CHAPTER 10 Majorizatio ad Matrix Iequalities 10.1/1 Fid two vectors x, y R 3 such that either x y or x y holds. Solutio: Take x = (1, 1 ad y = (2, 3. 10.1/2 Let y = (2, 1 R 2. Sketch the followig sets i the plae R 2 : {x R 2 : x y} ad {x R 2 : x w y}. Solutio: To be filled i; use xfig. 10.1/3 Let x, y R. Show that ( x = (x ad x y x y. Solutio: Let x = (x 1, x 2,..., x with x = (x 1, x 2,...,. The x ( x = ( x, x 1,..., x 1 =, x (x 1,..., x 1 =. x For the majorizatio iequality, we use Theorem 10.4(3: x y = x + ( y x + ( y = x y. 10.1/4 Let x = (x 1,..., x R. Show that x i = x i+1, i = 1, 2,...,. Solutio: x = (x 1, x 2,..., ad x x = (x, x 1,..., x 1. O the other had, x = (x 1, x 2,..., x. Apparetly, x i = x i+1. 10.1/5 Let x = (x 1, x 2,..., x ad y = (y 1, y 2,..., y be i R. Show that x w y (x 1, x 2,..., x k w (y 1, y 2,..., y k, k = 1, 2,...,. Solutio: : If x w y, the p x i p y i for every p k. Thus (x 1, x 2,..., x k w (y 1, y 2,..., y k, k = 1, 2,...,. : Set k =. The (x 1, x 2,..., x w (y 1, y 2,..., y. So x w y. 1
2 10.1/6 Let x = (x 1, x 2,..., x R. If x 1 x 2 x, show that 1 m m x i 1 x i for ay positive iteger m. Solutio: As k k m, k m, ( 1,..., 1 ( 1 m,..., 1 m, 0,..., 0. Theorem 10.5 yields 1 x i m 1 m x i. The claim is approved. 10.1/7 Let a 1, a 2,..., a be oegative umbers such that a i = 1. Show that ( 1, 1,..., 1 (a 1, a 2,..., a (1, 0,..., 0 ad that v (v 1, 0 (v 2, 0,..., 0 (1, 0,..., 0, where v k = ( 1 k, 1 k,..., 1 k with k copies of 1 k for k = 1, 2,...,. 1 m Solutio: By the previous theorem, m a i 1 a i for ay positive iteger m. Deote the average of a 1,..., a by v. The mv m a i. It follows that (v,..., v (a 1, a 2,..., a. Thus, whe a s add up to 1, v = 1 ad ( 1, 1,..., 1 (a 1, a 2,..., a. [Note: This ca also be prove by observig that ( 1, 1,..., 1 = (a 1, a 2,..., a S, where S is the doubly stochastic matrix all whose etries are 1. See Sectio 10.2.] That (a 1, a 2,..., a (1, 0,..., 0 is obvious as a s are oegative with summatio 1. To show (v p, 0 (v q, 0 for p q, it suffices to otice that k p k q. 10.1/8 Referrig to Theorem 10.6 (2, ca u v be replaced by u v? Solutio: No: (2, 1 (3, 1, (1, 0 (0, 1, but (2, 0 w (0, 1. 10.1/9 Let x, y R. Show that x y x y. If x w y, does it ecessarily follow that x w y (or y w x? Solutio: : Let x y. The k x i k y i for every k. Sice x i = y i, by subtractig, we have i= k+1 x i i= k+1 y i. Thus i= k+1 ( x i i= k+1 ( y i, that is, k ( x i k ( y i. As x i = y i, x y. : If x y, the above argumet shows x = ( x ( y = y. For weak majorizatio, this is ot true i geeral. Take x = (3, 0, 1, y = (4, 1, 2. The x w y, but either x w y or y w x. 10.1/10 Let x, y, z R. If x + y w z, show that x w z y. Solutio: If x + y w z, by Theorem 10.3(4, we have x = (x + y + ( y w z + ( y. By Problem 3, we kow that ( y = (y. Thus x w z y. Note: The last z i the origial problem should be z. Otherwise, it is ot true i geeral. Take x = (1, 2, y = (1, 2, ad z = (0, 2.
3 10.1/11 Let x, y R ad x y (compoetwise. Show that xp yp for ay permutatio matrix P ad cosequetly x y. Solutio: Sice y x, y x 0. Now P 0, we have (y xp 0, It follows that yp xp. Now let P be such a permutatio matrix that xp = (x p(1, x p(2,..., x p( = x. The (x p(1, x p(2,..., x p( (y p(1, y p(2,..., y p(. The iequality follows by observig that x 1 = x p(1 y p(1 y 1, x 2 mi{y p(1, y p(2 } y 2, ad for each k, x k = x p(k mi{y p(1,..., y p(k } y k. 10.1/12 Let e = (1, 1,..., 1 R. Fid all x R such that x e. Solutio: x = e. 10.1/13 Let x = (x 1, x 2,..., x R ad x = 1 (x 1 + x 2 + + x. Show that xe x, where e = (1, 1,..., 1 R. State the case of = 2. Solutio: If x = 0, i.e., x 1 + x 2 + + x = 0, the x 1 + + x k 0 for every k. It follows that xe = 0 x. Let x 0 ad deote x 0 = x 1 + +x. The xe = x 0 ( 1,..., 1. By Problem 7 (ad Problem 9 if x 0 < 0, ( 1,..., 1 1 x 0 x, which yields x 0 ( 1,..., 1 x. For = 2, ( x1+x2 2, x1+x2 2 (x 1, x 2 for all real umbers x 1 ad x 2. Note: The easiest way to show this is to use doubly stochastic matrix (Theorem 10.8 i Sectio 10.2: xe = xj, where J is the matrix all of whose etries are 1. 10.1/14 Let x, y R +. Show that (x, y w (x + y, 0. Solutio: Let u = (x, 0, v = (y, 0. The u + v = (x + y, 0, while u = (x, 0 ad v = (0, y. By Theorem 10.4(3, we have (x, y = u +v u+v = (x+y, 0. Sice (x, y ad (x, y cotai the same compoets except their order, the coclusio follows. Note: The compoets of x ad y eed be oegative, but they eed ot be decreasigly ordered i (x + y, 0, i.e., ot (x + y, 0. Of course, (x + y, 0 (x + y, 0. 10.1/15 Let x, y R. If x y ad if 0 α β 1, show that βx + (1 βy αx + (1 αy.
