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Research ArticlePositivity Betweenness and Strictness of Operator Means
Pattrawut Chansangiam
Department of Mathematics Faculty of Science King Mongkutrsquos Institute of Technology Ladkrabang Bangkok 10520 Thailand
Correspondence should be addressed to Pattrawut Chansangiam kcpattrakmitlacth
Received 5 March 2015 Accepted 7 May 2015
Academic Editor Sergei V Pereverzyev
Copyright copy 2015 Pattrawut Chansangiam This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
An operator mean is a binary operation assigned to each pair of positive operators satisfying monotonicity continuity fromabove the transformer inequality and the fixed-point property It is well known that there are one-to-one correspondencesbetween operator means operator monotone functions and Borel measures In this paper we provide various characterizationsfor the concepts of positivity betweenness and strictness of operator means in terms of operator inequalities operator monotonefunctions Borel measures and certain operator equations
1 Introduction
The concept of means a natural notion in mathematicsplays important roles in mathematics itself computer sci-ence statistics various branches in science engineering andeconomics This concept was developed since the ancientGreeks until the last century by many mathematicians (see[1]) Nowadays according to the definition of a mean forpositive real numbers in [1] a mean 119872 is defined to besatisfied by the following properties
(iii) strict if it is strict at both the right and the left
This paper focuses on means for positive operators ona Hilbert space Let 119861(H) be the algebra of bounded linearoperators on a Hilbert spaceH The set of positive operators
on H is denoted by 119861(H)+ Denote the spectrum of an
operator119883 by Sp(119883) For self-adjoint operators119860 119861 isin 119861(H)the partial order 119860 ⩽ 119861 indicates that 119861 minus 119860 isin 119861(H)
+ If119860 isin 119861(H)
+ is invertible then we write 119860 gt 0A starting point for the theory of operator means is the
presence of the notion of parallel sum in electrical networkanalysis (see [2]) A connection is a binary operation 120590
assigned to each pair of operators in 119861(H)+ such that the
following conditions are satisfied for all 119860 119861 119862119863 isin 119861(H)+
darr 119883 indicates that 119883119899is a decreasing sequence
(with respect to the partial order) and 119883119899converges
strongly to119883
This definition was modelled from significant properties ofthe parallel sum by Kubo and Ando in [3] Two trivialexamples are the left-trivial mean 120596
119897 (119860 119861) 997891rarr 119860 and the
right-trivial mean 120596119903 (119860 119861) 997891rarr 119861 See [4 Section 3] and
[5] for more information about operator connections Fromthe transformer inequality every connection is congruenceinvariant in the sense that for each 119860 119861 ⩾ 0 and 119862 gt 0 wehave
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2015 Article ID 851568 5 pageshttpdxdoiorg1011552015851568
2 Abstract and Applied Analysis
Amean in Kubo-Ando sense is a connection 120590with fixed-point property 119860120590119860 = 119860 for all 119860 ⩾ 0 The class of Kubo-Ando means cover many well-known means in practice forexample
(iv) logarithmic mean (119860 119861) 997891rarr 11986012
119891(119860minus12
119861119860minus12
)11986012
where the function 119891 is given by 119891(119909) = (119909minus 1) log119909for each 119909 isin R+ equiv [0infin) 119891(0) equiv 0 and 119891(1) equiv 1
A summary of Kubo-Ando theory is given in terms ofone-to-one correspondences between operator connectionson 119861(H)
+ operator monotone functions from R+ to R+and finite Borel measures on [0 1] Recall that a continuousfunction 119891 R+ rarr R+ is said to be operator monotone if
for all positive operators 119860 119861 isin 119861(H) and for all Hilbertspaces H This concept was introduced in [6] see also [7Chapter V] [4 Section 2] and [8] A connection 120590 on 119861(H)
+
can be characterized via operator monotone functions asfollows
Theorem 1 (see [3Theorem 32]) Given a connection 120590 thereis a unique operator monotone function 119891 R+ rarr R+ satis-fying
Moreover the map 120590 997891rarr 119891 is a bijection
We call 119891 the representing function of 120590 A connectionalso has a canonical characterization with respect to a Borelmeasure via a meaningful integral representation as follows
Theorem 2 Given a finite Borel measure 120583 on [0 1] thebinary operation
In this paper we provide various characterizations for theconcepts of positivity betweenness and strictness of operatormeans in terms of operator inequalities operator monotonefunctions Borel measures and certain operator equationsIt turns out that every mean satisfies the positivity propertyThe betweenness is a necessary and sufficient condition for aconnection to be a mean A mean is strict at the left (right)if and only if it is not the left-trivial mean (the right-trivialmean resp)
2 Positivity
We say that a connection 120590 satisfies the positivity property if
(5) rArr (3) (8) rArr (6) rArr (3) and (10) rArr (9) areclear Using the integral representations in Theorem 2 it isstraightforward to verify that the representing function of thezero connection 0 (119860 119861) 997891rarr 0 is the constant function119891 equiv 0and its associated measure is the zero measure Hence wehave the equivalences (3) hArr (9) hArr (11)
(9) rArr (10) Assume 119891 = 0 Suppose that there is 119886 gt 0such that119891(119886) = 0Then119891(119909) = 0 for all 119909 ⩽ 119886The concavityof 119891 implies that 119891(119909) = 0 for all 119909 ⩾ 119886 Hence 119891 = 0 acontradiction
(5) rArr (7) Assume (5) Let 119860 ⩾ 0 and 119861 gt 0 be such that119860120590119861 = 0 Then
0 = 11986112
(119861minus12
119860119861minus12
120590119868) 11986112 (14)
and 119861minus12
119860119861minus12
120590119868 = 0 Now (5) yields 119861minus12119860119861minus12
= 0 thatis 119860 = 0
(6) rArr (8) It is similar to (5) rArr (7)(10) rArr (1) Assume that 119891(119909) gt 0 for all 119909 gt 0 Since
Sp(119891(119860)) = 119891(Sp(119860)) by spectral mapping theorem we have119891(119860) gt 0 for all 119860 gt 0 Hence for each 119860 gt 0 and 119861 gt 0
119860120590119861 = 11986012
119891 (119860minus12
119861119860minus12
)11986012
gt 0 (15)
(10) rArr (4) Assume (10) Let119860 ⩾ 0 be such that119860120590119860 = 0Note that
119860120590119860 = lim120598darr0
119860120598120590119860120598= lim120598darr0
11986012120598
(119868120590119868) 11986012120598
= lim120598darr0
119891 (1) 119860120598= 119891 (1) 119860
(16)
here 119860120598equiv 119860 + 120598119868 Since 119891(1) gt 0 we have 119860 = 0
(10) rArr (5) Assume (10) Let119860 ⩾ 0 be such that119860120590119868 = 0Then 119892(119860) = 0 where 119892 is the representing function of thetranspose of 120590 We see that 119892(119909) gt 0 for 119909 gt 0The injectivityof functional calculus implies that 119892(120582) = 0 for all 120582 isin Sp(119860)We conclude that Sp(119860) = 0 that is 119860 = 0
(10) rArr (6) Assume (10) Let 119860 ⩾ 0 be such that 119868120590119860 = 0Then 119891(119860) = 0 By the injectivity of functional calculus wehave119891(120582) = 0 for all 120582 isin Sp(119860) Assumption (10) implies thatSp(119860) = 0 Thus 119860 = 0
Remark 5 It is not true that 120590 = 0 implies the condition thatfor all 119860 ⩾ 0 and 119861 ⩾ 0 119860120590119861 = 0 implies 119860 = 0 or 119861 = 0Indeed take 120590 to be the geometric mean and
119860 = (
1 00 0
)
119861 = (
0 00 1
)
(17)
3 Betweenness
We say that a connection 120590 satisfies the betweenness propertyif for each 119860 ⩾ 0 and 119861 ⩾ 0
