Calculation of the Reproducing Kernel on the Reproducing Kernel Space with Weighted Integral

Abstract

We provide a new definition for reproducing kernel space with weighted integral and present a method to construct and calculate the reproducing kernel for the space. The new reproducing kernel space is an enlarged reproducing kernel space, which contains the traditional reproducing kernel space. The proposed method of this paper is a universal method and is suitable for the case of that the weight is variable. Obviously, this new method will generalize a number of applications of reproducing kernel theory to many areas.

1. Introduction

A reproducing kernel is a basic tool for studying the spline interpolation of differential operators and is also the base of the reproducing kernel method, which were widely used in numerical analysis, genetic models, pattern analysis, and so forth. The concept of reproducing kernel is derived from the study of the integration equation, and paper [1] studied specially the reproducing kernels and presented its primary theory. From then on, the reproducing kernel theory and the reproducing kernel method have been studied by many authors [2–7].

Let 𝑊^𝑚2[0,𝑇] denote the function space on a finite interval [0,𝑇], 𝑊^𝑚2[0,𝑇] = {𝑓(𝑡),𝑡∈[0,𝑇],𝑓^(𝑚−1)(𝑡)∈𝐿²[0,𝑇]}, and this space becomes a reproducing kernel Hilbert space (RKHS) if we endow it with some inner product. This kind of the reproducing kernel space is the most popular space for solving the boundary value problems using reproducing kernel method. But in paper [8], the author firstly considered the reproducing kernel space with weighted integral 𝑊^22,𝜌[0,𝑇]={𝑢(𝑡),𝑡∈[0,𝑇],𝑢′(𝑡) is an absolute continuous real-valued function on [0,𝑇], ∫^𝑇0√𝑡(𝑢^(𝑡))²𝑑𝑡<+∞} and used it solving Volterra integral equation with weakly singular kernel. It is obvious that 𝑊²²[0,𝑇]⊆𝑊^22,𝜌[0,𝑇] and 𝑊^22,𝜌[0,𝑇] will be more widely applied.

In this paper, we are concerned with the reproducing kernel space with weighted integral 𝑊^𝑚2,𝛼[0,𝑇]={𝑢(𝑡),𝑡∈[0,𝑇],𝑢′(𝑡),…,𝑢^(𝑚−1) are absolute continuous real-valued functions on [0,𝑇],∫^𝑇0𝑡^𝛼(𝑢^(𝑚)(𝑡))²𝑑𝑡<+∞}, where 𝛼 is a constant and satisfies 0≤𝛼<1 (when 𝛼=0, 𝑊^𝑚2,𝛼[0,𝑇] is 𝑊^𝑚2[0,𝑇]). The method for computing the corresponding reproducing kernel is given.

2. Preliminaries

In order to get the main results of the paper, we introduce the method of Zhang for calculating the reproducing kernel of 𝑊^𝑚2[𝑎,𝑏] in a nutshell in this section.

The method of Zhang has very powerful system modeling capability. The idea is coming from the relationship between the Green function with reproducing kernel.

Set 𝐿=𝐷^𝑚+𝑎_𝑚−1𝐷^𝑚−1+⋯+𝑎₁𝐷+𝑎₀(𝑡),𝑡∈[𝑎,𝑏], where 𝑎_𝑗(𝑡)∈𝐶^𝑗[𝑎,𝑏] and Ker𝐿 = {𝑓∈𝑊^𝑚2[𝑎,𝑏]∶𝐿𝑓=0}.

Definition 2.1. 𝜑₁(𝑡),…,𝜑_𝑚(𝑡) are the basis in Ker𝐿. The 𝑖-th row of Wronskian matrix is (𝜑^1(𝑖−1)(𝑡),…,𝜑^{𝑚(𝑖−1)}(𝑡)), and the last line of its inverse matrix is (𝜑₁(𝑡),…,𝜑_𝑚(𝑡)). Call 𝜑₁(𝑡),…,𝜑_𝑚(𝑡) are the adjunct functions of 𝜑₁(𝑡),…,𝜑_𝑚(𝑡).