4 Solutio: Let β = α + γ, γ 0. Note that The compoets of βx + (1 βy are already i oicreasig order. For each k, (βx i + (1 βy i = Thus (αx i + (1 αy i γ( (y i x i. (βx i + (1 βy i (αx i + (1 αy i. Equality holds whe k = because x y. The proof is complete. 10.1/16 Let x, y R. Show that x y (x, z (y, z for all z R m. Solutio: Sufficiecy: Take z = 0. (x, 0 (y, 0 implies x y. Necessity: Let u = (x, z, v = (y, z. For k + m, let x 1,..., x p, z 1,..., z q, p + q = k, are the first k largest values of u. Thus u i = p q x i + z i p q y i + z i Equality holds whe k = m + because x y. v i. Note: If we are allowed to use Theorem 10.8, the x = yp for some doubly stochastic matrix P ad (x, z = (y, z (y, z. 10.1/17 Let x, y R ad z R m. Show that ( P 0 (x, z w (y, z x w y. Cosider the more geeral case. If (x, u (y, v for some u, v R m satisfyig u v, does it ecessarily follow that x y or x w y? Solutio: Let u = (x, z, v = (y, z. For each p, let y 1,..., y p, z 1,..., z q, are the first p + q largest values of v. The p+q v i = This implies that p q p+q y i + z i u i p y i 0 I p q x i + z i. p x i, i.e., x w y. Aswer to the secod part is o. Take x = (2, 0, u = (1, 1, y = (1, 1, (3, 3. The (x, u (y, v, u v, but x w y.
5 10.1/18 Let x, y R such that x w y. Show that (i there exists ỹ R which differs from y by at most oe compoet such that x ỹ; ad (ii there exist a, b R such that (x, a (y, b. Solutio: (i. Suppose y t = y. Let δ = y i x i. The x y δe t, where e t is the -vector whose tth compoet is 1 ad all other compoets are 0. Take ỹ = y δe t. (Oe checks that x y δe. (ii. Take a = mi{x 1,..., x } ad b = a + x i y i. 10.1/19 Let x, y, z R +. If 2x w y + z, show that (x, x w (y, z w (y + z, 0. Solutio: The secod w is Problem 14. For the first oe, let u = (x, x ad v = (y, z. We show that m u i m v i for m = 1, 2,..., 2. If m is eve, we write m = 2k, otherwise m = 2k+1, where k is some oegative iteger. For m = 2k, we have m u i = 2 x i For m = 2k + 1, we derive m (y + z i y i + z i v i. m k+1 u i = x i + x i 1 k+1 (y + z i 2 + 1 (y + z i 2 ( 1 k+1 ( k+1 y i 2 + z i + 1 y i 2 + z i ( = y i + + 1 2 (y k+1 + z k+1 ( y i + z i z i m + max{y k+1, z k+1 } v i. Equality apparetly holds whe m = 2. Thus (x, x (y, z. Note: All are ot eeded i the origial problem. 10.1/20 Let x = (x 1, x 2,..., x R ad α be a real umber such that x α x 1. Let β = x 1 +x 2 + +x. Show that (α, β α 1,..., β α 1 x. Solutio: Let y = (α, β α 1,..., β α 1 ad 1 < k.
6 β α 1 Case 1: α < β α 1. Sice x α, x 1 + +x 1 1 x 1 + +x k k (Problem 6. Thus k β α 1 x 1 + + x k. It is immediate that y x as equality y 1 + y 2 + + y = β = x 1 + x 2 + + x holds. Case 2: α β α 1. We eed to show that α+ k 1 1 (β α x 1 + +x k for every k. Observe that ( 1(x 2 + +x k (k 1(x 2 + +x by Problem 6. The majorizatio y x follows because y 1 + y 2 + + y k = α + k 1 (β α 1 = (1 k 1 1 α + k 1 1 β (1 k 1 1 x 1 + k 1 1 β = x 1 + k 1 1 (β x 1 = x 1 + k 1 1 (x 2 + + x x 1 + x 2 + + x k. 10.1/21 Let x, y R. If x y, show that ym x m y m+1 for some m. Solutio: If x = y, the there is othig to prove. Otherwise, y x cotais at least oe positive ad oe egative compoets. Note that x 1 y 1. Let m be the idex umber such that y i x i 0 for i = 1,..., m ad y m+1 x m+1 < 0, that is, y m+1 x m+1 is the first egative compoet i y x. The ym x m x m+1 > y m+1. 10.1/22 Let t R ad deote t + = max{t, 0}. For x, y R, if x w y, show that (x + 1,..., x+ w (y + 1,..., y+. Is the coverse true? Solutio: Deote x + = (x + 1,..., x+, y + = (y + 1,..., y+. If all x i are opositive, the x + = 0; we have othig to show. Suppose there are p positive x i. For every k p, take u = (1,..., 1, 0,..., 0 with k 1 s. By the secod result of Theorem 10.5, (x + i = x i = For k > p, x i u i y i u i = y i p p (x + i = (x + i (y + i (y + i. The coverse is false: x = (1, 1, y = (1, 2. (y + i.
