� EXTRACTA MATHEMATICAE Volumen 33, Número 2, 2018 instituto de investigación de matemáticas de la universidad de extremadura EXTRACTA MATHEMATICAE Vol. 37, Num. 2 (2022), 261 – 282 doi:10.17398/2605-5686.37.2.261 Available online October 26, 2022 Second derivative Lipschitz type inequalities for an integral transform of positive operators in Hilbert spaces S.S. Dragomir 1, 2 1 Mathematics, College of Engineering & Science Victoria University, PO Box 14428, Melbourne City 8001, Australia 2 DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences School of Computer Science & Applied Mathematics University of the Witwatersrand, Johannesburg, South Africa sever.dragomir@vu.edu.au , http://rgmia.org/dragomir Received July 6, 2022 Presented by M. Maestre Accepted October 5, 2022 Abstract: For a continuous and positive function w (λ), λ > 0 and µ a positive measure on (0,∞) we consider the following integral transform D (w,µ) (T) := ∫ ∞ 0 w (λ) (λ + T)−1 dµ (λ) , where the integral is assumed to exist for T a positive operator on a complex Hilbert space H. We show among others that, if A ≥ m1 > 0, B ≥ m2 > 0, then ‖D (w,µ) (B) −D (w,µ) (A) −D (D (w,µ)) (A) (B −A)‖ ≤‖B −A‖2 × { D(w,µ)(m2)−D(w,µ)(m1)−(m2−m1)D′(w,µ)(m1) (m2−m1)2 if m1 6= m2, 1 2 D′′ (w,µ) (m) if m1 = m2 = m, where D (D (w,µ)) is the Fréchet derivative of D (w,µ) as a function of operator and D′′ (w,µ) is the second derivative of D (w,µ) as a real function. We also prove the norm integral inequalities for power r ∈ (0, 1] and A, B ≥ m > 0, ∥∥∥∥∥ ∫ 1 0 ((1 − t) A + tB)r−1 dt− ( A + B 2 )r−1∥∥∥∥∥ ≤ 124 (1 −r) (2 −r) mr−3 ‖B −A‖2 and ∥∥∥∥Ar−1 + Br−12 − ∫ 1 0 ((1 − t) A + tB)r−1 dt ∥∥∥∥ ≤ 112 (1 −r) (2 −r) mr−3 ‖B −A‖2 . Key words: operator monotone functions, operator convex functions, operator inequalities, midpoint inequality, trapezoid inequality. MSC (2020): 47A63, 47A60. ISSN: 0213-8743 (print), 2605-5686 (online) © The author(s) - Released under a Creative Commons Attribution License (CC BY-NC 3.0) https://doi.org/10.17398/2605-5686.37.2.261 mailto:sever.dragomir@vu.edu.au http://rgmia.org/dragomir https://publicaciones.unex.es/index.php/EM https://creativecommons.org/licenses/by-nc/3.0/ 262 s.s. dragomir 1. Introduction Consider a complex Hilbert space (H,〈·, ·〉). An operator T on H is said to be positive (denoted by T ≥ 0) if 〈Tx,x〉 ≥ 0 for all x ∈ 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 on (0,∞) is said to be operator monotone if f(A) ≥ f(B) holds for any A ≥ B > 0. In 1934, K. Löwner [14] had given a definitive characterization of operator monotone functions as follows, see for instance [4, pp. 144 – 145]: Theorem 1. A function f : [0,∞) → R is operator monotone in [0,∞) if and only if it has the representation f (t) = f (0) + bt + ∫ ∞ 0 tλ t + λ dµ (λ) (1.1) where b ≥ 0 and a positive measure µ on [0,∞) such that∫ ∞ 0 λ 1 + λ dµ (λ) < ∞. We recall the important fact proved by Löwner and Heinz that states that the power function f : (0,∞) → R, f(t) = tα is an operator monotone function for any α ∈ [0, 1], see [12]. The function ln is also operator monotone on (0,∞). For other examples of operator monotone functions, see [10, 11]. Let