� EXTRACTA MATHEMATICAE Volumen 33, Número 2, 2018 instituto de investigación de matemáticas de la universidad de extremadura EXTRACTA MATHEMATICAE Vol. 34, Num. 1 (2019), 41 – 60 doi:10.17398/2605-5686.34.1.41 Available online March 4, 2019 Upper and lower bounds for the difference between the weighted arithmetic and harmonic operator means S.S. Dragomir 1,2 1 Mathematics, College of Engineering & Science, Victoria University PO Box 14428, Melbourne City, MC 8001, Australia sever.dragomir@vu.edu.au , http://rgmia.org/dragomir 2 School of Computer Science & Applied Mathematics, University of the Witwatersrand Private Bag 3, Johannesburg 2050, South Africa Received October 11, 2018 Presented by Mostafa Mbekhta Accepted February 4, 2019 Abstract: In this paper we establish some new upper and lower bounds for the difference between the weighted arithmetic and harmonic operator means under various assumption for the positive invertible operators A, B. Some applications when A, B are bounded above and below by positive constants are given as well. Key words: Young’s inequality, convex functions, arithmetic mean-Harmonic mean inequality, op- erator means, operator inequalities. AMS Subject Class. (2010): 47A63, 47A30, 15A60,.26D15; 26D10. 1. Introduction Throughout this paper A, B are positive invertible operators on a complex Hilbert space (H,〈·, ·〉). We use the following notations for operators A∇νB := (1 −ν) A + νB, the weighted operator arithmetic mean, A]νB := A 1/2 ( A−1/2BA−1/2 )ν A1/2, the weighted operator geometric mean and A!νB := ( (1 −ν) A−1 + νB−1 )−1 the weighted operator harmonic mean, where ν ∈ [0, 1]. When ν = 1 2 , we write A∇B, A]B and A!B for brevity, respectively. ISSN: 0213-8743 (print), 2605-5686 (online) https://doi.org/10.17398/2605-5686.34.1.41 mailto:sever.dragomir@vu.edu.au http://rgmia.org/dragomir https://www.eweb.unex.es/eweb/extracta/ https://creativecommons.org/licenses/by-nc/3.0/ 42 s.s. dragomir The following fundamental inequality between the weighted arithmetic, geometric and harmonic operator means holds A!νB ≤ A]νB ≤ A∇νB (1.1) for any ν ∈ [0, 1]. For various recent inequalities between these means we recommend the recent papers [3]-[6], [8]-[12] and the references therein. In the recent work [7] we obtained between others the following result: Theorem 1. Let A, B be positive invertible operators and M > m > 0 such that MA ≥ B ≥ mA. (1.2) Then for any ν ∈ [0, 1] we have rk (m,M) A ≤ A∇νB −A!