Week 3: Faces of convex sets

Transcription

1 Week 3: Faces of convex sets Conic Optimisation MATH515 Semester 018 Vera Roshchina School of Mathematics and Statistics, UNSW August 9, 018 Contents 1. Faces of convex sets 1. Minkowski theorem 3 3. Minimal faces 5 4. A couple of useful results 6 5. Structure of polyhedral sets 7 6. Exercises 8 References 10

2 Conic Optimisation/Week 3: Faces of convex sets 1 We define the faces of convex sets and using this notion study the structure of convex sets, focussing on compact and polyhedral sets. The results discussed here can be found in the classic textbooks [] and [3]. One may also wish to consult [1] for additional details. 1. Faces of convex sets Definition 1 (Face). Let C R n be a convex set. A convex set F C is called a face of C if for every x F and every y, z C such that x (y, z), we have y, z F. The fact that F is a face of C can be denoted by F C or F C. Observe that the set C itself is its own face. An empty set is also a face of C. A nonempty face F C such that F C is called proper. In Fig. 1 some faces and non-faces of a square and a disk are shown. Figure 1. Faces and non-faces: every vertex and every side (edge) of a square is a proper face; a subset shown in the second image is not a face, even though it is convex. The proper faces of a disk are the poins on its boundary; a segment of the boundary is not a face because it is not convex. Lemma. Let C R n be a convex set, and suppose that F C. Let E F. Then E F if and only if E C. Proof. It follows from the definition that if the set E is a face of C, then it is also a face of F. Conversely, suppose that E is a face of F, and let y, z C be such that (y, z) E. Since E F, we have (y, z) F. Hence y, z F since F is a face of C. But then we also have y, z E, since E is a face of F. Lemma 3. If C R n is a closed convex set, then every nonempty face F C is a closed convex set. Proof. Assume that this is not true: for some nonempty F C, where C is a closed convex subset of R n, there exists a point x cl F \ F. Since C is closed, we have x C. Since F is nonempty, there is some y ri F (y x). We know that all points on (x, y) are in the relative interior of F, and hence in F. Then by the definition of a face, the endpoints must also belong to F, and therefore x F, a contradiction. Zero-dimensional faces are called extreme points. For such special faces the definition can be restated as follows. Definition 4 (extreme point). Let C R n be a convex set. A point v C is called an extreme point of C if it does not lie in any open line segment joining two points of C. The set of all extreme points of a given convex set C R n is denoted by ext C. We can refine the statement of Lemma for the case of extreme points as follows. Corollary 5. Let C R n be a convex set, and let F C. Then ext F = ext C F. We have the following useful statement about faces.

3 018 Sem MATH515 UNSW Vera Roshchina Theorem 6 (Theorem 18.1 in [3]). Let C be a convex set, F C and D a convex subset of C. If ri D F, then D F. Proof. Let x ri D F. If z D and x z, then x ri D implies that there exists y D such that x (y, z). Since F C, we have z F. Corollary 7. Let C be a convex set and F C. Then F ri C if and only if F = C. Proof. Follows directly from Theorem 6. Corollary 8. Let C R n be a convex set, and let E, F C. If ri E ri F, then E = F. Proof. Follows from Theorem 6. See Exercise 4. Corollary 9. Let C be a convex set, and let F be a proper face of C. Then F rbd C, and dim F < dim C. Proof. Assume that F is a proper face of a convex set C R n. If ri C F, then C F by Theorem 6. Hence we conclude that F rbd C. To prove the assertion about dimension, observe that if dim F = dim C = d, then there are d + 1 affinely independent points in F (and hence in C) that affinely span aff C, and so aff F = aff C, and hence = ri F ri C, which effectively means that F ri C, and we arrive to the same contradiction. Proposition 10. Let C R n be convex. Then for every x C either x ri C or there exists a proper face F C of C such that x F. Proof. Let x C. If x ri C, there is nothing to prove. If x rbd C, define F := {[y, z] y, z C, x (y, z)}. First consider the case when F =. In this case there is no line segment [y, z] C such that x (y, z). Then x is an extreme point of C, and trivially x {x} C. Now consider the case F. It is evident that x F. We will show that F is a proper face of C. For any two line segments [y 1, z 1 ], [y, z ] C such that x (y 1, z 1 ), x (y, z ) we have [y 1, z 1 ], [y, z ] F by the definition of F. We will show that co{y 1, z 1, y, z } F. Take any point u co{y 1, z 1, y, z }. Observe that if x = u, then we are done. Otherwise, we have u = α 1 y 1 + β 1 z 1 + α y + β z, α 1, α, β 1, β 0, α 1 + β 1 + α + β = 1. On the other hand, since x (y 1, z 1 ) (y, z ), we have x = s 1 y 1 + t 1 z 1 = s y + t z, s 1 + t 1 = s + t = 1, s 1, t 1, s, t > 0. Let v λ := x + λ(x u) = (1 + λ)x λu.

