then λ1 x1 + · · · + λr xr is called a positive convex combination. For example, all convex combinations of distinct points x and y in R fulfil the segment [x, y] (see Definition 1.25), while all positive convex combinations of x and y fulfill the open segment (x, y). n
Theorem 2.3. A nonempty set X ⊂ Rn is convex if and only if it contains all convex combinations (equivalently, all positive convex combinations) of points from X.
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Proof. If X contains all positive convex combinations of its points, then, in particular, (1−λ)x+λy ∈ X whenever x, y ∈ X and 0 < λ < 1. Because [x, y] = {x} ∪ (x, y) ∪ {y}, the segment [x, y] lies in X for any choice of x, y ∈ X. Hence X is convex. Conversely, let X be convex. By induction on k, we are going to prove that X contains all convex combinations of k points from X. The case k = 1 is trivial (every point x ∈ X can be written as 1x), and the case k = 2 holds by the definition. Assuming that the statement is true for all positive integers k 6 r−1 (r > 2), choose a convex combination x = λ1 x1 + · · · + λr xr of points x1 , . . . , xr ∈ X. If λ1 = 1, then λ2 = · · · = λr = 0, and x = 1x1 ∈ X. Suppose that λ1 < 1. By the induction hypothesis, the convex combination λr λ2 x2 + · · · + xr y= 1 − λ1 1 − λ1 belongs to X. Since x can be written as x = λ1 x1 +(1−λ1 )y, the inclusion x ∈ X follows from the convexity of X. See Theorem 2.42 for an extension of Theorem 2.3 to the case of converging convex series. Theorem 2.4. A set X ⊂ Rn is convex if and only if µ1 X + · · · + µr X = (µ1 + · · · + µr )X for every choice of r > 2 and scalars µ1 , . . . , µr > 0.
(2.1)
Proof. Let X be convex. The inclusion (µ1 + · · · + µr )X ⊂ µ1 X + · · · + µr X holds for every set X ⊂ Rn and every choice of scalars µ1 , . . . , µr . Hence it remains to prove that µ1 X + · · · + µr X ⊂ (µ1 + · · · + µr )X provided µ1 , . . . , µr > 0. Since the statement is trivial when X = ∅, we may assume that X is nonempty. Excluding one more trivial case µ1 = · · · = µr = 0, we suppose that µ = µ1 + · · · + µr > 0. Put λi = µi /µ, 1 6 i 6 r. Then λ1 , . . . , λr > 0 and λ1 + · · · + λr = 1. According to Theorem 2.3, λ1 X + · · · + λr X = {λ1 x1 + · · · + λr xr : x1 , . . . , xr ∈ X} ⊂ X. Multiplying both sides of the latter inclusion by µ, we obtain µ1 X + · · · + µr X ⊂ (µ1 + · · · + µr )X. Conversely, let a nonempty set X ⊂ Rn satisfy the condition (2.1) for every choice of r > 2 and scalars µ1 , . . . , µr > 0. If x, y ∈ X and λ ∈ [0, 1], then (1 − λ)x + λy ∈ (1 − λ)X + λX = X, implying the convexity of X.
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Simplices Definition 2.5. Let {x1 , . . . , xr+1 }, r > 0, be an affinely independent set in Rn . The r-simplex ∆ = ∆(x1 , . . . , xr+1 ) with vertices x1 , . . . , xr+1 is defined as ∆ = {λ1 x1 + · · · + λr+1 xr+1 : λ1 , . . . , λr+1 > 0, λ1 + · · · + λr+1 = 1}. Equivalently, ∆ is the set of all convex combinations of x1 , . . . , xr+1 .
r x1
r x1
r x2
r x3 A A A A r Ar x1 x2
Fig. 2.2
r x3 C
J
CJ C J
C J
rX
Jr x1 XXXC r x2 x4
Simplices in R3 .
Theorem 2.6. Every r-simplex ∆(x1 , . . . , xr+1 ) ⊂ Rn is a convex set of dimension r, and aff ∆ = aff {x1 , . . . , xr+1 }. Proof. If x and y are points in ∆ = ∆(x1 , . . . , xr+1 ), then x = γ1 x1 + · · · + γr+1 xr+1 ,
γ1 + · · · + γr+1 = 1,
y = µ1 x1 + · · · + µr+1 xr+1 ,
µ1 + · · · + µr+1 = 1,
for certain nonnegative scalars γ1 , . . . , γr+1 and µ1 , . . . , µr+1 . Choose a scalar λ ∈ [0, 1], and let αi = (1 − λ)γi + λµi , 1 6 i 6 r + 1. Clearly, α1 , . . . , αr+1 > 0, and the equalities (1 − λ)x + λy = α1 x1 + · · · + αr+1 xr+1 ,
α1 + · · · + αr+1 = 1
show that (1 − λ)x + λy ∈ ∆. Hence ∆ is a convex set. Since convex combinations are particular cases of affine combinations, Corollary 1.52 implies the inclusions {x1 , . . . , xr+1 } ⊂ ∆ ⊂ aff {x1 , . . . , xr+1 }, and Theorem 1.50 gives aff ∆ = aff {x1 , . . . , xr+1 }. Furthermore, from Definition 1.75 and Theorem 1.60 one has dim ∆ = dim (aff ∆) = dim (aff {x1 , . . . , xr+1 }) = r.
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Combining Theorems 2.3 and 2.6 and Corollary 1.76, we obtain one more corollary. Corollary 2.7. For a nonempty convex set K ⊂ Rn , the following statements hold. (1) If {x1 , . . . , xr+1 } is an affinely independent subset of K, then the r-simplex ∆(x1 , . . . , xr+1 ) lies in K. (2) The dimension of K equals the maximum dimension of a simplex contained in K. (3) aff K = aff ∆ for every simplex ∆ ⊂ K whose dimension equals dim K. Algebra of Convex Sets Theorem 2.8. If F = {Kα } is a family of convex sets in Rn , then the following statements hold. (1) The intersection M = ∩ Kα is a convex set. α
(2) If the family F is nested, then the union N = ∪ Kα is a convex α set. Proof. (1) Since the statement is obvious when M is empty, we may suppose that M 6= ∅. If x, y ∈ M , then x, y ∈ Kα for every Kα ∈ F, which gives [x, y] ⊂ Kα by the convexity of Kα . Hence [x, y] ⊂ ∩ Kα = M , and α M is convex. (2) Excluding the trivial case N = ∅, we assume that N is nonempty. Let x and y be points in N . Then x ∈ Kγ and y ∈ Kµ for certain sets Kγ , Kµ ∈ F. By the hypothesis, one of the sets Kγ , Kµ contains the other. Let, for example, Kγ ⊂ Kµ . In this case, x, y ∈ Kµ , and [x, y] ⊂ Kµ by the convexity of Kµ . Thus [x, y] ⊂ Kµ ⊂ ∪ Kα = N , and N is a convex α set. Theorem 2.9. If K1 , . . . , Kr are convex sets in Rn and µ1 , . . . , µr are scalars, then the sum µ1 K1 + · · · + µr Kr is a convex set. Proof. Without loss of generality, we may suppose that all sets K1 , . . . , Kr are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 (if r = 1, then we can write µ1 K1 = µ1 K1 + µ2 o). Choose points x and y in µ1 K1 + µ2 K2 and a scalar λ ∈ [0, 1]. Then x = µ1 x1 + µ2 x2 and y = µ1 y1 + µ2 y2 for certain points x1 , y1 ∈ K1 and
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x2 , y2 ∈ K2 . By a convexity argument, (1 − λ)xi + λyi ∈ Ki , i = 1, 2. Finally, the inclusion (1 − λ)x + λy = µ1 ((1 − λ)x1 + λy1 ) + µ2 ((1 − λ)x2 + λy2 ) ∈ µ1 K1 + µ2 K2 shows that µ1 K1 + µ2 K2 is a convex set. Theorem 2.10. For a convex set K ⊂ Rn and a scalar µ ∈ (0, 1), there is a set X ⊂ Rn satisfying the condition µK = ∩ (z + K : z ∈ X). Proof. Since the case K = ∅ is obvious (let X = ∅), we assume that K is nonempty. It suffices to show that for every u ∈ Rn \ µK there is a point z ∈ Rn such that µK ⊂ z + K and u ∈ / z + K. The union of all such points z will form the desired set X. We consider apart the following cases. 1. Suppose that u+µ(x−u) ∈ µK for all x ∈ µK. Put z = (1−µ−1 )u. Then the implications x ∈ µK ⇒ u + µ(x − u) ∈ µK ⇒ x ∈ (1 − µ−1 )u + K = z + K, u∈ / µK ⇒ µu ∈ / (µ − 1)u + µK ⇒ u ∈ / (1 − µ−1 )u + K = z + K prove the statement. 2. Assume the existence of a point v ∈ µK such that u+µ(v−u) ∈ / µK. Since u + µ(v − u) ∈ [u, v], the convexity of µK shows that u + η(v − u) ∈ / µK for all η ∈ (0, µ). Hence there is a positive integer r such that u + (1 − µ)r (v − u) ∈ / µK. Assuming that r is the smallest integer with respect to the above property, one has w = u + (1 − µ)r−1 (v − u) ∈ µK. Let z = (1 − µ−1 )w. If x ∈ µK, then (1 − µ)w + µx ∈ µK, implying x ∈ µ−1 ((µ − 1)w + µK) = (1 − µ−1 )w + K = z + K. On the other hand, from µu + (1 − µ)w = u + (1 − µ)(w − u) = u + (1 − µ)r (v − u) ∈ / µK it follows that u∈ / µ−1 ((µ − 1)w + µK) = (1 − µ−1 )w + K = z + K. Exercises 2.5 and 2.6 give additional results on algebraic properties of convex sets. The next theorem shows that affine spans of convex sets can be describes in terms of affine combinations of two points (compare with Theorem 1.51 and Corollary 1.76). Theorem 2.11. The affine span of a nonempty convex set K ⊂ Rn can be expressed as aff K = {(1 − λ)x + λy : x, y ∈ K, λ ∈ R}.
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Proof. Because aff K is a plane containing K, Theorem 1.46 shows that the set L = {(1 − λ)x + λy : x, y ∈ K, λ ∈ R} lies in aff K. For the opposite inclusion, choose a point x ∈ aff K. By Theorem 1.51, x is an affine combination, x = λ1 x1 + · · · + λr xr , of certain points x1 , . . . , xr ∈ K. Omitting zero terms of the form 0xi , we may assume that all scalars λ1 , . . . , λr are distinct from zero. Since λ1 + · · · + λr = 1, at least one of the scalars λ1 , . . . , λr is positive. We suppose that all positive scalars among λ1 , . . . , λr are given by λ1 , . . . , λp , where 1 6 p 6 r. If p = r, then x is a convex combination of x1 , . . . , xr , and x ∈ K ⊂ aff K according to Theorem 2.3. Assume that p 6 r − 1 and let µ = λ1 + · · · + λp . Clearly, µ > 1. Put ( λi /µ if 1 6 i 6 p, µi = λi /(1 − µ) if p + 1 6 i 6 r. Obviously, u = µ1 x1 + · · · + µp xp
and v = µp+1 xp+1 + · · · + µr xr
are convex combinations of points from K. Therefore, u, v ∈ K according to Theorem 2.3. Since x can be written as x = (1 − µ)v + µu, it follows that x ∈ L. Summing up, aff K ⊂ L. A sharper version of Theorem 2.11 is given in Theorem 2.30. Theorem 2.12. If f : Rn → Rm is an affine transformation and K ⊂ Rn and M ⊂ Rm are convex sets, then both sets f (K) and f −1 (M ) are convex. Proof. Excluding the trivial cases K = ∅ and M ∩ rng f = ∅, we assume that both sets K and M ∩rng f are nonempty. Consequently, f −1 (M ) 6= ∅. Choose points x, y ∈ f (K) and a scalar λ ∈ [0, 1]. Let x0 and y0 be points in K satisfying the conditions f (x0 ) = x and f (y0 ) = y. Then (1 − λ)x0 + λy0 ∈ K because K is convex. By Theorem 1.86, (1 − λ)x + λy = (1 − λ)f (x0 ) + λf (y0 ) = f ((1 − λ)x0 + λy0 ) ∈ f (K), which shows the convexity of f (K). Similarly, if x, y ∈ f −1 (M ) and λ ∈ [0, 1], then f (x), f (y) ∈ M , and (1 − λ)f (x) + λf (y) ∈ M because of the convexity of M . Theorem 1.86 results in f ((1 − λ)x + λy) = (1 − λ)f (x) + λf (y) ∈ M ∩ rng f.
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Hence (1 − λ)x + λy ∈ f −1 (M ∩ rng f ) = f −1 (M ), and f −1 (M ) is convex. See Exercise 2.12 and references on page 114 for a family of mappings which preserve convexity. 2.2
Relative Interior of Convex Sets
Definition and Basic Properties Definition 2.13. A point c ∈ Rn is called relatively interior for a set X ⊂ Rn provided there is a scalar ρ > 0 such that Bρ (c) ∩ aff X ⊂ X. The set of relatively interior points of X is called the relative interior of X and denoted rint X (we let rint ∅ = ∅). A set X ⊂ Rn is said to be relatively open if rint X = X. Example. The relative interior of a plane L ⊂ Rn is the plane itself; in particular, rint {c} = {c} for every point c ∈ Rn . Example. If L ⊂ Rn is a plane of positive dimension and D is a closed halfplane of L given by D = {x ∈ L : x·c 6 γ},
c∈ / (sub L)⊥ ,
γ∈R
(see Corollary 1.38), then rint D coincides with the open halfplane E = {x ∈ L : x·c < γ},
c∈ / (sub L)⊥ ,
γ ∈ R.
Indeed, if D is given by D = V ∩ L, where V = {x ∈ Rn : x·c 6 γ} is the closed halfspace, Corollary 2.27 below shows that rint D = int V ∩ L = E. Similarly, the relative interior of a closed slab of L, expressed by F = {x ∈ L : γ 6 x·c 6 γ 0 },
c∈ / (sub L)⊥ ,
γ < γ0,
is the respective open slab (see Corollary 1.44). Remark. The difference between the relative interior and topological interior in Rn can be illustrated by the following example. The relative interior of a segment [x, y] in R2 is the open segment (x, y), while the topological interior of [x, y] is empty (indeed, no ball Bρ (z) ⊂ R2 centered at a point z ∈ [x, y] can lie in [x, y]).
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'$ Bρ (z) r
x
r r z y &%
Theorem 2.14. Let X ⊂ Rn be a set with nonempty relative interior. For any point c ∈ rint X, there is a scalar δ > 0 such that Bδ (c) ∩ aff X ⊂ rint X. Furthermore, if 0 < ε 6 δ, then aff (Bε (c) ∩ rint X) = aff (rint X) = aff X.
(2.2)
Proof. Choose a scalar ρ > 0 satisfying the condition Bρ (c) ∩ aff X ⊂ X and put δ = ρ/2. Let x be a point in Bδ (c) ∩ aff X. We first observe that Bδ (x) ⊂ Bρ (c). Indeed, if y ∈ Bδ (x), then ky − ck 6 ky − xk + kx − ck 6 ρ/2 + ρ/2 = ρ, implying the inclusion y ∈ Bρ (c). Consequently, Bδ (x) ∩ aff X ⊂ Bρ (c) ∩ aff X ⊂ X. Hence x ∈ rint X, and Bδ (c) ∩ aff X ⊂ rint X. If 0 < ε 6 δ, then Bε (c) ∩ aff X ⊂ Bδ (c) ∩ aff X ⊂ rint X. Now, a combination of Theorems 1.50 and 1.79 gives aff X = aff (Bε (c) ∩ aff X) ⊂ aff (rint X) ⊂ aff X. Therefore, the equalities (2.2) hold. Theorem 2.15. For sets X and Y in Rn , the following statements hold. (1) (2) (3) (4) (5) (6) (7)
If int X 6= ∅, then dim X = n and int X = rint X. int X ⊂ rint X ⊂ X. If X ⊂ Y and aff X = aff Y , then rint X ⊂ rint Y . rint (rint X) = rint X. If rint X ⊂ Y ⊂ X, then rint X = rint Y . If a translate of Y lies in aff X, then rint X + Y ⊂ rint (X + Y ). rint (X + v) = rint X + v for every point v ∈ Rn .
Proof. (1) Choose a point u ∈ int X and a scalar ρ > 0 such that Bρ (u) ⊂ X. Then aff X = Rn because of Rn = aff Bρ (u) ⊂ aff X ⊂ Rn . Hence dim X = n. In the latter case, int X = rint X. (2) If int X = ∅, then the inclusion int X ⊂ rint X is obvious. If int X 6= ∅, then statement (2) follows from (1).
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To exclude trivial cases, we suppose below that both sets rint X and rint Y are nonempty. (3) Choose a point x ∈ rint X and a scalar ρ > 0 satisfying the condition Bρ (x) ∩ aff X ⊂ X. Since aff X = aff Y , we have Bρ (x) ∩ aff Y = Bρ (x) ∩ aff X ⊂ X ⊂ Y. Hence x ∈ rint Y . (4) Because of rint X ⊂ X and aff (rint X) = aff X (see Theorem 2.14), statement (3) implies that rint (rint X) ⊂ rint X. For the opposite inclusion, choose a point x ∈ rint X. By the same lemma, there is a scalar δ > 0 such that Bδ (x) ∩ aff X ⊂ rint X. Thus Bδ (x) ∩ aff (rint X) = Bδ (x) ∩ aff X ⊂ rint X. Therefore, x ∈ rint (rint X). (5) We first observe that aff X = aff (rint X) = aff Y . Indeed, a combination of Theorems 1.50 and 2.14 gives aff X = aff (rint X) ⊂ aff Y ⊂ aff X. Now, statements (3) and (4) imply that rint X = rint (rint X) ⊂ rint Y ⊂ rint X. Summing up, rint X = rint Y . (6) Let z ∈ rint X + Y . Then z = x + y, where x ∈ rint X and y ∈ Y . Choose a scalar ρ > 0 satisfying the condition Bρ (x) ∩ aff X ⊂ X. Because aff (X + Y ) = aff X + Y = aff X + y (see Exercise 1.6) and Bρ (z) = Bρ (x) + y, one has Bρ (z) ∩ aff (X + Y ) = (Bρ (x) + y) ∩ (aff X + y) = Bρ (x) ∩ aff X + y ⊂ X + y ⊂ X + Y. Therefore, z ∈ rint (X + Y ), and rint X + Y ⊂ rint (X + Y ). (7) By statement (6) above, it suffices to show that rint (X + v) ⊂ rint X + v. Let u ∈ rint (X + v). Then u ∈ X + v, whence we can write u = x + v for a certain point x ∈ X. Choose a scalar δ > 0 such that Bδ (u) ∩ aff (X + v) ⊂ X + v. Consequently, Bδ (x) ∩ aff X = (Bδ (u) − v) ∩ (aff X + v − v) = Bδ (u) ∩ aff (X + v) − v ⊂ (X + v) − v = X. Hence x ∈ rint X and u = x + v ∈ rint X + v.
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Remark. The converse of statement (1) in Theorem 2.15 does not hold. Indeed, if X consists of n + 1 affinely independent points, then dim X = n (see Corollary 1.76), while int X = rint X = ∅. Nevertheless, it becomes true provided X is convex (see Corollary 2.18). Furthermore, the condition aff X = aff Y is essential in statement (3). For example, if X is a triangle in R2 and Y one of its sides, then Y ⊂ X, while rint X and rint Y are disjoint. Finally, the inclusion rint X + Y ⊂ rint (X + Y ) in statement (6) may be proper. Indeed, let X = [0, 1] ∪ Q and Y = [0, 1], where Q is the set of all rational numbers in R. Then rint X = (0, 1) and rint X + Y = (0, 2) 6= R = X + Y = rint (X + Y ). Relative Interior and Simplices We will need the following auxiliary statement. Theorem 2.16. Let {c1 , . . . , cr } be an affinely independent set in Rn and x, x1 , x2 , . . . be an infinite sequence of points in aff {c1 , . . . , cr }: x = λ1 c1 + · · · + λr cr , λ1 + · · · + λr = 1, (i)
(i)
(i) xi = λ1 c1 + · · · + λ(i) r cr , λ1 + · · · + λr = 1, i > 1. (i)
Then lim xi = x if and only if lim λj = λj for all 1 6 j 6 r. i→∞
i→∞
Proof. Since the case r = 1 is obvious, we assume that r > 2. By Theorem 1.60, the vectors b2 = c2 −c1 , . . . , br = cr −c1 are linearly independent. Put y = x − c1 and yi = xi − c1 , i > 2. Then y = λ2 b2 + · · · + λr br
(i)
and yi = λ2 b2 + · · · + λ(i) r br , i > 2.
Since kx − xi k = ky − yi k, one has xi → x if and only if yi → y. Further(i) more, yi → y if and only if λj → λj for all 2 6 j 6 r (see Exercise 0.7). In the latter case, (i)
(i)
lim λ1 = lim (1 − λ2 − · · · − λr(i) ) = 1 − λ2 − · · · − λr = λ1 .
i→∞
i→∞
Theorem 2.17. The relative interior of an r-simplex ∆ = ∆(x1 , . . . , xr+1 ) in Rn is the set of all positive convex combinations of x1 , . . . , xr+1 : rint ∆ = {λ1 x1 + · · · + λr+1 xr+1 : λ1 , . . . , λr+1 > 0, λ1 + · · · + λr+1 = 1}.
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Proof. We state first that the set M = {λ1 x1 + · · · + λr+1 xr+1 : λ1 , . . . , λr+1 > 0, λ1 + · · · + λr+1 = 1} lies in rint ∆. For this, choose a point x = λ1 x1 + · · · + λr+1 xr+1 ∈ M and put ε = min {λ1 , . . . , λr+1 }. Theorem 2.16 yields the existence of a scalar ρ > 0 with the following property: for every point u ∈ Bρ (x) ∩ aff ∆, its affine coordinates µ1 , . . . , µr+1 in the expression u = µ1 x1 + · · · + µr+1 xr+1 satisfy the inequalities |λi − µi | 6 ε, 1 6 i 6 r + 1. Consequently, µi > λi − ε > 0 for all i = 1, . . . , r + 1, which gives the inclusion u ∈ ∆. Hence Bρ (x)∩ aff ∆ ⊂ ∆, and x ∈ rint ∆. Conversely, let x = λ1 x1 + · · · + λr+1 xr+1 ∈ rint ∆. Then all scalars λ1 , . . . , λr+1 are nonnegative. Assume for a moment that at least one of them, say λ1 , equals zero. Choose a scalar ρ > 0, and put −1 1 γ = ρ kx1 k + kx2 k + · · · + kxr+1 k r and γ γ x2 + · · · + λr+1 + xr+1 . v = −γx1 + λ2 + r r Clearly, v is an affine combination of x1 , . . . , xr+1 . Therefore, v ∈ aff {x1 , . . . , xr+1 } = aff ∆ according to Theorem 2.6. Furthermore, v ∈ Bρ (x) because of γ γ kv − xk 6 γkx1 k + kx2 k + · · · + kxr+1 k = ρ. r r On the other hand, v ∈ / ∆ since its first affine coordinate, −γ, is negative. Summing up, v ∈ (Bρ (x) ∩ aff ∆) \ ∆, and Definition 2.13 shows that x∈ / rint ∆, contrary to the assumption. Hence all scalars λ1 , . . . , λr+1 are positive, which gives x ∈ M . Corollary 2.18. If K ⊂ Rn is a nonempty convex set, then rint K 6= ∅. Furthermore, rint ∆ ⊂ rint K for every simplex ∆ ⊂ K whose dimension equals dim K. Proof. Let m = dim K. By Corollary 2.7, K contains an m-simplex ∆0 such that aff ∆0 = aff K. A combination of Theorems 2.15 and 2.17 implies that ∅ 6= rint ∆0 ⊂ rint K. Similarly, if ∆ is an m-simplex in K, then aff ∆ = aff K according to Corollary 2.7, and Theorem 2.15 gives the inclusion rint ∆ ⊂ rint K.
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Theorem 2.19. Let K ⊂ Rn be a nonempty convex set of dimension m. A point x ∈ Rn belongs to rint K if and only if there is an m-simplex ∆ ⊂ K with the property x ∈ rint ∆. Proof. If x belongs to the relative interior of an m-simplex ∆ ⊂ K, then x ∈ rint K according to Corollary 2.18. Conversely, let x ∈ rint K. Since the case m = 0 is obvious (then K is a singleton, {u}, and ∆ = ∆(u) is the desired 0-simplex), we assume that m > 1. Choose a scalar ρ > 0 satisfying the inclusion Bρ (x) ∩ aff K ⊂ K. Because of dim (aff K) = dim K = m, the set S = aff K − x is an mdimensional subspace (see Theorem 1.2). Let b1 , . . . , bm be a basis for S such that kbi k = ρ/m, 1 6 i 6 m. Put bm+1 = −(b1 + · · · + bm ). Then bm+1 ∈ S and kbm+1 k 6 m (ρ/m) = ρ. Consequently, {b1 , . . . , bm+1 } ⊂ x + S = aff K. Let xi = x + bi , 1 6 i 6 m + 1. From kxm+1 − xk 6 ρ and kxi − xk = kbi k = ρ/m, 1 6 i 6 m, it follows that {x1 , . . . , xm+1 } ⊂ Bρ (x) ∩ aff K ⊂ K. By Corollary 1.63, the set {b1 , . . . , bm+1 } is affinely independent. Hence, its translate {x1 , . . . , xm+1 } also is affinely independent, as follows from Theorem 1.83. By Corollary 2.7, the m-simplex ∆(x1 , . . . , xm+1 ) lies in K. Because of x= =
1 1 m+1 ((m + 1)x + o) = m+1 ((m + 1)x + b1 + · · · + bm+1 ) 1 1 1 m+1 ((x + b1 ) + · · · + (x + bm+1 )) = m+1 x1 + · · · + m+1 xm+1 ,
Theorem 2.17 shows that x belongs to rint ∆(x1 , . . . , xm+1 ). The next result refines Theorem 2.19 (see also Exercise 2.7). Theorem 2.20. If K ⊂ Rn is a convex set of positive dimension m and u is a point in K, then a point x ∈ K \ {u} belongs to rint K if and only if K contains an m-simplex ∆ with vertex u such that x ∈ rint ∆. Proof. We observe that the basis b1 , . . . , bm for S in the proof of Theorem 2.19 was chosen arbitrarily. We modify the argument of that proof by letting ρ . b1 = δ(u − x), where δ = mku − xk As above, kb1 k = ρ/m and x=
1 m+1 x1
+ ··· +
1 m+1 xm+1 ,
(2.3)
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where x1 = x + b1 = (1 − δ)x + δu. Using this value of x1 in (2.3), we obtain x=
1−δ m+1 x
+
δ m+1 u
+
1 m+1 x2
+ ··· +
1 m+1 xm+1 .
Solving the latter equation for x, we express it as a positive convex combination x=
δ m+δ u
+
1 m+δ x2
+ ··· +
1 m+δ xm+1 .
Finally, combining Corollary 2.7 and Theorem 2.17, one has ∆(u, x2 , . . . , xm+1 ) ⊂ K
and x ∈ rint ∆(u, x2 , . . . , xm+1 ).
Geometric Properties of the Relative Interior Theorem 2.21. If K ⊂ Rn is a nonempty convex set, x ∈ rint K, and y ∈ K, then (1 − λ)x + λy ∈ rint K for all 0 < λ < 1. Consequently, the open segment (x, y) lies in rint K. Proof. The case x = y is trivial, we assume that x 6= y. Let z = (1 − λ)x + λy for a certain scalar 0 < λ < 1. Choose a scalar ρ > 0 satisfying the condition Bρ (x) ∩ aff K ⊂ K. We state that B(1−λ)ρ (z) ∩ aff K ⊂ K (see the figure below). Bρ (x) '$ ``` B(1−λ)ρ (z) ` `` ``` r `r r x z y &% λ 1 u − 1−λ y Indeed, let u be a point in B(1−λ)ρ (z) ∩ aff K. Put v = 1−λ (so that u = (1 − λ)v + λy). Then v ∈ aff K as an affine combination of 1 λ the points u, y ∈ aff K (see Theorem 1.46). Since x = 1−λ z − 1−λ y, one has 1 1 kx − vk = ku − zk 6 (1 − λ)ρ = ρ. 1−λ 1−λ
Hence v ∈ Bρ (x)∩ aff K ⊂ K, and u = (1−λ)v +λy ∈ K by the convexity of K. Thus B(1−λ)ρ (z) ∩ aff K ⊂ K, which gives z ∈ rint K. Corollary 2.22. If K ⊂ Rn is a convex set, then its relative interior is a convex set.
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Proof. The statement is trivial when K = ∅. If K 6= ∅, then the convexity of rint K follows from Theorem 2.21. Theorem 2.23. Let K ⊂ Rn be a nonempty convex set, and c be a point in rint K. Then rint K = M1 ∪ M2 ∪ · · · , where Mi is the homothetic copy of K with center c given by Mi = c + (1 −
1 i+1 )(K
− c)
i > 1.
Proof. Choose a point y ∈ rint K. By Theorem 2.24, there is a scalar γ > 1 such that the point z = (1 − γ)c + γy belongs to K. Obviously, y = (1 − γ −1 )c + γ −1 z and 0 < γ −1 < 1. Choose an integer i > 1 such 1 1 1 and put u = i+1 c + (1 − i+1 )z. Then y ∈ (c, u) and that γ −1 6 1 − i+1 u = c + (1 −
1 i+1 )(z
− c) ∈ Mi .
Since the set Mi is convex (see Theorem 2.9), we have y ∈ (c, u) ⊂ Mi . This argument proves the inclusion rint K ⊂ M1 ∪ M2 ∪ · · · . Conversely, if y ∈ Mi , then y = c + (1 −
1 i+1 )(z
− c) =
1 i+1 c
+ (1 −
1 i+1 )z
for a certain point z ∈ K. Therefore, y ∈ rint K according to Theorem 2.21. Hence M1 ∪ M2 ∪ · · · ⊂ rint K. The next theorem gives an important algebraic description of relatively interior points of a convex set. Theorem 2.24. For a nonempty convex set K ⊂ Rn and a point x ∈ Rn , the following conditions are equivalent. (1) x ∈ rint K. (2) For any point y ∈ aff K, there is a scalar 0 < λ < 1 such that (1 − λ)x + λy ∈ K. (3) For any point y ∈ aff K, there is a scalar γ > 1 satisfying the inclusion γx + (1 − γ)y ∈ K. (4) For any point y ∈ K, there is a scalar γ > 1 with the property γx + (1 − γ)y ∈ K. (5) There is a point y ∈ rint K and a scalar γ > 1 satisfying the inclusion γx + (1 − γ)y ∈ K. Proof. (1) ⇒ (2). Choose a scalar ρ > 0 such that Bρ (x) ∩ aff K ⊂ K. Let y ∈ aff K. Excluding the trivial case x = y, we suppose that x 6= y. Let a scalar δ satisfy the inequalities 0 < δ < min {ρ, kx − yk}. The line
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l = hx, yi lies in aff K (see Theorem 1.46) and meets the sphere Sδ (x) at two distinct points, u and v, as depicted below. Hence u, v ∈ Bδ (x) ∩ aff K ⊂ Bρ (x) ∩ aff K ⊂ K. By Lemma 1.26, one of the points u, v, say v, belongs to the open halfline (x, yi. Then v = (1 − λ)x + λy for a certain scalar λ > 0. From λkx − yk = kx − vk = δ < kx − yk it follows that 0 < λ < 1. '$ Bδ (x) r r r r u x v y &% (2) ⇒ (3). Choose a point y ∈ aff K, and let y 0 = 2x − y. Then y 0 belongs to aff K as an affine combination of x, y ∈ aff K (see Theorem 1.46). By condition (2), there is a scalar 0 < λ < 1 such that (1 − λ)x + λy 0 ∈ K. Put γ = 1 + λ. Then γ > 1 and γx + (1 − γ)y = (1 + λ)x + (−λ)(2x − y 0 ) = (1 − λ)x + λy 0 ∈ K. Since (3) ⇒ (4) ⇒ (5), it remains to show that (5) ⇒ (1). Choose a point y ∈ rint K (this is possible since rint K 6= ∅ according to Corollary 2.18). Since the case x = y is trivial, we assume that x 6= y. By the assumption, there is a scalar γ > 1 such that the point z = γx + (1 − γ)y belongs to K. Because x can be expressed as x = (1−γ −1 )y +γ −1 z, where 0 < γ −1 < 1, one has x ∈ (y, z) ⊂ rint K according to Theorem 2.21. Theorem 2.24 implies the corollary below. Corollary 2.25. Let K ⊂ Rn be a convex set of positive dimension. A point x ∈ Rn belongs to rint K if and only if any of the following conditions is satisfied. (1) For any y ∈ aff K \ {x}, there is a point u ∈ (x, y) ∩ K. (2) For any y ∈ aff K \ {x}, there is a point u ∈ K \ {x} such that x ∈ (u, y). (3) For any y ∈ K \ {x}, there is a point z ∈ K \ {x} such that x ∈ (y, z). (4) There are points y ∈ rint K \ {x} and z ∈ K \ {x} such that x ∈ (y, z).
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Remark. The condition dim K > 0 in Corollary 2.25 has a purpose to exclude from consideration empty intervals (x, x). Also, the convexity of K is essential in Theorem 2.24 and Corollary 2.25. Indeed, if K is the union of two circles and a closed segment centered at x, as depicted below, then x satisfies conditions (1)–(4) of the corollary, but is not relatively interior for K. r '$ '$ rx &% r&% Relative Interior and Algebra of Convex Sets Theorem 2.26. Let F = {Kα } be a family of convex sets in Rn whose relative interiors have a point in common. Then rint (∩ Kα ) ⊂ ∩ rint Kα . α
α
If, additionally, the family F is finite, then rint (∩ Kα ) = ∩ rint Kα . α
α
Proof. Since ∅ 6= ∩ rint Kα ⊂ ∩ Kα , Corollary 2.18 shows that rint (∩ Kα ) α
α
α
is nonempty. Choose points x ∈ rint (∩ Kα ) and y ∈ ∩ rint Kα . Then α α y ∈ ∩ Kα , and, by Theorem 2.24, there is a scalar γ > 1 such that the α
point z = γx + (1 − γ)y belongs to ∩ Kα . Since 0 < γ −1 < 1, Theorem 2.21 α gives x = (1 − γ −1 )y + γ −1 z ∈ rint Kα
for all
Kα ∈ F.
Therefore, x ∈ ∩ rint Kα . α
Suppose now that F is finite, and let F = {K1 , . . . , Kr }. By the above proved, it remains to verify the inclusion rint K1 ∩ · · · ∩ rint Kr ⊂ rint (K1 ∩ · · · ∩ Kr ). For this, choose points x ∈ rint K1 ∩ · · · ∩ rint Kr
and y ∈ K1 ∩ · · · ∩ Kr .
By Theorem 2.24, there are scalars γi > 1 such that zi = γi x + (1 − γi )y ∈ Ki ,
1 6 i 6 r.
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Put γ = min {γ1 , . . . , γr }. Then γ > 1 and γx + (1 − γ)y ∈ [y, z1 ] ∩ · · · ∩ [y, zr ] ⊂ K1 ∩ · · · ∩ Kr by the convexity of K1 , . . . , Kr . Therefore, Theorem 2.24 shows that x belongs to rint (K1 ∩ · · · ∩ Kr ). Remark. The inclusion rint (∩ Kα ) ⊂ ∩ rint Kα in Theorem 2.26 may be α
α
proper if the family {Kα } is infinite. Indeed, consider the closed intervals Ki = [0, 1 + 1i ], i > 1, on the real line R. Then rint (∩ Ki ) = rint [0, 1] = (0, 1) 6= (0, 1] = ∩ (0, 1 + 1i ) = ∩ rint Ki . i
i
i
n
Since rint L = L for every plane L ⊂ R , Theorem 2.26 implies the following corollary. Corollary 2.27. If a plane L ⊂ Rn meets the relative interior of a convex set K ⊂ Rn , then rint (K ∩ L) = rint K ∩ L. Theorem 2.28. Let F = {Kα } be a nested family of convex sets in Rn . Then the set ∪ rint Kα is convex and α
rint (∪ Kα ) ⊂ ∪ rint Kα . α
(2.4)
α
If, additionally, aff Kγ = aff (∪ Kα ) for every Kγ ∈ F, then α
rint (∪ Kα ) = ∪ rint Kα . α
(2.5)
α
Proof. First, we are going to show that the set ∪ rint Kα is convex. Exα cluding the trivial case ∪ Kα = ∅, we suppose that ∪ Kα is nonempty. Then α α ∪ rint Kα 6= ∅ according to Corollary 2.18. Let x, y be points in ∪ rint Kα . α α Then x ∈ rint Kβ and y ∈ rint Kδ for certain sets Kβ and Kδ from F. Assuming, for instance, that Kβ ⊂ Kδ , we obtain x ∈ Kδ . Consequently, Theorem 2.21 gives (x, y) ⊂ rint Kδ . Thus [x, y] = {x} ∪ (x, y) ∪ {y} ⊂ rint Kβ ∪ rint Kδ ⊂ ∪ rint Kα , α
which shows the convexity of ∪ rint Kα . α
To prove the inclusion (2.4), let x ∈ rint (∪ Kα ). Since the family F α
is nested, the family {aff Kα } of planes also is nested (see Theorem 1.50). By a dimension argument, the family {aff Kα } contains at most finitely many distinct planes. Hence there is a set Kδ ∈ F such that aff Kδ = aff (∪ Kα ). Choose a point y ∈ rint Kδ . Because the set ∪Kα is convex α
α
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(see Theorem 2.8) and y ∈ ∪Kα , Theorem 2.24 shows the existence of a α
scalar γ > 1 such that the point z = γx + (1 − γ)y belongs to ∪ Kα . α Therefore, z ∈ Kβ for a certain Kβ ∈ F. If Kβ ⊂ Kδ , then z ∈ Kδ , and the equality x = (1 − γ −1 )y + γ −1 z,
0 < γ −1 < 1,
(2.6)
together with Theorem 2.21, implies x ∈ rint Kδ ⊂ ∪ rint Kα . α Assume that Kδ ⊂ Kβ . By the choice of Kδ , Theorem 1.50 gives aff Kδ = aff Kβ = aff (∪ Kα ). Hence y ∈ rint Kδ ⊂ rint Kβ because of Theα
orem 2.15. As above, (2.6) and Theorem 2.21 give x ∈ rint Kβ ⊂ ∪ rint Kα . α
Summing up, the inclusion (2.4) holds. Finally, if aff Kγ = aff (∪ Kα ) for every Kγ ∈ F, then Theorem 2.15 α
implies that rint Kγ ⊂ rint (∪ Kα ) for all Kγ ∈ F. Hence the opposite to α
(2.4) inclusion holds. Theorem 2.29. If K1 , . . . , Kr are convex sets in Rn and µ1 , . . . , µr are scalars, then rint (µ1 K1 + · · · + µr Kr ) = µ1 rint K1 + · · · + µr rint Kr . Proof. Without loss of generality, we may assume that all sets K1 , . . . , Kr are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 (for r = 1, we can write µ1 K1 = µ1 K1 + µ2 o). By Corollary 2.18, all three sets rint K1 , rint K2 , and rint (µ1 K1 + µ2 K2 ) are nonempty. Let x ∈ rint (µ1 K1 + µ2 K2 ). Choose points yi ∈ rint Ki , i = 1, 2, and put y = µ1 y1 + µ2 y2 . By Theorem 2.24, there is a scalar γ > 1 such that z = γx + (1 − γ)y ∈ µ1 K1 + µ2 K2 . Hence z can be written as z = µ1 z1 + µ2 z2 , where zi ∈ Ki , i = 1, 2. Put xi = (1 − γ −1 )yi + γ −1 zi ,
i = 1, 2.
Since 0 < γ −1 < 1, Theorem 2.21 gives xi ∈ rint Ki , i = 1, 2. Therefore, x = (1 − γ −1 )y + γ −1 z = µ1 x1 + µ2 x2 ∈ µ1 rint K1 + µ2 rint K2 . Conversely, choose points x ∈ µ1 rint K1 + µ2 rint K2
and y ∈ µ1 K1 + µ2 K2 .
Then x = µ1 x1 + µ2 x2 and y = µ1 y1 + µ2 y2 for certain points xi ∈ rint Ki and yi ∈ Ki , i = 1, 2. By Theorem 2.24, there are scalars γi > 1 such that ui = γi xi + (1 − γi )yi ∈ Ki , i = 1, 2.
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Let γ = min {γ1 , γ2 }. Then γ > 1 and γ/γi ∈ [0, 1], i = 1, 2. Furthermore, 1 1 vi = γxi + (1 − γ)yi = γ ui + (1 − )yi + (1 − γ)yi γi γi γ γ = ui + 1 − yi ∈ [ui , yi ] ⊂ Ki , i = 1, 2. γi γi Thus γx + (1 − γ)y = µ1 v1 + µ2 v2 ∈ µ1 K1 + µ2 K2 .
(2.7)
Finally, Theorem 2.24 and (2.7) give x ∈ rint (µ1 K1 + µ2 K2 ). A sharper version of Theorem 2.29 for the case of large sums is considered in Exercise 2.9. Relative Interior and the Affine Structure The following result can be viewed as a refinement of Theorem 2.11. Theorem 2.30. If K ⊂ Rn is a nonempty convex set and x ∈ rint K, then aff K = {(1 − λ)x + λy : y ∈ rint K, λ ∈ R}. Proof. Since aff K is a plane containing K, Theorem 1.46 implies that the set L = {(1 − λ)x + λy : y ∈ rint K, λ ∈ R} lies in aff K. Conversely, let u ∈ aff K. By Theorem 2.24, there is a scalar λ ∈ (0, 1) for which the point z = (1−λ)x+λu belongs to K. Theorem 2.21 shows that the point y = 12 (x + z) lies in rint K. Hence u = (1 − λ−1 )x + λ−1 z = (1 − λ−1 )x + λ−1 (2y − x) = (1 − 2λ−1 )x + 2λ−1 y ∈ L. Theorem 2.31. If K1 , . . . , Kr are convex sets in Rn whose relative interiors have a point in common, then aff (K1 ∩ · · · ∩ Kr ) = aff K1 ∩ · · · ∩ aff Kr . Proof. From Theorem 1.50 it follows that aff (K1 ∩ · · · ∩ Kr ) ⊂ aff K1 ∩ · · · ∩ aff Kr . For the opposite inclusion, choose points x ∈ aff K1 ∩ · · · ∩ aff Kr
and y ∈ rint K1 ∩ · · · ∩ rint Kr .
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By Theorem 2.24, there are scalars γi > 1 with the property zi = (1 − γi )x + γi y ∈ Ki ,
1 6 i 6 r.
Put γ = min {γ1 , . . . , γr } and u = (1 − γ)x + γy. Obviously, γ > 1 and, by a convexity argument, u ∈ [y, z1 ] ∩ · · · ∩ [y, zr ] ⊂ K1 ∩ · · · ∩ Kr . γ 1 y − γ−1 u of Since x can be expressed as an affine combination x = γ−1 points y and u from K1 ∩ · · · ∩ Kr , Theorem 1.51 implies the inclusion x ∈ aff (K1 ∩ · · · ∩ Kr ).
Theorem 2.32. If K1 and K2 are convex sets in Rn whose relative interiors meet, then dim (K1 ∪ K2 ) = dim (K1 + K2 ) = dim K1 + dim K2 − dim (K1 ∩ K2 ). Proof. Because aff (K1 ∪ K2 ) is a translate of aff (K1 + K2 ) (see Theorem 1.56), one has dim (K1 ∪ K2 ) = dim (aff (K1 ∪ K2 )) = dim (aff (K1 + K2 )) = dim (K1 + K2 ). By Theorems 1.53 and 2.31, one has aff (K1 + K2 ) = aff K1 + aff K2 , aff (K1 ∩ K2 ) = aff K1 ∩ aff K2 . Finally, Theorem 1.7 gives dim (K1 + K2 ) = dim (aff (K1 + K2 )) = dim (aff K1 + aff K2 ) = dim (aff K1 ) + dim (aff K2 ) − dim (aff K1 ∩ aff K2 ) = dim (aff K1 ) + dim (aff K2 ) − dim (aff (K1 ∩ K2 )) = dim K1 + dim K2 − dim (K1 ∩ K2 ). The following corollary is important for the study of extreme faces of convex sets. Corollary 2.33. If a plane L ⊂ Rn of dimension m meets the relative interior of a convex set K ⊂ Rn , then dim K 6 n − m + dim (K ∩ L). In particular, if L is a hyperplane not containing K, then dim K = dim (K ∩ L) + 1.
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Proof. Since rint L = L, Theorem 2.32 gives dim K = dim (K ∪ L) − dim L + dim (K ∩ L) 6 n − m + dim (K ∩ L).
(2.8)
In particular, if L is a hyperplane not containing K, then dim (K ∪ L) = n, and (2.8) becomes as dim K = n − (n − 1) + dim (K ∩ L) = dim (K ∩ L) + 1. Remark. The condition L ∩ rint K 6= ∅ is essential in Corollary 2.33. Indeed, if K is the closed circle in R2 and L a line supporting K, then dim K = 2 2 − 1 + 0. Theorem 2.34. Let f : Rn → Rm be an affine transformation. For convex sets K ⊂ Rn and M ⊂ Rm , one has rint f (K) = f (rint K), rint f
−1
(M ) = f
−1
(rint (M ∩ rng f )).
(2.9) (2.10)
If, additionally, rint M ∩ rng f 6= ∅, then rint f −1 (M ) = f −1 (rint M ). Proof. Excluding the trivial cases K = ∅ and M ∩ rng f = ∅, we suppose that both sets K and M ∩rng f are nonempty. Consequently, f −1 (M ) 6= ∅. According to Theorem 2.12, the sets f (K) and f −1 (M ) are convex. For the equality (2.9), choose points x ∈ rint f (K) and y0 ∈ rint K. Then y = f (y0 ) ∈ f (rint K) ⊂ f (K). By Theorem 2.24, there is a scalar γ > 1 such that u = γx + (1 − γ)y ∈ f (K). Let a point u0 ∈ K satisfy the condition f (u0 ) = u. Put x0 = (1 − γ −1 )y0 + γ −1 u0 . Since 0 < γ −1 < 1, Theorem 2.21 yields x0 ∈ rint K. By the affinity of f , one has x = (1 − γ −1 )y + γ −1 u = f ((1 − γ −1 )y0 + γ −1 u0 ) = f (x0 ) ∈ f (rint K). Conversely, let x ∈ f (rint K) and y ∈ f (K). Then x = f (x0 ) and y = f (y0 ) for certain points x0 ∈ rint K and y0 ∈ K. By Theorem 2.24, there is a scalar γ > 1 such that u0 = γx0 + (1 − γ)y0 ∈ K. Consequently, γx + (1 − γ)y = γf (x0 ) + (1 − γ)f (y0 ) = f (u0 ) ∈ f (K). Hence x ∈ rint f (K) according to the same theorem. To prove the equality (2.10), let x ∈ rint f −1 (M ) = rint f −1 (M ∩ rng f ).
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Choose points z ∈ M ∩ rng f and z0 ∈ f −1 (M ∩ rng f ), with f (z0 ) = z. As above, there is a scalar γ > 1 such that γx + (1 − γ)z0 ∈ f −1 (M ∩ rng f ). Therefore, γf (x) + (1 − γ)z ∈ M ∩ rng f , implying f (x) ∈ rint (M ∩ rng f ). Hence x ∈ f −1 (rint (M ∩ rng f )). Conversely, let x ∈ f −1 (rint (M ∩rng f )) and z ∈ f −1 (M ∩rng f ). Then f (x) ∈ rint (M ∩ rng f )
and f (z) ∈ M ∩ rng f.
Theorem 2.24 gives the existence of a scalar γ > 1 with the property f (γx + (1 − γ)z) = γf (x) + (1 − γ)f (z) ∈ M ∩ rng f. Hence γx + (1 − γ)z ∈ f −1 (M ∩ rng f ), which gives x ∈ rint f −1 (M ∩ rng f ) = rint f −1 (M ). Now, assume that rint M ∩ rng f 6= ∅. According to Corollary 1.89, the set rng f is a plane, and Corollary 2.27 implies that rint M ∩ rng f = rint (M ∩ rng f ). Therefore, (2.9) shows that rint f −1 (M ) = f −1 (rint (M ∩ rng f )) = f −1 (rint M ∩ rng f ) = f −1 (rint M ).
2.3
Closure and Relative Boundary
Closure and Relative Interior Theorem 2.35. For a convex set K ⊂ Rn , its closure cl K is a convex set with the same affine span and the same dimension as K: aff (cl K) = aff K
and
dim (cl K) = dim K.
Proof. The statement is obvious when K = ∅. So, we assume that K 6= ∅. Because the closed unit ball B of Rn is convex (see example on page 72), Theorem 2.9 implies that every neighborhood Bρ (K) = ρB + K, ρ > 0, of K is convex. Theorem 2.8 shows that cl K is convex as the intersection of convex sets Bρ (K), ρ > 0 (see page 8). By Theorem 1.50, the inclusion K ⊂ cl K yields that aff K ⊂ aff (cl K). On the other hand, from K ⊂ aff K and the closedness of aff K (see Corollary 1.22) it follows that cl K ⊂ aff K. Hence aff (cl K) ⊂ aff K. Finally, the equality aff (cl K) = aff K gives dim (cl K) = dim K.
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The next result is a sharper version of Theorem 2.21. Theorem 2.36. If K ⊂ Rn is a nonempty convex set, x ∈ rint K, and y ∈ cl K, then (1 − λ)x + λy ∈ rint K for all 0 < λ < 1. Consequently, the open segment (x, y) lies in rint K. Proof. Since the case x = y is trivial, we assume that x 6= y. Choose a point z = (1−λ)x+λy, where 0 < λ < 1. By Theorem 2.14, there is a scalar δ > 0 such that Bδ (x) ∩ aff K ⊂ rint K. Because y ∈ cl K ⊂ aff K, the segment [x, y] lies in aff K. Let a point u ∈ K be such that ku−yk 6 1−λ λ δ, λ 1 z − 1−λ u (whence z = (1 − λ)v + λu, as depicted below). and put v = 1−λ From 1 λ λ x= z− y and kx − vk = ku − yk 6 δ 1−λ 1−λ 1−λ it follows that v ∈ Bδ (x). Furthermore, v ∈ aff K because v is an affine combination of the points u, z ∈ aff K (see Theorem 1.46). Hence v ∈ Bρ (x) ∩ aff K ⊂ rint K. By Theorem 2.21, z ∈ rint K. '$ Bδ (x) ru xr r (((r y ( ( ( ( z r v(( ( &% Theorem 2.37. Let K ⊂ Rn be a nonempty convex set, and c be a point in rint K. Then cl K = N1 ∩ N2 ∩ · · · , where Ni is the homothetic copy of K with center c given by Ni = c + (1 +
1 i+1 )(K
− c),
i > 1.
Proof. Let y ∈ cl K. Since the case y = c is obvious, we assume that y 6= c. Then [c, y) ⊂ rint K ⊂ K by Theorem 2.36. For every integer i > 1, there is a point z ∈ [c, y) ⊂ K so close to y that y ∈ [c, zi ), where 1 )(z − c). Because zi ∈ Ni and Ni is a convex set (see zi = c + (1 + i+1 Theorem 2.9), we have y ∈ Ni . Hence y ∈ N1 ∩ N2 ∩ · · · , which gives the inclusion K ⊂ N1 ∩ N2 ∩ · · · . Conversely, let y ∈ K1 ∩ K2 ∩ · · · . For every integer i > 1, there is 1 a point zi ∈ K such that y = c + (1 + i+1 )(zi − c). Expressing zi as i i zi = (1 − i+1 )c + i+1 y, we see that zi → y when i → ∞. Hence y ∈ cl K, implying the inclusion K1 ∩ K2 ∩ · · · ⊂ K. A relation between homothetic copies of a convex set and its neighborhoods is given in Theorem 2.56.
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Theorem 2.38. For a convex set K ⊂ Rn , one has cl (rint K) = cl K and rint (cl K) = rint K. Proof. Without loss of generality we may assume that K 6= ∅. Because the inclusion cl (rint K) ⊂ cl K is obvious, it suffices to prove the opposite one. Let x ∈ cl K and y ∈ rint K. By Theorem 2.36, the open segment (x, y) lies in rint K. Since the case x = y is trivial (then x ∈ rint K ⊂ cl (rint K)), we suppose that x 6= y. Put zλ = (1 − λ)x + λy, 0 < λ < 1. Clearly, zλ ∈ (x, y) ⊂ rint K and zλ → x as λ → 0. Hence x ∈ cl (rint K). Because aff K = aff (cl K) (see Theorem 2.35), the inclusion K ⊂ cl K and Theorem 2.15 give rint K ⊂ rint (cl K). Conversely, choose points x ∈ rint (cl K) and z ∈ rint K. Then u = γx + (1 − γ)z ∈ cl K for a certain γ > 1 (see Theorem 2.24). Therefore, x = (1 − γ −1 )z + γ −1 u ∈ rint K according to Theorem 2.36. Hence rint (cl K) ⊂ rint K. Corollary 2.39. If K1 and K2 are convex sets in Rn , then cl K1 = cl K2 if and only if rint K1 = rint K2 . Corollary 2.40. For a convex set K ⊂ Rn , the following conditions are equivalent. (1) K is a plane. (2) rint K is a plane. (3) cl K is a plane. Proof. (1) ⇔ (2). If K is a plane, then rint K = K, which shows that rint K also is a plane. Conversely, if rint K is a plane, then rint K ⊂ K ⊂ aff K = aff (rint K) = rint K according to Theorem 2.14. Hence K = rint K, and K is a plane. (1) ⇔ (3). If K is a plane, then cl K = K (see Corollary 1.22), which shows that cl K is a plane. Conversely, assume that cl K is a plane. Then Theorem 2.38 gives rint K = rint (cl K) = cl K, implying that rint K is a plane. By the above proved, K is a plane. The next theorem describes some local properties of convex sets. Theorem 2.41. If K ⊂ Rn is a nonempty convex set and c is a point in cl K, then Bρ (c) ∩ rint K 6= ∅ for every closed ball Bρ (c) ⊂ Rn , ρ > 0. Furthermore, aff (Bρ (c) ∩ rint K) = aff (Bρ (c) ∩ K) (2.11) = aff (Bρ (c) ∩ cl K) = aff K.
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Proof. First, we state that Bρ (c) ∩ rint K 6= ∅. Since the case K = {c} is obvious, we may assume that K has positive dimension. Then rint K also has positive dimension (see Theorem 2.14), implying that rint K 6= {c}. Choose points u ∈ rint K \ {c} and z ∈ (u, c) such that kz − ck < ρ. Then z ∈ rint K according to Theorem 2.36. By Theorem 2.14, there is a δ > 0 such that Bδ (z) ∩ aff K ⊂ rint K. Let 0 < ε < min{δ, ρ − kz − ck}. For every point x ∈ Bε (z), one has kx − ck 6 kx − zk + kz − ck 6 ε + kz − ck < ρ, implying the inclusion x ∈ Bρ (c). Consequently, Bε (z) ∩ rint K ⊂ Bρ (c) ∩ rint K. A combination of Theorem 2.14 (with z instead of c), and Theorems 1.50 and 2.35 gives aff K = aff (Bε (z) ∩ rint K) ⊂ aff (Bρ (c) ∩ rint K) ⊂ aff (Bρ (c) ∩ K) ⊂ aff (Bρ (c) ∩ cl K) ⊂ aff (cl K) = aff K. Summing up, the equalities (2.11) hold. Theorem 2.42. A nonempty convex set K ⊂ Rn contains every convergent series ∞ ∞ P P z= λi xi , where xi ∈ K, λi = 1, and λi > 0 for all i > 1. (2.12) i=1
i=1
Proof. Choose a point z of the form (2.12). Excluding zero multiples, we assume that all scalars λi are positive. Let X = {x1 , x2 , . . . } and L = aff X. Put r = dim L. Then X contains r + 1 affinely independent points (see Corollary 1.76). Without loss of generality, we may assume that these are x1 , . . . , xr+1 . So, L = aff {x1 , . . . , xr+1 }. Consider the convex set M = K ∩ L. We first state that z ∈ cl M . For this, we express z as ∞ k k P P P (k) λi xi = lim γk ηi xi z= λi xi = lim where γk =
j=1 λj
k→∞
k→∞ i=1
i=1
Pk
and
zk =
(k) ηi
i=1
= λi /γk for all 1 6 i 6 k. Clearly, every
(k) η1 x1
(k)
+ · · · + ηi xk ,
k > 1,
is a positive convex combination of points x1 , . . . , xk ∈ M . Hence all z1 , z2 , . . . belong to M (see Theorem 2.3). Since limk→∞ γk = 1, one has lim γk zk γk z k = lim = lim zk ∈ cl M. z = lim γk zk = k→∞ k→∞ k→∞ γk k→∞ lim γk k→∞
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To prove the inclusion z ∈ M , we write z=
∞ P
λi xi =
i=1
r+1 P i=1
= γr+1
r+1 P i=1
∞ P
λi xi +
λi xi
i=r+2
(r+1)
ηi
xi + (1 − γr+1 )
∞ P
δi xi ,
i=r+2
(r+1) ηi
where γr+1 and are defined as above, and δi = λi /(1 − γr+1 ) for all i > r + 2. Then z = γr+1 u + (1 − γr+1 )v, where u=
r+1 P i=1
(r+1)
ηi
xi ,
∞ P
v=
δi xi ,
0 < γr+1 < 1.
i=r+2
By the above argument, v ∈ cl M . We state that u ∈ rint M . Indeed, (r+1) since all scalars ηi are positive, Theorem 2.17 shows that u belongs to the relative interior of the simplex ∆ = ∆(x1 , . . . , xk+1 ). By the choice of x1 , . . . , xr+1 , we have aff ∆ = L = aff M . Now, Theorem 2.15 implies u ∈ rint ∆ ⊂ rint M . Finally, from the inclusions u ∈ rint M , v ∈ cl M and Theorem 2.36 it follows that the point z ∈ (u, v) belongs to rint M . Hence z ∈ M ⊂ K. Closure and Algebra of Convex Sets Theorem 2.43. If F = {Kα } is a family of convex sets in Rn , then cl (∩ Kα ) ⊂ ∩ cl Kα . α
(2.13)
α
Furthermore, cl (∩ Kα ) = ∩ cl Kα provided ∩ rint Kα 6= ∅. α
α
α
Proof. The convexity of ∩ cl Kα follows from Theorems 2.8 and 2.35. We α
observe that the inclusion (2.13) holds for any family {Kα } of sets, not necessarily convex. We state that ∩ cl Kα ⊂ cl (∩ Kα ) provided ∩ rint Kα 6= ∅. For this, α α α choose points x ∈ ∩ cl Kα and y ∈ ∩ rint Kα . By Theorem 2.36, one has α
α
(1 − λ)x + λy ∈ ∩ rint Kα α
for all
0 < λ < 1.
Therefore, x = lim ((1 − λ)x + λy) ∈ cl (∩ rint Kα ) ⊂ cl (∩ Kα ). λ→0
α
α
Remark. The second statement of Theorem 2.43 fails if ∩ rint Kα = ∅. α Indeed, let H be a hyperplane in Rn and W be an open halfspace determined by H. Then ∅ = cl (H ∩ W ) 6= cl H ∩ cl W = H.
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Since rint L = cl L = L for every plane L ⊂ Rn , Theorem 2.43 implies the following corollary. Corollary 2.44. If a plane L ⊂ Rn meets the relative interior of a convex set K ⊂ Rn , then cl (K ∩ L) = cl K ∩ L. Corollary 2.45. If F = {Kα } is a nested family of convex sets in Rn , then ∪ cl Kα is a convex set satisfying the inclusion α
∪ cl Kα ⊂ cl (∪ Kα ). α
(2.14)
α
Proof. The convexity of ∪ cl Kα follows from Theorems 2.8 and 2.35, and α
the inclusion (2.14) holds for any family {Kα } of sets, not necessarily convex. We will need the following corollary, which is a particular cases of a more general statement (see Exercise 0.10). Corollary 2.46. For convex sets K1 , K2 ⊂ Rn , the sum cl K1 + cl K2 is a convex set satisfying the inclusion cl K1 + cl K2 ⊂ cl (K1 + K2 ).
(2.15)
If at least one of the sets K1 , K2 is bounded or the planes aff K1 and aff K2 are independent, then the sum cl K1 + cl K2 is a closed set, and cl K1 + cl K2 = cl (K1 + K2 ).
(2.16)
Proof. The convexity of cl K1 + cl K2 follows from Theorems 2.9 and 2.35, and the remaining part of the statement easily derives from the properties of closures (see Exercise 0.10). y
6 K1 K2
Fig. 2.3
r
-
x
The sum of closed convex sets may be nonclosed.
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Remark. The inclusion (2.15) in Corollary 2.46 may be proper if both convex sets K1 and K2 are unbounded. Indeed, let K1 = {(x, y) : x > 0, xy > 1}
and K2 = {(x, 0) : x 6 0}.
Then both K1 and K2 are closed, while their sum K1 +K2 = {(x, y) : y > 0} is an open halfplane (see the picture above). Theorem 2.47. If K and M are convex sets and X is a nonempty bounded set in Rn such that K + X = M + X, then cl K = cl M . Proof. Without loss of generality, we may assume that both sets K and M are nonempty. By a symmetry argument, it suffices to prove the inclusion K ⊂ cl M . Choose points u ∈ K and x0 ∈ X. Since u + x0 ∈ K + X = M + X, there are points v1 ∈ M and x1 ∈ X such that u + x0 = v1 + x1 . Similarly, we select recursively points vi ∈ M and xi ∈ X such that u + xi−1 = vi + xi
for all
i > 1.
Adding the first r of these equalities, one has ru + x0 = v1 + · · · + vr + xr ,
r > 1.
Equivalently, u − 1r (v1 + · · · + vr ) = 1r (xr − x0 ),
r > 1.
By Theorem 2.3, wr = 1r (v1 + · · · + vr ) ∈ M for all r > 1. If δ = diam X, then kx0 − xr k 6 δ, and ku − wr k = 1r kx0 − xr k 6 δ/r → 0
when r → ∞.
Hence u = limr→∞ wr ∈ cl M , which gives the inclusion K ⊂ cl M . A similar to Theorem 2.47 statement is considered in Theorem 8.11. Theorem 2.48. Let f : Rn → Rm be an affine transformation and K ⊂ Rn and M ⊂ Rm be convex sets. Then f (cl K) is a convex set and f (cl K) ⊂ cl f (K), cl f
−1
(M ) = f
−1
(cl (M ∩ rng f )).
Furthermore, (1) f (cl K) = cl f (K) if f is one-to-one or K is bounded, (2) cl f −1 (M ) = f −1 (cl M ) provided rint M ∩ rng f 6= ∅.
(2.17) (2.18)
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Proof. Theorems 2.12 and 2.35 imply the convexity of f (cl K). The inclusions f (cl X) ⊂ cl f (X)
and
cl f −1 (Y ) ⊂ f −1 (cl (Y ∩ rng f ))
hold for any continuous function f : Rn → Rm and sets X ⊂ Rn and Y ⊂ Rm (see page 9). So, it remains to prove the opposite inclusion f −1 (cl (M ∩ rng f )) ⊂ cl f −1 (M ). Excluding the trivial case M ∩ rng f = ∅, we assume that M ∩ rng f is nonempty. Choose points x ∈ f −1 (cl (M ∩ rng f ))
and y ∈ rint (M ∩ rng f ).
Then f (x) ∈ cl (M ∩ rng f ). Let u ∈ f −1 (y) and vλ = (1 − λ)u + λx,
zλ = (1 − λ)y + λf (x),
0 < λ < 1.
Theorem 2.36 shows that zλ ∈ rint (M ∩ rng f ) for all 0 < λ < 1. Since f (vλ ) = (1 − λ)f (u) + λf (x) = zλ , Theorem 2.34 gives vλ ∈ f −1 (zλ ) ⊂ f −1 (rint (M ∩ rng f )) = rint f −1 (M ). Now, from Theorem 2.38 it follows x = lim vλ ∈ cl (rint f −1 (M )) = cl f −1 (M ). λ→1
Statement (1) of the theorem holds for any continuous function f and a set X ⊂ Rn (see page 9). For statement (2), we first observe that rng f is a plane (see Corollary 1.89). Then Corollary 2.44 gives cl M ∩ rng f = cl (M ∩ rng f ). Therefore, (2.18) implies that cl f −1 (M ) = f −1 (cl (M ∩ rng f )) = f −1 (cl M ∩ rng f ) = f −1 (cl M ). Remark. The inclusion f (cl K) ⊂ cl f (K) in Theorem 2.48 can be proper if f is not one-to-one and K is unbounded. Indeed, let f be the orthogonal projection of R2 on the x-axis, and K = {(x, y) : x > 0, xy > 1}. Then K is a closed convex set, while f (K) is the open halfline {(x, 0) : x > 0}.
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Closedness of Affine Images Given a convex set K ⊂ Rn and an affine transformation f : Rn → Rm , the closedness condition f (cl K) = cl f (K) plays an important role in convex analysis. Theorem 2.52 below shows that asymptotic planes can provide a useful tool in formulating such a condition for the case of any set X ⊂ Rn . We recall that the inf -distance between nonempty sets X and Y in Rn is defined by δ(X, Y ) = inf {kx − yk : x ∈ X, y ∈ Y }. Definition 2.49. A nonempty plane L ⊂ Rn is called asymptotic to a nonempty set X ⊂ Rn (or an asymptote of X) provided cl X ∩ L = ∅ and δ(X, L) = 0.
X L Fig. 2.4
An asymptotic plane L to a set X.
The next two lemmas show a relation of asymptotes with closedness of sums. Lemma 2.50. Let X ⊂ Rn be a nonempty set and S ⊂ Rn be a subspace. A translate L of S is an asymptote of X if and only if L ⊂ cl (X + S) \ (cl X + S). Proof. Let a translate L = c + S be an asymptote of X. We first state that L ∩ (cl X + S) = ∅. Indeed, assume for a moment the existence of a point z ∈ L ∩ (cl X + S). Then z = c + u = x + v, where x ∈ cl X and u, v ∈ S. Consequently, x = c + (u − v) ∈ c + S = L, contrary to the assumption cl X ∩ L = ∅. For the inclusion L ⊂ cl (X + S), choose a point z ∈ L. Then z = c + u, where u ∈ S. The condition δ(X, L) = 0 means the existence of a sequence
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x1 , x2 , . . . of points from X such that limi→∞ δ(xi , L) = 0. Denote by zi the orthogonal projection of xi on L. Clearly, δ(xi , L) = kxi − zi k (see Theorem 1.96). Because of zi ∈ L = c + S, we can write zi = c + ui for a certain ui ∈ S. Since xi + (z − zi ) = xi + (u − ui ) ∈ X + S, and since k(xi + z − zi ) − zk = kxi − zi k = δ(xi , L) → 0
as i → ∞,
one has z ∈ cl (X + S). Hence L ⊂ cl (X + S). Conversely, suppose that L ⊂ cl (X + S) \ (cl X + S). From L + S = L, one has (cl X ∩ L) + S ⊂ (cl X + S) ∩ (L + S) = (cl X + S) ∩ L = ∅. Therefore, cl X ∩ L = ∅. Now, let z ∈ L. The inclusion z ∈ L ⊂ cl (X + S) implies the existence of an infinite sequence z1 , z2 , . . . of points from X + S such that limi→∞ kzi − zk = 0. Expressing zi and z as zi = x + ui and z = c + u, where xi ∈ X and ui , u ∈ S, we obtain, because of c + u − ui ∈ c + S = L, that δ(xi , L) 6 kxi − (c + u − ui )k = kzi − zk → 0
as i → ∞.
Hence δ(X, L) = 0, and L is an asymptote of X. Lemma 2.51. If X ⊂ Rn is a nonempty set and S ⊂ Rn is a subspace, then a certain translate of S is an asymptote of X if and only if the sum cl X + S is not closed. Proof. First, we observe that cl (X + S) ⊂ cl (cl X + S) ⊂ cl (cl (X + S)) = cl (X + S), which gives the equality cl (X + S) = cl (cl X + S). If a certain translate L = c + S of S is an asymptote of X, then, by Lemma 2.50, L ⊂ cl (X + S) \ (cl X + S) = cl (cl X + S) \ (cl X + S), implying that the sum cl X + S is not closed. Conversely, suppose that the sum cl X + S is not closed. Choose a point a ∈ cl (cl X + S) \ (cl X + S) = cl (X + S) \ (cl X + S)
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and consider the plane M = a + S. We state that (cl X + S) ∩ M = ∅. Indeed, assuming the existence of a point z ∈ (cl X +S)∩M and expressing z as z = x+u = a+v, where x ∈ cl X and u, v ∈ S, we obtain the inclusion a = x + (u − v) ∈ cl X + S, which is impossible. Next, we state that M ⊂ cl (X + S). Indeed, let e ∈ M = a + S. Then e = a + u for a certain point u ∈ S. Because of a ∈ cl (X + S), there is a sequence of points a1 , a2 , . . . in X + S such that limi→∞ kai − ak = 0. Let ei = ai + u. Then ei ∈ (X + S) + S = X + S, lim kei − ek = lim kai − ak = 0.
i→∞
i→∞
Hence e ∈ cl (X + S), and M ⊂ cl (X + S). By Lemma 2.50, M is an asymptote of X. Theorem 2.52. Let f : Rn → Rm be an affine transformation expressed as f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. For every set X ⊂ Rn , one has f (cl X) ⊂ cl f (X). Furthermore, the following conditions are equivalent. (1) f (cl X) = cl f (X). (2) The sum cl X + null g is closed. (3) No translate of the subspace null g is an asymptote of X. Proof. Because of f (cl X) = a + g(cl X), we may assume that f is a linear transformation; that is, f = g. Without loss of generality, we may consider that X 6= ∅. For any set X ⊂ Rn , the inclusion f (cl X) ⊂ cl f (X) because f is continuous (see Exercise 0.6). (1) ⇒ (2). Condition (1) implies that the set f (cl X) is closed. Choose a subspace S ⊂ Rn complementary to null f and denote by Y the projection of cl X on S along null f . Then cl X + null f = Y + null f , which gives f (cl X) = f (Y ). Consequently, the set f (Y ) is closed. The mapping h : S → rng f defined by h(x) = f (x) is an invertible linear transformation (see Exercise 0.1). Clearly, h(Y ) = f (Y ). Furthermore, the set Y = h−1 (h(Y )) is closed by the continuity of h−1 : rng f → S and closedness of h(Y ). Since the subspaces S and null f are complementary, the sum Y + null f = cl X + null f is closed according to Theorem 2.48. (2) ⇒ (1). Repeating the above argument in the converse order, we obtain that the set f (cl X) is closed. Therefore, cl f (X) ⊂ cl f (cl X) = f (cl X) ⊂ cl f (X),
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implying that f (cl X) = cl f (X). The equivalence of conditions (2) and (3) follows from Lemma 2.51. Relative Boundary Definition 2.53. The relative boundary of a set X ⊂ Rn , denoted rbd X, is defined as rbd X = cl X \ rint X. Example. If L ⊂ Rn is a plane, then rbd L = cl L \ rint L = L \ L = ∅ (see Corollary 1.22 and example on page 78). Example. If L ⊂ Rn is a plane of positive dimension and D is a closed halfplane of L given by D = {x ∈ L : x·c 6 γ},
c∈ / (sub L)⊥ ,
γ∈R
(see Corollary 1.38), then rbd D is the (m − 1)-dimensional plane H ∩ L, where H is the hyperplane expressed by (1.3) (see Corollary 2.62 below). Similarly, the relative boundary of a closed slab of L, given by F = {x ∈ L : γ 6 x·c 6 γ 0 },
c∈ / (sub L)⊥ ,
γ < γ0,
coincides with the union of two parallel (m − 1)-dimensional planes H ∩ L and H 0 ∩ L, where H and H 0 are parallel hyperplanes, given by (1.3) and (1.4), respectively. Theorem 2.54. The relative boundary of a set X ⊂ Rn is a closed set. Furthermore, if X is convex, then rbd X = rbd (cl X) = rbd (rint X). Proof. Since the case rbd X = ∅ is obvious, we assume that rbd X is nonempty. Let x1 , x2 , . . . be a sequence of points in rbd X converging to a point x ∈ Rn . Then x ∈ cl X because of rbd X ⊂ cl X. We state that no set Bρ (x) ∩ aff X, ρ > 0, lies in X. Indeed, choose an index i large enough to satisfy the inequality kx − xi k 6 ρ/2. Since xi ∈ rbd X, there is a a point yi ∈ Bρ/2 (xi ) ∩ (aff X \ X). The inequalities kx − yi k 6 kx − xi k + kxi − yi k 6 ρ give yi ∈ Bρ (x). Therefore, Bρ (x) ∩ aff X 6⊂ X, and x ∈ / rint X. Summing up, x ∈ cl X \ rint X = rbd X, which confirms the closedness of rbd X.
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Assume that the set X is convex. Theorems 2.15 and 2.38 yield rbd (cl X) = cl (cl X) \ rint (cl X) = cl X \ rint X = rbd X, rbd (rint X) = cl (rint X) \ rint (rint X) = cl X \ rint X = rbd X. The next result describes an important property of the relative boundary of a convex set (see also the complementary Theorem 5.1 concerning the inclusion [x, zi ⊂ rint K). Theorem 2.55. Let K ⊂ Rn be a nonempty convex set, x ∈ rint K, and z ∈ aff K \ {x}. Then either [x, zi ⊂ rint K, or the halfline [x, zi contains a unique point u ∈ rbd K such that [x, u) ⊂ rint K and [x, zi \ [x, u] ⊂ aff K \ cl K.
K r x
r u
r z
-
Proof. Suppose that [x, zi 6⊂ rint K and choose a point v ∈ [x, zi \ rint K. We observe first that no point from [x, vi \ [x, v] belongs to cl K. Indeed, assuming the existence of a point y ∈ ([x, vi \ [x, v]) ∩ cl K, we would obtain the inclusion v ∈ (x, y) ⊂ rint K (see Theorem 2.36). Hence the set cl K ∩[x, zi is bounded, and, by the convexity of cl K, it is a closed segment, say [x, u]. We state that u ∈ rbd K. For if u belonged to rint K, then, according to Corollary 2.25, there would be a point y ∈ K ∩[x, zi satisfying the inclusion u ∈ (x, y), in contradiction with cl K ∩ [x, zi = [x, u]. Assume for a moment the existence of another point u0 ∈ [x, zi which belongs to rbd K. Then either u0 ∈ (x, u) or u ∈ (x, u0 ), and Theorem 2.36 shows that one of the points u, u0 belongs to rint K, which is impossible. Theorem 2.55 is used in establishing the following relation between homothetic copies of a convex set and its neighborhoods. Theorem 2.56. Let K ⊂ Rn be a nonempty convex set, c ∈ rint K, and ρ > 0 such that Bρ (c) ∩ aff K ⊂ K. For any scalar γ > 1, the homothetic copy Kγ = c + γ(K − c) contains the set B(γ−1)ρ (K) ∩ aff K, where B(γ−1)ρ (K) means the (γ − 1)ρ-neighborhood of K.
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Proof. Choose a point y ∈ B(γ−1)ρ (K) ∩ aff K. The case y ∈ K is obvious because of the inclusion K ⊂ Kγ for all γ > 1. Let y ∈ / K. Denote by u the nearest to y point in cl K. Then ky − uk 6 (γ − 1)ρ by the choice of y. If u ∈ [c, y], then kc − yk/kc − uk = (kc − uk + ku − yk)/kc − uk 6 (ρ + (γ − 1)ρ))/ρ = γ, implying the inclusion y ∈ Kγ . Assume that u ∈ / [c, y] and denote by L the 2-dimensional plane containing u, c, y. By Theorem 2.55, [c, y] meets rbd K at a single point, z. Denote by l the line in L which is tangent to the circle Bρ (c) ∩ L and meets [u, y] at a point u0 . Let v be the nearest to y point in l, and denote by w the point at which l supports Bρ (c) ∩ L. Clearly, kv − yk 6 ku0 − yk 6 ky − uk 6 (γ − 1)ρ, which gives kc − yk/kc − zk = kw − vk/kw − zk = (kc − wk + ky − vk)/kc − wk 6 (ρ + (γ − 1)ρ)/ρ = γ, again implying the inclusion y ∈ Kγ . Corollary 2.57. The relative boundary of a convex set K ⊂ Rn is empty if and only if K is a plane. Proof. If K is a plane, then rbd K = ∅ (see example on page 104). Suppose that K is not a plane. Then K 6= aff K and K 6= ∅. Choose points x ∈ rint K (see Corollary 2.18) and z ∈ aff K \ K. By Theorem 2.55, there is a point u ∈ [x, zi ∩ rbd K. Hence rbd K 6= ∅. Theorem 2.58. If K ⊂ Rn is a convex set distinct from a plane, then the following conditions are equivalent. (1) rbd K is a plane. (2) rbd K is a convex set. (3) There is an open halfplane E of aff K such that K is the union of E and a convex subset F of rbd E (possibly, F = ∅ or F = rbd E). Proof. (1) ⇒ (2). This part is trivial because every plane is a convex set. (2) ⇒ (1). According to Corollary 2.57, rbd K 6= ∅. If rbd K is a singleton, then, obviously, it is a plane. Suppose that rbd K is not a singleton
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and choose distinct points x, y ∈ rbd K. We state that hx, yi lies in rbd K. Assume for a moment the existence of a point z ∈ hx, yi \ rbd K. Since rbd K is convex, we have [x, y] ⊂ rbd K. So, z ∈ hx, yi \ [x, y]. Without loss of generality, we may suppose that y ∈ (x, z), as depicted below. Then z ∈ Rn \ cl K, because otherwise z ∈ cl K \ rbd K = rint K, implying the inclusion y ∈ (x, z) ⊂ rint K (see Theorem 2.36), which contradicts the assumption y ∈ rbd K. x rP P
y z r r PP Pr w PPP r v r u
Choose a point u ∈ rint K. By Theorem 2.55, there is a point v ∈ (u, z) which belongs to rbd K. By the assumption, [x, v] ⊂ rbd K. On the other hand, the segment [u, y] meets [x, v] at some point w (see Lemma 1.27 and the picture below). Theorem 2.36 shows that w ∈ rint K, in contradiction with w ∈ [x, v] ⊂ rbd K. Hence z ∈ rbd K, and hx, yi ⊂ rbd K. Finally, Theorem 1.46 shows that rbd K is a plane. (3) ⇒ (1). According to Corollary 1.38, the closure of E is a closed halfplane D of aff K such that D \ E is a plane. Since E ⊂ K ⊂ D, one has D = cl E ⊂ cl K ⊂ D, implying that rbd K = cl K \ rint K = D \ E. (1) ⇒ (3). Put L = aff K. Then cl K 6= L by Corollary 2.40. Let z ∈ L \ cl K and N = aff ({z} ∪ rbd K). Obviously, N ⊂ L. We state that cl K ⊂ N . Indeed, suppose cl K 6⊂ N . Then rint K 6⊂ N by Theorem 2.38. Choose a point y ∈ rint K \ N and consider the line l = hy, zi. We observe that l∩N = {z} (otherwise, l∩N would contain another point w, and y ∈ l = hw, zi ⊂ N by Lemma 1.26 and Theorem 1.46). On the other hand, the open segment (y, z) should contain a point of rbd K (see Theorem 2.55) which is distinct from z. The obtained contradiction proves the inclusion cl K ⊂ N . So, L = aff K ⊂ N , which gives L = N . Corollary 1.67 shows that dim (rbd K) = dim L − 1. The set L \ rbd K is the union of two open halfplanes of L, say E1 and E2 , as stated in Corollary 1.39. Without loss of generality, we may assume that z ∈ E1 . Since [u, z] ⊂ E1 for every point u ∈ E1 , and since each segment [x, z], with x ∈ rint K, should contain a point of rbd K (see Corollary 1.41), one has rint K ⊂ E2 . So,
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K ⊂ cl K ⊂ cl E2 , which shows that K = F ∪ E2 , where F = K ∩ rbd E2 is a convex subset of rbd E2 . Theorem 2.59. If F and K are nonempty convex sets in Rn , with F ⊂ cl K, then either F ⊂ rbd K or rint F ⊂ rint K. Furthermore, if F ⊂ rbd K, then aff F ∩ cl K ⊂ rbd K and dim F 6 dim K − 1. Proof. Suppose F 6⊂ rbd K. We observe first that rint F ∩ rint K 6= ∅, since otherwise rint F ⊂ rbd K, and a combination of Theorems 2.38 and 2.54 would give F ⊂ cl (rint F ) ⊂ rbd K. Let z ∈ rint F ∩ rint K. Choose a point x ∈ rint F . Then y = γx + (1 − γ)z ∈ F for a certain scalar γ > 1 (see Theorem 2.24). Since y ∈ F ⊂ cl K, Theorem 2.36 shows that the point x = (1 − γ −1 )z + γ −1 y belongs to rint K. Summing up, rint F ⊂ rint K. Now, let F ⊂ rbd K. Assume for a moment that aff F ∩ cl K 6⊂ rbd K. Then there is a point y ∈ aff F ∩ rint K. Let x ∈ rint F . By Theorem 2.24, a certain point z = (1 − λ)x + λy, 0 < λ < 1, belongs to F . On the other hand, z ∈ rint K because of x ∈ cl K (see Theorem 2.36), contradicting the assumption F ⊂ rbd K. Hence aff F ∩ cl K ⊂ rbd K. Finally, suppose that dim F = dim K. Then dim (aff F ) = dim (aff K), which, combined with the inclusion aff F ⊂ aff K, gives aff F = aff K (see Theorem 1.6). Therefore, rint F ⊂ rint K according to Theorem 2.15, in contradiction with rint F ⊂ F ⊂ rbd K. Hence dim F 6 dim K − 1. Corollary 2.60. If a hyperplane H ⊂ Rn meets a nonempty convex set K ⊂ Rn such that K 6⊂ H, then dim (H ∩ K) 6 dim K − 1. Proof. Clearly, the set F = H ∩ K is convex. If F ⊂ rbd K, then the inequality dim F 6 dim K − 1 follows from Theorem 2.59. Suppose that F 6⊂ rbd K. By the same theorem, rint F ⊂ rint K. Hence H meets rint K, and Corollary 2.33 gives dim (H ∩ K) = dim K − 1. Relative Boundary and Algebra of Convex Sets Theorem 2.61. Let F = {Kα } be a family of convex sets whose relative interiors have a point in common. If K = ∩ Kα , then α
∪ (rbd K ∩ rbd Kα ) = ∪ (cl K ∩ rbd Kα ) ⊂ rbd K. α
α
If, additionally, the family F is finite, then ∪ (rbd K ∩ rbd Kα ) = ∪ (cl K ∩ rbd Kα ) = rbd K. α
α
(2.19)
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Proof. By Theorem 2.26, rint K ⊂ ∩ rint Kα . Hence α
rint K ∩ rbd Kα = ∅ for all
Kα ∈ F,
implying that rbd K ∩ rbd Kα = (cl K \ rint K) ∩ rbd Kα = cl K ∩ rbd Kα . Hence ∪ (rbd K ∩ rbd Kα ) = ∪ (cl K ∩ rbd Kα ). α
α
Furthermore, since cl K ⊂ cl Kα for all Kα ∈ F, Theorem 2.26 gives ∪ (cl K ∩ rbd Kα ) = ∪ (cl K \ rint Kα ) = cl K \ (∩ rint Kα ) α
α
α
⊂ cl K \ rint (∩ Kα ) = cl K \ rint K = rbd K. α
If F is finite, then rint (∩ Kα ) = ∩ rint Kα (see Theorem 2.26), and α
α
(2.19) holds. Corollary 2.62. If a plane L ⊂ Rn meets the relative interior of a convex set K ⊂ Rn , then rbd (K ∩ L) = rbd K ∩ L. Furthermore, any side of the latter equality is empty if and only if K ∩ L is a plane which lies in rint K. Proof. Since rint L = cl L = L, Corollaries 2.27 and 2.44 give rbd (K ∩ L) = cl (K ∩ L) \ rint (K ∩ L) = (cl K ∩ L) \ (rint K ∩ L) = (cl K \ rint K) ∩ L = rbd K ∩ L. Furthermore, if K ∩L is a plane which lies in rint K, then rbd K ∩L = ∅. Conversely, assume that the set rbd K ∩ L is empty. First, we observe that L ⊂ rint K. Indeed, assume for a moment the existence of a point v ∈ L \ rint K. By the assumption, there is a point u ∈ rint K ∩ L. Clearly, u 6= v. Theorem 2.55 shows the existence of a point w ∈ rbd K ∩ [u, vi. Since w ∈ [u, vi ⊂ hu, vi ⊂ L (see Theorem 1.46), one has rbd K ∩ L 6= ∅, contrary to the assumption. Next, suppose that K ∩ L is not a plane. Then rbd (K ∩ L) 6= ∅ according to Corollary 2.57. Hence rbd K ∩ L 6= ∅ by the above proved, again contradicting the assumption. Theorem 2.63. For an affine transformation f : Rn → Rm and convex sets K ⊂ Rn and M ⊂ Rm , the following statements hold. (1) If K is bounded, then rbd f (K) ⊂ f (rbd K).
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(2) If f is one-to-one, then rbd f (K) = f (rbd K). (3) rbd f −1 (M ) = f −1 (rbd (M ∩ rng f )). (4) rbd f −1 (M ) = f −1 (rbd M ) provided rint M ∩ rng f 6= ∅. Proof. (1) If K is bounded, then Theorems 2.34 and 2.48 give rbd f (K) = cl f (K) \ rint f (K) = f (cl K) \ f (rint K) ⊂ f (cl K \ rint K) = f (rbd K). (2) Similarly, if f is one-to-one, then rbd f (K) = cl f (K) \ rint f (K) = f (cl K) \ f (rint K) = f (cl K \ rint K) = f (rbd K). (3) Again by Theorems 2.34 and 2.48, rbd f −1 (M ) = cl f −1 (M ) \ rint f −1 (M ) = f −1 (cl (M ∩ rng f )) \ f −1 (rint (M ∩ rng f )) = f −1 (cl (M ∩ rng f ) \ rint (M ∩ rng f )) = f −1 (rbd (M ∩ rng f )). (4) Since the set rng f is a plane (see Corollary 1.89), statement (3) above and Corollary 2.62 give rbd f −1 (M ) = f −1 (rbd (M ∩ rng f )) = f −1 (rbd M ∩ rng f ) = f −1 (rbd M ). Remark. The inclusion rbd f (K) ⊂ f (rbd K) in statement (1) of Theorem 2.63 does not hold if K is unbounded. Indeed, if f is the orthogonal projection of R2 on the x-axis and K = {(x, y) : xy > 1, x > 0}, then rbd f (K) = {o} 6⊂ {(x, 0) : x > 0} = f (rbd K). Exercises for Chapter 2 Exercise 2.1. Show the convexity of the following solid quadrics: solid ellipsoid, solid elliptic paraboloid, and solid elliptic hyperboloid, given, respectively, as Γ = {(x1 , . . . , xn ) ∈ Rn : a1 x21 + · · · + an x2n 6 1}, Φ = {(x1 , . . . , xn ) ∈ Rn : a1 x21 + · · · + an−1 x2n−1 6 xn }, Ω = {(x1 , . . . , xn ) ∈ Rn : a1 x21 + · · · + an−1 x2n−1 + 1 6 an x2n , xn > 0}, where a1 , . . . , an are positive scalars.
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Exercise 2.2. Show that a nonempty closed set X ⊂ Rn is convex if and only if for any pair of points x, y ∈ X there is a scalar 0 < λ < 1, possibly depending on x and y, such that (1 − λ)x + λy ∈ X. Exercise 2.3. Given a scalar 0 < λ0 < 1, show that a relatively open nonempty set X ⊂ Rn is convex if and only if (1 − λ0 )x + λ0 y ∈ X whenever x, y ∈ X. Exercise 2.4. Show that a nonempty convex set K ⊂ Rn is closed (respectively, relatively open) if and only if for every line l ⊂ aff K, the set K ∩ l is closed (respectively, relatively open). Exercise 2.5. (Grzybowski and Urba´ nski [106]) Show that for every convex set K ⊂ Rn and scalars λ, µ, one has λK + µK = (λ + µ)K +
|λ| + |µ| − |λ + µ| (K − K). 2
Exercise 2.6. (Grzybowski and Urba´ nski [106]) Let K, M ⊂ Rn be compact convex sets, and λ, µ, γ, δ be scalars, with λK + µM = γK + δM . Show that K=
λ−γ δ−µ M if λ 6= γ, or M = K if δ 6= µ. λ−γ δ−µ
Exercise 2.7. Let K ⊂ Rn be a nonempty convex set of dimension m, and ∆(x1 , . . . , xm+1 ) be an m-simplex in aff K. Show that a point z ∈ Rn belongs to rint K is and only if there is an m-simplex ∆(z1 , . . . , zm+1 ) ⊂ K positively homothetic to the simplex ∆(x1 , . . . , xm+1 ) such that z = 1 m+1 (z1 + · · · + zm+1 ). Exercise 2.8. Let K1 and K2 be convex sets in Rn such that a certain translate of K2 lies in aff K1 . Show that rint (K1 + K2 ) = rint K1 + K2 . Exercise 2.9. Let K1 , . . . , Kr be nonempty convex sets in Rn and µ1 , . . . , µr be scalars. Let also r > m = dim (µ1 K1 + · · · + µr Kr ). Show the existence of an index set I ⊂ {1, . . . , r}, card I 6 m, satisfying the condition P P rint (µ1 K1 + · · · + µr Kr ) = µi rint Ki + µi Ki . i∈I
i∈I /
Exercise 2.10. Show that for r-simplices ∆ and ∆0 in Rn , there is an invertible affine transformation f : Rn → Rn such that f (∆) = ∆0 .
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Exercise 2.11. (Starshaped set) Given a nonempty set X ⊂ Rn show that the intersection of all maximal convex subsets of X coincides with the set of points c ∈ X satisfying the property: [c, x] ⊂ X for all x ∈ X. Exercise 2.12. A projective transformation h : Rn → Rm is defined by h(x) = (a + g(x))/(α + x · c), where a ∈ Rm , g : Rn → Rm is a linear transformation, α is a scalar, and c is a nonzero vector in Rn . Show that the image h(K) of a convex set K lying in Rn \ {x ∈ Rn : α + x·c = 0} is a convex set. Notes for Chapter 2 Convex geometry. The first ideas on convexity date back to ancient Greece. For example, Archimedes (in his book “On the sphere and cylinder”, see [3, p. 404]) defines a line concave in the same direction by the following property: “... if every two points on it are taken either, all straight lines connecting the points fall on the same side of the line, or some fall on one and the same side while others fall on itself, but none on the other side.” Based on results of Euler, Cauchy, Schal¨ afly, and Steiner, convex geometry became an independent branch of mathematics around 1900, with notable contributions of Brunn [43, 44] and Minkowski [157, 158]. A comprehensive survey of convex geometry prior to 1934 is summarized in the well-known book of Bonnesen and Fenchel [29]. There are some articles on history of convex geometry, written by Fenchel [85] and Gruber [100, 101]. Intersections and unions of convex sets. Klee [134] posed the problem to describe those sets in Rn which are the intersections of decreasing infinite sequences of pairwise distinct convex sets. The problem is solved in [134] for dimensions two and three. In particular, a convex set in the plane is not the intersection of an infinite decreasing sequence of pairwise distinct convex sets if and only if it has the form S ∪ P ∪ Q, where S is an open convex m-gonal region (possibly, unbounded), P is the union of m open segments or halflines properly contained in the respective open sides of S, and every point of Q is a vertex of S which is an end point of two segments or halflines forming P . A similar problem on the union of convex sets is much easier: the convex sets in Rn which are not the unions of infinite increasing sequences of pairwise distinct convex sets are exactly the convex polytopes. Borovikov [32] showed that the intersection of a nested sequence of simplices in Rn also is a simplex; this statement was generalized by Eggleston, Gr¨ unbaum, and Klee [78] for the case of Hausdorff vector space. In this regard, Gruber [98] posed the problem to describe the families of closed convex sets in Rn which are closed under the intersections of nested sequences of their elements; he also proved in [98, 99] that the following two families have the above property: the family consisting of direct sums of simplices, simplicial cones and subspaces, and the family consisting of direct sums of parallelotopes, simplicial cones and subspaces
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(see also Lawrence [145]). Theorem 2.10 is proved by Jamison [120]. Topological properties of convex sets. Various elementary properties of relative interior and relative boundary of a convex set in n dimensions can be found in Steinitz [207]. Theorem 2.47 is attributed to R˚ adstr¨ om [178]. Klee [126] proved that for a proper subset X of Rn , the following conditions are equivalent: (a) X is a convex Fσ -set, (b) X is an orthogonal projection of a closed convex set, (c) X is an orthogonal projection of a convex Fσ -set. See also Yaksubaev [225] for condition (b) above, and Bromek and Kaniewski [36] for similar results for the case of convex cones. Answering a question of Klee [123], Larman [140] (for n = 3) and Preiss [175] (for all n > 3) showed that every convex Borel set in Rn can be obtained, starting from compact convex sets, by iteration of countable increasing unions or decreasing intersections. (We recall that a Borel set is a set in a topological space that can be formed from closed sets through the operations of countable union, countable intersection, and relative complement.) Theorem 2.42 is independently proved by Rubin and Wesler [188], Blackwell and Girshick [26, pp. 48–49], Cook and Webster [61]. Local convexity. A variety of sufficient conditions for a given set X ⊂ Rn to be convex are summarized in the surveys of Burago and Zalgaller [48] and ManiLevitska [152]. For instance, a closed connected set X ⊂ Rn is convex if and only if it is locally convex (see Nakajima [163] for n 6 3 and Tietze [215] for all n > 2). Similarly, if a connected nonempty set X ⊂ Rn is locally convex, then its relative interior is nonempty and convex (see Tamura [214]). We recall that a set X ⊂ Rn is called locally convex if for any point x ∈ X there is a ball Bρ (x) ⊂ Rn such that Bρ (x) ∩ X is convex (compare with Theorem 2.41). Universal convex sets. Mazur (see Mauldin [154, pp. 111–112]) posed the problem on the existence of a convex body K ⊂ R3 symmetric about the origin of R3 such that every planar centrally symmetric convex disc is affinely equivalent to the intersection of K with a certain 2-dimensional subspace. This problem was solved in the negative by Bessaga [23] and Gr¨ unbaum [103] (see Klee [127] for further references). Grz¸a´slewicz [105] showed the existence of a compact convex set K ⊂ Rn+2 such that every compact convex subset of the unit ball of Rn can be obtained as the intersection of K with a certain n-dimensional plane of Rn+2 . Starshaped sets. A subset X of a vector space E is said to be starshaped with respect to a point c ∈ X if the segment [x, c] belongs to X whenever x ∈ X. The kernel of X consists of all those points with respect to which X is starshaped. Brunn [46] proved that the kernel of a compact starshaped set X ⊂ Rn is a compact convex set. The statement of Exercise 2.11 is proved by Toranzos [216] (see also Smith [193]). Obviously, a set X ⊂ E is convex if and only if X is identical with its kernel.
