JENSEN S OPERATOR INEQUALITY FOR FUNCTIONS OF SEVERAL VARIABLES

Transcription

1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Voume 128, Number 7, Pages S Artice eectronicay pubished on February 16, 2000 JENSEN S OPERATOR INEQUALITY FOR FUNCTIONS OF SEVERAL VARIABLES HUZIHIRO ARAKI AND FRANK HANSEN Communicated by David R. Larson Abstract. The operator convex functions of severa variabes are characterized in terms of a non-commutative generaization of Jensen s inequaity, extending previous resuts of F. Hansen and G.K. Pedersen for functions of one variabe and of F. Hansen for functions of two variabes. 1. Introduction and the main resut Let I 1,...,I k be intervas in R and et f : I 1 I k R be an essentiay bounded rea function defined on the product of the intervas. Let x =x 1,...,x k be a k-tupe of bounded sefadjoint operators on Hibert spaces H 1,...,H k such that the spectrum of x i is contained in I i for i =1,...,k. We say that such a k-tupe is in the domain of f. If x i = λ i E i dλ i I i is the spectra resoution of x i for i =1,...,k, then we define fx =fx 1,...,x k = fλ 1,...,λ k E 1 dλ 1 E k dλ k I 1 I k as a bounded sefadjoint operator on H 1 H k. The above function f : I 1 I k R is said to be operator convex, if the operator inequaity fλx 1 +1 λy 1,...,λx k +1 λy k 1.1 λfx 1,...,x k +1 λfy 1,...,y k hods for any λ [0, 1], any Hibert spaces H 1,...,H k and any sefadjoint operators x 1,...,x k and y 1,...,y k on H 1,...,H k contained in the domain of f. The definition is meaningfu since aso the spectrum of λx i +1 λy i is contained in the interva I i for each i =1,...,k. If the inequaity 1.1 hods just for Hibert spaces of finite dimensions n 1,...,n k, then we say that f is matrix convex of order n 1,...,n k or just order n if n i = n for i =1,...,k. Definition 1.1. An -tupe of bounded inear operators a = a 1,...,a on a Hibert space H is caed a contraction row, if 1.2 a 1 a a a 1. Received by the editors August 30, Mathematics Subject Cassification. Primary 47A63; Secondary 47A80, 47Bxx c 2000 American Mathematica Society License or copyright restrictions may appy to redistribution; see

2 2076 HUZIHIRO ARAKI AND FRANK HANSEN It is caed a unitary row, if there exists a unitary operator U on the direct sum of copies of H such that a 1,...,a is the first row in the bock matrix representation of U. If a =a 1,...,a is a contraction row, it foows that the norm a i 1fori = 1,...,.A necessary but in genera not sufficient condition for an -tupe a 1,...,a to be a unitary row is that 1.3 a 1 a a a =1. A unitary row is in particuar a contraction row. If one of the operators in a contraction row is norma, then 1.3 is a sufficient condition for the row to be a unitary row; cf. Remark 2.3 beow. This resut was obtained in [5] for =2. In particuar, any -tupe of orthogona projections p 1,...,p withsump 1 + +p = 1 is a unitary row. This aso foows expicity, because the operator represented by the bock matrix p 1 p 2 p U = p j i mod +1 = p p 1 p i,j=1... p 2 p 3 p 1 is unitary on H C. Any extension of a unitary row of ength with zero operators to an m tupe a 1,...,a, 0,...,0 is a unitary row of ength m, since U U 0 = 0 1 m m is unitary on H C m, if U is unitary on H C. It is a non-trivia probem to determine whether a genera contraction row satisfying 1.3 is a unitary row. We are going to present Jensen s operator inequaity for functions of severa variabes. It generaizes previous resuts for functions of one variabe and for functions of two variabes; cf. [7] and [5]. Theorem 1.2. Let f be a continuous rea vaued function defined on I 1 I k where I i = [0,α i [ with α i for i = 1,...,k. The foowing statements are equivaent the statement iii is for each natura number 2: i f is operator convex and fr 1,...,r k 0 if r i =0for some i =1,...,k. ii For each natura number 2 and for each j =0, 1,..., 1 the operator inequaity 1.5 diag fa s x 11 1a s11,...,a s k k x ka sk k s =j mod a t 11 a t k k fx 1,...,x k a s11 a sk k t = s =j mod is vaid for a unitary rows a i =a 1i,...,a i of ength acting on a Hibert space H i for i =1,...,k and a k-tupes x 1,...,x k of sefadjoint operators in the domain of f acting on H 1,...,H k. The indices s, t are muti-indices of the form s =s 1,...,s k, where s i =1,..., for i =1,...,k with weight s = s s k. License or copyright restrictions may appy to redistribution; see

