# Some Properties of Furuta Type Inequalities and Applications

Authors: Jiangtao Yuan; Caihong Wang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

This work is to consider Furuta type inequalities and their applications. Firstly, some Furuta type inequalities underA≥B≥0 are obtained via Loewner-Heinz inequality; as an application, a proof of Furuta inequality is given without using the invertibility of operators. Secondly, we show a unified satellite theorem of grand Furuta inequality which is an extension of the results by Fujii et al. At the end, a kind of Riccati type operator equation is discussed via Furuta type inequalities.


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## Body

## 1. Introduction

Throughout this paper, an operatorT means a bounded linear operator on a Hilbert space. T≥0 and T>0 mean a positive operator and an invertible positive operator, respectively, (see [1, page 103]). The classical Loewner-Heinz inequality (L-H) is stated below (see [2, page 127]).Theorem 1 (Loewner-Heinz inequality (L-H)).
Letp∈[0,1]; then A≥B≥0 ensures
(1)Ap≥Bp.In general, (L-H) is not true forp>1. As a celebrated development of (L-H), Furuta provided a kind of order preserving operator inequality [2, page 129], the so-called Furuta inequality (FI).Theorem 2 (Furuta inequality (FI), [3]).
Letr≥0, p>0; then A≥B≥0 ensures
(2)(Br/2ApBr/2)(min⁡{1,p}+r)/(p+r)≥(Br/2BpBr/2)(min⁡{1,p}+r)/(p+r),(Ar/2ApAr/2)(min⁡{1,p}+r)/(p+r)≥(Ar/2BpAr/2)(min⁡{1,p}+r)/(p+r).Tanahashi proved that the outer exponentmin⁡{1,p}+r above is optimal; see [3] for related topics. In order to establish the order structure on Aluthge transform of nonnormal operators, the complete form of Furuta inequality was showed in [4].Theorem 3 (Complete form [4]).
Letq>0, r≥0, p>p0>0 and s(q)=min⁡{p,2p0+min⁡{q,r}}. Then A≥0 and B≥0 such that Aq≥Bq ensures
(3)(Ar/2Bp0Ar/2)(s(q)+r)/(p0+r)≥(Ar/2BpAr/2)(s(q)+r)/(p+r).We call the theorem above the complete form of Furuta inequality because the casep0=q=1 of it implies the essential part (p>1) of Furuta inequality by the Loewner-Heinz inequality for (1+r)/(s(1)+r)∈(0,1]. For convenience, we call Furuta inequality (Theorem 2) the original form of Furuta inequality.It is known that there are many applications of Furuta type inequalities; we cite [5–7].Based on Ito et al. [8] which is a continuation of [9], the equivalent relations between two operator inequalities are useful. For A≥0, A0 means the projection P(ker⁡A)⊥.Theorem 4 (see [8]).
Letr>0, 0≤p0<p, A≥0 and B≥0.(1)
Ifker⁡(ABp0/2)⊆ker⁡B, then, for each r, p0, and p, the following inequalities are equivalent to each other:
(4)(Bp/2ArBp/2)(p-p0)/(r+p)≥(Bp/2BrBp/2)(p-p0)/(r+p),(5)(Ar/2Bp0Ar/2)(p0+r)/(p0+r)≥(Ar/2BpAr/2)(p0+r)/(p+r).In particular, (4) implies (5) without condition ker⁡(ABp0/2)⊆ker⁡B.(2)
For eachr, p0, and p, the following inequalities are equivalent to each other:
(6)(Ap/2BrAp/2)(p-p0)/(r+p)≤(Ap/2ArAp/2)(p-p0)/(r+p),(Br/2Ap0Br/2)(p0+r)/(p0+r)≤(Br/2ApBr/2)(p0+r)/(p+r).It should be pointed out that (5) ensures (4) is not true without the condition ker⁡(ABp0/2)⊆ker⁡B [8, Remark 1]. Moreover, the proof of Theorem 4 is independent of (L-H).In Section2, some Furuta type inequalities under A≥B≥0 are proved via Loewner-Heinz inequality; as applications, we show alternate proofs of some well-known Furuta type inequalities (proofs of Theorems 10 and 2).In 1995, Furuta [10] proved the so-called grand Furuta inequality which is also an extension of Theorem 2.Theorem 5 (grand Furuta inequality [10]).