7 10.1/23 Give a example that Theorem 10.6 is ot valid for some x, y R. Solutio: Take x = (2, 2, y = (3, 1, u = ( 1, 3 (egative u or x = ( 2, 2, 2, y = ( 1, 2, 3, u = (3, 2, 1 (positive u. 10.1/24 Let x = (x 1,..., x, y = (y 1,..., y R. If x y, show that u i y i u i x i u i x i Solutio: For the first iequality, we show that u i y i, u = (u 1,..., u R. x y u i y i u i x i. Let t i = x i y i. The k t i = k x i k y i 0 for every k. u i x i u i y i = t i u i = t 1(u 1 u 2 + (t 1 + t 2 (u 2 u 3 + + (t 1 + + t 1 (u 1 u + (t 1 + + t u 0. By the first result of Theorem 10.4(4, u i x i u i x i, the first iequality of the problem follows immediately. The remaiig two iequalities are easy cosequeces the secod result of Theorem 10.4(4 ad the first result of Theorem 10.5, respectively. Note: The secod u i i the origial problem may be replaced by u i. 10.2/1 Fid a doubly stochastic matrix P such that (1, 2, 3 = (6, 0, 0P. Solutio: P = 1/6 1/3 1/2 1/3 2/3 0 1/2 0 1/2 10.2/2 Show that Q = (q ij is a doubly substochastic matrix if there exists a doubly stochastic matrix D = (d ij such that q ij d ij for all i, j. Solutio: We show that Q = (q ij is doubly substochastic (i.e., Q is oegative, eq e ad Qe T e T if ad oly if there exists a doubly stochastic matrix D = (d ij such that q ij d ij for all i, j. : If Q D etrywise, the eq ed = e ad Qe T De T = e T..
8 : Let Q be oegative, eq e ad Qe T e T. We show that there exists a doubly stochastic matrix D = (d ij such that q ij d ij for all i, j. If every row sum of Q is 1, the the sum of all etries of Q is, thus every colum sum is also 1, ad Q = D is doubly stochastic. Suppose that some row (say i sum of Q is less tha 1, the some colum (say j sum is less tha 1. Let h ij = 1 max{ q ik, q kj }, Q 1 = Q + h ij E ij, k=1 k=1 where E ij is the matrix with (i, j etry 1 ad 0 elsewhere. It is evidet that every rom (ad colum sum of Q 1 is o more tha 1 ad that Q 1 has oe more row or colum sum equal to 1. Repeat the process for Q 1 ad so o. I fiite steps, we obtai a doubly stochastic matrix D = Q + h ij E ij. 10.2/3 Let A = (a ij ad B = (b ij be matrices. Show that (a If A ad B are doubly substochastic, the C = (a ij b ij ad D = ( a ij b ij are also doubly substochastic. (b If A ad B are uitary, the E = ( a ij b ij is doubly substochastic, but F = ( a ij b ij is ot i geeral. Solutio: (a. Sice A is doubly substochastic, there is a doubly stochastic matrix G = (g ij such that a ij g ij for all i, j. Thus a ij b ij a ij g ij for all i, j, ad C = (a ij b ij is doubly substochastic. For D, otice that a ij b ij 1 2 (a ij + b ij, so D 1 2 (A + B (etrywise. Because D 0, ed 1 2 e(a+b e, ad 1 DeT 2 (A+BeT e T, D is doubly substochastic. (b. Sice a ij b ij 1 2 ( a ij 2 + b ij 2 ad sice A ad B are uitary (thus legth of each row or colum is 1, we have ee 1 2 e( a ij 2 + b ij 2 = 2( 1 e( aij 2 + e( b ij 2 = e. Likewise Ee T e T. Thus E is doubly substochastic. F = ( a ij b ij 1 is ot doubly substochastic i geeral. Take A = B =. ( 1 1 2 1 1 ( The A = B is uitary, but F = 1 1 1 2 1 1 is ot doubly substochastic. 10.2/4 Show that the followig matrix A satisfies (i ea e, Ae T e T ; (ii A Q is ever oegative for ay doubly stochastic matrix Q: A = 0 1 1 1 0 0 1 0 0, e = (1, 1, 1.
Solutio: ea = (2, 1, 1 e ad Ae T = (2, 1, 1 e T. If A Q were oegative, the q ij = 0 for all 2 i, j 3. This is impossible because each row ad colum sum of Q is 1. 10.2/5 Show that the followig doubly stochastic matrix caot be expressed as a product of T -trasforms: 1 1 0 2 2 1 1 2 0 2 1 1 2 2 0. Solutio: It suffices to show that the product of ay two 3 3 opermutatio doubly stochastic matrices has at least oe positive mai diagoal etry. Let A ad B be 3 3 doubly stochastic matrices, ot permutatio matrices. Let (a, b, c be the first row of A ad (x, y, z T be the first colum of B. The the (1, 1-etry of AB is ax + by + cz. If at least two umbers i a, b, c are positive ad two i x, y, z are positive, the ax + by + cz > 0. Otherwise, we may assume a = 1, b = c = 0 (permutatio may be applied if ecessary. The A has the form ( 1 0 0 0 d e 0 f g, i which d, e, f, g are all positive. The it is easy to see that the (2,2-etry or (3,3-etry of AB is positive. If two of x, y, z are zero, the argumet is proved i a similar way. 10.2/6 Let P be a square matrix. If both P ad its iverse P 1 are doubly stochastic, show that P is a permutatio matrix. Solutio: Let x 1,..., x ad y 1,..., y be oegative scalars such that x 1 + +x = 1 ad y 1 + +y = 1. The x 1 y 1 + +x y = 1 if ad oly if some x i y i = 1, i.e., x i = y i = 1, ad all other x j = y j = 0. Apply this to the kth row of P times the kth colum of P 1, k = 1,..., with P P 1 = I. Thus each row of P cotais a 1. Oe may also prove the assertio as follows: xp x ad x = (xp P 1 xp. So x xp x for all x, P has to be a permutatio. 10.2/7 Show each of the followig statemets. (a The (ordiary product of two doubly stochastic matrices is a doubly stochastic matrix. (b The (ordiary product of two doubly substochastic matrices is a doubly substochastic matrix. (c The Hadamard product of two doubly substochastic matrices is a doubly substochastic matrix. 9