Lemma 2.2. Assume 𝑔(𝑡,𝜏)=^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖̃(𝜏)(𝑡−𝜏)⁰⁺(2.1) and 𝛾₁,…,𝛾_𝑚 is a system of linear independent functions in Ker𝐿 and satisfies 𝛾_𝑘^𝑏𝑎𝑔(⋅,𝜏)𝑢(𝜏)𝑑𝜏=^𝑏𝑎𝛾_𝑘𝑔(⋅,𝜏)𝑢(𝜏)𝑑𝜏,(2.2) where 𝑒₁(𝑡),…,𝑒_𝑚(𝑡) are the dual basis of Ker𝐿 relative to 𝛾₁,…,𝛾_𝑚, and 𝑒₁̃(𝑡),…,𝑒_𝑚̃(𝑡) are the adjunct functions of 𝑒₁(𝑡),…,𝑒_𝑚(𝑡). Then for any functions 𝑓∈𝑊^𝑚2[𝑎,𝑏], they satisfy the form 𝑓(𝑡)=^𝑚_𝑖=1𝛾_𝑖𝑓𝑒_𝑖(𝑡)+^𝑏𝑎𝐺(𝑡,𝜏)⋅𝐿𝑓(𝜏)𝑑𝜏,(2.3) where 𝐺(𝑡,𝜏) is defined below 𝐺(𝑡,𝜏)=𝑔(𝑡,𝜏)−^𝑚_𝑖=1𝛾_𝑖𝑔(⋅,𝜏)𝑒_𝑖(𝑡)(2.4) and the expression is exclusive.

Lemma 2.3. 𝐿 is a linear differential operator. Assume 𝛾₁,…,𝛾_𝑚 are linear independent functions in Ker𝐿, satisfying (2.2). Then 𝑊^𝑚2[𝑎,𝑏] is a Hilbert space if the inner product is defined by the following form: (𝑓,ℎ)=^𝑚_𝑖=1𝛾_𝑖𝑓𝛾_𝑖ℎ+^𝑏𝑎𝐿𝑓(𝑡)𝐿ℎ(𝑡)𝑑𝑡,𝑓,ℎ∈𝑊^𝑚2[𝑎,𝑏].(2.5)

Lemma 2.4. Under the assumptions of Lemma 2.2 and the inner product (2.5), the Hilbert space 𝑊^𝑚2[𝑎,𝑏] is reproducing kernel Hilbert space with the reproducing kernel can be denoted by 𝐾(𝑡,𝜏)=^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏)+^𝑏𝑎𝐺(𝑡,𝑥)𝐺(𝜏,𝑥)𝑑𝑥.(2.6)

3. The New Method for Computing the Reproducing Kernel

It is known that the reproducing kernel of a reproducing kernel Hilbert space is existence and uniqueness. The reproducing kernel 𝐾 of a Hilbert space 𝐻 completely determines the space 𝐻.

This section discusses the method of calculating reproducing kernels for the following two cases. The first case is when the weight is constant. In the second case we deal with the general space 𝑊^𝑚2,𝛼[0,𝑇], where 0≤𝛼<1, and the result of this part is the main result of this paper.

3.1. Case 1: The Weight Is Constant

For general space 𝑊^𝑚2[0,𝑇], let 𝐿=𝐷^𝑚 be the linear differential operator of order 𝑚, and let 𝜆₁,𝜆₂,…,𝜆_𝑚 be the linear independent functions on Ker𝐿, where Ker𝐿 is defined by Ker𝐿={𝑓∈𝑊^𝑚2[0,𝑇],𝐿𝑓=0}. Let 𝑒₁,𝑒₂,…,𝑒_𝑚 be the dual basis of Ker𝐿 relative to 𝜆₁,𝜆₂,…,𝜆_𝑚. That means 𝐿𝑒_𝑖=0,𝜆_𝑖𝑒_𝑗=𝛿_𝑖𝑗,𝑖,𝑗=1,2,…,𝑚.(3.1) Let 𝐺 be the Green's function of 𝐿 and satisfy 𝐿_𝑡𝐺(𝑡,𝑠)=𝛿(𝑡−𝑠),𝜆_𝑖𝐺(⋅,𝑠)=0,𝑖=1,…,𝑚.(3.2) By the Lemma 2.4, 𝑊^𝑚2[0,𝑇] is a reproducing kernel Hilbert space if the inner product is defined by the following form: (𝑓,ℎ)₁=^𝑚_𝑖=1𝜆_𝑖𝑓𝜆_𝑖ℎ+^𝑇0𝐿𝑓(𝑡)𝐿ℎ(𝑡)𝑑𝑡,𝑓,ℎ∈𝑊^𝑚2[0,𝑇](3.3) and the reproducing kernel is 𝐾₁(𝑡,𝜏)=^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏)+^𝑇0𝐺(𝑡,𝑥)𝐺(𝜏,𝑥)𝑑𝑥.(3.4)

Let (𝑓,ℎ)₂=^𝑚_𝑖=1𝑎_𝑖𝜆_𝑖𝑓𝜆_𝑖ℎ+^𝑇0𝑏𝐿𝑓(𝑡)𝐿ℎ(𝑡)𝑑𝑡,𝑓,ℎ∈𝑊^𝑚2[0,𝑇],(3.5) where both 𝑎₁,𝑎₂,…,𝑎_𝑚 and 𝑏 are positive real numbers. The following proposition holds.

