J-selfadjointness of a class of high-order differential operators with transmission conditions

Abstract

This paper is to investigate the J-selfadjointness of a class of high-order complex coefficients differential operators with transmission conditions. Using the Lagrange bilinear form of J-symmetric differential equations, the definition of J-selfadjoint differential operators and the method of matrix representation, we prove that the operator is J-selfadjoint operator, and the eigenvectors and eigen-subspaces corresponding to different eigenvalues are C-orthogonal.

Keywords

High-order differential operator / transmission conditions / J-selfadjoint / C-orthogonal

1 Introduction and preliminaries

J-symmetric differential operator is a well-known class of asymmetric differential operators. Concerning the existence problem of J-selfadjoint extension of J-symmetric differential operator, Galindo [2] and Knowles [3] have successively proved that there exist J-selfadjoint expansions for any J-symmetric differential operator. The problem of analytical characterization of J-selfadjoint expansion domains of J-symmetric differential operator has been solved by Knowles [4], Race [9] and Shang [10] by making different restrictions on regular domains and using different methods. So far, the J-selfadjoint expansion problem of the J-symmetric differential operator has been basically solved.

A series of achievements have been made in the study of selfadjoint differential operators with transmission conditions (see [1, 7, 11, 13]), including selfadjoint expansion problems in Hilbert space, eigenvalue distribution, eigenfunction completeness and Green's function of this kind of operators, etc., and the selfadjointness of this type of operators on complete indefinite inner product spaces.

On the other hand, much less is known about the study of the problem in the case of non-symmetric differential operators with transmission conditions, so we consider J-selfadjoint expansion problems of higher-order J-symmetric differential operator with transmission conditions on the basis of second-order problems [5]. By establishing a Hilbert space H associated with the transmission condition, and the definition of the operator related to the problem given in H, we study J-selfadjointness of the operator and C-orthogonality of the eigenvectors and eigensubspaces of the operator corresponding to different eigenvalues.

Now, the basic concepts of C-operator, C-orthogonal, J-symmetric operator and J-selfadjoint operator and some conclusions are given below.

Definition 1.1　Let H be a separable Hilbert space, and (\cdot ,\cdot ) denote the inner product in H. Then an operator C in H is said to be a semilinear operator if

C(\alpha x+\beta y)=\overline{\alpha }x+ \overline{\beta } y,\mathrm{\forall }x,y\in H,

where \alpha and \beta are any complex numbers. Moreover, if C^2=I, then C is idempotent. If for arbitrary x and y in H, we have (x,y )=(Cy,Cx) holds, then C is isometric. An idempotent isometric semilinear operator in H is called the C-operator.

Definition 1.2　Let C be a C-operator in separable Hilbert space H. Then a closed densely defined linear operator T in H is said to be C-symmetric if

(CTf ,g)= (CTg ,f), \mathrm{\forall }f, g\in D( T).

That is, T\subset C{T}^{\ast } C, where D(T) is the domain of the operator T. T^{\ast } is a the adjoint of T. If T=C{T}^{\ast }C, then T is a C-selfadjoint operator.

Definition 1.3　Let H be a separable complex Hilbert space, C be a C-operator in H, (\cdot ,\cdot ) denote the inner product in H, and [\cdot ,\cdot ] denote the bilinear form in H,

(1.1)[x,y ]=(x,Cy), \mathrm{\forall } x,y\in H .

If [x,y]=(x ,Cy) =0, then x and y are said to be orthogonal or C-orthogonal with [\cdot ,\cdot ] and denoted it as x{\mathrm{\perp }}_{C}y. If for any x\in E and y\in F, [x,y ]=0 holds, then we say that the two subspaces E and F are C-orthogonal and denoted it as E{\mathrm{\perp }}_{C}F.

Remark 1.1 [12]　The bilinear form (1.1) is nondegenerate, that is, for any x\in H, [x,x] =0 if and only if x=0. Different from the conjugate bilinear form (\cdot ,\cdot ), bilinear form [\cdot ,\cdot ] is not necessarily negative, since for any \theta, we have [ {\mathrm{e}}^{\mathrm{i}\theta /2}x,{\mathrm{e}}^{\mathrm{i}\theta /2}x ]={\mathrm{e}}^{\mathrm{i}\theta }[x,x].