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The kernel of a star-shaped set X ⊂ Rn is called proper if its closure is different from the closure of X. L. Fejes T´ oth asked (see [173]) for a characterization of those convex sets in the plane which can be realized as proper kernels. Post [173] proved the following statements: (a) a strictly convex set K ⊂ R2 with regular boundary is a proper kernel if and only if bd K contains an arc C such that the set C \ K is countable (that is finite or denumerable), (b) a closed convex set K ⊂ R2 is a proper kernel if and only if it is neither a halfplane nor a slab between two parallel lines. Klee [131] established a much stronger result, by proving that a closed convex set K in a separable Banach space is a proper kernel of a starshaped set if and only if K contains no hyperplane. A convex set K ⊂ Rn containing a hyperplane is a proper kernel of a starshaped set if and only if the relative boundary of K contains a line which does not meet K (see Breen [35]). Convex combinations and distance. A point x in a normed vector space E is a convex combination, x = λ1 x1 + · · · + λr xr , of points x1 , . . . , xr ∈ E if and only if kx − zk 6 λ1 kx1 − zk + · · · + λr kxr − zk for all z ∈ E (see Bilyeu [24] for r = 2 and Wolfe [224] for r > 2). Convexity-preserving mappings. A mapping f : Rn → Rm is called convexity-preserving if the f -images of all convex sets in Rn are convex sets in Rm . As shown in Theorem 2.12, affine transformations are convexity-preserving. The converse statement is not true: for example, the mapping f in Rn , defined by f (x1 , x2 , . . . , xn ) = (x31 , 0, . . . , 0), is convexity-preserving but not affine. Walsh [221] proved that a one-to-one convexity-preserving mapping of R2 or R3 into itself is an affine transformation. This result was expanded by Meyer and Kay [156], who showed that a one-to-one and convexity-preserving mapping f of a real vector space V , dim V > 2, into another real vector space W is an affine transformation. Kuz0 minyh [139] observed that the one-to-one condition here cannot be dropped, by constructing a convexity-preserving mapping R2 → R2 which is onto and discontinuous at every point of R2 . Positively answering Alexandrov’s question, Kuz0 minyh [139] proved that a one-to-one mapping of a real vector space L, dim L > 2, into itself is an affine transformation if it maps every segment onto a convex set. Shaidenko-K¨ unzi [192] (correcting a proposition from [139]) established the following result: If K ⊂ Rn is a convex body which has at least n tangent hyperplanes in general position, and if H is the family of all (positive or negative) homothetic copies of K, then a bijection f : Rn → Rn which maps every convex body M ∈ H onto a convex set is an affine transformation. Ariyawansa, Davidon, and McKennon [4] showed that if D is an open connected subset of Rn , n > 2, and f : D → Rm is a continuous one-to-one convexitypreserving mapping, then there is a projective transformation h(x), expressed as h(x) = g(x)/ϕ(x), where g : Rn → Rm is an affine transformation and ϕ is a nonzero affine functional on Rn , such that h|D = f (see also Exercise 2.12).
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Homothetic convex sets. Rogers [185] showed that convex bodies K1 and K2 in Rn are positively homothetic if and only if all orthogonal projections of K1 and K2 on 2-dimensional planes are positively homothetic (see [199] for further references on this topic). Soltan [202] refined this statement by proving that compact (respectively, closed) convex sets K1 and K2 in Rn are homothetic provided for any given integer m, 2 6 m 6 n − 1 (respectively, 3 6 m 6 n − 1), the orthogonal projections of K1 and K2 on every m-dimensional plane of Rn are homothetic, where the homothety ratio may depend on the projection plane. Less attention in the literature is given to similar results involving planar sections of convex bodies, since they are viewed as dual forms of orthogonal projections with respect to polarity. Nevertheless, Rogers [185] (for the case when p1 ∈ int K1 and p2 ∈ int K2 ) and later Burton [49] proved of the following statement: if K1 , K2 ⊂ Rn are convex bodies and p1 , p2 ∈ Rn are some points such that for any pair of parallel 2-dimensional planes L1 , L2 through p1 , p2 , respectively, the intersections K1 ∩ L1 and K2 ∩ L2 are both empty or positively homothetic, then K1 and K2 are positively homothetic. Moreover, as proved by Burton and Mani [50], if no homothety is taking p1 to p2 transforms K1 onto K2 , then K1 and K2 are homothetic ellipsoids. For the case of unbounded convex sets, partial results in this direction are obtained by Soltan [199, 201]. Choquet simplices. In 1956 Choquet [55] defined a simplex (afterwards called a Choquet simplex ) as a convex set S in a linear space E of any dimension such that the intersection of any two homothetic copies of S, if nonempty, is again a homothetic copy of S, possibly degenerated into a point: (u + λS) ∩ (v + µS) = w + νS,
u, v, w ∈ E,
λ, µ, ν > 0.
By using the technique of measure representation (cf. Phelps [172]), it was shown later that finite-dimensional compact Choquet simplexes are precisely the simplices in the usual sense, i.e., they are the convex hulls of finitely many affinely independent points. Independently, Rogers and Shephard [186] proved that a convex body K in Rn is a simplex if and only if any nonempty intersection of K and a translate of K is a homothetic copy of K, possibly degenerated into a point: K ∩ (u + K) = v + λK,
u, v ∈ Rn ,
λ > 0.
Generalizing this statement, Soltan [196] proved that convex bodies K1 and K2 in Rn are homothetic simplices if and only if all n-dimensional intersections K1 ∩ (u + K2 ), u ∈ Rd , belong to at most countably many homothety classes of convex bodies. See the survey of Soltan [197] for further references and results on Choquet simplices.
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Convex Hulls
3.1
Algebraic Properties of Convex Hulls
Definition and Basic Properties Definition 3.1. For a given set X ⊂ Rn , the intersection of all convex sets containing X is called the convex hull of X and denoted conv X.
JJ J J
X Fig. 3.1
Q Q Q
conv X
The convex hull of a set X.
Theorem 2.8 shows that the convex hull of any set X ⊂ Rn exists and is the smallest convex set containing X. Furthermore, conv ∅ = ∅. Example. Let L ⊂ Rn be a plane of positive dimension m and F = {x ∈ L : γ 6 x·c 6 γ 0 } ⊂ Rn ,
c∈ / (sub L)⊥ ,
γ < γ0,
be a closed slab of L, as represented in Corollary 1.44. Then F is the convex hull of its relative boundary (see Theorem 3.15 for a more general statement), which is the union of parallel (m − 1)-dimensional planes M = {x ∈ L : x·c = γ}
and M 0 = {x ∈ L : x·c = γ 0 }.
Example. Every closed ball Bρ (c) ⊂ Rn is the convex hull of its boundary sphere Sρ (c) (see Exercise 3.2). 117
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Elementary properties of convex hulls are given in the theorem below. Theorem 3.2. For sets X and Y in Rn , the following statements hold. (1) (2) (3) (4) (5)
X ⊂ conv X, with X = conv X if and only if X is convex. conv (conv X) = conv X. conv X ⊂ aff X and aff (conv X) = aff X. conv X ⊂ conv Y if X ⊂ Y . conv X = conv Y if X ⊂ Y ⊂ conv X.
Furthermore, if {Xα } is a family of sets in Rn , then the statements below are true. (6) conv (∩ Xα ) ⊂ ∩ conv Xα . α
α
(7) ∪ conv Xα ⊂ conv (∪ Xα ). α
α
(8) ∪ conv Xα = conv (∪ Xα ) if the family {Xα } is nested. α
α
(9) conv (∪ conv Xα ) = conv (∪ Xα ). α
α
Proof. Let K(X) denote the family of all convex sets containing a given set X ⊂ Rn . The proofs of statements (1)–(9) derive from the following simple arguments. (a) (b) (c) (d)
conv X is the smallest element in K(X). K(X) is exactly the family of all convex sets containing conv X. If X ⊂ Y , then K(Y ) ⊂ K(X). The union of a nested family of convex sets is a convex set (see Theorem 2.8).
The next theorem gives an important description of convex hulls in terms of convex combinations of points. Theorem 3.3. The convex hull of a nonempty set X ⊂ Rn is the collection of all convex combinations of points from X: conv X = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 , . . . , λk > 0, λ1 + · · · + λk = 1}. Equivalently, conv X is the collection of all positive convex combinations of points from X. Proof. According to Theorem 2.8, conv X is a convex set containing X, and Theorem 2.3 implies that the set M = {λ1 x1 + · · · + λk xk : k > 1, x1 , . . . , xk ∈ X, λ1 , . . . , λk > 0, λ1 + · · · + λk = 1}
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lies in conv X. We state that M is a convex set. Indeed, choosing points x, y ∈ M and expressing them as convex combinations x = γ1 x 1 + · · · + γp x p
and y = µ1 y1 + · · · + µq yq
of certain points x1 , . . . , xp , y1 , . . . , yq from X, we see that (1 − λ)x + λy is a convex combination of these points for every choice of λ ∈ [0, 1]: (1 − λ)x + λy = (1 − λ)γ1 x1 + · · · + (1 − λ)γp xp + λµ1 y1 + · · · + λµq yq . Hence (1 − λ)x + λy ∈ M . Since X ⊂ M (every point x ∈ X can be written as 1x), the inclusions X ⊂ M ⊂ conv X and Theorem 3.2 give conv X = M . The second statement follows from the first one. Indeed, if a point x ∈ conv X is written as a convex combination x = λ1 x1 + · · · + λk xk of points x1 , . . . , xk ∈ X, then, eliminating all terms of the form 0xi , we obtain a positive convex combination. Corollary 3.4. For points x1 , . . . , xr ∈ Rn , one has conv {x1 , . . . , xr } = {λ1 x1 + · · · + λr xr : λ1 , . . . , λr > 0, λ1 + · · · + λr = 1}. Proof. By Theorem 3.3, a point x ∈ Rn belongs to conv {x1 , . . . , xr } if and only if it can be written as a convex combination x = λi1 xi1 + · · · + λit xit of certain points xi1 , . . . , xit from {x1 , . . . , xr }. If any of these points, say xip and xiq , coincide, then we replace λip xip + λiq xiq with (λip + λiq )xip . Since such a replacement can be performed at most finitely many times, one may assume that all points xi1 , . . . , xit are pairwise distinct. Finally, for every index i ∈ {1, . . . , r} \ {i1 , . . . , it }, we add 0xi to the right-hand side of x = λi1 xi1 + · · · + λit xit to express x as a convex combination of all points x1 , . . . , xr . A combination of Definition 2.5 and Corollary 3.4 shows that every rsimplex ∆(x1 , . . . , xr+1 ) ⊂ Rn is the convex hull of the set {x1 , . . . , xr+1 }. Theorem 3.5. The convex hull of a nonempty set X ⊂ Rn is the collection of all convex combinations (equivalently, of all positive convex combinations) of affinely independent points from X. Proof. By Theorem 3.3, a point x ∈ Rn belongs to conv X if and only if it can be written as a convex combination x = λ1 x1 + · · · + λk xk of certain
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points x1 , . . . , xk from X. Elimination all terms of the form 0xi , we assume that the scalars λ1 , . . . , λk are positive. We are going to show that x can be written as a convex combination of k − 1 of fewer points from the set {x1 , . . . , xk } provided x1 , . . . , xk are affinely dependent. So, let the set {x1 , . . . , xk } be affinely dependent. By the definition, there are scalars ν1 , . . . , νk , not all zero, such that ν1 x1 + · · · + νk xk = o
and ν1 + · · · + νk = 0.
Clearly, at least one of the scalar ν1 , . . . , νk is positive. Put t = min {λi /νi : νi > 0, 1 6 i 6 k}. Then λi − tνi > 0 for all 1 6 i 6 k, and λi − tνi = 0 for at least one index i ∈ {1, . . . , k}. Furthermore, (λ1 − tν1 ) + · · · + (λk − tνk ) = 1. Hence x = (λ1 − tν1 )x1 + · · · + (λk − tνk )xk is a convex combination of k − 1 or fewer points from X. Consecutively repeating this argument, we obtain a required expression of x. Remark. Although Theorem 3.5 looks similar to Theorems 1.59, convex hulls do not allow, in general, the existence of finite “convex bases” (compare with Theorem 1.68). For example, every point x in the convex hull of the unit circumference C = {(x, y) : x2 + y 2 = 1} of R2 is a convex combination of at most two points from C, while no finite subset X of C satisfies the condition conv X = conv C. The next result is a slight refinement of the well known Carath´eodory’s theorem. Theorem 3.6. For an m-dimensional set X ⊂ Rn , the following statements hold. (1) conv X is the collection of all positive convex combinations of m+1 or fewer affinely independent points from X, (2) conv X is the collection of all convex combinations of exactly m+1 affinely independent points from X, (3) conv X is the union of all m-simplices with vertices in X. Proof. (1) Since dim (aff X) = dim X = m, Corollary 1.61 shows that every affinely independent subset of X contains m + 1 or fewer points. Thus the statement follows from Theorem 3.5. (2) By statement (1) above, a point x ∈ Rn belongs to conv X if and only x is a convex combination x = λ1 x1 + · · · + λk xk of m + 1 or fewer
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affinely independent points x1 , . . . , xk ∈ X. Theorem 1.68 shows that the set {x1 , . . . , xk } can be expanded to an affinely independent subset {x1 , . . . , xm+1 } of X. Then x = λ1 x1 + · · · + λk xk + 0xk+1 + · · · + 0xm+1 is a desired expression of x. From Corollary 3.4 it follows that statement (3) is an equivalent form of statement (2). An extension of Theorem 3.6 to the case of finite subsets of conv X is considered in Exercise 3.6. As shown below, one point in the set X from Theorem 3.5 can be chosen in advance. Theorem 3.7. Given a nonempty set X ⊂ Rn and a point x0 ∈ X, the convex hull of X is the collection of all convex combinations of affinely independent subsets of X each containing x0 . Proof. By Theorem 3.5, a point x ∈ Rn belongs to conv X if and only if can be expressed as a positive convex combination x = λ1 x1 + · · · + λk xk of affinely independent points x1 , . . . , xk ∈ X. Without loss of generality, we suppose that the number k is this representation is minimum possible. If x0 ∈ Rn \ aff {x1 , . . . , xk }, then {x0 , x1 , . . . , xk } is affinely independent (see Corollary 1.63), and x = 0x0 +λ1 x1 +· · ·+λk xk is a desired expression. Let x0 ∈ aff {x1 , . . . , xk }. By Theorem 1.60, x0 can be uniquely written as an affine combination x0 = µ1 x1 + · · · + µk xk . Leaving aside the trivial case x = x0 (then x = 1x0 is the convex combination of x0 ), we assume that x 6= x0 . Then at least one of the scalars λ1 − µ1 , . . . , λk − µk is not zero, and the equality (λ1 − µ1 ) + · · · + (λk − µk ) = 0 shows that at least one of them is negative. Since all scalars µi + t(λi − µi ), 1 6 i 6 k, are positive for t = 1, a continuity argument shows the existence of a t0 > 1 such that µi + t0 (λi − µi ) > 0
for all
16i6k
and µi + t0 (λi − µi ) = 0 for at least one index i ∈ {1, . . . , k}. Therefore, the expression z = (1 − t0 )x0 + t0 x = (µ1 + t0 (λ1 − µ1 ))x1 + · · · + (µk + t0 (λk − µk ))xk is a convex combination of k − 1 or fewer points from {x1 , . . . , xk }. Let, for example, z = η1 x1 + · · · + ηk−1 xk−1 , η1 , . . . , ηk−1 > 0, η1 + · · · + ηk−1 = 1.
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Then −1 −1 −1 −1 x = (1 − t−1 0 )x0 + t0 z = (1 − t0 )x0 + t0 η1 x1 + · · · + t0 ηk−1 xk−1
is a convex combination of x0 , x1 , . . . , xk−1 (as depicted below for k = 3).
r x0
r x1 J r JJr z x J rh x3 hhhhhh J h Jr
x2
Finally, we observe that the set {x0 , x1 , . . . , xk−1 } is affinely independent. Indeed, if {x0 , x1 , . . . , xk−1 } were affinely dependent, then, by Theorem 3.5, x would be a positive convex combination of k − 1 or fewer points from {x0 , x1 , . . . , xk−1 }, contradicting the assumption on minimality of k. The following statement is a refinement of Theorem 3.6. Corollary 3.8. Let X ⊂ Rn be a nonempty set of dimension m, and x0 be a point in X. Then conv X coincides with the union of all m-simplices ∆(x0 , x1 , . . . , xm ), where x1 , . . . , xm ∈ X. Proof. According to Theorem 3.2, conv X contains every m-simplex with vertices x0 , x1 , . . . , xm ∈ X. Conversely, let x ∈ conv X. By Theorem 3.7, x can be written as a convex combination x = λ0 x0 + λ1 x1 + · · · + λk xk of affinely independent points x0 , x1 , . . . , xk from X. Theorem 1.68 shows that {x0 , x1 , . . . , xk } can be expanded to an affinely independent subset {x0 , x1 , . . . , xm } of X. Rewriting x as x = λ0 x0 + λ1 x1 + · · · + λk xk + 0xk+1 + · · · + 0xm , we obtain the inclusion x ∈ ∆(x0 , x1 , . . . , xm ). Iterative Construction of Convex Hulls Definition 3.9. For a nonempty set X ⊂ Rn and a positive integer r, denote by convr X the collection of all convex combination of r or fewer points from X: convr X = ∪ ( conv {x1 , . . . , xk } : k 6 r, x1 , . . . , xk ∈ X).
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Example. If X = {x1 , x2 , x3 , x4 } is an affinely independent set in R3 , then conv2 X is the union of all edges of the simplex ∆ = ∆(x1 , x2 , x3 , x4 ), and conv2 (conv2 X) = ∆. Indeed, every point x ∈ ∆ belongs to a closed segment with endpoints on [x1 , x2 ] and [x3 , x4 ], respectively, as depicted below. r x3
CJ
C Jr C J
x rC J Jr r
XXXr C x1 XXCCr x4 x2 A combination of Theorems 3.2 and 3.6 shows that X = conv1 X ⊂ conv2 X ⊂ · · · ⊂ convm+1 X = conv X
(3.1)
for any nonempty set X ⊂ Rn of dimension m. Theorem 3.10. For a nonempty set X ⊂ Rn and positive integers r and s, one has convr (convs X) = convrs X. Proof. Let z ∈ convr (convs X). Then z is a convex combination, z = λ1 y1 + · · · + λp yp , where y1 , . . . , yp ∈ convs X, p 6 r. Similarly, any yi , 1 6 i 6 p, can be written as a convex combination (i)
(i)
(i)
(i) (i) yi = µ1 x1 + · · · + µ(i) mi xmi , where x1 , . . . , xmi ∈ X, mi 6 s.
So, z is a convex combination of at most rs points from X: (1)
(1)
(1) z = λ1 µ1 x1 + · · · + λ1 µ(1) m1 xm1 + . . . (p)
(p)
(p) + λp µ1 x1 + · · · + λp µ(p) mp xmp ,
implying that z ∈ convrs X. Hence convr (convs X) ⊂ convrs X. Conversely, let z ∈ convrs X. Then z can be expressed as a convex combination z = η1 x1 + · · · + ηq xq of certain points x1 , . . . , xq ∈ X, q 6 rs. We may assume that q = rs (otherwise choose any points xq+1 , . . . , xrs in X and add 0xq+1 + · · · + 0xrs to η1 x1 + · · · + ηq xq ). Let λi = η(i−1)s+1 + η(i−1)s+2 + · · · + ηis ,
1 6 i 6 r.
For every λi = 6 0, 1 6 i 6 r, consider the convex combination η(i−1)s+1 η(i−1)s+2 ηis x(i−1)s+1 + x(i−1)s+2 + · · · + xis . yi = λi λi λi
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Then yi ∈ convs X. Also, choose any yi ∈ convs X if λi = 0, 1 6 i 6 r. Since z = λ1 y1 + · · · + λr yr is a convex combination of y1 , . . . , yr , one has z ∈ convr (convs X). Hence convrs X ⊂ convr (convs X). A combination of (3.1) and Theorem 3.10 gives the following result. Corollary 3.11. If X ⊂ Rn is a nonempty set of dimension m and r1 , . . . , rk are positive integers satisfying the inequality r1 · · · rk > m + 1, then convr1 (convr2 (. . . (convrk X) . . . )) = conv X. In particular, conv2 (conv2 (. . . (conv2 X) . . . )) = conv X, where the operation conv2 is consequently applied dlog2 (m + 1)e times. Algebra of Convex Hulls Theorem 3.12. For sets X1 , . . . , Xr ⊂ Rn and scalars µ1 , . . . , µr , one has conv (µ1 X1 + · · · + µr Xr ) = µ1 conv X1 + · · · + µr conv Xr . Proof. Excluding the trivial case when at least one of the sets X1 , . . . , Xr is empty, we assume that all these sets are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 (if r = 1, then we can write µ1 X1 = µ1 X1 + µ2 o). Since the set µ1 conv X1 + µ2 conv X2 is convex (see Theorem 2.9), the obvious inclusion µ1 X1 + µ2 X2 ⊂ µ1 conv X1 + µ2 conv X2 , and Theorem 3.2 give conv (µ1 X1 + µ2 X2 ) ⊂ µ1 conv X1 + µ2 conv X2 . For the opposite inclusion, choose an x ∈ µ1 conv X1 +µ2 conv X2 . Then x = µ1 x1 + µ1 x2 for certain points x1 ∈ conv X1 and x2 ∈ conv X2 . By Theorem 3.3, x1 and x2 can be written as convex combinations x1 = λ1 u1 + · · · + λp up
and x2 = γ1 v1 + · · · + γq vq
of certain points u1 , . . . , up ∈ X1 and v1 , . . . , vq ∈ X2 . Because µ1 ui + µ2 vj ∈ µ1 X1 + µ2 X2
for all
1 6 i 6 p and 1 6 j 6 q,
the equalities x = µ1 x1 + µ2 x2 =
p P q P i=1 j=1
λi γj (µ1 ui + µ2 vj ),
p P q P
λi γj = 1,
i=1 j=1
show that x is a convex combination of points from µ1 X1 + µ2 X2 . Theorem 3.3 gives x ∈ conv (µ1 X1 + µ2 X2 ).
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A sharper version of Theorem 3.12 for the case of large sums is considered in Exercise 3.9. The next theorem describes a “decomposition” of the convex hull of the union of sets (compare with Theorem 3.10). Theorem 3.13. For nonempty sets X1 , . . . , Xr in Rn , one has conv (X1 ∪ · · · ∪ Xr ) = {λ1 x1 + · · · + λr xr : x1 ∈ conv X1 , . . . , xr ∈ conv Xr , λ1 , . . . , λr > 0, λ1 + · · · + λr = 1}. Proof. Theorem 3.2 implies the inclusion conv X1 ∪ · · · ∪ conv Xr ⊂ conv (X1 ∪ · · · ∪ Xr ). This argument and Theorem 3.3 show that the set M = {λ1 x1 + · · · + λr xr : x1 ∈ conv X1 , . . . , xr ∈ conv Xr , λ1 , . . . , λr > 0, λ1 + · · · + λr = 1} lies in conv (X1 ∪ · · · ∪ Xr ). Conversely, let x ∈ conv (X1 ∪ · · · ∪ Xr ). By the same Theorem 3.3, x is expressible as a convex combination x = µ1 z1 + · · · + µk zk ,
where
z1 , . . . , zk ∈ X1 ∪ · · · ∪ Xr .
Renumbering the terms µ1 z1 , . . . , µk zk , we assume that {z1 , . . . , zk } is partitioned into subsets Y1 , . . . , Yr (some of them may be empty) such that Yi = {zpi−1 +1 , . . . , zpi } ⊂ Xi , 1 6 i 6 r, where p0 = 0, pr = k. Let λi = µpi−1 +1 + · · · + µpi , 1 6 i 6 r. Denote by I the set of all indices i ∈ {1, . . . , r} satisfying the condition: either Yi is empty or λi = 0. Let J = {1, . . . , r} \ I. For every i ∈ I, choose a point xi ∈ Xi , and for every i ∈ J, let µp +1 µp xi = i−1 zpi−1 +1 + · · · + i zpi . λi λi Then every xi is a convex combination of points from Xi , i ∈ J, and Theorem 3.3 shows that xi ∈ conv Xi for all 1 6 i 6 r. Finally, x = λ1 x1 + · · · + λr xr is a desired convex combination. Corollary 3.14. For nonempty convex sets K1 , . . . , Kr in Rn , one has conv (K1 ∪ · · · ∪ Kr ) = {λ1 x1 + · · · + λr xr : x1 ∈ K1 , . . . , xr ∈ Kr , λ1 , . . . , λr > 0, λ1 + · · · + λr = 1}. Theorem 3.15. For nonempty planes L1 and L2 in Rn , the following statements hold.
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(1) If L1 ∩ L2 6= ∅, then conv (L1 ∪ L2 ) = conv2 (L1 ∪ L2 ) = aff (L1 ∪ L2 ). (2) If L1 ∩ L2 = ∅ and G is the open slab of aff (L1 ∪ L2 ) determined by the parallel planes L01 = L1 + sub L2 and L02 = L2 + sub L1 , then conv (L1 ∪ L2 ) = conv2 (L1 ∪ L2 ) = L1 ∪ G ∪ L2 . Proof. (1) Let c ∈ L1 ∩ L2 . A combination of Theorem 1.56 and Corollary 3.14 shows that it suffices to prove the equality conv2 (L1 ∪ L2 ) = L1 + L2 − c. From Theorem 3.12 it follows that the latter equality can be rewritten as conv2 (S1 ∪ S2 ) = S1 + S2 , where S1 = L1 − c and S2 = L2 − c. Clearly, both S1 and S2 are subspaces. Because the subspace S1 + S2 is a convex set which contains S1 ∪ S2 , one has conv2 (S1 ∪ S2 ) ⊂ S1 + S2 . For the opposite inclusion, choose a point u ∈ S1 +S2 . Then u = u1 +u2 , where u1 ∈ S1 and u2 ∈ S2 . Consequently, 2u1 ∈ S1 and 2u2 ∈ S2 , which gives u = 12 (2u1 + 2u2 ) ∈ [2u1 , 2u2 ] ⊂ conv2 (S1 ∪ S2 ). (2) As above, Corollary 3.14 shows that it suffices to prove the equality conv2 (L1 ∪ L2 ) = L1 ∪ G ∪ L2 . Since all the sets L1 , G, L2 are convex, and since L1 and L2 belong to the opposite components of rbd G, we easily conclude that the set L1 ∪ G ∪ L2 is convex. Consequently, conv2 (L1 ∪ L2 ) ⊂ L1 ∪ G ∪ L2 . Conversely, since L1 ∪ L2 ⊂ conv2 (L1 ∪ L2 ), it suffices to prove the inclusion G ⊂ conv2 (L1 ∪ L2 ). For this, choose a point x ∈ G. Statement (1) of Theorem 1.56 implies that x ∈ hc1 , c2 i for suitable points c1 ∈ L1 and c2 ∈ L2 . Since G lies between L01 and L02 , we conclude that, in fact, x ∈ (c1 , c2 ) ⊂ conv2 (L1 ∪ L2 ). Theorem 3.16. For an affine transformation f : Rn → Rm and sets X ⊂ Rn and Y ⊂ Rm , one has conv f (X) = f (conv X), conv f
−1
(Y ) = f
−1
(conv (Y ∩ rng f )).
(3.2) (3.3)
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Proof. Excluding the trivial cases X = ∅ and Y ∩ rng f = ∅, we assume that both sets X and Y ∩ rng f are nonempty. Consequently, f −1 (Y ) 6= ∅. For the equality (3.2), choose a point x ∈ conv f (X). By Theorem 3.3, x can be written as a convex combination x = λ1 x1 + · · · + λp xp ,
x1 , . . . , xp ∈ f (X).
where
Let z1 , . . . , zp be points in X such that f (zi ) = xi for all 1 6 i 6 p. Put z = λ1 z1 + · · · + λp zp . Then z ∈ conv X by same theorem, and it follows that x = λ1 x1 + · · · + λp xp = λ1 f (z1 ) + · · · + λp f (zp ) = f (z) according to Theorem 1.87. Hence x = f (z) ∈ f (conv X), which proves the inclusion conv f (X) ⊂ f (conv X). Conversely, let x ∈ conv X. Similarly to the above, x can be written as a convex combination x = µ1 x1 + · · · + µq xq ,
where
x1 , . . . , xq ∈ X.
Then f (x) = µ1 f (x1 ) + · · · + µq f (xq ) according to Theorem 1.87, and Theorem 3.3 gives f (x) ∈ conv f (X). Hence f (conv X) ⊂ conv f (X). It remains to prove (3.3). Letting X = f −1 (Y ) in (3.2), one has f (conv f −1 (Y )) = conv f (f −1 (Y )) = conv (Y ∩ rng f ). Hence conv f −1 (Y ) ⊂ f −1 (f (conv f −1 (Y ))) = f −1 (conv (Y ∩ rng f )). For the opposite inclusion, let x ∈ f −1 (conv (Y ∩ rng f )). Then f (x) belongs to conv (Y ∩rng f ), and Theorem 3.3 implies that f (x) can be written as a convex combination f (x) = γ1 x1 + · · · + γr xr ,
where
x1 , . . . , xr ∈ Y ∩ rng f.
−1
Choose points z1 , . . . , zr ∈ f (Y ) such that f (zi ) = xi , 1 6 i 6 r, and put z = γ1 z1 + · · · + γr zr . Then z ∈ conv f −1 (Y ) and f (z) = γ1 f (z1 ) + · · · + γr f (zr ) = γ1 x1 + · · · + γr xr = f (x). Let ui = zi + (x − z), 1 6 i 6 r. Since ui is an affine combination, we have f (ui ) = f (zi ) + f (x) − f (z) = f (zi ) = xi . Hence ui ∈ f
−1
(xi ) ⊂ f −1 (Y ), 1 6 i 6 r. From the equalities x = ui − zi + z,
1 6 i 6 r,
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it follows that x can be written as a convex combination of u1 , . . . , ur : x = (γ1 + · · · + γr )x = γ1 (u1 − z1 + z) + · · · + γr (ur − zr + z) = (γ1 u1 + · · · + γr ur ) − (γ1 z1 + · · · + γr zr ) + (γ1 + · · · + γr )z = γ1 u1 + · · · + γr ur . Therefore, x ∈ conv f −1 (Y ). Summing up, f −1 (conv (Y ∩ rng f )) ⊂ conv f −1 (Y ). Remark. The equality (3.3) implies the inclusion conv f −1 (Y ) ⊂ f −1 (conv Y ), which may be proper. Indeed, let f : R2 → R2 be the orthogonal projection on the x-axis of R2 and Y = {(0, 1), (0, −1)}. Then conv f −1 (Y ) = f −1 (Y ) = ∅, while conv Y = {(0, y) : −1 6 y 6 1} and f −1 (conv Y ) is the y-axis.
3.2
Topological Properties of Convex Hulls
Convex Hulls and Closure Theorem 3.17. For a set X ⊂ Rn , one has conv (cl X) ⊂ cl (conv X). Furthermore, if X bounded, then conv X also is bounded and conv (cl X) = cl (conv X). Consequently, conv X is compact provided X is compact. Proof. Excluding the trivial case X = ∅, we assume that X is nonempty. Since X ⊂ conv X, one has cl X ⊂ cl (conv X). By Theorem 2.35, cl (conv X) is a convex set. Hence conv (cl X) ⊂ cl (conv X). Let X be bounded, and Bρ (c) ⊂ Rn be a ball containing X. Since Bρ (c) is a convex set (see example on page 72), Theorem 3.2 implies the inclusion conv X ⊂ Bρ (c). Hence conv X is a bounded set. For the equality conv (cl X) = cl (conv X), it remains to show, by the above proved, that cl (conv X) ⊂ conv (cl X). Let x ∈ cl (conv X). Then x = limi→∞ xi for a sequence of points x1 , x2 , . . . from conv X. Put m = dim X. According to Theorem 3.6, every xi can be written as a convex combination (i) (i)
(i)
(i)
xi = λ1 y1 + · · · + λm+1 ym+1 ,
(i)
(i)
y1 , . . . , ym+1 ∈ X,
i > 1.
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(1)
(2)
Consider the m + 1 sequences of points yk , yk , . . . from X and m + (1) (2) 1 sequences of scalars λk , λk , . . . from [0, 1], 1 6 k 6 m + 1. By a compactness argument, there is an infinite sequence of indices i1 , i2 , . . . such that all 2m + 2 subsequences (i )
(i )
yk 1 , yk 2 , . . . ,
(i )
(i )
λk 1 , λk 2 , . . . ,
1 6 k 6 m + 1,
converge. Put (i )
yk = lim yk s s→∞
(i )
and λk = lim λk s , s→∞
1 6 k 6 m + 1.
Then y1 , . . . , ym+1 ∈ cl X, λ1 , . . . , λm+1 > 0, λ1 + · · · + λm+1 = 1. Furthermore, (i ) (i )
(i )
(i )
s s x = lim xis = lim (λ1 s y1 s + · · · + λm+1 ym+1 )
s→∞
s→∞
= λ1 y1 + · · · + λm+1 ym+1 , which shows that x is a convex combination of y1 , . . . , ym+1 . Hence x belongs to conv (cl X), and the inclusion cl (conv X) ⊂ conv (cl X) is proved. If X is compact, then the equality conv X = conv (cl X) = cl (conv X) implies the closedness of conv X. Since conv X is bounded, it also is compact. Remark. If a set X ⊂ Rn is closed but not compact, then conv X may be nonclosed. For example, the set X = {(x, y) : y = 1/|x|} ⊂ R2 is closed, while its convex hull is the open halfplane {(x, y) : y > 0}. The following particular result gives a closedness condition for the convex hull of a pair of planes. Theorem 3.18. For nonempty planes L1 and L2 in Rn , the set conv (L1 ∪ L2 ) is closed if and only if any of the following two conditions holds: (a) L1 ∩ L2 6= ∅,
(b) L1 is a translate of L2 .
Proof. If L1 ∩ L2 6= ∅, then, by Theorem 3.15, conv (L1 ∪ L2 ) coincides with the plane aff (L1 ∪ L2 ), which is a closed set (see Corollary 1.22). If L1 and a translate of L2 , then, by the same theorem, conv (L1 ∪ L2 ) is the closed slab of aff (L1 ∪ L2 ) bounded by L1 and L2 , which also is a closed set (see Corollary 1.44).
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Conversely, let conv (L1 ∪ L2 ) be a closed set. If L1 ∩ L2 = ∅, then, by Theorem 3.15, conv (L1 ∪ L2 ) = L1 ∪ G ∪ L2 , where G is the open slab of aff (L1 ∪ L2 ) bounded by the parallel planes L01 = L1 + sub L2 and L02 = L2 + sub L1 . Since the closure of conv (L1 ∪ L2 ) equals L01 ∪ G ∪ L02 , one has L1 = L01 and L2 = L2 . The latter is possible if and only if sub L1 = sub L2 , that is, when L1 is a translate of L2 . Convex Hulls and Relative Interior Theorem 3.19. For a set X in Rn , one has conv (rint X) ⊂ rint (conv X). Furthermore, the set conv X is relatively open provided X is relatively open. Proof. By Theorem 3.2, aff X = aff (conv X). The inclusion X ⊂ conv X and Theorem 2.15 imply that rint X ⊂ rint (conv X). Since rint (conv X) is convex (see Corollary 2.22), one has conv (rint X) ⊂ rint (conv X). If X is relatively open (that is, X = rint X), then, by the above proved, conv X = conv (rint X) ⊂ rint (conv X) ⊂ conv X. Hence rint (conv X) = conv X, and conv X is relatively open. Theorem 3.20. Let X ⊂ Rn be a nonempty set. A point x ∈ Rn belongs to rint (conv X) if and only if x is a positive convex combination of finitely many points x1 , . . . , xk ∈ X such that aff {x1 , . . . , xk } = aff X. Proof. Put m = dim X. Assume first that x ∈ rint (conv X). According to Corollary 1.76, X contains m + 1 affinely independent points y1 , . . . , ym+1 with the property aff X = aff {y1 , . . . , ym+1 }. Let ∆ = ∆(y1 , . . . , ym+1 )
and y =
1 m+1 (y1
+ · · · + ym+1 ).
A combination of Theorems 2.15 and 2.17 gives y ∈ rint ∆ ⊂ rint (conv X). Since the case x = y is obvious, we assume that x and y are distinct. Theorem 2.24 shows the existence of a scalar γ > 1 such that the point z = γx + (1 − γ)y belongs to conv X. By Theorem 3.3, z can be written as a convex combination z = λ1 z1 +· · ·+λp zp of certain points z1 , . . . , zp ∈ X. Excluding zero scalars, we suppose that all λ1 , . . . , λp are positive. Then x = (1 − γ −1 )y + γ −1 z =
(3.4) 1 − γ −1 1 − γ −1 y1 + · · · + ym+1 + γ −1 λ1 z1 + · · · + γ −1 λp zp m+1 m+1
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is a positive convex combination of y1 , . . . , ym+1 , z1 , . . . , zp . Theorem 1.50 and the inclusions {y1 , . . . , ym+1 } ⊂ {y1 , . . . , ym+1 , z1 , . . . , zp } ⊂ aff X = aff {y1 , . . . , ym+1 } show that the affine span of these points equals aff X. Hence (3.4) gives a desired expression of x. Conversely, let x = λ1 x1 + · · · + λk xk be a positive convex combination of points x1 , . . . , xk ∈ X satisfying the condition aff {x1 , . . . , xk } = aff X. According to Theorem 1.68, the set {x1 , . . . , xk } contains m + 1 affinely independent points whose affine span equals aff X. Without loss of generality, we may suppose that these are exactly x1 , . . . , xm+1 . If k = m + 1, then conv {x1 , . . . , xm+1 } is the m-simplex ∆ = ∆(x1 , . . . , xm+1 ), and a combination of Theorems 2.17 and 2.15 gives x ∈ rint ∆ ⊂ rint (conv X). Let k > m + 2, and put λ = λ1 + · · · + λm+1 . Clearly, 0 < λ < 1, and λm+1 λm+2 λk λ1 x1 + · · · + xm+1 , z = xm+2 + · · · + xk y= λ λ 1−λ 1−λ are convex combinations of points from X. Hence y, z ∈ conv X according to Theorem 3.3. Since y ∈ rint ∆ ⊂ rint (conv X), Theorem 2.21 implies the inclusion x = λy + (1 − λ)z ∈ rint (conv X). Remark. Unlike Theorem 3.5, the points x1 , . . . , xk from Theorem 3.20 cannot be chosen affinely independent. Indeed, let X = {x1 , x2 , x3 , x4 } be the set of points in R2 depicted below. Clearly, o ∈ int (conv X) and X is affinely dependent. At the same time, no proper subset Y of X satisfies the condition o ∈ int (conv Y ). x3 r Q Q Q Qr r r x1 Q x4 o Q Q Q r x2 The next corollary can be viewed as a generalization of Theorem 2.17. Corollary 3.21. For a finite set X = {x1 , . . . , xr } in Rn , one has rint (conv X) = {λ1 x1 + · · · + λr xr : λ1 , . . . , λr > 0, λ1 + · · · + λr = 1}.
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Proof. Theorem 3.20 shows that every positive convex combination of points x1 , . . . , xr belongs to rint (conv X). Conversely, choose a point x in rint (conv X). Since the point y = 1r (x1 + · · · + xr ) is a convex combination of x1 , . . . , xr and whence belongs to conv X, there is a scalar γ > 1 such that z = γx + (1 − γ)y ∈ conv X (see Theorem 2.24). By Corollary 3.4, z can be written as a convex combination z = µ1 x1 + · · · + µr xr . Consequently, x = (1 − γ −1 )y + γ −1 z 1 − γ −1 1 − γ −1 = + γ −1 µ1 x1 + · · · + + γ −1 µr xr r r is a positive convex combination of x1 , . . . , xr . Algebra of Convex Hulls and Relative Interior A combination of Theorems 2.29 and 3.12 implies the following statement. Corollary 3.22. For sets X1 , . . . , Xr ⊂ Rn and scalars µ1 , . . . , µr , rint (conv (µ1 X1 + · · · + µr Xr )) = µ1 rint (conv X1 ) + · · · + µr rint (conv Xr ). The result below complements Theorem 3.13. Theorem 3.23. For nonempty sets X1 , . . . , Xr ⊂ Rn , the relative interior of conv (X1 ∪ · · · ∪ Xr ) is the collection of all positive convex combinations of the form λ1 x1 + · · · + λr xr , where xi ∈ rint (conv Xi ) for all 1 6 i 6 r. (3.5) Consequently, rint (conv (X1 ∪ · · · ∪ Xr )) ⊂ conv (rint (conv X1 ) ∪ · · · ∪ rint (conv Xr )). Proof. Denote by M the set of all points of the form (3.5). Then the statement can be rewritten as M = rint (conv (X1 ∪ · · · ∪ Xr )).
(3.6)
First, we establish the convexity of M . Indeed, let x = γx1 + · · · + γxr
and y = µ1 y1 + · · · + µr yr
be points from M , expressed in the form (3.5). For any 0 < λ < 1, the point (1 − λ)x + λy = η1 z1 + · · · + ηr zr is a positive convex combination, where (1 − λ)γi λµi ηi = (1 − λ)γi + λµi and zi = xi + yi , 1 6 i 6 r. ηi ηi
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By Theorem 2.21, zi ∈ (xi , yi ) ⊂ rint (conv Xi ),
1 6 i 6 r,
which shows that (1 − λ)x + λy ∈ M . Summing up, M is a convex set. Next, we will prove the inclusion M ⊂ rint (conv (X1 ∪ · · · ∪ Xr )). For this, choose a point x ∈ M , expressed in the form (3.5). By Theorem 3.20, (i) (i) (i) (i) each xi is a positive convex combination xi = µ1 y1 + · · · + µpi ypi , where (i)
(i)
y1 , . . . , yp(i) ∈ Xi and aff {y1 , . . . , yp(i) } = aff Xi , 1 6 i 6 r. i i Clearly, (1) (1)
(r) (r)
(1) (r) (r) x = λ1 µ1 y1 + · · · + λ1 µ(1) p1 yp1 + · · · + λr µ1 y1 + · · · + λr µpr ypr
is a positive convex combination of all points from the set (1)
(r)
Y = {y1 , . . . , yp(1) , . . . , y1 , . . . , yp(r) }. 1 r Furthermore, Theorems 1.50 and 3.2 give (1)
(r)
aff Y = aff (aff {y1 , . . . , yp(1) } ∪ · · · ∪ aff {y1 , . . . , yp(r) }) 1 r = aff (aff X1 ∪ · · · ∪ aff Xr ) = aff (X1 ∪ · · · ∪ Xr ) = aff (conv (X1 ∪ · · · ∪ Xr )). Therefore, x ∈ rint (conv (X1 ∪ · · · ∪ Xr )) according to Theorem 3.20. Next, we state that conv (X1 ∪ · · · ∪ Xr ) ⊂ cl M . For this, let x ∈ conv (X1 ∪ · · · ∪ Xr ). To prove the inclusion x ∈ cl M , it suffices to show that for every ρ > 0 there is a point z ∈ M such that kx − zk < ρ. By Theorem 3.13, x can be expressed as a convex combination x = γ1 x 1 + · · · + γr x r ,
where
x1 ∈ conv X1 , . . . , xr ∈ conv Xr .
Renumbering the sets X1 , . . . , Xr , we assume that all positive scalars γi are placed before all zero scalars: γ1 , . . . , γ s > 0
and γs+1 = · · · = γr = 0,
1 6 s 6 r.
Consider first the case s = r. Since conv Xi ⊂ cl (rint (conv Xi )) due to Theorem 2.38, there are points zi ∈ rint (conv Xi ) satisfying the conditions kxi − zi k 6 ρ for all 1 6 i 6 r. Let z = γ1 z1 + · · · + γr zr . Then z ∈ M and kx − zk 6 γ1 kx1 − z1 k + · · · + γr kxr − zr k 6 ρ. Now, assume that s < r and put µ = ρ2 ,
δ = max {kx1 k, . . . , kxr k},
ρ 0 < ε < min γ1 , . . . , γs , 4δs .
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As above, choose points zi ∈ rint (conv Xi ) such that kxi − zi k 6 µ for all 1 6 i 6 r. Then kzi k 6 kzi − xi k + kxi k 6 µ + δ, 1 6 i 6 r. Consider the positive convex combination z = (γ1 − ε)z1 + · · · + (γs − ε)zs +
εs εs zs+1 + · · · + zr . r−s r−s
Clearly, z ∈ M and kx − zk 6 kγ1 x1 − (γ1 − ε)z1 k + · · · + kγs xs − (γs − ε)zs k εs εs + kzs+1 k + · · · + kzr k r−s r−s = k(γ1 − ε)(x1 − z1 ) + εx1 k + · · · + k(γs − ε)(xs − zs ) + εxs k εs kzs+1 k + · · · + kzr k + r−s 6 (γ1 − ε)kx1 − z1 k + · · · + (γs − ε)kxs − zs k εs + ε kx1 k + · · · + kxs k + kzs+1 k + · · · + kzr k r−s εs 6 (γ1 + · · · + γs − εs)µ + εsδ + (r − s)(δ + µ) r−s ρ ρ δs = ρ. = µ + 2εδs < + 2 2 4δs Summing up, x ∈ cl M , and conv (X1 ∪ · · · ∪ Xr ) ⊂ cl M . Finally, Theorem 2.38 gives M ⊂ rint (conv (X1 ∪ · · · ∪ Xr )) ⊂ rint (cl M ) = rint M ⊂ M, implying (3.6). Theorem 3.24. For convex sets K1 , . . . , Kr ⊂ Rn , one has rint (conv (K1 ∪ · · · ∪ Kr )) ⊂ conv (rint K1 ∪ · · · ∪ rint Kr ).
(3.7)
If the relative interiors of K1 , . . . , Kr have a point in common, then rint (conv (K1 ∪ · · · ∪ Kr )) = conv (rint K1 ∪ · · · ∪ rint Kr ).
(3.8)
Proof. The inclusion (3.7) immediately follows from Theorem 3.23. Assume that the relative interiors of K1 , . . . , Kr have a point c in common. Due to (3.7), it suffices to prove that conv (rint K1 ∪ · · · ∪ rint Kr ) ⊂ rint (conv (K1 ∪ · · · ∪ Kr )). First, we state that c ∈ rint (conv (K1 ∪ · · · ∪ Kr )). Indeed, let x be any point in conv (K1 ∪ · · · ∪ Kr ). By Theorem 3.13, x is a convex combination,
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x = λ1 x1 + · · · + λr xr , of some points xi ∈ Ki , 1 6 i 6 r. Theorem 2.24 shows the existence of scalars γi > 1 such that the point γi c + (1 − γi )xi belongs to Ki , 1 6 i 6 r. Replacing each γi with γ = min{γ1 , . . . , γr }, a convexity argument implies the inclusions yi = γc + (1 − γ)xi ∈ Ki for all 1 6 i 6 r. Finally, with y = λ1 y1 + · · · + λr yr , we obtain the equality y = γc + (1 − γ)x. Since 0 < γ −1 < 1, Theorem 2.24 gives c = (1 − γ −1 )x + γ −1 y ∈ rint (conv (K1 ∪ · · · ∪ Kr )). Next, choose any point z ∈ conv (rint K1 ∪ · · · ∪ rint Kr ). As above, z is a convex combination, z = µ1 z1 + · · · + µr zr , of some points zi ∈ rint Ki , 1 6 i 6 r. By Theorem 2.24, there are scalars ηi > 1 such that the point ui = ηi zi + (1 − ηi )c belongs to Ki , 1 6 i 6 r. As above, we may suppose that η1 = · · · = ηr = η. Let u = µ1 u1 + · · · + µr ur . Then u ∈ conv (K1 ∪ · · · ∪ Kr ), and Theorem 2.24 gives z = (1 − η −1 )c + η −1 u ∈ rint (conv (K1 ∪ · · · ∪ Kr )). Remark. The inclusion (3.7) may be proper. Indeed, if K1 = {(0, y) : 0 6 y 6 1} and K2 = {(1, y) : 0 6 y 6 1}, then rint (conv (K1 ∪ K2 )) = {(x, y) : 0 < x < 1, 0 < y < 1}, conv (rint K1 ∪ rint K2 ) = {(x, y) : 0 < x < 1, 0 6 y 6 1}. The next corollary follows from Theorems 2.34 and 3.16. Corollary 3.25. For an affine transformation f : Rn → Rm and sets X ⊂ Rn and Y ⊂ Rm , one has rint (conv f (X)) = f (rint (conv X)), rint (conv f −1 (Y )) = f −1 (rint (conv (Y ∩ rng f ))).
Steinitz’s Theorem If X ⊂ Rn is a nonempty set, then a combination of Theorem 3.20 and Corollary 3.21 shows that a point x ∈ Rn belongs to rint (conv X) if and only if there is a finite subset Y of X such that x ∈ rint (conv Y ) and aff Y = aff X. Theorem 3.26, which is attributed to Steinitz, sharpens this statement by establishing bounds on the cardinality of Y . The picture below, with X = {x1 , x2 , x3 , x4 , x5 } and x ∈ int (conv X), illustrates two possible cases described in this theorem for m = n = 2 (where Y = {x1 , x2 , x4 } corresponds to the first case, and Y = {x1 , x2 , x3 , x4 } does to the second one).
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xr 3 !r x4 !!
! r ! r
r ! x x@ x5 2 @
r @
x1
xr 3 @ r x@ 2
r @ @r x x4
r x5
@ @r x1
Theorem 3.26. Let X ⊂ Rn be a set of positive dimension m. A point x ∈ Rn belongs to rint (conv X) if and only if there is a subset Y of X such that x ∈ rint (conv Y ),
m + 1 6 card Y 6 2m,
aff Y = aff X.
(3.9)
Furthermore, if Y is a minimal subset of X satisfying (3.9), then either card Y 6 2m − 1 or Y consists of 2m points collinear in pairs with x. Proof. If a subset Y of X satisfies the conditions (3.9), then Theorem 3.20 and Corollary 3.21 imply the inclusion x ∈ rint (conv X). Conversely, let x ∈ rint (conv X). By the same argument, there is a finite subset Z ⊂ X with the properties x ∈ rint (conv Z) and aff Z = aff X. We are going to choose in Z a subset Y satisfying (3.9). Because the cases m 6 1 are obvious, we suppose that m > 2. Denote by F the family of planes in Rn , each of dimension m − 1 or less, such that every plane L ∈ F contains x and is the affine span of points from {x}∪Z. Since the set Z is finite, the family F also is finite. Choose in aff X a line l through x which does not lie in any plane L ∈ F. According to Theorem 3.17, conv Z is a compact set, and from Theorem 2.55 it follows that the intersection of l and conv Z is a segment [y, y 0 ], where y, y 0 ∈ rbd (conv Z) and x ∈ (y, y 0 ). So, x = (1 − η)y + ηy 0 for a scalar η ∈ (0, 1). By Theorem 3.6, y is a positive convex combination, y = λ1 x1 + · · · + λk xk , of m + 1 or fewer affinely independent points x1 , . . . , xk ∈ Z. According to Corollary 3.21, y ∈ rint (conv {x1 , . . . , xk }). We state that k = m and x ∈ / aff {x1 , . . . , xm }. Indeed, assume for a moment that k = m + 1. Then dim (aff {x1 , . . . , xm+1 }) = m according to Corollary 1.66. Since aff {x1 , . . . , xm+1 } ⊂ aff Z
and
dim (aff Z) = m,
Theorem 1.6 implies that aff {x1 , . . . , xm+1 } = aff Z. Consequently, Theorem 2.15 gives y ∈ rint (conv {x1 , . . . , xm+1 }) ⊂ rint (conv Z), in contradiction with y ∈ rbd (conv Z). Hence k 6 m.
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If k 6 m − 1 or x belonged to aff {x1 , . . . , xk }, then the plane L = aff {x, x1 , . . . , xk } would be a member of F, and l = hx, yi ⊂ L, contrary to the choice of l. Summing up, y = λ1 x1 + · · · + λm xm , where the set {x, x1 , . . . , xm } is affinely independent (see Corollary 1.63). Similarly, y 0 is a positive convex combination, y 0 = µ1 x01 + · · · + µm x0m , of points x01 , . . . , x0m ∈ Z such that the set {x, x01 , . . . , x0m } is affinely independent. Put Y = {x1 , . . . , xm , x01 , . . . , x0m }. Clearly, card Y 6 2m and aff Y = aff X. Since x is expressible as a positive convex combination x = (1 − η)y + ηy 0 = (1 − η)λ1 x1 + · · · + (1 − η)λm xm + ηµ1 x01 + · · · + ηµm x0m , Theorem 3.20 implies that x ∈ rint (conv Y ). Finally, card Y > m + 1 by Corollary 1.76. Now, suppose that Y is a minimal subset of X satisfying the conditions (3.9), with card Y = 2m. Then, with the notation above, y and y 0 belong, respectively, to the relative interiors of the simplices ∆ = ∆(x1 , . . . , xm )
and
∆0 = ∆(x01 , . . . , x0m ).
We observe that both ∆ and ∆0 lie in rbd (conv Z). Indeed, assume for a moment that ∆ contains a point u ∈ rint (conv Z). According to Theorem 2.24, there is a scalar γ > 1 satisfying the condition v = γy + (1 − γ)u ∈ ∆ ⊂ conv Z, and Theorem 2.21 gives y = (1 − γ −1 )u + γ −1 v ∈ rint (conv Z), in contradiction with y ∈ rbd (conv Z). The above argument holds for every choice of y in rint ∆ and the corresponding point y 0 in rint ∆0 . Clearly, y → y 0 is a one-to-one correspondence between rint ∆ and rint ∆0 . Allowing some scalars λi , µj from y = λ1 x1 + · · · + λm xm
and y 0 = µ1 x01 + · · · + µm x0m ,
take zero values, we expand this correspondence to a bijection f : ∆ → ∆0 (see Theorem 2.16). We state that f ({x1 , . . . , xm }) ⊂ {x01 , . . . , x0m }. Indeed, assume for a moment that f (x1 ) ∈ / 0 as a convex combination f (x1 ) = α1 x1 + scalars, say α1 and α2 , are not 0. Put
(3.10)
{x01 , . . . , x0m }, and express f (x1 ) · · · + αm x0m , where at least two
z1 = (α1 + α2 )x01 + α3 x03 + · · · + αm x0m , z2 = (α1 + α2 )x02 + α3 x03 + · · · + αm x0m .
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Then z1 and z2 are distinct points in ∆0 . Hence their pre-images u1 = f −1 (z1 ) and u2 = f −1 (z2 ) are distinct points in ∆. Because of α1 α2 f (x1 ) = z1 + z2 ∈ (z1 , z2 ), α1 + α2 α1 + α2 the point x1 should belong to (u1 , u2 ), as depicted below. u1r PP PPr r r PP x x1 f (x1 ) P PP PP r r z u2 P 1 PP rz2 PP
Let x1 = (1 − λ)u1 + λu2 , where 0 < λ < 1. Since both u1 and u2 belong to ∆, they can be expressed as convex combinations u1 = β1 x1 + · · · + βm xm
and u2 = γ1 x1 + · · · + γm xm .
Consequently, x1 = (1 − λ)u1 + λu2 = ((1 − λ)β1 + λγ1 )x1 + · · · + ((1 − λ)βm + λγm )xm . Because x1 = 1x1 + 0x2 + · · · + 0xm is the only way to express x1 as an affine combination of points x1 , . . . , xm (see Theorem 1.60), one has (1 − λ)β1 + λγ1 = 1,
(1 − λ)βi + λγi = 0,
2 6 i 6 m.
(3.11)
Solving (3.11) for βi and γi , we obtain β1 = γ1 = 1 and βi = γi = 0 for all 2 6 i 6 m. Thus u1 = u2 , contrary to the assumption. So, the inclusion (3.10) holds. This argument implies that the set Y can be partitioned into pairs {xi , x0j }, each of them collinear with x. An extension of Theorem 3.26 to the case of finite (or even compact) subsets of rint (conv X) is considered in Exercises 3.7 and 3.8. The following lemma will be of use in the next two theorems. Lemma 3.27. Let {x1 , . . . , xm } ⊂ Rn be a linearly independent set, and let y = λ1 x1 + · · · + λm xm , with exactly r nonzero scalars among λ1 , . . . , λm , 0 6 r 6 m. For negative scalars α1 , . . . , αm , there is a subset Y of the set Z = {x1 , α1 x1 , . . . , xm , αm xm } satisfying the conditions o ∈ rint (conv ({y} ∪ Y )),
card Y = 2m − r,
dim ({y} ∪ Y ) = m.
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Proof. Since the case r = 0 is trivial (put Y = Z), we suppose that r > 1. Renumbering the vectors x1 , . . . , xm , we assume that namely the scalars λ1 , . . . , λr are distinct from zero. Replacing, if necessary, each term λi xi with (αi λi )(αi−1 xi ) (and, respectively, xi with αi−1 xi ), one can suppose that all scalars λ1 , . . . , λr are negative. Suppose first that r = m. Dividing both parts of the equality o = y − λ1 x1 − · · · − λm xm by the positive scalar γ = 1 − λ1 − · · · − λm , we express o as a positive convex combination of the points y, x1 , . . . , xm . So, we can put Y = {x1 , . . . , xm }. Assume now that r 6 m − 1 and let x0i = αi xi , r + 1 6 i 6 m. Adding the equalities o = y − λ1 x1 − · · · − λr xr ,
o = −αi xi + x0i ,
r + 1 6 i 6 m,
we obtain o = y − (λ1 x1 + · · · + λr xr + αr+1 xr+1 + · · · + αm xm ) + x0r+1 + · · · + x0m . Dividing both parts of the latter equality by the positive scalar γ = m − r + 1 − (λ1 + · · · + λr + αr+1 + · · · + αm ), we express o as a positive convex combination of the points y, x1 , . . . , xm , x0r+1 , . . . , x0m . Put Y = {x1 , . . . , xm , x0r+1 , . . . , x0m }. From Theorem 1.60 it follows that the set {o, x1 , . . . , xm } is affinely independent. Therefore, Theorem 3.20 gives o ∈ rint (conv ({y} ∪ Y ))
and
dim ({y} ∪ Y ) = m.
The next statement complements Theorem 3.26. Theorem 3.28. For a set X ⊂ Rn of positive dimension m, the following conditions are equivalent. (1) There is a point x ∈ rint (conv X) such that every m-dimensional subset Y ⊂ X with the property x ∈ rint (conv Y ) has 2m or more points. (2) There is a point x ∈ Rn and lines l1 , . . . , lm through x such that: (a) l1 , . . . , lm are independent, (b) X ⊂ l1 ∪ · · · ∪ lm , (c) each of l1 , . . . , lm contains points of X on both sides of x.
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Proof. (1) ⇒ (2). Let a point x ∈ Rn satisfy condition (1). Translating X on −x, we assume that x = o. By Theorem 3.26, X contains a subset Z of 2m points which can be partitioned into pairs {xi , x0i } such that o ∈ (xi , x0i ) and the lines li = hxi , x0i i are independent, 1 6 i 6 m. It remains to show that X ⊂ l1 ∪ · · · ∪ lm . For this, choose a point y ∈ X. Because the case y = o is obvious, we may assume that y 6= o. Since {x1 , . . . , xm } is a basis for span X = aff X, the vector y is uniquely expressed as a linear combination y = λ1 x1 + · · · + λm xm . Let r be the number of nonzero scalars among λ1 , . . . , λm . Clearly, r > 1 because of y 6= o. By Lemma 3.27, there is a subset Y of Z such that o ∈ rint (conv ({y} ∪ Y )),
card Y = 2m − r,
dim ({y} ∪ Y ) = m.
By the assumption, 2m 6 card ({y} ∪ Y ). Therefore, 2m − 1 6 card Y = 2m − r, which gives r 6 1. Hence r = 1 and y is a scalar multiple of a vector xi , i ∈ {1, . . . , m}; that is, y ∈ li . Summing up, X ⊂ l1 ∪ · · · ∪ lm . (2) ⇒ (1). Let a point x ∈ Rn and lines l1 , . . . , lm through x satisfy conditions (a)–(c). Choose an m-dimensional subset Y of X \ {x} with the property z ∈ rint (conv Y ), and consider the partition Y = Y1 ∪ · · · ∪ Ym ,
where
Yi = Y ∩ li ,
1 6 i 6 m.
If at least one of the sets Y1 , . . . , Ym were empty, then Y would lie in the sum of fewer than m lines l1 , . . . , lm , which is impossible because of dim Y = m. Assume for a moment that at least one of the sets Y1 , . . . , Ym , say Y1 , lies on one side of x on the line l1 . Denote by D the closed halfplane of aff (l1 ∪ · · · ∪ lm ) which is determined by the plane L = aff (l2 ∪ · · · ∪ lm ) and contains Y1 . Clearly, x ∈ L, Y ⊂ D, and Y 6⊂ L. By Theorem 2.59, rint (conv Y ) ⊂ rint D, implying that rint (conv Y ) ∩ L = ∅. The last is in contradiction with x ∈ rint (conv Y ). Hence every set Yi contains a pair of points which lie on opposite sides of x. Summing up, Y contains at least 2m points. The following proposition is analogous to Theorem 3.7 Theorem 3.29. Let X ⊂ Rn be a set of positive dimension m, and let x0 ∈ X. A point x ∈ Rn belongs to rint (conv X) if and only if there is a subset Y ⊂ X such that x ∈ rint (conv ({x0 } ∪ Y )),
m 6 card Y 6 2m − 1,
aff ({x0 } ∪ Y ) = aff X.
(3.12)
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Proof. If a subset Y of X satisfies the conditions (3.12), then a combination of Theorem 3.20 and Corollary 3.21 implies the inclusion x ∈ rint (conv X). Conversely, let x ∈ rint (conv X). By Theorem 3.26, X contains a subset Z such that x ∈ rint (conv Z),
m + 1 6 card Z 6 2m,
aff Z = aff X.
Without loss of generality, we assume that Z is minimal. If card Z 6 2m − 1, then we put Y = Z \ {x0 }. Consequently, m 6 card Y 6 2m − 1
and
aff ({x0 } ∪ Y ) = aff Z = aff X
according to Theorem 1.50. Furthermore, Theorem 2.15 gives x ∈ rint (conv Z) ⊂ rint (conv ({x0 } ∪ Y )). Assume that card Z = 2m. Translating X on −x, we assume that x = o. By Theorem 3.26, Z consists of m disjoint pairs {xi , x0i } such that o ∈ (xi , x0i ) and the lines li = hxi , x0i i, 1 6 i 6 m, are independent. Excluding the trivial case x0 = o, we suppose that x0 6= o. Since {x1 , . . . , xm } is a basis for span X, the vector x0 is uniquely expressed as a linear combination x0 = λ1 x1 + · · · + λm xm . Let r be the number of nonzero scalars among λ1 , . . . , λm . Clearly, r > 1 because of x0 6= o. By Lemma 3.27, there is a subset Y of Z such that o ∈ rint (conv ({x0 } ∪ Y )), card Y = 2m − r, aff ({x0 } ∪ Y ) = aff X. Furthermore, Theorem 1.50 shows that card Y = 2m − r 6 2m − 1
and
aff ({x0 } ∪ Y ) = aff X.
By Corollary 1.76, one has m + 1 6 card ({x0 } ∪ Y ). Therefore, m 6 card Y . Exercises for Chapter 3 Exercise 3.1. Let ∆ = ∆(x1 , . . . , xr+1 ) be an r-simplex in Rn . For every proper subset X of {x1 , . . . , xr+1 }, denote by ∆1 and ∆2 the simplices with the vertex sets X and Y = {x1 , . . . , xr+1 } \ X, respectively. Show that ∆ = conv2 (∆1 ∪ ∆2 ) = ∪ ([u1 , u2 ] : u1 ∈ ∆1 , u2 ∈ ∆2 ). Exercise 3.2. Show that every closed ball Bρ (c) ⊂ Rn is the convex hull of its boundary sphere Sρ (c). Furthermore, Bρ (c) = conv2 Sρ (c).
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Exercise 3.3. Let Q = {(x1 , . . . , xn ) : 0 6 xi 6 1, i = 1, . . . , n} be the unit cube of Rn and X be the set of points (x1 , . . . , xn ) in Q with all irrational coordinates x1 , . . . , xn . Show that conv X = conv2 X = int Q. Exercise 3.4. Let a convex set K ⊂ Rn and points x, y ∈ Rn \ K satisfy the conditions x ∈ conv (y ∪ K) and y ∈ conv (x ∪ K). Show that x = y. Exercise 3.5. (Radon [177]) Show that a set X ⊂ Rn is affinely independent if and only if it does not contain disjoint subsets whose convex hulls meet. Exercise 3.6. (Reay [179, Lemma 4.1]) Let X ⊂ Rn be a set of positive dimension m and Y be a subset of conv X of cardinality r > 2. Show the existence of a set Z ⊂ X of cardinality rm such that Y ⊂ conv Z. Exercise 3.7. (Reay [179, Lemma 4.6]) Let X ⊂ Rn be a set of positive dimension m and Y be a subset of rint (conv X) of cardinality r > 2. Show the existence of a set Z ⊂ X of cardinality (r +1)m such that Y ⊂ rint (conv Z) and aff Z = aff X. Exercise 3.8. Let X ⊂ Rn and Y be a compact subset of rint (conv X). Show the existence of a finite subset Z of X such that Y ⊂ rint (conv Z) and aff Z = aff X. Exercise 3.9. Let X1 , . . . , Xr be nonempty sets in Rn , µ1 , . . . , µr ∈ R, and x ∈ conv (µ1 X1 + · · · + µr Xr ). Let r > m = dim (µ1 X1 + · · · + µr Xr ). Show the existence of an index set I = I(x) ⊂ {1, . . . , r}, card I 6 m, and points zi ∈ Xi , i ∈ {1, . . . , r} \ I, such that P P x∈ µi zi + µi conv Xi . i∈I /
i∈I
Notes for Chapter 3 Carath´ eodory’s theorem and its variations. The concept of convex hull is originated by Carath´eodory [51, 52], who defined it as the smallest convex set containing a compact set in Rn . Statement (2) of Theorem 3.6 was proved in [52, p. 200] for the case of compact sets X ⊂ Rn , using the support properties of convex sets and induction on n. The extension of Theorem 3.6 to arbitrary sets in Rn , as well as Theorems 3.3 and 3.5, are attributed to Steinitz [205, Part I, § 10]. Corollary 3.8 is proved by Reay [179, Lemma 4.1]. Cook [60] established the following analogue of Carath´eodory’s theorem for the case of convex combinations with linear constraints. Baker [19] showed that if a point x belongs to the convex
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hull of a finite set X ⊂ Rn such that card X = k and dim X = m, then there are k − m or more m-simplices with vertices in X that contain x. Fenchel [82] (for n = 3) and Stoelinga [209, Theorem 25] (for all n > 3) proved that every point in the convex hull of an n-dimensional connected set can be expressed as a convex combination of n points from X. Bunt [47, p. 23] (see also Kramer [138]) further observed that connectedness of X can be replaced by the weaker condition “has n or fewer components.” Hanner and R˚ adstr¨ om [111] defined a set M ⊂ Rn as convexly connected if there is no hyperplane H ⊂ Rn such that H ∩ M = ∅ and M has points in both open halfspaces determined by H. They proved that if a set X in Rn is compact and has n or fewer convexly connected components, then every point x ∈ X can be expressed as a convex combination of n points from X. Similar results on convex hulls of special curves in Rn are given by Derry [72], Danielyan and Movsisyan [64]. Tverberg [218] proved that if X is the union of convex sets in Rn , all meeting a fixed hyperplane, then every point x ∈ conv X is a convex combination of n points from X. See Danzer, Gr¨ unbaum, and Klee [66], Eckhoff [77], and Reay [179] for additional references on Carath´eodory-type theorems. Iterative construction of convex hulls. The property of convr described in Theorem 3.10 was discovered by Lepin [150] and independently by Klee (see [30]). A particular case r1 = · · · = rk = 2 and m = n of Corollary 3.11, which implies the inequality k 6 dlog2 (n + 1)e, is well-known due to Brunn [45] (also Abe, Kubota, and Yoneguchi [1], Bonnesen and Fenchel [29, p. 10], Danzer, Gr¨ unbaum, and Klee [66, pp. 116–117], and Straszewicz [212, pp. 29–36]). Topological properties of convex hulls. Theorem 3.18 is due to Bair [17]. Theorem 3.26 is proved by Steinitz [206, § 20] in terms of positive hulls. By using positive convex combinations of points, this theorem was proved by Robinson [183] for the case of convex polytopes (see also Blumenthal [27, pp. 197–198]); other proofs can be found in Bonnice and Klee [30], Gustin [108], Peterson [170], and Valentine [219, pp. 41–42]. Theorem 3.28 is due to Reay [179, p. 14]. Exercise 3.7 is a weaker version of a result of Reay [179, Lemma 4.6]: If X ⊂ Rn is a set of positive dimension m, and Y ⊂ rint (conv X) is a set of cardinality r > 2, then there is a set Z ⊂ X of cardinality rm such that Y ⊂ rint (conv Z) and aff Z = aff X. Given a set X ⊂ Rn , denote by intr X the union of relative interiors of all r-simplices contained in X; equivalently, a point x belongs to intr X if and only if there is an r-dimensional plane L ⊂ Rn such that x ∈ rint (L ∩ X). Bonnice and Klee [30] proved the following result, which becomes Carath´eodory’s theorem when r = 0 and Steinitz’s theorem when r = n: For a set X ⊂ Rn , an integer r (0 6 r 6 n), and a point x ∈ intr (conv X), there is a subset Y ⊂ X such that x ∈ intr (conv Y ) and card Y 6 max {2r, n + 1}. A simplified proof of this assertion is given by Reay [180]. Furthermore, Bonnice and Reay [31] showed that if m is the highest dimension of a simplex with vertices in X that contains x in its relative interior, then r + 1 + b(r − 1)/mc 6 card Y 6 max {2r − m + 1, min {2m + 1, n + 2}}
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for r 6 m, and r + 1 6 card Y 6 m + 1 for m 6 r − 1. Convex hull and large sums. Exercise 3.9 is a refinement of a lemma of Shapley and Folkman (see Starr [203]), which states that for every point x in conv (X1 + · · · + Xr ), where X1 , . . . , Xr are compact subset of Rn and r > n + 1, there is an index set I ⊂ {1, . . . , r} such that card I 6 n and x∈
P i∈I
conv Xi +
P
Xi .
i∈I /
Their proof was gradually simplified and expanded to the case of arbitrary sets by Artstein [6], Cassels [53], Howe [117], Starr [204], and Zhou [227]. Lawrence and Soltan [146] gave the following refinement of this statement: If X1 , . . . , Xr are nonempty sets in Rn , then for every point x ∈ conv (X1 + · · · + Xr ), there is an index set I ⊂ {1, . . . , r} with card I 6 n and nonempty subsets Yi ⊂ Xi , 1 6 i 6 r, such that x∈
P i∈I
conv Yi +
P i∈I /
Yi ,
P
card Yi 6 n + card I,
card Yi = 1, i ∈ / I.
i∈I
Shapley-Folkman’s lemma is used in an asymptotic approximation of the set conv (X1 + · · · + Xr ) by the sum of compact sets X1 , . . . , Xr ⊂ Rn , with applications in mathematical economics (see, e.g., Arrow and Hahn [5, pp. 392–400], Ekeland and Temam [80, Apendix 1], Ichiishi [119, pp. 24–25], Schneider [191, Capter 3] for additional references and results). For the case r = n + 1 and X1 = · · · = Xr = X, Shapley-Folkman’s lemma gives X + n conv X = conv X + n conv X (see also Borwein and O’Brien [33]). This equality is closely related to the function c(X) = inf {λ > 0 : X + λ conv X is convex}, studied by Schneider [189, 191]. Convex closure. The correspondence h : X → conv X can be considered as a closure mapping on the family of all subsets of Rn , since it has the following obvious properties: X ⊂ hX, hhX = hX, hX ⊂ hY if X ⊂ Y . Koenen [137] proved that hchchX = chchX, where cZ means the complement Rn \ Z of a set Z. Therefore, applying the operations h and c in any order to a given set X ⊂ Rn , one can obtain 10 or fewer distinct sets. Similar problems on semigroups generated by the mappings h, c, i = chc, and f = h∩hc are studied by Soltan [195, § 7]: if m(ϕ, η, . . . ) denotes the number of elements in the semigroup generated by mappings ϕ, η, . . . on the family of all subsets of Rn , then m(c, f ) = 6, m(f, i) = 7, m(h, i) = 5, m(f, h) = 5, m(f, h, i) = 12, m(c, f, h) = 22.
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Convex Cones and Conic Hulls
4.1
Convex Cones
Definition and Characteristic Properties Definition 4.1. A nonempty set C ⊂ Rn is called a cone with apex s ∈ Rn provided s + λ(x − s) ∈ C whenever x ∈ C and λ > 0. A convex cone is a cone which is a convex set. s a
Fig. 4.1
C
A convex cone with apex s.
Setting λ = 0 in Definition 4.1, we observe that s ∈ C. One can reformulate this definition, saying that a nonempty set C ⊂ Rn is a cone with apex s if and only if every halfline [s, xi lies in C whenever x ∈ C \{s}. Hence a cone with apex s is either the singleton {s} or a union of closed halflines with common endpoint s. Obviously, the condition λ > 0 can be replaces with the combination s ∈ C and λ > 0. Some authors use only positive scalars λ in Definition 4.1, thus allowing the apex s be outside the cone C, which also may be empty. The condition λ > 0 makes easier the study of the relative interior of convex cones and conic hulls, while the condition λ > 0 is more suitable for dealing with the closure of these sets.
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Example. Every nonempty plane L ⊂ Rn is a convex cone, and each point s ∈ L is its apex. Indeed, this is obvious if L = {s}. If dim L > 1 and x is a point in L \ {s}, then [s, xi ⊂ hs, xi ⊂ L (see remark on page 38). Example. Every closed halfplane D in Rn is a convex cone, and each point s ∈ rbd D is its apex. Indeed, let x be a point in D\{s}, and h be the closed halfline through x with endpoint s. If x ∈ rbd D, then h ⊂ hs, xi ⊂ rbd D because rbd D is a plane (see Theorem 2.58). If x ∈ D \ rbd D, then h ⊂ D according to Corollary 1.41. Example. The nonnegative octant of Rn , O = {(x1 , . . . , xn ) ∈ Rn : x1 > 0, . . . , xn > 0}, is a convex cone, and the origin o is its only apex (see Theorem 4.6 for a more general statement). Theorem 4.2. For a cone C ⊂ Rn with apex s, the following statements hold. (1) C = s + λ(C − s) for all λ > 0. (2) conv C is a convex cone with apex s. Proof. (1) Choose a scalar λ > 0. Then s + λ(C − s) = {s + λ(x − s) : x ∈ C} ⊂ C. Conversely, for any x ∈ C, the point z = s + λ−1 (x − s) belongs to C. Hence x = s + λ(s + λ−1 (x − s) − s) = s + λ(z − s) ∈ s + λ(C − s), which gives the opposite inclusion C ⊂ s + λ(C − s). (2) Let x ∈ conv C and λ > 0. By Theorem 3.3, x can be written as a convex combination x = µ1 x1 + · · · + µk xk ,
where
x1 , . . . , xk ∈ C.
Since s + λ(xi − s) ∈ C for all 1 6 i 6 k, the same theorem gives the inclusion s + λ(x − s) = s + λ(µ1 (x1 − s) + · · · + µk (xk − s)) = µ1 (s + λ(x1 − s)) + · · · + µk (s + λ(xk − s)) ∈ conv C. Hence conv C is a convex cone with apex s.
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Theorem 4.3. A convex set K ⊂ Rn is a convex cone with apex s ∈ K if and only if there is a positive scalar λ 6= 1 such that K = s + λ(K − s). Proof. If K is a convex cone with apex s, then, due to Theorem 4.2, K = s + λ(K − s) for all λ > 0. Conversely, assume the existence of a positive scalar λ 6= 1 such that K = s + λ(K − s). Since the case K = ∅ is obvious, we assume that K is nonempty. Rewriting this condition as K − s = λ(K − s), we easily obtain by induction on r that λ−r (K − s) = K − s = λr (K − s) Let x ∈ K and µ > 0. µ > 0. Choose integers positive if λ > 1, and p the convexity of K − s,
for all
r > 1.
Since the case µ = 0 is obvious, we assume that p, q such that λp < µ < λq (p is negative and q is is positive and q is negative if 0 < λ < 1). Due to one has
µ(x − s) ∈ (λp (x − s), λq (x − s)) ∈ K − s. So, s + µ(x − s) ∈ s + (λp (x − s), λq (x − s)) ⊂ s + (K − s) = K, which shows that K is a cone with apex s. The result below is analogous to Theorems 1.47 and 2.3. Theorem 4.4. For a nonempty set X ⊂ Rn and a point s ∈ Rn , the following conditions are equivalent. (1) X is a convex cone with apex s. (2) s + λ(x − s) + µ(y − s) ∈ X whenever x, y ∈ X and λ, µ > 0. (3) X contains all linear combinations s+λ1 (x1 −s)+· · ·+λk (xk −s), where k > 1, x1 , . . . , xk ∈ X, and λ1 , . . . , λk > 0. (4) µ1 X + · · · + µr X = (µ1 + · · · + µr − 1)s + X for all r > 1 and all scalars µ1 , . . . , µr > 0. Proof. (1) ⇒ (3). Let x1 , . . . , xk ∈ X and λ1 , . . . , λk > 0. Since the case λ1 = · · · = λk = 0 is obvious, we assume that λ = λ1 + · · · + λk > 0. By Theorem 2.3, X contains the convex combination λk λ1 x1 + · · · + xk . z= λ λ Because X is a cone with apex s, one has s + λ1 (x1 − s) + · · · + λk (xk − s) λ λk λ1 + · · · + λk 1 =s+λ x1 + · · · + xk − s λ λ λ = s + λ(z − s) ∈ X.
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Condition (3) implies (2); so, it remains to show that (2) ⇒ (1). Choosing x = y and λ = µ = η/2, one has s + η(x − s) ∈ X whenever x ∈ X and η > 0, which is equivalent to Definition 4.1. Hence X is a cone with apex s. To prove that X is a convex set, let x, y ∈ X and λ ∈ [0, 1]. Then (1 − λ)x + λy = s + (1 − λ)(x − s) + λ(y − s) ∈ X. (1) ⇔ (4). Let X be a convex cone with apex s, and µ1 , . . . , µr be nonnegative scalars. Since X − s is a convex cone with apex o, a combination of Theorems 2.4 and 4.2 gives µ1 (X − s) + · · · + µr (X − s) = (µ1 + · · · + µr )(X − s) = X − s. Consequently, (µ1 X −µ1 s) + · · · + (µr X −µr s) = X −s, which is equivalent to condition (4). Conversely, if X satisfies condition (4), then s + µ1 (X − s) + · · · + µr (X − s) = X for all r > 1 and all µ1 , . . . , µr > 0. Therefore, X contains all positive combinations s + µ1 (x1 − s) + · · · + µr (xr − s) of points x1 , . . . , xr from X, and condition (3) shows that X is a convex cone with apex s. Simplicial Cones Definition 4.5. Let {s, x1 , . . . , xr } be an affinely independent set in Rn . The simplicial cone (more exactly, simplicial r-cone) Cs (x1 , . . . , xr ) with apex s generated by x1 , . . . , xr is defined by Cs (x1 , . . . , xr ) = {s + λ1 (x1 − s) + · · · + λr (xr − s) : λ1 , . . . , λr > 0}.
r s
x2 r r r s x1
r x1
Fig. 4.2
x2 r x1 r r P s PPP r x3 P
Simplicial cones in R3 .
Example. The nonnegative octant O of Rn (see page 146) is a simplicial cone because it can be expressed as O = Co (e1 , . . . , en ), where e1 , . . . , en is the standard basis for Rn .
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Remark. For a point x ∈ Cs (x1 , . . . , xr ), the scalars λ1 , . . . , λr in the expression x = s + λ1 (x1 − s) + · · · + λr (xr − s)
(4.1)
are uniquely determined by x. Indeed, since x can be written as an affine combination x = (1 − λ1 − · · · − λr )s + λ1 x1 + · · · + λr xr of affinely independent points s, x1 , . . . , xr , the uniqueness of λ1 , . . . , λr follows from Theorem 1.60. Consequently, a point x ∈ aff {s, x1 , . . . , xr } belongs to Cs (x1 , . . . , xr ) if and only if all scalars λ1 , . . . , λr in the expression (4.1) are nonnegative. Theorem 4.6. Every simplicial r-cone Cs (x1 , . . . , xr ) ⊂ Rn is a convex cone of dimension r, and s is its only apex. Furthermore, aff Cs (x1 , . . . , xr ) = aff {s, x1 , . . . , xr }.
(4.2)
Proof. Let x ∈ Cs (x1 , . . . , xr ) and η > 0. By the definition, x = s + λ1 (x1 − s) + · · · + λr (xr − s) for certain scalars λ1 , . . . , λr > 0, which gives s + η(x − s) = s + λ1 η(x1 − s) + · · · + λr η(xr − s) ∈ Cs (x1 , . . . , xr ). Hence Cs (x1 , . . . , xr ) is a cone with apex s. Choose points y, z ∈ Cs (x1 , . . . , xr ) and a scalar µ ∈ [0, 1]. Then y and z can be written as y = s + λ1 (x1 − s) + · · · + λr (xr − s), λ1 , . . . , λr > 0, z = s + γ1 (x1 − s) + · · · + γr (xr − s), γ1 , . . . , γr > 0. Let µi = (1 − µ)λi + µγi , 1 6 i 6 r. Clearly, µ1 , . . . , µr > 0 and (1 − µ)y + µz = s + µ1 (x1 − s) + · · · + µr (xr − s) ∈ Cs (x1 , . . . , xr ), which shows that Cs (x1 , . . . , xr ) is a convex set. Next, assume that a point s0 = s + α1 (x1 − s) + · · · + αr (xr − s),
α1 , . . . , αr > 0,
is an apex of Cs (x1 , . . . , xr ). Then s0 + 2(s − s0 ) ∈ Cs (x1 , . . . , xr ). Clearly, s0 + 2(s − s0 ) = s + (−α1 )(x1 − s) + · · · + (−αr )(xr − s), implying that −α1 , . . . , −αr > 0 (see remark on page 149). Hence α1 = · · · = αr = 0, and s0 = s.
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The obvious inclusions {s, x1 , . . . , xr } ⊂ Cs (x1 , . . . , xr ) ⊂ aff {s, x1 , . . . , xr } and Theorem 1.50 give (4.2). Finally, Theorem 1.60 shows that dim Cs (x1 , . . . , xr ) = dim (aff {s, x1 , . . . , xr }) = r. A combination of Corollary 1.76 and Theorems 4.4 and 4.6 yields a frequently used statement, given below. Corollary 4.7. For a convex cone C ⊂ Rn of positive dimension with apex s, the following statements hold. (1) If x1 , . . . , xr are points in C such that the set {s, x1 , . . . , xr } is affinely independent, then the simplicial cone Cs (x1 , . . . , xr ) lies in C. (2) The dimension of C equals the maximum dimension of a simplicial cone with apex s contained in C. (3) If m = dim C, then aff C = aff Cs (x1 , . . . , xm ) for every simplicial m-cone Cs (x1 , . . . , xm ) contained in C. Algebra of Convex Cones Theorem 4.8. For a family {Cα } of convex cones in Rn with a common apex s, the following statements hold. (1) The intersection ∩ Cα is a convex cone with apex s. α
(2) The set conv ( ∪ Cα ) is the smallest convex cone with apex s conα taining ∪ Cα . α
(3) If the family {Cα } is nested, then the union ∪ Cα is a convex cone α with apex s. Proof. Clearly, the union and intersection of a family of cones with common apex s are cones with apex s. This argument and Theorem 2.8 imply all statements of the theorem. Remark. The union and the intersection of a nested family of convex cones with distinct apices are not, in general, convex cones. For example, the union of closed halfplanes {(x, y) : x > k1 }, k > 1, and the intersection of convex cones Ck = {(x, y) : y > 0} ∪ {(x, 0) : x > k}, k > 1, coincide with the open halfplane {(x, y) : y > 0} of R2 , which is not a convex cone.
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Theorem 4.9. If C1 , . . . , Cr are convex cones in Rn with apices s1 , . . . , sr , respectively, and if µ1 , . . . , µr are scalars, then µ1 C1 +· · ·+µr Cr is a convex cone with apex µ1 s1 + · · · + µr sr . Furthermore, µ1 C1 + · · · + µr Cr = µ1 s1 + · · · + µr sr + conv (µ1 (C1 − s1 ) ∪ · · · ∪ µr (Cr − sr )). Proof. An induction argument shows that the proof can be reduced to the case r = 2 (for r = 1, we can write µ1 C1 = µ1 C1 + µ2 o). By Theorem 2.9, the sum µ1 C1 + µ2 C2 is a convex set. Choose a point x ∈ µ1 C1 + µ2 C2 . Then x = µ1 x1 + µ2 x2 , where x1 ∈ C1 and x2 ∈ C2 . For a scalar λ > 0, one has (µ1 s1 + µ2 s2 ) + λ(x − (µ1 s1 + µ2 s2 )) = µ1 (s1 + λ(x1 − s1 )) + µ2 (s2 + λ(x2 − s2 )) ∈ µ1 C1 + µ2 C2 , which shows that µ1 C1 + µ2 C2 is a cone with apex µ1 s1 + µ2 s2 . To prove the equality µ1 C1 + µ2 C2 = µ1 s1 + µ2 s2 + conv (µ1 (C1 − s1 ) ∪ µ2 (C2 − s2 )),
(4.3)
let C10 = C1 − s1 and C20 = C2 − s2 . Then C10 and C20 are convex cones with common apex o, and (4.3) can be rewritten as µ1 C10 + µ2 C20 = conv (µ1 C10 ∪ µ2 C20 ).
(4.4)
If x ∈ µ1 C10 +µ2 C20 , then x = µ1 x1 +µ2 x2 , where x1 ∈ C10 and x2 ∈ C20 . Since 2x1 ∈ C10 and 2x2 ∈ C20 , Corollary 3.14 gives x = 21 (µ1 (2x1 ) + µ2 (2x2 )) ∈ conv (µ1 C10 ∪ µ2 C20 ). Conversely, if x ∈ conv (µ1 C10 ∪ µ2 C20 ), then, by Corollary 3.14, x can be expressed as a convex combination x = λ1 x1 + λ2 x2 ,
where
x1 ∈ µ1 C10 and x2 ∈ µ2 C20 .
Consequently, x = λ1 x1 + λ2 x2 ∈ µ1 (λ1 C10 ) + µ2 (λ2 C20 ). If both λ1 and λ2 are positive, then λ1 C10 = C10 and λ2 C20 = C20 , implying the inclusion x ∈ µ1 C10 + µ2 C20 . If, for instance, λ1 = 1 and λ2 = 0, then x = x1 + o ∈ µ1 C10 + µ2 o ⊂ µ1 C10 + µ2 C20 . Summing up, the equality (4.4) holds.
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Corollary 4.10. If C ⊂ Rn is a convex cone with apex s, then for any point r ∈ Rn and a scalar µ, the set r + µC is a convex cone with apex r + µs. Furthermore, r + (µ − 1)s + C if µ > 0, r + µC = {r} if µ = 0, r + (µ + 1)s − C if µ < 0. Proof. By Theorem 4.9, the set r + µC is a convex cone with apex r + µs. The second statement follows from the equalities r + µC = r + µs + µ(C − s) and µ(C − s) = sgn (µ)(C − s), where the latter one derives from Theorem 4.2. Theorem 4.11. If C ⊂ Rn is a convex cone with apex s, then aff C = s + (C − C) = conv (C ∪ (2s − C)). Proof. Consider the convex cone C 0 = C − s with apex o. We state that span C 0 = C 0 − C 0 = conv (C 0 ∪ −C 0 ). (4.5) Indeed, let x ∈ span C 0 . According to Theorem 2.11, x = (1 − λ)y + λz, where y, z ∈ C 0 and λ ∈ R. Therefore, (y + λz) − (λy) ∈ C 0 − C 0 if λ > 0, x= (1 − λ)y − (−λz) ∈ C 0 − C 0 if λ < 0, which gives the inclusion span C 0 ⊂ C 0 − C 0 . Similarly, if x ∈ C 0 −C 0 , then x = y−z for certain y, z ∈ C 0 . Expressing x as a convex combination x = 21 (2y) + 12 (−2z), we obtain the inclusion x ∈ conv (C 0 ∪ −C 0 ). Hence C 0 − C 0 ⊂ conv (C 0 ∪ −C 0 ). Next, if x ∈ conv (C 0 ∪ −C 0 ), then Corollary 3.14 shows that x can be written as x = (1 − λ)y + λz for certain y, z ∈ C 0 ∪ −C 0 and λ ∈ [0, 1]. If both y and z belong to C 0 , then x ∈ C 0 ⊂ span C 0 by the convexity of C 0 . Similarly, x ∈ span (−C 0 ) = span C 0 provided y, z ∈ −C 0 . Suppose that y ∈ C 0 and z ∈ −C 0 . Then x = (1 − λ)y − λ(−z) ∈ span C 0 . Summing up, conv (C 0 ∪ −C 0 ) ⊂ span C 0 , and the equalities (4.5) hold. Finally, a combination of Theorems 1.50 and 1.53 gives aff C = aff (s + C 0 ) = s + aff C 0 = s + span C 0 , while (4.5) and Theorem 3.12 result in s + span C 0 = s + conv (C 0 ∪ −C 0 ) = conv (C ∪ (2s − C)). Theorem 4.12. Let f : Rn → Rm be an affine transformation and C ⊂ Rn and D ⊂ Rm be convex cones with apices s and r, respectively, where r ∈ rng f . Then the sets f (C) and f −1 (D) are convex cones with apices f (s) and t ∈ f −1 (r), respectively.
