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Spectral Analysis of the Bounded Linear Operator in the Reproducing Kernel Space
We first introduce some related definitions of the bounded linear operator in the reproducing kernel space . Then we show spectral analysis of and derive several property theorems.
It is well known that spectral analysis of linear operators  is an important topic in functional analysis. For example, the matrix eigenvalue in linear algebra and eigenvalue problems for differential equation have been discussed emphatically. Two major reasons are as follows. Firstly, spectral analysis arises from vibration frequency problems and the stability theory of system. Secondly, spectral analysis comes from the need of discussing the structure of the operator and solving the corresponding equation by using the eigenvalue and spectral theorem. Also, spectral analysis can be used to study the structure of the solution for homogeneous or nonhomogeneous differential system and the normalized form of matrix which can be obtained clearly by matrix eigenvalues.
So far, the spectral decomposition method  has become a central topic in the theory of spectral analysis of linear operators. This method has been successfully applied in Hilbert space and perfect spectral decomposition theorem . In recent years, the spectral decomposition method has been developed into spectral theorems of spectral operators and decomposable operators in Banach space [4, 5].
To our knowledge, reproducing kernel space has been applied in many fields, such as linear systems [6–8], nonlinear systems [9–11], operator equation, stochastic processes, wavelet transform, signal analysis, and pattern recognition [12–17]. Since the reproducing kernel space is a Hilbert space, this paper will apply the theory of spectral analysis for linear operator in the reproducing kernel space and derive some useful conclusions.
The paper is organized as follows. In Section 2, we introduce some related definitions for the eigenvalue of the bounded linear operator in the reproducing kernel space . In Section 3, the regular point and the spectral point of bounded linear operator in are given. In Section 4, we show spectral analysis of the bounded linear operator and also establish several theorems. Section 5 ends this paper with a brief conclusion.
2. Related Definitions
Definition 1. Let be an abstract set, the reproducing kernel space, and the bounded linear operator space. with , if there exists nonvanishing vector , such that Then is called an eigenvalue of and is called the eigenvector of according to , where denotes the identity operator.
Definition 2. and all eigenvectors and zero vector of compose the eigenvector space which is denoted by .
Obviously, is a linear closed subspace of .
Definition 3. Denote the dimension of by ; it is called the multiplicity of eigenvalue . That is, is the number of vectors of maximum linear independence.
Example 4. Let be a binary function on , , with then is the eigenvalue of if and only if the following integral equation has nonzero solution: If , is a linear independence vector system, then (3) can be converted into the equivalent equation Thus we have the following results.(a)If , then (4) has nonzero solution if and only if and , . It follows that, for the eigenvalue of , the eigenvector space is infinite-dimensional.(b)If , then the solution of (4) can be denoted by where () are constants.
Summing up the above results, we can see that eigenvalues of (3) and (5) are equivalent, where () are undetermined coefficients. In addition, in order to solve the eigenvector, we just need to solve () in (5).
3. Regular Point and Spectral Point
In Definitions 1–3, we introduce the eigenvalue, eigenvector, eigenvector space, and of for the homogeneous equation (1). However, for many problems in mathematics and physics, we just need to solve the following nonhomogeneous equation: where is a given operator, is a given vector, and is an unknown vector. In order to discuss this problem, we need to introduce the following definitions and theorems.
Definition 5. Let , , , and , where denotes the domain of and denotes the range of values of . If the inverse operator of exists and is linearly bounded, then is called a regular operator.
Let be a linear operator; and ; if exists, then and , where and are, respectively, identity operators of subspace and . Inversely, if there exists a linear operator , such that and , then exists and . In fact, ; if , then . Hence, is invertible. Since , then ; we have such that . That is, .
Summing up the above disscusion, . Hence, we have . Particularly, when and is a bounded linear operator, we can derive the following results.
Theorem 6. Let , ; then is a regular operator if and only if , such that and .
Theorem 7. Let , ; if is a regular operator, then is also a regular operator and .