Theorem 3.1. Using the above hypothesis, 𝑊^𝑚2[0,𝑇] is a reproducing kernel Hilbert space if it has been endowed with the inner product (3.5) and the reproducing kernel is 𝐾₂(𝑡,𝜏)=^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏)𝑎_𝑖+^𝑇01𝑏𝐺(𝑡,𝑥)𝐺(𝜏,𝑥)𝑑𝑥.(3.6)

Proof. Let 𝐿=√𝑏𝐿=√𝑏𝐷^𝑚, and 𝜆̃_𝑖𝑓=√𝑎_𝑖𝜆_𝑖𝑓.
It is obvious that 𝜆̃₁,𝜆̃₂,…,𝜆̃_𝑚 are also the linear independent functions on Ker𝐿. From Lemma 2.3, we have that 𝑊^𝑚2[0,𝑇] is a Hilbert space if the inner product is defined by (𝑓,ℎ)_𝐿=^𝑚_𝑖=1𝜆̃_𝑖=𝑓𝜆̃_𝑖ℎ+^𝑇0𝐿𝑓(𝑡)𝐿ℎ(𝑡)𝑑𝑡_𝑚_𝑖=1𝑎_𝑖𝜆_𝑖𝑓𝜆_𝑖ℎ+^𝑇0𝑏𝐿𝑓(𝑡)𝐿ℎ(𝑡)𝑑𝑡=(𝑓,ℎ)₂.(3.7) Next, we will proof 𝐾₂(𝑡,𝑠) is the reproducing kernel of the space 𝑊^𝑚2[0,𝑇] with the inner product (⋅,⋅)₂.
𝐾₁(𝑡,𝑠) is the reproducing kernel of the space 𝑊^𝑚2[0,𝑇] with the inner product (⋅,⋅)₁. In particular, 𝐾₁(𝑡,𝑠) is contained in 𝑊^𝑚2[0,𝑇]. So 𝐾₂(𝑡,𝑠) is also contained in 𝑊^𝑚2[0,𝑇].
For any 𝑓∈𝑊^𝑚2[0,𝑇], 𝑓(𝑠),𝐾₂(𝑡,𝑠)₂=^𝑚_𝑖=1𝑎_𝑖𝜆_𝑖𝑓𝜆_𝑖𝐾₂(𝑡,⋅)+^𝑇0𝑏𝑓^(𝑚)(𝑠)𝜕^𝑚𝜕𝑠^𝑚𝐾₂(𝑡,𝑠)𝑑𝑠.(3.8) From (3.1) and (3.2), we have 𝑓(𝜏),𝐾₂(𝑡,𝜏)₂=^𝑚_𝑖=1𝑎_𝑖𝜆_𝑖𝑓𝜆_𝑖^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏)𝑎_𝑖+^𝑇0𝑏𝑓^(𝑚)(𝜏)𝜕^𝑚𝜕𝜏^𝑚^𝑇01𝑏𝐺(𝑡,𝑥)=𝐺(𝜏,𝑥)𝑑𝑥𝑑𝜏_𝑚_𝑖=1𝜆_𝑖𝑓𝜆_𝑖^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏)+^𝑇0𝑓^(𝑚)(𝜏)𝜕^𝑚𝜕𝜏^𝑚^𝑇0𝐺(𝑡,𝑥)𝐺(𝜏,𝑥)𝑑𝑥𝑑𝜏.(3.9) Similarly, from (3.1) and (3.2), we obtain 𝑓(𝑡)=𝑓(𝜏),𝐾₁(𝑡,𝜏)₁=^𝑚_𝑖=1𝜆_𝑖𝑓𝜆_𝑖^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏)+^𝑇0𝑓^(𝑚)(𝑠)𝜕^𝑚𝜕𝜏^𝑚^𝑇0𝐺(𝑡,𝑥)𝐺(𝜏,𝑥)𝑑𝑥𝑑𝜏.(3.10) So 𝑓(𝑡)=(𝑓(𝜏),𝐾₂(𝑡,𝜏))₂ holds.
The proof is complete.