3 iii JENSEN S OPERATOR INEQUALITY 2077 For each j =0, 1,..., 1 the operator inequaity 1.6 diag fp s11x 1 p s11,...,p sk kx k p sk k s =j mod p t11 p tk kfx 1,...,x k p s11 p sk k t = s =j mod is vaid for a partitions of unity p 1i + + p i =1on H i by orthogona projections for each i =1,...,k and a k-tupes x 1,...,x k of sefadjoint operators in the domain of f acting on H 1,...,H k. The index notation s and t for rows and coumns of the bock matrices are the same as in statement ii above. Remark 1.3. The statements in ii andiii extended to = 1 are triviay true with equaity for any f. To get i fromiii for any fixed 2, it is enough to assume the inequaity in iii for j = 0 and identica partitions of the unity acting on the same infinite dimensiona Hibert space H for i =1,...,k. Coroary 1.4. Let f be a continuous rea vaued function defined on I 1 I k where I i =[0,α i [ with α i for i =1,...,k. If f is operator convex and fr 1,...,r k 0 if r i =0for some i =1,...,k, then the operator inequaity 1.7 diag fa s x 11 1a s11,...,a s k k x ka sk k s =j mod m a t 11 a t k kfx 1,...,x k a s11 a sk k t = s =j modm is vaid for a natura numbers 1, m>,j=0, 1,...,m 1, a contraction rows a i =a 1i,...,a i of ength acting on a Hibert space H i for i =1,...,k and a k-tupes x 1,...,x k of sefadjoint operators in the domain of f acting on H 1,...,H k. Proof. We appy Lemma 2.1 to construct unitary rows of ength m from the given contraction rows a i =a 1i,...,a i and consider Jensen s operator inequaity for k and m. Take j =0, 1,...,m 1 and consider the submatrix corresponding to indices s, t taking vaues ony in {1,...,} k and having weights j mod m. QED Remark 1.5. If we set k =1,=1andm = 2 in Coroary 1.4, then we get the inequaity fa xa a fxa, vaid for a contractions a and sefadjoint x in the domain of f. This is Jensen s operator inequaity for functions of one variabe [7]. If we set k =2,=1, and m = 2 in the coroary and choose projections p and q as contraction rows of ength 1, then we get Auja s inequaity [2] fpxp, qyq p qfx, yp q which characterizes separatey operator convex functions of two variabes. Remark 1.6. If we put = k and j = 0 in ii of Theorem 1.2 and take the submatrix corresponding to coinciding indices s 1 = = s k = i for i =1,...,k, License or copyright restrictions may appy to redistribution; see

4 2078 HUZIHIRO ARAKI AND FRANK HANSEN then we obtain the inequaity k diag fa i1x 1 a i1,...,a ikx k a ik i=1 1.8 a i1 a ik fx 1,...,x k a j1 a jk k i,j=1 for a unitary rows a i =a 1i,...,a ki acting on a Hibert space H i for i =1,...,k and a k-tupes x 1,...,x k of sefadjoint operators in the domain of f acting on H 1,...,H k. Remark 1.7. If we et a the operators in 1.8 act on the same Hibert space and ony consider one common unitary row, which with a change of notation we denote a 1,...,a k, then we obtain the inequaity k diag fa i x 1a i,...,a i x ka i i=1 1.9 k a i a i fx 1,...,x k a j a j i,j=1. This is for k = 2 the generaization of Jensen s operator inequaity to functions of two variabes obtained in [5], except that in the reference it was obtained ony for unitary rows of a certain type. It appears that the inequaity 1.9 is too weak to impy operator convexity of f except in the cases k =1, 2 which were aready setted in the iterature [7, 5]. 2. Preiminaries Lemma 2.1. A contraction row a 1,...,a of ength on a Hibert space H can for any m>be extended to a unitary row of ength m. More specificay, the row a 1,...,a,a, where 1/2 a = 1 a i a i, as we as a 1,...,a,a,0,...,0 for any number of zeros are unitary rows. Proof. The bock matrix with operators on H as its eements a 1... a 0 0 A = i=1 is by the assumption a contraction on H C. Indeed, the norm AA = a i a i 1. The 2 2 bock matrix A 1 AA U = 1/2 1 A A 1/2 A is unitary [3, Soution 177]. When expicity written out, this bock matrix spits as a direct sum of two unitary maps a bock matrix of size +1andtheidentity matrix of size 1, one of which has the first row a 1,...,a,a. QED License or copyright restrictions may appy to redistribution; see