Letp≥1, -1≤t<0, r≥-t and s≥1. If A≥B≥0 with A>0; then
(7)A1+t+r≥(Ar/2(At/2BpAt/2)sAr/2)(1+t+r)/((p+t)s+r).Fujii et al. proved some satellite theorems of grand Furuta inequality.Theorem 6 (see [11]).
Letp≥1, -1≤t<0, r≥-t and s≥1. If A≥B≥0 with A>0; then
(8)(Ar/2(At/2BpAt/2)sAr/2)(1+t+r)/((p+t)s+r)≤A(r+t)/2BA(r+t)/2.Theorem 7 (see [12]).
Letp≥1, -1≤t<0, r≥-t and s≥1. If A≥B≥0 with A>0; then
(9)(Ar/2(At/2BpAt/2)sAr/2)(1+t+r)/((p+t)s+r)≤Ar/2(At/2BpAt/2)(1+t)/(p+t)Ar/2.Theorems6 and 7 are extensions of Theorem 5.In Section3, we will show a unified satellite theorem which is an extension of Theorems 6 and 7 via the complete forms of Furuta inequality with negative powers.Lastly, it is known that Riccati type operator equationsK=THT-TC-C*T relate to control theory closely and have been studied extensively [13]. Pedersen and Takesaki [14] developed the special kind of Riccati equation K=THT as a useful tool for the noncommutative Radon-Nikodym theorem.Yuan and Gao [15] discussed the Riccati type equation:
(10)Kp=Hδ/2T1/2(T1/2Hδ+rT1/2)(p-δ)/(δ+r)T1/2Hδ/2.In Section4, as a continuation of [15, 16], we will consider the Riccati type equation:
(11)Kp=Hδ/2T1/2(T1/2Hδ+rT1/2)(1/w)-1T1/2Hδ/2
via Furuta type inequalities.


## 2. Furuta Type Inequalities under the OrderA≥B≥0

Reference [17] proved a kind of equivalent relations which can be regarded as a parallel result to Theorem 4.Theorem 8 (see [17]).
Letr>0, 0<p0<p, A≥0 and B≥0. If ker⁡(ABp0/2)⊆ker⁡B, then, for each r, p0 and p, the following inequalities are equivalent to each other:
(12)(Bp0/2ArBp0/2)(p-p0)/(r+p0)≥(resp.≤)(Bp0/2BrBp0/2)(p-p0)/(r+p0),(13)(Ar/2Bp0Ar/2)(p+r)/(p0+r)≥(resp.≤)(Ar/2BpAr/2)(p+r)/(p+r).
In particular, (12) implies (13) without condition ker⁡(ABp0/2)⊆ker⁡B.The proof of Theorem8 is different from Theorem 4 and independent of (L-H).In this section, we consider some Furuta type inequalities under the orderA≥B≥0. As applications, alternate proofs of some Furuta type inequalities are given (proofs of Theorems 10 and 2). Especially, we prove (FI) without using the invertibility of operators.Theorem 9.
Let1≥r>0, A≥B≥0. (1)
For eachp0>0 and p with p0<p≤2p0+r, the following inequalities hold and they are equivalent to each other:
(14)(Bp0/2ArBp0/2)(p-p0)/(r+p0)≥(Bp0/2BrBp0/2)(p-p0)/(r+p0),(15)(Ar/2Bp0Ar/2)(p+r)/(p0+r)≥(Ar/2BpAr/2)(p+r)/(p+r).(2)
For eachp0>0 and p with p0<p≤2p0+r, the following inequalities hold:
(16)(Ap0/2BrAp0/2)(p-p0)/(r+p0)≤(Ap0/2ArAp0/2)(p-p0)/(r+p0),(17)(Br/2Ap0Br/2)(p+r)/(p0+r)≤(Br/2ApBr/2)(p+r)/(p+r).(3)
Ifp>1, then
(18)A1+r≥Ar/2BAr/2≥(Ar/2BpAr/2)(1+r)/(p+r),(19)B1+r≤Br/2ABr/2≤(Br/2ApBr/2)(1+r)/(p+r).Proof.
