Linear, affine, and convex sets and hulls In the sequel, unless otherwise specified, X will denote a real vector space.

Lines and segments. Given two points x, y ∈ X, we define ←→ xy = {x + t(y − x): t ∈ R} = {(1 − t)x + ty : t ∈ R} , [x, y] = {x + t(y − x): t ∈ [0, 1]} = {(1 − t)x + ty : t ∈ [0, 1]} .

The sets ←→xy and [x, y] are, respectively, the line passing through x and y, and the (closed) segment with endpoints x, y. Observe that they reduce to a singleton whenever x = y. We shall use also the following notation for non-closed segments:

(x, y] = [x, y] \{x}, [x, y) = [x, y] \{y}, (x, y) = [x, y] \{x, y}.

Linear, affine, and convex sets. A set A ⊂ X is called: • linear if A is a vector subpace of X (i.e., A is nonempty, and αx + βy ∈ A whenever x, y ∈ A, α, β ∈ R); • affine if the line passing through any two points of A is entirely contained in A (i.e., (1 − t)x + ty ∈ A whenever x, y ∈ A, t ∈ R); • convex if any segment with endpoints in A is contained in A (i.e., (1 − t)x + ty ∈ A whenever x, y ∈ A, t ∈ [0, 1]). Obviously, each linear set is affine, and each affine set is convex. Moreover, any translate of an affine (convex, respectively) set is affine (convex, resp.).

2 Example 0.1. A linear set in R is either the singleton {0}, or a line containing 0, 2 or the whole R . 2 2 An affine set in R is either ∅, or a singleton, or a line, or R . 2 There is a large variety of convex sets in R ... Proposition 0.2. Let A be a set in a vector space X. (a) A is linear if and only if A is affine and contains 0. (b) A is affine if and only if either A = ∅ or A is a translate of a linear set. (Moreover, the linear set is unique in this case.)

Proof. Exercise. 

By Proposition 0.2(b), the following definition is justified. The dimension and the codimension of an affine set A ⊂ X is defined as, respectively, the dimension and the codimension of the (unique) linear translate of A.

1 2

Hyperplanes. A set H ⊂ X is a hyperplane in X if it is an affine set of codimension 1. Equivalently, a hyperplane is any maximal proper affine subset of X. Proposition 0.3. A set H ⊂ X is a hyperplane if and only if it is of the form −1 H = ϕ (α) where ϕ: X → R is a nonzero linear functional and α ∈ R.

Proof. “⇒” Fix any x0 ∈ H. By Proposition 0.2, the set Y = H − x0 is a of codimension 1 in X. Fix any v0 ∈ X \ Y . Then each x ∈ X has a unique representation of the form x = yx + txv0 where yx ∈ Y , tx ∈ R. The mapping ϕ(x) := tx is linear and satisfies

x ∈ H ⇔ x − x0 ∈ Y ⇔ ϕ(x − x0) = 0 ⇔ ϕ(x) = ϕ(x0).

Thus we can put α = ϕ(x0). “⇐” It is easy to see that ϕ−1(α) is a translate of the kernel ϕ−1(0). We have to −1 −1 show that ϕ (0) has codimension 1. Fix some v0 ∈ X \ ϕ (0) (this is possible since ϕ 6≡ 0). By substituting v0 by its appropriate multiple, we can suppose that ϕ(v0) = 1. Then every x ∈ X can be written in the form −1 x = [x − ϕ(x)v0] + ϕ(x)v0 ∈ ϕ (0) + Rv0 since ϕ(x − ϕ(x)v0) = 0.  Corollary 0.4. Let H be a hyperplane in X. Then X can be written as a disjoint union X = H ∪ H+ ∪ H− in such a way that if x ∈ H+ and y ∈ H− then [x, y] ∩ H is a singleton. (The sets H+,H− are the algebraically open halfspaces generated by H.) Proof. Exercise. Hint: take ϕ, α as in Proposition 0.3 and consider H+ = {x ∈ X : − ϕ(x) > α}, H = {x ∈ X : ϕ(x) < α}. 