Definition 1.4　Let J be a conjugate-linear involution from complex Hilbert space H into itself, i.e., J is surjective and satisfies

(Jx, Jy)= (y,x ),J^{2}x=x,

for all x and y in H. Then J is called a conjugation operator in L^2.

Definition 1.5　A densely defined linear operator T in H is said to be J-symmetric (C-symmetric) if

(1.2)(Tx, Jy)= (x,J Ty), \mathrm{\forall }x, y\in D( T),

where D(T) is the domain of T. This is equivalent to requiring

(Tx, \overline{y} )=(x,\overline{Ty} ).

Lemma 1.1 [6]　 T is J-symmetric (C-symmetric) if and only if

T\subset J{T}^{\ast }J(orJT J\subset T^{\ast }).

Obviously, the J-symmetric operators are special C-symmetric operators.

Definition 1.6　If

T=J{T}^{\ast }J(\text{or}JTJ=T^{\ast }),

then T is said to be J-selfadjoint (C-selfadjoint).

Lemma 1.2 [6]　 J{T}^{\ast }J is a closed linear operator.

Lemma 1.3 [6]　 If T is J-symmetric, then \overline{T} is also J-symmetric.

2 A new Hilbert space and a new linear operator

We consider here the ordinary form of 2nth-order linear differential expression

(2.1)l(y)=\sum _{k=0}^n(-1{)}^{n-k}(p_k(x) y^{(n-k)}{)}^{(n-k)}, x\in \mathcal{M}=[-1, 0)\cup (0,1 ],

with boundary conditions

(2.2)A{C}_y(-1)+B{C}_y(1) =0,

and transmission conditions at the interior discontinuous point

(2.3)C\cdot C_y(0-)+D\cdot C_y(0+)=0 ,

where the functions p_0^{-1}(x),p_1(x) ,\cdots ,p_n(x) are complex-valued and measurable over \mathcal{M}, the limits \underset{x\to 0\pm }{lim}{p}_k(x) =p_k(0\pm )(k =0,1, 2,\cdots ,n ) exist. A=(a_{ij} ),B=(b_{ij} ),C=(c_{ij} ),D=(d_{ij} )(i,j =1,2, \cdots ,2n) are 2n\times 2n complex non-singular matrix, and detC={\rho }^n,\rho >0 ,detD={\theta }^n,\theta >0.

Let

(2.4)\theta A{Q}_{2n} A^T=\rho B{Q}_{2n}{B}^T, \theta C{Q}_{2n} C^T=\rho D{Q}_{2n}{D}^T,

where Q is a 2n\times 2n matrix as follows:

Q_{2n}=\left(\begin{array}{cccccc}& & & & & -1\\ & & & & \cdots & \\ & & & -1& & \\ & & 1& & & \\ & \cdots & & & & \\ 1& & & & & \end{array}\right).

Clearly Q has the property

det{Q}_{2n}\ne 0,Q_{2n}^T= -Q_{2n}=Q_{2n}^{-1},[Q_{2n}^{-1}{]}^T=[Q_{2n}^T {]}^{-1}=-Q_{2n}^{-1}.

Definition 2.1 [10]　For the quasi-derivatives of a function y relative to expression (2.1), we introduce the functions y^{[k]}(x) (k=0,1,\cdots ,2n) defined by the formulae as follows:

\begin{array}{rl}& y^{[k]}(x)=\frac{{\mathrm{d}}^{k}y}{\mathrm{d}{x}^k},k=0,1,\cdots ,n-1,\\ & y^{[n]}(x)=p_0\frac{{\mathrm{d}}^{n}y}{\mathrm{d}{x}^n},\\ & y^{[n+k]}(x)=p_k\frac{{\mathrm{d}}^{n-k}y}{\mathrm{d}{x}^{n-k}}-\frac{\mathrm{d}}{\mathrm{d}x}(y^{[n+k-1]}),k=1,\cdots ,n.\end{array}

Then l(y) in (2.1) may be simply written by l(y)= y^{[2n]}.