(1) Sincep0<p≤2p0+r, 1≥(p-p0)/(r+p0)>0 follows. By A≥B≥0 and (L-H) for 1≥r>0 and 1≥(p-p0)/(r+p0)>0, we have
(20)Ar≥Br,Bp0/2ArBp0/2≥Bp0/2BrBp0/2,(Bp0/2ArBp0/2)(p-p0)/(r+p0)≥(Bp0/2BrBp0/2)(p-p0)/(r+p0).
Hence, (14) holds. Since A≥B≥0, ker⁡(ABp0/2)⊆ker⁡B follows. So, the equivalency follows by Theorem 8.
(2) Similar to the proof of (14), we have
(21)Ar≥Br,Ap0/2BrAp0/2≤Ap0/2ArAp0/2,(Ap0/2BrAp0/2)(p-p0)/(r+p0)≤(Ap0/2ArAp0/2)(p-p0)/(r+p0).
Hence, (16) holds. Since (12) implies (13) without kernel condition, (17) follows by (16).
(3) By (15), there exists the function d(t)=t+r defined on (0,∞) satisfying [18, Lemma 2.6(1)]. Hence, case 1=p0<p of   [18, Lemma 2.6(2)] implies
(22)(Ar/2BAr/2)(min⁡{p,2+r}+r)/(1+r)≥(Ar/2BpAr/2)(min⁡{p,2+r}+r)/(p+r).
So (18) holds by A≥B≥0 and (L-H) for (1+r)/(min⁡{p,2+r}+r)∈(0,1). It is easy to prove (19) in a similar way.As prompt applications, we show alternate proofs of some Furuta type inequalities.Theorem 10 (see [19]).
Letα>0, β0>0, A≥0, B≥0. For δ such that -β0<δ≤α, if
(23)(Bβ0/2AαBβ0/2)(δ+β0)/(α+β0)≥(resp.≤)Bδ+β0,
then
(24)(Bβ/2AαBβ/2)(δ+β)/(α+β)≥(resp.≤)Bδ+β,
where β≥β0. Moreover, for each δ′>-α, the function
(25)f(β)=(Aα/2BβAα/2)(δ′+α)/(β+α)
is decreasing (resp., increasing) for β≥max⁡{β0,δ′}.Proof.
It is enough to prove the case≥ because the case ≤ can be proved in a similar manner. Denote (23) by A1≥B1; that is,
(26)A1=(Bβ0/2AαBβ0/2)(δ+β0)/(α+β0),B1=Bδ+β0.
For 1≥r1>0, p1≥1, by (19) of Theorem 9, we have
(27)(B1r1/2A1pB1r1/2)(1+r1)/(p1+r1)≥B11+r1.
By putting p1=(α+β0)/(δ+β0), the inequality above becomes
(28)(B(β0+(δ+β0)r1)/2AαB(β0+(δ+β0)r1)/2)(δ+β0+(δ+β0)r1)/(α+β0+(δ+β0)r1)≥Bδ+β0+(δ+β0)r1.
This implies that (24) holds for β0≤β≤β1=β0+(δ+β0). Denote A2=(Bβ1/2AαBβ1/2)(δ+β1)/(α+β1) and B2=Bδ+β1; repeating this process, (24) holds for β≥β0.
