HERMITE-HADAMARD TYPE INEQUALITIES FOR OPERATOR GEOMETRICALLY CONVEX FUNCTIONS

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1 HERMITE-HADAMARD TYPE INEQUALITIES FOR OPERATOR GEOMETRICALLY CONVEX FUNCTIONS A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR Abstract In this paper, we introduce the concept of operator geometrically convex functions for positive linear operators and prove some Hermite- Hadamard type inequalities for these functions As applications, we obtain trace inequalities for operators which give some refinements of previous results Introduction and preliminaries Let A be a sub-algebra of B(H stand for the commutative C -algebra of all bounded linear operators on a complex Hilbert space H with inner product, An operator A A is positive and write A if Ax, x for all x H Let A + stand for all strictly positive operators in A Let A be a self-adjoint operator in A The Gelfand map establishes a - isometrically isomorphism Φ between the set C(Sp(A of all continuous functions defined on the spectrum of A, denoted Sp(A, and the C -algebra C (A generated by A and the identity operator H on H as follows: For any f, g C(Sp(A and any α, β C we have: Φ(αf + βg = αφ(f + βφ(g; Φ(fg = Φ(fΦ(g and Φ( f = Φ(f ; Φ(f = f := sup t Sp(A f(t ; Φ(f = H and Φ(f = A, where f (t = and f (t = t, for t Sp(A with this notation we define f(a = Φ(f for all f C(Sp(A and we call it the continuous functional calculus for a self-adjoint operator A If A is a self-adjoint operator and f is a real valued continuous function on Sp(A, then f(t for any t Sp(A implies that f(a, ie f(a is a positive operator on H Moreover, if both f and g are real valued functions on Mathematics Subject Classification 47A63, 5A6, 47B5, 47B, 6D5 Key words and phrases Hermite-Hadamard inequality, Operator geometrically convex function, Trace inequality

2 A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR Sp(A then the following important property holds: f(t g(t for any t Sp(A implies that f(a g(a, ( in the operator order of B(H, see [8] Let I be an interval in R Then f : I R is said to be convex function if f(λa + ( λb λf(a + λf(b for a, b I and λ [, ] The following inequality holds for any convex function f defined on R ( a + b b f(a + f(b (b af f(xdx (b a, a, b R ( a It was firstly discovered by Hermite in 88 in the journal Mathesis (see [] But this result was nowhere mentioned in the mathematical literature and was not widely known as Hermite s result [4] Beckenbach, a leading expert on the history and the theory of convex functions, wrote that this inequality was proven by Hadamard in 893 [] In 974, Mitrinovič found Hermites note in Mathesis [] Since (3 was known as Hadamards inequality, the inequality is now commonly referred as the Hermite- Hadamard inequality [4] Definition [] A continuous function f : I R + R + is said to be geometrically convex function (or multiplicatively convex function if for a, b I and λ [, ] f(a λ b λ f(a λ f(b λ The author of [8] established the Hermite-Hadamard type inequalities for geometrically convex functions as follows: Theorem Let f : I R + R + be a geometrically convex function and a, b I with a < b If f L [a, b], then f( b ( ab ab f(tf dt ln b ln a t t a b ln b ln a a f(b f(a ln f(b ln f(a f(a + f(b f(t dt t

3 HERMITE-HADAMARD TYPE INEQUALITY FOR OPERATOR G-CONVEX 3 By changing variables t = a λ b λ we have ln b ln a b a f(t dt = t f(a λ b λ dλ Remark It is well-known that for positive numbers a and b min{a, b} G(a, b = ab L(a, b = b a a + b A(a, b = max{a, b} ln b ln a The author of [9] mentioned the following inequality, but here we provide a short proof which gives a refinement for above theorem Theorem 3 Let f be a geometrically convex function defined on I a subinterval of R + Then, we have f( b ( ab ab f(tf dt f(af(b ln b ln a t t for a, b I Proof Since f is geometrically convex function, we can write f( ( ab = f (a λ b λ (a λ b λ f(a λ b λ f(a λ b λ f(a λ f(b λ f(a λ f(b λ for all λ [, ] So, we have f( ab Integrate (3 over [, ], we have f( ab a = f(af(b f(a λ b λ f(a λ b λ f(af(b (3 ln b ln a b a f(tf t ( ab dt f(af(b t Lemma 4 [, Page 56] Suppose that I is a subinterval of R + and f : I (, is a geometrically convex function Then F = log f exp : log(i R is a convex function Conversely, if J is an interval ( for which exp(j is a subinterval of R + and F : J R is a convex function, then f = exp F log : exp(j R +