The maximal operator L_{MT} generated by T is defined by

\begin{array}{l}D(L_{MT})=\{f\in H|f_1^{[2n-1]}\in A{C}_{loc}[-1,0),f_2^{[2n-1]}\in A{C}_{loc}(0,1],l(f)\in H,\\ C\cdot C_f(0-)+D\cdot C_f(0+)=0\},\\ L_{MT}f=l(f),\mathrm{\forall }f\in D(L_{MT}).\end{array}

The minimal operator L_{0T} generated by T is defined by

\begin{array}{l}D(L_{0T})=\{f\in D(L_{MT})|C_f(-1)=C_f(1)=0\},\\ L_{0T}f=l(f),\mathrm{\forall }f\in D(L_{0T}).\end{array}

The operator T is defined by

D(T)=\{f\in D (L_{MT} )|A{C}_f(-1)+B{C}_f(1)=0\}, Tf=l(f ).

Definition 2.2 [10]　For any y, z in D(T), we have

(2.5)l(y)z-y l(z)=\frac{\mathrm{d}}{\mathrm{d}x}[y, \overline{z} ],

where

(2.6)\begin{array}{rl}[y,z](x)& =W(y,\overline{z};x)\\ & =\sum _{k=1}^n\{y^{[k-1]}(x){\overline{z}}^{[2n-k]}(x)-y^{[2n-k]}(x){\overline{z}}^{[k-1]}(x)\}\\ & =R_{\overline{z}}(x)Q_{2n}{C}_y(x).\end{array}

the row vector R_{\overline{z}} (x) and the column vector C_y(x) in (2.6) are given by

R_{\overline{z}} (x)=(\overline{z}(x) ,{\overline{z}}^{[1]}(x),\cdots , {\overline{z}}^{[2n-1]}(x)), C_y(x) =(y(x) ,y^{[1]}(x ),\cdots , y^{[2n-1]}(x) {)}^T.

(2.5) is the so-called Lagrange's identity, (2.6) is called the Lagrange bilinear form corresponding to T.

For any y,z in D(L_{MT}), the limits [y,\overline{z}](a) =\underset{x\to a^{+}}{lim}[y,\overline{z}] and [y,\overline{z}](b )=\underset{x\to b^{-}}{lim}[y,\overline{z}] exist and we have

{\int }_a^{b}l(y)z\mathrm{d}x-{\int }_a^{b}y l(z)\mathrm{d}x=[ y,\overline{z}{]}_a^b,

where [y,\overline{z}{]}_a^b=[y ,\overline{z}]( b)-[y,\overline{z}](a ).

Lemma 2.1 [8]　 For any a, b in , and a set of complex constants {\alpha }_0, {\alpha }_1,\cdots , {\alpha }_{2n-1},{\beta }_0, {\beta }_1,\cdots ,{\beta }_{2n-1}, there exists a function y\in D(L_{MT} ), satisfying

y^{[k]}(a)={\alpha }_k,y^{[k]}(b)= {\beta }_k, k=0,1,2,\cdots ,2n-1.

In order to investigate problem (2.1)−(2.3), we define a new inner product in L^2(\mathcal{M}) as

(f,g )={\int }_{-1}^0{f}_1\overline{g_1}\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^1{f}_2\overline{g_2}\mathrm{d}x ,\mathrm{\forall } f,g\in L^2(\mathcal{M}),

where f_1(x)=f(x) {|}_{[-1,0)},f_2(x )=f(x) {|}_{(0,1]}, similarly f_1^{[k]}(x) =f^{[k]}(x ){|}_{[-1,0)} ,f_2^{[k]}(x)=f^{[k]}(x){|}_{(0,1]}, k=1,2,\cdots ,2n-1. It is easy to verify that H=(L^2(\mathcal{M}),(\cdot ,\cdot )) is a new Hilbert space with the inner product. By Definition 1.3, we have

\begin{array}{rl}[f,g]=(f,Cg)& ={\int }_{-1}^0{f}_1\overline{C{g}_1}\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^1{f}_2\overline{C{g}_2}\mathrm{d}x\\ & ={\int }_{-1}^0{f}_1{g}_1\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^1{f}_2{g}_2\mathrm{d}x,\mathrm{\forall }f,g\in L^2(\mathcal{M}).\end{array}

The following is then an immediate consequence of the properties of the inner product and Definition 1.3,

(2.7)\begin{array}{rl}& (\alpha f+\beta g,z)=\alpha (f,z)+\beta (g,z),(z,\alpha f+\beta g)=\overline{\alpha }(z,f)+\overline{\beta }(z,g),\\ & [\alpha f+\beta g,z]=\alpha [f,z]+\beta [g,z],[z,\alpha f+\beta g]=\alpha [z,f]+\beta [z,g],\end{array}

for any \alpha , \beta \in \mathbb{C} and f, g,z\in L^2(\mathcal{M}), where \mathbb{C} denotes the set of complex numbers.

