Volume 2013, Issue 1 124510
Research Article
Open Access

Some Operator Inequalities on Chaotic Order and Monotonicity of Related Operator Function

Changsen Yang

Corresponding Author

Changsen Yang

College of Mathematics and Information Science, Henan Normal University, Xinxiang 453002, China htu.cn

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Yanmin Liu

Yanmin Liu

College of Mathematics and Information Science, Henan Normal University, Xinxiang 453002, China htu.cn

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First published: 16 May 2013
Academic Editor: Yisheng Song

Abstract

We will discuss some operator inequalities on chaotic order about several operators, which are generalization of Furuta inequality and show monotonicity of related Furuta type operator function.

1. Introduction

An operator T is said to be positive (denoted by T ≥ 0) if (Tx, x) ≥ 0 for all vectors x in a Hilbert space, and T is said to be strictly positive (denoted by T > 0) if T is positive and invertible.

Theorem LH [Löwner-Heinz inequality, denoted by (LH) briefly] If AB ≥ 0 holds, then AαBα for any α ∈ [0,1].

This was originally proved in [1, 2] and then in [3]. Although (LH) asserts that AB ≥ 0 ensures AαBα for any α ∈ [0,1], unfortunately AαBα does not always hold for α > 1. The following result has been obtained from this point of view.

Theorem F (Furuta inequality). If AB ≥ 0, then for each r ≥ 0,

  • (i)

    (Br/2ApBr/2) 1/q ≥ (Br/2BpBr/2) 1/q,

  • (ii)

    (Ar/2ApAr/2) 1/q ≥ (Ar/2BpAr/2) 1/q

hold for p ≥ 0 and q ≥ 1 with (1 + r)qp + r.

The original proof of Theorem F is shown in [4], an elementary one-page proof is in [5], and alternative ones are in [6, 7]. We remark that the domain of the parameters p, q, and r in Theorem F is the best possible for the inequalities (i) and (ii) under the assumption AB ≥ 0; see [8].

We write AB if log A ≥ log B for A, B > 0, which is called the chaotic order.

Theorem A. For A, B > 0, the following (i) and (ii) hold:

  • (i)

    AB holds if and only if Ar ≥ (Ar/2BpAr/2) r/(p+r) for p, r ≥ 0;

  • (ii)

    AB holds if and only if for any fixed δ ≥ 0, FA,B(p, r) = Ar/2(Ar/2BpAr/2) (δ + r)/(p+r)Ar/2 is a decreasing function of pδ and r ≥ 0.

(i) in Theorem A is shown in [9, 10], an excellent proof in [11], a proof in the case p = r in [12], (ii) in [9, 10], and so forth.

Lemma B (see [11].)Let A be a positive invertible operator, and let B be an invertible operator. For any real number λ,

()

Definition 1. Let An, An−1, …, A2, A1,   B ≥ 0, r1, r2, …, rn ≥ 0, and p1, p2, …, pn ≥ 0 for a natural number n.

Let be defined by

()
For example,
()
Let q[n] be defined by
()
For example,
()
For the sake of convenience, we define
()
and these definitions in (6) may be reasonable by (2) and (4).

Lemma 2. For An, An−1, …, A2, A1,  B ≥ 0 and any natural number n, we have

  • (i)

    ,

  • (ii)

    q[n] = q[n − 1]pn + rn.

Proof. (i) and (ii) can be easily obtained by definitions (2) and (4).

2. Basic Results Associated with and q[n]

We will give some operator inequalities on chaotic order, and Theorem 5 is further extension of Theorem  3.1 in [13].

Lemma 3. If AB, for p ≥ 0 and r ≥ 0, then A ≫ (Ar/2BpAr/2) 1/(p+r).

Proof. Since AB, we can obtain the following inequality.

Ar ≥ (Ar/2BpAr/2) r/(p+r) holds for p ≥ 0 and r ≥ 0 by (i) of Theorem A.

Take the logarithm on both sides of the previous inequality; that is,

()
therefor we have
()

Theorem 4. If AnAn−1 ≫ ⋯≫A2A1B and r1, r2, …, rn ≥ 0, p1, p2, …, pn ≥ 0 for a natural number n. Then the following inequality holds:

()
where and q[n] are defined in (2) and (4).

Proof. We will show (9) by mathematical induction. In the case n = 1.

Since  A1B  implies

()
holds for any p1 ≥ 0 and r1 ≥ 0 by Lemma 3, whence (9) for n = 1.

Assume that (9) holds for a natural number k (1 ≤ k < n). We will show that (9) holds r1, r2, …, rk, rk+1 ≥ 0 and p1, p2, …, pk, pk+1 ≥ 0 for k + 1.

Put D = Ak+1,   E = Ak, and , and (9) holds for n = k implying

()
Equation (11) yields the following by Lemma 3, for r ≥ 0 and p ≥ 0:
()
that is,
()
Put r = rk+1,   p = q[k]pk+1 in (13), then by (ii) of Lemma 2, the exponential power 1/(p + r) of the right hand side of (13) can be written as follows:
()
and we have the following desired (15) by (12) and (13):
()
so that (15) shows that (9) holds for k + 1.