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Proof. As shown in Theorem 2.12, both sets f (C) and f −1 (D) are convex. So, it suffices to prove that the sets are cones. For this, choose a point x ∈ f (C) and a scalar λ > 0. Then x = f (x0 ) for a certain point x0 ∈ C. Since s + λ(x0 − s) is an affine combination which lies in C, Theorem 1.86 gives f (s) + λ(x − f (s)) = f (s + λ(x0 − s)) ∈ f (C). Therefore, f (C) is a cone with apex f (s). Similarly, if x is a point in f −1 (D) and λ > 0, then for every point t ∈ f −1 (r), one has f (t + λ(x − t)) = r + λ(f (x) − r) ∈ D. Hence t + λ(x − t) ∈ f −1 (D), and f −1 (D) is a cone with apex t. The Apex Set of a Convex Cone Definition 4.13. The apex set of a cone C ⊂ Rn , denoted ap C, is the set of all apices of C. Example. If L ⊂ Rn is a nonempty plane, then ap L = L (see example on page 145). Example. If D is a closed halfplane in Rn , then ap D = rbd D (see example on page 146). Example. The apex set of a simplicial cone Cs (x1 , . . . , xr ) is the singleton {s}, as follows from Theorem 4.6. Theorem 4.14. The apex set of any cone C ⊂ Rn is a plane. Proof. The statement is obvious if ap C is a singleton. So, let ap C contain more than one point. According to Theorem 1.46, it suffices to show that for distinct points s1 , s2 ∈ ap C, the line l = hs1 , s2 i lies in ap C. J Jr u r J x Jr z r Jr s1 s2
Ez l
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First, we observe that l ⊂ C because l can be expressed as the union of halflines hs1 , s2 ] and [s1 , s2 i, both lying in C due to s1 , s2 ∈ ap C. Since the case l = C is obvious, we may suppose that l 6= C. Choose a point z ∈ C \ l and consider the 2-dimensional plane Lz = aff {s1 , s2 , z}. Denote by Ez the open halfplane of Lz determined by l and containing z (see the figure above). Theorem 1.71 shows that Ez = {λ1 s1 + λ2 s2 + λ3 z : λ1 + λ2 + λ3 = 1, λ3 > 0}. We state that Ez ⊂ C. Indeed, let x ∈ Ez . By the above argument, x can be written as an affine combination x = α1 s1 + α2 s2 + α3 z, α1 + α2 + α3 = 1, α3 > 0. From α1 + α2 = 1 − α3 < 1 it follows that at least one of the scalars α1 , α2 is less than 1. If α1 < 1, then the point α3 α3 α3 s2 + z, > 0, u= 1− 1 − α1 1 − α1 1 − α1 belongs to the halfline [s2 , zi, which entirely lies in C because of s2 ∈ ap C. Since x can be expressed as x = (1 − (1 − α1 ))s1 + (1 − α1 )u,
1 − α1 > 0,
we conclude that x ∈ [s1 , ui ⊂ C because of s1 ∈ ap C. Similarly, if α2 < 1, the inclusion x ∈ C follows from x = (1 − (1 − α2 ))s2 + (1 − α2 )w ∈ [s2 , wi, where w is the point in [s1 , zi given by α3 α3 s1 + z, w = 1− 1 − α2 1 − α2
1 − α2 > 0,
α3 > 0. 1 − α2
Finally, let s be a point in l. Then s = (1 − µ)s1 + µs2 for a certain scalar µ. Because every point y from the open halfline (s, zi can be written as an affine combination y = (1 − η)s + ηz = (1 − η)(1 − µ)s1 + (1 − η)µs2 + ηz, η > 0, one has (s, zi ⊂ Ez ⊂ C by the above proved. Hence l ⊂ ap C, and ap C is a plane. Theorem 4.15. For a convex cone C ⊂ Rn and an apex s of C, the following statements hold. (1) ap C is the largest plane among all planes containing s and lying in C.
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(2) u − s + ap C ⊂ C for any point u ∈ C. (3) ap C = C ∩ (2s − C). Proof. (1) Due to Theorem 4.14, it suffices to prove that every line l through s which lies in C also lies in ap C. This statement is obvious provided ap C = C. So, we may assume that ap C 6= C. Choose points s1 ∈ l \ {s} and z ∈ C \ {s1 }. Since the inclusion [s1 , zi ⊂ C is trivial when z ∈ l, we will suppose that z ∈ C \ l. Put s2 = 2s − s1 . Clearly, s2 ∈ l and s = (s1 + s2 )/2. Denote by Ez the open halfplane of aff {s1 , s2 , z} which is determined by l and contains z. According to Theorem 1.71, Ez = {λ1 s1 + λ2 s2 + λ3 z : λ1 + λ2 + λ3 = 1, λ3 > 0}. We state that Ez ⊂ C. Indeed, let x be a point in Ez . Then x = α1 s1 + α2 s2 + α3 z, α1 + α2 + α3 = 1, α3 > 0. If α1 6 α2 , then α1 6 21 (α1 + α2 ) = 21 (1 − α3 ) < 12 . In this case, γ = α3 /(1 − 2α1 ) ∈ [0, 1], and the point u = (1 − γ) s2 + γz belongs to C because of the convexity of C. Consequently, x = 2α1 s + (1 − 2α1 )((1 − γ) s2 + γz) = 2α1 s + (1 − 2α1 ) u ∈ [s, ui ⊂ C since s ∈ ap C (see the figure below). r x zr a u
aaa
r a aar r
r s1 s s2
Ez l
Similarly, if α2 6 α1 , then 1 − 2α2 > 0 and x = 2α2 s + (1 − 2α2 ) w ∈ [s, wi ⊂ C, where w is a point in [s1 , z] given by α3 ∈ [0, 1]. 1 − 2α2 Finally, [s1 , zi ⊂ Ez ⊂ C, which gives s1 ∈ ap C. Hence l ⊂ ap C. (2) Choose a point x in u − s + ap C. Then w = (1 − µ)s1 + µz, with µ =
x0 = x − u + s ∈ ap C ⊂ C
and v 0 = s + 2(x0 − s) ∈ C
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because of s ∈ ap C. Similarly, u0 = s + 2(u − s) ∈ C. Therefore, x = 21 (2u + 2x0 − 2s) = 12 (u0 + v 0 ) ∈ [u0 , v 0 ] ⊂ C by the convexity of C. Hence (u − s) + ap C ⊂ C. (3) If x ∈ ap C, then x ∈ C and x0 = x + 2(s − x) ∈ C, which gives x = 2s − x0 ∈ 2s − C. Hence x ∈ C ∩ (2s − C). Conversely, let x ∈ C ∩ (2s − C). Since the case x = s is obvious, we assume that x 6= s. Then x ∈ C \ {s} and x = 2s − x0 , where x0 ∈ C \ {s}. Both halflines [s, xi and [s, x0 i lie in C since s ∈ ap C. Consequently, the line l = hx, x0 i lies in C, and x ∈ l ⊂ ap C by statement (1) above. Theorem 4.16. For any convex cone C ⊂ Rn , the following statement hold. (1) The set C 0 = C \ ap C is convex. (2) If L is a nonempty plane in ap C, then the set CL = C 0 ∪ L is a convex cone with ap CL = L. Proof. Both statements are trivial when ap C = C. Suppose that C 6= ap C. Then C 0 6= ∅. (1) Choose points x, y ∈ C 0 . Then [x, y] ⊂ C by the convexity of C. Assume for a moment that [x, y] 6⊂ C 0 . Then the open segment (x, y) contains a point s ∈ ap C, and hx, yi = hx, s] ∪ [s, yi ⊂ C. According to Theorem 4.15, the line hx, yi lies in ap C, in contradiction with the choice of x and y. Hence [x, y] ⊂ C 0 , implying the convexity of C 0 . (2) For the convexity of CL , choose any points x, y ∈ CL . If both x and y belong to C 0 , then [x, y] ⊂ C 0 ⊂ CL by the proved above. Similarly, if both x and y belong to L, then [x, y] ⊂ L ⊂ CL . Suppose that x ∈ L and y ∈ C 0 . Then x ∈ ap C and (x, yi ⊂ C 0 ⊂ CL . Summing up, the set CL is convex. As above, (s, xi ⊂ C 0 ⊂ CL for any choice of points s ∈ L and x ∈ C 0 . This argument gives the inclusion L ⊂ ap CL . Conversely, if s ∈ ap CL and x ∈ C 0 , then (s, xi lies in C 0 , which shows the inclusion ap CL ⊂ ap C. Since C 0 ∩ ap C = ∅, one has ap CL ⊂ L. Topological Properties of Convex Cones Theorem 4.17. If Cs (x1 , . . . , xr ) is a simplicial cone in Rn , then rint Cs (x1 , . . . , xr ) = {s + λ1 (x1 − s) + · · · + λr (xr − s) : λ1 , . . . , λr > 0}.
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Proof. First, we state that every positive combination of the form x = s + λ1 (x1 − s) + · · · + λr (xr − s) lies in rint Cs (x1 , . . . , xr ). Indeed, rewriting x as an affine combination x = (1 − λ1 − · · · − λr ) s + λ1 x1 + · · · + λr xr and using Theorem 4.6, we see that x ∈ aff {s, x1 , . . . , xr } = aff Cs (x1 , . . . , xr ). Put ε = min {λ1 , . . . , λr+1 }. Theorem 2.16 implies the existence of a scalar ρ > 0 such that for every point u = s + µ1 (x1 − s) + · · · + µr (xr − s) ∈ Bρ (x) ∩ aff Cs (x1 , . . . , xr ), its affine coordinates 1 − µ1 − · · · − µr , µ1 , . . . , µr relative to s, x1 , . . . , xr satisfy the inequalities |(1 − λ1 − · · · − λr ) − (1 − µ1 − · · · − µr )| 6 ε, |λi − µi | 6 ε, i = 1, . . . , r. Hence µi > λi − ε > 0, i = 1, . . . , r, which gives u ∈ Cs (x1 , . . . , xr ). Definition 2.13 shows that x ∈ rint Cs (x1 , . . . , xr ). Conversely, let x ∈ rint Cs (x1 , . . . , xr ). Then x can be written as x = s + λ1 (x1 − s) + · · · + λr (xr − s), λ1 , . . . , λr > 0. Assume for a moment that at least one of the scalars λ1 , . . . , λr , say λ1 , equals 0. From Theorem 2.16 it follows that for a given ρ > 0, there is a sufficiently small δ > 0 such that the point y = s − δ(x1 − s) + λ2 (1 + δ)(x2 − s) + · · · + λr (1 + δ)(xr − s) belongs to Bρ (x). Since y is an affine combination of s, x1 , . . . , xr , one has y ∈ aff {s, x1 , . . . , xr } \ Cs (x1 , . . . , xr ) (see remark on page 149). Therefore, Bρ (x) ∩ aff {s, x1 , . . . , xr } 6⊂ Cs (x1 , . . . , xr ), contradicting the choice of x. Hence all scalars λ1 , . . . , λr are positive. Theorem 4.18. For a convex cone C ⊂ Rn of positive dimension m and an apex s of C, the following statements hold. (1) If Cs (x1 , . . . , xm ) is a simplicial m-cone lying in C, then rint Cs (x1 , . . . , xm ) ⊂ rint C.
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(2) A point x ∈ Rn \ {s} belongs to rint C if and only if there is a simplicial m-cone Cs (x1 , . . . , xm ) ⊂ C containing x in its relative interior. Proof. Statement (1) follows from Theorem 2.15 and Corollary 4.7. For the second statement, let x ∈ rint C. A combination of Theorems 2.17 and 2.20 shows that x can be written as a positive convex combination, x = λ0 s + λ1 x1 + · · · + λm xm , of affinely independent points s, x1 , . . . , xm ∈ C. Then x = s + λ1 (x1 − s) + · · · + λm (xm − s), and Theorem 4.17 gives the inclusion x ∈ rint Cs (x1 , . . . , xm ). Conversely, let x belong to the relative interior of a simplicial m-cone Cs (x1 , . . . , xm ) ⊂ C. Because aff Cs (x1 , . . . , xm ) = aff C (see Corollary 4.7), the inclusion x ∈ rint C follows from Theorem 2.15. x3r r x1
r r x s@ r x@ 2@
Remark. Unlike Theorem 2.20, none of the points x1 , . . . , xm in statement (2) of Theorem 4.18 can be chosen in advance. Indeed, let s = o and X = {x1 , x2 , x3 }, where x1 = (−1, 0), x2 = (1, −1), and x3 = (1, 1), as depicted above. Clearly, the point x = (1, 0) belongs to cones X = R2 but does not belong to a simplicial cone of the form Cs (x1 , xi ), i = 1, 2. Corollary 4.19. For a convex cone C ⊂ Rn , the following conditions are equivalent. (1) (2) (3) (4)
C is a plane. ap C = C. ap C ⊂ rint C. ap C ∩ rint C 6= ∅.
Proof. (1) ⇔ (2). This part of the proof follows from Theorem 4.14 and example on page 153. (2) ⇒ (3). This part is true because C = ap C = rint C for any nonempty plane C ⊂ Rn . Since condition (3) implies (4), it remains to show that (4) ⇒ (2). Let s be a point in ap C ∩ rint C. Leaving aside the trivial case C = {s}, choose
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a point x ∈ C \ {s}. By Corollary 2.25, there is a point u ∈ C such that s ∈ (u, x). Then hs, xi = hu, s] ∪ [s, xi ⊂ C, and Theorem 4.15 shows the inclusion x ∈ ap C. Consequently, C ⊂ ap C, which gives C = ap C. Corollary 4.20. For a convex cone C ⊂ Rn , the following conditions are equivalent. (1) (2) (3) (4)
C is not a plane. ap C 6= C. ap C ⊂ rbd C. ap C ∩ rbd C 6= ∅.
Furthermore, ap C = rbd C if and only if C is a closed halfplane of aff C. Proof. The equivalence of statements (1)–(4) follows from Corollary 4.19. If C is a closed halfplane, then ap C = rbd C (see example on page 146). Conversely, suppose that ap C = rbd C. By Theorem 4.14, ap C is a plane, and Theorem 2.58 shows that cl C is a closed halfplane. The inclusion rbd C = ap C ⊂ C implies that C itself is a closed halfplane. Theorem 4.21. If C ⊂ Rn is a convex cone, then all three sets, cl C,
C1 = ap C ∪ rint C,
C2 = ap (cl C) ∪ rint C,
are convex cones such that C1 ⊂ C and C2 ⊂ cl C. Furthermore, ap C = ap C1 ⊂ ap (cl C) = ap C2 .
(4.6)
Proof. Since both statements of the theorem are obvious when C is a plane, we further assume that C is distinct from a plane. Then ap C ⊂ rbd C according to Corollary 4.20. We divide our argument into several steps. 1. First, we state that all three sets cl C, C1 , and C2 are convex. Indeed, the convexity of cl C follows from Theorem 2.35. For the convexity of C1 , choose points x, y ∈ C1 . If both x and y belong either to ap C or to rint C, then, respectively, [x, y] ⊂ ap C (see Theorem 4.14) or [x, y] ⊂ rint C (see Corollary 2.22). If x ∈ ap C and y ∈ rint C, then (x, y) ⊂ rint C due to ap C ⊂ rbd C and Theorem 2.21, implying the inclusion [x, y] ⊂ ap C ∪ rint C = C1 . Similarly, let x, y ∈ C2 . If x, y ∈ ap (cl C) or x, y ∈ rint C, then [x, y] ⊂ ap (cl C) by Theorem 4.14, or [x, y] ⊂ rint C by Corollary 2.22. Let x ∈ ap (cl C) and y ∈ rint C. Since cl C is not a plane, one has
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(x, y) ⊂ rint C due to ap (bd C) ⊂ rbd (cl C) = rbd C and Theorem 2.36; whence [x, y] ⊂ ap (cl C) ∪ rint C = C2 . The inclusions C1 ⊂ C and C2 ⊂ cl C are obvious. 2. We are going to show that ap C = ap C1 . Indeed, let s ∈ ap C and x ∈ C1 \ {s}. If x ∈ ap C, then [s, xi ⊂ hs, xi ⊂ ap C because ap C is a plane (see Theorem 4.14). If x ∈ rint C, then [s, xi ⊂ C because s ∈ ap C; furthermore, for any z ∈ (s, xi, there is a point u ∈ (s, xi such that z ∈ (x, u), implying the inclusion z ∈ rint C (see Theorem 2.21). Hence (s, xi ⊂ rint C, and [s, xi ⊂ ap C ∪ rint C = C1 . Summing up, s ∈ ap C1 , and ap C ⊂ ap C1 . For the opposite inclusion, we observe that rint C ⊂ C1 ⊂ cl C, which gives rint C1 = rint C (see Corollary 2.39). Since C is not a plane, C1 also is distinct from a plane, and Corollary 4.20 implies ap C1 ⊂ C1 \ rint C1 = (ap C ∪ rint C) \ rint C = ap C. 3. Next, we are going to prove the inclusion ap C ⊂ ap (cl C). Equivalently, that [s, xi ⊂ cl C for every choice of points s ∈ ap C and x ∈ cl C \ {s}. Indeed, let z ∈ [s, xi. Since the case z = s is obvious, we assume that z 6= s. Then z = s + λ(x − s) for a certain scalar λ > 0. If ρ > 0, then the inclusion x ∈ cl C shows the existence of a point x0 ∈ Bρ/λ (x) ∩ C. Put z 0 = s + λ(x0 − s). Then kz − z 0 k = λkx − x0 k 6 ρ and z 0 ∈ [s, z 0 i ⊂ C because s ∈ ap C. Therefore, every ball Bρ (z), ρ > 0, meets C, which gives z ∈ cl C. Hence [s, xi ⊂ cl C. Thus s ∈ ap (cl C), and the inclusion ap C ⊂ ap (cl C) holds. 4. Finally, we state that ap (cl C) = ap C2 . As above, let s ∈ ap (cl C) and x ∈ C2 \{s}. If x ∈ ap (cl C), then [s, xi ⊂ hs, xi ⊂ ap (cl C) according to Theorem 4.14. Let x ∈ rint C. Then [s, xi ⊂ cl C because s ∈ ap (cl C). For any z ∈ (s, xi, there is a point u ∈ (s, xi such that z ∈ (x, u), implying the inclusion z ∈ rint C (see Theorem 2.36). Hence [s, xi ⊂ ap (cl C) ∪ rint C = C2 , which shows that s ∈ ap C2 . Thus ap (cl C) ⊂ ap C2 .
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To prove the opposite inclusion, we observe that rint C ⊂ C2 ⊂ cl C, which gives rint C2 = rint C according to Corollary 2.39. Since C is not a plane, C2 also is distinct from a plane, and Corollary 4.20 gives ap C2 ⊂ C2 \ rint C2 = (ap (cl C) ∪ rint C) \ rint C = ap (cl C). The fact that all three sets cl C, C1 , and C2 are cones follows from (4.6) and the assumption ap C 6= ∅. Remark. All four convex cones from Theorem 4.21 may be pairwise distinct. Indeed, let C = {(x, y, z) : y > |x|} ∪ {(x, x, 0) : x > 0}. Then cl C = {(x, y, z) : y > |x|} and rint C = {(x, y, z) : y > |x|}. Furthermore, ap C = {o} and ap (cl C) = {(0, 0, z) : z ∈ R}, which shows that C, C1 , C2 , and cl C are pairwise distinct. Clearly, the set ap (cl C) ∪ C is not a convex cone. The inclusion ap C ⊂ C ∩ ap (cl C) may be proper. For instance, if C = {(x, y) : y < 0} ∪ {(x, 0) : x > 0}, then ap C = {o} and ap (cl C) is the x-axis of R2 , implying that ap C 6= C ∩ ap (cl C). Analyzing the proof of Theorem 4.21, we derive the corollary below. Corollary 4.22. For a convex cone C ⊂ Rn with apex s and a point x in cl C \ {s}, the following statements hold. (1) The halfline [s, xi lies in rbd C if and only if x ∈ rbd C. (2) The halfline (s, xi lies in rint C if and only if x ∈ rint C. 4.2
Conic Hulls
Definition and Basic Properties Definition 4.23. For a given set X ⊂ Rn , the intersection of all convex cones with apex s ∈ Rn containing X is called the conic hull with apex s of X and denoted cones X. Theorem 4.8 shows for any set X ⊂ Rn and a point s ∈ Rn , the conic hull cones X exists and is the smallest convex cone with apex s containing X. Furthermore, cones ∅ = {s}.
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cones X
s r Fig. 4.3
X
The conic hull with apex s of a set X.
The next result describes elementary properties of conic hulls. Theorem 4.24. For sets X and Y in Rn and a point s ∈ Rn , the following statements hold. (1) X ⊂ cones X, and X = cones X if and only if X is a convex cone with apex s. (2) cones X = cones ({s} ∪ X). (3) (r − s) + cones X = coner (r − s + X) for every point r ∈ Rn . (4) cones (cones X) = cones X. (5) cones X ⊂ cones Y if X ⊂ Y . (6) cones X = cones Y if X ⊂ Y ⊂ cones X. Furthermore, if {Xα } is a family of sets in Rn and s ∈ Rn , then the statements below are true. (7) cones (∩ Xα ) ⊂ ∩ cones Xα . α
α
(8) ∪ cones Xα ⊂ cones (∪ Xα ). α
α
(9) ∪ cones Xα = cones (∪ Xα ) if the family {Xα } is nested. α
α
(10) cones (∪ cones Xα ) = cones (∪ Xα ). α
α
Proof. Let C(X) denote the family of all convex cones with apex s containing a given set X ⊂ Rn . The proofs of statements (1)–(10) derive from the following arguments. (a) cones X is the smallest element in C(X). (b) C(X) is exactly the family of convex cones with apex s containing cones X. (c) If X ⊂ Y , then C(Y ) ⊂ C(X). (d) The union of a nested family of convex cones with apex s is a convex cone with apex s (see Theorem 4.8). The theorem below gives an important description of conic hulls in terms of nonnegative combinations of points.
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Theorem 4.25. For a nonempty set X ⊂ Rn and a point s ∈ Rn , one has cones X = {s + λ1 (x1 − s) + · · · + λk (xk − s) : k > 1, x1 , . . . , xk ∈ X, λ1 , . . . , λk > 0}. Proof. Since cones X is a convex cone with apex s containing X, Theorem 4.4 implies that the set C = {s + λ1 (x1 − s) + · · · + λk (xk − s) : k > 1, x1 , . . . , xk ∈ X, λ1 , . . . , λk > 0} lies in cones X. We state that C is a convex cone with apex s. Indeed, if x ∈ C, then x = s + λ1 (x1 − s) + · · · + λk (xk − s) for certain points x1 , . . . , xk ∈ X and scalars λ1 , . . . , λk > 0. For a scalar µ > 0, one has s + µ(x − s) = s + µλ1 (x1 − s) + · · · + µλk (xk − s) ∈ C, implying that C is a cone with apex s. For the convexity of C, choose points u, w ∈ C. Then u and w can be expressed as nonnegative combinations u = s + γ1 (u1 − s) + · · · + γp (up − s), u1 , . . . , up ∈ X, w = s + µ1 (w1 − s) + · · · + µq (wq − s), w1 , . . . , wq ∈ X. For a scalar λ ∈ [0, 1], one has (1 − λ)u + λw = s + (1 − λ)γ1 (u1 − s) + · · · + (1 − λ)γp (up − s) + λµ1 (w1 − s) + · · · + λµq (wq − s) ∈ C. Hence C is a convex set. Since X ⊂ C (every point x ∈ X can be written as x = s + 1(x − s)), the inclusions X ⊂ C ⊂ cones X and Theorem 4.24 give cones X = C. Corollary 4.26. For points s, x1 , . . . , xr ∈ Rn , one has cones {x1 , . . . , xr } = {s + λ1 (x1 − s) + · · · + λr (xr − s) : λ1 , . . . , λr > 0}. Proof. By Theorem 4.25, a point x ∈ Rn belongs to cones {x1 , . . . , xr } if and only if it can be expressed as x = s + λi1 (xi1 − s) + · · · + λit (xit − s),
(4.7)
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where λi1 , . . . , λit > 0 and xi1 , . . . , xit ∈ {x1 , . . . , xr }. We may assume that all points xi1 , . . . , xit are pairwise distinct. Indeed, if any of these points, say xip and xiq , coincide, then we replace λip (xip − s) + λiq (xiq − s)
with
(λip + λiq )(xip − s).
Next, for every index i ∈ {1, . . . , r} \ {i1 , . . . , it }, we add 0(xi − s) to the right-hand side of (4.7) to obtain a desired expression of x. A combination of Definition 4.5 and Corollary 4.26 shows that every rsimplex ∆(x1 , . . . , xr+1 ) ⊂ Rn is the convex hull of the set {x1 , . . . , xr+1 }. Theorem 4.27. Let s ∈ Rn and X ⊂ Rn be a set such that X \ {s} 6= ∅. Then cones X consists of s and all positive combinations x = s + λ1 (x1 − s) + · · · + λk (xk − s),
x1 , . . . , xk ∈ X,
k > 1,
where {s, x1 , . . . , xk } is affinely independent. Proof. By Theorem 4.25, a point x ∈ Rn belongs to cones X if and only if it can be written as x = s + λ1 (x1 − s) + · · · + λk (xk − s),
(4.8)
where x1 , . . . , xk ∈ X and λ1 , . . . , λk > 0. Since the case x = s is obvious, we suppose that x 6= s. Also, we may assume that (4.8) involves a minimum possible number of points x1 , . . . , xk and that all scalars λ1 , . . . , λk are positive. First, we are going to show that {x1 , . . . , xk } is affinely independent. Indeed, assume for a moment that {x1 , . . . , xk } is affinely dependent; that is, there are scalars ν1 , . . . , νk , not all zero, such that ν1 x1 + · · · + νk xk = o
and ν1 + · · · + νk = 0.
Clearly, at least one of the scalar ν1 , . . . , νk is positive. Put t = min {λi /νi : νi > 0, 1 6 i 6 k}. Then λi − tνi > 0 for all 1 6 i 6 k, and λi − tνi = 0 for at least one i ∈ {1, . . . , k}. Since x = s + (λ1 − tν1 )(x1 − s) + · · · + (λk − tνk )(xk − s), we obtain a contradiction with the minimality of k. Next, we state that {s, x1 , . . . , xk } is affinely independent. By the above proved, it suffices to show that s ∈ / aff {x1 , . . . , xk }, as follows from Corollary 1.63. Assume for a moment that s ∈ aff {x1 , . . . , xk }. Then s can be
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written as an affine combination s = µ1 x1 +· · ·+µk xk (see Corollary 1.52). Consequently, o = µ1 (x1 − s) + · · · + µk (xk − s). Clearly, at least one of the scalars µ1 , . . . , µk is positive. Put α = min {λi /µi : µi > 0, 1 6 i 6 k}. Then λi − αµi > 0 for all 1 6 i 6 k, and λi − αµi = 0 for at least one i ∈ {1, . . . , k}. Thus x = s + (λ1 − αµ1 )(x1 − s) + · · · + (λk − αµk )(xk − s), in contradiction with the minimality of k. Similarly to Theorem 3.7, we prove one more variation of Theorem 4.27. Theorem 4.28. Let X ⊂ Rn be a nonempty set, x0 ∈ X, and s ∈ Rn . Then cones X consists of all nonnegative combinations s + λ0 (x0 − s) + λ1 (x1 − s) + · · · + λk (xk − s), where {x0 , x1 , . . . , xk } is an affinely independent subset of X. Proof. By Theorem 4.25, a point x belongs to cones X if and only if it can be written as x = s + λ1 (y1 − s) + · · · + λm (ym − s), where y1 , . . . , ym ∈ X and λ1 , . . . , λm > 0. Since the case x = s is obvious, we suppose that x 6= s. Then λ = λ1 + · · · + λm > 0. Put y = λ−1 (x − s) = λ−1 λ1 (y1 − s) + · · · + λ−1 λm (ym − s). Clearly, y is a convex combination of y1 − s, . . . , ym − s. By Theorem 3.3, y ∈ conv (X − s) = conv X − s, or y + s ∈ conv X. Theorem 3.7 shows the existence of an affinely independent subset {x0 , x1 , . . . , xk } of X such that y + s can be written as a convex combination y + s = µ0 x0 + µ1 x1 + · · · + µk xk . Equivalently, y = µ0 (x0 − s) + µ1 (x1 − s) + · · · + µk (xk − s). Finally, x = s + λy = s + λµ0 (x0 − s) + λµ1 (x1 − s) + · · · + λµk (xk − s) is a desired expression of x.
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Remark. Theorem 4.28 becomes a particular case of Theorem 4.27 when x0 = s. If x0 6= s, then the set {s, x0 , x1 , . . . , xk } in Theorem 4.27 may be affinely dependent. Indeed, Let s = o and X = {x0 , x1 , x2 } ⊂ R2 , where x0 = (−1, 0), x1 = (1, −1), and x2 = (1, 1), as depicted below. Clearly, coneo X = R2 , and no interior point of the cone C = coneo {x1 , x2 } can be written as a nonnegative combination of an affinely independent subset of {o} ∪ X, which contains both o and x0 . x2r r x0
r C o@ r x@ 1@
Affine Spans, Conic Hulls, and Convex Hulls The following result can be considered as an analog of Theorem 3.6. Theorem 4.29. For a set X ⊂ Rn of positive dimension m and a point s ∈ Rn , the following statements hold. (1) If s ∈ / aff X, then cones X consists of s and all positive combinations x = s + λ1 (x1 − s) + · · · + λk (xk − s),
1 6 k 6 m + 1,
where x1 , . . . , xk ∈ X and {s, x1 , . . . , xk } is affinely independent. (2) If s ∈ aff X, then cones X consists of s and all positive combinations x = s + λ1 (x1 − s) + · · · + λk (xk − s),
1 6 k 6 m,
where x1 , . . . , xk ∈ X and {s, x1 , . . . , xk } is affinely independent. (3) If s ∈ / aff X, then cones X is the union of simplicial (m + 1)-cones Cs (x1 , . . . , xm+1 ), where x1 , . . . , xm+1 ∈ X. (4) If s ∈ aff X, then cones X is the union of simplicial m-cones Cs (x1 , . . . , xm ), where x1 , . . . , xm ∈ X. Proof. Corollary 1.78 shows that dim ({s} ∪ X) 6 m + 1, with the equality if and only if s ∈ / aff X. Since k + 1 is the maximum cardinality of affinely independent subsets of a k-dimensional set Z ⊂ Rn (see Corollary 1.76), statements (1) and (2) follows from Theorem 4.27. (3) Let s ∈ / aff X. If {x1 , . . . , xm+1 } is an affinely independent subset of X, then {s, x1 , . . . , xm+1 } also is affinely independent (see Corollary 1.63).
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By Corollary 4.7, the simplicial cone Cs (x1 , . . . , xm+1 ) lies in cones X. Conversely, let x ∈ cones X. Since the case x = s is obvious, we assume that x 6= s. Theorem 4.27 implies that x can be expressed as a positive combination x = s + λ1 (x1 − s) + · · · + λk (xk − s),
x1 , . . . , xk ∈ X,
where the set {s, x1 , . . . , xk } of X is affinely independent. Theorem 1.68 shows that {x1 , . . . , xk } can be expanded to an affinely independent subset {x1 , . . . , xm+1 } of X. Because {s, x1 , . . . , xm+1 } is affinely independent, the equality x = s + λ1 (x1 − s) + · · · + λk (xk − s) + 0(xk+1 − s) + · · · + 0(xm+1 − s), implies the inclusion x ∈ Cs (x1 , . . . , xm+1 ). (4) The case s ∈ aff X is similar. Various relations between affine spans, conic hulls, and convex hulls are given in the next theorem (compare with Theorem 4.24). Theorem 4.30. For a nonempty set X ⊂ Rn and a point s ∈ Rn , the following statements hold. (1) conv X ⊂ cones X. (2) conv (cones X) = cones (conv X) = cones X. (3) If X 6= {s} and Y denotes the union of all halflines [s, xi, x ∈ X, then cones X = conv Y . (4) cones X = {s + λ(x − s) : x ∈ conv X, λ > 0}. (5) cones X = conv {s + λ(x − s) : x ∈ X, λ > 0}. (6) aff (cones X) = aff ({s} ∪ X). (7) cones X ⊂ aff X if and only if s ∈ aff X. (8) cones X is a plane if and only if cones X = aff X. (9) If s ∈ / aff X, then conv X = aff X ∩ cones X. Proof. Statement (1) is trivial. Since the set cones X is convex, one has conv (cones X) = cones X. The equality cones (conv X) = cones X follows from the inclusions X ⊂ conv X ⊂ cones X and Theorem 4.24. (3) Clearly, Y is a cone with apex s and X ⊂ Y . Theorem 4.2 shows that conv Y is a convex cone with the same apex. Due to the inclusions X ⊂ conv Y ⊂ cones X, Theorem 4.24 gives cones X = cones (conv Y ) = conv Y.
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(4) Put C = {s + λ(x − s) : x ∈ conv X, λ > 0}. Then C ⊂ cones X by statement (2) above and Theorem 4.25. Conversely, if z ∈ cones X, then z = s + λ1 (x1 − s) + · · · + λk (xk − s) for appropriate x1 , . . . , xk ∈ X and λ1 , . . . , λk > 0. Let λ = λ1 + · · · + λk . Since the case λ = 0 is obvious, assume that λ > 0 and consider the convex combination λk λ1 λk λ1 (xk − s) = x1 + · · · + xk . x = s + (x1 − s) + · · · + λ λ λ λ Then x ∈ conv X (see Theorem 3.3), and z = s + λ(x − s) ∈ C. (5) Let D = {s + λ(x − s) : x ∈ X, λ > 0}. From Theorem 4.25 it follows that D ⊂ cones X. By a convexity argument, one has conv D ⊂ cones X. For the opposite inclusion, choose a point u ∈ cones X. Then u = s + λ1 (x1 − s) + · · · + λk (xk − s) for certain x1 , . . . , xk ∈ X and λ1 , . . . , λk > 0. Since s + kλi (xi − s) ∈ D
for all
1 6 i 6 k,
u can be written as a convex combination of points from D: u = k1 (s + kλ1 (x1 − s)) + · · · + k1 (s + kλk (xk − s)). Hence u ∈ conv D, and cones X ⊂ conv D. (6) The inclusions {s} ∪ X ⊂ cones X ⊂ aff ({s} ∪ X) and Theorem 1.50 give aff (cones X) = aff ({s} ∪ X). (7) Assume first that s ∈ aff X, and choose a point y ∈ cones X. If y = s, then y ∈ aff X. Suppose that y 6= s. By statement (4) above, y = s + λ(x − s), where x ∈ conv X \ {s} and λ > 0. Since x ∈ conv X ⊂ aff X, one has y ∈ [s, xi ⊂ hs, xi ⊂ aff X according to Theorem 1.46. Conversely, if cones X ⊂ aff X, then s ∈ cones X ⊂ aff X. (8) If cones X is a plane, then aff X ⊂ cones X because aff X is the smallest plane in Rn containing X. The opposite inclusion, cones X ⊂ aff X, follows from statement (7).
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Conversely, if cones X = aff X, then cones X is a plane. (9) The inclusion conv X ⊂ aff X ∩ cones X follows from conv X ⊂ aff X (see Theorem 3.2) and statement (1) above. For the opposite inclusion, choose a point x ∈ aff X ∩ cones X. By Theorem 4.25, we can write x = s + λ1 (x1 − s) + · · · + λk (xk − s), where x1 , . . . , xk ∈ X and λ1 , . . . , λk > 0. Since x 6= s (because s ∈ / aff X), at least one of the scalars λ1 , . . . , λk is positive. Put λk λ1 x1 + · · · + xk . λ λ Clearly, u is a convex combination of x1 , . . . , xk , which gives u ∈ conv X. Moreover, x = s + λ(u − s). Assuming that λ 6= 1, we would obtain λ = λ1 + · · · + λk
and u =
λ 1 x− u ∈ hx, ui ⊂ aff X, 1−λ 1−λ contradicting the assumption s ∈ / aff X. Hence λ = 1, which shows that x = u. s=
The next theorem shows that the conic hull of a convex set can be defined locally. Theorem 4.31. For a nonempty convex set K ⊂ Rn and a ball Bρ (s) ⊂ Rn , the following statements hold. (1) If s ∈ K, then cones K = cones (Bρ (s) ∩ K). (2) If s ∈ cl K, then cones (rint K) = cones (Bρ (s) ∩ rint K). Proof. Theorem 4.24 gives the inclusions cones (Bρ (s) ∩ K) ⊂ cones K, cones (Bρ (s) ∩ rint K) ⊂ cones (rint K). For the opposite inclusions, choose a point y in cones K (respectively, in cones (rint K)). By Theorem 4.30, y = s+λ(x−s) for a certain scalar λ > 0 and point x in K (respectively, in rint K \ {s}). One has [s, x] ⊂ K by the convexity of K (respectively, (s, x) ⊂ rint K according to Theorem 2.36). Choose a point z ∈ (s, x) such that kz − sk 6 ρ. Then z ∈ Bρ (s) ∩ K (respectively, z ∈ Bρ (s) ∩ rint K). Expressing z as z = (1 − µ)s + µx for a certain 0 < µ < 1, we can write x = (1 − µ−1 )s + µ−1 z, which gives y = s + λ(x − s) = s + λ((1 − µ−1 )s + µ−1 z − s) = s + λµ−1 (z − s). Thus y ∈ cones (Bρ (s) ∩ K), (respectively, y ∈ cones (Bρ (s) ∩ rint K).
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Algebra of Conic Hulls Theorem 4.32. For nonempty sets X1 , . . . , Xr in Rn and a point s ∈ Rn , one has cones (X1 ∪ · · · ∪ Xr ) = {s + λ1 (x1 − s) + · · · + λr (xr − s) : x1 ∈ conv X1 , . . . , xr ∈ conv Xr , λ1 , . . . , λr > 0}. Proof. Theorem 4.24 implies the inclusion cones X1 ∪ · · · ∪ cones Xr ⊂ cones (X1 ∪ · · · ∪ Xr ). This argument and Theorem 4.25 show that the set C = {s + λ1 (x1 − s) + · · · + λr (xr − s) : x1 ∈ conv X1 , . . . , xr ∈ conv Xr , λ1 , . . . , λr > 0} lies in cones (X1 ∪ · · · ∪ Xr ). Conversely, let x ∈ cones (X1 ∪ · · · ∪ Xr ). By the same Theorem 4.25, x is expressible as a nonnegative combination x = s + µ1 (z1 − s) + · · · + µk (zk − s), z1 , . . . , zk ∈ X1 ∪ · · · ∪ Xr . Renumbering the terms µ1 (z1 − s), . . . , µk (zk − s), we assume that the set {z1 , . . . , zk } is partitioned into subsets Y1 , . . . , Yr (some of them may be empty) such that Yi = {zpi−1 +1 , . . . , zpi } ⊂ Xi , 1 6 i 6 r, where p0 = 0, pr = k. Let λi = µpi−1 +1 + · · · + µpi , 1 6 i 6 r. Denote by I the set of all indices i ∈ {1, . . . , r} satisfying the property: either Yi is empty or λi = 0. Let J = {1, . . . , r} \ I. For every i ∈ I, choose a point xi ∈ Xi , and for every i ∈ J, let µp +1 µp xi = i−1 zpi−1 +1 + · · · + i zpi . λi λi Clearly, every point xi is a convex combination of points from Xi , i ∈ J, and Theorem 3.3 shows that xi ∈ conv Xi for all 1 6 i 6 r. Finally, x = s + λ1 (x1 − s) + · · · + λr (xr − s) is a desired representation. Corollary 4.33. For nonempty convex sets K1 , . . . , Kr in Rn and a point s ∈ Rn , one has cones (K1 ∪ · · · ∪ Kr ) = {s + λ1 (x1 − s) + · · · + λr (xr − s) : x1 ∈ K1 , . . . , xr ∈ Kr , λ1 , . . . , λr > 0}. Equivalently, cones (K1 ∪ · · · ∪ Kr ) = ∪ (cones {x1 , . . . , xr } : x1 ∈ K1 , . . . , xr ∈ Kr ).
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Theorem 4.34. If X ⊂ Rn is a set, r and s are points in Rn , and µ is a scalar, then coner+µs (r + µX) = r + µ cones X. Furthermore, r + (µ − 1)s + cones X coner+µs (r + µX) = {r} r + (µ + 1)s − cones X
if if if
µ > 0, µ = 0, µ < 0.
(4.9)
Proof. Since the case X = ∅ is obvious, we may assume that X is nonempty. Clearly, r + µX ⊂ r + µ cones X. By Theorem 4.9, r + µ cones X is a convex cone with apex r + µs, and Theorem 4.24 gives coner+µs (r + µX) ⊂ r + µ cones X. Conversely, let u ∈ r + µ cones X. Then u = r + µy, where y ∈ cones X. By Theorem 4.25, y can be written as y = s + λ1 (x1 − s) + · · · + λk (xk − s) for certain points x1 , . . . , xk ∈ X and scalars λ1 , . . . , λk > 0. By the same theorem, the equality u = (r + µs) + λ1 ((r + µx1 ) − (r + µs)) + . . . + λk ((r + µxk ) − (r + µs)) implies the inclusion u ∈ coner+µs (r + µX). Hence r + µ cones X ⊂ coner+µs (r + µX). The equalities (4.9) follow from Corollary 4.10. Theorem 4.35. For sets X1 , . . . , Xr ⊂ Rn , points s1 , . . . , sr ∈ Rn , and scalars µ1 , . . . , µr , r > 2, one has coneµ1 s1 +···+µr sr (µ1 X1 + · · · + µr Xr ) ⊂ µ1 cones1 X1 + · · · + µr conesr Xr ⊂ coneµ1 s1 +···+µr sr (µ1 ({s1 } ∪ X1 ) + · · · + µr ({sr } ∪ Xr )). In particular, if si ∈ Xi for all 1 6 i 6 r, then coneµ1 s1 +···+µr sr (µ1 X1 + · · · + µr Xr ) = µ1 cones1 X1 + · · · + µr conesr Xr .
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Proof. An induction argument shows that the proof can be reduced to the case r = 2 (for r = 1, we can write µ1 X1 = µ1 X1 +µ2 o). Assume that both sets X1 and X2 are nonempty (otherwise the statement is trivial). Since µ1 X1 + µ2 X2 ⊂ µ1 cones1 X1 + µ2 cones2 X2 , and since µ1 cones1 X1 +µ2 cones2 X2 is a convex cone with apex µ1 s1 +µ2 s2 (see Theorem 4.9), we obtain from Theorem 4.24 that coneµ1 s1 +µ2 s2 (µ1 X1 + µ2 X2 ) ⊂ µ1 cones1 X1 + µ2 cones2 X2 . For the second inclusion, choose a point x ∈ µ1 cones1 X1 +µ2 cones2 X2 . Then x = µ1 x1 + µ2 x2 , where x1 ∈ cones1 X1 and x2 ∈ cones2 X2 . By Theorem 4.25, x1 and x2 can be expressed as nonnegative combinations x1 = s1 + γ1 (u1 − s1 ) + · · · + γp (up − s1 ), u1 , . . . , up ∈ X1 , x2 = s2 + β1 (v1 − s2 ) + · · · + βq (vq − s2 ), v1 , . . . , vq ∈ X2 . For all 1 6 i 6 p and 1 6 j 6 q, one has µ1 ui + µ2 s2 , µ1 s1 + µ2 vj ∈ µ1 ({s1 } ∪ X1 ) + µ2 ({s2 } ∪ X2 ). Therefore, the equality x = (µ1 s1 + µ2 s2 ) + γ1 ((µ1 u1 + µ2 s2 ) − (µ1 s1 + µ2 s2 )) + . . . + γp ((µ1 up + µ2 s2 ) − (µ1 s1 + µ2 s2 )) + β1 ((µ1 s1 + µ2 v1 ) − (µ1 s1 + µ2 s2 )) + . . . + βq ((µ1 s1 + µ2 vq ) − (µ1 s1 + µ2 s2 )) and Theorem 4.25 give the inclusion x ∈ coneµ1 s1 +µ2 s2 (µ1 ({s1 } ∪ X1 ) + µ2 ({s2 } ∪ X2 )). Theorem 4.36. If f : Rn → Rm is an affine transformation and s ∈ Rn and r ∈ rng f , then for sets X ⊂ Rn and Y ⊂ Rm , one has conef (s) f (X) = f (cones X), coneu f where u is a point in f
−1
−1
(Y ) = f
−1
(coner (Y ∩ rng f )),
(4.10) (4.11)
(r).
Proof. Excluding the trivial cases X = ∅ and Y ∩ rng f = ∅, we assume that both sets X and Y ∩ rng f are nonempty. Consequently, f −1 (Y ) 6= ∅. For the equality (4.10), choose a point x ∈ conef (s) f (X). By Theorem 4.25, there are points x1 , . . . , xp ∈ f (X) and scalars λ1 , . . . , λp > 0 such that x = f (s) + λ1 (x1 − f (s)) + · · · + λp (xp − f (s)).
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Choose zi ∈ X such that f (zi ) = xi for all 1 6 i 6 p. Put z = s + λ1 (z1 − s) + · · · + λp (zp − s). Then z ∈ cones X by the same theorem. Since z is an affine combination of s, z1 , . . . , zp , Theorem 1.87 gives x = f (s) + λ1 (x1 − f (s)) + · · · + λp (xp − f (s)) = f (s) + λp (f (z1 ) − f (s)) + · · · + λp (f (zp ) − f (s)) = f (s + λ1 (z1 − s) + · · · + λp (zp − s)) = f (z). Hence x = f (z) ∈ f (cones X), which proves the inclusion conef (s) f (X) ⊂ f (cones X). Conversely, let x ∈ cones X. According to Theorem 4.25, x = s + µ1 (x1 − s) + · · · + µq (xq − s), where x1 , . . . , xq ∈ X and µ1 , . . . , µq > 0. As above, x is an affine combination of s, x1 , . . . , xq , which gives f (x) = f (s) + µ1 (f (x1 ) − f (s)) + · · · + µq (f (xq ) − f (s)). Therefore, f (x) ∈ conef (s) f (X) by the same theorem. Hence f (cones X) ⊂ conef (s) f (X). It remains to prove (4.11). Letting X = f −1 (Y ) in (4.10), we obtain f (coneu f −1 (Y )) = conef (u) f (f −1 (Y )) = coner (Y ∩ rng f ). Hence coneu f −1 (Y ) ⊂ f −1 (f (coneu f −1 (Y ))) = f −1 (coner (Y ∩ rng f )). Conversely, let x ∈ f −1 (coner (Y ∩ rng f )). Then f (x) ∈ coner (Y ∩ rng f ), and Theorem 4.25 implies that f (x) can be written as f (x) = r + γ1 (x1 − r) + · · · + γk (xk − r) for certain points x1 , . . . , xk ∈ Y ∩ rng f and scalars γ1 , . . . , γk > 0. Choose zi ∈ f −1 (Y ) such that f (zi ) = xi , 1 6 i 6 k, and put z = u + γ1 (z1 − u) + · · · + γk (zk − u). Then z ∈ coneu f −1 (Y ) and f (z) = f (u) + γ1 (f (z1 ) − f (u)) + · · · + γk (f (zk ) − f (u)) = r + γ1 (x1 − r) + · · · + γk (xk − r) = f (x).
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Let γ = γ1 + · · · + γk . The case γ = 0 is obvious, because f (x) = r and x ∈ f −1 (Y ) ⊂ coneu f −1 (Y ). Assume that γ > 0, and let vi = zi + γ −1 (x − z), 1 6 i 6 k. Since vi is an affine combination of xi , x, z, Theorem 1.87 gives f (vi ) = f (zi ) + γ −1 f (x) − γ −1 f (z) = f (zi ) = xi . Hence vi ∈ f −1 (xi ) ⊂ f −1 (Y ), 1 6 i 6 k. From the equalities x = z + (x − z) = u + γ1 (z1 − u) + · · · + γk (zk − u) + (γ1 + · · · + γk )γ −1 (x − z) = u + γ1 (z1 + γ −1 (x − z) − u) + · · · + γk (zk + γ −1 (x − z) − u) = u + γ1 (v1 − u) + · · · + γk (vk − u) and Theorem 4.25 it follows that x ∈ coneu f −1 (Y ). Conic Hulls and Closure Theorem 4.37. For any set X ⊂ Rn and a point s ∈ Rn , one has cones (cl X) ⊂ cl (cones X).
(4.12)
Furthermore, if X is bounded and s ∈ / cl (conv X), then cones (cl X) = cl (cones X).
(4.13)
Proof. Since the case X = ∅ is obvious, we assume that X is nonempty. Clearly, cl X ⊂ cl (cones X). Since cl (cones X) is a convex cone with apex s (see Theorem 4.21), the inclusion (4.12) holds. Suppose X is bounded and s ∈ / cl (conv X). To prove (4.13), it suffices to show that cl (cones X) ⊂ cones (cl X). Choose a point x ∈ cl (cones X). Then x = limi→∞ xi for a certain sequence of points x1 , x2 , . . . from cones X. By Theorem 4.30, xi = s + γi (zi − s) for appropriate points zi ∈ conv X and scalars γi > 0, i > 1. Since conv X is bounded, a certain subsequence z10 , z20 , . . . of z1 , z2 , . . . converges to a point z ∈ cl (conv X). Clearly, z 6= s. Denote by x01 , x02 , . . . and γ10 , γ20 , . . . the respective subsequences of x1 , x2 , . . . and γ1 , γ2 , . . . , respectively. The sequence γ10 , γ20 , . . . converges to a scalar γ > 0 due to kx0i − sk kx − sk . = i→∞ kzi0 − sk kz − sk
lim γi0 = lim
i→∞
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Thus x = lim x0i = lim (s + γi0 (zi0 − s)) i→∞
i→∞
= s + γ(z − s) ∈ cones (cl (conv X)). According to Theorem 3.17, cl (conv X) = conv (cl X). This argument and Theorem 4.30 give x ∈ cones (cl (conv X)) = cones (conv (cl X)) = cones (cl X). Hence cl (cones X) ⊂ cones (cl X). Remark. The assumptions in the second statement of Theorem 4.37 are essential. Indeed, for the unbounded closed convex set K = {(x, y) : x > 0, xy > 1} ⊂ R2 , the set coneo X = {o} ∪ {(x, y) : x, y > 0} is not closed. Similarly, for the compact convex set K = {(x, y) : (x2 +(y−1)2 6 1} ⊂ R2 , which contains o, the set coneo K equals {o} ∪ {(x, y) : y > 0}. Corollary 4.38. Every simplicial cone Cs (x1 , . . . , xr ) ⊂ Rn is a closed set. Proof. Let X = {x1 , . . . , xr }. Since the set {s}∪X is affinely independent (see Definition 4.5), Corollary 1.63 shows that s ∈ / aff X. Because conv X ⊂ aff X (see Theorem 3.2) and the plane aff X is closed (see Corollary 1.22), one has s ∈ / cl (conv X). Clearly, the set X is bounded. So, the statement follows from Theorem 4.37. Corollary 4.39. For a finite set X ⊂ Rn and a point s ∈ Rn , the conic hull cones X is a closed set. Proof. The statement is obvious when X = ∅. Assume that X is nonempty, and let m = dim X. By Theorem 4.29, cones X is the union of simplicial cones of the form Cs (x1 , . . . , xk ), where k 6 m + 1 and x1 , . . . , xk ∈ X. Since X is finite, the family of such cones if finite. Hence cones X is closed as a finite union of closed sets (see Corollary 4.38). Conic Hulls and Relative Interior Theorem 4.40. For a nonempty set X ⊂ Rn and a point s ∈ Rn , the following conditions are equivalent. (1) s ∈ rint (conv X). (2) s ∈ rint (cones X).
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(3) cones (rint (conv X)) = aff X. (4) cones X = aff X. (5) cones X is a plane. Proof. Excluding the trivial case X = {s}, we assume that X \ {s} = 6 ∅. (1) ⇒ (2). Let s ∈ rint (conv X). Then s ∈ aff X, and cones X ⊂ aff X according Theorem 4.30. The inclusions conv X ⊂ cones X ⊂ aff X = aff (conv X) and Theorem 1.50 give aff (conv X) = aff (cones X). Theorem 2.15 implies s ∈ rint (conv X) ⊂ rint (cones X). (1) ⇒ (3). If s ∈ rint (conv X), then s ∈ aff X, and Theorem 4.30 gives the inclusion cones (rint (conv X)) ⊂ aff X. Conversely, let x ∈ aff X. Since the case x = s is trivial, we may assume that x 6= s. By Corollary 2.25, there is a point z ∈ rint (conv X) such that z ∈ (s, x). Hence x ∈ [s, zi ⊂ cones (rint (conv X). (2) ⇒ (4). If s ∈ rint (cones X), then Corollary 4.19 shows that cones X is a plane, and Theorem 4.30 implies that cones X = aff X. (3) ⇒ (4). If cones (rint (conv X)) = aff X, then s ∈ aff X, and Theorem 4.30 gives aff X = cones (rint (conv X)) ⊂ cones (conv X) = cones X ⊂ aff X. Hence cones X = aff X. The equivalence of conditions (4) and (5) is proved in Theorem 4.30. (4) ⇒ (1). Choose a point x ∈ conv X \ {s}. Then [s, xi ⊂ cones X = aff X, which implies the inclusion hs, xi ⊂ cones X. Let y be a point in hs, xi such that s ∈ (x, y). Because cones X = cones (conv X), there is a point z ∈ conv X such that y ∈ [s, zi. Clearly, s ∈ (x, z). Finally, s ∈ rint (conv X) according to Corollary 2.25. Theorem 4.41. For a set X ⊂ Rn and a point s ∈ Rn , one has cones (rint (conv X)) = {s} ∪ rint (cones X).
(4.14)
Proof. Since the case X ⊂ {s} is obvious, we assume that X \ {s} = 6 ∅. Next, we suppose that s ∈ / rint (cones X), since otherwise both sides of (4.14) coincide with aff X (see Theorem 4.40). Under these assumptions, we state that rint (conv X) ⊂ rint (cones X). Indeed, let x ∈ rint (conv X). To prove the inclusion x ∈ rint (cones X), it suffices to show that for every u ∈ cones X \ {x} there is a point w ∈
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cones X such that x ∈ (u, w), as follows from Corollary 2.25. Because the case u ∈ [s, xi is trivial, we assume that u ∈ / [s, xi. By Theorem 4.30, there is a point z ∈ conv X such that u ∈ [s, zi. Due to x ∈ rint (conv X), one can find a point y ∈ conv X with the property x ∈ (y, z). If z ∈ [s, u], then the halflines [u, xi and [s, yi have a common point, w (see Lemma 1.27 and the picture below). Consequently, w ∈ [s, yi ⊂ cones (conv X) and x ∈ (u, w). If u ∈ [s, z], then we can choose a point r ∈ [x, y] so close to x that the halflines [s, ri and [u, xi meet at a point, say w. As above, w ∈ [s, ri ⊂ cones (conv X) and x ∈ (u, w). Summing up, rint (conv X) ⊂ rint (cones X). ry HHw r H rx H H HHr r r s z u
rw r y HH r H rx rH H HHr r r s z u
We return to the equality (4.14). From Theorem 4.21 it follows that {s} ∪ rint (cones X) is a convex cone with apex s. Hence the inclusion rint (conv X) ⊂ rint (cones X) gives cones (rint (conv X)) ⊂ {s} ∪ rint (cones X). For the opposite inclusion, choose a point x ∈ rint (cones X). Then x ∈ cones X, and Theorem 4.30 shows that the convex set K = [s, xi ∩ conv X is nonempty. Choose a point y ∈ rint K. We intend to show that y ∈ rint (conv X). For this, it suffices to prove that every u ∈ conv X \ {y} there is a point w ∈ conv X such that y ∈ (u, w), as described in Corollary 2.25. If u ∈ K, then K is one-dimensional, and the inclusion y ∈ rint K implies the existence of points r, t ∈ K \ {y} such that y ∈ (r, t). By Lemma 1.26, either y ∈ (u, t) or y ∈ (r, u); so w may be chosen in {r, t}. u r r 0 r r r y r ry rt P P x s PP r PrP w P p r P qPPP P
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Assume that u ∈ / K. Then u ∈ / [s, xi, as depicted above. Because u ∈ conv X ⊂ cones X and x ∈ rint (cones X), Corollary 2.25 shows the existence of a point p ∈ cones X such that x ∈ (u, p). Since cones X = cones (conv X), there is a point q ∈ conv X such that p ∈ [s, qi, as follows from Theorem 4.30. Obviously, [u, q] ⊂ conv X, and the point y 0 = [s, xi ∩ [u, q] belongs to K. If y 0 = y, then put w = q. Let y 0 6= y. Then K 6= {y}, and, as above, there are points r, t ∈ K \ {y} such that y ∈ (r, t). Then the halfline [u, yi meets one of the segments [r, q] and [t, q]. Assume, for example, that [u, yi meets [t, q], and denote by w their common point. Clearly, y ∈ (u, w) and w ∈ [t, q] ⊂ conv X. Summing up, y ∈ rint (conv X)
and x ∈ [s, yi ⊂ cones (rint (conv X)).
Hence {s} ∪ rint (cones X) ⊂ cones (rint (conv X)). Steinitz-Type Results The following result is analogous to Theorem 3.20. Theorem 4.42. Let s ∈ Rn and X ⊂ Rn be a set such that X \{s} = 6 ∅. A point x ∈ Rn belongs to rint (cones X) if and only if there are finitely many points x1 , . . . , xk ∈ X and scalars λ1 , . . . , λk > 0 satisfying the conditions x = s + λ1 (x1 − s) + · · · + λk (xk − s), aff {x1 , . . . , xk } = aff X.
(4.15)
Proof. Let x ∈ rint (cones X). If x = s, then Theorem 4.40 implies the inclusion s ∈ rint (conv X). In this case, according to Theorem 3.20, there are finitely many points x1 , . . . , xk ∈ X such that aff {x1 , . . . , xk } = aff X and s is their positive convex combination, s = λ1 x1 + · · · + λk xk . Clearly, s = s + λ1 (x1 − s) + · · · + λk (xk − s). Assume that x 6= s. Theorem 4.41 gives x ∈ cones (rint (conv X)), and Theorem 4.30 implies the existence of a point z ∈ rint (conv X) such that x = s + λ(z − s) for a certain scalar λ > 0. As above, Theorem 3.20 shows the existence of finitely many points x1 , . . . , xk ∈ X such that aff {x1 , . . . , xk } = aff X and z is their positive convex combination, z = µ1 x1 + · · · + µk xk . Put λi = λµi , 1 6 i 6 k. Then λ1 , . . . , λk > 0 and x = s + λ(z − s) = s + λ1 (x1 − s) + · · · + λk (xk − s). Conversely, suppose x satisfies the conditions (4.15). Let λ = λ1 + · · · + λk
and u = λ−1 (λ1 x1 + · · · + λk xk ).
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Clearly, u is a positive convex combination of x1 , . . . , xk , and Theorem 3.20 shows that u ∈ rint (conv X). If s = u, Theorem 4.40 implies that cones X is a plane; hence x ∈ cones X = rint (cones X). If s 6= u, then Theorem 4.41 gives x = s + λ(u − s) ∈ cones (rint (conv X)) \ {s} = rint (cones X). An important particular case of Theorem 4.42, combined with Corollary 3.21, is given in the statement below. Corollary 4.43. For a finite set X = {x1 , . . . , xr } ⊂ Rn and a point s ∈ Rn , one has rint (cones X) = {s + λ1 (x1 − s) + · · · + λr (xr − s) : λ1 , . . . , λr > 0}. The next result is an analog of Steinitz’s Theorem 3.26 for the case of conic hulls. Theorem 4.44. Let X ⊂ Rn be a set of positive dimension m and s ∈ Rn . A point x ∈ Rn belongs to rint (cones X) if and only if there is a subset Y of X such that x ∈ rint (cones Y ),
m + 1 6 card Y 6 2m,
aff Y = aff X.
Proof. Let x ∈ rint (cones X). Then x ∈ cones (rint (conv X)) according to Theorem 4.41. Assume first that x = s. Then Theorem 4.40 implies the equality cones X = aff X. By Theorem 1.68, X contains an affine basis Y for aff X. Clearly, card Y = m + 1 6 2m, and x = s ∈ aff X = aff Y = rint (aff Y ) = rint (cones Y ). Suppose that x 6= s. From Theorem 4.30 it follows that x = s+λ(z−s), where λ > 0 and z ∈ rint (conv X). Theorem 3.26 implies the existence of a set Y ⊂ X such that z ∈ rint (conv Y ),
m + 1 6 card Y 6 2m,
and
aff Y = aff X.
Finally, x ∈ cones (rint (conv Y )) \ {s} = rint (cones Y ) according to Theorem 4.41. The converse statement follows from Theorem 4.42.
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Remark. If 2m is the minimum cardinality of a set Y ⊂ X satisfying the hypothesis of Theorem 4.44, then, unlike Theorem 3.26, Y cannot be partitioned into m pairs of points collinear with x. For example, let x = (3, 0, 0) and X = {x1 , x2 , x3 , x4 }, where x1 = (1, 1, 0), x2 = (1, −1, 0), x3 = (2, 0, 2), x4 = (2, 0, −2). Then coneo X = {(x, y, z) : x > |y| + |z|} and x ∈ int (coneo X). Clearly, no proper subset Y of X satisfies the condition x ∈ int (coneo Y ), and X cannot be partitioned into two pairs of points collinear with x. Corollary 4.45. Let X ⊂ Rn be a set and s be a point in Rn such that cones X is a plane of positive dimension m. Then s ∈ aff X and X contains a subset Y which satisfies the following conditions: cones Y = aff Y = aff X
and
m + 1 6 card Y 6 2m.
Proof. Theorem 4.40 implies that cones X = aff X and s ∈ rint (cones X). Theorem 4.44 shows the existence of a set Y ⊂ X satisfying the conditions x ∈ rint (cones Y ),
m + 1 6 card Y 6 2m,
aff Y = aff X.
Again by Theorem 4.40, cones Y is a plane. Hence cones Y = aff Y . Exercises for Chapter 4 Exercise 4.1. Let L ⊂ Rn be a plane of positive dimension m with an affine basis c1 , . . . , cm+1 and s be a point in the relative interior of the simplex ∆(c1 , . . . , cm+1 ). Show that L = cones {c1 , . . . , cm+1 } = conv (h1 ∪ · · · ∪ hm+1 ), where hi = [s, ci i is the halfline though ci with endpoint s, 1 6 i 6 m + 1. Exercise 4.2. Let C = Cs (x1 , . . . , xr ) be a simplicial cone with apex s. For a proper subset X of {x1 , . . . , xr }, denote by C1 and C2 the simplicial cones with apex s generated by X and Y = {x1 , . . . , xr } \ X, respectively. Show that C1 ∩ C2 = {s} and C = conv2 (C1 ∪ C2 ) = ∪ ([u1 , u2 ] : u1 ∈ C1 , u2 ∈ C2 ). Exercise 4.3. Let C = Cs (x1 , . . . , xr ) and C 0 = Cs0 (x01 , . . . , x0r ) be simplicial r-cones in Rn . Show the existence of an invertible affine transformation f : Rn → Rn with the property f (C) = C 0 .
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Exercise 4.4. Let L ⊂ Rn be a proper plane, s be a point in Rn \ L, and L0 be a translate of L containing s. Show that cones L is the union of {s} and the open halfplane of the plane M = aff ({s} ∪ L) which is determined by L0 and contains L. Exercise 4.5. (Bair [13]) Show that a nonempty convex set K ⊂ Rn satisfies the equality K = a + µK, where a 6= o and µ > 0, if and only if K is either a cone or contains a translate of the line ho, ai. Exercise 4.6. (Rockafellar [184, p. 22]) For a pair of nonempty disjoint convex sets K1 and K2 in Rn , the umbra of K1 with respect to K2 is defined by U (K1 , K2 ) = ∩ (1 − γ)x + γK1 : γ > 1 and x ∈ K2 , and the penumbra of K1 with respect to K2 is given by P (K1 , K2 ) = ∪ (1 − γ)x + γK1 : γ > 1 and x ∈ K2 . Show that both set U (K1 , K2 ) and P (K1 , K2 ) are convex. Notes for Chapter 4 Convex cones. Various facts about convex cones and their apices can be traced in the works of Minkowski [158, pp. 131–229] and Steinitz [206]. Sung and Tam [213] showed that for a convex cone C ⊂ Rn with apex s ∈ Rn , the following conditions are equivalent. (a) C is an Fσ -set. (b) C = cones K for a certain compact convex set K ⊂ Rn . (c) C is a countable union of a nested family of closed convex cones with common apex s. Conic hulls. The conic hull coneo X is often called the positive hull of a set X ⊂ Rn and is denoted pos X. The first statement of Theorem 4.29 can be found in Pforr [171] and Stoer and Witzgall [210]. Corollary 4.43 is proved by Fischer [86] for the case s = o. Positive bases. Following Steinitz [206], a set X of vectors in Rn is called positively independent provided no vector x ∈ X belongs to pos (X \ {x}). A set X ⊂ Rn is call a positive basis for Rn provided X is positively independent and pos X = Rn . As proved by Steinitz [206, p. 15] (also, by Gale [90] and Davis [67]), every positive basis X for Rn satisfies the inequalities n + 1 6 card X 6 2n. Here the equality card X = n + 1 holds if and only if X consists of the vectors of a linear basis for Rn together with a sum of negative multiples of these vectors, and the equality card X = 2n holds if and only if X consists of the vectors of
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a linear basis for Rn together with their negative multiples. Reay [179] proved that a positive basis X for Rn admits a partition into pairwise disjoint subsets X = X1 ∪ · · · ∪ Xr such that: (a) card X1 > · · · > card Xr > 2, (b) pos (X1 ∪ · · · ∪ Xi ) is a subspace of dimension card X1 + · · · + card Xi − i for all 1 6 i 6 r. See McKinney [155] for an algebraic description of positive bases for Rn , who also studied the notion of strong positive independence of a set X ⊂ Rn , given by the condition pos U ∩ pos V = {o} for all pairs of disjoint subsets U and V of X. Hansen and Klee [112], generalizing results from [155], showed that a set X ⊂ Rn is strongly positively independent if and only if X = X0 ∪ X1 ∪ · · · ∪ Xr such that: (a) X0 is linearly independent, (b) the subspaces span X1 , . . . , span Xr are independent, (c) card Xi = dim (span Xi ) + 1 for all 1 6 i 6 r. A set X is called a strongly positive basis for Rn if X is strongly positively independent and pos X = Rn . If X is a strongly positive basis for Rn then X = X1 ∪ · · · ∪ Xk such that conditions (b) and (c) above are satisfied (see McKinney [155]). Giving a positive basis c1 , . . . , cr for Rn , a representation x = α1 c1 +· · ·+αr cr of a given vector x ∈ Rn as a positive combination of c1 , . . . , cr is called minimal provided it involves the minimum possible number of nonzero coefficients α1 , . . . , αr . Reay [181], expanding results of Bonnice and Klee [30] and McKinney [155], showed that a positive basis c1 , . . . , cr for Rn is strongly positive if and only if, for every vector x ∈ Rn its minimal representation as a positive combination of c1 , . . . , cr is unique.
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Recession and Normal Directions
5.1
Recession Cones
Unbounded Convex Sets and Halflines Theorem 5.1. Let the closure of a convex set K ⊂ Rn contain a closed halfline h with endpoint x. Then the following statements hold. (1) If x ∈ rint K, then h ⊂ rint K. (2) If x ∈ rbd K, then either h ⊂ rbd K or h \ {x} ⊂ rint K. (3) For points u ∈ cl K and v ∈ rint K, the halflines h0 = (u − x) + h
and
h00 = (v − x) + h
lie in cl K and rint K, respectively. Proof. (1) Let x ∈ rint K. Assume for a moment that h 6⊂ rint K. By Theorem 2.55, there is a unique point u ∈ h ∩ rbd K with the property h \ [x, u] ⊂ Rn \ cl K, in contradiction with the assumption h ⊂ cl K. (2) Let x ∈ rbd K. Suppose that h 6⊂ rbd K, and choose a point y ∈ h ∩ rint K. Clearly, the halfline g = (y − x) + h lies in h. Furthermore, g ⊂ rint K by the above proved. Since (x, y) ⊂ rint K (see Theorem 2.36), one has h \ {x} = (x, y) ∪ g ⊂ rint K. x r
w wλ r !r ! J ! J !!! r J vλ !! Jr r!! u z
h
h0
(3) We first observe that u is the endpoint of h0 . To prove the inclusion h ⊂ cl K, choose a point z ∈ h0 . Excluding the trivial case z = u, we 0
183
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assume that z 6= u. Put w = x − u + z. Then w ∈ (x − u) + h0 = h ⊂ cl K and x 6= w. Let 1 λ vλ = (1 − λ)w + λz, wλ = w− x, 0 < λ < 1. 1−λ 1−λ Clearly, w ∈ (x, wλ ) and wλ ∈ [x, wi = h (see Lemma 1.26). From vλ = w + λ(z − w) = w + λ(u − x) 1 λ = (1 − λ) w− x + λu 1−λ 1−λ = (1 − λ)wλ + λu and the convexity of cl K (see Theorem 2.35) it follows that vλ ∈ [u, wλ ] ⊂ cl K, as depicted above. Finally, z = lim ((1 − λ)w + λz) = lim vλ ∈ cl K. λ→1
λ→1
0
Summing up, h ⊂ cl K. Since v ∈ rint K ⊂ cl K, the halfline h00 lies in cl K, as proved above. Finally, statement (1) yields the inclusion h00 ⊂ rint K. Corollary 5.2. Let K ⊂ Rn be a convex set and l be a line in cl K. If x ∈ l, then for points u ∈ cl K and v ∈ rint K, the lines l0 = (u − x) + l and l00 = (v − x) + l lie in cl K and rint K, respectively. Proof. Express l as the union of halflines h1 and h2 with common endpoint x. Then l0 = ((u − x) + h1 ) ∪ ((u − x) + h2 ),
l00 = ((v − x) + h1 ) ∪ ((v − x) + h2 ).
The inclusions l0 ⊂ cl K and l00 ⊂ rint K follow from Theorem 5.1. Theorem 5.3. For a convex set K ⊂ Rn , the following conditions are equivalent. (1) K is unbounded. (2) For every point x ∈ cl K, there is a closed halfline with endpoint x which lies in cl K. (3) For every point x ∈ rint K, there is a closed halfline with endpoint x which lies in rint K. Proof. (1) ⇒ (2). Choose a point x ∈ cl K and a sequence of points x1 , x2 , . . . in K \ {x} such that limi→∞ kxi − xk = ∞. Put xi − x ui = x + , i > 1. (5.1) kxi − xk
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Clearly, kui −xk = 1 for all i > 1, which shows that the set {u1 , u2 , . . . } lies on the sphere S1 (x) of radius 1 centered at x. Since S1 (x) is compact, the sequence u1 , u2 , . . . contains a subsequence which converges to a point u ∈ S1 (x). Without loss of generality, we may assume that the sequence u1 , u2 , . . . itself converges to u. If an integer i0 is such that kxi − xk > 1 for all i > i0 , then, by (5.1), ui can be written as the convex combination 1 1 x+ xi , i > i0 . ui = 1 − kxi − xk kxi − xk Hence ui ∈ [xi , x] ⊂ cl K for all i > i0 by the convexity of cl K (see Theorem 2.35). Summing up, u = limi→∞ ui ∈ cl K. We state that the halfline [x, ui lies in cl K. Indeed, choose a point y in [x, ui. Excluding the trivial case y = x, we conclude that y = (1−λ)x+λu for a certain scalar λ > 0. Clearly, kx − yk = λkx − uk = λ. Since limi→∞ kxi − xk = ∞, there is an integer i1 (> i0 ) such that kxi − xk > λ for all i > i1 . Put yi = (1 − λ)x + λui . From yi ∈ [x, ui i = [x, xi i (see Lemma 1.26) and the inequality kyi − xk = λ < kxi − xk,
i > i1 ,
it follows that yi ∈ [xi , x] ⊂ cl K for all i > i1 . Because of lim yi = lim ((1 − λ)x + λui ) = (1 − λ)x + λu = y,
i→∞
i→∞
the point y is in cl K. Summing up, [x, ui ⊂ cl K. (2) ⇒ (3). This part of the proof follows directly from Theorem 5.1. (3) ⇒ (1). Let a halfline [x, zi lie in rint K. Choose a scalar ρ > 0 and a point ρ . y = (1 − λ)x + λz, where λ > kx − zk Then y ∈ [x, zi ⊂ rint K ⊂ K and kx − yk = kx − (1 − λ)x − λzk = λkx − zk > ρ. Hence K is unbounded. Remark. It might happen that for an unbounded convex set K ⊂ Rn and a point x ∈ K, no halfline with endpoint x lies in K. Indeed, if K ⊂ R2 is given by K = {o} ∪ {(x, y) : 0 < y < 1}, then no halfline h ⊂ R2 with endpoint o lies in K. The next result gives a slight generalization of Theorem 5.3. Theorem 5.4. Let K ⊂ Rn be a unbounded convex set and x and v be points in cl K and Rn , respectively. The following statements hold.
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(1) If an unbounded sequence x1 , x2 , . . . of points in K is such that the unit vectors xi − v zi = , i > 1, (5.2) kxi − vk converge to a vector z, then the halfline [x, x + zi lies in cl K. (2) Every unbounded sequence of points in K contains a subsequence, say x1 , x2 , . . . , such that the respective sequence (5.2) converges. Proof. (1) The assumption on the unboundedness of x1 , x2 , . . . is equivalent to the condition limi→∞ kxi k = ∞. From the inequalities kxi k − kxk 6 kxi − xk 6 kxi k + kxk, kxi k − kvk 6 kxi − vk 6 kxi k + kvk
(5.3)
it follows that lim kxi − xk = lim kxi − vk = ∞ and
i→∞
i→∞
lim
i→∞
kxi − vk = 1. kxi − xk
Therefore, x −v v−x xi − x i = lim + i→∞ kxi − xk i→∞ kxi − xk kxi − xk kx − vk x − v v−x i i = lim + i→∞ kxi − xk kxi − vk kxi − xk lim
= 1z + o = z. Hence the vectors u1 , u2 , . . . from (5.1) converge to x + z. As shown in the proof of Theorem 5.3, the halfline [x, x + zi lies in cl K. (2) Let y1 , y2 , . . . be an unbounded sequence of points in K. By an argument from the proof of Theorem 5.3, y1 , y2 , . . . contains a subsequence x1 , x2 , . . . such that the respective sequence (5.1) converging to a point u ∈ cl K. As above, from the inequalities (5.3) it follows that lim kxi − xk = lim kxi − vk = ∞
i→∞
i→∞
and
lim
i→∞
kxi − xk = 1. kxi − vk
Let z = u − x. Then x −x x−v xi − v i = lim + lim i→∞ kxi − vk i→∞ kxi − vk kxi − vk kx − xk x − x x−v i i = lim + i→∞ kxi − vk kxi − xk kxi − vk = 1(u − x) + o = z.
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Recession Cones Definition 5.5. The recession cone of a nonempty set X ⊂ Rn , denoted rec X, is defined by rec X = {e ∈ Rn : λe + x ∈ X whenever λ > 0 and x ∈ X}. We put rec ∅ = ∅. Remark. Given a vector e ∈ Rn , the scalar multiples λe, λ > 0, fulfil the halfline [o, ei (or the singleton {o} if e = o). This argument shows that the recession cone of a nonempty set is either {o} or the union of closed halflines with common endpoint o.
A A
X o
Fig. 5.1
rec X
The recession cone of a set X.
Example. If C ⊂ Rn is a cone with apex s, then rec C = C − s. In particular, the recession cone of a nonempty plane L ⊂ Rn is its characteristic subspace sub L. Example. If M is a closed slab between parallel hyperplanes H and H 0 (see Definition 1.33), then rec M is the hypersubsace which is parallel to both H and H 0 . Indeed, from Theorem 1.34 it follows that for points x ∈ M and e ∈ Rn , the halfline h = {x + λe : λ > 0} lies in M if and only if h is parallel to both H and H 0 . This argument shows that e ∈ rec M if and only if the halfline h0 = {λe : λ > 0} lies in rec M . A similar argument holds for the open slab between H and H 0 Example. If L ⊂ Rn is a plane of positive dimension and F = {x ∈ L : γ 6 x·c 6 γ 0 } is a closed slab of L (see Corollary 1.44), then Theorem 5.7 and the above example show that rec F is a subspace of dimension dim L − 1, given by rec F = {x ∈ sub L : x·c = 0}. A similar statement holds for an open slab of L.
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Theorem 5.6. If K is a nonempty convex set in Rn , then the following statements hold. (1) rec K = {e ∈ Rn : e + K ⊂ K}. (2) rec K is the largest among all convex cones C ⊂ Rn with apex o which satisfy the inclusion C +K ⊂ K, or, equivalently, the equality C + K = K. (3) rec K = ∩ (K − x : x ∈ K). (4) Given a point u ∈ K, one has rec K ⊂ ∩ (λ(K − u) : λ > 0).
(5.4)
Furthermore, rec K = ∩ (λ(K − u) : λ > 0) provided K is closed or relatively open. (5) rec K ⊂ sub K, and rec K = sub K if and only if K is a plane. Proof. (1) If e ∈ rec K, then 1e + x ∈ K for all x ∈ K, which implies the inclusion e + K ⊂ K. Conversely, let e be a point in Rn satisfying the condition e + K ⊂ K. By induction on r > 1, we easily conclude that re + K ⊂ K for all r > 1. Choose a point x ∈ K, a scalar λ > 0, and an integer r0 such that λ 6 r0 . Then λe ∈ [o, r0 e], and λe + x ∈ [o, r0 e] + x = [x, r0 e + x] ⊂ K by the convexity of K. Hence e ∈ rec K according to Definition 5.5. (2) First, we state that rec K is a convex cone with apex o. Indeed, let e ∈ rec K and µ > 0. For a scalar λ > 0 and a point x ∈ K, we have λ(µe) + x = (λµ)e + x ∈ K, which shows the inclusion µe ∈ rec K. For the convexity of rec K, choose points b and e in rec K and a scalar γ ∈ [0, 1]. Let x be a point in K. Then λb + x ∈ K and λe + x ∈ K for all λ > 0. Since K is a convex set, Theorem 2.4 yields λ((1 − γ)b + γe) + x = (1 − γ)(λb + x) + γ(λc + x) ∈ (1 − γ)K + γK = K. Hence (1 − γ)b + γe ∈ rec K, giving the convexity of rec K. If C is a convex cone with apex o satisfying the inclusion C + K ⊂ K, then, for any e ∈ C, λ > 0, and x ∈ K, one has λe + x ∈ C + K ⊂ K, implying that e ∈ rec K. Hence C ⊂ rec K, and rec K is the largest cone among all convex cones C with apex o that satisfy the condition C+K ⊂ K. Since o + x = x for every x ∈ K, we have rec K + K = K.
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(3) If e ∈ rec K, then e + x ∈ K, or, equivalently, e ∈ K − x for every point x ∈ K. Hence e ∈ ∩ (K − x : x ∈ K). Conversely, suppose that a vector e does not belong to ∩ (K − x : x ∈ K). Then e ∈ / K − x0 for a certain x0 ∈ K. Consequently, e + x0 ∈ / K, implying that e ∈ / rec K. (4) According to statement (2) above, rec K is a convex cone with apex o and u + rec K ⊂ K. Theorem 4.2 shows that λ rec K = rec K for all λ > 0. Therefore, the inclusion rec K ⊂ K − u gives rec K = λ rec K ⊂ λ(K − u),
λ > 0.
Hence rec K ⊂ ∩ (λ(K − u) : λ > 0). Next, we state that rec K = ∩ (λ(K − u) : λ > 0) provided K is closed or relatively open. Due to the above argument, it suffices to prove the inclusion ∩ (λ(K − u) : λ > 0) ⊂ rec K. Let e ∈ ∩ (λ(K − u) : λ > 0). Then e/λ ∈ K − u for all λ > 0. Since the case e = o is obvious, we assume that e 6= o. The points e/λ, λ > 0, fulfill the closed halfline h = [o, ei, which lies in K − u. Equivalently, u + h ⊂ K. From Theorem 5.1 it follows that x + h ⊂ K for all x ∈ K. Consequently, e ∈ rec K. (5) Let e ∈ rec K. Then e + x ∈ K whenever x ∈ K. According to Corollary 1.55, one has e ∈ K − x ⊂ aff K − x = span (K − K) = sub K. If K is a plane, then rec K = K − K = sub K (see example on page 187). Conversely, let rec K = sub K. Then rec K is a translate of aff K (see Corollary 1.55), and a combination of Theorem 1.2 and statement (2) above gives K = rec K + K = aff K, implying that K is a plane. Remark. The inclusion (5.4) may be proper if K is neither closed nor relatively open. Indeed, let K = {o} ∪ {(x, y) : 0 < y < 1}. Then rec K = {o}, while ∩ (λ(K −u) : λ > 0) is the x-axis of R2 for any point u ∈ K \{o}. Algebra of Recession Cones Theorem 5.7. For a family F = {Kα } of convex sets in Rn satisfying the condition ∩ Kα 6= ∅, one has α
∩ rec Kα ⊂ rec (∩ Kα ). α
α
(5.5)
Furthermore, if every element of the family F is closed or relatively open, then ∩ rec Kα = rec (∩ Kα ). α
α
(5.6)
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If, additionally, the family F is nested, then the family {rec Kα } is nested in the same order, and ∪ rec Kα ⊂ rec (∪ Kα ). α
(5.7)
α
Proof. Choose a point e ∈ ∩ rec Kα . Given a set Kα ∈ F, the inclusion α λe + x ∈ Kα holds for all λ > 0 and x ∈ Kα . Hence λe + x ∈ ∩ Kα for α
all λ > 0 and x ∈ ∩ Kα , implying that e ∈ rec (∩ Kα ). Hence the inclusion α
α
(5.5) holds. Now, assume that every set from F is either closed or relatively open. To prove the equality (5.6), it suffices, by the above argument, to show that rec (∩ Kα ) ⊂ ∩ rec Kα . Since this inclusion is obvious when α
α
rec (∩ Kα ) = {o}, we may assume that rec (∩ Kα ) contains a closed halfline α α h with endpoint o. Choose a point x ∈ ∩ Kα . Then x + h ⊂ ∩ Kα . In parα α ticular, x+h ⊂ Kα for every Kα ∈ F. If z is a point in Kα , then z+h ⊂ Kα whenever Kα is closed or relatively open (see Theorem 5.1). This argument shows that h ⊂ rec Kα for every Kα ∈ F. Hence h ⊂ ∩ rec Kα , and the α
inclusion rec (∩ Kα ) ⊂ ∩ rec Kα is established. α α Finally, suppose additionally that the family F is nested. Choose any sets Kβ , Kγ ∈ F and assume, for example, that Kβ ⊂ Kγ . We state that rec Kβ ⊂ rec Kγ . Indeed, let e ∈ rec Kβ . If e = o, then the inclusion e ∈ rec Kγ is true. Suppose e 6= o and denote by h the halfline [o, ei. Given a point x ∈ Kβ , the halfline x + h lies in Kβ . If z ∈ Kγ , then z − x + h ⊂ Kγ whenever Kγ is closed or relatively open (see Theorem 5.1). The same theorem gives the inclusion h ⊂ rec Kγ . Hence rec Kβ ⊂ rec Kγ , In a similar way, we obtain the inclusion (5.7). Corollary 5.8. If a plane L ⊂ Rn meets a convex set K ⊂ Rn , then rec K ∩ sub L ⊂ rec (K ∩ L). Furthermore, if K is closed or relatively open, then rec K ∩ sub L = rec (K ∩ L). Proof. The proof follows from Theorem 5.7 and the fact that rec L coincides with the characteristic subspace sub L of L (see example on page 187). Remark. The inclusions in Corollary 5.8 may be proper. Indeed, let K = {(0, 1)} ∪ {(x, y) : −1 < y < 1} Then rec K = {o} = 6 rec (K ∩ L) = L.
and L = {(x, 0) : x ∈ R}.
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Theorem 5.9. If K ⊂ Rn is a convex set, a is a point in Rn , and µ is a scalar, then rec (a + µK) = µ rec K = sgn (µ) rec K.
(5.8)
Proof. Since both cases K = ∅ and µ = 0 are obvious, we may assume that K is nonempty and µ 6= 0. One has e ∈ rec (a + µK) if and only if λe + (a + µx) ∈ a + µK whenever λ > 0 and x ∈ K. Equivalently, λ(e/µ) + x ∈ K for all λ > 0 and x ∈ K. In other words, e/µ ∈ rec K, or e ∈ µ rec K. Summing up, rec (a + µK) = µ rec K. Since rec K is a convex cone with apex o, Theorem 4.2 gives rec K if µ > 0 µ rec K = {o} if µ = 0 = sgn (µ) rec K. −rec K if µ < 0 Theorem 5.10. For nonempty convex sets K1 , . . . , Kr ⊂ Rn and scalars µ1 , . . . , µr , one has µ1 rec K1 + · · · + µr rec Kr ⊂ rec (µ1 K1 + · · · + µr Kr ). If, additionally, the planes aff K1 , . . . , aff Kr are independent, then µ1 rec K1 + · · · + µr rec Kr = rec (µ1 K1 + · · · + µr Kr ). Proof. The case r = 1 follows from Theorem 5.9, and an induction argument shows that the proof can be reduced to the case r = 2. Choose points e1 ∈ µ1 rec K1 and e2 ∈ µ2 rec K2 . Let x be a point in µ1 K1 + µ2 K2 . Then x can be expressed as x = µ1 x1 + µ2 x2 for certain points x1 ∈ K1 and x2 ∈ K2 . For a scalar λ > 0, one has λ(µ1 e1 + µ2 e2 ) + x = µ1 (λe1 + x1 ) + µ2 (λe2 + x2 ) ∈ µ1 K1 + µ2 K2 , which implies the inclusion µ1 e1 + µ2 e2 ∈ rec (µ1 K1 + µ2 K2 ). Hence µ1 rec K1 + µ2 rec K2 ⊂ rec (µ1 K1 + µ2 K2 ). Assume now that aff K1 and aff K2 are independent planes. Choose points u1 ∈ K1 and u2 ∈ K2 , and put K10 = K1 − u1
and K20 = K2 − u2 .
Then S1 = aff K10 = aff K1 − u1
and S2 = aff K20 = aff K2 − u2
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are independent subspaces. Furthermore, Theorem 5.9 gives rec K10 = rec K1 ⊂ S1
and
rec K20 = rec K2 ⊂ S2 .
By the above proved, it suffices o show that rec (µ1 K10 + µ2 K20 ) ⊂ µ1 rec K10 + µ2 rec K20 . Let e ∈ rec (µ1 K10 + µ2 K20 ). By Theorem 5.6, e ∈ sub (µ1 K10 + µ2 K20 ) = span (µ1 K10 + µ2 K20 ) ⊂ span K10 + span K20 = S1 + S2 . Hence e = e1 + e2 for certain vectors e1 ∈ S1 and e2 ∈ S2 . Choose points x1 ∈ K10 and x2 ∈ K20 . Then (λe1 + µ1 x1 ) + (λe2 + µ2 x2 ) = λe + (µ1 x1 + µ2 x2 ) ∈ µ1 K10 + µ2 K20 ,
λ > 0.
Since λe1 + µ1 x1 ∈ S1 and λe2 + µ2 x2 ∈ S2 , and the sum µ1 K10 + µ2 K20 is direct (see page 4), one has λe1 + µ1 x1 = u1 ∈ µ1 K10
and λe2 + µ2 x2 = u2 ∈ µ2 K20
for all λ > 0. Therefore, e1 ∈ rec (µ1 K10 ) = µ1 rec K10
and e2 ∈ rec (µ2 K20 ) = µ2 rec K20 ,
implying the inclusion e = e1 + e2 ∈ µ1 rec K10 + µ2 rec K20 . Remark. The inclusion in Theorem 5.10 may be proper if the planes aff K1 , . . . , aff Kr are not independent. Indeed, let K1 = {(x, y) : y > x2 }
and K2 = {(x, y) : y 6 −x2 }.
Then rec K1 and rec K2 are opposite halflines: rec K1 = {(0, y) : y > 0}
and
rec K2 = {(0, y) : y 6 0},
which gives rec K1 + rec K2 = {(0, y) : y ∈ R}. On the other hand, K1 + K2 = R2 , implying that rec (K1 + K2 ) = R2 . Theorem 5.11. For a linear transformation g : Rn → Rm and convex sets K ⊂ Rn and M ⊂ Rm , one has g(rec K) ⊂ rec g(K)
and
rec g −1 (M ) = g −1 (rec (M ∩ rng f )).
Furthermore, if g is one-to-one, then g(rec K) = rec g(K).
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Proof. Excluding the trivial cases K = ∅ and M ∩ rng g = ∅, we assume that both sets K and M ∩ rng g are nonempty. Consequently, g −1 (M ) 6= ∅. First, we will prove the inclusion g(rec K) ⊂ rec g(K). Let e ∈ g(rec K), x ∈ g(K), and λ > 0. Choose points e0 ∈ rec K and x0 ∈ K such that g(e0 ) = x and g(x0 ) = x. Then λe0 + x0 ∈ K, which gives the inclusion λe + x = g(λe0 + x0 ) ∈ g(K),
λ > 0.
Hence e ∈ rec g(K), and g(rec K) ⊂ rec g(K). With K = g −1 (M ), we obtain, by the above proved, g(rec g −1 (M )) ⊂ rec g(g −1 (M )) = rec (M ∩ rng g). Thus, rec g −1 (M ) ⊂ g −1 (rec (M ∩ rng g)). Conversely, let e ∈ g −1 (rec (M ∩ rng g)). Choose a point x ∈ g −1 (M ) and a scalar scalar λ > 0. Let e0 = g(e) and x0 = g(x). Then e0 ∈ rec (M ∩ rng g) and x0 ∈ M ∩ rng g, which gives λe0 + x0 ∈ M ∩ rng g
for all
λ > 0.
Therefore, λe + x ∈ g −1 (λe0 + x0 ) ∈ g −1 (M ) So, e ∈ rec g
−1
for all
λ > 0.
(M ), and g −1 (rec (M ∩ rng g)) ⊂ rec g −1 (M ).
Finally, we state that g(rec K) = rec g(K) provided g is one-to-one. By the above proved, it suffices to show the inclusion rec g(K) ⊂ g(rec K). Choose points e ∈ rec g(K) and x ∈ g(K). Then e + x ∈ g(K). Hence there are points u, v ∈ K such that g(u) = x and g(v) = e + x. So, e = g(v) − x = g(v − u). We state that the vector c = v − u belongs to rec K. Indeed, for y ∈ K and λ > 0, we have g(λc + y) = λg(c) + g(y) = λe + g(y) ∈ g(K). Thus λc + y ∈ K because g is one-to-one. This argument gives c ∈ rec K, implying that e = g(c) ∈ g(rec K). Remark. The inclusion g(rec K) ⊂ rec g(K) in Theorem 5.11 may be proper. Indeed, if K = {(x, y) : y > x2 } and g is the orthogonal projection of R2 on the x-axis, then rec K = {(0, y) : y > 0}
and g(rec K) = {o}.
On the other hand, g(K) = {(x, 0) : x ∈ R} and rec g(K) = g(K).
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Recession Cones and Topology Theorem 5.12. If C ⊂ Rn is a convex cone, then rec (cl C) = cl (rec C). Proof. Let s be an apex of C. By Theorem 4.21, cl C is a convex cone with apex s. Since rec D = D − s for any cone D ⊂ Rn with apex s (see example on page 187), one has rec (cl C) = cl C − s = cl (C − s) = cl (rec C). Theorem 5.13. For a convex set K ⊂ Rn , one has rec K ⊂ rec (rint K) = rec (cl K) = cl (rec (cl K)). If K 6= ∅, then K is unbounded if and only if rec (cl K) 6= {o}. Proof. Since the case K = ∅ is obvious, we assume that K is nonempty. The inclusion rec K ⊂ rec (rint K) is trivial when rec K = {o}. Let rec K 6= {o}, and choose in rec K a closed halfline h with endpoint o. If x ∈ K, then, x + h ⊂ K according to Theorem 5.6. For a point y ∈ rint K, the halfline (y − x) + h lies in rint K (see Theorem 5.3), which implies the inclusion h ⊂ rec (rint K). Hence rec K ⊂ rec (rint K). In a similar way, Theorem 5.3 shows that a halfline h ⊂ Rn with apex o lies in rec (rint K) if and only if it lies in rec (cl K). Hence rec (rint K) = rec (cl K). The cone rec (cl K) is closed as the intersection of closed sets cl K − x, x ∈ cl K (see Theorem 5.6). Hence rec (cl K) = cl (rec (cl K)). The second statement follows from Theorem 5.3. Remark. The inclusion rec K ⊂ rec (rint K) may be proper. Indeed, for an unbounded convex set K = {o} ∪ {(x, y) : 0 < y < 1} ⊂ R2 . one has rec K = {o}, while rec (rint K) is the x-axis of R2 . Theorems 5.1 and 5.6 imply the following corollary. Corollary 5.14. For a nonempty convex set K ⊂ Rn , the following statements hold. (1) If x is a point in cl K, then rec (cl K) is the largest among all convex cones C ⊂ Rn with apex o which satisfy the inclusion x + C ⊂ cl K. (2) If x is a point in rint K, then rec (rint K) is the largest among all convex cones C ⊂ Rn with apex o which satisfy the inclusion x + C ⊂ rint K.