3 The \mathit{J}-selfadjointness of operator \mathit{T}

In this section, we consider problems (2.1)−(2.3) by investigation operator T, and show that the operator T is J-selfadjoint operator, and the eigenvectors and eigen-subspaces corresponding to different eigenvalues are C-orthogonal.

Lemma 3.1　 The domain D(T) of the operator T is dense in H.

Proof　Let {\tilde{C}}_0^{\mathrm{\infty }} be the set of all functions defined on [-1, 0)\cup (0,1 ], such that

\eta (x)=\{\begin{array}{l}{\eta }_1(x),x\in [-1,0);\\ {\eta }_2(x),x\in (0,1].\end{array}

For {\eta }_1(x)\in C_0^{\mathrm{\infty }}[- 1,0) ,{\eta }_2(x) \in C_0^{\mathrm{\infty }}(0,1 ], clearly, {\tilde{C}}_0^{\mathrm{\infty }}\subset D(T). For any f\in H,

f(x)=\{\begin{array}{l}{f}_1(x),x\in [-1,0);\\ f_2(x),x\in (0,1].\end{array}

and for any \epsilon >0, from the well-known fact that C_0^{\mathrm{\infty }}(a,b) is dense in the Hilbert space L^2(a,b), it follows that the set is dense in the Hilbert space L^2[-1, 0). Then there exists g_1(x)\in C_0^{\mathrm{\infty }}[- 1,0), such that

Similarly, there exists g_2(x)\in C_0^{\mathrm{\infty }}(0,1], such that

Let

g(x)=\{\begin{array}{l}{g}_1(x),x\in [-1,0);\\ g_2(x),x\in (0,1].\end{array}

For any f\in H and \epsilon > 0, there exists g(x)\in {\tilde{C}}_0^{\mathrm{\infty }}, such that

(3.1)

(3.1) means that {\tilde{C}}_0^{\mathrm{\infty }} is dense in the Hilbert space H, and therefore D(T) is dense in H.

Theorem 3.1　 The operator T is J-selfadjoint in H.

Proof　Let f and g be arbitrary elements of D(T). By partial integrations we obtain

(3.2)\begin{array}{rl}(Tf,\overline{g})& ={\int }_{-1}^{0}l(f)g\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^{1}l(f)g\mathrm{d}x\\ & ={\int }_{-1}^{0}fl(g)\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^{1}fl(g)\mathrm{d}x+[f,\overline{g}{]}_{-1}^0+\frac{\theta }{\rho }[f,\overline{g}{]}_0^1\\ & =(f,\overline{Tg})+W(f,g;0-)-W(f,g;-1)+\frac{\theta }{\rho }W(f,g;1)-\frac{\theta }{\rho }W(f,g;0+).\end{array}

Since f and g satisfy the boundary conditions (2.2) and (2.4), by (2.6), it follows that

(3.3)\begin{array}{rl}W(f,g;1)& =R_g(1)Q_{2n}{C}_f(1)\\ & =(C_g(1){)}^T{Q}_{2n}{C}_f(1)\\ & =(-B^{-1}A\cdot C_g(-1){)}^T{Q}_{2n}(-B^{-1}A\cdot C_f(-1))\\ & =R_g(-1)[(B^{-1}A{)}^T{Q}_{2n}{B}^{-1}A]C_f(-1)\\ & =R_g(-1)\frac{\rho }{\theta }{Q}_{2n}{C}_f(-1)\\ & =\frac{\rho }{\theta }W(f,g;-1).\end{array}

From the transmission conditions (2.3) and (2.4), we get

(3.4)\begin{array}{rl}W(f,g;0+)& =R_g(0+)Q_{2n}{C}_f(0+)\\ & =(C_g(0+){)}^T{Q}_{2n}{C}_f(0+)\\ & =(-D^{-1}C\cdot C_g(0-){)}^T{Q}_{2n}(-D^{-1}C\cdot C_f(0-))\\ & =R_g(0-)[(D^{-1}C{)}^T{Q}_{2n}{D}^{-1}C]C_f(0-)\\ & =R_g(0-)\frac{\rho }{\theta }{Q}_{2n}{C}_f(0-)\\ & =\frac{\rho }{\theta }W(f,g;0-).\end{array}

Further, putting (3.3) and (3.4) into (3.2), we get the required equality (Tf, \overline{g} )=(f,\overline{Tg} ), so T is J-symmetric.