ByTheorem 4 every mean enjoys the positivity propertyIn fact the betweenness property is a necessary and sufficientcondition for a connection to be a mean
Theorem 6 The following statements are equivalent for aconnection 120590 with representing function 119891
(1) 120590 is a mean(2) 120590 satisfies the betweenness property(3) for all 119860 ⩾ 0 119860 ⩽ 119868 rArr 119860 ⩽ 119860120590119868 ⩽ 119868(4) for all 119860 ⩾ 0 119868 ⩽ 119860 rArr 119868 ⩽ 119868120590119860 ⩽ 119860(5) for all 119905 ⩾ 0 1 ⩽ 119905 rArr 1 ⩽ 119891(119905) ⩽ 119905(6) for all 119905 ⩾ 0 119905 ⩽ 1 rArr 119905 ⩽ 119891(119905) ⩽ 1(7) for all 119860 ⩾ 0 and 119861 ⩾ 0 119860 ⩽ 119861 rArr 119860 ⩽ 119860120590119861 ⩽
119861(8) for all 119860 ⩾ 0 119860 ⩽ 119868 rArr 119860 ⩽ 119860120590119868 ⩽ 1(9) for all 119860 ⩾ 0 119868 ⩽ 119860 rArr 1 ⩽ 119868120590119860 ⩽ 119860(10) the only solution 119883 gt 0 to the equation 119883120590119883 = 119868 is
119883 = 119868(11) for all 119860 gt 0 the only solution 119883 gt 0 to the equation
Therefore 120590 is a mean byTheorem 3(7) rArr (9) rArr (1) It is similar to (7) rArr (8) rArr (1)(1) rArr (11) Let 119860 gt 0 Consider 119883 gt 0 such that 119883120590119883 =
119860 Then by the congruence invariance of 120590 we have
119883 = 11988312
(119868120590119868)11988312
= 119883120590119883 = 119860 (21)
(2) rArr (5) If 119905 ⩾ 1 then 119868 ⩽ 119868120590(119905119868) ⩽ 119905119868 which is 119868 ⩽
119891(119905)119868 ⩽ 119905119868 that is 1 ⩽ 119891(119905) ⩽ 119905(5) rArr (1) We have 119891(1) = 1(2) rArr (6) rArr (1) It is similar to (2) rArr (5) rArr (1)
Remark 7 For a connection 120590 and 119860 119861 ⩾ 0 the operators119860 119861 and119860120590119861 need not be comparableThe previous theoremtells us that if 120590 is a mean then the condition 0 ⩽ 119860 ⩽ 119861
guarantees the comparability between 119860 119861 and 119860120590119861
4 Strictness
We consider the strictness of Kubo-Ando means as that forscalar means in [1]
4 Abstract and Applied Analysis
Definition 8 A mean 120590 on 119861(H)+ is said to be
(i) strict at the left if for each 119860 gt 0 and 119861 gt 0
(iii) strict if it is strict at both the right and the left
In order to prove the next two lemmas recall thefollowing facts if 119891 R+ rarr R+ is operator monotone then
(i) 119891 is operator concave and hence concave in usualsense (see [9] or [4 Corollary 254])
(ii) 119892(119909) = 119909119891(119909) is convex in usual sense (see [3 Lemma52])
(iii) ℎ(119909) = 119909119891(119909) is operator monotone on (0infin) (see[9] or [4 Corollary 256])
Lemma 9 If 119891 R+ rarr R+ is an operator monotone functionsuch that 119891 is a constant on an interval [119886 119887] with 119886 lt 119887 then119891 is a constant on R+
Proof Assume that 119891(119909) = 119896 for all 119886 ⩽ 119909 ⩽ 119887 The case119886 = 0 is done by using the monotonicity and concavity of 119891Consider the case 119886 gt 0 The monotonicity and concavity of119891 imply that 119891(119909) = 119896 for all 119909 ⩾ 119887 If 119896 = 0 then 119891(119909) = 119896
on [0 119886] by the monotonicity of 119891 Consider the case 119896 gt 0and suppose there is an 1199090 isin [0 119886) such that 119891(1199090) lt 119896 Thenthe slope of the line segment joining the point (1199090 1199090119891(1199090))and the point (119886 119886119891(119886)) is greater than 119896 This contradicts theconvexity of the function 119909119891(119909)
Lemma 10 If 119891 R+ rarr R+ is an operator monotone func-tion such that 119891(119909) = 119898119909 + 119888 for some 119898 gt 0 and 119888 ⩾ 0 on aninterval [119886 119887] with 119886 lt 119887 then 119891(119909) = 119898119909 + 119888 on R+
Proof If there is 1199090 gt 0 such that 119891(1199090) = 0 then 119891 equiv 0 byLemma 9 Suppose that 119891(119909) gt 0 for all 119909 gt 0 For simplicityassume that 119891(119909) = 119909 for all 119886 ⩽ 119909 ⩽ 119887 Then the functionℎ(119909) = 119909119891(119909) is operator monotone on (0infin) and henceon R+ by continuity Note that ℎ(119909) = 1 on [119886 119887] Lemma 9implies that ℎ(119909) = 1 on R+ that is 119891(119909) = 119909 on R+
Theorem 11 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the left(2) 120590 is not the left-trivial mean(3) for all 119860 ⩾ 0 119868120590119860 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119860120590119868 = 119860 rArr 119860 = 119868(5) for all 119860 gt 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119868120590119860 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119868120590119860 ⩽ 119868 rArr 119860 ⩽ 119868
(8) for all 119860 gt 0 119860 ⩽ 119860120590119868 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119860120590119868 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 gt 0 and 119861 ⩾ 0 119860 ⩽ 119860120590119861 rArr 119860 ⩽ 119861(11) for all 119860 gt 0 and 119861 ⩾ 0 119860120590119861 ⩽ 119860 rArr 119861 ⩽ 119860(12) 119891 is not the constant function 119909 997891rarr 1(13) for all 119909 ⩾ 0 119891(119909) = 1 rArr 119909 = 1(14) for all 119909 ⩾ 0 119891(119909) ⩾ 1 rArr 119909 ⩾ 1(15) for all 119909 ⩾ 0 119891(119909) ⩽ 1 rArr 119909 ⩽ 1(16) 120583 is not the Dirac measure at 0
Proof It is clear that (5) rArr (1) and each of (1)(4) and (6)ndash(11) implies (2) Also each of (13)ndash(15) implies(12)
(2) rArr (3) Let119860 ⩾ 0 be such that 119868120590119860 = 119868Then 119891(119860) = 119868
and hence 119891(120582) = 1 for all 120582 isin Sp(119860) Suppose that 120572 equiv
inf Sp(119860) lt 119903(119860) where 119903(119860) is the spectral radius of119860 Then119891(119909) = 1 for all 119909 isin [120572 119903(119860)] It follows that 119891 equiv 1 on R+ byLemma 9This contradicts assumption (2) We conclude that120572 = 119903(119860) that is Sp(119860) = 120582 for some 120582 ⩾ 0 Suppose nowthat 120582 lt 1 Since 119891(1) = 1 we have that 119891 is a constant onthe interval [120582 1] Again Lemma 9 implies that 119891 equiv 1 onR+a contradiction Similarly 120582 gt 1 gives a contradiction Thus120582 = 1 which implies 119860 = 119868
(2) rArr (4) Let 119860 gt 0 be such that 119860120590119868 = 119860 Then119892(119860) = 119860 where 119892 is the representing function of thetranspose of 120590 Hence 119892(119909) = 119909 for all 119909 isin Sp(119860) Supposethat 120572 equiv inf Sp(119860) lt 119903(119860) Then 119892(119909) = 119909 for all 119909 isin
[120572 119903(119860)] It follows that119892(119909) = 119909 onR+ by Lemma 10 Hencethe transpose of 120590 is the right-trivial mean This contradictsassumption (2) We conclude that 120572 = 119903(119860) that is Sp(119860) =
120582 for some 120582 ⩾ 0The same argument as in (2) rArr (3) yields119860 = 119868
(3) rArr (5) Use the congruence invariance of 120590(2) rArr (6) Assume that 120590 is not the left-trivial mean Let
119860 ⩾ 0 be such that 119868120590119860 ⩽ 119868 Then 119891(119860) ⩾ 119868 The spectralmapping theorem implies that 119891(120582) ⩾ 1 for all 120582 isin Sp(119860)Suppose that there exists 119905 isin Sp(119860) such that 119905 lt 1 Since119891(119905) ⩽ 119891(1) = 1 we have 119891(119905) = 1 It follows that 119891(119909) = 1for 119905 ⩽ 119909 ⩽ 1 By Lemma 9 119891 equiv 1 onR+ a contradiction Weconclude that 120582 ⩾ 1 for all 120582 isin Sp(119860) that is 119860 ⩾ 119868
(2) rArr (7) It is similar to (2) rArr (6)(6) rArr (8) Assume (6) Let 119860 gt 0 be such that 119860 ⩽ 119860120590119868
Then
119860 ⩽ 11986012
(119868120590119860minus1)119860
12 (24)
which implies 119868 ⩽ 119868120590119860minus1 By (6) we have I ⩽ 119860
minus1 or 119860 ⩽ 119868(7) rArr (9) It is similar to (6) rArr (8)(6) rArr (10) Use the congruence invariance of 120590(7) rArr (11) Use the congruence invariance of 120590(2) hArr (12) hArr (16) Note that the representing function
