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

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1 Lnear, affne, and convex sets and hulls In the sequel, unless otherwse specfed, X wll denote a real vector space. Lnes and segments. Gven two ponts x, y X, we defne 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, respectvely, the lne passng through x and y, and the (closed) segment wth endponts x, y. Observe that they reduce to a sngleton whenever x = y. We shall use also the followng notaton for non-closed segments: (x, y] = [x, y] \ {x}, [x, y) = [x, y] \ {y}, (x, y) = [x, y] \ {x, y}. Lnear, affne, and convex sets. A set A X s called: lnear f A s a vector subpace of X (.e., A s nonempty, and αx + βy A whenever x, y A, α, β R); affne f the lne passng through any two ponts of A s entrely contaned n A (.e., (1 t)x + ty A whenever x, y A, t R); convex f any segment wth endponts n A s contaned n A (.e., (1 t)x + ty A whenever x, y A, t [0, 1]). Obvously, each lnear set s affne, and each affne set s convex. Moreover, any translate of an affne (convex, respectvely) set s affne (convex, resp.). Example 0.1. A lnear set n R 2 s ether the sngleton {0}, or a lne contanng 0, or the whole R 2. An affne set n R 2 s ether, or a sngleton, or a lne, or R 2. There s a large varety of convex sets n R 2... Proposton 0.2. Let A be a set n a vector space X. (a) A s lnear f and only f A s affne and contans 0. (b) A s affne f and only f ether A = or A s a translate of a lnear set. (Moreover, the lnear set s unque n ths case.) Proof. Exercse. By Proposton 0.2(b), the followng defnton s justfed. The dmenson and the codmenson of an affne set A X s defned as, respectvely, the dmenson and the codmenson of the (unque) lnear translate of A. 1

2 2 Hyperplanes. A set H X s a hyperplane n X f t s an affne set of codmenson 1. Equvalently, a hyperplane s any maxmal proper affne subset of X. Proposton 0.3. A set H X s a hyperplane f and only f t s of the form H = ϕ 1 (α) where ϕ: X R s a nonzero lnear functonal and α R. Proof. Fx any x 0 H. By Proposton 0.2, the set Y = H x 0 s a lnear subspace of codmenson 1 n X. Fx any v 0 X \ Y. Then each x X has a unque representaton of the form x = y x + t x v 0 where y x Y, t x R. The mappng ϕ(x) := t x s lnear and satsfes x H x x 0 Y ϕ(x x 0 ) = 0 ϕ(x) = ϕ(x 0 ). Thus we can put α = ϕ(x 0 ). It s easy to see that ϕ 1 (α) s a translate of the kernel ϕ 1 (0). We have to show that ϕ 1 (0) has codmenson 1. Fx some v 0 X \ ϕ 1 (0) (ths s possble snce ϕ 0). By substtutng v 0 by ts approprate multple, we can suppose that ϕ(v 0 ) = 1. Then every x X can be wrtten n the form snce ϕ(x ϕ(x)v 0 ) = 0. x = [x ϕ(x)v 0 ] + ϕ(x)v 0 ϕ 1 (0) + Rv 0 Corollary 0.4. Let H be a hyperplane n X. Then X can be wrtten as a dsjont unon X = H H + H n such a way that f x H + and y H then [x, y] H s a sngleton. (The sets H +, H are the algebracally open halfspaces generated by H.) Proof. Exercse. Hnt: take ϕ, α as n Proposton 0.3 and consder H + = {x X : ϕ(x) > α}, H = {x X : ϕ(x) < α}. Convex and affne combnatons. Defnton 0.5. Let A X. An affne combnaton of elements of A s any fnte sum of the form (1) λ x where x A, λ R, n 1 λ j = 1. A convex combnaton of elements of A s any fnte sum of the form (1) wth λ 0 ( = 1,..., n). Proposton 0.6. Every convex/affne/lnear set n a vector space X s closed under makng convex/affne/lnear combnatons of ts elements. Proof. The lnear part s well known from Lnear Algebra. Let C be a convex set and x = n λ x be a convex combnaton of elements of C. We want to prove that x C. Let us proceed by nducton wth respect to n. For n = 1, we have x = x 1 C. Now, suppose that the case n = k holds, and consder the case n = k + 1, that s, x = k+1 λ x wth x C, λ 0,

