Sum of squared logarithms - An inequality relating positive definite matrices and their matrix logarithm

Transcription

1 Sum of squared logarithms - An inequality relating positive definite matrices and their matrix logarithm arxiv: v1 [math.ca] 24 Jan 201 Mircea Bîrsan 12 Patrizio Neff 1 and Johannes Lankeit 1 1 Lehrstuhl für Nichtlineare Analysis und Modellierung Fakultät für Mathematik Universität Duisburg-Essen Germany 2 Department of Mathematics University A.I. Cuza of Iaşi Romania August 2018 Abstract Let y 1 y 2 y a 1 a 2 a (0 ) be such that y 1 y 2 y = a 1 a 2 a and y 1 +y 2 +y a 1 +a 2 +a y 1 y 2 +y 2 y +y 1 y a 1 a 2 +a 2 a +a 1 a. (logy 1 ) 2 +(logy 2 ) 2 +(logy ) 2 (loga 1 ) 2 +(loga 2 ) 2 +(loga ) 2. This can also be stated in terms of real positive definite -matrices P 1 P 2 : If their determinants are equal detp 1 = detp 2 then trp 1 trp 2 and trcofp 1 trcofp 2 = logp 1 2 F logp 2 2 F where log is the principal matrix logarithm and P 2 F = ij=1 P2 ij denotes the Frobenius matrix norm. Applications in matrix analysis and nonlinear elasticity are indicated. Key words: matrix logarithm elementary symmetric polynomials ineqality characteristic polynomial positive definite matrices means AMS-Classification: 26D05 26D07 To whom correspondence should be addressed. patrizio.neff@uni-due.de 1

2 1 Introduction Convexity is a powerful source for obtaining new inequalities see e.g. [1 8]. In applications coming from nonlinear elasticity we are faced however with variants of the squared logarithm function see the last section. The function (log(x)) 2 is neither convex nor concave. Nevertheless the sum of squared logarithm inequality holds. We will proceed as follows: In the first section we will give several equivalent formulations of the inequality for example in terms of the coefficients of the characteristic polynomial (Theorem 1) in terms of elementary symmetric polynomials (Theorem ) in terms of means (Theorem 5) or in terms of the Frobenius matrix norm (Theorem 7). A proof of the inequality will be given in Section 2 and some counterexamples for slightly changed variants of the inequality are discussed in Section. In the last section an application of the sum of squared logarithms inequality in matrix analysis and in the mathematical theory of nonlinear elasticity is indicated. 2 Formulations of the problem All theorems in this section are equivalent. Theorem 1. For n = 2 or n = let P 1 P 2 R n n be positive definite real matrices. Let the coefficients of the characteristic polynomials of P 1 and P 2 satisfy trp 1 trp 2 and trcofp 1 trcofp 2 and detp 1 = detp 2. logp 1 2 F logp 2 2 F. For n = we will now give equivalent formulations of this statement. The case n = 2 can be treated analogously. For its proof see Remark 15. By orthogonal diagonalization of P 1 and P 2 the inequalities can be rewritten in terms of the eigenvalues y 1 y 2 y and a 1 a 2 a respectively. Theorem 2. Let the real numbers a 1 a 2 a > 0 and y 1 y 2 y > 0 be such that y 1 +y 2 +y a 1 +a 2 +a y 1 y 2 +y 2 y +y 1 y a 1 a 2 +a 2 a +a 1 a y 1 y 2 y = a 1 a 2 a. (1) (logy 1 ) 2 +(logy 2 ) 2 +(logy ) 2 (loga 1 ) 2 +(loga 2 ) 2 +(loga ) 2. (2) 2

