On the Positive Operator Solutions to an Operator Equation

Abstract

The necessary conditions and the sufficient condition for the existence of positive operator solutions to the operator equation 𝑋^𝑠+𝐴^∗𝑋^−𝑡𝐴=𝑄 are established. An iterative method for obtaining the positive operator solutions is proposed.

1. Introduction

Let ℬ(ℋ) be the set of all bounded linear operators on the Hilbert space ℋ. In this paper, we consider the nonlinear operator equation 𝑋^𝑠+𝐴^∗𝑋^−𝑡𝐴=𝑄, (1) 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 in ℬ(ℋ), the following facts are well known.(1)If 𝑃≥𝑄>0, then 𝑃⁻¹≤𝑄⁻¹.(2)Denote 𝜆_max(𝑃)=max{𝜆∶𝜆∈𝜎(𝑃),𝑃>0},𝜆_min(𝑃)=min{𝜆∶𝜆∈𝜎(𝑃),𝑃>0}. Then 𝜆_min(𝑃)𝐼≤𝑃≤𝜆_max(𝑃)𝐼.(3)If the sequence {𝑋_𝑛}^+∞_𝑛=1 of positive operator is monotonically increasing and has upper bound, that is, 𝑋_𝑘≤𝑋_𝑘+1≤𝐶₁, or is monotonically decreasing and has lower bound, that is, 𝐶₂≤𝑋_𝑘+1≤𝑋_𝑘, 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. Let 𝐴,𝐵∈ℬ(ℋ). If 𝐴≥𝐵≥0, then ‖𝐴‖≥‖𝐵‖.

Lemma 2. Let 𝐴 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)^𝑡𝐴𝑄⁻¹𝐴^∗≤𝑋≤^𝑠𝑄 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 𝑋^𝑠≤𝑄; that is, 𝑋≤^𝑠𝑄. From 𝐴^∗𝑋^−𝑡𝐴≤𝑄, it follows that 𝑄^−1/2𝐴^∗𝑋^−𝑡/2𝑋^−𝑡/2𝐴𝑄^−1/2𝑋≤𝐼,_−𝑡/2𝐴𝑄^−1/2𝑄^−1/2𝐴^∗𝑋^−𝑡/2≤𝐼,𝐴𝑄⁻¹𝐴^∗≤𝑋^𝑡. (2) That is, 𝑋≥^𝑡𝐴𝑄⁻¹𝐴^∗.
(2) From (1), we have 𝑋^𝑡+𝑠+𝑋^𝑡/2𝐴^∗𝑋^−𝑡𝐴𝑋^𝑡/2=𝑋^𝑡/2𝑄𝑋^𝑡/2≤𝜆_max(𝑄)𝑋^𝑡. (3) Then 𝑟(𝐴)²=𝑟²𝑋^𝑡/2𝐴𝑋^−𝑡/2≤𝑋^𝑡/2𝐴𝑋^−𝑡/22≤𝜆_max(𝑄)𝑋^𝑡−𝑋^𝑡+𝑠. (4) According to the spectral decomposition of 𝑋, we have 𝜆_max(𝑄)𝑋^𝑡−𝑋^𝑡+𝑠=_𝜎(𝑋)𝜆_max(𝑄)𝜆^𝑡−𝜆^𝑡+𝑠𝑑𝐸𝜆. (5) Denote 𝜆_max(𝑄)=‖𝑄‖=𝑏. Then 𝜆_max(𝑄)𝑋^𝑡−𝑋^𝑡+𝑠≤max_{𝜆∈0,𝑠√𝑏}𝑏𝜆^𝑡−𝜆^𝑡+𝑠=𝑏𝑡𝑠+𝑡^𝑡/𝑠𝑏𝑠.𝑠+𝑡 (6) Therefore, 𝑟(𝐴)≤(𝑏𝑡/(𝑠+𝑡))^𝑡/2𝑠(𝑏𝑠/(𝑠+𝑡))^1/2.

Theorem 4. If 𝐴 is invertible and 𝐴^∗𝑋^−𝑡𝐴≤𝑄−(𝐴𝑄⁻¹𝐴^∗)^𝑠/𝑡 for all 𝑋∈[^𝑡𝐴𝑄⁻¹𝐴^∗,^𝑠𝑄], then (1) has a positive operator solution.