10 (d The Kroecker product of two doubly substochastic matrices is a doubly substochastic matrix. (e The covex combiatio of fiite doubly stochastic matrices is a doubly stochastic matrix. (f The covex combiatio of fiite doubly substochastic matrices is a doubly substochastic matrix. Solutio: (a Let A ad B be doubly stochastic matrices. The AB is oegative, ad e(ab = (eab = eb = e ad (ABe T = A(Be T = Ae T = e T. So AB is a doubly stochastic matrix. (b Let A ad B be doubly substochastic matrices. The AB is oegative, ad e(ab = (eab eb e ad (ABe T = A(Be T Ae T e T. So AB is a doubly substochastic matrix. (c If A ad B are doubly substochastic, the A ad B are oegative ad b ij 1 for all i, j. Thus A B A ad A B is doubly substochastic. (d Let A ad B be doubly substochastic matrices (of possibly differet sizes, say, m ad. The e m+ (A B = (e m e (A B = (e m A (e B e m e, where e k is the all 1 row vector of k 1 s. Similarly, (A Be T m+ e T m+. (e Let P = t 1 P 1 + + t P be a covex combiatio of doubly stochastic matrices P 1,..., P, where t i are oegative addig up to 1. The ep = t 1 ep 1 + + t ep = (t 1 + + t e = e. Similarly, P e T = e T. Thus P is a doubly stochastic matrix. (f Similar to the proof for (e. 10.2/8 Show that ay square submatrix of a doubly stochastic matrix is doubly substochastic ad that every doubly substochastic matrix ca be regarded as a square submatrix of a doubly stochastic matrix. Solutio: The first part is obvious. Let Q be a doubly substochastic matrix. The there is a doubly stochastic matrix P such that Q P (etrywise ad is doubly stochastic. ( Q P Q P Q Q 10.2/9 A square (0, 1-matrix is called sub-permutatio matrix if each row ad each colum have at most oe 1. Show that a matrix is doubly substochastic if ad oly if it is a covex combiatio of fiite subpermutatio matrices.
11 Solutio: If a matrix is a covex combiatio of fiite sub-permutatio matrices, it is obviously doubly substochastic. Coversely, if Q is doubly substochastic, the there are matrices A, B, C such that ( Q A B C is doubly stochastic. By Theorem 5.21, every doubly stochastic matrix is a covex combiatio of permutatio matrices. It follows that Q is a covex combiatio of sub-permutatio matrices. 10.2/10 Let A = (a ij be a doubly stochastic matrix. Show that etries of A ca be chose from differet rows ad colums so that their product is positive; that is, there exists a permutatio p such that a ip(i > 0. [Hit: Use the Frobeius Köig theorem.] Solutio: If the product of the etries o every diagoal is zero, the by the Frobeius Köig theorem (Theorem 5.20, A has a r s zero submatrix, r + s = + 1. We may assume, after permutig rows ad colums if ecessary, that A = ( B C D 0, where B is r ( s ad D is ( r s. Sice A is doubly stochastic, each row sum of B is 1 ad the sum of all etries of B is r. As the sum of all etries of B ad all etries of C is s, r s. This cotradicts r + s = + 1. ( 10.2/11 A matrix A of order 2 is said to be reducible if P AP T = B 0 C D for some permutatio matrix P, where B ad D are some square matrices; A is irreducible if it is ot reducible. A matrix A is said to be decomposable if P AQ = for some permutatio matrices P ( B C 0 D ad Q, where B ad D are some square matrices; A is idecomposable if it is ot decomposable. Prove each of the followig statemets. (a If A is a oegative idecomposable matrix of order, the the etries of A 1 are all positive. (b The product of two oegative idecomposable matrices is idecomposable. (c The product of two oegative irreducible matrices eed ot be irreducible. Solutio: (a Let B be a oegative matrix. If Bx > 0 for all oegative vectors x 0, the B > 0. Otherwise, say, b ij = 0. Settig x = e T i = (0,..., 0, 1, 0,..., 0 T i Bx gives the jth colum of B which cotais b ij = 0, cotradictig Bx > 0. Now for A, sice A is idecomposable (o row of A is etirely zero, Ay > 0 wheever y > 0. We show that A 1 x > 0 for all oegative vectors x 0. If x > 0, the above argumet says Ax > 0, so A 2 x = A(Ax > 0. Iductively, A 1 x > 0. Let x 0 be a oegative vector
12 havig some zero compoets. We first show that the umber of positive compoets of Ax is strictly more tha that of x, that is, the umber of positive compoets of x is icreased after multiplyig by A. Suppose that P (Ax = ( 0 u ad QT x = ( 0 v, where P ad Q are permutatio matrices, ad u ad v are positive vectors of s ad t compoets, respectively. We show that s > t. Suppose that s t ad partitio P AQ = ( A 1 A 2 A 3 A 4, i which A1 ad A 4 are square matrices of sizes t ad t, respectively. The P Ax = P AQQ T x = ( A 1 A 2 A 3 A 4 ( 0 v = ( 0 u. Thus A 2 v = 0. Sice v > 0, A 2 = 0 ad A is decomposable, a cotradictio to the assumptio that A is idecomposable. Deote by p(z the umber of the positive compoets of a oegative vector z. The above argumet says that 1 p(x p(ax p(a 2 x p(a 1 x. It s impossible for all strict iequalities to hold; so p(a k x = p(a k+1 x for some k. This happes oly if A k x > 0. It follows that A k+1 x > 0, A k+2 x > 0,..., A 1 x > 0. (b. Let A ad B be idecomposable matrices. If AB is decomposable, let P ad Q be permutatio matrices such that P ABQ = ( D C E 0, where C is q q. Take x = (0, u T, where u > 0 with q compoets. The p(p ABQx p(x. However, from the discussios i (a we kow that p(p ABQx = p(abqx p(bqx > p(qx = p(x, a cotradictio. So AB is idecomposable. (c. A = ( 0 1 1 0 is irreducible, but A2 = I 2 is reducible. 10.2/12 Let x, y R +. Show that x w y if ad oly if x is a covex combiatio of the vectors yq 1, yq 2,..., yq m, where Q 1, Q 2,..., Q m are sub-permutatio matrices. Solutio: This is a combiatio of Theorem 10.10 ad Problem 9. If x w y, by Theorem 10.10, there exists a doubly substochastic matrix S such that x = ys. By Problem 9, S is a fiite covex combiatio of sub-permutatio matrices, say Q 1, Q 2,..., Q m. Thus x is a covex combiatio of the vectors yq 1, yq 2,..., yq m. Coversely, if x = t 1 yq 1 + t 2 yq 2 + + t m yq m. The x = yq, where Q = t 1 Q 1 + t 2 Q 2 + + t m Q m is doubly substochastic. So x w y.