5 JENSEN S OPERATOR INEQUALITY 2079 In fact, the unitary matrix corresponding to a unitary row of ength +1appearing in the proof of Lemma 2.1 can be computed expicity as a 1 a a a 1 a a a 1. 1 A A 1/2. = v 11 v 1 a , a v 1 v a where v ij = a i 1 + a 1 a j δ ij for i, j =1,...,. Coroary 2.2. Let a 1,...,a be a contraction row and suppose that dim ker a i = dim ker a i for at east one i =1,...,. Then the row is a unitary row, if and ony if 1.3 hods. Proof. The necessity of 1.3 is immediate as mentioned earier. To prove the sufficiency of 1.3 we may without oss of generaity assume that dim ker a = dim ker a. Let a = u a be the poar decomposition of a. Then u is a partia isometry with u u and uu being the sefadjoint projections on ker a and ker a. By assumption, there exists a partia isometry v with v v and vv being the sefadjoint projections on ker a and ker a. Therefore w = u+v is unitary, and since the support of a is ker a we obtain a = w a. Set b j = w a j for j =1,...,k 1. Then by b j b j = 1 w 1 a j a j w = w a a w = a 2. j=1 j=1 Hence b 1,...,b 1, a is a unitary row by Lemma 2.1, thus being the first row of a unitary bock matrix V. Finay, a 1,...,a 1,a =wb 1,...,wb 1,w a is the first row of the unitary bock matrix w E V, where E denotes the identity matrix of order. QED Remark 2.3. If one of the operators in a contraction row is norma and 1.3 is satisfied, then the contraction row is a unitary row. This foows from Coroary 2.2, because the kerne of a norma operator coincides with the kerne of its adjoint. The root of unity β = e 2πi/ is a simpe root of the poynomia X 1. In the foowing we make repeated use of the identities β =1, β = β 1, and 1 + β j + β 2j + + β 1j =0forj =1,..., 1. We consider the operator R i on H i C given by the diagona bock matrix representation R i =diagβ p p=1,...,. For i =1,...,k and any unitary row a i =a 1i,...,a i acting on a Hibert space H i with associated unitary U i on H i C we set U mi = U i R m i = U i diagβ pm p=1,..., which defines a unitary operator on H i C. Reative to these unitary operators we define the average Ui X i ofanoperatorx i acting on H i C by setting Ui X i = 1 Umi X iu mi. License or copyright restrictions may appy to redistribution; see

