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

Authors: Changsen Yang; Yanmin Liu
Journal: Abstract and Applied Analysis
 (2013)

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## 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.


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## Body

## 1. Introduction

An operatorT 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]
IfA≥B≥0 holds, then Aα≥Bα for any α∈[0,1].This was originally proved in [1, 2] and then in [3]. Although (LH) asserts that A≥B≥0 ensures Aα≥Bα for any α∈[0,1], unfortunately Aα≥Bα does not always hold for α>1. The following result has been obtained from this point of view.Theorem F (Furuta inequality).
IfA≥B≥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 forp≥0 and q≥1 with (1+r)q≥p+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 A≥B≥0; see [8].We writeA≫B if logA≥logB for A,B>0, which is called the chaotic order.Theorem A.
ForA,B>0, the following (i) and (ii) hold:(i)
A≫B holds if and only if Ar≥(Ar/2BpAr/2)r/(p+r) for p,r≥0;(ii)
A≫B holds if and only if for any fixed δ≥0,FA,B(p,r)=A-r/2(Ar/2BpAr/2)(δ+r)/(p+r)A-r/2 is a decreasing function of p≥δ 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]).
LetA be a positive invertible operator, and let B be an invertible operator. For any real number λ,
(1)(BAB*)=BA1/2(A1/2B*BA1/2)λ-1A1/2B*.Definition 1.
LetAn,An-1,…,A2,A1,B≥0, r1,r2,…,rn≥0, and p1,p2,…,pn≥0 for a natural number n.
LetCAi,B[n] be defined by
(2)CAi,B[n]=Anrn/2{An-1rn-1/2[⋯A3r3/2{A2r2/2(A1r1/2Bp1A1r1/2)p2A2r2/2}p3×A3r3/2⋯{A2r2/2(A1r1/2Bp1A1r1/2)p2A2r2/2}p3]An-1rn-1/2}pnAnrn/2.
For example,
(3)CAi,B[2]=A2r2/2(A1r1/2Bp1A1r1/2)p2A2r2/2,CAi,B[4]=A4r4/2{A3r3/2[A2r2/2(A1r1/2Bp1A1r1/2)p2A2r2/2]p3×A3r3/2[A2r2/2(A1r1/2Bp1A1r1/2)p2A2r2/2]p3}p4A4r4/2.
Let q[n] be defined by
(4)q[n]={⋯[(p1+r1)p2+r2]p3+⋯+rn-1}pn+rn.
For example,
(5)q[1]=p1+r1,q[2]=(p1+r1)p2+r2,q[4]={[(p1+r1)p2+r2]p3+r3}p4+r4.
For the sake of convenience, we define
(6)CAi,B[0]=B,q[0]=1,
and these definitions in (6) may be reasonable by (2) and (4).Lemma 2.
ForAn,An-1,…,A2,A1,B≥0 and any natural number n, we have(i)
CAi,B[n]=Anrn/2CAi,B[n-1]pnAnrn/2,(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 withCAi,B[n] and q[n]