B(H) be the Banach algebra of bounded linear operators on a complex Hilbert space H. The absolute value of an operator A is the positive operator |A| defined as |A| := (A∗A)1/2. It is known that [3] in the infinite-dimensional case the map f(A) := |A| is not Lipschitz continuous on B(H) with the usual operator norm, i.e., there is no constant L > 0 such that ‖|A|− |B|‖≤ L‖A−B‖ for any A, B ∈B(H). However, as shown by Farforovskaya in [7, 8] and Kato in [13], the following inequality holds ‖|A|− |B|‖≤ 2 π ‖A−B‖ ( 2 + log ( ‖A‖ + ‖B‖ ‖A−B‖ )) (1.2) second derivative lipschitz type inequalities 263 for any A,B ∈B(H) with A 6= B. If the operator norm is replaced with Hilbert-Schmidt norm ‖C‖HS := (tr C∗C) 1/2 of an operator C, then the following inequality is true [1] ‖|A|− |B|‖HS ≤ √ 2‖A−B‖HS (1.3) for any A,B ∈B(H). The coefficient √ 2 is best possible for a general A and B. If A and B are restricted to be selfadjoint, then the best coefficient is 1. It has been shown in [3] that, if A is an invertible operator, then for all operators B in a neighborhood of A we have ‖|A|− |B|‖≤ a1 ‖A−B‖ + a2 ‖A−B‖2 + O ( ‖A−B‖3 ) (1.4) where a1 = ∥∥A−1∥∥‖A‖ and a2 = ∥∥A−1∥∥ + ∥∥A−1∥∥3 ‖A‖2 . In [2] the author also obtained the following Lipschitz type inequality ‖f (A) −f (B)‖≤ f ′ (a)‖A−B‖ (1.5) where f is an operator monotone function on (0,∞) and A,B ≥ a > 0. One of the problems in perturbation theory is to find bounds for ‖f (A) −f (B)‖ in terms of ‖A−B‖ for different classes of measurable func- tions f for which the function of operator can be defined. For some results on this topic, see [5, 9] and the references therein. We have the following integral representation for the power function when t > 0, r ∈ (0, 1], see for instance [4, p. 145] tr = sin (rπ) π t ∫ ∞ 0 λr−1 λ + t dλ. (1.6) Observe that for t > 0, t 6= 1, we have∫ u 0 dλ (λ + t) (λ + 1) = ln t t− 1 + 1 1 − t ln ( u + t u + 1 ) for all u > 0. By taking the limit over u →∞ in this equality, we derive ln t t− 1 = ∫ ∞ 0 dλ (λ + t) (λ + 1) , 264 s.s. dragomir which gives the representation for the logarithm ln t = (t− 1) ∫ ∞ 0 dλ (λ + 1) (λ + t) (1.7) for all t > 0. Motivated by these representations, we introduce, for a continuous and positive function w(λ), λ > 0, the following integral transform D(w,µ)(t) := ∫ ∞ 0 w(λ) λ + t dµ(λ), t > 0 , (1.8) where µ is a positive measure on (0,∞) and the integral (1.8) exists for all t > 0. For µ the Lebesgue usual measure, we put D(w)(t) := ∫ ∞ 0 w (λ) λ + t dλ, t > 0 . (1.9) If we take µ to be the usual Lebesgue measure and the kernel wr(λ) = λ r−1, r ∈ (0, 1], then tr−1 = sin (rπ) π D (wr) (t) , t > 0 . (1.10) For the same measure, if we take the kernel wln(λ) = (λ + 1) −1, t > 0, we have the representation ln t = (t− 1)D (wln) (t), t > 0 . (1.11) Assume that T > 0, then by the continuous functional calculus for selfad- joint operators, we can define the positive operator D (w,µ) (T) := ∫ ∞ 0 w (λ) (λ + T) −1 dµ (λ) , (1.12) where w and µ are as above. Also, when µ is the usual Lebesgue measure, then D (w) (T) := ∫ ∞ 0 w (λ) (λ + T) −1 dλ, (1.13) for T > 0. second derivative lipschitz type inequalities 265 In this paper, we