# On the Positive Operator Solutions to an Operator Equation

Authors: Kai-Fan Yang; Fang-An Deng; Hong-Ke Du
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

The necessary conditions and the sufficient condition for the existence of positive operator solutions to the operator equationXs+A*X-tA=Q are established. An iterative method for obtaining the positive operator solutions is proposed.


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## Body

## 1. Introduction

Letℬ(ℋ) be the set of all bounded linear operators on the Hilbert space ℋ. In this paper, we consider the nonlinear operator equation
(1)Xs+A*X-tA=Q,
where A,Q∈ℬ(ℋ), Q>0, and X is an unknown operator in ℬ(ℋ). Both s and t are positive integers.This type of equation often arises from many areas such as dynamic programming [1], control theory [2, 3], stochastic filtering and statistics, and so forth [4, 5]. In the recent years, for matrices, (1) has been considered by many authors (see [6–13]), and different iterative methods for computing the positive definite solutions to (1) are proposed in finite-dimensional space. The case s=t=1 has been extensively studied by several authors. In this paper, we extend the study of the operator equation (1) from a finite-dimensional space to an infinite-dimensional Hilbert space. We derive some necessary conditions for the existence of positive solutions to the operator equation (1). Moreover, conditions under which the operator equation (1) has positive operator solutions are obtained. Based on Banach’s fixed-point principle, we obtain the positive operator solution to the operator equation (1). First, we introduce some notations and terminologies, which are useful later. For A∈ℬ(ℋ), if (Ax,x)≥0 for all x∈ℋ, then A is said to be a positive operator and is denoted by A≥0. If A is a positive operator and invertible, then denote A>0. For A and B in ℬ(ℋ), A≥B means that A-B is a positive operator. For A∈ℬ(ℋ), A*, ω(A), σ(A), and r(A) denote the adjoint, the radius of numerical range, the spectrum, and the spectral radius of A, respectively.For positive operators inℬ(ℋ), the following facts are well known.(1)
IfP≥Q>0, then P-1≤Q-1.(2)
Denoteλmax(P)=max{λ:λ∈σ(P),P>0},λmin(P)=min{λ:λ∈σ(P),P>0}. Then λmin(P)I≤P≤λmax(P)I.(3)
If the sequence{Xn}n=1+∞ of positive operator is monotonically increasing and has upper bound, that is, Xk≤Xk+1≤C1, or is monotonically decreasing and has lower bound, that is, C2≤Xk+1≤Xk, then this sequence is convergent to a positive limit operator, where C1,C2∈ℬ(ℋ) are given operators.


