ISRN Mathematical AnalysisVolume 2013 (2013), Article ID 145606, 4 pageshttp://dx.doi.org/10.1155/2013/145606

Research Article

On the Positive Operator Solutions to an Operator Equation

Kai-Fan Yang,¹ Fang-An Deng,¹ and Hong-Ke Du²

¹School of Mathematics and Computer Science, Shaanxi University of Technology, Shaanxi 723001, China²College of Mathematics and Information Science, Shaanxi Normal University, Xi'an 710062, China

Received 3 May 2013; Accepted 9 June 2013

Academic Editors: M. Lindstrom and K. A. Lurie

Copyright © 2013 Kai-Fan Yang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

The necessary conditions and the sufficient condition for the existence of positive operator solutions to the operator equation are established. An iterative method for obtaining the positive operator solutions is proposed.

1. Introduction

Let be the set of all bounded linear operators on the Hilbert space . In this paper, we consider the nonlinear operator equation where , , and is an unknown operator in . Both and 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 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 , if for all , then is said to be a positive operator and is denoted by . If is a positive operator and invertible, then denote . For and in , means that is a positive operator. For , , , , and denote the adjoint, the radius of numerical range, the spectrum, and the spectral radius of , respectively.

For positive operators in , the following facts are well known.(1)If , then .(2)Denote . Then .(3)If the sequence of positive operator is monotonically increasing and has upper bound, that is, , or is monotonically decreasing and has lower bound, that is, , then this sequence is convergent to a positive limit operator, where are given operators.

2. Main Results and Proofs

In order to prove our main result, we begin with some lemmas as follows.

Lemma 1. Let . If , then .

Lemma 2. Let and be self-adjoint operators in . If , then, for every , one has .

In this section, we give our main results and proofs.

Theorem 3. If the operator equation (1) has a positive operator solution , then and , where .

Proof. (1) If the operator equation (1) has a positive operator solution , then . Hence, we obtain ; that is, . From , it follows that That is, .(2) From (1), we have Then According to the spectral decomposition of , we have Denote . Then Therefore, .

Theorem 4. If is invertible and for all , then (1) has a positive operator solution.

Proof. Let ; then is continuous for ; ; and . So maps into itself. Furthermore, is continuous on because ; hence, has a fixed point in ; that is, there exists such that ; this implies that (1) has a positive operator solution.

Theorem 5. If the operator equation (1) has a positive operator solution , then and when , where and is a solution to the equation in .

Proof. We consider the sequence Let be a positive solution to (1). Then ; that is, . Assuming that , then Hence, for all . It is straightforward to check that the sequence is monotonically decreasing; also, is bounded below; hence, is convergent. Denote and let ; then ; that is, is a solution to the equation . Let . Then Thus, it follows that .Since Then the equation may have two solutions; one of these solutions is in the interval .In order to prove that the limit of the sequence is in , we assume that (obviously ); then Therefore, for each . Since , then and . The proof is completed.

Consider the following iterative sequence:

We will prove the following theorem.

Theorem 6. If the operator equation (1) has a positive operator solution , then it has a maximal one . Moreover, the sequence in (12) for is monotonically decreasing and converges to , where is defined in Theorem 5.

Proof. Consider the iterative sequence (12) with . According to Theorem 4, we have for any positive operator solution to (1).Suppose that . Then Hence, hold for each . According to the definition of and the monotonicity of the function on , we have for all , where . We compute Using inequality (14) and equality (15), we obtain .Assuming that , then Therefore, for all ; that is, the sequence is monotonically decreasing. Hence, converges to the positive operator solution to (1). Since for any positive operator solution , it follows that is the maximal solution.

Theorem 7. Let , and are solutions to the equation in and , respectively. Then there are the following conclusions. (i)If , then the sequence in (12) is monotonically increasing and converges to a positive operator solution to (1).(ii)If , then the sequence in (12) is monotonically decreasing and converges to the maximal positive operator solution to (1).(iii)If and , then the sequence in (12) converges to the unique solution , where and are defined in Theorem 5.

Proof. Consider the function . It is monotonically increasing on and monotonically decreasing on , and Since , then, for each such that , the inequalities are satisfied.(i) Let . We will prove that the operator sequence in (12) is monotonically increasing and is bounded above.According to the fourth inequality in (18), we compute Assuming that , then Therefore, for all ; that is, the sequence is monotonically increasing. Obviously . We suppose that ; from the first inequality in (18), we obtain Hence, converges to a positive operator solution to (1). Since for all , we have .(ii) Let . From the proving procession of Theorem 6, the sequence (12) is monotonically decreasing. Since and supposing that , it follows that By induction principle, hold for each . Hence, is convergent. From Theorem 6, converges to such that .(iii) Considering the sequence in (12) for , that is, and supposing that , then, for , Hence, , for all . Now, we considering , and , it follows that in (12) is a Cauchy sequence in the Banach space . Hence, it has a limit in , and is a unique solution to (1) in . According to Theorem 6, it is the maximal solution, that is, . The proof is completed.

Acknowledgments

This paper is supported by the NSF of China Grant no. 11171197 and by the Natural Science Foundation of Shaanxi Educational Committee Grant no. 12JK0884.

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