On Symplectic Self-Adjointness of Hamiltonian Operator Matrices

Abstract

The symmetry of the spectrum and the completeness of the eigenfunction system of the Hamiltonian operator matrix have important applications in the symplectic Fourier expansion method in elasticity. However, the symplectic self-adjointness of Hamiltonian operator matrices is important to the characterization of the symmetry of the point spectrum. Therefore, in this paper, the symplectic self-adjointness of infinite dimensional Hamiltonian operators is studied by using the spectral method of unbounded block operator matrices, and some sufficient conditions of the symplectic self-adjointness of infinite dimensional Hamiltonian operators are obtained. In addition, the necessary and sufficient conditions are also investigated for some special infinite dimensional Hamiltonian operators.

1. Introduction

The theory of linear operators in Hilbert space, especially the spectral theory of self-adjoint operators is one of the most important achievements in the field of mathematical science in the 20th century. By studying the spectral theory of linear operators, we can not only understand the structure of operators themselves, but also describe the energy change, stability, constructive of the solution, etc. At present, the self-adjoint operator theory has formed a relatively complete framework. The method of the separation of variables (also known as the Fourier series method) for solving partial differential equations is based on this theory. In mathematical physics equations, the traditional method of the separation of variable is effective to solve partial differential equations which can be transformed into the Strum–Liouville problem after separating variables, such as vibrating string equation and harmonic equations. However, for the equations that cannot be transformed into the Strum–Liouville problem after separating the variables, the method of the separation of variable is useless [1]. In elasticity, the symplectic approach (i.e., the Hamiltonian system approach) was first applied in the early 1990s by Professor Wanxie Zhong [1,2,3], who extended the traditional method of separating variables and established a new method of solving elasticity problems, namely the symplectic Fourier series expansion method of Hamiltonian operator matrix eigenfunction systems. The Hamiltonian system is an important branch in dynamical systems, and has various applications in our daily life. It is, by now, almost beyond dispute that all real physical processes with negligible dissipations, no matter whether they are classical, quantum, or relativistic or of finite or infinite degrees of freedom, can always be cast in a suitable Hamiltonian form [4,5]. In addition, the Hamiltonian operator matrices (see Definition 2 below) have a number of important applications in optimal control, elasticity, mechanics of composites, fracture problems, and evolution equations [6,7,8,9,10]. It is worth noting that the symmetry of the point spectrum with respect to the imaginary axis is an important theoretical basis that is used in the method of infinite dimensional Hamiltonian operators to solve practical problems [9]. However, the symmetry of the spectrum is based on the symplectic self-adjointness of Hamiltonian operators. In general, the point spectra of an infinite dimension Hamiltonian operator are not necessarily symmetric with respect to the imaginary axis. For instance, let

$$H=\left(\begin{array}{cc}A& 0\\ 0& -A^{*}\end{array}\right):\mathcal{H}=\mathcal{D}\left(A\right)\times \mathcal{D}\left(A^{*}\right)\subset X\times X\to X\times X$$

be a Hamiltonian operator, where $X=L^2$$[0,+\infty )$, $A=\frac{d}{dt}$ and $\mathcal{D}\left(A\right)=\left\{x\right(t)\in X:x(t)$ are absolutely continuous and $x^{\prime }\left(t\right)\in X,x\left(0\right)=0\}.$ Then, $A^{*}=-\frac{d}{dt},\mathcal{D}\left(A^{*}\right)=\{x\left(t\right)\in X:x\left(t\right)$ is absolutely continuous and $x^{\prime }\left(t\right)\in X\}.$

It can be known by calculation that

$$\begin{array}{ccc}\hfill {\sigma }_p\left(H\right)& =& \{\lambda \in \mathbb{C}:Re(\lambda )<0\},\hfill \\ \hfill {\sigma }_r\left(H\right)& =& \{\lambda \in \mathbb{C}:Re(\lambda )>0\},\hfill \\ \hfill {\sigma }_c\left(H\right)& =& \{\lambda \in \mathbb{C}:Re(\lambda )=0\}.\hfill \end{array}$$

where ${\sigma }_p\left(H\right),{\sigma }_r\left(H\right)$ and ${\sigma }_c\left(H\right)$ are the point spectrum, residual spectrum and continuous spectrum of the infinite dimensional Hamiltonian operator, respectively.