It remains to show that if (Tf, \overline{g} )=(f,\omega ) for all f\in D(T), then g\in D(T) and \overline{l(g)}= \omega. For an arbitrary f\in {\tilde{C}}_0^{\mathrm{\infty }}\subset D(T), (Tf,\overline{g}) =(f, \omega ) holds. According to normal Sturm-Liouville theory and Definition 1.6, we have g_1^{[2n-1]}\in A{C}_{loc}[-1,0),g_2^{[2n-1]}\in A{C}_{loc}(0,1 ], \overline{l(g)}\in H and \omega = \overline{l(g)}.

Next, we prove g\in D( T), that is, A{C}_g(-1)+B{C}_g(1)=0 and C\cdot C_g(0- )+D\cdot C_g(0+)=0 holds.

By (Tf, \overline{g} )=(f,\omega )=( f,\overline{Tg}), we have

(Tf, \overline{g} )={\int }_{-1}^{0}fl( g)\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^{1}f l(g)\mathrm{d}x.

On the other hand,

\begin{array}{rl}(Tf,\overline{g})=& {\int }_{-1}^{0}fl(g)\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^{1}fl(g)\mathrm{d}x\\ & +W(f,g;0-)-W(f,g;-1)+\frac{\theta }{\rho }W(f,g;1)-\frac{\theta }{\rho }W(f,g;0+).\end{array}

So

(3.5)W(f, g;0-)-W(f,g ;-1)+\frac{\theta }{\rho }W(f,g ;1)- \frac{\theta }{\rho } W(f, g;0+)=0.

By Lemma 2.1, there exists f_1,f_2, \cdots ,f_{2n} \in D(T ), such that

f_i(0+)=f_i^{[1]}(0+)=\cdots = f_i^{[2n-1]}(0+) =0,i =1,2, \cdots ,2n.

For such f_i\in D(T)(i =1,2, \cdots ,2n) satisfing transmission conditions (2.3), we have W(f, g;0-)=0. Then from (3.5) we have \theta W(f,g;1)=\rho W (f,g ;-1), that is

(3.6)\theta R_g(1)Q_{2n} C_{f_i}(1 )=\rho R_g(-1)Q_{2n} C_{f_i}(- 1),i =1,2, \cdots ,2n.

Let

\left(\begin{array}{cccc}{f}_1(1)& f_2(1)& \cdots & f_{2n}(1)\\ f_1^{[1]}(1)& f_2^{[1]}(1)& \cdots & f_{2n}^{[1]}(1)\\ ⋮& ⋮& & ⋮\\ f_1^{[2n-1]}(1)& f_2^{[2n-1]}(1)& \cdots & f_{2n}^{[2n-1]}(1)\end{array}\right)=F (1),

\left(\begin{array}{cccc}{f}_1(-1)& f_2(-1)& \cdots & f_{2n}(-1)\\ f_1^{[1]}(-1)& f_2^{[1]}(-1)& \cdots & f_{2n}^{[1]}(-1)\\ ⋮& ⋮& & ⋮\\ f_1^{[2n-1]}(-1)& f_2^{[2n-1]}(-1)& \cdots & f_{2n}^{[2n-1]}(-1)\end{array}\right)= F(-1).

Then it is equal to \theta R_g(1)Q_{2n} F(1)=\rho R_g(-1)Q_{2n} F(-1).

Let Q_{2n}F(1)=B^T,\rho Q_{2n}F(-1)=-\theta A^T. Then R_g(1)B^T= -R_g(-1)A^T, which is equal to A{C}_g(-1)+B{C}_g(1)=0. So g satisfies the boundary conditions (2.2).

On the other hand, by (3.6), A{C}_{f_i}( -1)+B{C}_{f_i}(1)=0(i =1,2, \cdots ,2n). Moreover, we have \theta A Q_{2n}{A}^T= \rho B{Q}_{2n} B^T. Therefore A,B satisfy conditions (2.4).