13 10.2/13 Let A be a oegative matrix. Show that A is doubly substochastic if ad oly if Ax w x for all colum vectors x R +. Solutio: : This is Theorem 10.10. : Take x = e T to be the colum vector with all compoets equal to 1. The Ae T w e T. So every row sum of A is o more tha 1, i.e., Ae T e T. Set x = e T i with ith compoet equal to 1 ad 0 elsewhere. The a i w e T i, where a i deotes the ith colum of A, i = 1, 2,...,. Thus every colum sum of A is o more tha 1, i.e., ea e. So A is doubly substochastic. 10.2/14 Give a example that x R +, y R, x = ys for some substochastic matrix S, but x w y does ot hold. Solutio: x = ( 1 3, 0, y = (1, 1, S = 1 3 ( 2 1 1 1. 10.2/15 Let x, y R. Show that for ay real umbers a ad b, x y (ax 1 +b, ax 2 +b,..., ax +b (ay 1 +b, ay 2 +b,..., ay +b. Solutio: Let x = ys, where S is doubly stochastic. Let e R be the row vector of all oes. Note that es = e. We have (ax 1 + b, ax 2 + b,..., ax + b = ax + be = ays + bes = (ay + bes ay + be. 10.2/16 Let T = ( 1 a a b 1 b, S = ( a a b b, 0 < a < 1, 0 < b < 1. Show that T = I + 1 r S, r = 1 (a + b, 1 r for every positive iteger. Fid T as. Solutio: Observe that S 2 = ts, where t = (a + b. So S = t 1 S for all 2. Note that T = I + S, 1 + t = r. We have T 2 = (I + S 2 = I + 2S + S 2 = I + 2S + ts = I + (1 + rs. T 3 = T 2 T = [I + (1 + rs](i + S = I + (1 + r + r 2 S. T = I + (1 + r + r 2 + + r 1 S = I + 1 r 1 r S. Sice 0 < a + b < 2, 1 < r < 1. Thus T I + 1 1 r S = 1 Note: Oe may also compute T by biomial theorem. a a. a+b ( b b
14 10.3/1 Let x, y R. If x y (compoetwise, show that x w y. Solutio: Let x j1 x j2 x j be the arragemet of the compoets of x i decreasig order; that is, x i = x j i. The k x i = k x j i k y j i k y i for every k. Thus x w y. 10.3/2 Let α > 1. Show that f(t = t α is strictly icreasig ad strictly covex o R + ad that g(t = t α is strictly covex o R. Solutio: For α > 1 ad t > 0, f (t = αt α 1 > 0 ad f (t = α(α 1t α 2 > 0, so f is strictly icreasig ad strictly covex o R +. For g, g is the same as f o R +, so it is covex o R +. If t < 0, the g(t = ( t α ad g (t = α( t α 1 ( 1 ad g (t = α(α 1( t α 2 > 0. So g is strictly covex o R. 10.3/3 Show that f(t = e αt, α > 0, is strictly icreasig ad strictly covex o R ad that g(t = t is strictly cocave ad icreasig o R +. Solutio: For α > 0, f (t = αe αt > 0. So f is strictly icreasig o R. Sice f (t = α 2 e αt > 0, f is strictly covex o R. g (t = 1 2 t > 0 for t > 0, so g is icreasig o R +. g (t = 1 4 t 3/2 < 0, so it is strictly cocave o R +. 10.3/4 Show that f(t = l( 1 t 1 is covex o (0, 1 2 but ot o ( 1 2, 1. Solutio: f (t = 1 2t (t 2 t 2 > 0 for t (0, 1 2 ad < 0 for t ( 1 2, 1. 10.3/5 The followig iequalities are of fudametal importace. They ca be show i various ways. Oe way is to use iductio; aother way is to use Jese iequality with covex fuctios (f(x = l x, say. (a Use Jese iequality to show the geeral arithmetic mea geometric mea iequality: if all a i 0, p i > 0 ad p i = 1, a pi i p i a i. (b Use (a to show the Hölder iequality: if p, q > 1 ad 1 p + 1 q = 1, ( 1/p ( 1/q a i b i a i p b i q for complex umbers a 1,..., a, b 1,..., b.