Proof. Since , , and by taking conjugate on both sides of the above formulas, we obtain In view of Theorem 6, we can see that is a regular operator and . The proof is complete.
Definition 8. Let , , ; denotes the complex number field.(1) is a regular operator, that is, is a one-to-one linear operator from to . In addition, the inverse operator is a linear bounded operator. Then is called a regular point of . All regular points compose the regular set of , which is denoted by .(2)If is not a regular point, then is called a spectral point of . All spectral points compose the spectral set of , which is denoted by .
In view of Definition 8, we have . Then we have the following property results.
Lemma 9. Let be a bounded linear operator in reproducing kernel space , ; then is a regular point of if and only if , there exists a solution of , which satisfies , where m is a positive constant.
Proof. Since , then , such that . In addition, in view of the boundedness of and the Cauchy-Schwartz inequality, we have
Let ; then .
Since , we have . Next, we will prove that is one-to-one. In fact, ; if , , then namely, the image of is 0. Hence, ; that is, . Therefore, is one-to-one and exists. Furthermore, since , we have that is, Hence, exists and is a bounded linear operator. Summing up the above, is a regular operator, where is a regular point of .
Lemma 9 shows that ; when is a continuous linear operator and is the regular point of , has a unique solution . Furthermore, the continuity of depends on the right term. In other words, if are column vectors and , then .
Lemma 10. Let be a bounded linear operator in the reproducing kernel space , . If is not the eigenvalue of and , one has , . That is, is invertible.
Proof. Otherwise, the invertible operator can convert the nonvanishing vector to nonvanishing vector. Hence, there exists such that . That is, is not the eigenvalue of .
When is a finite dimension space and is not the eigenvalue of , we can derive that is an invertible mapping. Obviously, . In fact, let be the basis of ; then is a linear independent system in and also a basis of . Therefore, . In view of the inverse operator Theorem, is bounded. It follows that .
So, the proof of the theorem is complete.
Lemma 10 shows that regular point and spectral point are absolutely opposite for finite dimension normed spaces. That is, spectral point of can only be an eigenvalue in finite dimension normed space. This is entirely consistent with the conclusion of the theory of linear algebra. But if is an infinite-dimensional space and is not the eigenvalue of , then may not be a regular point of , so far as is not a map from to .
For example, let , has only zero solutions. Hence, has not eigenvalue. That is, zero is not the eigenvalue. However, the range of values is all functions of the from for . This shows that the spectral point is complex in infinite-dimensional space for the operator .
Now, we will classify the spectral set by three situations.(a)If is not one-to-one, then is called point spectral of ; the set of point spectral is denoted by .(b)If is one-to-one and is dense in , then is called continuous spectral of ; the set of continuous spectral is denoted by .(c)If is one-to-one and is not dense in , then is called residual spectral of ; the set of residual spectral is denoted by .
Obviously, , , and are mutually disjoint sets and .
4. Spectral Analysis
Let , , , , , , such that ; that is, . In view of the completeness of , there exists , such that Hence, converges in the sense of and the limit is denoted by .
Let ; then For , , we have If , then Namely, and .
For , one obtains
Summing up the above parts, we have the following theorems.
Theorem 11. Let , , then one has the following.(1)Consider .(2)Consider .(3)When , .
Theorem 12. Let , ; if , then one has the following.(1) if and only if is a regular point of .(2)When , .(3)When , .
Proof. , since , if and only if . Replacing by in Theorem 11, we have
Furthermore, we have
Then we obtain
It follows that is a regular point of and with .
In addition, when , we have . In view of (2) of Theorem 11, we have The proof is complete.
Theorem 13. Let , ; then one has the following.(1) is an open set.(2)When is nonempty, ; if , then is a regular point of and , where .(3) is a closed set.(4)Consider .
Proof. (1) If , the conclusion is obvious. If , then
where is a bounded linear operator in the reproducing kernel space . We use instead of in Theorem 11, such that
That is, when , exists and is bounded. Hence, when , ; that is, is an open set.
(2) If is nonempty, , let . In view of (2) of Theorem 11 and (1) of Theorem 13, we have
(3) Since and (1), is a closed set.