4 4 A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR is geometrically convex function Theorem 5 Let f be a geometrically convex function defined on [a, b] such that < a < b Then, we have f( ( ab f(a 3 4 b 4 f(a 4 b 3 4 ( b log f(t exp dt log b log a a t f( ab 4 f(a 4 f(b f(af(b for a, b I Proof Let f : [a, b] R be a geometrically convex function So, by Lemma 4 we have is convex Then, by [3, Remark 93] ( log a + log b F F (x = log f exp(x : [log a, log b] R ( F ( 3 log a + log b 4 log b + F ( log a + 3 log b F (xdx log b log a log a ( ( log a + log b F (log a + F (log b F + F (log a + F (log b By definition of F, we obtain log f exp(log ab ( ( ( log f exp log a 3 4 b 4 + log f exp log b log f exp(xdx log b log a log a ( ( log f exp log a b + log f exp(log a + log f exp(log b 4 log a 3 4 b 4 log f exp(log a + log f exp(log b

5 HERMITE-HADAMARD TYPE INEQUALITY FOR OPERATOR G-CONVEX 5 It follows that log f( ab ( log f(a b 4 + log f(a 4 b 4 log b log f exp(xdx log b log a log a ( ( log f a log f(a + log f(b b + log f(a + log f(b Since exp(x is increasing, we have f( ( ab f(a 3 4 b 4 f(a 4 b 3 4 ( log b exp log f(exp(xdx log b log a log a f( ab 4 f(a 4 f(b f(af(b Using change of variable t = exp(x to obtain the desired result the desired result The author of [, p 58] showed that every polynomial P (x with nonnegative coefficients is a geometrically convex function on [, More generally, every real analytic function f(x = n= c nx n with non-negative coefficients is geometrically convex function on (, R where R denotes the radius of convergence This gives some different examples of geometrically convex function It is easy to show that exp(x is geometrically convex function In this paper, we introduce the concept of operator geometrically convex functions and prove the Hermite-Hadamard type inequalities for these class of functions These results lead us to obtain some inequalities for trace functional of operators Inequalities for operator geometrically convex functions In this section, we prove Hermite-Hadamard type inequality for operator geometrically convex function In [5] Dragomir investigated the operator version of the Hermite-Hadamard inequality for operator convex functions Let f : I R be an operator convex function on the interval I then, for any self-adjoint operators A and B with

6 6 A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR spectra in I, the following inequalities holds ( 3 A + B 4 f f(ta + ( tbdt ( 4 [ ( ( ] 3A + B A + 3B f + f ( 4 4 f (( ta + tb dt [ ( ] A + B f(a + f(b f + (3 f(a + f(b, (4 for the first inequality in above, see [6] To give operator geometrically convex function definition, we need following lemmas Lemma [, Lemma 3] Let A and B be two operators in A +, and f a continuous function on Sp(A Then, AB = BA implies that f(ab = Bf(A Since f(t = t λ is continuous function for λ [, ] and A is a commutative C -algebra, we have A λ B = BA λ Moreover, by applying above lemma for f(t = t λ again, we have A λ B λ = B λ A λ, for operators A and B in A + It means A λ and B λ commute together whenever A and B commute Lemma Let A and B be two operators in A + Then is convex {A λ B λ : λ } Proof We know that {λa + ( λb : λ } is convex for arbitrary operator A and B So, {λ log A + ( λ log B : λ } is convex Since A and B are commutative and knowing that e f is convex when f is convex, we have So, A λ B λ is convex for λ e (λ log A+( λ log B = e λ log A ( λ log B e = A λ B λ Lemma 3 [8, Theorem 53] Let A and B be in a Banach algebra such that AB = BA Then Sp(AB Sp(A Sp(B

