Norm inequalities related to the Heinz means

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1 Gao and Ma Journal of Inequalities and Applications (08) 08:303 R E S E A R C H Open Access Norm inequalities related to the Heinz means Fugen Gao * and Xuedi Ma * Correspondence: gaofugen08@6.com College of Mathematics and Information Science, Henan Normal University, Xinxiang, China Abstract Let (I, ) be a two-sided ideal of operators equipped with a unitarily invariant norm. We generalize the results of Kapil s, using a new contractive map in I to obtain a norm inequality. And we give a new inequality, which is a comparison between the Heinz means and other related inequalities; moreover, we will obtain some correlative conclusions. Keywords: Norm inequalities; Contractive maps; Unitarily invariant norm Introduction and preliminaries In this paper, let B(H) denote the algebra of all bounded linear operators on a complex separable Hilbert space, B(H) + denote the cone of positive operators and K(H)denotethe ideal of compact operators in B(H). And (I, ) is a two-sided ideal of B(H) equipped with a unitarily invariant norm. We shall denote this by I instead of (I, )for convenience. For any compact operator A K(H), let S (A), S (A),... be the eigenvalues of A = (A A) arranged in decreasing order. If A M n, which is the algebra of all n n matrices over C, wetakes k (A) =0fork > n. A unitarily invariant norm in K(H) isamap : K(H) [0, ] givenby A = g(s(a)), A K(H), where g is a symmetric gauge function, cf. [6, 9]. The Schatten p-norms A p =( Sp (A)) p for p aresignificant examples of the unitarily invariant norm, and is a special unitarily invariant norm which is the Hilbert Schmidt norm defined, for A M n, as follows: A = i, a i = tr ( A A ). It is well known that the arithmetic geometric mean inequality ab a + b for positive numbers a and b has been generalized in various directions. Bhatia et al. [] haveobtainedtheresultthatifa, B, andx are n n matrices with A and B are positive definite, then A XB AX + XB. The Author(s) 08. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Gao and MaJournal of Inequalities and Applications (08) 08:303 Page of 7 The Heinz means H v (a, b)= a v b v + a v b v is an interpolation between the arithmetic and geometric means for a, b 0andv [0, ]. It is easy to see that H v is symmetric and convex function for v [0, ] and attains its minimum at v = ;thuswehave ab Hv (a, b) a + b. Thematrixversionhasbeenprovedin[]: if A, B and X are positive definite, then for every unitarily invariant norm the function g(v)= A v XB v + A v XB v is convex on [0, ], and attains its minimum at v =.Thuswehave A XB A v XB v + A v XB v AX + XB. The Heron means is defined by F α (a, b)=( α) ab + α a + b for 0 α, see []. This family is the linear interpolation between the geometric and the arithmetic mean. Clearly, F α F β whenever α β. Bhatia and Davis have proved that the inequality ( ) ( ) ( α)a XB AX + XB + α ( β)a XB AX + XB + β is always true for 0 α β, β,andthisrestrictiononβ is necessary in []. The Heinz means and the Heron means satisfy the inequality H v (a, b) F α(v) (a, b) for 0 v. And α(v) = 4(v v ), this is a convex function, its minimum value is α( ) = 0, and its maximum value is α(0) = α() =. In [0] the authors have presented the result that A v XB v + A v XB v ( ) ( α)a XB AX + XB + α for 4 v 3 4 and α [, ] and they used the properties of contractive map on I to prove it. A different version of the Heinz inequality, A α XB α A α XB α α AX XB, for α [0, ] was proved by Bhatia and Davis [3] in 995.

3 Gao and MaJournal of Inequalities and Applications (08) 08:303 Page 3 of 7 A further generalization, namely +t A α XB α + A α XB α AX + XB + ta XB, was proved for t [, ], and α [ 4, 3 ]. For more details refer to [3]and[4]. 4 Contractive maps on I play a key role to prove the inequalities, more details refer to [4, 5, 7, 8, ]. This in turn proves the corresponding Schur multiplier maps to be contractive. We use the ideals to establish the fact that some special maps on I are contractive. Our paper consists of three parts. In the second part, we will give a new norm inequality whichisprovedbythepropertiesofcontractivemapsoni.inthethirdpart,wewillpresent a inequality related to the Heinz means and we notice that this inequality is a comparison between the Heinz means and the geometric mean. Our results are extensions of some previous conclusions about norm inequalities. Norm inequalities with contractive maps Let L X, R Y denote the left and right multiplication maps on B(H), respectively, that is, L X (T)=XT and R Y (T)=TY and we have e L X+R Y (T)=e X Te Y. () Neeb []hasprovedthat(l X ±R Y ) sin(l X ±R Y )isanexpansivemaponi by using the Weierstrass factorization theorem. We use X and Y to denote two selfadoint operators in B(H)andD = L X R Y. The following proposition is a result for a contractive map in I; for more details refer to [0]. Proposition. ([0, Proposition.4]) Let 0 r sors r 0, then each of the following operator maps is a contraction on I. () (+t) cosh( r D) t+cosh(sd) () t+cosh(rd) t+cosh(sd) for t. for t. (3) cosh(rd) cosh(sd). In particular r (4) s sinh(rd) r sinh(sd) cosh(rd) cosh(sd) dr = sd. The case r =0would become sinh(rd) rd cosh(sd) sinh(sd). is a contraction. Corollary. ([0, Corollary.5]) The following maps are contractive on I. (t+) cosh((v )D) () for t+cosh D 4 v 3, and t. 4 cosh(rd) () cosh(s D)+cosh(s D) for 0 r max{ s +s, s s } or min{ s +s, s s } r 0. (s (3) +s ) sinh(rd) r(sinh(s D)+sinh(s D)) for 0 r s +s or s +s r 0. Corollary.3 Let 0 v and t. Then the map (t+) cosh((v )D) t+cosh(d) is contractive on I. Proof This map is a special case of () of Proposition. when s =. From the condition 0 v, we can obtain v. Letting r =v,wegetthedesiredresult. With these above results about contractive maps in I, we obtain a new norm inequality which derives from the Heinz means and other related inequalities. Let R be an invertible operator in B(H) +, then there exists a selfadoint operator S B(H) suchthatr = e S.To avoid repetitions, we denote the two invertible operators A and B in B(H) + by e X and