3.2. Case 2: The Weight Is Variable

In this case, we construct the inner product of the space 𝑊^𝑚2,𝛼[0,𝑇], and calculate the corresponding reproducing kernel.

Define 𝐿₁=𝑡^𝛼/2𝐷^𝑚. From the definition of the space 𝑊^𝑚2,𝛼[0,𝑇], we know that 𝐿₁∶𝑊^𝑚2,𝛼[0,𝑇]→𝐿²[0,𝑇] (mapping the space 𝑊^𝑚2,𝛼[0,𝑇] to the square integrable space on [0,𝑇]).

Under the hypothesis of Case 1, 𝜆₁,𝜆₂,…,𝜆_𝑚 are also the linear independent functions on Ker𝐿₁, where Ker𝐿₁ is defined by Ker𝐿₁={𝑓∈𝑊^𝑚2,𝛼[0,𝑇],𝐿₁𝑓=0}, and 𝑒₁,𝑒₂,…,𝑒_𝑚 is also the dual basis of Ker𝐿₁ relative to 𝜆₁,𝜆₂,…,𝜆_𝑚. 𝑒₁̃(𝑡),…,𝑒_𝑚̃(𝑡) are the adjunct functions of 𝑒₁(𝑡),…,𝑒_𝑚(𝑡).

Similar to Case 1, define an algorithm (⋅,⋅)₃ as the following form (𝑓,ℎ)₃=^𝑚_𝑖=1𝜆_𝑖𝑓𝜆_𝑖ℎ+^𝑇0𝐿₁𝑓(𝑡)𝐿₁ℎ(𝑡)𝑑𝑡,𝑓,ℎ∈𝑊^𝑚2,𝛼[0,𝑇](3.11)

Theorem 3.2. Under the above assumption, (⋅,⋅)₃ is the inner product of the space 𝑊^𝑚2,𝛼[0,𝑇].

If act in accordance with the four basic rules of the inner product, the proof of this proposition is easy. So one overlaps the proof.

Divide the space 𝑊^𝑚2,𝛼[0,𝑇] into two parts Ker𝐿₁ and (Ker𝐿₁)^⟂, where Ker𝐿₁ is the linear space of order 𝑚. From the results in [9], one has the following proposition.

Proposition 3.3. Under the above assumption, Ker𝐿₁ is the reproducing kernel Hilbert space with the inner product below (𝑓,ℎ)₄=^𝑚_𝑖=1𝜆_𝑖𝑓𝜆_𝑖ℎ𝑓,ℎ∈Ker𝐿₁(3.12) and the corresponding reproducing kernel is 𝐾₄(𝑡,𝜏)=^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏).(3.13)

Let 𝑔₁(𝑡,𝜏)=1𝜏^𝑚𝛼/2_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖̃(𝜏)(𝑡−𝜏)⁰⁺=⎧⎪⎨⎪⎩𝜏1^𝑚𝛼/2𝐺0,∑_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖̃(𝜏),𝑡𝑡<≥𝜏𝜏,₁,(𝑡,𝜏)=𝑔₁(𝑡,𝜏)−^𝑚_𝑖=1𝜆_𝑖𝑔₁(⋅,𝜏)𝑒_𝑖(𝑡).(3.14)

It is obvious that 𝐺₁(𝑡,𝜏)=(1/𝜏^𝛼/2)𝐺(𝑡,𝜏).

The following theorem holds.

Theorem 3.4. 𝐺₁ is the Green's function of 𝐿₁ and for any 𝑓∈𝑊^𝑚2[0,𝑇], 𝑢(𝑡)=𝐿₁𝑓(𝑡), satisfies 𝐿_1𝑡^𝑇0𝜆𝐺₁(𝑡,𝜏)𝑢(𝜏)𝑑𝜏=𝑢(𝑡),_𝑖𝐺₁(⋅,𝜏)=0,𝑖=1,2,…,𝑚.(3.15)

Proof. For any 𝑖=1,2,⋅,𝑚, we have 𝜆_𝑖𝐺₁(⋅,𝜏)=𝜏^𝛼/2𝜆_𝑖𝐺(⋅,𝜏).(3.16) From (3.2), 𝜆_𝑖𝐺₁(⋅,𝜏)=𝜏^𝛼/2𝜆_𝑖𝐺(⋅,𝜏)=0,𝑖=1,2,⋅,𝑚.(3.17) Then from the results in [9, 10], for any 𝑓∈𝑊^𝑚2[0,𝑇], we obtain 𝐿_𝑡^𝑇0𝐺(𝑡,𝜏)𝐿𝑓(𝜏)𝑑𝜏=𝐿𝑓.(3.18) So 𝐿_1𝑡^𝑇0𝐺₁(𝑡,𝜏)𝑢(𝜏)𝑑𝜏=𝑡^𝛼/2𝐿_𝑡^𝑇01𝜏^𝛼/2=𝑡_𝛼/2𝐺(𝑡,𝜏)𝜏^𝛼/2𝐿𝑓(𝜏)𝑑𝜏𝐿𝑓(𝑡)=𝐿₁𝑓(𝑡)=𝑢(𝑡).(3.19)
The proof is complete.