From the perspective of a general block operator matrix, the infinite dimensional Hamiltonian operator satisfies

$${\left(JH\right)}^{*}\supset JH,$$

where $J=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$ is the unit symplectic operator matrix and satisfies $J^{*}=J^{-1}=-J$.

Since the operator J can induce the symplectic structure, the infinite dimensional Hamiltonian operator is a symplectic symmetric operator. Furthermore, if

$${\left(JH\right)}^{*}=JH,$$

we call the infinite dimensional Hamiltonian operator a symplectic self-adjoint Hamiltonian operator.

The spectra of the symplectic self-adjoint Hamiltonian operators are symmetric with respect to the imaginary axis. In addition, the point spectra are symmetric with respect to the imaginary axis when the residual spectrum of it is an empty set. In [11,12], the symplectic self-adjointness of infinite dimensional Hamiltonian operators is studied by using linear operator perturbation theory and Schur decomposition, and sufficient conditions for symplectic self-adjointness of infinite dimensional Hamiltonian operators are obtained. However, the Schur decomposition method is only applicable to some special infinite dimensional Hamiltonian operators, such as diagonally dominant, off-diagonally dominant, etc. It is useless for more general infinite dimensional Hamiltonian operators. For instance, let $T,S$ be self-adjoint operators on a Hilbert space X, satisfying $\mathcal{D}\left(T\right)⫋\mathcal{D}\left(S\right)$. We also let

$$A=\left(\begin{array}{cc}0& S\\ T& 0\end{array}\right),B=\left(\begin{array}{cc}T& 0\\ 0& S\end{array}\right),C=\left(\begin{array}{cc}S& 0\\ 0& T\end{array}\right),$$

then, the infinite dimensional Hamiltonian operator

$$H=\left(\begin{array}{cc}A& B\\ C& -A^{*}\end{array}\right),$$

is neither diagonally dominant nor off-diagonally dominant; therefore, we cannot use Schur decomposition. In view of this, we attempt to establish some sufficient conditions for a Hamiltonian operator matrix to be symplectic self-adjoint by using spectral methods of block operator matrices and obtain some sufficient and necessary conditions of symplectic self-adjointness for some special infinite dimensional Hamiltonian operators.

Next, we will introduce the symbols used in this paper and give some important concepts and conclusions.

2. Preliminaries

In this paper, let $X,X_i(i=1,2)$ be complex Hilbert spaces. We denote by $\mathcal{D}\left(T\right)$ and $R\left(T\right)$ the domain and range of T, respectively. For a closed densely defined operator $T:X\to X$, if $T-\lambda I$ is injective and surjective, then $\lambda$ belongs to the resolvent set $\rho \left(T\right)$, denoted by $\lambda \in \rho \left(T\right)$. The complementary set of the resolvent set $\rho \left(T\right)$ is the spectrum of T, i.e.,

$$\sigma \left(T\right)=\mathbb{C}\setminus \rho \left(T\right).$$

When $\lambda \in \sigma \left(T\right)$, there are three kind of situations:

(i) $T-\lambda I$ is not injective; then, we call $\lambda$ a point spectrum of $T.$ The whole point spectra of T are denoted by ${\sigma }_p\left(T\right)$;

(ii) $T-\lambda I$ is injective, but $\overline{R(T-\lambda I)}\ne X$; then, we call $\lambda$ a residual spectrum of $T.$ The whole residual spectra of T are denoted by ${\sigma }_r\left(T\right)$;

(iii) $T-\lambda I$ is injective, $\overline{R(T-\lambda I)}=X$, but $R(T-\lambda I)\ne X$; then, we call $\lambda$ q continuous spectrum of T. The whole continuous spectra of T are denoted by ${\sigma }_c\left(T\right)$.