Next choose f_1,f_2, \cdots ,f_{2n} \in D(T ), such that

f_i(- 1)=f_i^{[1]}(-1)=\cdots = f_i^{[2n-1]}(-1)=0,i=1,2,\cdots ,2n,

thus W(f,g ;-1)=0. Then from (3.5) we have \theta W(f,g;0+)=\rho W(f, g;0-), that is

\theta R_g(0+)Q_{2n}{C}_{f_i}( 0+)= \rho R_g(0-)Q_{2n}{C}_{f_i}(0- ),i= 1,2,\cdots ,2n.

Let

\left(\begin{array}{cccc}{f}_1(0+)& f_2(0+)& \cdots & f_{2n}(0+)\\ f_1^{[1]}(0+)& f_2^{[1]}(0+)& \cdots & f_{2n}^{[1]}(0+)\\ ⋮& ⋮& & ⋮\\ f_1^{[2n-1]}(0+)& f_2^{[2n-1]}(0+)& \cdots & f_{2n}^{[2n-1]}(0+)\end{array}\right)=F(0+) ,

\left(\begin{array}{cccc}{f}_1(0-)& f_2(0-)& \cdots & f_{2n}(0-)\\ f_1^{[1]}(0-)& f_2^{[1]}(0-)& \cdots & f_{2n}^{[1]}(0-)\\ ⋮& ⋮& & ⋮\\ f_1^{[2n-1]}(0-)& f_2^{[2n-1]}(0-)& \cdots & f_{2n}^{[2n-1]}(0-)\end{array}\right)= F(0- ).

Then

(3.7)\theta R_g(0+)Q_{2n}F(0+)= \rho R_g(0-)Q_{2n}F( 0-).

Let Q_{2n}F(0+)= D^T,\rho Q_{2n}F(0-)=-\theta C^T. Then R_g(0+)D^T= -R_g(0-)C^T, that is C\cdot C_g(0-)+D\cdot C_g(0+)=0. So g satisfies the transmission conditions (2.3).

On the other hand, by (3.7), C\cdot C_{f_i}(0- )+D\cdot C_{f_i}(0 +)=0 (i=1 ,2,\cdots , 2n). Moreover, we have \theta C{Q}_{2n} C^T=\rho D{Q}_{2n}{D}^T. Therefore C,D satisfy conditions (2.4).

From the above discussion, we can know g\in D(T), and T is J-selfadjoint operator.

Theorem 3.2　 Let {\lambda }_1 and {\lambda }_2 be two different eigenvalues of the operator T. Then the corresponding eigenvectors {\chi }_1(x) and {\chi }_2(x) are C-orthogonal.

Proof　Let

{\lambda }_1=\{\begin{array}{l}{\lambda }_{11},x\in [-1,0);\\ {\lambda }_{12},x\in (0,1],\end{array} {\lambda }_2= \{\begin{array}{l}{\lambda }_{21},x\in [-1,0);\\ {\lambda }_{22},x\in (0,1],\end{array}

{\chi }_1=\{\begin{array}{l}{\chi }_{11},x\in [-1,0);\\ {\chi }_{12},x\in (0,1],\end{array} {\chi }_2= \{\begin{array}{l}{\chi }_{21},x\in [-1,0);\\ {\chi }_{22},x\in (0,1].\end{array}

Then T {\chi }_1={\lambda }_1{\chi }_1,T {\chi }_2={\lambda }_2{\chi }_2, and we have

\begin{array}{rl}{\lambda }_1[{\chi }_1,{\chi }_2]& ={\lambda }_1({\chi }_1,C{\chi }_2)\\ & ={\lambda }_{11}{\int }_{-1}^0{\chi }_{11}{\chi }_{21}\mathrm{d}x+{\lambda }_{12}\cdot \frac{\theta }{\rho }{\int }_0^1{\chi }_{12}{\chi }_{22}\mathrm{d}x\\ & ={\int }_{-1}^{0}T{\chi }_1\cdot {\chi }_{21}\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^{1}T{\chi }_1\cdot {\chi }_{22}\mathrm{d}x\\ & =(T{\chi }_1,C{\chi }_2)\\ & =({\chi }_1,T^{\ast }C{\chi }_2).\end{array}