show among others that, if A ≥ m1 > 0, B ≥ m2 > 0, then ‖D (w,µ) (B) −D (w,µ) (A) −D (D (w,µ)) (A) (B −A)‖ ≤‖B −A‖2×   D(w,µ)(m2)−D(w,µ)(m1)−(m2−m1)D′(w,µ)(m1) (m2−m1)2 if m1 6= m2, 1 2 D′′ (w,µ) (m) if m1 = m2 = m, where D (D(w,µ)) is the Fréchet derivative of D (w,µ) as a function of oper- ator and D′′ (w,µ) is the second derivative of D (w,µ) as a real function. We also prove the norm integral inequalities for power r ∈ (0, 1] and A,B ≥ m > 0, ∥∥∥∥∥ ∫ 1 0 ((1 − t) A + tB)r−1dt − ( A + B 2 )r−1 ∥∥∥∥∥ ≤ 1 24 (1 −r) (2 −r) mr−3 ‖B −A‖2 and ∥∥∥∥∥A r−1 + Br−1 2 − ∫ 1 0 ((1 − t) A + tB)r−1dt ∥∥∥∥∥ ≤ 1 12 (1 −r) (2 −r) mr−3 ‖B −A‖2 . 2. Preliminary results We have the following representation of the Fréchet derivative: Lemma 1. For all A > 0, D (D (w,µ)) (A) (V ) = − ∫ ∞ 0 w (λ) (λ + A) −1 V (λ + A) −1 dµ (λ) (2.1) for all V ∈ S (H), the class of all selfadjoint operators on H. Proof. By the definition of D (w,µ) we have for t in a small open interval 266 s.s. dragomir around 0 that D(w,µ) (A + tV ) −D (w,µ) (A) = ∫ ∞ 0 w (λ) [ (λ + A + tV ) −1 − (λ + A)−1 ] dµ (λ) = ∫ ∞ 0 w (λ) [ (λ + A + tV ) −1 (λ + A−λ−A− tV ) (λ + A)−1 ] dµ (λ) = −t ∫ ∞ 0 w (λ) [ (λ + A + tV ) −1 V (λ + A) −1 ] dµ (λ) . Therefore, lim t→0 D (w,µ) (A + tV ) −D (w,µ) (A) t = − lim t→0 ∫ ∞ 0 w (λ) [ (λ + A + tV ) −1 V (λ + A) −1 ] dµ (λ) = − ∫ ∞ 0 w (λ) [ (λ + A) −1 V (λ + A) −1 ] dµ (λ) and the identity (2.1) is obtained. The second Fréchet derivative can be represented as follows: Lemma 2. For all A > 0, D2(D(w,µ))(A)(V,V ) = 2 ∫ ∞ 0 w (λ) (λ + A) −1 V (λ + A) −1 V (λ + A) −1 dµ (λ) (2.2) for all V ∈ S (H). Proof. We have by the definition of the Fréchet second derivative that D2 (D (w,µ)) (A) (V,V ) = lim t→0 D (D (w,µ)) (A + tV ) (V ) −D (D (w,µ)) (A) (V ) t . Observe, by (2.1), that we have for t in a small open interval around 0 D (D (w,µ)) (A + tV ) (V ) = − ∫ ∞ 0 w (λ) (λ + A + tV ) −1 V (λ + A + tV ) −1 dµ (λ) , second derivative lipschitz type inequalities 267 which gives that D (D (w,µ)) (A + tV ) (V ) −D (D (w,µ)) (A) (V ) = − ∫ ∞ 0 w (λ) (λ + A + tV ) −1 V (λ + A + tV ) −1 dµ (λ) + ∫ ∞ 0 w (λ) (λ + A) −1 V (λ + A) −1 dµ (λ) = ∫ ∞ 0 w(λ)× [ (λ+A)−1V (λ+At)−1 − (λ+A+tV )−1V (λ+A+tV )−1 ] dµ(λ). Define for λ ≥ 0 and t as above, Ut,λ := (λ + A) −1 V (λ + A) −1 − (λ + A + tV )−1 V (λ + A + tV )−1 . If we multiply both sides of Ut,λ with λ + A + tV , the we get (λ + A + tV )Ut,λ (λ + A + tV ) = (λ + A + tV ) (λ + A) −1 V (λ + A) −1 (λ + A + tV ) −V = ( 1 + tV (λ + A) −1 ) V ( 1 + t (λ + A) −1 V ) −V = ( V + tV (λ + A) −1 V )( 1 + t (λ + A) −1 V ) −V = V + tV (λ + A) −1 V + tV (λ + A) −1 V + t2V (λ + A) −1 V (λ + A) −1 V −V = 2tV (λ + A) −1 V + t2V (λ + A) −1 V (λ + A) −1 V = t [ 2V (λ + A) −1 V + tV (λ + A) −1 V (λ + A) −1 V ] . (2.3) If we multiply the equality by (λ + A + tV ) −1 both sides, we get for t 6= 0 Ut,λ t = (λ+A+tV ) −1 [ 2V (λ+A) −1 V +tV (λ+A) −1 V (λ+A) −1 V ] × (λ + A + tV )−1 . (2.4) If we take the limit over t → 0 in, then we get lim t→0 ( Ut,λ t ) = 2 (λ + A) −1 V (λ + A) −1 V (λ + A) −1 . 