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Remark. Corollary 5.14 does not hold if we replace cl K or rint K with K. For example, let K = {(x, y) : 0 < y < 1} ∪ {(x, 1) : x > 0} ∪ {(x, 0) : x 6 0}, as depicted below. Clearly, rec K = {o}. On the other hand, for every point x ∈ K, there is a horizontal halfline h with endpoint x (the direction of h depends on x) such that h ⊂ K. r K r Theorem 5.15. If K is a nonempty convex set in Rn , then rec (cl K) is the set of limits of all converging sequences λ1 x1 , λ2 x2 , . . . , where x1 , x2 , . . . belong to K and λ1 , λ2 , . . . are nonnegative scalars tending to 0. Proof. We state first that every point e ∈ rec (cl K) can be expressed in the described above form. This fact is obvious if e = o (choose a point z ∈ K and put xi = z and λi = 1i for all i > 1). So, we may suppose that e 6= o. Then the halfline [o, ei lies in rec (cl K). Let x ∈ rint K. Then [x, x + ei ⊂ cl K, and Theorem 5.1 shows that [x, x + ei ⊂ rint K. Therefore x + ie ∈ K for all i > 1. Finally, e = limi→∞ 1i (x + ie). Conversely, suppose e = limi→∞ λi xi , where x1 , x2 , . . . ∈ K and nonnegative scalars λ1 , λ2 , . . . satisfy the condition limi→∞ λi = 0. If the sequence x1 , x2 , . . . is bounded, then e = lim λi xi = o ∈ rec (cl K). i→∞
Suppose that the sequence x1 , x2 , . . . is unbounded and that e 6= o. Eliminating all zero terms, we may assume that λi xi 6= o for all i > 1. Let c = e/kek and ci = xi /kxi k, i > 1. Then c=
e limi→∞ λi xi λi xi = = lim = lim ci , kek limi→∞ λi kxi k i→∞ λi kxi k i→∞
and Theorem 5.4 (with v = o) implies that the halfline [u, u + ci lies in cl K for every choice of u in cl K. Therefore, e ∈ [o, ci ⊂ rec (cl K). The result below (which expands Theorem 4.37) demonstrates that recessions cones are useful in describing closures of conic hulls. Theorem 5.16. If K ⊂ Rn is a convex set and s ∈ Rn \ cl K, then cl (cones K) = cones (cl K) ∪ (s + rec (cl K)).
(5.9)
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Proof. Excluding the trivial case K = ∅ (then both sides of (5.9) are equal to {s}), we assume that K is nonempty. If K is bounded, then rec (cl K) = {o} (see Theorem 5.13), and Theorem 4.37 confirms the equality (5.9). Hence one can assume that K is unbounded. We state first that cones (cl K) ∪ (s + rec (cl K)) ⊂ cl (cones K).
(5.10)
Since cones (cl K) ⊂ cl (cones K) according to Theorem 4.37, it suffices to prove the inclusion s + rec (cl K) ⊂ cl (cones K). For this, choose a point x ∈ s + rec (cl K). Because the case x = s is obvious, we may suppose that x 6= s. Then the halfline h = [o, x − si lies in rec (cl K). Let u be a point in rint K. By the definition of rec (cl K), the halfline h0 = u + h lies in cl K, and Theorem 5.1 gives h0 ⊂ rint K. Put v = u + (x − s), and let xλ = (1 − λ)v + λx,
vλ =
λ 1 v− u, 0 < λ < 1. 1−λ 1−λ
Then v ∈ h0 , which gives vλ ∈ h0 for all 0 < λ < 1, as depicted below. u v vλ r r u+h !r ! J ! ! J !! ! Jr xλ ! Jr r!! s+h s x From xλ = v + λ(x − v) = v + λ(s − u) 1 λ = (1 − λ) v− u + λs 1−λ 1−λ = (1 − λ)vλ + λs it follows that xλ ∈ [s, vλ i ⊂ cones K. Since x = limλ→1 xλ , we conclude that x ∈ cl (cones K). Summing up, the inclusion (5.10) holds. For the opposite inclusion, it suffices to show that cl (cones K) \ cones (cl K) ⊂ s + rec (cl K).
(5.11)
Because (5.11) is obvious when cl (cones K) = cones (cl K), we will assume that cones (cl K) is a proper subset of cl (cones K) (see Theorem 4.37). Choose a point x ∈ cl (cones K) \ cones (cl K) and a sequence of points
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x1 , x2 , . . . in cones K which converges to x. Then xi ∈ [s, zi i for certain points zi ∈ K, i > 1. We state that the sequence z1 , z2 , . . . is unbounded. Indeed, assume for a moment that z1 , z2 , . . . is bounded. Then it contains a subsequence zi1 , zi2 , . . . converging to a point z ∈ cl K. Clearly, x 6= z due to the assumption x ∈ / cones (cl K). By a continuity argument, x = lim xij ∈ [s, lim zij i = [s, zi ⊂ cones (cl K), j→∞
j→∞
again contradicting the choice of x. Thus z1 , z2 , . . . must be unbounded. The inclusion xi ∈ [s, zi i shows that xi − s = λi (zi − s) for a certain scalar λi > 0. Since the sequence x1 − s, x2 − s, . . . is bounded, one has kxi − sk = 0. lim λi = lim i→∞ i→∞ kzi − sk Therefore, x − s = lim (xi − s) = lim λi (zi − s) = lim λi zi , i→∞
i→∞
i→∞
and Theorem 5.15 gives x − s ∈ rec (cl K), or x ∈ s + rec (cl K). Summing up, the inclusion (5.11) holds, and the equality (5.9) is proved. Corollary 5.17. If K ⊂ Rn is a nonempty convex set and s is a point in Rn \ cl K, then the equality cl (cones K) = cones (cl K) holds if and only if s + rec (cl K) ⊂ cones (cl K). Remark. Theorem 5.16 is not true if s ∈ rbd K. Indeed, for the compact convex set K = {(x, y) : (x2 + (y − 1)2 6 1} ⊂ R2 , which contains o, the set coneo K equals {o} ∪ {(x, y) : y > 0}.
5.2
Linearity Spaces
Definition and Basic Properties Definition 5.18. The linearity space of a nonempty set X in Rn , denoted lin X, is defined by lin X = {e ∈ Rn : λe + x ∈ X whenever λ ∈ R and x ∈ X}. We put lin ∅ = ∅. Example. If C ⊂ Rn is a convex cone, then lin C is the characteristic subspace of the plane ap C (see Corollary 1.55 and Theorem 4.14). In particular, the linearity space of an m-dimensional plane L ⊂ Rn coincides with the characteristic subspace sub L.
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F
Fig. 5.2
r o
lin F
The linearity space of a slab F .
Example. If L ⊂ Rn is a plane of positive dimension and F is a closed halfplane or a slab of L, given, respectively, as F = {x ∈ L : x·c 6 γ}
or F = {x ∈ L : γ 6 x·c 6 γ 0 },
where c ∈ / (sub L)⊥ and γ < γ 0 (see Corollaries 1.38 and 1.44), then lin F is a subspace of dimension dim L − 1, given by lin F = {x ∈ sub L : x·c = 0} (see example on page 187). A similar statement holds for all other types of slab (see Definition 1.33). Theorem 5.19. For a nonempty convex set K ⊂ Rn , the following statements hold. (1) lin K = {e ∈ Rn : ±e + K ⊂ K} = {e ∈ Rn : e + K = K}. (2) lin K is the largest among all subspaces S of Rn satisfying the inclusion S + K ⊂ K, or, equivalently, the equality S + K = K. (3) lin K = rec K ∩ (−rec K) = lin (rec K) = ap (rec K). (4) lin K ⊂ sub K, and lin K = sub K if and only if K is a plane. Proof. (1) First, we are going to prove that the set M = {e ∈ Rn : ±e + K ⊂ K} equals lin K. Indeed, if e ∈ lin K, then ±e + x ∈ K for all x ∈ K, which implies the inclusion ±e + K ⊂ K. Hence lin K ⊂ M . Conversely, let e ∈ M . By induction on r > 1, we easily conclude that ±re + K ⊂ K for all r > 1. Choose a point x ∈ K, a scalar λ > 0, and an integer r0 such that |λ| 6 r0 . Then λe ∈ [−r0 e, r0 e], and λe + x ∈ [−r0 e, r0 e] + x = [x − r0 e, r0 e + x] ⊂ K by the convexity of K. Hence e ∈ lin K according to Definition 5.5. Next, we state that M coincides with the set M 0 = {e ∈ Rn : e + K = K}. Indeed, for every e ∈ M , one has K = o + K = e + (−e + K) ⊂ e + K ⊂ K,
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implying that e + K = K. Hence e ∈ M 0 . Conversely, let e ∈ M 0 . Then e + K = K. Furthermore, −e + K = −e + (e + K) = o + K = K, which gives the inclusion e ∈ M . (2) To show that lin K is a subspace, choose vectors b, c ∈ lin K and a scalar α ∈ R. Let x be a point in K. Then γb + x ∈ K and βc + x ∈ K for all γ, β ∈ R. Since K is a convex set, we have λ(αb + c) + x = 21 (2αλb + x) + 21 (2λc + x) ∈ K,
λ ∈ R.
Hence αb + c ∈ lin K, and lin K is a subspace. If S is a subspace satisfying the inclusion S + K ⊂ K, then, for any e ∈ S, λ ∈ R, and x ∈ K, one has λe + x ∈ S + K ⊂ K, implying that e ∈ lin K. Hence S ⊂ rec K, and lin K is the largest among all subspaces satisfying the condition S +K ⊂ K. Since o+x = x for every point x ∈ K, we have lin K + K = K. (3) The equality lin K = rec K ∩ (−rec K) follows from the definitions. Furthermore, lin (rec K) = rec (rec K) ∩ (−rec (rec K)) = rec K ∩ (−rec K) = lin K. Finally, lin (rec K) = ap (rec K) because of Theorem 4.15. Statement (4) follows from (3) above and Theorem 5.6. The next theorem gives a way to reduce the study of convex sets to those with zero linearity. The method is based on a decomposition of the convex set into a direct sum of its linearity space and a convex set of smaller dimension. For example, the closed halfplane D = {(x, y) : x > 1} is the direct sum of its linearity space (which is the y-axis) and the halfline h = {(x, 0) : x > 1}, as depicted below. lin D D r
o
r
h
Theorem 5.20. If K ⊂ Rn is a nonempty convex set and S ⊂ Rn is a subspace complementary to lin K, then K can be expressed as the direct sum K = (K ∩ S) ⊕ lin K,
where
lin (K ∩ S) = {o}.
Furthermore, rec K = (rec K ∩ S) ⊕ lin K = rec (K ∩ S) ⊕ lin K.
(5.12)
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Proof. Choose a point x ∈ K and consider the plane L = x + lin K. Since L and S are complementary planes, Theorem 1.11 shows that L ∩ S is a singleton, say {z}. By Theorem 5.19, L ⊂ K. Hence z ∈ L ∩ S ⊂ K ∩ S. Because x + lin K = z + lin K (see Theorem 1.2), we have x ∈ x + lin K = z + lin K ⊂ (K ∩ S) ⊕ lin K. Conversely, let x ∈ (K ∩ S) ⊕ lin K. Then x = u + e such that u ∈ K ∩ S and e ∈ lin K. By Theorem 5.19, x = u + e ∈ u + lin K ⊂ K. Assume for a moment that lin (K ∩ S) 6= {o}. Then lin K is a proper subset of the subspace T = lin (K ∩ S) ⊕ lin K. Given a point x ∈ K ∩ S, we have x + T = x + lin (K ∩ S) ⊕ lin K = (x + lin (K ∩ S)) ⊕ lin K ⊂ (K ∩ S) ⊕ lin K = K, in contradiction with the maximality of lin K (see Theorem 5.19). Using Theorem 5.19 and (5.12), with rec K instead of K, we obtain rec K = (rec K ∩ S) ⊕ lin (rec K) = (rec K ∩ S) ⊕ lin K. Finally, for the equality rec K = rec (K ∩ S) ⊕ lin K, it suffices to show that rec K ∩ S = rec (K ∩ S). Indeed, let e ∈ rec K ∩ S. Choose a point x ∈ K ∩ S. Then λe + x ∈ K for all λ > 0. Similarly, λe + x ∈ S whenever λ > 0, because S is a subspace. Hence λe + x ∈ K ∩ S for all λ > 0, implying the inclusion e ∈ rec (K ∩ S). Conversely, let e ∈ rec (K ∩ S) and x ∈ K. Because of K = (K ∩ S) ⊕ lin K, we can write x = u + c, where u ∈ K ∩ S and c ∈ lin K. Therefore, λe + x = (λe + u) + c ∈ (K ∩ S) ⊕ lin K = K, implying that e ∈ rec K. Clearly, e ∈ S because e + u ∈ K ∩ S for every u ∈ K ∩ S. Summing up, e ∈ rec K ∩ S. The next result provides a useful generalization of Theorem 5.20. Theorem 5.21. For a nonempty convex set K ⊂ Rn , a subspace V of lin K, and a complementary to V subspace W of Rn , one has K = (K ∩ W ) ⊕ V. Furthermore, rec K = (rec K ∩ W ) ⊕ V = rec (K ∩ W ) ⊕ V .
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Proof. Let T = lin K ∩ W , and denote by S a subspace of W complementary to T . Clearly, lin K = T ⊕ V,
W = S ⊕ T,
Rn = lin K ⊕ S.
Also, K + T = K (see Theorem 5.19). Now Theorem 5.20 gives K = (K ∩ S) ⊕ lin K = (K ∩ S) ⊕ (T ⊕ V ) = (K + T ) ∩ (S + T ) ⊕ V = (K ∩ W ) ⊕ V. Replacing K with rec K in the above equality, we obtain rec K = (rec K ∩ W ) ⊕ V. For the equality rec K = rec (K ∩ W ) ⊕ V , it suffices to show that rec K ∩ W = rec (K ∩ W ). Indeed, let e ∈ rec K ∩ W and x ∈ K ∩ W . Then λe + x ∈ K for all λ > 0. Similarly, λe + x ∈ W whenever λ > 0 because W is a subspace. Hence λe + x ∈ K ∩ W for all λ > 0, implying the inclusion e ∈ rec (K ∩ W ). Conversely, let e ∈ rec (K ∩ W ). Choose a point x ∈ K. Then x = u + c, where u ∈ K ∩ W and c ∈ V . Therefore, λe + x = (λe + e) + c ∈ (K ∩ W ) ⊕ V = K, which gives e ∈ rec K. Furthermore, e ∈ W since e + u ∈ K ∩ W for every point u ∈ K ∩ W . Summing up, e ∈ rec K ∩ W . Algebra of Linearity Spaces Theorem 5.22. For a family F = {Kα } of convex sets in Rn satisfying the condition ∩ Kα 6= ∅, one has α
∩ lin Kα ⊂ lin (∩ Kα ). α
α
Furthermore, if every set Kα ∈ F is closed or relatively open, then ∩ lin Kα = lin (∩ Kα ). α
α
If, additionally, the family F is nested, then the family {lin Kα } is nested in the same order and ∪ lin Kα ⊂ lin (∪ Kα ). α
α
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Proof. By Theorems 5.7 and 5.19, ∩ lin Kα = ∩ (rec Kα ∩ (−rec Kα )) = (∩ rec Kα ) ∩ (∩ (−rec Kα )) α
α
α
α
⊂ rec (∩Kα ) ∩ (−rec (∩Kα )) = lin (∩Kα ). α
α
α
The other two statements are obtained in a similar way. Corollary 5.23. If a plane L ⊂ Rn meets a convex set K ⊂ Rn , then lin K ∩ sub L ⊂ lin (K ∩ L). Furthermore, if K is closed or relatively open, then lin K ∩ sub L = lin (K ∩ L). Proof. The proof follows from Theorem 5.22 and the fact that lin L coincides with the characteristic subspace sub L of L (see example on page 197). Remark. The inclusions in Corollary 5.23 may be proper. Indeed, let K = {(0, 1)} ∪ {(x, y) : −1 < y < 1}
and L = {(x, 0) : x ∈ R}.
Then lin K = {o} = 6 lin (K ∩ L) = L. Theorem 5.24. For a nonempty convex set K ⊂ Rn , a point a ∈ Rn , and a nonzero scalar µ, one has lin (a + µK) = lin K. Furthermore, if K = (K ∩ S) ⊕ lin K, where S is a subspace of Rn complementary to lin K, then a + µK = (c + µ(K ∩ S)) ⊕ lin K, where c is the projection of a on S along lin K. Proof. By Theorem 5.9, lin (a + µK) = rec (a + µK) ∩ (−rec (a + µK)) = µ rec K ∩ (−µ rec K) = µ lin K = lin K. Since S and lin K are complementary subspaces, a can be written as a = c + e where c is the projection of a on S along lin K and e = a − c ∈ lin K. In this case, a + µK = a + µ((K ∩ S) ⊕ lin K) = (c + µ(K ∩ S)) ⊕ (e + µ lin K) = (c + µ(K ∩ S)) ⊕ lin K.
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Theorems 5.10 and 5.19 imply the following corollary. Corollary 5.25. For nonempty convex sets K1 , . . . , Kr ⊂ Rn and scalars µ1 , . . . , µr , one has µ1 lin K1 + · · · + µr lin Kr ⊂ lin (µ1 K1 + · · · + µr Kr ). If, additionally, the planes aff K1 , . . . , aff Kr are independent, then µ1 lin K1 + · · · + µr lin Kr = lin (µ1 K1 + · · · + µr Kr ). Remark. The inclusion in Corollary 5.25 may be proper. Indeed, if K1 = {(x, y) : x, y > 0} and K2 = {(x, y) : x, y 6 0}, then lin K1 = lin K2 = {o}. On the other hand, K1 + K2 = R2 , implying that lin (K1 + K2 ) = R2 . Theorem 5.26. If g : Rn → Rm is a linear transformation and K ⊂ Rn and M ⊂ Rm are convex sets, then g(lin K) ⊂ lin g(K) and lin g −1 (M ) = g −1 (lin (M ∩ rng g)). If, additionally, g is one-to-one, then g(lin K) = lin g(K). Proof. Excluding the trivial cases K = ∅ and M ∩ rng g = ∅, we assume that both sets K and M ∩ rng g are nonempty. Consequently, g −1 (Y ) 6= ∅. By Theorems 5.11 and 5.19, g(lin K) = g(rec K ∩ (−rec K)) ⊂ g(rec K) ∩ g(−rec K) ⊂ rec g(K) ∩ (−rec g(K)) = lin g(K). The same theorems imply that lin g −1 (M ) = rec g −1 (M ) ∩ (−rec g −1 (M )) = g −1 (rec (M ∩ rng g)) ∩ (−g −1 (rec (M ∩ rng g))) = g −1 (rec (M ∩ rng g) ∩ (−rec (M ∩ rng g))) = g −1 (lin (M ∩ rng g)). If g is one-to-one, then, according to Theorem 5.11, we have g(lin K) = g(rec K ∩ (−rec K)) = g(rec K) ∩ g(−rec K) = rec g(K) ∩ (−rec g(K)) = lin g(K). Remark. The inclusion g(lin K) ⊂ lin g(K) in Theorem 5.26 may be proper even if g(rec K) = rec g(K). Indeed, let K = {(x, y) : y > |x|} and g be the orthogonal projection of R2 on the x-axis. Then rec K = K and g(K) is the x-axis, which gives g(rec K) = g(K) = rec g(K). On the other hand, lin K = {o}, while lin g(K) is the x-axis, showing that g(lin K) = {o} = 6 lin g(K).
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Linearity Spaces and Topology Theorem 5.27. If K ⊂ Rn is a convex set, then lin K ⊂ lin (cl K) = lin (rint K). Proof. By Theorem 5.13, lin K = rec K ∩ (−rec K) ⊂ rec (cl K) ∩ (−rec (cl K)) = lin (cl K). Similarly, the same theorem gives lin (cl K) = lin (rint K). Remark. The inclusion lin K ⊂ lin (cl K) in Theorem 5.27 may be proper. Indeed, if K = {o} ∪ {(x, y) : 0 < y < 1}, then lin K = {o}, while lin (cl K) is the x-axis of R2 . Corollary 5.14 implies one more result. Corollary 5.28. For a nonempty convex set K ⊂ Rn , the following statements hold. (1) If x is a point in cl K, then lin (cl K) is the largest among all subspaces S ⊂ Rn satisfying the inclusion x + S ⊂ cl K. (2) If x is a point in rint K, then lin (rint K) is the largest among all subspaces S ⊂ Rn satisfying the inclusion x + S ⊂ rint K. Theorem 5.29. If K ⊂ Rn is a nonempty convex set and s is a point in Rn \ cl K, then lin (cl (cones K)) = lin (cl K).
(5.13)
Proof. Since cl K ⊂ cl (cones K), the last statement of Theorem 5.22 shows that lin (cl K) ⊂ lin (cl K ∪ cl (cones K)) = lin (cl (cones K)). Hence it suffices to prove the opposite inclusion lin (cl (cones K)) ⊂ lin (cl K). For this, choose a vector e ∈ lin (cl (cones K)). The case e = o is obvious; whence we assume that e 6= o. Because of lin (cl (cones K)) = rec (cl (cones K)) ∩ (−rec (cl (cones K))), both vectors e and −e belong to rec (cl (cones K)). Theorem 5.6 shows that s + e and s − e belong to cl (cones K). Let y1 , y2 , . . . and z1 , z2 , . . . be sequences of points in cones K such that lim yi = s + e
i→∞
and
lim zi = s − e.
i→∞
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Put yi = s + λi (ui − s),
zi = s + µi (vi − s),
i > 1,
(5.14)
where ui , vi ∈ K and λi , µi > 0 for all i > 1. Choosing certain subsequences of y1 , y2 , . . . and z1 , z2 , . . . , we assume that both sequences λ1 , λ2 , . . . and µ1 , µ2 , . . . converge (possibly, to ∞). We state that lim λi = lim µi = 0.
i→∞
i→∞
Indeed, assume, for example, that limi→∞ λi > γ > 0. Then 0<
1 < γ −1 for all i > 1. λi + µi
Consider the sequence of convex combinations si =
λi µi ui + vi , λi + µi λi + µi
i > 1.
Then all s1 , s2 , . . . belong to K by the convexity of K. Using (5.14), we express si as si =
(yi − s) + (zi − s) yi + (λi − 1)s zi + (µi − 1)s + =s+ . λi + µi λi + µi λi + µi
Since limi→∞ ((yi − s) + (zi − s)) = e − e = o, one has lim si = s + lim
i→∞
i→∞
(yi − s) + (zi − s) = s. λi + µi
Thus s ∈ cl K, contrary to the assumption. Hence limi→∞ λi = limi→∞ µi = 0. Finally, the equalities e = lim (yi − s) = lim λi (ui − s) = lim λi ui , i→∞
i→∞
i→∞
−e = lim (zi − s) = lim µi (vi − s) = lim µi vi i→∞
i→∞
i→∞
and Theorem 5.15 show that both e and −e belong to rec (cl K). Hence e ∈ lin (cl K), which proves the inclusion lin (cl (cones K)) ⊂ lin (cl K). Remark. Theorem 5.29 does not hold if s ∈ cl K. Indeed, let K be the closed unit ball in R2 . Clearly, lin K = {o}. If s ∈ int K, then cones K = R2 and lin (cones K) = R2 . If s ∈ bd K, say s = (1, 0), then cl (cones K) = {(x, y) : x 6 1} and lin (cl (cones K)) is the y-axis.
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A combination of Theorem 2.29 and Corollaries 2.27, 2.44, and 2.46 implies the following result. Corollary 5.30. Let a nonempty convex set K ⊂ Rn be expressed as the direct sum K = (K ∩S)⊕ lin K, where S ⊂ Rn is a subspace complementary to lin K. Then rint K = rint (K ∩ S) ⊕ lin K = (rint K ∩ S) ⊕ lin K, cl K = cl (K ∩ S) ⊕ lin K = (cl K ∩ S) ⊕ lin K, rbd K = rbd (K ∩ S) ⊕ lin K = (rbd K ∩ S) ⊕ lin K. A similar argument, combined with Theorem 5.21, gives one more statement. Corollary 5.31. Let a nonempty convex set K ⊂ Rn be expressed as the direct sum K = (K ∩ W ) ⊕ V , where V is a subspace of lin K and W ⊂ Rn is a subspace complementary to V . Then rint K = rint (K ∩ W ) ⊕ V = (rint K ∩ W ) ⊕ V, cl K = cl (K ∩ W ) ⊕ V = (cl K ∩ W ) ⊕ V, rbd K = rbd (K ∩ W ) ⊕ V = (rbd K ∩ W ) ⊕ V.
5.3
Normal and Barrier Cones
Metric Projections Definition 5.32. Given a nonempty set X ⊂ Rn and a point a ∈ Rn , we will say that a point c ∈ X is a nearest to a point in X provided ka − ck = inf{ka − xk : x ∈ X}. Example. If L is a nonempty plane in Rn , then every point a ∈ Rn has a unique nearest point in L, which is the orthogonal projection of a on L (see Theorem 1.96). Theorem 5.33. Let K ⊂ Rn be a nonempty convex set and a be a point in Rn . Then cl K contains a unique nearest to a point c. Furthermore, if a ∈ Rn \ cl K, then K lies in the closed halfspace V = {x ∈ Rn : (x − c)·(a − c) 6 0}.
(5.15)
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Proof. Since the case a ∈ cl K is obvious (put c = a), we assume that a ∈ Rn \ cl K. The continuous function δa (x) = kx − ak (see Exercise 0.5) achieves its minimum value on cl K at a certain point c ∈ cl K. Put e = a − c and γ = c·e. Then the set V from (5.15) can be expressed as V = {x ∈ Rn : x·e 6 γ}, implying that V is a closed halfspace whose boundary hyperplane H = {x ∈ Rn : x·e = γ} contains c (see Definition 1.28 the picture above). r aQ B Q B Q W Q B Q r BBr u c r V
K
We state that K ⊂ V . Indeed, assume for a moment that a certain point u ∈ K belongs to the opposite open halfspace W = {x ∈ Rn : x·e > γ}. Then u · e > γ, or (u − c) · e > 0. By the convexity of cl K, all points xλ = (1 − λ)c + λu, 0 6 λ 6 1, belong to cl K. From the equalities ka − xλ k2 = ke − λ(u − c)k2 = (e − λ(u − c))·(e − λ(u − c)) = kek2 + λ(λku − ck2 − 2 e·(u − c)) it follows that ka − xλ k < kek = ka − ck
if
0 < λ < 2 e·(u − c)/ku − ck2 .
Since this argument contradicts the choice of c, no point of K lies in W . Hence K ⊂ V . For the uniqueness of c, assume for a moment the existence of another nearest to a point c1 ∈ cl K. Then c1 ∈ V . Similarly to the above argument, cl K belongs to the closed halfspace V1 = {x ∈ Rn : (x − c1 )·(a − c1 ) 6 0}. Consequently, (c1 − c)·(a − c) 6 0
and
(c − c1 )·(a − c1 ) 6 0.
Adding these two inequalities, we obtain kc − c1 k2 = (c − c1 )·(c − a + a − c1 ) = (c − c1 )·(c − a) + (c − c1 )·(a − c1 ) 6 0, which results in c = c1 .
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Theorem 5.33 gives a base to the following definition. Definition 5.34. Let K ⊂ Rn be a nonempty convex set. The mapping pK : Rn → cl K defined for every x ∈ Rn by pK (x) = z, where z is the (unique) nearest to x point of cl K, is called the metric projection on cl K. Theorem 5.35. For a nonempty convex set K ⊂ Rn , the metric projection pK satisfies the Lipschitz condition kpK (x) − pK (y)k 6 kx − yk
whenever
x, y ∈ Rn .
Consequently, pK is a continuous function on Rn . Proof. Choose points x, y ∈ Rn . Then (5.15) gives (letting a = x and then a = y) that (pK (y) − pK (x))·(x − pK (x)) 6 0, (pK (x) − pK (y))·(y − pK (y)) 6 0.
(5.16)
With u = x − pK (x) and v = y − pK (y), the inequalities (5.16) can be rewritten as u·(pK (x) − pK (y)) > 0,
v·(pK (x) − pK (y)) 6 0.
Therefore, (u − v)·(pK (x) − pK (y)) > 0, which implies kx − yk2 = k(u − v) + (pK (x) − pK (y))k2 = ku − vk2 + 2(u − v)·(pK (x) − pK (y)) + kpK (x) − pK (y)k2 > kpK (x) − pK (y)k2 . Hence kpK (x) − pK (y)k 6 kx − yk. Lemma 5.36. Let K ⊂ Rn be a nonempty convex set. For a point x ∈ Rn and a scalar λ > 0, one has pK (pK (x) + λ(x − pK (x))) = pK (x). Proof. Given an x ∈ Rn and λ > 0, let y = pK (x) + λ(x − pK (x)). Then (5.15) gives (pK (y) − pK (x))·(x − pK (x)) 6 0, (pK (x) − pK (y))·(y − pK (y)) 6 0. Consequently, 0 6 kpK (y) − pK (x)k2 = (pK (y) − pK (x))·(pK (y) − pK (x)) = (pK (y) − pK (x))·(y − pK (x) + pK (y) − y) = (pK (y) − pK (x))·(y − pK (x)) + (pK (x) − pK (y))·(y − pK (y)) = λ(pK (y) − pK (x))·(x − pK (x)) + (pK (x) − pK (y))·(y − pK (y)) 6 0. Hence kpK (y) − pK (x)k = 0, and pK (y) = pK (x).
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Theorem 5.37. For a nonempty convex set K ⊂ Rn and a point z ∈ cl K, the normal set NK (z) = {x ∈ Rn : pK (x) = z} is a closed convex cone with apex set L = z + (sub K)⊥ , where sub K is the characteristic subspace of K. Furthermore, NK (z) = L if and only if z ∈ rint K.
NK (v) NK (u)
ru
rv K
Fig. 5.3
Normal sets NK (u) and NK (v) of a convex set K.
Proof. Lemma 5.36 shows that for a point x ∈ NK (z) and a scalar λ > 0, the point z + λ(x − z) belongs to NK (z). Hence NK (z) is a cone with apex z. To show the convexity of NK (z), choose points x1 , x2 ∈ NK (z) and a scalar µ ∈ [0, 1]. From Theorem 5.33 it follows that for every point y ∈ K, one has ((1 − µ)x1 + µx2 − z)·(y − z) = (1 − µ)(x1 − z)·(y − z) + µ(x2 − z)·(y − z) 6 0, which gives the inclusion (1−µ)x1 +µx2 ∈ NK (z). Hence NK (z) is a convex cone. Furthermore, if x1 , x2 , . . . is a sequence of points in NK (z) converging to a point x0 ∈ Rn , then, by Theorem 5.35, pK (x0 ) = limi→∞ pK (xi ) = z, implying that x0 ∈ NK (z). Thus NK (z) is a closed set. We state now that the plane L = z + (sub K)⊥ is the apex set of NK (z). First, we observe that L ⊂ NK (z). Indeed, if u ∈ L, then u−z ∈ (sub K)⊥ . Because o is the nearest to u−z point in sub K (see page 6), z is the nearest to u point in z+sub K = aff K. Consequently, z is the nearest to u point in cl K, which gives u ∈ NK (z). Since L is a plane containing z, Theorem 4.15 shows that L ⊂ ap (NK (z)). Conversely, choose a point u ∈ ap NK (z). Since the case u = z is obvious, we assume that u 6= z. Let u0 = 2z − u. Then u0 ∈ ap NK (z). By Theorem 5.33, for every x ∈ K, one has (u − z)·(x − z) 6 0
and
(u0 − z)·(x − z) 6 0.
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Thus (u − z)·(x − z) = −(u0 − z)·(x − z) > 0, which gives (u − z) · (x − z) = 0. Hence K lies in the hyperplane H = (u − z)·(x − z) = 0. Therefore, sub K ⊂ H − H. From here, u − z ∈ (H − H)⊥ ⊂ (sub K)⊥ , implying the inclusion u ∈ z + (sub K)⊥ = L. Now, suppose that z ∈ rint K. Assume for a moment the existence of a point u ∈ NK (z) \ L. Then u − z ∈ / (sub K)⊥ and the orthogonal projection of u − z on sub K is a nonzero point, say v. Clearly, w = v + z is the orthogonal projection of u on z + sub K = aff K such that w 6= z. By Theorem 2.24, there is a a scalar λ ∈ (0, 1) with the property y = (1 − λ)z + λw ∈ K. Then ky − wk = (1 − λ)kz − wk, which gives ku − yk = (ku − wk2 + ky − wk2 )1/2 < (ku − wk2 + kz − wk2 )1/2 = ku − zk, contrary to the chose of z. Summing up, NK (z) = L. Conversely, let z ∈ rbd K. We state that NK (z) ∩ (aff K \ cl K) 6= ∅. For this, choose points ui ∈ aff K \ cl K such that kui − zk 6 1/i for all i > 1. Put zi = pK (ui ), and let vi = zi + (ui − zi )/kui − zi k, i > 1. Then vi ∈ aff K \ cl K, since otherwise ui ∈ (vi , zi ) ⊂ K. Furthermore, kvi − zi k = 1, and pK (vi ) = zi according to Lemma 5.36. From Theorem 5.35 we conclude that kzi − zk = kpK (ui ) − pK (z)k 6 kui − zk 6 1/i. Hence limi→∞ zi = z. The inequalities kvi k 6 kvi − zi k + kzi − zk + kzk 6 1 + 1/i + kzk, i > 1, show that the sequence v1 , v2 , . . . is bounded. Therefore, it contains a convergent subsequence. Without loss of generality, we assume that v1 , v2 , . . . itself converges to a point v. Clearly, v ∈ aff K\cl K because of δ(v, K) = 1. Finally, the continuity of pK gives pK (v) = lim pK (vi ) = lim zi = z, i→∞
i→∞
which implies the inclusion v ∈ NK (z) ∩ (aff K \ cl K). On the other hand, L ∩ aff K = (z + (sub K)⊥ ) ∩ (z + sub K) = z + (sub K)⊥ ∩ sub K = z + {o} = {z}. Hence NK (z) 6= L. Corollary 5.38. For a proper convex set K ⊂ Rn and a point u ∈ Rn , the metric projection pK (u) belongs to rbd K if and only if u ∈ Rn \ (rint K + (sub K)⊥ ), where sub K is the characteristic subspace of K.
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Normal Cones and Barrier Cones Definition 5.39. For a nonempty convex set K ⊂ Rn , the union nor K = ∪ (NK (z) − z : z ∈ cl K) is called the normal cone of K. Example. The normal cone of a nonempty plane L ⊂ Rn is the subspace (sub L)⊥ . Indeed, since rint L = L, one has NK (z) = z + (sub L)⊥ for every point z ∈ L (see Theorem 5.37). In particular, if L is a hyperplane given as L = {x ∈ Rn : x·c = γ}, then nor L = {λc : λ ∈ R}. Example. If V = {x ∈ Rn : x·c 6 γ} is a closed halfspace in Rn , then nor V = {λc : λ > 0} (compare with Definition 1.28). Theorem 5.40. For a nonempty convex set K ⊂ Rn , the following statements hold. (1) (2) (3) (4)
nor K is a cone with apex o, which contains (sub K)⊥ . nor K = (sub K)⊥ if and only if K is a plane. If K is not a plane, then nor K = ∪ (NK (z) − z : z ∈ rbd K). nor K = nor (cl K) = nor (rint K).
Proof. (1) Given a point z ∈ cl K, the normal set NK (z) is a closed convex cone with apex set z + (sub K)⊥ (see Theorem 5.37). According to Corollary 4.10, the set NK (z) − z is a convex cone with apex set (sub K)⊥ . Hence nor K, as the union of such cones, contains (sub K)⊥ . (2) From Theorem 5.37 it follows that the equality nor K = (sub K)⊥ holds if and only if rbd K = ∅. The latter condition is satisfied if and only if K is a plane (see Corollary 2.57). (3) According to Corollary 2.57, rbd K 6= ∅ provided K is not a plane. According to Theorem 5.37, every normal set NK (z), z ∈ cl K, contains z + (sub K)⊥ . So, in the definition of nor K we may consider only points z ∈ rbd K. (4) This statement follows from the fact that for every point z ∈ cl K, the normal sets NK (z), NrintK (z), and NclK (z) coincide. Remark. Unlike apices of cones NK (z), the apex set of nor K can be larger than a translate of (sub K)⊥ . For example, if K is the unit circle of the xy-plane in R3 , then (sub K)⊥ is the z-axis, while nor K = R3 .
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Furthermore, nor K may be neither closed, nor convex, even if K is closed. Indeed, if the convex set K ⊂ R3 is given by 1 1 1 1 , 1−x , 1+y , 1−y } , (5.17) K = {(x, y, z) : |x| + |y| 6 1, z > max 1+x then nor K is the union of the x- and y-axes and the open halfspace {(x, y, z) : z < 0}. The normal cone of a convex set K ⊂ R2 is always convex. Definition 5.41. The barrier cone of a nonempty convex set K ⊂ Rn , denoted bar K, is defined by bar K = {e ∈ Rn : ∃ γ = γ(e) ∈ R such that x·e 6 γ for all x ∈ K}. Example. If L is a proper plane in Rn , then bar L = nor L = (sub L)⊥ , as shown in Exercise 5.6. Theorem 5.42. For a nonempty convex set K ⊂ Rn , the following statements hold. (1) bar K is a convex cone with apex o. (2) nor K ⊂ bar K. (3) bar K = bar (cl K) = bar (rint K). Proof. (1) Let a vector e ∈ bar K and a scalar γ satisfy the inequality x·e 6 γ whenever x ∈ K. Then (µx)·e 6 µγ for all µ > 0, which shows that µe ∈ bar K. Hence bar K is a cone with apex o. For the convexity of bar K, choose vectors e1 , e2 ∈ bar K and a scalar λ ∈ [0, 1]. If γi satisfies the inequality x·ei 6 γi for all x ∈ K, i = 1, 2, then x·((1 − λ)e1 + λe2 ) = (1 − λ)x·e1 + λx·e2 6 (1 − λ)γ1 + λγ2 , implying that (1 − λ)e1 + λe2 ∈ bar K. Hence bar K is a convex set. (2) Let e ∈ nor K. Since the case e = o is obvious, we assume that e 6= o. According to Definition 5.39, e ∈ NK (z) − z for a certain point z ∈ cl K. Hence pK (e + z) = z. By Theorem 5.33, K lies in the closed halfspace V = {x ∈ Rn : x·e 6 γ}, where γ = z·e. Hence e ∈ bar K. (3) From Definition 5.41 it follows that a nonzero vector e ∈ Rn belongs to bar K if and only if a certain closed halfspace V = {x ∈ Rn : x·e 6 γ} contains K. Since V is a closed set (see Theorem 1.29), K lies in V if and only if each of the sets cl K and rint K lies in V . Consequently, e ∈ bar K if and only if each of the inclusions e ∈ bar (cl K) and e ∈ bar (rint K) holds.
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Remark. A barrier cone of a closed convex set may be nonclosed. For example, if K = {(x, y) : y > x2 }, then nor K = bar K = {o} ∪ {(x, y) : y < 0}. Also, the cones bar K and nor K may be distinct. For example, if K ⊂ R2 is a closed convex set bounded by a pair of asymptotic horizontal lines l1 and l2 , as depicted below, then bar K = {(x, y) : x 6 0} = 6 nor K = {o} ∪ {(x, y) : x < 0}. l2 K l1 Polar Cones Definition 5.43. For a nonempty set X ⊂ Rn , the set X ◦ = {u ∈ Rn : u·x 6 0 for all x ∈ X} is called the polar cone of X.
PP PP
PP PP Pr X X◦ oB B B B Fig. 5.4
The polar cone of a set X.
Example. Given a nonzero vector e ∈ Rn , the polar cone {e}◦ is the closed halfspace V = {x ∈ Rn : x·e 6 0}. Furthermore, V ◦ is the closed halfline [o, ei = {λe : λ > 0}. Example. If S ⊂ Rn is a subspace, then S ◦ = S ⊥ . Indeed, the inclusion S ⊥ ⊂ S ◦ is trivial. Conversely, if e ∈ S ◦ , then, for every vector x ∈ S, the inequalities e·x 6 0 and e·(−x) 6 0 give e·x = 0, implying the inclusion e ∈ S⊥.
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The theorem below describes some general properties of polar set. Theorem 5.44. For nonempty sets X and Y in Rn , the following statements hold. (1) (2) (3) (4) (5) (6) (7) (8) (9)
Y ◦ ⊂ X ◦ provided X ⊂ Y . (X ∪ Y )◦ = X ◦ ∩ Y ◦ . X ⊥ ⊂ X ◦ = (cl X)◦ . X ◦ is a closed convex cone with apex o. X ◦ = (cl (coneo X))◦ . (X ◦ )◦ = cl (coneo X). If X is a convex cone with apex o, then (X ◦ )◦ = cl X. X ◦ 6= Rn if and only if X 6= {o}. X ◦ 6= {o} if and only if coneo X 6= Rn .
Proof. Statements (1) and (2) and the inclusion X ⊥ ⊂ X ◦ from statement (3) immediately follow from the definitions (see page 5 for X ⊥ ). For the equality X ◦ = (cl X)◦ , we first observe that (cl X)◦ ⊂ X ◦ due to statement (1) above. Conversely, let u ∈ X ◦ . If u = o, then the inclusion u ∈ (cl X)◦ is trivial. Assume that u 6= o. From Definition 5.43 it follows that X lies in the closed halfspace V = {x ∈ Rn : u · x 6 0}. Since V is a closed set (see Theorem 1.29), one has cl X ⊂ V . Consequently, u ∈ (cl X)◦ , and the inclusion X ◦ ⊂ (cl X)◦ holds. (4) If u ∈ X ◦ and λ > 0, then (λu)· x 6 0 whenever x ∈ X. Hence λu ∈ X ◦ , implying that X ◦ is a cone with apex o. For the convexity of X ◦ , choose vectors u1 , u2 ∈ X ◦ and a scalar λ ∈ [0, 1]. Then ((1 − λ)u1 + λu2 )·x = (1 − λ)u1 ·x + λu2 ·x 6 0 ◦
for all
x ∈ X,
◦
implying that (1 − λ)u1 + λu2 ∈ X . Hence X is a convex set. If a sequence of points u1 , u2 , . . . from X ◦ converges to a point u ∈ Rn , then u·x = lim ui ·x 6 0 i→∞ ◦
for all
x ∈ X,
◦
implying the inclusion u ∈ X . Thus X is closed. (5) We observe that (cl (coneo X))◦ ⊂ X ◦ because of the inclusion X ⊂ cl (coneo X) and statement (3) above. So, it remains to prove the opposite inclusion. For this, choose a point u ∈ X ◦ . If u = o, then the inclusion u ∈ (cl (coneo X))◦ is obvious. Let u 6= o, and let v ∈ coneo X. By Theorem 4.25, v can be expressed as a nonnegative combination v = λ1 v1 + · · · + λk vk of certain points v1 , . . . , vk ∈ X. From u·v = λ1 u·v1 + · · · + λk u·vk 6 0,
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it follows that v belongs to the halfspace V = {x ∈ Rn : x·u 6 0}. Hence coneo X ⊂ V , which gives the inclusion cl (coneo X) ⊂ V . Thus u·x 6 0 for all x ∈ cl (coneo X), and Definition 5.43 shows that u ∈ (cl (coneo X))◦ . (6) Let x ∈ cl (coneo X) and u ∈ X ◦ . Then u ∈ (cl (coneo X))◦ by statement (1) above, which gives u·x 6 0. Since this argument holds for any choice of u ∈ X ◦ , we conclude that x ∈ (X ◦ )◦ . So, it remains to prove the opposite inclusion (X ◦ )◦ ⊂ cl (coneo X). Assume for a moment the existence of a point v ∈ (X ◦ )◦ \ cl (coneo X). By Theorem 5.33, there is a point z ∈ cl (coneo X) such that the closed halfspace V = {x ∈ Rn : (x − z)·(v − z) 6 0} contains cl (coneo X). We state that o belongs to the boundary hyperplane H = {x ∈ Rn : (x − z)·(v − z) = 0}
(5.18)
of V . Indeed, this fact is trivial if z = o because of z ∈ H. Let z 6= o. Then 2z ∈ cl (coneo X) since cl (coneo X) is a convex cone with apex o (see Theorem 4.21). From z ∈ (o, 2z) ⊂ V it follows that [o, 2z] ⊂ H, according to Corollary 1.32. The inclusion o ∈ H and (5.18) give z ·(v − z) = 0. Hence V can be expressed as V = {x ∈ Rn : x · (v − z) 6 0}. This argument and the inclusion cl (coneo X) ⊂ V show that v − z ∈ (cl (coneo X))◦ = X ◦ . On the other hand, v·(v − z) = (v − z + z)·(v − z) = kv − zk2 + z·(v − z) = kv − zk2 > 0 contrary to the assumption v ∈ (X ◦ )◦ . (7) This statement follows from statement (6) above. (8) Suppose that X 6= {o} and choose a nonzero point x0 ∈ X. Then X ◦ lies in the closed halfspace V = {u ∈ Rn : u·x0 6 0}, implying that X ◦ 6= Rn . If X = {o}, then X ◦ = Rn . (9) Suppose that X ◦ 6= {o} and choose a nonzero point u0 ∈ X. Then X lies in the closed halfspace V = {x ∈ Rn : x · u0 6 0}, implying that coneo X ⊂ V . Therefore, coneo X 6= Rn . If X ◦ = {o}, then statement (6) above gives that cl (coneo X) = (X ◦ )◦ = ({o})◦ = Rn . According to Corollary 2.40, coneo X = Rn .
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The theorem below gives a generalization of decomposition of Rn as the orthogonal sum of a subspace S ⊂ Rn and its orthogonal complement S ⊥ . Theorem 5.45. Let C ⊂ Rn be a convex cone with apex o, and let x ∈ Rn . Vectors y ∈ cl C and z ∈ C ◦ are the metric projections of x on cl C and C ◦ , respectively, if and only if y and z are orthogonal and satisfy the equality x = y + z. xq PP PPq z P PP q yPPP Pq cl C C◦ o B B B B
Proof. Assume first that vectors y ∈ cl C and z ∈ C ◦ are orthogonal and satisfy the equality x = y + z. Then (x − y)·y = (x − z)·z = y·z = 0. Choose a point u ∈ cl C. Since (cl C)◦ = C ◦ , one has u·(x − y) = u·z 6 0. Therefore, kx − uk2 − kx − yk2 = (x − u)·(x − u) − (x − y)·(x − y) = ku − yk2 + 2(x − y)·y − 2u·(x − y) > ku − yk2 > 0. Hence kx − uk > kx − yk, implying that y is the metric projection of x on cl C. Similarly, if v ∈ C ◦ , then (x − z)·v = y·v 6 0, which gives kx − vk2 − kx − zk2 = (x − v)·(x − v) − (x − z)·(x − z) = kv − zk2 + 2(x − z)·z − 2(x − z)·v > kv − zk2 > 0. Therefore, z is the metric projections of x on C ◦ . Conversely, let y be the metric projection of x on cl C. We observe that o is the nearest to x − y point in cl C. Indeed, let u be any point in cl C. Then u + y ∈ cl C because cl C is a convex cone with apex o (see Theorem 4.21). This argument gives k(x − y) − uk = kx − (u + y)k > kx − yk = k(x − y) − ok, where the equality holds if and only if u = o. Therefore, Theorem 5.33 implies that (x − y)·v = ((x − y) − o)·(v − o) 6 0
for all
v ∈ cl C.
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Equivalently, x − y ∈ (cl C)◦ = C ◦ . Next, we state that y and x − y are orthogonal. Since this is true when y = o or x = y, we may suppose that both vectors y and x − y are distinct from o. Because y is the metric projection of x on cl C, and since λy ∈ cl C for all λ > 0, we have kx − λyk > kx − yk, with equality if and only if λ = 1. Hence λ = 1 is the double zero of the quadratic function kx − λyk2 − kx − yk2 = λ2 y·y − 2λ x·y + 2x·y − y·y, which happens if and only if x·y = y·y, or (x − y)·y = 0. Finally, let z = x − y. Then x = y + z, z ∈ C ◦ , and z ·y = 0. By the above proved, z is the metric projection of x on C ◦ . Theorem 5.46. If C1 , . . . , Cr ⊂ Rn are convex cones with common apex o, then (C1 ∪ · · · ∪ Cr )◦ = C1◦ ∩ · · · ∩ Cr◦ , (cl C1 ∩ · · · ∩ cl Cr )◦ = cl (conv (C1◦ ∪ · · · ∪ Cr◦ )).
(5.19)
Proof. The first equality follows from Theorem 5.44. Since both sets cl C1 ∩ · · · ∩ cl Cr
and
cl (conv (C1◦ ∪ · · · ∪ Cr◦ ))
are closed convex cone with apex o (see Theorem 4.8), a combination of Theorems 5.44 and 4.9 gives (cl C1 ∩ · · · ∩ cl Cr )◦ = ((C1◦ )◦ ∩ · · · ∩ (Cr◦ )◦ )◦ = ((C1◦ ∪ · · · ∪ Cr◦ )◦ )◦ = cl (coneo (C1◦ ∪ · · · ∪ Cr◦ )) = cl (conv (C1◦ ∪ · · · ∪ Cr◦ )). Polar Decomposition Theorem 5.47. If C ⊂ Rn is a convex cone with apex o, then C ⊥ = (cl C)⊥ = ap C ◦
and
ap C ⊂ ap (cl C) = (C ◦ )⊥ .
Consequently, the following statements hold. (1) span C = (ap C ◦ )⊥ and span C ◦ = (ap (cl C))⊥ . (2) dim (ap C ◦ ) = n − dim C and dim C ◦ = n − dim (ap (cl C)). (3) dim C ◦ = n if and only if ap (cl C) = {o}, and dim C = n if and only if ap C ◦ = {o}. (4) C ◦ is a subspace if and only if C is a subspace.
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Proof. Since X ⊥ = (cl X)⊥ for any nonempty set X ⊂ Rn , it suffices to show that C ⊥ = ap C ◦ . The inclusion C ⊥ ⊂ ap C ◦ holds because C ⊥ is a subspace which lies in the convex cone C ◦ (see Theorem 5.44) and because ap C ◦ is the largest subspace lying in C ◦ (see Theorem 4.15). Conversely, let x ∈ ap C ◦ . Then −x ∈ ap C ◦ , which gives x·u 6 0 and (−x)·u 6 0 for all u ∈ C. Thus x·u = 0 for all u ∈ C, implying that x ∈ C ⊥ . Hence ap C ◦ ⊂ C ⊥ . For the second statement, we first observe that ap C ⊂ ap (cl C) according to Theorem 4.21. By the argument above, ap (cl C) = ap (C ◦ )◦ = (C ◦ )⊥ . Statement (1) follows from the above argument and the fact that (X ⊥ )⊥ = span X
and
span X ⊕ X ⊥ = Rn
for any nonempty set X ⊂ Rn . Statement (2) obviously follows from statement (1), and property (3) is an important particular case of (2). (4) A convex cone C ⊂ Rn is a subspace if and only if C = ap C, which is equivalent to the equality C ◦ = ap C ◦ ; that is, if C ◦ is a subspace. The next theorem gives a simultaneous representation of a convex cone and its polar. We illustrate this representations with the following example. lz
h−
aa C aa o aar aa C◦ a
h+ aa a
lx
Example. Let a convex cone C ⊂ R3 be given by C = {o} ∪ {(0, y, z) : y > 0}. Its polar cone is C ◦ = {(x, y, 0) : y 6 0}. If lx and lz denote the x- and z-axes, and h+ and h− stand for the nonnegative and nonpositive parts of the y-axis, then cl C = h+ ⊕ lz and C ◦ = h− ⊕ lx , as depicted above. Clearly, ap C = {o},
ap (cl C) = lz ,
ap (C ◦ ) = lx .
In terms of Theorem 5.48, span C is the yz-plane, and L is the y-axis. Then cl C ∩ L = h+ and C ◦ ∩ L = h− , confirming (5.20).
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Theorem 5.48. Let C ⊂ Rn be a convex cone with apex o and L be the orthogonal complement of ap (cl C) within span C. Then cl C = (cl C ∩ L) ⊕ ap (cl C)
and
C ◦ = (C ◦ ∩ L) ⊕ ap C ◦ .
(5.20)
Furthermore, if C 6= ap C, then C ◦ 6= ap C ◦ and a point e ∈ Rn belongs to rint C ◦ if and only if x·e < 0 for all nonzero vectors x ∈ cl C ∩ L. Proof. Since cl C is a convex cone with apex o (see Theorem 4.21), a comparison of Theorems 4.15 and 5.19 shows that ap (cl C) = lin (cl C). Denote by S the orthogonal complement of lin (cl C) in Rn . Then L ⊂ S and span (cl C) ∩ S = span C ∩ S = L. Therefore, Theorem 5.20 gives cl C = (cl C ∩ S) ⊕ lin (cl C) = (cl C ∩ span (cl C) ∩ S) ⊕ lin (cl C) = (cl C ∩ L) ⊕ ap (cl C). Due to Theorem 5.47, ap C ◦ = (cl C)⊥ = (span (cl C))⊥ = (lin (cl C) ⊕ L)⊥ = (lin (cl C))⊥ ∩ L⊥ = S ∩ L⊥ . From the obvious identity L⊥ = lin (cl C) ⊕ (S ∩ L⊥ ), we obtain that L⊥ = lin (cl C) ⊕ (S ∩ L⊥ ) = ap (cl C) ⊕ ap C ◦ . Denote by M and T the orthogonal complements of ap C ◦ within span C ◦ and Rn , respectively. Similarly to the above argument, span C ◦ ∩ T = M, C ◦ = (C ◦ ∩ M ) ⊕ ap C ◦ , M ⊥ = ap C ◦ ⊕ ap (cl C). Therefore, L = M and C ◦ = (C ◦ ∩ L) ⊕ ap C ◦ . For the second statement, the condition C 6= ap C implies that C is not a subspace; whence C ◦ also is not a subspace (see Theorem 5.47). Therefore, C ◦ 6= ap C ◦ . Assume first that a point e ∈ Rn belongs to rint C ◦ . According to Corollary 5.30, rint C ◦ = (rint C ◦ ∩ T ) ⊕ lin C ◦ = (rint C ◦ ∩ L) ⊕ lin C ◦ = (rint C ◦ ∩ L) ⊕ ap C ◦ . Hence we can write e = u + v, where u ∈ rint C ◦ ∩ L and v ∈ ap C ◦ = (span C)⊥ . Let x be a nonzero vector in cl C ∩ L. Since u + x ∈ L, Theorem 2.24 shows the existence of an ε > 0 such that u + εx = (1 − ε)u + ε(u + x) ∈ rint C ◦ ∩ L.
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Then x·(u + εx) 6 0, implying that x·e = x·(u + v) = x·u 6 −εkxk2 < 0. Conversely, suppose that a vector e ∈ Rn satisfies the condition x·e < 0 for all nonzero vectors x ∈ cl C ∩ L. Clearly, e ∈ (cl C)◦ = C ◦ . Assume for a moment that e ∈ / rint C ◦ . Then e ∈ rbd C ◦ . Choose a point u ∈ rint C ◦ . By Theorem 2.55, [u, ei \ [u, e] ⊂ span C ◦ \ C ◦ . Let ei = (− 1i )u + (1 + 1i )e,
i > 2.
Clearly, ei ∈ / C ◦ because of ei ∈ [u, ei \ [u, e]. By the definition of C ◦ , there is a unit vector xi ∈ cl C such that xi ·ei > 0. Because of (5.20), we can write xi = vi + wi , where vi ∈ cl C ∩ L and wi ∈ ap (cl C). By Theorem 5.47, wi ·ei = 0, which gives vi ·ei = (xi − wi )·ei = xi ·ei > 0, i > 2. Clearly, vi0 ·ei > 0, where vi0 = vi /kvi k for all i > 2. Replacing the sequence v20 , v30 , . . . with a converging subsequence, we assume that v20 , v30 , . . . itself tends to a unit vector v 0 ∈ cl C ∩ L. Therefore, v 0 ·e = lim vi0 ·ei > 0, i→∞
in contradiction with the choice of e. Hence e ∈ rint C ◦ . A combination of Theorems 5.47 and 5.48 implies the following corollary, which is useful for the study of support properties of convex sets. Corollary 5.49. If C ⊂ Rn is a proper convex cone with apex o, then for every nonzero vector c ∈ rint C ◦ , the (n − 1)-dimensional subspace S = {x ∈ Rn : x·c = 0} satisfies the condition S ∩ cl C = ap (cl C). Proof. A combination of Theorems 5.44 and 5.47 shows that ap (cl C) = (C ◦ )⊥ ⊂ {c}⊥ = S. Hence ap (cl C) ⊂ S ∩ cl C. For the opposite inclusion, choose a point u ∈ S ∩ cl C. If L is the orthogonal complement of ap (cl C) within span C, then Theorem 5.48 implies that u can be written as u = x + v, where x ∈ cl C ∩ L and v ∈ ap (cl C). Because of u ∈ S, one has x·e = (x + v)·c = u· c = 0. The same theorem shows that x = o. Therefore, u = v ∈ ap (cl C), and the inclusion S ∩ cl C ⊂ ap (cl C) holds.
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Polarity of Barrier, Normal, and Recession Cones Theorem 5.50. For a nonempty convex set K ⊂ Rn , one has rint (rec (cl K))◦ ⊂ nor K ⊂ bar K ⊂ (rec (cl K))◦ .
(5.21)
Consequently, rec (cl K) = (nor K)◦ = (bar K)◦ . Proof. According to Theorems 5.40 and 5.42, nor K = nor (cl K) and bar K = bar (cl K). So, we may assume that K is closed. Then (5.21) becomes as rint (rec K)◦ ⊂ nor K ⊂ bar K ⊂ (rec K)◦ . With S = (lin K)⊥ , we have K = (K ∩ S) ⊕ lin K (see Theorem 5.20). Furthermore, S = (lin K)◦ because lin K is a subspace (see example on page 213). Translating K on a certain vector, we assume that o ∈ rint K. First, we are going to prove that rint (rec K)◦ ⊂ nor K. Since this inclusion is obvious when (rec K)◦ = {o}, we may suppose that (rec K)◦ 6= {o}. Choose a nonzero vector e ∈ rint (rec K)◦ . Our goal is to find a point z ∈ K such that pK (e + z) = z. This argument will give e + z ∈ NK (z), or, equivalently, e ∈ NK (z) − z ⊂ nor K. For the latter, we state the existence of a closed halfspace Vγ = {x ∈ Rn : x·e 6 γ},
γ ∈ R,
which contains K∩S. Indeed, assume for a moment that no such a halfspace contains K ∩ S. Then, for every integer i > 1, there is a point xi ∈ K ∩ S such that xi ·e > i. From the inequalities kxi kkek > xi ·e > i it follows that the sequence x1 , x2 , . . . is unbounded. By Theorem 5.4, there is a subsequence x01 , x02 , . . . of x1 , x2 , . . . with the following property: the unit vectors ci = x0i /kx0i k tend to a unit vector c ∈ K ∩ S such that the halfline h = [o, ci lies in K ∩ S. A combination of Theorems 5.1 and 5.6 gives the inclusion c ∈ h ⊂ rec (K ∩ S) = rec K ∩ S. If M = span (rec K), then the subspace L = M ∩S is the orthogonal complement of lin K within M . Clearly, rec K ∩ S = rec K ∩ L, and Theorem 5.48 shows that c·e < 0. On the other hand, since the vectors c1 , c2 , . . . belong to Rn \V0 , the limit point c belongs to the closed halfspace {x ∈ Rn : x·e > 0}, contrary to c·e < 0. The obtained contradiction shows that K ∩ S ⊂ Vγ for a certain scalar γ.
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Next, we state that the intersection of K ∩ S with every hyperplane Hγ = {x ∈ Rn : x·e = γ},
γ ∈ R,
is compact. Indeed, this is trivial if K ∩ S is bounded. Suppose that K ∩ S is unbounded. By Theorem 5.48, c·e < 0 for every nonzero point c ∈ rec K ∩ S, implying that H0 ∩ rec (K ∩ S) = {o}. Theorem 5.54 shows that the intersection of K ∩ S with every Hγ is compact. Choose a pair of parallel hyperplanes Hµ and Hγ , µ < γ, both meeting K ∩ S, and denote by F the intersection of K ∩ S with the closed slab between Hµ and Hγ . We state that F is compact. Indeed, assume for a moment that F is unbounded. By Theorem 5.3, F contains a halfline h, and Theorem 1.34 shows that h lies in a hyperplane Hβ , µ 6 β 6 γ. Hence the set Hβ ∩ (K ∩ S) is unbounded, contrary to the above argument. Therefore, F is compact. Since the linear functional ϕ(x) = x·e is continuous (see Exercise 0.4), it attains on F a maximum value α at a certain point z ∈ F . Clearly, the halfspace Vα contains K ∩ S such that z ∈ Hα . We also observe that K ⊂ Vα . Indeed, let x be a point in K. Since K = (K ∩ S) ⊕ lin K, we can write x = y + c, where y ∈ K ∩ S and c ∈ lin K. Because of c·e = 0 (S and lin K are orthogonal subspaces), we have x·c = y ·e 6 α. Hence x ∈ Vα , and K ⊂ Vα . Since z is the nearest to z + e point in Vα (e is a normal vector of Vα ), it is the nearest to z + e point in K. Therefore, pK (e + z) = z, as desired. The inclusion nor K ⊂ bar K is proved in Theorem 5.42. So, it remains to show that bar K ⊂ (rec K)◦ . Equivalently, that c · e 6 0 whenever c ∈ bar K and e ∈ rec K. Choose a scalar γ such that x · c 6 γ for all x ∈ K. Let u ∈ K. Then λe + u ∈ K for all λ > 0. Therefore, (λe + u)·c 6 γ for all λ > 0, which is possible only if c·e 6 0. Since the set (rec (cl K))◦ is closed (see Theorem 5.44), we obtain from Theorem 5.50 and Corollary 2.39 the following statement. Corollary 5.51. For a nonempty convex set K ⊂ Rn , one has cl (nor K) = cl (bar K) = (rec (cl K))◦ , rint (nor K) = rint (bar K) = rint (rec (cl K))◦ . Line-Free Convex Sets and Bounded Sections Definition 5.52. A convex set K ⊂ Rn is called line-free if it contains no line.
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A combination of Theorems 5.19 and 5.29 implies the corollary below. Corollary 5.53. For a convex set K ⊂ Rn , the following conditions are equivalent. (1) (2) (3) (4) (5)
K is line-free. cl K is line-free. rec (cl K) is line-free. lin (cl K) = {o}. For every point s ∈ Rn \ cl K, the conic hull cones K is line-free.
Conditions (1)–(5) remain equivalent if we replace cl K with rint K. Remark. If K ⊂ Rn is a line-free convex set, then Theorems 5.13 and 5.27 show that rec K is line-free and lin K = {o}. Neither of these conditions is sufficient for K to be line-free. Indeed, the convex set K = {o} ∪ {(x, y) : 0 < y < 1} ⊂ R2 contains a line, while rec K = lin K = {o}. Theorem 5.54. Let K ⊂ Rn be a nonempty convex set and L ⊂ Rn be a plane. If a translate of L meets K along a bounded set, then rec K ∩ sub L = {o}. Furthermore, the following conditions are equivalent. (1) (2) (3) (4) (5) (6)
There is a translate of L which meets cl K along a bounded set. There is a translate of L which meets rint K along a bounded set. Every section of cl K by a translate of L is bounded. Every section of K by a translate of L is bounded. Every section of rint K by a translate of L is bounded. rec (cl K) ∩ sub L = rec (rint K) ∩ sub L = {o}.
Proof. If a translate L0 of L meets K along a bounded set, then, according to Theorem 5.13, rec (K ∩ L0 ) = {o}. Consequently, Corollary 5.8 gives {o} ⊂ rec K ∩ sub L = rec K ∩ sub L0 ⊂ rec (K ∩ L0 ) = {o}. Hence rec K ∩ sub L = {o}. For the equivalence of conditions (1)–(5), we observe that the implications (3) ⇒ (1) and (3) ⇒ (4) ⇒ (5) ⇒ (2) are obvious, and the statement (1) ⇒ (6) follows from the above argument. So, it it remains to show that (6) ⇒ (3) and (2) ⇒ (1). (6) ⇒ (3). Suppose that rec (cl K) ∩ sub L = {o}. If a translate L0 of L meets cl K, then Corollary 5.8 gives rec (cl K ∩ L0 ) = rec (cl K) ∩ rec L0 = rec (cl K) ∩ rec L = {o}.
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Consequently, Theorem 5.13 implies that cl K ∩ L0 is bounded. The case rec (rint K) ∩ sub L = {o} is similar (see also Theorem 5.13). (2) ⇒ (1). Assume for a moment that a translate L0 of L meets cl K along an unbounded set. According to Theorem 5.3, the set cl K ∩ L0 contains a halfline h with apex z, say. Choose a point y ∈ rint K. By Theorem 5.1, the halfline y −z +h lies in rint K. Since y −z +h ⊂ y −z +L, the set rint K ∩ (y − z + L) is unbounded, contrary to condition (2). Corollary 5.55. Let K ⊂ Rn be a proper convex set. If dim (lin (cl K)) = r, 0 6 r 6 n − 1, then there is a plane L ⊂ Rn of dimension n − r − 1 such that any translate of L meets K along a bounded set. In particular, if K is line-free, then there is a hyperplane H ⊂ Rn such that any translate of H meets K along a bounded set. Proof. By Corollary 5.49, for every nonzero vector c ∈ rint (rec (cl K))◦ , the (n − 1)-dimensional subspace S = {x ∈ Rn : x · c = 0} satisfies the condition S ∩ rec (cl K) = lin (cl K). If a subspace L ⊂ S complements lin (cl K), that is S = L ⊕ lin (cl K), then dim L = n − r − 1, and Theorem 5.54 shows that any translate of L meets K along a bounded set. The second statement follows from the fact that lin (cl K) = {o} provided K is line-free (see Corollary 5.53). Remark. If every translate of a hyperplane H ⊂ Rn meets a convex set K ⊂ Rn along a bounded set, then K is not necessarily line-free (see Exercise 5.7). Corollary 5.56. Let C ⊂ Rn be a line-free convex cone with apex s, which is distinct from {s}. For any vector e ∈ rint (C − s)◦ and a scalar γ < s·e, the hyperplane H = {x ∈ Rn : x·e = γ} meets C along a bounded set such that C = cones (H ∩ C) = ∪ ([s, xi : x ∈ H ∩ C). Proof. A combination of Theorems 5.12 and 5.44 gives (rec (cl C))◦ = (cl (rec C))◦ = (rec C)◦ = (C − s)◦ . Therefore, e ∈ rint (C − s)◦ = rint (rec (cl C))◦ . Corollary 5.49 shows that the (n − 1)-dimensional subspace S = {x ∈ Rn : x · e = 0} satisfies the condition S ∩ rec C = {s}. Choose a scalar γ < s·e and let µ = γ − s·e. Clearly, µ < 0. Since C is distinct from {s}, it contains a point u which is distinct from s. Then the
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cone C − s contains u − s which is distinct from o. One has (u − s)·e < 0 according to Theorem 5.48. Put µ (u − s). v =s+ (u − s)·e Then v ∈ C due to
µ (u−s)·e
> 0. Furthermore,
v·e = s·e +
µ (u − s)·e = γ. (u − s)·e
The latter equality shows that v belongs to the hyperplane H = {x ∈ Rn : x·e = γ}. Since H is a translate of S, Theorem 5.54 implies that the set H ∩ C is bounded. We observe that the inclusion ∪ ([s, xi : x ∈ H ∩ C) ⊂ C is obvious. For the opposite inclusion, choose any point u ∈ C. Repeating the above argument, we will find a point v ∈ H ∩ C such that u = s + λ(v − s) for a suitable scalar λ > 0. Hence u ∈ [s, vi ⊂ ∪ ([s, xi : x ∈ H ∩ C). Exercises for Chapter 5 Exercise 5.1. Let K ⊂ Rn be a nonempty convex set and C ⊂ Rn be a convex cone with apex o. Given a point u ∈ K, show that cl (K + C) = cl (conv (K ∪ (u + C)). Exercise 5.2. (Bair [16]) Let K ⊂ Rn be a nonempty convex set distinct from a plane. Show that rec (cl K) = ∩ (cones (cl K) − s : s ∈ rbd K). Exercise 5.3. Show that for a convex set K ⊂ Rn of positive dimension m, the following statements hold. (1) dim (lin K) = m if and only if K is a plane. (2) dim (lin K) = m − 1 if and only if K is either a halfplane or a plane slab (closed, open, or semi-open). Exercise 5.4. Show that for a convex set K ⊂ Rn , the following conditions are equivalent. (1) rbd K is disconnected. (2) rbd K lies in the union of two distinct parallel planes of the same dimension. (3) K is the union of an open slab G of aff K and two convex subsets of rbd G which belong to distinct planes bounding G.
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Exercise 5.5. Let K ⊂ Rn be a convex set distinct from a plane and z be a point in rint K. Show that if u is a nearest to z point in rbd K, then K lies in the closed halfspace V = {x ∈ Rn : (x − u)·(z − u) > 0}. Exercise 5.6. For a nonempty plane L in Rn , one has nor L = bar L = (sub L)⊥ . Exercise 5.7. Let K ⊂ Rn be a convex set and H ⊂ Rn be a hyperplane which meets cl K along a bounded set. Show that either K is line-free, or lin (cl K) is one-dimensional and cl K = (sub H ∩ cl K) ⊕ lin (cl K). Notes for Chapter 5 Recession cones. The concept of recession cone goes back to Steinitz [207, § 26]. Theorems 5.1 and 5.3, and part (2) of Theorem 5.6 can be found in Stoker [211]. The last statement of Corollary 5.53 is contributed to Dragomirescu [75]. Uniqueness of metric projection. The following statement, usually attributed to Motzkin, complements Theorem 5.33. A nonempty closed set X ⊂ Rn is convex provided it satisfies the following condition: for every point a ∈ Rn \ X, there is exactly one nearest to a point in X (see Motzkin [162] for n = 2 and Jessen [121] for all n > 2). Normal and barrier cones. The set NK (x) is often called the normal cone of K at x. For barrier cones, see Rockafellar [184, Section 14]. Theorem 5.45 is due to Moreau [160]. Plane slabs. The statements of Exercises 5.3 and 5.4 can be found, respectively, in Goberna, Jornet, Rodr´ıguez [94] and Steinitz [207, § 26]. Recession cones and closeness conditions. Rockafellar [184, Section 9] (see also Dieudonn´e [74]) obtained the following results: (1) If K ⊂ Rn is a convex set and f : Rn → Rm is a linear transformation such that rec (cl K) ∩ null f ⊂ lin (cl K), then f (cl K) = cl f (K). (2) If K1 and K2 are nonempty convex sets in Rn , satisfying the condition rec (cl K1 ) ∩ (−rec (cl K2 )) ⊂ lin (cl K1 ) ∩ lin (cl K2 ), then cl (K1 + K2 ) = cl K1 + cl K2 . Some of these results are expanded by Auslender and Teboulle [8] to the case of non-convex sets.
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Chapter 6
Support and Separation Properties
6.1
Bounds, Supports, and Asymptotes
Bounding Planes We recall that sub X stands for the characteristic subspace of a nonempty set X ⊂ Rn (see Definition 1.54). Furthermore, if H = {x ∈ Rn : x· c = γ} is a hyperplane in Rn , then sub H = {x ∈ Rn : x·c = 0} and (sub H)⊥ = {λc : λ ∈ R}. Definition 6.1. We say that a plane L ⊂ Rn properly bounds a nonempty convex set K ⊂ Rn if L ∩ rint K = ∅. Furthermore, L bounds K if either K ⊂ L or L properly bounds K.
L K
Fig. 6.1
A plane L properly bounds the convex set K.
Example. A plane L ⊂ Rn bounds a closed ball Bρ (z) ⊂ Rn if and only if ρ 6 δ(z, L) = inf{kz − xk : x ∈ L} (see also Exercise 6.1). Theorem 6.2. If K ⊂ Rn is a nonempty convex set and L is a proper plane in Rn bounding (respectively, properly bounding) K, then there is a hyperplane H ⊂ Rn containing L and bounding (respectively, properly bounding) K. 227
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Proof. Since the case K ⊂ L is obvious, we assume that L ∩ rint K = ∅. Excluding one more trivial case, namely, when L is a hyperplane, we suppose that dim L 6 n − 2. We intend to enlarge L to a plane L0 ⊂ Rn such that dim L0 = dim L + 1
and
L0 ∩ rint K = ∅.
A consecutive repetition of this enlargement procedure will give a desired hyperplane. Translating both K and L on a certain vector, we assume that L is a subspace. Denote by T the orthogonal complement of L (put T = Rn if L = {o}), and let p : Rn → T be the orthogonal projection on T . By Theorems 2.12 and 2.34, the set M = p(K) is convex and rint M = p(rint K). Furthermore, o ∈ / rint M , since otherwise L ∩ rint K 6= ∅. Consider the convex cone C = coneo M . From Theorem 4.40 it follows that C is not a plane. So, rbd C 6= ∅ according to Corollary 2.57. Choose a nonzero point x ∈ rbd C. We state that the line l = ho, xi does not meet rint C. Indeed, [o, xi ⊂ rbd C by Corollary 4.22. Suppose for a moment the existence of a point u ∈ h−x, o) ∩ rint C. Then, according to Theorem 2.36, (o, x) ⊂ (u, x) ⊂ rint C, contrary to [o, xi ⊂ rbd C. Thus l ∩ rint C = ∅. Consequently, Theorem 4.41 implies that l ∩ rint M = ∅. Finally, let L0 = L + l. Then L0 is a subspace of dimension dim L + 1 which contains L and bounds K; indeed, if L0 ∩ rint K 6= ∅, then we would have l ∩ rint M = p(L0 ) ∩ p(rint K) = p(L0 ∩ rint K) 6= ∅, contrary to the above argument. Remark. Theorem 6.2 does not hold if we replace the wording “properly bounding” with that of “disjoint.” Indeed, let K = {(x, y) : y > 0} ∪ {(x, 0) : x > 0},
L = {o}.
Then L ∩ K = ∅, while every line through L meets K. Another example: let K = {(x, y, z) : x > 0, y > 0, z > 0, xy > z 2 },
L = {(0, y, 1) : y ∈ R}.
Then K is a closed convex cone with apex o, and the line L does not meet K. On the other hand, it is easy to see that every 2-dimensional plane through L meets K.
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The next result complements Theorem 6.2. We recall that sets X and Y in Rn are strongly disjoint provided δ(X, Y ) > 0 (see page 9). Theorem 6.3. If a nonempty convex set K ⊂ Rn and a plane L ⊂ Rn are strongly disjoint, then there is a hyperplane H ⊂ Rn containing L and strongly disjoint from K. Proof. Excluding the trivial case when L a hyperplane itself, we assume that L is a plane with dim L 6 n − 2. Translating both K and L on a certain vector, we suppose that L is a subspace. Let T be the orthogonal complement of L, and f : Rn → T be the orthogonal projection. Put M = f (K). Theorem 2.12 shows that M is a convex set. Choose a scalar ρ > 0 such that Bρ (L) ∩ Bρ (K) = ∅ (see Exercise 0.9). Clearly, f (Bρ (L)) = Bρ (o) ∩ T
and f (Bρ (K)) = Bρ (M ) ∩ T,
which gives Bρ (o) ∩ Bρ (M ) = ∅. Equivalently, o ∈ / cl M . According to Theorem 5.33, cl M has a unique nearest to o point, say c, such that cl M lies in the closed halfspace V = {x ∈ Rn : (x − c)·(o − c) 6 0}. The hyperplane H = {x ∈ Rn : x·c = 0} lies in the open halfspace Rn \ V , and δ(M, H) = δ(c, H) = kck > 0 (see Theorem 1.97). Since c ∈ T , one has T ⊂ {c}⊥ = H
and δ(K, H) = δ(M, H) > 0.
Thus H is a desired hyperplane. Corollary 6.4. If K is a proper convex set in Rn , then for every point a ∈ Rn \ rint K (respectively, a ∈ Rn \ cl K), there is a hyperplane H ⊂ Rn through a properly bounding K (respectively, strongly disjoint from K). Theorem 6.5. For a hyperplane H ⊂ Rn and a nonempty convex set K ⊂ Rn , the following statements hold. (1) H bounds K if and only if K lies in a closed halfspace determined by H. (2) H properly bounds K if and only if K lies in a closed halfspace determined by H such that K 6⊂ H. (3) H and K are strongly disjoint if and only if a certain ρ-neighborhood Bρ (K) of K lies in a closed halfspace determined by H. (4) There is a translate of H bounding K (respectively, strongly disjoint from K) if and only if (sub H)⊥ ∩ bar K 6= {o}.
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Proof. (1) Since the case K ⊂ H is obvious, we may assume that H ∩ rint K = ∅. Therefore, statement (1) directly follows from statement (2). (2) Assume that H properly bounds K. Then K 6⊂ H, since otherwise rint K ⊂ K ⊂ H, contrary to the assumption. Choose a point u ∈ rint K \ H and denote by V the closed halfspace determined by H and containing u. We state that K ⊂ V . Indeed, assume for a moment that the complementary open halfspace Rn \ V contains a point v ∈ K. By Theorem 1.31, the open segment (u, v) meets H at a single point w, and Theorem 2.21 implies that w ∈ rint K, in contradiction with the assumption H ∩ rint K = ∅. Therefore, K ⊂ V . Conversely, let K lie in a closed halfspace V of Rn determined by H such that K 6⊂ H. According to Theorem 2.59, either rint K ⊂ int V or K ⊂ H. Since the latter is impossible by the assumption, we have rint K ⊂ int V . Therefore, H ∩ rint K = ∅, as desired. (3) If H and K are strongly disjoint, then Bρ (K) ∩ H = ∅ for a certain scalar ρ > 0. From statement (2) it follows that Bρ (K) lies in a closed halfspace determined by H. Conversely, if a certain neighborhood Bρ (K) of K lies in a closed halfspace determined by H, then δ(H, K) > ρ (see Theorem 1.97). Hence H and K are strongly disjoint. (4) By Theorem 1.17, H can be expressed as H = {x ∈ Rn : x·c = γ}, where c is a nonzero vector of Rn and γ a scalar. A translate of H bounds K is and only if there is a scalar µ such that one of the inequalities µ 6 inf {x·c : x ∈ K}
and µ > sup {x·c : x ∈ K}
holds. In other words, one of the normal vectors c, −c of H belongs to bar K (see Definition 5.41). Hence (sub H)⊥ ∩ bar K 6= {o}. Conversely, suppose that the subspace (sub H)⊥ ∩bar K is distinct from {o}. Choose a nonzero vector c ∈ (sub H)⊥ ∩ bar K. By the definition of bar K, there is a scalar γ ∈ R such that K lies in the closed halfspace V = {x ∈ Rn : x·c 6 γ}. Let ε > 0 and H 0 = {x ∈ Rn : x·c = γ + ε}. According to Corollary 1.18, H 0 is a translate of H. Finally, H 0 and K are strongly disjoint, because δ(H 0 , K) > ε (see Theorem 1.97). Support Planes Definition 6.6. We say that a plane L ⊂ Rn properly supports a nonempty convex set K if L meets cl K such that L ∩ rint K = ∅. Furthermore, L supports K if either K ⊂ L or L properly supports K.
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Fig. 6.2
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K
A plane L properly supports the convex set K.
Example. Every vertical line l = {(x, y, z) : x = cos α, y = sin α, z ∈ R} properly supports the closed (respectively, open) unit ball of R3 . Example. A hyperplane H = {x ∈ Rn : x · c = γ} properly supports a closed (respectively, open) unit ball of Rn if and only if kck = |γ| (see Exercise 6.1 for a more general statement). The result below shows that a support property of convex sets is a local one. Theorem 6.7. Let K ⊂ Rn be a nonempty convex set, c be a point in cl K, and Bρ (c) be a closed ball in Rn . A plane L ⊂ Rn through c supports (respectively, properly supports) K if and only if L supports (respectively, properly supports) Bρ (c) ∩ K. Proof. Since the “if” part of the statement is trivial, it remains to prove the “only if” part. So, let L support Bρ (c) ∩ K. If Bρ (c) ∩ K ⊂ L, then K ⊂ aff K = aff (Bρ (c) ∩ K) ⊂ L according to Theorems 2.41 and 1.50. Hence L supports K. Suppose now that L properly supports Bρ (c) ∩ K. Then Bρ (c) ∩ K 6⊂ L. Assume for a moment that L does not support K properly. Then L ∩ rint K 6= ∅. Choose a point u ∈ L ∩ rint K \ {c}. By Theorem 2.36, (c, u) ⊂ rint K. Choose a point v ∈ (c, u) so close to c that kc − vk < ρ. Then v ∈ L and v ∈ Uρ (c) ∩ rint K = rint (Bρ (c) ∩ K) (see Theorem 2.26). Since the latter inclusion contradicts the assumption L ∩ rint (Bρ (c) ∩ K) = ∅, it follows that L properly supports K. Theorem 6.8. A proper plane L ⊂ Rn supporting (respectively, properly supporting) a nonempty convex set K ⊂ Rn lies within a hyperplane which supports (respectively, properly supports) K.
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Proof. If L contains K, then every hyperplane H containing L also contains K and, therefore, supports K. If L properly supports K, then L properly bounds K, and Theorem 6.2 implies the existence of a hyperplane H containing L and properly bounding K. In the latter case, the hyperplane H properly supports K because of ∅ 6= L ∩ cl K ⊂ H ∩ cl K. Corollary 6.9. If K ⊂ Rn is a convex set distinct from a plane and F is a nonempty convex subset of rbd K, then the plane L = aff F properly supports K. Consequently, F lies in a hyperplane properly supporting K. Proof. According to Theorem 2.59, L ∩ K ⊂ L ∩ cl K ⊂ rbd K. Hence L ∩ rint K = ∅. Since L meets cl K, we obtain that L properly supports K. The second statement follows from Theorem 6.8. Theorem 6.5 implies the following corollary. Corollary 6.10. For a proper convex set K ⊂ Rn and a hyperplane H ⊂ Rn , the following statements hold. (1) H supports K if and only if H ∩ cl K 6= ∅ and one of the closed halfspaces determined by H contains K. (2) H properly supports K if and only if H supports K such that K 6⊂ H. Corollary 6.11. A nonempty convex set K ⊂ Rn has a properly supporting hyperplane if and only if K is not a plane. Proof. If a hyperplane H ⊂ Rn properly supports K, then H ∩rint K = ∅, which means that H meets cl K along rbd K. Hence rbd K 6= ∅, and Corollary 2.57 shows that K is not a plane. Conversely, let K be distinct from a plane. Then rbd K 6= ∅ by the same Corollary 2.57. Choose a point x ∈ rbd K. Since {x} is a 0-dimensional plane properly supporting K, Theorem 6.8 shows the existence of a hyperplane through x properly supporting K. Theorem 6.12. Let K ⊂ Rn be a proper convex set and H ⊂ Rn be a hyperplane. A certain translate of H supports K if and only if (sub H)⊥ ∩ nor K 6= {o}. Furthermore, H properly supports K if and only if (sub H)⊥ ∩ (nor K \ (sub K)⊥ ) 6= ∅.
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Proof. By Theorem 1.17, H can be expressed as H = {x ∈ Rn : x·c = γ}, where c is a nonzero vector and γ a scalar. Suppose that a certain translate H 0 of H supports K. Choose a point z ∈ H 0 ∩ cl K. Translating both K and H 0 on −z, we may put z = o. Consequently, aff K = sub K and H 0 = sub H. According Corollary 1.18, H 0 = {x ∈ Rn : x·c = 0} and c ∈ (H 0 )⊥ . The assumption that H 0 supports K means that either K ⊂ H 0 or H 0 properly supports K. If K ⊂ H 0 , then pK (c) = o because o is the nearest to c point in H 0 . The latter means that c ∈ nor K (see Definition 5.39). Hence (sub H)⊥ ∩ nor K 6= {o}. Let H 0 properly support K. Then H 0 ∩ rint K = ∅ and, according to Theorem 6.5, K lies in one one of the closed halfspaces determined by H 0 : V1 = {x ∈ Rn : x·c 6 0}
and V2 = {x ∈ Rn : x·c > 0}.
Assuming that K ⊂ V1 (the case K ⊂ V2 is similar), we conclude that pK (c) = o because o the nearest to c point in V1 . Equivalently, c ∈ nor K. Furthermore, since H 0 ∩ rint K = ∅, the subspace sub K does not lie in H 0 ; whence c ∈ / (sub K)⊥ . Summing up, c ∈ (sub H)⊥ ∩ (nor K \ (sub K)⊥ ). Conversely, suppose the existence of a nonzero vector c ∈ (sub H)⊥ ∩ nor K. By the definition of nor K, there is a point z ∈ cl K such that z is a nearest to c + z point in cl K. According to Theorem 5.33, K lies in the closed halfspace V = {x ∈ Rn : (x − z)·c 6 0}, and Theorem 6.5 shows that H bounds K. Since z ∈ H ∩ cl K, the hyperplane H supports K. Assume, additionally, that c ∈ nor K \(sub K)⊥ . We state that K 6⊂ H. Indeed, suppose for a moment that K ⊂ H. Then aff K ⊂ H, implying the inclusion sub K ⊂ sub H (see Theorem 1.13). Consequently, c ∈ (sub H)⊥ ⊂ (sub K)⊥ , contrary to the assumption. Summing up, H properly supports K. Asymptotic Planes We recall (see Definition 2.49) that a nonempty plane L ⊂ Rn is asymptotic to a nonempty set X ⊂ Rn (or an asymptote of X) provided cl X ∩ L = ∅ and δ(X, L) = 0, where δ(X, L) is the inf -distance between X and L. Corollary 6.13. Every nonempty plane bounding a nonempty convex set K ⊂ Rn either is strongly disjoint from K, or supports K, or is asymptotic to K. Remark. Unlike Theorems 6.3 and 6.8, an asymptotic plane to a convex
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set K does not necessarily lie within an asymptotic plane of higher dimension. Indeed, let K ⊂ R3 be the sum of the closed unit ball B1 (z) centered at z = (0, 0, 1) and the plane convex set N = {(x, y, 0) : x > 0, xy > 1}. The x- and y-axis of R3 are asymptotic lines to K, while every 2-dimensional plane containing any of these lines meets K. Another example: the convex set M = {(x, y, z) : x > 0, xy > 1, z > (x + y)2 } ⊂ R3 has two asymptotic planes (the xz- and yz-coordinate planes), neither containing an asymptotic line of M . Theorem 6.14. Let K ⊂ Rn be a nonempty convex set and L ⊂ Rn be a proper plane of dimension m bounding K. If M ⊂ Rn is an (m + 1)dimensional plane containing L and meeting rint K, then M contains a translate of L which either supports K or is asymptotic to K. Proof. Let K 0 = K ∩ M . By Corollary 2.27, rint K 0 = rint K ∩ M . Hence L ∩ rint K 0 = ∅. Choose a hyperplane H ⊂ Rn such that H ∩ L = M (see Theorem 1.35). Then H ∩ rint K 0 = ∅, and H bounds K 0 according to Theorem 6.5. If H is given by H = {x ∈ Rn : x·c = γ}, then we may assume that K 0 lies within the closed halfspace V = {x ∈ Rn : x·c 6 γ}. Let β = sup {x·c : x ∈ K 0 }; also, put H 0 = {x ∈ Rn : x·c = β},
V 0 = {x ∈ Rn : x·c 6 β},
L0 = H 0 ∩ M.
Clearly, K 0 ⊂ V 0 ∩ M and δ(H 0 , K 0 ) = 0 (see Theorem 1.97). Hence H 0 either supports K 0 or is asymptotic to K 0 . Therefore, L0 either supports K or is asymptotic to K. A combination of Theorems 6.14, 6.5, and 6.12 implies the corollary below. Corollary 6.15. For a proper convex set K ⊂ Rn and a hyperplane H ⊂ Rn , the following statements hold. (1) If H bounds K, then there is a translate of H which either supports K or is asymptotic to K. (2) A translate of H is asymptotic to K if and only if there is a normal vector of H which belongs to bar K \ nor K. Remark. Exercises 6.4–6.7 that the results of this section can be formulated in terms of subplanes of the affine span of a convex set K ⊂ Rn .
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Containing and Supporting Halfspaces
Containing Halfspaces A combination of Corollary 6.4 and Theorem 6.5 implies the following corollary. Corollary 6.16. Given a proper convex set K ⊂ Rn , the following statements hold. (1) For any point a ∈ Rn \ cl K, there is a closed halfplane V ⊂ Rn which contains K but not a. (2) A translate of a closed halfspace V ⊂ Rn contains K if and only if nor V ⊂ bar K. (3) If F is the family of all closed halfspaces containing K, then bar K is the union of {o} and all normal vectors to the halfspaces from F. Theorem 6.17. For a proper convex set K ⊂ Rn , the following statements hold. (1) If F is the family of all closed halfspaces of Rn containing K, then cl K = ∩ (V : V ∈ F).
(6.1)
(2) Every family F of closed halfspaces of Rn satisfying the condition (6.1) contains a countable subfamily satisfying the same condition. Proof. (1) Clearly, it suffices to prove the inclusion ∩ (V : V ∈ F) ⊂ cl K.
(6.2)
For this, choose a point v ∈ Rn \ cl K. If z is the nearest to v point in K, then Theorem 5.33 shows that cl K lies the closed halfspace V = {x ∈ Rn : (x − z)·(v − z) 6 0}.
(6.3)
Since v ∈ / V , the inclusion (6.2) holds. (2) From (6.1) it follows that Rn \ cl K = Rn \ (∩ (V : V ∈ F)) = ∪ (Rn \ V : V ∈ F). Because every set Rn \ V is open, Lindel¨of’s theorem (see page 7) implies the existence of a countable subfamily G ⊂ F with the property Rn \ cl K = ∪ (Rn \ V : V ∈ G).
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Hence cl K = Rn \ (Rn \ cl K) = Rn \ (∪ (Rn \ V : V ∈ G)) = ∩ (V : V ∈ G). Although Theorem 6.17 shows the existence of a countable subfamily G satisfying the condition cl K = ∩ (V : V ∈ G), it provides no description of G. The theorem below describes such a family. Theorem 6.18. Let K ⊂ Rn be a proper convex set, Φ be a dense subset of R, and E be a dense subset of S ∩ bar K, where S is the unit sphere of Rn . Then cl K is the intersection of countably many closed halfspaces of the form Vγ (e) = {x ∈ Rn : x·e 6 γ}, e ∈ E, γ ∈ Φ.
(6.4)
Proof. Denote by F be the family of all closed halfspaces Vγ (e) of the form (6.4), which contain K. According to Theorem 6.17, it suffices to show that F satisfies condition (6.1). Equivalently, that for every point u ∈ Rn \ cl K, there is a halfspace Vγ (e) ∈ F containing K but not u. Denote by z the nearest to u point in cl K. By Theorem 5.33, the closed halfspace V = {x ∈ Rn : (x − z)·(u − z) 6 0} contains K such that its boundary hyperplane H supports K. We can express V as u−z and µ1 = z·e1 . V = {x ∈ Rn : x·e1 6 µ1 }, where e1 = ku − zk Clearly, e1 ∈ S ∩ nor K ⊂ S ∩ bar K and u·e1 > µ1 . Let β1 = u·e1 . Our goal is to replace V1 with a certain halfspace Vγ (e) ∈ F satisfying the conditions K ⊂ Vγ (e) and u ∈ / Vγ (e). We assume first that the cone bar K is one-dimensional. Then bar K is either the halfline [o, e1 i or the line ho, e1 i. In either case, e1 ∈ E ∩ bar K, implying that Vγ (e1 ) is a desired halfspace for any choice of γ ∈ (µ1 , β1 )∩Φ. Suppose now that dim (bar K) = r > 2. Choose r − 1 unit vectors e2 , . . . , er in rint (bar K) such that the set {e1 , . . . , er } is linearly independent. Since bar K is a convex cone with apex o (see Theorem 5.42), it contains the simplicial cone C = Co (e1 , . . . , er ). Theorem 4.18 gives rint C ⊂ rint (bar K). Because rint (bar K) ⊂ nor K (see Theorem 5.50), there are points z2 , . . . , zr ∈ rbd K such that zi is the nearest to ei + zi point in cl K. By Theorem 5.33, K lies in each of the closed halfspaces Vi = {x ∈ Rn : x·ei 6 µi },
where
µi = zi ·ei ,
2 6 i 6 r.