Remark 3.5. If acting in accordance with the process of the paper [9], we have 𝐾₃(𝑡,𝜏)=^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏)+^𝑇0𝐺₁(𝑡,𝑥)𝐺₁(𝜏,𝑥)𝑑𝑥.(3.20) But 𝐾₃(𝑡,𝜏) is not the reproducing kernel of 𝑊^𝑚2[0,𝑇], since 𝐾₃(𝑡,𝜏)∉𝑊^𝑚2[0,𝑇].

Now, we will give an important property of the arbitrary element of 𝑊^𝑚2,𝛼[0,𝑇].

Theorem 3.6. For any 𝑓∈𝑊^𝑚2,𝛼[0,𝑇], 𝐿₁𝑓(𝑡)=𝑢(𝑡). Then there are some real constant 𝑐₁,𝑐₂,…,𝑐_𝑚, satisfying 𝑓(𝑡)=^𝑚_𝑖=1𝑐_𝑖𝑒_𝑖(𝑡)+^𝑇0^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖̃(𝜏)𝑢(𝜏)𝑑𝜏,(3.21) and the expression is exclusive.

Proof. 𝐿₁=𝑡^𝛼/2𝐷^𝑚 is a linear mapping, and 𝐿₁∶𝑊^𝑚2,𝛼[0,𝑇]→𝐿²[0,𝑇] is a homomorphic mapping. For any 𝑢(𝑡)∈𝐿²[0,𝑇], we have a function ℎ(𝑥) satisfies ℎ(𝑡)=^𝑇0^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖̃(𝜏)𝑢(𝜏)𝑑𝜏.(3.22) Because 𝐿₁ℎ(𝑡)=𝑢(𝑡)∈𝐿²[0,𝑇], ℎ(𝑡)∈𝑊^𝑚2,𝛼[0,𝑇] holds. So 𝐿₁∶𝑊^𝑚2,𝛼[0,𝑇]→𝐿²[0,𝑇] is a surjective homomorphism.
Ker𝐿₁=Span{𝑒₁,𝑒₂,…,𝑒_𝑚} is a linear system and the dimension of the system is 𝑚. So from the knowledge of the group homomorphism, we have 𝐿₁∶𝑊^𝑚2,𝛼[0,𝑇]/Ker𝐿₁→𝐿²[0,𝑇](3.23) is isomorphic.
On the one hand for any ℎ(𝑡)∈𝑊^𝑚2,𝛼[0,𝑇]/Ker𝐿₁, there exists the exclusive 𝑢(𝑡), satisfying 𝐿₁ℎ(𝑡)=𝑢(𝑡). On the other hand for the 𝑢(𝑡)∈𝐿²[0,𝑇], ℎ₀(𝑡)=∫^𝑇0(∑^𝑚𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖̃(𝜏))𝑢(𝜏)𝑑𝜏 satisfies ℎ₀(𝑡)∈𝑊^𝑚2,𝛼[0,𝑇]/Ker𝐿₁ and 𝐿₁ℎ₀(𝑡)=𝑢(𝑡). So for ℎ(𝑡)∈𝑊^𝑚2,𝛼[0,𝑇]/Ker𝐿₁, that holds ℎ(𝑡)=^𝑇0^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖̃(𝜏)𝐿₁ℎ(𝜏)𝑑𝜏.(3.24)
For any 𝑓(𝑡)∈𝑊^𝑚2,𝛼[0,𝑇], 𝑓(𝑡)=𝑓₀(𝑡)+ℎ(𝑡) holds, where 𝑓₀(𝑡)∈Ker𝐿₁ and ℎ(𝑡)∈𝑊^𝑚2,𝛼[0,𝑇]/Ker𝐿₁. At the same time, the decomposition is exclusive because of the orthogonality between Ker𝐿₁ and 𝑊^𝑚2,𝛼[0,𝑇]/Ker𝐿₁.
Furthermore, Ker𝐿₁=Span{𝑒₁,𝑒₂,…,𝑒_𝑚}, so 𝑓₀(𝑡)=∑^𝑚𝑖=1𝑐_𝑖𝑒_𝑖(𝑡), where 𝑐₁,𝑐₂,…,𝑐_𝑚 are real numbers, and the expression is exclusive.
So for any 𝑓(𝑡)∈𝑊^𝑚2,𝛼[0,𝑇], we have 𝑓(𝑡)=𝑓₀(𝑡)+ℎ(𝑡)=^𝑚_𝑖=1𝑐_𝑖𝑒_𝑖(𝑡)+^𝑇0^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖̃(𝜏)𝑢(𝜏)𝑑𝜏,(3.25)
where 𝑐₁,𝑐₂,…,𝑐_𝑚 are real numbers, and the the expression is exclusive.
The proof is complete.