Convex and affine combinations. Definition 0.5. Let A ⊂ X. An affine combination of elements of A is any finite sum of the form n X Pn (1) λixi where xi ∈ A, λi ∈ R, 1 λj = 1. i=1

A of elements of A is any finite sum of the form (1) with λi ≥ 0 (i = 1, . . . , n). Proposition 0.6. Every convex/affine/linear set in a vector space X is closed under making convex/affine/linear combinations of its elements. Proof. The “linear part” is well known from Linear Algebra. Pn Let C be a and x = i=1 λixi be a convex combination of elements of C. We want to prove that x ∈ C. Let us proceed by induction with respect to n. For n = 1, we have x = x1 ∈ C. Now, suppose that the case n = k holds, Pk+1 and consider the case n = k + 1, that is, x = i=1 λixi with xi ∈ C, λi ≥ 0, 3

Pk+1 1 λj = 1. If λk+1 = 1 then necessarily x = xk+1 ∈ C. Suppose λk+1 6= 1. Then Pk s := 1 λj = 1 − λk+1 6= 0. We can write " k # X λi x = (1 − λ ) x + λ x . k+1 s i k+1 k+1 i=1 Since the sum in the quare brackets belongs to C by our induction assumption, x belongs to C. The “affine part” can be proved in the same way. The only difference is that we start with indexing the points xi in such a way that λk+1 6= 1 (if this is not possible, we are in a trivial case).  It is easy to see that the set of all convex combinations of elements of {x, y} is the segment [x, y], and the set of all affine combinations of elements of {x, y} is the line ←→xy. Observation 0.7. A convex/affine combination of convex/affine combinations of elements of A is a convex/affine combination of elements of A. Fact 0.8. Let X be a normed linear space, x, y, z ∈ X, z ∈ (x, y). Then kz−yk kx−zk kx − yk = kx − zk + kz − yk and z = kx−yk x + kx−yk y . Proof. We have z = (1 − t)x + ty for some t ∈ (0, 1). Then z − x = t(y − x) and kz−xk kz−yk z − y = (1 − t)(x − y). Passing to norms we get t = ky−xk and 1 − t = kx−yk . Now the two formulas easily follow. 

Hulls. It is an easy but important observation that the intersection of any family of linear/affine/convex sets is again a linear/affine/convex set. (The same does not hold for unions, but does holds for linearly ordered unions.) Given a set A ⊂ X, the intersection of all linear sets containing A is the linear hull of A, denoted by span(A). Analogously, we can define the affine hull of A as the intersection of all affine sets containing A, and the of A as the intersection of all convex sets containing A. The affine and the convex hull of A will be denoted by aff(A) and conv(A) , respectively. Obviously, A is linear if and only if span(A) = A; A is affine if and only if aff(A) = A; A is convex if and only if conv(A) = A. It is a well known fact that the linear hull of a set A coincides with the set of all linear combinations of elements of A. The following theorem states that analogous properties hold for convex hulls and for affine hulls as well. Theorem 0.9. Let A be a set in a vector space X. Then conv(A) = {x ∈ X : x is a convex combination of elements of A} , aff(A) = {x ∈ X : x is an affine combination of elements of A} . 4

Proof. Let us prove the first formula (the second one is analogous). By Obser- vation 0.7, the set C of all convex combinations of points of A is convex; thus conv(A) ⊂ C. On the other hand, any point x ∈ C, being a convex combination of points of conv(A), belongs to conv(A) by Proposition 0.6.  Let A be an affine set in a vector space X of a finite dimension d and let x ∈ aff(A). By translation, we can suppose that 0 ∈ A. In this case, x belongs to the linear hull of A, and hence it is a of at most d elements of A:

d X x = λixi where λi ∈ R, xi ∈ A. i=1 Since 0 ∈ A, we can write x as an affine combination of d + 1 points of A:

d X Pn x = λ0 · 0 + λixi with λ0 = 1 − 1 λj. i=1 Thus we have proved that, in an d-dimensional vector space, every point of the affine hull of a set is an affine combination of d + 1 or fewer points of A. The following important theorem shows that a similar result holds for convex hulls as well. Theorem 0.10 (Carath´eodory). Let A be a subset of a d-dimensional vector space X. Then conv(A) = {x ∈ X : x is a convex combination of d + 1 or fewer points of A} . Proof. By Theorem 0.9, it suffices to show that every point of the form n n X X x = λixi where λi ∈ R, xi ∈ A, λj = 1 , i=0 0 (a convex combination of n + 1 points of A) is a convex combination of d + 1 or fewer points of A. If n ≤ d, there is nothing to prove. Let n > d. By translation, we can (and do) suppose that x0 = 0. Since the set {x1, . . . , xn} is linearly dependent, there exist real numbers α1, . . . , αn, not all of them null, such that n X (2) αixi = 0 . i=1

