Convex Sets and Functions


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1 B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line segment connecting x nd y belongs to C Exmple B2 The convex subsets of R, +, ) re the intervls of R Regulr polygons re convex subsets of R 2 Definition B3 Let U be subset of rel liner spce L, +, ) A convex combintion of U is n element of L of the form 1 x k x k, where x 1,,x k U, i for 1 i k, nd k = 1 If the conditions i re dropped, we hve n ffine combintion of U In other words, x is n ffine combintion of U if there exist 1,, k R such tht x = 1 x k x k, for x 1,,x k U, nd k i = 1 Definition B4 Let U be subset of rel liner spce L, +, ) A subset {x 1,,x n } is ffinely dependent if = 1 x n x n such tht t lest one of the numbers 1,, n is nonzero nd i = If no such ffine combintion exists, then x 1,,x n re ffinely independent Theorem B5 The set U = {x 1,,x n } is ffinely independent if nd only if the set V = {x 1 x n,x n 1 x n } is linerly independent Proof Suppose tht U is ffinely independent but V is linerly dependent; tht is, = b 1 x 1 x n ) + + b n 1 x n 1 x n ) such tht not ll numbers b i re This implies b 1 x b n 1 x n 1 n 1 b i )x n =, which contrdicts the ffine independence of U
2 574 B Convex Sets nd Functions Conversely, suppose tht V is linerly independent but U is not ffinely independent In this cse, = 1 x n x n such tht t lest one of the numbers 1,, n is nonzero nd i = This implies n = 1 i, so = 1 x 1 x n ) + + n 1 x n 1 x n ) Observe tht t lest one of the numbers 1,, n 1 must be distinct from becuse otherwise we would hve 1 = = n 1 = n = This contrdicts the liner independence of V, so U is ffinely independent Exmple B6 Let x 1 nd x 2 be two elements of the liner spce R 2, +, ) The line tht psses through x 1 nd x 2 consists of ll x such tht x x 1 nd x x 2 re colliner; tht is, x x 1 ) + bx x 2 ) = for some, b R such tht + b Thus, we hve x = 1 x x 2, where = + b + b + b = 1, so x is n ffine combintion of x 1 nd x 2 It is esy to see tht the segment of line contined between x 1 nd x 2 is given by convex combintion of x 1 nd x 2 ; tht is, by n ffine combintion 1 x x 2 such tht 1, 2 Theorem B7 If C is convex subset of rel liner spce L, +, ), then C contins ll convex liner combintions of C Proof The proof is by induction on k 2 nd is left to the reder Theorem B8 The intersection of ny collection of convex sets of liner spce L, +, ) is convex set Proof Let C = {C i i I} be collection of convex sets nd let C = C Suppose tht x 1,,x k C, i for 1 i k, nd k = 1 Since x 1,,x k C i, it follows tht 1 x k x k C i for every i I Thus, 1 x k x k C, which proves the convexity of C Corollry B9 The fmily of convex sets of liner spce L, +, ) is closure system on PL) Proof This sttement follows immeditely from Theorem B8 by observing tht the set L is convex Corollry B9 llows us to define the convex hull of subset U of L s the closure K conv U) of U reltive to the closure system of the convex subsets of L If U R n consists of n+1 points such tht no point is n ffine combintion of the other n points, then K conv U) is n ndimensionl simplex in L Exmple B1 A twodimensionl simplex is defined strting from three points x 1,x 2,x 3 in R 2 such tht none of these points is n ffine combintion of
3 B Convex Sets nd Functions 575 the other two no point is colliner with the others two) Thus, the twodimensionl symplex generted by x 1,x 2,x 3 is the full tringle determined by x 1,x 2,x 3 In generl, n ndimensionl simplex is the convex hull of set of n + 1 points x 1,,x n+1 in R n such tht no point is n ffine combintion of the remining n points Let S be the ndimensionl simplex generted by the points x 1,,x n+1 in R n nd let x S If x S, then x is convex combintion of x 1,,x n,x n+1 In other words, there exist 1,, n, n+1 such tht 1,, n, n+1, 1), +1 i = 1, nd x = 1 x n x n + n+1 x n+1 The numbers 1,, n, n+1 re the bricentric coordintes of x reltive to the simplex S nd re uniquely determined by x Indeed, if we hve x = 1 x n x n + n+1 x n+1 = b 1 x b n x n + b n+1 x n+1, nd i b i for some i, this implies 1 b 1 )x n b n )x n + n+1 b n+1 )x n+1 =, which contrdicts the ffine independence of x 1,,x n+1 The next sttement plys centrl role in the study of convexity We reproduce the