3 k+1 1 λ j = 1. If λ k+1 = 1 then necessarly x = x k+1 C. Suppose λ k+1 1. Then s := k 1 λ j = 1 λ k+1 0. We can wrte [ k ] λ x = (1 λ k+1 ) s x + λ k+1 x k+1. Snce the sum n the quare brackets belongs to C by our nducton assumpton, x belongs to C. The affne part can be proved n the same way. The only dfference s that we start wth ndexng the ponts x n such a way that λ k+1 1 (f ths s not possble, we are n a trval case). It s easy to see that the set of all convex combnatons of elements of {x, y} s the segment [x, y], and the set of all affne combnatons of elements of {x, y} s the lne xy. Observaton 0.7. A convex/affne combnaton of convex/affne combnatons of elements of A s a convex/affne combnaton of elements of A. Fact 0.8. Let X be a normed lnear space, x, y, z X, z (x, y). Then x y = x z + z y and z = z y x y x + x z x y y. Proof. We have z = (1 t)x + ty for some t (0, 1). Then z x = t(y x) and z y = (1 t)(x y). Passng to norms we get t = z x z y y x and 1 t = x y. Now the two formulas easly follow. 3 Hulls. It s an easy but mportant observaton that the ntersecton of any famly of lnear/affne/convex sets s agan a lnear/affne/convex set. (The same does not hold for unons, but does holds for lnearly ordered unons.) Gven a set A X, the ntersecton of all lnear sets contanng A s the lnear hull of A, denoted by span(a). Analogously, we can defne the affne hull of A as the ntersecton of all affne sets contanng A, and the convex hull of A as the ntersecton of all convex sets contanng A. The affne and the convex hull of A wll be denoted by aff(a) and conv(a), respectvely. Obvously, A s lnear f and only f span(a) = A; A s affne f and only f aff(a) = A; A s convex f and only f conv(a) = A. It s a well known fact that the lnear hull of a set A concdes wth the set of all lnear combnatons of elements of A. The followng theorem states that analogous propertes hold for convex hulls and for affne hulls as well. Theorem 0.9. Let A be a set n a vector space X. Then conv(a) = {x X : x s a convex combnaton of elements of A}, aff(a) = {x X : x s an affne combnaton of elements of A}.

4 4 Proof. Let us prove the frst formula (the second one s analogous). By Observaton 0.7, the set C of all convex combnatons of ponts of A s convex; thus conv(a) C. On the other hand, any pont x C, beng a convex combnaton of ponts of conv(a), belongs to conv(a) by Proposton 0.6. Let A be an affne set n a vector space X of a fnte dmenson d and let x aff(a). By translaton, we can suppose that 0 A. In ths case, x belongs to the lnear hull of A, and hence t s a lnear combnaton of at most d elements of A: x = d λ x where λ R, x A. Snce 0 A, we can wrte x as an affne combnaton of d + 1 ponts of A: x = λ d λ x wth λ 0 = 1 n 1 λ j. Thus we have proved that, n an d-dmensonal vector space, every pont of the affne hull of a set s an affne combnaton of d + 1 or fewer ponts of A. The followng mportant theorem shows that a smlar result holds for convex hulls as well. Theorem 0.10 (Carathéodory). Let A be a subset of a d-dmensonal vector space X. Then conv(a) = {x X : x s a convex combnaton of d + 1 or fewer ponts of A}. Proof. By Theorem 0.9, t suffces to show that every pont of the form x = λ x where λ R, x A, λ j = 1, =0 (a convex combnaton of n + 1 ponts of A) s a convex combnaton of d + 1 or fewer ponts of A. If n d, there s nothng to prove. Let n > d. By translaton, we can (and do) suppose that x 0 = 0. Snce the set {x 1,..., x n } s lnearly dependent, there exst real numbers α 1,..., α n, not all of them null, such that (2) α x = 0. Snce (2) remans true f we change the sgn of all α s, we can suppose that n 1 α 0. Observe that the set I = { {1,..., n} : α > 0} s nonempty. (Indeed, otherwse we would have n 1 α < 0 snce not all a s are null.) For each t > 0, we have x = λ x t α x = (λ tα )x. Observe that all coeffcents n the last sum wll be nonnegatve provded t λ α each I. Choose k I so that λ k α k = mn{ λ α 0 for : I}. Then, for t = λ k α k, we have

5 λ tα 0 for each 1 n, λ k tα k = 0, and n 1 (λ j tα j ) = n 1 λ j t n 1 α j n 1 λ j 1. Snce x = [ 1 n 1 (λ j tα j ) ] 0 + (λ tα )x, we have wrtten x as a convex combnaton of less than n + 1 ponts of A. So, we have proved that, n any convex combnaton x of more than d + 1 ponts, an approprate change of coeffcents allows us to throw out one of the ponts wthout changng x. Now, the proof follows by repeatng ths procedure untll we arrve to at most d + 1 ponts. Theorem Let X be a normed space of a fnte dmenson, K X a compact set. Then conv(k) s compact. Proof. Let d = dm(x). Denote Λ = {λ = (λ ) d 0 [0, 1]d+1 : d 0 λ = 1}, and defne F : Λ K d+1 X by F (λ, x 0,..., x d ) = d =0 λ x. By the Carathéodory theorem, we have k conv(k) = {F (λ, x 0,..., x d ) : λ Λ, x K} = F (Λ K d+1 ). The last set s compact snce F s contnuous and Λ K d+1 s a compact metrc space. Corollary In any normed space, the convex hull of a fnte set s compact. (Indeed, we can restrct ourselves to a fnte-dmensonal subspace, and apply the above theorem.) The next example shows that the assumpton on the dmenson of the space n Theorem 0.11 cannot be omtted. Example Consder the Hlbert space l 2 and a set K = { en n : n N} {0}, where e n s the n-th vector of the standard orthonormal bass of l 2. The set K s compact snce en n 0. We clam that conv(k) s not compact snce t s not closed. Frst, the ponts x n := ( n 1 2 j ) 1 n 2 (e /) (n N) belong to K. Second, the sequence {x n } converges n l 2 to the pont x = (x n ) wth x n = 2 n (1/n) for each n (Exercse: prove ths!). Thrd, observe that every element of conv(k) has a fnte support; thus x / conv(k). However, we shall see n a moment that, f the normed space s complete, the closedness s the unque thng whch can prevent the convex hull of a compact set from beng compact. Defnton Let X be a normed space, A X. The closed convex hull of A s the ntersecton of all closed convex sets contanng A, and t s denoted by conv(a). Observaton Let X be a normed space, A X. Then conv(a) = conv(a). 5

