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Some Integral Inequalities for Operator Monotonic Functions on Hilbert Spaces Received May 6, 2020; Accepted June 24, 2020

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Some Integral Inequalities for Operator Monotonic Functions on Hilbert Spaces Received May 6, 2020; Accepted June 24, 2020 Spec. Matrices 2020; 8:172–180 Research Article Open Access Silvestru Sever Dragomir* Some integral inequalities for operator monotonic functions on Hilbert spaces https://doi.org/10.1515/spma-2020-0108 Received May 6, 2020; accepted June 24, 2020 Abstract: Let f be an operator monotonic function on I and A, B 2 SAI (H) , the class of all selfadjoint op- erators with spectra in I. Assume that p : [0, 1] ! R is non-decreasing on [0, 1]. In this paper we obtained, among others, that for A ≤ B and f an operator monotonic function on I, Z1 Z1 Z1 0 ≤ p (t) f ((1 − t) A + tB) dt − p (t) dt f ((1 − t) A + tB) dt 0 0 0 1 ≤ [p (1) − p (0)] [f (B) − f (A)] 4 in the operator order. Several other similar inequalities for either p or f is dierentiable, are also provided. Applications for power function and logarithm are given as well. Keywords: Operator monotonic functions, Integral inequalities, Čebyšev inequality, Grüss inequality, Os- trowski inequality MSC: 47A63, 26D15, 26D10. 1 Introduction Consider a complex Hilbert space (H, h·, ·i). An operator T is said to be positive (denoted by T ≥ 0) if hTx, xi ≥ 0 for all x 2 H and also an operator T is said to be strictly positive (denoted by T > 0) if T is positive and invertible. A real valued continuous function f (t) on (0, ∞) is said to be operator monotone if f (A) ≥ f (B) holds for any A ≥ B > 0. In 1934, K. Löwner [7] had given a denitive characterization of operator monotone functions as follows: Theorem 1. A function f : (0, ∞) ! R is operator monotone in (0, ∞) if and only if it has the representation ∞ Z t f (t) = a + bt + dm (s) t + s 0 where a 2 R and b ≥ 0 and a positive measure m on (0, ∞) such that Z∞ dm (s) < ∞. t + s 0 *Corresponding Author: Silvestru Sever Dragomir: Mathematics, College of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science & Applied Mathemat- ics, University of the Witwatersrand, Johannesburg, South Africa E-mail: [email protected], http://rgmia.org/dragomir Open Access. © 2020 Silvestru Sever Dragomir, published by De Gruyter. This work is licensed under the Creative Commons Attribution alone 4.0 License. Some integral inequalities for operator monotonic functions Ë 173 We recall the important fact proved by Löwner and Heinz that states that the power function f : (0, ∞) ! R, α f (t) = t is an operator monotone function for any α 2 [0, 1] . In [3], T. Furuta observed that for αj 2 [0, 1] , j = 1, ..., n the functions 0 1−1 n n α X α X − j g (t) := @ t− j A and h (t) = 1 + t−1 j=1 j=1 are operator monotone in (0, ∞). Let f(t) be a continuous function (0, ∞) ! (0, ∞). It is known that f(t) is operator monotone if and only if g(t) = t/f (t) =: f *(t) is also operator monotone, see for instance [3] or [8]. Consider the family of functions dened on (0, ∞) and p 2 [−1, 2] \ f0, 1g by p p − 1 t − 1 fp (t) := p tp−1 − 1 and t f0 (t) := ln t, 1 − t t − 1 f1 (t) := (logarithmic mean). ln t We also have the functions of interest 2t p f−1 (t) = (harmonic mean), f (t) = t (geometric mean). 