νB ≤ RK (m,M) A, (1.3) where r = min{ν, 1 −ν}, R = max{ν, 1 −ν} and the bounds K (m,M) and k (m,M) are given by K(m,M) (1.4) :=   (m− 1)2 (m + 1)−1 if M < 1, max { (m− 1)2 (m + 1)−1 , (M − 1)2 (M + 1)−1 } if m ≤ 1 ≤ M, (M − 1)2 (M + 1)−1 if 1 < m, and k (m,M) :=   (M − 1)2 (M + 1)−1 if M < 1, 0 if m ≤ 1 ≤ M, (m− 1)2 (m + 1)−1 if 1 < m. (1.5) In particular, 1 2 k (m,M) A ≤ A∇B −A!B ≤ 1 2 K (m,M) A. (1.6) Let A, B positive invertible operators and positive real numbers m, m′, M, M ′ such that the condition 0 < mI ≤ A ≤ m′I < M ′I ≤ B ≤ MI holds. Put h := M m and h′ := M ′ m′ , then for any ν ∈ [0, 1] we have [7] r ( h′ − 1 )2 ( h′ + 1 )−1 A ≤ A∇νB −A!νB ≤ R (h− 1)2 (h + 1)−1 A, (1.7) upper and lower bounds 43 where r = min{ν, 1 −ν}, R = max{ν, 1 −ν} and, in particular, 1 2 ( h′ − 1 )2 ( h′ + 1 )−1 A ≤ A∇B −A!B ≤ 1 2 (h− 1)2 (h + 1)−1 A. (1.8) Let A, B positive invertible operators and positive real numbers m, m′, M, M ′ such that the condition 0 < mI ≤ B ≤ m′I < M ′I ≤ A ≤ MI holds. Then for any ν ∈ [0, 1] we also have [7] r ( h′ − 1 )2 ( h′ + 1 )−1 ( h′ )−1 A ≤ A∇νB −A!νB ≤ R (h− 1)2 (h + 1)−1 h−1A, (1.9) and, in particular, 1 2 ( h′ − 1 )2 ( h′ + 1 )−1 ( h′ )−1 A ≤ A∇B −A!B ≤ 1 2 (h− 1)2 (h + 1)−1 h−1A. (1.10) Motivated by the above facts, in this paper we establish some new upper and lower bounds for the difference A∇νB−A!νB for ν ∈ [0, 1] under various assumption for the positive invertible operators A, B. Some applications when A, B are bounded above and below by positive constants are given as well. A graphic comparison for upper bounds is provided as well. 2. Min and max bounds The following lemma is of interest in itself. Lemma 1. For any a, b > 0 and ν ∈ [0, 1] we have ν (1 −ν) (b−a)2 max{b,a} ≤ Aν (a,b) −Hν (a,b) ≤ ν (1 −ν) (b−a)2 min{b,a} , (2.1) where Aν (a,b) and Hν (a,b) are the scalar weighted arithmetic mean and harmonic mean, respectively, namely Aν (a,b) := (1 −ν) a + νb and Hν (a,b) := ab (1 −ν) b + νa . 44 s.s. dragomir In particular, 1 4 (b−a)2 max{b,a} ≤ A (a,b) −H (a,b) ≤ 1 4 (b−a)2 min{b,a} , (2.2) where A (a,b) := a + b 2 and H (a,b) := 2ab b + a . Proof. Consider the function ξν : (0,∞) → (0,∞) defined by ξν(x) = 1 −ν + νx− x (1 −ν) x + ν , where ν ∈ [0, 1]. Then ξν (x) = (1 −ν + νx) [(1 −ν) x + ν] −x (1 −ν) x + ν = (1 −ν)2 x + ν (1 −ν) x2 + ν (1 −ν) + ν2x−x (1 −ν) x + ν = ν (1 −ν) x2 − 2ν (1 −ν) x + ν (1 −ν) (1 −ν) x + ν = ν (1 −ν) (x− 1)2 (1 −ν) x + ν , (2.3) for any x > 0 and ν ∈ [0, 1]. If we take in the definition of ξν, x = b a > 0, then we have ξν ( b a ) = 1 a [Aν (a,b) −Hν (a,b)] . From the equality (2.3) we also have ξν ( b a ) = ν (1 −ν) (b−a)2 aAν (b,a) . Therefore, we have the equality Aν (a,b) −Hν (a,b) = ν (1 −ν) (b−a)2 Aν (b,a) (2.4) for any a, b > 0 and ν ∈ [0, 1]. upper and lower bounds 45 Since for any a, b > 0 and ν ∈ [0, 1] we have min{a,b}≤ Aν (b,a) ≤ max{a,b} then ν (1 −ν) (b−a)2 max{a,b} ≤ ν (1 −ν) (b−a)2 Aν (b,a) ≤ ν (1 −ν) (b−a)2 min{a,b} (2.5) and by (2.4) we get the desired result (2.1). Remark 1. We show that there is no constant K1 > 1 and K2 < 1 such that ν (1 −ν) (b−a)2 max{b,a} ≤ Aν (a,b) −Hν (a,b) ≤ ν (1 −ν) (b−a)2 min{b,a} , (2.6) for some ν ∈ (0, 1) and any a, b > 0. Assume that there exist K1, K2 > 0 such that K1ν (1 −ν) (b−a)2 max{b,a} ≤ Aν (a,b) −Hν (a,b) ≤ K2ν (1 −ν) (b−a)2 min{b,a} , (2.7) for some ν ∈ (0, 1) and any a, b > 0. Let ε > 0 and write the inequality (2.7) for a > 0 and b = a + ε to get, via (2.4) that K1ν (1 −ν) ε2 a + ε ≤ ν (1 −ν) ε2 (1 −ν) ε + a ≤ K2ν (1 −ν) ε2 a . (2.8) If we divide by ν (1 −ν) ε2 > 0 in (2.8), then we get K1 1 a + ε ≤ 1 (1 −ν) ε + a ≤ K2 1 a , (2.9) for any a > 0 and ε > 0. By letting ε → 0+ in (2.9), we get K1 ≤ 1 ≤ K2 and the statement is proved. 46 s.s. dragomir We have the following operator double inequality: Theorem 2. Let A, B be positive invertible operators and M > m > 0 such that the condition (1.2). Then for any ν ∈ [0, 1] we have ν (1 −ν) c (m,M) A ≤ ν (1 −ν) max{M, 1} (B −A) A−1 (B −A) ≤ A∇νB −A!νB ≤ ν (1 −ν) min{m, 1} (B −A) A−1 (B −A) ≤ ν (1 −ν) C (m,M) A, (2.10) where c (m,M) :=   (M − 1)2 if M < 1, 0 if m ≤ 1 ≤ M, (m−1)2 M if 1 < m, and C (m,M) :=   (m−1)2 m if M < 1, 1 m max { (m− 1)2 , (M − 1)2 } if m ≤ 1 ≤ M, (M − 1)2 if 1 < m. In particular, 1 4 c (m,M) A ≤ 1 4 max{M, 1} (B −A) A−1 (B −A) ≤ A∇B −A!B ≤ 1 4 min{m, 1} (B −A) A−1 (B −A) ≤ 1 4 C (m,M) A. (2.11) Proof. If we write the inequality (2.1) for a = 1 and b = x, then we get ν (1 −ν) (x− 1)2 max{x, 1} ≤ 1 −ν + νx− ( (1 −ν) + νx−1 )−1 ≤ ν (1 −ν) (x− 1)2 min{x, 1} (2.12) for any x > 0 and for any ν ∈ [0, 1]. upper and lower bounds 47 If x ∈ [m,M] ⊂ (0,∞), then max{x, 1} ≤ max{M, 1} and min{m, 1} ≤ min{x, 1} and by (2.12) we get ν (1 −ν) minx∈[m,M] (x− 1) 2 max{M, 1} ≤ ν (1 −ν) (x− 1)2 max{M, 1} ≤ 1 −ν + νx− ( (1 −ν) + νx−1 )−1 ≤ ν (1 −ν) (x− 1)2 min{m, 1} ≤ ν (1 −ν) maxx∈[m,M] (x− 1) 2 min{m, 1} (2.13) for any x ∈ [m,M] and for any ν ∈ [0, 1]. Observe that min x∈[m,M] (x− 1)2 =   (M − 1)2 if M < 1, 0 if m ≤ 1 ≤ M, (m− 1)2 if 1 < m, and max x∈[m,M] (x− 1)2 =   (m− 1)2 if M < 1, max { (m− 1)2 , (M − 1)2 } if m ≤ 1 ≤ M, (M − 1)2 if 1 < m. Then minx∈[m,M] (x− 1) 2 max{M, 1} =   (M − 1)2 if M < 1, 0 if m ≤ 1 ≤ M, (m−1)2 M if 1 < m, = c (m,M) and maxx∈[m,M](x− 1)2 min{m, 1} =   (m−1)2 m if M < 1, 1 m max { (m− 1)2 , (M − 1)2 } if m ≤ 1 ≤ M, (M − 1)2 if 1 < m, = C (m,M) . 