of the left-trivial mean is the constant function 119891 equiv 1 Itsassociated measure is the Dirac measure at 0
(2) rArr (13) Assume (2) Let 119909 ⩾ 0 be such that 119891(119909) = 1Suppose that 119909 = 1 It follows that 119891(119909) = 1 for all 119909
lying between 119909 and 1 Lemma 9 implies that 119891 equiv 1 on R+contradicting assumption (2)
Abstract and Applied Analysis 5
(2) rArr (14) (15) Modify the argument in the proof (2) rArr(13)
Theorem 12 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the right(2) 120590 is not the right-trivial mean(3) for all 119860 ⩾ 0 119860120590119868 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119868120590119860 = 119860 rArr 119860 = 119868(5) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 = 119861 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119860120590119868 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119860120590119868 ⩽ 119868 rArr 119860 ⩽ 119868(8) for all 119860 gt 0 119860 ⩽ 119868120590119860 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119868120590119860 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 ⩾ 0 and 119861 gt 0 119861 ⩽ 119860120590119861 rArr 119861 ⩽ 119860(11) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 ⩽ 119861 rArr 119860 ⩽ 119861(12) 119891 is not the identity function 119909 997891rarr 119909(13) 120583 is not the associated measure at 1
Proof Replace 120590 by its transpose in the previous theo-rem
We immediately get the following corollaries
Corollary 13 A mean is strict if and only if it is nontrivial
Corollary 14 Let 120590 be a nontrivial mean For each119860 gt 0 and119861 gt 0 the following statements are equivalent
Remark 16 (i) It is not true that if120590 is not the left-trivialmeanthen for all 119860 ⩾ 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861 Indeedtake 120590 to be the geometric mean 119860 = 0 and
119861 = (
0 00 1
) (25)
The case of right-trivial mean is just the same
(ii) The assumption of invertibility of 119860 or 119861 inCorollary 14 cannot be omitted as a counter example in (i)shows Also the invertibility of119860 or 119861 in Corollary 15 cannotbe omitted Consider the geometric mean and
119860 = (
1 00 0
)
119861 = (
0 00 1
)
(26)
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author is supported by KingMongkutrsquos Institute of Tech-nology Ladkrabang Research Fund Grant no KREF045710
References
[1] G Toader and S Toader Greek means and the arithmetic-geometric mean [dissertation] Victoria University 2005
[2] J Anderson and R J Duffin ldquoSeries and parallel addition ofmatricesrdquo Journal of Mathematical Analysis and Applicationsvol 26 pp 576ndash594 1969
[3] F Kubo and T Ando ldquoMeans of positive linear operatorsrdquoMathematische Annalen vol 246 no 3 pp 205ndash224 197980
[4] F Hiai ldquoMatrix analysis matrix monotone functions matrixmeans and majorizationrdquo Interdisciplinary Information Sci-ences vol 16 no 2 pp 139ndash248 2010
[5] PChansangiamandWLewkeeratiyutkul ldquoCharacterizations ofconnections for positive operatorsrdquo Southeast Asian Bulletin ofMathematics vol 37 no 5 pp 645ndash657 2013
[6] K Lowner ldquoUber monotone matrixfunktionenrdquo Mathematis-che Zeitschrift vol 38 no 1 pp 177ndash216 1934
[7] R BhatiaMatrix analysis vol 169 of Graduate Texts in Mathe-matics Springer New York NY USA 1997
[8] F Hiai and K Yanagi Hilbert Spaces and Linear OperatorsMakino 1995
[9] F Hansen and G K Pedersen ldquoJensenrsquos inequality for operatorsand Lownerrsquos theoremrdquoMathematische Annalen vol 258 no 3pp 229ndash241 1982
[10] M Fiedler and V Ptak ldquoA new positive definite geometricmean of two positive definite matricesrdquo Linear Algebra and ItsApplications vol 251 pp 1ndash20 1997
Amean in Kubo-Ando sense is a connection 120590with fixed-point property 119860120590119860 = 119860 for all 119860 ⩾ 0 The class of Kubo-Ando means cover many well-known means in practice forexample
(iv) logarithmic mean (119860 119861) 997891rarr 11986012
119891(119860minus12
119861119860minus12
)11986012
where the function 119891 is given by 119891(119909) = (119909minus 1) log119909for each 119909 isin R+ equiv [0infin) 119891(0) equiv 0 and 119891(1) equiv 1
A summary of Kubo-Ando theory is given in terms ofone-to-one correspondences between operator connectionson 119861(H)
+ operator monotone functions from R+ to R+and finite Borel measures on [0 1] Recall that a continuousfunction 119891 R+ rarr R+ is said to be operator monotone if
for all positive operators 119860 119861 isin 119861(H) and for all Hilbertspaces H This concept was introduced in [6] see also [7Chapter V] [4 Section 2] and [8] A connection 120590 on 119861(H)
+
can be characterized via operator monotone functions asfollows
Theorem 1 (see [3Theorem 32]) Given a connection 120590 thereis a unique operator monotone function 119891 R+ rarr R+ satis-fying
Moreover the map 120590 997891rarr 119891 is a bijection
We call 119891 the representing function of 120590 A connectionalso has a canonical characterization with respect to a Borelmeasure via a meaningful integral representation as follows
Theorem 2 Given a finite Borel measure 120583 on [0 1] thebinary operation
In this paper we provide various characterizations for theconcepts of positivity betweenness and strictness of operatormeans in terms of operator inequalities operator monotonefunctions Borel measures and certain operator equationsIt turns out that every mean satisfies the positivity propertyThe betweenness is a necessary and sufficient condition for aconnection to be a mean A mean is strict at the left (right)if and only if it is not the left-trivial mean (the right-trivialmean resp)
2 Positivity
We say that a connection 120590 satisfies the positivity property if
(5) rArr (3) (8) rArr (6) rArr (3) and (10) rArr (9) areclear Using the integral representations in Theorem 2 it isstraightforward to verify that the representing function of thezero connection 0 (119860 119861) 997891rarr 0 is the constant function119891 equiv 0and its associated measure is the zero measure Hence wehave the equivalences (3) hArr (9) hArr (11)
(9) rArr (10) Assume 119891 = 0 Suppose that there is 119886 gt 0such that119891(119886) = 0Then119891(119909) = 0 for all 119909 ⩽ 119886The concavityof 119891 implies that 119891(119909) = 0 for all 119909 ⩾ 119886 Hence 119891 = 0 acontradiction
(5) rArr (7) Assume (5) Let 119860 ⩾ 0 and 119861 gt 0 be such that119860120590119861 = 0 Then
0 = 11986112
(119861minus12
119860119861minus12
120590119868) 11986112 (14)
and 119861minus12
119860119861minus12
120590119868 = 0 Now (5) yields 119861minus12119860119861minus12
= 0 thatis 119860 = 0
(6) rArr (8) It is similar to (5) rArr (7)(10) rArr (1) Assume that 119891(119909) gt 0 for all 119909 gt 0 Since
Sp(119891(119860)) = 119891(Sp(119860)) by spectral mapping theorem we have119891(119860) gt 0 for all 119860 gt 0 Hence for each 119860 gt 0 and 119861 gt 0
119860120590119861 = 11986012
119891 (119860minus12
119861119860minus12
)11986012
gt 0 (15)
(10) rArr (4) Assume (10) Let119860 ⩾ 0 be such that119860120590119860 = 0Note that
119860120590119860 = lim120598darr0
119860120598120590119860120598= lim120598darr0
11986012120598
(119868120590119868) 11986012120598
= lim120598darr0
119891 (1) 119860120598= 119891 (1) 119860
(16)
here 119860120598equiv 119860 + 120598119868 Since 119891(1) gt 0 we have 119860 = 0
(10) rArr (5) Assume (10) Let119860 ⩾ 0 be such that119860120590119868 = 0Then 119892(119860) = 0 where 119892 is the representing function of thetranspose of 120590 We see that 119892(119909) gt 0 for 119909 gt 0The injectivityof functional calculus implies that 119892(120582) = 0 for all 120582 isin Sp(119860)We conclude that Sp(119860) = 0 that is 119860 = 0
(10) rArr (6) Assume (10) Let 119860 ⩾ 0 be such that 119868120590119860 = 0Then 119891(119860) = 0 By the injectivity of functional calculus wehave119891(120582) = 0 for all 120582 isin Sp(119860) Assumption (10) implies thatSp(119860) = 0 Thus 119860 = 0