6 2080 HUZIHIRO ARAKI AND FRANK HANSEN Lemma 2.4. For any bounded operator x i on H i, consider the bock matrix x i D x i = as an operator on H i C. Then for i =1,...,k we have Ui D x i = diag a pi x ia pi p=1 for any unitary row a i =a 1i,...,a i on H i with associated unitary U i. Proof. The entry at the pth row and qth coumn of YRi m is just β mq y pq for any operator Y acting on H i C withbockmatrixrepresentationy =y pq p,q=1,...,. Likewise, the entry [Ri m YRi m ] pq = β q pm y pq and consequenty 1 Ri m YRi m is a diagona bock matrix composed of the diagona of Y. Since the diagona of Ui D x i U i is given by a 1i x ia 1i,...,a i x ia i, the statement foows. QED 3. Proof of Jensen s operator inequaity Proof of Theorem 1.2. i ii: We consider unitary rows a i =a 1i,...,a i for i =1,...,k with associated unitaries U 1,...,U k. The k k bock matrix diag fa s 11x 1 a s11,...,a s k kx k a sk k is naturay identified with a 11x 1a 11 f... a 1x 1a 1 s 1,...,s k =1 a 1kx k a 1k,...,... a kx k a k under the bock matrix representation of tensor products. Appying Lemma 2.4 and the convexity of f, we thus obtain diag fa s x 11 1a s11,...,a s k k x ka sk k s 1,...,s k =1 = f U1 D x 1,..., Uk D x k 1 = 1 1 f Um1D x 1 U m1,...,umkd x k U mk U m1 U mk f D x 1,...,D x k U m1 U mk U m1 U mk D k fx 1,...,x k U m1 U mk. License or copyright restrictions may appy to redistribution; see

7 JENSEN S OPERATOR INEQUALITY 2081 The ast inequaity is a consequence of the assumption that fr 1,...,r k 0 whenever r i =0forsomei =1,...,k. The matrix eements of each term of the ast sum are cacuated as [ U m1 U mk D k ] fx 1,...,x k U m1 U mk ts = β m s t [ ] U 1 U k D k fx 1,...,x k U 1 U k The average over m gives rise to 1 β m s t = δ s t and this automaticay spits the above inequaity into parts. Takej = 0, 1,..., 1 and et Q j denote the projection on the subspace corresponding to the indices s =s 1,...,s k ofafixedweight s = s s k = j mod. It foows that Q j diag fa s x 11 1a s11,...,a s k k x ka sk k Q j s 1,...,s k =1 Q j U 1 U k D k fx 1,...,x k U 1 U k Q j. The eft-hand side is a diagona matrix with non-trivia diagona entries for s = j mod andd kfx 1,...,x k is a diagona matrix which has its ony non-trivia eement at 1k =1,...,1. Because [U 1 U k Q j ] 1ks = a s11 a sk k, we concude that the above inequaity, when restricted to the image subspace of Q j, is Jensen s operator inequaity 1.5. ii iii : Obvious speciaization. iii i: We sha ony use iii for j =0. The operator V defined by the bock matrix 1 β β 2 β 1 1 β 2 β 4 β 2 1 V =v pq p,q=1 =.... 1/2...., 1 β 1 β 2 1 β where v pq = 1/2 β pq 1, is unitary. We introduce the foowing bock matrices with operator eements X i =diagx ji j=1,..., and Z i = V X i V for 1 = 1,...,k. The diagona eements of Z i are given by [Z i ] jj = and consequenty we obtain s=1 v sj v sj x si = x 1i + + x i P j Z i P j = x 1i + + x i P j ts. License or copyright restrictions may appy to redistribution; see

8 2082 HUZIHIRO ARAKI AND FRANK HANSEN for i =1,...,k and j =1,...,, where P i is the projection given by the bock matrix P i =diagδ ji j=1,..., and δ is Kronecker s symbo. The statement iii of Theorem 1.2 now entais the inequaity 3.1 diag fp s1 Z 1 P s1,...,p sk Z k P sk s =0 mod P t1 P tk fz 1,...,Z k P s1 P sk t = s =0 = P t1 V P tk V fx 1,...,X k VP s1 VP sk t = s =0 mod which we sha examine in more detai. The bock matrix VP j has vanishing matrix eements except for the jth coumn which is equa to the jth coumn of V. We can therefore cacuate the matrix eements [VP s1 VP sk ] pq =[VP s1 ] p1q 1 [VP sk ] pk q k = δ q1s 1 1/2 β p1s1 1 δ qk s k 1/2 β p ks k 1 = k/2 δ qs β p s p for p =p 1,...,p k,q=q 1,...,q k andp 1,...,p k,q 1,...,q k =1,...,.We observe in particuar that the ony nonzero coumn in VP s1 VP sk is the sth coumn. The right hand side of 3.1 is a k 1 k 1 bock matrix where each bock is a k k bock matrix of operators. Let us cacuate the p, q entryofthis k k bock matrix. [ P t1 V P tk V ] fx 1,...,X k VP s1 VP sk pq [ = P t1 V P tk V ] fx 1,...,X k pu k/2 δ qs β u s u = k = k δ pt δ qs δ pt β u u t fx u11,...,x uk kδ qs β u s u fx u11,...,x uk kβ u s t where the cacuation was faciitated because fx 1,...,X k is a diagona matrix with uth diagona entry fx u11,...,x uk k. We discard a rows and coumns with indices different from t and s in each bock in 3.1. Since this operation preserves the inequaity, we obtain x x 1 f,..., x 1k + + x k E k 1 k fx u11,...,x uk k β u s t t = s =0 mod where E k 1 denotes the identity matrix of order k 1. If we set Π u = k 1 β u s t t = s =0 mod, License or copyright restrictions may appy to redistribution; see