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Choose a vector e ∈ S ∩ rint C. Then e = λ1 e1 + · · · + λr er for certain positive scalars λ1 , . . . , λr (see Theorem 4.17). Let µ = λ1 µ1 + · · · + λr µr . We state that the halfspace Vµ (e) = {x ∈ Rn : x·e 6 µ} contains the set M = V1 ∩ · · · ∩ Vr . Indeed, if x ∈ M , then x· ei 6 µi for all 1 6 i 6 r, which gives x·e = x·(λ1 e1 + · · · + λr er ) 6 λ1 µ1 + · · · + λr µr = µ, implying the inclusion K ⊂ M ⊂ Vµ (e). Choose a scalar δ > 0 (see Exercise 0.7) such that for every point e ∈ Bδ (e1 ) ∩ S ∩ rint C, the scalars λ1 , λ2 , . . . , λr in the expression e = λ1 e1 + · · · + λr er are so close to 1, 0, . . . , 0, respectively, that they satisfy the inequality λ1 (u·e1 − µ1 ) > λ2 (µ2 − u·e2 ) + · · · + λr (µr − u·er ). Equivalently, u·e = u·(λ1 e1 + · · · + λr er ) > λ1 µ1 + · · · + λr µr = µ, which shows that u ∈ / Vµ (e). Because E is dense in S ∩ rint C, we may assume that e ∈ E. Finally, let β = u·e and choose a scalar γ ∈ (µ, β) ∩ Φ. Then the halfspace Vγ (e) ∈ F contains K but not u. Corollary 6.19. Let E be a dense subset of the unit sphere S of Rn and Φ be a dense subset of R. For every line-free convex set K ⊂ Rn , its closure is the intersection of countably many closed halfspaces Vγ (e) = {x ∈ Rn : x·e 6 γ}, e ∈ E, γ ∈ Φ, each containing K. Proof. Let K ⊂ Rn be a line-free convex set. The case K = ∅ is obvious. So, we may assume that K is nonempty. Since the cone rec (cl K) is linefree (see Corollary 5.53), a combination of Theorems 5.47 and 5.50 shows that the cone bar K is n-dimensional. Then E ∩ int (bar K) is dense in S ∩ int (bar K), and the corollary follows from Theorem 6.18. Normal directions provide a suitable tool for describing recession cones and linearity spaces of closed convex sets. Theorem 6.20. Let K ⊂ Rn be a proper convex set; also, let E ⊂ Rn and Φ ⊂ R be nonempty sets. If cl K is the intersection of a family F of closed halfspaces of the form Vγ (e) = {x ∈ Rn : x·e 6 γ}, e ∈ E, γ ∈ Φ,
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then rec (cl K) and lin (cl K) are, respectively, the intersections of the families of closed halfspaces and (n − 1)-dimensional subspaces of the form V0 (e) = {x ∈ Rn : x·e 6 0}, e ∈ E, H0 (e) = {x ∈ Rn : x·e = 0}, e ∈ E. Proof. Let C = ∩ (V0 (e) : e ∈ E). First, we are going to prove the inclusion rec (cl K) ⊂ C. Since o ∈ C, the case when rec (cl K) = {o} is obvious. Hence we may suppose that rec (cl K) 6= {o}. Choose a nonzero vector c ∈ rec (cl K) and a point u ∈ K. Then the halfline [u, c + ui lies in cl K. Therefore, [u, c + ui ⊂ Vγ (e) for any halfspace Vγ (e) ∈ F. Equivalently, (λc + u) · e 6 γ for all λ > 0. As easy to see, the latter is possible only if c·e 6 0. Hence c ∈ V0 (e) for all e ∈ E, which gives c ∈ C. Summing up, rec (cl K) ⊂ C. Conversely, let c ∈ C. Choose a vector e ∈ E and a point u ∈ cl K. Then c· e 6 0 and u ∈ Vγ (e). Since (u + λc) · e 6 u · e 6 γ if λ > 0, it follows that u + λc ∈ Vγ (e). Thus u + λc ∈ ∩ (Vγ (e) : Vγ (e) ∈ F) = cl K
for all
λ > 0,
which gives c ∈ rec (cl K). The second statement of the theorem follows from the equalities lin (cl K) = rec (cl K) ∩ (−rec (cl K), H0 (e) = V0 (e) ∩ (−V0 (e)) for all e ∈ E. Support Halfspaces Definition 6.21. We say that a closed halfspace V ⊂ Rn supports a convex set K ⊂ Rn provided V contains K such that the boundary hyperplane H of V supports K. Furthermore, V properly supports K if K ⊂ V and H properly supports K. Theorem 6.12 implies the following corollary. Corollary 6.22. Given a proper convex set K ⊂ Rn , the following statements hold. (1) For any point a ∈ Rn \ cl K, there is a closed halfplane V ⊂ Rn which supports K and does not contain a. (2) A translate of a closed halfspace V of Rn supports (respectively, properly supports) K if and only if nor V ⊂ nor K (respectively, (nor V \ {o}) ⊂ nor K \ (sub K)⊥ ).
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(3) If F is the family of all closed halfspaces supporting (respectively, properly supporting) K, then nor K \ {o} (respectively, nor K \ (sub K)⊥ ), is the set of all normal vectors of the halfspaces from F. Theorem 6.23. Let K ⊂ Rn be a convex set distinct from a plane and E be a dense subset of S ∩ nor K (respectively, of (S ∩ nor K) \ (sub K)⊥ ), where S is the unit sphere of Rn . Then there is a countable family F of halfspaces Vγ (e) = {x ∈ Rn : x·e 6 γ}, e ∈ E, supporting (respectively, properly supporting) K such that cl K = ∩ (Vγ (e) : Vγ (e) ∈ F).
(6.5)
Proof. According to Theorem 6.18, cl K is the intersection of a certain countable family G of closed halfspaces of the form Vµ (e) = {x ∈ Rn : x · e 6 µ}, e ∈ E. A combination of Theorem 5.33 and Definition 5.39 shows that every halfspace Vµ (e) can be translated into a position Vγ (e) = {x ∈ Rn : x · e 6 γ} supporting K. Let F denote the family of these translates. Theorem 6.12 shows that the normal vector of a halfspace Vγ (e) ∈ F satisfies the inclusion S ∩ nor K (respectively, e ∈ (S ∩ nor K) \ (sub K)⊥ ) if Vγ (e) properly supports K. Finally, (6.5) follows from the obvious inclusions K ⊂ Vγ (e) ⊂ Vµ (e). In a similar way, Corollary 6.19 implies the following result. Corollary 6.24. Let E be a dense subset of the unit sphere S of Rn . For every line-free convex set K ⊂ Rn , there there is a countable family F of halfspaces Vγ (e) = {x ∈ Rn : x·e 6 γ}, e ∈ E, properly supporting K such that (6.5) holds. Proof. Let K ⊂ Rn be a line-free convex set. The case K = ∅ is obvious. So, we may assume that K is nonempty. Because rec (cl K) is line-free (see Corollary 5.53), a combination of Theorems 5.47 and 5.50 shows that the cone nor K is n-dimensional. Then E ∩ ((S ∩ nor K) \ (sub K)⊥ ) is dense in (S ∩ nor K) \ (sub K)⊥ . Finally, Theorem 6.23 shows the existence of a desired family F.
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Theorem 6.25. Let K ⊂ Rn be a convex set distinct from a plane and F be the family of all closed halfspaces properly supporting K. Then cl K = ∩ (V : V ∈ F), rint K = ∩ (int V : V ∈ F), rbd K = ∪ (rbd K ∩ bd V : V ∈ F).
(6.6)
Proof. Since K is not a plane, its relative boundary is nonempty (see Corollary 2.57). According to Corollary 6.9, every point x from rbd K belongs to a hyperplane H ⊂ Rn properly supporting K. By Theorem 6.5, a closed halfspace V determined by H contains K. So, V ∈ F, and the family F is nonempty. We state first that cl K = ∩ (V : V ∈ F). Obviously, it suffices to prove the inclusion ∩ (V : V ∈ F) ⊂ cl K. For this, choose points x ∈ Rn \ cl K and y ∈ rint K. First, assume that x ∈ aff K \ cl K. Then the open segment (x, y) meets rbd K at exactly one point z (see Theorem 2.55). By the above argument, there is a closed halfspace V ∈ F such that z belongs to the boundary hyperplane of V . Then x ∈ / V , since otherwise (x, y) ⊂ bd V (see Theorem 6.5). In the latter case, y ∈ bd V , contradicting the choice of V. Now assume that x ∈ Rn \ aff K. Let z ∈ rbd K. Choose a point u ∈ (x, y). Then u ∈ / aff K because (x, y) is disjoint from aff K. Since z is the only common point of aff K and the line l = hu, zi, we have l ∩ rint K = ∅. Theorem 6.8 gives the existence of a hyperplane H ⊂ Rn which contains l and properly supports K. Let V denote the closed halfspace determined by H and containing K. Then V ∈ F and x ∈ / V by the above argument. For the second equality, rint K = ∩ (int V : V ∈ F), choose x ∈ rint K and V ∈ F. Since bd V ∩ rint K = ∅, we have rint K ⊂ int V (see Theorem 2.59). Thus rint K ⊂ ∩ (int V : V ∈ F). Conversely, let x ∈ Rn \rint K. If x ∈ Rn \ cl K, then, by the above proved, x ∈ / V for a certain halfspace V ∈ F. Suppose that x ∈ rbd K. Denote by V a halfspaces from F properly supporting K at x. Then x ∈ / int V , implying the inclusion ∩ (int V : V ∈ F) ⊂ rint K. Finally, since the inclusion ∪ (rbd K ∩ bd V : V ∈ F) ⊂ rbd K is obvious, it suffices to verify the opposite inclusion. Let x ∈ rbd K. We state the existence of a halfspace V0 ∈ F such that x ∈ bd V0 . Indeed, if x belonged to the interiors of all halfspaces from F, then, by the above proved, x would belong to rint K. Hence x ∈ rbd K ∩ bd V0 for a certain V0 ∈ F.
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Remark. In Theorem 6.25, the equality rint K = ∩ (int V : V ∈ F) does not hold if F means the family of all closed halfspaces supporting K (not necessarily properly). For example, let K = {(0, y) : y > 0} be the positive halfline of the y-axis of R2 . A closed halfplane V ⊂ R2 properly supports K if and only if it can be expressed as V = {(x, y) : y > ax}, where a ∈ R. There are two more closed halfplanes, V0 = {(x, y) : x 6 0} and V1 = {(x, y) : x 6 0}, both supporting K, but not properly. With F 0 = F ∪ {V0 , V1 }, one has rint K = {(0, y) : y > 0} = 6 ∅ = ∩ (int V : V ∈ F 0 ). Similarly, one cannot state the existence of a countable (even a proper) subfamily of the family F from Theorem 6.25 which satisfies the last two equalities (6.6). Indeed, let K be the closed unit circle of R2 , and F be the family of all closed halfplanes of R2 supporting K. Let V 0 be the closed halfplane supporting K at the point e = (1, 0). If G = F \ {V 0 }, then K = ∩ (V : V ∈ G),
int K ∪ {e} = ∩ (int V : V ∈ G),
bd K \ {e} = ∪ (rbd K ∩ bd V : V ∈ G). 6.3
Separation by Hyperplanes and Slabs
Separation of Sets by Hyperplanes Definition 6.26. We say that a hyperplane H ⊂ Rn separates nonempty sets X1 and X2 provided X1 and X2 lie in the opposite closed halfspaces determined by H. The hyperplane H properly separates X1 and X2 if X1 ∪ X2 6⊂ H. Finally, H strongly separates X1 and X2 provided there is a scalar ρ > 0 such that H separates the ρ-neighborhoods Bρ (X1 ) and Bρ (X2 ) of these sets. Theorem 6.27. For nonempty sets X1 and X2 in Rn , the following statements hold. (1) X1 and X2 are separated by a hyperplane if and only if there is a nonzero vector c ∈ Rn such that sup {x·c : x ∈ X1 } 6 inf {x·c : x ∈ X2 }.
(6.7)
(2) X1 and X2 are properly separated by a hyperplane if and only if there is a nonzero vector c ∈ Rn satisfying both (6.7) and inf {x·c : x ∈ X1 } < sup {x·c : x ∈ X2 }.
(6.8)
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(3) X1 and X2 are strongly separated by a hyperplane if and only if there is a nonzero vector c ∈ Rn such that sup {x·c : x ∈ X1 } < inf {x·c : x ∈ X2 }.
(6.9)
Proof. (1) Let X1 and X2 be separated by a hyperplane H ⊂ Rn . According to Theorem 1.17, H can be expressed as H = {x ∈ Rn : x·c = γ},
(6.10)
n
where c ∈ R is a nonzero vector and γ ∈ R. Denote by V1 and V2 the closed halfspaces determined by H (see Definition 1.28). We can write V1 = {x ∈ Rn : x·c 6 γ}
and V2 = {x ∈ Rn : x·c > γ}.
(6.11)
If X1 ⊂ V1 and X2 ⊂ V2 , then sup {x·c : x ∈ X1 } 6 γ 6 inf {x·c : x ∈ X2 },
(6.12)
and if X1 ⊂ V2 and X2 ⊂ V1 , then replace c and γ with −c and −γ, respectively, to obtain (6.12). Conversely, let X1 and X2 satisfy the condition (6.7). Choose a scalar γ such that (6.12) holds. Then the closed halfspaces V1 and V2 from (6.11) contain X1 and X2 , respectively. Hence X1 and X2 are separated by H. (2) Let X1 and X2 be properly separated by a hyperplane H ⊂ Rn , which is given by (6.10). As above, we may assume that the inequalities (6.12) holds. Since X1 ∪ X2 6⊂ H, at least one of the inequalities inf {x·c : x ∈ X1 } 6 γ 6 sup {x·c : x ∈ X2 }
(6.13)
is strict, implying the condition (6.8). Conversely, if both conditions (6.7) and (6.8) hold, then X1 and X2 are separated by the hyperplane H of the form (6.10), with γ satisfying the inequalities (6.12). Clearly, (6.8) shows that X1 ∪ X2 6⊂ H. (3) Assume the existence of a scalar ρ > 0 such that the ρ-neighborhoods Bρ (X1 ) and Bρ (X2 ) are separated by a hyperplane H ⊂ Rn . Without loss of generality, we may suppose that H is given by (6.10), and V1 and V2 are given by (6.12) such that Bρ (Xi ) ⊂ Vi , i = 1, 2. Choose a point u ∈ X1 . Then u−ρc/kck ∈ Bρ (u) ⊂ V1 , implying the inequality (u−ρc/kck)·c > γ, or u·c > γ + ρkck. Similarly, v ·c 6 γ − ρkck for all v ∈ X2 . Hence the inequality (6.9) holds. Conversely, assume the existence of a nonzero vector c ∈ Rn such the inequality (6.9) holds. Let γ be a scalar satisfying sup {x·c : x ∈ X1 } < γ < inf {x·c : x ∈ X2 }.
(6.14)
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Since the linear functional ϕ(x) = x·c is continuous on Rn (see Exercise 0.4), there is a scalar ρ > 0 such that x·c < γ for all x ∈ Bρ (X1 ) and x·c > γ for all x ∈ Bρ (X2 ). This argument implies that the hyperplane (6.10) separates the sets Bρ (X1 ) and Bρ (X2 ). Corollary 6.28. For nonempty sets X1 and X2 in Rn and a nonzero vector c ∈ Rn , the following statements take place. (1) If the inequality (6.7) holds, then a hyperplane H of the form (6.10) separates X1 and X2 if and only if (6.12) is satisfied. (2) If both inequalities (6.7) and (6.8) hold, then a hyperplane H of the form (6.10) properly separates X1 and X2 if and only if (6.12) is satisfied. (3) If the inequality (6.9) holds, then a hyperplane H of the form (6.10) strongly separates X1 and X2 if and only if (6.14) is satisfied. One more corollary show that a separation of sets by a hyperplane can be reduced to the case of separation of the zero vector and a set. Corollary 6.29. Let X1 and X2 be nonempty sets in Rn , c be a nonzero vector, and α1 = sup{x·c : x ∈ X1 }. Then the following statements take place. (1) If the inequality (6.7) holds, then a hyperplane H of the form (6.10) separates X1 and X2 if and only if the hyperplane H 0 = {x ∈ Rn : x·c = γ − α1 } (6.15) separates o from X2 − X1 . (2) If both inequalities (6.7) and (6.8) hold, then a hyperplane H of the form (6.10) properly separates X1 and X2 if and only if the hyperplane (6.15) properly separates o and X2 − X1 . (3) If the inequality (6.9) holds, then a hyperplane H of the form (6.10) strongly separates X1 and X2 if and only if the hyperplane (6.15) strongly separates o and X2 − X1 . Proof. Statements (1)–(3) follows from Corollary 6.28 and the following arguments: (a) the inequality (6.13) holds if and only if 0 6 γ − α1 6 inf {x·c : x ∈ X2 } − sup {x·c : x ∈ X1 } = inf{x·c : x ∈ X2 − X1 }; (b) the inequality (6.14) holds if and only if 0 < γ − α1 < inf {x·c : x ∈ X2 } − sup {x·c : x ∈ X1 } = inf{x·c : x ∈ X2 − X1 }.
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Separation of Convex Sets by Hyperplanes Theorem 6.30. Nonempty convex sets K1 and K2 in Rn are properly separated by a hyperplane if and only if rint K1 ∩ rint K2 = ∅. Proof. First, assume that K1 and K2 are properly separated by a hyperplane H ⊂ Rn . Denote by V1 and V2 the closed halfspaces determined by H such that K1 ⊂ V1 and K2 ⊂ V2 . Since K1 ∪ K2 6⊂ H, we may suppose that K1 6⊂ H. Then H ∩ rint K1 = ∅ (see Theorem 2.59), implying the inclusion rint K1 ⊂ int V1 . Hence rint K1 ∩ rint K2 ⊂ int V1 ∩ V2 = ∅. Conversely, suppose that rint K1 ∩ rint K2 = ∅. Consider the convex set K = K2 − K1 . Since rint K = rint K2 − rint K1 (see Theorem 2.29), we obtain that o ∈ / rint K. According to Theorem 6.2, there is a hyperplane H ⊂ Rn through o such that H ∩ rint K = ∅. Theorem 6.5 shows that H properly bounds K; that is, K lies in a closed halfspace V determined by H such that rint K ⊂ int V . Equivalently, the hyperplane H properly separates o and K2 − K1 . According to Corollary 6.29, a translate of H properly separates K1 and K2 . Corollary 6.31. Nonempty convex sets K1 and K2 in Rn are separated by a hyperplane if and only if any of the following conditions holds. (1) K1 ∪ K2 lies in a hyperplane. (2) rint K1 ∩ rint K2 = ∅. Theorem 6.32. If K1 and K2 are nonempty convex sets in Rn , then the following conditions are equivalent. (1) K1 and K2 are strongly separated by a hyperplane. (2) δ(K1 , K2 ) > 0. (3) o ∈ / cl (K1 − K2 ). Proof. The equivalence of conditions (2) and (3) is obvious. So, it remains to prove that (1) ⇔ (2). Assume first that K1 and K2 are strongly separated by a hyperplane H ⊂ Rn . Equivalently, there is a scalar ρ > 0 such that the ρ-neighborhoods Bρ (K1 ) and Bρ (K2 ) of these sets lie in the distinct closed halfspaces, say V1 and V2 , determined by H. Suppose that Bρ (K1 ) ⊂ V1 and Bρ (K2 ) ⊂ V2 . Since Bρ/2 (o) ⊂ Uρ (o) = int Bρ (o), we have Bρ/2 (Ki ) = Bρ/2 + Ki ⊂ int Bρ + Ki = int (Bρ + Ki ) = int Bρ (Ki ) ⊂ int Vi , i = 1, 2.
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Therefore, Bρ/2 (K1 ) ∩ Bρ/2 (K2 ) ⊂ int V1 ∩ int V2 = ∅, and the inequality δ(K1 , K2 ) > 0 holds. Conversely, suppose δ(K1 , K2 ) > 0. Then there is a scalar ρ > 0 such that Bρ (K1 ) ∩ Bρ (K2 ) = ∅ (see Exercise 0.9). Theorem 6.30 shows that Bρ (K1 ) and Bρ (K2 ) are separated by a hyperplane; hence K1 and K2 are strongly separated. Separation of Convex Sets by Slabs We recall (see Definition 1.33) that a closed slab in Rn is a set M = {x ∈ Rn : γ 6 x·c 6 γ 0 },
(6.16)
where c ∈ Rn is a nonzero vector and γ < γ 0 . The opposite closed halfspaces determined by M are defined as V = {x ∈ Rn : x·c 6 γ}
and V 0 = {x ∈ Rn : γ 0 6 x·c}.
(6.17)
Definition 6.33. We say that a closed slab M ⊂ Rn separates nonempty subsets X1 and X2 of Rn provided X1 and X2 lie in the opposite closed halfspaces determined by M . Lemma 6.34. For nonempty sets X1 and X2 in Rn , the following statements hold. (1) X1 and X2 are strongly separated by a hyperplane H ⊂ Rn if and only if they are separated by a closed slab whose boundary hyperplanes are parallel to H. (2) X1 and X2 are separated by a closed slab M ⊂ Rn if and only if o and X2 − X1 are separated by a certain translate of M . Proof. (1) Suppose first that X1 and X2 are separated by a hyperplane H. Let H be given by (6.10), where c is a nonzero vector and γ is a suitable scalar. Without loss of generality, we may assume that the inequalities (6.14) holds. Choosing any scalar γ 0 in the open interval between γ and inf{x · c : x ∈ X2 }, we conclude that the closed slab of the form (6.16) separates X1 and X2 . Conversely, if a closed slab of the form (6.16) separates X1 and X2 , then, according to Corollary 6.28, for any scalar δ ∈ (γ, γ 0 ), the hyperplane G = {x ∈ Rn : x·c = δ} strongly separates X1 and X2 .
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(2) Suppose first that X1 and X2 are separated by a closed slab M of the form (6.16). Without loss of generality, we may assume that X1 ⊂ V and X2 ⊂ V 0 , where V and V 0 are closed halfspaces of the form (6.17). Corollary 6.29 implies that o and X2 − X1 lie, respectively, in the opposite closed halfspaces W = {x ∈ Rn : x·c 6 0}
and W 0 = {x ∈ Rn : γ 0 − γ 6 x·c}.
Since W and W 0 are translates of V and V 0 on the same vector, we conclude that the slab N = {x ∈ Rn : 0 6 x·c 6 γ 0 − γ} is a translate of M and separates o and X2 − X1 . The converse statement is proved in a similar way. In what follows, by the width of a slab we will mean the distance between its boundary hyperplanes. So, if a slab M ⊂ Rn is given by (6.16), then its width equals (γ 0 − γ)/kck (see Theorem 1.97). Theorem 6.35. Let nonempty convex sets K1 and K2 in Rn be strongly disjoint. Then there is a unique slab M ⊂ Rn of width δ(K1 , K2 ) separating K1 and K2 , and the width of any other slab separating these sets is less then δ(K1 , K2 ). Furthermore, if e is the metric projection of o on cl (K2 − K1 ), then nor M = span {e}.
K1
oq q e
M
cl (K2 − K1 ) K2 Proof. Let δ = δ(K1 , K2 ). The equality δ = inf{kx1 − x2 k : x1 ∈ K1 , x2 ∈ K2 } can be rewritten as δ = inf{ko − xk : x ∈ K2 − K1 }, which shows that δ = δ(o, K2 − K1 ). Since the set K2 − K1 is convex (see Theorem 2.9), Theorem 5.33 implies the existence of a unique point e ∈ cl (K2 − K1 ) nearest to o and such that the closed halfspace V = {x ∈ Rn : (x − e)·e > 0}
(6.18)
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contains K2 − K1 . Let β = e· e. From the above argument it follows that the closed slab N = {x ∈ Rn : 0 6 x·e 6 β}
(6.19)
separates o and K2 − K1 . Furthermore, since e is a normal vector of both hyperplanes H0 = {x ∈ Rn : x·e = 0}
and Hβ = {x ∈ Rn : x·e = β},
and since o ∈ H0 and e ∈ Hβ , the width of N equals δ(H0 , Hβ ) = kek = δ and e is a normal vector of N . Finally, Lemma 6.34 shows that a certain translate of N separates K1 and K2 . Next, choose a closed slab M ⊂ Rn separating K1 and K2 . Denote by ε the width of M . According to Lemma 6.34, a certain translate M 0 of M separates o and K2 − K1 . Clearly, the width of M 0 equals ε. Let M 0 = {x ∈ Rn : µ 6 x·b 6 µ0 }, where b ∈ Rn is a nonzero vector and µ < µ0 . Since o and K2 − K1 lie in the opposite closed halfspaces determined by M 0 , the closed segment [o, e] meets the boundary hyperplanes H = {x ∈ Rn : x·b = µ}
and H 0 = {x ∈ Rn : x·b = µ0 }
of M 0 at some points u ∈ H and u0 ∈ H 0 , as depicted below. If v denotes the orthogonal projection of u on H 0 , then ε = ku − vk 6 ku − u0 k 6 kek = δ, where the equality ε = δ holds if and only if [u, u0 ] = [o, e], which occurs if and only if M 0 coincides with the closed slab N defined by (6.19). Hence M has width δ only if M is a translate of N ; otherwise, the width of M is less than δ. qo @ u @r @ @ u0 r r @ v @q
H H0 e
Finally, assume for a moment the existence of distinct slabs M1 and M2 of width δ, both separating K1 and K2 . By the above argument, these slabs are translates of N . So, they can be expressed as M1 = {x ∈ Rn : γ1 6 x·e 6 γ10 },
M2 = {x ∈ Rn : γ2 6 x·e 6 γ20 }.
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Let γ0 = min{γ1 , γ2 } and γ00 = max{γ10 , γ20 }. Then the slab M0 = {x ∈ Rn : γ0 6 x·b 6 γ00 } separates K1 and K2 , and the width δ0 of M0 equals (γ00 − γ0 )/kek, which is greater than δ = (γ10 − γ1 )/kek = (γ20 − γ2 )/kek. By the above argument, the slab N0 = {x ∈ Rn : 0 6 x·b 6 γ00 − γ0 } separates o and K2 − K1 . Consequently, kek = ko − ek > δ0 , contrary to kek = δ. The obtained contradiction shows the uniqueness of a closed slab of width δ separating K1 and K2 . The next theorem provides a sufficient condition for a separation of convex sets by slabs. We will need the following auxiliary lemma. Lemma 6.36. Let nonempty convex sets K1 and K2 in Rn satisfy the condition cl (K1 ) ∩ cl (K2 ) = ∅. Then rec (cl K1 ) ∩ rec (cl K2 ) 6= {o} provided any of the following two conditions holds. (1) There are unbounded sequences x1 , x2 , . . . and y1 , y2 , . . . in K1 and K2 , respectively, such that lim inf {kxj − yj k : j > i} < ∞.
i→∞
(2) δ(K1 , K2 ) = 0. Proof. (1) Let unbounded sequences x1 , x2 , . . . ∈ K1 and y1 , y2 , . . . ∈ K2 satisfy condition (1) of the lemma. Then lim inf {kxj − yj k : j > i} 6 α
i→∞
for a certain scalar α > 0. Consider the unit vectors ui = xi /kxi k, i > 1. Choosing, if necessary, a certain subsequence, we assume that the sequence u1 , u2 , . . . itself converges to a unit vector u. Then u ∈ rec (cl K1 ) according to Theorem 5.15. The same theorem and the equalities y − x xi yi i i = lim + =o+u=u lim i→∞ i→∞ kxi k kxi k kxi k show that u ∈ rec (cl K2 ). (2) Suppose that δ(K1 , K2 ) = 0 and choose sequences x1 , x2 , . . . ∈ K1 and y1 , y2 , . . . ∈ K2 satisfying the condition limi→∞ kxi − yi k = 0. We observe first that both set X = {x1 , x2 , . . . } and Y = {y1 , y2 , . . .} must
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be unbounded. Indeed, assume for a moment that X is bounded. Then X contains a subsequence x01 , x02 , . . . converging to a point x ∈ cl X. If y10 , y20 , . . . is the respective subsequence from Y , then the inequalities kx − yi0 k 6 kx − x0i k + kx0i − yi0 k,
i > 1,
show that the set Y 0 = {y10 , y20 , . . . } is bounded. Choose in Y 0 a subsequence y100 , y200 , . . . converging to a point y ∈ cl Y . If x001 , x002 , . . . is the respective subsequence from x01 , x02 , . . . , then kx − yk = lim kx00i − yi00 k = lim kxi − yi k = 0, i→∞
i→∞
implying that x = y. The latter is in contradiction with the assumption cl K1 ∩ cl K2 = ∅. Hence both sequences x1 , x2 , . . . and y1 , y2 , . . . are unbounded, and rec (cl K1 ) ∩ rec (cl K2 ) 6= {o} by the above proved. Theorem 6.37. If K1 and K2 are nonempty convex sets in Rn such that cl K1 ∩ cl K2 = ∅ and neither K1 nor K2 admits an asymptotic plane, then δ(K1 , K2 ) > 0. Proof. We will prove this statement by induction on n. The case n = 1 is trivial. Assume that the statement holds for all n 6 m − 1, m > 2, and consider convex sets K1 and K2 in Rm which satisfy the hypothesis. Suppose for a moment that δ(K1 , K2 ) = 0. By Lemma 6.36, the cone rec (cl K1 ) ∩ rec (cl K2 ) is distinct from {o}; so, it contains a halfline h with apex o. We state that no line parallel to h meets both cl K1 and cl K2 . Indeed, assume for a moment the existence of a line l which is parallel to h and meets cl K1 and cl K2 at points x1 and x2 , respectively. Clearly, one of the halflines h1 = x1 + h and h2 = x2 + h contains the other. Since h1 ⊂ cl K1 and h2 ⊂ cl K2 , we obtain that cl K1 ∩ cl K2 6= ∅, contrary to the assumption. Choose a new orthonormal basis c1 , . . . , cm for Rm such that h = [o, cm i. Let p : Rm → Rm−1 = span {c1 , . . . , cm−1 } be the orthogonal projection of Rm on Rm−1 . Put K10 = p(K1 ) and K20 = p(K2 ). Clearly, null p = ho, cm i. Because neither K1 nor K2 admits an asymptotic line parallel to h, Theorem 2.52 implies that cl K10 = p(cl K1 ) and cl K20 = p(cl K2 ). Since no line parallel to h meets both cl K1 and cl K2 , one has cl K10 ∩ cl K20 = p(cl K1 ) ∩ p(cl K2 ) = ∅.
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We state that neither K10 nor K20 admits an asymptotic plane. Indeed, if L ⊂ Rm−1 were an asymptotic plane for Ki0 , then the plane L + ho, cm i would be an asymptotic plane for Ki , which is impossible. By the induction hypothesis, δ(K10 , K20 ) > 0. Obviously, δ(K1 , K2 ) > δ(K10 , K20 ) > 0. Remark. The condition that neither K1 nor K2 admits an asymptotic plane is essential in Theorem 6.37. Indeed, let K1 = {(x, y) : x > 0, xy > 1}
and K2 = {(x, y) : y 6 0}.
Then K1 and K2 are disjoint closed convex sets, while δ(K1 , K2 ) = 0. Nearest Pairs of Points Given nonempty sets X1 and X2 in Rn , we will say that points x1 ∈ X1 and x2 ∈ X2 form a nearest pair provided kx1 − x2 k = δ(X1 , X2 ). Theorem 6.38. Let nonempty convex sets K1 and K2 in Rn be such that cl K1 ∩ cl K2 = ∅
and
δ(K1 , K2 ) = kc1 − c2 k
for certain points c1 ∈ cl K1 and c2 ∈ cl K2 . If H1 and H2 are hyperplanes through c1 and c2 , respectively, both orthogonal to the segment [c1 , c2 ], then the closed slab determined by H1 and H2 separates K1 and K2 .
K1 r
c1
H1
rc2
H2
K2 Proof. Clearly, x1 6= x2 . According to Theorem 5.33, the halfspaces V1 = {x ∈ Rn : (x − c1 )·(c2 − c1 ) 6 0} V2 = {x ∈ Rn : (x − c2 )·(c2 − c1 ) > 0} contain K1 and K2 , respectively. So, the hyperplanes H1 = {x ∈ Rn : (x − c1 )·(c2 − c1 ) = 0} H2 = {x ∈ Rn : (x − c2 )·(c2 − c1 ) = 0} support K1 and K2 at c1 and c2 , respectively (see the picture above). Consequently, the closed slab between H1 and H2 separates K1 and K2 .
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If at least one of nonempty sets X1 and X2 in Rn is bounded and cl X1 ∩ cl X2 = ∅, then δ(X1 , X2 ) = kx1 − x2 k for certain points x1 ∈ cl K1 and x2 ∈ cl K2 (see Exercise 0.8). The theorem below refines this result for the case of convex sets. Theorem 6.39. If nonempty convex sets K1 and K2 in Rn satisfy the condition rec (cl K1 ) ∩ rec (cl K2 ) = {o}, then δ(K1 , K2 ) = kc1 − c2 k for certain points c1 ∈ cl K1 and c2 ∈ cl K2 . Proof. Since the case cl K1 ∩ cl K2 6= ∅ is obvious, we assume that K1 and K2 are disjoint. Let sequences u1 , u2 , . . . ∈ K1 and v1 , v2 , . . . ∈ K2 satisfy the condition limi→∞ kui − vi k = δ(K1 , K2 ). We state that at least one of these sequences is bounded. Indeed, if both u1 , u2 , . . . and v1 , v2 , . . . were unbounded, then, according to Lemma 6.36, we would have rec (cl K1 ) ∩ rec (cl K2 ) 6= {o}. Assume, for example, that u1 , u2 , . . . is bounded. Then there is a subsequence up1 , up2 , . . . of u1 , u2 , . . . which converges to a point c1 ∈ cl K1 . Choose an index r such that kui − vi k 6 δ(K1 , K2 ) + 1 for all i > r and kupi − c1 k 6 1 for all pi > r. Then kc1 − vpi k 6 kc1 − vpi k + kvpi − upi k 6 δ(K1 , K2 ) + 2 for all pi > r, implying that the sequence vp1 , vp2 , . . . is bounded. Therefore, there is a subsequence vq1 , vq2 , . . . of vp1 , vp2 , . . . which converges to a point c2 ∈ cl K2 . Therefore, kc1 − c2 k = lim kuqi − vqi k = δ(K1 , K2 ). i→∞
Exercises for Chapter 6 Exercise 6.1. Show that a hyperplane H = {x ∈ Rn : x·c = γ} bounds (respectively, strongly bounds) a ball Bρ (z) if and only if ρkck 6 |γ − z·c| (respectively, ρkck < |γ − z ·c|). Consequently, H supports Bρ (z) if and only if ρkck = |γ − z·c|. Exercise 6.2. Let u be a boundary point of a ball Bρ (z) ⊂ Rn . Show that H = {x ∈ Rn : (x − u)·(u − z) = 0} is the only hyperplane through u which supports Bρ (z). Furthermore, H ∩ Bρ (z) = {u}. Exercise 6.3. Let L1 and L2 be nonempty planes in Rn which meet each other. Show the existence of a scalar γ > 0 satisfying the following condition: if a ∈ L1 and c and e are the orthogonal projections of a on L1 ∩ L2 and L2 , respectively, then ka − ck 6 γka − ek.
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Exercise 6.4. Let K ⊂ Rn be a nonempty convex set and L ⊂ Rn be a proper plane. Show that L bounds (respectively, properly bounds) K if and if any of the following two conditions is satisfied: (a) L ∩ aff K = ∅, (b) L ∩ aff K 6= ∅ and the plane L ∩ aff K bounds (respectively, properly bounds) K. Exercise 6.5. Let K ⊂ Rn be a nonempty convex set and L ⊂ Rn be a proper plane. Show that L is strongly disjoint from K if and if any of the following two conditions is satisfied: (a) L ∩ aff K = ∅, (b) L ∩ aff K 6= ∅ and the plane L ∩ aff K is strongly disjoint from K. Exercise 6.6. Let K ⊂ Rn be a nonempty convex set and L ⊂ Rn be a proper plane. Show that L supports (respectively, properly supports) K if and if L∩aff K 6= ∅ and the plane L∩aff K supports (respectively, properly supports) K. Exercise 6.7. Let K ⊂ Rn be a nonempty convex set and L ⊂ Rn be a proper plane. Show that L is asymptotic to K if and only if L ∩ aff K 6= ∅ and the plane L ∩ aff K is asymptotic to K. Exercise 6.8. Let K ⊂ Rn be a convex set distinct from a plane and F be a family of closed halfspaces whose intersection is cl K. Show that F contains a subfamily G such that V ∩ aff K is a closed halfplane of aff K for any choice of V ∈ G, and cl K = ∩ (V ∩ aff K : V ∈ G). Exercise 6.9. Let K ⊂ Rn be a convex set distinct from a plane and X be a dense subset of rbd K. For every point x ∈ X, let Dx be a closed halfplane of aff K properly supporting K at x. Show that the family D = {Dx : x ∈ X} contains a countable subfamily whose intersection is cl K. Notes for Chapter 6 Bounding hyperplanes and halfspaces. Theorem 6.2 belongs to Rockafellar [184, Theorem 11.2]. Theorem 6.3 (when L is a singleton) can be found in Carath´eodory [51]. We will say that a family F of closed halfspaces of Rn is dense provided for every affinely independent set {x0 , x1 , . . . , xn } ⊂ Rn there is a halfspace V ∈ F such that x1 , . . . , xn ∈ V and x0 ∈ / V . Kirsch [122] proved that, given a dense family F of closed halfspaces in Rn , every line-free closed convex set K ⊂ Rn is the intersection of closed halfspaces from F which contain K. Since every closed halfspace can be expressed as the intersection of denumerably many open halfspaces. Kirsch’s result also holds for dense families of open halfspaces.
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Support hyperplanes and halfspaces. Minkowski [157, §16] introduced the notion of support hyperplane of a set in Rn and showed that every point of a bounded convex hypersurface S ⊂ Rn , containing o in its interior, belongs to a hyperplane supporting S. His proof uses a description of S as the “unit sphere” with respect to a certain real-valued function on Rn , called the Minkowski distance nowadays (see [29] for further references). An extension of Minkowski’s support theorem to the case of n-dimensional convex sets, possibly unbounded, was given by Steinitz [207, pp. 6–8]. Corollary 6.9 refines a result of Rockafellar [184, Theorem 11.6]. For a variety of results characterizing convex sets in Rn by support (or locally support) properties, see Bonnesen and Fenchel [29, p. 6–7]. Soltan and Vasiloi [194] proved that a closed convex set K ⊂ Rn of dimension n is a convex cone provided any three support hyperplanes of K meet. Asymptotic planes. Given an m-dimensional closed convex set K ⊂ Rn , m > 2, let α(K) denote the set of integers j between 1 and m − 1 such that K admits an j-dimensional asymptotic plane. When K admits no boundary halfline, then α(K) = ∅ or α(K) = {1, . . . , m − 1}; when K is a convex cone, then α(K) = ∅ or α(K) = {1, . . . , m − 2}. Klee [128] posed the question to describe the sets α(K). The answer to this question was given in Klee [132] and Goossens [97]: For every set J ⊂ {1, . . . , m − 1}, m > 2, there is an m-dimensional closed convex set K ⊂ Rn such that α(K) = J. Separation by hyperplanes and slabs. Separation properties of convex sets go up to Minkowski (see [158], p. 141), who showed that a pair of convex bodies in R3 is separated by a plane provided their interiors do not meet. Various generalizations of this result, predominantly for the case of infinite-dimensional vector spaces, are summarized by Klee in [133]. Separation of finitely many convex sets are reviewed by Deumlich, Elster, and Nehse [73]. Theorem 6.30 is attributed to Fenchel (see [84], Chapter II, Theorem 28); it became widely known due to Rockafellar [184, Theorem 11.3], who gave another proof. Klee [133] provides various references related to Theorems 6.32 and 6.37. From Dieudonn´e [74] it follows that disjoint closed convex sets K1 and K2 in Rn are strongly separated by a hyperplane provided rec K1 ∩ rec K2 = {o}. Theorem 6.35 is attributed to Dax [69]. Various separation results. Nieuwenhuis [165], defined convex sets K1 and K2 in Rn be strongly cone separated if there if there are points z1 and z2 and a simplicial n-cone C with apex o such that K1 ⊂ z1 − C,
K2 ⊂ z2 + C,
(z1 − C) ∩ (z2 + C) = ∅.
As proved in [165], nonempty closed convex sets K1 and K2 in Rn are strongly cone separated if and only if both are line-free and rec K1 ∩ rec K2 = {o}. A particular case of Zizler’s [229] result states that for every pair of disjoint compact convex sets K1 and K2 in Rn there are disjoint balls B1 and B2 such that K1 ⊂ B1 and K2 ⊂ B2 . Evenly convex set. According to Fenchel [83], a subset of Rn is called evenly
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convex if it is an intersection of open halfspaces. As follows from Theorem 6.17, every open or closed convex set in Rn is evenly convex. Various properties of evenly convex sets in Rn are studied by Goberna, Jornet, Rodr´ıguez [95], and Klee, Maluta, Zanco [135]. The intersection of all evenly convex sets containing a given set X ⊂ Rn is called the evenly convex hull of X (see Goberna and Rod´ıguez [96] for some details). Minimal intersection basis for convex sets. A family F of convex sets in a vector space E is called an intersection basis provided every convex set in E is the intersection of some elements from F. Following Hammer [109], a semispace at a point x ∈ E is a maximal convex set in E \ {x}. As proved in [109] (see also [110]), the family B of sets consisting of E and all semispaces in E is the smallest intersection basis in E. The structure of semispaces in Rn can be described as follows (see Hammer [109, 110]). For a point c ∈ Rn , let {c} = L0 ⊂ L1 ⊂ · · · ⊂ Ln−1 ⊂ Ln = Rn be a strictly increasing sequence of planes in Rn , and E1 , . . . , En be a sequence of sets such that Ei is an open halfplane of Li determined by Li−1 , 1 6 i 6 n. Then the union E1 ∪ · · · ∪ En is a semispace at x, and every semispace at c is constructed in such a way (see also Leichtweiß [148, § 1]). Convex halfspaces. A nonempty convex set F in a vector space E is called a convex halfspace provided its complement E \ F is a convex set. Hammer [109] (see also Tukey [217] for the case of normed spaces) showed that a pair of disjoint convex sets K, M in E can be separated by complementary convex halfspaces F, G: K ⊂ F, M ⊂ G, F ∩ G = ∅, F ∪ G = E. Some results about convex halfspaces in E can be found in P´ ales [167]. The structure of convex halfspaces in Rn is explicitly given by Lassak [144] (see also Mart´ınez-Legaz and Singer [153]): for every pair of complementary convex halfspaces F and G of Rn , there is a nested sequence of sets E1 ⊂ · · · ⊂ En , used in the above description of semispaces, such that one of F and G equals Er ∪ Er+1 ∪ · · · ∪ En for a certain integer r ∈ {1, . . . , n}. Bair [9–12] (see also Bair and Fourneau [18, Section III.1]) studied the structure of finitely many convex sets which partition a vector space E.
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The Extreme Structure of Convex Sets
7.1
Extreme Faces
Definition, Examples, and Basic Properties Definition 7.1. Let K ⊂ Rn be a convex set. A nonempty convex set F ⊂ K is called an extreme face of K if for any points x, y ∈ K and a scalar 0 < λ < 1, the inclusion (1 − λ)x + λy ∈ F implies that x, y ∈ F . The empty set ∅ is assumed to be an extreme face of K. An extreme face F of K is called proper provided ∅ 6= F 6= K. For example, the planar closed convex set K ⊂ R2 depicted below has the following extreme faces: ∅, every singleton from the closed arc bd K \ (a, c), the segment [a, c], and K itself.
K r a
r c
We can reformulate Definition 7.1, saying that a nonempty convex subset F of the convex set K ⊂ Rn is an extreme face of K provided a segment [x, y] ⊂ K, x 6= y, entirely lies in F whenever the respective open segment (x, y) meets F . Example. The extreme faces of a closed ball Bρ (c) ⊂ Rn are ∅, all singletons from its boundary sphere Sρ (c), and the ball itself (see Exercise 7.1). Example. The extreme faces of an r-simplex ∆ = ∆(x1 , . . . , xr+1 ) ⊂ Rn are ∅ and all simplices ∆(z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr+1 } (see Exercise 7.2). 255
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Example. The extreme faces of a simplicial cone Cs (x1 , . . . , xr ) ⊂ Rn are ∅, {s}, and all simplicial cones Cs (z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr } (see Exercise 7.3). Example. The extreme faces of a plane L ⊂ Rn are ∅ and L itself. If dim L = m > 0 and D is a closed halfplane of L, then ∅, rbd D, and D itself are the extreme faces of D. If G is a closed slab of L, then the extreme faces of G are ∅, two (m − 1)-dimensional planes determining G, and G itself (see Exercise 7.4). Theorem 7.2. If K ⊂ Rn is a convex set and F is a convex subset of K, then F is an extreme face of K if and only if the following two conditions are satisfied: K ∩ aff F = F and the set K \ F is convex. Proof. Since both cases F = ∅ and F = K are obvious, we may assume that F is a proper subset of K. Let F be an extreme face of K. To show that K ∩ aff F = F , it suffices to prove the inclusion K ∩aff F ⊂ F (clearly, because the opposite inclusion is obvious). Choose a point x ∈ K ∩ aff F , and let y ∈ rint F (the latter is possible because rint F 6= ∅ according to Corollary 2.18). By Theorem 2.24, there is a scalar 0 < λ < 1 such that (1 − λ)x + λy ∈ F . Since F is an extreme face of K, one has x, y ∈ F . Therefore, K ∩ aff F ⊂ F . Next, Definition 7.1 implies that F ∩ [x, z] = ∅ whenever x, z ∈ K \ F . Because [x, z] ⊂ K by the convexity of K, we conclude that [x, z] ⊂ K \ F . Hence the set K \ F is convex. Conversely, suppose a convex subset F of K satisfies the condition K ∩ aff F = F such that the set K \ F is convex. Choose any points x, y ∈ K and a scalar 0 < λ < 1 with z = (1 − λ)x + λy ∈ F . At least one of the points x, y should lie in F (otherwise [x, y] ⊂ K \ F by the convexity of K \ F ). Let, for example, x ∈ F . Then y = (1 − λ−1 )x + λ−1 z ⊂ aff F (see Theorem 1.46), implying the inclusion y ∈ K ∩ aff F = F . Summing up, F is an extreme face of K. Theorem 7.3. For a convex set K ⊂ Rn , the following statements hold. (1) The intersection of any family of extreme faces of K is an extreme face of K. (2) If F is an extreme face of K and G is an extreme face of F , then G is an extreme face of K. (3) If F is an extreme faces of K and M is a convex subset of K such that F ⊂ M , then F is an extreme face of M .
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Proof. Since all three statements are obvious for the case of empty sets, we may assume that all sets involved are nonempty. (1) Let F = {Fα } be a family of extreme faces of K; put F = ∩ Fα . α Since the case F = ∅ is obvious, we may assume that F is nonempty. Chose points x, y ∈ K and a scalar 0 < λ < 1 such that (1 − λ)x + λy ∈ F . Then (1 − λ)x + λy ∈ Fα for all Fα ∈ F. Since every set Fα ∈ F is an extreme face of K, we have x, y ∈ Fα . Hence both x and y belong to F = ∩ Fα . α Therefore, F is an extreme face of K. (2) Let F be an extreme face of K, and G be an extreme face of F . Choose x, y ∈ K and a scalar 0 < λ < 1 such that z = (1 − λ)x + λy ∈ G. Then z ∈ F , and hence both x and y belong to F . Since G is an extreme face of F , we obtain that x, y ∈ G. Thus G is an extreme face of K. (3) Let x, y ∈ M and 0 < λ < 1 such that z = (1 − λ)x + λy ∈ F . Since x, y ∈ K and F is an extreme face of K, both x and y belong to F . Hence F is an extreme face of M . Theorem 7.4. For a convex set K ⊂ Rn , the following statements hold. (1) If F is an extreme face of K, then F = K ∩ cl F ; in particular, F is closed provided K is closed. (2) If F is an extreme face of K and M is a convex subset of K such that F ∩ rint M 6= ∅, then M ⊂ F . (3) If F is an extreme face of K and M is a convex subset of F , then the set (K \ F ) ∪ M is convex. (4) If an extreme face F of K meets rint K, then F = K. (5) Every proper extreme face F of K lies in K ∩ rbd K and dim F 6 dim K − 1. (6) Distinct extreme faces of K have disjoint relative interiors. (7) If F is an extreme face of K and X is a subset of K, then conv (F ∩ X) = F ∩ conv X. Proof. All seven statements are obvious for the case of empty sets; so, we may assume that all sets involved are nonempty. (1) Since the inclusion F ⊂ K ∩ cl F is obvious, it remains to prove that K ∩ cl F ⊂ F . Indeed, choose a point x ∈ K ∩ cl F , and let y ∈ rint F (the latter is possible because rint F 6= ∅ according to Corollary 2.18). Because the plane aff F is closed, we have cl F ⊂ aff F . Now, Theorem 2.24 implies the existence of a scalar 0 < λ < 1 such that (1 − λ)x + λy ∈ rint F ⊂ F . Hence x ∈ F because F is an extreme face of K.
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(2) Let x ∈ M . Choose a point z ∈ F ∩ rint M . By Theorem 2.24, there is a scalar 0 < λ < 1 such that (1 − λ)x + λy ∈ M . Definition 7.1 implies that x ∈ F . Hence M ⊂ F . (3) Choose points x and y in (K \F )∪M . If both x and y belong to one of the sets K \ F or M , then [x, y] lies in this set by a convexity argument (K \F is convex according to Theorem 7.2). Let, for example, x ∈ K\F and y ∈ M . We observe that (x, y) ⊂ K \ F . Indeed, assuming the existence of a point z ∈ F ∩ (x, y), we would obtain that y ∈ hx, zi ⊂ aff F , contrary to K ∩ aff F = ∅ by the same theorem. Hence [x, y] = [x, y) ∪ {y} ⊂ (K \ F ) ∪ M, implying the convexity of (K \ F ) ∪ M . Statements (4) and (5) follows from statement (2) above. Furthermore, if a proper extreme face F of K lies in K ∩ rbd K, then Theorem 2.59 implies that dim F 6 dim K − 1. (6) If F and G are extreme faces of K such that rint F ∩ rint G 6= ∅, then statement (4) above gives F = G. (7) Since the set F ∩conv X is convex, the inclusion F ∩X ⊂ F ∩conv X implies that conv (F ∩ X) ⊂ F ∩ conv X. For the opposite inclusion, choose a point x ∈ F ∩ conv X. By Theorem 3.3, x can be written as a convex combination x = λ1 x1 + · · · + λk xk of certain points x1 , . . . , xk from X. Excluding all zero multiples in this expression, we assume that the scalars λ1 , . . . , λk are positive. According to Corollary 3.21, x belongs to the relative interior of the convex set M = conv {x1 , . . . , xk } ⊂ K. The inclusion x ∈ F ∩ rint M and statement (3) above imply that M ⊂ F . Consequently, x1 , . . . , xk ∈ F , and x1 , . . . , xk ∈ F ∩ X. Finally, Theorem 3.3 shows that x ∈ conv (F ∩ X), confirming the inclusion F ∩ conv X ⊂ conv (F ∩ X). Remark. If F is an extreme face of a convex set K ⊂ Rn , then cl F is not necessarily an extreme face of cl K. For example, if K = {o} ∪ {(x, y) : y > 0}, then {o} is an extreme face of K, while the smallest extreme face of cl K = {(x, y) : y > 0} containing o is the x-axis of R2 . The next theorem shows that we can reduce the study of extreme faces of arbitrary convex sets to those with trivial linearity. Theorem 7.5. Let a nonempty convex set K ⊂ Rn be expressed as K = (K ∩ S) ⊕ lin K, where S ⊂ Rn is a subspace complementary to lin K. A nonempty set F ⊂ Rn is an extreme face of K if and only if F = E ⊕ lin K, where E is an extreme face of K ∩ S.
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Proof. Let F be an extreme face of K. Clearly, F ⊂ p(F ) ⊕ lin K, where p is the projection on S along lin K. We state that F = p(F )⊕ lin K. Indeed, assume for a moment the existence of a point x ∈ (p(F ) ⊕ lin K) \ F . Then p(x) ∈ p(F ); so, there is a point y ∈ F \ {x} such that p(x) = p(y). This argument shows that both x and y belong to the plane L = y + lin K ⊂ K. Choose a point z ∈ L such that y ∈ (x, z). Then x, z ∈ F because F is an extreme face of K, a contradiction with the choice of x. Next, we state that p(F ) is an extreme face of K ∩ S. Indeed, let x, y ∈ K ∩ S and 0 < λ < 1 be such that z = (1 − λ)x + λy ∈ p(F ). Then z ∈ F because p(F ) ⊂ F . Since F is an extreme face of K, both x and y belong to F , and hence x, y ∈ p(F ). Summing up, p(F ) is an extreme face of K ∩ S. Conversely, let E be an extreme face of K ∩ S. Put F = E ⊕ lin K. Choose points x, y ∈ K and a scalar 0 < λ < 1 such that z = (1−λ)x+λy ∈ F . Then p(z) = (1 − λ)p(x) + λp(y) ∈ E, where p is the projection on S along lin K. Since E is an extreme face of K ∩ S, we have p(x), p(y) ∈ E. Therefore, x, y ∈ F , implying that F is an extreme face of K.
The next result shows that extreme faces of a convex set can be defined locally. Theorem 7.6. Let K ⊂ Rn be a nonempty convex set and Bρ (o) ⊂ Rn be a closed ball of radius ρ > 0 centered at o. A nonempty subset F of K is an extreme face of K if and only if F is an extreme face of Bρ (F ) ∩ K, where Bρ (F ) = Bρ (o) + F is the ρ-neighborhood of F . Proof. If F is an extreme face of K, then, according to Theorem 7.3, F is an extreme face of the convex subset Bρ (F ) ∩ K of K. Conversely, let F be an extreme face of the set M = Bρ (F ) ∩ K. Choose points x, y ∈ K and a scalar 0 < λ < 1 such that the point z = (1 − λ)x + λy belongs to F . Assume for a moment that at least one of x, y does not belong to F . Let, for example, x ∈ / F . Then x ∈ / cl F according to Theorem 7.4. So, we can 0 choose a point x ∈ [x, z) \ cl F such that x0 ∈ Bρ (F ). Clearly, z ∈ (x0 , y). Let y 0 = y if y ∈ cl F ; otherwise choose a point y 0 ∈ [y, z) \ F such that y 0 ∈ Bρ (F ). As above, z ∈ (x0 , y 0 ). Furthermore, [x0 , y 0 ] ⊂ M by the convexity of M . Since F is an extreme face of M , one should have x0 ∈ F , contrary to the choice of x0 . The obtained contradiction shows that F is an extreme face of K.
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Generated Extreme Faces Theorem 7.3 gives a base to the definition below. Definition 7.7. Let K ⊂ Rn be a convex set and X be a subset of K. The intersection of all extreme faces of K containing X is called the extreme face of K generated by X and denoted FK (X). The corollary below follows from the definitions. Corollary 7.8. Given a convex set K ⊂ Rn and subsets X and Y of K, the following statements hold. (1) X ⊂ FK (X), with X = FK (X) if and only if X is an extreme face of K. In particular, FK (∅) = ∅ and FK (K) = K. (2) FK (X) is the smallest extreme face of K which contains X. (3) FK (FK (X)) = FK (X). (4) FK (conv X) = FK (X). (5) FK (X) ⊂ FK (Y ) if X ⊂ Y . The next result deals with extreme faces generated by points. Theorem 7.9. For a nonempty convex set K ⊂ Rn and a point x ∈ K, the following statements hold. (1) FK (x) is the union of x and all points y ∈ K \ {x} such that x ∈ (y, z) for a certain z ∈ K. (2) FK (x) is the largest among all convex subsets C of K satisfying the condition x ∈ rint C. (3) FK (x) ⊂ K ∩ rbd K if and only if x ∈ rbd K. (4) For a point y ∈ K, one has FK (x) = FK (y) if and only if y ∈ rint FK (x). (5) If F is an extreme face of K, then F = FK (x) if and only if x ∈ rint F . (6) FK (x) is the unique extreme face of K containing x in its relative interior. Proof. (1) Denote by G the set consisting of x and all points y ∈ K \ {x} such that x ∈ (y, z) for a certain z ∈ K. To prove the equality G = FK (x), it suffices to show that G is the smallest extreme face of K containing x. First, we will prove that G is a convex set. Indeed, let v1 , v2 ∈ G. Since the case v1 = v2 = x is obvious, we may assume, for example, that v1 6= x. Then x ∈ (v1 , w1 ) for a certain point w1 ∈ K. Clearly, [v1 , w1 ] ⊂ G. If
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v2 ∈ [v1 , w1 ], then [v1 , v2 ] ⊂ [v1 , w1 ] ⊂ G. Hence it remains to consider the case when v2 ∈ / [v1 , w1 ]. As above, x ∈ (v2 , w2 ) for a certain point w2 ∈ K. Given a point v ∈ [v1 , v2 ], there is a point w ∈ [w1 , w2 ] such that x ∈ (v, w) (see the picture below). Since both v and w lie in K, we have [v1 , v2 ] ⊂ G, implying the convexity of G. w1 r PP PP r Pr v r w x PPP PP P r r w2 v1 v rP2 P
Next, we are going to show that G is an extreme face of K. Indeed, let y, z be points in K such that the open segment (y, z) meets G at a point u. If u = x, then the inclusions y, z ∈ G follow from the definition of G. Let u 6= x. Then x ∈ (u, w) for a certain point w ∈ G. According to Lemma 1.27, there are points y 0 ∈ (z, w) and z 0 ∈ (y, w) such that x ∈ (y, y 0 ) ∩ (z, z 0 ). Since y 0 , z 0 ∈ K by the convexity of K, we have y, z ∈ G according to the definition of G. Summing up, G is an extreme face of K. Finally, Definition 7.1 shows that G lies in every extreme face of K which contains x. Summing up, G = FK (x). (2) The inclusion x ∈ rint FK (x) follows from statement (1) above and Corollary 2.25. Let C be a convex subset of K such that x ∈ rint C. If y ∈ C \ {x}, then, by Corollary 2.25, there is a point z ∈ C such that x ∈ (y, z). Hence y ∈ FK (x), and C ⊂ FK (x). (3) If FK (x) ⊂ K ∩ rbd K, then the inclusion x ∈ rbd K is obvious. Conversely, let x ∈ rbd K. Assume for a moment the existence of a point y ∈ FK (x) ∩ rint K. By statement (2) above, there is a point z ∈ K such that x ∈ (y, z), and Theorem 2.21 implies the inclusion x ∈ rint K, in contradiction with the choice of x. Hence FK (x) ⊂ K ∩ rbd K. (4) If FK (x) = FK (y), then y ∈ rint FK (x) = rint FK (y) by statement (2) above. Conversely, let y ∈ rint FK (x). Then FK (x) and FK (y) share a common relatively interior point, y. By Theorem 7.4, FK (x) = FK (y). (5) If F = FK (x), then x ∈ rint FK (x) = rint F . Conversely, let x ∈ rint F . Then FK (x) ⊂ F as the smallest extreme face of K that contains x. On the other hand, F ⊂ FK (x), since FK (x) is the largest convex subset of K that contains x in its relative interior.
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(6) The last statement follows from Theorem 7.4 and statement (2) above. Corollary 7.10. A nonempty convex set K ⊂ Rn has only trivial extreme faces if and only if K is relatively open. The following theorem shows that the study of generated extreme faces FK (X) can be reduced to those generated by points. Theorem 7.11. If X is a nonempty subset of a convex set K ⊂ Rn , then FK (X) = FK (u) for every point u ∈ rint (conv X). Proof. Let G = ∪ (FK (x) : x ∈ conv X) and u ∈ rint (conv X). We are going to show that G = FK (u). Since FK (u) ⊂ G, it suffices to prove the opposite inclusion. Indeed, let y be a point in G. If y ∈ conv X, then u ∈ (y, z) for a certain point z ∈ conv X (see Theorem 2.21), which gives y ∈ FK (u). Assume that y ∈ G \ conv X. Then y ∈ FK (z) for a point z ∈ conv X. Equivalently, there is a point p ∈ K such that z ∈ (p, y). By Theorem 2.21 again, there is a point q ∈ conv X with the property u ∈ (q, z). According to Lemma 1.27, there is a point v ∈ (p, q) such that u ∈ (v, y). Therefore, y ∈ FK (u). Finally, the inclusion X ⊂ G = FK (u) and Theorem 7.9 imply that FK (u) is the smallest extreme face of K which contains X. Hence FK (X) = FK (u). Corollary 7.12. A convex set K ⊂ Rn and its “proper” relative boundary K ∩ rbd K are disjoint unions of relative interiors of extreme faces of K. Furthermore, each relatively open convex subset of K lies in the relative interior of a certain extreme face of K. Proof. From Theorem 7.9 it follows that every point of K belongs to the relative interior of an extreme face of K. As shown in Theorem 7.4, the relative interiors of distinct extreme faces of K are disjoint. Consequently, K is a disjoint union of relative interiors of its extreme faces. Since any extreme face F of K, whose relative interior contains a point from K ∩ rbd K entirely lies in K ∩ rbd K (see Theorem 7.9), the above argument shows that K ∩ rbd K is a disjoint union of relative interiors of extreme faces of K. If Q is a relatively open convex subset of K and u is a point in Q, then, according to Theorem 7.11, Q ⊂ FK (u). Since u ∈ Q ∩ rint FK (u), Theorem 2.26 shows that Q = rint Q ⊂ rint FK (u).
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Extreme Faces and Algebra of Convex Sets Theorem 7.13. Let F = {Kα } be a family of convex sets in Rn , and let K = ∩ Kα . The following statements hold. α
(1) If Fα is an extreme face of Kα , Kα ∈ F, then the set F = ∩ Fα α is an extreme face of K. (2) If F is finite, say F = {K1 , . . . , Kr }, then every extreme face G of K can be expressed as G = FK 1 (G) ∩ . . . FK r (G), where FK i (G) is the extreme face of Ki generated by G, 1 6 i 6 r. Proof. (1) Since the case F = ∅ is trivial, we may assume that F is nonempty. Choose points x, y ∈ K such that the open segment (x, y) meets F . Then x, y ∈ Kα and (x, y) ∩ Fα 6= ∅ for every Kα ∈ F. Since Fα is an extreme face of Kα , one has x, y ∈ Fα . Hence x, y ∈ ∩ Fα = F , α and F is an extreme face of K. (2) Because the inclusion G ⊂ FK 1 (G) ∩ · · · ∩ FK r (G) is obvious, it remains to prove the opposite one. Choose a point z ∈ rint G. By Theorem 7.11, G = FK (z)
and FK i (G) = FK i (z)
for all
1 6 i 6 r.
So, it suffices to show that FK 1 (z)∩· · ·∩FK r (z) ⊂ FK (z). Indeed, let x be a point in FK 1 (z)∩· · ·∩FK r (z). Since the case x = z is trivial, we assume that x 6= z. Choose points yi ∈ FK i such that z ∈ (x, yi ), 1 6 i 6 r. Because the segment (x, y) = (x, y1 ) ∩ · · · ∩ (x, yr ) lies in K1 ∩ · · · ∩ Kr = K and contains z, we obtain x ∈ FK (z). Remark. The second statement of Theorem 7.13 does not hold if the family F is infinite. Indeed, consider the closed intervals Ki = [0, 1 + 1i ], i > 1. Then K = K1 ∩ K2 ∩ · · · = [0, 1], implying that FK (1) = {1}. On the other hand, FK i (1) = [0, 1 + 1i ], which gives FK 1 (1) ∩ FK 2 (1) ∩ · · · = [0, 1] 6= FK (1). A combination of Theorem 7.13 and Corollary 2.33 derives the corollary below. Corollary 7.14. Let K ⊂ Rn be a convex set and L ⊂ Rn be a plane which meets K. If F is an extreme face of K, then F ∩ L is an extreme face of K ∩ L. Conversely, if G is an extreme face of K ∩ L, then G = FK (G) ∩ L.
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Theorem 7.15. For convex sets K1 , . . . , Kr in Rn and nonzero scalars µ1 , . . . , µr , let K = µ1 K1 + · · · + µr Kr . Then the following statements hold. (1) Any extreme face F of K can be expressed as F = µ1 F1 +· · ·+µr Fr , where every Fi is an extreme face of Ki , 1 6 i 6 r. (2) If, additionally, the extreme face F of K is nonempty and bounded, then the faces F1 , . . . , Fr above are nonempty and bounded, and are uniquely determined by F . Proof. (1) Without loss of generality, we may assume that F 6= ∅. An induction argument shows that the proof can be reduced to the case r = 2 and µ1 µ2 6= 0 (for r = 1, we can write µ1 K1 = µ1 K1 + µ2 o). We state that the sets F1 = {x1 ∈ K1 : ∃ x2 ∈ K2 such that µ1 x1 + µ2 x2 ∈ F }, F2 = {x2 ∈ K2 : ∃ x1 ∈ K1 such that µ1 x1 + µ2 x2 ∈ F } are desired extreme faces. Clearly, both F1 and F2 are nonempty because of F 6= ∅. By a symmetry argument, it suffices to prove that F1 is an extreme face of K1 . To show the convexity of F1 , choose points x1 , x2 ∈ F1 and a scalar λ ∈ [0, 1]. Then there are points y1 , y2 ∈ K2 such that both z1 = µ1 x1 + µ2 y1 and z2 = µ1 x2 + µ2 y2 belong to F . By the convexity of the sets K1 , K2 , and F , we have (1 − λ)x1 + λx2 ∈ K1 ,
(1 − λ)y1 + λy2 ∈ K2 ,
(1 − λ)z1 + λz2 ∈ F.
Since µ1 ((1 − λ)x1 + λx2 ) + µ2 ((1 − λ)y1 + λy2 ) = (1 − λ)z1 + λz2 , we conclude that (1 − λ)x1 + λx2 ∈ F1 , and hence F1 is convex. To prove that F1 is an extreme face of K1 , choose points u, v ∈ K1 and a scalar 0 < λ < 1 such that (1 − λ)u + λv ∈ F1 . Then there is a point y ∈ K2 with the property µ1 ((1 − λ)u + λv) + µ2 y ∈ F . Equivalently, (1 − λ)(µ1 u + µ2 y) + λ(µ1 v + µ2 y) ∈ F. Since both µ1 u+µ2 y and µ1 v +µ2 y belong to K, and since F is an extreme face of K, both µ1 x1 +µ2 y and µ1 x2 +µ2 y belong to F . Hence x1 , x2 ∈ F1 , and F1 is an extreme face of K1 . It remains to prove the equality F = µ1 F1 + µ2 F2 . First, we observe that the inclusion F ⊂ µ1 F1 + µ2 F2 follows from the definition of F1 and F2 . Conversely, let x = µ1 x1 + µ2 x2 , where x1 ∈ F1 and x2 ∈ F2 . Then
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there are points y1 ∈ K2 and y2 ∈ K1 such that both µ1 x1 + µ2 y1 and µ1 y2 + µ2 x2 are in F . By the convexity of F , the point z = 12 (µ1 x1 + µ2 y1 ) + 12 (µ1 y2 + µ2 x2 ) lies in F . Put x = µ1 x1 + µ2 x2 and y = µ1 y2 + µ2 y1 . Clearly, x, y ∈ K and z = 12 (x + y). Therefore, x, y ∈ F because F is an extreme face of K. Summing up, µ1 F1 + µ2 F2 ⊂ F . (2) Let F be bounded. Suppose that F = µ1 F10 + µ2 F20 , where F10 and 0 F2 are extreme faces of K1 and K2 , respectively. From the definition of F1 and F2 it follows that F10 ⊂ F1 and F20 ⊂ F2 . Therefore, F = µ1 F10 + µ2 F20 ⊂ µ1 F1 + µ2 F20 ⊂ F, which gives µ1 F10 + µ2 F20 = µ1 F1 + µ2 F20 . Since all sets µ1 F1 , µ1 F10 , and µ2 F20 are bounded, Theorem 2.47 shows that cl (µ1 F1 ) = cl (µ2 F 0 ). Hence cl F1 = cl F10 . As a result, F1 = K1 ∩ cl F1 = K1 ∩ cl F10 = F10 (see Theorem 7.4). Similarly, F2 = F20 . Remark. It might be that the sum F1 + F2 of extreme faces F1 and F2 of convex sets K1 and K2 , respectively, is not an extreme face of K1 + K2 . Indeed, if K1 = K2 = B, where B is the unit ball of Rn , then for a boundary point x of B, the sets {x} and {−x} are 0-dimensional extreme faces of B, while {x} + {−x} = {o} is not an extreme face of K1 + K2 = 2B. Theorem 7.16. For an affine transformation f : Rn → Rm and convex sets K ⊂ Rn and M ⊂ Rm , the following statements hold. (1) If G is an extreme face of M , then f −1 (G) is an extreme face of f −1 (M ). (2) If G is a convex subset of M ∩ rng f such that f −1 (G) is an extreme face of f −1 (M ), then G is an extreme face of M ∩ rng f . (3) A subset G of f (K) is an extreme face of f (K) if and only if f −1 (G) ∩ K is an extreme face of K. (4) If f is one-to-one, then a set F is an extreme face of K if and only if f (F ) is an extreme face of f (K). Proof. Since the statement is obvious if the respective sets are empty, we may assume that all involved sets are nonempty. Furthermore, the convexity of these sets follows from Theorem 2.12.
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(1) Choose points x, y ∈ f −1 (M ) and a scalar 0 < λ < 1 such that z = (1 − λ)x + λy ∈ f −1 (G). Then f (z) ∈ f (f −1 (G)) ⊂ G. Furthermore, f (z) = (1 − λ)f (x) + λf (y)
and f (x), f (y) ∈ f (f −1 (M )) ⊂ M.
Since G is an extreme face of M , one has f (x), f (y) ∈ G. Hence x, y ∈ f −1 (G), implying that f −1 (G) is an extreme face of f −1 (M ). (2) Choose points x, y ∈ M ∩ rng f and a scalar 0 < λ < 1 such that z = (1−λ)x+λy ∈ G. Let x0 , y0 be points in f −1 (M ) such that f (x0 ) = x and f (y0 ) = y. Put z0 = (1 − λ)x0 + λy0 . Then f (z0 ) = z, implying that z0 ∈ f −1 (G). Hence x0 , y0 ∈ f −1 (G), which gives the inclusions x, y ∈ G. Summing up, G is is an extreme face of M ∩ rng f . (3) Let G be an extreme face of f (K). Choose points x, y ∈ K, a scalar 0 < λ < 1 such that z = (1 − λ)x + λy ∈ f −1 (G) ∩ K. Then f (z) = (1 − λ)f (x) + λf (y) ∈ G. Consequently, f (x), f (y) ∈ G because G is an extreme face of f (K). Hence x, y ∈ f −1 (G) ∩ K, implying that f −1 (G) ∩ K is an extreme face of K. Conversely, let f −1 (G) ∩ K be an extreme face of K. Choose points x, y ∈ f (K) and a scalar 0 < λ < 1 such that the point z = (1 − λ)x + λy belongs to G. Let x0 , y0 be points in K with f (x0 ) = x and f (y0 ) = y. Put z0 = (1 − λ)x0 + λy0 . Then z0 ∈ K by the convexity of K. From f (z0 ) = (1 − λ)x + λy ∈ G we conclude that z0 ∈ f −1 (G) ∩ K. Since f −1 (G) ∩ K is an extreme face of K, one has x0 , y0 ∈ f −1 (G) ∩ K. Thus both x = f (x0 ) and y = f (y0 ) belong to f (f −1 (G)) ⊂ G, implying that G is an extreme face of f (K). (4) Let F be an extreme face of K. Choose points x, y ∈ f (K) and a scalar 0 < λ < 1 such that the point z = (1 − λ)x + λy belongs to f (F ). Let x0 and y0 be the points in K such that f (x0 ) = x and f (y0 ) = y. Put z0 = (1 − λ)x0 + λy0 . Then z0 = f −1 (z) ∈ F because f (z0 ) = z and f is one-to-one. Since F is an extreme face of K, we have x0 , y0 ∈ F , and hence x, y ∈ f (F ). Summing up, f (F ) is an extreme face of f (K). The converse statement follows from statement (3) above. Remark. The assumption that f is one-to-one is essential in statement (4) of Theorem 7.16. Indeed, if K = {(x, y) : x2 + y 2 6 1} is the unit circle of R2 and f is the orthogonal projection of R2 on the x-axis, then every singleton from bd K is a 0-dimensional extreme face of K, while the only 0-dimensional extreme faces of the segment f (K) = {(x, 0) : −1 6 x 6 1} are its endpoints.
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Extreme Representations
Compact Convex Sets Reformulating Definition 7.1 for the case of singletons, we say that x is an extreme point of a convex set K ⊂ Rn if x ∈ K and, for any y, z ∈ K and a scalar 0 < λ < 1, the equality x = (1 − λ)y + λz implies that x = y = z. In what follows, ext K denotes the set of all extreme points of K (put ext ∅ = ∅). Theorem 7.2 shows that a point x ∈ Rn is an extreme point of a convex set K ⊂ Rn if and only if x ∈ K and the set K \ {x} is convex. Example. The extreme points of a closed ball Bρ (c) ⊂ Rn are exactly the points of the boundary sphere Sρ (c) (see Exercise 7.1). Example. The extreme points of an r-simplex ∆(x1 , . . . , xr+1 ) ⊂ Rn are exactly its vertices, x1 , . . . , xr+1 (see Exercise 7.2). Example. The only extreme point of a simplicial cone Cs (x1 , . . . , xr ) ⊂ Rn is its apex s (see Exercise 7.3). Theorem 7.17. Every compact convex set K ⊂ Rn is the convex hull of its extreme points: K = conv (ext K). Proof. Since the case K = ∅ is obvious, we may assume that K is nonempty. Clearly, conv (ext K) ⊂ K because of the inclusion ext K ⊂ K and the convexity of K. By induction on m = dim K, we are going prove the opposite inclusion K ⊂ conv (ext K). The case m = 0 is trivial: K is a singleton, and ext K = K. If dim K = 1, then K is a closed segment, say [x, z]; in this case, ext K = {x, z} and K = conv (ext K). Assume that the statement holds for all compact convex sets of dimension r 6 m − 1 (m > 2), and let K ⊂ Rn be a compact convex set of dimension m. Choose a point x ∈ K and a line l through x. Since K is compact, the set K ∩ l is a closed segment [u, v] (possibly, u = v). From Corollary 2.25 it follows that both v and w belong to rbd K. Theorem 7.9 shows that the generated extreme faces FK (u) and FK (v) also lie in rbd K, and Theorem 7.4 implies that these faces are compact convex sets of dimension less than m. By the induction hypothesis, FK (u) ⊂ conv (ext FK (u))
and FK (v) ⊂ conv (ext FK (v)).
Furthermore, Theorem 7.3 shows that ext FK (u) ∪ ext FK (v) ⊂ ext K.
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Finally, using Theorem 3.2, we conclude x ∈ [u, v] ⊂ conv (FK (u) ∪ FK (v)) ⊂ conv (conv (ext FK (u)) ∪ conv (ext FK (v))) = conv (ext FK (u) ∪ ext FK (v)) ⊂ conv (ext K). Hence K ⊂ conv (ext K). The next result shows the minimality of ext K. Theorem 7.18. If K ⊂ Rn is a compact convex set and X is a subset of K, then K = conv X if and only if ext K ⊂ X. Proof. Let K = conv X. Assume for a moment the existence of a point x ∈ ext K \ X. Then X ⊂ ext K \ {x} ⊂ K \ {x}. Since K \ {x} is a convex set, one has K = conv X ⊂ K \ {x}, a contradiction. Conversely, if ext K ⊂ X, then, by Theorem 7.17, K = conv (ext K) ⊂ conv X ⊂ K, which gives K = conv X. Remark. The equality K = conv (ext K) does not imply that the convex set K ⊂ Rn is bounded or closed. For example, for the unbounded and nonclosed convex set K = {(x, y) : x > 0, y > x2 } ⊂ R2 , one has ext K = {(x, y) : x > 0, y = x2 }
and K = conv (ext K).
A description of closed convex sets K ⊂ Rn satisfying the condition K = conv (ext K) is given in Corollary 7.26. A combination of Theorems 3.6 and 7.17 implies the following corollary. Corollary 7.19. If K ⊂ Rn is a compact convex set of dimension m, then every point u ∈ K is a positive convex combination of m + 1 or fewer affinely independent extreme points of K. Line-Free Closed Convex Sets We recall (see Definition 5.52) that a convex set K ⊂ Rn is line-free if it contains no line. Theorem 7.20. For a nonempty closed convex set K ⊂ Rn , the following statements hold.
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(1) ext K 6= ∅ if and only if K is line-free. (2) K has a unique extreme point, say x, if and only if K is a line-free convex cone with apex x. Proof. (1) Suppose that ext K 6= ∅ and choose a point x ∈ ext K. Assume for a moment that K contains a line l. Then the line l0 which is parallel to l and contains x entirely lies in K (see Corollary 5.2). Choosing points y, z ∈ l0 \ {x} such that x ∈ (y, z), we conclude that x ∈ / ext K. The obtained contradiction shows that K must be line-free. Conversely, let K be line-free. We will show the existence of an extreme point of K by induction on m = dim K. If m = 0, then K is a singleton and ext K = K. If dim K = 1, then K is either a segment [x, z] or a halfline [x, zi; in either case x is an extreme point of K. Assume that the statement holds for all line-free closed convex set of dimension r 6 m − 1, and let K ⊂ Rn be a line-free closed convex set of dimension m. Choose a point z ∈ rbd K. By Theorem 7.9, the generated extreme face FK (z) lies in rbd K, and Theorem 7.4 implies that FK (z) is a closed convex set of dimension m − 1 or less. Since FK (z) is line-free, it has, by the induction hypothesis, an extreme point x. Finally, Theorem 7.3 shows that x is an extreme point of K. (2) Suppose that K has exactly one extreme point, x. As shown above, K is line-free. By induction on m = dim K, we are going to prove that K is a cone with apex x. The case m = 0 is clear because K = {x} is a trivial cone with apex x. Let dim K = 1. Then K is either a segment [x, z] or a halfline [x, zi. Since [x, z] has two extreme points, K must be [x, zi, which is a cone with apex x. Assume that the statement holds for all closed convex set of dimension r 6 m − 1, and let K ⊂ Rn be a closed convex set of dimension m with a unique extreme point, x. Choose a point y ∈ K \ {x} and consider the generated extreme face FK (y). As above, FK (y) is a line-free closed convex set, so it has at least one extreme point z, which also is an extreme point of K. Hence x = z, which shows that x is the only extreme point of FK (y). By the induction hypothesis, FK (y) is a cone with apex x. Therefore, [x, yi ⊂ FK (y) ⊂ K. Since y was chosen arbitrarily in K \ {x}, it follows that K is a cone with apex x. Conversely, let K be a line-free convex cone with apex x. If K = {x}, then x is the only extreme point of K. Suppose that K 6= {x}. Then no point y ∈ K \ {x} is extreme because it belongs to a nontrivial open interval (x, yi ⊂ K. It remains to show that x is an extreme point of K.
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Indeed, if x belonged to an open line interval (u, w), where u, w ∈ K \{x}, then hu, wi = hu, x] ∪ [x, wi ⊂ K, contrary to the assumption that K is line-free. The next result shows that every extreme point of a line-free closed convex set can be included into a “cap” of K of an arbitrary small diameter. Theorem 7.21. Let K ⊂ Rn be a nonempty line-free closed convex set. For a point u ∈ ext K and a scalar δ > 0, there is an open halfspace W ⊂ Rn such that u ∈ K ∩ W ⊂ Bδ (u). Proof. Let M = (B2δ (u) ∩ K) \ Uδ (u). Clearly, M is a compact set which lies in the convex set K \ {u} (see Theorem 7.2). According to Theorem 3.17, conv M is a compact convex set. Furthermore, u ∈ / conv M because of the inclusion conv M ⊂ K \ {u}. By Theorem 6.32, there is a hyperplane H ⊂ Rn strongly separating u and conv M . Denote by W the open halfspace determined by H and containing u. We state that K ∩ W ⊂ Bδ (u). Indeed, assume for a moment that K ∩ W contains a point v ∈ / Bδ (u). The segment [u, v] lies in K by the convexity of K. Choose a point w ∈ [u, v] such that δ 6 ku − wk 6 2δ. Then w ∈ M ∩ W , contrary to the condition conv M ∩ W = ∅. Definition 7.22. An extreme halfline (also called an extreme ray) of a convex set K ⊂ Rn is a halfline h ⊂ K which is an extreme face of K. In what follows, extr K denotes the union of extreme halflines of K (put extr K = ∅ if K has no extreme halflines). Example. The extreme halflines of a simplicial cone Cs (x1 , . . . , xr ) in Rn are exactly the halflines [s, xi i, 1 6 i 6 r (see Exercise 7.3). Remark. If K ⊂ Rn is a closed convex set, then Theorems 7.3 and 7.4 show that every extreme halfline h of K is closed, and the endpoint of h is an extreme point of K. The next lemma will be used in the proof of Theorem 7.24. Lemma 7.23. For a nonempty closed convex set K ⊂ Rn , the following conditions are equivalent. (1) There is a line l such that K is the union of parallel to l segments [u, v], where u, v ∈ rbd K. (2) K is the convex hull of rbd K.
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(3) K is neither a plane nor a closed halfplane of aff K. Proof. The statement (1) ⇒ (2) is trivial. (2) ⇒ (3). The set K cannot be a plane, since otherwise rbd K = ∅ (see Corollary 2.57). Similarly, K cannot be a closed halfplane of aff K, since otherwise rbd K would be a proper subplane of aff K (see Theorem 2.58), which would give K 6= rbd K = conv (rbd K). (3) ⇒ (1). By Corollary 2.57 and Theorem 2.58, the set rbd K is nonempty and nonconvex. Hence there are distinct points x, y ∈ rbd K such that rint K ∩ (x, y) 6= ∅. Choose a point w ∈ rint K ∩ (x, y) and consider the line l = hx, yi. We state that K ∩ l = [x, y]. Indeed, since the inclusion [x, y] ⊂ K ∩l is obvious, it remains to prove the opposite inclusion. Assume for a moment the existence of a point p ∈ K ∩ l \ [x, y]. Then one of the points x, y, say x, belongs to (p, w). In this case, x ∈ rint K according to Theorem 2.21, which contradicts the assumption x ∈ rbd K. Hence K ∩ l ⊂ [x, y]. Next, we state that for a given point z ∈ K, the line l0 = (z − x) + l through z also meets K along a closed segment. Indeed, if K ∩ l0 were unbounded, then a halfline h with endpoint z would lie in K ∩ l0 , implying that the halfline h0 = (x−z)+ h would lie in K ∩l (see Theorem 5.1), which is impossible due to K ∩ l = [x, y]. Hence K ∩ l0 is a closed segment, say [u, v]. From Corollary 2.25 it follows that both u and v belong to rbd K. Since z ∈ [u, v], condition (1) holds. Theorem 7.24. For a closed convex set K ⊂ Rn , the following conditions are equivalent. (1) K is line-free. (2) K = conv (ext K ∪ extr K). (3) K = conv (ext K) + rec K. Proof. Since the case K = ∅ is obvious, we may assume that K is nonempty. (1) ⇒ (2). The inclusion conv (ext K ∪ extr K) ⊂ K is obvious. So, we are going to prove the opposite inclusion K ⊂ conv (ext K ∪ extr K). This will be done by induction on m = dim K. The cases m = 0 and m = 1 are simple: if m = 0, then K is a singleton and K = ext K; if m = 1, then K is either a closed segment (which is the convex hull of its endpoints), or a closed halfline, which is an extreme halfline itself. Assume that the statement holds for all r 6 m − 1, m > 2, and let K ⊂ Rn be a line-free closed convex set of dimension m. Then K is neither
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a plane nor a closed halfplane. Choose a point x ∈ K. By Lemma 7.23, there is a line l through x such that K ∩ l is a segment, [y, z], where both y and z belong to rbd K. Theorem 7.9 shows that both generated extreme faces FK (u) and FK (v) lie in rbd K, and Theorem 7.4 implies that these faces are closed convex sets of dimension less than m. Clearly, FK (u) and FK (v) are line-free. By the induction hypothesis, FK (y) ⊂ conv (ext FK (y) ∪ extr FK (y)), FK (z) ⊂ conv (ext FK (z) ∪ extr FK (z)). According to Theorem 7.3, the extreme points and the extreme halflines of both sets FK (y) and FK (z) are also extreme points and extreme halflines of K. Hence FK (y) ∪ FK (z) ⊂ conv (ext K ∪ extr K), which gives x ∈ conv {y, z} ⊂ conv (FK (y) ∪ FK (z)) ⊂ conv (ext K ∪ extr K). Hence K ⊂ conv (ext K ∪ extr K). (2) ⇒ (3). Since conv (ext K) ⊂ K, the inclusion conv (ext K)+rec K ⊂ K follows from Theorem 5.6. Hence it remains to prove the opposite inclusion. Choose a point x ∈ K = conv (ext K ∪ extr K). Theorem 3.3 shows that x can be written as a convex combination x = λ1 x1 + · · · + λp xp + λp+1 zp+1 + · · · + λq zq , where x1 , . . . , xp ∈ ext K and zp+1 , . . . , zq ∈ extr K \ ext K. If [ui , vi i is an extreme halfline of K containing zi , then Theorem 5.6 gives zi = ui + γi wi , where γi > 0 and wi = vi − ui ∈ rec K, p + 1 6 i 6 q. Hence x = z + w, where z = λ1 x1 + · · · + λp xp + λp+1 up+1 + · · · + λq uq is a convex combination of points from ext K, and w = λp+1 γp+1 wp+1 + · · · + λq γq wq is a positive combination of points from rec K. Clearly, z ∈ conv (ext K), and w ∈ rec K because rec K is a convex cone with apex o. Summing up, x ∈ conv (ext K) + rec K. (3) ⇒ (1). Assume for a moment that K contains a line l. Then, for a given point x ∈ K, the line l0 which is parallel to l and contains x entirely
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lies in K (see Corollary 5.2). Choosing points y, z ∈ l0 \ {x} such that x ∈ (y, z), we conclude that x ∈ / ext K. Hence ext K = ∅. Since the endpoint of an extreme halfline is an extreme point of K, it follows that extr K = ∅. Therefore, K = conv (ext K) + rec K = ∅, contrary to the assumption K 6= ∅. Hence K must be line-free. The next result expands Theorem 7.18 to the case of line-free closed convex sets. We will say that a set X ⊂ Rn is coterminal with a halfline h = {u + λv : λ > 0} provided sup {λ : u + λv ∈ X} = ∞. Theorem 7.25. For a line-free closed convex set K ⊂ Rn and a subset X of K, the following statements hold. (1) K = conv X if and only if ext K ⊂ X and X is coterminal with every extreme halfline of K. (2) K = conv X if and only if ext K ⊂ X and h = conv (h ∩ X) for every extreme halfline of K. (3) K = conv X + rec K if and only if ext K ⊂ X. Proof. (1) Let K = conv X. Assume for a moment the existence of a point x ∈ ext K \ X. Then X ⊂ ext K \ {x} ⊂ K \ {x}. Since K \ {x} is a convex set, one has K = conv X ⊂ K \ {x}, a contradiction. Thus ext K ⊂ K. Next, assume that X is not coterminal with an extreme halfline h of K. Then there is a point v ∈ h such that the open halfline h0 ⊂ h with endpoint v contains no point from X. Consider the set M = (K \h)∪[u, v], where u is the endpoint of h. It is easy to see that M contains X and is a proper convex subset of K. Hence conv X ⊂ M 6= K, a contradiction. Conversely, let a subset X of K contain ext K and be coterminal with every extreme halfline of K. Then conv X contains ext K ∪ extr K. By Theorem 7.24, K = conv (ext K ∪ extr K) ⊂ conv X ⊂ K, which gives K = conv X. (1) ⇔ (2). Assume first that ext K ⊂ X and X is coterminal with every extreme halfline of K. Let h be an extreme halfline of K. Then the endpoint of h belongs to X as an extreme point of K, and the condition of coterminality implies that h = conv (h ∩ X). Conversely, if an extreme halfline h of K satisfies the condition h = conv (h ∩ X), then, obviously, X is coterminal with h. (3) Let K = conv X + rec K. Assume for a moment the existence of a point x ∈ ext K \ X. Since x ∈ K, one can write x = u + v, where u ∈
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conv X and v ∈ rec K. We state that v 6= o. Indeed, suppose v = o. Then x = u ∈ conv X. On the other hand, X ⊂ ext K \ {x} ⊂ K \ {x}. Since the set K \ {x} is convex, one has x ∈ conv X ⊂ K \ {x}, a contradiction. Hence v 6= o. This argument shows that the halfline [o, vi lies in rec K, and Theorem 5.6 gives the inclusion [u, xi = [u, u + vi = u + [o, vi ⊂ K. Because x is not the endpoint of the halfline [u, xi, it cannot be extreme in K, contrary to the assumption x ∈ ext K \ X. Hence ext K ⊂ X. Conversely, let ext K ⊂ X. Then conv (ext K) ⊂ conv X. A combination of Theorems 5.6 and 7.24 gives K = conv (ext K) + rec K ⊂ conv X + rec K ⊂ K + rec K = K. Hence K = conv X + rec K. The corollary below, which refines Theorem 7.17, follows from Theorems 7.24 and 7.25. Corollary 7.26. A closed convex set K ⊂ Rn is the convex hull of ext K if and only if K is line-free and has no extreme halflines. In the latter case, given a set X ⊂ Rn , one has K = conv X if and only if ext K ⊂ X. For the next result, we will need one more notation (compare with Definition 7.34). Namely, for a convex set K ⊂ Rn , denote by ext1 K the union of all nonempty extreme faces of K of dimension at most one. So, if K is line-free, then ext1 K is the union of all extreme points, all extreme segments, and all extreme halflines of K. Theorem 7.27. If K ⊂ Rn is a line-free closed convex set of dimension m, then every point u ∈ K is a positive convex combination of m or fewer affinely independent points from ext1 K. Proof. Choose a point u ∈ K. Since the case dim K = 0 is obvious, we assume that dim K > 1. We state the existence of a hyperplane H ⊂ Rn through u such that H ∩ K is bounded and K 6⊂ H. For this, assume first that K is bounded. Since dim K > 1, there is a point z ∈ K \ {u}. Then the hyperplane H = {x ∈ Rn : (x − u)·(u − z) = 0} meets K along the bounded set H ∩ K such that z ∈ / H. Hence K 6⊂ H. Suppose now that K is unbounded. Then the recession cone of K contains a halfline h with endpoint o (see Theorem 5.13). By the same
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theorem and Corollary 5.53, the recession cone of K is closed and linefree. Theorem 5.47 shows that the polar cone (rec K)◦ is n-dimensional, and Corollary 5.49 derives that for any nonzero vector z ∈ (rec K)◦ , the (n − 1)-dimensional subspace S = {x ∈ Rn : x·z = 0} supports rec K such that S ∩ rec K = {o}. Consider the hyperplane H = u + S. Theorem 5.54 shows that the convex set M = H ∩ K is bounded. Since the halfline u + h lies in K (see Theorem 5.6) and (u + h) ∩ H = {u} + (h ∩ S) = {u}, it follows that K 6⊂ H. Since K 6⊂ H, Corollary 2.60 gives the inequality dim M 6 dim K − 1. By Corollary 7.19, u can be expressed as a positive convex combination of affinely independent points x1 , . . . , xp ∈ ext M , p 6 m. Consider the generated extreme faces FK (xi ), 1 6 i 6 p. Since xi ∈ rint FK (xi ) and FK (xi ) ∩ H = {xi }, Corollary 2.33 shows that dim FK (xi ) 6 1 for all 1 6 i 6 p. Hence the points x1 , . . . , xp belong to ext1 K, as desired. Closed Convex Sets A combination of Theorems 5.10, 5.20, and 7.24 implies the following result. Corollary 7.28. Let K ⊂ Rn be a nonempty closed convex set. If S ⊂ Rn is a subspace complementary to lin K, then K = conv (ext (K ∩ S) ∪ extr (K ∩ S)) ⊕ lin K, K = (conv (ext (K ∩ S)) + rec (K ∩ S)) ⊕ lin K, K = conv (ext (K ∩ S)) + rec K. Definition 7.29. Let K ⊂ Rn be a nonempty convex set. A nonempty extreme face F of K ⊂ Rn is called planar (respectively, halfplanar ) if F is a plane (respectively, F is a halfplane). Example. If K ⊂ R3 is a convex cone, given by K = {(x, y, z) : y > |x|}, then ap K = lin K = {(0, 0, z) : z ∈ R} is the only planar extreme face of K, and the halfplanes D1 = {(x, x, z) : x > 0}
and D2 = {(x, −x, z) : x 6 0}
are the halfplanar extreme faces of K. Clearly, the planar extreme faces of a line-free closed convex set K ⊂ Rn are its extreme points, and the halfplanar extreme faces of K are its extreme halflines.