Remark 5 It is not true that 120590 = 0 implies the condition thatfor all 119860 ⩾ 0 and 119861 ⩾ 0 119860120590119861 = 0 implies 119860 = 0 or 119861 = 0Indeed take 120590 to be the geometric mean and
119860 = (
1 00 0
)
119861 = (
0 00 1
)
(17)
3 Betweenness
We say that a connection 120590 satisfies the betweenness propertyif for each 119860 ⩾ 0 and 119861 ⩾ 0
ByTheorem 4 every mean enjoys the positivity propertyIn fact the betweenness property is a necessary and sufficientcondition for a connection to be a mean
Theorem 6 The following statements are equivalent for aconnection 120590 with representing function 119891
(1) 120590 is a mean(2) 120590 satisfies the betweenness property(3) for all 119860 ⩾ 0 119860 ⩽ 119868 rArr 119860 ⩽ 119860120590119868 ⩽ 119868(4) for all 119860 ⩾ 0 119868 ⩽ 119860 rArr 119868 ⩽ 119868120590119860 ⩽ 119860(5) for all 119905 ⩾ 0 1 ⩽ 119905 rArr 1 ⩽ 119891(119905) ⩽ 119905(6) for all 119905 ⩾ 0 119905 ⩽ 1 rArr 119905 ⩽ 119891(119905) ⩽ 1(7) for all 119860 ⩾ 0 and 119861 ⩾ 0 119860 ⩽ 119861 rArr 119860 ⩽ 119860120590119861 ⩽
119861(8) for all 119860 ⩾ 0 119860 ⩽ 119868 rArr 119860 ⩽ 119860120590119868 ⩽ 1(9) for all 119860 ⩾ 0 119868 ⩽ 119860 rArr 1 ⩽ 119868120590119860 ⩽ 119860(10) the only solution 119883 gt 0 to the equation 119883120590119883 = 119868 is
119883 = 119868(11) for all 119860 gt 0 the only solution 119883 gt 0 to the equation
Therefore 120590 is a mean byTheorem 3(7) rArr (9) rArr (1) It is similar to (7) rArr (8) rArr (1)(1) rArr (11) Let 119860 gt 0 Consider 119883 gt 0 such that 119883120590119883 =
119860 Then by the congruence invariance of 120590 we have
119883 = 11988312
(119868120590119868)11988312
= 119883120590119883 = 119860 (21)
(2) rArr (5) If 119905 ⩾ 1 then 119868 ⩽ 119868120590(119905119868) ⩽ 119905119868 which is 119868 ⩽
119891(119905)119868 ⩽ 119905119868 that is 1 ⩽ 119891(119905) ⩽ 119905(5) rArr (1) We have 119891(1) = 1(2) rArr (6) rArr (1) It is similar to (2) rArr (5) rArr (1)
Remark 7 For a connection 120590 and 119860 119861 ⩾ 0 the operators119860 119861 and119860120590119861 need not be comparableThe previous theoremtells us that if 120590 is a mean then the condition 0 ⩽ 119860 ⩽ 119861
guarantees the comparability between 119860 119861 and 119860120590119861
4 Strictness
We consider the strictness of Kubo-Ando means as that forscalar means in [1]
4 Abstract and Applied Analysis
Definition 8 A mean 120590 on 119861(H)+ is said to be
(i) strict at the left if for each 119860 gt 0 and 119861 gt 0
(iii) strict if it is strict at both the right and the left
In order to prove the next two lemmas recall thefollowing facts if 119891 R+ rarr R+ is operator monotone then
(i) 119891 is operator concave and hence concave in usualsense (see [9] or [4 Corollary 254])
(ii) 119892(119909) = 119909119891(119909) is convex in usual sense (see [3 Lemma52])
(iii) ℎ(119909) = 119909119891(119909) is operator monotone on (0infin) (see[9] or [4 Corollary 256])
Lemma 9 If 119891 R+ rarr R+ is an operator monotone functionsuch that 119891 is a constant on an interval [119886 119887] with 119886 lt 119887 then119891 is a constant on R+
Proof Assume that 119891(119909) = 119896 for all 119886 ⩽ 119909 ⩽ 119887 The case119886 = 0 is done by using the monotonicity and concavity of 119891Consider the case 119886 gt 0 The monotonicity and concavity of119891 imply that 119891(119909) = 119896 for all 119909 ⩾ 119887 If 119896 = 0 then 119891(119909) = 119896
on [0 119886] by the monotonicity of 119891 Consider the case 119896 gt 0and suppose there is an 1199090 isin [0 119886) such that 119891(1199090) lt 119896 Thenthe slope of the line segment joining the point (1199090 1199090119891(1199090))and the point (119886 119886119891(119886)) is greater than 119896 This contradicts theconvexity of the function 119909119891(119909)
Lemma 10 If 119891 R+ rarr R+ is an operator monotone func-tion such that 119891(119909) = 119898119909 + 119888 for some 119898 gt 0 and 119888 ⩾ 0 on aninterval [119886 119887] with 119886 lt 119887 then 119891(119909) = 119898119909 + 119888 on R+
Proof If there is 1199090 gt 0 such that 119891(1199090) = 0 then 119891 equiv 0 byLemma 9 Suppose that 119891(119909) gt 0 for all 119909 gt 0 For simplicityassume that 119891(119909) = 119909 for all 119886 ⩽ 119909 ⩽ 119887 Then the functionℎ(119909) = 119909119891(119909) is operator monotone on (0infin) and henceon R+ by continuity Note that ℎ(119909) = 1 on [119886 119887] Lemma 9implies that ℎ(119909) = 1 on R+ that is 119891(119909) = 119909 on R+
Theorem 11 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the left(2) 120590 is not the left-trivial mean(3) for all 119860 ⩾ 0 119868120590119860 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119860120590119868 = 119860 rArr 119860 = 119868(5) for all 119860 gt 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119868120590119860 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119868120590119860 ⩽ 119868 rArr 119860 ⩽ 119868
(8) for all 119860 gt 0 119860 ⩽ 119860120590119868 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119860120590119868 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 gt 0 and 119861 ⩾ 0 119860 ⩽ 119860120590119861 rArr 119860 ⩽ 119861(11) for all 119860 gt 0 and 119861 ⩾ 0 119860120590119861 ⩽ 119860 rArr 119861 ⩽ 119860(12) 119891 is not the constant function 119909 997891rarr 1(13) for all 119909 ⩾ 0 119891(119909) = 1 rArr 119909 = 1(14) for all 119909 ⩾ 0 119891(119909) ⩾ 1 rArr 119909 ⩾ 1(15) for all 119909 ⩾ 0 119891(119909) ⩽ 1 rArr 119909 ⩽ 1(16) 120583 is not the Dirac measure at 0
Proof It is clear that (5) rArr (1) and each of (1)(4) and (6)ndash(11) implies (2) Also each of (13)ndash(15) implies(12)
(2) rArr (3) Let119860 ⩾ 0 be such that 119868120590119860 = 119868Then 119891(119860) = 119868
and hence 119891(120582) = 1 for all 120582 isin Sp(119860) Suppose that 120572 equiv
inf Sp(119860) lt 119903(119860) where 119903(119860) is the spectral radius of119860 Then119891(119909) = 1 for all 119909 isin [120572 119903(119860)] It follows that 119891 equiv 1 on R+ byLemma 9This contradicts assumption (2) We conclude that120572 = 119903(119860) that is Sp(119860) = 120582 for some 120582 ⩾ 0 Suppose nowthat 120582 lt 1 Since 119891(1) = 1 we have that 119891 is a constant onthe interval [120582 1] Again Lemma 9 implies that 119891 equiv 1 onR+a contradiction Similarly 120582 gt 1 gives a contradiction Thus120582 = 1 which implies 119860 = 119868
(2) rArr (4) Let 119860 gt 0 be such that 119860120590119868 = 119860 Then119892(119860) = 119860 where 119892 is the representing function of thetranspose of 120590 Hence 119892(119909) = 119909 for all 119909 isin Sp(119860) Supposethat 120572 equiv inf Sp(119860) lt 119903(119860) Then 119892(119909) = 119909 for all 119909 isin
[120572 119903(119860)] It follows that119892(119909) = 119909 onR+ by Lemma 10 Hencethe transpose of 120590 is the right-trivial mean This contradictsassumption (2) We conclude that 120572 = 119903(119860) that is Sp(119860) =
120582 for some 120582 ⩾ 0The same argument as in (2) rArr (3) yields119860 = 119868
(3) rArr (5) Use the congruence invariance of 120590(2) rArr (6) Assume that 120590 is not the left-trivial mean Let
119860 ⩾ 0 be such that 119868120590119860 ⩽ 119868 Then 119891(119860) ⩾ 119868 The spectralmapping theorem implies that 119891(120582) ⩾ 1 for all 120582 isin Sp(119860)Suppose that there exists 119905 isin Sp(119860) such that 119905 lt 1 Since119891(119905) ⩽ 119891(1) = 1 we have 119891(119905) = 1 It follows that 119891(119909) = 1for 119905 ⩽ 119909 ⩽ 1 By Lemma 9 119891 equiv 1 onR+ a contradiction Weconclude that 120582 ⩾ 1 for all 120582 isin Sp(119860) that is 119860 ⩾ 119868