15 (c Use (b to show the Mikowski iequality: if 1 p <, Solutio: ( 1/p ( 1/p ( 1/p a i + b i p a i p + b i p for complex umbers a 1,..., a, b 1,..., b. (a If some a i = 0, the iequality obviously holds. Assume that all a i > 0. Sice f(t = l t is strictly covex o (0,, by Jese s iequality, we have ( l p i a i p i ( l a i. So ( l p i a i l a pi i = l ( The iequality follows as l t is mootoic. Equality occurs if ad oly if a 1 = = a. (b Give oegative real umbers a, b, set x 1 = a p ad x 2 = b q. With p 1 = 1/p ad p 2 = 1/q i (a, we have a pi i ab = x p1 1 xp2 2 p 1x 1 + p 2 x 2 = 1 p ap + 1 q bq ad equality occurs if ad oly if a p = b q. If all a i or all b i are 0, there is othig to prove. Assumig otherwise, we have ( a i p 1/p ( b i q 1/q âˆb 0. Set â i = a i /â, ˆb i = b i /ˆb. The â i p = 1, ˆb i q = 1. Thus â i ˆbi â i ˆb ( 1 i p â i p + 1 q ˆb i q = 1 p + 1 q = 1. The desired iequality Hölder iequality follows at oce. For the equality case, equality holds if ad oly if â i p = ˆb i q ad âi ˆb i = â i ˆb i which meas â i ˆbi = c â i ˆbi for a uit complex c ad all i. Equivaletly, there exist some oegative umber t ad uit complex umber u such that b i q = t a i p ad a i b i = u a i b i for all i..
16 (c If p = 1, equality holds. So we let p > 1. We may also assume that ot all a i are 0 ad ot all b i are 0. Let q be such that 1/p + 1/q = 1. Note that a i + b i p a i + b i p 1 a i + a i + b i p 1 b i. Applyig Hölder iequality to each term o the right, we have a i + b i p = ( 1/q a i + b i (p 1q ( 1/p ( 1/p a i p + b i p ( 1/q a i + b i p ( 1/p ( 1/p a i p + b i p. Now dividig both sides by ( a i + b i p 1/q reveals the the Mikowski iequality. Equality occurs if ad oly if (i all a i are 0; or (ii b i = ta i for all i ad some t 0; or (iii p = 1 ad for each i, either a i = 0 or b i = t i a i for some t i 0. 10.3/6 Show that (i if f(t is covex o R, the f 1 (x = f(x 1 + + f(x ad f 3 (x = f 1 (f 2 (x, where f 2 (x = (f(x 1,..., f(x, are Schurcovex o R ; ad (ii if g is symmetric ad covex o R, the g is Schur-covex. Solutio: (i Let f(t be covex o R. Let x y, where x, y R. Theorem 10.13(1 says f(x i f(y i, i.e., f 1 (x f 2 (y; that is, f 1 is Schur-covex. Now for f 3 (x, from the proof of Theorem 10.12, f 2 (x f 2 (ya for some doubly stochastic matrix A. It follows that f 1 (f 2 (x f 1 (f 2 (ya f 1 (f 2 (y because f 2 (ya f 2 (y ad f 1 is Schur-covex. (ii Let g be symmetric ad covex o R. For x y, where x, y R, we write x = ya, where A is a doubly stochastic matrix. The g(x = g(ya. O the other had, A is a covex combiatio of permutatio matrices (Theorem 5.21. Write A = t 1 P 1 + + t k P k, where t i are positive ad have sum 1, ad P i
17 are permutatio matrices. The, sice g is covex ad symmetric, ( ( g(x = g(ya = g y t i P i = g t i yp i t i g (yp i = t i g(y = g(y. Note: The origial problem eeds to be reworded a bit as Schurcovexity is ot defied for vector-valued fuctios; i additio, g eeds to be symmetric. 10.3/7 Let f(t be a positive fuctio defied o a iterval I R. If l f(t is (strictly covex o I, show that F (x = f(x i is (strictly Schur-covex o I = {(x 1,..., x : x 1,..., x I} R. Solutio: For a, b I ad oegative α, β, α + β = 1, the covexity of l f says l f(αa + βb α l f(a + β l f(b. Now let G(x = l F (x = l =1 f(x i = l f(x i. The G(x is covex because G(αx + βy = l f(αx i + βy i α l f(x i + β = αg(x + βg(y. l f(y i Let x y, where x, y R. We write x = ya, where A is a doubly stochastic matrix. The G(x = G(yA. O the other had, A is a covex combiatio of permutatio matrices (Theorem 5.21. Write A = t 1 P 1 + + t k P k, where t i are positive ad have sum 1, ad P i are permutatio matrices. The G(x = G(yA t k G(yP k = t k G(y = G(y. It follows that F (x F (y, i.e., F is Schur-covex. If f is strict, the G is strict, ad thus F is strict. Note: If f(t is covex should read if l f(t is covex. Otherwise, take f(x = x 1. The f(x is oegative ad cotiuous o the iterval I = ( 2, 2. For x = (0, 0 y = (1, 1, F (x = 1 > 0 = F (y. Chage oegative cotiuous fuctio i the origial problem to positive fuctio.