6 6 Recall that a metrc space (M, d) s totally bounded (or precompact) f, for each ε > 0, t contans a fnte ε-net, that s, a fnte set F ε such that d(x, F ε ) < ε for each x M. It s a well known fact that M s compact f and only f M s complete and totally bounded. Exercse Let (M, d) be a metrc space, A M. (a) A s totally bounded f and only f A s totally bounded. (b) A s totally bounded f and only f, for each ε > 0, there exsts a compact set K M such that d(x, K) < ε for each x A. Theorem Let X be a normed lnear space, A X a totally bounded set. (a) conv(a) s totally bounded. (b) If X s a Banach space, then conv(a) s compact. Proof. (a) Fx ε > 0. There exsts a fnte set A 0 A such that, for each a A, there exsts y a A 0 wth a y a < ε. The set conv(a 0 ) s compact by Corollary Now, f x conv(a), we can wrte x = n λ a where a A, λ 0, n 1 λ j = 1. The pont c = n λ y a belongs to conv(a 0 ) and t satsfes x c n λ a y a < ε n λ = ε. By Exercse 0.16(b), conv(a) s totally bounded. To show (b) t suffces to observe that conv(a) s complete (snce t s closed and X s complete) and totally bounded (by (a) above and Exercse 0.16(a)). Now, let us consder convex hulls of fntely many convex sets. We can see the part (b) of Theorem 0.18 as a generalzaton of the fact that a convex hull of a fnte set s compact. Theorem Let X be a normed space. Let C 1,..., C n be convex subsets of X. (a) conv(c 1... C n ) = { n λ x : x C, λ 0, n 1 λ j = 1}. (b) If each C s compact, then conv(c 1... C n ) s compact. (c) If C 1 s closed and bounded, and the sets C 2,..., C n are compact, then conv(c 1... C n ) s closed. Proof. (a) The ncluson s obvous. To see the reverse one, consder x conv(c 1... C n ) and wrte t as a convex combnaton of elements y k (k = 1,..., K) of C 1... C n. Snce each of y k s belongs to some C, we can group them wth respect to whch set they belong. Thus x can be wrtten n the form x = λ k y k, k J where J s are parwse dsjont, n 1 J = {1,..., K}, K 1 λ k = 1, and x k C whenever k J.

7 For every fxed {1,..., n}, denote µ = k J λ k. If µ = 0, fx an arbtrary x C. If µ > 0, denote x = λ k k J µ y k and observe that x C. Now, we have x = µ x and µ = 1. (b) follows n the same way as n the proof of Theorem Indeed, usng (a), we can wrte conv(c 1... C n ) = F (Λ C 1... C n ) where Λ = {λ [0, 1] n : n 1 λ = 1} and F (λ, x 1,..., x n ) = n λ x. Thus conv(c 1... C n ) s compact snce t s a contnuous mage of a compact metrc space. (c) Let {x m } conv(c 1... C n ) be a sequence convergng to some x X. By (a), each x m can be wrtten as a convex combnaton x m = λ (m) where C. Usng the fact that [0, 1], C 2,..., C n are compact, we can pass to subsequences to assure that λ (m) λ [0, 1] (1 n) and c C (2 n) as m +. Observe that n 1 λ = 1. Let us consder two cases. If λ 1 = 0, we have λ (m) (snce C 1 s bounded) and hence x = lm x m = lm λ (m) m m = λ c conv(c 2... C n ) conv(c 1... C n ). =2 If λ 1 > 0, we can wrte =2 ( 1 = 1 λ (m) x m 1 =2 λ (m) Thus 1 1 λ (x n =2 λ c ) =: c 1 C 1 (snce C 1 s closed). Consequently, x = lm x m = λ c conv(c 1... C n ). m Corollary In any normed lnear space, conv(c F ) s closed whenever C s closed, bounded, and convex, and F s fnte. Example The assumpton that C s bounded n Corollary 0.19 cannot be omtted. To see ths, consder a lne C n the plane, and a pont x 0 / C. Then conv(c {x 0 }) s not closed. (Why?) ). 7