(4) In view of (1) of Theorem 12 and , we have , which means that .
The proof is complete.
Definition 14. Let , , ; is called the spectral radius of .
From the purpose of solving equations, spectral radius has the following meanings.(1)For , due to the fact that is a regular point of , then for any , has a unique solution .(2)For , it cannot guarantee this equation has a solution for any . In many practical problems, in order to calculate the spectral range, one needs to estimate the spectral radius. In terms of (4) of Theorem 13, we can get . In practical terms, this estimate is convenient, but it is imprecise.
Theorem 15. Let , ; then .
Proof. In terms of (4) of Theorem 13, . Hence, one only needs to prove that . For , one obtains
Consider , . If , then is a regular point of . In addition, since , then Laurent expansions of are established, where .
Let , ; we have Let , ; then
In terms of the resonance Theorem, must be bounded. It follows that there exists a positive constant , such that and . Namely, Let ; then The proof is complete.
Theorem 16. Let , ; then .
Proof. If , in view of the properties of the reproducing kernel space, ; hence, , where the unit element is denoted by . In terms of the functional extension Theorem, , such that . In addition, and ; when , we have
in terms of the assumption that and Theorem 15; when , one obtains
Therefore, when , we have
That is, is bounded. In terms of the Liouville Theorem, must be a constant, so we have .
The proof is complete.
Definition 17. If , , , then is called a generalized nilpotent operator.
Definition 17 is the finite-dimensional space concept nilpotent operator in the infinite-dimensional space to promote. In the spectral theory of operators, generalized nilpotent operator is a kind of important operator.
In terms of Theorem 16 and the spectral radius theorem, one can obtain that the generalized nilpotent operator has only a spectral point . For example, let , , , In terms of the property of , one obtains Note that we have This shows that is a generalized nilpotent operator; spectral point is not the eigenvalue of .
Definition 18. Let , , ; if there exists , such that , then is called an approximate spectral point. All approximate spectral points are denoted by ; the other spectral point is called remainder spectral point. All the remainder spectral points are denoted by .
Theorem 19. Let , ; then(1),(2), and ,(3) is an open set,(4), where denotes the boundary of ,(5) is a nonempty closed set.
Proof. (1) If , then there exists nonzero element of , such that
Without loss of the generality, let ; we choose , , and then and ; namely, ; this shows that .
(2) In terms of Definition 18, one obtains and .
(3) If , then . Hence, , such that When , , we have It shows that for any which satisfies it is impossible to be an approximate spectral point of . Hence, if one can prove that when , is not a regular point of , then . That is, is an inner point of , so is an open set.
Now, we prove that when , . But not vice versa, , ; then . Note that ; if , then In view of (2) of Theorem 13, let be a regular point of ; if , then In a particular case, let ; note that then . This is a contradiction with . It follows that is an open set.
(4) Since is a closed set, when , . In addition, , , so we have .
(5) Since , then is a closed set. Furthermore, since , , this shows that and are all nonempty sets.
The proof is complete.
This paper first introduces the eigenvalue, eigenvector, eigenvector space, and of the bounded linear operator in the reproducing kernel space . Then we show some definitions and properties of the regular operator. The regular set and spectral set of bounded linear operator are also introduced. From the solvability of the equation, we show the spectral classification and give three conditions. Finally, we introduce the spectral analysis of the bounded linear operator . It includes the definitions of spectral radius, nilpotent operator, approximate spectral point, and remainder spectral point. We also establish some property theorems of the bounded linear operator in the reproducing kernel space .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
This work is partially supported by the National Science Foundation of China (11271100, 11301113, and 71303067), Harbin Science and Technology Innovative Talents Project of Special Fund (2013RFXYJ044), China Postdoctoral Science Foundation funded Project (Grant no. 2013M541400), the Heilongjiang Postdoctoral Fund (Grant no. LBH-Z12102), and the Fundamental Research Funds for the Central Universities (Grant no. HIT. HSS. 201201).
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