Proof. Let Φ(𝑋)=(𝑄−𝐴^∗𝑋^−𝑡𝐴)^1/𝑠; then Φ is continuous for 𝑋∈[^𝑡𝐴𝑄⁻¹𝐴^∗,^𝑠𝑄]; Φ(𝑋)≤^𝑠𝑄; and Φ(𝑋)≥[𝑄−(𝑄−(𝐴𝑄⁻¹𝐴^∗)^𝑠/𝑡)]^1/𝑠=(𝐴𝑄⁻¹𝐴^∗)^1/𝑡. So Φ maps [^𝑡𝐴𝑄⁻¹𝐴^∗,^𝑠𝑄] into itself. Furthermore, Φ is continuous on [^𝑡𝐴𝑄⁻¹𝐴^∗,^𝑠𝑄] because 𝑋>0; hence, Φ has a fixed point in [^𝑡𝐴𝑄⁻¹𝐴^∗,^𝑠𝑄]; 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 𝜆_min(𝑇^∗𝑇)≤(𝑡/(𝑡+𝑠))^𝑡/𝑠(𝑠/(𝑡+𝑠)) and 𝑋≤^𝑠𝛼𝑄 when 𝜆_min(𝑇^∗𝑇)≠1, where 𝑇=𝑄^{−𝑡/2𝑠}𝐴𝑄^−1/2 and 𝛼 is a solution to the equation 𝛼^𝑡/𝑠(1−𝛼)=𝜆_min(𝑇^∗𝑇) in [𝑡/(𝑡+𝑠),1].

Proof. We consider the sequence 𝛼₀=1,𝛼_𝑘+1−𝜆=1_min𝑄^{−𝑡/2𝑠}𝐴𝑄^−1/2∗𝑄^{−𝑡/2𝑠}𝐴𝑄^−1/2𝛼_𝑘^𝑡/𝑠,𝑘=0,1,2,…. (7)
Let 𝑋 be a positive solution to (1). Then 𝑋^𝑠=𝑄−𝐴^∗𝑋^−𝑡𝐴≤𝑄=𝛼₀𝑄; that is, 𝑋≤^𝑠𝛼₀𝑄. Assuming that 𝑋≤^𝑠𝛼_𝑘𝑄, then 𝑋^𝑠=𝑄−𝐴^∗𝑋^−𝑡𝐴𝐴≤𝑄−^∗𝑄^−𝑡/𝑠𝐴𝛼_𝑘^𝑡/𝑠=𝑄^1/2𝑄𝐼−^−1/2𝐴^∗𝑄^{−𝑡/2𝑠}𝑄^{−𝑡/2𝑠}𝐴𝑄^−1/2𝛼_𝑘^𝑡/𝑠𝑄^1/2≤𝑄^1/2×𝜆1−^min𝑄^{−𝑡/2𝑠}𝐴𝑄^−1/2∗𝑄^{−𝑡/2𝑠}𝐴𝑄^−1/2𝛼_𝑘^𝑡/𝑠×𝑄^1/2=𝛼_𝑘+1𝑄. (8) Hence, 𝑋≤^𝑠𝛼_𝑛𝑄 for all 𝑛=0,1,2,…. It is straightforward to check that the sequence {𝛼_𝑛} is monotonically decreasing; also, {𝛼_𝑛} is bounded below; hence, {𝛼_𝑛} is convergent. Denote 𝑇=𝑄^{−𝑡/2𝑠}𝐴𝑄^−1/2 and let lim_𝑛→∞𝛼_𝑛=𝛼; then 𝛼=1−𝜆_min(𝑇^∗𝑇)/𝛼^𝑡/𝑠; that is, 𝛼 is a solution to the equation 𝛼^𝑡/𝑠(1−𝛼)=𝜆_min(𝑇^∗𝑇). Let 𝑓(𝑥)=𝑥^𝑡/𝑠(1−𝑥). Then max_[]𝑥∈0,1𝑓(𝑥)𝑡=𝑓=𝑡𝑡+𝑠𝑡+𝑠^𝑡/𝑠𝑠.𝑡+𝑠 (9) Thus, it follows that 𝜆_min(𝑇^∗𝑇)≤(𝑡/(𝑡+𝑠))^𝑡/𝑠(𝑠/(𝑡+𝑠)).
Since 𝜆_min𝑇^∗𝑇≤𝑡𝑡+𝑠^𝑡/𝑠𝑠.𝑡+𝑠 (10) Then the equation 𝛼^𝑡(1−𝛼)=𝜆_min(𝑇^∗𝑇) may have two solutions; one of these solutions is in the interval [𝑡/(𝑡+𝑠),1].
In order to prove that the limit 𝛼 of the sequence {𝛼_𝑛} is in [𝑡/(𝑡+𝑠),1], we assume that 𝛼_𝑖≥𝑡/(𝑡+𝑠) (obviously 𝛼₀=1>𝑡/(𝑡+𝑠)); then 𝛼_𝑖+1𝜆=1−^min𝑇^∗𝑇𝛼_𝑖^𝑡/𝑠≥1−𝜆_min𝑇^∗𝑇𝑡+𝑠𝑡^𝑡/𝑠𝑡≥1−𝑡+𝑠^𝑡/𝑠𝑠𝑡+𝑠𝑡+𝑠𝑡^𝑡/𝑠=𝑡.𝑡+𝑠 (11) Therefore, 𝛼_𝑖≥𝑡/(𝑡+𝑠) for each 𝑖=1,2,…. Since 𝑋≤^𝑠𝛼_𝑛𝑄, then 𝑋≤^𝑠𝛼𝑄 and 𝛼∈[𝑡/(𝑡+𝑠),1]. The proof is completed.