4 Conic Optimisation/Week 3: Faces of convex sets 3 We have explicitly (observing that x = 1x + 1x) ( s1 v λ = (1 + λ) y 1 + t 1 z 1 + s y + t ) z λ (α 1 y 1 + β 1 z 1 + α y + β z ) ( ) ( ) (1 + λ)s1 (1 + λ)t1 = λα 1 y 1 + λβ 1 z 1 ( ) ( ) (1 + λ)s (1 + λ)t + λα y + λβ z. It is evident that there exists a sufficiently small λ > 0 such that all four coefficients are nonnegative. It is also clear that they sum to 1. We hence have v λ C, and x = λ v λ + λ 1 + λ u, so by the definition of F we have u [u, v λ ] F. Now if we take any v 1, v F, there are y 1, z 1, y, z such that x (y 1, z 1 ) (y, z ) and v 1 [y 1, z 1 ], v [y, z ]. Then any w [v 1, v ] co{y 1, z 1, y, z } F, hence, F is convex. Likewise, if there is some u (v 1, v ) F such that v 1, v C, then by the definition of F we have v 1 [y 1, z 1 ], v [y, z ], and x (y 1, z 1 ) (y, z ). We then know that [v 1, v ] co{y 1, z 1, y, z } F, hence F is a face of C. It remains to show that F is a proper face of C. Assume that there is y F ri C. Then by the definition of F we have x (y, z) for some z C. We then must have x ri C, a contradiction. Hence, F ri C =, and therefore F C, hence F is a proper face of C. We can refine the preceding statement into the following powerful result. Theorem 11 (Theorem 18. in [3]). Let C R n be a convex set. Then C is the disjoint union of the relative interiors of its faces, i.e. C = {ri F, F C}, ri F ri E = E, F C such that E F. Proof. We will first show that for every x C there exists a face F C such that x ri F, and then demonstrate that this face is unique. Let x C. If x ri C, then we let F := C. Otherwise x rbd C, and by Proposition 10 there exists a proper face F C such that x F. We can apply Proposition 10 to x and F again, deducing that either x ri F or there is a yet strictly smaller face E F C such that x E. We can continue this process until we reach such a face of G C that x ri G. This process is finite because on each step we identify a nonempty face of dimension smaller than the dimension of the preceding face (see Corollary 9). It follows from Corollary 8 that the face F C such that x ri F is unique.. Minkowski theorem Extreme points play a special role in the structure of convex sets due to the Minkowski theorem. This theorem asserts that every compact convex set is completely characterised by its extreme points. Theorem 1 (Minkowski theorem). A compact convex set C R n is the convex hull of its extreme points.

5 4 018 Sem MATH515 UNSW Vera Roshchina Before we prove this theorem, it is worth observing that the set of extreme points of a compact convex set may not be closed. The classic example of the convex hull of a circle and a transversal line segment (see Fig. ) demonstrates this. Figure. An example of a convex set C for which ext C is not closed: this set is the convex hull of a circle and a line segment orthogonal to the plane of the circle Proof of the Minkowski theorem. Let C R n be a compact convex set. We will proceed by induction in the dimension of the set C. Observe that for dim C = 0 the set C is a singleton, and hence ext C = C, so the statement holds trivially. Now suppose that for any set C R n of dimension d = dim C, 0 d m we have C = co ext C. We will show that then it is true for d = dim C = m + 1. Assuming that dim C = d = m + 1, pick any point x C. By Theorem 11 there exists a unique face F C such that x ri F. If F is a proper face of C, then dim F < dim C by Corollary 9 and by the inductive assumption we have x co ext F. By Corollary 5 we have ext F ext C, hence x co ext C. It remains to consider the case when x ri C. Since dim C = m + 1 > 0, the set C is not a singleton, and there exists another point y C, y x. Consider the line x + λ(y x) passing through x and y. We will next show that this line intersects the relative boundary of the set C at precisely two points lying on different sides of x. Let λ := inf{λ x + λ(y x) C}, λ + := sup{λ x + λ(y x) C}. Since the set C is compact (hence bounded), both the infimum λ and the supremum λ + are finite. Since the set C is closed, x := x + λ (y x) C, x + := x + λ + (y x) C. It is evident that λ + > 0 (since x+1 (y x) = y C). Moreover, x+λ(y x) aff C for all λ, and since x ri C, there must be a sufficiently small λ < 0 such that x+λ (y x) C, which means that λ λ < 0. It is evident from the definition of the relative interior and the fact that x + λ(y x) aff C that x and x + are both in the relative boundary of C, so there exist proper faces F, F + C such that x F and x + F +. By Corollary 9 we have dim F, dim F + < dim C = m + 1, hence by the inductive assumption and also from Corollary 5 x co ext F ext C, x + co ext F + ext C. Since x is a convex combination of x and x +, and both x and x + can be represented as convex combinations of points from ext C, it is a technical exercise to verify that x is also a convex combination of points from ext C. Corollary 13. Let C R n be a nonempty compact convex set. Then ext C.