3 The elementary symmetric polynomials see e.g. [9 p.178] e 0 (y 1 y 2 y ) = 1 e 1 (y 1 y 2 y ) = y 1 +y 2 +y e 2 (y 1 y 2 y ) = y 1 y 2 +y 1 y +y 2 y e (y 1 y 2 y ) = y 1 y 2 y are known to have the Schur-concavity property (i.e. e k is Schur-convex) [1 5] see (16). It is possible to express the problem in terms of these elementary symmetric polynomials as follows: Theorem. Let a 1 a 2 a > 0 and y 1 y 2 y > 0 satisfy e 1 (y 1 y 2 y ) e 1 (a 1 a 2 a ) e 2 (y 1 y 2 y ) e 2 (a 1 a 2 a ) e (y 1.y 2 y ) = e (a 1 a 2 a ). e 1 ((logy 1 ) 2 (logy 2 ) 2 (logy ) 2 ) e 1 ((loga 1 ) 2 (loga 2 ) 2 (loga ) 2 ). Because y 1 y 2 y = a 1 a 2 a > 0 we have y 1 y 2 +y 2 y +y 1 y a 1 a 2 +a 2 a +a 1 a y 1 y 2 y a 1 a 2 a Thus we obtain Theorem 4. Let the real numbers a 1 a 2 a > 0 and y 1 y 2 y > 0 be such that y 1 +y 2 +y a 1 +a 2 +a 1 y y y 1 a a a y 1 y 2 y = a 1 a 2 a. () (logy 1 ) 2 +(logy 2 ) 2 +(logy ) 2 (loga 1 ) 2 +(loga 2 ) 2 +(loga ) 2. (4) The conditions () are also simple expressions in terms of arithmetic harmonic and geometric and quadratic mean A(y 1 y 2 y ) = y 1 +y 2 +y H(y 1 y 2 y ) = 1 y y y G(y 1 y 2 y ) = 1 y 1 y 2 y Q(y 1 y 2 y ) = (y2 1 +y2 2 +y2 )

4 Theorem 5. Let a 1 a 2 a > 0 and y 1 y 2 y > 0. A(y 1 y 2 y ) A(a 1 a 2 a ) H(a 1 a 2 a ) H(y 1 y 2 y ) ( reverse! ) and G(y 1 y 2 y ) = G(a 1 a 2 a ) imply Q(logy 1 logy 2 logy ) Q(loga 1 loga 2 loga ) We denote by and arrive at a i =: d 2 i y i =: x 2 i. Theorem 6. Let the real numbers d i and x i be such that d 1 d 2 d > 0 x 1 x 2 x > 0 and x 2 1 +x2 2 +x2 d2 1 +d2 2 +d2 x 2 1x 2 2 +x 2 2x 2 +x 2 1x 2 d 2 1d 2 2 +d 2 2d 2 +d 2 1d 2 x 1 x 2 x = d 1 d 2 d. (5) (logx 1 ) 2 +(logx 2 ) 2 +(logx ) 2 (logd 1 ) 2 +(logd 2 ) 2 +(logd ) 2. (6) If we again view x i and d i as eigenvalues of positive definite matrices an equivalent formulation of the problem can be given in terms of their Frobenius matrix norms: Theorem 7. For n {2} let P 1 P 2 R n n be positive definite real matrices. Let P 1 2 F P 2 2 F and P 1 2 F P 1 2 and detp F 1 = detp logp 1 2 F logp 2 2 F. Let us reconsider the formulation from Theorem 5. If we denote c i := loga i z i := logy i from H(a 1 a 2 a ) H(y 1 y 2 y ) we obtain e z 1 +e z 2 +e z e c 1 +e c 2 +e c. Theorem 8. Let the real numbers c 1 c 2 c and z 1 z 2 z be such that e z 1 +e z 2 +e z e c 1 +e c 2 +e c e z 1 +e z 2 +e z e c 1 +e c 2 +e c z 1 +z 2 +z = c 1 +c 2 +c. (7) z 2 1 +z2 2 +z2 c2 1 +c2 2 +c2. (8) 4