Clearly, ${\sigma }_p\left(T\right)$, ${\sigma }_r\left(T\right)$ and ${\sigma }_c\left(T\right)$ are mutually disjoint and according to the density and closeness of the range of $T-\lambda I$, we can further divide the point spectra and residual spectra as follows:

$$\begin{array}{ccc}\hfill {\sigma }_{p,1}\left(T\right)& =& \{\lambda \in {\sigma }_p\left(T\right):R(T-\lambda I)=X\};\hfill \\ \hfill {\sigma }_{p,2}\left(T\right)& =& \{\lambda \in {\sigma }_p\left(T\right):\overline{R(T-\lambda I)}=X,R(T-\lambda I)\mathrm{is} \mathrm{not} \mathrm{closed}\};\hfill \\ \hfill {\sigma }_{p,3}\left(T\right)& =& \{\lambda \in {\sigma }_p\left(T\right):\overline{R(T-\lambda I)}\ne X,R(T-\lambda I)\mathrm{is} \mathrm{closed}\};\hfill \\ \hfill {\sigma }_{p,4}\left(T\right)& =& \{\lambda \in {\sigma }_p\left(T\right):\overline{R(T-\lambda I)}\ne X,R(T-\lambda I)\mathrm{is} \mathrm{not} \mathrm{closed}\};\hfill \\ \hfill {\sigma }_{r,1}\left(T\right)& =& \{\lambda \in {\sigma }_r\left(T\right):R(T-\lambda I)\mathrm{is} \mathrm{closed}\};\hfill \\ \hfill {\sigma }_{r,2}\left(T\right)& =& \{\lambda \in {\sigma }_r\left(T\right):R(T-\lambda I)\mathrm{is} \mathrm{not} \mathrm{closed}\}.\hfill \end{array}$$

The detailed proof of the following properties of point spectra, residual spectra and continuous spectra can be found in [13].

Lemma 1.

Let T be a closed densely defined operator in Hilbert space X; then,

(1) $\lambda \in {\sigma }_{p,1}\left(T\right)$ if and only if $\overline{\lambda }\in {\sigma }_{r,1}\left(T^{*}\right)$;

(2) $\lambda \in {\sigma }_{p,2}\left(T\right)$ if and only if $\overline{\lambda }\in {\sigma }_{r,2}\left(T^{*}\right)$;

(3) $\lambda \in {\sigma }_{p,3}\left(T\right)$ if and only if $\overline{\lambda }\in {\sigma }_{p,3}\left(T^{*}\right)$;

(4) $\lambda \in {\sigma }_{p,4}\left(T\right)$ if and only if $\overline{\lambda }\in {\sigma }_{p,4}\left(T^{*}\right)$;

(5) $\lambda \in {\sigma }_c\left(T\right)$ if and only if $\overline{\lambda }\in {\sigma }_c\left(T^{*}\right)$;

(6) $\lambda \in \sigma \left(T\right)$ if and only if $\overline{\lambda }\in \sigma \left(T^{*}\right)$.

Our notion of an operator matrix is taken from [13].

Definition 1.

Let $X_1,X_2$ be Hilbert spaces and consider the linear operators $A:\mathcal{D}\left(A\right)\subset X_1\to X_1,B:\mathcal{D}\left(B\right)\subset X_2\to X_1,C:\mathcal{D}\left(C\right)\subset X_1\to X_2$ and $D:\mathcal{D}\left(D\right)\subset X_2\to X_2$. Then, the matrix $\mathcal{A}$ = $\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ is called a block operator matrix in the product space $X_1\times X_2$, where $A,B,C$ and D are closable operators with a dense domain. For any ${\left(\begin{array}{cc}{x}_1& x_2\end{array}\right)}^T\in \mathcal{D}\left(\mathcal{A}\right)=(\mathcal{D}\left(A\right)\cap \mathcal{D}\left(C\right))\times (\mathcal{D}\left(B\right)\cap \mathcal{D}\left(D\right))$,

$\mathcal{A}\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{c}A{x}_1+B{x}_2\\ C{x}_1+D{x}_2\end{array}\right)$.

We use the following definition of a Hamiltonian operator matrix.

Definition 2

([13]). Let X be a complex Hilbert space. The closed densely defined block operator matrix

$$H=\left(\begin{array}{cc}A& B\\ C& -A^{*}\end{array}\right):\mathcal{D}\subset X\times X\to X\times X$$

is called a Hamiltonian operator, where A is a densely defined closed operator, and $B,C$ are symmetric operators. In addition, if $B,C$ are non-negative self-adjoint operators, then we call H a non-negative Hamiltonian operator. In particular, if $C=0$, then we call H an upper triangular infinite dimensional Hamiltonian operator.

The following Lemma will play an important role in the proof of our main theorems.

Lemma 2.