## 2. Main Results and Proofs

In order to prove our main result, we begin with some lemmas as follows.Lemma 1.
LetA,B∈ℬ(ℋ). If A≥B≥0, then ∥A∥≥∥B∥.Lemma 2.
LetA and B be self-adjoint operators in ℬ(ℋ). If A≤B, then, for every T∈ℬ(ℋ), one has T*AT≤T*BT.In this section, we give our main results and proofs.Theorem 3.
If the operator equation (1) has a positive operator solution X, then (1)AQ-1A*t≤X≤Qs and (2)r(A)≤(bt/(s+t))t/2s(bs/(s+t))1/2, where b=∥Q∥.Proof.
(1) If the operator equation (1) has a positive operator solution X, then 0≤A*X-tA≤Q. Hence, we obtain Xs≤Q; that is, X≤Qs. From A*X-tA≤Q, it follows that
(2)Q-1/2A*X-t/2X-t/2AQ-1/2≤I,X-t/2AQ-1/2Q-1/2A*X-t/2≤I,AQ-1A*≤Xt.
That is, X≥AQ-1A*t.
(2) From (1), we have
(3)Xt+s+Xt/2A*X-tAXt/2=Xt/2QXt/2≤λmax(Q)Xt.
Then
(4)r(A)2=r2(Xt/2AX-t/2)≤∥Xt/2AX-t/2∥2≤∥λmax(Q)Xt-Xt+s∥.
According to the spectral decomposition of X, we have
(5)λmax(Q)Xt-Xt+s=∫σ(X)(λmax(Q)λt-λt+s)dEλ.
Denote λmax(Q)=∥Q∥=b. Then
(6)∥λmax(Q)Xt-Xt+s∥≤maxλ∈(0,bs]|bλt-λt+s|=(bts+t)t/s(bss+t).
Therefore, r(A)≤(bt/(s+t))t/2s(bs/(s+t))1/2.Theorem 4.
IfA is invertible and A*X-tA≤Q-(AQ-1A*)s/t for all X∈[AQ-1A*t,Qs], then (1) has a positive operator solution.Proof.
LetΦ(X)=(Q-A*X-tA)1/s; then Φ is continuous for X∈[AQ-1A*t,Qs]; Φ(X)≤Qs; and Φ(X)≥[Q-(Q-(AQ-1A*)s/t)]1/s=(AQ-1A*)1/t. So Φ maps [AQ-1A*t,Qs] into itself. Furthermore, Φ is continuous on [AQ-1A*t,Qs] because X>0; hence, Φ has a fixed point in [AQ-1A*t,Qs]; that is, there exists X≥0 such that (Q-A*X-tA)1/s=X; this implies that (1) has a positive operator solution.Theorem 5.
If the operator equation (1) has a positive operator solution X, then λmin(T*T)≤(t/(t+s))t/s(s/(t+s)) and X≤α~Qs when λmin(T*T)≠1, where T=Q-t/2sAQ-1/2 and α~ is a solution to the equation αt/s(1-α)=λmin(T*T) in [t/(t+s),1].Proof.
We consider the sequence(7)α0=1,αk+1=1-λmin((Q-t/2sAQ-1/2)*Q-t/2sAQ-1/2)αkt/s,k=0,1,2,….
LetX be a positive solution to (1). Then Xs=Q-A*X-tA≤Q=α0Q; that is, X≤α0Qs. Assuming that X≤αkQs, then
(8)Xs=Q-A*X-tA≤Q-A*Q-t/sAαkt/s=Q1/2[I-Q-1/2A*Q-t/2sQ-t/2sAQ-1/2αkt/s]Q1/2≤Q1/2×[1-λmin((Q-t/2sAQ-1/2)*Q-t/2sAQ-1/2)αkt/s]×Q1/2=αk+1Q.
Hence, X≤αnQs for all n=0,1,2,…. It is straightforward to check that the sequence {αn} is monotonically decreasing; also, {αn} is bounded below; hence, {αn} is convergent. Denote T=Q-t/2sAQ-1/2 and let limn→∞αn=α~; then α~=1-λmin(T*T)/α~t/s; that is, α~ is a solution to the equation αt/s(1-α)=λmin(T*T). Let f(x)=xt/s(1-x). Then
(9)maxx∈[0,1]f(x)=f(tt+s)=(tt+s)t/s(st+s).
Thus, it follows that λmin(T*T)≤(t/(t+s))t/s(s/(t+s)).
Since(10)λmin(T*T)≤(tt+s)t/s(st+s).
Then the equation αt(1-α)=λmin(T*T) may have two solutions; one of these solutions is in the interval [t/(t+s),1].
In order to prove that the limitα~ of the sequence {αn} is in [t/(t+s),1], we assume that αi≥t/(t+s) (obviously α0=1>t/(t+s)); then
(11)αi+1=1-λmin(T*T)αit/s≥1-λmin(T*T)(t+st)t/s≥1-(tt+s)t/s(st+s)(t+st)t/s=tt+s.
Therefore, αi≥t/(t+s) for each i=1,2,…. Since X≤αnQs, then X≤α~Qs and α~∈[t/(t+s),1]. The proof is completed.Consider the following iterative sequence:(12)X0s=γQ,Xk+1s=Q-A*Xk-tA,k=0,1,…,γ>0.We will prove the following theorem.Theorem 6.
If the operator equation (1) has a positive operator solution X, then it has a maximal one XL. Moreover, the sequence {Xi} in (12) for γ∈[α~,1] is monotonically decreasing and converges to XL, where α~ is defined in Theorem 5.Proof.
Consider the iterative sequence (12) with γ∈[α~,1]. According to Theorem 4, we have Xs≤α~Q≤γQ=X0s for any positive operator solution X to (1).
Suppose thatXis≥Xs. Then
(13)Xi+1s=Q-A*Xi-tA≥Q-A*X-tA=Xs.
