Fundamental inequalities and Mond-Pe carić method

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1 Chapter Fundamental inequalities and Mond-Pecarić method In this chapter, we have given a very brief and rapid review of some basic topics in Jensen s inequality for positive linear maps and the Kantorovich inequality for several types. We present some basic ideas and the viewpoints of the Mond-Pecarić method for convex functions.. Classical Jensen s inequality In this section, we introduce a classical Jensen s inequality associated with a convex function, and naturally extend it to an operator version. First we introduce some notations. If a complex vector space H having the inner product is complete with respect to the distance d(x y) =kx yk dened by the norm kxk := (x x) =,thenh is called a Hilbert space. A linear operator A on a Hilbert space H is said to be bounded if kak := supfkaxk : kxk x Hg < : Then kak is said to be the operator norm of A. The adjoint operator A of A is dened by (Ax y) =(x A y)forx y H. Then it follows that kak = ka k = ka Ak =.Inanalgebra of all linear operators (H! H) on a Hilbert space H with the operator norm, we denote by

2 FUNDAMENTAL INEQUALITIES AND MOND-PECARIĆ METHOD B(H) a semi-algebra of all bounded (i.e., continuous) linear operators on H. The spectrum of an operator A is the set Sp(A) =f C : A H is not invertible in B(H)g: The spectrum Sp(A) is nonempty and compact. A bounded linear operator A on a Hilbert space H is said to be selfadjoint if A = A. An operator A B(H) is selfadjoint if and only if (Ax x) R for every vector x H. We denote by B h (H) a semi-space of all selfadjoint operators in B(H). B(H) A H V x () x y M a b () c d H V Ax () x y Figure.: Graphic chart of space B(H) We introduce a partial order in B h (H) as follows: Denition. An operator A B h (H) is positive semi-denite, (simply, positive) and we write A, if(ax x) for every vector x H. An operator A B(H) is positive if and only if A = B B for some operator B B(H). For operators A B B h (H) we write A B (or B A) if B A, i.e., (Bx x) (Ax x) for every vector x H. We call it the operator order. In particular, for some scalars m and M, we write m H A M H if m (Ax x) M for every unit vector x H. Notice that for a selfadjoint operator A, Sp(A) [m M] implies m H A M H. A positive semi-denite operator A B h (H) is positive denite (strictly positive) and we write A >, if there is a real number m > such that A m H. We denote by B + (H) the set of all positive operators and B ++ (H) the set of all strictly positive operators (or positive invertible operators) in B h (H). ThesetB + (H) is the convex cone contained in B h (H). Now, we review the continuous functional calculus. A rudimentary functional calculus for an operator A can be dened as follows: For a polynomial p(t) = P k j= jt j,dene p(a) = H + A + A + + k A k :

3 . CLASSICAL JENSEN S INEQUALITY 3 The mapping p! p(a) is a homomorphism from the algebra of polynomials to the algebra of operators. The extension of this map to larger algebras of functions is really signicant in operator theory. LetA be a selfadjointoperatoron a HilbertspaceH. Then the Gelfand map establishes a -isometrically isomorphism between the set C(Sp(A)) of all continuous functions on Sp(A) andc -algebra C (A) generated by A and the identity operator H on H as follows: For f g C(Sp(A)) and C (i) ( f + g) =(f )+(g). (ii) (fg)=(f )(g)and(f )=(f ). (iii) k(f )k = kf k(:= sup tsp(a) jf (t)j). (iv) (f )= H and (f )=A,wheref (t) =andf (t) =t. With this notation, we dene f (A) =(f ) for all f C(Sp(A)) and we call it the continuous functional calculus for a selfadjoint operator A. This map is an extension of p(a) for a polynomial p. The continuous functional calculus is applicable. For example, if A is a positive operator and f = (t) = p t,then A = = f = (A). If A is a selfadjoint operator and f (t) is a real valued continuous function on Sp(A) such that f (t) on Sp(A), then f (A), i.e., f (A) is a positive operator. Moreover, if g(t) is a real valued continuous function on Sp(A) such that f (t) g(t) on Sp(A), then f (A) g(a). Next, we shall introduce a spectral decomposition theorem for selfadjoint, bounded linear operators on a Hilbert space H. For the sake of convenience, we recall the following well known diagonalization of Hermitian matrices in matrix theory. If A is a Hermitian k k matrix, then there exists a unitary matrix U (i.e., U U = UU = k ) such that A = U U (.) where =diag( k )andthe i ( R) are the eigenvalues of A. If we put E = U diag( )U E = U diag( )U E k = U diag( )U then (.) can be rewritten as follows: A = E + (E E )++ k (E k E k )= j E j (.)