(2) rArr (7) It is similar to (2) rArr (6)(6) rArr (8) Assume (6) Let 119860 gt 0 be such that 119860 ⩽ 119860120590119868
Then
119860 ⩽ 11986012
(119868120590119860minus1)119860
12 (24)
which implies 119868 ⩽ 119868120590119860minus1 By (6) we have I ⩽ 119860
minus1 or 119860 ⩽ 119868(7) rArr (9) It is similar to (6) rArr (8)(6) rArr (10) Use the congruence invariance of 120590(7) rArr (11) Use the congruence invariance of 120590(2) hArr (12) hArr (16) Note that the representing function
of the left-trivial mean is the constant function 119891 equiv 1 Itsassociated measure is the Dirac measure at 0
(2) rArr (13) Assume (2) Let 119909 ⩾ 0 be such that 119891(119909) = 1Suppose that 119909 = 1 It follows that 119891(119909) = 1 for all 119909
lying between 119909 and 1 Lemma 9 implies that 119891 equiv 1 on R+contradicting assumption (2)
Abstract and Applied Analysis 5
(2) rArr (14) (15) Modify the argument in the proof (2) rArr(13)
Theorem 12 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the right(2) 120590 is not the right-trivial mean(3) for all 119860 ⩾ 0 119860120590119868 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119868120590119860 = 119860 rArr 119860 = 119868(5) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 = 119861 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119860120590119868 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119860120590119868 ⩽ 119868 rArr 119860 ⩽ 119868(8) for all 119860 gt 0 119860 ⩽ 119868120590119860 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119868120590119860 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 ⩾ 0 and 119861 gt 0 119861 ⩽ 119860120590119861 rArr 119861 ⩽ 119860(11) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 ⩽ 119861 rArr 119860 ⩽ 119861(12) 119891 is not the identity function 119909 997891rarr 119909(13) 120583 is not the associated measure at 1
Proof Replace 120590 by its transpose in the previous theo-rem
We immediately get the following corollaries
Corollary 13 A mean is strict if and only if it is nontrivial
Corollary 14 Let 120590 be a nontrivial mean For each119860 gt 0 and119861 gt 0 the following statements are equivalent
Remark 16 (i) It is not true that if120590 is not the left-trivialmeanthen for all 119860 ⩾ 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861 Indeedtake 120590 to be the geometric mean 119860 = 0 and
119861 = (
0 00 1
) (25)
The case of right-trivial mean is just the same
(ii) The assumption of invertibility of 119860 or 119861 inCorollary 14 cannot be omitted as a counter example in (i)shows Also the invertibility of119860 or 119861 in Corollary 15 cannotbe omitted Consider the geometric mean and
119860 = (
1 00 0
)
119861 = (
0 00 1
)
(26)
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author is supported by KingMongkutrsquos Institute of Tech-nology Ladkrabang Research Fund Grant no KREF045710
References
[1] G Toader and S Toader Greek means and the arithmetic-geometric mean [dissertation] Victoria University 2005
[2] J Anderson and R J Duffin ldquoSeries and parallel addition ofmatricesrdquo Journal of Mathematical Analysis and Applicationsvol 26 pp 576ndash594 1969
[3] F Kubo and T Ando ldquoMeans of positive linear operatorsrdquoMathematische Annalen vol 246 no 3 pp 205ndash224 197980
[4] F Hiai ldquoMatrix analysis matrix monotone functions matrixmeans and majorizationrdquo Interdisciplinary Information Sci-ences vol 16 no 2 pp 139ndash248 2010
[5] PChansangiamandWLewkeeratiyutkul ldquoCharacterizations ofconnections for positive operatorsrdquo Southeast Asian Bulletin ofMathematics vol 37 no 5 pp 645ndash657 2013
[6] K Lowner ldquoUber monotone matrixfunktionenrdquo Mathematis-che Zeitschrift vol 38 no 1 pp 177ndash216 1934
[7] R BhatiaMatrix analysis vol 169 of Graduate Texts in Mathe-matics Springer New York NY USA 1997
[8] F Hiai and K Yanagi Hilbert Spaces and Linear OperatorsMakino 1995
[9] F Hansen and G K Pedersen ldquoJensenrsquos inequality for operatorsand Lownerrsquos theoremrdquoMathematische Annalen vol 258 no 3pp 229ndash241 1982
[10] M Fiedler and V Ptak ldquoA new positive definite geometricmean of two positive definite matricesrdquo Linear Algebra and ItsApplications vol 251 pp 1ndash20 1997
(5) rArr (3) (8) rArr (6) rArr (3) and (10) rArr (9) areclear Using the integral representations in Theorem 2 it isstraightforward to verify that the representing function of thezero connection 0 (119860 119861) 997891rarr 0 is the constant function119891 equiv 0and its associated measure is the zero measure Hence wehave the equivalences (3) hArr (9) hArr (11)
(9) rArr (10) Assume 119891 = 0 Suppose that there is 119886 gt 0such that119891(119886) = 0Then119891(119909) = 0 for all 119909 ⩽ 119886The concavityof 119891 implies that 119891(119909) = 0 for all 119909 ⩾ 119886 Hence 119891 = 0 acontradiction
(5) rArr (7) Assume (5) Let 119860 ⩾ 0 and 119861 gt 0 be such that119860120590119861 = 0 Then
0 = 11986112
(119861minus12
119860119861minus12
120590119868) 11986112 (14)
and 119861minus12
119860119861minus12
120590119868 = 0 Now (5) yields 119861minus12119860119861minus12
= 0 thatis 119860 = 0
(6) rArr (8) It is similar to (5) rArr (7)(10) rArr (1) Assume that 119891(119909) gt 0 for all 119909 gt 0 Since
Sp(119891(119860)) = 119891(Sp(119860)) by spectral mapping theorem we have119891(119860) gt 0 for all 119860 gt 0 Hence for each 119860 gt 0 and 119861 gt 0
119860120590119861 = 11986012
119891 (119860minus12
119861119860minus12
)11986012
gt 0 (15)
(10) rArr (4) Assume (10) Let119860 ⩾ 0 be such that119860120590119860 = 0Note that
119860120590119860 = lim120598darr0
119860120598120590119860120598= lim120598darr0
11986012120598
(119868120590119868) 11986012120598
= lim120598darr0
119891 (1) 119860120598= 119891 (1) 119860
(16)
here 119860120598equiv 119860 + 120598119868 Since 119891(1) gt 0 we have 119860 = 0
(10) rArr (5) Assume (10) Let119860 ⩾ 0 be such that119860120590119868 = 0Then 119892(119860) = 0 where 119892 is the representing function of thetranspose of 120590 We see that 119892(119909) gt 0 for 119909 gt 0The injectivityof functional calculus implies that 119892(120582) = 0 for all 120582 isin Sp(119860)We conclude that Sp(119860) = 0 that is 119860 = 0
(10) rArr (6) Assume (10) Let 119860 ⩾ 0 be such that 119868120590119860 = 0Then 119891(119860) = 0 By the injectivity of functional calculus wehave119891(120582) = 0 for all 120582 isin Sp(119860) Assumption (10) implies thatSp(119860) = 0 Thus 119860 = 0
Remark 5 It is not true that 120590 = 0 implies the condition thatfor all 119860 ⩾ 0 and 119861 ⩾ 0 119860120590119861 = 0 implies 119860 = 0 or 119861 = 0Indeed take 120590 to be the geometric mean and
119860 = (
1 00 0
)
119861 = (
0 00 1
)
(17)
3 Betweenness
We say that a connection 120590 satisfies the betweenness propertyif for each 119860 ⩾ 0 and 119861 ⩾ 0
ByTheorem 4 every mean enjoys the positivity propertyIn fact the betweenness property is a necessary and sufficientcondition for a connection to be a mean
Theorem 6 The following statements are equivalent for aconnection 120590 with representing function 119891
(1) 120590 is a mean(2) 120590 satisfies the betweenness property(3) for all 119860 ⩾ 0 119860 ⩽ 119868 rArr 119860 ⩽ 119860120590119868 ⩽ 119868(4) for all 119860 ⩾ 0 119868 ⩽ 119860 rArr 119868 ⩽ 119868120590119860 ⩽ 119860(5) for all 119905 ⩾ 0 1 ⩽ 119905 rArr 1 ⩽ 119891(119905) ⩽ 119905(6) for all 119905 ⩾ 0 119905 ⩽ 1 rArr 119905 ⩽ 119891(119905) ⩽ 1(7) for all 119860 ⩾ 0 and 119861 ⩾ 0 119860 ⩽ 119861 rArr 119860 ⩽ 119860120590119861 ⩽
119861(8) for all 119860 ⩾ 0 119860 ⩽ 119868 rArr 119860 ⩽ 119860120590119868 ⩽ 1(9) for all 119860 ⩾ 0 119868 ⩽ 119860 rArr 1 ⩽ 119868120590119860 ⩽ 119860(10) the only solution 119883 gt 0 to the equation 119883120590119883 = 119868 is