5 In order to prove Theorem 8 one can assume without loss of generality that Thus we have the equivalent formulation z 1 +z 2 +z = c 1 +c 2 +c = 0. (9) Theorem 9. Let the real numbers c 1 c 2 c and z 1 z 2 z be such that e z 1 +e z 2 +e z e c 1 +e c 2 +e c e z 1 +e z 2 +e z e c 1 +e c 2 +e c z 1 + z 2 + z = c 1 + c 2 + c = 0. (10) z z2 2 + z2 c2 1 + c2 2 + c2. (11) Let us prove that Theorem 8 can be reformulated as Theorem 9. Indeed let us assume that Theorem 9 is valid and show that the statement of Theorem 8 also holds true. We denote by s the sum s = z 1 +z 2 +z = c 1 +c 2 +c and we designate z i = z i s c i = c i s (i = 12). the real numbers z i and c i satisfy the hypotheses of Theorem 9 and we obtain z z2 2 + z2 c2 1 + c2 2 + c2. This inequality is equivalent to ) 2 ( z i s which by virtue of the condition (7) reduces to Thus Theorem 8 is also valid. By virtue of the logical equivalence ( c i s ) 2 z 2 1 +z2 2 +z2 c2 1 +c2 2 +c2. (A B C) ( C A B) for any statements ABC we can formulate the inequality (11) (i.e. Theorem 9) in the following equivalent manner: Theorem 10. Let the real numbers c 1 c 2 c and z 1 z 2 z be such that z 1 +z 2 +z = c 1 +c 2 +c = 0 and z 2 1 +z2 2 +z2 < c2 1 +c2 2 +c2. (12) one of the following inequalities holds: e z 1 +e z 2 +e z < e c 1 +e c 2 +e c or e z 1 +e z 2 +e z < e c 1 +e c 2 +e c. (1) 5

6 We use the statement of Theorem 10 for the proof. Before continuing let us show that our new inequality is not a consequence of majorization and Karamata s inequality [4]. Consider z = (z 1...z n ) R n + and c = (c 1...c n ) R n + arranged already in decreasing order z 1 z 2... z n and c 1 c 2... c n. If k k n n z i c i (1 k n 1) z i = c i (14) we say that z majorizes c denoted by z c. The following result is well known [4 5][2 p.89]. If f : R R is convex then n n z c f(z i ) f(c i ). (15) A function g : R n R which satisfies z c g(z 1...z n ) g(c 1...c n ) (16) is called Schur-convex. In Theorem 8 the convex function to be considered would be f(t) = t 2. Do conditions (7) (upon rearrangement of zc R + if necessary) yield already majorization z c? This is not the case as we explain now. Let the real numbers z 1 z 2 z and c 1 c 2 c be such that e z 1 +e z 2 +e z e c 1 +e c 2 +e c e z 1 +e z 2 +e z e c 1 +e c 2 +e c z 1 +z 2 +z = c 1 +c 2 +c. These conditions do not imply the majorization z c (17) z 1 c 1 z 1 +z 2 c 1 +c 2 z 1 +z 2 +z = c 1 +c 2 +c. (18) Therefore ourinequality (i.e. z 2 1 +z2 2 +z2 c2 1 +c2 2 +c2 )doesnotfollowfrommajorization in disguise. Indeed let z 1 = z 2 = z = and c 1 = c 2 = c = 1 we have z 1 > z 2 > z and c 1 > c 2 > c together with e z 1 +e z 2 +e z = > = e c 1 +e c 2 +e c e z 1 +e z 2 +e z = > = e c 1 +e c 2 +e c z 1 +z 2 +z = c 1 +c 2 +c = 0 but the majorization inequalities (18) are not satisfied since z 1 < c 1. 6