For eachδ′>-α, β1≥max⁡{β0,δ′}, by (24) and (L-H),
(29)(Bβ1/2AαBβ1/2)(δ+β1)/(α+β1)≥Bδ+β1,(Bβ1/2AαBβ1/2)v/(α+β1)≥Bv,
where 0<v≤δ+β0. This together with Theorem 8 and (L-H) deduce that
(30)(Aα/2Bβ1Aα/2)(v+β1+α)/(β1+α)≥(Aα/2Bv+β1Aα/2)(v+β1+α)/(v+β1+α),(Aα/2Bβ1Aα/2)(δ′+α)/(β1+α)≥(Aα/2Bβ2Aα/2)(δ′+α)/(β2+α),
where β2=v+β1∈(β1,δ+β0+β1]. So, the monotonicity of the function f(β) holds.It should be pointed out that, ifδ=0 and 0<β<β0, the assertion that (23) ensures (24) is not true [15, Theorem 2.8].Theorem 11 (see [15]).
Given any positive numbersp, r, p1, and r1 with r1>r, there exist invertible positive operators H and K such that
(31)Hr1≥(Hr1/2Kp1Hr1/2)r1/(p1+r1),aHr≱(Hr/2KpHr/2)r/(p+r),
where a is an arbitrary positive number.Alternate Proof of Theorem2. The case r≥0 and 0<p≤1 of Theorem 2 follows by (L-H) directly. Theorem 9(3) means the case 0<r≤1 and p>1 of Theorem 2; this together with Theorem 10 implies the case r>0 and p>1 of Theorem 2. So, the proof is complete.The proof above says that the original form of Furuta inequality (Theorem2) is a composition of (L-H), Theorems 9 and 10. The proof here is independent of the invertibility of the operators A and B.


## 3. A Unified Satellite Theorem of Grand Furuta Inequalities

Denote∶=(At/2BpAt/2)1/(p+t), where p+t≠0.Theorem 12.
Let-1≤t<0, p≥1, p~≥1, C≥A≥B≥0 with A>0.(1)
Ifr≥-t, s≥1, 1≤p0≤p and 1≤p0≤p~, then
(32)(Ar/2(At/2BpAt/2)sAr/2)(p0+t+r)/((p+t)s+r)≤A(r+t)/2Bp0A(r+t)/2≤Ar/2(At/2Bp~At/2)(p0+t)/(p~+t)Ar/2.(2)
Ifr≥-t and s≥1, then
(33)(Cr/2(At/2BpAt/2)sCr/2)(1+t+r)/((p+t)s+r)≤Cr/2(At/2BpAt/2)(1+t)/(p+t)Cr/2.The caseC=A of Theorem 12(2) is just Theorem 7. The special case p0=1 of Theorem 12(1) implies the result below.Corollary 13.
Let-1≤t<0, p≥1, p~≥1, A≥B≥0 with A>0. If r≥-t and s≥1, then
(34)(Ar/2(At/2BpAt/2)sAr/2)(1+t+r)/((p+t)s+r)≤A(r+t)/2BA(r+t)/2≤Ar/2(At/2Bp~At/2)(1+t)/(p~+t)Ar/2.It is obvious that the special casep=p~ of Corollary 13 is a unified result of Theorems 6 and 7; that is, it is an extension of Theorems 6 and 7. So, we call Theorem 12 a unified satellite theorem of grand Furuta inequality (Theorem 5).In order to give a proof, we prepare some results in advance.Lemma 14 (see [18]).
Let-1≤t<0, p≥1 and s≥1. Then C≥A≥B≥0 with A>0 ensures that the function
(35)f(s)=(C-t/2(At/2BpAt/2)sC-t/2)1/((p+t)s-t)
is decreasing for s≥1. In particular,
(36)C≥(C-t/2(At/2BpAt/2)sC-t/2)1/((p+t)s-t).Lemma 15 (see [18]).
Let-1≤t<0, p≥1, r≥-t, 1≤s0<s and δ=min⁡{(p+t)s,2(p+t)s0}. Then C≥A≥B≥0 with A>0 ensures
(37)(Cr/2D(p+t)s0Cr/2)(δ+r)/((p+t)s0+r)≥(Cr/2D(p+t)sCr/2)(δ+r)/((p+t)s+r).Lemma 16.