If $H=\left(\begin{array}{cc}A& B\\ C& -A^{*}\end{array}\right):\mathcal{D}\left(H\right)\subset X\times X\to X\times X$ is infinite dimensional Hamiltonian operator and J is the unite symplectic operator matrix. Then, ${\sigma }_r\left(H\right)=Ø$ if and only if ${\sigma }_r\left(JHJ\right)=Ø$.

Proof. 

Let $\mathfrak{J}=iJ$, then we know from $J^{*}=J^{-1}=-J$ that $\mathfrak{J}={\mathfrak{J}}^{-1}={\mathfrak{J}}^{*},$ and

$$JHJ=\mathfrak{J}(-H)\mathfrak{J}.$$

So, $JHJ$ is similar to that of $-H$. Therefore, they have the same spectral properties, i.e.,

$${\sigma }_r\left(JHJ\right)={\sigma }_r(-H).$$

which implies that ${\sigma }_r\left(H\right)=Ø$ if and only if ${\sigma }_r\left(JHJ\right)=Ø$. □

3. Main Results

Theorem 1.

If $H=\left(\begin{array}{cc}A& B\\ C& -A^{*}\end{array}\right):\mathcal{D}\left(H\right)\subset X\times X\to X\times X$ is an infinite dimensional Hamiltonian operator and satisfies $\rho \left(H\right)\ne Ø$ and ${\sigma }_r\left(H\right)=Ø$, then $JH={\left(JH\right)}^{*}$.

Proof. 

It is easily seen that ${\left(JH\right)}^{*}\supseteq JH$ always holds. To prove the converse is also true, we need only show that $\mathcal{D}\left(H^{*}\right)\subset \mathcal{D}\left(JHJ\right)$.

As $\rho \left(H\right)\ne Ø$, then for any $\lambda \in \rho \left(H\right)$, according to part (6) of Lemma 1, we have $\overline{\lambda }\in \rho \left(H^{*}\right)$. According to the definition of a resolvent set, $H^{*}-\overline{\lambda }I$ is injective and ${(H^{*}-\overline{\lambda }I)}^{-1}$ is bounded.

On the other hand, from $JHJ\subset H^{*}$, we have

$$H^{*}{\mid }_{\mathcal{D}\left(JHJ\right)}=JHJ.$$

Thus, $JHJ-\overline{\lambda }I$ is injective and ${(JHJ-\overline{\lambda }I)}^{-1}$ is a bounded operator. So,

$$\overline{\lambda }\in {\sigma }_{r1}\left(JHJ\right)\bigcup \rho \left(JHJ\right).$$

From Lemma 2, if ${\sigma }_r\left(H\right)=Ø$, then ${\sigma }_{r1}\left(JHJ\right)=Ø$, which implies that $\overline{\lambda }\in \rho \left(JHJ\right)$.

For any $x^{*}\in \mathcal{D}\left(H^{*}\right)$, there exists $x\in \mathcal{D}\left(JHJ\right)$ such that

$$(JHJ-\overline{\lambda }I)x=(H^{*}-\overline{\lambda }I)x^{*}.$$

In view of $(H^{*}-\overline{\lambda }I){|}_{\mathcal{D}\left(JHJ\right)}=JHJ-\overline{\lambda }I$, we can obtain that

$$(H^{*}-\overline{\lambda }I)(x-x^{*})=0.$$

It is clear from $H^{*}-\overline{\lambda }I$ being an injective operator that $x^{*}=x\in \mathcal{D}\left(JHJ\right)$. This show that

$$\mathcal{D}\left(H^{*}\right)\subseteq \mathcal{D}\left(JHJ\right),$$

which implies $JH\supseteq {\left(JH\right)}^{*}$. □

Remark 1.

From the proof of Theorem 1, we can see that the condition ${\sigma }_r\left(H\right)=Ø$ can be replaced with ${\sigma }_{r,1}\left(H\right)=Ø$ or ${\sigma }_{r,1}\left(JHJ\right)=Ø$.

For an unbounded block operator matrix, it is difficult to calculate its residual spectra. For this reason, we will give a criterion for a symplectic of a Hamiltonian operator matrix which does not use the residual spectra.

Theorem 2.