6 Conic Optimisation/Week 3: Faces of convex sets 5 Proof. The proof follows directly from the Minkowski theorem (Theorem 1): since C = co ext C, the set of extreme points ext C has to be nonempty. Notice that the assumption on compactness is crucial here. For the set of extreme points to be nonempty and for the Minkowski theorem to hold it is essential that the set should be both closed and bounded, as shown in the next two examples. Example 14. Consider the open unit ball, C := B = B 1 (0) = {x R n x < 1}. Then ext C =. Indeed, observe that ri C = int C = C, and hence the only nonempty face of C is the set C itself, so there are no extreme points. Example 15. Consider a closed half-space C := {(x, y) R y 0}. Observe that this set is closed but unbounded. We have ri C = int C = {(x, y) R y > 0}, and F := {(x, y) R y = 0} is the only proper face of C. There are no extreme points. 3. Minimal faces Recall that for every point of a convex set there exists a face in this set that contains this point in its relative interior (see Theorem 11). This face is the minimal face of the set C that contains the point. We can in fact go a little further and define a more general notion of a minimal face. Definition 16 (Minimal face). Let C R n be a convex set and let S C. The minimal face F min (S, C) of C containing S is a face F C such that S F and there is no smaller face E C, E F such that S E. In the case when S = {x} is a singleton, we will abuse the notation writing F min (x, C) for F min ({x}, C). We will also write F min omitting S and C whenever S and C are clear from the context. The minimal face is well-defined: it always exists (although it may coincide with the set C itself) and is unique. Theorem 17. Let C R n be a convex set and let S C. Then the minimal face F min (S, C) is well-defined and unique; moreover, F min (S, C) = {F F C, S F }. Proof. To prove the statement, it is sufficient to show that E = {F F C, S F } is a face of C: both the minimality and the uniqueness follow from its definition. Since F is the intersection of convex sets, it is also convex. Now let y, z C be such that there is some point x (y, z) E. By the definition of E for every F C such that S F we have y, z F. Then we must have [y, z] E, again, by the definition. We deduce that E is a face of C. We have the following properties of the minimal face (see e.g. [4, Proposition..5])

7 6 018 Sem MATH515 UNSW Vera Roshchina Proposition 18. Let C be a nonempty convex set, = S C, and F C. Then the following statements are equivalent. (i) F = F min (S, C), (ii) S F and co S ri F, (iii) ri co S ri F. Moreover, we have (iv) F min (S, C) = F min (s, C) for any s ri co S. Proof. (iii) (ii): since S, there exists x ri co S ri F. Hence co S ri F. Furthermore, (iii) implies that S F by Theorem 6. (ii) (i): Fix any F C such that S F. Since S F, and S ri F, we have F ri F, hence, by Theorem 6 F F. Hence, by Theorem 17 F = F min (S, C). (i) (iii): by the definition of the minimal face, we have S F, and by the convexity of F ri co S co S F. Assume that (iii) is not true. Then there exists x ri co S \ ri F. By Proposition 10 there exists a subface E F C (E F ) such that x ri E, so ri co S E, and then by Theorem 6 we have co S E, a contradiction with the minimality of F. To show (iv), observe that from (i) (iii) we have ri{s} = {s} ri co S ri F min (S, C) for any s ri co S. Then from (iii) (i) we deduce that F min (S, C) = F min (s, C). In the context of conic optimisation it makes sense to talk about chains of faces. Definition 19. A subset of faces F of a convex set C R n is a chain if for every two faces E, F F, E F either E is a proper subset of F or F is a proper subset of E. It is evident from Lemma and Corollary 9 that any chain of faces must be finite, moreover, the longest possible chain of nonempty proper faces has length at most dim C. 4. A couple of useful results Lemma 0 (Lemma 5.39 in [1]). Let P, Q R n be nonempty convex compact sets, and let C = P + Q. Then every face of C is the Minkowski sum of faces of P and Q. More precisely, F C F P P, F Q Q such that F = F P + F Q. Proof. Let F be a nonempty face of C. We construct two sets F P := {x P y Q, x + y F }, F Q := {y Q x P, x + y F }. Both F P and F Q are nonempty since F is nonempty. First we show that F = F P + F Q. It is obvious that F F P + F Q, and it remains to show the reverse inclusion. For that, pick an arbitrary x F P, y F Q. We will next show that z = x + y F. By the definition of F P and F Q there exist u P and v Q such that x + v F and y + u F. If x = u or y = v, there is nothing to prove, as in this case z = u + v F. Otherwise, by the convexity of F we have z = x + v + y + u F. At the same time, notice that x + y P + Q C; likewise, u + v P + Q C, and z (x + y, u + v). Since F is a face of C, this yields z = x + y F. It remains to show that both F P and F Q are faces of P and Q respectively. First note that both are closed compact sets, and that F Q Q and F P P.

