FASCICULIMATHEMATICI Nr 62 2019

S. S. Dragomir

SOME REVERSES OF HOLDER¨ VECTOR OPERATOR INEQUALITY

Abstract.In this paper we obtain some new reverses of H¨older vector inequality for positive operators on Hilbert spaces. Key words: Young’s inequality, H¨older operator inequality, arithmetic mean-geometric mean inequality. AMS Mathematics Subject Classification: 47A63, 47A30, 26D15, 26D10, 15A60.

1. Introduction

Throughout this paper A, B are positive invertible operators on a com- plex Hilbert space (H, h·, ·i) . We use the following notation ν 1/2  −1/2 −1/2 1/2 A]νB := A A BA A ,

1 the weighted geometric mean. When ν = 2 we write A]B for brevity. In [6] the authors obtained the following result:

q p p 1/p q 1/q (1) B ]1/pA x, x ≤ hA x, xi hB x, xi ! 1/p m1 M1 q p ≤ λ p; q−1 , q−1 B ]1/pA x, x M2 m2 for any x ∈ H, where 0 < m1I ≤ A ≤ M1I, 0 < m2I ≤ B ≤ M2I, p, q > 1 1 1 with p + q = 1,I is the identity operator and " #p 1 M p − mp λ (p; m, M) := p1/pq1/q (M − m)1/p (mM p − Mmp)1/q for 0 < m < M. In particular, one can obtain from (1) the following noncommutative version of Greub-Rheinboldt inequality

1/2 1/2 m m + M M (2) A2]B2x, x ≤ A2x, x B2x, x ≤ √1 2 1 2 A2]B2x, x 2 m1m2M1M2 36 S. S. Dragomir for any x ∈ H. Moreover, if A and B are replaced by C1/2 and C−1/2 in (2), then we get the Kantorovich inequality [20]

1/2 m + M hCx, xi1/2 C−1x, x ≤ √ , x ∈ H with kxk = 1, 2 mM provided mI ≤ C ≤ MI for some 0 < m < M. For various related inequalities, see [5]-[12] and [16]- [17]. In this paper, by making use of some recent Young’s type inequalities outlined below, we establish some reverses and a refinement of H¨older’s inequality for the positive operators A, B

q p p 1/p q 1/q B ]1/pA x, x ≤ hA x, xi hB x, xi , x ∈ H

1 1 where p, q > 1 with p + q = 1. The famous Young inequality for scalars says that if a, b > 0 and ν ∈ [0, 1], then

(3) a1−νbν ≤ (1 − ν) a + νb with equality if and only if a = b. The inequality (3) is also called ν-weighted arithmetic-geometric mean inequality. We recall that Specht’s ratio is defined by [21]

 1 h h−1   1  if h ∈ (0, 1) ∪ (1, ∞) (4) S (h) := e ln h h−1   1 if h = 1.

1
It is well known that limh→1 S (h) = 1, S (h) = S h > 1 for h > 0, h 6= 1. The function is decreasing on (0, 1) and increasing on (1, ∞). The following inequality provides a refinement and a multiplicative re- verse for Young’s inequality ar a (5) S a1−νbν ≤ (1 − ν) a + νb ≤ S a1−νbν, b b where a, b > 0, ν ∈ [0, 1], r = min {1 − ν, ν}. The second inequality in (5) is due to Tominaga [22] while the first one is due to Furuichi [7]. We consider the Kantorovich’s constant defined by

(h + 1)2 (6) K (h) := , h > 0. 4h Some reverses of Holder¨ vector operator . . . 37

The function K is decreasing on (0, 1) and increasing on [1, ∞) ,K (h) ≥ 1 1
for any h > 0 and K (h) = K h for any h > 0. The following multiplicative refinement and reverse of Young inequality in terms of Kantorovich’s constant holds a a (7) Kr a1−νbν ≤ (1 − ν) a + νb ≤ KR a1−νbν b b where a, b > 0, ν ∈ [0, 1], r = min {1 − ν, ν} and R = max {1 − ν, ν}. The first inequality in (7) was obtained by Zou et al. in [23] while the second by Liao et al. [19]. Kittaneh and Manasrah [14], [15] provided a refinement and an additive reverse for Young inequality as follows: √ √ 2 √ √ 2 (8) r a − b ≤ (1 − ν) a + νb − a1−νbν ≤ R a − b where a, b > 0, ν ∈ [0, 1], r = min {1 − ν, ν} and R = max {1 − ν, ν}. The 1 case ν = 2 reduces (8) to an identity. In the recent paper [1] we obtained the following reverses of Young’s inequality as well:

(9) 0 ≤ (1 − ν) a + νb − a1−νbν ≤ ν (1 − ν)(a − b) (ln a − ln b) and (1 − ν) a + νb h  a i (10) 1 ≤ ≤ exp 4ν (1 − ν) K − 1 , a1−νbν b where a, b > 0, ν ∈ [0, 1]. In [2] we obtained the following inequalities that improve the correspond- ing results of Furuichi and Minculete from [9] 1 (11) ν (1 − ν) (ln a − ln b)2 min {a, b} ≤ (1 − ν) a + νb − a1−νbν 2 1 ≤ ν (1 − ν) (ln a − ln b)2 max {a, b} 2 and " # 1  min {a, b} 2 (1 − ν) a + νb (12) exp ν (1 − ν) 1 − ≤ 2 max {a, b} a1−νbν " # 1 max {a, b} 2 ≤ exp ν (1 − ν) − 1 2 min {a, b} for any a, b > 0 and ν ∈ [0, 1]. 38 S. S. Dragomir

2. Some reverse inequalities

We have the following reverse of H¨older’sinequality: Theorem 1. Let A and B be two positive invertible operators, p, q > 1 1 1 with p + q = 1 and m, M > 0 such that (13) mpBq ≤ Ap ≤ M pBq.

Then for any x ∈ H we have the inequality M p (14) hApx, xi1/p hBqx, xi1/q ≤ S Bq] Apx, x . m 1/p

t Proof. Assume that ν ∈ (0, 1) . Let a, b ∈ [t, T ] ⊂ (0, ∞), then T ≤ a T t T a  t
a
t
T
b ≤ t with T < 1 < t . If b ∈ T , 1 then S b ≤ S T = S t . If a T  a
T
b ∈ 1, t then also S b ≤ S t . Therefore for any a, b ∈ [t, T ] we have by Tominaga’s inequality (5) that T  (15) (1 − ν) a + νb ≤ S a1−νbν. t Now, if C is an operator with tI ≤ C ≤ TI then for p > 1 we have tpI ≤ p p 1 C ≤ T I. Using the functional calculus we get from (15) for ν = p that    p 1 1 T 1− 1 1 − d + Cp ≤ S d p C, p p t namely, the vector inequality,    p 1 1 T 1− 1 (16) 1 − d + hCpy, yi ≤ S d p hCy, yi , p p t for any y ∈ H, kyk = 1 and d ∈ [tp,T p]. Since d := hCpy, yi ∈ [tp,T p] for any y ∈ H, kyk = 1, hence by (16) we have    p 1 1 T 1− 1 1 − hCpy, yi + hCpy, yi ≤ S hCpy, yi p hCy, yi , p p t that is equivalent to  p T 1− 1 hCpy, yi ≤ S hCpy, yi p hCy, yi t and to T p (17) hCpy, yi ≤ Sp hCy, yip t Some reverses of Holder¨ vector operator . . . 39 for any y ∈ H, kyk = 1. z If z ∈ H with z 6= 0, then by taking y = kzk in (17) we get T p hCpz, zi kzk2p−2 ≤ Sp hCz, zip t

1 for any z ∈ H, and by taking the power p we get T p (18) hCpz, zi1/p hz, zi1/q ≤ S hCz, zi t for any z ∈ H. − q p Now, from (13) by multiplying both sides with B 2 we have m I ≤ 1 q q  q q  p − 2 p − 2 p 1 − 2 p − 2 B A B ≤ M I and by taking the power p we get mI ≤ B A B ≤ MI. 1  − q p − q  p By writing the inequality (18) for C = B 2 A B 2 , t = m, T = M q and z = B 2 x, with x ∈ H, we have

1/p 1/q D − q p − q q q E D q q E B 2 A B 2 B 2 x, B 2 x B 2 x, B 2 x  p  1  M  − q p − q  p q q ≤ S B 2 A B 2 B 2 x, B 2 x , m namely

 p  1  p 1/p q 1/q M q  − q p − q  p q hA x, xi hB x, xi ≤ S B 2 B 2 A B 2 B 2 x, x m for any x ∈ H, and the inequality (14) is proved. 

Remark 1. We observe, for A and B two positive invertible operators, that the condition (13) is equivalent to following condition

1  − q p − q  p (19) mI ≤ B 2 A B 2 ≤ MI.