We will give some operator inequalities on chaotic order, and Theorem5 is further extension of Theorem  3.1 in [13].Lemma 3.
IfA≫B, for p≥0 and r≥0, then A≫(Ar/2BpAr/2)1/(p+r).Proof.
SinceA≫B, we can obtain the following inequality.
A
r
≥
(
A
r
/
2
B
p
A
r
/
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,(7)logAr≥log(Ar/2BpAr/2)r/(p+r),
therefor we have
(8)A≫(Ar/2BpAr/2)1/(p+r).Theorem 4.
IfAn≫An-1≫⋯≫A2≫A1≫B and r1,r2,…,rn≥0, p1,p2,…,pn≥0 for a natural number n. Then the following inequality holds:
(9)An≫CAi,B[n]1/q[n],
where CAi,B[n] and q[n] are defined in (2) and (4).Proof.
We will show (9) by mathematical induction. In the case n=1.
SinceA1≫B  implies
(10)A1≫(A1r1/2Bp1A1r1/2)r1/(p1+r1)
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.
PutD=Ak+1,E=Ak, and F=CAi,B[k]1/q[k], and (9) holds for n=k implying
(11)D≫E≫F>0.
Equation (11) yields the following by Lemma 3, for r≥0 and p≥0:(12)D≫(Dr/2FpDr/2)1/(p+r),
that is,
(13)Ak+1≫(Ak+1r/2CAi,B[k]p/q[k]Ak+1r/2)1/(p+r).
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:
(14)1p+r=1q[k]pk+1+rk+1=1q[k+1],
and we have the following desired (15) by (12) and (13):
(15)Ak+1≫{Ak+1rk+1/2(CAi,B[k])pk+1Ak+1rk+1/2}1/q[k+1]=CAi,B[k+1]1/q[k+1],
so that (15) shows that (9) holds for k+1.Theorem 5.
IfAn≫An-1≫⋯≫A2≫A1≫B and r1,r2,…,rn≥0 for a natural number n. For any fixed δ≥0, let p1,p2,…,pn be satisfied by
(16)p1≥δ,p2≥δ+r1p1+r1,⋮pk≥δ+r1+r2+⋯+rk-1q[k-1],⋮pn≥δ+r1+r2+⋯+rn-1q[n-1].
The operator function Ik(pk,rk) for any natural number k such that 1≤k≤n is defined by
(17)Ik(pk,rk)=Ak-rk/2CAi,B[k](δ+r1+r2+⋯+rk)/q[k]Ak-rk/2.
Then the following inequality holds:
(18)Ak-1rk-1/2Ik-1(pk-1,rk-1)Ak-1rk-1/2≥Ik(pk,rk)
for every natural number k such that 1≤k≤n, where CAi,B[n] and q[n] are defined in (2) and (4).Proof.
SinceCAi,B[0]=B, q[0]=1 in (6), we may define I0(p0,r0)=Bδ for p0=r0=0.
BecauseA1≫B, then for any fixed δ≥0,
(19)Bδ≥A1-r1/2(A1r1/2Bp1A1r1/2)(δ+r1)/(p1+r1)A1-r1/2forp1≥δ,r1≥0,
since FA1,B(δ,r0)≥FA1,B(p1,r1) holds by (ii) of Theorem A. And (19) can be expressed as
(20)Bδ=A0r0/2I0(p0,r0)A0r0/2≥I1(p1,r1).
We can apply Theorem 4, and we have the following (21) for any natural number k such that 1≤k≤n:
(21)Ak+1≫Ak≫CAi,B[k]1/q[k].
Since X≫Y implies that Xt≫Yt holds for any t≥0, (21) ensures
(22)Ak+1δ+r1+r2+⋯+rk≫CAi,B[k](δ+r1+r2+⋯+rk)/q[k].
Putting A=Ak+1δ+r1+r2+⋯+rk, B1=CAi,B[k](δ+r1+r2+⋯+rk)/q[k] and applying (19) for δ=1 and A≫B1, we have
(23)B1≥A-r/2(Ar/2B1pAr/2)(1+r)/(p+r)A-r/2
holds for p≥1 and r≥0.
Puttingrk+1=r(δ+r1+r2+⋯+rk) in (23), then (23) can be rewritten by
(24)B1≥Ak+1-rk+1/2(Ak+1rk+1/2CAi,B[k]((δ+r1+r2+⋯+rk)/q[k])p×Ak+1rk+1/2)(1+r)/(p+r)Ak+1-rk+1/2.
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
(25)Akrk/2Ik(pk,rk)Akrk/2=B1=CAi,B[k](δ+r1+r2+⋯+rk)/q[k]≥Ak+1-rk+1/2×(Ak+1rk+1/2CAi,B[k]((δ+r1+r2+⋯+rk)/q[k])pAk+1rk+1/2)(1+r)/(p+r)×Ak+1-rk+1/2=Ak+1-rk+1/2CAi,B[k+1](δ+r1+r2+⋯+rk+rk+1)/(q[k+1])Ak+1-rk+1/2=Ik+1(pk+1,rk+1),
and we have (18) for k such that 1≤k≤n by (25) and (20) since (20) means (18) for k=1.Corollary 6.
IfAn≫An-1≫⋯≫A2≫A1≫B 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:(26)Bδ≥A1-r1/2(A1r1/2Bp1A1r1/2)(δ+r1)/(p1+r1)A1-r1/2≥A1-r1/2A2-r2/2×[A2r2/2(A1r1/2Bp1A1r1/2)p2A2r2/2](δ+r1+r2)/((p1+r1)p2+r2)×A2-r2/2A1-r1/2⋮≥A1-r1/2A2-r2/2A3-r3/3⋯An-1-rn-1/2An-rn/2×CAi,B[n](δ+r1+r2+⋯+rn)/q[n]×An-rn/2An-1-rn-1/2⋯A3-r3/3A2-r2/2A1-r1/2,
where CAi,B[n], q[n], and Ik(pk,rk)(1≤k≤n) are defined in (2), (4), and (17).Proof.
Applying (18) of Theorem 5 for k such that 1≤k≤n, we have
(27)Bδ=Ar0/2I0(p0,r0)Ar0/2≥I1(p1,r1)=A1-r1/2(A1r1/2Bp1A1r1/2)(δ+r1)/(p1+r1)A1-r1/2≥A1-r1/2I2(p2,r2)A1-r1/2=A1-r1/2A2-r2/2[A2r2/2(A1r1/2Bp1A1r1/2)p2×A2r2/2(A1r1/2Bp1A1r1/2)p2](δ+r1+r2)/((p1+r1)p2+r2)×A2-r2/2A1-r1/2⋮≥A1-r1/2A2-r2/2A3-r3/3⋯An-1-rn-1/2In(pn,rn)×An-1-rn-1/2⋯A3-r3/3A2-r2/2A1-r1/2=A1-r1/2A2-r2/2A3-r3/3⋯An-1-rn-1/2An-rn/2×CAi,B[n](δ+r1+r2+⋯+rn)/q[n]×An-rn/2An-1-rn-1/2⋯A3-r3/3A2-r2/2A1-r1/2.