119883 = 119868(11) for all 119860 gt 0 the only solution 119883 gt 0 to the equation
Therefore 120590 is a mean byTheorem 3(7) rArr (9) rArr (1) It is similar to (7) rArr (8) rArr (1)(1) rArr (11) Let 119860 gt 0 Consider 119883 gt 0 such that 119883120590119883 =
119860 Then by the congruence invariance of 120590 we have
119883 = 11988312
(119868120590119868)11988312
= 119883120590119883 = 119860 (21)
(2) rArr (5) If 119905 ⩾ 1 then 119868 ⩽ 119868120590(119905119868) ⩽ 119905119868 which is 119868 ⩽
119891(119905)119868 ⩽ 119905119868 that is 1 ⩽ 119891(119905) ⩽ 119905(5) rArr (1) We have 119891(1) = 1(2) rArr (6) rArr (1) It is similar to (2) rArr (5) rArr (1)
Remark 7 For a connection 120590 and 119860 119861 ⩾ 0 the operators119860 119861 and119860120590119861 need not be comparableThe previous theoremtells us that if 120590 is a mean then the condition 0 ⩽ 119860 ⩽ 119861
guarantees the comparability between 119860 119861 and 119860120590119861
4 Strictness
We consider the strictness of Kubo-Ando means as that forscalar means in [1]
4 Abstract and Applied Analysis
Definition 8 A mean 120590 on 119861(H)+ is said to be
(i) strict at the left if for each 119860 gt 0 and 119861 gt 0
(iii) strict if it is strict at both the right and the left
In order to prove the next two lemmas recall thefollowing facts if 119891 R+ rarr R+ is operator monotone then
(i) 119891 is operator concave and hence concave in usualsense (see [9] or [4 Corollary 254])
(ii) 119892(119909) = 119909119891(119909) is convex in usual sense (see [3 Lemma52])
(iii) ℎ(119909) = 119909119891(119909) is operator monotone on (0infin) (see[9] or [4 Corollary 256])
Lemma 9 If 119891 R+ rarr R+ is an operator monotone functionsuch that 119891 is a constant on an interval [119886 119887] with 119886 lt 119887 then119891 is a constant on R+
Proof Assume that 119891(119909) = 119896 for all 119886 ⩽ 119909 ⩽ 119887 The case119886 = 0 is done by using the monotonicity and concavity of 119891Consider the case 119886 gt 0 The monotonicity and concavity of119891 imply that 119891(119909) = 119896 for all 119909 ⩾ 119887 If 119896 = 0 then 119891(119909) = 119896
on [0 119886] by the monotonicity of 119891 Consider the case 119896 gt 0and suppose there is an 1199090 isin [0 119886) such that 119891(1199090) lt 119896 Thenthe slope of the line segment joining the point (1199090 1199090119891(1199090))and the point (119886 119886119891(119886)) is greater than 119896 This contradicts theconvexity of the function 119909119891(119909)
Lemma 10 If 119891 R+ rarr R+ is an operator monotone func-tion such that 119891(119909) = 119898119909 + 119888 for some 119898 gt 0 and 119888 ⩾ 0 on aninterval [119886 119887] with 119886 lt 119887 then 119891(119909) = 119898119909 + 119888 on R+
Proof If there is 1199090 gt 0 such that 119891(1199090) = 0 then 119891 equiv 0 byLemma 9 Suppose that 119891(119909) gt 0 for all 119909 gt 0 For simplicityassume that 119891(119909) = 119909 for all 119886 ⩽ 119909 ⩽ 119887 Then the functionℎ(119909) = 119909119891(119909) is operator monotone on (0infin) and henceon R+ by continuity Note that ℎ(119909) = 1 on [119886 119887] Lemma 9implies that ℎ(119909) = 1 on R+ that is 119891(119909) = 119909 on R+
Theorem 11 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the left(2) 120590 is not the left-trivial mean(3) for all 119860 ⩾ 0 119868120590119860 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119860120590119868 = 119860 rArr 119860 = 119868(5) for all 119860 gt 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119868120590119860 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119868120590119860 ⩽ 119868 rArr 119860 ⩽ 119868
(8) for all 119860 gt 0 119860 ⩽ 119860120590119868 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119860120590119868 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 gt 0 and 119861 ⩾ 0 119860 ⩽ 119860120590119861 rArr 119860 ⩽ 119861(11) for all 119860 gt 0 and 119861 ⩾ 0 119860120590119861 ⩽ 119860 rArr 119861 ⩽ 119860(12) 119891 is not the constant function 119909 997891rarr 1(13) for all 119909 ⩾ 0 119891(119909) = 1 rArr 119909 = 1(14) for all 119909 ⩾ 0 119891(119909) ⩾ 1 rArr 119909 ⩾ 1(15) for all 119909 ⩾ 0 119891(119909) ⩽ 1 rArr 119909 ⩽ 1(16) 120583 is not the Dirac measure at 0
Proof It is clear that (5) rArr (1) and each of (1)(4) and (6)ndash(11) implies (2) Also each of (13)ndash(15) implies(12)
(2) rArr (3) Let119860 ⩾ 0 be such that 119868120590119860 = 119868Then 119891(119860) = 119868
and hence 119891(120582) = 1 for all 120582 isin Sp(119860) Suppose that 120572 equiv
inf Sp(119860) lt 119903(119860) where 119903(119860) is the spectral radius of119860 Then119891(119909) = 1 for all 119909 isin [120572 119903(119860)] It follows that 119891 equiv 1 on R+ byLemma 9This contradicts assumption (2) We conclude that120572 = 119903(119860) that is Sp(119860) = 120582 for some 120582 ⩾ 0 Suppose nowthat 120582 lt 1 Since 119891(1) = 1 we have that 119891 is a constant onthe interval [120582 1] Again Lemma 9 implies that 119891 equiv 1 onR+a contradiction Similarly 120582 gt 1 gives a contradiction Thus120582 = 1 which implies 119860 = 119868
(2) rArr (4) Let 119860 gt 0 be such that 119860120590119868 = 119860 Then119892(119860) = 119860 where 119892 is the representing function of thetranspose of 120590 Hence 119892(119909) = 119909 for all 119909 isin Sp(119860) Supposethat 120572 equiv inf Sp(119860) lt 119903(119860) Then 119892(119909) = 119909 for all 119909 isin
[120572 119903(119860)] It follows that119892(119909) = 119909 onR+ by Lemma 10 Hencethe transpose of 120590 is the right-trivial mean This contradictsassumption (2) We conclude that 120572 = 119903(119860) that is Sp(119860) =
120582 for some 120582 ⩾ 0The same argument as in (2) rArr (3) yields119860 = 119868
(3) rArr (5) Use the congruence invariance of 120590(2) rArr (6) Assume that 120590 is not the left-trivial mean Let
119860 ⩾ 0 be such that 119868120590119860 ⩽ 119868 Then 119891(119860) ⩾ 119868 The spectralmapping theorem implies that 119891(120582) ⩾ 1 for all 120582 isin Sp(119860)Suppose that there exists 119905 isin Sp(119860) such that 119905 lt 1 Since119891(119905) ⩽ 119891(1) = 1 we have 119891(119905) = 1 It follows that 119891(119909) = 1for 119905 ⩽ 119909 ⩽ 1 By Lemma 9 119891 equiv 1 onR+ a contradiction Weconclude that 120582 ⩾ 1 for all 120582 isin Sp(119860) that is 119860 ⩾ 119868
(2) rArr (7) It is similar to (2) rArr (6)(6) rArr (8) Assume (6) Let 119860 gt 0 be such that 119860 ⩽ 119860120590119868
Then
119860 ⩽ 11986012
(119868120590119860minus1)119860
12 (24)
which implies 119868 ⩽ 119868120590119860minus1 By (6) we have I ⩽ 119860
minus1 or 119860 ⩽ 119868(7) rArr (9) It is similar to (6) rArr (8)(6) rArr (10) Use the congruence invariance of 120590(7) rArr (11) Use the congruence invariance of 120590(2) hArr (12) hArr (16) Note that the representing function
of the left-trivial mean is the constant function 119891 equiv 1 Itsassociated measure is the Dirac measure at 0
(2) rArr (13) Assume (2) Let 119909 ⩾ 0 be such that 119891(119909) = 1Suppose that 119909 = 1 It follows that 119891(119909) = 1 for all 119909
lying between 119909 and 1 Lemma 9 implies that 119891 equiv 1 on R+contradicting assumption (2)
Abstract and Applied Analysis 5
(2) rArr (14) (15) Modify the argument in the proof (2) rArr(13)
Theorem 12 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the right(2) 120590 is not the right-trivial mean(3) for all 119860 ⩾ 0 119860120590119868 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119868120590119860 = 119860 rArr 119860 = 119868(5) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 = 119861 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119860120590119868 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119860120590119868 ⩽ 119868 rArr 119860 ⩽ 119868(8) for all 119860 gt 0 119860 ⩽ 119868120590119860 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119868120590119860 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 ⩾ 0 and 119861 gt 0 119861 ⩽ 119860120590119861 rArr 119861 ⩽ 119860(11) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 ⩽ 119861 rArr 119860 ⩽ 119861(12) 119891 is not the identity function 119909 997891rarr 119909(13) 120583 is not the associated measure at 1