Consider the following iterative sequence: 𝑋^𝑠₀=𝛾𝑄,𝑋^𝑠_𝑘+1=𝑄−𝐴^∗𝑋_𝑘^−𝑡𝐴,𝑘=0,1,…,𝛾>0. (12)

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 𝑋_𝐿. Moreover, the sequence {𝑋_𝑖} in (12) for 𝛾∈[𝛼,1] is monotonically decreasing and converges to 𝑋_𝐿, where 𝛼 is defined in Theorem 5.

Proof. Consider the iterative sequence (12) with 𝛾∈[𝛼,1]. According to Theorem 4, we have 𝑋^𝑠≤𝛼𝑄≤𝛾𝑄=𝑋^𝑠₀ for any positive operator solution 𝑋 to (1).
Suppose that 𝑋^𝑠_𝑖≥𝑋^𝑠. Then 𝑋^𝑠_𝑖+1=𝑄−𝐴^∗𝑋_𝑖^−𝑡𝐴≥𝑄−𝐴^∗𝑋^−𝑡𝐴=𝑋^𝑠. (13)
Hence, 𝑋_𝑖≥𝑋 hold for each 𝑖. According to the definition of 𝛼 and the monotonicity of the function 𝑓(𝑥)=𝑥^𝑡/𝑠(1−𝑥) on [𝑡/(𝑡+𝑠),1], we have 𝛾^𝑡/𝑠1−𝛾𝐼≤𝛼^𝑡/𝑠(1−𝛼)𝐼=𝜆_min𝑇^∗𝑇𝐼≤𝑇^∗𝑇 (14) for all 𝛾∈[𝛼,1], where 𝑇=𝑄^{−𝑡/2𝑠}𝐴𝑄^−1/2. We compute 𝑋^𝑠₁=𝑄−𝐴^∗𝛾𝑄^−𝑡/𝑠𝐴=𝑄^1/2𝐼−𝑄^−1/2𝐴^∗𝛾𝑄^−𝑡/𝑠𝐴𝑄^−1/2𝑄^1/2=𝑄^1/2𝐼−𝛾^−𝑡/𝑠𝑇^∗𝑇𝑄^1/2. (15) Using inequality (14) and equality (15), we obtain 𝑋^𝑠₁≤𝛾𝑄=𝑋^𝑠₀.
Assuming that 𝑋^𝑠_𝑖≤𝑋^𝑠_𝑖−1, then 𝑋^𝑠_𝑖+1=𝑄−𝐴^∗𝑋_𝑖^−𝑡𝐴≤𝑄−𝐴^∗𝑋^−𝑡_𝑖−1𝐴=𝑋^𝑠_𝑖. (16) Therefore, 𝑋^𝑠_𝑖+1≤𝑋^𝑠_𝑖 for all 𝑖=0,1,…; that is, the sequence {𝑋^𝑠_𝑛} is monotonically decreasing. Hence, {𝑋^𝑠_𝑛} converges to the positive operator solution 𝑋_𝐿 to (1). Since 𝑋^𝑠_𝐿≥𝑋^𝑠 for any positive operator solution 𝑋, it follows that 𝑋_𝐿 is the maximal solution.

