Lemma 5.22.If C is a nonempty convex set in a finite-dimensional affine spaceA, thenri(C)6=∅. Moreover,dim(C) = dim(ri(C)).
Proof. Let{xi}k1⊂C be an affine basis for aff(C). Consider the simplices S:=
k X
i=1
tixi:
k
X
1
ti= 1, ti>0, i= 1, . . . , k
⊂C,
Sk:=
(t1, . . . , tk)∈Rk:
k
X
1
ti= 1, ti>0, i= 1, . . . , k
. Since{xi}k1 is affinely independent, the map
T(t1, . . . , tk) =
k
X
i
tixi
is a one-to-one onto affine transformation from the affine space aff(C) to the hyperplane H := {t ∈Rk : Pk
1ti = 1}, hence a homoeomorphism between the two affine spaces. Evidently,Sk is open inH, and thusS=T(Sk) is open in aff(C). This proves that ri(C)6=∅.
Note that we also have dim(C) =k−1 = dim(ri(C)). ut Combining Corollary 5.21 and Lemma 5.22, we immediately obtain the following important result for finite-dimensional convex sets.
6∅, and
ri(C) = ri(C) = rai(C) and C= ri(C) = ac(C).
5.5 Facial Structure of Convex Sets 129 A point x∈C is called an extreme point of C if{x} is a face ofC. We denote the set of extremal points ofC byext(C).
A vectord∈Eis called an extreme directionofCif there is a pointp∈C such that the rayp+R+d:={p+td:t≥0} is a face ofC.
A vectord∈E is called a recession direction of C if there exists a point p∈Csuch that the rayp+R+dstays inC. The set of all recession directions ofC is called the recession coneofC,
recC:={d∈Rn:there existsp∈C such that p+td∈C for allt≥0}. Lemma 5.25.Let F andC be two convex sets such that F ⊂C. ThenF is a face ofC if and only if C\F is a convex set.
This is an easy consequence of the definition of a face, and can be used as an alternative definition of a face.
Remark 5.26.It is easy to see that the union of a nested set of faces of C is a face (the nestedness condition is needed only to ensure the convexity of the union), and the same is true for the intersection of any set of faces. The face of a face is a face, that is, ifF2⊆F1⊆CwithF2a face ofF1andF1a face of C, thenF2is a face ofC. Also, ifF is a face ofC, then ext(F) = ext(C)∩F. A face actually satisfies a stronger property given below. This can be used to show, among other things, that a convex setCcan be written as a disjoint union of relative interiors of different faces ofC; see Rockafellar [228], Theorem 18.2.
Lemma 5.27.Let C be a convex set in a vector spaceE,F a face of C, and D a convex subset ofC. Ifri(D)∩F 6=∅, then D⊆F.
Proof. Pickz∈ri(D)∩F. Ifx∈D, then z∈ri(D) implies that there exists y∈D such thatz∈(x, y). SinceF is a face ofC, we havex∈F. ut As a first step toward the decomposition of a closed convex set, we char- acterize its affine faces.
Lemma 5.28.Let C be a closed convex set in a vector spaceE. If06=d∈E is a recession direction ofC, thenq+R+d⊆C for everyq∈C.
Consequently,C+ recC=C.
Proof. Suppose that p+R+d ⊆ C, and let q ∈ C, q 6= p. It is easy to see geometrically that the convex hull ofqandp+Ris the union of sets{q}and [p, q) +R, whose closure is the set [p, q] +R; seeFigure 5.4.SinceCis closed,
we haveq+R⊆C. ut
Corollary 5.29.Let C be a closed, convex set in a vector space E. If C contains an affine subspace K := p+L, where p ∈ C and L is a linear subspace ofE, thenC=C+L.
p
q
C
Fig. 5.4.Recession direction in a convex set.
linear spaceE. ThenC contains an affine faceF.
The affine face F is unique, is a maximal affine subspace of C, and has the formF =p+LC, wherep∈C is arbitrary and
LC:={x∈E:C+x=C} is a linear subspace of E.
Moreover,C=C+LC.