Letr<0, A>0 and B≥0. Then the following assertion (1) implies (2).(1)
There exists an increasing functiond(p):(-r,∞)→(0,∞) such that, for each p0>-r, if p0<p≤p0+d(p0), then
(38)(Ar/2Bp0Ar/2)(p+r)/(p0+r)≥(resp.≤)(Ar/2BpAr/2)(p+r)/(p+r).(2)
The functiond(p):(-r,∞)→(0,∞) in (1) satisfies that, for each p0>-r, if p0<p, then
(39)(Ar/2Bp0Ar/2)(min⁡{p,p0+d(p0)}+r)/(p0+r)≥(resp.≤)(Ar/2BpAr/2)(min⁡{p,p0+d(p0)}+r)/(p+r).Lemma16 is a complement to [18, Lemma 2.6].Proof.
It is sufficient to prove the case≥ for the case ≤ can be proved in a similar manner. For each p0>0 and p0<p, if p≤p0+d(p0), then (2) follows by (1) immediately. Suppose that pn<p≤pn+1=pn+d(pn) for some positive integer n and p1=p0+d(p0). By (1), for k=0,1,…,n-1, we have
(40)(Ar/2BpkAr/2)(pk+1+r)/(pk+r)≥(Ar/2Bpk+1Ar/2)(pk+1+r)/(pk+1+r),(Ar/2BpnAr/2)(p+r)/(pn+r)≥(Ar/2BpAr/2)(p+r)/(p+r).
Noting that ((p1+r)/(pk+1+r))∈[0,1] and ((p1+r)/(p+r))∈[0,1], these together with (L-H) deduce that
(41)(Ar/2Bp0Ar/2)(p1+r)/(p0+r)≥(Ar/2Bp1Ar/2)(p1+r)/(p1+r)≥⋯≥(Ar/2BpnAr/2)(p1+r)/(pn+r)≥(Ar/2BpAr/2)(p1+r)/(p+r).
Therefore, the function d in (1) satisfies (2).Lemma 17.
Let0>r≥-1, p>p0>-r and β=min⁡{p,2p0+r}; then A≥B≥0 with A>0 ensures
(42)(Ar/2Bp0Ar/2)(β+r)/(p0+r)≤(Ar/2BpAr/2)(β+r)/(p+r).Proof.
Firstly, we prove the casep≤2p0+r of Lemma 17. By [10, Lemma 1], (42) is equivalent to
(43)Ar/2Bp0/2(Bp0/2ArBp0/2)(p-p0)/(p0+r)Bp0/2Ar/2≤Ar/2Bp0/2Bp-p0Bp0/2Ar/2.
On the other hand, (Bp0/2ArBp0/2)(p-p0)/(p0+r)≤Bp-p0 holds by Loewner-Heinz inequality for ((p-p0)/(p0+r))∈(0,1]. So (42) holds for p≤2p0+r.
Now, it is proved that (42) holds when -r<p0<p≤2p0+r. Meanwhile, it is easy to see that the increasing function d(t)=t+r satisfies (1) of Lemma 16, so (42) holds for -r<p0<p.Proof of Theorem12.