If we assume that

(20) rBq ≤ Ap ≤ RBq, for r, R > 0, then by (14) we have the inequality R (21) hApx, xi1/p hBqx, xi1/q ≤ S Bq] Apx, x r 1/p for any x ∈ H. 40 S. S. Dragomir

Corollary 1. Let A and B be two positive invertible operators and m, M > 0 such that

1 (22) mI ≤ B−1A2B−1
2 ≤ MI.

Then for any x ∈ H we have the inequality

2! 1/2 1/2 M  (23) A2x, x B2x, x ≤ S A2]B2x, x , m where 1/2 A2]B2 = A A−1B2A−1
A = B2]A2.

Now, by taking A = C1/2 and B = C−1/2, then the condition (22) be- comes

(24) mI ≤ C ≤ MI and by (23) we get

2! 1/2 M  hCx, xi1/2 C−1x, x ≤ S , m for any x ∈ H with kxk = 1. Corollary 2. Assume that A and B satisfy the conditions

(25) m1I ≤ A ≤ M1I, m2I ≤ B ≤ M2I for some 0 < m1 < M1 and 0 < m2 < M2. Then we have  p  q p 1/p q q 1/q M1 M2 q p (26) hA x, xi hB x, B xi ≤ S B ]1/pA x, x , m1 m2 for any x ∈ H. In particular, we have

2! 1/2 1/2 M M  (27) A2x, x B2x, x ≤ S 1 2 A2]B2x, x , m1m2 for any x ∈ H. Proof. We have from (25) that

p p p m1I ≤ A ≤ M1 I. Some reverses of Holder¨ vector operator . . . 41

Then p −q p −q − q p − q p −q p −q m1M2 I ≤ m1B ≤ B 2 A B 2 ≤ M1 B ≤ M1 m2 I, which implies that

q 1 q − q q − p  − p −  p p m1M2 I ≤ B 2 A B 2 ≤ M1m2 I.

q q − p − p Now, on using the inequality (14) for m = m1M2 and M = M1m2 , we get the desired result (26).  Using Kantorovich’s constant (6) we also have: Theorem 2. With the assumptions of Theorem 1 we have the inequality n o  p max 1 , 1 M (28) hApx, xi1/p hBqx, xi1/q ≤ K p q Bq] Apx, x m 1/p for any x ∈ H. Proof. Assume that ν ∈ (0, 1) and R = max {1 − ν, ν}. Let a, b ∈ t a T t T a  t
[t, T ] ⊂ (0, ∞), then T ≤ b ≤ t with T < 1 < t . If b ∈ T , 1 then R a
R t
R T
a T  R a
R T
K b ≤ K T = K t . If b ∈ 1, t then also K b ≤ K t . Therefore for any a, b ∈ [t, T ] we have by inequality (7) that T  (29) (1 − ν) a + νb ≤ KR a1−νbν. t Now, if C is an operator with tI ≤ C ≤ TI then for p > 1 we have tpI ≤ p p 1 C ≤ T I. Using the functional calculus we get from (29) for ν = p that   n o  p 1 1 max 1 , 1 T 1− 1 1 − d + Cp ≤ K p q d p C, p p t namely, the vector inequality,   n o  p 1 1 max 1 , 1 T 1− 1 1 − d + hCpy, yi ≤ K p q d p hCy, yi , p p t for any y ∈ H, kyk = 1 and d ∈ [tp,T p]. Now, by employing a similar argument to the one in the proof of Theo- rem 1 we deduce the desired result (28). The details are omitted. 

Corollary 3. With the assumptions of Corollary 1 we have

" 2!#1/2 1/2 1/2 M  (30) A2x, x B2x, x ≤ K A2]B2x, x , m for any x ∈ H. 42 S. S. Dragomir

We also have: Corollary 4. With the assumptions of Corollary 2 we have (31) hApx, xi1/p hBqx, Bqxi1/q p q n 1 1 o M  M   max p , q 1 2 q p ≤ K B ]1/pA x, x , m1 m2 for any x ∈ H. In particular, we have

" 2!#1/2 1/2 1/2 M M  (32) A2x, x B2x, x ≤ K 1 2 A2]B2x, x , m1m2 for any x ∈ H.

