# On the Positive Operator Solutions to an Operator EquationX−A∗X−tA=Q

Authors: Kaifan Yang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, the positive operator solutions to operator equationX−A∗X−tA=Q (t > 1) are studied in infinite dimensional Hilbert space. Firstly, the range of norm and the spectral radius of the solution to the equation are given. Secondly, by constructing effective iterative sequence, it gives some conditions for the existence of positive operator solutions to operator equation X−A∗X−tA=Q (t > 1). The relations of these operators in the operator equation are given.


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## Body

## 1. Introduction

Operator algebra has played an important role in the subject of functional analysis; it has been considered by many authors. At the same time, operator equation is one of the hottest topics in operator theory. The researches on the positive solutions to operator equations began in 1990s; it has been applied to many fields such as dynamic programming [1], stochastic filtering, control theory [2, 3], and statistics [4]. In recent years, operator equation attained a great development and many scholars put into studying different kinds of operator equations (see [5–11]).In this paper, letH be infinite dimensional Hilbert space and B (H) denote the set of all bounded linear operators on H; we will consider nonlinear operator equation.(1)X−A∗X−tA=Q,in H; here, A,Q∈BH with Q > 0, A∗ is the adjoint of A.In the past few years, many authors used different iterative methods for computing the positive definite solutions to equation (1) in finite dimensional space. In this paper, we extend the study of operator equation (1) from finite dimensional space to infinite dimensional Hilbert space. Some necessary conditions for the existence of positive operator solutions to operator equation (1) are derived. Furthermore, conditions under which operator equation (1) has positive operator solutions are obtained.ForA∈BH, A∗,σA,A denote the adjoint, the spectrum, and the norm of A, respectively. If Ax,x≥0 for all x∈H, then A is said to be a positive operator and denoted by A≥0. For positive operators in BH, the following conclusions are obvious:(1)
ForP ≥ Q > 0, we have P−1≤Q−1.(2)
For positive operatorP,λminPI≤P≤λmaxPI, where(2)λminP=minλ:λ∈σP,P>0,λmaxP=maxλ:λ∈σP,P>0.