By the cases0=1 of Lemma 15 and (L-H) for ((p0+t+r)/(δ+r))∈[0,1],
(44)(Cr/2(At/2BpAt/2)sCr/2)(p0+t+r)/((p+t)s+r)≤(Cr/2(At/2BpAt/2)Cr/2)(p0+t+r)/((p+t)+r).(1)
For1≤p0≤p, Theorem 3 and (L-H) deduce that
(45)(A(r+t)/2BpA(r+t)/2)(p0+t+r)/(p+r+t)≤A(r+t)/2Bp0A(r+t)/2.Meanwhile, for-1≤t<0 and 1≤p0≤p~, Lemma 17 and (L-H) imply
(46)At/2Bp0At/2≤(At/2Bp~At/2)(p0+t)/(p~+t).Hence,(1) follows by the case C=A of (44), (45), and (46).(2)
By (L-H), (44), f(1)≤C, Theorem 3 and Lemma 17 ensure
(47)(Cr/2(At/2BpAt/2)sCr/2)(1+t+r)/((p+t)s+r)≤(Cr/2(At/2BpAt/2)Cr/2)(1+t+r)/((p+t)+r)hhhhby(44)and(L-H)=(C(r+t)/2(C-t/2At/2BpAt/2C-t/2)(1/p)pcccccccc×C(r+t)/2(At/2)(1/p)p)(1+r+t)/(p+r+t)=(C(r+t)/2(f(1))pC(r+t)/2)(1+r+t)/(p+r+t)≤(C(r+t)/2(f(1))1C(r+t)/2)(1+r+t)/(1+r+t)cccccbyf(1)≤C,Theorem3and(L-H)=Cr/2Ct/2(f(1))1Ct/2Cr/2≤Cr/2(Ct/2(f(1))pCt/2)(1+t)/(p+t)Cr/2cccccbyf(1)≤C,Lemma17and(L-H)=Cr/2(At/2BpAt/2)(1+t)/(p+t)Cr/2.The f(s) above is the same as the function f(s) in Lemma 14.


## 4. Riccati Type Operator Equations

Yuan and Gao [15] discussed the Riccati type equation:
(48)Kp=Hδ/2T1/2(T1/2Hδ+rT1/2)(p-δ)/(δ+r)T1/2Hδ/2.Theorem 18 (see [15]).
LetH≥0, K≥0 and assume that ker⁡H={0}.(1)
The following statements are equivalent for eachp>0, r>0 and p≥δ≥0.
(a)
aHδ+r≥(Hr/2KpHr/2)(δ+r)/(p+r) for some a≥0.(b)
There exists a unique operatorT≥0 that satisfies ∥T∥≤a and (48).If in additionalH is invertible, (1) holds for p≥δ>-r.(2)
If there existsT≥0 satisfying (48) for fixed p>0, r>0 and p≥δ≥0, then, for p1≥p and r1≥r, there exists T1≥0 satisfying
(49)Kp1=Hδ/2T11/2(T11/2Hδ+r1T11/2)(p1-δ)/(δ+r1)T11/2Hδ/2.One of the applications of Riccati equation (48) is to show that the inclusion relations among class A(p,r) operators are strict [15, Theorem 3.1]. Recently, there are some developments on operator equations including the following equation (see [16, 20]):
(50)Kp=Hδ/2T1/2(T1/2Hδ+rT1/2)(1/w)-1T1/2Hδ/2.Obviously, the special casew=(δ+r)/(p+r) of (50) is just (48).Theorem 19 (see [16]).
LetH≥0, K≥0 and assume that ker⁡H={0}. The following statements are equivalent for each p>0, r>0, δ≥0 and 0<w≤1.(1)
aHδ+r≥(Hr/2KpHr/2)w for some a≥0.(2)
There exists a unique operatorT≥0 which satisfies ∥T∥≤a and (50).
If in additionalH is invertible, the condition δ≥0 can be replaced with δ∈ℛ where ℛ means the set of all real numbers, and if H and K are both invertible, the conditions δ≥0 and 0<w≤1 can be replaced with δ∈ℛ and w≠0.The casew=(δ+r)/(p+r) of Theorem 19 is a generalization of Theorem 18(1). In this section, we give a generalization of Theorem 18(2).Lemma 20.
Letα>0, β0>0, α′>0, β0′>0, A≥0, B≥0. For δ and δ′ such that δ∈(-β0,∞) and δ′∈(-β0′,α′], if
(51)(Bβ0/2AαBβ0/2)(δ′+β0′)/(α′+β0′)≥(resp.≤)Bδ+β0,
then, for β≥β0,
(52)(Bβ/2AαBβ/2)(δ′+β′)/(α′+β′)≥(resp.≤)Bδ+β,
where β′=((β(δ′+β0′)+β0′δ-β0δ′)/(δ+β0))(≥β0′).The caseα=α′, β0=β0′ and δ=δ′ of Lemma 20 implies Theorem 10. The case α=α′, β0=β0′ and δ=δ′=0 of Lemma 20 implies Yanagida's result [21, Proposition 4].Proof.