3. Exponential reverses

We have: Theorem 3. With the assumptions of Theorem 1 we have the inequality (33) hApx, xi1/p hBqx, xi1/q  4  M p  ≤ exp K − 1 Bq] Apx, x pq m 1/p for any x ∈ H. Proof. Assume that ν ∈ (0, 1). Let a, b ∈ [t, T ] ⊂ (0, ∞), then by the inequality (10) we have   T   (34) (1 − ν) a + νb ≤ a1−νbν exp 4ν (1 − ν) K − 1 . t Now, if C is an operator with tI ≤ C ≤ TI then for p > 1 we have tpI ≤ p p 1 C ≤ T I. Using the functional calculus we get from (34) for ν = p that      p  1 1 4 T 1− 1 1 − d + Cp ≤ exp K − 1 d p C, p p pq t namely, the vector inequality,      p  1 1 4 T 1− 1 1 − d + hCpy, yi ≤ exp K − 1 d p hCy, yi , p p pq t for any y ∈ H, kyk = 1 and d ∈ [tp,T p]. Now, by employing a similar argument to the one in the proof of Theorem 1 we deduce the desired result (33). The details are omitted.  Some reverses of Holder¨ vector operator . . . 43

We have: Corollary 5. With the assumptions of Corollary 1 we have

" 2# ! 1/2 1/2 M  (35) A2x, x B2x, x ≤ exp K − 1 A2]B2x, x , m for any x ∈ H. Corollary 6. With the assumptions of Corollary 2 we have

(36) hApx, xi1/p hBqx, Bqxi1/q    p  q  4 M1 M2 q p ≤ exp K − 1 B ]1/pA x, x , pq m1 m2 for any x ∈ H. In particular, we have

" 2# ! 1/2 1/2 M M  (37) A2x, x B2x, x ≤ exp K 1 2 − 1 A2]B2x, x , m1m2 for any x ∈ H. Finally, we have the following reverse of H¨older’sinequality as well: Theorem 4. With the assumptions of Theorem 1 we have the inequality " # 1 M p 2 (38) hApx, xi1/p hBqx, xi1/q ≤ exp − 1 Bq] Apx, x 2pq m 1/p for any x ∈ H. Proof. If a, b ∈ [t, T ] ⊂ (0, ∞) and since

max {a, b} T 0 < − 1 ≤ − 1, min {a, b} t hence max {a, b} 2 T 2 − 1 ≤ − 1 . min {a, b} t Therefore, by (12) we get " # 1 T 2 (39) (1 − ν) a + νb ≤ a1−νbν exp ν (1 − ν) − 1 , 2 t for any a, b ∈ [t, T ] and ν ∈ (0, 1). 44 S. S. Dragomir

Now, if C is an operator with tI ≤ C ≤ TI then for p > 1 we have p p p 1 t I ≤ C ≤ T I. Using the functional calculus we get from (34) for ν = p that   "  p 2# 1 1 1 T 1− 1 1 − d + Cp ≤ exp − 1 d p C, p p 2pq t namely, the vector inequality,   "  p 2# 1 1 1 T 1− 1 1 − d + hCpy, yi ≤ exp − 1 d p hCy, yi , p p 2pq t for any y ∈ H, kyk = 1 and d ∈ [tp,T p]. Now, by employing a similar argument to the one in the proof of Theorem 1 we deduce the desired result (39). The details are omitted.  We have: Corollary 7. With the assumptions of Corollary 1 we have   2 !2 1/2 1/2 1 M  (40) A2x, x B2x, x ≤ exp − 1 A2]B2x, x 8 m  for any x ∈ H. If mI ≤ C ≤ MI for some m, M with 0 < m < M, then by (40) we get   2 !2 1/2 1 M  (41) hCx, xi1/2 C−1x, x ≤ exp − 1 kxk2 , 8 m  for any x ∈ H. Corollary 8. With the assumptions of Corollary 2 we have hApx, xi1/p hBqx, xi1/q "  p  q 2# 1 M1 M2 q p ≤ exp − 1 B ]1/pA x, x , 2pq m1 m2 for any x ∈ H. In particular, we have

1/2 1/2 (42) A2x, x B2x, x    2 !2 1 M1M2 2 2 ≤ exp  − 1  A ]B x, x , 8 m1m2 for any x ∈ H. Some reverses of Holder¨ vector operator . . . 45

Acknowledgement. The author would like to thank the anonymous referee for valuable suggestions that have been implemented in the final version of the paper.

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S. S. Dragomir Mathematics, College of Engineering & Science Victoria University, PO Box 14428 Melbourne City, MC 8001, Australia or School of Computer Science & Applied Mathematics University of the Witwatersrand Private Bag 3, Johannesburg 2050, South Africa e-mail: [email protected]

Received on 08.10.2019 and, in revised form, on 29.10.2019.