7 Proof of the inequality Of course we may assume without loss of generality that c 1 c 2 c and z 1 z 2 z (and the same for a i d i x i y i ). The proof begins with the crucial Lemma 11. Let the real numbers a b c and x y z be such that the inequality a+b+c = x+y +z = 0 a 2 +b 2 +c 2 = x 2 +y 2 +z 2. (19) is satisfied if and only if the relation holds or equivalently if and only if holds. Proof. Let us denote by r := and we find e a +e b +e c e x +e y +e z (20) a x (21) c z (22) 2 (a2 +b 2 +c 2 ) > 0. from (19) it follows b+c = a b 2 +c 2 = 2 r2 a 2 y +z = x y 2 +z 2 = 2 r2 x 2 ( b = 1 2 a+ (r2 a 2 ) ) ( c = 1 2 a (r2 a 2 ) ) ( y = 1 2 x+ (r2 x 2 ) ) ( z = 1 2 x (r2 x 2 ) ). (2) In view of (19) and a b c x y z one can show that ax [ r 2 r] by [ r 2 r ] [ r ] cz r. (24) 2 2 Indeed let us verify the relations (24). We have r 2 a r 1 6 (a2 +b 2 +c 2 ) a 2 2 (a2 +b 2 +c 2 ) b 2 +c 2 5a 2 and a 2 2(b 2 +c 2 ) b 2 +(a+b) 2 5a 2 and (b+c) 2 2(b 2 +c 2 ) 4a 2 2ab 2b 2 0 and b 2 +c 2 2bc 2(a b)(2a+b) 0 and (b c) 2 0 7

8 which hold true since a b and 2a+b a+b+c = 0. Similarly we have r 2 b r b 2 r2 4b (a2 +b 2 +c 2 ) 5b 2 a 2 +c 2 5b 2 a 2 +(a+b) 2 2a 2 +2ab 4b 2 0 2(a b)(a+2b) 0 which holds true since a b and a+2b a+b+c = 0. Also we have r c r r 2 c 2 r (a2 +b 2 +c 2 ) c (a2 +b 2 +c 2 ) 2(a 2 +b 2 ) c 2 and 5c 2 a 2 +b 2 2(a 2 +b 2 ) (a+b) 2 and 5(a+b) 2 a 2 +b 2 (a b) 2 0 and 4a 2 +10ab+4b 2 0 (a b) 2 0 and 2(a+2b)(2a+b) 0 which hold true since a+2b a+b+c = 0 and 2a+b a+b+c = 0. One can show in the same way that x [ r 2 r] y [ r 2 r ] [ r ] z r so that (24) has been 2 2 verified. We prove now that the inequality (21) holds if and only if (22) holds. Indeed using (2) 24 and (24) we get c z a (r 2 a 2 ) x (r 2 x 2 ) a ( r + 1 ( a) ) 2 x ( r r + 1 ( x) ) 2 a x r since the function t t+ (1 t 2 ) is decreasing for t [ 1 2 1]. Let us prove next that the inequalities (20) and (21) are equivalent. To accomplish this we introduce the function f : [ r 2 r] R by ( ) ( ) f(x) = e x x+ (r +e 2 x 2 ) /2 x (r +e 2 x 2 ) /2. (25) Taking into account (2) and (24) 1 the inequality (20) can be written equivalently as f(a) f(x) (26) which is equivalent to a x since the function f defined by (25) is monotone increasing on [ r 2 r] as we show next. To this aim we denote by cosϕ := x r [ 1 x ) 2 1] i.e. ϕ := arccos( [ 0 π ]. r 8