Let $H=\left(\begin{array}{cc}A& B\\ C& -A^{*}\end{array}\right):\mathcal{D}\left(H\right)\subset X\times X\to X\times X$ be an infinite dimensional Hamiltonian operator satisfying $\rho \left(H\right)\cap iR\ne Ø$, where $iR$ denotes the union of all of pure imaginary numbers and 0. Then, $JH={\left(JH\right)}^{*}$.

Proof. 

Let $\lambda \in \rho \left(H\right)\cap iR$; following the same methods as in Lemma 2, we can show that $-\lambda \in \rho \left(JHJ\right)\cap \rho \left(H^{*}\right)$. Then, for any $x^{*}\in \mathcal{D}\left(H^{*}\right)$, there exists $x\in \mathcal{D}\left(JHJ\right)$ such that

$$(JHJ+\lambda I)x=(H^{*}+\lambda I)x^{*}.$$

In view of $(H^{*}+\lambda I){|}_{\mathcal{D}\left(JHJ\right)}$=$JHJ+\lambda I$, we can obtain that

$$(H^{*}+\lambda I)(x-x^{*})=0.$$

It is clear from $H^{*}+\lambda I$ being an injective operator that we have $x^{*}=x\in \mathcal{D}\left(JHJ\right)$. Thus,

$$\mathcal{D}\left(H^{*}\right)\subseteq \mathcal{D}\left(JHJ\right).$$

This complete the proof of $JH\supseteq {\left(JH\right)}^{*}$. The opposite sense is always true. □

From the above two theorem conditions, we found that it is very easy to judge the symplectic self-adjointness of a Hamiltonian operator, and the common condition is that its resolvent set is not empty. In addition, the invertibility is consistent with the existence of bounded inverses under the condition of symplectic self-adjointness. So, we have a great interest in whether the Hamiltonian operator is a symplectic self-adjoint operator under the condition $\rho \left(H\right)=Ø$. The following example can give an answer.

Example 1.

Let H be an infinite dimensional Hamiltonian operator on $L^2[0,1]\times L^2[0,1]$ given by

$$H=\left(\begin{array}{cc}A& 0\\ 0& -A^{*}\end{array}\right),$$

where $A=i\frac{d}{dt},\mathcal{D}\left(A\right)=\{x\in L^2[0,1]:x^{\prime }\in L^2[0,1],x\left(0\right)=x\left(1\right)=0\}$. From [1], we know that $\sigma \left(A\right)=\mathbb{C}$ and ${\left(A^{*}\right)}^{*}=A.$ Then,

$$H^{*}=\left(\begin{array}{cc}{A}^{*}& 0\\ 0& -{\left(A^{*}\right)}^{*}\end{array}\right)=\left(\begin{array}{cc}{A}^{*}& 0\\ 0& -A\end{array}\right)=JHJ.$$

On the other hand, we can obtain from $\rho \left(A\right)=Ø$ that $\rho (-A^{*})=Ø$; therefore, $\rho \left(H\right)=\rho \left(A\right)\cap \rho (-A^{*})=Ø.$

Example 1 illustrates that $\rho \left(H\right)\ne Ø$ is not a sufficient condition for the symplectic self-adjointness of Hamiltonian operators.

In the following, we use another example to illustrate the validity of Theorem 2.

Example 2.

Let $X=L^2\left([0,1]\right)$ and H be the infinite dimensional Hamiltonian operator on $X\times X$, given by

$$H=\left(\begin{array}{cc}A& |A^{*}|\\ {\left|A\right|}^2& -A^{*}\end{array}\right),$$

where $A=i\frac{d}{dt},\mathcal{D}\left(A\right)=\{x\in X:x\left(t\right)$ is absolutely continuous, $x^{\prime }\in X,x\left(0\right)=0\}$ and $\left|A\right|={\left(A^{*}A\right)}^{\frac{1}{2}}.$ Then, $JH={\left(JH\right)}^{*}.$

Proof. 