268 s.s. dragomir Therefore, by the properties of limit under the sign of integral, we get lim t→0 D (D (w,µ)) (A + tV ) (V ) −D (D (w,µ)) (A) (V ) t = ∫ ∞ 0 w (λ) lim t→0 ( Ut,λ t ) dµ (λ) = 2 ∫ ∞ 0 w (λ) (λ + A) −1 V (λ + A) −1 V (λ + A) −1 dµ (λ) and the representation (2.2) is obtained. Remark 1. One may ask if the above integral representation can be ex- tended for higher derivative. The author thinks that is possible, however the calculations are more difficult to perform and are not presented here. We have the following representation for the transform D(w,µ): Theorem 2. For all A,B > 0 we have D (w,µ) (B) = D (w,µ) (A) − ∫ ∞ 0 w (λ) (λ + A) −1 (B −A) (λ + A)−1 dµ (λ) + 2 ∫ 1 0 (1 − t) [∫ ∞ 0 w (λ) (λ + (1 − t) A + tB)−1 (B −A) ×(λ + (1 − t) A + tB)−1(B −A)(λ + (1 − t)A + tB)−1 dµ(λ) ] dt. (2.5) Proof. We use the Taylor’s type formula with integral remainder, see for instance [6, p. 112], f (E) = f (C) + D (f) (C) (E −C) + ∫ 1 0 (1 − t) D2 (f) ((1 − t) C + tE) (E −C,E −C) dt (2.6) that holds for functions f which are of class C2 on an open and convex subset O in the Banach algebra B (H) and C,E ∈O. If we write (2.6) for D (w,µ) and A,B > 0, we get D (w,µ) (B) = D (w,µ) (A) + D (D (w,µ)) (A) (B −A) + ∫ 1 0 (1 − t) D2 (D (w,µ)) ((1 − t) A + tB) (B −A,B −A) dt and by the representations (2.1) and (2.2) we obtain the desired result (2.5). second derivative lipschitz type inequalities 269 3. Main results We have the following Lipschitz type inequality: Theorem 3. Assume that A ≥ m1 > 0, B ≥ m2 > 0, then ‖D (w,µ) (B) −D (w,µ) (A) −D (D (w,µ)) (A) (B −A)‖ (3.1) ≤‖B −A‖2×   D(w,µ)(m2)−D(w,µ)(m1)−(m2−m1)D′(w,µ)(m1) (m2−m1)2 if m1 6= m2 , 1 2 D′′ (w,µ) (m) if m1 = m2 = m. Proof. From (2.5) we get ‖D (w,µ) (B) −D (w,µ) (A) −D (D (w,µ)) (A) (B −A)‖ ≤ 2 ∫ 1 0 (1 − t) [∫ ∞ 0 w (λ) ∥∥∥(λ + (1 − t) A + tB)−1 (B −A) (3.2) ×(λ + (1 − t) A + tB)−1 (B −A) (λ + (1 − t) A + tB)−1 ∥∥∥dµ (λ) ]dt ≤ 2‖B −A‖2 ∫ 1 0 (1 − t) (∫ ∞ 0 w (λ) ∥∥∥(λ + (1 − t) A + tB)−1∥∥∥3 dµ (λ))dt. Assume that m2 > m1. Then (1 − t) A + tB + λ ≥ (1 − t) m1 + tm2 + λ, which implies that ((1 − t) A + tB + λ)−1 ≤ ((1 − t) m1 + tm2 + λ)−1 , and ∥∥∥((1 − t) A + tB + λ)−1∥∥∥3 ≤ ((1 − t) m1 + tm2 + λ)−3 (3.3) for all t ∈ [0, 1] and λ ≥ 0. 270 s.s. dragomir Therefore, by integrating (3.3) we derive∫ 1 0 (1 − t) (∫ ∞ 0 w (λ) ∥∥∥(λ + (1 − t) A + tB)−1∥∥∥3 dµ (λ))dt ≤ ∫ 1 0 (1 − t) (∫ ∞ 0 w (λ) ((1 − t) m1 + tm2 + λ)−3 dµ (λ) ) dt (3.4) = 1 (m2 −m1)2 ∫ 1 0 (1 − t) [∫ ∞ 0 w (λ) ((1 − t) m1 + tm2 + λ)−1 (m2 −m1) × ((1 − t) m1 + tm2 + λ)−1 (m2 −m1) × ((1 − t) m1 + tm2 + λ)−1 dµ(λ) ] dt. From (2.5) we have for m2 > m1 that D (w,µ) (m2) −D (w,µ) (m1) + (m2 −m1) ∫ ∞ 0 w (λ) (λ + m1) −2 dµ (λ) = 2 ∫ 1 0 (1 − t) [∫ ∞ 0 w (λ) ((1 − t) m1 + tm2 + λ)−1 (m2 −m1) × ((1 − t) m1 + tm2 + λ)−1 (m2 −m1) (3.5) × ((1 − t) m1 + tm2 + λ)−1 dµ(λ) ] dt. Also ∫ ∞ 0 w (λ) (λ + m1) −2 dµ (λ) = −D′ (w,µ) (m1) , and then by (3.5) we get 1 2 (m2 −m1)2 [ D (w,µ) (m2) −D (w,µ) (m1) − (m2 −m1)D′ (w,µ) (m1) ] (3.6) = 1 (m2 −m1)2 ∫ 1 0 (1 − t) [∫ ∞ 0 w (λ) ((1 − t) m1 + tm2 + λ)−1 (m2 −m1) × ((1 − t)m1 + tm2 + λ)−1(m2 −m1)((1 − t)m1 + tm2 + λ)−1dµ(λ) ] dt. By utilizing (3.2) and (3.4) – (3.6) we derive (3.1). The case m2 < m1 goes in a similar way and we also obtain (3.1). second derivative lipschitz type inequalities 271 Assume