9 the function (25) can be written as f(x) = h(rϕ) where h : (0 ) [ 0 π h(rϕ) = e rcosϕ +e rcos(ϕ+2π/) +e rcos(ϕ 2π/). ] R (27) We have to show that h(rϕ) is decreasing with respect to ϕ [ 0 π ]. We compute the first derivative [ ϕ h(rϕ) = r e rcosϕ sinϕ+e rcos(ϕ+2π/) sin(ϕ+ 2π )+ercos(ϕ 2π/) sin(ϕ ]. 2π ) The function (28) has the same sign as the function F(rϕ) := 1 r e rcosϕ h(rϕ) (29) ϕ i.e. the function F : (0 ) [ 0 π ] R given by (28) F(rϕ) = sinϕ e r sin(ϕ+π/) sin(ϕ+ 2π ) er sin(ϕ π/) sin(ϕ 2π ). (0) In order to show that F(rϕ) 0 for all (rϕ) (0 ) [ 0 π ] we remark that lim F(rϕ) = 0 for fixed ϕ [ 0 π ] and we compute rց0 r F(rϕ) = ] [e r sin(ϕ+π/) sin(ϕ+ π )sin(ϕ+ 2π ) er sin(ϕ π/) sin(ϕ π )sin(ϕ 2π ) = [ e r sin(ϕ+π/) 2( 1 ) cos(2ϕ+π)+cos π e r sin(ϕ π/) 1 ( ) ] cos(2ϕ π)+cos π 2 ( 1 ) [ = cos2ϕ+ e r sin(ϕ+π/) e sin(ϕ π/)] r since ϕ [ 0 π ] implies cos2ϕ 1 and sin(ϕ+ π 2 ) sin(ϕ π ). Consequently the function F(r ϕ) is decreasing with respect to r and for any (r ϕ) (0 ) [ 0 π ] we have that F(rϕ) lim rց0 F(rϕ) = 0. (1) From (29) and (1) it follows that h(rϕ) is decreasing with respect to ϕ [ 0 π ]. This means that f(x) is increasing as a function of x [ r 2 r] i.e. the relation (26) is indeed equivalent to a x and the proof is complete. 9

10 Consequence 12. Let the real numbers a b c and x y z be such that a+b+c = x+y +z = 0 a 2 +b 2 +c 2 = x 2 +y 2 +z 2. one of the following inequalities holds: or e a +e b +e c e x +e y +e z (2) e a +e b +e c e x +e y +e z. () The inequalities (2) and () are satisfied simultaneously if and only if a = x b = y and c = z. Proof. According to Lemma 11 the inequality (2) is equivalent to while the inequality () is equivalent to a x (4) a x. (5) Since one of the relations (4) and (5) must hold we have proved that one of the inequalities (2) and () is satisfied. They are simultaneously satisfied if and only if both (4) and (5) hold true i.e. a = x (and consequently b = y c = z). Consequence 1. Let the real numbers a b c and x y z be such that a+b+c = x+y +z = 0 a 2 +b 2 +c 2 = x 2 +y 2 +z 2 and e a +e b +e c = e x +e y +e z. we have a = x b = y and c = z. Proof. Since by hypothesis e a +e b +e c e x +e y +e z holds we can apply the Lemma 11 to deduce a x and c z. On the other hand by virtue of the inverse inequality e x + e y + e z e a + e b + e c and Lemma 11 we obtain x a and z c. In conclusion we get a = x c = z and b = y. Proof of Theorem 10. In order to prove (1) we define the real numbers c 2 1 t i = kz i (i = 12) where k = +c2 2 +c2 > 1. (6) z1 2 +z2 2 +z 2 we have t 1 +t 2 +t = c 1 +c 2 +c = 0 and t 2 1 +t2 2 +t2 = c2 1 +c2 2 +c2. (7) 10