Proof. We first demonstrate the existence of an affine face by induction on the dimension ofC. Suppose dim(C) =nand that we have proved the existence of affine faces for convex sets with dimension less than n. Clearly, we may assume thatC is full-dimensional, that is,E has dimensionn. We may also assume that∂C6=∅; otherwiseC=E(C is both closed and open), in which case the lemma is obviously true. Letp∈∂C. By Theorem 6.8, there exists a support hyperplaneH :=H(a,α)atpsuch thatC⊆H¯−. ThenD:=C∩H is a convex set inE with dimension less thann, so by the induction hypothesis, Dcontains an affine faceF. We claim thatF is a face ofC. By Remark 5.26, this will follow if we can show thatDis a face ofC. Suppose thatx, y∈Cwith z∈(x, y)∩D. We have ha, xi ≤α,ha, yi ≤ α, but ha, zi=α; consequently, ha, xi=α=ha, yi, which proves thatD is a face ofC.
Next, we show that an affine face F ⊆Cis a maximal affine subset ofC.
LetM be a maximal affine subset of C containingF. Ifx∈M, pickz ∈F and consider the line passing throughxandz. This line is contained inM, so that z∈(x, y) for somey ∈M; sinceF is a face of C, we have x∈F. This proves thatF =M.
IfF =p+Lis an affine face ofC, Corollary 5.29 implies thatC=C+L.
Consider the setLC defined in the statement of the theorem. Obviously,L⊆ Theorem 5.30.LetCbe a nonempty, closed, convex set in a finite-dimensional
5.5 Facial Structure of Convex Sets 131 LC. We claim thatLC is a linear subspace. It will then follow thatL=LC. It is easily seen thatLC is convex, closed under addition, and that LC=−LC. Thus, ifx∈LC, thenkx∈LC for every integerk >0. The convexity of LC
implies thattx∈LC for every t >0, and consequently for everyt∈R. ut The linear subspaceLC defined above is called thelineality subspaceofC, and its dimension thelineality ofC.
Since affine faces are points if and only ifLC={0}, we have the following corollary.
Corollary 5.31.A nonempty, closed, convex set in a finite-dimensional lin- ear space contains an extreme point if and only if it is line-free, that is, it contains no whole lines.
Definition 5.32.Two linear subspaces L, M of a vector space E are called complementaryif E =L+M and L∩M ={0}, and we denote this by the notation
E=L⊕M.
IfC⊆E is a convex set such thatC=C1+C2, where C1 andC2 are convex subsets ofLandM, respectively, we write
C=C1⊕C2.
Lemma 5.33.LetCbe a nonempty, closed, convex set in a finite-dimensional linear spaceE, and letM be a linear subspace ofE complementary toLC.
The setC can decomposed as
C= ˆC⊕LC, whereCˆ is a line-free, closed convex set inM.
Proof. Define ˆC:=C∩M. Ifx∈C, we can writex=l+m, wherel∈LCand m∈M; then we havem∈C, sinceˆ m∈M, andm=x−l∈C+LC=Cby virtue of Theorem 5.30. This shows that ˆC6=∅,x=m+l∈C+Lˆ C, and hence C⊆LC+ ˆC. The reverse inclusion also holds, sinceLC+ ˆC⊆LC+C=C, so we haveC=LC+ ˆC.
Suppose that ˆC contains a line q+{td: t ∈R} =q+Rd, where q ∈Cˆ and 06=d∈M. Then
q+LC⊂q+ (Rd+LC)⊆C+LC=C,
whereq+ (Rd+LC) is an affine subset ofCstrictly containingq+LC, which by Theorem 5.30 contradicts the maximality of the affine subspace q+LC.
This proves that ˆC is line-free. ut
Lemma 5.34.Let C be a closed convex set C in a finite-dimensional linear spaceE. The relative boundaryrbd(C) :=C\ri(C)ofC is convex if and only ifC is either an affine subspace ofE or the intersection of an affine subspace with a closed half-space.
Proof. IfCis either an affine subset or the intersection of an affine subspace with a closed half-space, then it is clear that rbd(C) is convex.
To prove the converse, assume without loss of generality that C is full- dimensional. If the boundary ofC is empty, thenC=E(C is both open and closed), and we are done. Otherwise, Theorem 6.16 implies that there exists a support hyperplaneH such that ∂C ⊆H, C does not lie entirely onH, and C⊆H¯+, where ¯H+ is one of the two closed half-spaces boundingH.
We claim that C = ¯H+. Pick p ∈ C\H; then p ∈ H+ := int( ¯H+). If xis in H+ but not inC, then the line segment [x, p] must contain a point w ∈ ∂C, a contradiction because w ∈ H+; consequently, int(C) = H+ and
C= ¯H+. ut
Lemma 5.35.LetCbe a nonempty, closed, convex set in a finite-dimensional linear spaceE. If C is not an affine subspace or the intersection of an affine subspace with a closed half-space, then every point in the relative interior of C lies on a line segment whose endpoints lie on the relative boundary of C;
consequentlyC= co(rbd(C)).