18 10.3/8 Give a example of covex, oicreasig fuctio f(t for which (f(x 1,..., f(x w (f(y 1,..., f(y is ot true eve if x w y. Solutio: Let f(x = (x 3 2. The f is covex o R +. For x = (2, 1 w y = (3, 1, f(x = (1, 4 w f(y = (0, 4. 10.3/9 Let x, y R. Prove or disprove: (a x w y x w y, i.e., ( x 1,..., x w ( y 1,..., y. (b x w y x 2 w y 2, i.e., (x 2 1,..., x 2 w (y 2 1,..., y 2. (c x w y x 2 w y 2, i.e., (x 2 1,..., x 2 w (y 2 1,..., y 2. (d x y x 3 w y 3, i.e., (x 3 1,..., x 3 w (y 3 1,..., y 3. (e x y x 3 w y 3, i.e., ( x 1 3,..., x 3 w ( y 1 3,..., y 3. (f x w y e x w e y, i.e., (e x1,..., e x w (e y1,..., e y. Solutio: (a False. x = ( 1, 0 w (0, 0 = y but x w y. (b True. By Theorem 10.2(2, x z y for some z 0. It follows that x 2 = x 2 z 2 w y 2 = y 2 by Corollary 10.1(2. (c False. x = ( 1, 0 w (0, 0 = y but x 2 w y 2 (d False. x = ( 0.1, 0.1 (0, 0.2 = y but x 3 = ( 0.001, 0.001 w (0, 0.008 = y 3. If x, y R +, the it is true.. (e True. f(t = t 3 is covex. Use Theorem 10.12. (f True. f(t = e t is icreasig ad covex. Use Theorem 10.12. 10.3/10 Let x, y R. Show that x y w x y. Solutio: By Theorem 10.4.(3, x y = x + ( y x y. By Corollary 10.1(1, takig the absolute values gives x y w x y. 10.3/11 Let x, y R +, x y. Show that i=k x i i=k y i for each k. Solutio: Let f(t = t. The f(t is covex o R +. By Theorem 10.12, f(x w f(y, i.e., x w y. It is sufficiet to ote that ( x i = x i+1 ad that k ( x i = i= k+1 x i. 10.3/12 Show that the followig fuctios are Schur-covex o R. (a f(x = max i x i. (b g(x = x i p, p 1. ( 1/p, (c h(x = x i p p 1.
19 (d p(x = 1 x i, where all x i > 0. Solutio: (a If x y, the x w y (Corollary 10.1. It follows that the largest compoet of x is o more tha that of y, i.e., max i x i max i y i. So f is Schur-covex. (b If p = 1, x y x w y (Corollary 10.1, so g(x g(y. For p > 1, t p is covex o R (Problem 2. By Theorem 10.12, x p w y p. Cosequetly, g(x g(y. (c This is immediate from (b by takig the pth roots of both sides. (d 1 t, t > 0, is a covex fuctio. For x y, by Theorem 10.12, ( 1 1 x 1,..., x w ( 1 y 1,..., 1 y. As a result p(x p(y. 10.3/13 Let x, y R +. If x y, show that x i y i ad that the strict iequality holds if y is ot a permutatio of x. Show by example that this is ivalid if x or y cotais oegative compoets. Solutio: If some x i = 0 the x = 0 ad x y = 0. I this case we have othig to show. Without loss of geerality, we may assume that all x i ad all y i are positive. Let f(t = l t. The f(t is covex. By Theorem 10.13, we have l x i l y i. Thus l x i l y i, or l x i l y i, which yields that x i y i. For the equality case, let x = ya, where A is doubly stochastic but ot a permutatio matrix. The x i = ya i = j=1 a jiy j, where A i is the ith colum of A. Sice l t is strictly covex, we have ( l x i = l a ji y j a ji l y j, i = 1,...,, j=1 i which at least oe strict iequality holds (as A is ot a permutatio. Now takig the sum over i, we have l x i > l y i. The the strict iequality for product holds. If x ad y are allowed to have egative compoets, the the coclusio is ot true i geeral. Take x = (1, 0, 1, y = (3, 1, 2. Note: if both x ad y cotai zero compoets, equality holds eve x is ot a permutatio of y. 10.3/14 Let x, y R +. Show that the sum iequalities k x i k y i (k imply the product iequalities k x i k y i (k. j=1
20 Solutio: We may assume that all compoets of x ad y are positive (otherwise use cotiuity argumet. Note that the set of iequalities k x i k y i (k is equivalet to sayig y w x. The fuctio f(t = l( t is icreasig ad covex o (, 0. By Theorem 10.12, f( y w f( x; that is, (l y 1,..., l y w (l x 1,..., l x, which imply that or l k k y i l k k x i y i. x i 10.3/15 Let x, y, z be the three iterior agles of ay triagle. Show that 0 < si x + si y + si z 3 2 3. Solutio: ( π 3, π 3, π 3 (x, y, z (π, 0, 0. By Theorem 10.13(1 with the covex fuctio f(t = si t, we obtai the iequalities. 10.3/16 Let a 1,..., a ad b 1,..., b be positive umbers. Show that ( a 1 b 1 (,..., a a1 b w,..., a b 1 b ( a 1 w b,..., a. b 1 Solutio: This is immediate from Theorem 10.16 (Equ. (10.5 by takig x = (a 1,..., a ad y = ( 1 b 1,..., 1 b. 10.3/17 Let x 1,..., x be positive umbers such that x 1 + +x = 1. Prove (a 1 x i x i ( 1; x i 1 x i ( 1. (b 1+x i x i ( + 1; (+1 x i 1+x i 1 2. (c 1+x i 1 x i (d x i l 1 x i l. (+1 1 ; 1 1 x i 1+x i ( 1 +1. Solutio: Note that ( 1,..., 1 (x 1,..., x (1, 0,..., 0. Applyig the first part of Theorem 10.12 to each of the followig covex fuctios (which ca be verified by takig the 2d derivatives o (0, 1 results i the desired iequalities.
21 (a f(t = 1 t t ; g(t = t 1 t. (b f(t = 1+t t ; g(t = t 1+t. (c f(t = 1+t 1 t 1 t ; g(t = 1+t. (d f(t = t l 1 t = t l t. 10.3/18 Let A be a positive semidefiite matrix, > 1. Show that tr e A e tr A + ( 1 with equality if ad oly if rak (A 1. Solutio: It is easy to see that equality holds whe rak (A = 0 or 1. Let rak (A > 1. Let λ 1,..., λ be the eigevalues of A. The the iequality is the same as e λi e λi + ( 1. Note that f(t = e t 1 is a strictly covex fuctio with f(0 0. The coclusio follows from a applicatio of Theorem 10.11 to f(t. 10.3/19 Let x, y R +. If x log y ad x y, show that x = y. Solutio: Sice x log y, y cotais o more tha 0s tha x (if ay, we may assume that the compoets of y are all positive. O the other had, x y, so the compoets of x are also all positive. If x y, the l x is ot a permutatio of l y. x log y says that l x l y. Applyig Theorem 10.14 to l x l y with f(t = e t, we see that f(l x i < f(l y i; that is, x i < y i. This cotradicts the assumptio x y. 10.3/20 For real umber t, deote t + = max{t, 0}. Let x, y R. Show that (a x y if ad oly if (x i t + (y i t + for all t R ad x i = y i. (b x y if ad oly if x i t y i t for all t R. Solutio: (a. For ay fixed t R, f(x = (x t + is a covex fuctio. By Theorem 10.13(1, (x i t + (y i t +. Coversely, for each k, take t = y k. The (y i y k = (y i y k + (x i y k + (x i y k + (x i y k.