11 If we apply the Consequence 12 for the numbers c 1 c 2 c and t 1 t 2 t then we obtain that e t 1 +e t 2 +e t e c 1 +e c 2 +e c or (8) e t 1 +e t 2 +e t e c 1 +e c 2 +e c. In what follows let us show that e z 1 +e z 2 +e z < e t 1 +e t 2 +e t. (9) 2 Using the notations ρ := (z2 1 +z2 2 +z) 2 and cosζ := z 1 ρ [ 1 ( 2 1] z1 ) i.e. ζ := arccos [ 0 π ] ρ 2 we have kρ := (t2 1 +t 2 2 +t 2 ) and cosζ = t 1. With the help of the function h kρ defined in (27) we can write the inequality (9) in the form e ρcosζ +e ρcos(ζ+2π/) +e ρcos(ζ 2π/) < e kρcosζ +e kρcos(ζ+2π/) +e kρcos(ζ 2π/) or h(ρζ) < h(kρζ) (ρζ) (0 ) [ 0 π ] k > 1. (40) The relation (40) asserts that the function h defined in (27) is increasing with respect to the first variable r (0 ). To show this we compute the derivative r h(rϕ) = ercosϕ cosϕ+e rcos(ϕ+2π/) cos(ϕ+ 2π )+ercos(ϕ 2π/) cos(ϕ 2π ). (41) By virtue of the Chebyshev s sum inequality we deduce from (41) that h(rϕ) > 0. (42) r Indeed the Chebyshev s sum inequality [2 2.17] asserts that: if a 1 a 2... a n and b 1 b 2... b n then n ( n )( n n a k b k a k b k ). k=1 k=1 Inourcase wederivethefollowingresult: foranyrealnumbers xyz suchthatx+y+z = 0 the inequality k=1 xe x +ye y +ze z 1 (x+y +z)(ex +e y +e z ) = 0 (4) holds true with equality if and only if x = y = z = 0. 11

12 Applying the result (4) to the function (41) we deduce the relation (42). This means that h(r ϕ) is an increasing function of r i.e. the inequality (40) holds and hence we have proved (9). One can show analogously that the inequality e z 1 +e z 2 +e z < e t 1 +e t 2 +e t (44) is also valid. From (8) (9) and (44) it follows that the assertion (1) holds true. Thus the proof of Theorem 10 is complete. Since the statements of the Theorems 8 and 10 are equivalent we have proved also the inequality (8). Remark 14. The inequality (8) becomes an equality if and only if z i = c i i = 12. Proof. Indeed assumethatz 2 1+z 2 2+z 2 = c 2 1+c 2 2+c 2. wecanapplytheconsequence 12 and we deduce that e z 1 +e z 2 +e z e c 1 +e c 2 +e c or e z 1 +e z 2 +e z e c 1 +e c 2 +e c. (45) Taking into account (7) 12 in conjunction with (45) we find e z 1 +e z 2 +e z = e c 1 +e c 2 +e c or e z 1 +e z 2 +e z = e c 1 +e c 2 +e c. (46) By virtue of (46) we can apply the Consequence 1 to derive z 1 = c 1 and consequently z 2 = c 2 z = c. Let us prove the following version of the inequality (6) for two pairs of numbers d 1 d 2 and x 1 x 2 : Remark 15. If the real numbers d 1 d 2 > 0 and x 1 x 2 > 0 are such that then the inequality x 2 1 +x2 2 d2 1 +d2 2 and x 1 x 2 = d 1 d 2 = 1 (47) holds true. Note that the additional condition is automatically fulfilled. (logx 1 ) 2 +(logx 2 ) 2 (logd 1 ) 2 +(logd 2 ) 2 (48) 1 x x d 2 1 d