4 4 FUNDAMENTAL INEQUALITIES AND MOND-PECARIĆ METHOD where E j = E j E j and E =. Iff (t) is a real valued continuous function on the spectrum Sp(A), then f (A) may be dened by f (A) = f ( j )E j : (.3) This result can be generalized to selfadjoint operators on a Hilbert space H. Let A be a selfadjoint operator on a Hilbert space H and f (t) a real valued continuous function dened on an interval [m M], where m =inf kxk= (Ax x)andm =max kxk= (Ax x). Then A can be expressed as follows: A = Z M m de (.4) where fe : Rg is a family of projections such that E E if, E + = E, E =ande = H. Since a selfadjoint operator A on a Hilbert space H is an extension of a selfadjoint matrix, (.4) can be naturally considered as an extension of (.). Therefore, we have an extension of (.3) under the above situation as follows: f (A) = Z M m f ( )de : (.5) Next, we shall introduce a classical Jensen s inequality as an inequality associated with a convex function: Theorem. (CLASSICAL JENSEN S INEQUALITY) If f (t) is a convex function on an interval [m M] for some scalars m < M, then for every x x x k [m M] and every positive real numbers t t P k t k with t j =, X k t j x j A t j f (x j ): (.6) Proof. Since f (t) is convex, then for each point (s f (s)) there exists a real number l such that l(x s)+f (s) f (x) for all x [m M]: (.7) Put s = P k t jx j [m M], then it follows from (.7) that l(x j s )+f (s ) f (x j ) for j = k: Multiplying this inequality with t j R+ and summing of j we have t j (l(x j s )+f (s )) t j f (x j ):

5 . CLASSICAL JENSEN S INEQUALITY 5 Since t j (l(x j s )+f (s )) = X k t j x j s t j A+ f (s )=f (s ) we have a desired inequality. We rephrase it under matrix situation. If we put A x... x n A and x p t p. A tn then a classical Jensen s inequality (.6) in Theorem. is expressed as f ((Ax x)) (f (A)x x) for every unit vector x. The following theorem is an operator version of Theorem. (classical Jensen s inequality). Theorem. Let A B h (H) be a selfadjoint operator with Sp(A) [m M] for some scalars m < M. If f(t) is a convex function on [m M],then for every unit vector x H. f ((Ax x)) (f (A)x x) (.8) Proof. If we put s =(Ax x), then m s M. Foragiven >, there exist a straight line l(t) such that (i) l(t) f (t)forallt [m M] and (ii) l(s) f (s). Then (i) implies l(a) f (A). Hence we have (f (A)x x) (l(a)x x) =l(s) f (s) for every unit vector x H. Since is an arbitrary, we have (f (A)x x) f ((Ax x)). The following theorem is a multiple operator version of Theorem.: Theorem.3 Let A j B h (H) be selfadjoint operators with Sp(A j ) [m M] (j = k) for some scalars m < M. Let x x x k H be any nite number of vectors such that P k kx jk =.Iff(t) is a convex function on [m M],then X k (A j x j x j ) A (f (A j )x j x j ): (.9)