Since T is a J-selfadjoint operator, we get

\begin{array}{rl}({\chi }_1,T^{\ast }C{\chi }_2)& =({\chi }_1,JT{\chi }_2)\\ & ={\int }_{-1}^0{\chi }_{11}\cdot \overline{JT{\chi }_{21}}\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^1{\chi }_{12}\cdot \overline{JT{\chi }_{22}}\mathrm{d}x\\ & ={\int }_{-1}^0{\chi }_{11}\cdot {\lambda }_{21}{\chi }_{21}\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^1{\chi }_{12}\cdot {\lambda }_{22}{\chi }_{22}\mathrm{d}x\\ & ={\lambda }_2\left({\int }_{-1}^0{\chi }_{11}{\chi }_{21}\mathrm{d}x+\frac{\theta }{\rho }{\int }_0^1{\chi }_{12}{\chi }_{22}\mathrm{d}x\right)\\ & ={\lambda }_2[{\chi }_1,{\chi }_2].\end{array}

Therefore {\lambda }_1[{\chi }_1,{\chi }_2]={\lambda }_2[{\chi }_1, {\chi }_2]. Since {\lambda }_1\ne {\lambda }_2, it follows that

(3.8)[{\chi }_1, {\chi }_2]=({\chi }_1, C{\chi }_2)=0.

By Definition 1.3, (3.8) means that {\chi }_1 and {\chi }_2 are C-orthogonal.

Theorem 3.3　 Let {\lambda }_1 and {\lambda }_2 be two different eigenvalues of the operator T. Then the corresponding subspace \text{Ker}(T-{\lambda }_{1}I{)}^{m_1} and \text{Ker}(T-{\lambda }_{2}I{)}^{m_2} are C-orthogonal. That is

(3.9)\text{Ker}(T -{\lambda }_{1}I {)}^{m_1}{\mathrm{\perp }}_C\text{Ker}(T-{\lambda }_{2}I {)}^{m_2},

where m_1 and m_2 are the nonnegative integers.

Proof　If m_1= m_2=1, then (3.9) holds by Theorem 3.2. Let

{\lambda }_1=\{\begin{array}{l}0,x\in [-1,0);\\ 0,x\in (0,1],\end{array}{\lambda }_2= \{\begin{array}{l}{\lambda }_{21},x\in [-1,0);\\ {\lambda }_{22},x\in (0,1],\end{array}

where {\lambda }_{21}\ne 0,{\lambda }_{22}\ne 0, and their corresponding eigenvectors as

{\chi }_1=\{\begin{array}{l}{\chi }_{11},x\in [-1,0);\\ {\chi }_{12},x\in (0,1],\end{array} {\chi }_2= \{\begin{array}{l}{\chi }_{21},x\in [-1,0);\\ {\chi }_{22},x\in (0,1].\end{array}

Then

(T- {\lambda }_{1}I{)}^j{\chi }_1=T^j{\chi }_1=\{\begin{array}{l}{T}^j{\chi }_{11},x\in [-1,0);\\ T^j{\chi }_{12},x\in (0,1],\end{array}j =0,1, \cdots ,m_1,

(T- {\lambda }_{2}I{)}^k{\chi }_2=\{\begin{array}{l}(T-{\lambda }_{21}I{)}^k{\chi }_{21},x\in [-1,0);\\ (T-{\lambda }_{22}I{)}^k{\chi }_{22},x\in (0,1],\end{array} k=0,1,\cdots ,m_2,

and

\begin{array}{l}\text{Ker}(T-{\lambda }_{1}I{)}^{m_1}=\text{span}\{{\chi }_1,T{\chi }_1,T^2{\chi }_1,\cdots ,T^{m_1}{\chi }_1\}={\epsilon }_1,\\ \text{Ker}(T-{\lambda }_{2}I{)}^{m_2}=\text{span}\{{\chi }_2,(T-{\lambda }_{2}I){\chi }_2,(T-{\lambda }_{2}I{)}^2{\chi }_2,\cdots ,(T-{\lambda }_{2}I{)}^{m_2}{\chi }_2\}={\epsilon }_2,\end{array}

where T^{m_1}{\chi }_1\ne 0,(T-{\lambda }_{2}I{)}^{m_2}{\chi }_2\ne 0, but T^{m_1+1}{\chi }_1=0,(T -{\lambda }_{2}I {)}^{m_2+1}{\chi }_2=0.