## 2. Main Results and Proofs

Lemma 1 (see [12]).
LetA,B∈BH. If A≥B, then A≥B.Lemma 2 (see [12]).
LetA and B be self-adjoint operators in BH. If A≤B, then for any T∈BH, we have T∗AT≤T∗BT.Proposition 1 (see [12]).
IfA∈BH is normal, then the C∗−algebra generated by A is commutative.Theorem 1.
If the operator equation (1) has a positive operator solution X, then(3)Q≤X≤Q1+Q−t/2AQ−1/22.Proof.
If the operator equation (1) has a positive operator solution X, from A∗X−tA>0, we can obtain X≥Q. From Lemma 2, it is easy to see X≥Q. From equation (1),(4)X=Q+A∗X−tA≤Q+A∗Q−tA=Q1/2I+Q−1/2A∗Q−tAQ−1/2Q1/2.
We can obtainX≤QI+Q−1/2A∗Q−tAQ−1/2=Q1+Q−t/2AQ−t/22. That is, Q≤X≤Q1+Q−t/2AQ−1/22. The theorem is proved.Theorem 2.
If A is invertible, then equation (1) has a positive operator solution.Proof.
LetϕX=Q+A∗X−tA, then ϕ is continuous for any X∈Q,Q+A∗Q−tA. Clearly, ϕX≥Q for any X, combination with the proof of Theorem 1, we have ϕX=Q+A∗X−tA≤Q+A∗Q−tA, that is, ϕX∈Q,Q+A∗Q−tA; this implies ϕ is a mapping to itself on Q,Q+A∗Q−tA. By the fixed point theorem , ϕ has a fixed point X0 in Q,Q+A∗Q−tA such that ϕX0=X0, i.e., X0=Q+A∗X0−tA, that is, X0 is a positive operator solution to equation (1).Theorem 3.
LetA∈BH. The operator equationX−A∗X−tA=Q has a positive invertible operator solution X if and only if A has the factor decompositionA=W∗Wt/2B, where W, B satisfy W∗W−B∗B=Q and W is invertible.Proof.
If operator equation (1) has a positive operator solution X, let W=X, then W∗=W and W is invertible, so X=W∗W. According to equation (1), we have(5)W∗W−W∗W−t/2A∗W∗W−t/2A=Q.
LetW∗W−t/2A=B, then A=W∗Wt/2B. From equality (5), we obtain W∗W−B∗B=Q. Conversely, if A has the factor decomposition A=W∗Wt/2B, W, B satisfy W∗W−B∗B=Q and W is invertible. Let X=W∗W, then(6)X−A∗X−tA=W∗W−W∗Wt/2B∗W∗W−tW∗Wt/2B=W∗W−B∗B=Q.
That is,X is a positive operator solution to equation (1).
In [6], it proves that if A is not invertible and then X is a positive solution to X+A∗X−2A=I, then λmaxX=1. In finite dimensional space, if A is not invertible, then NA≠0, but in infinite dimensional space, if A is not invertible, NA maybe a null space; the lemma in [6] is not held in infinite dimensional space, but the following conclusion holds.Theorem 4.
IfX−A∗X−tA=Q has positive operator solution X, thenX=Q if and only if A is not bounded below.Proof.
From Theorem1, we know X≥Q for any positive operator solution to equation (1).
Necessary. If X−A∗X−tA=Q has positive operator solution X and X=Q, then ωX=γX=X=Q, hence there exists unit sequence xnn=1+∞ such thatXsxn,xn⟶bn⟶∞. For any unit vector x∈H, we have(7)Xx,x=A∗X−tA+Qx,x=Qx,x+A∗X−tAx,x,hence X−tAxn,Axn⟶0. On the other hand, X−tAxn,Axn=X−t/2Axn2≥Axn2/Xt/22.
Hence,Axn⟶0, therefore A is not bounded below.
Sufficient. Assume X>Q for any positive operator solution X, then X−QI is nonnegative and invertible, then for any unit vector x∈H, there exists constant δ>0 such that X−QIx,x≥δx2. Since X=A∗X−tA+Q and Q≤QI, we can conclude that(8)δx2≤X−QIx,x≤X−Qx,x=A∗X−tAx,x=X−t/2Ax2≤Xt/22Ax2,that is, Ax≥δX−t/2−1x; this illustrates that A is bounded below; it is a contradiction, so X=Q.Theorem 5.
IfX is a positive operator solution to equation (1), then(9)λmaxX−λmaxQλmintX≤λmaxA∗A≤λmaxX−λminQλmaxtX.Proof.
From0<λminXI≤X≤λmaxXI, we have Q+A∗A/λmaxtX≤Q+A∗X−tA≤Q+A∗A/λmintX, that is,(10)Q+A∗AλmaxtX≤X≤Q+A∗AλmintX.
From the first inequality of (10), we have A∗A/λmaxtX≤λmaxX−Q≤λmaxX−λminQ, that is, λmaxA∗A≤λmaxX−λminQλmaxtX.
From the second inequality of (10), we have λmaxX≤λmaxQ+A∗A/λmintX≤λmaxQ+λmaxA∗A/λmintX, that is,λmaxA∗A≥λmaxX−λmaxQλmintX. Therefore, λmaxX−λmaxQλmintX≤λmaxA∗A≤λmaxX−λminQλmaxtX.Theorem 6.
If A is normal,t=2m and A, Q, t satisfy tA2at−1/Q2t<1, then equation (1) has positive operator solution, where m is the positive integer and a=Q+A∗Q−tA.Proof.
Consider the sequence of positive operatorsXnn=0+∞,(11)X0=Q,Xk+1=Q+A∗Xk−tA,k=0,1,2,….
According to the iteration sequence (11), Xi is in the C∗−algebra generated by A and Q. Because A is normal, in accordance with Proposition 1, for any n=0,1,2,…, we have AXn=XnA,Xn+1Xn=XnXn+1. Since X1=Q+A∗X0−tA≥Q=X0and X1−t≤X0−t, it is easy to see X2=Q+A∗X1−tA≤Q+A∗X0−tA=X1; this implies Q=X0≤X2≤X1=Q+A∗Q−tA. Successive analogy: we can prove(12)Q=X0≤X2≤X4≤…≤X5≤X3≤X1=Q+A∗Q−tA,therefore the subsequence X2nn=0+∞ and X2n+1n=0+∞ both converge to positive operators. At the same time, for all nonnegative integers i, we have Q≤Xi≤Q+A∗Q−tA.
Q≤Xi≤Q+A∗Q−tA. Denote Q+A∗Q−tA=a, then(13)X2k+1−X2k=A∗X2k−tA−A∗X2k−1−tA≤A2X2k−tX2k−1t−X2ktX2k−1−t≤A2Q−2tX2k−1t−X2kt,(14)X2k−1t−X2kt=X2k−1t/2X2k−1t/2−X2kt/2+X2k−1t/2−X2kt/2X2kt/2≤X2k−1t/2−X2kt/2X2k−1t/2+X2kt/2≤2at/2X2k−1t/2−X2kt/2.
In the same way, we have(15)X2k−1t/2−X2kt/2≤2at/4X2k−1t/4−X2kt/4,…
Successive analogy:(16)X2k−1t−X2kt≤2at/22at/4⋯2at/2mX2k−1t/2m−X2kt/2m=2mat−1X2k−1−X2k.
Therefore,X2k+1−X2k≤tat−1A2Q−2tX2k−1−X2k. Combined with the condition tA2at−1/Q2t<1, we can know that subsequence X2nn=0+∞ and X2n+1n=0+∞ converge to the same positive operator, which is the positive operator solution to equation (1).



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