On the Positive Operator Solutions to an Operator Equation

Abstract

In this paper, the positive operator solutions to operator equation (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 (t > 1). The relations of these operators in the operator equation are given.

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, let H be infinite dimensional Hilbert space and B (H) denote the set of all bounded linear operators on H; we will consider nonlinear operator equation.in H; here, 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.

For, denote the adjoint, the spectrum, and the norm of A, respectively. If for all , then A is said to be a positive operator and denoted by . For positive operators in , the following conclusions are obvious:(1)For P ≥ Q > 0, we have .(2)For positive operator P,, where

2. Main Results and Proofs

Lemma 1 (see [12]). Let . If , then .

Lemma 2 (see [12]). Let A and B be self-adjoint operators in . If , then for any , we have .

Proposition 1 (see [12]). If is normal, then the algebra generated by A is commutative.

Theorem 1. If the operator equation (1) has a positive operator solution X, then

Proof. If the operator equation (1) has a positive operator solution X, from , we can obtain . From Lemma 2, it is easy to see . From equation (1),We can obtain . That is, . The theorem is proved.

Theorem 2. If A is invertible, then equation (1) has a positive operator solution.

Proof. Let , then is continuous for any . Clearly, for any X, combination with the proof of Theorem 1, we have , that is, ; this implies is a mapping to itself on . By the fixed point theorem , has a fixed point in such that , i.e., , that is, is a positive operator solution to equation (1).

Theorem 3. Let . The operator equation has a positive invertible operator solution X if and only if A has the factor decomposition, where W, B satisfy and W is invertible.

Proof. If operator equation (1) has a positive operator solution X, let , then and W is invertible, so . According to equation (1), we haveLet , then . From equality (5), we obtain . Conversely, if A has the factor decomposition , W, B satisfy and W is invertible. Let , thenThat 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 , then . In finite dimensional space, if A is not invertible, then , but in infinite dimensional space, if A is not invertible, maybe a null space; the lemma in [6] is not held in infinite dimensional space, but the following conclusion holds.

Theorem 4. If has positive operator solution X, then if and only if A is not bounded below.

Proof. From Theorem 1, we know for any positive operator solution to equation (1).
Necessary. If has positive operator solution X and , then , hence there exists unit sequence such that. For any unit vector , we havehence . On the other hand, .
Hence, , therefore A is not bounded below.
Sufficient. Assume for any positive operator solution X, then is nonnegative and invertible, then for any unit vector , there exists constant such that . Since and , we can conclude thatthat is, ; this illustrates that A is bounded below; it is a contradiction, so .

Theorem 5. If X is a positive operator solution to equation (1), then

Proof. From , we have , that is,From the first inequality of (10), we have , that is, .
From the second inequality of (10), we have , that is,. Therefore, .

Theorem 6. If A is normal, and A, Q, t satisfy , then equation (1) has positive operator solution, where m is the positive integer and .

Proof. Consider the sequence of positive operators ,According to the iteration sequence (11), is in the algebra generated by A and Q. Because A is normal, in accordance with Proposition 1, for any , we have . Since and , it is easy to see ; this implies . Successive analogy: we can provetherefore the subsequence and both converge to positive operators. At the same time, for all nonnegative integers i, we have .
. Denote , thenIn the same way, we haveSuccessive analogy:Therefore, . Combined with the condition , we can know that subsequence and converge to the same positive operator, which is the positive operator solution to equation (1).

Data Availability

This paper is a theoretical analysis without data.

Conflicts of Interest

The author declares no conflicts of interest.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (11301318) and the Natural Science Foundation of Shaanxi Educational Committee Grant (18JK0162). This work was also supported by the NSF of China.