6 6 FUNDAMENTAL INEQUALITIES AND MOND-PECARIĆ METHOD Proof. If we put à A... A k A and x x. x k A then we have Sp(Ã) [m M], k xk =and P k (A jx j x j )=(à x x). It follows from Theorem. that f ((à x x)) (f (Ã) x x) and hence we have (.9). As a special case of Theorem., we have the following Hölder-McCarthy inequality. Theorem.4 (HÖLDER-MCCARTHY INEQUALITY) Let A B h (H) be a positive operator on a Hilbert space H. Then (i) (A r x x) (Ax x) r for all r > and every unit vector x H. (ii) (A r x x) (Ax x) r for all < r < and every unit vector x H. (iii) If A is invertible, then (A r x x) (Ax x) r for all r < and every unit vector x H. Proof. Since the power function f (t) =t r is convex for r > orr <, and concave for < r <, this theorem follows from Theorem... Operator convexity In this section, we consider another operator version of a classical Jensen s inequality (.6) in Theorem.. We rephrase it under another matrix situation. If we put A x p t... A and V A. p x n tn then a classic Jensen s inequality is expressed as f (V AV) V f (A)V: The formulation offers a fresh insight into the noncommutative case. Its noncommutative version is considered in various way. We shall start with the following denition. Denition. A real valued continuous function f (t) on an interval I is said to be operator convex (resp. operator concave)if f (( )A + B) ( )f (A)+ f (B) (.)

7 . OPERATOR CONVEXITY 7 (resp. f (( )A + B) ( )f (A)+ f (B)) (.) for all [ ] and for every selfadjoint operator A and B on a Hilbert space H whose spectra are contained in I. Also, the condition (.) can be replaced by the more special condition A + B f (A)+f(B) f : (.) Notice that a function f is operator concave if f is operator convex. A real valued continuous function f (t) on an interval I is said to be operator monotone if it is monotone with respect to the operator order, i.e., A B with Sp(A) Sp(B) I implies f (A) f (B): Before we present basic examples of such functions, we prove some lemmas needed later. Lemma.5 If A B + (H) is positive, then X AX for every X B(H). Proof. For every vector x H,(X AXx x) =(AXx Xx). Lemma.6 If A B h (H) is selfadjoint and U is unitary, i.e. U U = UU = H,then f (U AU) =U f (A)U foreveryf C(Sp(A)). Proof. Put B = U AU, thenb is selfadjoint and Sp(B) = Sp(A). Since B m = U A m U for every integer m, we have p(b) =U p(a)u for every polynomial p(t). Since there exist polynomials fp j g such that kf p j k 7! asj!for a given f C(Sp(A)), we have kf (U AU) U f (A)Uk kf (U AU) p j (U AU)k + kp j (U AU) U p j (A)Uk + ku p j (A)U U f (A)Uk 7! as j!and so f (U AU) =U f (A)U. Lemma.7 If A B(H) and f C([ kak ]),thenaf(a A)=f (AA )A. Proof. Since A(A A) n =(AA ) n A for every integer n, we have Ap(A A)=p(AA )A for every polynomial p(t). Since there exist polynomials fp j g such that kf p j k 7! as j!for a given f C([ kak ]), we obtain Af(A A)=f (AA )A. Now, we study basic examples of such functions. Example. The function f (t) = + t is operator monotone on every interval for all R and. It is operator convex for all R.