It is enough to prove the case≥ because the case ≤ can be proved in a similar manner. Denote (51) by A1≥B1; that is,
(53)A1=(Bβ0/2AαBβ0/2)(δ′+β0′)/(α′+β0′),B1=Bδ+β0.
For r1>0, p1≥1, by (FI) (Theorem 2), we have
(54)(B1r1/2A1p1B1r1/2)(1+r1)/(p1+r1)≥B11+r1.
By putting p1=((α′+β0′)/(δ′+β0′))(≥1), the inequality above becomes
(55)(B(β0+(δ+β0)r1)/2Aαcccc×B(β0+(δ+β0)r1)/2)(δ′+β0′+(δ′+β0′)r1)/(α′+β0′+(δ′+β0′)r1)≥Bδ+β0+(δ+β0)r1.
Denote β=β0+(δ+β0)r1 and β′=β0′+(δ′+β0′)r1; then β≥β0 and β′=((β(δ′+β0′)+β0′δ-β0δ′)/(δ+β0))≥β0′, so that (52) holds by the inequality above.Theorem 21.
LetH≥0, K≥0 and assume that ker⁡H={0}. For each p>0, r>0 and δ≥0, if there exists T≥0 satisfying the equation
(56)Kp=Hδ/2T1/2(T1/2Hδ+rT1/2)(p′-δ′)/(δ′+r′)T1/2Hδ/2,
where p′>0, r′>0 and -r′<δ′≤p′. Then, for r1≥r and r1′=((r1(δ′+r′)+r′δ-rδ′)/(δ+r))(≥r′), there exists T1≥0 satisfying
(57)Kp=Hδ/2T11/2(T11/2Hδ+r1T11/2)(p′-δ′)/(δ′+r1′)T11/2Hδ/2.
If H is invertible, the condition δ≥0 can be replaced with δ>-r.Proof.
By the assumption, (1) of Theorem19 holds for some a>0; that is,
(58)(a1/(δ+r(1-w))H)δ+r≥((a1/(δ+r(1-w))H)r/2cccccccccc×Kp(a1/(δ+r(1-w))H)r/2)w.
So, the following holds by Lemma 20:
(59)(a1/(δ+r(1-w))H)δ+r1≥((a1/(δ+r(1-w))H)r1/2vvvvvv×Kp(a1/(δ+r(1-w))H)r1/2)(δ′+r1′)/(p′+r1′),
where r1′=((r1(δ′+r′)+r′δ-rδ′)/(δ+r))(≥r′); that is,
(60)a(δ+r1(1-w1))/(δ+r(1-w))Hδ+r1≥(Hr1/2KpHr1/2)w1,
where w1=((δ′+r1′)/(p′+r1′))∈(0,1]. Hence, (57) is solvable by Theorem 19.The result below is the caser=r′ and δ=δ′ of Theorem 21.Corollary 22.
LetH≥0, K≥0 and assume that ker⁡H={0}. For each p>0, r>0 and δ≥0, if there exists T≥0 satisfying the equation
(61)Kp=Hδ/2T1/2(T1/2Hδ+rT1/2)(p′-δ)/(δ+r)T1/2Hδ/2,
where p′>0 and p′≥δ. Then, for r1≥r, there exists T1≥0 satisfying
(62)Kp=Hδ/2T11/2(T11/2Hδ+r1T11/2)(p′-δ)/(δ+r1)T11/2Hδ/2.
If H is invertible, the condition δ≥0 can be replaced with δ>-r.It is obvious that Corollary22 is a generalization of the case p1=p of Theorem 18(2).



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*Source: 457367-2014-04-13.xml*