Now just prove that the subspaces {\epsilon }_1 and {\epsilon }_2 are C-orthogonal, or orthogonal by [\cdot ,\cdot ]. First, we show that (T- {\lambda }_{2}I{)}^{m_2}{\chi }_2 and {\epsilon }_1 are C-orthogonal. For 0⩽j⩽m_1, we have

\begin{array}{rl}& {\lambda }_2^{m_1+1-j}[T^j{\chi }_1,(T-{\lambda }_{2}I{)}^{m_2}{\chi }_2]\\ & ={\int }_{-1}^0{T}^j{\chi }_{11}\cdot {\lambda }_{21}^{m_1+1-j}(T-{\lambda }_{21}I{)}^{m_2}{\chi }_{21}\mathrm{d}x\\ & +\frac{\theta }{\rho }{\int }_0^1{T}^j{\chi }_{12}\cdot {\lambda }_{22}^{m_1+1-j}(T-{\lambda }_{22}I{)}^{m_2}{\chi }_{22}\mathrm{d}x\\ & ={\int }_{-1}^0{T}^j{\chi }_{11}\cdot T^{m_1+1-j}(T-{\lambda }_{21}I{)}^{m_2}{\chi }_{21}\mathrm{d}x\\ & +\frac{\theta }{\rho }{\int }_0^1{T}^j{\chi }_{12}\cdot T^{m_1+1-j}(T-{\lambda }_{22}I{)}^{m_2}{\chi }_{22}\mathrm{d}x\\ & =[T^j{\chi }_1,T^{m_1+1-j}(T-{\lambda }_{2}I{)}^{m_2}{\chi }_2]\\ & =[T^{m_1+1}{\chi }_1,(T-{\lambda }_{2}I{)}^{m_2}{\chi }_2]\\ & =0.\end{array}

Since {\lambda }_2\ne 0, we have

[T^j{\chi }_1,(T -{\lambda }_{2}I {)}^{m_2}{\chi }_2]=0, 0⩽j⩽m_1.

Thus (T-{\lambda }_{2}I{)}^{m_2}{\chi }_2{\mathrm{\perp }}_C{\epsilon }_1.

For some k(0⩽ k⩽m_2- 1), if [ T^j{\chi }_1, (T- {\lambda }_{2}I{)}^{m_2-k}{\chi }_2]=0(0 ⩽j⩽m_1), by (2.7), we obtain

\begin{array}{rl}{\lambda }_2[T^j{\chi }_1,(T-{\lambda }_{2}I{)}^{m_2-k-1}{\chi }_2]& =[T^j{\chi }_1,{\lambda }_2(T-{\lambda }_{2}I{)}^{m_2-k-1}{\chi }_2]\\ & +[T^j{\chi }_1,(T-{\lambda }_{2}I{)}^{m_2-k}{\chi }_2]\\ & =[T^j{\chi }_1,{\lambda }_2(T-{\lambda }_{2}I{)}^{m_2-k-1}{\chi }_2+(T-{\lambda }_{2}I{)}^{m_2-k}{\chi }_2]\\ & =[T^j{\chi }_1,({\lambda }_{2}I+T-{\lambda }_{2}I)(T-{\lambda }_{2}I{)}^{m_2-k-1}{\chi }_2]\\ & =[T^j{\chi }_1,T(T-{\lambda }_{2}I{)}^{m_2-k-1}{\chi }_2].\end{array}

Again, as know by the J-selfadjointness of T, we get

[T^j{\chi }_1,T( T-{\lambda }_{2}I {)}^{m_2-k-1}{\chi }_2]=[ T^{j+1}{\chi }_1,(T- {\lambda }_{2}I{)}^{m_2-k-1} {\chi }_2].

Thus {\lambda }_2^{m_1+1-j} [T^j{\chi }_1,(T -{\lambda }_{2}I {)}^{m_2-k-1}{\chi }_2]=[ T^{m_1+1}{\chi }_1,(T -{\lambda }_{2}I {)}^{m_2-k-1}{\chi }_2]=0.

Therefore, for 0⩽j⩽m_1, [T^j{\chi }_1,(T -{\lambda }_{2}I {)}^{m_2-k-1}{\chi }_2]=0 holds, which indicates that for 0⩽j⩽m_1,0 ⩽k⩽m_2, we have

[T^j{\chi }_1,(T -{\lambda }_{2}I {)}^{m_2-k} {\chi }_2]=0.

Thus {\epsilon }_2{\mathrm{\perp }}_C{\epsilon }_1.

In conclusion, we complete the proof.

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