22 So k y i k x i. With equality for k =, we have x y. We use (a to prove (b. is easy because f(x = x t is covex for every t. To show, takig t to be a value smaller tha ay of x i ad y i, we see x i y i. Similarly, we have the reverse iequality by settig t to be a large value. We arrive at x i = y i. Now, otice that t + t = 2t +. We have (x i t + = 1 2 1 2 ( (xi t + x i t ( (yi t + y i t = (y i t +. 10.3/21 Let f(t be a real-valued fuctio defied o a iterval I R. Show that if f(t is icreasig ad strictly covex, the f(t is strictly icreasig. Solutio: Let α I ad δ be ay positive umber such that α+δ I. We eed to show that f(α < f(α+δ. Suppose that f(α = f(α+δ. The f(α + 1 2 δ = f( 1 2 α + 1 2 (α + δ < 1 2 (f(α + f(α + δ = f(α, a cotradictio to f beig icreasig. Note: The origial problem is replaced by the ew oe above. Note that f(t = t is covex ad strictly icreasig but ot strictly covex. Theorem 10.14 does ot apply: (0, 0 (1, 1, 0 + 0 1 + ( 1. 10.3/22 Let x, y R +. If x w y, show that x m w y m for all itegers m 1, where z m = (z m 1,..., z m for z = (z 1,..., z R. If x m y m for some iteger m > 1, show that x = y. Solutio: f(t = t m is strictly covex ad icreasig o R +. By Theorem 10.12, f(x w f(y, i.e., x m w y m. Let x m y m for some positive iteger m > 1. If x y, amely, x is ot a permutatio of y, by Theorem 10.14, xm i < ym i. This cotradicts the assumptio x m y m (which implies all compoet sum idetity. Note: Chage for all itegers m 1 to for some iteger m > 1. 10.3/23 Let g be a differetiable fuctio o a iterval I R. Show that (a g is covex if ad oly if g( a+b 2 1 2 (g(a + g(b for all a, b I. (b g is liear if ad oly if g( a+b 2 = 1 2 (g(a + g(b for all a, b I. (c g(x g(y wheever x y, x, y R, if ad oly if g is liear.
23 Solutio: (a is obvious. For, let x, y I ad z = t 1 x + t 2 y, where t 1 + t 2 = 1 ad t 1, t 2 > 0. We first show by iductio that the covexity holds for all t 1, t 2 of the forms p/2, q/2. Suppose the iequality holds for, we show the case of + 1. Assume that p < q, p + q = 2 +1. The q > 2. Let r = q 2. Write z = p 2 +1 x + q 2 +1 y = 1 ( p 2 2 x + r 2 y + y It follows that g(z 1 ( ( px + ry g 2 2 + g(y p q g(x + g(y. 2+1 2+1 Sice the the set of t of the form p/2 is dese i (0, 1 ad from the cotiuity of g, we see that g is covex. [A similar solutio is see i the book Iequalities by Hardy et al, pp. 71 72.] (b is obvious. For, differetiatig g( a+b 2 = 1 2 (g(a+g(b with respect to a, we get 1 2 g ( a+b 2 = 1 2 g (a, or g ( a+b 2 = g (a for all b I. Thus g (a is costat ad g is liear. (c If g is liear, say g(t = at+b, the x y implies (ax 1,..., ax (ay 1,..., ay (ote that this is ot true for w, ad (ax 1 + b,..., ax + b (ay 1 + b,..., ay + b, i.e., g(x g(y. Coversely, observe that ( a+b (a, b. If > 2, add ( 2 2, a+b 2 0s to make them vectors i R. It follows that ( a + b 2g 2 + ( 2g(0 = g(a + g(b + ( 2g(0. Thus 2g( a+b 2 = g(a + g(b. By (b, g is liear. 10.3/24 If x, y, u, v R +, show that x wlog u, y wlog v x y wlog u v. Solutio: We first observe that x y log x y. This is because l x + l y l x + l y, that is, l(x y l(x y. I a similar way, oe may prove that if x wlog u the x y wlog u y. Thus x y wlog u y wlog u v, as desired. 10.3/25 Let a, b R ad all compoets of a ad b be positive. Show that a i b i r + s 2 rs a i b i,
24 where r ad s are the umbers such that r ai b i s > 0, i = 1,...,. [Hit: Use the Katorovich iequality for A = diag( a1 b 1,..., a b.] Solutio: Let A = diag( a1 b 1,..., a b. The the smallest eigevalue of A is λ = mi i { ai b i } ad largest eigevalue of A is λ 1 = mi i { ai b i }. The Katorovich iequality states that for ay colum vector x C, (x Ax(x A 1 x (λ 1 + λ 2 4λ 1 λ (x x 2. Put x = ( a 1 b 1,..., a b T. It follows that ( a 2 i ( b 2 i (λ 1 + λ 2 ( 2. a i b i 4λ 1 λ Takig the square roots ad by the Cauchy-Schwarz iequality, a i b i a 2 i b 2 i (λ 1 + λ 2 a i b i. λ 1 λ Note that λ1+λ 2 λ 1λ r+s 2 rs as log as r λ 1 ad λ s > 0.