4 Gao and MaJournal of Inequalities and Applications (08) 08:303 Page 4 of 7 e Y,respectively,whereX and Y in B(H) are selfadoint. The corresponding operator map L X R Y is denoted by D. Theorem.4 Let v [0, ] and α [, ). Supposed that A and B are any two invertible operators in B(H) +, X I, then A v XB v + A v XB v ( α)a XB + α A 3 XB + A XB 3. () Proof Put A XB = T and use () we can notice that A v XB v + A v XB v = A v A XB B v + A v A XB B v = A v TB v + A v TB v = e(v )(L X R Y ) T + e (v )(L X R Y ) T = cosh ( (v )D ) T. (3) And we also have ( α)a XB + α A 3 XB + A XB 3 =( α)t + α AA XB B + A A XB B =( α)t + α ATB + A TB =( α)t + α e(l X R Y ) T + e (L X R Y ) T = [ ( α)+αcosh(d) ] T. (4) Setting = α in (4) and applying Corollary.3, we can obtain ()from(3)and(4). +t 3 Norm inequality related to the Heinz means According to the above results, we set H v (a, b)= av b v + a v b v, G v (a, b)=( v)a b + v a 3 b + a b 3 Then we have the following result, which is a interpolation of H v and G v..

5 Gao and MaJournal of Inequalities and Applications (08) 08:303 Page 5 of 7 Theorem 3. Let a, b >0and v [,].Then ( H (a, b) ) v H v (a, b) G v (a, b). (5) H (a, b) Proof Without loss of generality, we can suppose that a =. Then the inequality (5) can be simplified to b v + b v To prove this inequality, let ( ) b v [ ( v) b + v b + b 3 ]. +b f (v)=log ( b v + b v) ( v) log b +b log[ ( v) b + v ( b + b 3 )]. Calculations show that f (v)= 4bv b v log b (b + b 3 b ) + (b v + b v ) [( v)b + v(b + b 3 )]. We can notice that f (v) 0for v, thus f is convex on [,].Thenwehavef (v) max{f ( ), f ()}. By calculation, we have f ( ) ( b 4 ( + b) ) = log b + b + b 0 3 because of b 4 (+b) b + b + b 3 (b >0),whichequals b 4 ( + b) b + b + b 3,that is, 8( + b)b 3 6b + b 4 +4b 3 +4b +. Setting b = x (x >0),thuswehavex 8 +4x 6 8x 5 +6x 4 8x 3 +4x + 0, that is, (x ) (x 6 +x 5 +7x 4 +4x 3 +7x +x +) 0, which is always true when x >0,thenwe get the desired results. In the same way, ( ) +b f () = log 0 b + b 3 +b because of (b >0),whichequals+b b b +b 3 + b 3,thatis(b )(b 3 ) 0, whichisalwaystruewhenb >0.Hencethevaluesoff ( )andf() are both less than 0, that is, f (v) 0. Then we complete the proof. Remark 3. The inequality which we obtained from Theorem 3. can also be written as H v (a, b) ( ) ab v G v, a + b and we notice that it is a new comparison between the Heinz means and the geometric means.

6 Gao and MaJournal of Inequalities and Applications (08) 08:303 Page 6 of 7 Corollary 3.3 Let A, B M n +, X M n and v. If there are two positive numbers m, Msuchthatm A, B M, then A v XB v + A v XB v ( ) m + M v ( v)a XB + v A 3 XB + A XB 3 mm. Proof Let A = U diag(λ i )U and B = V diag(μ )V be the spectral decompositions of A and B, respectively. Letting U XV = Y,wehave A v XB v + A v XB v = U diag(λv i )Y diag(μ v )+diag(λ v i )Y diag(μ v ) = U [λv i μ v + λ v i μ v ] [y i] V, (6) where denotes the Schur product. Applying (5) we have A v XB v + A v XB v = ( λ v i μ v + λ v i i, i, ( λi + μ λ i λ μ v ) y i ) (v ) ( ( v)λ i μ V + v λ 3 i μ + λ i μ 3 ) y i ( ) m + M (v ) ( v)a XB + v A 3 XB + A XB 3 mm, (7) wherewehaveusedthefactthatm λ i, μ M to obtain the last inequality. Funding This research is supported by the National Natural Science Foundation of China (7054), the Natural Science Foundation of the Department of Education, Henan Province (6A0003), (9A000), and the Program for Graduate Innovative Research of Henan Normal University (No.YL070). Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed to each part of this work equally, and they all read and approved the final manuscript. Publisher s Note Springer Nature remains neutral with regard to urisdictional claims in published maps and institutional affiliations. Received: 9 July 08 Accepted: 8 October 08 References. Bhatia, R.: Interpolating the arithmetic geometric mean inequality and its operator version. Linear Algebra Appl. 43, (006). Bhatia, R., Davis, C.: More matrix forms of the arithmetic geometric mean inequality. SIAM J. Matrix Anal. Appl. 4, 3 36 (993) 3. Bhatia, R., Davis, C.: A Cauchy Schwarz inequality for operators with applications. Linear Algebra Appl. 3/4, 9 9 (995)

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