Theorem 7. Let 𝜆_min(𝑇^∗𝑇)<𝜆_max(𝑇^∗𝑇)≤(𝑡/(𝑡+𝑠))^𝑡/𝑠(𝑠/(𝑡+𝑠)), and 𝛽₁,𝛽₂ are solutions to the equation 𝛽^𝑡/𝑠(1−𝛽)=𝜆_max(𝑇^∗𝑇) in [0,𝑡/(𝑡+𝑠)] and [𝑡/(𝑡+𝑠),1], respectively. Then there are the following conclusions.(i)If 𝛾∈[𝛽₁,𝛽₂], then the sequence {𝑋_𝑛} in (12) is monotonically increasing and converges to a positive operator solution 𝑋_𝛾∈[^𝑠𝛾𝑄,^𝑠𝛼𝑄] to (1).(ii)If 𝛾∈[𝛼,1], then the sequence {𝑋_𝑛} in (12) is monotonically decreasing and converges to the maximal positive operator solution 𝑋_𝐿∈[^𝑠𝛽₂𝑄,^𝑠𝛼𝑄] to (1).(iii)If 𝛾∈(𝛽₂,𝛼) and (𝑡/𝑠)‖𝐴‖²≤(𝛽₂𝜆_min(𝑄))^{(𝑠+𝑡)/𝑠}, then the sequence {𝑋_𝑛} in (12) converges to the unique solution 𝑋_𝐿∈[^𝑠𝛽₂𝑄,^𝑠𝛼𝑄], where 𝛼 and 𝑇 are defined in Theorem 5.