9 JENSEN S OPERATOR INEQUALITY 2083 then the inequaity can be written as x x 1 f,..., x 1k + + x k E k 1 1 fx u11,...,x uk kπ u. It is an easy cacuation to show that the matrices Π u are sefadjoint projections. Because of β u s t = k δ ts, it foows that u 1,,u k =1 Π u = E k 1. Due to t = s we note that Π u =Π u+jkmod for j =0, 1,..., 1 wherejk = j,...,jandu + jkmod are different indices for different j. We introduce an equivaence reation u v by setting u v, if u v = jkmod for any natura number j and denote the equivaence cass of u by u. Each equivaence cass contains exacty distinct members of the form u + jk forj =0, 1,..., 1. We denote the common projection corresponding to each equivaence cass u by Π u and obtain Π u = E k 1. u Hence the projections Π u are mutuay orthogona, and setting u =1k weobtain in particuar x x 1 f,..., x 1k + + x k Π 1k 1 fx 11,...,x 1k + + fx 1,...,x k Π 1k. Therefore f is operator convex. The rest of statement i foows by examining diagona eements in 3.1, which at t for t =0mod gives fp t1 Z 1 P t1,...,p tk Z k P tk P t1 P tk fz 1,...,Z k P t1 P tk. If we ook at the sth diagona entry of this inequaity for s t, then x x 1 f δ t1s 1,..., x 1k + + x k δ tk s k 0 and we concude that fr 1,...,r k 0ifr i =0forateastonei =1,...,k. QED References [1] T. Ando. Concavity of certain maps of positive definite matrices and appications to Hadamard products. Linear Agebra App., 26: , MR 80f:15023 [2] MR 94h:15008 J.S. Auja. Matrix convexity of functions of two variabes. Linear Agebra and Its Appications, 194: , MR 94h:15008 [3] P.R. Hamos. A Hibert space probem book. Van Nostrand, MR 34:8178 [4] F. Hansen. An operator inequaity. Math. Ann., 246: , MR 82a:46065 License or copyright restrictions may appy to redistribution; see

10 2084 HUZIHIRO ARAKI AND FRANK HANSEN [5] F. Hansen. Jensen s operator inequaity for functions of two variabes. Proc. Amer. Math. Soc., 125: , MR 97i:47027 [6] F. Hansen. Operator convex functions of severa variabes. Pub.Res.Inst.Math.Sci., 33: , MR 99b:47022b; MR 99b:47022a [7] F. Hansen and G.K. Pedersen. Jensen s inequaity for operators and Löwner s theorem. Math. Ann., 258: , MR 83g:47020 [8] A. Korányi. On some casses of anaytic functions of severa variabes. Trans Amer. Math. Soc., 101: , MR 25:226 [9] G.K. Pedersen. Extreme n-tupes of eements in C -agebras. Bu. London Math. Soc., 19: , MR 88m:46070 Department of Mathematics, Facuty of Science and Technoogy, The Science University of Tokyo, Noda, Chiba-ken , Japan E-mai address: araki@ma.noda.sut.ac.jp Institute of Economics, University of Copenhagen, Studiestraede 6, 1455 Copenhagen K, Denmark E-mai address: frank.hansen@econ.ku.dk License or copyright restrictions may appy to redistribution; see