Then similar to the Theorem 3.4, we have the following theorem.

Theorem 3.7. 𝐺₁ is the Green's function of 𝐿₁ and for any 𝑓∈𝑊^𝑚2,𝛼[0,𝑇], 𝐿₁𝑓(𝑡)=𝑢(𝑡), satisfies 𝐿_1𝑡^𝑇0𝐺₁(𝑡,𝜏)𝑢(𝜏)𝑑𝜏=𝑢(𝑡).(3.26)

Proof. For any 𝑓∈𝑊^𝑚2,𝛼[0,𝑇], 𝐿₁𝑓(𝑡)=𝑢(𝑡), 𝐿_1𝑡^𝑇0𝐺₁(𝑡,𝜏)𝑢(𝜏)𝑑𝜏=𝑡^𝛼/2𝐿_𝑡^𝑇01𝜏^𝛼/2=𝑡_𝛼/2𝐺(𝑡,𝜏)𝜏^𝛼/2𝐿𝑓(𝜏)𝑑𝜏𝐿_𝑡^𝑇0𝐺(𝑡,𝜏)𝐿𝑓(𝜏)𝑑𝜏.(3.27)
From Theorem 3.6, we have 𝐿𝑓(𝑡)=𝐿_𝑡^𝑇0𝐺(𝑡,𝜏)𝐿𝑓(𝜏)𝑑𝜏.(3.28)
Thus, we know that (3.26) is true.
The proof is complete.

Theorem 3.8. Under the above hypothesis and the inner product (⋅,⋅)₃, 𝑊^𝑚2,𝛼[0,𝑇] is the Hilbert space.

Proof. The norm of the space is denoted by ‖𝑢‖₃=√(𝑢,𝑢)₃, where 𝑢∈𝑊^𝑚2,𝛼[0,𝑇].
Suppose that {𝑓_𝑛}^∞𝑛=1 is a Cauchy sequence in 𝑊^𝑚2,𝛼[0,𝑇], that is, ‖‖𝑓_𝑛+𝑝−𝑓_𝑛‖‖²³→0(𝑛→∞).(3.29) From Theorems 3.6 and 3.1, we have ‖‖𝑓_𝑛+𝑝−𝑓_𝑛‖‖²³=^𝑚_𝑖=1𝜆_𝑖𝑓_𝑛+𝑝−𝜆_𝑖𝑓_𝑛²+^𝑇0𝐿₁𝑓_𝑛+𝑝(𝑡)−𝑓_𝑛(𝑡)²𝑑𝑡→0(𝑛→∞).(3.30) By the completeness of Ker𝐿₁ and 𝐿²[0,𝑇], there exist a real number 𝑟_𝑖, (𝑖=1,2,…,𝑚) and a real function ℎ∈𝐿²[0,𝑇], such that lliimm_{𝑛→∞𝑛→∞}𝜆_𝑖𝑓_𝑛=𝑟_𝑖,𝑖=1,2,…,𝑚,^𝑇0𝐿₁𝑓_𝑛(𝑡)−ℎ(𝑡)²=0.(3.31)
Set 𝑓₀(𝑡)=∑^𝑚𝑖=1𝑟_𝑖𝑒_𝑖(𝑡)+∫^𝑇0𝐺₁(𝑡,𝜏)ℎ(𝜏)𝑑𝜏. It follows that 𝑓₀∈𝑊^𝑚2,𝛼[0,𝑇] and ‖‖𝑓_𝑛−𝑓₀‖‖²³→0(𝑛→∞).(3.32) So 𝑊^𝑚2,𝛼[0,𝑇] is complete. Namely, 𝑊^𝑚2,𝛼[0,𝑇] is Hilbert space.
The proof is complete.