8 Conic Optimisation/Week 3: Faces of convex sets 7 Let x F P, and pick any interval [a, b] P such that x (a, b). By the definition of C, for an arbitrary y F Q we have a + y, b + y C. At the same time, x + y F P + F Q = F and x+y (a+y, b+y). From F C we have [a+y, b+y] F, hence, a+y, b+y F, and therefore a, b F P. This shows that F P is a face of P. The proof for F Q is identical. It follows from the previous lemma that every face of a closed convex set C must contain its lineality space. Lemma 1. For a convex set C R n and F C we have Proof. Left as an exercise (see Exercise 10). F = aff F C. 5. Structure of polyhedral sets Recall that the sets defined by a finite number of linear inequalities are called polyhedral. Generally speaking, a system of linear inequalities need not be consistent, so a polyhedral set may be empty. Likewise, a polyhedral set is not necessarily bounded. Theorem. Let P be a polyhedral set, P = {x a i, x b i, i I}, where I is a finite index set (for example, I = {1,,..., m}). Then F is a nonempty face of P if and only if it can be represented as a nonempty polyhedral set (1) F I0 = {x a i, x = b i i I 0, a i, x b i I \ I 0 }, where I 0 I. Proof. Observe that for every index I 0 I the set F I0 is a face of P. Indeed, let x F I0, and choose any y, z P such that x (y, z). For every i I 0 we have for some α (0, 1) a i, x = (1 α) a i, y + α a i, z = b i, which is only possible when a i, y = a i, z = b i, hence, y, z F I0, and F I0 is a face of P. Let F be an arbitrary nonempty face of P. We will show that F = F I0 for some I 0 I. Let x 0 ri F, and let I 0 = {i I a i, x 0 = b i }. Now observe that x 0 ri F. Indeed, there exists a sufficiently small ε > 0 such that for all x in B ε (x 0 ), B ε (x 0 ) = {x x x 0 ε}, we have a i, x < b i for i I \ I 0. At the same time, for any y aff F I0 we have p p y = λ j x j, λ j = 1, x j F I0, and hence a i, y = j=1 j=1 p λ j a i, x j = j=1 p λ j b i = b i i I 0. Therefore, B ε (x 0 ) aff F I0 F I0, and x 0 ri F I0. Since x 0 ri F ri F I0, from Corollary 8 we have F = F I0. j=1

9 8 018 Sem MATH515 UNSW Vera Roshchina Corollary 3. Every face of a polyhedral set is a polyhedral set. Proof. Evident from the representation (1). As a consequence of Theorem, we have the following statement. Theorem 4. A polyhedral set P has a finite number of different faces. Moreover, the extreme points of polyhedral sets are nonempty finite intersections where I I. {e} = i I {x a i, x = b i }, Proof. Finiteness follows from the finiteness of the power set of I, and the last representation follows directly from Theorem discarding the redundant inequalities. Observe that there is a finite number of sets that can be represented as in Theorem 4, hence, the number of extreme points (vertices) of a polyhedral set is finite. Also we have the following result directly from the Minkowski theorem. Lemma 5. A bounded polyhedron is a convex hull of a finite number of points (i.e. it is a convex polytope). For polyhedral sets it is conventional to call the extreme points vertices, the onedimensional faces edges, and n 1 dimensional faces facets. 6. Exercises Exercise 1 (Assignment, 5%). Consider the following optimisation problem: min c, x s.t. x C, where c R n, and C R n is a compact convex set. Prove that the set of optimal solutions to this problem { } S := x R n c, x = min c, x x C is a face of C. Exercise. Identify all proper faces of the convex set C = {(x, y) x 1, y 0}. Solution. We can rewrite the definition of C as C = {(x, y) x 1, x 1, y 0}. This is a polyhedral set, we can obtain all candidates for the faces from Theorem, selecting the nonempty sets, as follows. F 1 = {(x, y) x = 1, x 1, y 0} = {(x, y) x = 1, y 0}, F = {(x, y) x 1, x = 1, y 0} = {(x, y) x = 1, y 0}, F 3 = {(x, y) x 1, x 1, y = 0}, F 4 = {(x, y) x = 1, x = 1, y 0} =, F 5 = {(x, y) x = 1, x 1, y = 0} = {(1, 0)}, F 6 = {(x, y) x 1, x = 1, y = 0} = {( 1, 0)}, F 7 = {(x, y) x = 1, x = 1, y = 0} =,