Proof Replace 120590 by its transpose in the previous theo-rem
We immediately get the following corollaries
Corollary 13 A mean is strict if and only if it is nontrivial
Corollary 14 Let 120590 be a nontrivial mean For each119860 gt 0 and119861 gt 0 the following statements are equivalent
Remark 16 (i) It is not true that if120590 is not the left-trivialmeanthen for all 119860 ⩾ 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861 Indeedtake 120590 to be the geometric mean 119860 = 0 and
119861 = (
0 00 1
) (25)
The case of right-trivial mean is just the same
(ii) The assumption of invertibility of 119860 or 119861 inCorollary 14 cannot be omitted as a counter example in (i)shows Also the invertibility of119860 or 119861 in Corollary 15 cannotbe omitted Consider the geometric mean and
119860 = (
1 00 0
)
119861 = (
0 00 1
)
(26)
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author is supported by KingMongkutrsquos Institute of Tech-nology Ladkrabang Research Fund Grant no KREF045710
References
[1] G Toader and S Toader Greek means and the arithmetic-geometric mean [dissertation] Victoria University 2005
[2] J Anderson and R J Duffin ldquoSeries and parallel addition ofmatricesrdquo Journal of Mathematical Analysis and Applicationsvol 26 pp 576ndash594 1969
[3] F Kubo and T Ando ldquoMeans of positive linear operatorsrdquoMathematische Annalen vol 246 no 3 pp 205ndash224 197980
[4] F Hiai ldquoMatrix analysis matrix monotone functions matrixmeans and majorizationrdquo Interdisciplinary Information Sci-ences vol 16 no 2 pp 139ndash248 2010
[5] PChansangiamandWLewkeeratiyutkul ldquoCharacterizations ofconnections for positive operatorsrdquo Southeast Asian Bulletin ofMathematics vol 37 no 5 pp 645ndash657 2013
[6] K Lowner ldquoUber monotone matrixfunktionenrdquo Mathematis-che Zeitschrift vol 38 no 1 pp 177ndash216 1934
[7] R BhatiaMatrix analysis vol 169 of Graduate Texts in Mathe-matics Springer New York NY USA 1997
[8] F Hiai and K Yanagi Hilbert Spaces and Linear OperatorsMakino 1995
[9] F Hansen and G K Pedersen ldquoJensenrsquos inequality for operatorsand Lownerrsquos theoremrdquoMathematische Annalen vol 258 no 3pp 229ndash241 1982
[10] M Fiedler and V Ptak ldquoA new positive definite geometricmean of two positive definite matricesrdquo Linear Algebra and ItsApplications vol 251 pp 1ndash20 1997
(iii) strict if it is strict at both the right and the left
In order to prove the next two lemmas recall thefollowing facts if 119891 R+ rarr R+ is operator monotone then
(i) 119891 is operator concave and hence concave in usualsense (see [9] or [4 Corollary 254])
(ii) 119892(119909) = 119909119891(119909) is convex in usual sense (see [3 Lemma52])
(iii) ℎ(119909) = 119909119891(119909) is operator monotone on (0infin) (see[9] or [4 Corollary 256])
Lemma 9 If 119891 R+ rarr R+ is an operator monotone functionsuch that 119891 is a constant on an interval [119886 119887] with 119886 lt 119887 then119891 is a constant on R+
Proof Assume that 119891(119909) = 119896 for all 119886 ⩽ 119909 ⩽ 119887 The case119886 = 0 is done by using the monotonicity and concavity of 119891Consider the case 119886 gt 0 The monotonicity and concavity of119891 imply that 119891(119909) = 119896 for all 119909 ⩾ 119887 If 119896 = 0 then 119891(119909) = 119896
on [0 119886] by the monotonicity of 119891 Consider the case 119896 gt 0and suppose there is an 1199090 isin [0 119886) such that 119891(1199090) lt 119896 Thenthe slope of the line segment joining the point (1199090 1199090119891(1199090))and the point (119886 119886119891(119886)) is greater than 119896 This contradicts theconvexity of the function 119909119891(119909)
Lemma 10 If 119891 R+ rarr R+ is an operator monotone func-tion such that 119891(119909) = 119898119909 + 119888 for some 119898 gt 0 and 119888 ⩾ 0 on aninterval [119886 119887] with 119886 lt 119887 then 119891(119909) = 119898119909 + 119888 on R+
Proof If there is 1199090 gt 0 such that 119891(1199090) = 0 then 119891 equiv 0 byLemma 9 Suppose that 119891(119909) gt 0 for all 119909 gt 0 For simplicityassume that 119891(119909) = 119909 for all 119886 ⩽ 119909 ⩽ 119887 Then the functionℎ(119909) = 119909119891(119909) is operator monotone on (0infin) and henceon R+ by continuity Note that ℎ(119909) = 1 on [119886 119887] Lemma 9implies that ℎ(119909) = 1 on R+ that is 119891(119909) = 119909 on R+
Theorem 11 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the left(2) 120590 is not the left-trivial mean(3) for all 119860 ⩾ 0 119868120590119860 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119860120590119868 = 119860 rArr 119860 = 119868(5) for all 119860 gt 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119868120590119860 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119868120590119860 ⩽ 119868 rArr 119860 ⩽ 119868
(8) for all 119860 gt 0 119860 ⩽ 119860120590119868 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119860120590119868 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 gt 0 and 119861 ⩾ 0 119860 ⩽ 119860120590119861 rArr 119860 ⩽ 119861(11) for all 119860 gt 0 and 119861 ⩾ 0 119860120590119861 ⩽ 119860 rArr 119861 ⩽ 119860(12) 119891 is not the constant function 119909 997891rarr 1(13) for all 119909 ⩾ 0 119891(119909) = 1 rArr 119909 = 1(14) for all 119909 ⩾ 0 119891(119909) ⩾ 1 rArr 119909 ⩾ 1(15) for all 119909 ⩾ 0 119891(119909) ⩽ 1 rArr 119909 ⩽ 1(16) 120583 is not the Dirac measure at 0
Proof It is clear that (5) rArr (1) and each of (1)(4) and (6)ndash(11) implies (2) Also each of (13)ndash(15) implies(12)
(2) rArr (3) Let119860 ⩾ 0 be such that 119868120590119860 = 119868Then 119891(119860) = 119868
and hence 119891(120582) = 1 for all 120582 isin Sp(119860) Suppose that 120572 equiv
inf Sp(119860) lt 119903(119860) where 119903(119860) is the spectral radius of119860 Then119891(119909) = 1 for all 119909 isin [120572 119903(119860)] It follows that 119891 equiv 1 on R+ byLemma 9This contradicts assumption (2) We conclude that120572 = 119903(119860) that is Sp(119860) = 120582 for some 120582 ⩾ 0 Suppose nowthat 120582 lt 1 Since 119891(1) = 1 we have that 119891 is a constant onthe interval [120582 1] Again Lemma 9 implies that 119891 equiv 1 onR+a contradiction Similarly 120582 gt 1 gives a contradiction Thus120582 = 1 which implies 119860 = 119868
(2) rArr (4) Let 119860 gt 0 be such that 119860120590119868 = 119860 Then119892(119860) = 119860 where 119892 is the representing function of thetranspose of 120590 Hence 119892(119909) = 119909 for all 119909 isin Sp(119860) Supposethat 120572 equiv inf Sp(119860) lt 119903(119860) Then 119892(119909) = 119909 for all 119909 isin
[120572 119903(119860)] It follows that119892(119909) = 119909 onR+ by Lemma 10 Hencethe transpose of 120590 is the right-trivial mean This contradictsassumption (2) We conclude that 120572 = 119903(119860) that is Sp(119860) =
120582 for some 120582 ⩾ 0The same argument as in (2) rArr (3) yields119860 = 119868
(3) rArr (5) Use the congruence invariance of 120590(2) rArr (6) Assume that 120590 is not the left-trivial mean Let
119860 ⩾ 0 be such that 119868120590119860 ⩽ 119868 Then 119891(119860) ⩾ 119868 The spectralmapping theorem implies that 119891(120582) ⩾ 1 for all 120582 isin Sp(119860)Suppose that there exists 119905 isin Sp(119860) such that 119905 lt 1 Since119891(119905) ⩽ 119891(1) = 1 we have 119891(119905) = 1 It follows that 119891(119909) = 1for 119905 ⩽ 119909 ⩽ 1 By Lemma 9 119891 equiv 1 onR+ a contradiction Weconclude that 120582 ⩾ 1 for all 120582 isin Sp(119860) that is 119860 ⩾ 119868