1 + t 1/2 In [2] the authors showed that fp is operator monotone for 1 ≤ p ≤ 2. In the same category, we observe that the function t − 1 gp (t) := tp − 1 is an operator monotone function for p 2 (0, 1], [3]. It is well known that the logarithmic function ln is operator monotone and in [3] the author obtained that the functions 1 1 f t = t 1 + t ln 1 + , g t = ( ) ( ) t ( ) 1 (1 + t) ln 1 + t are also operator monotone functions on (0, ∞) . Let f be an operator monotonic function on an interval of real numbers I and A, B 2 SAI (H) , the class of all selfadjoint operators with spectra in I. Assume that p : [0, 1] ! R is non-decreasing on [0, 1]. In this paper we obtain, among others, that for A ≤ B and f an operator monotonic function on I, Z1 Z1 Z1 0 ≤ p (t) f ((1 − t) A + tB) dt − p (t) dt f ((1 − t) A + tB) dt 0 0 0 1 ≤ [p (1) − p (0)] [f (B) − f (A)] 4 in the operator order. Several other similar inequalities for either p or f is dierentiable, are also provided. Applications for power function and logarithm are given as well. 2 Main Results For two Lebesgue integrable functions h, g : [a, b] ! R, consider the Čebyšev functional: b b b Z Z Z 1 1 1 C (h, g) := h(t)g(t)dt − h(t)dt g(t)dt. (2.1) b − a b − a b − a a a a 174 Ë Silvestru Sever Dragomir It is well known that, if h and g have the same monotonicity on [a, b] , then b b b Z Z Z 1 1 1 h(t)g(t)dt ≥ h(t)dt g(t)dt, (2.2) b − a b − a b − a a a a which is known in the literature as Čebyšev’s inequality. In 1935, Grüss [4] showed that 1 jC (h, g)j ≤ (M − m)(N − n) , (2.3) 4 provided that there exists the real numbers m, M, n, N such that m ≤ h (t) ≤ M and n ≤ g (t) ≤ N for a.e. t 2 [a, b] . (2.4) 1 The constant 4 is best possible in (2.1) in the sense that it cannot be replaced by a smaller quantity. Let f be a continuous function on I. If (A, B) 2 SAI (H) , the class of all selfadjoint operators with spectra in I and t 2 [0, 1] , then the convex combination (1 − t) A + tB is a selfadjoint operator with the spectrum in I showing that SAI (H) is a convex set in the Banach algebra B (H) of all bounded linear operators on H. By the continuous functional calculus of selfadjoint operator we also conclude that f ((1 − t) A + tB) is a selfadjoint operator in B (H) . For A, B 2 SAI (H) , we consider the auxiliary function '(A,B) : [0, 1] ! B (H) dened by '(A,B) (t) := f ((1 − t) A + tB) . (2.5) For x 2 H we can also consider the auxiliary function '(A,B);x : [0, 1] ! R dened by '(A,B);x (t) := '(A,B) (t) x, x = hf ((1 − t) A + tB) x, xi . (2.6) Theorem 2. Let A, B 2 SAI (H) with A ≤ B and f an operator monotonic function on I. If p : [0, 1] ! R is monotonic nondecreasing on [0, 1] , then Z1 Z1 Z1 0 ≤ p (t) f ((1 − t) A + tB) dt − p (t) dt f ((1 − t) A + tB) dt (2.7) 0 0 0 1 ≤ [p (1) − p (0)] [f (B) − f (A)] . 4 If p : [0, 1] ! R is monotonic nonincreasing on [0, 1] , then Z1 Z1 Z1 0 ≤ p (t) dt f ((1 − t) A + tB) dt − p (t) f ((1 − t) A + tB) dt (2.8) 0 0 0 1 ≤ [p (0) − p (1)] [f (B) − f (A)] . 4 Proof. Let 0 ≤ t1 < t2 ≤ 1 and A ≤ B. Then (1 − t2) A + t2B − (1 − t1) A − t1B = (t2 − t1)(B − A) ≥ 0 and by operator monotonicity of f we get f ((1 − t2) A + t2B) ≥ f ((1 − t1) A + t1B) , which is equivalent to '(A,B);x (t2) = hf ((1 − t2) A + t2B) x, xi ≥ hf ((1 − t1) A + t1B) x, xi = '(A,B);x (t1) Some integral inequalities for operator monotonic functions Ë 175 that shows that the scalar function '(A,B);x : [0, 1] ! R is monotonic nondecreasing for A ≤ B and for any x 2 H. If we write the inequality (2.2) for the functions p and '(A,B);x we get Z1 Z1 Z1 p (t) hf ((1 − t) A + tB) x, xi dt ≥ p (t) dt hf ((1 − t) A + tB) x, xi dt, 0 0 0 which can be written as 0 1 0 1 * Z1 + * Z1 Z1 + @ p (t) f ((1 − t) A + tB) dtA x, x ≥ @ p (t) dt f ((1 − t) A + tB)A dtx, x 0 0 0 for x 2 H, and the rst inequality in (2.7) is obtained. We also have that hf (A) x, xi = '(A,B);x (0) ≤ '(A,B);x (t) = hf ((1 − t) A + tB) x, xi ≤ '(A,B);x (1) = hf (B) x, xi and p (0) ≤ p (t) ≤ p (1) for all t 2 [0, 1] . By writing Grüss’ inequality for the functions '(A,B);x and p, we get Z1 Z1 Z1 0 ≤ p (t) hf ((1 − t) A + tB) x, xi dt − p (t) dt hf ((1 − t) A + tB) x, xi dt 0 0 0 1 ≤ [p (1) − p (0)] [hf (B) x, xi − hf (A) x, xi] 4 for x 2 H and the second inequality in (2.7) is obtained. A continuous function g : SAI (H) ! B (H) is said to be Gâteaux dierentiable in A 2 SAI (H) along the direction B 2 B (H) if the following limit exists in the strong topology of B (H) g (A + sB) − g (A) rgA (B) := lim 2 B (H) .
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