13 Proof. Since x 1 x 2 = d 1 d 2 = 1 and d 1 d 2 > 0 x 1 x 2 > 0 we have x 1 1 d 1 1 and logx 1 = logx 2 0 logd 1 = logd 2 0 so that the inequality (48) is equivalent to logx 1 logd 1 i.e. we have to show that x 1 d 1. Indeed if we insert x 2 = 1 x 1 and d 2 = 1 d 1 into the inequality (47) 1 then we find x x 2 1 d d 2 1 which means that x 1 d 1 since the function t t t 2 is increasing for t [1 ). This completes the proof. Alternative proof of Remark 15. Let x = d = 1. (47) implies x x2 2 + x2 d 2 1 +d 2 2 +d 2 and x 1 x 2 x = d 1 d 2 d = 1 as well as x 2 1 x2 2 +x2 2 x2 +x2 1 x2 = 1+x2 2 +x2 1 1+d2 2 +d2 1 = d2 1 d2 2 +d2 2 d2 +d2 1 d2 (49) because x 2 1 x2 2 = 1 = d2 1 d2 2 and Theorem 6 provides the assertion. 4 Some counterexamples for weakened assumptions Example 16. Unlike in the 2D case in Remark 15 for two triples of numbers the second condition (18) 2 of Theorem 2 namely y 1 y 2 +y 2 y +y 1 y a 1 a 2 +a 2 a +a 1 a cannot be removed. Let y 1 = e 6 y 2 = 1y = e 6 a 1 = e 4 a 2 = e 4 a = e 8. y 1 y 2 y = a 1 a 2 a = 1 and but y 1 +y 2 +y > e 6 > e 2 e 4 > e 4 > a 1 +a 2 +a (logy 1 ) 2 +(logy 2 ) 2 +(logy ) 2 = < = (loga 1 ) 2 +(loga 2 ) 2 +(loga ) 2. Example 17. The condition y 1 y 2 y = a 1 a 2 a cannot be weakened to y 1 y 2 y a 1 a 2 a. Indeed let y 2 = y = a 1 = a 2 = 1 y 1 = e a = e 2. y 1 +y 2 +y = e e 2 = a 1 +a 2 +a y 1 y 2 +y 1 y +y 2 y = e+e+1 1+e 2 +e 2 = a 1 a 2 +a 1 a +a 2 a y 1 y 2 y = e e 2 = a 1 a 2 a. 1

14 But nevertheless (logy 1 ) 2 +(logy 2 ) 2 +(logy ) 2 = < = (loga 1 ) 2 +(loga 2 ) 2 +(loga ) 2. A counterexample for the two variable case can be constructed analogously. Example 18. Even with an analogous condition the inequality (4) does not hold for n = 4 numbers (without further assumptions). Indeed let y 1 = ey 2 = y = e 7 y 4 = e 15 a 1 = a 2 = e 6 a = e 7 a 4 = e 19. y 1 y 2 y y 4 = a 1 a 2 a a 4 = 1. Also y 1 +y 2 +y +y 4 = e+e 7 +e 7 +e 15 > 0+e 7 +2e 6 +e 19 = a 1 +a 2 +a +a 4. Furthermore and y 1 y 2 +y 1 y +y 1 y 4 +y 2 y +y 2 y 4 +y y 4 = e 8 +e 8 +e 14 +e 14 +e 8 +e 8 a 1 a 2 +a 1 a +a 1 a 4 +a 2 a +a 2 a 4 +a a 4 = e 12 +e 1 +e 1 +e 1 +e 1 +e 12. Since e 2 > 2e+1 we have e 14 > e 1 +e 1 +e 12 and therefore y 1 y 2 +y 1 y +y 1 y 4 +y 2 y +y 2 y 4 +y y 4 a 1 a 2 +a 1 a +a 1 a 4 +a 2 a +a 2 a 4 +a a 4. Nevertheless for the sum of squared logarithms the reverse inequality holds true. (logy 1 ) 2 +(logy 2 ) 2 +(logy ) 2 +(logy 4 ) 2 = = 24 < 482 = = (loga 1 ) 2 +(loga 2 ) 2 +(loga ) 2 +(loga 4 ) 2 Example 19. The inequality (4) does not remain true either if the function log(y) is replaced by its linearization (y 1). Indeed let y 1 = 9 y 2 = 5 y = 1 a 45 1 = 10 a 2 = 1 a = y 1 +y 2 +y > 14 > 11.1 = a 1 +a 2 +a and y 1 y 2 +y 1 y +y 2 y > = a 1 a 2 +a 1 a +a 2 a. But ( ) 2 44 (y 1 1) 2 +(y 2 1) 2 +(y 1) 2 = ( ) 2 9 < 81 < = (a 1 1) 2 +(a 2 1) 2 +(a 1)