Proof. Consider the function 𝑓(𝑥)=𝑥^𝑡/𝑠(1−𝑥),𝑥∈[0,1]. It is monotonically increasing on [0,𝑡/(𝑡+𝑠)] and monotonically decreasing on [𝑡/(𝑡+𝑠),1], and max_[]𝑥∈0,1𝑓(𝑥)𝑡=𝑓=𝑡𝑡+𝑠𝑡+𝑠^𝑡/𝑠𝑠.𝑡+𝑠 (17) Since 𝜆_max(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 𝛾₂^𝑡/𝑠1−𝛾₂𝐼≤𝜆_min𝑇^∗𝑇≤𝑇^∗𝑇≤𝜆_max𝑇^∗𝑇≤𝛾₁^𝑡/𝑠1−𝛾₁𝐼 (18) are satisfied.
(i) Let 𝛾∈[𝛽₁,𝛽₂]. We will prove that the operator sequence {𝑋_𝑛} in (12) is monotonically increasing and is bounded above.
According to the fourth inequality in (18), we compute 𝑋^𝑠₁=𝑄−𝐴^∗𝛾𝑄^−𝑡/𝑠𝐴=𝑄^1/2𝐼−𝛾^−𝑡/𝑠𝑄^−1/2𝐴^∗𝑄^−𝑡/𝑠𝐴𝑄^−1/2𝑄^1/2=𝑄^1/2𝐼−𝛾^−𝑡/𝑠𝑇^∗𝑇𝑄^1/2≥𝛾𝑄=𝑋^𝑠₀. (19)
Assuming that 𝑋^𝑠_𝑖≥𝑋^𝑠_𝑖−1, then 𝑋^𝑠_𝑖+1=𝑄−𝐴^∗𝑋_𝑖^−𝑡𝐴≥𝑄−𝐴^∗𝑋^−𝑡_𝑖−1𝐴=𝑋^𝑠_𝑖. (20) Therefore, 𝑋^𝑠_𝑖+1≥𝑋^𝑠_𝑖 for all 𝑖=0,1,…; that is, the sequence {𝑋^𝑠_𝑛} is monotonically increasing. Obviously 𝑋^𝑠₀=𝛾𝑄≤𝛽₂𝑄≤𝛼𝑄. We suppose that 𝑋^𝑠_𝑖≤𝛼𝑄; from the first inequality in (18), we obtain 𝑋^𝑠_𝑖+1=𝑄−𝐴^∗𝑋_𝑖^−𝑡𝐴≤𝑄−𝐴^∗(𝛼𝑄)^−𝑡/𝑠𝐴≤𝛼𝑄. (21)
Hence, {𝑋^𝑠_𝑛} converges to a positive operator solution 𝑋^𝑠_𝛾 to (1). Since 𝑋_𝑖∈[^𝑠𝛾𝑄,^𝑠𝛼𝑄] for all 𝑖=0,1,…, we have 𝑋_𝛾∈[^𝑠𝛾𝑄,^𝑠𝛼𝑄].
(ii) Let 𝛾∈[𝛼,1]. From the proving procession of Theorem 6, the sequence (12) is monotonically decreasing. Since 𝑋₀=^𝑠𝛾𝑄≥^𝑠𝛽₂𝑄 and supposing that 𝑋^𝑠_𝑖≥𝛽₂𝑄, it follows that 𝑋^𝑠_𝑖+1=𝑄−𝐴^∗𝑋_𝑖^−𝑡𝐴≥𝑄−𝐴^∗𝛽₂𝑄^−𝑡/𝑠𝐴=𝑄^1/2𝐼−𝛽₂^−𝑡/𝑠𝑄^−1/2𝐴^∗𝑄^−𝑡/𝑠𝐴𝑄^−1/2𝑄^1/2≥𝑄^1/2𝐼−𝛽₂^−𝑡/𝑠𝜆_max𝑇^∗𝑇𝐼𝑄^1/2=𝛽₂𝑄. (22) By induction principle, 𝑋^𝑠_𝑖≥𝛽₂𝑄 hold for each 𝑖=0,1,2,…. Hence, {𝑋^𝑠_𝑛} is convergent. From Theorem 6, {𝑋^𝑠_𝑛} converges to 𝑋^𝑠_𝐿 such that 𝑋_𝐿∈[^𝑠𝛽₂𝑄,^𝑠𝛾𝑄].
(iii) Considering the sequence {𝑋^𝑠_𝑛} in (12) for 𝛾∈(𝛽₂,𝛼), that is, 𝑋^𝑠₀∈(𝛽₂𝑄,𝛼𝑄) and supposing that 𝑋^𝑠_𝑖∈(𝛽₂𝑄,𝛼𝑄), then, for 𝑋^𝑠_𝑖+1, 𝑋^𝑠_𝑖+1=𝑄−𝐴^∗𝑋_𝑖^−𝑡𝐴<𝑄^1/2𝐼−𝛽₂^−𝑡/𝑠𝑄^−1/2𝐴^∗𝑄^−𝑡/𝑠𝐴𝑄^−1/2𝑄^1/2≤𝑋𝛼𝑄,_𝑠𝑖+1=𝑄−𝐴^∗𝑋_𝑖^−𝑡𝐴>𝑄^1/2𝐼−𝛼^−𝑡/𝑠𝑄^−1/2𝐴^∗𝑄^−𝑡/𝑠𝐴𝑄^−1/2𝑄^1/2≥𝛽₂𝑄. (23) Hence, 𝑋^𝑠_𝑖∈(𝛽₂𝑄,𝛾𝑄), for all 𝑖=0,1,2,…. Now, we considering ‖𝑋^𝑠_𝑖+1−𝑋^𝑠_𝑖‖, 𝑋^𝑠_𝑖+1−𝑋^𝑠_𝑖=𝑄−𝐴^∗𝑋_𝑖^−𝑡𝐴−𝑄−𝐴^∗𝑋^−𝑡_𝑖−1𝐴≤1𝑠_𝑠1𝛽₂𝜆_min(𝑄)^𝑠−1×𝑡‖𝐴‖²_𝑠𝛽₂𝜆_min(𝑄)^𝑡+1𝑋_𝑖−𝑋_𝑖−1=𝑡𝑠‖𝐴‖²_𝑠1𝛽₂𝜆_min(𝑄)^𝑡+𝑠×𝑋_𝑖−𝑋_𝑖−1, (24) and (𝑡/𝑠)‖𝐴‖²≤(𝛽₂𝜆_min(𝑄))^{(𝑠+𝑡)/𝑠}, it follows that {𝑋_𝑛} in (12) is a Cauchy sequence in the Banach space [^𝑠𝛽₂𝑄,^𝑠𝛼𝑄]. Hence, it has a limit 𝑋_𝛾 in [^𝑠𝛽₂𝑄,^𝑠𝛼𝑄], and 𝑋_𝛾 is a unique solution to (1) in [^𝑠𝛽₂𝑄,^𝑠𝛼𝑄]. According to Theorem 6, it is the maximal solution, that is, 𝑋_𝛾=𝑋_𝐿. 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.