We easily know from the conditions that A is invertible, and thus $A^{*}$ is invertible, which implies that $A^{*}A$ is invertible. As

$$H=\left(\begin{array}{cc}A& |A^{*}|\\ {\left|A\right|}^2& -A^{*}\end{array}\right)$$

$$=\left(\begin{array}{cc}I& A{\left(A^{*}A\right)}^{-1}\\ 0& I\end{array}\right)\left(\begin{array}{cc}0& |A^{*}|+A{\left(A^{*}A\right)}^{-1}{A}^{*}\\ {\left|A\right|}^2& 0\end{array}\right)\left(\begin{array}{cc}I& \overline{-{\left(A^{*}A\right)}^{-1}{A}^{*}}\\ 0& I\end{array}\right)$$

$$=\left(\begin{array}{cc}I& {\left(A^{*}\right)}^{-1}\\ 0& I\end{array}\right)\left(\begin{array}{cc}0& |A^{*}{|+I|}_{D\left(A^{*}\right)}\\ {\left|A\right|}^2& 0\end{array}\right)\left(\begin{array}{cc}I& -A^{-1}\\ 0& I\end{array}\right)\triangleq RTS,$$

we obtain that

$$H^{*}=\left(\begin{array}{cc}I& 0\\ -{\left(A^{*}\right)}^{-1}& I\end{array}\right)\left(\begin{array}{cc}0& {\left(A^{*}A\right)}^{*}\\ \left(\right|A^{*}{|+I|}_{D\left(A^{*}\right)}{)}^{*}& 0\end{array}\right)\left(\begin{array}{cc}I& 0\\ A^{-1}& I\end{array}\right).$$

Using the self-adjointness of $A^{*}A$ and Corollary 3.5.1 in [11], we know that $|A^{*}{|+I|}_{\mathcal{D}\left({\mathcal{A}}^{*}\right)}$ is a self-adjoint operator. Hence,

$$H^{*}=\left(\begin{array}{cc}{A}^{*}& {\left|A\right|}^2\\ |A^{*}|& -A\end{array}\right)=JHJ.$$

On the other hand, we need to prove that $\rho \left(H\right)\cap iR\ne Ø.$

Since $R,S$ are invertible, we obtain that the invertibility of H is equivalent to $|A^{*}{|+I|}_{\mathcal{D}\left(A^{*}\right)}.$

In fact,

$$\parallel \left(\right|A^{*}{|+I|}_{\mathcal{D}\left(A^{*}\right)})x\parallel \ge \parallel x\parallel ,\forall x\in \mathcal{D}\left(A^{*}\right),$$

which implies that $|A^{*}{|+I|}_{\mathcal{D}\left(A^{*}\right)}$ is bounded below. Then, $0\in {\sigma }_r\left(\right|A^{*}{|+I|}_{\mathcal{D}\left(A^{*}\right)})\bigcup \rho (|A^{*}{|+I|}_{\mathcal{D}\left(A^{*}\right)})$.

As $|A^{*}{|+I|}_{\mathcal{D}\left(A^{*}\right)}$ is a self-adjoint operator, according to part (1) of Lemma 1.2.4 in [11], we can obtain that $|A^{*}{|+I|}_{\mathcal{D}\left(A^{*}\right)}$ is invertible and so

$$\rho \left(H\right)\cap iR\ne Ø.$$

□

Next, we will discuss when an infinite dimensional Hamiltonian operator is a symplectic self-adjoint operator without the restriction $\rho \left(H\right)\ne Ø$.

Theorem 3.

Let $H=\left(\begin{array}{cc}A& B\\ C& -A^{*}\end{array}\right):\mathcal{D}\left(H\right)\subset X\times X\to X\times X$ be an infinite dimensional Hamiltonian operator. If ${\sigma }_{p,1}\left(H\right)\cap iR\ne Ø$, then $JH={\left(JH\right)}^{*}$.

Proof. 

As noted earlier, it is clear that $JH\subseteq {\left(JH\right)}^{{}^{*}}$. As one should expect, the argument of ${\left(JH\right)}^{*}\subseteq JH$ is somewhat parallel to that of the proof of Theorems 1 and 2.

For $\lambda \in {\sigma }_{p,1}\left(H\right)\cap iR$, by the definition of ${\sigma }_{p,1}\left(H\right)$, we obtain that

$$JHJ+\lambda I=J(H-\lambda )J$$

is surjective. Furthermore, it is known from Lemma 1 that $\overline{\lambda }=-\lambda \in {\sigma }_{r,1}\left(H^{*}\right).$ So, $H^{*}+\lambda I$ is injective.