## 3. Monotonicity Property on Operator Functions

We would like to emphasize that the condition of Theorem7 is stronger than Theorem 5, and moreover when we discuss monotonicity property on operator functions, we can only apply Theorem 7.Theorem 7.
IfAn≫An-1≫⋯≫A2≫A1≫B and r1,r2,…,rn≥0, p1,p2,…,pn≥0 for a natural number n. Then the following inequality holds:
(28)Anrn≥CAi,B[n]rn/q[n],
where CAi,B[n] and q[n] are defined in (2) and (4).Proof.
We will show (28) by mathematical induction. In the case n=1.
SinceA1≫B  implies
(29)A1≥(A1r1/2Bp1A1r1/2)r1/(p1+r1)
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:
(30)Akrk≥CAi,B[k]rk/q[k],
hence we have Ak+1≫Ak≫CAi,B[k]1/q[k], and (i) of Theorem A ensures
(31)Ak+1r≥(Ak+1r/2CAi,B[k]p/q[k]Ak+1r/2)r/(p+r)forp,r≥0.
Putting r=rk+1 and p=q[k]pk+1, then we have the following inequality:
(32)Ak+1rk+1≥(Ak+1rk+1/2CAi,B[k]pk+1Ak+1rk+1/2)rk+1/(q[k]pk+1+rk+1)=CAi,B[k+1]rk+1/q[k+1],
so that (32) shows (28) for k+1.Theorem 8.
IfAn≫An-1≫⋯≫A2≫A1≫B and r1,r2,…,rn≥0 for a natural number n. For any fixed δ≥0, let p1,p2,…,pn be satisfied by (16).
Then(33)In(pn,rn)=An-rn/2CAi,B[n](δ+r1+r2+⋯+rn)/q[n]An-rn/2
is a decreasing function of both rn≥0 and pn which satisfies
(34)pn≥δ+r1+r2+⋯+rn-1q[n-1],
where CAi,B[n] 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):
(35)q[n]=q[n-1]pn+rn≥δ+r1+r2+⋯+rn-1+rn
because the inequality in (35) follows by (ii) of Lemma 2, and the inequality follows by
(36)q[n-1]pn≥δ+r1+r2+⋯+rn-1
obtained by (34).
(a) Proof of the result thatIn(pn,rn) is a decreasing function of pn.
Without loss of generality, we can assume thatpn>0. We can obtain the following inequality by (28) and by (i) of Lemma 2:
(37)Anrn≥CAi,B[n]rn/q[n]=(Anrn/2CAi,B[n-1]pnAnrn/2)rn/q[n]=Anrn/2CAi,B[n-1]pn/2×(CAi,B[n-1]pn/2AnrnCAi,B[n-1]pn/2)(rn-q[n])/q[n]×CAi,B[n-1]pn/2Anrn/2,
and (37) implies
(38)(CAi,B[n-1]pn/2AnrnCAi,B[n-1]pn/2)(q[n]-rn)/q[n]≥CAi,B[n-1]pn.
Put α=ω/pn∈[0,1] for pn≥ω≥0, then we raise each side of (38) to the power α=ω/pn∈[0,1], then