48 s.s. dragomir Using the inequality (2.13) we have ν (1 −ν) c (m,M) ≤ ν (1 −ν) (x− 1)2 max{M, 1} ≤ 1 −ν + νx− ( (1 −ν) + νx−1 )−1 ≤ ν (1 −ν) (x− 1)2 min{m, 1} ≤ ν (1 −ν) C (m,M) (2.14) for any x ∈ [m,M] and for any ν ∈ [0, 1]. If we use the continuous functional calculus for the positive invertible op- erator X with mI ≤ X ≤ MI, then we have from (2.14) that ν (1 −ν) c (m,M) I ≤ ν (1 −ν) max{M, 1} (X − I)2 ≤ (1 −ν) I + νX − ( (1 −ν) I + νX−1 )−1 ≤ ν (1 −ν) min{m, 1} (X − I)2 ≤ ν (1 −ν) C (m,M) I (2.15) for any ν ∈ [0, 1]. If we multiply (1.2) both sides by A−1/2 we get MI ≥ A−1/2BA−1/2 ≥ mI. By writing the inequality (2.15) for X = A−1/2BA−1/2 we obtain ν(1−ν)c(m,M)I ≤ ν(1 −ν) max{M, 1} ( A−1/2BA−1/2 − I )2 (2.16) ≤ (1 −ν)I + νA−1/2BA−1/2 −A−1/2 ( (1 −ν)A−1 + νB−1 )−1 A−1/2 ≤ ν(1 −ν) min{m, 1} ( A−1/2BA−1/2 − I )2 ≤ ν(1 −ν)C(m,M)I for any ν ∈ [0, 1]. upper and lower bounds 49 If we multiply the inequality (2.16) both sides with A1/2, then we get ν (1 −ν) c (m,M) A ≤ ν (1 −ν) max{M, 1} A1/2 ( A−1/2BA−1/2 − I )2 A1/2 ≤ (1 −ν) A + νB − ( (1 −ν) A−1 + νB−1 )−1 ≤ ν (1 −ν) min{m, 1} A1/2 ( A−1/2BA−1/2 − I )2 A1/2 ≤ ν (1 −ν) C (m,M) A, (2.17) and since A1/2 ( A−1/2 BA−1/2 − I )2 A1/2 = A1/2 ( A−1/2 (B −A) A−1/2 )2 A1/2 = A1/2A−1/2 (B −A) A−1/2A−1/2 (B −A) A−1/2A1/2 = (B −A) A−1 (B −A) , then by (2.17) we get the desired result (2.10). When the operators A and B are bounded above and below by constants we have the following result as well: Corollary 1. Let A, B be two positive operators and m, m′, M, M ′ be positive real numbers. Put h := M m and h′ := M ′ m′ . (i) if 0 < mI ≤ A ≤ m′I < M ′I ≤ B ≤ MI, then ν (1 −ν) (h′ − 1)2 h A ≤ ν (1 −ν) h (B −A) A−1 (B −A) ≤ A∇νB −A!νB ≤ ν (1 −ν) (B −A) A−1 (B −A) ≤ ν (1 −ν) (h− 1)2 A, (2.18) and, in particular, (h′ − 1)2 4h A ≤ 1 4h (B −A) A−1 (B −A) ≤ A∇B −A!B ≤ 1 4 (B −A) A−1 (B −A) ≤ 1 4 (h− 1)2 A. (2.19) 50 s.s. dragomir (ii) if 0 < mI ≤ B ≤ m′I < M ′I ≤ A ≤ MI, then ν (1 −ν) ( h′ − 1 h′ )2 A ≤ ν (1 −ν) (B −A) A−1 (B −A) ≤ A∇νB −A!νB ≤ ν (1 −ν) h (B −A) A−1 (B −A) ≤ ν (1 −ν) (h− 1)2 h A (2.20) and, in particular, 1 4 ( h′ − 1 h′ )2 A ≤ 1 4 (B −A) A−1 (B −A) ≤ A∇B −A!B ≤ 1 4 h (B −A) A−1 (B −A) ≤ (h− 1)2 4h A. (2.21) Proof. We observe that h, h′ > 1 and if either of the condition (i) or (ii) holds, then h ≥ h′. If (i) is valid, then we have A < h′A = M ′ m′ A ≤ B ≤ M m A = hA, (2.22) while, if (ii) is valid, then we have 1 h A ≤ B ≤ 1 h′ A < A. (2.23) If we use the inequality (2.10) and the assumption (i), then we get (2.18). If we use the inequality (2.10) and the assumption (ii), then we get (2.20). 