For any $x^{*}\in \mathcal{D}\left(H^{*}\right)$, there exists $x\in \mathcal{D}\left(JHJ\right)$ such that

$$(JHJ+\lambda I)x=(H^{*}+\lambda I)x^{*}.$$

Due to $H^{*}{|}_{\mathcal{D}\left(JHJ\right)}=JHJ$, we can obtain that

$$(H^{*}+\lambda I)(x-x^{*})=0.$$

From the injectivity of $H^{*}+\lambda I$, we note that $x^{*}=x\in \mathcal{D}\left(JHJ\right)$, which implies that $JH={\left(JH\right)}^{*}$.

Although, in general for a Hamiltonian operator, $\rho \left(H\right)\ne Ø$ is not a necessary and sufficient condition, combining Lemma 4.3.1 and Lemma 4.4.2 in [13] and Theorem 2, we can obtain a necessary and sufficient condition for some special Hamiltonian operators. □

Theorem 4.

Let $H=\left(\begin{array}{cc}A& B\\ C& -A^{*}\end{array}\right):\mathcal{D}\left(H\right)\subset X\times X\to X\times X$ be a Hamiltonian operator. If there exists a constant $M>0$ such that

$$|(By,y)+(Cx,x)|\ge M\Vert \left(\begin{array}{c}x\\ y\end{array}\right){\Vert }^2,forany{\left(\begin{array}{cc}x& y\end{array}\right)}^T\in \mathcal{D}\left(H\right)$$

is fulfilled, then $JH={\left(JH\right)}^{*}$ if and only if $0\in \rho \left(H\right)$.

Proof. 

Suppose that the Hamiltonian operator H is symplectic self-adjoint and there exists a constant $M>0,$ such that

$$|(By,y)+(Cx,x)|\ge M\Vert \left(\begin{array}{c}x\\ y\end{array}\right){\Vert }^2,forany{(x,y)}^T\in \mathcal{D}\left(H\right).$$

Then, from Lemma 4.3.1 and Lemma 4.4.2 in [13], we obtain that

$$0\in \rho \left(H\right).$$

So,

$$\rho \left(H\right)\cap iR\ne Ø.$$

The converse is clear from Theorem 2. □

Corollary 1.

Let $H=\left(\begin{array}{cc}A& B\\ C& -A^{*}\end{array}\right):\mathcal{D}\left(H\right)\subset X\times X\to X\times X$ be a non-negative Hamiltonian operator and one of the following conditions be fulfilled.

(1) $0\in \rho \left(B\right)\cap \rho \left(C\right)$;

(2) $0\in \rho \left(A\right)$ and $A^{-1}{B}^{\frac{1}{2}}$, ${\left(A^{*}\right)}^{-1}{C}^{\frac{1}{2}}$ are bounded linear operators.

Then, $JH={\left(JH\right)}^{*}$ if and only if $0\in \rho \left(H\right)$.

Proof. 

Suppose that $0\in \rho \left(H\right)$. By the special case of Theorem 2, $JH={\left(JH\right)}^{*}$ is trivial. Conversely, if $JH={\left(JH\right)}^{*}$, it can be asserted that $0\notin {\sigma }_{r,1}\left(H\right)$.

In fact, if we assume that $0\in {\sigma }_{r,1}\left(H\right)$, we may employ Lemma 1 to obtain that

$$0\in {\sigma }_{p,1}\left(H^{*}\right)={\sigma }_{p,1}(-H).$$

which is a contradiction with $H^{*}=JHJ$.

Moreover, from Theorem 4.3.7 in [13], we know that when a non-negative Hamiltonian operator H satisfies one of the conditions (1) or (2), H is invertible and its inverse is bounded, i.e.,

$$0\in \rho \left(H\right)\cup {\sigma }_{r,1}\left(H\right).$$

In consideration of $0\notin {\sigma }_{r,1}\left(H\right)$, we can obtain $0\in \rho \left(H\right)$. □

4. Conclusions

Through the research in this article, we have given some necessary and sufficient conditions for symplectic self-adjointness of infinite-dimension Hamiltonian operators. The research method in this article is extensive and new, and it starts from the most basic theory to characterize the conditions for symplectic self-adjointness of infinite-dimension Hamiltonian operators that do not satisfy diagonal dominance or skew-diagonal dominance, which further develops the research on symplectic self-adjointness of infinite-dimension Hamiltonian operators. However, some conclusions have relatively restrictive conditions, and there is a lack of specific examples for the main conclusions. These issues will be further optimized in future research.