Proof. Again, we may assume that C is full-dimensional. Since ∂C is not convex by Lemma 5.34, there exist two points x, y ∈ ∂C such that [x, y]
intersects int(C). It follows from Lemma 5.28 that the line passing through any point of int(C) and parallel to [x, y] must intersectC in a line segment.
u t Theorem 5.36.A nonempty, line-free, closed, convex set C in a finite- dimensional linear spaceEis the convex hull of its extreme points and extreme rays, that is, any pointx∈C has a representation
x=
k
X
i=1
λivi+
l
X
j=1
µjdj, (5.5)
where{vi}k1 and{dj}l1are extreme points and extreme directions ofC, respec- tively,{λi}k1,{µj}l1 are nonnegative, andPk
i=1λi = 1.
Proof. We use induction on the dimension of C. Suppose dim(C) = n and that we have proved the theorem for convex sets with dimension less thann.
First, suppose thatx∈rbd(C) =C\ri(C). By Theorem 6.8, there exists a support hyperplane H at x such that C ⊆ H¯+. Then D := C∩H is a convex set inE with dimension less thann, so by the induction hypothesis, xhas a representation (5.5), where{vi}k1 and{dj}l1 are extreme points and extreme directions ofD, respectively. Since Dis a face ofC(see the proof of Theorem 5.30),{vi}k1 and{dj}l1 are also extreme points and directions ofC, respectively.
If x ∈ ri(C), then Lemma 5.35 implies that x ∈ (y, z) for two points y, z∈∂C. Sinceyandzboth have representations in the form (5.5), so doesx.
u t
5.5 Facial Structure of Convex Sets 133 Theorem 5.37.LetCbe a closed convex setCin a finite-dimensional vector spaceE. The setC can be decomposed as
C=LC⊕C,ˆ
whereLC is a linear subspace andCˆ is a line-free convex set lying in a linear subspace complementary toLC. The set of extreme points of Cˆ is nonempty, and
Cˆ = rec( ˆC) + co(ext( ˆC)).
Moreover, we also have the decompositions
recC=LC⊕rec( ˆC) and C= co(ext( ˆC)) + rec(C).
Proof. The theorem follows immediately from Lemma 5.33 and Theorem 5.36.
u t We also have the following classical result.
Theorem 5.38. (Minkowski)A compact, convex set in a finite-dimensional linear space is the convex hull of its extreme points.
Finally, we include the following two interesting results relating the faces of a sum set to the faces of its summands.
Lemma 5.39.Let C = C1 +C2, where C1 and C2 are nonempty convex sets. If F is a face of C, then there exist facesFi of Ci, i= 1,2, such that F =F1+F2.
Proof. Define the sets
F1={x∈C1: ∃y∈C2, x+y∈F}, F2={x∈C2: ∃x∈C1, x+y∈F}. It is easy to verify thatF1andF2are convex sets. We claim thatF1 is a face of C1. Let x∈F1 such that x+y ∈ F for some y ∈ C2. If x∈(u1, u2) for someu1, u2∈C1, thenx+y∈(u1+y, u2+y), and sinceF is a face ofC, we have [u1+y, u2+y]⊆F. It follows from the definition ofF1thatu1, u2∈F1, proving thatF1is a face ofC1. Similarly,F2 is a face ofC2.
It follows from the definition ofF1 andF2thatF ⊆F1+F2. To prove the reverse inclusion F1+F2 ⊆F, letx1 ∈F1 and y2 ∈ F2; it suffices to show thatx1+y2∈F. If y1∈C2 andx2∈C1are such that xi+yi∈F, i= 1,2, then (x1+y1)/2 + (x2+y2)/2 = (x1+y2)/2 + (x2+y1)/2∈F; sinceF is a
face ofC, we havex1+y2∈F. ut
Lemma 5.40.LetC=C1+C2, whereC1 andC2are nonempty convex sets.
If z is an extreme point of C, thenz has a unique representation z=x+y, wherex∈C1,y∈C2. Moreover, in this representationxis an extreme point ofC1 andy is an extreme point of C2.
Proof. It follows from Lemma 5.39 thatz=x+y, wherexandy are extreme points ofC1 and C2, respectively. Ifz =x+y, where x∈C1, y ∈C2, then we havez= (x+y)/2 + (x+y)/2, and becausez is an extreme point of C, x+y=x+y =z. Comparing this equation with z=x+y givesx=xand
y=y. ut