Hence,Xi≥X hold for each i. According to the definition of α~ and the monotonicity of the function f(x)=xt/s(1-x) on [t/(t+s),1], we have
(14)γt/s(1-γ)I≤α~t/s(1-α~)I=λmin(T*T)I≤T*T
for all γ∈[α~,1], where T=Q-t/2sAQ-1/2. We compute
(15)X1s=Q-A*(γQ)-t/sA=Q1/2[I-Q-1/2A*(γQ)-t/sAQ-1/2]Q1/2=Q1/2(I-γ-t/sT*T)Q1/2.
Using inequality (14) and equality (15), we obtain X1s≤γQ=X0s.
Assuming thatXis≤Xi-1s, then
(16)Xi+1s=Q-A*Xi-tA≤Q-A*Xi-1-tA=Xis.
Therefore, Xi+1s≤Xis for all i=0,1,…; that is, the sequence {Xns} is monotonically decreasing. Hence, {Xns} converges to the positive operator solution XL to (1). Since XLs≥Xs for any positive operator solution X, it follows that XL is the maximal solution.Theorem 7.
Letλmin(T*T)<λmax(T*T)≤(t/(t+s))t/s(s/(t+s)), and β1,β2 are solutions to the equation βt/s(1-β)=λmax(T*T) in [0,t/(t+s)] and [t/(t+s),1], respectively. Then there are the following conclusions.(i)
Ifγ∈[β1,β2], then the sequence {Xn} in (12) is monotonically increasing and converges to a positive operator solution Xγ∈[γQs,α~Qs] to (1).(ii)
Ifγ∈[α~,1], then the sequence {Xn} in (12) is monotonically decreasing and converges to the maximal positive operator solution XL∈[β2Qs,α~Qs] to (1).(iii)
Ifγ∈(β2,α~) and (t/s)∥A∥2≤(β2λmin(Q))(s+t)/s, then the sequence {Xn} in (12) converges to the unique solution XL∈[β2Qs,α~Qs], where α~ and T are defined in Theorem 5.Proof.
Consider the functionf(x)=xt/s(1-x),x∈[0,1]. It is monotonically increasing on [0,t/(t+s)] and monotonically decreasing on [t/(t+s),1], and
(17)maxx∈[0,1]f(x)=f(tt+s)=(tt+s)t/s(st+s).
Since λmax(T*T)≤(t/(t+s))t/s(s/(t+s)), then, for each γ1,γ2 such that 0<β1≤γ1≤β2<α~≤γ2≤1, the inequalities
(18)γ2t/s(1-γ2)I≤λmin(T*T)≤T*T≤λmax(T*T)≤γ1t/s(1-γ1)I
are satisfied.
(i) Letγ∈[β1,β2]. We will prove that the operator sequence {Xn} in (12) is monotonically increasing and is bounded above.
According to the fourth inequality in (18), we compute
(19)X1s=Q-A*(γQ)-t/sA=Q1/2(I-γ-t/sQ-1/2A*Q-t/sAQ-1/2)Q1/2=Q1/2(I-γ-t/sT*T)Q1/2≥γQ=X0s.
Assuming thatXis≥Xi-1s, then
(20)Xi+1s=Q-A*Xi-tA≥Q-A*Xi-1-tA=Xis.
Therefore, Xi+1s≥Xis for all i=0,1,…; that is, the sequence {Xns} is monotonically increasing. Obviously X0s=γQ≤β2Q≤α~Q. We suppose that Xis≤α~Q; from the first inequality in (18), we obtain
(21)Xi+1s=Q-A*Xi-tA≤Q-A*(α~Q)-t/sA≤α~Q.
Hence,{Xns} converges to a positive operator solution Xγs to (1). Since Xi∈[γQs,α~Qs] for all i=0,1,…, we have Xγ∈[γQs,α~Qs].
(ii) Letγ∈[α~,1]. From the proving procession of Theorem 6, the sequence (12) is monotonically decreasing. Since X0=γQs≥β2Qs and supposing that Xis≥β2Q, it follows that
(22)Xi+1s=Q-A*Xi-tA≥Q-A*(β2Q)-t/sA=Q1/2(I-β2-t/sQ-1/2A*Q-t/sAQ-1/2)Q1/2≥Q1/2(I-β2-t/sλmax(T*T)I)Q1/2=β2Q.
By induction principle, Xis≥β2Q hold for each i=0,1,2,…. Hence, {Xns} is convergent. From Theorem 6, {Xns} converges to XLs such that XL∈[β2Qs,γQs].
(iii) Considering the sequence{Xns} in (12) for γ∈(β2,α~), that is, X0s∈(β2Q,α~Q) and supposing that Xis∈(β2Q,α~Q), then, for Xi+1s,
(23)Xi+1s=Q-A*Xi-tA<Q1/2(I-β2-t/sQ-1/2A*Q-t/sAQ-1/2)Q1/2≤α~Q,Xi+1s=Q-A*Xi-tA>Q1/2(I-α~-t/sQ-1/2A*Q-t/sAQ-1/2)Q1/2≥β2Q.
Hence, Xis∈(β2Q,γQ), for all i=0,1,2,…. Now, we considering ∥Xi+1s-Xis∥,
(24)∥Xi+1s-Xis∥=∥(Q-A*Xi-tA)-(Q-A*Xi-1-tA)∥≤1s1(β2λmin(Q)s)s-1×t∥A∥2(β2λmin(Q)s)t+1∥Xi-Xi-1∥=ts∥A∥21(β2λmin(Q)s)t+s×∥Xi-Xi-1∥,
and (t/s)∥A∥2≤(β2λmin(Q))(s+t)/s, it follows that {Xn} in (12) is a Cauchy sequence in the Banach space [β2Qs,α~Qs]. Hence, it has a limit Xγ in [β2Qs,α~Qs], and Xγ is a unique solution to (1) in [β2Qs,α~Qs]. According to Theorem 6, it is the maximal solution, that is, Xγ=XL. The proof is completed.



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