Theorem 5. If AnAn−1 ≫ ⋯≫A2A1B and r1, r2, …, rn ≥ 0 for a natural number n. For any fixed δ ≥ 0, let p1, p2, …, pn be satisfied by

()
The operator function Ik(pk, rk) for any natural number k such that 1 ≤ kn is defined by
()
Then the following inequality holds:
()
for every natural number k such that 1 ≤ kn, where and q[n] are defined in (2) and (4).

Proof. Since , q[0] = 1 in (6), we may define I0(p0, r0) = Bδ for p0 = r0 = 0.

Because A1B, then for any fixed δ ≥ 0,

()
since holds by (ii) of Theorem A. And (19) can be expressed as
()
We can apply Theorem 4, and we have the following (21) for any natural number k such that 1 ≤ kn:
()
Since XY implies that XtYt holds for any t ≥ 0, (21) ensures
()
Putting , and applying (19) for δ = 1 and AB1, we have
()
holds for p ≥ 1 and r ≥ 0.

Putting rk+1 = r(δ + r1 + r2 + ⋯+rk) in (23), then (23) can be rewritten by

()
Putting p = (q[k]pk+1)/(δ + r1 + r2 + ⋯+rk) ≥ 1, since pk+1 ≥ (δ + r1 + r2 + ⋯+rk)/q[k] in (16), then we have
()
and we have (18) for k such that 1 ≤ kn by (25) and (20) since (20) means (18) for k = 1.

Corollary 6. If AnAn−1 ≫ ⋯≫A2A1B and r1, r2, …, rn ≥ 0 for a natural number n. For any fixed δ ≥ 0, let p1, p2, …, pn be satisfied by (16).

Then the following inequalities hold:

()
where , q[n], and Ik(pk, rk)  (1 ≤ kn) are defined in (2), (4), and (17).

Proof. Applying (18) of Theorem 5 for k such that 1 ≤ kn, we have

()

3. Monotonicity Property on Operator Functions

We would like to emphasize that the condition of Theorem 7 is stronger than Theorem 5, and moreover when we discuss monotonicity property on operator functions, we can only apply Theorem 7.

Theorem 7. If AnAn−1 ≫ ⋯≫A2A1B and r1, r2, …, rn ≥ 0, p1, p2, …, pn ≥ 0 for a natural number n. Then the following inequality holds:

()
where and q[n] are defined in (2) and (4).

Proof. We will show (28) by mathematical induction. In the case n = 1.

Since  A1B  implies

()
holds for any, p1 ≥ 0 and r1 ≥ 0 by (i) of Theorem A, whence (28) for n = 1.

Assume that (28) holds for a natural number k (1 ≤ k < n). We will show (28) for r1, r2, …, rk+1 ≥ 0 and p1, p2, …, pk,   pk+1 ≥ 0 for k + 1.

We can obtain the following inequality from the hypothesis (28) for the case n = k:

()
hence we have , and (i) of Theorem A ensures
()
Putting r = rk+1 and p = q[k]pk+1, then we have the following inequality:
()
so that (32) shows (28) for k + 1.

Theorem 8. If AnAn−1 ≫ ⋯≫A2A1B and r1, r2, …, rn ≥ 0 for a natural number n. For any fixed δ ≥ 0, let p1, p2, …, pn be satisfied by (16).

Then

()
is a decreasing function of both rn ≥ 0 and pn which satisfies
()
where and q[n] are defined in (2) and (4).

Proof. Since the condition (16) with δ ≥ 0 suffices (28) in Theorem 7, we have the following inequality by Theorem 7; see (28).

We state the following important inequality (35) for the forthcoming discussion which is the inequality in (16):

()
because the inequality in (35) follows by (ii) of Lemma 2, and the inequality follows by
()
obtained by (34).

(a) Proof of the result that In(pn, rn) is a decreasing function of pn.

Without loss of generality, we can assume that pn > 0. We can obtain the following inequality by (28) and by (i) of Lemma 2:

()
and (37) implies
()
Put α = ω/pn ∈ [0,1] for pnω ≥ 0, then we raise each side of (38) to the power α = ω/pn ∈ [0,1], then
()

Whence we have

()
and the last inequality holds by LH because (39) and (δ + r1 + r2 + ⋯+rn)/(q[n − 1](pn + ω) + rn) ∈ [0,1] which is ensured by (35) and q[n] + q[n − 1]ω = q[n − 1](pn + ω) + rnq[n] by (4), so that In(pn, rn) is a decreasing function of pn.

(b) Proof of the result that In(pn, rn) is a decreasing function of rn.

Without loss of generality, we can assume that rn > 0. Raise each side of (28) to the power μ/rn ∈ [0,1] for rnμ ≥ 0 by LH, then

()
We state the following inequality by (ii) of Lemma 3 and (35):
()
Then we have
()
and the last inequality holds by LH because (41) and
()
so that Ik(pk, rk) is a decreasing function of rn.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (1127112; 11201127), Technology and Pioneering project in Henan Province (122300410110).

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