Since (2) remains true if we change the sign of all αi’s, we can suppose that Pn 1 αi ≥ 0. Observe that the set I = {i ∈ {1, . . . , n} : αi > 0} is nonempty. Pn (Indeed, otherwise we would have 1 αi < 0 since not all ai’s are null.) For each t > 0, we have n n n X X X x = λixi − t αixi = (λi − tαi)xi . i=1 i=1 i=1 Observe that all coefficients in the last sum will be nonnegative provided t ≤ λi for αi each i ∈ I. Choose k ∈ I so that λk = min{ λi : i ∈ I}. Then, for t = λk , we have αk αi αk 5

Pn Pn Pn λi −tαi ≥ 0 for each 1 ≤ i ≤ n, λk −tαk = 0, and 1 (λj −tαj) = 1 λj −t 1 αj ≤ Pn 1 λj ≤ 1. Since n  Pn  X x = 1 − 1 (λj − tαj) · 0 + (λi − tαi)xi , i=1 i6=k we have written x as a convex combination of less than n + 1 points of A. So, we have proved that, in any convex combination x of more than d + 1 points, an appropriate change of coefficients allows us to throw out one of the points without changing x. Now, the proof follows by repeating this procedure untill we arrive to at most d + 1 points.  Theorem 0.11. Let X be a normed space of a finite dimension, K ⊂ X a compact set. Then conv(K) is compact.

d d+1 Pd Proof. Let d = dim(X). Denote Λ = {λ = (λi)0 ∈ [0, 1] : 0 λi = 1}, and define d+1 Pd F :Λ × K → X by F (λ, x0, . . . , xd) = i=0 λixi. By the Carath´eodory theorem, we have d+1 conv(K) = {F (λ, x0, . . . , xd): λ ∈ Λ, xi ∈ K} = F (Λ × K ). The last set is compact since F is continuous and Λ × Kd+1 is a compact metric space.  Corollary 0.12. In any normed space, the convex hull of a finite set is compact. (Indeed, we can restrict ourselves to a finite-dimensional subspace, and apply the above theorem.) The next example shows that the assumption on the dimension of the space in Theorem 0.11 cannot be omitted.

en Example 0.13. Consider the Hilbert space `2 and a set K = { n : n ∈ N} ∪ {0}, where en is the n-th vector of the standard orthonormal basis of `2. The set K is en compact since n → 0. We claim that conv(K) is not compact since it is not closed. Pn −j −1 Pn −i First, the points xn := ( 1 2 ) i=1 2 (ei/i)(n ∈ N) belong to K. Second, −n the sequence {xn} converges in `2 to the point x = (xn) with xn = 2 (1/n) for each n (Exercise: prove this!). Third, observe that every element of conv(K) has a finite support; thus x∈ / conv(K). However, we shall see in a moment that, if the normed space is complete, the closedness is the unique thing which can prevent the convex hull of a compact set from being compact. Definition 0.14. Let X be a normed space, A ⊂ X. The closed convex hull of A is the intersection of all closed convex sets containing A, and it is denoted by conv(A). Observation 0.15. Let X be a normed space, A ⊂ X. Then conv(A) = conv(A) . 6

Recall that a metric space (M, d) is totally bounded (or precompact) if, for each ε > 0, it contains a finite ε-net, that is, a finite set Fε such that d(x, Fε) < ε for each x ∈ M. It is a well known fact that M is compact if and only if M is complete and totally bounded.

Exercise 0.16. Let (M, d) be a metric space, A ⊂ M. (a) A is totally bounded if and only if A is totally bounded. (b) A is totally bounded if and only if, for each ε > 0, there exists a compact set K ⊂ M such that d(x, K) < ε for each x ∈ A.

Theorem 0.17. Let X be a normed linear space, A ⊂ X a totally bounded set. (a) conv(A) is totally bounded. (b) If X is a , then conv(A) is compact.