proof given in [59] Theorem B11 Crthéodory s Theorem) If U is subset of R n, then for every x K conv U) we hve x = +1 ix i, where x i U, i for 1 i n + 1, nd +1 i = 1 Proof Consider x K conv U) We cn write x = p+1 ix i, where x i U, i for 1 i p + 1, nd p+1 i = 1 Let p be the smllest number which llows this kind of expression for x We prove the theorem by showing tht p n Suppose tht p n+1 Then, the set {x 1,,x p+1 } is ffinely dependent, so there exist b 1,, b p+1 not ll zero such tht = p+1 b ix i nd p+1 b i = Without loss of generlity, we cn ssume b p+1 > nd p+1 b p+1 i b i for ll i such tht 1 i p nd b i > Define i c i = b i ) p+1 b i b p+1 for 1 i p We hve c i = i p+1 b p+1 b i = 1 Furthermore, c i for 1 i p Indeed, if b i, then c i i ; if b i >, then c i becuse p+1 b p+1 i b i for ll i such tht 1 i p nd b i > Thus, we hve
4 576 B Convex Sets nd Functions c i x i = i ) p b i x i = b p i x i = x, which contrdicts the choice of p A finite set of points P in R 2 is convex polygon if no member p of P lies in the convex hull of P {p} Theorem B12 A finite set of points P in R 2 is convex polygon if nd only if no member p of P lies in twodimensionl simplex formed by three other members of P Proof The rgument is strightforwrd nd is left to the reder s n exercise Theorem B13 Rdon s Theorem) Let P = {x i R n 1 i n + 2} be set of n + 2 points in R n Then, there re two disjoint subsets R nd Q of P such tht K conv R) K conv Q) Proof Since n+2 points in R n re ffinely dependent, there exist 1,, n+2 not ll equl to such tht n+2 i x i = B1) nd +2 i = Without loss of generlity, we cn ssume tht the first k numbers re positive nd the lst n + 2 k re not Let = k i > nd let b j = j for 1 j k Similrly, let c l = l for k + 1 l n + 2 Equlity B1) cn now be written s k b j x j = j=1 n+2 l=k+1 c l x l Since the numbers b j nd c l re nonnegtive nd k j=1 b j = +2 l=k+1 c l = 1, it follows tht K conv {x 1,,x k }) K conv {x k+1,,x n+2 }) Theorem B14 Klein s Theorem) If P R 2 is set of five points such tht no three of them re colliner, then P contins four points tht form convex qudrilterl Proof Let P = {x i 1 i } If these five points form convex polygon, then ny four of them form convex qudrilterl If exctly one point is in the interior of convex qudrilterl formed by the remining four points, then the desired conclusion is reched Suppose tht none of the previous cses occur Then, two of the points, sy x p,x q, re locted inside the tringle formed by the remining points x i,x j,x k Note tht the line x p x q intersects two sides of the tringle x i x j x k,
5 B Convex Sets nd Functions 577 x i x q x p x k x j Fig B1 A fivepoint configurtion in R 2 sy x i x j nd x i x k see Figure B1) Then x p x q x k x j is convex qudrilterl A function f : R R is convex if its grph on n intervl is locted below the chord determined by the endpoints of the intervl More formlly, we hve the following definition Definition B15 A function f : R R is convex if ftx + 1 t)y) tfx) + 1 t)fy) for every x, y Domf) nd t [, 1] The function g : R R is concve if g is convex Theorem B16 If f : R R is convex function nd < b c, then fb) f) b fc) f) c Proof Since < b c, we cn write b = t + 1 t)c, where t = c b c, 1] The convexity of f yields the inequlity fb) c b c f) + b c fc), which is esily seen to be equivlent with the desired inequlity A similr result follows Theorem B17 If f : R R is convex function nd b < c, then fc) f) c fc) fb) c b
6 578 B Convex Sets nd Functions Proof The rgument is similr to the proof of Theorem B16 Corollry B18 Let f : R R be convex function nd let p, q, p, q be four numbers such tht p p < q q We hve the inequlity fq) fp) q p fq ) fp ) q p B2) Proof By Theorem B16 pplied to the numbers p, q, q, we hve fq) fp ) q p fq ) fp ) q p Similrly, by pplying Theorem B17 to p, p, q, we obtin fq) fp) q p fq) fp ) q p The inequlity of the corollry cn be obtined by combining the lst two inequlities From Corollry B18, it follows tht if f : R R is convex nd differentible everywhere, then its derivtive is n incresing function The converse is lso true; nmely, if f is differentible everywhere nd its derivtive is n incresing function, then f is convex Indeed, let, b, c be three numbers such tht < b < c By the men vlue theorem, there is p, b) nd