3. Bounds in term of Kantorovich’s constant We consider the Kantorovich’s constant defined by K (h) := (h + 1) 2 4h , h > 0. (3.1) The function K is decreasing on (0, 1) and increasing on [1,∞), K(h) ≥ 1 for any h > 0 and K(h) = K( 1 h ) for any h > 0. upper and lower bounds 51 Observe that for any h > 0 K(h) − 1 = (h− 1)2 4h = K ( 1 h ) − 1. Observe that K ( b a ) − 1 = (b−a)2 4ab for a,b > 0. Since, obviously ab = min{a,b}max{a,b} for a,b > 0, then we have the following version of Lemma 1: Lemma 2. For any a, b > 0 and ν ∈ [0, 1] we have 4ν (1 −ν) min{a,b} [ K ( b a ) − 1 ] ≤ Aν (a,b) −Hν (a,b) ≤ 4ν (1 −ν) max{a,b} [ K ( b a ) − 1 ] . (3.2) For positive invertible operators A, B we define A∇∞B := 1 2 (A + B) + 1 2 A1/2 ∣∣∣A−1/2 (B −A) A−1/2∣∣∣A1/2, A∇−∞B := 1 2 (A + B) − 1 2 A1/2 ∣∣∣A−1/2 (B −A) A−1/2∣∣∣A1/2. If we consider the continuous functions f∞, f−∞ : [0,∞) → [0,∞) defined by f∞ (x) = max{x, 1} = 1 2 (x + 1) + 1 2 |x− 1| , f−∞ (x) = max{x, 1} = 1 2 (x + 1) − 1 2 |x− 1| , then, obviously, we have A∇±∞B = A1/2f±∞ ( A−1/2BA−1 ) A1/2. (3.3) If A and B are commutative, then A∇±∞B = 1 2 (A + B) ± 1 2 |B −A| = B∇±∞A. 52 s.s. dragomir Theorem 3. Let A, B be positive invertible operators and M > m > 0 such that the condition (1.2) holds. Then we have 4ν (1 −ν) g (m,M) A∇−∞B ≤ A∇νB −A!νB ≤ 4ν (1 −ν) G (m,M) A∇∞B, (3.4) where g (m,M) :=   K (M) − 1 if M < 1, 0 if m ≤ 1 ≤ M, K (m) − 1 if 1 < m, G (m,M) :=   K (m) − 1 if M < 1, max{K (m) ,K (M)}− 1 if m ≤ 1 ≤ M, K (M) − 1 if 1 < m. In particular, g (m,M) A∇−∞B ≤ A∇B −A!B ≤ G (m,M) A∇∞B. (3.5) Proof. From (3.2) we have for a = 1 and b = x that 4ν (1 −ν) min{1,x} [K (x) − 1] ≤ 1 −ν + νx− ( (1 −ν) + νx−1 )−1 ≤ 4ν (1 −ν) max{1,x} [K (x) − 1] (3.6) for any x > 0. From (3.6) we then have 4ν (1 −ν) f−∞(x) min x∈[m,M] [K (x) − 1] ≤ 1 −ν + νx− ( (1 −ν) + νx−1 )−1 ≤ 4ν (1 −ν) f∞ (x) max x∈[m,M] [K (x) − 1] (3.7) for any x ∈ [m,M]. Observe that max x∈[m,M] [K (x) − 1] =   K (m) − 1 if M < 1, max{K (m) ,K (M)}− 1 if m ≤ 1 ≤ M, K (M) − 1 if 1 < m, = G (m,M) upper and lower bounds 53 and min x∈[m,M] [K (x) − 1] =   K (M) − 1 if M < 1, 0 if m ≤ 1 ≤ M, K (m) − 1 if 1 < m. = g (m,M) . Therefore by (3.7) we get 4ν (1 −ν) f−∞ (x) g (m,M) ≤ 1 −ν + νx− ( (1 −ν) + νx−1 )−1 ≤ 4ν (1 −ν) f∞ (x) G (m,M) (3.8) for any x ∈ [m,M] and ν ∈ [0, 1]. If we use the continuous functional calculus for the positive invertible op- erator X with mI ≤ X ≤ MI, then we have from (3.8) that 4ν (1 −ν) f−∞ (X) g (m,M) ≤ (1 −ν) I + νX − ( (1 −ν) + νX−1 )−1 ≤ 4ν (1 −ν) f∞ (X) G (m,M) (3.9) for any x ∈ [m,M] and ν ∈ [0, 1]. By writing the inequality (3.9) for X = A−1/2BA−1/2 we obtain 4ν (1 −ν) f−∞ ( A−1/2BA−1/2 ) g (m,M) (3.10) ≤ (1 −ν) I + νA−1/2BA−1/2 −A−1/2 ( (1 −ν) A−1 + νB−1 )−1 A−1/2 ≤ 4ν (1 −ν) f∞ ( A−1/2BA−1/2 ) G (m,M) for any ν ∈ [0, 1]. If we multiply (3.10) both sides by A1/2 we get 4ν (1 −ν)A1/2f−∞ ( A−1/2BA−1/2 ) A1/2g (m,M) ≤ (1 −ν) A + νB − ( (1 −ν) A−1 + νB−1 )−1 ≤ 4ν (1 −ν) A1/2f∞ ( A−1/2BA−1/2 ) A1/2G (m,M) for any ν ∈ [0, 1], which, by (3.3) produces the desired result (3.4). We have: 54 s.s. dragomir Corollary 2. Let A, B be two positive operators and m, m′, M, M ′ be positive real numbers. Put h := M m and h′ := M ′ m′ . If either of the conditions (i) or (ii) from Corollary 1 holds, then 4ν (1 −ν) [ K ( h′ ) − 1 ] A∇−∞B ≤ A∇νB −A!νB (3.11) ≤ 4ν (1 −ν) [K (h) − 1] A∇∞B. In particular,[ K ( h′ ) − 1 ] A∇−∞B ≤ A∇B −A!B ≤ [K (h) − 1] A∇∞B. (3.12) Proof. If (i) is valid, then we have A < h′A = M ′ m′ A ≤ B ≤ M m A = hA. By using the inequality (3.4) we get (3.11). If (ii) is valid, then we have 1 h A ≤ B ≤ 1 h′ A < A. By using the inequality (3.4) we get 4ν (1 −ν) [ K ( 1 h′ ) − 1 ] A∇−∞B ≤ A∇νB −A!νB ≤ 4ν (1 −ν) [ K ( 1 h ) − 1 ] A∇∞B, and since K ( 1 h′ ) = K (h′) and K ( 1 h ) = K (h), the inequality (3.11) is also obtained. 4. Further bounds The following result also holds: Theorem 4. Let A, B be positive invertible operators and M > m > 0 such that the condition (1.2) holds. Then we have pν (m,M) A ≤ A∇νB −A!νB ≤ Pν (m,M) A (4.1) upper and lower bounds 55 for any ν ∈ [0, 1], where pν(m,M) :=   ν(1−ν)(M−1)2 (1−ν)M+ν if M < 1, 0 if m ≤ 1 ≤ M, ν(1−ν)(m−1)2 (1−ν)m+ν if 1 < m, Pν (m,M) :=   ν(1−ν)(m−1)2 (1−ν)m+ν if M < 1, max { ν(1−ν)(m−1)2 (1−ν)m+ν , ν(1−ν)(M−1)2 (1−ν)M+ν } if m ≤ 1 ≤ M, ν(1−ν)(M−1)2 (1−ν)M+ν if 1 < m. Proof. Consider the function ξν : (0,∞) → (0,∞) defined by ξν (x) = 1 −ν + νx− x (1 −ν) x + ν , where ν ∈ [0, 1]. Taking the derivative, we have ξ′ν(x) = ν − (1 −ν)x + ν −x(1 −ν) [(1 −ν)x + ν]2 = ν [(1 −ν)x + ν]2 − 1 [(1 −ν)x + ν]2 = ν(1 −ν)(x− 1) [(1 −ν)x + ν + 1] [(1 −ν)x + ν]2 for any x ≥ 0 and ν ∈ [0, 1]. This shows that the function is decreasing on [0, 1] and increasing on (1,∞). We have ξν (0) = 1 −ν, ξν (1) = 0 and limx→∞ξν (x) = ∞. Since, by (2.3) ξν (x) = ν (1 −ν) (x− 1)2 (1 −ν) x + ν , x ≥ 0, then for [m,M] ⊂ [0,∞) we have min x∈[m,M] ξν (x) =   ν(1−ν)(M−1)2 (1−ν)M+ν if M < 1, 0 if m ≤ 1 ≤ M, ν(1−ν)(m−1)2 (1−ν)m+ν if 1 < m, = pν (m,M) 56 s.s. dragomir and max x∈[m,M] ξν(x) =   ν(1−ν)(m−1)2 (1−ν)m+ν if M < 1, max { ν(1−ν)(m−1)2 (1−ν)m+ν , ν(1−ν)(M−1)2 (1−ν)M+ν } if m ≤ 1 ≤ M, ν(1−ν)(M−1)2 (1−ν)M+ν if 1 < m, = Pν (m,M) . Therefore pν (m,M) ≤ 1 −ν + νx− ( (1 −ν) + νx−1 )−1 ≤ Pν (m,M) (4.2) for any x ∈ [m,M] and ν ∈ [0, 1]. If we use the continuous functional calculus for the positive invertible operator X with mI ≤ X ≤ MI, then we have from (4.2) that p (m,M) I ≤ (1 −ν) I + νX − ( (1 −ν) I + νX−1 )−1 ≤ Pν (m,M) I (4.3) for any ν ∈ [0, 1]. If we multiply (1.2) both sides by A−1/2 we get MI ≥ A−1/2BA−1/2 ≥ mI. By writing the inequality (4.3) for X = A−1/2BA−1/2 we obtain p(m,M)I ≤ (1 −ν)I + νA−1/2BA−1/2 −A−1/2 ( (1 −ν)A−1 + νB−1 )−1 A−1/2 ≤ Pν (m,M) I (4.4) for any ν ∈ [0, 1]. If we multiply (4.4) both sides by A1/2 we get p (m,M) A ≤ (1 −ν) A + νB − ( (1 −ν) A−1 + νB−1 )−1 ≤ Pν (m,M) A for any ν ∈ [0, 1]. upper and lower bounds 57 Remark 2. If we consider p (m,M) :=   (M−1)2 2(M+1) if M < 1, 0 if m ≤ 1 ≤ M, (m−1)2 2(m+1) if 1 < m, P (m,M) :=   (m−1)2 2(m+1) if M < 1, max { (m−1)2 2(m+1) , (M−1)2 2(M+1) } if m ≤ 1 ≤ M, (M−1)2 2(M+1) if 1 < m, then by (4.1) we have p (m,M) A ≤ A∇B −A!B ≤ P (m,M) A, (4.5) provided that A, B are positive invertible operators and M > m > 0 are such that the condition (1.2) holds. Corollary 3. Let A, B be two positive operators and m, m′, M, M ′ be positive real numbers. Put h := M m and h′ := M ′ m′ . (i) if 0 < mI ≤ A ≤ m′I < M ′I ≤ B ≤ MI, then for any ν ∈ [0, 1] ν (1 −ν) (h′ − 1)2 (1 −ν) h′ + ν A ≤ A∇νB −A!νB ≤ ν (1 −ν) (h− 1)2 (1 −ν) h + ν A (4.6) and, in particular, (h′ − 1)2 2 (h′ + 1) A ≤ A∇B −A!B ≤ (h− 1)2 2 (h + 1) A. (4.7) (ii) if 0 < mI ≤ B ≤ m′I < M ′I ≤ A ≤ MI, then for any ν ∈ [0, 1] ν (1 −ν) (h′ − 1)2 h′ (1 −ν + νh′) A ≤ A∇νB −A!νB ≤ ν (1 −ν) (h− 1)2 h (1 −ν + νh) A (4.8) 58 s.s. dragomir and, in particular, (h′ − 1)2 2h′ (1 + h′) A ≤ A∇B −A!B ≤ (h− 1)2 2h (1 + h) A. (4.9) Proof. We observe that h, h′ > 1 and if either of the condition (i) or (ii) holds, then h ≥ h′. If (i) is valid, then we have A < h′A = M ′ m′ A ≤ B ≤ M m A = hA, while, if (ii) is valid, then we have 1 h A ≤ B ≤ 1 h′ A < A. If we use the inequality (4.1) and the assumption (i), then we get (4.6). If we use the inequality (4.1) and the assumption (ii), then we get (4.8). 5. A comparison We observe that an upper bound for the difference A∇νB −A!νB as pro- vided in (1.3) is B1 (ν,m,M) A := max{ν, 1 −ν}×   (m−1)2 m+1 A if M < 1, max { (m−1)2 m+1 , (M−1)2 M+1 } A if m ≤ 1 ≤ M, (M−1)2 M+1 A if 1 < m while the one from (2.10) is B2 (ν,m,M) A := ν (1 −ν)×   (m−1)2 m A if M < 1, 1 m max { (m− 1)2, (M − 1)2 } A if m ≤ 1 ≤ M, (M − 1)2 A if 1 < m, where A, B are positive invertible operators and M > m > 0 such that the condition (1.2) holds. upper and lower bounds 59 We consider for x = m ∈ (0, 1) and y = ν ∈ [0, 1] the difference D1 (x,y) = max{y, 1 −y} (x− 1)2 x + 1 −y (1 −y) (x− 1)2 x that has the 3D plot on the box [0.3, 0.6] × [0, 1] depicted in Figure 1 show- ing that it takes both positive and negative values, meaning that neither of the bounds B1 (ν,m,M) A and B2 (ν,m,M) A is better in the