7 HERMITE-HADAMARD TYPE INEQUALITY FOR OPERATOR G-CONVEX 7 Let A and B be two positive operators in A with spectra in I Now, Lemma and functional calculus [8, Theorem 3 (c] imply that for λ Sp(A λ B λ Sp(A λ Sp(B λ = Sp(A λ Sp(B λ I Definition 4 A continuous function f : I R + R + is said to be operator geometrically convex if f(a λ B λ f(a λ f(b λ for A, B A + such that Sp(A, Sp(B I Now, we are ready to prove Hermite-Hadamard type inequality for operator geometrically convex functions Theorem 5 Let f be an operator geometrically convex function Then, we have log f( AB log f(a t B t dt log f(af(b (5 for t and A, B A + such that Sp(A, Sp(B I Proof Since f is operator geometrically convex function, we have f( AB f(af(b Let replace A and B by A t B t and A t B t respectively, we obtain f( AB f(a t B t f(a t B t (6 It is well-known that log t is operator monotone function on (, (see [7], ie, log t is operator monotone function if log A log B when A B So, by above inequality, we have log f( AB log f(a t B t f(a t B t = log ( f(a t B t f(a t B t = ( log f(a t B t + log f(a t B t Therefore, log f( AB ( log f(a t B t + log f(a t B t Integrate above inequality over [, ], we can write the following log f( ABdt ( log f(a t B t dt + log f(a t B t dt = log f(a t B t dt (7

8 8 A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR The last equality above follows by knowing that Hence, from (7, we have log f(a t B t dt = log f( AB log f(a t B t dt log f(a t B t dt This proved left inequality of (5 On the other hand, we have f(a t B t f(a t f(b t It follows that So, log f(a t B t log f(a t f(b t = log f(a t + log f(b t = t log f(a + ( t log f(b log f(a t B t t log f(a + ( t log f(b (8 Now, integrate of (8 on [, ], we have log f(a t B t dt This completes the proof = log f(a t log f(adt + tdt + log f(b = (log f(a + log f(b = log f(af(b ( t log f(bdt ( tdt We should mention, when f is operator geometrically convex function, then we have So, we have f( AB = f( A t B t A t B t f(a t B t f(a t B t f(a t f(b t f(a t f(b t = f(af(b f( AB f(a t B t f(a t B t f(af(b Integrate above inequality over [, ], we obtain f( AB f(at B t f(a t B t dt f(af(b,

9 HERMITE-HADAMARD TYPE INEQUALITY FOR OPERATOR G-CONVEX 9 for t and A, B A + such that Sp(A, Sp(B I Let A, B A and A B, by continuous functional calculus [8, Theorem 3 (b], we can easily obtain exp(a exp(b This means exp(t is operator monotone on [, for A, B A On the other hand, like the classical case, the arithmetic-geometric mean inequality holds for operators as following A ( A BA ν A ( νa + νb, ν [, ] (9 with respect to operator order for positive non-commutative operator in B(H Whenever, A and B commute together, then inequality (9 reduces to A ν B ν ( νa + νb, ν [, ] ( Since exp(t is an operator monotone function, by above inequality we have exp ( A ν B ν exp (( νa + νb = exp(( νa exp(νb = exp(a ν exp(b ν, for A, B A + and ν [, ] So, in this case exp(t is an operator geometrically convex function on [, Let replace f in Theorem 5 by exp(t as an operator geometrically convex function, we have So, log exp( AB log exp(a t B t dt log exp(a exp(b AB = log (exp(a exp(b = (log exp(a + log exp(b A t B t dt A + B, ( for A, B A + Here, we mention some remarks for operator geometrically convex functions Remark f(x = x is geometrically convex function for usual operator norms since the following hold f(a α B α = A α B α A α B α = f(a α f(b α Above inequality is a special case of McIntosh inequality

10 A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR Remark 3 If f(t is an operator geometrically convex function, then so is g(t = tf(t for α [, ] and A, B A + g(a α B α = A α B α f(a α B α A α B α f(a α f(b α A α f(a α B α f(b α = g(a α g(b α Remark 4 Operator geometrically convex functions is an algebra with some complication of operators spectra To see this we make use of the following inequality A α B α + C α D α (A + C α + (B + D α ( for A, B, C, D A + Let f and g be operator geometrically convex functions First, we prove that f + g is an operator geometrically convex function (f + g(a α B α = f(a α B α + g(a α B α f(a α f(b α + g(a α g(b α (f(a + g(a α + (f(b + g(b α = ((f + g(a α + ((f + g(b α for A, B A + In the last inequality above we applied ( Second, we show that mf is an operator geometrically convex function for a scalar m (mf(a α B α mf(a α f(b α = (mf(a α (mf(b α for A, B A + Third, h = fg is an operator geometrically convex function for A, B A + h(a α B α = f(a α B α g(a α B α f(a α f(b α g(a α g(b α = f(a α g(a α f(b α g(b α = h(a α h(b α Let {e i } i I be an orthonormal basis of H, we say that A B(H is trace class if A := i I A e i, e i < (3