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Theorem 7.30. Let K ⊂ Rn be a nonempty closed convex set expressed as K = (K ∩ S) ⊕ lin K, where S is a subspace of Rn complementary to lin K. The following statements hold. (1) An extreme face F of K is planar if and only if F = x + lin K, where x ∈ ext (K ∩ S). (2) An extreme face F of K is halfplanar if and only if F = h + lin K, where h is an extreme halfline of K ∩ S. Proof. According to Theorem 7.5, an extreme face F of K can be expressed as F = G ⊕ lin K, where G is an extreme face of K ∩ S. (1) If F = x + lin K, where x ∈ ext (K ∩ S), then F is a plane, which is an extreme face of K by the above argument. Conversely, let F a planar extreme face of K. Choose a point x ∈ G. Then x + lin K ⊂ G ⊕ lin K = F. On the other hand, Corollary 5.28 implies the opposite inclusion F ⊂ x + lin K. (2) If F = h + lin K, where h is an extreme halfline of K ∩ S, then F is a halfplane (see Theorem 1.42), which is an extreme face of K by the above argument. Conversely, let F be a halfplanar extreme face of K. Theorem 1.42 shows that F = g + L, where g is a halfline with endpoint o and L is a plane. Denote by h the projection of g on S along lin K. Since F = G ⊕ lin K, it is easy to see that G is a closed halfline, h, with endpoint o (otherwise G = {o} and L is a translate of lin K). Then, by the above argument, F = h + lin K. The corollary below follows from Theorem 7.20 and the argument in the proof of Theorem 7.30. Corollary 7.31. Let K ⊂ Rn be a nonempty closed convex set. An extreme face F of K is planar if and only if F is a translate of lin K. Furthermore, K has exactly one planar extreme face, F , if and only if K is a convex cone with ap K = F . The next result expands condition (2) of Theorem 7.24. Theorem 7.32. A nonempty closed convex set K ⊂ Rn is the convex hull of the union of its planar and halfplanar extreme faces. Similarly, K is the sum of rec K and the convex hull of the union of its planar extreme faces.
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Proof. Express K as a direct sum K = (K ∩ S) ⊕ lin K, where S is a subspace complementary to lin K. By Corollary 7.28, K = conv (ext (K ∩ S) ∪ extr (K ∩ S)) ⊕ lin K = conv ∪ (x + lin K : x ∈ ext (K ∩ S)) ∪ h + lin K : h is an extreme halfline of K ∩ S) . This argument and Theorem 7.30 give K = conv ∪ (F : F is a planar or a halfplanar extreme face of K) . Similarly, Theorems 5.20 and 7.24 give rec K = rec (K ∩ S) ⊕ lin K, K ∩ S = conv (ext (K ∩ S)) + rec (K ∩ S), which implies K = (conv (ext (K ∩ S)) + rec (K ∩ S)) ⊕ lin K = conv ∪ (x + lin K : x ∈ ext (K ∩ S)) + rec (K ∩ S) ⊕ lin K = conv ∪ (F : F is a planar extreme face of K) + rec K. The next result expands Theorem 7.25 to the case of arbitrary closed convex sets. Theorem 7.33. For a nonempty closed convex set K ⊂ Rn and a subset X of K, the following statements hold. (1) K = conv X if and only if F = conv (F ∩ X) for every planar or halfplanar extreme face F of K. (2) K = conv X + rec K if and only if F = conv (F ∩ X) for every planar extreme face F of K. Proof. (1) Suppose that K = conv X, and choose an extreme face F of K. By Theorem 7.4, one has F = F ∩ K = F ∩ conv X = conv (F ∩ X). Conversely, suppose that X satisfies the hypothesis of statement (1). By Theorem 7.32, K = conv ∪ (F : F is a planar or halfplanar extreme face of K) = conv ∪ (conv (F ∩ X) : F is a planar or halfplanar extreme face of K) ⊂ conv (conv X) = conv X ⊂ K, implying the equality K = conv X. The proof of statement (2) of the theorem is similar.
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r-Extreme Sets Definition 7.34. Let K be a nonempty convex set in Rn and r be an integer satisfying the inequalities 0 6 r 6 dim K. The union of all nonempty extreme faces of K of dimension r or less, is called the r-extreme set of K and denoted extr K. The elements of extr K are called r-extreme points of K. Clearly, ext0 K = ext K. Another terminology used for the r-extreme set is the r-skeleton. Theorem 7.35. If K ⊂ Rn is a nonempty convex set of dimension m and an integer r satisfies the inequalities 0 6 r 6 m, then the following statements hold. (1) ext0 K ⊂ ext1 K ⊂ · · · ⊂ extm−1 K = K ∩ rbd K ⊂ extm K = K. (2) The set K \ extr K is convex. Furthermore, K \ extr K is the union of relative interiors of extreme faces of K, each of dimension at least r + 1. (3) For a point x ∈ K, the following conditions are equivalent. (a) (b) (c) (d)
x ∈ extr K. dim FK (x) 6 r. No (r + 1)-simplex ∆ ⊂ K satisfies the inclusion x ∈ rint ∆. x ∈ extr (Bρ (x) ∩ K) for every ball Bρ (x) ⊂ Rn , ρ > 0.
Proof. (1) The inclusions ext0 K ⊂ ext1 K ⊂ · · · ⊂ extm−1 K ⊂ K ∩ rbd K and the equality extm K = K follow from definitions and Theorem 7.4. The opposite inclusion, K∩ rbd K ⊂ extm−1 K, is a consequence of Theorem 7.9. (2) Let Fr denote the family of all extreme faces of K, each of dimension r or less. By the definition, extr K = ∪ (F : F ∈ Fr ). The equality K \ extr K = K \ (∪ (F : F ∈ Fr )) = ∩ (K \ F : F ∈ Fr ) shows that K \ extr K is convex as the intersection of convex sets K \ F , F ∈ Fr (see Theorem 7.2). The second part of statement (2) follows from Corollary 7.12. (3) (a) ⇒ (b). Let x ∈ extr K. Then there is an extreme face F of K of dimension r or less such that x ∈ F . By Corollary 7.8, FK (x) is the smallest extreme face of K containing x. Hence dim FK (x) 6 dim F 6 r.
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(b) ⇒ (c). Let dim FK (x) 6 r. Choose a simplex ∆ ⊂ K such that x ∈ rint ∆. Since FK (x) is an extreme face of K, Theorem 7.4 implies the inclusion ∆ ⊂ FK (x). Hence dim ∆ 6 dim FK (x) 6 r. (c) ⇒ (a). Let x be a point in K such that no (r + 1)-simplex from K contains x in its relative interior. Put p = dim FK (x). According to Theorem 7.9, x ∈ rint FK (x). Therefore, there is a simplex ∆ ⊂ FK (x) such that x ∈ rint ∆ and dim ∆ = dim FK (x) (see Theorem 2.19). By the assumption, dim ∆ 6 r. Consequently, p 6 r, implying the inclusion x ∈ extr K. (a) ⇔ (d). Assume first that x ∈ extr K. By condition (c) above, no (r+ 1)-simplex ∆ ⊂ Bρ (x) ∩ K contains x in its relative interior. Consequently, x ∈ extr (Bρ (x) ∩ K). Conversely, let x ∈ extr (Bρ (x)∩K). Assume for a moment the existence of an (r + 1)-simplex ∆ = ∆(x1 , . . . , xr+2 ) ⊂ K satisfying the inclusion x ∈ rint ∆. By Theorem 2.17, x can be written as a positive convex combination x = λ1 x1 + · · · + λr+2 xr+2 . Let δ = max{kx − xi k : 1 6 i 6 r + 2} and ρ x0i = x + (xi − x), 1 6 i 6 r + 2. δ A combination of Theorems 1.92 and 1.83 shows that x01 , . . . , x0r+2 are affinely independent. Furthermore, r+2 r+2 r+2 P P P P ρ ρ r+2 λi xi − λi x = x + (x − x) = x, λi x0i = λi x + δ δ i=1 i=1 i=1 i=1 0 0 0 and Theorem 2.17 gives x ∈ rint ∆ , where ∆ = ∆(x1 , . . . , x0r+2 ). Since {x01 , . . . , x0r+2 } ⊂ ∆, it follows that ∆0 = conv {x01 , . . . , x0r+2 } ⊂ ∆ ⊂ K. Similarly, the inequalities ρ kx − x0i k 6 kx − xi k 6 ρ, 1 6 i 6 r + 2, δ imply that ∆0 ⊂ Bρ (x). Summing up, ∆0 ⊂ Bρ (x) ∩ K. Consequently, x∈ / extr (Bρ (x) ∩ K), contrary to the assumption on x. The following corollary is a consequence of Theorem 7.5. Corollary 7.36. Let a nonempty convex set K ⊂ Rn of dimension m be expressed as a direct sum K = (K ∩ S) ⊕ lin K, where S is a subspace complementary to lin K. If dim (lin K) = p, then extr K = ∅ for all 0 6 r 6 p − 1, and extp+q K = extq (K ∩ S) ⊕ lin K, 0 6 q 6 m − p.
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The following lemma is used in the proof of Theorem 7.38. Lemma 7.37. Let K ⊂ Rn be a convex set of dimension m > 2 and F be an (m−1)-dimensional convex subset of rbd K. For any point z ∈ rint F , there is a closed ball Bρ (z) ⊂ Rn such that Bρ (z)∩ rbd K ⊂ rint F . Consequently, the set rbd K \ rint F is closed. Proof. By Theorem 2.59, aff F ∩ cl K ⊂ rbd K; so, aff F ∩ rint K = ∅. Choose a point s ∈ rint K. Then s ∈ / aff F , and the set {s} ∪ F is m-dimensional (see Corollary 1.78). The inclusion {s} ∪ F ⊂ K gives aff ({s} ∪ F ) ⊂ aff K, and Theorem 1.6 shows that aff ({s} ∪ F ) = aff K. Furthermore, Theorem 4.30 implies that aff (cones F ) = aff K. Let z ∈ rint F . By Theorem 4.41, z ∈ rint (cones F ), and Theorem 2.14 shows the existence of a scalar ρ > 0 such that Bρ (z) ∩ aff K = Bρ (z) ∩ aff (cones F ) ⊂ rint (cones F ).
(7.1)
Finally, let x ∈ Bρ (z) ∩ rbd K. By the above argument, x ∈ Bρ (z) ∩ aff K ⊂ rint (cones F ) = cones (rint F ) \ {s}. Hence there is a point u ∈ rint F such that x ∈ (s, ui (see Theorem 4.30). According to Theorem 2.55, the halfline (s, ui meets rbd K at a unique point. Since both x and u belong to (s, ui ∩ rbd K, we conclude that x = u ∈ rint F . Theorem 7.38. For a convex set K ⊂ Rn of dimension m > 2, one has extm−1 K = K ∩ cl (extm−1 K), extm−2 K = K ∩ cl (extm−2 K). Consequently, both sets extm−1 K and extm−2 K are closed provided K is closed. Proof. By Theorem 7.35, one has extm−1 K = K ∩ rbd K. Because the set rbd K is closed (see Theorem 2.54), we have extm−1 K ⊂ K ∩ cl (extm−1 K) ⊂ K ∩ cl (rbd K) = K ∩ rbd K = extm−1 K. Hence extm−1 K = K ∩ cl (extm−1 K). Since the inclusion extm−2 K ⊂ K ∩ cl (extm−2 K) is obvious, it remains to prove the opposite inclusion. Assume for a moment the existence of a point x ∈ K ∩ cl (extm−2 K) \ extm−2 K.
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Then dim FK (x) > m − 1 according to Theorem 7.35. On the other hand, the inclusion x ∈ cl (extm−2 K) ⊂ rbd K and a combination of Theorems 7.4 and 7.9 shows that FK (x) is a convex subsets of rbd K and dim FK (x) 6 m − 1. So dim FK (x) = m − 1. By Lemma 7.37, there is a ball Bρ (x) such that Bρ (x) ∩ rbd K ⊂ rint FK (x). From here we obtain that x cannot belong to cl (extm−2 K), a contradiction. Remark. Theorem 7.38 does not hold for extr K if r 6 m − 3. Indeed, let K be the convex hull of the compact set X = {u, v} ∪ {(x, y, 0) : (x − 1)2 + y 2 = 1} ⊂ R3 , where u = (0, 0, 1) and v = (0, 0, −1). Every point (x, y, 0), with (x − 1)2 + y 2 = 1 and x 6= 0, is an extreme point of K. Hence the origin o belongs to cl (ext0 K). On the other hand, o ∈ (u, v); so, o ∈ / ext0 K. The next result generalizes Theorems 7.24 and 7.25 and Corollary 7.26. Theorem 7.39. For a nonempty closed convex set K ⊂ Rn of dimension m and an integer 0 6 r 6 m, the following statements hold. (1) If dim (lin K) 6 r, then K = conv (extr+1 K). (2) K = conv (extr K) if and only if dim (lin K) 6 r and K has no extreme halfplane of dimension greater than r. Proof. Let K = (K ∩ S) ⊕ lin K, where S is a subspace complementary to lin K and K ∩ S is a line-free closed convex set (see Theorem 5.20). (1). Put p = dim (lin K), where p 6 r. Theorem 7.24 gives that K ∩ S = conv (ext (K ∩ S) ∪ extr (K ∩ S)). By the definitions, ext (K ∩ S) ∪ extr (K ∩ S) ⊂ ext1 (K ∩ S), Therefore, Corollary 7.36 shows that K = (K ∩ S) ⊕ lin K ⊂ conv (ext1 (K ∩ S)) ⊕ lin K = conv (extp+1 K) ⊂ conv (extr+1 K). (2). Suppose first that dim (lin K) 6 r and K has no extreme halfplane of dimension greater than r. Then, in the notation above, K ∩ S has no extreme halflines, and Corollary 7.26 gives K = (K ∩ S) ⊕ lin K ⊂ conv (ext0 (K ∩ S)) ⊕ lin K = conv (extp K) ⊂ conv (extr K).
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Conversely, let K = conv (extr K). We state first that dim (lin K) 6 r. Indeed, suppose for a moment that dim (lin K) > r+1. Then every extreme face F of K, expressed as F = E ⊕ lin K (see Theorem 7.5), is at least (r + 1)-dimensional. In this case extr K = ∅ (see Corollary 7.36), and K = conv (extr K) = ∅, in contradiction with the assumption K 6= ∅. Next, we observe that no extreme halfplane of K has dimension greater than r. Indeed, let D be an extreme halfplane of K. As above, D = (D ∩ S)⊕ lin K. Since K ∩S is a line-free closed convex set, its extreme face D∩S is a closed halfline. Therefore, dim D = dim (lin K) + 1 6 r + 1. We state that dim D 6 r. Suppose that dim D = r + 1. In this case, dim (lin K) = r and D ∩ S is a halfline. By Corollary 7.36, extr K = ext (K ∩ S) ⊕ lin K, implying that (K ∩ S) ⊕ lin K = K = conv (extr K) = conv (ext (K ∩ S)) ⊕ lin K. Hence K ∩ S = conv (ext (K ∩ S)), which is impossible (see Corollary 7.26). The obtained contradiction shows that dim D 6 r. The next result provides a generalization of Theorem 7.27. Theorem 7.40. Let K ⊂ Rn be a closed convex set of dimension m, with dim (lin K) = r. Then every point x ∈ K is a positive convex combination of m − r or fewer affinely independent points from extr+1 K. Proof. Let K = (K ∩ S) ⊕ lin K, where S is a subspace complementary to lin K and K ∩ S is a line-free closed convex set of dimension m − r (see Theorem 5.20). Choose a point x ∈ K and express it as x = y + z, where y ∈ K ∩ S and z ∈ lin K. By Theorem 7.27, y can be written as a positive convex combination y = λ1 y1 + · · · + λp yp , where p 6 m − r and y1 , . . . , yp are affinely independent points in ext (K ∩ S) ∪ extr (K ∩ S). Let xi = yi + z, 1 6 i 6 p. Clearly, the set {x1 , . . . , xp } is affinely independent and xi ∈ yi + lin K for all 1 6 i 6 p. Since ext (K ∩ S) ∪ extr (K ∩ S) ⊂ ext1 (K ∩ S), Corollary 7.36 implies that x1 , . . . , xp ∈ ext1 K ⊕ lin K = extr+1 K, and the equality x = y + z = λ1 y1 + · · · + λp yp + z = λ1 (y1 + z) + · · · + λp (yp + z) = λ1 x1 + · · · + λp xp shows that x is a positive convex combination of p (6 m − r) affinely independent points from extr+1 K.
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Exercises for Chapter 7 Exercise 7.1. Show that the extreme faces of a closed ball Bρ (c) ⊂ Rn are ∅, all singletons from the boundary sphere Sρ (c), and the ball itself. Exercise 7.2. Let ∆ = ∆(x1 , . . . , xr+1 ) be an r-simplex in Rn . Show that the extreme faces of ∆ are ∅ and all simplices ∆(z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr+1 }. Exercise 7.3. Let C = Cs (x1 , . . . , xr ) be a simplicial cone in Rn . Show that the extreme faces of C are ∅, {s}, and all simplicial cones of the form Cs (z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr }. Exercise 7.4. Let K ⊂ Rn be a closed convex set. Show that the following statements hold. (1) K has exactly two extreme faces if and only if K is a plane. (2) K has exactly three extreme faces if and only if K is a closed halfplane. (3) K has exactly four extreme faces if and only if K is a closed plane slab. Exercise 7.5. (Bj¨ orck [25]) Show that a nonempty set X ⊂ Rn is the set of extreme points of a compact convex set in Rn if and only if its satisfies the following the conditions. (1) cl X ⊂ conv X. (2) cl X is compact. (3) X ∩ conv Y ⊂ Y for all subsets Y of X. Exercise 7.6. Let K ⊂ Rn be a closed convex set. Show that ext K is a Gδ -sets (equivalently, K \ ext K is an Fσ -set). Exercise 7.7. Let K and M be nonempty convex sets in Rn . Show that a point x ∈ K + M is an extreme point of K + M if and only if it is uniquely expressible as x = y + z, where y and z are extreme points of K and M , respectively. Notes for Chapter 7 Extreme faces. Surveys on the extreme structure of convex sets in Rn can be found in Schneider [190, 191]. Theorem 7.9 and Corollary 7.12, belong to Dubins [76] (see also Bourbaki [34, p. TVS II.87]). Theorem 7.5 implies the following
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statement of Lommatzsch [151]: given a closed convex set K ⊂ Rn , the minimum dimension of an extreme face of K equals the dimension of lin K. Some necessary and sufficient conditions for boundedness of extreme points of a convex set are discussed in Batson [21]. Price [176] showed that if K1 , K2 , . . . is a sequence of compact convex sets in Rn which converges to a compact convex set K, then for every point x ∈ ext K there is a sequence of points xi ∈ ext Ki , i > 1, such that x = limi→∞ xi . Dubins [76] (see also Klee [130] and Pranger [174]) proved that if K is a compact convex set in Rn and L is a plane such that K ∩ L 6= ∅, then every extreme point of K ∩ L is a convex combination of n + 1 − dim L or fewer extreme points of K. Klee and Martin [136] proved the following statement. If K is a convex body in Rn , then the function x 7→ FK (x), x ∈ bd K, is lower semicontinuous almost everywhere in the sense of category and upper semicontinuous almost everywhere in the sense of measure. As shown by Reiter and Stavrakas [182], Debs [71], and Papadopoulou [168,169], the following conditions are equivalent for a convex body K ⊂ Rn : (a) the face-function x 7→ FK (x) is lower semicontinuous, (b) the metric space of all extreme faces of a convex body K ⊂ Rn is compact (in the Hausdorff metric), (c) all sets extr K, r = 0, 1, . . . , dim K, are closed, (d) the mapping (x, y) → (x + y)/2 from K × K on K is open. The statement of Exercise 7.6 is proved by Klee [125] (see also Phelps [172, p. 5]). A generalization to the case of extr K belongs to Schneider [189, p. 66] . Soltan and Vasiloi [194] proved that a closed convex set K in Rn is a convex cone provided every two extreme faces of K meet. Extreme structure of the sum of convex sets. The first part of Theorem 7.15 goes back to Minkowski [158, p. 181], where it is proved for the case of compact convex sets in R3 , with a simplified proof and a generalization to Rn by Fujiwara [89] (see also Dalla [63], Husain and Tweddle [118], Roy [187]). The second part of Theorem 7.15 belongs to Husain and Tweddle [118]. Given convex sets K1 , K1 ⊂ Rn , some conditions when the sum of points x1 ∈ ext K1 and x2 ∈ ext K2 is an extreme point of K1 + K2 are given by Bair [15]. The following result is proved in Soltan [200] (see Husain and Tweddle [118] for the case of compact sets). Let a line-free closed convex set A ⊂ Rn be the sum of closed convex sets B and C. Then both sets B0 = B +rec A and C0 = C +rec A are closed and satisfy the following conditions: (a) for every point a ∈ ext A there are unique points b ∈ ext B0 and c ∈ ext C0 such that a = b + c, (b) the sets extC B = {x ∈ ext B : ∃ y ∈ ext C such that x + y ∈ ext A}, extB C = {x ∈ ext C : ∃ y ∈ ext B such that x + y ∈ ext A}
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are dense in ext B0 and ext C0 , respectively, (c) extC B = ext B and extB C = ext C if both B and C are compact. Topological properties of extreme points. An result of Collier [59] states that given a convex body K ⊂ R3 , every component of cl (ext K) \ ext K is a subset of a 1-dimensional extreme face of K. Klee [125] posed (for n = 3) the problem to “find a useful and simple characterization of the class G of subsets of the unit square K in Rn such that there is a homeomorphism of K onto the boundary of a convex body K ⊂ Rn mapping a given set X ∈ G onto ext K.” He also mentioned that every member of G is a Gδ -set, and that G includes every closed subset of K which has at least n + 1 points. Collier [58] partially solved this problem by proving that given a subset X of a compact 0-dimensional metric space Z, there is a homeomorphism of Z into the boundary of a convex body K ⊂ Rn , n > 3, mapping X onto ext K if and only if X is a Gδ -set with at least n + 1 points. For n = 3, this result was improved by Bronshte˘ın [38], who showed that given a Gδ -subset M ⊂ R, with at least 4 points, there is a topological embedding ϕ : R → R3 such that ext (conv ϕ(R)) = ϕ(M ). Two more results of Bronste˘ın [37,39,40] are related to more restricted classes of sets as follows. (a) If M is a bounded, locally compact subset of Rn with at least n + 3 points, then there is a topological embedding ϕ : cl M → Rn+2 such that ext (conv ϕ(cl M )) = ϕ(M ). (b) If M is a bounded subset of Rn with cl M \ M countable, then there is a topological embedding ϕ : cl M → Rn+4 such that ext (conv ϕ(cl M )) = ϕ(M ). One–skeletons of convex bodies. Balinski [20] showed that given an ndimensional polytope P ⊂ Rn , the graph formed by the vertices and edges of P is n-connected; that is, it has at least n + 1 vertices and no two vertices can be separated by removing fewer than n vertices. Larman and Rogers [142] established the following direct generalization of this statement: If b and c are distinct exposed points of a convex body K ⊂ Rn , then there are n simple arcs S1 , . . . , Sn in the 1-skeleton ext1 K, each joining b and c in that Si ∩ Sj = {b, c} when 1 6 i < j 6 n. By analogy with the simplex algorithm for convex polytopes, Larman and Rogers [143] conjectured that for a non-constant linear functional ϕ on Rn and an extreme point c of a convex body K ⊂ Rn there is a path S initiated at c such that ϕ strictly increases along C from ϕ(c) to its maximum value on K. This conjecture was disproved by Gallivan and Larman [92] and by Gallivan and Gardner [91] under a stronger assumption that c is an exposed point of K. Sharpening a result of Larman and Rogers [143], Gallivan proved that if a non-constant linear functional ϕ attains its maximum value on a convex body K ⊂ Rn over the whole of an r-dimensional extreme face of K, then there are r + 1 paths in the one-skeleton of K, disjoint except at their end-points, along each of which ϕ strictly increases to its maximum value on K. A survey on various results and open problems about one-skeletons of convex bodies is given by
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Larman [141]. Extreme representations. Theorem 7.17 is proved by Minkowski [158, p. 160] for the case n = 3 and expanded by Steinitz [207, p. 16] to all n > 3. Lemma 7.23 belongs to Steinitz [207, p. 8], while the equivalence of statements (1) and (2) in Theorem 7.24 and statement (1) of Theorem 7.25 are proved by Klee [124]. The equivalence of statements (1) and (3) in Theorem 7.24 is explicitly shown by Gr¨ unbaum [104, p. 25]. Also, the first statement of Theorem 7.32 is proved by Klee [126, Proposition 2.5]. If K ⊂ Rn is a line-free closed convex, then the set B = {x ∈ K : K ∩ (x − rec K) = {x}} is the smallest subset of K (not necessary convex or closed) such that K = B + rec K (see He and Sun [113]). Zamfirescu [226] obtained an extensions of Minkowski and Klee results for the case of nonclosed convex sets sets. Danielyan, Movsisyan, and Tatalyan [65] posed the following problem: If K ⊂ Rn is a compact convex set, r, s > 2, and n1 > n2 > · · · > ns > 1 are integers such that n1 +· · ·+ns = r, then for every r-extreme point x of K there are distinct ni -extreme points xi of K, 1 6 i 6 s, such that x is a convex combination of x1 , . . . , xs . This problem is confirmed by Lawrence and Soltan [146] for the case of convex polytopes. Their paper also contains the following Carath´eodory-type results concerning a family {K1 , . . . , Kr } of nonempty line-free closed convex sets in Rn . (a) For every point z ∈ K1 +· · ·+Kr , there are nonempty extreme faces Fi of Ki , 1 6 i 6 r, such that z ∈ F1 + · · · + Fr and dim F1 + · · · + dim Fr 6 n. (b) For every point z ∈ conv (K1 ∪ · · · ∪ Kr ), there is an index set I ⊂ {1, . . . , r}, with |I| 6 n + 1, and nonempty P extreme faces Fi of Ki , i ∈ I, such that z ∈ conv (∪ Fi : i ∈ I) and (dim P Fi : i ∈ I) 6 n. If, additionally, all K1 , . . . , Kr are compact, then (dim Fi : i ∈ I) 6 n + 1 − |I|.
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The Exposed Structure of Convex Sets
8.1
Exposed Faces
Definition, Examples, and Basic Properties Definition 8.1. Let K ⊂ Rn be a convex set. An exposed face of K is a subset G ⊂ K satisfying any of the following conditions: (i) G = H ∩ K for a certain hyperplane H ⊂ Rn supporting K, (ii) G = ∅ or G = K. The exposed face G is called proper provided ∅ 6= G 6= K. Example. The exposed faces of a closed ball Bρ (c) ⊂ Rn are the empty set, all singletons from the boundary sphere Sρ (c), and the ball itself (see Exercise 8.1). Example. The exposed faces of an r-simplex ∆ = ∆(x1 , . . . , xr+1 ) ⊂ Rn are exactly its extreme faces: the empty set and all simplices ∆(z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr+1 } (see Exercise 8.2). Example. The exposed faces of a simplicial cone Cs (x1 , . . . , xr ) ⊂ Rn coincide with its extreme faces: the empty set, the singleton {s}, and all simplicial cones Cs (z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr } (see Exercise 8.3). Example. The exposed faces of a plane L ⊂ Rn are ∅ and L itself. If dim L = m > 0 and D is a closed halfplane of L, then ∅, rbd D, and D are the exposed faces of D. If G is a closed slab of L, then the exposed faces of G are ∅, two (m − 1)-dimensional planes determining G, and G itself (see Exercise 8.4). 287
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Theorem 8.2. Every exposed face of a convex set K ⊂ Rn also is an extreme face of K. Proof. Let G be an exposed face of K. Excluding the trivial cases, we assume that G is a proper subset of K. Then there is a hyperplane H ⊂ Rn supporting K such that H ∩ K = G. Denote by V the closed halfspace determined by H and containing K. Without loss of generality, we can write V = {x ∈ Rn : x·c 6 γ} for a nonzero vector c ∈ Rn and a scalar γ ∈ R. Suppose that points x, y ∈ K and a scalar 0 < λ < 1 are such that z = (1 − λ)x + λy ∈ G. Then x, y ∈ V and z ∈ H, giving γ = z·c = (1 − λ)x·c + λ y·c 6 (1 − λ)γ + λγ = γ. Therefore, x·c = y ·c = γ, which gives the inclusion x, y ∈ H ∩ K = G. Hence G is an extreme face of K. The following example shows that a closed convex set K ⊂ Rn may have proper extreme faces which are not exposed. Example. Let K ⊂ R2 be the convex hull of two unit circles B1 (o) and B1 (c), where c = (2, 0) (see the picture below). Put x = (0, 1). Then x is an extreme but not an exposed point of K. Indeed, if a line l through x supports K, then l should be horizontal and K ∩ l = [x, z] 6= {x}, where z = (2, 1). xr zr '$ '$ ' $ l &% &% & % K Theorem 8.2 allows to derive some properties of exposed faces from Theorems 7.3 and 7.4, as shown in the following corollary. Corollary 8.3. For a convex set K ⊂ Rn , the following statements hold. (1) If G is an exposed faces of K and M is a convex subset of K such that G ⊂ M , then G is an exposed face of M . (2) If G is an exposed face of K, then G = K ∩ cl G; in particular, G is closed provided K is closed. (3) If G is an exposed face of K and M is a convex subset of K such that G ∩ rint M 6= ∅, then M ⊂ G.
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(4) If G is an exposed face of K and M is a convex subset of G, then the set (K \ G) ∪ M is convex. (5) If an exposed face G of K meets rint K, then G = K. (6) Every proper exposed face of K lies in K ∩ rbd K and has dimension less than dim K. (7) Distinct exposed faces of K have disjoint relative interiors. (8) If G is an exposed face of K and X is a subset of K, then conv (G ∩ X) = G ∩ conv X. Remark. Example on page 288 shows that the property of being an exposed face is not hereditary (compare with statement (2) of Theorem 7.3). Indeed, {z} is an exposed face of [x, z], and [x, z] is an exposed face of K, while {z} is not an exposed face of K. Corollary 8.4. If F is a nonempty (respectuvely, proper) exposed face of a convex set K ⊂ Rn and a hyperplane H ⊂ Rn satisfies the condition H ∩ K = F , then H supports (respectuvely, properly supports) K. Proof. Since the case F = K is obvious, we assume that F is a proper exposed face of K. Then ∅ 6= H ∩ K = F ⊂ rbd K (see Corollary 8.3), implying that H ∩ cl K 6= ∅ and H ∩ rint K = ∅. Consequently, H properly supports K. Theorem 8.5. Let K ⊂ Rn be a convex set and F be an extreme face of K. There is a sequence of sets F = Gr ⊂ Gr−1 ⊂ · · · ⊂ G1 ⊂ G0 = K such that every Gi is a proper exposed face of Gi−1 , 1 6 i 6 r − 1. Proof. Excluding the trivial cases, we assume that F is a proper subset of K. Let G0 = K. Theorem 7.4 shows that F ⊂ G0 ∩ rbd G0 , and Corollary 6.9 implies the existence of a hyperplane H1 which contains F and properly supports G0 . Put G1 = G0 ∩H1 . Then G1 is a proper exposed face of G0 . Furthermore, G1 ⊂ G0 ∩ rbd G0 because of H1 ∩ rint K = ∅. By Theorem 2.59, dim G1 6 dim G0 − 1. Since F ⊂ G1 , Theorem 7.3 implies that F is an extreme face of G1 . If F = G1 , the proof is done. If F 6= G1 , then F ⊂ G1 ∩ rbd G1 by Theorem 7.4, and we can apply to G1 and F the above argument. Repeating this procedure finitely many times, we obtain the desired conclusion.
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Corollary 8.6. If K ⊂ Rn is a convex set of positive dimension m, then every (m − 1)-dimensional extreme face of K is its exposed face. The next theorem shows that we can reduce the study of exposed faces of arbitrary convex sets to those with trivial linearity. Theorem 8.7. If a convex set K ⊂ Rn is expressed as K = (K ∩S)⊕ lin K, where S is a subspace complementary to lin K, then a set G ⊂ Rn is a proper exposed face of K if and only if G = E ⊕ lin K, where E is a proper exposed face of K ∩ S. Proof. Since the case lin K = {o} is trivial, we may assume that lin K is distinct from {o}. Let G be a proper exposed face of K, and H ⊂ Rn be a hyperplane such that H ∩ K = G. By Theorem 8.2, G is a proper extreme face of K, and Theorem 7.5 implies that G = E ⊕ lin K, where E is a proper extreme face of K ∩ S. For every point x ∈ E, one has x + lin K ⊂ G ⊂ H. Hence H = (H ∩ S) ⊕ lin K, and H ∩ K = ((H ∩ S) ⊕ lin K) ∩ ((K ∩ S) ⊕ lin K) = (H ∩ K ∩ S) ⊕ lin K. Comparing this expression with that of H ∩ K = G = E ⊕ lin K, we obtain H ∩ (K ∩ S) = E. Furthermore, if V is a closed halfspace determined by H and containing K, then, V also contains K ∩ S. Hence H supports K ∩ S, implying that E is a proper exposed face of K ∩ S. Conversely, let E be a proper exposed face of K ∩ S, and let H be a hyperplane supporting K ∩ S such that E = H ∩ (K ∩ S). Since S does not lie in H (otherwise E = K ∩S), the plane H ∩S has dimension dim S−1 (see Theorem 1.19). If V is a closed halfspace determined by H and containing K, then V ∩S is a closed halfplane of S determined by H ∩S and containing K ∩ S. Put G = E ⊕ lin K and H 0 = (H ∩ S) ⊕ lin K. Clearly, H 0 is a hyperplane with the property H 0 ∩ E = K such that the closed halfspace V 0 = (V ∩ S) ⊕ lin K contains K. Hence H 0 properly supports K, and E is a proper exposed face of K. The following result shows that exposed faces of a convex set can be defined locally. Theorem 8.8. Let K ⊂ Rn be a convex set and Bρ (o) ⊂ Rn be a ball of radius ρ > 0 centered at o. A convex subset F of K is an exposed face of K if and only if F is an exposed face of Bρ (F )∩K, where Bρ (F ) = Bρ (o)+F is the ρ-neighborhood of F .
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Proof. Since the cases F = ∅ and F = K is obvious, we may assume that ∅ 6= F 6= K. Put M = Bρ (F ) ∩ K. Let F be a proper exposed face of K. Clearly, F ⊂ Bρ (F ) ∩ K = M . Choose a hyperplane H ⊂ Rn supporting K such that F = H ∩ K. Then F = F ∩ Bρ (F ) = H ∩ K ∩ Bρ (F ) = H ∩ M. Since H ∩rint K = ∅ and Bρ (F ) meets rint K (see Theorem 2.41), it follows that H properly supports M . Therefore, F is a proper exposed face of M . Conversely, supposed that F is a proper exposed face of M , and let H be a hyperplane supporting M such that F = H ∩ M . We state that F = H ∩K. Indeed, assume for a moment the existence of a point x ∈ H ∩K \F . Let z ∈ F . Then [x, z] ⊂ H ∩K by a convexity argument. The line segment [x, z] meets cl F along a closed line segment [x, u]. Clearly, [x, u] ⊂ F and u ∈ (x, z) due to F = K ∩ cl F (see Corollary 8.3). Choosing a point v ∈ (u, z) such that ku − vk 6 ρ, one has v ∈ Bρ (F ) ∩ K \ F , contrary to the assumption F = H ∩M . Hence F = H ∩K. Since F ⊂ rbd M ⊂ rbd K, we conclude that H ∩ rint K = ∅. Summing up, H properly supports K, and F is an exposed face of K. Exposed Faces and Algebra of Convex Sets The following theorem is similar to statement (1) of Theorem 7.3. Theorem 8.9. Let F = {Kα } be a family of convex sets in Rn , with an exposed face Gα chosen in every Kα ∈ F. Then the set G = ∩ Gα is an α exposed face of the convex set K = ∩ Kα . α
Proof. Excluding the trivial cases, we suppose that ∅ 6= G 6= K and Gα 6= Kα for all Kα ∈ F. Denote by H = {Hα } the family of hyperplanes in Rn such that Gα = Hα ∩ Kα for all Kα ∈ F. By Corollary 8.4, every hyperplane Hα properly supports Kα . Denote by Vα the closed halfspace determined by Hα and containing Kα . According to Theorem 1.4, the set L = ∩ Hα is a plane. Furthermore, α
G = ∩ Gα = (∩ Hα ) ∩ (∩ Kα ) = L ∩ K. α
α
α
Put m = dim L. From Theorem 1.20 it follows that H contains a subfamily {H1 , . . . , Hn−m } of hyperplanes whose intersection is L. By Theorem 1.17, we can write Hi = {x ∈ Rn : x · ci = γi }, where ci ∈ Rn is a nonzero vector and γi a scalar, 1 6 i 6 n − m. We may suppose that Vi = {x ∈ Rn : x·ci 6 γi }
for all
16i6n−m
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(otherwise, replace ci with −ci ). Clearly, the closed convex set P = V1 ∩ · · · ∩ Vn−m contains K. Let z ∈ G, and consider the subspaces S = L − z,
Si = Hi − z,
1 6 i 6 n − m.
Obviously, S = S1 ∩ · · · ∩ Sn−m , which gives ⊥ S ⊥ = (S1 ∩ · · · ∩ Sn−m )⊥ = S1⊥ + · · · + Sn−m = span {c1 , . . . , cn−m }.
Hence dim (span {c1 , . . . , cn−m }) = n − dim S = n − m, implying that the set {c1 , . . . , cn−m } is linearly independent. Put c=
1 n−m (c1
+ · · · + cn−m )
and γ =
1 n−m (γ1
+ · · · + γn−m ).
Also, let H = {x ∈ Rn : x·c = γ}
and V = {x ∈ Rn : x·c 6 γ}.
We observe that P ⊂ V . Indeed, if x ∈ P , then x · ci 6 γi for all 1 6 i 6 n − m, which gives x·c =
1 n−m
x·(c1 + · · · + cn−m ) 6
1 n−m (γ1
+ · · · + γn−m ) = γ.
Similarly, L ⊂ H. Since z ∈ G ⊂ L ⊂ H and K ⊂ P ⊂ V , the hyperplane H supports K. Next, we state that H ∩ P = L. By the above argument, it suffices to show the inclusion H ∩ P ⊂ L. Indeed, if x ∈ H ∩ P , then ci · x 6 γi for all 1 6 i 6 n − m, and γ = c·x =
1 n−m (c1
+ · · · + cn−m )·x 6
1 n−m (γ1
+ · · · + γn−m ) = γ.
Hence ci·x = γi for all 1 6 i 6 n−m, which gives x ∈ H1 ∩· · ·∩Hn−m = L. Finally, from G = L ∩ K = (H ∩ P ) ∩ K = H ∩ (P ∩ K) = H ∩ K it follows that G is an exposed face of K. Remark. Unlike extreme faces (see Theorem 7.13), an exposed face of the intersection of convex sets K1 and K2 in Rn may be not expressible as the intersection of exposed faces G1 ⊂ K1 and G2 ⊂ K2 . Indeed, let K1 and K2 be convex sets in the plane, as depicted below. Then {u} is a 0-dimensional exposed face of K1 ∩ K2 = [u, v], while [u, v] is the only exposed face of K1 (and K2 ) containing u.
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K1
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Theorem 8.10. For convex sets K1 , . . . , Kr in Rn and nonzero scalars µ1 , . . . , µr , let K = µ1 K1 + · · · + µr Kr . Then the following statements hold. (1) Any exposed face G of K can be expressed as G = µ1 G1 +· · ·+µr Gr , where every Gi is an exposed face of Ki , 1 6 i 6 r. (2) If, additionally, the exposed face G is nonempty and bounded, then the faces G1 , . . . , Gr above are nonempty and bounded, and are uniquely determined by G. Proof. (1) Since the cases K = Rn and G = ∅ are obvious, we assume that K 6= Rn and G 6= ∅. Let H ⊂ Rn be a hyperplane supporting K such that H ∩ K = G. Then K lies in a closed halfspace, say V , determined by H. Without loss of generality, we may assume that V = {x ∈ Rn : x·c 6 γ}, where c is a nonzero vector in Rn and γ is a certain scalar. Choose a point z ∈ G. Then z = µ1 z1 + · · · + µr zr for certain points zi ∈ Ki , 1 6 i 6 r. Put Hi = {x ∈ Rn : x·c = γi }, where γi = zi ·c, 1 6 i 6 r. Consequently, γ = z·c = (µ1 z1 + · · · + µr zr )·c = µ1 γ1 + · · · + µr γr . We are going to show that Hi supports Ki . For this, consider the closed halfspaces ( {x ∈ Rn : x·c 6 γi } if µi > 0, Vi = 16i6r {x ∈ Rn : x·c > γi } if µi < 0, We state that Ki ⊂ Vi for all 1 6 i 6 r. Indeed, assume the existence of a point yj ∈ Kj \ Vj for some j = 1, . . . , r and put ε = µj (yj ·c − γj ). Clearly, ε > 0. Since the linear functional ϕ(x) = x·c is continuous on Rn (see Exercise 0.4), we can choose points ui ∈ Ki so close to zi that |µi (ui ·c − γi )| < ε/r, Let y =
P
1 6 i 6 r.
µi ui + µj yj . Then y ∈ K and
i6=j
y·c =
P i6=j
µi ui ·c + µj yj ·c >
P i6=j
(µi γi − ε/r) + µj γj + ε > γ,
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in contradiction with K ⊂ V . Since zi ∈ Hi ∩ Ki , the hyperplane Hi supports Ki . Hence the set Gi = Hi ∩ Ki is a exposed face of Ki , 1 6 i 6 r. Finally, we state that G = µ1 G1 + · · · + µr Gr . Indeed, if x ∈ G, then x = µ1 x1 + · · · + µr xr for certain points xi ∈ Ki , 1 6 i 6 r. By the above proved, xi ∈ Vi , which gives µi xi ·c 6 µi γi . Therefore, r r P P γ = x·c = µi xi ·c 6 µi γi = γ, i=1
i=1
implying that xi · c = γi . Hence xi ∈ Hi ∩ Ki = Gi for all 1 6 i 6 r, and G ⊂ µ1 G1 + · · · + µr Gr . Conversely, if x ∈ µ1 G1 + · · · + µr Gr and x = µ1 x1 +· · ·+µr xr for certain points xi ∈ Gi , 1 6 i 6 r, then x ∈ K and x·c = γ, implying that x ∈ H. So, x ∈ G, giving µ1 G1 + · · · + µr Gr ⊂ G. (2) Suppose that G is bounded. According to Theorem 8.2, all sets G, G1 , . . . , Gr are extreme faces of K, K1 , . . . , Kr , respectively. Now, Theorem 7.15 implies that G1 , . . . , Gr are uniquely determined by G. The next result is analogous to Theorem 2.47. Theorem 8.11. If K and M are convex sets and X is a nonempty compact set in Rn such that K + X = M + X, then K = M . Proof. Since the case K = M = ∅ is obvious, we assume that both sets K and M are nonempty. By Theorem 3.17, the set Y = conv X is compact. Furthermore, Theorem 3.12 gives K + Y = K + conv X = conv (K + X) = conv (M + X) = M + conv X = M + Y. From Theorem 2.47 it follows that cl K = cl M . We will prove the equality K = M by induction on m = dim (K + Y ). The case m = 0 is trivial: all three sets K, M and Y are singletons. Let m = 1. The 1-dimensional convex set K + Y may be a line, a halfline, or a segment. We consider each of these possibilities separately. 1. If K + Y is a line, l, then K is a line parallel to l (because Y is bounded). In this case, from cl K = cl M it follows that K = M . 2. Suppose that K + Y is a halfline h with endpoint u. Then cl K is a closed halfline, which is a translate of cl h. Furthermore, u = x + y, where x is an endpoint of cl K and y is the relatively boundary point of Y . Clearly, the inclusion u ∈ K + Y = M + Y holds if and only if both K and M are closed halflines with endpoint x. Therefore, K = M . 3. Suppose that K + Y is a segment. Then cl K = cl M = [u, v] (possibly, u = v). As above, each of the points u, v either belongs to both K and M , or misses both of them. This argument shows that K = M .
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Let the induction statement be true for all m 6 r − 1. Suppose that convex set K, M and a compact convex set Y satisfy the conditions K +Y = M + Y such that dim (K + Y ) = r. As above, cl K = cl M . If both cl K and cl M are planes, then K = M . Hence, we may assume that K and M are not planes. In this case, rbd K = rbd M 6= ∅ (see Corollary 2.57). According to Corollary 2.39, one has rint K = rint M . Hence it suffices to show that K ∩ rbd K = M ∩ rbd M . Let x ∈ K ∩ rbd K. Choose a closed halfspace V ⊂ Rn which properly supports K at x (see Corollary 6.9). Denote by H the boundary hyperplane of V . Let K0 = H ∩ K and M0 = H ∩ M . Also, let V 0 be a translate of V supporting Y . Put Y 0 = H 0 ∩ Y , where H 0 is the boundary hyperplane of V 0 . By Theorem 8.10, one has K0 + Y 0 = M0 + Y 0 . Obviously, K0 and M0 are convex set, and Y 0 is compact. Furthermore, K0 + Y 0 is a proper exposed face of the r-dimensional convex set K + Y , implying that dim (K0 + Y 0 ) 6 r − 1 (see Corollary 8.3). By the induction hypothesis, K0 = M0 . Hence x ∈ M0 = H ∩ M ⊂ M ∩ rbd M . Similarly, M ∩ rbd M ⊂ K ∩ rbd K. Summing up, K = rint K ∪ (K ∩ rbd K) = rint M ∪ (M ∩ rbd M ) = M. Corollary 8.12. If a nonempty compact convex set K ⊂ Rn is the sum of convex sets K1 and K2 , then both sets K1 and K2 are compact. Proof. According to Corollary 2.46, K = K1 + K2 ⊂ K1 + cl K2 ⊂ cl K1 + cl K2 ⊂ cl (K1 + K2 ) = cl K = K. So, K1 + cl K2 = cl K1 + cl K2 . The set K2 , as a summand of K, must be bounded. Therefore, cl K2 is compact. Theorem 8.11 gives K1 = cl K1 . Similarly, K1 is bounded and K2 = cl K2 . Hence both K1 and K2 are compact. Theorem 8.13. For an affine transformation f : Rn → Rm and convex sets K ⊂ Rn and M ⊂ Rm , the following statements hold. (1) If G is an exposed face of M , then f −1 (G) is an exposed face of f −1 (M ). (2) If G is a convex subset of M ∩ rng f such that f −1 (G) is an exposed face of f −1 (M ), then G is an exposed face of M ∩ rng f . (3) A subset G of f (K) is an exposed face of f (K) if and only if f −1 (G) ∩ K is an exposed face of K. (4) If f is one-to-one, then a set F is an exposed face of K if and only if f (F ) is an exposed face of f (K).
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Proof. Without loss of generality, we may assume that f is a linear transformation. Indeed, if f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation, then f (X) = a + g(X) and f −1 (Y ) = f −1 (Y ∩ rng f ) = g −1 ((Y − a) ∩ rng g) = g −1 (Y − a). for all sets X ⊂ Rn and Y ⊂ Rm . (1) Excluding the trivial cases, we assume that G is a proper subset of M . Then G = H ∩ M for a certain hyperplane H ⊂ Rm properly supporting M . Denote by V the closed halfspace of Rm determined by H and containing M . Let G0 = G ∩ rng f,
H 0 = H ∩ rng f,
M 0 = M ∩ rng f,
V 0 = V ∩ rng f.
Since rng f is a plane, the set H 0 also is a plane. Clearly, G0 = H 0 ∩ M 0 . Without loss of generality, we may suppose that G0 is a proper subset of M 0 . Then rng f 6⊂ H 0 , since otherwise G0 = G ∩ rng f = (H ∩ M ) ∩ rng f = M ∩ rng f = M 0 . Therefore, V 0 is a closed halfplane of rng f determined by H 0 and containing M 0 (see Definition 1.36). Hence H 0 properly supports M 0 . According to Theorem 2.59, this argument gives H 0 ∩ rint M 0 = ∅. Due to f −1 (G) = f −1 (G0 ) = f −1 (H 0 ) ∩ f −1 (M 0 ) = f −1 (H) ∩ f −1 (M ) it suffices to show that f −1 (H 0 ) is a hyperplane in Rn properly supporting f −1 (M 0 ). Indeed, Corollary 1.89 shows that f −1 (H 0 ) is a plane, and Theorem 2.34 implies that f −1 (H 0 ) ∩ rint f −1 (M 0 ) = f −1 (H 0 ∩ rint M 0 ) = ∅. Hence the plane f −1 (H 0 ) properly supports f −1 (M 0 ). We state that f −1 (H 0 ) is a hyperplane of Rn . Indeed, Theorem 1.19 shows that dim H 0 = dim (rng f ) − 1. Next, the subspace L = H 0 − H 0 lies in rng f and is a translate of H 0 (see Theorem 1.2). Hence f −1 (L) is a translate of f −1 (H 0 ). Then (see Exercise 0.2) dim f −1 (H 0 ) = dim f −1 (L) = dim L + dim (null f ) = dim (rng f ) − 1 + dim (null f ) = n − 1, which shows that f −1 (H 0 ) is a hyperplane of Rn . (2) Without loss of generality, we suppose that f −1 (G) is a proper exposed face of f −1 (M ). Let H ⊂ Rn be a hyperplane properly supporting f −1 (M ) such that f −1 (G) = H ∩ f −1 (M ). Clearly, translates of null f lie
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in f −1 (G) and H, respectively. Choose a subspace S ⊂ Rn complementary to null f . Then f −1 (M ) = M 0 ⊕ null f,
f −1 (G) = G0 ⊕ null f,
H = H 0 ⊕ null f,
where G0 and M 0 are convex subset of S and H 0 is a plane in S. Clearly, G0 = H 0 ∩ M 0 ,
M = f (M 0 ),
G = f (G0 ),
f (H) = f (H 0 ).
Since the linear transformation h : S → rng f defined by h(x) = f (x) is invertible on S (see page 5), we have G = f (G0 ) = f (H 0 ∩ M 0 ) = f (H 0 ) ∩ f (M 0 ) = f (H 0 ) ∩ M. Furthermore, dim f (H 0 ) = dim H 0 = dim S − 1 = dim (rng f ) − 1. Choose a hyperplane L of Rm with the property L∩rng f = f (H 0 ). Clearly, L supports M ∩ rng f such that G = L ∩ (M ∩ rng f ). Summing up, G is a proper exposed face of M ∩ rng f . (3) Let G be an exposed face of f (K). Without loss of generality, we suppose that G is a proper subset of f (K). Then there is a hyperplane L of Rm which properly supports f (K) such that G = L ∩ f (K). Consequently, L ∩ rint f (K) = ∅. Since K ⊂ f −1 (f (K)), the equality f −1 (G) = f −1 (L) ∩ f −1 (f (K)) gives f −1 (G) ∩ K = f −1 (L) ∩ f −1 (f (K)) ∩ K = f −1 (L) ∩ K. As shown in the proof of statement (1) above, the hyperplane f −1 (L) = f −1 (L ∩ rng ) supports f −1 (K). Hence f −1 (L) also supports f −1 (G) ∩ K, implying that f −1 (G) ∩ K is an exposed face of K. Conversely, let the set C = f −1 (G) ∩ K be a proper exposed face of K. Denote by H the hypersubspace of Rn with the property C = H ∩ K. We state that H can be replaced by another hypersubspace H 0 ⊂ Rn such that C = H 0 ∩ K and null f ⊂ H 0 . Since the case null f ⊂ H is obvious, we assume that null f 6⊂ H. Let L = aff C and express both L and H as direct sums L = L0 + L ∩ null f
and H = H0 + H ∩ null f,
where L0 ⊂ H0 . Clearly, dim L0 6 dim H0 6 dim (rng f ) − 1.
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Put M = L + null f . Then dim M = dim L0 + dim(null f ) 6 n − 1. Denote by S the subspace which is complementary to null f and contains L0 . Choose in S a subspace S0 of dimension n − 1 − dim (null f ) and put H 0 = M + S0 . Clearly, H 0 is a hypersubspace and C = H 0 ∩ K. Since the linear transformation h : S → rng f defined by h(x) = f (x) is invertible on S, we have G = f (C) = f (H 0 ∩ K) = f (H 0 ) ∩ f (K). Furthermore, dim f (H 0 ) = dim H 0 = dim S − 1 = dim (rng f ) − 1. Choose a hypersubspace L of Rm with the property f (H 0 ) = L ∩ rng f . Clearly, G = L ∩ f (K), and, similarly to the above argument, L ∩ rint f (K) = ∅. Summing up, G is a proper exposed face of f (K). (4) Let M = f (K) and E = f (G). Then K = f −1 (M ) and G = f −1 (E) since f is one-to-one. According to statements (1)–(3) above, G is an exposed face of K if and only if f (G) is an exposed face of f (K). Remark. Example on page 266 shows that the assumption on f to be one-to-one is essential in Theorem 8.13. Generated Exposed Faces Theorem 8.9 gives a base to the following definition. Definition 8.14. Let K ⊂ Rn be a convex set and X be a subset of K. The intersection of all exposed faces of K containing X is called the exposed face of K generated by X and denoted GK (X). The corollary below follows from the definitions. Corollary 8.15. For a convex set K ⊂ Rn and subsets X and Y of K, the following statements hold. (1) X ⊂ FK (X) ⊂ GK (X), with X = GK (X) if and only if X is an exposed face of K. In particular, GK (∅) = ∅ and GK (K) = K. (2) GK (X) is the smallest exposed face of K which contains X. (3) GK (GK (X)) = GK (X). (4) GK (conv X) = GK (X). (5) GK (X) ⊂ GK (Y ) if X ⊂ Y .
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Corollary 8.16. For a nonempty convex set K ⊂ Rn and a point u ∈ K, the following statements hold. (1) If u ∈ rbd K, then GK (u) ⊂ K ∩ rbd K. (2) If G is an exposed face of K and u ∈ rint G, then G = FK (u) = GK (u). Proof. (1) By the assumption, the 0-dimensional plane {u} properly supports K. Theorem 6.8 shows the existence of a hyperplane H ⊂ Rn which contains u and properly supports K. Then H ∩ rint K = ∅, implying that the exposed face H ∩ K of K lies in K ∩ rbd K. Consequently, GK (u) ⊂ H ∩ K ⊂ K ∩ rbd K. (2) If u ∈ rint G, then Theorem 7.4 gives G ⊂ FK (u). Since the inclusions FK (u) ⊂ GK (u) ⊂ G follow from the definitions, we conclude that G = FK (u) = GK (u). Remark. Unlike extreme faces (see Theorem 7.9), an exposed face G of a convex set K can be generated by a point from rbd G. Indeed, in example on page 288, the exposed face G = [x, z] is generated by the point x, which belongs to rbd G. The same example shows that x does not belong to the relative interior of an exposed face of K (compare with Corollary 7.12). Corollary 8.17. A convex set K ⊂ Rn has no proper exposed faces if and only if K is relatively open. Consequently, planes are the only nonempty closed convex sets without proper exposed faces. Proof. Corollary 8.16 implies that K has no proper exposed faces if and only if rbd K = ∅, which is exactly the definition of a relatively open set. The second statement is a direct consequence of Corollary 2.57.
8.2
Exposed Representations
Compact Convex Sets Reformulating Definition 8.1 for the case of singletons, we say that x is an exposed point of the convex set K ⊂ Rn if there is a hyperplane H ⊂ Rn such that H ∩ K = {x}. In what follows, exp K denotes the set of all exposed points of K (put exp ∅ = ∅). Theorem 8.2 shows that exp K ⊂ ext K.
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Example. The exposed points of a closed ball Bρ (c) ⊂ Rn are exactly the points of the boundary sphere Sρ (c) (see Exercise 8.1). Example. The exposed points of an r-simplex ∆ = ∆(x1 , . . . , xr+1 ) ⊂ Rn are exactly its vertices, x1 , . . . , xr+1 (see Exercise 8.2). Example. The only exposed point of a simplicial cone Cs (x1 , . . . , xr ) in Rn is its apex s (see Exercise 8.3). Lemma 8.18. If K ⊂ Rn is a compact convex set and W ⊂ Rn is an open halfspace which meets K, then W contains an exposed point of K. Proof. Suppose that W is expressed as W = {x ∈ Rn : x · c > γ} for a certain nonzero vector c and a scalar γ (see Definition 1.28). Choose a point u ∈ K ∩ W . Then µ = u·c > γ. Put d = diam K. Let a scalar λ > 0 satisfy the inequality d2 + 2λ(γ − µ) < 0. Put v = u − λc. The continuous function δv (x) = kx − vk (see Exercise 0.5) attains its maximum value, ρ, on K at a certain point z ∈ K. Equivalently, K ⊂ Bρ (v) and z ∈ Sρ (v). We state that z ∈ W . This is trivial if K ⊂ W . So, we assume that K 6⊂ W . For a point y ∈ K \ W , one has y·c 6 γ < µ, which gives ky − vk2 = ky − u + λck2 = ky − uk2 + 2λ(y − u)·c + λ2 kck2 6 d2 + 2λ(γ − µ) + λ2 kck2 < λ2 kck2 = ku − vk2 6 kz − vk2 = ρ2 . Hence ky − vk < kz − vk, which shows that y 6= z. Therefore, z cannot be in K \ W . Since z belongs to the sphere Sρ (v), the hyperplane H = {x ∈ Rn : (x − v)·(z − v) = 0} supports Bρ (v) such that H ∩ Bρ (v) = {z} (see Exercise 6.2). Hence {z} ⊂ H ∩ K ⊂ H ∩ Bρ (v) = {z}. Consequently, H ∩ K = {z}, and z is an exposed point of K. Theorem 8.19. If K ⊂ Rn is compact convex set, then K = conv (cl (exp K)) = cl (conv (exp K)). Proof. The statement is obvious when K is empty; so, we assume that K 6= ∅. Since the set exp K is bounded, the equality conv (cl (exp K)) = cl (conv (exp K))
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follows from Theorem 3.17. It remains to prove that K = cl (conv (exp K)). The inclusion cl (conv (exp K)) ⊂ K is obvious; hence it suffices to prove the opposite inclusion. Assume for a moment that K does not lie in cl (conv (exp K)) and choose a point u ∈ K \ cl (conv (exp K)). According to Theorem 5.33, there is a closed halfspace V ⊂ Rn containing cl (conv (exp K)) such that u belongs to the opposite open halfspace W = Rn \ V . On the other hand, W contains an exposed point of K (see Lemma 8.18), contrary to the choice of V . Hence K ⊂ cl (conv (exp K)). Corollary 8.20. For a compact convex set K ⊂ Rn , one has exp K ⊂ ext K ⊂ cl (exp K). Furthermore, a subset X of K satisfies any of the condition cl (conv X) = K
and
conv (cl X) = K
(8.1)
if and only if exp K ⊂ cl X. Consequently, both equalities (8.1) hold provided X is a dense subset of exp K. Proof. According to Theorem 7.17, K = conv (ext K). A combination of Theorems 7.18 and 8.19 gives ext K ⊂ cl (exp K). Since any subset X of K is bounded, the equivalence of conditions cl (conv X) = K and conv (cl X) = K follows from Theorem 3.17. Finally, Theorem 7.18 shows that conv (cl X) = K if and only if ext K ⊂ cl X, which is equivalent to the inclusion exp K ⊂ ext K ⊂ cl (cl X) = cl X. Corollary 8.21. If K ⊂ Rn is a nonempty compact convex set of dimension m, then every point x ∈ rint K is a positive convex combination of m + 1 or fewer affinely independent points from exp K. Proof. According to Theorem 8.19, K = cl (conv (exp K)). Therefore, Corollary 2.39 gives rint K = rint (cl (conv (exp K))) = rint (conv (exp K)). Finally, Theorem 3.6 shows that every point x ∈ rint K is a positive convex combination of m + 1 or fewer affinely independent points from exp K. Line-Free Closed Convex Sets Theorem 8.22. For a nonempty closed convex set K ⊂ Rn , the following statements hold.
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(1) exp K 6= ∅ if and only if K is line-free. (2) K has a unique exposed point if and only if K is a line-free convex cone. Proof. (1) If exp K 6= ∅, then from the inclusion exp K ⊂ ext K and Theorem 7.20 it follows that K is line-free. Conversely, let K be line-free. Then rec K is a line-free closed convex cone (see Corollary 5.53) and ap (rec K) = {o}. Choose in rint (nor K) a unit vector c. Since rint (nor K) = rint (rec K)◦ (see Corollary 5.51), we obtain from Theorem 5.48 that the hypersubspace S = {x ∈ Rn : x·c = 0} satisfies the condition S ∩ rec K = {o}. Because c ∈ nor K, Theorem 6.12 shows the existence of a translate, H = {x ∈ Rn : x·c = µ}, of S which supports K at a certain point u ∈ rbd K (so that u·c = µ and x·c 6 µ for all x ∈ K). Due to S = H − H, Theorem 5.54 shows that H ∩ K is a bounded set. Choose a scalar γ < µ and consider the halfspace V = {x ∈ Rn : x·c > γ}. The set K ∩ V is closed as the intersection of closed sets K and V . We state that K ∩ V is compact. Indeed, assume for a moment that K ∩ V is unbounded. By Theorem 5.3, K ∩ V contains a halfline h with endpoint u. Since h lies between the hyperplanes H and H 0 = {x ∈ Rn : x · c = γ}, Theorem 1.34 shows that h is parallel to H. Furthermore, h ⊂ H because the endpoint u of h belongs to H. Hence h ⊂ K ∩ H, implying the unboundedness of K ∩ H, contrary to the above argument. Summing up, K ∩ V is compact. Let d = diam (K ∩ V ). Choose a scalar λ > 0 such that d2 + 2λ(γ − µ) < 0. Put v = u − λc. The continuous function δv (x) = kx − vk (see Exercise 0.5) attains on the compact set K ∩ V its maximum value, ρ, at a certain point z ∈ K ∩ V . We state that z belongs to the open halfspace W = int V . Indeed, for a point y ∈ K ∩ H 0 , one has y ·c = γ < µ, which gives ky − vk2 = ky − u + λck2 = ky − uk2 + 2λ(y − u)·c + λ2 kck2 6 d2 + 2λ(γ − µ) + λ2 kck2 < λ2 kck2 = ku − vk2 6 kz − vk2 = ρ2 . Thus ky − vk < ρ, implying that y 6= z. Hence z ∈ W . Choose an ε > 0 such that Bε (z) ⊂ W and put M = Bε (z) ∩ K. By the above argument, z ∈ M ⊂ K ∩ V ⊂ Bρ (v) such that the point z belongs to the boundary sphere of Bρ (v). The hyperplane H = {x ∈ Rn : (x − v)·(z − v) = 0}
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supports Bρ (v) such that H ∩ Bρ (v) = {z} (see Exercise 6.2). Hence H ∩ M = {z}, and z is an exposed point of M . Finally, Theorem 8.8 shows that z is an exposed point of K. (2) Let K have a unique exposed point u. By Theorem 8.2, u ∈ ext K. We state that u is the only extreme point of K. Indeed, assume for a moment the existence of another point v ∈ ext K. Choose a scalar ρ such that 0 < ρ < ku − vk and consider the compact convex set M = Bρ (v) ∩ K. According to Corollary 8.20, there is a point w ∈ exp M satisfying the condition kv − wk 6 ρ/2. Let H be a hyperplane supporting M such that H ∩ M = {w}. Since w ∈ / bd Bρ (v), Theorem 2.61 shows that w ∈ rbd K. Clearly, Bρ/2 (w) ∩ M = Bρ/2 (w) ∩ K
and H ∩ (Bρ/2 (w) ∩ M ) = {w}.
Hence w is an exposed point of Bρ/2 (w) ∩ K, and Theorem 8.8 shows that w ∈ exp K, contrary to the assumption on the uniqueness of u. Hence u is the only extreme point of K, and Theorem 7.20 implies that K is a convex cone with apex u. Conversely, suppose that K is a line-free convex cone. By Theorem 7.20, K has a unique extreme point u, which is the apex of K. Since the case K = {u} is obvious (then u is a unique exposed point of K), we may assume that K 6= {u}. Consider the cone C = K − u. Then ap C = {o}, and dim C ◦ = n (see Theorem 5.47). Choose a nonzero vector e ∈ rint C ◦ . Corollary 5.49 shows that the (n − 1)-dimensional subspace S = {x ∈ Rn : x · e = 0} supports C such that S ∩ C = {o}. Consequently, the hyperplane H = u + S supports K such that H ∩ K = {u}, implying that u is an exposed point of K. Since every exposed point of K is its extreme point (see Theorem 8.2), and since K has a unique extreme point (see Theorem 7.20), it follows that K has a unique exposed point. The following theorem shows that the family of hyperplanes supporting a given line-free closed convex set at its exposed points is sufficiently large. Theorem 8.23. Let K ⊂ Rn be a nonempty line-free closed convex set and H be the family of hyperplanes in Rn all supporting K such that H ∩ K is a singleton for any choice of H ∈ H. Then the set of normal vectors of the hyperplanes from H is dense in the cone nor K. Consequently, K is the intersection of closed halfspaces containing K and determined by the hyperplanes from H. Proof. For the first statement, it suffices to show that the set of unit normal vectors of the hyperplanes from H is dense in S ∩ rint (nor K),
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where S is the unit sphere of Rn . Choose a unit vector c ∈ rint (nor K). Let the scalars µ, γ, λ, the closed halfspace V , and the point u ∈ K ∩ V be as in the proof of Theorem 8.22. Put v = u − λc and d = diam (K ∩ V ). Denote by z a furthest from v point in K ∩ V . Let ρ = kv − zk, e = (z − v)/kz − vk,
and H = {x ∈ Rn : x·e = ρ}.
Then e is a unit normal vector of the hyperplane H, and H supports K such that {z} = H ∩ K (see the proof of Theorem 8.22). We state that e → c as λ → ∞. First, we observe that |λ − ρ| = ku − vk − kz − vk 6 ku − zk 6 d, which gives λ ρ = lim = 1. λ→∞ ρ λ→∞ λ lim
Next, ku − zk2 = k(u − v) − (z − v)k2 = kλc − ρek2 = λ2 + ρ2 − 2λρ c·e. Therefore, lim kc − ek2 = lim (kck2 + kek2 − 2 c·e)
λ→∞
λ→∞
ku − zk2 − λ2 − ρ2 λ→∞ λρ 2 λ ρ ku − zk − lim − lim = 0. = 2 + lim λ→∞ ρ λ→∞ λ λ→∞ λρ Hence the set of unit normal vectors of the hyperplanes from H is dense in nor K. The second statement follows from Theorem 6.23. = 2 + lim
Definition 8.24. An exposed halfline (also called an exposed ray) of a convex set K ⊂ Rn is a halfline h ⊂ K which is an exposed face of K. In what follows, expr K denotes the union of exposed halflines of K (we put expr K = ∅ if K has no exposed halflines). Remark. Theorem 8.2 shows that every exposed halfline of a convex set K ⊂ Rn is its extreme halfline. Consequently, the endpoint of an exposed halfline of K is an extreme point of K, but not necessarily an exposed point of K. Indeed, let K = B + h ⊂ R2 , where B is the unit ball of R2 and h is the positive halfline of the x-axis. Then both halflines h1 = {(x, −1) : x > 0}
and h2 = {(x, 1) : x > 0}
are exposed halflines of K, while their endpoints, (0, −1) and (0, 1), are extreme but not exposed points of K.
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The following theorem expands Corollary 8.20. Theorem 8.25. If K ⊂ Rn is a line-free closed convex set, then ext K ⊂ cl (exp K)
and
ext K ∪ extr K ⊂ cl (exp K ∪ expr K).
Proof. Since the case K = ∅ is obvious, we assume that K is nonempty. By Theorem 7.20, ext K 6= ∅. Choose a point x ∈ ext K and consider the compact convex set N = Bρ (x) ∩ K, where Bρ (x) is a closed ball of radius ρ > 0 centered at x. From Theorem 7.6 it follows that x ∈ ext N . By Corollary 8.20, ext N ⊂ cl (exp N ). Hence there is a point z ∈ exp N such that kx − zk < ρ/2. From the inclusion Bρ/2 (z) ∩ K ⊂ Bρ (x) ∩ K = N and Corollary 8.3 it follows that z ∈ exp (Bρ/2 (z) ∩ K). Now, Theorem 8.8 gives z ∈ exp K. Since ρ > 0 was chosen arbitrarily, one has x ∈ cl (exp K). Summing up, ext K ⊂ cl (exp K). For the second inclusion, it suffices to prove that extr K ⊂ cl (exp K ∪ expr K).
(8.2)
Choose a point x ∈ extr K. Since the case x ∈ cl (exp K) is obvious, we may suppose that x is not in cl (exp K). Then there is a ball Bδ (x) ⊂ Rn disjoint from cl (exp K). Due to ext K ⊂ cl (exp K), one has Bδ (x) ∩ ext K = ∅. The inclusion x ∈ extr K \ ext K means that the generated extreme face FK (x) is one-dimensional. If K = FK (x), then extr K = expr K = K. So, we may assume that dim K > 2. Choose in FK (x) an open interval (y, z) such that x ∈ (y, z). Let u ∈ rint K, and let H ⊂ Rn be a hyperplane which contains {u, x} but not [y, z]. We observe that x is an extreme point of the closed convex set M = H ∩ K. Indeed, assume for a moment that x belongs to an open segment (r, s) ⊂ M . Then the convex quadrangle Q = conv {r, s, y, z} lies in K such that x ∈ rint Q, which gives dim FK (x) > 2, contrary to x ∈ extr K. Because ext M ⊂ cl (exp M ), there is a point v ∈ exp M such that kx − vk 6 δ. Furthermore, v ∈ / ext K by the choice of Bδ (x). Hence there is an open segment (c, e) ⊂ K which contains v. Clearly, (c, e) does not lie in H, since otherwise v ∈ (c, e) ⊂ H ∩ K = M , contrary to v ∈ exp M . Since v ∈ exp M , there is a hyperplane G ⊂ Rn satisfying the condition G ∩ M = {v}. Obviously, G properly supports M , and G 6= H, implying that the plane G ∩ H is (n − 2)-dimensional (see Theorem 1.19). Furthermore, (G ∩ H) ∩ K = (G ∩ H) ∩ M = {v}.
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According to Theorem 6.8, there is a hyperplane L ⊂ Rn containing G ∩ H and properly supporting K. Then (c, e) ⊂ L (see Corollary 1.32). We state that K ∩ L is a subset of the line hc, ei. Indeed, assume for a moment the existence of a point w ∈ (K ∩L)\hc, ei. Clearly, w ∈ / H due to (K ∩ L) ∩ H = {v}. Hence w belongs an open halfspace W determined by H. Since c and e belong to distinct open halfspaces determined by H, one of these points, say c, does not belong to W . By Theorem 1.31, the open segment (c, w) meets H at a unique point, v 0 . One has v 0 ∈ H ∩K = M and v 6= v 0 due to w ∈ / hc, ei. The latter contradicts the condition L∩M = {v}. Summing up, the exposed face K ∩ L of K is 1-dimensional. Hence the inclusion v ∈ rint (K ∩ L) implies that dim GK (v) = 1. Therefore, v ∈ expr K. Since kx − vk 6 δ, the inclusion (8.2) holds. Remark. For a line-free closed convex set K ⊂ Rn , n > 3, the inclusion extr K ⊂ cl (expr K) does not generally hold (compare with Theorem 8.39 for the case r = 1). For example, let √ M = {(x, y, z) : x2 + y 2 6 2x z, z > 0}. Geometrically, every plane z = a > 0 meets M along the circle {(x, y, a) : (x − a)2 + y 2 6 a2 }. It is easy to see that M is a line-free closed convex set and h = {(0, 0, z) : z > 0} is the only exposed halfline of M . Let e = (0, 1, 0), M 0 = e + M , and K = conv (M ∪ M 0 ). Then the halflines h and h0 = e + h are the only extreme halflines of the line-free closed convex set K. Neither h nor h0 is an exposed halfline of K because the coordinate x-plane is the only plane supporting K at h or h0 . Consequently, extr K = h ∪ h0 and expr K = ∅. Theorem 8.26. For a closed convex set K ⊂ Rn , the following conditions are equivalent. (1) K is line-free. (2) K = conv (cl (exp K ∪ expr K)) = cl (conv (exp K ∪ expr K)). (3) K = conv (cl (exp K)) + rec K = cl (conv (exp K)) + rec K. Proof. Since the case K = ∅ is obvious, we may assume that K is nonempty. (1) ⇒ (2). Because of the inclusion exp K ∪ expr K ⊂ K, the convexity and closedness of K give conv (cl (exp K ∪ expr K)) ⊂ K,
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while a combination of Theorems 7.24 and 8.25 implies the opposite inclusion: K = conv (ext K ∪ extr K) ⊂ conv (cl (exp K ∪ expr K)). The second equality, K = cl (conv (exp K ∪ expr K)), follows from the above argument and Theorem 3.17: K = conv (cl (exp K ∪ expr K)) ⊂ cl (conv (exp K ∪ expr K)) ⊂ K. (1) ⇒ (3). According to Theorems 3.17 and 8.25, one has K = conv (ext K) + rec K ⊂ conv (cl (exp K)) + rec K ⊂ cl (conv (exp K)) + rec K ⊂ K + rec K = K, implying both equalities of (3). (2) ⇒ (1) and (3) ⇒ (1). If K contained a line, then ext K = ∅ (see the proof of Theorem 7.24), implying that exp K = expr K = ∅, which is impossible. Hence K must be line-free. Corollary 8.27. If a line-free closed convex set K ⊂ Rn has no exposed halflines, then K = conv (cl (exp K)) = cl (conv (exp K)). Remark. The converse to Corollary 8.27 statement is not true provided n > 3 (compare with Corollary 7.26). Indeed, if K = conv (cl (exp K)), then K should be line-free, but may contain exposed halflines. For example, let K be the line-free closed convex set given by √ K = {(x, y, z) : (x − z)2 + y 2 6 z, z > 0}. It is easy to see that h = {(0, 0, z) : z > 0} is the unique exposed halfline of K and exp K = {o} ∪ (bd K \ h). At the same time, K = conv (cl (exp K)). For the next result, we recall (see page 273) that a set X ⊂ Rn is coterminal with a halfline h = {u + λv : λ > 0} provided sup {λ : u + λv ∈ X} = ∞. Theorem 8.28. For a line-free closed convex set K ⊂ Rn and a subset X of K, the following statements hold. (1) K = conv (cl X) if and only if exp K ⊂ cl X and cl X is coterminal with every exposed halfline of K. (2) K = conv (cl X) if and only if exp K ⊂ cl X and h = conv (h ∩ cl X) for every exposed halfline of K.
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(3) Any of the equalities K = conv (cl X) + rec K
and
K = cl (conv X) + rec K
is satisfied if and only if exp K ⊂ cl X. Proof. Since the case K = ∅ is obvious, we let K 6= ∅. (1) Assume that K = conv (cl X). Then exp K ⊂ ext K ⊂ cl X and every exposed halfline of K, being an extreme face of K, is coterminal with cl X (see Theorem 7.25). Conversely, suppose that exp K ⊂ cl X and cl X is coterminal with every exposed halfline of K. Then conv (cl X) contains exp K ∪ expr K. By Theorem 8.26, K = conv (cl (exp K ∪ expr K)) ⊂ conv (cl (cl X)) = conv (cl X) = K, which implies that K = conv (cl X). (1) ⇔ (2). Assume first that exp K ⊂ cl X and cl X is coterminal with every exposed halfline of K. Let h be an exposed halfline of K, with endpoint u. Then u ∈ ext K ⊂ cl (exp K) ⊂ cl X, and the condition of coterminality implies that h = conv (h ∩ cl X). Conversely, if an exposed halfline h of K satisfies the condition h = conv (h ∩ cl X), then, obviously, cl X is coterminal with h. (3) If K = conv (cl X) + rec K, then Theorem 7.25 gives exp K ⊂ ext K ⊂ cl X. Conversely, suppose that exp K ⊂ cl X. In this case ext K ⊂ cl (exp K) ⊂ cl X, and Theorem 7.25 implies that K = conv (cl X) + rec K. Now, assume that K = cl (conv X) + rec K. To show the inclusion exp K ⊂ cl X, choose a point u ∈ exp K and a scalar δ > 0. Then u ∈ ext K, and Theorem 7.21 shows the existence of an open halfspace W ⊂ Rn such that u ∈ K ∩ W ⊂ Bδ (u). We state that X meets K ∩ W . Indeed, assume for a moment that X entirely lies in the closed halfspace V = Rn \W . Then cl (conv X) ⊂ K ∩ V . It is easy to see that the cone x + rec K lies in K ∩ V for any choice of x ∈ K ∩ V (see Theorem 5.6). This argument gives K = cl (conv X) + rec K ⊂ K ∩ V , contrary to u ∈ K \ V . Hence X meets K ∩ W . Since δ is chosen arbitrarily, it follows that u ∈ cl X. Summing up, exp K ⊂ cl X. Conversely, if exp K ⊂ cl X, then, by Theorem 3.17 and the above argument, K = conv (cl X) + rec K ⊂ cl (conv X) + rec K ⊂ K, which gives K = cl (conv X) + rec K.
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Remark. In statements (1) and (2) of Theorem 8.28, the equality K = conv (cl X) cannot be replaced with K = cl (conv X). Indeed, let K = {(x, y) : x, y > 0}
and X = {o} ∪ {(x, y) : x > 0, xy = 1}.
Then K = cl (conv X), while cl X is not coterminal with any of the exposed halflines of K. Corollary 8.29. If K ⊂ Rn is a nonempty line-free closed convex set of dimension m, then every point x ∈ rint K is a positive convex combination of m + 1 or fewer affinely independent points from exp K ∪ expr K. Proof. According to Theorem 8.26, K = cl (conv (exp K ∪ expr K)), and Corollary 2.39 gives rint K = rint (cl (conv (exp K ∪ expr K)) ⊂ conv (exp K ∪ expr K). Finally, Theorem 3.6 shows that every point x ∈ rint K is a positive convex combination of m + 1 or fewer affinely independent points from exp K ∪ expr K. Closed Convex Sets A combination of Theorems 5.10, 5.20, and 8.26 implies the following result. Corollary 8.30. Let K ⊂ Rn be a proper closed convex set. If S ⊂ Rn is a subspace complementary to lin K, then K = conv (cl (exp (K ∩ S) ∪ expr (K ∩ S))) ⊕ lin K, K = cl (conv (exp (K ∩ S) ∪ expr (K ∩ S))) ⊕ lin K, K = (conv (cl (exp (K ∩ S))) + rec (K ∩ S)) ⊕ lin K, K = (cl (conv (exp (K ∩ S))) + rec (K ∩ S)) ⊕ lin K, K = conv (cl (exp (K ∩ S))) + rec K, K = cl (conv (exp (K ∩ S))) + rec K. Definition 8.31. Let K ⊂ Rn be a nonempty convex set. A nonempty exposed face F of K ⊂ Rn is called planar (respectively, halfplanar ) if F is a plane (respectively, F is a halfplane). Example. If K ⊂ R3 is a convex cone, given by K = {(x, y, z) : y > |x|}, then ap K = lin K = {(0, 0, z) : z ∈ R} is the only planar exposed face of K, and the halfplanes D1 = {(x, x, z) : x > 0}
and D2 = {(x, −x, z) : x 6 0}
are the halfplanar exposed faces of K.
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Clearly, the planar exposed faces of a line-free closed convex set K ⊂ Rn are its exposed points, and the halfplanar exposed faces of K are its exposed halflines. Theorem 8.32. Let K ⊂ Rn be a nonempty closed convex set expressed as K = (K ∩ S) ⊕ lin K, where S is a subspace of Rn complementary to lin K. The following statements hold. (1) An exposed face F of K is planar if and only if F = x + lin K, where x ∈ exp (K ∩ S). (2) An exposed face F of K is halfplanar if and only if F = h + lin K, where h is an exposed halfline of K ∩ S. Proof. According to Theorem 8.7, an exposed face F of K can be expressed as F = G ⊕ lin K, where G is an exposed face of K ∩ S. (1) If F = x + lin K, where x ∈ exp (K ∩ S), then F is a plane, which is an exposed face of K by the above argument. Conversely, let F a planar exposed face of K. Choose a point x ∈ G. Then x + lin K ⊂ G ⊕ lin K = F. On the other hand, Corollary 5.28 implies the opposite inclusion F ⊂ x + lin K. (2) If F = h + lin K, where h is an exposed halfline of K ∩ S, then F is a halfplane (see Theorem 1.42), which is an exposed face of K by the above argument. Conversely, let F be a halfplanar exposed face of K. Theorem 1.42 shows that F = g + L, where g is a halfline with endpoint o and L is a plane. Denote by h the projection of g on S along lin K. Since F = G ⊕ lin K, it is easy to see that G is a closed halfline, h, with endpoint o (otherwise G = {o} and F is a translate of lin K). Then, by the above argument, F = h + lin K. The corollary below follows from Theorem 8.22 and the argument in the proof of Theorem 8.32. Corollary 8.33. Let K ⊂ Rn be a nonempty closed convex set. An exposed face F of K is planar if and only if F is a translate of lin K. Furthermore, K has exactly one such exposed face, F , if and only if K is a convex cone with ap K = F . Remark. Every nonempty exposed face of a closed convex set K ⊂ Rn contains a translate of lin K (see Theorem 8.7). Therefore, planar exposed faces of K are minimal among all nonempty exposed faces of K. Example
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on page 288 shows that the converse statement is not true: the exposed face [x, z] is minimal, but is not a singleton. The following result expands Theorem 8.26 to the case of closed convex sets in Rn . Theorem 8.34. Each nonempty closed convex set K ⊂ Rn is the closure of the convex hull (or the convex hull of the closure) of the union of its planar and halfplanar exposed faces. Similarly, K is the sum of rec K and the closure of the convex hull (or the convex hull of the closure) of the union of its planar exposed faces. Proof. Express K as a direct sum K = (K ∩ S) ⊕ lin K, where S is a subspace complementary to lin K. By Corollary 8.30, K = conv (cl (exp (K ∩ S) ∪ expr (K ∩ S))) ⊕ lin K = conv (cl (∪ (x + lin K : x ∈ exp (K ∩ S)) ∪ (h + lin K : h is an exposed halfline of K ∩ S))). By Theorem 8.32, every planar exposed face of K is expressible as x+lin K, x ∈ exp K, and every exposed halfplane of K is expressible as h + lin K, where h is an exposed halfline of K ∩ S. Hence K = conv (cl (∪ (G : G is a planar or a halfplanar exposed face of K))). Similarly, K = cl (conv (∪ (G : G is a planar or a halfplanar exposed face of K))). For the second statement, observe first that the equalities K ∩ S = conv (cl (exp (K ∩ S))) + rec (K ∩ S), rec K = rec (K ∩ S) ⊕ lin K give K = conv (cl (exp (K ∩ S)) + rec (K ∩ S)) ⊕ lin K = conv (cl (∪ (x + lin K : x ∈ exp (K ∩ S)))) + rec (K ∩ S) ⊕ lin K = conv (cl (∪ (G : G is a planar exposed face of K))) + rec K. Similarly, K = cl (conv (∪ (G : G is a planar exposed face of K))) + rec K. A combination of Theorems 8.28 and 8.34 implies the corollary below. Corollary 8.35. For a closed convex set K ⊂ Rn and a subset X of K, the following statements hold.
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(1) K = conv (cl X) if and only if G = conv (G∩cl X) for every planar or halfplanar exposed face of K. (2) Any of the equalities K = conv (cl X) + rec K
and
K = cl (conv X) + rec K
is satisfied if and only if G = conv (G ∩ cl X) for every planar exposed face of K. r-Exposed Sets Definition 8.36. Let K be a nonempty convex set in Rn and r be an integer satisfying the inequalities 0 6 r 6 dim K. The union of all exposed faces of K of dimension r or less is called the r-exposed set of K and denoted expr K. The elements of expr K are called r-exposed points of K. Clearly, exp0 K = exp K. Another terminology used for the r-extreme set is the r-skeleton. Theorem 8.37. If K ⊂ Rn is a nonempty convex set of dimension m and r is an integer satisfying the inequalities 0 6 r 6 m, then the following statements hold. (1) expr K ⊂ extr K. (2) exp0 K ⊂ exp1 K ⊂ · · · ⊂ expm−1 K = K ∩ rbd K ⊂ expm K = K. (3) The set K \ expr K is convex and contains the union of relative interiors of exposed faces of K, each of dimension at least r + 1. (4) For a point x ∈ K, the following conditions are equivalent. (a) x ∈ expr K. (b) dim GK (x) 6 r. (c) x ∈ expr (Bρ (x) ∩ K) for every ball Bρ (x) ⊂ Rn , ρ > 0. Proof. (1) Let x be a point in expr K. Then x belongs to an exposed face G of K such that dim G 6 r. Since every exposed face of K also is an extreme face of K (see Theorem 8.2), we conclude that x ∈ extr K. Therefore expr K ⊂ extr K. (2) This part follows from the definition. (3) Let Gr denote the family of all exposed faces of K, each of dimension at most r. By the definition, expr K = ∪ (G : G ∈ Gr ). The equality K \ expr K = K \ (∪ (G : G ∈ Gr )) = ∩ (K \ G : G ∈ Gr )
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shows that K \ expr K is convex as the intersection of convex sets K \ G, G ∈ Gr (see Theorem 7.2). For the second part of (3), let a point u ∈ K belong to the relative interior of an exposed face G of K. Then G = GK (u) according to Corollary 8.16. Hence any exposed face of K which contains u also contains G. So, no exposed face of dimension r or less may contain u. In other words, u ∈ K \ expr K (4) (a) ⇔ (b). Let x ∈ expr K. Then there is an exposed face G of K of dimension r or less such that x ∈ G. By Theorem 8.9, GK (x) is the smallest exposed face of K containing x. Hence dim GK (x) 6 dim G 6 r. Conversely, if dim GK (x) 6 r, then GK (x) is an exposed face of K which contains x and has dimension r or less. Therefore x ∈ expr K. (a) ⇔ (c). Let x ∈ expr K and G be an exposed face of K which contains x and has dimension r or less. By Theorem 8.9, Bρ (x) ∩ G is an exposed face of Bρ (x) ∩ K. Clearly, dim (Bρ (x) ∩ G) 6 dim G 6 r, which shows the inclusion x ∈ expr (Bρ (x) ∩ K). Conversely, let x ∈ expr (Bρ (x) ∩ K) for a certain ball Bρ (x) ⊂ Rn . Choose an exposed face G of Bρ (x)∩K which contains x and has dimension r or less. Let H be a hyperplane supporting Bρ (x) ∩ K such that G = H ∩ (Bρ (x) ∩ K). Then H supports K (see Theorem 6.7). Hence H ∩ K is an exposed face of K containing x. Furthermore, from Theorem 2.41 it follows that dim (H ∩ K) = dim (Bρ (x) ∩ (H ∩ K)) = dim G 6 r. Hence x ∈ expr K. The corollary below is a consequence of Theorem 8.7. Corollary 8.38. Let a convex set K ⊂ Rn of dimension m be expressed as a direct sum K = (K ∩ S) ⊕ lin K, where S is a subspace complementary to lin K. If dim (lin K) = p, then expr K = ∅ for all 0 6 r 6 p − 1, and expp+q K = expq (K ∩ S) ⊕ lin K,
0 6 q 6 m − p.
A generalization of Theorem 8.25 to the case of r-extreme and r-exposed points is given in the next theorem. Theorem 8.39. Let K ⊂ Rn be a closed convex set of dimension m. Then extr K ⊂ cl (expr K)
for all
0 6 r 6 m.
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Proof. The statement is trivial for r = m − 1 and r = m because of extm−1 K = expm−1 K = rbd K
and
extm K = expm K = K
(see Theorems 7.35 and 8.37). So, one may assume that r 6 m − 2. We will prove the statement by induction on r. Let r = 0. If K contains a line, then ext0 K = exp0 K = ∅ (see Corollaries 7.36 and 8.38). If K is line-free, then ext0 K ⊂ cl (exp0 K), as shown in Theorem 8.25. Suppose that the statement holds for all s 6 r − 1, where r > 1, and choose a point x ∈ extr K (the case extr K = ∅ is trivial). Assume for a moment that x ∈ / cl (expr K). Then there is a ball Bδ (x) ⊂ Rn which is disjoint from cl (expr K). Consequently, Bδ (x) ∩ extr−1 K = ∅ because of extr−1 K ⊂ cl (expr−1 K) ⊂ cl (expr K). The inclusion x ∈ extr K \ extr−1 K means that the generated extreme face FK (x) is r-dimensional. If K = FK (x), then extr K = expr K = K. So, we may assume that K 6= FK (x) and dim K > r + 1. As a proper subset of K, the extreme face FK (x) lies in rbd K (see Theorem 7.4). Let L = aff FK (u). Then dim L = r 6 m − 2 and L ∩ rint K = ∅, as follows from Theorem 7.2. Choose a point u ∈ rint K and a plane N ⊂ Rn through {x, u} which is complementary to L. Then dim N = n − r, as shown in Theorem 1.10. We observe that x is an extreme point of the closed convex set M = K ∩ N . Indeed, assume for a moment that x belongs to an open segment (r, s) ⊂ M . Since x ∈ rint FK (x) (see Theorem 7.9), Theorem 3.24 gives x ∈ rint FK (x) ∪ (r, s) ⊂= rint (conv (FK (x) ∪ [r, s])). Consequently, Theorem 7.4 shows that conv (FK (x) ∪ [r, s]) ⊂ FK (x), contrary to the condition FK (x) ∩ [r, s] ⊂ L ∩ N = {x}. Due to the inclusion ext0 M ⊂ cl (exp0 M ), there is a point v ∈ exp0 M such that kx−vk 6 δ. Let H ⊂ Rn be a hyperplane satisfying the condition H ∩ M = {v}. Clearly, H properly supports M . Hence N 6⊂ H, and the plane H ∩ N is (n − r − 1)-dimensional (see Theorem 1.19). Furthermore, (H ∩ N ) ∩ K = (H ∩ N ) ∩ M = {v}. Because H ∩ K is an exposed face of K containing v, and because v ∈ rint FK (v), Corollary 8.3 implies that FK (v) ⊂ H ∩ K. Furthermore, dim (H ∩ K) > dim GK (v) > r + 1
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due to the assumption v ∈ / expr K. Put Q = aff (H ∩ K). Then dim Q = dim (H ∩ K) > r + 1. Both planes H ∩ N and Q lie in H. Hence their sum lies in the hyperplane 2H. A combination of Theorems 1.6 and 1.7 gives dim (N ∩ Q) = dim (H ∩ N ) + dim Q − dim (H ∩ N + Q) > dim (H ∩ N ) + dim Q − dim (2H) = n − r − 1 + dim Q − (n − 1) = dim Q − r. Because (N ∩ Q) ∩ K ⊂ (H ∩ N ) ∩ K = {v}, there is a plane T ⊂ Q of dimension dim Q − 1 which contains N ∩ Q and supports H ∩ K = K ∩ Q. By the above, dim T − dim (N ∩ Q) 6 (dim Q − 1) − (dim Q − r) = r − 1. Finally, since T contains the exposed face H ∩ K of K, one has FK (v) ⊂ T . Because v ∈ rint FK (v) and (N ∩ Q) ∩ FK (v) ⊂ (N ∩ Q) ∩ K = {v}, Theorem 2.32 gives dim FK (v) = dim (FK (v) ∪ (N ∩ Q)) − dim (N ∩ Q) + dim (FK (v) ∩ (N ∩ Q)) 6 dim T − dim (N ∩ Q) + dim {v} 6 r − 1. Hence v ∈ Bδ (x) ∩ extr−1 K, contrary to the assumption. Summing up, x ∈ cl (expr K), and the inclusion extr K ⊂ cl (expr K) holds. The next result generalizes Theorem 8.26 and Corollary 8.27 (compare with Theorem 7.39). Theorem 8.40. For a nonempty closed convex set K ⊂ Rn of dimension m and an integer 0 6 r 6 m, the following statements hold. (1) If dim (lin K) 6 r, then K = conv (cl (expr+1 K)) = cl (conv (expr+1 K)). (2) If dim (lin K) 6 r and K has no exposed halfplane of dimension greater than r, then K = conv (cl (expr K)) = cl (conv (expr K)).