8 8 FUNDAMENTAL INEQUALITIES AND MOND-PECARIĆ METHOD Example. If f g are operator monotone, and if are positive real numbers, then f + g is also operator monotone. If f n are operator monotone and f n (t)! f (t) as n!, then f is also operator monotone. Example.3 The function f (t) =t on [ ) is not operator monotone though it is monotone increasing. As a matter of fact, if we put A = and B = then A B and A 6 B since A B = Example.4 The function f (t) =t is operator convex on every interval. To see it, for any selfadjoint operators A and B, A + B A + B = 4 (A + B AB BA) = 4 (A B) : This shows that the function f (t) =t + t + is operator convex for all R,. Example.5 The function f (t) =t 3 on [ ) is not operator convex though it is convex on [ ). In fact, if we put A = and B = then we have A 3 + B 3 A + B 3 = : Example.6 The function f (t) = t is operator convex on ( ) and g(t) = t is operator monotone on ( ). In fact, for any positive invertible operators A and B A + B A + B = A + B 4(A(A + B )B) = A + B 4B (A + B ) A = (A + B B )(A + B ) (A (A + B )) = (A B )(A + B ) (A B ) : The last inequality holds by Lemma.5.

9 . OPERATOR CONVEXITY 9 This fact shows that f (t) = t is operator convex. Next, let A B. Then H A BA. Taking inverse both sides, we have I H A B A and hence A B. Therefore it follows that A B and hence g(t) = t is operator monotone on ( ). To relate this, we introducethe following famouslöwner-heinz inequality established in 934. Theorem.8 (LÖWNER-HEINZ INEQUALITY) Let A and B be positive operators on a HilbertspaceH.IfA B,thenA r B r for all r [ ]. We need some elementary results in operator theory in order to prove it. The spectral radius of an operator A is dened as r(a) =maxfj j : Sp(A)g: Notice that r(a) kak and r(a) =kak if A is a selfadjoint operator. Moreover, it follows that r(ab) =r(ba) foralla B B(H), since Sp(AB)nfg = Sp(BA)nfg. Also, if A is positive, then A H if and only if r(a). An operator A is a contraction (kak ) if and only if A A H. Proof of Theorem.8. Let A B. Suppose that A is invertible. Put = fr R : A r B r g: Then the set is closed since r! A r B r are norm continuous and obviously. The hypothesis A B ensures. Therefore, to prove [ ] is sufcient to show that r s implies r+s. If r, then H A r B r A r = B r A r r B A r and hence r B A r : By the same argument, if s, then s B A s. So, we have A (r+s) 4 B r+s (r+s) A 4 = r A (r+s) 4 B r+s (r+s) A 4 by A (r+s) 4 B r+s (r+s) A 4 is positive = r A rs 4 A (r+s) 4 B r+s (r+s) A 4 A sr 4 by r(st) =r(ts) = r A s r+s r B A A s r+s r B A by r(x) k r B A r s B A s :

10 FUNDAMENTAL INEQUALITIES AND MOND-PECARIĆ METHOD Therefore we have A (r+s) 4 B r+s A (r+s) 4 H and hence A r+s B r+s i:e: r + s : This fact shows the theorem under the assumption that A is invertible. Suppose that A is not invertible. For each >, A + H is invertible and A + H B. Therefore it follows from above argument that (A + H ) r B r for all r : By letting!, we have the desired inequality A r B r. Now, we go back to Jensen s inequality. We show some characterizations of operator convexity and operator monotonicity based on the ideas due to Hansen-Pedersen. This leads to some conditions equivalent to Jensen s inequality. Theorem.9 (JENSEN S OPERATOR INEQUALITY) Let H and K be Hilbert space. Let f be a real valued continuous function on an interval I. Let A and A j be selfadjoint operators on H with spectra contained in I (j = k). Then the following conditions are mutually equivalent: (i) f is operator convex on I. (ii) f (C AC) C f (A)C for every A B h (H) and isometry C B(K H), i.e., C C = K. (iii) f (C AC) C f (A)C foreveryab h (H) and isometry C B(H). P k P (iv) f C j A k jc j C j f (A j)c j for every A j B h (H) and C j B(KH) with P k C j C j = K (j = k). P (v) f k P C j A k jc j C j f (A j)c j for every A j B h (H) and C j B(H) with P k C j C j = H (j = k). (vi) f P k P P P k ja j P j P j f (A j )P j for every A j B h (H) and projection P j B h (H) k with P j = H (j = k). Proof. (i) ) (ii): Let X = B B h (K) with (B) I and A B B h (H K) for some selfadjoint operator U = C D C V = C D C B(K H H K)