11 HERMITE-HADAMARD TYPE INEQUALITY FOR OPERATOR G-CONVEX The definition of A does not depend on the choice of the orthonormal basis {e i } i I We denote by B (H the set of trace class operators in B(H We define the trace of a trace class operator A B (H to be Tr(A := i I Ae i, e i, (4 where {e i } i I an orthonormal basis of H Note that this coincides with the usual definition of the trace if H is finitedimensional We observe that the series (4 converges absolutely The following result collects some properties of the trace: Theorem 6 We have (i If A B (H then A B (H and Tr(A = Tr(A; (5 (ii If A B (H and T B(H, then AT, T A B (H and Tr(AT = Tr(T A and Tr(AT A T ; (6 (iii Tr( is a bounded linear functional on B (H with Tr = ; (iv If A, B B (H then Tr(AB = Tr(BA For the theory of trace functionals and their applications the reader is referred to [5] For A, B we have Tr(AB Tr(A Tr(B Also, since f(t = t is monotone we have Tr(AB Tr(A Tr(B (7 for positive operator A and B in B(H We know that f(t = Tr(t is operator geometrically convex function [7, p53], ie Tr(A t B t Tr(A t Tr(B t for t and positive operators A, B B (H For commutative case, we have Tr(AB Tr( AB Tr(A Tr(B, since (Tr(AB Tr(AB

12 A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR Moreover, by Theorem 5 we can write log Tr( AB log Tr(A t B t dt log Tr(A Tr(B = (log Tr(A + log Tr(B Let replace A and B by A and B in above inequality, respectively By applying commutativity of algebra and knowing that Tr(A (Tr A for positive operator A, we have log Tr(AB log Tr(A t B ( t dt log (Tr(A Tr(B 3 More results on trace functional class for product of operators In this section we prove some trace functional class inequalities for operators which are not necessarily commutative We consider the wide class of unitarily invariant norms Each of these norms is defined on an ideal in B(H and it will be implicitly understood that when we talk of T, then the operator T belongs to the norm ideal associated with Each unitarily invariant norm is characterized by the invariance property UT V = T for all operators T in the norm ideal associated with and for all unitary operators U and V in B(H For p <, the Schatten p-norm of an operator A B (H defined by A p = (Tr A p /p These Schatten p-norms are unitarily invariant In [], Bhatia and Davis proved the following inequality A XB r AA X r XBB r (3 for all operators A, B, X and r As we know, A = Tr A From (3 for p =, we have So, by inequality (3, we can write A XB r AA X r XBB r (3 (Tr A XB r Tr( AA X r Tr( XBB r, (33 for all operators A, B B (H, X B(H and r Let, X = I in above inequality, we have (Tr A B r Tr( AA r Tr( BB r Moreover, let r = in inequality (33, we have Tr(A XB (Tr A XB Tr( AA X Tr( XBB

13 HERMITE-HADAMARD TYPE INEQUALITY FOR OPERATOR G-CONVEX 3 Put X instead of X and applying the property of trace we have Tr(AB X Tr( AA X Tr( X BB, (34 for all A, B B (H and X B(H Let X = I in (34 Corollary 3 Let A, B B (H Then Tr(AB Tr(AA Tr(BB (35 In [6, Theorem 5], Dragomir proved the following inequality for X B(H, A, B B (H and α [, ] Tr(AB X Tr ( A X α ( Tr B X ( α Here, we give a generalization for above inequality when α R Theorem 3 Let X B (H, A, B B(H and α R Then Tr(AB X Tr ( A X α ( Tr B X ( α (36 Proof Let replace A and B in Corollary 3 with X α A and X ( α B, where α R It follows that Tr(AB X Tr ( X α AA X α Tr ( X ( α BB X ( α = Tr (AA X α X α Tr (BB X ( α X ( α = Tr ( A X α ( Tr B X ( α So, we have Tr(AB X Tr ( A X α ( Tr B X ( α Let A = B = I in Theorem 3, we have Tr(X Tr ( X α Tr ( X ( α, for X B (H and α R Above inequality is a refinement for [6, Inequality (3] Also, let X B (H and normal operators A, B B(H For α R, we have Tr(AB X Tr ( A X α ( Tr B X ( α