Proof. (a) Fix ε > 0. There exists a finite set A0 ⊂ A such that, for each a ∈ A, there exists ya ∈ A0 with ka − yak < ε. The set conv(A0) is compact by Corollary 0.12. Pn Pn Now, if x ∈ conv(A), we can write x = i=1 λiai where ai ∈ A, λi ≥ 0, 1 λj = 1. Pn The point c = i=1 λiyai belongs to conv(A0) and it satisfies Pn Pn kx − ck ≤ i=1 λikai − yai k < ε i=1 λi = ε . By Exercise 0.16(b), conv(A) is totally bounded. To show (b) it suffices to observe that conv(A) is complete (since it is closed and X is complete) and totally bounded (by (a) above and Exercise 0.16(a)). 

Now, let us consider convex hulls of finitely many convex sets. We can see the part (b) of Theorem 0.18 as a generalization of the fact that a convex hull of a finite set is compact.

Theorem 0.18. Let X be a normed space. Let C1,...,Cn be convex subsets of X. Pn Pn (a) conv(C1 ∪ ... ∪ Cn) = { i=1 λixi : xi ∈ Ci, λi ≥ 0, 1 λj = 1}. (b) If each Ci is compact, then conv(C1 ∪ ... ∪ Cn) is compact. (c) If C1 is closed and bounded, and the sets C2,...,Cn are compact, then conv(C1 ∪ ... ∪ Cn) is closed. Proof. (a) The inclusion “⊃” is obvious. To see the reverse one, consider x ∈ conv(C1∪...∪Cn) and write it as a convex combination of elements yk (k = 1,...,K) of C1 ∪ ... ∪ Cn. Since each of yk’s belongs to some Ci, we can group them with respect to which set they belong. Thus x can be written in the form

n   X X x =  λkyk , i=1 k∈Ji

Sn PK where Ji’s are pairwise disjoint, 1 Ji = {1,...,K}, 1 λk = 1, and xk ∈ Ci whenever k ∈ Ji. 7

For every fixed i ∈ {1, . . . , n}, denote µ = P λ . If µ = 0, fix an arbitrary i k∈Ji k i x ∈ C . If µ > 0, denote x = P λk y and observe that x ∈ C . Now, we have i i i i k∈Ji µi k i i n n X X x = µixi and µi = 1 . i=1 i=1 (b) follows in the same way as in the proof of Theorem 0.11. Indeed, using (a), we can write conv(C1 ∪ ... ∪ Cn) = F (Λ × C1 × ... × Cn) n Pn Pn where Λ = {λ ∈ [0, 1] : 1 λi = 1} and F (λ, x1, . . . , xn) = i=1 λixi. Thus conv(C1 ∪ ... ∪ Cn) is compact since it is a continuous image of a compact metric space.

(c) Let {xm} ⊂ conv(C1 ∪ ... ∪ Cn) be a sequence converging to some x ∈ X. By (a), each xm can be written as a convex combination n X (m) (m) (m) xm = λi ci where ci ∈ Ci . i=1

Using the fact that [0, 1],C2,...,Cn are compact, we can pass to subsequences to (m) (m) assure that λi → λi ∈ [0, 1] (1 ≤ i ≤ n) and ci → ci ∈ Ci (2 ≤ i ≤ n) as Pn m → +∞. Observe that 1 λi = 1. Let us consider two cases. (m) (m) If λ1 = 0, we have λ1 c1 → 0 (since C1 is bounded) and hence n n X (m) (m) X x = lim xm = lim λ c = λici ∈ conv(C2 ∪...∪Cn) ⊂ conv(C1 ∪...∪Cn) . m m i i i=2 i=2 If λ1 > 0, we can write n ! (m) 1 X (m) (m) c = x − λ c . 1 (m) m i i λ1 i=2 Thus c(m) → 1 (x − Pn λ c ) =: c ∈ C (since C is closed). Consequently, 1 λi i=2 i i 1 1 1 n X x = lim xm = λici ∈ conv(C1 ∪ ... ∪ Cn) . m i=1  Corollary 0.19. In any normed linear space, conv(C ∪ F ) is closed whenever C is closed, bounded, and convex, and F is finite. Example 0.20. The assumption that C is bounded in Corollary 0.19 cannot be omitted. To see this, consider a line C in the plane, and a point x0 ∈/ C. Then conv(C ∪ {x0}) is not closed. (Why?)