Theorem 3.9. Under the above hypothesis and the inner product (⋅,⋅)₃, 𝑊^𝑚2,𝛼[0,𝑇] is the reproducing kernel Hilbert space, and the reproducing kernel is 𝐾₃(𝑡,𝜏)=^𝑚_𝑖=1𝑒_𝑖(𝑡)𝑒_𝑖(𝜏)+^𝑇0𝐺₁(𝑡,𝑥)𝐺₁(𝜏,𝑥)𝑑𝑥.(3.33)

Proof. From Theorem 3.8 and Proposition 3.3, we only need to demonstrate that 𝐾₅(𝑡,𝜏)=^𝑇0𝐺₁(𝑡,𝑥)𝐺₁(𝜏,𝑥)𝑑𝑥.(3.34) is the reproducing kernel of (Ker𝐿₁)^⟂, where the inner product is defined by (𝑓,ℎ)₅=^𝑇0𝐿₁𝑓(𝑡)𝐿₁ℎ(𝑡)𝑑𝑡,𝑓,ℎ∈Ker𝐿₁^⟂.(3.35) From Theorem 3.7, 𝐿₁𝐾₅(𝑡,𝜏)≠0, so 𝐾₅(𝑡,𝜏)∈(Ker𝐿₁)^⟂.
For any ℎ∈(Ker𝐿₁)^⟂, ℎ(𝜏),𝐾₅(𝑡,𝜏)₅=^𝑇0𝐿₁ℎ(𝜏)𝐿₁^𝑇0𝐺₁(𝑡,𝑥)𝐺₁(𝜏,𝑥)𝑑𝑥𝑑𝜏.(3.36) From Theorem 3.7, ℎ(𝜏),𝐾₅(𝑡,𝜏)₅=^𝑇0𝐿₁ℎ(𝜏)𝐺₁(𝑡,𝜏)𝑑𝜏.(3.37) Furthermore, from the definition of the (Ker𝐿₁)^⟂, we have ^𝑇0𝐿₁ℎ(𝜏)𝐺₁(𝑡,𝜏)𝑑𝜏=^𝑚_𝑖=1𝜆_𝑖ℎ𝑒_𝑖(𝑡)+^𝑇0𝐿₁ℎ(𝜏)𝐺₁(𝑡,𝜏)𝑑𝜏.(3.38) Finally, from the Theorem 3.6, ℎ(𝑡)=^𝑚_𝑖=1𝜆_𝑖ℎ𝑒_𝑖(𝑡)+^𝑇0𝐿₁ℎ(𝜏)𝐺₁(𝑡,𝜏)𝑑𝜏.(3.39) So ℎ(𝜏),𝐾₅(𝑡,𝜏)₅=ℎ(𝑡).(3.40)
The proof is complete.

4. Example

Example 4.1. We consider the space mentioned in the introduction 𝑊^22,1/2[0,1]={𝑢(𝑡),𝑡∈[0,1],𝑢′(𝑡) is an absolute continuous real-valued function on [0,1], ∫¹⁰√𝑡(𝑢^(𝑡))²𝑑𝑡<+∞}. Let 𝐿=𝐷², 𝜆₁𝑢=𝑢(0), and 𝜆₂𝑢=𝑢^(0). Using Theorems 3.8 and 3.9, 𝑊^22,1/2[0,1] is endowed with the inner product: (𝑢,𝑣)₃=𝑢(0)𝑣(0)+𝑢^(0)𝑣^(0)+¹⁰√𝑥𝑢^(𝑥)𝑣^(𝑥)𝑑𝑥(4.1) and the corresponding reproducing kernel is 𝐾(𝑥,𝑦)=⎧⎪⎨⎪⎩11++𝑥𝑥𝑦𝑦++44𝑥𝑥3𝑦^3/2_3/2−41𝑦5^5/2,𝑦≤𝑥,𝑦3−41𝑥5^5/2,𝑦>𝑥.(4.2) This result is in accord with Theorem 2.1 of [8].
If 𝜆₁𝑢=𝑢(0) and 𝜆₂𝑢=𝑢(1) and the inner product of 𝑊^22,1/2[0,1] is given by (𝑢,𝑣)₃=𝑢(0)𝑣(0)+𝑢(1)𝑣(1)+¹⁰√𝑥𝑢^(𝑥)𝑣^(𝑥)𝑑𝑥,(4.3) using the method of this paper, the reproducing kernel of the this space is 𝐾(𝑥,𝑦)=1−𝑥−𝑦+46𝑥𝑦15+4(𝑦−155)𝑥𝑦^3/2+4(𝑥−5)𝑥^3/2𝑦−⎧⎪⎨⎪⎩4𝑦_3/215(𝑦−5𝑥)4𝑥_3/215,𝑦≤𝑥,(𝑥−5𝑦)15,𝑦>𝑥.(4.4)