14 4 A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR In [4, Theorem 3], F M Dannan proved that if S i and T i (i =,,, n are positive definite matrices, then we have ( ( n n ( n Tr S i T i Tr Tr (37 Si Ti i= i= i= Moreover, if S i T i, (i =,,, n Then ( n ( n Tr S i T i Tr S i T i i= Tr So, i= ( n ( ( n n Tr S i T i Tr Si i= Tr ( n Ti i= Si Tr Ti i= i= i= ( n Here, we prove inequality (37 for arbitrary operators Theorem 33 Let S i and T i (i =,,, n be arbitrary operators in B (H Then, ( n Tr ( n ( n S i Ti Tr S i Si Tr T i Ti (38 i= i= S S S n T T T n Proof Let A = and B = So, we have n i= S iti AB =, i= n i= S isi AA =,

15 HERMITE-HADAMARD TYPE INEQUALITY FOR OPERATOR G-CONVEX 5 n i= T iti BB = Put A and B in inequality (35, by property of trace, we obtain the desired result Corollary 34 Let S i and T i (i =,,, n be positive operators in B (H Then, we have ( ( n n ( n Tr S i T i Tr Tr i= Proof By Theorem 33 for positive operators S i and T i, we obtain ( n Tr ( n ( n S i T i Tr Si Tr Ti i= Since S i and T i are positive operators, we have Tr(S i T i It follows that Tr( n i= S it i because Tr ( n i= S it i = n i= Tr(S it i So, ( ( n n ( n Tr S i T i Tr Tr i= i= i= i= S i S i References [] EF Beckenbach, Convex functions, Bull Amer Math Soc 54 ( [] R Bhatia, C Davis, A Cauchy-Schwarz inequality for operators with applications, Linear Algebra Appl 3/4 ( [3] R Bhatia, Matrix Analysis, GTM 69, Springer-Verlag, New York, 997 [4] F M Dannan, Matrix and Operator Inequalities, J Ineq Pure and Appl Math ( Article 34 [5] S S Dragomir, Hermite-Hadamards type inequalities for operator convex functions, Applied Mathematics and Computation 8 ( [6] S S Dragomir, Some Inequalities for Trace Class Operators Via a Kato s Result, Preprint RGMIA Res Rep Coll 7 (4, Article 5 [Online [7] R A Horn, C R Johnson, Matrix Analysis, Cambridge University Press, [8] İ, İscan, Some new Hermite-Hadamard type inequalities for geometrically convex functions, Mathematics and Statistics, ( ( [9] İ, İscan, On some new Hermite-Hadamard type inequalities for s-geometrically convex functions, International Journal of Mathematics and Mathematical Sciences, vol4, Article ID 639, 8 pages i= i= i= T i T i

16 6 A TAGHAVI, V DARVISH, H M NAZARI, S S DRAGOMIR [] DS Mitrinović, IB Lacković, Hermite and convexity, Aequationes Math 8 ( [] M Nagisa, M Ueda and S Wada, Commutativity of operators, Nihonkai Math J 7 (6-8 [] C P Niculescu, Convexity according to the geometric mean, Math Inequal Appl 3 ( (, [3] C P Niculescu and L E Persson, Convex Functions and their Applications: A Contemporary Approach, Springer, 4 [4] JE Pečarić, F Proschan, YL Tong, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press Inc, San Diego, 99 [5] B Simon, Trace Ideals and Their Applications, Cambridge University Press, Cambridge, 979 [6] A Taghavi, V Darvish, H M Nazari, S S Dragomir, Some inequalities associated with the Hermite-Hadamard inequalities for operator h-convex functions, RGMIA Research Report Collection, 8 (5, Article ( [7] X Zhan, Matrix Inequalities, Springer-Verlag, Berlin, [8] K Zhu, An introduction to operator algebras, CRC Press, 993 Department of Mathematics,, Faculty of Mathematical Sciences,, University of Mazandaran,, P O Box ,, Babolsar, Iran address: taghavi@umzacir, vahiddarvish@vueduau, mnazari@stuumzacir Mathematics, School of Engineering and Science Victoria University,, PO Box 448, Melbourne City, MC 8, Australia School of Computational and Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg 5, South Africa address: severdragomir@vueduau