15 5 Conjecture for arbitrary n The structure of the inequality in dimensions n = 2 and n = and extensive numerical sampling strongly suggest that the inequality holds for all n N if the n corresponding conditions are satisfied more precisely in terms of the elementary symmetric polynomials Conjecture 20. Let n N and y i a i > 0 for i = 1...n. If for all i = 1...n 1 we have e i (y 1...y n ) e i (a 1...a n ) and e n (y 1...y n ) = e n (a 1...a n ) then 6 Applications n (logy i ) 2 n (loga i ) 2. The investigation in this paper has been motivated by some recent applications. The new sum of squared logarithm inequality is one of the fundamental tools in deducing a novel optimality result in matrix analysis and the conditions in the form () had been deduced in the course of that work. Optimality in the matrix problem suggested the sum of squared logarithm inequality. Indeed based on the present result in [6] it has been shown that for all invertible Z C and for any definition of the matrix logarithm as possibly multivalued solution X C of expx = Z it holds min Q Q=I logq Z 2 F = logu p Z 2 = F logh 2 F min Q Q=I symlogq Z 2 F = symlogu p Z 2 = F logh 2 F (50) where symx = 1 2 (X+X ) is the Hermitian part of X C and U p is the unitary factor in the polar decomposition of Z into unitary and Hermitian positive definite matrix H Z = U p H. (51) This result (50) generalizes the fact that for any complex logarithm and for all z C\{0} min log C [e iϑ z] 2 = log R z 2 ϑ ( ππ] min Relog C [e iϑ z] 2 = log R z 2. (52) ϑ ( ππ] The optimality result (50) can now also be viewed as another characterization of the unitary factor in the polar decomposition. In addition in a forthcoming contribution [7] we use (50) to calculate the geodesic distance of the isochoric part of the deformation 15

16 gradient F detf 1 on SL(R) to the effect that SL( R)) to SO( R)) in the canonical left-invariant Riemannian metric dist 2 geod ( F SO(R)) = detf 1 dev log F T F 2 F. (5) where dev X = X 1 trx I is the orthogonal projection of X R to trace free matrices. Thereby we provide a rigorous geometric justification for the preferred use of the Hencky-strain measure log F T F in nonlinear elasticity and plasticity theory []. References 2 F [1] K. Guan Schur-convexity of the complete elementary symmetric functions J. Inequalities Appl. (2006) doi: /jia/2006/ [2] G.H. Hardy J.E. Littlewood and G. Pólya Inequalities The University Press 194. [] H. Hencky Über die Form des Elastizitätsgesetzes bei ideal elastischen Stoffen Z. Techn. Physik 9 (1928) [4] J. Karamata Sur une inégalité relative aux fonctions convexes Publ. Math. Univ. Belgrad 1 (192) [5] A.R. Khan N. Latif and J. Pečarić Exponential convexity for majorization J. Inequalities Appl. 105 (2012) 1 1. [6] P. Neff and Y. Nagatsukasa The unitary polar factor Q = U p minimizes log(q A) 2 and sym(log(q A) 2 for the Frobenius- and spectral matrix norm in three dimensions submitted (201). [7] P. Neff and F. Osterbrink Appropriate strain measures: The Hencky shear strain energy devlog F T F 2 measures the geodesic distance of the isochoric part of the deformation gradient F SL() to SO() in the canonical left invariant Riemannian metric on SL(). in preparation (201). [8] I. Roventa A note on Schur-concave functions J. Inequalities Appl. 159 (2012) 1 9. [9] J.M. Steele The Cauchy-Schwarz Master Class: An introduction to the art of mathematical inequalities Cambridge University Press Cambridge