q b, c) such tht f p) = fb) f) b Since f p) f q), we obtin nd f q) = fc) fb) c b fb) f) b fc) fb), c b which implies fb) c b c f) + b c fc); tht is, the convexity of f Thus, if f is twice differentible everywhere nd its second derivtive is nonnegtive everywhere, then it follows tht f is convex Clerly, under the sme conditions of differentibility s bove, if the second derivtive is nonpositive everywhere, then f is concve The functions listed in the Tble B1, defined on the set R, provide exmples of convex or concve) functions Theorem B19 Jensen s Theorem) Let f be function tht is convex on n intervl I If t 1,,t n [, 1] re n numbers such tht t i = 1, then
7 B Convex Sets nd Functions 579 n ) f t i x i t i fx i ) for every x 1,, x n I Proof The rgument is by induction on n, where n 2 The bsis step, n = 2, follows immeditely from Definition B15 Suppose tht the sttement holds for n, nd let u 1,, u n, u n+1 be n + 1 numbers such tht +1 u i = 1 We hve fu 1 x u n 1 x n 1 + u n x n + u n+1 x n+1 ) = f u 1 x u n 1 x n 1 + u n + u n+1 ) u ) nx n + u n+1 x n+1 u n + u n+1 By the inductive hypothesis, we cn write fu 1 x u n 1 x n 1 + u n x n + u n+1 x n+1 ) ) un x n + u n+1 x n+1 u 1 fx 1 ) + + u n 1 fx n 1 ) + u n + u n+1 )f u n + u n+1 Next, by the convexity of f, we hve ) un x n + u n+1 x n+1 u n f fx n ) + u n+1 fx n+1 ) u n + u n+1 u n + u n+1 u n + u n+1 Combining this inequlity with the previous inequlity gives the desired conclusion Of course, if f is concve function nd t 1,,t n [, 1] re n numbers such tht t i = 1, then n ) f t i x i t i fx i ) B3) Tble B1 Exmples of convex or concve functions Function Second Convexity Derivtive Property x r for rr 1)x r 2 concve for r < 1 r > convex for r 1 ln x 1 x 2 concve xln x 1 x convex e x e x convex
8 58 B Convex Sets nd Functions Exmple B2 We sw tht the function fx) = ln x is concve Therefore, if t 1,, t n [, 1] re n numbers such tht t i = 1, then n ) ln t i x i t i lnx i This inequlity cn be written s n ) n ln t i x i ln x ti i, or equivlently n t i x i n x ti i, for x 1,,x n, ) In the specil cse where t 1 = = t n = 1 n, we hve the inequlity tht reltes the rithmetic to the geometric verge on n positive numbers: x x n n n x i ) 1 n B4) Let w = w 1,,w n ) R n be such tht w i = 1 For r, the wweighted men of order r of sequence of n positive numbers x = x 1,, x n ) R n > is the number n ) 1 r µ r w x) = w i x r i Of course, µ r w x) is not defined for r = ; we will give s specil definition µ wx) = lim r µ r wx) We hve lim r lnµr w x) = lim r = lim r n = = ln ln w ix r i r w ix r i lnx i w ix r i w i lnx i n x wi i
9 B Convex Sets nd Functions 581 Thus, if we define µ w x) = n xwi i, the weighted men of order r becomes function continuous everywhere with respect to r For w 1 = = w n = 1 n, we hve µ 1 nx 1 x n w x) = x 2 x n + + x 1 x n 1 the hrmonic verge of x), µ wx) = x 1 x n ) 1 n the geometric verge of x), µ 1 wx) = x x n n the rithmetic verge of x) Theorem B21 If p < r, we hve µ p wx) µ r wx) Proof There re three cses depending on the position of reltive to p nd r In the first cse, suppose tht r > p > The function fx) = x r p is convex, so by Jensen s inequlity pplied to x p 1,, xp n, we hve which implies ) r p w i x p i n w i x r i, ) 1 p w i x p i w i x r i ) 1 r, which is the inequlity of the theorem If r > > p, the function fx) = x r p is gin convex becuse f x) = r r p p )x 1 r p 2 Thus, the sme rgument works s in the previous cse Finlly, suppose tht > r > p Since < r p < 1, the function fx) = x r p is concve Thus, by Jensen s inequlity, ) r p w i x p i n w i x r i Since 1 r <, we obtin gin ) 1 p w i x p i w i x r i ) 1 r
10
11 C Useful Integrls nd Formuls C1 Euler s Integrls The integrls B, b) = Γ) = x 1 1 x) b 1 dx, x 1 e x dx, re known s Euler s integrl of the first type nd Euler s integrl of the second type, respectively We ssume here tht nd b re positive numbers to ensure tht the integrls re convergent Replcing x by 1 x yields the equlity B, b) = which shows tht B is symmetric Integrting B, b) by prts, we obtin B, b) = = 1 x 1 1 x) b 1 dx 1 x) b 1 d x 1 = x 1 x) 1 b) = b 1 = b 1 1 x) 1 x) b 1 dx = Bb, ), + b 1 x 1 x) b 2 dx x 1 1 x) b 2 dx b 1 b 1 B, b 1) B, b), x 1 1 x) b 1 dx
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