that m2 = m1 > 0. Let � > 0. Then B + � ≥ m + � > m. By the first inequality for m2 = m + � and m1 = m, we have ‖D (w,µ) (B + �) −D (w,µ) (A) −D (D (w,µ)) (A) (B + �−A)‖ (3.7) ≤‖B + �−A‖2 1 �2 [ D (w,µ) (m + �) −D (w,µ) (m) − �D′ (w,µ) (m) ] . By Taylor’s expansion theorem with the Lagrange remainder we have D (w,µ) (m + �) −D (w,µ) (m) − �D′ (w,µ) (m) = 1 2 �2D′′ (w,µ) (ζ�) with m + � > ζ� > m. Therefore lim �→0+ 1 �2 [ D (w,µ) (m + �) −D (w,µ) (m) − �D′ (w,µ) (m) ] = 1 2 D′′ (w,µ) (m) and by taking the limit � → 0+ in (3.7) then we get∥∥D (w,µ) (B) −D (w,µ) (A)−D (D (w,µ)) (A) (B −A) ∥∥ ≤ 1 2 ‖B −A‖2 D′′ (w,µ) (m) and the second part of (3.1) is proved. The case of operator monotone function is as follows: Corollary 1. Assume that f : [0,∞) → R is an operator monotone function with f (0) = 0. If A ≥ m1 > 0, B ≥ m2 > 0, then∥∥f (B) B−1 −(2 −A−1B)A−1f (A) −A−1D (f) (A) (B −A)∥∥ (3.8) ≤‖B −A‖2 ×   1 (m2−m1)2 [ f(m2) m2 − f(m1) m1 − (m2 −m1) f′(m1)m1−f(m1) m21 ] if m1 6= m2 , 1 2 f′′(m)m2−2mf′(m)+2f(m) m3 if m1 = m2 = m. Proof. We denote by ` the identity function ` (t) = t, t > 0. By `−1 we denote the function `−1(t) = t−1, t > 0. Using these notations we have D(`,µ)(t) = f(t) t − b, t > 0 , where b ≥ 0 and µ is a positive measure on (0,∞). 272 s.s. dragomir The derivative of this function is D′ (`,µ) (t) = f ′ (t) t−f (t) t2 , t > 0 , and the second derivative D′′ (`,µ) (t) = (f ′ (t) t−f (t))′ t2 − 2t (f ′ (t) t−f (t)) t4 = (f ′′ (t) t + f ′ (t) −f ′ (t)) t2 − 2t (f ′ (t) t−f (t)) t4 = f ′′ (t) t3 − 2t2f ′ (t) + 2tf (t) t4 = f ′′ (t) t2 − 2tf ′ (t) + 2f (t) t3 . We have D (w,µ) (B) −D (w,µ) (A) −D ( `−1f ) (A)(B −A) = f (B) B−1 −f (A) A−1 − [ D ( `−1 ) (A) (B −A) f (A) + `−1 (A) D (f) (A) (B −A) ] = f (B) B−1 −f (A) A−1 + A−1 (B −A) A−1f (A) −A−1D (f) (A) (B −A) , since, by using the definition of the Fréchet derivative, D ( `−1 ) (A) (B −A) = −A−1 (B −A) A−1. Also D (w,µ) (m2)−D (w,µ) (m1) − (m2 −m1)D′ (w,µ) (m1) = f (m2) m2 − f (m1) m1 − (m2 −m1) f ′ (m1) m1 −f (m1) m21 . By making use of (3.1) we deduce (3.8). We consider the representation obtained from (1.9) for the operator T > 0 and the power r ∈ (0, 1], Tr−1 = D (w̃r) (T) with the kernel w̃r(λ) := sin(rπ) π λr−1, r ∈ (0, 1]. second derivative lipschitz type inequalities 273 From (3.1) we get for A ≥ m1 > 0, B ≥ m2 > 0 and r ∈ (0, 1] that∥∥∥∥Br−1 −Ar−1 + ∫ ∞ 0 λr−1 (λ + A) −1 (B −A) (λ + A)−1 dλ ∥∥∥∥ (3.9) ≤‖B −A‖2 ×   (1−r)(m2−m1)mr−21 −m r−1 1 +m r−1 2 (m2−m1)2 if m1 6= m2 , 1 2 (1 −r) (2 −r) mr−3 if m1 = m2 = m. We have the following error bounds for operator Jensen’s gap related to the n-tuple of positive operators A = (A1, . . . ,An) and probability density n-tuple p = (p1, . . . ,pn), J (A,p,D (w,µ)) := n∑ k=1 pkD (w,µ) (Ak) −D (w,µ) ( n∑ k=1 pkAk ) . Theorem 4. Assume that Ai ≥ m > 0 for i ∈ {1, . . . ,n} and consider the probability density n-tuple p = (p1, . . . ,pn), then ∥∥J (A,p,D (w,µ)) ∥∥ ≤ 1 2 D′′ (w,µ) (m) n∑ k=1 pk ∥∥∥∥∥Ak − n∑ j=1 pjAj ∥∥∥∥∥ 2 ≤ 1 2 D′′ (w,µ) (m) n∑ k=1 n∑ j=1 pjpk ‖Ak −Aj‖ 2 ≤ 1 2 D′′ (w,µ) (m) max k,j∈{1,...,n} ‖Ak −Aj‖ 2 . (3.10) Proof. From (3.1) we get∥∥∥∥∥D(w,µ) (Ak) −D (w,µ) ( n∑ j=1 pjAj ) −D (D (w,µ)) ( n∑ j=1 pjAj )( Ak − n∑ j=1 pjAj )∥∥∥∥∥ ≤ 1 2 D′′(w,µ) (m) ∥∥∥∥∥Ak − n∑ j=1 pjAj ∥∥∥∥∥ 2 (3.11) for all k ∈{1, . . . ,n}. 