Example 4.2. We consider the space 𝑊^42,𝛼[0,1]={𝑢(𝑡),𝑡∈[0,1], 𝑢⁽³⁾(𝑡) is an absolute continuous real-valued function on [0,1], ∫¹⁰𝑡^𝛼(𝑢⁽⁴⁾(𝑡))²𝑑𝑡<+∞}. Let 𝐿=𝐷⁴, 𝜆₁𝑢=𝑢(0), 𝜆₂𝑢=𝑢(1), 𝜆₃𝑢=𝑢′(0) and 𝜆₄𝑢=𝑢′(1). The inner product is given by (𝑢,𝑣)₃=𝑢(0)𝑣(0)+𝑢(1)𝑣(1)+𝑢′(0)𝑣′(0)+𝑢′(1)𝑣′(1)+¹⁰𝑥^𝛼𝑢⁽⁴⁾(𝑥)𝑣⁽⁴⁾(𝑥)𝑑𝑥(4.5) Similar to Example 4.1, we can compute the reproducing kernel of the reproducing kernel space 𝑊^42,𝛼[0,1] is 𝐾(𝑥,𝑦)+=((−𝑥1−+1)𝑥²)𝑥𝑥(−₂1+𝑦)²𝑦+𝑥²(−3+2𝑥)𝑦²(−3+2𝑦)−2(−(−11++𝑥)𝑦𝑥)𝑦²₂+(−1+𝑥)²(1+2𝑥)(−1+𝑦)²(1+2𝑦)(3+3𝛼(−1+𝑦)−8𝑦)𝑦²−720−1764𝛼+1106𝑥2₂4𝛼²−735𝛼³+175𝛼⁴−21𝛼⁵+𝛼⁶(−3+2𝑥)(1+𝑎(−1+𝑦)−3𝑦)𝑦²−𝑟−(𝑥5,0𝑦4)0−+𝑟1(𝑦3,0𝑥6)8+𝛼−1𝑅𝑅3((1𝑥𝑦3,,2𝑥𝑦𝛼))²,,+𝑦𝑦67≤>69𝑥𝑥𝛼,,³−1960𝛼⁴+322𝛼⁵−28𝛼⁶+𝛼⁷(4.6) where 𝑟(𝑥=,𝑦𝑥)_(4−𝛼)𝑦²−3−42+55𝛼−14𝛼²+𝛼³𝑥−3−14+23𝛼−10𝛼²+𝛼³𝑥²(−2+𝑦)+𝑥_(4−𝛼)6(−7+𝛼)(−2+𝛼)(−1+𝛼)360−342𝛼+119𝛼²−18𝛼³+𝛼⁴𝑦²−210+107𝛼−18𝛼²+𝛼³𝑦+−6+11𝛼−6𝛼²+𝛼³𝑥³(−3+2𝑦)𝑅(=𝑥,𝑦𝑦)_(4−𝛼)6(−7+𝛼)(−2+𝛼)(−1+𝛼)360−342𝛼+119𝛼²−18𝛼³+𝛼⁴,(−7+𝛼)(−6+𝛼)(−5+𝛼)𝑥³−3(−7+𝛼)(−6+𝛼)(−1+𝛼)𝑥²𝑦+𝑦6(−7+𝛼)(−6+𝛼)(−5+𝛼)(−4+𝛼)(−3+𝛼)(−2+𝛼)(−1+𝛼)_(4−𝛼)3(−7+𝛼)(−2+𝛼)(−1+𝛼)𝑥𝑦²−(−3+𝛼)(−2+𝛼)(−1+𝛼)𝑦³6(−7+𝛼)(−6+𝛼)(−5+𝛼)(−4+𝛼)(−3+𝛼)(−2+𝛼)(−1+𝛼).(4.7)

5. Conclusion

In this paper, we have proposed a method to compute the reproducing kernel on the reproducing kernel space with weighted integral. Theorems 3.8 and 3.9 are the most important theorems of the paper. To our best knowledge, Theorem 3.6 is the first results about the component of the space 𝑊^𝑚2,𝛼[0,𝑇]. From the example, we know that the reproducing kernel space of [8] is just one space of the 𝑊^𝑚2,𝛼[0,𝑇], and the proposed method of this paper is a universal method.

Acknowledgment

The work is supported by NSF of China under Grant no. 10971226.