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Proof. Let K = (K ∩ S) ⊕ lin K, where S is a subspace complementary to lin K (see Theorem 5.20). (1) Let p = dim (lin K), where p 6 r. Since K ∩ S is a line-free closed convex set, Theorem 8.26 gives K ∩ S = conv (cl (exp (K ∩ S) ∪ expr (K ∩ S))). By the definitions, exp (K ∩ S) ∪ expr (K ∩ S) ⊂ exp1 (K ∩ S), Therefore, Corollary 8.38 shows that K = (K ∩ S) ⊕ lin K ⊂ conv (cl (exp1 (K ∩ S))) ⊕ lin K = conv (cl (expp+1 K)) ⊂ conv (cl (expr+1 K)) ⊂ K. Hence, K = conv (cl (expr+1 K)). Similarly, K = cl (conv (expr+1 K)). (2) If K has no exposed halfplane of dimension greater than r, then Theorem 8.7 shows that the set K ∩ S has no exposed faces of positive dimension. By Corollary 8.27, K ∩ S = conv (cl (exp0 (K ∩ S))) = cl (conv (exp0 (K ∩ S))). Therefore, Corollary 8.38 shows that K = (K ∩ S) ⊕ lin K ⊂ conv (cl (exp0 (K ∩ S))) ⊕ lin K = conv (cl (expp K)) ⊂ conv (cl (expr K)) ⊂ K. Hence, K = conv (cl (expr K)). Similarly, K = cl (conv (expr K)). The next result provides a generalization of Corollary 8.29. Theorem 8.41. Let K ⊂ Rn be a nonempty closed convex set of dimension m, with dim (lin K) = r. Then every point x ∈ rint K is a positive convex combination of m−r+1 or fewer affinely independent points from expr+1 K. Proof. Let K = (K ∩ S) ⊕ lin K, where S is a subspace complementary to lin K (see Theorem 5.20). Then K ∩ S is a line-free closed convex set of dimension m − r. Choose a point x ∈ rint K and express it as x = y + z, where y ∈ rint (K ∩ S) and z ∈ lin K (see Corollary 5.31). By Corollary 8.29, y can be written as a positive convex combination y = λ1 y1 + · · · + λp yp , where p 6 m − r + 1 and y1 , . . . , yp are affinely independent points in exp (K ∩ S) ∪ expr (K ∩ S). Let xi = yi + z, 1 6 i 6 p. Clearly, the set
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{x1 , . . . , xp } is affinely independent and xi ∈ yi + lin K for all 1 6 i 6 p. Since exp (K ∩ S) ∪ expr (K ∩ S) ⊂ exp1 (K ∩ S), Corollary 8.38 implies that x1 , . . . , xp ∈ exp1 K ⊕ lin K = expr+1 K, and the equality x = y + z = λ1 y1 + · · · + λp yp + z = λ1 (y1 + z) + · · · + λp (yp + z) = λ1 x1 + · · · + λp xp shows that x is a positive convex combination of p (6 m − r + 1) affinely independent points from expr+1 K. Exercises for Chapter 8 Exercise 8.1. Show that the exposed faces of the closed ball Bρ (c) ⊂ Rn are ∅, all singletons from the boundary sphere Sρ (c), and the ball itself. Exercise 8.2. Let ∆ = ∆(x1 , . . . , xr+1 ) be an r-simplex in Rn . Show that the exposed faces of ∆ are ∅ and all simplices ∆(z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr+1 }. Exercise 8.3. Show that the exposed faces of a simplicial cone Cs (x1 , . . . , xr ) in Rn are the empty set, {s}, and all simplicial cones Cs (z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr }. Exercise 8.4. Let K ⊂ Rn be a closed convex set. Show that the following statements hold. (1) K exactly two exposed faces if and only if K is a plane. (2) K has exactly three exposed faces if and only if K is a closed halfplane. (3) K has exactly four exposed faces if and only if K is a plane closed slab. Notes for Chapter 8 Generated exposed faces. Brown [42] showed that for every convex body K ⊂ Rn , n 6 5, and an exposed face G of K, either there is an exposed point x of K that belongs to rbd G, or there is a point x ∈ rbd G such that GK (x) = G. It is an open question whether this statement holds for all n > 5.
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Exposed structure of the sum of convex sets. The following result is proved in Soltan [200]. Let a line-free closed convex set A ⊂ Rn be the sum of closed convex sets B and C. Then both sets B0 = B + rec A and C0 = C + rec A are closed and satisfy the following conditions: (a) for every point a ∈ exp A there are unique points b ∈ exp B0 and c ∈ exp C0 such that a = b + c, (b) the sets expC B = {x ∈ exp B : ∃ y ∈ exp C such that x + y ∈ exp A}, expB C = {x ∈ exp C : ∃ y ∈ exp B such that x + y ∈ exp A} are dense in exp B0 and exp C0 , respectively. De Wilde [70] obtained some separation results that involve exposed points of line-free closed convex sets. Topological properties of exposed faces. Klee [125] gave an example of a convex body K ⊂ R3 such that exp K is not a Gδ -set, and of a convex body M ⊂ R3 such that all sets bd M \ ext M , ext M \ exp M , and exp M are dense in bd M . Corson [62], negatively answering a question from [125], gave an example of a convex body K ⊂ R3 such that exp K is of the first category. Choquet, Corson, and Klee [56] showed that for every closed convex set K ⊂ Rn , the set exp K is the union of an Fσ -set, a Gδ -set, and n − 2 sets each of which is the intersection of an Fσ -set and a Gδ -set. Holick´ y and Laczkovich [116], negatively answering a question from [56], gave an example of a convex body K ⊂ R3 for which exp K is not the intersection of an Fσ -set and a Gδ -set. Theorem 8.39 is proved by Asplund [7] for the case of compact convex sets. Reiter and Stavrakas [182] proved that the metric space of all exposed faces of a convex body K ⊂ Rn is compact (in the Hausdorff metric) if and only if all sets expr K are compact, 0 6 r 6 dim K − 1. Exposed representations. Theorem 8.19 is proved by Straszewicz [212]. The equality K = cl (conv (exp K ∪ expr K)) in Theorem 8.26 is due to Klee [125]. Soltan [202] refined Theorem 8.19 by showing that every compact convex set K ⊂ Rn is the closure of the convex hull of its exposed diameters. We recall that an exposed diameter of K is a chord [a, c] of K such that there are parallel hyperplanes Ha and Hc satisfying the conditions K ∩Ha = {a} and K ∩Hc = {c} (see, e. g., the survey [198]). A point x of a nonempty compact convex set K ⊂ Rn is called bare provided there is a closed ball Bρ (c) ⊂ Rn containing K such that x belongs to the boundary sphere of Bρ (c). Clearly, every bare point of K is its exposed point, but not converse. Berberian [22] and Orland [166] (for n = 2) proved that a nonempty compact convex set K ⊂ Rn is the closed convex hull of its bare points.
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Polyhedra
9.1
Intersections of Halfspaces
Representations with Halfspaces Definition 9.1. A set P ⊂ Rn is called a polyhedron if either P = Rn or P the intersection of finitely many closed halfspaces, say V1 , . . . , Vr : P = V1 ∩ · · · ∩ Vr .
(9.1)
In the latter case, we say that the family F = {V1 , . . . , Vr } represents P . A bounded convex polyhedron is called a polytope. The empty set ∅ is a polyhedron.
@ @ @ P @ @ Fig. 9.1
P HH H HH H
A polytope and a polyhedron in the plane.
Example. Every plane L in Rn is a polyhedron. Indeed, eliminating the trivial cases L = ∅ and L = Rn , suppose that L is a proper plane. Then L is the intersection of n − dim L + 1 closed halfspaces (see Exercise 9.1). Example. Every closed halfplane D in Rn is a polyhedron. Since the case when D is a halfspace is obvious, we may assume that dim D = m 6 n − 1. 319
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Then D is the intersection of the plane aff D and a certain closed halfspace. Because aff D has dimension m (see Corollary 1.80), it can be expressed as the intersection of n − m + 1 closed halfspaces (see the above example). Hence D is the intersection of n − m + 2 closed halfspaces. Example. Every simplex ∆ = ∆(x1 , . . . , xr+1 ) in Rn is a polytope. Indeed, if r = 0, then ∆ = {x1 } is the intersection of n + 1 closed halfspaces (see the first example above). Suppose that r > 1. Then a combination of Definition 2.5 and Theorem 1.71 shows that ∆ is the intersection of r + 1 closed halfplanes Di = {λ1 x1 + · · · + λr+1 xr+1 : λ1 + · · · + λr+1 = 1, λi > 0}, 1 6 i 6 r + 1. If r = n, then every Di is a halfspace, and ∆ is the intersection of n + 1 closed halfspaces. Let 1 6 r 6 n−1. Then every Di is the intersection of the r-dimensional plane aff ∆ and a certain closed halfspace Vi , 1 6 i 6 r + 1. So, we can write ∆ = aff ∆ ∩ V1 ∩ · · · ∩ Vr . Since aff ∆ is the intersection of n−r+1 closed halfspaces (see the first example above), ∆ is the intersection of (n − r + 1) + (r + 1) = n + 2 closed halfspaces. Example. Every simplicial cone C = Cs (x1 , . . . , xr ) in Rn is a polyhedron. Indeed, from Definition 4.5 and Theorem 1.71 it follows that C is the intersection of r closed halfplanes Di = {(1 − λ1 − · · · − λr )s + λ1 x1 + · · · + λr xr : λi > 0}, 1 6 i 6 r. If r = n, then every Di is a halfspace, and C is the intersection of n closed halfspaces. Let 1 6 r 6 n − 1. Then every Di is the intersection of the r-dimensional plane aff C and a certain closed halfspace Vi , 1 6 i 6 r. So, C = aff C ∩ V1 ∩ · · · ∩ Vr . Since the r-dimensional plane aff C is the intersection of n − r + 1 closed halfspaces (see the first example above), C is the intersection of (n − r + 1) + r = n + 1 closed halfspaces. Corollary 9.2. If P1 , . . . , Pr are polyhedra in Rn , then their intersection is a polyhedron. Consequently, a plane section of a polyhedron P ⊂ Rn is again a polyhedron. Proof. By the definition, each Pi is the intersection of a finite family Fi of closed halfspaces, 1 6 i 6 r. Then P1 ∩ · · · ∩ Pr is a polyhedron as the intersection of the finite family F1 ∪ · · · ∪ Fr of closed halfspaces. If L ⊂ Rn is a subspace meeting P , then P ∩ L is a polyhedron because L is a polyhedron (see the example above).
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Corollary 9.3. Every polyhedron P ⊂ Rn is a closed convex set. Proof. Excluding the trivial cases P = ∅ and P = Rn , we assume that P is a proper subset of Rn . Then P is the intersection of finitely many closed halfspaces. Since every closed halfspace is a closed convex set (see Theorem 1.29 and example on page 72), and since the intersection of every family of closed convex sets is again a closed convex set (see, respectively, page 8 and Theorem 2.8), it follows that P is closed and convex. Theorem 9.4. Let a proper polyhedron P ⊂ Rn be represented by a family F = {V1 , . . . , Vr } of closed halfspaces. Then F can be partitioned into two subfamilies, say G = {V1 , . . . , Vs } and H = {Vs+1 , . . . , Vr }, such that the following statements hold (let s = 0 if P is a plane and s = r if dim P = n). (1) Every set Di = Vi ∩ aff P , 1 6 i 6 s, is a closed halfplane of aff P , and P = D1 ∩ · · · ∩ Ds = V1 ∩ · · · ∩ Vs ∩ aff P.
(9.2)
(2) Furthermore, rint P = rint D1 ∩ · · · ∩ rint Ds = int V1 ∩ · · · ∩ int Vs ∩ aff P,
(9.3)
rbd P = (P ∩ rbd D1 ) ∪ · · · ∪ (P ∩ rbd Ds ) = (P ∩ bd V1 ) ∪ · · · ∪ (P ∩ bd Vs ).
(9.4)
(3) Every halfspace Vi , s + 1 6 i 6 r, contains aff P , and aff P = Vs+1 ∩ · · · ∩ Vr .
(9.5)
Proof. First, we consider the special cases when P is a plane or dim P = n. If P is a plane, then the equalities (9.2), (9.3), and (9.4) are excluded because of s = 0, and (9.5) becomes (9.1). Similarly, if dim P = n, then (9.5) is excluded because of s = r. Furthermore, aff P = Rn , Di = Vi , 1 6 i 6 r, and (9.2) becomes (9.1). The equalities (9.3) and (9.4) hold due to int P = int (V1 ∩ · · · ∩ Vr ) = int V1 ∩ · · · ∩ int Vr , bd P = bd (V1 ∩ . . . Vr ) = (P ∩ bd V1 ) ∪ · · · ∪ (P ∩ bd Vs ), and (9.4) is excluded since s = r. So, we may assume that P is not a plane and dim P 6 n − 1. (1) Denote by G the family of all halfspaces from F which do not contain aff P . We observe that G = 6 ∅, since otherwise P ⊂ aff P ⊂ V1 ∩ · · · ∩ Vr = P,
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contrary to the assumption P 6= aff P . Without loss of generality, we may suppose that G = {V1 , . . . , Vs }, 1 6 s 6 r. Then every set Di = Vi ∩ aff P , 1 6 i 6 s, is a closed halfplane of aff P (see Definition 1.36). Clearly, P ⊂ D1 ∩ · · · ∩ Ds = V1 ∩ · · · ∩ Vs ∩ aff P. For the opposite inclusion, choose a point z ∈ aff P \ P . Then z ∈ aff P \ Vi for a certain halfspace Vi ∈ F. Therefore, Vi ∈ G, which gives V1 ∩ · · · ∩ Vs ∩ aff P ⊂ P. Summing up, the equalities (9.2) hold. (2) For all 1 6 i 6 s, the inclusions P ⊂ Di ⊂ aff P give aff P = aff Di (see Theorem 1.50); therefore, rint P ⊂ rint Di according to Theorem 2.15. Since rint P 6= ∅, a combination of Theorem 2.26 and Corollary 1.38 implies rint P = rint (D1 ∩ · · · ∩ Ds ) = rint D1 ∩ · · · ∩ rint Ds = (int V1 ∩ aff P ) ∩ · · · ∩ (int Vs ∩ aff P ) = int V1 ∩ · · · ∩ int Vs ∩ aff P. The equalities (9.4) follow from (9.3) (see Theorem 2.61 and Corollary 2.62): rbd P = (P ∩ rbd D1 ) ∪ · · · ∪ (P ∩ rbd Ds ) = (P ∩ (bd V1 ∩ aff P )) ∪ · · · ∪ (P ∩ (bd Vs ∩ aff P ) = (P ∩ bd V1 ) ∪ · · · ∪ (P ∩ bd Vs ). (3) First, we state that the family H = F \ G is nonempty. For this, choose a point u ∈ rint P . There should be at least on halfspace Vi ∈ F with the property u ∈ bd Vi . Indeed, otherwise u ∈ int V1 ∩ · · · ∩ int Vr = int (V1 ∩ · · · ∩ Vr ) = int P, contrary to the assumption dim P 6 n−1 (see Corollary 2.18). If a halfspace Vi ∈ F satisfies the condition u ∈ bd Vi , then Theorem 2.59 shows that P ⊂ bd Vi . Because bd Vi is a hyperplane, one has aff P ⊂ bd Vi ⊂ Vi . Summing up, H 6= ∅. Equivalently, s < r. Denote by H0 the family of all halfspaces Vi from H with the property u ∈ bd Vi . As shown above, H0 6= ∅, and aff P ⊂ bd Vi for every Vi ∈ H0 . Renumbering the elements of H, we suppose that H0 = {Vs+1 , . . . , Vt }, where s + 1 6 t 6 r. Then aff P ⊂ bd Vs+1 ∩ · · · ∩ bd Vt ⊂ Vs+1 ∩ · · · ∩ Vt . We state that aff P = bd Vs+1 ∩ · · · ∩ bd Vt = Vs+1 ∩ · · · ∩ Vt .
(9.6)
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Assume for a moment the existence of a point v ∈ (Vs+1 ∩ · · · ∩ Vt ) \ aff P. Then [u, v] ⊂ Vs+1 ∩ · · · ∩ Vt by a convexity argument, while (u, v) ∩ aff P = ∅ (see Theorem 1.31). Furthermore, u ∈ int Vi for every halfspace Vi ∈ F \ H0 . Indeed, u ∈ int Vi for all 1 6 i 6 s because of (9.3), and u ∈ int Vi for all i > t + 1 (if t 6 r − 1) by the choice of H0 . Therefore, there is a scalar ρ > 0 such that Bρ (u) ⊂ Vi for all Vi ∈ F \ H0 . Choose a point w ∈ (u, v) such that ku − wk 6 ρ. By the above argument, w ∈ ∩ (Vi : Vi ∈ H0 ) ∩ Bρ (u) ⊂ ∩ (Vi : Vi ∈ H0 ) ∩ (Vi : Vi ∈ F \ H0 ) = V1 ∩ · · · ∩ Vr = P, in contradiction with (u, v) ∩ aff P = ∅. Summing up, the equalities (9.6) hold. Finally, if t 6 r − 1, then every halfspace Vi , t + 1 6 i 6 r, contains aff P , which gives aff P = Vs+1 ∩ · · · ∩ Vt = Vs+1 ∩ · · · ∩ Vr . Remark. We observe that an infinite family of closed halfplanes whose intersection is a polyhedron may not contain a finite subfamily with this property. For example, the triangle P = {(x, y) : x > 0, y > 0, x + y 6 1} ⊂ R2 is the intersection of the infinite family F of closed halfplanes V = {(x, y) : x > 0}, V 0 = {(x, y) : y > 0}, Vk = {(x, y) : y 6
k+1 k (1
− x)},
k > 1.
Clearly, F does not contain a minimal subfamily. Corollary 9.5. A proper set P ⊂ Rn is a polyhedron if and only if P is either a plane or a finite intersection of closed halfplanes of aff P . Proof. If a proper set P ⊂ Rn is a polyhedron, then, according to Theorem 9.4, P is either a plane or a finite intersection of closed halfplanes of aff P . Conversely, let P be either a plane or a finite intersection of closed halfplanes of aff P . If P is a plane, then P is a polyhedron as a finite intersection of closed halfspaces (see Exercise 9.1). Suppose that P is the intersection of closed halfplanes D1 , . . . , Dp of aff P . If dim P = n, then aff P = Rn and these halfplanes are halfspaces, implying that P is a polyhedron. Let dim P 6 n − 1. Every halfplane Di can be expressed as
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Di = Vi ∩ aff P , where Vi is a certain closed halfspace, 1 6 i 6 p. Denote by {Vp+1 , . . . , Vq } a family of closed halfspaces expressing aff P (see Exercise 9.1). Then P = D1 ∩ · · · ∩ Dp = (V1 ∩ · · · ∩ Vp ) ∩ aff P = (V1 ∩ · · · ∩ Vp ) ∩ (Vp+1 ∩ · · · ∩ Vq ), implying that P is a polyhedron. Irreducible Representations Definition 9.6. A family F = {V1 , . . . , Vr } of closed halfspaces representing a proper polyhedron P ⊂ Rn is called irreducible provided either r = 1, or r > 2 and P 6= V1 ∩ · · · ∩ Vi−1 ∩ Vi+1 ∩ · · · ∩ Vr
for all
1 6 i 6 r.
Similarly, if P is not a plane, then a family D = {D1 , . . . , Ds } of closed halfplanes of aff P representing P is called irreducible if either s = 1, or s > 2 and P 6= D1 ∩ · · · ∩ Di−1 ∩ Di+1 ∩ · · · ∩ Ds
for all
1 6 i 6 s.
Remark. If a family F = {V1 , . . . , Vr } of closed halfspaces represents a proper polyhedron P ⊂ Rn , then, by a finiteness argument, F contains an irreducible subfamily. Theorem 9.7. Let a proper polyhedron P ⊂ Rn be represented by a family F = {V1 , . . . , Vr } of closed halfspaces. If F is partitioned into subfamilies, say G = {V1 , . . . , Vs } and H = {Vs+1 , . . . , Vr }, satisfying conditions (1)–(3) of Theorem 9.4, then the following statements hold (let s = 0 if P is a plane and s = r if dim P = n). (1) F is irreducible if and only if both families G and H are irreducible under the conditions (9.2) and (9.5), respectively. (2) If P is not a plane, then G is irreducible if and only if either s = 1, or s > 2 and bd Vi ∩ rint (V1 ∩ · · · ∩ Vi−1 ∩ Vi+1 ∩ · · · ∩ Vr ) 6= ∅, 1 6 i 6 s. (9.7) Equivalently, the family D = {D1 , . . . , Ds } of closed halfplanes of aff P , where Di = Vi ∩ aff P , 1 6 i 6 s, is irreducible if and only if either s = 1, or s > 2 and rbd Di ∩ rint (D1 ∩ · · · ∩ Di−1 ∩ Di+1 ∩ · · · ∩ Ds ) 6= ∅ for all 1 6 i 6 s.
(9.8)
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(3) If dim P 6 n − 1, then H is irreducible if and only if aff P = bd Vs+1 ∩ · · · ∩ bd Vr = Vs+1 ∩ · · · ∩ Vr
(9.9)
and no set Ci = Vs+1 ∩ · · · ∩ Vi−1 ∩ Vi+1 ∩ · · · ∩ Vr , s + 1 6 i 6 r, is a plane. Proof. (1) This part of the proof follows from Theorem 9.4 and the method the subfamilies G and H were chosen. (2) Let G be irreducible. Suppose for a moment that s > 2 and at least one of the halfspaces from G, say V1 , does not satisfy the condition (9.7). Then rint (V2 ∩ · · · ∩ Vs ) ⊂ int V1 , and Theorem 2.38 gives V2 ∩ · · · ∩ Vs = cl (rint (V2 ∩ · · · ∩ Vs )) ⊂ cl (int V1 ) = V1 . Consequently, P = V2 ∩ · · · ∩ Vs ∩ aff P , contrary to the irreducibly of G. Conversely, if s > 2 and G satisfies the conditions (9.7), then no halfspace Vi , 1 6 i 6 s, can be omitted in the representation P = V1 ∩ · · · ∩ Vr , implying that G is irreducible. An equivalent statement for the halfplanes Di follows from the relations rbd Di = bd Vi ∩ aff P , 1 6 i 6 s (see Corollary 2.62). (3) Suppose first that H is irreducible. Then (9.9) follows from the proof of statement (3) of Theorem 9.4. From (9.7) one has Ci 6⊂ Vi for all s + 1 6 i 6 r. Assume for a moment that a certain set Ci , s + 1 6 i 6 r, is a plane. Then Ci contains a point u ∈ int Vi (see Corollary 1.37). On the other hand, (9.9) implies that u ∈ Ci ∩ int Vi ⊂ Ci ∩ Vi = aff P = bd Vs+1 ∩ · · · ∩ bd Vr ⊂ bd Vi , a contradiction. Conversely, suppose that (9.9) holds, and that no set Ci , s + 1 6 i 6 r, is a plane. Since Vs+1 ∩ · · · ∩ Vr is a plane, one has Ci 6= Vs+1 ∩ · · · ∩ Vr = aff P,
s + 1 6 i 6 r.
Therefore, the family H is irreducible. Theorem 9.8. If a polyhedron P ⊂ Rn of positive dimension is not a plane, then there is a unique family of closed halfplanes of aff P irreducibly representing P .
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Proof. Let G = {D1 , . . . , Ds } and G 0 = {D10 , . . . , Dt0 } be families of closed halfplanes of aff P both irreducibly representing P : P = D1 ∩ · · · ∩ Ds = D10 ∩ · · · ∩ Dt0 . By a symmetry argument, it suffices to show that G ⊂ G 0 . Choose a halfplane Di ∈ G. Clearly, rbd Di is a plane of dimension m − 1, where m = dim P (see example on page 104). Consider the convex set Ki = rbd Di ∩P . We state that Ki ⊂ rbd P
and
aff Ki = rbd Di .
Indeed, the inclusion Ki ⊂ rbd P follows from (9.4). Let Ci = D1 ∩ · · · ∩ Di−1 ∩ Di+1 ∩ · · · ∩ Ds . The inclusion P ⊂ Ci ⊂ aff P gives aff Ci = aff Pi (see Theorem 1.50). Furthermore, Ki = rbd Di ∩ P = (rbd Di ∩ Di ) ∩ Ci = rbd Di ∩ Ci . Since the condition (9.8) can be rewritten as rbd Di ∩ rint Ci 6= ∅, Theorem 2.31 gives aff Ki = rbd Di ∩ aff Ci = rbd Di ∩ aff P = rbd Di . Consequently, dim Ki = dim (aff Ki ) = dim (rbd Di ) = m − 1. Now, choose a point u ∈ rint Ki . Since u ∈ Ki ⊂ rbd P , the equality rbd P = (P ∩ rbd D10 ) ∪ · · · ∪ (P ∩ rbd Dt0 ) implies the existence of a halfplane Dj0 ∈ G 0 such that u ∈ P ∩ rbd Dj0 . By Theorem 2.59, Ki ⊂ rbd Dj0 . Because dim Ki = dim (rbd Dj0 ) = m − 1, Corollary 1.77 shows that rbd Di = aff Ki = rbd Dj0 . Finally, since both closed halfplanes Di and Dj0 contain P and are bounded by the same (m − 1)-dimensional subplane rbd Di = rbd Dj0 of aff P , Corollary 1.38 gives Di = Dj0 . So, Di ∈ G 0 , implying the inclusion G ⊂ G0. Remark. A similar to Theorem 9.8 statement does not hold in the case of irreducible representations by halfspaces. For example, the halfplane P = {(x, y, 0) : x > 0} ⊂ R3 can be irreducibly expressed by closed halfspaces in distinct ways. For instance, P = V1 ∩ V2 ∩ V3
and P = V1 ∩ V2 ∩ V30 ,
where V1 and V2 are given by the inequalities z > 0 and z 6 0, respectively; V3 = {(x, y, z) : z 6 x}
and V30 = {(x, y, z) : z 6 −x}.
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Theorem 9.9. If a proper plane L ⊂ Rn of dimension m is irreducibly represented by a family H = {V1 , . . . , Vp } of closed halfspaces, then n − m + 1 6 p 6 2(n − m). Proof. Since the family H is irreducible, one has L = bd V1 ∩· · ·∩bd Vp according to Theorem 9.7. Choose a vector u ∈ L and translate all V1 , . . . , Vp on the same vector −u. Then L becomes an m-dimensional subspace and every Vi can be written as Vi = {x ∈ Rn : x·ci 6 0} = {ci }◦ , 1 6 i 6 p. By Theorem 5.46, L = V1 ∩ · · · ∩ Vp = {c1 }◦ ∩ · · · ∩ {cp }◦ = {c1 , . . . , cp }◦ , and a combination of Theorem 5.44 and Corollary 4.39 gives L⊥ = L◦ = ({c1 , . . . , cp }◦ )◦ = cl (coneo {c1 , . . . , cp }) = coneo {c1 , . . . , cp }. Hence coneo {c1 , . . . , cp } is an (n−m)-dimensional subspace. Corollary 4.45 shows the existence of a set Y ⊂ {c1 , . . . , cp } such that coneo Y = coneo {c1 , . . . , cp } and n − m + 1 6 card Y 6 2(n − m). Let Y = {ci1 , . . . , cir }, where r = card Y and 1 6 i1 < · · · < ir 6 p. As above, L = (L◦ )◦ = (coneo Y )◦ = (coneo {ci1 , . . . , cir })◦ = {ci1 , . . . , cir }◦ = {ci1 }◦ ∩ · · · ∩ {cir }◦ = Vi1 ∩ · · · ∩ Vir . Since the representation L = V1 ∩ · · · ∩ Vp is irreducible, one has {V1 , . . . , Vp } = {Vi1 , . . . , Vir }, which gives the equality Y = {c1 , . . . , cp }. Hence p = r. A combination of Theorems 9.7, 9.8, and 9.9 implies the following corollary. Corollary 9.10. For a proper polyhedron P ⊂ Rn of positive dimension m, there is a nonnegative integer s such that every family F = {V1 , . . . , Vr } of closed halfspaces irreducibly representing P is uniquely partitioned into subfamilies G and H, say G = {V1 , . . . , Vs } and H = {Vs+1 , . . . , Vr } (let s = 0 if P is a plane and s = r if dim P = n), satisfying conditions (1)–(3) of Theorem 9.7. Furthermore, n − m + 1 6 r − s 6 2(n − m). Remark. There is no upper bound on each of the numbers s and r in Corollary 9.10. For example, if P ⊂ R2 is a convex polygon with s sides, then s = r and every irreducible family of closed halfplanes representing P consists of exactly s members.
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Properties of Polyhedra
Extreme Faces of Polyhedra Definition 9.11. Let P ⊂ Rn be a proper polyhedron distinct from a plane and F = {V1 , . . . , Vr } be a family of closed halfspaces irreducibly representing P . If the subfamily G = {V1 , . . . , Vs } of F satisfies conditions (1) and (2) of Theorem 9.7, then the sets Qi = bd Vi ∩ P,
1 6 i 6 s,
are called the facets of P . q
Vi
Qi P q P P q @ @ Fig. 9.2
A facet Qi of the polyhedron P .
In terms of Definition 9.11, every facet Qi of P equals rbd Di ∩ P , where Di = Vi ∩ aff P is a closed halfspace, 1 6 i 6 s. A combination of Theorems 9.7 and 9.8 shows that the family of facets of a polyhedron P ⊂ Rn is uniquely determined and does not depend on the choice of an irreducible family F of closed halfspaces representing P . Example. As follows from example on page 320, the facets of an r-simplex ∆(x1 , . . . , xr+1 ) ⊂ Rn , r > 1, are exactly the (r − 1)-simplices ∆i = ∆(x1 , . . . , xi−1 , xi+1 , . . . , xr+1 ), 1 6 i 6 r + 1. Example. From example on page 320 we conclude that the facets of a simplicial cone Cs (x1 , . . . , xr ) ⊂ Rn , r > 2, are exactly the simplicial cones Cs (x1 , . . . , xi−1 , xi+1 , . . . , xr ), 1 6 i 6 r. Theorem 9.12. Let P ⊂ Rn be a polyhedron of positive dimension m, distinct from a plane. Then every facet of P is a polyhedron. Furthermore, a set F is a facet of P if and only if F is an (m − 1)-dimensional extreme face of P . Proof. Let F = {V1 , . . . , Vr } be an irreducible family of closed halfspaces representing P , and let Qi = bd Vi ∩ P , 1 6 i 6 s, be the facets of P .
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Denote by Vi0 the opposite to Vi closed halfspace determined by the hyperplane bd Vi , 1 6 i 6 s. Clearly, bd Vi = Vi ∩ Vi0 , which gives Qi = bd Vi ∩ P = (Vi ∩ Vi0 ) ∩ (V1 ∩ · · · ∩ Vr ). Therefore, Qi is a polyhedron as a finite intersection of closed halfspaces. For the second statement, suppose first that F is a facet of P , say F = Qi . Since the hyperplane bd Vi properly supports P along Qi , it follows that Qi is an exposed face of P . Consequently, Qi is a proper extreme face of P , as shown in Theorem 8.2. Furthermore, according to (9.7) and to Corollary 2.33, dim Qi = dim P − 1 = m − 1. Conversely, suppose that F is an (m − 1)-dimensional extreme face of P . By Corollary 8.6, F is an exposed face of P . Then there is a hyperplane H ⊂ Rn satisfying the condition H ∩ P = F . One of the closed halfspaces, say V , determined by H properly supports P along F . Clearly, V does not contain aff P ; so, V ∩ aff P is a closed halfplane of aff P . As shown in the proof of Theorem 9.8, there is an index 1 6 i 6 s such that V ∩ aff P = Vi ∩ aff P . Therefore, F = bd V ∩ P = (bd V ∩ aff P ) ∩ P = rbd (V ∩ aff P ) ∩ P = rbd (Vi ∩ aff P ) ∩ P = (bd Vi ∩ aff P ) ∩ P = bd Vi ∩ P = Qi . Hence F is a facet of P . Theorem 9.13. Let P ⊂ Rn be a polyhedron of positive dimension m, distinct from a plane. Then the following statements hold. (1) Any proper extreme face of P lies within a facet of P . (2) Every extreme face of P is a polyhedral set. (3) Given proper extreme faces Fj and Fk of P such that Fj ⊂ Fk and dim Fj = j < dim Fk = k, there is a sequence of extreme faces Fi of P satisfying the conditions Fj ⊂ Fj+1 ⊂ · · · ⊂ Fk
and
dim Fi = i,
j 6 i 6 k.
(4) Every proper extreme face F of P coincides with the intersection of all facets of P containing F . Proof. Let F = {V1 , . . . , Vr } be an irreducible family of closed halfspaces representing P such that Qi = bd Vi ∩ P , 1 6 i 6 s, are the facets of P . (1) Choose a proper extreme face F of P . Then F ⊂ rbd P according to Theorem 7.4. Let x be a point in rint F . Then x ∈ rbd P , and Theorem 9.4 shows the existence of a halfspace Vi , 1 6 i 6 s, such that x ∈ bd Vi ∩ P .
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Then F ⊂ bd Vi (see Theorem 2.59), implying the inclusion F ⊂ bd Vi ∩P = Qi . The proofs of statements (2)–(4) use induction on m = dim P . Let m = 1. Then P is either a closed segment, [a, b], or a closed halfline with an endpoint a. Then {a} and {b} (respectively, {a}) are all proper extreme faces of P . In this case, all three statements (2)–(4) hold. Assume that statements (2)–(4) hold for all m 6 r − 1 (2 6 r 6 n), and let P ⊂ Rn be a polyhedron of dimension r distinct from a plane. (2) Let F be an extreme face of P . If F = ∅ or F = P , then P is a polyhedron. Assume that F is a proper extreme face of P . By statement (1) above, there is a facet Qi of P containing F . According to Theorem 9.12, Qi is a polyhedral set. Furthermore, Qi is an extreme face of P of dimension r − 1. Since F is an extreme face of Qi (see Theorem 7.3), the induction hypothesis implies that F is a polyhedron. (3) The case k − j = 1 is trivial. So, we assume that k − j > 2. Since Fk is a polyhedron, and since Fj is an extreme face of Fk , statement (1) above shows the existence of a facet Q of Fk containing Fj . Let Fk−1 = Q. By the induction hypothesis, there is a sequence Fj ⊂ Fj+1 ⊂ · · · ⊂ Fk−1 of extreme faces of Fk such that dim Fi = i for all j 6 i 6 k − 1. According to Theorem 7.3, all sets Fj , . . . , Fk−1 are extreme faces of P . So, the sequence Fj ⊂ Fj+1 ⊂ · · · ⊂ Fk is a desired one. (4) Let F be a proper extreme face of P . By statement (1) above, F lies within a certain facet Qi of P . By the induction hypothesis, F is the intersection of all facets of Qi containing F . We state that every facet of Qi is the intersection of two facets of P . Indeed, consider the closed halfplanes Ej = bd Vi ∩ Vj ,
j 6= i,
1 6 i 6 r.
Form Qi = bd Vi ∩ P it follows that Qi = bd Vi ∩ (V1 ∩ · · · ∩ Vi−1 ∩ Vi+1 ∩ · · · ∩ Vr ) = E1 ∩ · · · ∩ Ei−1 ∩ Ei+1 ∩ · · · ∩ Er . Choose in {E1 , . . . , Ei−1 , Ei+1 , . . . , Er } a subfamily, say {E10 , . . . , Ep0 }, of closed halfplanes of bd Vi irreducibly representing Qi . Then rbd Ek0 ∩ Qi , 1 6 k 6 p, are the facets of Qi . Consequently, Qi can be expressed as the intersection of two facets of P : rbd Ek0 ∩ Qi = rbd (bd Vi ∩ Vk0 ) ∩ (bd Vi ∩ P ) = (bd Vi ∩ bd Vk0 ) ∩ (bd Vi ∩ P ) = (bd Vi ∩ P ) ∩ (bd Vk0 ∩ P ). Consequently, F is the intersection of all facets of P containing F .
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Corollary 9.14. For every polyhedron P ⊂ Rn , the families of extreme faces and exposed faces of P coincide. Proof. Since every exposed face of P is its extreme face (see Theorem 8.2), it remains to prove the converse statement. Let F be an extreme face of P . Excluding the trivial cases F = ∅ and F = P , we assume that F is a proper subset of P . By Theorem 9.13, F is the intersection of facets of P containing F . Since every facet of P is an exposed face of P , Theorem 8.9 implies that F is an exposed face of P . Polyhedrality and Finiteness Conditions Theorem 9.15. For a nonempty closed convex set K ⊂ Rn , the following conditions are equivalent. (1) K is a polyhedron. (2) K has at most finitely many extreme faces. (3) K has at most finitely many exposed faces. Proof. Excluding the trivial cases, we assume that K is a proper subset of Rn of positive dimension. If K is a plane, then K is a polyhedron (see example on page 319) without proper extreme or exposed faces. So, we suppose that K is not a plane. (1) ⇒ (2). If K is a polyhedron, then K has at most finitely many facets. By Theorem 9.13, every proper extreme face of K is an intersection of its facets. Consequently, K has at most finitely many extreme faces. (2) ⇒ (3). This part is obvious because every exposed face of K is its extreme face (see Theorem 8.2). (3) ⇒ (1). Let G1 , . . . , Gr , r > 1, be all proper exposed faces of K (such faces exist because K is not a plane; see Corollary 8.17). Denote by Vi a closed halfspace containing K such that its boundary hyperplane Hi satisfies the condition Gi = K ∩ Hi , 1 6 i 6 r. Since each hyperplane Hi properly supports K, no halfspace Vi contains aff K, 1 6 i 6 r. We state that K coincides with the polyhedron P = V1 ∩ · · · ∩ Vr ∩ aff K. Because of the obvious inclusion K ⊂ P , it suffices to show that P ⊂ K. Assume for a moment the existence of a point z ∈ P \ K. Choose a point x ∈ rint K. Due to z ∈ P ⊂ aff K, the open segment (x, z) contains a point u ∈ rbd K (see Theorem 2.55). By Corollary 8.15, there is a proper
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exposed face G of K which contains u. By the assumption, G = Gj for a certain j ∈ {1, . . . , r}. Thus, [x, z] ⊂ Vj such that u ∈ Hj . In this case, [x, z] ⊂ Hj (see Corollary 1.32). Hence x ∈ K ∩ Hj = Gj ⊂ rbd K, contrary to the choice of x. Summing up, P ⊂ K. The next result refines Theorem 9.15 by dealing with planar and halfplanar faces only (compare with Theorems 7.32 and 8.34). Theorem 9.16. For a nonempty closed convex set K ⊂ Rn , the following conditions are equivalent. (1) K is a polyhedron. (2) K has at most finitely many planar and halfplanar extreme faces. (3) If S ⊂ Rn is a subspace complementary to lin K, then K ∩ S has at most finitely many extreme points and extreme halflines. (4) K has at most finitely many planar and halfplanar exposed faces. (5) If S ⊂ Rn is a subspace complementary to lin K, then K ∩ S has at most finitely many exposed points and exposed halflines. Proof. Since the equivalence of conditions (1)–(5) is obvious if K is a plane, we may assume that K is not a plane. (1) ⇒ (2). This part follows from Theorem 9.15. (2) ⇒ (3). If S ⊂ Rn is a subspace complementary to lin K, then the convex set K ∩ S is closed and line-free (see Theorem 5.20 and Corollary 5.53). By Theorem 7.5, a subset F of K is an extreme face of K if and only if it can be expressed as F = E ⊕ lin K, where E is an extreme face of K ∩ S. Clearly, F is a planar or halfplanar extreme face of K if and only if so is E for K ∩ S. Hence condition (2) implies that K ∩ S has at most finitely many planar and halfplanar extreme faces. Finally, since K ∩ S is line-free, its planar extreme faces are points and its halfplanar extreme faces are halflines. (3) ⇒ (1). First, we state that the set K ∩ S is polyhedral. By Theorem 9.15, it suffices to show that K ∩ S has at most finitely many extreme faces. For this, choose an extreme face E of K ∩ S. Then E is a line-free closed convex set (see Theorem 7.4), and, by Theorem 7.24, E is the convex hull of the set ext E ∪ extr E. Furthermore, ext E ⊂ ext (K ∩ S) and extr E ⊂ extr (K ∩ S) (see Theorem 7.3). Since the sets ext (K ∩ S) and extr (K ∩ S) are finite, there are at most finitely many sets of the form conv (X ∪ Y ), where X ⊂ ext (K ∩ S) and Y is a family of halflines from extr (K ∩ S). In conclusion, K ∩ S has at most finitely many extreme faces.
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Because every extreme face of K can be expressed in the form F = E ⊕ lin K, where E is an extreme face of K ∩ S (see Theorem 5.20), the family of extreme faces of K is finite. (2) ⇒ (4) and (3) ⇒ (5). These statements follow from the fact that every planar exposed face (respectively, every halfplanar exposed face) of K is its planar extreme face (respectively, every halfplanar extreme face). The latter statement derives from Theorems 7.30 and 8.32 (4) ⇒ (5). This part is analogous to that of (2) ⇒ (3), with a reference on Theorem 8.7 instead of Theorem 7.5. (5) ⇒ (3). Since K ∩ S has at most finitely many exposed points, Theorem 8.39 shows that K ∩S has at most finitely many extreme points. It remains to show that K ∩S has at most finitely many extreme halflines. For this, we observe first that K∩S has at most finitely many exposed segments. Indeed, since the endpoints of each exposed segment of K ∩ S are extreme points of K∩S, their finiteness implies the finiteness of the family of exposed segments of K ∩ S. From Theorem 8.39 it follows that the set ext1 (K ∩ S) lies in the closure of exp1 (K ∩S), which is a finite union of extreme segments and extreme halflines of K ∩S. Hence ext1 (K ∩S) = exp1 (K ∩S), implying that K ∩ S has at most finitely many extreme halflines.
Polyhedra as Convex Hulls and Sums We will need the following auxiliary lemma. Lemma 9.17. Let L ⊂ Rn be a plane of positive dimension m, M ⊂ L be a plane of dimension m − 1, and D be a closed halfplane of L determined by M . Then D = conv (h ∪ M ), where h is a closed halfline with endpoint a ∈ M , which meets D \ M . Proof. We first observe that one may put a = o. Indeed, according to Theorem 3.12, D = conv (h ∪ M ) if and only if D − a = conv (h ∪ M ) − a = conv ((h − a) ∪ (M − a)). By Theorem 1.42, D = h + M . Since h ∪ M ⊂ D, and since D is a convex set (see example on page 72), one has conv (h ∪ M ) ⊂ D. For the opposite inclusion, choose a point x ∈ D. By the above, x = u + v, where u ∈ h and v ∈ M . Since 2u ∈ h and 2v ∈ M , one has x = 21 (2u + 2v) ∈ conv (h ∪ M ). Summing up, D = conv (h ∪ M ). Theorem 9.18. A closed convex set K ⊂ Rn is a polyhedron if and only if K is the convex hull of finitely many points and closed halflines.
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Proof. Because the case K = ∅ is obvious, we may suppose that K is nonempty. Assume first that K is a polyhedron. By Theorem 9.16, K has at most finitely many planar and halfplanar extreme faces. If K is line-free, then these are extreme points and extreme halflines. According to Theorem 7.24, K is the convex hull of finitely many points and closed halflines. Suppose that the polyhedron K is not line-free, and let r = dim (lin K). We state that K is the convex hull of finitely many closed halflines. Choose a basis c1 , . . . , cr for lin K, and let er+1 = −(e1 + · · · + er ). Then {e1 , . . . , er+1 } is an affine basis for lin K (see Corollary 1.63). Consider the closed halflines hi = {λci : λ > 0}, 1 6 i 6 r + 1. Exercise 4.1 shows that lin K = conv (h1 ∪ · · · ∪ hr+1 ). Choose a subspace S ⊂ Rn complementary to lin K. By Corollary 9.2, the set K ∩ S is a polyhedron. Theorem 9.16 shows that K ∩ S has at most finitely many extreme points, say x1 , . . . , xp , and at most finitely many extreme halflines, say g1 , . . . , gq . Let zj denote the endpoint of gj , 1 6 j 6 q. From Theorem 7.5 it follows that Fj = xj + lin K, 1 6 j 6 p, are the planar extreme faces of K, and Gj = conv ((zj + lin K) ∪ gj ), 1 6 j 6 q, are the extreme halfplanes of K. By the above argument, Fj = xj + lin K = xj + conv (h1 ∪ · · · ∪ hr+1 ) = conv (h1 (xj ) ∪ · · · ∪ hr+1 (xj )), where hi (xj ) = xj + hi , 1 6 i 6 r + 1, are halflines with common endpoint xj . Similarly (see Lemma 9.17), Gj = conv ((zj + conv (h1 ∪ · · · ∪ hr+1 ) ∪ gj ) = conv (h1 (zj ) ∪ · · · ∪ hr+1 (zj ) ∪ gj ), 1 6 j 6 q, where hi (zj ) = zj + hi , 1 6 i 6 r + 1, are halflines with common endpoint zj . Finally, a combination of Theorems 7.32 and 3.2 gives K = conv (F1 ∪ · · · ∪ Fp ∪ G1 ∪ · · · ∪ Gq ) = conv
p
∪ conv (h1 (xj ) ∪ · · · ∪ hr+1 (xj ))
j=1
q ∪ ∪ conv (gj ∪ h1 (zj ) ∪ · · · ∪ hr+1 (zj )) j=1
= conv
p
∪ h1 (xj ) ∪ · · · ∪ hr+1 (xj )
j=1
q ∪ ∪ gj ∪ h1 (zj ) ∪ · · · ∪ hr+1 (zj ) , j=1
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implying that K is the convex hull of finitely many closed halflines. Conversely, suppose that the closed convex set K is the convex hull of certain points x1 , . . . , xp and certain closed halflines h1 , . . . , hq . Denote by zi the endpoint of hi , 1 6 i 6 q, and let X = {x1 , . . . , xp } ∪ h1 ∪ · · · ∪ hq . If F is a halfplanar extreme face of K, then, by Theorem 7.33, F = conv (F ∩ X). It is easy to see that if a halfline hj meets F , then either hj ⊂ F or hj ∩ F = {zj }. Consequently, the set F ∩ X is the union of a subset of {x1 , . . . , xp } and some of the halflines h1 , . . . , hq . Because there are at most finitely many sets of the form F ∩ X, the family of planar and halfplanar extreme faces of K is finite. According to Theorem 9.16, K is a polyhedron. Remark. The closedness of K in Theorem 9.18 is essential. For example, the convex hull of the point u = (1, 0) and the halfline h = {(x, 0) : x > 0} is not closed; whence this convex hull is not a polyhedral set. Theorem 9.19. A convex set K ⊂ Rn is a polyhedron if and only if there are finite sets X and Y in Rn such that K = conv X + coneo Y . Proof. Because the case K = ∅ is obvious, we may suppose that K is nonempty. Assume first that K is a polyhedron. By Theorem 9.18, K can be expressed as the convex hull of certain points, say x1 , . . . , xp , and certain closed halflines, say h1 , . . . , hq . Denote by zi the endpoint of hi , and let h0i = hi −zi , 1 6 i 6 q. Clearly, h01 , . . . , h0q are closed halflines with common endpoint o. Put X = {x1 , . . . , xp , z1 , . . . , zq }
and C = conv (h01 ∪ · · · ∪ h0q ).
We state that K = conv X + C. Indeed, observe first that conv X ⊂ K by the convexity of K. Next, Theorem 5.1 implies the inclusion x + h0i ⊂ K for all x ∈ conv X and all 1 6 i 6 q. Hence x + C ⊂ K by a convexity argument. Therefore, conv X + C = ∪ (x + C : x ∈ conv X) ⊂ K. For the opposite inclusion, choose a point u ∈ K. By the assumption on K, one can express u as a convex combination u = λ1 v1 + · · · + λr vr + µ1 w1 + · · · + µs ws ,
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where v1 , . . . , vr ∈ X and w1 , . . . , ws ∈ h1 ∪ · · · ∪ hq . If wi ∈ hm(i) , where 1 6 m(i) 6 q, then put wi0 = wi − zm(i) . In these terms, ws0 ∈ h0m(i) and u = (λ1 v1 + · · · + λr vr + µ1 zm(1) + · · · + µs zm(s) ) + (µ1 w10 + · · · + µs ws0 ) ∈ conv X + coneo {w10 , . . . , wq0 } ⊂ conv X + conv (h01 ∪ · · · ∪ h0q ) = conv X + C. Summing up, K ⊂ conv X + C. Finally, choosing nonzero points yi ∈ hi , 1 6 i 6 q, and letting Y = {y1 , . . . , yq }, we obtain K = conv X + conv (h01 ∪ · · · ∪ h0q ) = conv X + coneo {y1 , . . . , yq } = conv X + coneo Y. Conversely, suppose that K = conv X + coneo Y for certain finite sets X and Y in Rn . First, we state that K is closed. Indeed, coneo Y is closed by Corollary 4.39. Therefore, K is closed as the sum of the compact set conv X and the closed set coneo Y (see Theorem 3.17 and Corollary 2.46). Let X = {x1 , . . . , xp } and Y = {y1 , . . . , yq }. Excluding the trivial cases, we suppose that all points y1 , . . . , yq are distinct from o. Denote by gj the closed halfline with endpoint o containing yj , and let hij = xi + gj for all 1 6 i 6 p and 1 6 j 6 q. Since gj ⊂ coneo Y , one has hij ⊂ conv X + coneo Y = K. We state that K equals the set M = conv ( ∪ (hij : 1 6 i 6 p, 1 6 j 6 q)). The inclusion M ⊂ K holds because of the convexity of K and the above argument. Hence it suffices to prove that K ⊂ M . For this, choose a point u ∈ K. By the assumption on K, we can write u = x+y, where x ∈ conv X and y ∈ coneo Y . The points x and y are expressible, respectively, as a convex combination of certain points x1 , . . . , xp ∈ X and a nonnegative combination of points y1 , . . . , yq ∈ coneo Y : x = λx1 + · · · + λr xr
and y = µ1 y1 + · · · + µq yq .
Without loss of generality, we may assume that all scalars λ1 , . . . , λp are positive and µ1 + · · · + µq > 0. Then u can be written as p P q P µ1 + · · · + µq λi µj (xi + yj ). (9.10) u= λi p i=1 j=1 µ1 + · · · + µq
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Clearly, every point zij = xi +
µ1 + · · · + µq yj , λi p
1 6 i 6 p,
1 6 j 6 q,
belongs to the halfline hij , and u=
q p P P
λi µj zij µ + · · · + µq 1 i=1j=1
is their convex combination. Hence u ∈ M , and K ⊂ M . Finally, Theorem 9.18 implies that K is a polyhedron. Theorem 9.20. The following statements hold. (1) A bounded convex set in Rn is a polytope if and only if it is the convex hull of finitely many points. (2) A convex cone in Rn is polyhedral if and only if it is the convex hull of finitely many closed halflines with common endpoint. (3) A convex set in Rn is a polyhedron if and only if it is the sum of a polytope and a polyhedral cone with apex o. (4) A convex set in Rn is a polyhedron if and only if it is the closure of the convex hull of finitely many points and closed halflines. (5) A convex set in Rn is a polyhedron if and only if it is the closure of the convex hull of a polytope and a polyhedral cone. Proof. Without loss of generality, we may assume that convex sets K ⊂ Rn involved in statements (1)–(5) are nonempty. Furthermore, we assume that K is expressed as K = (K ∩ S) ⊕ lin K, where S ⊂ Rn is a subspace complementary to lin K (see Theorem 5.20). Statement (1) is a particular cases of Theorem 9.19 (corresponding to the cases Y = {o}). (2) Let K ⊂ Rn be a convex cone. Assume first that K is polyhedral. Then K ∩ S is a line-free polyhedral cone, with a unique apex, say s. Then s is the only extreme point of K (see Theorem 7.20), and Theorem 9.16 shows that K ∩S has at most finitely many extreme halflines, say h1 , . . . , hp . Furthermore, each of the halflines g1 , . . . , gq has s as the endpoint. According to Theorem 7.30, every set Fj = gj + lin K, 1 6 j 6 q, is an extreme halfplane of K, and Theorem 7.32 shows that K = conv (F1 ∪ · · · ∪ Fp . It was shown in the proof of Theorem 9.18 that lin K can be expressed as the convex hull of closed halfline, say h1 , . . . , hp with common endpoint o.
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Finally, Theorem 3.2 gives K = conv (F1 ∪ · · · ∪ Fq ) = conv ((g1 + lin K) ∪ · · · ∪ (gq + lin K)) = conv ((g1 ∪ (s + lin K)) ∪ · · · ∪ (gq ∪ (s + lin K))) q = conv ∪ (gj ∪ conv ((s + h1 ) ∪ · · · ∪ (s + hp ))) = conv
j=1 q
∪ (gj ∪ (s + h1 ) ∪ · · · ∪ (s + hp )) .
j=1
(3) By Theorem 9.19, a convex set K ⊂ Rn is a polyhedron if and only if there are finite sets X and Y in Rn such that K = conv X + coneo Y . The above argument shows that conv X is a polytope, and coneo Y is a polyhedral cone. (4) If a convex set K ⊂ Rn is a polyhedron, then Theorem 9.18 shows that K can be expressed as the convex hull of finitely many points and closed halflines. Conversely, suppose that K is the closure of the convex hull of a certain finite set X = {x1 , . . . , xp } of points and a finite family of closed halflines h1 , . . . , hq : K = cl (conv (X ∪ h1 ∪ · · · ∪ hq )). We state that K is a polyhedron. Indeed, let zi denote the endpoint of hi , and put h0i = hi − zi , 1 6 i 6 q. Clearly, the halflines h01 , . . . , h0q have o as a common endpoint, and conv (h01 ∪ · · · ∪ h0q ) = coneo {y1 , . . . , yq }, where yi is a nonzero point in h0i , 1 6 i 6 q. Since hi = zi + h0i ⊂ K, Theorem 5.1 shows that xi + h0j ⊂ K for all 1 6 i 6 p and 1 6 j 6 q. Consequently, with Z = {x1 , . . . , xp , z1 , . . . , zq }, one has X ∪ h1 ∪ · · · ∪ hq ⊂ Z + (h01 ∪ · · · ∪ h0q ) ⊂ K. By a convexity argument, conv (X ∪ h1 ∪ · · · ∪ hq ) ⊂ conv (Z + (h01 ∪ · · · ∪ h0q )) = conv Z + conv (h01 ∪ · · · ∪ h0q ) = conv Z + coneo {y1 , . . . , yq } ⊂ K. According to Theorem 9.19, the set M = conv Z + coneo {y1 , . . . , yq } is a polyhedron. Therefore, M is closed, which gives K = cl (conv (X ∪ h1 ∪ · · · ∪ hq )) ⊂ M ⊂ K. Hence K = M , implying that K is a polyhedron. (5) By statement (3) above, K is a polyhedron if and only if it is the sum of a certain polytope Q and a polyhedral cone C with apex o. If u is a point in Q, then, as shown in Exercise 5.1, K = Q + C = cl (conv (Q ∪ (u + C))).
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Algebra of Polyhedra Theorem 9.21. If P1 , . . . , Pr ⊂ Rn are polyhedra, then the set Q = cl (conv (P1 ∪ · · · ∪ Pr )) also is a polyhedron. Proof. Expressing every polyhedron Pi as the convex hull of a finite set Xi and a finitely many closed halflines hi1 , . . . , hiqi (see Theorem 9.18), we obtain, by Theorem 3.2, that Q = cl (conv (P1 ∪ · · · ∪ Pr )) r
= cl (conv ( ∪ conv (Xi ∪ hi1 ∪ · · · ∪ hiqi ))) i=1 r
= cl (conv ( ∪ Xi ∪ hi1 ∪ · · · ∪ hiqi )). i=1
According to Theorem 9.20, Q is a polyhedron. Theorem 9.22. For polyhedra P1 , . . . , Pr ⊂ Rn and scalars µ1 , . . . , µr , the set P = µ1 P1 + · · · + µr Pr is a polyhedron. Proof. Without loss of generality, we may assume that all sets P1 , . . . , Pr are nonempty. An induction argument shows that the proof can be reduced to the case r = 2 and µ1 µ2 6= 0 (if, for instance, r = 1 or µ2 = 0, we can write µ1 P1 + µ2 o). By Theorem 9.18, Pi = conv Xi + coneo Yi , for certain finite sets Xi and Yi in Rn , i = 1, 2. With Yi0 = Yi ∪ {o}, one has coneo Yi0 = coneo Yi , i = 1, 2. By Theorems 3.12 and 4.35 µ1 P1 + µ2 P2 = (conv (µ1 X1 ) + coneo (µ1 Y1 )) + (conv (µ2 X2 ) + coneo (µ2 Y2 )) = (conv (µ1 X1 ) + conv (µ2 X2 )) + (coneo (µ1 Y10 ) + coneo (µ2 Y20 )) = conv (µ1 X1 + µ2 X2 ) + coneo (µ1 Y10 + µ2 Y20 ). Since both sets µ1 X1 + µ2 X2 and µ1 Y10 + µ2 Y20 are finite, Theorem 9.18 shows that the set µ1 P1 + µ2 P2 is a polyhedron. Corollary 9.23. No polyhedron P ⊂ Rn has asymptotic planes. Proof. Let L ⊂ Rn be a plane of positive dimension. Since L is a polyhedron (see example on page 319), Theorem 9.22 shows that P + L also is a polyhedron. In particular, P + L is closed, and Lemma 2.51 shows that L cannot be asymptotic for P .
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Theorem 9.24. If f : Rn → Rm is an affine transformation and P ⊂ Rn and Q ⊂ Rm are polyhedra, then both sets f (P ) and f −1 (Q) are polyhedra. Proof. Let f (x) = z + g(x), where z ∈ Rm and g : Rn → Rm is a linear transformation. By Theorem 9.18, P = conv X + coneo Y for certain finite sets X and Y in Rn . A combination of Theorems 3.16 and 4.36 gives f (P ) = z + g(P ) = z + g(conv X + coneo Y ) = z + conv g(X) + coneo g(Y ). By Theorem 9.18, the set T = conv g(X) + coneo g(Y ) is a polyhedron; so f (P ) also is a polyhedron as a translate of T . Since the subspace rng g is a polyhedron (see example on page 319), the set T = Q ∩ rng g also is a polyhedron. Clearly, g −1 (Q) = g −1 (T ). By Theorem 9.19, T = conv X+coneo Y for certain finite sets X = {x1 , . . . , xp } and Y = {y1 , . . . , yq } in rng g. Choose in Rn sets X 0 = {x01 , . . . , x0p } and Y 0 = {y10 , . . . , yq0 } such that g(x0i ) = xi and g(yj0 ) = yj for all 1 6 i 6 p and 1 6 j 6 q. By Theorem 9.19 again, the set T 0 = conv X 0 + coneo Y 0 is polyhedral. A combination of Theorems 3.16 and 4.36 gives g(T 0 ) = g(conv X 0 + coneo Y 0 ) = conv g(X 0 ) + coneo g(Y 0 ) = conv X + coneo Y = T. Finally, according to Theorem 9.22, the set g −1 (Q) = g −1 (T ) = T 0 + null g is a polyhedron. Polyhedrality of Associated Cones Theorem 9.25. If P is a proper polyhedral set and s is a point in P , then the cone cones P is polyhedral. If, additionally, P is represented by a finite family F of closed halfspaces and G is the family of all halfspaces V ∈ F satisfying the condition s ∈ bd V , then cones P = ∩(V : V ∈ G). Proof. Since the statement is obvious when P is a singleton, we may assume that P 6= {s}. If s ∈ rint P , then cones P = aff P (see Theorem 4.40), implying that cones P is a polyhedron. Suppose that s ∈ rbd P . Then G = 6 ∅, as follows from Theorem 9.4. Let Q = ∩(P : P ∈ G). Given a halfspace V ∈ G, the obvious inclusion P ⊂ V and Theorem 1.31 show that [s, xi ⊂ V for every point x ∈ P . Therefore, Theorem 4.30 gives cones P = ∪([s, xi : x ∈ P \ s) ⊂ V,
V ∈ G.
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Consequently, cones P ⊂ Q. For the opposite inclusion, Q ⊂ cones P , choose a point x ∈ Q. Then [s, x] ⊂ Q by a convexity argument. Since the inclusion x ∈ P is trivial when x = s, we assume that x 6= s. If G = F, then Q = P ⊂ cones P . Suppose that G 6= F. Then s ∈ int V for every halfspace V ∈ F \ G. By the finiteness of F \ G, there is a point z ∈ (s, x) so close to x that z ∈ V for every V ∈ F \ G. Hence z ∈ Q ∩ (V : V ∈ F \ G) = P , and x ∈ [s, zi ⊂ cones P . Summing up, Q ⊂ cones P . Theorem 9.26. If P is a proper polyhedral set and s is a point in Rn \ P , then cl (cones P ) is a polyhedral cone. Proof. By Theorem 9.18, P can be expressed as the convex hull of finitely many points, say x1 , . . . , xp , and finitely many closed halflines, say h1 , . . . , hq . Let zj be the endpoint of hj , 1 6 j 6 q. Denote by h0j the translate of hj on s − zj (so that s is the endpoint of h0j ), and put h00j = [s, zi i, 1 6 j 6 q. It is easy to see that hj ⊂ cones (h0j ∪ h00j ). From Theorem 5.1 it follows that h0j ∪ h00j ⊂ cl (cones P ),
1 6 j 6 q.
Clearly, every halfline gi = [s, xi i lines in cones P , 1 6 i 6 p. Using Theorems 4.24 and 4.30, we obtain cones P = cones (conv ({x1 , . . . , xp } ∪ h1 ∪ · · · ∪ hp )) = cones ({x1 , . . . , xp } ∪ h1 ∪ · · · ∪ hp ) p
q
i=1 p
j=1 q
i=1 p
j=1 q
⊂ cones (( ∪ gi ) ∪ ( ∪ cones (h0j ∪ h00j ))) = cones (( ∪ gi ) ∪ ( ∪ h0j ∪ h00j )) = conv (( ∪ gi ) ∪ ( ∪ h0j ∪ h00j )) i=1
j=1
⊂ cl (cones P ). By Theorem 9.20, the set p
q
i=1
j=1
Q = conv (( ∪ gi ) ∪ ( ∪ h0j ∪ h00j )) is a polyhedral cone with apex s. Therefore, cl (cones P ) = Q, which shows that cl (cones P ) is a polyhedral cone.
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Theorem 9.27. Given a nonempty polyhedron P ⊂ Rn , its recession, barrier, and normal cones are polyhedral. Furthermore, if P is represented by a finite family of closed halfspaces Vi = {x ∈ Rn : x·ci 6 γi },
1 6 i 6 p,
then p
rec K = ∩ {x ∈ Rn : x·ci 6 0},
(9.11)
bar K = nor K = coneo {c1 , . . . , cp }.
(9.12)
i=1
Proof. The equality (9.11) follows from Theorem 6.20. Hence the recession cone rec K is polyhedral as the intersection of finitely many closed halfspaces. For the second statement, we first observe that all halfspaces Wi = {x ∈ Rn : x·ci 6 0},
1 6 i 6 p,
are closed convex cones with apex o. Since Wi◦ = hi = [o, ci i, Corollary 5.51 and Theorem 5.46 give cl (bar K) = (rec K)◦ = (W1 ∩ · · · ∩ Wp )◦ = cl (conv (h1 ∪ · · · ∪ hp )) = cl (coneo {c1 , . . . , cp }) = coneo {c1 , . . . , cp } because coneo {c1 , . . . , cp } is a closed set (see Corollary 4.39). The obvious inclusion {c1 , . . . , cp } ⊂ bar K and the convexity of bar K (see Theorem 5.42), imply that coneo {c1 , . . . , cp } ⊂ bar K. Hence bar K = coneo {c1 , . . . , cp }, and bar K is a polyhedron according to Theorem 9.20. Since K has no asymptotic planes (see Corollary 9.23), Corollary 6.15 implies that bar K \ nor K = ∅. This argument and the inclusion nor K ⊂ bar K shows that nor K = bar K. Exercises for Chapter 9 Exercise 9.1. Show that every proper plane L ⊂ Rn of dimension m is an intersection of n − m + 1 closed halfspaces. Furthermore, each such representation of L is irreducible. Exercise 9.2. Let a proper plane L ⊂ Rn be the intersection of a family F of closed halfspaces, each containing L in its boundary hyperplane. Show that L = ∩ (bd V : V ∈ F).
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Exercise 9.3. Let L ⊂ Rn be a plane of positive dimension and F ⊂ Rn be a closed (respectively, open) halfplane. Show that the sum L + F is either a plane or a closed (respectively, open) halfplane. Similarly, if L ∩ F 6= ∅, then the intersection L ∩ F is either a plane or a closed (respectively, open) halfplane. Exercise 9.4. Let f : Rn → Rm be an affine transformation. Show that for a closed (respectively, open) halfplane F ⊂ Rn , its image f (F ) is either a plane or a closed (respectively, open) halfplane. Similarly, if G ⊂ Rm is a closed (respectively, open) halfplane, with G ∩ rng f 6= ∅, then f −1 (G) is either a plane or a closed (respectively, open) halfplane. Exercise 9.5. (Bair [13]) Show that if a convex polytope P ⊂ Rn is the sum of convex sets A and B, then both A and B are convex polytopes. Exercise 9.6. (Bair [14]) Show that if a convex polyhedron P ⊂ Rn is the sum of convex sets A and B such that B is bounded, then A is a polyhedron. Exercise 9.7. Given a nonempty bounded convex set K ⊂ Rn , a point c ∈ rint K, and a scalar ε > 0, show the existence of a polytope P ⊂ Rn such that c ∈ rint P and P ⊂ K ⊂ (1 + ε)P − εc. If, additionally, K is closed, then the vertices of P can be chosen in exp K. Notes for Chapter 9 Polyhedra and polytopes. General (possibly nonconvex) polyhedra (or polytopes, in other terminology) in Rn are recursively defined as complexes of lowerdimensional polyhedra, while convex polyhedra were introduced as bounded intersections of finitely many closed halfspaces (see, e. g., Steinitz [208, Part IV]). Alternatively, Minkowski [158, p. 136] (for n = 3) and Weyl [223] (for all n > 3) defined a convex polyhedron in Rn as the set of all convex combinations of finitely many given points (equivalently, as the convex hull of these points). The intersections (possibly, unbounded) of finitely many closed halfspaces were introduced for geometric interpretations of solution sets of systems of linear inequalities. Such intersections were called polyhedral sets (see, e. g., Motzkin [161, § 5]). Afterwards, bounded polyhedral sets were called polytopes (or, convex polytopes); they were defined either as convex hulls of finitely many points, or as compact convex sets with at most finitely many extreme points. Theorem E3 from [161] states that every polyhedral set in Rn can be represented as the sum of a subspace, a polytope, and a line-free polyhedral cone. Representations of polyhedra. Every n-dimensional polytope with f facets is a plane section of an (f − 1)-dimensional simplex (see Davis [68] and Gr¨ unbaum [104, p. 71]), while every n-dimensional polytope with v vertices is a
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parallel projection of an (v − 1)-dimensional simplex (see Gr¨ unbaum [104, p. 72]). Naumann [164] showed that every n-dimensional polytope with f facets is a section of a 2n (n + 1)f -dimensional cube. Epifanov [81] proved that for every ndimensional polyhedron P with m facets in Rn , there is an (n+m−1)-dimensional vector space E containing Rn as a subspace and a solid polyhedral cone C ⊂ E with m facets and pairwise orthogonal (n + m − 2) faces such that P = E ∩ C. Davis [68] showed that each line-free polyhedral cone is affine-equivalent to the intersection of the positive orthant with a linear subspace, and is affine-equivalent to a parallel projection of the positive orthant. Corollary 9.23 and Theorem 6.37 imply the following result of Motzkin [161, Theorem G5]: A pair of disjoint polyhedra in Rn can be strongly separated by ˇ a hyperplane (see also Cernikov [54] for an extension of this statement to any vector spaces). Characteristic properties of polyhedra. Bol [28] showed that a compact convex set K ⊂ R3 is a polytope if and only if all projections of K on 2-dimensional planes are polygons. Mirkil [159] proved that a closed convex cone C ⊂ Rn is polyhedral if and only if all projections of C on 2-dimensional planes are closed. Klee [126] showed that a convex set K ⊂ Rn is polyhedral if and only if any of the following statements holds: (a) all sections of K by 2-dimensional planes through a given point u ∈ rint K are polygonal, (b) all sections of K by 3-dimensional planes through a given point u ∈ Rn are polyhedral, (c) all projections on 3-dimensional planes of Rn are polyhedral. If, additionally, K is bounded, then K is a polytope is and only if all its projections on 2-dimensional planes are polyhedral. Approximations by polyhedra. The statement of Exercise 9.7 belongs to Minkowski [158, p. 139]. For the case of unbounded convex sets, a similar statement is proved by Klee [126]: If K ⊂ Rn is a closed convex set and ε > 0, then there are boundedly polyhedral sets P and Q such that P ⊂ K ⊂ Bε (P ) and K ⊂ Q ⊂ Bε (K). We recall that a set X ⊂ Rn is boundedly polyhedral provided its intersection with every polytope is polyhedral. Hoffman [115] proved that if a polyhedron P ⊂ Rn is the union of closed convex sets K1 , . . . , Kr , there there are polyhedra Pi ⊂ Ki , 1 6 i 6 r, such that P = P1 ∪ · · · ∪ Pr . Dually, if a polyhedral set P ⊂ Rn is the intersection of closed convex sets K1 , . . . , Kr , then there are polyhedra P1 , . . . , Pr such that Ki ⊂ Pi , 1 6 i 6 r and P = P1 ∩ · · · ∩ Pr (see Eggleston [79]). If K1 , . . . , Kr ⊂ Rn are pairwise disjoint convex sets, then there are convex polyhedra P1 , . . . , Pr ⊂ Rn of which no two have common interior points and such that Ki ⊂ Pi for all 1 6 i 6 r (Stoelinga [209, Theorem 31]).
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Chapter 0 Exercise 0.1. The linearity of h is obvious. We state that h is onto. Indeed, let u ∈ rng f and x ∈ Rn be such that f (x) = u. Since Rn = S ⊕ null f , one can write x = x0 + y, where x0 ∈ S and y ∈ null f . Then h(x0 ) = f (x0 ) = f (x − y) = f (x) = u, implying the equality rng h = rng f . Furthermore, h is invertible because dim S = n − dim (null f ) = dim (rng f ). For the equality f −1 (x) = h−1 (x) + null f , choose a point u ∈ f −1 (x). Then u = u0 + y, where u0 ∈ S and y ∈ null f . From x = f (u) = f (u0 ) = h(u0 ) it follows that u0 = h−1 (x) and u = u0 + y ∈ h−1 (x) + null f . Conversely, if u ∈ h−1 (x) + null f and u = u0 + y, where u0 = h−1 (x) ∈ S and y ∈ null f , then u ∈ f −1 (x) because of f (u) = f (u0 ) = f (h−1 (x)) = h(h−1 (x)) = x. Exercise 0.2. Choose in L a subspace N complementary to L∩null f , and let S ⊂ Rn be a subspace containing N and complementary to L ∩ null f . Then f (L) = f (N ⊕ (L ∩ null f )) = f (N ) + {o} = f (N ). As shown in Exercise 0.1, the mapping h : S → rng f , defined by h(x) = f (x) is an invertible linear transformation. Hence dim f (L) = dim f (N ) = dim h(N ) = dim N = dim L − dim (L ∩ null f ). Similarly, from f −1 (M ) = f −1 (M ∩ rng f ) = h−1 (M ∩ rng f ) ⊕ null f it follows that dim f −1 (M ) = dim (M ∩ rng f ) + dim (null f ). 345
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Exercise 0.3. Let f be a projection in Rn . Put S = rng f and T = null f . So, every vector x ∈ Rn is uniquely expressible as x = y + z such that y ∈ S, z ∈ T , and f (x) = y. Then f (y) = y, which gives f 2 (x) = f (f (x)) = f (y) = y = f (x). Conversely, assume that a linear transformation f : Rn → Rn satisfies the condition f 2 = f . Let S = rng f . We state that S = {y ∈ Rn : f (y) = y}. Indeed, if f (y) = y, then y ∈ rng f . Conversely, if y ∈ rng f , then y = f (x) for a certain vector x ∈ Rn . Therefore, f (y) = f (f (x)) = f (x) = y. Put T = {x − f (x) : x ∈ Rn }. We claim that T = null f . Indeed, from f (x − f (x)) = f (x) − f (f (x)) = o if follows that T ⊂ null f . Conversely, if x ∈ null f , then x = x − f (x) ∈ T . It remains to show that the subspaces S and T are complementary. First, we observe that S + T = Rn because every vector x ∈ Rn can be written as x = f (x) + (x − f (x)) ∈ S + T. Next, choose a vector y ∈ S ∩ T . Then f (y) = y and y = z − f (z) for a certain vector z ∈ Rn . This gives y = f (y) = f (z − f (z)) = f (z) − f (f (z)) = o, implying the equality S ∩ T = {o}. Hence the sum S + T = Rn is direct, and f is a projection on S along T . Exercise 0.4. Choose a vector x0 ∈ Rn and an infinite sequence of vectors x1 , x2 , . . . converging to x0 . From |ϕ(xi ) − ϕ(x0 )| = |xi ·c− x0 ·c| = |(xi − x0 )·c| 6 kxi − x0 kkck it follows that limi→∞ ϕ(xi ) = ϕ(x0 ). So, ϕ is continuous on Rn . Exercise 0.5. The triangle inequality kx − yk 6 kx − zk + kz − yk gives kx − yk − kz − yk 6 kx − zk.
(9.13)
Similarly, the inequality ky − zk 6 ky − xk + kx − zk implies −kx − zk 6 ky − xk − ky − zk = kx − yk − kz − yk. (9.14) Combining (9.13) and (9.14), we obtain kx − yk − kz − yk 6 kx − zk. Choose a vector x0 ∈ Rn and an infinite sequence of vectors x1 , x2 , . . . converging to x0 . From |δc (xi ) − δc (x0 )| = kxi − ck − kx0 − ck 6 kxi − x0 k it follows that limi→∞ δc (xi ) = δc (x0 ). Hence δc (x) is continuous on Rn .
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Exercise 0.6. Let f (ek ) = (a1k , . . . , amk ), 1 6 k 6 n, where e1 , . . . , en is the standard basis for Rn . For every vector x = (x1 , . . . , xn ), one has f (x) = f (x1 e1 + · · · + xn en ) = x1 f (e1 ) + · · · + xn f (en ) = (y1 , . . . , ym ), where yi =
n P
aik xk , 1 6 i 6 m. Let α =
k=1
n m P P i=1 k=1
n
a2ik
1/2
. We state
that kf (x)k 6 αkxk for all x ∈ R . Indeed, using the Cauchy-Schwarz inequality (β1 γ1 + · · · + βn γn )2 6 (β12 + · · · + βn2 )(γ12 + · · · + γn2 ), we obtain n P
yi2 =
aik xk
2
6
k=1
n P
n P
a2ik
k=1
x2k =
k=1
n P
a2ik kxk2 .
k=1
Hence kf (x)k2 =
m P i=1
m P n P
yi2 6
a2ik kxk2 = α2 kxk2 .
i=1 k=1
For the continuity of f , choose a vector x0 ∈ Rn and a sequence x1 , x2 , . . . converging to x0 . From kf (xi ) − f (x0 )k = kf (xi − x0 )k 6 αkxi − x0 k it follows that limi→∞ f (xi ) = f (x0 ). So, f is continuous on Rn . (i)
Exercise 0.7. Assume first that limi→∞ λj = λj , 1 6 j 6 r. Put γ = max {kb1 k, . . . , kbr k}. Given an ε > 0, there is an index i0 such that (i) |λj − λj | < ε/(γr) for all j = 1, . . . , r and i > i0 . Hence kxi − xk = k
r P j=1
(i)
(λj − λj )bj k 6
r P j=1
(i)
|λj − λj | kbj k 6
ε γr = ε γr
for all i > i0 , which gives limi→∞ xi = x. Conversely, suppose that limi→∞ xi = x. Put ui = xi − x and choose an orthonormal basis e1 , . . . , er for the subspace span {b1 , . . . , br }. We can write r r P P (i) ui = µk ek , i > 1, and bj = ajk ek , 1 6 j 6 r. k=1
k=1
The r × r-matrix A = (ajk ) is invertible because the set {b1 , . . . , br } is linearly independent. From limi→∞ ui = 0 and the equality r r P 2 P (i) (i) (i) 2 + · · · + µ(i) kui k2 = µk ek · µk ek = µ1 r k=1
k=1
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(i)
it follows that limi→∞ µk = 0 for all 1 6 k 6 r. Rewriting ui as ui = =
r P
(i)
j=1 r P k=1
(λj − λj )bj = P r j=1
r P
(i)
(λj − λj )
j=1
r P
ajk ek
k=1
(i) (λj − λj )ajk ek
one has (i)
µk =
r P j=1
(i)
(λj − λj )ajk ,
1 6 k 6 r, i > 1.
(9.15)
(i)
Since A is invertible, (9.15) is solvable for λj − λj : (i)
λj − λj =
r P k=1
(i)
bjk µk ,
1 6 j 6 r, i > 1.
Therefore, (i)
lim (λj − λj ) = lim (
i→∞
r P
(i)
i→∞ k=1
bjk µk ) =
r P k=1
(i)
bjk lim µk = 0, i→∞
1 6 j 6 r.
(i)
Hence limi→∞ λj = λj for all 1 6 j 6 r. Exercise 0.8. Since the case X ∩Y 6= ∅ is obvious, we may assume that X and Y are disjoint. Choose sequences x1 , x2 , . . . and y1 , y2 , . . . in the sets X and Y , respectively, such that limi→∞ kxi −yi k = δ(X, Y ). Because X is bounded, there is a subsequence xk1 , xk2 , . . . of x1 , x2 , . . . which converges to a point x ∈ cl X. Let an index j be such that kxi − yi k 6 δ(X, Y ) + 1 for all i > j and kxki − xk 6 1 for all ki > j. Then kx − yki k 6 kx − xki k + kxki − yki k 6 δ(X, Y ) + 2 for all ki > j, implying that the sequence yk1 , yk2 , . . . is bounded. Therefore, there is a subsequence ym1 , ym2 , . . . of yk1 , yk2 , . . . which converges to a point y ∈ cl Y . Finally, kx − yk = lim kxmi − ymi k = δ(X, Y ). i→∞
Exercise 0.9. (1) ⇒ (2). Assume for a moment that o ∈ cl (X −Y ). Then limi→∞ zi = o for a certain sequence of points zi ∈ X − Y , i > 1. Writing zi = xi − yi , where xi ∈ X and yi ∈ Y , one has δ(X, Y ) 6 lim kxi − yi k = 0, i→∞
in contradiction with the assumption δ(X, Y ) > 0.
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(2) ⇒ (3). The condition o ∈ / cl (X −Y ) implies the existence of a scalar ρ > 0 such that B2ρ (o) ∩ (X − Y ) = ∅. We state that Bρ (X) ∩ Bρ (X) = ∅. Indeed, assume for a moment the existence of a point z in Bρ (X) ∩ Bρ (X). Then one can find points x ∈ X and y ∈ Y such that kx − zk 6 ρ and ky − zk 6 ρ. Therefore, kx − yk 6 kx − zk + kz − yk 6 2ρ. Hence x − y ∈ B2ρ (o), contrary to B2ρ (o) ∩ (X − Y ) = ∅. (3) ⇒ (1). Assume for a moment that δ(X, Y ) = 0. Then kx − yk 6 2ρ for certain points x ∈ X and y ∈ Y . Let z = (x + y)/2. Then kz − xk 6 ρ and kz−yk 6 ρ, implying the inclusion z ∈ Bρ (X)∩Bρ (Y ), in contradiction with Bρ (X) ∩ Bρ (Y ) = ∅. Exercise 0.10. Since the case when at least one of the sets X and Y is empty is obvious, we may assume that both X and Y are nonempty. For the inclusion cl X + cl Y ⊂ cl (X + Y ), choose a point z ∈ cl X + cl Y . Then z = x + y for some points x ∈ cl X and y ∈ cl Y , and we can write x = limi→∞ xi and y = limi→∞ yi for certain sequences x1 , x2 , . . . and y1 , y2 , . . . from X and Y , respectively. Therefore, z = x + y = lim xi + lim yi = lim (xi + yi ) ∈ cl (X + Y ). i→∞
i→∞
i→∞
(1) Assume that at least one of the sets X and Y , say X, is bounded. By the above proved, it suffices to verify the inclusion cl (X + Y ) ⊂ cl X + cl Y . Choose a point z ∈ cl (X + Y ). Then z = limi→∞ zi for a certain sequence z1 , z2 , . . . from X + Y . Every zi can be written as zi = xi + yi , where xi ∈ X and yi ∈ Y , i > 1. Since X is bounded, the sequence x1 , x2 , . . . contains a subsequence xk1 , xk2 , . . . converging to a point x ∈ cl X. From lim yki = lim (zki − xki ) = lim zki − lim xki = z − x
i→∞
i→∞
i→∞
i→∞
it follows that z − x ∈ cl Y . Hence z = x + (z − x) ∈ cl X + cl Y . (2) Assume now that the sum X + Y is direct. In other words, there are independent subspaces S and T such that X ⊂ S and Y ⊂ T . Since X +Y ⊂ S +T , and since the subspace S +T is closed, one has cl (X +Y ) ⊂ S + T . As above, it suffices to show that cl (X + Y ) ⊂ cl X + cl Y . Let z ∈ cl (X + Y ). Then z = limi→∞ zi , where all z1 , z2 , . . . are in X + Y . Furthermore, zi = xi + yi , i > 1, with x1 , x2 , . . . in X and y1 , y2 , . . . in Y . Let p = dim S and q = dim T . Choose vectors b1 , . . . , bp+q such that b1 , . . . , bp and bp+1 , . . . , bp+q form bases for S and T , respectively. Then z = λ1 b1 + · · · + λp+q bp+q ,
(i)
(i)
zi = λ1 b1 + · · · + λp+q bp+q ,
i > 1.
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Clearly, (i)
xi = λ1 b1 + · · · + λ(i) p bp ,
(i)
(i)
yi = λp+1 bp+1 + · · · + λp+q bp+q ,
i > 1.
(i)
According to Exercise 0.7, limi→∞ λj = λj for all 1 6 j 6 p + q. By the same exercise, limi→∞ xi = x and limi→∞ yi = y, where x = λ1 b1 + · · · + λp bp
and y = λp+1 bp+1 + · · · + λp+q bp+q .
Finally, z = lim zi = lim (xi +yi ) = lim xi + lim yi = x+y ∈ cl X +cl Y . i→∞
i→∞
i→∞
i→∞
Exercise 0.11. Express X as a union of a countable nested sequence X1 ⊂ X2 ⊂ · · · of compact sets. Also, let Y = Y1 ∪Y2 ∪· · · be a similar representation of Y . By Exercise 0.10, all sets Xi + Yi , i > 1, are compact. Because X1 + Y1 ⊂ X2 + Y2 ⊂ · · ·
and X + Y = (X1 + Y1 ) ∪ (X2 + Y2 ) ∪ · · ·
we conclude that X + Y is an Fσ -set. Since f (X) is the union of the nested sequence f (X1 ) ⊂ f (X2 ) ⊂ · · · , and since every set f (Xi ), i > 1, is compact as a continuous image of a compact set (see Exercise 0.6), we obtain that f (X) is an Fσ -set. Chapter 1 Exercise 1.1. If L is a hyperplane, then its complement consists of two open halfspaces; so, Rn \ L is not connected. Suppose now that L is not a hyperplane. Let a subspace S ⊂ Rn be complementary to L. Clearly, dim S = n − dim L > 2. Choose points x and y in Rn \ L, and denote by x0 and y0 , respectively, their projections on S along L. By Theorem 1.11, the set L ∩ S is a singleton, z. Clearly, one can choose a simple polygonal path P with end points x and y which does not contain z. The polygonal path [x, x0 ] ∪ P ∪ [y0 , y] connects x and y within Rn \ L. Hence Rn \ L is connected. Exercise 1.2. If one of the planes, say Lγ ∈ F, contains all the others, then L = Lγ . Conversely, assume that L is a plane. Since the case L = ∅ is obvious, we assume that L 6= ∅. We are going to prove the inclusion L ∈ F by induction on m = dim L. The case m = 0 is obvious. Let m > 1. If m = 1, then at least one Lα ∈ F coincides with L (otherwise, each Lα would be a singleton and ∪ Lα would be a countable subset of L). α
Suppose that the inclusion L ∈ F holds for all m 6 k − 1 (k > 2), and let dim L = k. Assume for a moment that L ∈ / F. Choose in L a plane M of dimension k − 1. Then M = ∪ (M ∩ Lα ) is a countable union of α
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planes M ∩ Lα (see Theorem 1.4). By the induction hypothesis, there is a plane Lβ ∈ F such that M ∩ Lβ = M , which gives M = Lβ ∈ F (indeed, if M 6= Lβ then L = Lβ ∈ F, contrary to the assumption). Let l ⊂ L be a line which meets M at a single point, z. For a given point x ∈ l, consider the plane Mx = (x − z) + M . Clearly, all planes Mx lie in L, and Mx ∩ My = ∅ whenever x 6= y. Every plane Mx is a member of F by the above argument. Since F is countable, the family {Mx } should also be countable, which is impossible. The obtained contradiction shows that L ∈ F. Exercise 1.3. If X is an m-dimensional plane, then, according to Theorem 1.79, the set Bρ (c) ∩ X is an m-dimensional ball. Conversely, assume the existence of a scalar ρ > 0 such that the set Bρ (c) ∩ X is an mdimensional ball for every choice of a point c ∈ X. Suppose for a moment that X is not a plane. Then L = aff X 6= X. Choose points u ∈ X and v ∈ L \ X. By Theorem 1.46, the line hu, vi lies in L. Denote by (w, v) the largest open segment in [u, v] which is disjoint with X (possibly, w = v), and choose a point s ∈ [u, w] ∩ X such that ks − wk 6 ρ/2. By the assumption, the set Bρ (s) ∩ X is an m-dimensional ball. Hence there is an m-dimensional plane M through s such that Bρ (s) ∩ X = Bρ (s) ∩ M . On the other hand, L = aff (Bρ (s) ∩ X) because Bρ (s) ∩ X is an m-dimensional subset of L (see Corollary 1.77). This argument and Theorem 1.79 give L = aff (Bρ (s) ∩ X) = aff (Bρ (s) ∩ M ) = M. Consequently, Bρ (s) ∩ X = Bρ (s) ∩ M = Bρ (s) ∩ L. Furthermore, v 6= w, since otherwise v ∈ Bρ (s) ∩ L = Bρ (s) ∩ X ⊂ X. Now, choose a point r ∈ (v, w) such that kr − wk 6 ρ/2. Then kr − sk 6 ks − wk + kw − rk 6 ρ
and r ∈ hu, vi ⊂ L,
implying that r ∈ Bρ (s) ∩ L = Bρ (s) ∩ X ⊂ X, contrary to (w, v) ∩ X = ∅. Summing up, X is a plane. If, additionally, the set X is closed, then v 6= w in the above argument. Hence every scalar ρ = ρ(w) > 0 satisfies the desired conclusion. Exercise 1.4. Let H = {x ∈ Rn : x · c = γ}, where c 6= o and γ is a scalar (see Theorem 1.17). Without loss of generality, assume that a belongs to the open halfspace W = {x ∈ Rn : x · c < γ}. Denote by u the point at which h meets H. Then a · c < γ = u · c, and Lemma 1.26
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shows that h = [a, ui. Consequently, h0 = (b − a) + [a, ui = [b, u0 i, where u0 = b − a + u. Clearly, b·c = (a − u)·c + u0 ·c < u0 ·c.
(9.16)
Every point x ∈ [b, u0 i can be written as x = (1 − λ)b + λu0 , λ > 0, which gives x·c = (1 − λ)b·c + λu0 ·c, λ > 0.
(9.17)
Assume first that b ∈ W . Then b·c < γ, and h0 meets H at the point γ − b·c x0 = (1 − λ0 )b + λ0 u0 ∈ [b, u0 i, where λ0 = 0 > 0. (u − b)·c Indeed, x0 ·c =
u0 ·c − γ γ − b·c 0 b·c + 0 u ·c = γ. 0 (u − b)·c (u − b)·c
If b ∈ / W , then b·c > γ (since b ∈ / H), and a combination of (9.16) and (9.17) shows that x·c > γ for all points x = (1 − λ)b + λu0 , where λ > 0. Hence h0 ∩ H = ∅. Exercise 1.5. Let m = dim L. Denote by Mα the (m − 1)-dimensional plane which determines a halfplane Fα ∈ F (see Corollary 1.39). Corollary 1.38 shows that Fα can be described in one of the following ways: Fα = {x ∈ L : x·cα 6 γα },
Fα = {x ∈ L : x·cα < γα },
(9.18)
where cα is a nonzero vector in Nα = sub L ∩ (sub Mα )⊥ . It is easy to see that Nα is a 1-dimensional subspace. (1) Assume first that U is a halfplane and denote by the (m − 1)dimensional plane which determines U . Corollary 1.40 shows that each plane Mα is a translate of M . In this case, all subspaces Nα coincide. Hence all cα are scalar multiples of the same nonzero vector, say c. This argument shows that the family F is nested. Conversely, suppose that the family F is nested. By Corollary 1.40, the planes {Mα } are pairwise parallel. Hence they have a common normal vector c ∈ sub L. Without loss of generality, we may assume that each Fα has one of the forms Fα = {x ∈ L : x·c 6 µα },
Fα = {x ∈ L : x·c < µα }.
With µ = sup µα , the set U has one of the forms U = {x ∈ Rn : x·c 6 µ},
U = {x ∈ Rn : x·c < µ}.
Hence U is a halfplane of L. Part (2) of the exercise is similar.
(9.19)
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Exercise 1.6. Suppose that a translate Z = Y + c of Y lies in aff X. Then aff Z ⊂ aff X. Choose a point y ∈ Y and let z = y + c. By Theorem 1.53, Z − z ⊂ aff X − z = aff (X − z). Since o = z − z ∈ Z − z ⊂ aff (X − z), the plane aff (X − z) is a subspaces (see Theorem 1.2). Theorem 1.2 gives aff (X − z) + aff Z = aff (X − z) + Z = aff X. From Theorems 1.2 and 1.53 it follows aff (X + Y ) = aff (X − z + Z + y) = aff (X − z) + aff Z + y = aff X + y, aff X + Y = aff (X − z) + Z + y = aff X + y. Consequently, aff (X + Y ) = aff X + Y = aff X + y. Exercise 1.7. Let Li = aff Xi , 1 6 i 6 r. By Theorem 1.53, aff (µ1 X1 + · · · + µr Xr ) = µ1 L1 + · · · + µr Lr . Theorem 1.2 gives Li = ai + Si , where Si is a subspace, 1 6 i 6 r. Since µ1 L1 + · · · + µr Lr = (µ1 a1 + · · · + µr ar ) + (µ1 S1 + · · · + µr Sr ), one has dim (µ1 S1 + · · · + µr Sr ) = m. Let F = {µ1 S1 , . . . , µr Sr }. Choose recursively a maximal subset H = {µi1 Si1 , . . . , µik Sik } ⊂ F such that µij+1 Sij+1 6⊂ µi1 Si1 + · · · + µij Sij ,
1 6 j 6 k − 1.
Clearly, k 6 m < r and µi1 Si1 + · · · + µik Sik = µ1 S1 + · · · + µr Sr , since otherwise we could further enlarge H. Let I = {i1 , . . . , ik }. Then card I 6 m and aff (µ1 X1 + · · · + µr Xr ) = µ1 L1 + · · · + µr Lr = (µ1 a1 + · · · + µr ar ) +
P
µi Si
i∈I
=
P
µi ai +
P i∈I /
µi (ai + Si )
i∈I
i∈I /
=
P
µi ai +
P i∈I
µi Li =
P i∈I /
µi ai +
P i∈I
µi aff Xi .