(2) rArr (7) It is similar to (2) rArr (6)(6) rArr (8) Assume (6) Let 119860 gt 0 be such that 119860 ⩽ 119860120590119868
Then
119860 ⩽ 11986012
(119868120590119860minus1)119860
12 (24)
which implies 119868 ⩽ 119868120590119860minus1 By (6) we have I ⩽ 119860
minus1 or 119860 ⩽ 119868(7) rArr (9) It is similar to (6) rArr (8)(6) rArr (10) Use the congruence invariance of 120590(7) rArr (11) Use the congruence invariance of 120590(2) hArr (12) hArr (16) Note that the representing function
of the left-trivial mean is the constant function 119891 equiv 1 Itsassociated measure is the Dirac measure at 0
(2) rArr (13) Assume (2) Let 119909 ⩾ 0 be such that 119891(119909) = 1Suppose that 119909 = 1 It follows that 119891(119909) = 1 for all 119909
lying between 119909 and 1 Lemma 9 implies that 119891 equiv 1 on R+contradicting assumption (2)
Abstract and Applied Analysis 5
(2) rArr (14) (15) Modify the argument in the proof (2) rArr(13)
Theorem 12 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the right(2) 120590 is not the right-trivial mean(3) for all 119860 ⩾ 0 119860120590119868 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119868120590119860 = 119860 rArr 119860 = 119868(5) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 = 119861 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119860120590119868 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119860120590119868 ⩽ 119868 rArr 119860 ⩽ 119868(8) for all 119860 gt 0 119860 ⩽ 119868120590119860 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119868120590119860 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 ⩾ 0 and 119861 gt 0 119861 ⩽ 119860120590119861 rArr 119861 ⩽ 119860(11) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 ⩽ 119861 rArr 119860 ⩽ 119861(12) 119891 is not the identity function 119909 997891rarr 119909(13) 120583 is not the associated measure at 1
Proof Replace 120590 by its transpose in the previous theo-rem
We immediately get the following corollaries
Corollary 13 A mean is strict if and only if it is nontrivial
Corollary 14 Let 120590 be a nontrivial mean For each119860 gt 0 and119861 gt 0 the following statements are equivalent
Remark 16 (i) It is not true that if120590 is not the left-trivialmeanthen for all 119860 ⩾ 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861 Indeedtake 120590 to be the geometric mean 119860 = 0 and
119861 = (
0 00 1
) (25)
The case of right-trivial mean is just the same
(ii) The assumption of invertibility of 119860 or 119861 inCorollary 14 cannot be omitted as a counter example in (i)shows Also the invertibility of119860 or 119861 in Corollary 15 cannotbe omitted Consider the geometric mean and
119860 = (
1 00 0
)
119861 = (
0 00 1
)
(26)
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author is supported by KingMongkutrsquos Institute of Tech-nology Ladkrabang Research Fund Grant no KREF045710
References
[1] G Toader and S Toader Greek means and the arithmetic-geometric mean [dissertation] Victoria University 2005
[2] J Anderson and R J Duffin ldquoSeries and parallel addition ofmatricesrdquo Journal of Mathematical Analysis and Applicationsvol 26 pp 576ndash594 1969
[3] F Kubo and T Ando ldquoMeans of positive linear operatorsrdquoMathematische Annalen vol 246 no 3 pp 205ndash224 197980
[4] F Hiai ldquoMatrix analysis matrix monotone functions matrixmeans and majorizationrdquo Interdisciplinary Information Sci-ences vol 16 no 2 pp 139ndash248 2010
[5] PChansangiamandWLewkeeratiyutkul ldquoCharacterizations ofconnections for positive operatorsrdquo Southeast Asian Bulletin ofMathematics vol 37 no 5 pp 645ndash657 2013
[6] K Lowner ldquoUber monotone matrixfunktionenrdquo Mathematis-che Zeitschrift vol 38 no 1 pp 177ndash216 1934
[7] R BhatiaMatrix analysis vol 169 of Graduate Texts in Mathe-matics Springer New York NY USA 1997
[8] F Hiai and K Yanagi Hilbert Spaces and Linear OperatorsMakino 1995
[9] F Hansen and G K Pedersen ldquoJensenrsquos inequality for operatorsand Lownerrsquos theoremrdquoMathematische Annalen vol 258 no 3pp 229ndash241 1982
[10] M Fiedler and V Ptak ldquoA new positive definite geometricmean of two positive definite matricesrdquo Linear Algebra and ItsApplications vol 251 pp 1ndash20 1997
(2) rArr (14) (15) Modify the argument in the proof (2) rArr(13)
Theorem 12 Let 120590 be a mean with representing function 119891
and associated measure 120583 Then the following statements areequivalent
(1) 120590 is strict at the right(2) 120590 is not the right-trivial mean(3) for all 119860 ⩾ 0 119860120590119868 = 119868 rArr 119860 = 119868(4) for all 119860 gt 0 119868120590119860 = 119860 rArr 119860 = 119868(5) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 = 119861 rArr 119860 = 119861(6) for all 119860 ⩾ 0 119868 ⩽ 119860120590119868 rArr 119868 ⩽ 119860(7) for all 119860 ⩾ 0 119860120590119868 ⩽ 119868 rArr 119860 ⩽ 119868(8) for all 119860 gt 0 119860 ⩽ 119868120590119860 rArr 119860 ⩽ 119868(9) for all 119860 gt 0 119868120590119860 ⩽ 119860 rArr 119868 ⩽ 119860(10) for all 119860 ⩾ 0 and 119861 gt 0 119861 ⩽ 119860120590119861 rArr 119861 ⩽ 119860(11) for all 119860 ⩾ 0 and 119861 gt 0 119860120590119861 ⩽ 119861 rArr 119860 ⩽ 119861(12) 119891 is not the identity function 119909 997891rarr 119909(13) 120583 is not the associated measure at 1
Proof Replace 120590 by its transpose in the previous theo-rem
We immediately get the following corollaries
Corollary 13 A mean is strict if and only if it is nontrivial
Corollary 14 Let 120590 be a nontrivial mean For each119860 gt 0 and119861 gt 0 the following statements are equivalent
Remark 16 (i) It is not true that if120590 is not the left-trivialmeanthen for all 119860 ⩾ 0 and 119861 ⩾ 0 119860120590119861 = 119860 rArr 119860 = 119861 Indeedtake 120590 to be the geometric mean 119860 = 0 and
119861 = (
0 00 1
) (25)
The case of right-trivial mean is just the same
(ii) The assumption of invertibility of 119860 or 119861 inCorollary 14 cannot be omitted as a counter example in (i)shows Also the invertibility of119860 or 119861 in Corollary 15 cannotbe omitted Consider the geometric mean and
119860 = (
1 00 0
)
119861 = (
0 00 1
)
(26)
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper
Acknowledgment
The author is supported by KingMongkutrsquos Institute of Tech-nology Ladkrabang Research Fund Grant no KREF045710
References
[1] G Toader and S Toader Greek means and the arithmetic-geometric mean [dissertation] Victoria University 2005
[2] J Anderson and R J Duffin ldquoSeries and parallel addition ofmatricesrdquo Journal of Mathematical Analysis and Applicationsvol 26 pp 576ndash594 1969
[3] F Kubo and T Ando ldquoMeans of positive linear operatorsrdquoMathematische Annalen vol 246 no 3 pp 205ndash224 197980
[4] F Hiai ldquoMatrix analysis matrix monotone functions matrixmeans and majorizationrdquo Interdisciplinary Information Sci-ences vol 16 no 2 pp 139ndash248 2010
[5] PChansangiamandWLewkeeratiyutkul ldquoCharacterizations ofconnections for positive operatorsrdquo Southeast Asian Bulletin ofMathematics vol 37 no 5 pp 645ndash657 2013
[6] K Lowner ldquoUber monotone matrixfunktionenrdquo Mathematis-che Zeitschrift vol 38 no 1 pp 177ndash216 1934
[7] R BhatiaMatrix analysis vol 169 of Graduate Texts in Mathe-matics Springer New York NY USA 1997
[8] F Hiai and K Yanagi Hilbert Spaces and Linear OperatorsMakino 1995
[9] F Hansen and G K Pedersen ldquoJensenrsquos inequality for operatorsand Lownerrsquos theoremrdquoMathematische Annalen vol 258 no 3pp 229ndash241 1982
[10] M Fiedler and V Ptak ldquoA new positive definite geometricmean of two positive definite matricesrdquo Linear Algebra and ItsApplications vol 251 pp 1ndash20 1997