case 0 < m < M < 1. Figure 1: Plot of difference D1(x,y) Figure 2: Plot of difference D2(x,y) 60 s.s. dragomir We consider for x = M ∈ (1,∞) and y = ν ∈ [0, 1] the difference D2 (x,y) = max{y, 1 −y} (x− 1)2 x + 1 −y (1 −y) (x− 1)2 that has the 3D plot on the box [1, 3]×[0, 1] depicted in Figure 2 showing that it takes both positive and negative values, meaning that neither of the bounds B1 (ν,m,M) A and B2 (ν,m,M) A is better in the case 1 < m < M < ∞. Similar conclusions may be derived for lower bounds, however the details are left to the interested reader. References [1] S.S. Dragomir, Bounds for the normalised Jensen functional, Bull. Austral. Math. Soc. 74 (3) (2006), 417 – 478. [2] S.S. Dragomir, A note on Young’s inequality, preprint RGMIA Res. Rep. Coll. 18 (2015), Art. 126, http://rgmia.org/papers/v18/v18a126.pdf. [3] S.S. Dragomir, Some new reverses of Young’s operator inequality, preprint RGMIA Res. Rep. Coll. 18 (2015), Art. 130, http://rgmia.org/papers/v18/v18a130.pdf. [4] S.S. Dragomir, On new refinements and reverses of Young’s opera- tor inequality, preprint RGMIA Res. Rep. Coll. 18 (2015), Art. 135, http://rgmia.org/papers/v18/v18a135.pdf. [5] S.S. Dragomir, Some inequalities for operator weighted geomet- ric mean, preprint RGMIA Res. Rep. Coll. 18 (2015), Art. 139, http://rgmia.org/papers/v18/v18a139.pdf. [6] S.S. Dragomir, Some reverses and a refinement of Hölder opera- tor inequality, preprint RGMIA Res. Rep. Coll. 18 (2015), Art. 147, http://rgmia.org/papers/v18/v18a147.pdf. [7] S.S. Dragomir, Some inequalities for weighted harmonic and arithmetic operator means, preprint RGMIA Res. Rep. Coll. 19 (2016), Art. 5, http://rgmia.org/papers/v19/v19a05.pdf. [8] S. Furuichi, Refined Young inequalities with Specht’s ratio, J. Egyptian Math. Soc. 20 (2012), 46 – 49. [9] S. Furuichi, On refined Young inequalities and reverse inequalities, J. Math. Inequal. 5 (2011), 21 – 31. [10] W. Liao, J. Wu, J. Zhao, New versions of reverse Young and Heinz mean inequalities with the Kantorovich constant, Taiwanese J. Math. 19 (2) (2015), 467 – 479. [11] M. Tominaga, Specht’s ratio in the Young inequality, Sci. Math. Japon. 55 (2002), 583 – 588. [12] G. Zuo, G. Shi, M. Fujii, Refined Young inequality with Kantorovich constant, J. Math. Inequal. 5 (2011), 551 – 556. http://rgmia.org/papers/v18/v18a126.pdf http://rgmia.org/papers/v18/v18a130.pdf http://rgmia.org/papers/v18/v18a135.pdf http://rgmia.org/papers/v18/v18a139.pdf http://rgmia.org/papers/v18/v18a147.pdf http://rgmia.org/papers/v19/v19a05.pdf Introduction Min and max bounds Bounds in term of Kantorovich's constant Further bounds A comparison