11 . OPERATOR CONVEXITY where D = p H CC. Since C D = p K C CC = B h (H K) anddc = C p K C C =B h (K H), it follows that both U and V are unitary operators of K H onto H K. Then U XU = C AC C AD DAC DAD + CBC and V XV = C AC C AD DAC DAD + CBC : So, we have C AC D AD + CBC = U XU + V XV : Hence, it follows from the operator convexity of f andlemma.6that f (C AC) C f (D AD + CBC ) = f AC D AD + CBC U XU + V XV = f f (U XU)+f (V XV) = U f (X)U + V f (X)V C = f (A)C D f (A)D + Cf(B)C : Thus we have f (C AC) C f (A)C by seeing the (,)-components. (ii) (iv): X = Let B@ A A... CA B h (H H) C = B@ C C. CA B(K H H): A k C k Then C C = K and hence it follow from (ii) that f ( (iv) (vi): Obviously. C j A j C j )=f ( C X C) C f (X) C = C j f (A j )C j : p p and U = p t p t.thenuis a unitary operator t t (vi) (i): Let A and B be selfadjoint operators with spectrum in I and let t. Let X = A B, P = H on H H. Thus we have f (( t)a + tb) f (ta +(t)b) = f (PU XUP +( HH P)U XU( HH P)) Pf(U XU)P +( HH P)f (U XU)( HH P) = PU f (X)UP +( HH P)U f (X)U( HH P) ( t)f (A)+tf(B) = : tf(a)+(t)f (B)

12 FUNDAMENTAL INEQUALITIES AND MOND-PECARIĆ METHOD Hence f is operator convex on I by seeing the (,)-components. Therefore, we proved the implications (i) (ii) (iv) (vi) (i). To complete the proof, we need the implication (iii) (v) because it is non-trivial in (i) (ii) (iii) (v) (vi) (i). (iii) (v): We onlyshowthecase ofk =, which is essential. Let X = A A B@ A... C CA B and C = CA : H Then C is isometry in B(H H ), i.e., C C = HH. Hence it follows from (iii) f (C A C + C A C ) f (A ) = f (C XC) C f (X)C = C f (A )C + C f (A )C f (A ) A :... Thus we have f (C A C + C A C ) C f (A )C + C f (A )C by seeing the (,)- components. By Theorem.9, we show the following Hansen-Pedersen type Jensen s inequality. Theorem. (HANSEN-PEDERSEN-JENSEN S INEQUALITY) Let I be an interval containing and let f be a real valued continuous function dened on I. Let A and A j be selfadjoint operators on H with spectra contained in I (j = k). Then the following conditions are mutually equivalent: (i) f is operator convex on I and f (). (ii) f (C AC) C f (A)C for every A B h (H) and contraction C B(H), i.e., C C H. P P k (iii) f ( C k j A j C j ) C j f (A j )C j for every A B h (H) and C j B(H) with P k C j C j H (iv) f (PAP) Pf(A)P for every A B h (H) and projection P. Proof. (i) (ii): Suppose that f is operator convex and f (). For every contraction C, put D = p H C C. Since C C + D D = H, it follows from (v) of Theorem.9 that f (C AC) = f (C AC + D D) C f (A)C + D f ()D = C f (A)C by f ()