274 s.s. dragomir If we multiply this inequality by pk ≥ 0 and sum over k from 1 to n, then we get n∑ k=1 ∥∥∥∥∥pkD (w,µ) (Ak) −pkD (w,µ) ( n∑ j=1 pjAj ) (3.12) −D (D (w,µ)) ( n∑ j=1 pjAj ) pk ( Ak − n∑ j=1 pjAj )∥∥∥∥∥ ≤ 1 2 D′′ (w,µ) (m) n∑ k=1 pk ∥∥∥∥∥Ak − n∑ j=1 pjAj ∥∥∥∥∥ 2 . By making use of the triangle inequality for norms, we also have n∑ k=1 ∥∥∥∥∥pkD (w,µ) (Ak) −pkD (w,µ) ( n∑ j=1 pjAj ) −D (D (w,µ)) ( n∑ j=1 pjAj ) pk ( Ak − n∑ j=1 pjAj )∥∥∥∥∥ ≥ ∥∥∥∥∥ n∑ k=1 pkD (w,µ) (Ak) − n∑ k=1 pkD (w,µ) ( n∑ j=1 pjAj ) (3.13) −D (D (w,µ)) ( n∑ j=1 pjAj )( n∑ k=1 pkAk − n∑ j=1 pjAj )∥∥∥∥∥ = ∥∥∥∥∥ n∑ k=1 pkD (w,µ) (Ak) −D (w,µ) ( n∑ j=1 pjAj )∥∥∥∥∥. By utilizing (3.12) and (3.13) we deduce the first part of (3.10). The rest is obvious. Remark 2. From (3.10) we can obtain the following norm inequalities for power r ∈ (0, 1], second derivative lipschitz type inequalities 275 ∥∥∥∥∥∥ n∑ k=1 pkA r−1 k − ( n∑ k=1 pkAk )r−1∥∥∥∥∥∥ (3.14) ≤ 1 2 (1 −r) (2 −r) mr−3 n∑ k=1 pk ∥∥∥∥∥Ak − n∑ j=1 pjAj ∥∥∥∥∥ 2 ≤ 1 2 (1 −r) (2 −r) mr−3 n∑ k=1 n∑ j=1 pjpk ∥∥Ak −Aj∥∥2 ≤ 1 2 (1 −r) (2 −r) mr−3 max k,j∈{1,...,n} ∥∥Ak −Aj∥∥2, where Ai ≥ m > 0 for i ∈ {1, . . . ,n} and the probability density n-tuple p = (p1, . . . ,pn). 4. Midpoint and trapezoid inequalities We have the following midpoint norm inequality: Theorem 5. If A,B ≥ m > 0 for some constant m, then∥∥∥∥ ∫ 1 0 D (w,µ) ((1 − t) A + tB) dt−D (w,µ) ( A + B 2 )∥∥∥∥ (4.1) ≤ 1 24 D′′ (w,µ) (m)‖B −A‖2 . Proof. From (3.1) we have for all t ∈ [0, 1] and A,B ≥ m > 0,∥∥∥∥∥D (w,µ) ((1 − t)A + tB) −D (w,µ) ( A + B 2 ) −D (D (w,µ)) ( A + B 2 )( (1 − t) A + tB − A + B 2 )∥∥∥∥ ≤ 1 2 D′′ (w,µ) (m) ∥∥∥∥(1 − t) A + tB − A + B2 ∥∥∥∥2 (4.2) = 1 2 D′′ (w,µ) (m) ( t− 1 2 )2 ‖B −A‖2 . 276 s.s. dragomir If we integrate this inequality, we get∫ 1 0 ∥∥∥∥D (w,µ) ((1 − t) A + tB) −D (w,µ) ( A + B 2 ) −D (D (w,µ)) ( A + B 2 )( (1 − t) A + tB − A + B 2 )∥∥∥∥dt ≤ 1 2 D′′ (w,µ) (m)‖B −A‖2 ∫ 1 0 ( t− 1 2 )2 dt (4.3) = 1 24 D′′ (w,µ) (m)‖B −A‖2 . Using the properties of norm and integral, we also have∫ 1 0 ∥∥∥∥D (w,µ) ((1 − t) A + tB) −D (w,µ) ( A + B 2 ) −D (D (w,µ)) ( A + B 2 )( (1 − t) A + tB − A + B 2 )∥∥∥∥dt ≥ ∥∥∥∥ ∫ 1 0 D (w,µ) ((1 − t) A + tB) dt−D (w,µ) ( A + B 2 ) (4.4) − (∫ 1 0 ( t− 1 2 ) dt ) D (D (w,µ)) ( A + B 2 ) (B −A) ∥∥∥∥ = ∥∥∥∥ ∫ 1 0 D (w,µ) ((1 − t) A + tB) dt−D (w,µ) ( A + B 2 )∥∥∥∥ . By employing (4.3) and (4.4) we derive the desired result (4.1). Corollary 2. Assume that f : [0,∞) → R is an operator monotone function with f (0) = 0. If A,B ≥ m > 0, then∥∥∥∥∥ ∫ 1 0 ((1 − t) A + tB)−1 f ((1 − t) A + tB) dt− ( A + B 2 )−1 f ( A + B 2 )∥∥∥∥∥ ≤ f ′′ (m) m2 − 2mf ′ (m) + 2f (m) 24m3 ‖B −A‖2 . (4.5) The proof follows by (4.1) for D (`,µ) (t) = f (t) t − b, t > 0 , where b ≥ 0 and µ is a positive measure on (0,∞). second derivative lipschitz type inequalities 277 Remark 3. If A,B ≥ m > 0, then for r ∈ (0, 1] we get by (4.5) that∥∥∥∥∥ ∫ 1 0 ((1 − t) A + tB)r−1 dt− ( A + B 2 )r−1∥∥∥∥∥ ≤ 1 24 (1 −r) (2 −r) mr−3 ‖B −A‖2 . (4.6) The trapezoid norm inequality will be our concern from now on. For a