10 Conic Optimisation/Week 3: Faces of convex sets 9 hence, the faces F 1, F, F 3, F 5 and F 6 are the proper faces of C. Exercise 3 (Assignment 5%). Identify all proper faces of the nonnegative orthant in R 3, defined as R 3 + = {(x, y, z) R 3 x 0, y 0, z 0}. Exercise 4. Prove Corollary 8 using Theorem 6: let E, F C. If ri E ri F, show that E = F. Solution. From ri E ri F we have ri E F. Since F is a face of C, from Theorem 6 we have E F. Likewise, we obtain and F E, hence, E = F. Exercise 5 (Assignment 5%). Let F C, where both F and C are convex sets. Is it true that F is a face of C if and only if C \ F is a convex set? Exercise 6. Prove that if D is a convex subset of the relative boundary of a non-empty convex set C R n, then dim D < dim C. Solution. If D rbd C, then D ri C =, and by Proposition 18 the minimal face F = F min (D, C) is a proper face of C. By Corollary 9 dim D dim F < dim C. Exercise 7. Let C = C 1 + C, where C 1 and C are nonempty convex sets in R n. Show that if x is an extreme point of C, then x = y + z, where y and z are some extreme points of C 1 and C, respectively. Solution. This follows directly from Lemma 0, however we can also prove this without relying on this lemma. Suppose that x = y +z, y C 1, z C, but, say, y is not an extreme point of C. Then it must be in the relative interior of a face F C 1, so there are some points y 1, y C such that y (y 1, y ). From here we have x 1 := y 1 + z, x := y + z, x 1 x, x 1, x C and x (x 1, x ). Since x is an extreme point, hence, a face of C, we have x 1, x {x}, which is contradictory. Exercise 8 (Assignment 5%). Prove that the Minkowski sum of two polytopes is a polytope. Exercise 9. Consider k sites (points) v 1, v,..., v k R n. A Voronoi cell V i of v i is the set of points nearest to v i, i.e. V i := {x R n x v i x v j Show that Voronoi cells are polyhedral. Solution. Observe that the set defined by the inequality is the same as the set defined by x v i x v j () x v i x v j. We have j i}. (3) x v i = x v i, x v i = x x, v i + v i,

11 Sem MATH515 UNSW Vera Roshchina Using (3), the equation () becomes x x, v i + v i x x, v j + v j ; cancelling out x and regrouping the terms, we have x, v j v i v j v i, a linear inequality. Since the set V i is defined by a finite number of such inequalities, we conclude that this set is polyhedral. Exercise 10. Prove Lemma 1: for a convex set C R n and a face F C, F we have F = aff F C. Solution. Observe that F aff F C. It remains to show the converse. Assume the contrary and let y aff F C \ F. Since F, there exists x ri F, x y. Since y aff F, there is some z aff{x, y} C such that x (y, z). This means that y, z F by the definition of a face, a contradiction with our choice of y. References [1] Osman Güler. Foundations of optimization, volume 58 of Graduate Texts in Mathematics. Springer, New York, 010. [] Jean-Baptiste Hiriart-Urruty and Claude Lemaréchal. Fundamentals of convex analysis. Grundlehren Text Editions. Springer-Verlag, Berlin, 001. Abridged version of Convex analysis and minimization algorithms. I and II. [3] R. Tyrrell Rockafellar. Convex analysis. Princeton Mathematical Series, No. 8. Princeton University Press, Princeton, N.J., [4] Yuen-Lam Cheung (Viris Voronin). Preprocessing and reduction for semidefinite programming via facial reduction: Theory and practice. PhD thesis, University of Waterloo, 013.