13 . OPERATOR CONVEXITY 3 and hence we have (ii). (ii) (iii): Put X and C as in the proof (ii) (iv) of Theorem.9, then C C H and hence we have (iii). (iii) (iv): obviously. (iv) (i): Under the same situation in the proof (iv) (i) of Theorem.9, we have f (( t)a + tb) f () PU f (X)UP = = f (PU XUP) ( t)f (A)+tf(B) : Hence f is operator convex and f (). Theorem. Let f C([ ). Iff(t) for all t [ ), then conditions (i) (vi) in Theorems.9 are again equivalent to the following condition (vii) f is an operator monotone function. Proof. Suppose that f is operator convex. Let A B B(H), A B. Then for any < < we can write Since f is operator convex, we have B = A +( ) (B A): f ( B) f (A)+( )f (B A) Since f (X) is positive for every positive operator X, it follows that f ( B) f (A). Letting tend to, we have f (B) f (A). Hence f is operator monotone. Conversely, suppose that f is operator monotone. Let C B(H) beanisometry. Consider the unitary operator U on H H given by where D = p H CC. We put and note that X = U XU = U = Choose now a constant > andset Y = C D C A B h (H H) C AC C AD DAC DAD : C AC + H H :

14 4 FUNDAMENTAL INEQUALITIES AND MOND-PECARIĆ METHOD where is a positive constant to be xed later. We observe that Y U XU = H C AD DAC H DAD F for H H F where F = C AD. Furthermore let, H, then H F F H H DAD = k k +(F )+(F )+ kk k k kfkk kkk + kk for kfk : For a sufciently large we thus obtain U XU Y and consequently the operator monotonicity of f implies U f (X)U = f (U XU) f (Y) or written as matrices C f (A)C Df(A)C C f (A)D Df(A)D f (C AC + H ) : f ( H ) In particular we have C f (A)C f (C AC+ H ). Letting tend to, we get the conclusion of the theorem. Corollary. Let f be a real valued continuous function mapping the positive half line [ ) into itself. Then f is operator monotone if and only if f is operator concave. Theorem.3 Let f C([ r)) and r. Then the following conditions are mutually equivalent. (i) f is operator convex and f (). (ii) The function t 7! f (t) t is operator monotone on ( r). Proof. Suppose that f is operator convex. Let A B B h (H) be selfadjoint operators with Sp(A) Sp(B) ( r) anda B. ThenA and B are invertible. If we put C = B = A =, then CC = B = AB = H and kck. Since A = C BC, it follows from (ii) in Theorem. that f (A) =f (C BC) C f (B)C = A = B = f (B)B = A =

15 . OPERATOR CONVEXITY 5 and hence A = f (A)A = B = f (B)B =. Therefore we have A f (A) B f (B) and f (t)=t is operator monotone. Conversely, suppose that f (t)=t is operator monotone on ( r). Since f (t)=t f ()= for < t < r, wehavef (t) (f ()=)t. Letting t 7!, we have f (). We will show that f satises the condition (iv) of Theorem.. Let P be any projection and let A be any positive operator with spectrum in ( r) and Sp(( + )A) ( r) fora sufciently small >. Put P = P + H and X = P A.SinceP ( + ) H,wehave A P A ( + )A. Since f A P A = f A P A A P A t = f (X X )(X X ) = X = X f (X X )X X X f (X X )X it follows from the operator monotonicity of f (t)=t that Therefore we obtain X f (X X )X = f t f t A P A (( + )A) = (+) A f (( + )A)A : f (X X ) ( + ) X A f (( + )A)A X =(+) P f (( + )A)P : Let!. This gives X X! A PA and hence we have f (APA) Pf(A)P as desired. From the previous theorem we obtain the following corollary. Corollary.4 Let f C([ )) and f >. The function f is operator monotone if and only if the function t=f (t) is operator monotone. Proof. Suppose that f is operator monotone. Since f is operator convex, it follows from Theorem.3 that f (t)=t is operator monotone on ( ). Hence t=f (t) = (f (t)=t) is operator monotone on ( ). By the continuity of f, we have the desired result. Conversely, suppose that t=f (t) is operator monotone. If we put g(t) =t=f (t), then g(t) and by Theorem. the operator monotonicity of g(t) implies the operator convexity of g(t) andg(). It follows from Theorem.3 that g(t)=t = =f (t) is operator monotone on ( ) and this fact is equivalent to the operator monotonicity of f (t).