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Exercise 1.8. Let f be a homothety. Then f (x) = a + µx for a certain a ∈ Rn and µ 6= 0. Choose a line l ⊂ Rn . We can write l = c + S, where c ∈ Rn and S is a 1-dimensional subspace of Rn . From f (l) = a + µl = (a + µc) + µS = (a + µc) + S = µc + l it follows that f (l) is a translate of l. Conversely, assume that f is a mapping which takes every line on a parallel one. Excluding the trivial case when f is the identity map, we assume the existence of a point u ∈ Rn such that f (u) 6= u. Choose a line l ⊂ Rn through u which does not contain f (u). Then the line f (l), being parallel to l, must be disjoint from l. Let a point v ∈ l \ {u} be such that f (u) 6= f (v). Then l = hu, vi and f (l) = hf (u), f (v)i according to Lemma 1.26. f (v) v (((r( v r r f (v) r( c( ((( r ( h hhhrh r r u hhh rhh u f (u) f (u) 1. Suppose first that the lines hu, f (u)i and hv, f (v)i are parallel (see the picture above). Then v = u + f (v) − f (u), or f (u) − u = f (v) − v. Hence the translate t(x) = a + x, where a = f (u) − u, takes u and v on f (u) and f (v), respectively. We state that f = t. Indeed, choose any point x ∈ Rn \ {u, v}. 1a. Let x ∈ / hu, vi. Since f (u) 6= f (v), the point f (x) is distinct from at least one of them. Assume, for example, that f (x) 6= f (u). By the assumption, hf (u), f (x)i is parallel to hu, xi. Hence f (v) ∈ / hf (u), f (x)i, implying that f (x) 6= f (v). Because hf (v), f (x)i is parallel to hv, xi, the triangle ∆(f (x), f (u), f (v)) equals a + ∆(x, u, v). Consequently, f (x) = a + x = t(x). 1b. Let x ∈ hu, vi. Choose a point z ∈ Rn which does not belong to the union of lines hu, vi and hu, f (u)i. Then f (z) = a + z, as proved in 1a. Since the lines hu, f (u)i and hz, f (z)i are parallel, and since x ∈ / hu, zi, we can apply the argument above (with u, z instead of u, v). This gives f (x) = a + x = t(x). 2. Suppose now that the lines hu, f (u)i and hv, f (v)i are not parallel. Then they meet at a single point, c. Since hf (u), f (v)i and hu, vi are parallel, there is a scalar µ 6= 0 such that f (u) − f (v) = µ(u − v). It is easy to see (see the above picture) that f (u) − c = µ(u − c)
and f (v) − c = µ(v − c),
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which implies that the homothety h(x) = c + µ(x − c) takes u and v on f (u) and f (v), respectively. We state that f = h. For this, observe first that f (c) = c. Indeed, the parallel lines hu, f (u)i and hf (u), f (f (u))i coincide because they share the point f (u). Similarly, the lines hv, f (v)i and hf (v), f (f (v))i coincide. Consequently, {f (c)} = hf (u), f (f (u))i ∩ hf (v), f (f (v))i = hu, f (u)i ∩ hv, f (v)i = {c}. Finally, let x be a point in Rn \ {c}. Then x does not belong to at least one of the lines hc, ui and hc, vi. Suppose that x ∈ / hc, ui (the case x ∈ / hc, vi is similar). Similarly to the argument above, the lines hc, xi and hc, f (x)i coincide, implying the inclusion f (x) ∈ hc, xi. Hence f (x) 6= f (u). Since the lines hu, xi and hf (u), f (x)i are parallel, the triangles ∆(c, u, x) and ∆(c, f (u), f (x)) are homothetic. Therefore, f (x) − c = µ(x − c), or f (x) = c + µ(x − c) = h(x), as desired. Exercise 1.9. The properties of norm (see page 6) give k(1 − λ)x + λy − zk = k(1 − λ)(x − z) + λ(y − z)k 6 (1 − λ)kx − zk + λky − zk.
(9.20)
Suppose that x 6= y and 0 < λ < 1. Choosing suitable Cartesian coordinates, we suppose that x = (0, 0, . . . , 0), y = (1, 0, . . . , 0), and z = (z1 , z2 , 0, . . . , 0). Then (9.20) becomes as ((z1 − λ)2 + z22 )1/2 6 (1 − λ)(z12 + z22 )1/2 + λ((z1 − 1)2 + z22 )1/2 .
(9.21)
If z ∈ hx, yi then z2 = 0, and (9.21) becomes as |z1 − λ| 6 (1 − λ)|z1 | + λ|z1 − 1|, with equality if and only if either z1 6 0 or z1 > 1. Suppose that z ∈ / hx, yi. Then z2 6= 0. Squaring both parts of (9.21) and simplifying the expressions, we obtain ((z1 − 1)z1 + z22 )2 6 ((z1 − 1)2 + z22 )(z12 + z22 ), which is strict because of z2 6= 0.
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Chapter 2 Exercise 2.1. (1) Solid ellipsoid. If y, z ∈ Γ, then y = (y1 , . . . , yn ) and z = (z1 , . . . , zn ) such that a1 y12 + · · · + an yn2 6 1,
a1 z12 + · · · + an zn2 6 1.
For every scalar λ ∈ [0, 1], one has (1 − λ)y + λz = ((1 − λ)y1 + λz1 , . . . , (1 − λ)yn + λzn ). Using the obvious inequalities ((1 − λ)yi + λzi )2 6 (1 − λ)yi2 + λzi2 ,
1 6 i 6 n,
(9.22)
we obtain a1 ((1 − λ)y1 + λz1 )2 + · · · + an ((1 − λ)yn + λzn )2 6 (1 − λ)(a1 y12 + · · · + an yn2 ) + λ(a1 z12 + · · · + an zn2 ) 6 (1 − λ) + λ = 1. Hence (1 − λ)y + λz ∈ Γ, which shows that Γ is convex. (2) Solid elliptic paraboloid. If y, z ∈ Φ, then y = (y1 , . . . , yn ) and z = (z1 , . . . , zn ) such that 2 a1 y12 + · · · + an−1 yn−1 6 yn ,
2 a1 z12 + · · · + an−1 zn−1 6 zn .
For every scalar λ ∈ [0, 1], one has (1 − λ)y + λz = ((1 − λ)y1 + λz1 , . . . , (1 − λ)yn + λzn ). From (9.22) we obtain a1 ((1 − λ)y1 + λz1 )2 + · · · + an−1 ((1 − λ)yn−1 + λzn−1 )2 2 2 6 (1 − λ)(a1 y12 + · · · + an−1 yn−1 ) + λ(a1 z12 + · · · + an−1 zn−1 )
6 (1 − λ)yn + λzn . Hence (1 − λ)y + λz ∈ Φ, which shows that Φ is convex. (3) Solid elliptic hyperboloid. If y, z ∈ Ω, then y = (y1 , . . . , yn ) and z = (z1 , . . . , zn ) such that yn , zn > 0 and 2 a1 y12 + · · · + an−1 yn−1 + 1 6 an yn2 ,
2 a1 z12 + · · · + an−1 zn−1 + 1 6 an zn2 .
For every scalar λ ∈ [0, 1], one has (1 − λ)y + λz = ((1 − λ)y1 + λz1 , . . . , (1 − λ)yn + λzn ). Clearly, (1 − λ)yn + λzn > 0. From the Cauchy-Schwarz inequality (α1 β1 + · · · + αn βn )2 6 (α12 + · · · + αn2 )(β12 + · · · + βn2 )
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it follows that a1 y1 z1 + · · · + an−1 yn−1 zn−1 + 1 2 2 6 [(a1 y12 + · · · + an−1 yn−1 + 1)(a1 z12 + · · · + an−1 zn−1 + 1)]1/2 1/2 6 (an yn2 )(an zn2 ) = an yn zn .
This inequality gives a1 ((1 − λ)y1 + λz1 )2 + · · · + an−1 ((1 − λ)y1 + λz1 )2 + 1 2 = (1 − λ)2 (a1 y12 + · · · + an−1 yn−1 + 1)
+ 2(1 − λ)λ(a1 y1 z1 + · · · + an−1 yn−1 zn−1 + 1) 2 + λ2 (a1 z12 + · · · + an−1 zn−1 + 1)
6 (1 − λ)2 an yn2 + 2(1 − λ)λ an yn zn + λ2 an zn2 = an ((1 − λ)yn + λzn )2 . Hence (1 − λ)y + λz ∈ Ω, which shows that Ω is convex. Exercise 2.2. Since the “if” part is obvious, it remains to prove the “only if” part. To show the convexity of X, choose points x, y ∈ X. Assume for a moment that [x, y] 6⊂ X. Then [x, y] \ X is a relatively open subset of [x, y]; so, it can be expressed as a disjoint union of countably many open subintervals of [x, y]. Let (u, v) be one of these subintervals. Then u, v ∈ X because X is closed. By the assumption, X contains a point w = (1 − λ)u + λv, with 0 < λ < 1. Then w ∈ (u, v), contradicting the choice of (u, v). Hence X is a convex set. Exercise 2.3. Choose a pair of points x, y ∈ X, and denote by Φ the set of scalars 0 < η < 1 such that (1 − η)x + ηy ∈ X. Clearly, {0, λ0 , 1} ⊂ Φ. We state that (1 − λ0 )η1 + λ0 η2 ∈ Φ for any choice of η1 , η2 ∈ Φ. Indeed, if z1 = (1 − η1 )x + η1 y
and z2 = (1 − η2 )x + η2 y,
then z1 , z2 ∈ X, which gives (1 − λ0 )z1 + λ0 z2 ∈ X. Consequently, the inclusion (1 − (1 − λ0 )η1 − λ0 η2 )x + ((1 − λ0 )η1 + λ0 η2 )y = (1 − λ0 )z1 + λ0 z2 ∈ X shows that (1 − λ0 )η1 + λ0 η2 ∈ Φ. This property of Φ obviously implies that Φ is dense in [0, 1]. Therefore, the set X ∩ [x, y] is dense in [x, y]. Since X is relatively open, one has [x, y] ⊂ X. Hence X is convex.
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Exercise 2.4. If K is closed, then for every line l ⊂ Rn , the set K ∩ l is closed. Conversely, let K ⊂ Rn be a convex set such that for every line l ⊂ Rn , the set K ∩l is closed. Choose a point z ∈ rint K. For a given point x ∈ cl K consider the line l = hz, xi. By Theorem 2.36, K ∩ l contains the open segment (z, x). Furthermore, [x, z] ⊂ K ∩ l ⊂ K because K ∩ l is closed. Hence x ∈ K, and K is closed. If K ⊂ Rn a relatively open convex set and l is a line meeting K, then Corollary 2.27 implies that K ∩ l = rint K ∩ l = rint (K ∩ l). Hence K ∩ l is relatively open. Conversely, let a convex set K ⊂ Rn be such that for every line l the intersection K ∩ l is relatively open. To prove that K is relatively open, it suffices to show that every point x ∈ K belongs to rint K. Since the case K = {x} is obvious, we may assume that dim K > 1. Let y ∈ K be a point distinct from x, and l be the line through x and y. Since the relatively open set K ∩ l contains both x and y, there is a scalar γ > 1 such that (1 − γ)y + γx ∈ K ∩ l ⊂ K. By Theorem 2.24, x ∈ rint K. Exercise 2.5. Put M = (λ + µ)K + then, by Theorem 2.4,
|λ|+|µ|−|λ+µ| (K 2
− K). If λ + µ > 0,
M = λK + µK + 0(K − K) = λK + µK if λ > 0 and µ > 0, M = (λ + µ)K − µ(K − K) = λK + µK if λ > −µ > 0, M = (λ + µ)K − λ(K − K) = λK + µK if µ > −λ > 0. If λ + µ 6 0, then replace K with −K, µ with −λ, and λ with −µ in the above argument. Exercise 2.6. We consider only the case λ > γ, since all other cases, λ < γ, µ > δ, or µ < δ are similar. By Theorem 2.47, we obtain from λK + µM = γK + δM that (λ − γ)K + µM = δM . Similarly, 1) (λ − γ)K = (δ − µ)M if δ > µ, δ−µ M if δ = µ, 2) (λ − γ)K = {o} and K = {o} = λ−γ 3) (λ − γ)K + (µ − δ)M = {o} and K = M = {o} if δ < µ. Summing up, K =
δ−µ λ−γ M .
Exercise 2.7. Assume first that z ∈ rint K. By the definition, there is a scalar ρ > 0 such that Bρ (z) ∩ aff K ⊂ K. Put y=
1 m+1 (y1
+ · · · + ym+1 ),
δ = max {ky − y1 k, . . . , ky − ym+1 k},
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and consider the positive homothety ( (z − y) + x f (x) = u + ρ/δ(x − u), u =
359
if δ 6 ρ, δ δ−ρ z
−
ρ δ−ρ y
if δ > ρ.
Let zi = f (yi ), 1 6 i 6 m + 1. As easily seen, z1 , . . . , zm+1 ∈ aff K
and z =
1 m+1 (z1
+ · · · + zm+1 ).
By Theorem 1.83, the points z1 , . . . , zm+1 are affinely independent. Clearly, the simplices ∆(y1 , . . . , ym+1 ) and ∆(z1 , . . . , zm+1 ) are homothetic. If δ 6 ρ, then zi = (z − y) + yi and kz − zi k = ky − yi k 6 ρ, 1 6 i 6 m + 1. If δ > ρ, then zi = z + ρδ (yi − y) and ρ kz − zi k = ky − yi k < ky − yi k 6 ρ, 1 6 i 6 m + 1. δ In either case, z1 , . . . , zm+1 ∈ Bρ (z). Hence ∆(z1 , . . . , zm+1 ) = conv {z1 , . . . , zm+1 } ⊂ Bρ (z) ∩ aff K ⊂ K. Conversely, let ∆ = ∆(z1 , . . . , zm+1 ) ⊂ K be a simplex such that z = 1 m+1 (z1 + · · · + zm+1 ). By Theorem 2.17, z ∈ rint ∆. Since aff ∆ = aff K, Theorem 2.15 gives z ∈ rint ∆ ⊂ rint K. Exercise 2.8. One has rint K1 + K2 ⊂ rint (K1 + K2 ) according to Theorem 2.15. The opposite inclusion, rint (K1 + K2 ) = rint K1 + rint K2 ⊂ rint K1 + K2 , follows from Theorem 2.29. Exercise 2.9. Choose points zi ∈ µi Ki and put Ki0 = µi Ki −zi , 1 6 i 6 r. Then o ∈ Ki0 and o ∈ K10 + · · · + Kr0 , which gives, by Theorem 1.50, aff (K10 + · · · + Kr0 ) = span (K10 + · · · + Kr0 ), aff Ki0 = span Ki0 , 1 6 i 6 r. Clearly, there is an index set I ⊂ {1, . . . , r} such that card I 6 m and P span (K10 + · · · + Kr0 ) = span Ki0 . i∈I
By Theorems 2.29 and 2.15, rint (K10 + · · · + Kr0 ) = rint K10 + · · · + rint Kr0 P P 0 ⊂ rint Ki0 + Ki ⊂ rint (K10 + · · · + Kr0 ). i∈I
i∈I /
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Hence, rint (K10 + · · · + Kr0 ) =
P
rint Ki0 +
i∈I
P
Ki0 . Finally,
i∈I /
rint (µ1 K1 + · · · + µr Kr ) = (z1 + · · · + zr ) + rint (K10 + · · · + Kr0 ) P P 0 = (z1 + · · · + zr ) + rint Ki0 + Ki i∈I
=
P
µi rint Ki +
i∈I
P
i∈I /
µi Ki .
i∈I /
Exercise 2.10. Let ∆ = ∆(x1 , . . . , xr+1 ) and ∆0 = ∆(x01 , . . . , x0r+1 ). According to Theorem 1.68, both affinely independent sets {x1 , . . . , xr+1 } and {x01 , . . . , x0r+1 } can be completed, respectively, to affine bases {x1 , . . . , xn+1 } and {x01 , . . . , x0n+1 } for Rn . By Theorem 1.85, there is an invertible affine transformation f : Rn → Rn such that f (xi ) = x0i for all 1 6 i 6 n + 1. We state that f (∆) = ∆0 . Indeed, if u ∈ ∆, then x can be written as a convex combination u = λ1 x1 + · · · + λr+1 xr+1 , and Theorem 1.87 gives f (u) = λ1 f (x1 ) + · · · + λr+1 f (xr+1 ) = λ1 x01 + · · · + λr+1 x0r+1 ∈ ∆0 . Hence f (∆) ⊂ ∆0 . Conversely, if v 0 ∈ ∆0 , then v 0 can be expressed as a convex combination v 0 = µ1 x01 + · · · + µr+1 x0r+1 . With v = µ1 x1 + · · · + µr+1 xr+1 , we have v ∈ ∆ and f (v) = v 0 . Summing up, f (∆) = ∆0 . Exercise 2.11. Let F = Cα be the family of all maximal convex subsets of X. From Zorn’s lemma (see page 1) it follows that every point x ∈ X belongs to a set Cα ∈ F. Suppose first that a point c belongs to ∩ Cα . If x ∈ X and Cα is an element of F containing x, then [c, x] ⊂ Cα ⊂ X. Conversely, let [c, x] ⊂ X for all x ∈ X. Assume for a moment the existence of a set Cγ ∈ F such that c ∈ / Cγ . It is easy to see that the set C = ∪([c, u] : u ∈ Cγ ) is a convex set which lies in X due to the assumption on c. Since Cγ is a proper subset of C, we obtain the contradiction with the choice of Cγ . Summing up, c ∈ Cγ , and c ∈ ∩ Cα . Exercise 2.12. Clearly, f (x) = a + g(x) and ϕ(x) = α + x·c are affine transformations. It suffices to show that the h-image of every segment [x, z] ⊂ Rn is the segment [h(x), h(z)]. Indeed, choose a point y = (1 − λ)x + λz ∈ [x, z],
0 6 λ 6 1,
and let µ = λϕ(z)/((1 − λ)ϕ(x) + λϕ(z)). Then 0 < µ < 1 because both
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ϕ(x) and ϕ(z) are either positive or negative. Therefore, (1 − λ)f (x) + λf (z) (1 − λ)ϕ(x) + λϕ(z) (1 − λ)ϕ(x) f (x) λϕ(z) f (z) = + (1 − λ)ϕ(x) + λϕ(z) ϕ(x) (1 − λ)ϕ(x) + λϕ(z) ϕ(z)
h(y) =
= (1 − µ) h(x) + µ h(z) ∈ [h(x), h(z)]. Hence h([x, z]) ⊂ [h(x), h(z)]. Conversely, let u = (1 − δ) h(x) + δ h(z), where 0 < δ < 1. Put γ = δϕ(x)/(δϕ(x) + (1 − δ)ϕ(z)). Then 0 < γ < 1 and v = (1 − γ)x + γz ∈ (x, z). As above, h(v) = u, which gives [h(x), h(z)] ⊂ h([x, z]). Chapter 3 Exercise 3.1. Without loss of generality, we may put X = {x1 , . . . , xt }, where t 6 r. Since ∆ is a convex set (see Theorem 2.6), the set K = ∪ ([u1 , u2 ] : u1 ∈ ∆1 , u2 ∈ ∆2 ) lies in ∆. Conversely, let u ∈ ∆. Then u can be written as an affine combination u = λ1 x1 + · · · + λr+1 xr+1 . Put λ = λ1 + · · · + λt . If λ = 0 or λ = 1, then the inclusion u ∈ K is obvious. Suppose 0 < λ < 1 and consider the convex combinations u1 =
λt λ1 x1 + · · · + xt λ λ
and u2 =
λt+1 λr+1 xt+1 + · · · + xr+1 . 1−λ 1−λ
Clearly, u1 ∈ ∆1 , u2 ∈ ∆2 , and u = λu1 + (1 − λ)u2 ∈ [u1 , u2 ]. Hence ∆ ⊂ K. Exercise 3.2. As shown in example on page 72, the ball Bρ (c) is a convex set, which gives conv2 Sρ (c) ⊂ Bρ (c). Conversely, let u ∈ Bρ (c). If u = c, then choose any point v ∈ Sρ (c), and let w = 2c−u. From kw −ck = ku− ck = ρ it follows that w ∈ Sρ (c). Furthermore, c = 21 (u+w) ∈ conv2 Sρ (c). Suppose that u 6= c and put u1 = c −
ρ (u − c), ku − ck
u2 = c +
ρ (u − c). ku − ck
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Then ku1 − ck = ku2 − ck = ρ, implying the inclusions u1 , u2 ∈ Sρ (c). With λ = (ρ + ku − ck)/(2ρ), one has ρ − ku − ck ρ (c − (u − c)) 2ρ ku − ck ρ ρ + ku − ck (c + (u − c)) + 2ρ ku − ck ρ − ku − ck ρ + ku − ck = u1 + u2 = (1 − λ)u1 + λu2 . 2ρ 2ρ
u=
Hence Bρ (c) ⊂ conv2 Sρ (c). Exercise 3.3. The set int Q is convex as the intersection of open halfspaces Wi = {(x1 , . . . , xn ) : xi > 0}, Wi0 = {(x1 , . . . , xn ) : xi < 1}, i 6 i 6 n, (see example on page 72 and Theorem 2.8). Then X ⊂ conv2 X ⊂ conv X ⊂ int Q. So, it suffices to show that int Q ⊂ conv2 X. For this, choose any point u = (u1 , . . . , un ) ∈ int Q. Reordering the coordinate vectors of Rn , we may assume that all u1 , . . . , uk are rational and all uk+1 , . . . , un are irrational (possibly, k = 0 or k = n). If k = 0, then u ∈ X ⊂ conv2 X. Let k > 1. Choose an irrational number ξ > 0 so small that 0 < ui − ξ and ui + ξ < 1 for all 1 6 i 6 k. Let x = (u01 , . . . , u0k , uk+1 , . . . , un )
and z = (u001 , . . . , u00k , uk+1 , . . . , un ),
where u0i = ui + ξ and u00i = ui − ξ, 1 6 i 6 k. Clearly, x, z ∈ X and u = (x + z)/2. Hence u ∈ conv2 X. Exercise 3.4. According to Corollary 3.14, the inclusion x ∈ conv (y ∪ K) gives x = (1 − λ)y + λu for a certain point u ∈ K and a scalar λ ∈ [0, 1]. Similarly, y ∈ conv (x ∪ K) gives y = (1 − µ)z + µx for a certain z ∈ K and µ ∈ [0, 1]. Assume for a moment that x 6= y. Then λ 6= 0 and µ 6= 1, which implies 0 6 (1 − λ)µ < 1. From y = (1 − µ)z + µx = (1 − µ)z + µ((1 − λ)y + λu), it follows that (1 − (1 − λ)µ)y = (1 − µ)z + λµu. Therefore, y=
λµ 1−µ z+ u 1 − µ + λµ 1 − µ + λµ
is a convex combination of z and u. Hence y ∈ [z, u] ⊂ K, in contradiction with the hypothesis. Consequently, x = y.
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Exercise 3.5. We will prove the contrapositive statement: A finite set X ⊂ Rn is affinely dependent if and only if it it contains disjoint subsets whose convex hulls meet. Let a set X = {x1 , . . . , xr } be affinely dependent. This means the existence of scalars ν1 , . . . , νr , not all zero, such that ν1 x1 + · · · + νr xr = o,
ν1 + · · · + νr = 0.
Put I = {i : νi > 0} and J = {i : νi < 0}. Then both sets I and J are nonempty and partition the set {1, . . . , r}. Furthermore, P P νi xi = (−νj )xj . (9.23) i∈I
i∈J
P
P Let ν = i∈I νi . Clearly, ν > 0 and i∈J (−νj ) = ν. Dividing both parts of (9.23) by ν, we obtain P (−νj ) P νi xi = xj . (9.24) ν ν i∈J i∈I As easily seen, left-hand part of (9.24) is a convex combination of points from the set Y = {xi : i ∈ I}; similarly, right-hand side of (9.24) is a convex combination of points from the set Z = {xj : j ∈ J}. Hence Y ∩ Z = ∅ and conv Y ∩ conv Z 6= ∅. Conversely, let X = Y ∪ Z be a partition of X such that conv Y ∩ conv Z 6= ∅. Choose a point x ∈ conv Y ∩ conv Z. By Theorem 3.3, x can be expressed as convex combinations, x = λ1 y1 + · · · + λk yk = µ1 z1 + · · · + µm zm , of certain points y1 , . . . , yk ∈ Y and z1 , . . . , zm ∈ Z. Definition 1.58 and the equalities λ1 y1 + · · · + λk yk − µ1 z1 − · · · − µm zm = o, λ1 + · · · + λk − µ1 − · · · − µm = 0 show that the set {y1 , . . . , yk , z1 , . . . , zm } is affinely dependent. Hence X is affinely dependent. Exercise 3.6. Let Y = {y1 , . . . , yr }. We proceed by induction on r > 2. Let r = 2. By Theorem 3.7, y1 can be expressed as a convex combination y1 = λ1 x1 + · · · + λp xp + λp+1 y2 , there x1 , . . . , xp ∈ X, p 6 m. Similarly, there are q (6 m) points z2 , . . . , zq+1 ∈ X such that y2 can be written as a convex combination y2 = µ1 y1 + µ2 z2 + · · · + µq+1 zq+1 . Both scalars λp+1 and µ1 are distinct from 1 because of y1 6= y2 . Solving the equality y1 = λ1 x1 + · · · + λp xp + λp+1 y2 = λ1 x1 + · · · + λp xp + λp+1 (µ1 y1 + µ2 z2 + · · · + µq+1 zq+1 )
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for y1 , we obtain the convex combination λ1 λp y1 = x1 + · · · + xp 1 − λp+1 µ1 1 − λp+1 µ1 λp+1 µ2 λp+1 µq+1 + z2 + · · · + zq+1 . 1 − λp+1 µ1 1 − λp+1 µ1 Similarly, y2 can be expressed as a convex combination λ1 µ1 λp µ1 y2 = x1 + · · · + xp 1 − λp+1 µ1 1 − λp+1 µ1 µq+1 µ2 + z2 + · · · + zq+1 . 1 − λp+1 µ1 1 − λp+1 µ1 Hence {y1 , y2 } lies in the convex hull of 2m or fewer points from X. Assume that the statement holds for all r 6 s, and let Y = {y1 , . . . , ys+1 } be a set of s + 1 points in conv X, s > 2. By the induction hypothesis, there is a set Z0 ⊂ X of cardinality sm such that y1 , . . . , ys ∈ conv Z0 . Choose a point z0 ∈ Z0 . By Theorem 3.7, there are t (6 m) points u1 , . . . , ut ∈ X such that ys+1 can be written as a convex combination ys+1 = α1 u1 + · · · + αt ut + αt+1 z0 . Therefore, ys+1 ∈ conv {u1 , . . . , ut , z0 }. Put Z = Z0 ∪ {u1 , . . . , ut }. Clearly, Z ⊂ X, card Z 6 (s + 1)m, and Y ⊂ conv Z. Exercise 3.7. Let Y = {y1 , . . . , yr }. For every point yi there is a finite subset Zi of X such that yi ∈ rint (conv Zi ) and aff X = aff Zi , 1 6 i 6 r (see Theorem 3.20 and Corollary 3.21). Put Z = Z1 ∪ · · · ∪ Zr . Then aff Z = aff X, and Theorem 2.15 gives r
r
i=1
i=1
Y ⊂ ∪ rint (conv Zi ) ⊂ rint ( ∪ conv Zi ) r
⊂ rint (conv ( ∪ Zi )) = rint (conv Z). i=1
As in the proof of Theorem 3.26, we can choose a point v ∈ rbd (conv Z) which be written as a positive convex combination v = λ1 x1 + · · · + λm xm of m affinely independent points x1 , . . . , xm ∈ Z. For a given point yi , denote by wi the point of intersection of the open halfline (v, yi i with rbd (conv Z). By Theorem 3.6, wi can be written as a convex combination w = µ1 z1 + · · · + µk zk of affinely independent points z1 , . . . , zk ∈ Z, where k 6 m + 1. Excluding all zero multiples 0xi , we assume that the scalars µ1 , . . . , µk are positive. By Theorem 3.20, wi ∈ rint (conv {z1 , . . . , zk }). We state that k 6 m. Indeed, if k = m + 1, then aff {z1 , . . . , zm+1 } = aff Z and, according Theorem 2.15, wi ∈ rint (conv {z1 , . . . , zm+1 }) ⊂ rint (conv Z),
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in contradiction to the condition wi ∈ rbd (conv Z). Hence k 6 m. Since yi ∈ (v, wi ), Theorem 3.24 (for r = 2) gives the inclusion yi ∈ rint (conv {x1 , . . . , xm , z1 , . . . , zk }). With V = {x1 , . . . , xm } and Vi = {z1 , . . . , zk }, we have yi ∈ rint (conv (V ∪ Vi )),
V ∪ Vi ⊂ Z,
card (V ∪ Vi ) 6 2m.
By Theorems 3.23, 2.15 and 3.2, r
r
Y ⊂ ∪ rint (conv (V ∪ Vi )) ⊂ rint ( ∪ conv (V ∪ Vi )) i=1
i=1
r
⊂ rint (conv (V ∪ ( ∪ Vi ))). i=1
r
Finally, card (V ∪ ( ∪ Vi )) 6 (r + 1)m. i=1
Exercise 3.8. Since the case Y = ∅ is obvious (the existence of a finite set Z ⊂ X with aff Z = aff X is proved in Theorem 1.68), we may assume that Y 6= ∅. Let m = dim X. From Theorem 3.20 and Corollary 3.21 it follows that for every point x ∈ Y there is a finite set Zx ⊂ X such that x ∈ rint (conv Zx ) and aff Zx = aff X. Since the set Y is compact, the relatively open cover {rint (conv Zx ) : x ∈ Y } of Y contains a finite subcover {rint (conv Zxi ) : 1 6 i 6 r}. Let Z = Zx1 ∪ · · · ∪ Zxr . Then aff Z = aff X, and a combination of Theorems 2.15 and 3.2 gives r
r
Y ⊂ ∪ rint (conv Zxi ) ⊂ rint ( ∪ conv Zxi ) i=1
i=1
r
⊂ rint (conv ( ∪ Zxi )) = rint (conv Z). i=1
Exercise 3.9. By Theorem 3.12, x = µ1 x1 + · · · + µr xr , where xi ∈ conv Xi , 1 6 i 6 r. Without loss of generality, we may assume that all scalars µ1 , . . . , µr are distinct from zero. Theorem 3.6 shows that each xi (i) (i) (i) (i) is a convex combination, xi = λ1 z1 + · · · + λpi zpi , where (i)
(i)
Zi = {z1 , . . . , zp(i) } ⊂ Xi , pi 6 m + 1, 1 6 i 6 r. i Furthermore, choose Z1 , . . . , Zr such that the sum p1 + · · · + pr is minimal (i) (i) possible. Then all coefficients λ1 , . . . , λpi , 1 6 i 6 r, are distinct from zero. Renumbering X1 , . . . , Xr , we suppose that p1 > · · · > pr . We state that Zm+1 is a singleton. Suppose for a moment that this is not the case. Then each of Z1 , . . . , Zm+1 contains two or more points, and (i) xi 6= z1 , 1 6 i 6 m + 1. Since the vectors (1)
(m+1)
µ1 (x1 − z1 ), . . . , µm+1 (xm+1 − z1
)
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belong to the m-dimensional subspace sub (µ1 X1 + · · · + µr Xr ), they are linearly dependent. Hence there are scalars γ1 , . . . , γm+1 , not all zero, such that (1)
(m+1)
γ1 µ1 (x1 − z1 ) + · · · + γm+1 µm+1 (xm+1 − z1
) = o.
For a scalar t ∈ R, put (i)
yi (t) = xi + γi t(xi − z1 ),
1 6 i 6 m + 1.
Then (i)
yi (t) = (1 + γi t)xi − γi tz1 (i) (i)
(i)
(i) = (1 + γi t)(λ1 z1 + · · · + λ(i) pi zpi ) − γi tz1 (i)
(i)
(i) (i)
= ((1 + γi t)λ1 − γi t)z1 + (1 + γi t)λ2 z2 + . . . (i) + (1 + γi t)λ(i) pi zpi ,
where (i)
(i)
(1 + γi t)λ1 − γi t + (1 + γi t)λ2 + · · · + (1 + γi t)λ(i) pi = 1. Since yi (0) = xi , there is a scalar t0 6= 0 such that all scalars (i)
(i)
(1 + γi t0 )λ1 − γi t0 , (1 + γi t0 )λ2 , . . . , (1 + γi t0 )λ(i) pi , 1 6 i 6 m + 1, are nonnegative and at least one of them is zero. In other words, yi (t0 ) ∈ conv Zi , 1 6 i 6 m + 1, and there is an index i ∈ {1, . . . , m + 1} such that yi (t0 ) belongs to the convex hull of a proper subset of Zi . Because of x = µ1 y1 (t0 ) + · · · + µm+1 ym+1 (t0 ) + µm+2 xm+2 + · · · + µr xr , we obtained a contradiction with the minimality of p1 + · · · + pr . Hence Zm+1 is a singleton. Consequently, all Zm+2 , . . . , Zr are singletons by the assumption pm+1 > · · · > pr . Chapter 4 Exercise 4.1. By Corollary 3.4, ∆(c1 , . . . , cm+1 ) = conv {c1 , . . . , cm+1 }. A combination of Theorems 4.30 and 4.40 gives L = aff {c1 , . . . , cm+1 } = cones {c1 , . . . , cm+1 } = conv (h1 ∪ · · · ∪ hm+1 ). Exercise 4.2. Without loss of generality, we may put X = {x1 , . . . , xt }, where t 6 r − 1. Assume for a moment that the set C1 ∩ C2 contains a point u 6= s. Then u can be expressed as nonnegative combinations u = s + λ1 (x1 − s) + · · · + λt (xt − s) = s + λt+1 (xt+1 − s) + · · · + λr (xr − s),
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where at least one of the scalars λ1 , . . . , λr is not zero. Consequently, λ1 x1 + · · · + λt xt − λt+1 xt+1 − · · · − λr xr − (λ1 + · · · + λt − λt+1 − · · · − λr )s = o, contrary to the affine independence of {s, x1 , . . . , xr } (see Definition 4.5). Hence C1 ∩ C2 = {s}. For the second statement, we first observe that C is a convex set according to Corollary 3.14. Hence the set K = ∪ ([u1 , u2 ] : u1 ∈ C1 , u2 ∈ C2 ) lies in C. Conversely, let u ∈ C. Then u can be written as a nonnegative combination u = s + λ1 (x1 − s) + · · · + λr (xr − s). Let u1 = s + 2λ1 (x1 − s) + · · · + 2λt (xt − s), u2 = s + 2λt+1 (xt+1 − s) + · · · + 2λr (xr − s). Clearly, u1 ∈ C1 , u2 ∈ C2 , and u = (u1 + u2 )/2. Hence C ⊂ K. Exercise 4.3. According to Theorem 1.68, both affinely independent sets {s, x1 , . . . , xr } and {s0 , x01 , . . . , x0r } can be completed, respectively, to affine bases {s, x1 , . . . , xn } and {s0 , x01 , . . . , x0n } for Rn . By Theorem 1.85, there is an invertible affine transformation f : Rn → Rn such that f (s) = s0 and f (xi ) = x0i for all 1 6 i 6 n. We state that f (C) = C 0 . Indeed, if u ∈ C, then x can be written as an affine combination u = s + λ1 (x1 − s) + · · · + λr (xr − s), and Theorem 1.87 gives f (u) = f (s) + λ1 (f (x1 ) − f (s)) + · · · + λr (f (xr ) − f (s)) = s0 + λ1 (x01 − s0 ) + · · · + λr (x0r − s0 ) ∈ C 0 . Hence f (C) ⊂ C 0 . Conversely, if v 0 ∈ C 0 , then v 0 can be expressed as an affine combination v 0 = s0 + µ1 (x01 − s0 ) + · · · + µr (x0r − s0 ). With v = s + µ1 (x1 − s) + · · · + µr (xr − s), we have v ∈ C and f (v) = v 0 . Summing up, f (C) = C 0 . Exercise 4.4. Let m = dim L. By Corollary 1.67, the plane M has dimension m + 1. Clearly, L0 ⊂ M and L ∩ L0 = ∅. Hence L lies in an open halfplane, say E, of M determined by L0 (see Corollary 1.40). For any point x ∈ L, the open halfline (s, xi lies in E, as follows from Corollary 1.41. Hence cones L = ∪([s, xi : x ∈ L) ⊂ {s} ∪ E. For the opposite inclusion, we observe first that, by Theorem 1.35 and Corollary 1.38, there is a nonzero vector c ∈ Rn \ (sub L)⊥ and a scalar γ 0 satisfying the conditions L0 = {x ∈ M : x·c} = γ 0
and E = {x ∈ M : x·c < γ 0 }.
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Clearly, L = {x ∈ M : x·c} = γ, where γ < γ 0 . Now, let u ∈ E, λ = (γ − γ 0 )/(u·c − γ 0 ),
and v = s + λ(u − s).
Then λ > 0, and the equality v· c = (s + λ(u − s))·c = γ 0 +
γ − γ0 (u·c − γ 0 ) = γ u·c − γ 0
shows that v ∈ L. Hence v ∈ cones L, and the inclusion E ⊂ cones L holds. Since s ∈ cones L, one has {s} ∪ E ⊂ cones L. Exercise 4.5. Let a convex set K ⊂ Rn satisfy the equality K = a + µK, with a 6= o and µ > 0. If µ 6= 1, then K=
1 1−µ
a + µ(K −
1 1−µ
a) = s + µ(K − s),
where s = (1 − µ)−1 a, which shows that K is a convex cone with apex s (see Theorem 4.2). If µ = 1, then, by induction on r, we obtain that K = ±ra + K for all r > 1, and a convexity argument shows that K contains a translate of the line ho, ai. The proof of the converse statement is similar. Exercise 4.6. From Theorem 4.30 it follows that each penumbra P (K1 , z), z ∈ K2 , coincides with conez K1 . Therefore, the umbra U (K1 , K2 ) is a convex set, as the intersection of convex sets P (K1 , z), z ∈ K2 (see Theorem 2.8). For the convexity of P (K1 , K2 ), choose any points x, y ∈ P (K1 , K2 ) and a scalar λ ∈ [0, 1]. Then x = (1−η)x2 +ηx1 and y = (1−θ)y2 +θy1 for some points x1 , y1 ∈ K1 , x2 , y2 ∈ K2 and scalars η, θ > 1. Put µ = (1−λ)η +λθ. Clearly, µ > 1. By Theorem 2.4, (1 − λ)x + λy = (1 − λ)((1 − η)x2 + ηx1 ) + λ((1 − θ)y2 + θy1 ) ∈ (1 − λ)((1 − η)K2 + ηK1 ) + λ((1 − θ)K2 + θK1 ) ⊂ ((1 − λ)(1 − η) + λ(1 − θ))K2 + ((1 − λ)η + λθ)K1 = (1 − µ)K2 + µK1 = ∪ ((1 − µ)x + µK1 : x ∈ K2 ) ⊂ ∪ ((1 − γ)x + γK1 : γ > 1 and x ∈ K2 ) = P (K1 , K2 ), which proves the convexity of P (K1 , K2 ).
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Chapter 5 Exercise 5.1. Let x be a point in conv (K ∪ (u + C). Since both sets K and u + C are convex, Corollary 3.14 shows that x can be written as x = (1 − λ)y + λz, where y ∈ K, z ∈ u + C, and 0 6 λ 6 1. Furthermore, z = u + v, where v ∈ C. By the convexity of K, the point (1 − λ)y + λu lies in K, and λv ∈ C since C is a cone with apex o. Consequently, x = ((1 − λ)y + λu) + λv ∈ K + C. Hence conv (K ∪ (u + C)) ⊂ K + C, and cl (conv (K ∪ (u + C))) ⊂ cl (K + C). For the opposite inclusion, choose a point x ∈ K + C. Then x can be written as x = y + z, where y ∈ K and z ∈ C. Since the case z = o is obvious, we may assume that z 6= o. Put h = [o, zi. Because the halfline u + h lies in u + C, it also lies in conv (K ∪ (u + C)). From Theorem 5.1 it follows that x = y + z ∈ y + h ⊂ cl (conv (K ∪ (u + C))). Summing up, K + C ⊂ cl (conv (K ∪ (u + C))), and cl (K + C) ⊂ cl (conv (K ∪ (u + C))). Exercise 5.2. One has rbd K 6= ∅ because K is not a plane (see Corollary 2.57). According Theorem 5.6, one has s + rec (cl K) ⊂ cl K ⊂ cones (cl K) for every point s ∈ rbd K. Hence rec (cl K) ⊂ ∩ (cones (cl K) − s : s ∈ rbd K). Conversely, suppose that a halfline h with endpoint o does not lie in rec (cl K). Then there is a point x ∈ cl K such that the halfline h0 = x + h does not lie in cl K. Clearly, h0 contains a point u ∈ rbd K such that cl K ∩ (h0 \ [x, u]) = ∅. Furthermore, u + h 6⊂ cones (cl K). Therefore, h 6⊂ ∩ (cones (cl K) − x : x ∈ rbd K), implying the inclusion ∩ (cones (cl K) − x : x ∈ rbd K) ⊂ rec (cl K). Exercise 5.3. (1) If K is a plane, then lin K = sub K (see example on page 197), and dim (lin K) = m. Conversely, assume that dim (lin K) = m. Choose a point x ∈ K. Then x + lin K ⊂ K according to Theorem 5.19. Consequently, x + lin K ⊂ aff K. Since dim (aff K) = m, Theorem 1.6 implies the equality x + lin K = aff K. Hence K = aff K, and K is a plane.
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(2) If K is a halfplane of a slab of a plane L ⊂ Rn , then dim (lin K) = m − 1 (see example on page 202). Conversely, assume that dim (lin K) = m − 1. Since the case m = 1 is obvious (K is either a halfline or a segment), we may suppose that m > 2. Express K as K = (K ∩ S) ⊕ lin K (see Theorem 5.20). Then dim (K ∩ S) = dim K − dim (lin K) = 1, which shows that K ∩ S is either a halfline or a segment. Therefore, K is either a halfplane or a slab (compare with Theorem 1.42). Exercise 5.4. Because the implications (3) ⇒ (2) ⇒ (1) are obvious, it remains to show that (1) ⇒ (3). Since rbd K = rbd (cl K), we may assume that the set K is closed. Under this assumption, (3) becomes equivalent to the statement that K is a closed slab. Choose a subspace S ⊂ Rn complementary to lin K and express K as K = (K ∩ S) ⊕ lin K, where lin (K ∩ S) = {o} (see Theorem 5.20). Then rbd K = rbd (K ∩ S) ⊕ lin K according to Corollary 5.30. Furthermore, K ∩ S is line-free (see Corollary 5.53). First, we observe that dim (K ∩S) > 1. Indeed, if dim (K ∩S) = 0, then K ∩ S is a singleton and K is translate of lin K, implying that rbd K = ∅. Next, suppose that dim (K ∩ S) = 1. Then K ∩ S is either a closed segment, a closed halfline, or a line. Under this assumption, rbd (K ∩ S) is disconnected (namely, consists of two points) if and only if K ∩ S is a segment. Therefore, rbd K is disconnected (namely, consists of two translates of lin K) if and only if K is a closed slab. Finally, we are going to show that the set rbd K is connected provided dim (K ∩ S) > 2. For this, we will show that any points u, v ∈ rbd (K ∩ S) can be connected by a continuous curve which lies in rbd (K ∩ S). Indeed, this is trivial if [u, v] ⊂ rbd (K ∩ S). Alternatively, let (u, v) ⊂ rint (K ∩ S) (see Theorem 2.59). Choose in S a 2-dimensional plane L which contains (u, v). From K ∩ L = (K ∩ S) ∩ L it follows that the 2-dimensional closed convex set K ∩ L is line-free. It is easy to see that u and v are endpoints of a continuous arc Γ ⊂ rbd (K ∩ L). Since rbd (K ∩ L) = rbd (K ∩ S) ∩ L (see Corollary 2.62), we conclude that Γ ⊂ rbd (K ∩ S). Now, let x and z be points in rbd K. Denote by x0 and z0 the projections of x and z, respectively, on S along lin K. Clearly, [x, x0 ] ⊂ rbd K and [z, z0 ] ⊂ rbd K. By the above proved, x0 and z0 are connected by a continuous arc Γ ⊂ rbd (K ∩ S) = rbd K ∩ S. Summing up, the continuous path [x, x0 ] ∪ Γ ∪ [z0 , z] joins x and z in rbd K. Exercise 5.5. One has rbd K 6= ∅ because K is not a plane (see Corol-
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lary 2.57). The existence of a nearest to z point u ∈ rbd K follows from the continuity of the function δz (x) = kx − zk on Rn and the closedness of rbd K (see Corollary 2.57). Assume for a moment that K 6⊂ V and choose a point v ∈ K \ V . Denote by L the 2-dimensional plane spanned by {u, v, z}. Let F (respectively, G) be the open halfplane of L determined by the line l = hu, vi and missing z (respectively, determined by the line m = hu, zi and missing v). Put Q = F ∩ G. We state that K ∩ Q = ∅. Indeed, assume for a moment the existence of a point x ∈ K ∩ Q. According to Theorem 1.31, the segment [x, v] meets the line m at a point y ∈ F . Furthermore, y ∈ K by the convexity of K. The latter is impossible because, according to Theorem 2.55, the open halfline [z, ui \ [z, u] must be disjoint from K. Hence K ∩ Q = ∅. Denote by h the halfline with endpoint z which meets the line l and is perpendicular to it. Since h meets Q, we have h 6⊂ K. By Theorem 2.59, h meets rbd K at a point c ∈ L \ F . Denote by w the point at which h meets l. Then c ∈ [w, z], which gives kc − zk 6 kw − zk < ku − zk, in contradiction with the choice of u. The obtained contradiction proves the inclusion K ⊂ P . Exercise 5.6. It suffices to prove the inclusions nor L ⊂ bar L ⊂ (sub L)⊥ ⊂ nor L. Indeed, the first inclusion is proved in Theorem 5.42. Let e ∈ bar L. Then there is a scalar γ such that L lies within the halfspace V = {x ∈ Rn : x·e 6 γ}. By Theorem 1.30, L is parallel to the hyperplane H = {x ∈ Rn : x·e = γ}, and Theorem 1.13 shows that the characteristic subspace sub L lies in the hypersubspace S = {x ∈ Rn : x·e = 0}. Hence e ∈ S ⊥ ⊂ (sub L)⊥ . Now, let c ∈ (sub L)⊥ . Then o is the nearest to c point in sub L. If u is a point in L, then L = u + sub L (see Theorem 1.2), and, by the above proved, u is the nearest to c + u point in L. Hence c ∈ nor L. Exercise 5.7. Suppose that K is not line-free. Then lin (cl K) 6= {o} according to Corollary 5.53. We observe that sub H ∩ lin (cl K) = {o}. Otherwise, choosing a point x ∈ rint K and a 1-dimensional subspace l ⊂ sub H ∩ lin (cl K), one would have x + l ⊂ (x + sub H) ∩ rint (cl K) = H ∩ rint K ⊂ H ∩ K (see Corollary 5.14), contrary to the assumption that H ∩ K is bounded. Hence the subspace lin (cl K) has dimension one, implying that sub H is complementary to lin (cl K). Finally, Theorem 5.20 gives cl K = (sub H ∩ cl K) ⊕ lin (cl K).
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Chapter 6 Exercise 6.1. Clearly, H bounds (respectively, strongly bounds) the ball Bρ (z) if and only if ρ 6 δ(z, H) (respectively, ρ < δ(z, H)). Since δ(z, H) = |γ − z·c|/kck (see Theorem 1.97), H bounds (respectively, strongly bounds) Bρ (z) if and only if ρkck 6 |γ − z·c| (respectively, ρkck < |γ − z·c|). Exercise 6.2. If y ∈ Bρ (z), then ρ2 > ky − zk2 = k(y − u) + (u − z)k2 = ky − uk2 + 2(y − u)·(u − z) + ku − zk2 = ky − uk2 + 2(y − u)·(u − z) + ρ2 . Hence (y − u)·(u − z) 6 0, and the equality (y − u)·(u − z) = 0 holds if and only if y = u. In other words, Bρ (z) lies in the closed halfspace V = {x ∈ Rn : (x − u)·(u − z) 6 0}, and u is the only point of Bρ (z) which belongs to H. Exercise 6.3. Choosing a point u ∈ L1 ∩ L2 and translating the planes L1 and L2 on −u, we may assume that both L1 and L2 are subspaces. If a ∈ L1 ∩ L2 , then a = c = e, implying that ka − ck 6 γka − ek for any choice of γ > 0. Assume that a ∈ L1 \ L2 . Then c 6= a 6= e. Let T be the orthogonal complement of L1 ∩ L2 in Rn , and put T1 = L1 ∩ T and T2 = L2 ∩ T . Then L1 = T1 ⊕ (L1 ∩ L2 )
and L2 = T2 ⊕ (L1 ∩ L2 ).
Every point a ∈ L1 is uniquely expressible as a = a1 + c, where a1 and c are the orthogonal projections of a on T1 and L1 ∩L2 , respectively. Clearly, a1 6= o because a ∈ S1 \ L2 . Denote by f the orthogonal projection of Rn on L2 . Put e1 = f (a1 ). Since f is a linear transformation, one has e = f (a) = f (a1 + c) = f (a1 ) + f (c) = e1 + c. Hence a − a1 = c = e − e1 , which gives a − e = a1 − e1 . Consequently, ka − ck/ka − ek = ka1 k/ka1 − e1 k. Let a0 = a1 /ka1 k be the unit vector in the direction of a1 . Then ka1 k/ka1 − e1 k = ka0 k/ka0 − e0 k = 1/ka0 − e0 k, where e0 = e1 /ka1 k. Since T1 ∩ T2 = (L1 ∩ T ) ∩ (L2 ∩ T ) = (L1 ∩ L2 ) ∩ T = {o}
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the unit sphere E of T1 is disjoint from T2 . Because E is compact, there is a scalar ρ > 0 such that δ(E, T2 ) > ρ (see page 9). The inclusions a0 ∈ E and e0 ∈ T2 give ka0 − e0 k > δ(E, T2 ) > ρ. Let γ = 1/ρ. Then ka − ck/ka − ek = ka1 k/ka1 − e1 k = 1/ka0 − e0 k 6 1/ρ = γ. Exercise 6.4. Let a proper plane L bound K. If K ⊂ L, then K ⊂ L ∩ aff K, and condition (b) is satisfied. Hence we may assume that L properly bounds K. Suppose that L ∩ aff K 6= ∅. From (L ∩ aff K) ∩ rint K = L ∩ (aff K ∩ rint K) = L ∩ rint K = ∅ it follows that the plane L ∩ aff K properly bounds K. Conversely, assume that a proper plane L satisfies one of the conditions (a) and (b). If L ∩ aff K = ∅, then L ∩ rint K ⊂ L ∩ aff K = ∅, implying that L properly bounds K. Similarly, if L ∩ aff K 6= ∅ and K ⊂ L, then K ⊂ L ∩ aff K. Suppose that L ∩ aff K properly bounds K. Then from L ∩ rint K = L ∩ (aff K ∩ rint K) = (L ∩ aff K) ∩ rint K = ∅ if follows that L properly bounds K. Exercise 6.5. Let a proper plane L be strongly disjoint from K. Assume that L∩aff K 6= ∅. Since δ(K, L∩aff K) > δ(K, L) > 0, the plane L∩aff K is strongly disjoint from K. Conversely, assume that a proper plane L satisfies one of the conditions (a) and (b). If L ∩ aff K = ∅, then δ(K, L) > δ(aff K, L) > 0 (see Theorem 1.98), implying that L is strongly disjoint from K. Suppose that L ∩ aff K 6= ∅ and L ∩ aff K is strongly disjoint from K. Choose in K a sequence of points x1 , x2 , . . . satisfying the equality limi→∞ δ(xi , L) = δ(K, L). Let ui and vi be the orthogonal projections of xi on L and L ∩ aff K, respectively. Then δ(xi , L) = kxi − ui k
and δ(xi , L ∩ aff K) = kxi − vi k i > 1.
Since there is a scalar α > 0 satisfying the inequalities kxi − vi k 6 αkxi − ui k,
i > 1,
(see Exercise 6.3), we obtain that 0 < δ(K, L ∩ aff K) 6 inf kxi − vi k 6 α inf kxi − ui k = δ(K, L), i>1
implying that L is strongly disjoint from K.
i>1
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Exercise 6.6. Let a plane L support K. If K ⊂ L, then K ⊂ aff K = L ∩ aff K implying that L ∩ aff K 6= ∅ and the plane L ∩ aff K supports K. Assume that L properly supports K. Then ∅ 6= L ∩ cl K ⊂ L ∩ aff K. Furthermore, from (L ∩ aff K) ∩ cl K = L ∩ (aff K ∩ cl K) = L ∩ cl K 6= ∅, (L ∩ aff K) ∩ rint K = L ∩ (aff K ∩ rint K) = L ∩ rint K = ∅ it follows that the plane L ∩ aff K properly supports K. Conversely, suppose that L∩aff K 6= ∅ and the plane L∩aff K supports K. Then ∅ 6= (L ∩ aff K) ∩ cl K ⊂ L ∩ cl K. If K ⊂ L ∩ aff K, then K ⊂ L, implying that L supports K. Assume that K 6⊂ L ∩ aff K. Then from L ∩ rint K = L ∩ (aff K ∩ rint K) = (L ∩ aff K) ∩ rint K = ∅ if follows that L properly supports K. Exercise 6.7. Let a plane L be asymptotic to K. Then L ∩ aff K 6= ∅, since otherwise δ(L, K) > δ(L, aff K) > 0 (see Theorem 1.98). Next, (L ∩ aff K) ∩ cl K = L ∩ (aff K ∩ cl K) = L ∩ cl K = ∅. Hence, to show that L ∩ aff K is asymptotic to K, it remains to prove the equality δ(K, L ∩ aff K) = 0. Choose in K a sequence of points x1 , x2 , . . . satisfying the equality limi→∞ δ(xi , L) = δ(K, L) = 0. Let ui and vi be the orthogonal projections of xi on L and L ∩ aff K, respectively. Then δ(xi , L) = kxi − ui k
and δ(xi , L ∩ aff K) = kxi − vi k i > 1.
Since there is a scalar α > 0 satisfying the inequalities kxi − vi k 6 αkxi − ui k,
i > 1,
(see Exercise 6.3), we obtain that δ(K, L ∩ aff K) 6 inf kxi − vi k 6 α inf kxi − ui k = δ(K, L) = 0, i>1
i>1
implying that L ∩ aff K is asymptotic to K. Exercise 6.8. Let G be the family of all halfspaces from F which do not contain aff K. The family F is nonempty, since otherwise cl K ⊂ aff K ⊂ ∩(V : V ∈ F) = cl K, contrary to the assumption that K is not a plane (see Corollary 2.40). Clearly, V ∩ aff K is a closed halfplane of aff K for any choice of V ∈ G, and cl K = cl K ∩ aff K = (∩(V : V ∈ F)) ∩ aff K = ∩(V ∩ aff K : V ∈ F) = ∩(V ∩ aff K : V ∈ G).
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Exercise 6.9. We observe first that a properly supporting halfplane Dx exists for every point x ∈ X. Indeed, since {x} is a convex subset of rbd K, Corollary 6.9 implies the existence of a hyperplane H through x properly supporting K. Then one of the closed halfspaces determined by H, say V , properly contains K. Hence aff K 6⊂ V , and the set Dx = V ∩ aff K is a closed halfplane of aff K properly supporting K at x. As in the proof of Theorem 6.23, it suffices to show that ∩ (Dx : Dx ∈ D) ⊂ cl K. Equivalently, that for any point u ∈ aff K \ cl K, there is a halfplane Dx ∈ D such that u ∈ / Dx . Choose a point v ∈ rint K. Then the open segment (u, v) meets rbd K at a unique point z (see Theorem 2.55). Since z = (1 − λ)u + λv for a certain scalar 0 < λ < 1, we can write v = (1 − λ−1 )u + λ−1 z. According to Theorem 2.14, there is a scalar ρ > 0 such that Bρ (v) ∩ aff K ⊂ rint K. Because X is dense in rbd K, one can choose a point x ∈ X satisfying the inequality kx − zk < λρ. Put w = (1 − λ−1 )u + λ−1 x. Then w ∈ aff K because w is an affine combination of u and x. Furthermore, from kv−wk = kλ−1 (x−z)k < ρ it follows that w ∈ Bρ (v)∩aff K ⊂ rint K. Finally, we state that u ∈ / Dx . Indeed, assume for a moment that u ∈ Dx . Since Dx properly supports K, one has w ∈ rint K ⊂ rint Dx , and Corollary 1.41 gives x = (1−λ)u+λw ∈ (u, w) ⊂ rint Dx , in contradiction with the assumption x ∈ rbd Dx . Chapter 7 Exercise 7.1. Theorem 7.4 shows that every extreme point of Bρ (c) belongs to Sρ (c). Conversely, let z ∈ Sρ (c). We state that z is an extreme point of Bρ (c). Indeed, assume for a moment that z = (1 − λ)u + λv for certain distinct points u, v ∈ Bρ (c) and a scalar 0 < λ < 1. From the inequalities ρ = kz − ck = k(1 − λ)u + λv − ck 6 (1 − λ)ku − ck + λkv − ck 6 ρ it follows that kz − ck = k(1 − λ)u + λv − ck = (1 − λ)ku − ck + λkv − ck. According to Exercise 1.9, c belongs to hu, vi such that either u ∈ [c, v] or v ∈ [u, c]. Each of these cases is impossible because of ku − ck = kv − ck = ρ. Hence z ∈ ext Bρ (c), ext Bρ = (c)Sρ (c). It remains to show that every extreme face F of Bρ (c) containing at least two points coincides with Bρ (c). Let u and v be distinct points in F .
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We state first that (u, v) ⊂ Uρ (c). Indeed, choose a point w ∈ (u, v) and express it as w = (1 − λ)u + λv, where 0 < λ < 1. If at least one of the points u, v, say u, satisfies the inequality ku − ck < ρ, then kw − ck = k(1 − λ)(u − c) + λ(v − c)k 6 (1 − λ)ku − ck + λkv − ck < ρ, implying the inclusion w ∈ Uρ (c). Assume that both points u, v belong to the sphere Sρ (c). By the above proved, kw − ck < ρ, again giving w ∈ Uρ (c). Summing up, (u, v) ⊂ Uρ (c). Finally, F ∩ Uρ (c) 6= ∅ because of (u, v) ⊂ F , and Theorem 7.4 gives F = Bρ (c) Exercise 7.2. First, we are going to show that every simplex ∆0 = ∆(z1 , . . . , zt ) whose vertices z1 , . . . , zt are chosen in {x1 , . . . , xr+1 } is an extreme face of ∆. Renumbering, if necessary, the vertices x1 , . . . , xr+1 , we assume that zi = xi for all 1 6 i 6 t. Let x, y ∈ ∆ and 0 < λ < 1 be such that z = (1 − λ)x + λy ∈ ∆0 . Then x and y can be written as convex combinations of x1 , . . . , xr+1 , and z can be written as a convex combination of x1 , . . . , xt : x = α1 x1 + · · · + αr+1 xr+1 , y = β1 x1 + · · · + βr+1 xr+1 , z = γx1 + · · · + γt xt . On the other hand, z = ((1 − λ)α1 + λβ1 )x1 + · · · + ((1 − λ)αr+1 + λβr+1 )xr+1 . Since the points x1 , . . . , xr+1 are affinely independent, the representation of each point from ∆ as a convex combination of x1 , . . . , xr+1 is unique. Thus (1 − λ)αi + λβi = γi , 1 6 i 6 t, (1 − λ)αi + λβi = 0, t + 1 6 i 6 r + 1. Because the scalars α1 , . . . , αr+1 , β1 , . . . , βr+1 are nonnegative, the last equalities imply that αi = βi = 0 for all t + 1 6 i 6 r + 1. Hence both x and y lie in ∆0 , i. e., ∆0 is an extreme face of ∆. Conversely, let F be an extreme face of ∆. Choose a point x ∈ F . Since x ∈ ∆, it can be expressed as x = λ1 x1 + · · · + λr+1 xr+1 , λ1 , . . . , λr+1 > 0, λ1 + · · · + λr+1 = 1. We state that if in this expression a scalar λi , 1 6 i 6 r + 1, is positive, then xi ∈ F . Indeed, this is trivial if λi = 1; then λj = 0 for all j 6= i and 1xi = x ∈ F . Let 0 < λi < 1 and put λi−1 λi+1 λr+1 λ1 x1 + · · · + xi−1 + xi+1 + · · · + xr+1 z= 1 − λi 1 − λi 1 − λi 1 − λi
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Since the numbers λi−1 λi+1 λr+1 λ1 ,..., , ,..., 1 − λi 1 − λi 1 − λi 1 − λi are nonnegative and their sum equals 1, z is a convex combination of the vertices of ∆. Hence z ∈ ∆. Furthermore, from x = λi xi + (1 − λi )z and Definition 7.1 it follows that xi , z ∈ F . Let z1 , . . . , zt be the vertices of ∆ that lie in F . By the above argument, F ⊂ ∆(z1 , . . . , zt ). On the other hand, since F is convex, we have ∆(z1 , . . . , zt ) ⊂ F . Summing up, F = ∆(z1 , . . . , zt ). Exercise 7.3. First, we are going to show that s is an extreme point of Cs . Indeed, assume for a moment that s = (1 − λ)u + λw for certain points u, w ∈ Cs and a scalar 0 < λ < 1. According to Definition 4.5, we can write u = s + α1 (x1 − s) + · · · + αr (xr − s),
α1 , . . . , αr > 0,
w = s + β1 (x1 − v) + · · · + βr (xr − s),
β1 , . . . , βr > 0.
Then s = s + ((1 − λ)α1 + λβ1 )(x1 − s) + · · · + ((1 − λ)αr + λβr )(xr − s), or ((1 − λ)α1 + λβ1 )(x1 − s) + · · · + ((1 − λ)αr + λβr )(xr − s) = o. (9.25) By Theorem 1.60, the set {x1 − s, . . . , xr − s} is linearly independent, so (9.25) gives (1 − λ)α1 + λβ1 = · · · = (1 − λ)αr + λβr = 0. Since all scalars αi , βi , 1 6 i 6 r, are nonnegative and 0 < λ < 1, we have α1 = · · · = αr = β1 = · · · = βr = 0, which shows that s = u = w; i. e., s is an extreme point of C. Next, we are going to show that every simplicial cone C 0 = Cs (z1 , . . . , zt ) whose vertices z1 , . . . , zt are chosen in {x1 , . . . , xr } is an extreme face of Cs . Renumbering, if necessary, the vertices x1 , . . . , xr , we assume that zi = xi for all 1 6 i 6 t. Let x, y ∈ Cs and 0 < λ < 1 be such that z = (1 − λ)x + λy ∈ C 0 . Then x, y, and z can be written as x = s + α1 (x1 − s) + · · · + αr (xr − s), α1 , . . . , αr > 0, y = s + β1 (x1 − s) + · · · + βr (xr − s), β1 , . . . , βr > 0, z = s + γ1 (x1 − s) + · · · + γs (xs − s), γ1 , . . . , γt > 0.
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On the other hand, z = (1 − λ)x + λy gives z = s + ((1 − λ)α1 + λβ1 )(x1 − s) + · · · + ((1 − λ)αr + λβr )(xr − s). Since the points v, x1 , . . . , xr are affinely independent, the representation of each point from C 0 as a nonnegative combination of s, x1 − s, . . . , xr − s is unique. Comparing the above expressions for z, we have (1 − λ)αi + λβi = γi , 1 6 i 6 t, (1 − λ)αi + λβi = 0, t + 1 6 i 6 r. Because the scalars α1 , . . . , αr , β1 , . . . , βr are nonnegative, the last r − t − 1 equalities imply that αi = βi = 0 for all t + 1 6 i 6 r. Hence both x and y lie in C 0 , i. e., C 0 is an extreme face of C. Conversely, let F be an extreme face of Cs . First, we observe that for every point x ∈ F \ {s} the entire halfline [s, xi lies in F . Indeed, [s, x] ⊂ F by the convexity of F . If λ > 1, then x ∈ [s, s + λ(x − s)). Since F is an extreme face of Cs , we obtain that s + λ(x − s) ∈ F . Hence [s, xi = {s + λ(x − s) : λ > 0} ⊂ F , implying that F is a convex cone with apex s. Choose a point x ∈ F . Since x ∈ Cs , it can be written as x = s + λ1 (x1 − s) + · · · + λr (xr − s), λ1 , . . . , λr > 0. We state that if in this expression a scalar λi , 1 6 i 6 r, is positive, then xi ∈ F . Indeed, consider the points y = s + 2λi (xi − s), z = s + 2λ1 (x1 − s) + · · · + 2λr (xr − s) − 2λi (xi − s). Since y, z ∈ Cs and x = (y + z)/2 ∈ (y, z), we conclude that both y and z belong to F . By the above, xi = s + λ−1 i λi (xi − s) ∈ F . Let z1 , . . . , zt be all points from {x1 , . . . , xr } that lie in F . By the above argument, we have F ⊂ Cs (z1 , . . . , zt ). On the other hand, since F is a convex cone with apex v, we have Cs (z1 , . . . , zt ) ⊂ F . Summing up, F = Cs (z1 , . . . , zt ). Exercise 7.4. First, we observe that every nonempty convex set K ⊂ Rn has two (trivial) extreme faces: ∅ and K. (1) According to Corollary 7.10, K has exactly two extreme faces if and only if K is relatively open, which happens if and only if rbd K = ∅, or, equivalently, if and only if K is a plane (see Corollary 2.57). (2) If K is a closed halfplane, then it has exactly three extreme faces: ∅, rbd K, and K. Conversely, assume that K has exactly three extreme faces.
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By the above proved, K is not a halfplane and rbd K 6= ∅. According to Theorem 7.9, every point x ∈ rbd K belongs to the generated extreme face FK (x) ⊂ rbd K. Thus assumption (2) shows that FK (x) = FK (y) for any points x, y ∈ rbd K. This argument proves that rbd K is a convex set, implying that K is a halfspace (see Theorem 2.58). (3) If K is a closed plane slab, then K has exactly four extreme faces: ∅, two boundary planes, and K. Conversely, assume that K has exactly four extreme faces. By the above proved, K has nonempty relative boundary, which is not a convex set. Hence there are distinct points u, v ∈ rbd K such that [u, v] 6⊂ rbd K. Consider the generated extreme faces FK (u) and FK (v). By Theorem 7.9, FK (u)∪FK (v) ⊂ rbd K. If the set rbd K \(FK (u)∪ FK (v)) contained a point w, then FK (w) would be a new extreme face of K. Hence rbd K = FK (u) ∪ FK (v). Similarly, if FK (u) ∩ FK (v) 6= ∅, then, according to Theorem 7.13, FK (u) ∩ FK (v) would be a new extreme face of K. Summing up, rbd K is a disjoint union of two sets, which shows that K is a closed slab (see Exercise 5.4). Exercise 7.5. Let K ⊂ Rn be a compact convex set and X = ext K. From cl X ⊂ K = conv X we conclude that cl X is compact and that cl X ⊂ conv X. Assume for a moment the existence of a set Y ⊂ X such that X ∩ conv Y 6⊂ Y . Then we can choose a point z ∈ (X ∩ conv Y ) \ Y . In particular, z ∈ X \ Y , implying that Y ⊂ X \ {z} ⊂ K \ {z}. Since z is an extreme point of K, the set K \ {z} is convex. Hence conv Y ⊂ K \ {z}, i. e., z ∈ / conv Y , a contradiction. Conversely, let a set X ⊂ Rn satisfy conditions (1)–(3). Put M = conv X. By condition (1), X is bounded. Then conv X also is bounded. A combination of Theorem 3.17 and condition (2) implies that cl M = cl (conv X) = conv (cl X) ⊂ conv (conv X) = conv X = M. Hence M is compact. Theorem 7.17 implies that ext M ⊂ X. Assume for a moment that ext M 6= X, i. e., that there is a point z ∈ X \ ext M . Put Y = X \ {z}. Since ext M ⊂ Y , we have conv Y = M . At the same time, condition (3) implies that X = X ∩ M = X ∩ conv Y ⊂ Y , contradicting the inclusion z ∈ X. Hence ext M = X. Exercise 7.6. For every positive integer r > 1, let Xr denote the set of midpoints of all segments of length 1/r contained in K. We state that every set Xr , r > 1, is closed. Indeed, let u1 , u2 , . . . be a sequence of points in Xr which converges to a point u0 . Then one can find an index j such that
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kui −u0 k 6 1 for all i > j. Every point ui is the center of a certain segment [vi , wi ] ⊂ K of length 1/r. From the inequality kvi − u0 k 6 kvi − ui k + kui − u0 k 6 1/r + 1,
i > j,
it follows that the sequence vj , vj+1 , . . . is bounded. Hence it contains a convergent subsequence. Without loss of generality, we may suppose that vj , vj+1 , . . . itself converges to a certain point v0 . By a similar argument, we assume that the sequence wj , wj+1 , . . . converges to a certain point w0 . Then v0 , w0 ∈ K by the closedness of K. Consequently, [v0 , w0 ] ⊂ K by the convexity of K. Furthermore u0 = lim ui = lim (vi + wi )/2 = (v0 + w0 )/2 i→∞
i→∞
and kv0 − w0 k = limi→∞ kvi − wi k = 1/r. Summing up, the set Xr is closed. It is easy to see that a point x belongs to K \ ext K if and only if it belongs to a certain set Xr . Thus K \ ext K = X1 ∪ X2 ∪ · · · is and Fσ -set. Hence ext K is a Gδ -set. Exercise 7.7. If x is an extreme point of K + M then {x} is a 0dimensional extreme face of K + M . By Theorem 7.15, we can write {x} = F + G, where F and G are uniquely defined extreme faces of K and M , respectively. Obviously, both F and G are singletons: F = {y} and G = {z}, where y and z are extreme points of K and M , respectively. Conversely, let a point x ∈ K + M be uniquely expressed as x = y + z with y ∈ ext K and z ∈ ext M . Assume for a moment that x is not an extreme point of K +M . Then there are distinct points x1 , x2 ∈ K +M and a scalar 0 < λ < 1 such that x = (1−λ)x1 +λx2 . We can write x1 = y1 +z1 and x2 = y2 + z2 for certain points y1 , y2 ∈ K and z1 , z2 ∈ M . Then y1 6= y2 or z1 6= z2 since otherwise x1 = x2 . Let, for example, y1 6= y2 . Put v = (1 − λ)y1 + λy2 and w = (1 − λ)z1 + λz2 . Then v ∈ K and w ∈ M by a convexity argument. Obviously, x = v + w. The assumed uniqueness of this representation gives y = v and z = w. Hence y = (1 − λ)y1 + λy2 for distinct points y1 , y2 ∈ K and a scalar 0 < λ < 1. The latter is in contradiction with the choice y ∈ ext K. Hence x ∈ ext (K + M ). Chapter 8 Exercise 8.1. Corollary 8.3 implies that every exposed point of Bρ (c) belongs to Sρ (c). Conversely, let z ∈ Sρ (c). Exercise 6.1 shows that H = {x ∈ Rn : (x − z)·(z − c) = 0}
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is the only hyperplane through z which supports Bρ (c) such that H ∩ Bρ (z) = {z}. Hence z ∈ exp Bρ (c). Summing up, Sρ (c) = exp Bρ (c). It remains to prove that every exposed face G of Bρ (c) containing at least two points coincides with Bρ (c). Indeed, according to Theorem 8.2, G is an extreme face of Bρ (c), and Exercise 7.1 shows that G = Bρ (c). Exercise 8.2. If G is an exposed face of the simplex ∆ = ∆(x1 , . . . , xr+1 ), then, by Theorem 8.2, G is an extreme face of ∆, and Exercise 7.2 shows that either G = ∅ or G = ∆(z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr+1 }. Conversely, choose a simplex ∆0 = ∆(z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr+1 }. We are going to prove that ∆0 is an exposed face of ∆. Since the case ∆0 = ∆ is obvious, we assume that ∆0 is a proper subset of ∆. Let L = aff {z1 , . . . , zt }. Because ∆0 is an extreme face of ∆, Theorem 7.2 gives ∆ ∩ L = ∆0 . This argument and Exercise 3.1 show that L is disjoint from ∆00 = ∆(zt+1 , . . . , zr+1 ). Furthermore, L and ∆00 are strongly disjoint because L is closed (see page 9 and Corollary 1.22) and ∆00 is compact (see Theorem 3.17). According to Theorem 6.3, there is a hyperplane H ⊂ Rn containing L and strongly disjoint from ∆00 . Finally, we state that H ∩ ∆ = ∆0 (which implies that ∆0 is an exposed face of ∆). Indeed, assume for a moment the existence of a point u ∈ (H ∩ ∆) \ ∆0 . By the above argument, u ∈ / ∆00 . Therefore, u can be written as a convex combination u = (1 − λ)u0 + λu00 of certain points u0 ∈ ∆0 and u00 ∈ ∆00 , where 0 < λ < 1 (see Exercise 3.1). In this case, u00 = λ−1 u + (1 − λ−1 )u0 ∈ hu, u0 i ⊂ H, contrary to H ∩ ∆00 = ∅. Hence H ∩ ∆ = ∆0 . Exercise 8.3. If G is an exposed face of the simplicial cone C = Cs (x1 , . . . , xr ), then, by Theorem 8.2, G is an extreme face of C, and Exercise 7.3 shows that either G = ∅, or G = {s}, or G = ∆(z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr }. Conversely, choose a simplicial cone C 0 = Cs (z1 , . . . , zt ), where {z1 , . . . , zt } is a nonempty subset of {x1 , . . . , xr }. We are going to prove that C 0 is an exposed face of C. Since the case C 0 = C is obvious, we assume that C 0 is a proper subset of C. Let L = aff {z1 , . . . , zt }. Because C 0 is an extreme face of C, Theorem 7.2 gives C ∩ L = C 0 . This argument and Exercise 4.2 show that L is disjoint from the simplex ∆ = ∆(zt+1 , . . . , zr ). Furthermore, implies that L and ∆ are strongly disjoint because L is closed (see Corollary 1.22) and ∆ is compact (see page 9 and Theorem 3.17). Ac-
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cording to Theorem 6.3, there is a hyperplane H ⊂ Rn containing L and strongly disjoint from ∆. Finally, we state that H ∩ C = C 0 (which implies that C 0 is an exposed face of C). Indeed, assume for a moment the existence of a point u ∈ (H ∩ C) \ C 0 . By the above argument, u ∈ / ∆. Therefore, u can be written 0 as a convex combination u = (1 − λ)u + λu00 of certain points u0 ∈ C 0 and u00 ∈ C 00 = Cs (zt+1 , . . . , zr ), where 0 < λ < 1 (see Exercise 4.2). In this case, u00 = λ−1 u + (1 − λ−1 )u0 ∈ hu, u0 i ⊂ H, contrary to H ∩ ∆ = ∅. Hence H ∩ C = C 0 . Exercise 8.4. Since every exposed face of a convex set is its extreme face (see Theorem 8.2) all three statements of the exercise follow from Exercise 7.4. Chapter 9 Exercise 9.1. Translating L on a certain vector, we suppose that L is a subspace. Choose a basis b1 , . . . , bm for L and a basis bm+1 , . . . , bn for L⊥ . Let bn+1 = −(bm+1 + · · · + bn ). We state that L is the intersection of n − m + 1 closed halfspaces Vi = {x ∈ Rn : x·bi 6 0},
m + 1 6 i 6 n + 1.
Since the inclusion L ⊂ Vm+1 ∩ · · · ∩ Vn+1 is obvious, it remains to prove the opposite inclusion. For this, choose a point x ∈ Vm+1 ∩ · · · ∩ Vn+1 ⊂ L. Let x = ξ1 b1 + · · · + ξn bn . The inequalities x·bi 6 0, m + 1 6 i 6 n + 1 imply that ξm+1 6 0, . . . , ξn+1 6 0, ξm+1 + · · · + ξn+1 > 0. Hence ξm+1 = · · · = ξn+1 = 0, and x ∈ L. Consequently, Vm+1 ∩ · · · ∩ Vn+1 ⊂ L. From Theorem 9.9 it follows that every family of n − m + 1 closed halfspaces whose intersection is L should be irreducible. Exercise 9.2. The equality L = ∩ (bd V : V ∈ F) follows from L ⊂ ∩ (bd V : V ∈ F) ⊂ ∩ (V : V ∈ F) = L. Exercise 9.3. Let L = c + S, where c ∈ Rn and S is a subspace. Assume that the halfplane F is closed (the case when F is open is similar). By Theorem 1.42, F = M + h, where M is a plane and h is a halfline with endpoint o, which is not parallel to M . Express M as M = a + T , where a ∈ Rn and T is a subspace. From the same theorem and the equality L + F = (a + c) + (S + T ) + h
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it follows that L+F is either a plane if h ⊂ S+T , or a halfplane if h 6⊂ S+T . Suppose that L ∩ F 6= ∅ and choose a point v ∈ L ∩ F . Theorem 1.30 implies that the plane T 0 = v + T lies in F . If L ∩ F ⊂ T 0 , then the inclusions L ∩ T 0 ⊂ L ∩ F ⊂ L ∩ T 0 show that L ∩ F = L ∩ T 0 ; whence L ∩ F is a plane as the intersection of the planes L and T 0 . Assume that L ∩ F 6⊂ T 0 and choose a point w ∈ (L ∩ F ) \ T 0 . Then the line l = hv, wi lies in aff F and is not parallel to the plane M = u+T . From Theorem 1.15 it follows that l ∩ M 6= ∅. Choose a point e ∈ l ∩ M . Theorem 1.2 gives L = e + S and M = e + T . Consequently, F = e + T + [o, v − ei, according to Theorem 1.42. Therefore, L ∩ F = (e + S) ∩ (e + T + [o, v − ei) = e + S ∩ T + [o, v − ei, and Theorem 1.42 implies that L ∩ F is a halfplane. Exercise 9.4. Let f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. For any sets X ⊂ Rn and Y ⊂ Rn , one has f (X) = a + g(X),
f −1 (Y ) = g −1 (Y − a),
Y ∩ rng f = (Y − a) ∩ rng g + a. Since a translate of a plane (respectively, of a halfplane) also is a plane (respectively, a halfplane), it suffices to prove that the statements hold for g. Furthermore, we will assume that both halfplanes F and G are closed, because the case when they are open is similar. By Theorem 1.42, the closed halfplane F ⊂ Rn can be expressed as F = M + h, where M is a plane and h is a halfline with endpoint o, which is not parallel to M . Express M as L = c + T , where c ∈ Rn and T is a subspace. Then g(F ) = g(c + T + h) = g(c) + g(T ) + g(h). Let h = [o, bi = {λb : λ > 0}, where b is a nonzero vector in h. From g(h) = g({λ b : λ > 0}) = {λ g(b) : λ > 0} it follows that g(h) is a closed halfline with endpoint o if g(b) 6= o, or the singleton {o} if g(b) = o. Since g(T ) is a subspace, Theorem 1.42 shows that g(c) + g(T ) + g(h) is either a closed halfplane if g(h) 6⊂ g(T ), or the plane g(c) + g(T ) if g(h) ⊂ g(T ). Let G ⊂ Rm be a closed halfplane with G ∩ rng g 6= ∅. By Exercise 9.4, the set H = G ∩ rng g is either a plane or a closed halfplane. If H is a plane, then Corollary 1.89 shows that the set g −1 (G) = g −1 (H) is a plane. Assume that H is a closed halfplane. Then H = e + N + h, where e ∈ H, N is a subspace of Rm , and h ⊂ rng g is a closed halfline with endpoint o which does not lie in N (see Theorem 1.42). Clearly, g −1 (G) = g −1 (H) = g −1 (e) + g −1 (N ) + g −1 (h).
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By Corollary 1.89, g −1 (e) is a plane and g −1 (N ) is a subspace; so, their sum g −1 (e)+g −1 (N ) is a plane (see Theorem 1.4). If h = [o, ci for a certain nonzero point c ∈ h, and if c0 is a point in Rn satisfying the conditions f (c0 ) = c, then Theorem 1.42 and the obvious equality g −1 (h) = g −1 ([o, ci) = [o, c0 i + null g imply that g −1 (h) is a closed halfplane. Finally, g −1 (G) is a closed halfplane as the sum of the plane g −1 (e) + g −1 (N ) and the closed halfplane g −1 (h) (see Exercise 9.4). Exercise 9.5. Since the polytope P is compact, Corollary 8.12 shows that both sets A and B are compact. Suppose that P is irreducibly represented by a family F = {V1 , . . . , Vr } of closed halfspaces. Let Vi = {x ∈ Rn : x·ci 6 γi },
1 6 i 6 r.
Because the hyperplane bd Vi supports P , there is a point ui ∈ bd Vi ∩ P . From P = A + B we deduce the existence of points ai ∈ A and bi ∈ B such that ui = ai + bi . Put αi = ai ·ci and βi = bi ·ci . Clearly, αi + βi = (ai + bi )·ci = γi . Put Vi0 = {x ∈ Rn : x·ci 6 αi }
and Vi00 = {x ∈ Rn : x·ci 6 βi }.
Then Vi0 + Vi00 = Vi . We state that A ⊂ Vi0 and B ⊂ Vi00 . Indeed, assuming the existence of a point a ∈ A \ Vi0 , we would obtain that a + bi ∈ (A + B) \ Vi because of (a + bi )·ci > αi + βi , contrary to the inclusion A + B = P ⊂ Vi . Let A0 = V10 ∩ · · · ∩ Vr0 and B0 = V100 ∩ · · · ∩ Vr00 . From P = A + B ⊂ A0 + B ⊂ A0 + B 0 ⊂ P it follows that A + B = A0 + B. Theorem 2.47 gives A = A0 . Similarly, B = B0 . Summing up, both A and B are polytopes. Exercise 9.6. From the obvious inclusions P = A + B ⊂ A + cl B ⊂ cl (A + B) = P it follows that P = A + cl B. Suppose that P is irreducibly represented by a family F = {V1 , . . . , Vr } of closed halfspaces. Let Vi = {x ∈ Rn : x·ci 6 αi },
1 6 i 6 r.
Since cl B is compact, the continuous function ϕi (x) = x · ci attains a maximum value, βi , on cl B.
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Because P is closed, there is a point xi ∈ P such that ϕi (xi ) = αi . Let xi = ai + bi , where ai ∈ A and bi ∈ cl B. Then ϕ(bi ) 6 βi , and γi = sup{ϕ(x) : x ∈ A} > ϕ(ai ) = ϕ(xi − bi ) = ϕ(xi ) − ϕ(bi ) > αi − βi . On the other hand, if yi is a point in cl B with ϕi (yi ) = βi , then x + yi ∈ P for every point x ∈ A, implying that ϕi (x) + βi = ϕi (x) + ϕi (yi ) = ϕ(x + yi ) 6 αi . Therefore, γi = sup{ϕ(x) : x ∈ A} 6 αi − βi , which gives αi + βi = γi . Put Vi0 = {x ∈ Rn : x·ci 6 βi }
and Vi00 = {x ∈ Rn : x·ci 6 γi }.
Then Vi0 + Vi00 = Vi . By the above argument, A ⊂ Vi00 and B ⊂ Vi0 . Let A0 = V100 ∩ · · · ∩ Vr00 and B0 = V10 ∩ · · · ∩ Vr0 . From P = A + B ⊂ A0 + B ⊂ A0 + B 0 ⊂ P it follows that A + B = A0 + B. Theorem 2.47 gives A = A0 . Hence A is a polyhedron. Exercise 9.7. Since c ∈ rint K, there is a scalar ρ > 0 with the property Bρ (c) ∩ aff K ⊂ K. Because K is bounded, there is a finite set X such that X ⊂ K ⊂ Bερ (X). Adding to X finitely many points from K, we assume that aff X = aff K. By Theorem 3.12, the convex polytope P = conv X satisfies the inclusions P ⊂ K ⊂ Bερ (P ). From Theorem 2.56 it follows that the homothetic copy c + (1 + ε)(P − c) of P contains the set Bερ (P ) ∩ aff P . Summing up, P ⊂ K ⊂ (1 + ε)P − εc.
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Author Index
Abe, Y., 143 Aboubabdullah, D., 70 Alexandrov, A., 114 Archimedes, 112 Ariyawansa, K. A., 114 Arrow, K. J., 144 Artstein, Z., 144 Asplund, E., 318 Auslender, A., 226 Bair, J., 143, 181, 225, 254, 284, 343 Baker, M. J. C., 142 Balinski, M. L., 285 Batson, R. G., 284 Berberian, S. K., 318 Bessaga, C., 113 Bilyeu, R. G., 114 Bj¨ orck, G., 283 Blackwell, D., 113 Blumenthal, L. M., 143 Bol, G., 344 Bonnesen, T., 112, 143, 253 Bonnice, W. E., 143, 182 Borovikov, V. A., 112 Borwein, J. M., 144 Bourbaki, N., 283 Breen, M., 114 Bromek, T., 113 Bronshte˘ın, E. M., 285 Brown, A. L., 317 Brunn, H., 112, 113, 143
Brøndsted, A., vi Bunt, L. N. H., 143 Burago, Yu. D., 113 Burton, G. R., 115 Carath´eodory, C., 120, 142, 252 Cassels, J. W. S., 144 Cernikov, S. N., 344 Choquet, G., 115, 318 Chubarev, A., 70 Collier, J. B., 285 Cook, W. D., 113, 142 Corson, H. H., 318 Dalla, L., 284 Danielyan, E. A., 143 Danzer, L., 143 Davidon, W. C., 114 Davis, C., 181, 343, 344 Dax, A., 253 De Wilde, M., 318 Debs, G., 284 Derry, D., 143 Deumlich, R., 253 Dieudonn´e, J., 226, 253 Dragomirescu, M., 226 Dubins, L. E., 283, 284 Eckhoff, J., 143 Eggleston, H. G., 112, 344 Ekeland, I., 144 399
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Lectures on Convex Sets
Elster, K.-H., 253 Epifanov, G. V., 344 Fejes T´ oth, L., 114 Fenchel, W., 112, 143, 253 Fischer, T., 181 Florenzano, M., v Folkman, J., 144 Fourneau, R., 254 Frenkel, J., 70 Fujiwara, M., 284 Gale, D., 181 Gallivan, S., 285 Gardner, R. G., v, 285 Girshick, M. A., 113 Goberna, M. A., 226, 254 Goossens, P., 253 Gr¨ unbaum, B., vi, 112, 113, 143, 286, 343, 344 Gruber, P. M., v, 112 Grz¸a´slewicz, R., 113 Grzybowski, J., 111 Guler, O., v Gustin, W., 143 Hahn, F. H., 144 Hammer, P. C., 254 Hanner, O., 143 Hansen, W., 182 He, Y., 286 Hiriart-Urruty, J.-B., v Hoffman, A. J., 344 Holick´ y, P., 318 Howe, R., 144 Husain, T., 284 Ichiishi, T., 144
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Klee, V. L., 70, 112–114, 143, 182, 253, 254, 284–286, 318, 344 Koenen, W., 144 Kramer, H., 143 Kubota, T., 143 Kuz0 minyh, A. V., 114 Laczkovich, M., 318 Larman, D. G., 113, 285, 286 Lassak, M., 254 Lawrence, J., 113, 144 Lay, S. R., v Le Van, C., v Leichtweiß, K., 254 Lemar´echal, C., v Lenz, H., 70 Lepin, A. A., 143 Lommatzsch, K., 284 Maluta, E., 254 Mani-Levitska, P., 113, 115 Mart´ınez-Legaz, J.-E., 254 Martin, M., 284 Mauldin, R. D., 113 Mazur, S., 113 McKennon, K. D., 114 McKinney, R. L., 182 Meyer, W., 114 Minkowski, H., 112, 181, 253, 284, 286, 343, 344 Mirkil, H., 344 Moreau, J.-J., 226 Motzkin, T., 226, 343, 344 Movsisyan, G. S., 143 Nakajima, S., 113 Naumann, H., 344 Nehse, R., 253 Nieuwenhuis, J. W., 253
Jamison, R. E., 113 Jessen, B., 226 Jornet, V., 226, 254
O’Brien, R. C., 144 Orland, G. H., 318
Kaniewski, J., 113 Kay, D. C., 114 Kirsch, A., 252
P´ ales, Z., 254 Papadopoulou, S., 284 Peterson, B. B., 143
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Author Index
Pforr, E.-A., 181 Phelps, R. R., 115, 284 Pinelis, I., 70 Post, K. A., 114 Pranger, W., 284 Preiss, D, 113 Price, G. B., 284 Radon, J., 142 Radstr¨ om, H., 113, 143 Reay, J. R., 142, 143, 182 Reiter, H. B., 284, 318 Robinson, C. V., 143 Rockafellar, R. T., v, 181, 226, 252, 253 Rodr´ıguez, M. M. L., 226, 254 Rogers, C. A., 115, 285 Roy, A. K., 284 Rubin, H., 113 Schneider, R., v, 144, 283, 284 Sha˘ıdenko-K¨ unzi, A. V., 114 Shapley, L., 144 Shephard, G. C., 115 Singer, I., 254 Smith, C. R., 113 Soltan, P., 253, 284 Soltan, V., 115, 144, 284, 318 Starr, R. M., 144 Stavrakas, N. M., 284, 318 Steinitz, E., 113, 142, 143, 181, 226, 253, 286, 343 Stoelinga, T. G. D., 143, 344 Stoer, J., v, 181 Stoker, J. J., 226 Straszewicz, S., 143, 318 Sun, J., 286 Sung, C. H., 181
Tam, B. S., 181 Tamura, T., 113 Teboulle, M., 226 Temam, R., 144 Tietze, H., 113 Toranzos, F. A., 113 Tukey, J. W., 254 Tverberg, H., 143 Tweddle, I., 284 Urba´ nski, R., 111 Valentine, F. A., 143 Vasiloi, A., 253, 284 Veblen, O., 70 Walsh, J. L., 114 Webster, R. J., v, 113 Wesler, O., 113 Weyl, H., 343 Whitehead, J. H. C., 70 Witzgall, C., v, 181 Wolfe, D., 114 Yaksubaev, K. D., 113 Yoneguchi, H., 143 Zalgaller, V. A., 113 Zamfirescu, T., 286 Zanco, C., 254 Zhou, L., 144 Ziegler, G. M., vi Zizler, V., 253
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401
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Subject Index
ball
face
closed unit –, 7 closed –, 7 open unit –, 7 open –, 7 basis, 4 affine –, 49 positive –, 181 strongly positive –, 182
exposed –, 287 extreme –, 255 generated exposed –, 298 generated extreme –, 260 halfplanar exposed –, 309 halfplanar extreme –, 275 planar exposed –, 309 planar extreme –, 275 proper exposed –, 287 proper extreme –, 255 facet, 328 family of sets disjoint –, 1 irreducible –, 324 nested –, 1 pairwise disjoint –, 1 functional affine –, 57 linear –, 5
combination of points/vectors affine –, 37 convex –, 72 linear –, 4 positive convex –, 72 cone, 145 apex of –, 145 apex set of –, 153 barrier –, 212 convex –, 145 normal –, 211 polar –, 213 recession –, 187 simplicial –, 148 coordinates, 2 affine –, 51 De Morgan’s Laws, 1 distance, 6 inf –, 9 dot product, 5
halfline closed –, 1, 25 exposed –, 304 extreme –, 270 open –, 1, 25 halfplane closed –, 34 open –, 34 halfspace closed –, 28 403
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9508-Lectures on Convex Sets
Lectures on Convex Sets
open –, 28 proper support –, 238 support –, 238 homothety, 64 contraction, 64 expansion, 64 negative –, 64 positive –, 64 hull conic –, 161 convex –, 117 hyperplane, 20 properly separating –, 241 separating –, 241 strongly separating –, 241 interval, 1 closed –, 1 open –, 1 semi-open –, 1 line, 13, 25 mapping, 2 continuous –, 9 inverse –, 2 invertible –, 2 one-to-one –, 2 onto –, 2 range of –, 2 plane, 11 asymptotic –, 101, 233 bounding –, 227 dimension of –, 13 proper bounding –, 227 proper support –, 230 proper –, 11 support –, 230 planes complementary –, 15 independent –, 15 parallel –, 17 point, 2, 13 r-exposed –, 312 r-extreme –, 278
closure –, 8 exposed –, 299 extreme –, 267 interior –, 7 nearest –, 206 relatively interior –, 78 polyhedron, 319 polytope, 319 projection affine –, 65 linear –, 5 metric –, 208 orthogonal –, 6, 65 scalar, 1 segment closed –, 25 open –, 26 semi-open –, 25 set Fσ –, 9 Gδ –, 9 ρ-neighborhood of –, 7 f -image of –, 2 r-exposed –, 312 r-extreme –, 278 affinely dependent –, 45 affinely independent –, 45 boundary of –, 8 bounded –, 8 closed –, 8 closure of –, 8 compact –, 8 connected –, 9 convex –, 71 coterminal –, 273 countable –, 1 dense –, 8 denumerable –, 1 diameter of –, 8 dimension of –, 55 interior of –, 7 inverse f -image of –, 2 line-free –, 222 linearity space of –, 197 linearly dependent –, 4
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Subject Index
linearly independent –, 4 open –, 7 orthogonal complement of –, 5 orthonormal –, 6 positively independent –, 181 relative boundary of –, 104 relative interior of –, 78 relatively open –, 78 scalar multiple of –, 3 sets (Minkowski) sum of –, 3 intersection of –, 1 orthogonal –, 5 set difference of –, 1 union of –, 1 simplex, 74 slab closed plane –, 37 closed –, 32 open plane –, 37 open –, 32 separating –, 245 span, 4 affine –, 40 sphere, 7 subset proper –, 1 subspace, 3 characteristic – of a plane, 12 characteristic – of a set, 42 dimension of –, 4 hyper-, 4 trivial –, 3 zero –, 3 subspaces complementary –, 4 direct sum of –, 4 independent –, 3
theorem Heine-Borel –, 8 Lindel¨ of’s –, 7 transformation affine –, 57 linear –, 4 null space of –, 5 translate, 3, 64 value absolute –, 1 sign –, 1 vector, 2 length of –, 6 norm of –, 6 normal –, 22, 28 unit –, 6 vector space Rn , 2 vectors orthogonal –, 5 Zorn’s lemma, 1
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