(39)(CAi,B[n-1]pn/2AnrnCAi,B[n-1]pn/2)((q[n]-rn)ω)/(q[n]pn)≥CAi,B[n-1]ω.
Whence we have(40)In(pn,rn)=An-rn/2(Anrn/2CAi,B[n-1]pnAnrn/2)(δ+r1+r2+⋯+rn)/q[n]An-rn/2=An-rn/2×{Anrn/2)(q[n]+q[n-1]ω)/q[n](Anrn/2CAi,B[n-1]pn×Anrn/2)(q[n]+q[n-1]ω)/q[n]}(δ+r1+r2+⋯+rn)/(q[n]+q[n-1]ω)×An-rn/2=An-rn/2{Anrn/2CAi,B[n-1]pn/2×(CAi,B[n-1]pn/2Anrn×CAi,B[n-1]pn/2)(q[n-1]ω)/q[n]×CAi,B[n-1]pn/2Anrn/2}(δ+r1+r2+⋯+rn)/(q[n]+q[n-1]ω)×An-rn/2byLemmaB=An-rn/2{Anrn/2CAi,B[n-1]pn/2×(CAi,B[n-1]pn/2Anrn×CAi,B[n-1]pn/2)((q[n]-rn)ω)/(q[n]pn)×CAi,B[n-1]pn/2×Anrn/2}(δ+r1+r2+⋯+rn)/(q[n]+q[n-1]ω)An-rn/2≥An-rn/2(Anrn/2CAi,B[n-1]pn/2CAi,B[n-1]ω×CAi,B[n-1]pn/2Anrn/2)(δ+r1+r2+⋯+rn)/(q[n-1](pn+ω)+rn)×An-rn/2=In(pn+ω,rn),
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)+rn≥q[n] by (4), so that In(pn,rn) is a decreasing function of pn.
(b) Proof of the result thatIn(pn,rn) is a decreasing function of rn.
Without loss of generality, we can assume thatrn>0. Raise each side of (28) to the power μ/rn∈[0,1] for rn≥μ≥0 by LH, then
(41)Anμ≥(Anrn/2CAi,B[n-1]pnAnrn/2)μ/q[n].
We state the following inequality by (ii) of Lemma 3 and (35):
(42)q[n]-(δ+r1+r2+⋯+rn)=q[n-1]pn+rn-(δ+r1+r2+⋯+rn)=q[n-1]pn-(δ+r1+r2+⋯+rn-1)≥0.
Then we have
(43)In(pn,rn)=An-rn/2CAi,B[n](δ+r1+r2+⋯+rn)/q[n]An-rn/2=An-rn/2(Anrn/2CAi,B[n-1]pnArn/2)(δ+r1+r2+⋯+rn)/q[n]An-rn/2=CAi,B[n-1]pn/2×(CAi,B[n-1]pn/2Anrn×CAi,B[n-1]pn/2)(δ+r1+r2+⋯+rn-q[n])/q[n]CAi,B[n-1]pn/2=CAi,B[n-1]pn/2×{×CAi,B[n-1]pn/2)(q[n]+μ)/q[n](CAi,B[n-1]pn/2Anrn×CAi,B[n-1]pn/2)(q[n]+μ)/q[n]}(δ+r1+r2+⋯+rn-q[n])/(q[n]+μ)×CAi,B[n-1]pn/2=CAi,B[n-1]pn/2×{CAi,B[n-1]pn/2Anrn/2×(Anrn/2CAi,B[n-1]pnAnrn/2)μ/q[n]Anrn/2×CAi,B[n-1]pn/2}(δ+r1+r2+⋯+rn-q[n])/(q[n]+μ)×CAi,B[n-1]pn/2≥CAi,B[n-1]pn/2×{CAi,B[n-1]pn/2Anrn+μ×CAi,B[n-1]pn/2}(δ+r1+r2+⋯+rn-q[n])/(q[n]+μ)×CAi,B[n-1]pn/2=In(pn,rn+μ),
and the last inequality holds by LH because (41) and
(44)δ+r1+r2+⋯+rn-q[n]q[n]+μ=-q[n]-(δ+r1+r2+⋯+rn)q[n]+μ∈[-1,0],
so that Ik(pk,rk) is a decreasing function of rn.



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