continuous function f on (0,∞) and A,B > 0 we consider the auxiliary function fA,B : [0, 1] → R defined by fA,B (t) := f ((1 − t) A + tB) , t ∈ [0, 1]. We have the following representations of the derivatives: Lemma 3. Assume that the operator function generated by f is twice Fréchet differentiable in each A > 0, then for B > 0 we have that fA,B is twice differentiable on [0, 1], dfA,B (t) dt = D (f) ((1 − t) A + tB) (B −A) , (4.7) d2fA,B (t) dt2 = D2 (f) ((1 − t) A + tB) (B −A,B −A) (4.8) for t ∈ [0, 1], where in 0 and 1 the derivatives are the right and left derivatives. Proof. We prove only for the interior points t ∈ (0, 1). Let h be in a small interval around 0 such that t + h ∈ (0, 1). Then for h 6= 0, fA,B (t + h) −f (t) h = f ((1 − (t + h)) A + (t + h) B) −f ((1 − t) A + tB) h = f ((1 − t) A + tB + h (B −A)) −f ((1 − t) A + tB) h and by taking the limit over h → 0, we get dfA,B (t) dt = lim h→0 fA,B (t + h) −f (t) h = lim h→0 [ f ((1 − t) A + tB + h (B −A)) −f ((1 − t) A + tB) h ] = D (f) ((1 − t) A + tB) (B −A) , which proves (4.7). 278 s.s. dragomir Similarly, 1 h [ dfA,B (t + h) dt − dfA,B (t) dt ] = D (f) ((1 − (t + h)) A + (t + h) B) (B −A) −D (f) ((1 − t) A + tB) (B −A) h = D (f) ((1 − t) A + tB + h (B −A)) (B −A) −D (f) ((1 − t) A + tB) (B −A) h and by taking the limit over h → 0, we get d2fA,B (t) dt2 = lim h→0 { 1 h [ dfA,B (t + h) dt − dfA,B (t) dt ]} = D2 (f) ((1 − t) A + tB) (B −A,B −A) , which proves (4.8). For the transform D (w,µ) (t) defined in the introduction, we consider the auxiliary function D (w,µ)A,B (t) : = D (w,µ) ((1 − t) A + tB) = ∫ ∞ 0 w (λ) (λ + (1 − t) A + tB)−1 dµ (λ) where A,B > 0 and t ∈ [0, 1]. Corollary 3. For all A,B > 0 and t ∈ [0, 1], dD (w,µ)A,B (t) dt = D (D (w,µ)) ((1 − t) A + tB) (B −A) (4.9) = − ∫ ∞ 0 w (λ) (λ + (1 − t) A + tB)−1 (B −A) × (λ + (1 − t) A + tB)−1 dµ (λ) and d2D (w,µ)A,B (t) dt2 = D2 (D (w,µ)) ((1 − t) A + tB) (B −A,B −A) (4.10) = 2 ∫ ∞ 0 w (λ) (λ + (1 − t) A + tB)−1 (B −A) × (λ + (1 − t) A + tB)−1 (B −A) (λ + (1 − t) A + tB)−1 dµ (λ) . second derivative lipschitz type inequalities 279 We observe that if f (t) = D (w,µ) (t), t > 0, in Lemma 3, then by the representations from Lemma 1 and Lemma 2 we obtain the desired equalities (4.9) and (4.10). We have the following identity for the trapezoid rule: Lemma 4. For all A,B > 0 we have the identity D (w,µ) (A) + D (w,µ) (B) 2 − ∫ 1 0 D (w,µ) ((1 − t) A + tB) dt = ∫ 1 0 t (1 − t) [∫ ∞ 0 w (λ) (λ + (1 − t) A + tB)−1 (B −A) (4.11) × (λ + (1 − t) A + tB)−1 (B −A) (λ + (1 − t) A + tB)−1 dµ (λ) ] dt. Proof. Using integration by parts for the Bochner integral, we have 1 2 ∫ 1 0 t (1 − t) d2D (w,µ)A,B (t) dt2 dt = 1 2 [ t (1 − t) dD (w,µ)A,B (t) dt ∣∣∣∣1 0 − ∫ 1 0 (1 − 2t) dD (w,µ)A,B (t) dt dt ] = ∫ 1 0 ( t− 1 2 ) dD (w,µ)A,B (t) dt dt = ( t− 1 2 ) D (w,µ)A,B (t) ∣∣∣∣1 0 − ∫ 1 0 D (w,µ)A,B (t) dt = 1 2 [ D (w,µ)A,B (1) + D (w,µ)A,B (0) ] − ∫ 1 0 D (w,µ)A,B (t) dt, that gives the identity D (w,µ) (A) + D (w,µ) (B) 2 − ∫ 1 0 D (w,µ) ((1 − t) A + tB) dt = 1 2 ∫ 1 0 t (1 − t) d2D (w,µ)A,B (t) dt2 dt. (4.12) 280 s.s. dragomir By (4.12) we have 1 2 ∫ 1 0 t (1 − t) d2D (w,µ)A,B (t) dt2 dt (4.13) = ∫ 1 0 t (1 − t) [∫ ∞ 0 w (λ) (λ + (1 − t) A + tB)−1 (B −A) × (λ + (1 − t) A + tB)−1 (B −A) (λ + (1 − t) A + tB)−1 dµ (λ) ] dt. By making use of (4.10) and (4.13) we deduce (4.11). We can state now the corresponding trapezoid norm inequality: Theorem 6. If A,B ≥ m > 0 for some constant m, then∥∥∥∥D (w,µ) (A) + D (w,µ) (B)2 − ∫ 1 0 D (w,µ) ((1 − t) A + tB) dt ∥∥∥∥ ≤ 1 12 D′′ (w,µ) (m)‖B −A‖2 . (4.14) Proof. By taking the norm in (4.11), we obtain∥∥∥∥D (w,µ) (A) + D (w,µ) (B)2 − ∫ 1 0 D (w,µ) ((1 − t) A + tB) dt ∥∥∥∥ (4.15) ≤‖B −A‖2 ∫ 1 0 t (1 − t) (∫ ∞ 0 w (λ) ∥∥∥(λ + (1 − t) A + tB)−1∥∥∥3 dµ (λ))dt. Since A,B ≥ m > 0, then for λ ≥ 0 and t ∈ [0, 1], λ + (1 − t) A + tB ≥ λ + m, which implies that (λ + (1 − t) A + tB)−1 ≤ (λ + m)−1 . This implies that ∥∥∥(λ + (1 − t) A + tB)−1∥∥∥3 ≤ (λ + m)−3 for λ ≥ 0 and t ∈ [0, 1]. second derivative lipschitz type inequalities 281 By multiplying this inequality by t (1 − t) w (λ) ≥ 0 and integrating we get∫ 1 0 t (1 − t) (∫ ∞ 0 w (λ) ∥∥∥(λ + (1 − t) A + tB)−1∥∥∥3 dµ (λ))dt ≤ (∫ 1 0 t (1 − t) dt )(∫ ∞ 0 w (λ) (λ + m) −3 dµ (λ) ) = 1 6 ∫ ∞ 0 w (λ) (λ + m) −3 dµ (λ) . (4.16) Taking the derivative over t twice in (1.8), we get D′′ (w,µ) (t) := 2 ∫ ∞ 0 w (λ) (λ + t) 3 dµ (λ) , t > 0 , that gives ∫ ∞ 0 w (λ) (λ + m) −3 dµ (λ) = 1 2 D′′ (w,µ) (m) and by (4.15) and (4.16) we derive (4.14). Corollary 4. Assume that f : [0,∞) → R is an operator monotone function with f (0) = 0. If A,B ≥ m > 0, then∥∥∥∥A−1f (A) + B−1f (B)2 − ∫ 1 0 ((1 − t) A + tB)−1 f ((1 − t) A + tB) dt ∥∥∥∥ ≤ f ′′ (m) m2 − 2mf ′ (m) + 2f (m) 12m3 ‖B −A‖2 . (4.17) The proof follows by (4.14) for D (`,µ) (t) = f (t) t − b, t > 0 , where b ≥ 0 and µ is a positive measure on (0,∞). Remark 4. If A,B ≥ m > 0, then for r ∈ (0, 1] we get by (4.5) that∥∥∥∥Ar−1 + Br−12 − ∫ 1 0 ((1 − t) A + tB)r−1 dt ∥∥∥∥ ≤ 1 12 (1 −r) (2 −r) mr−3 ‖B −A‖2 . (4.18) 282 s.s. dragomir Acknowledgements The author would like to thank the anonymous referee for valu- able comments that have been implemented in the final version of the manuscript. References [1] H. Araki, S. Yamagami, An inequality for Hilbert-Schmidt norm, Comm. Math. Phys. 81 (1981), 89 – 96. [2] R. Bhatia, First and second order perturbation bounds for the operator ab- solute value, Linear Algebra Appl. 208/209 (1994), 367 – 376. [3] R. 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Nikolskaya, Modulus of continuity of oper- ator functions, Algebra i Analiz 20 (3) (2008), 224 – 242; translation in St. Petersburg Math. J. 20 (3) (2009) 3, 493 – 506. [10] J.I. Fujii, Y. Seo, On parametrized operator means dominated by power ones, Sci. Math. 1 (1998), 301 – 306. [11] T. Furuta, Precise lower bound of f(A) − f(B) for A > B > 0 and non- constant operator monotone function f on [0,∞), J. Math. Inequal. 9 (1) (2015), 47 – 52. [12] E. Heinz, Beiträge zur Störungsteorie der Spektralzerlegung, Math. Ann. (in German) 123 (1951), 415 – 438. [13] T. Kato, Continuity of the map S → |S| for linear operators, Proc. Japan Acad. 49 (1973), 143 – 162. [14] K. Löwner, Über monotone MatrixFunktionen, Math. Z. (in German) 38 (1934) 177 – 216. Introduction Preliminary results Main results Midpoint and trapezoid inequalities