On the Equality A = A₁A₂ for Linear Relations

Abstract

Assume that A, $A_1$, and $A_2$ are three selfadjoint linear relations (multi-valued linear operators) in a certain complex Hilbert space. In this study, conditions are presented for the multi-valued operator equality $A=A_1{A}_2$ to hold when the inclusion $A\subset A_1{A}_2$ is assumed to be satisfied. The present study is strongly motivated by the invalidity of a classical result from A. Devinatz, A. E. Nussbaum, and J. von Neumann in the general case of selfadjoint linear relations. Two types of conditions for the aforementioned equality to hold are presented. Firstly, a condition is given in terms of the resolvent sets of the involved objects, which does not depend on the product structure of the right-hand side, $A_1{A}_2$. Secondly, a condition is also presented where the structure of the right-hand side is taken into account. This one is based on the notion of the $\mathfrak{L}$-stability of a linear operator under linear subspaces. It should be mentioned that the classical Devinatz–Nussbaum–von Neumann theorem is obtained as a particular case of one of the main results.

1. Introduction

Assume that $\mathfrak{H}$ is a complex Hilbert space and A, $A_1$, and $A_2$ are three selfadjoint linear operators in $\mathfrak{H}$. A classical result from A. Devinatz, A. E. Nussbaum, and J. von Neumann says that if $A\subset A_1{A}_2$, then, in fact, one has equality, i.e., $A=A_1{A}_2$, cf. [1].

This result is strongly connected to the commutativity of selfadjoint operators and related topics; see, for instance, [2,3,4,5,6] and the references therein.

The concept of a linear relation in a Hilbert space generalizes the concept of a linear operator to that of a multi-valued operator. This mathematical object goes back at least to R. Arens who presented a systematic study in [7]. Since then, it has proved very useful in different areas of mathematics and applied mathematics and it has been studied in various contexts; cf. [8,9].

More specifically, the theory of selfadjoint linear relations in Hilbert spaces plays an important role in modeling phenomena in quantum physics, quantum chemistry, engineering, and control theory. There are situations when many objects (selfadjoint operators or relations) are involved in such a mathematical model and it is important to decide if their product is also a selfadjoint object. In the present study, the answer is obtained with the help of a third object, namely a selafadjoint relation which is assumed to be a subrelation of their product. In the context of linear operators, some related results can be found in [10,11].

The main aim in this study is to generalize the result of Devinatz, Nussbaum, and von Neumann in the context of linear relations in complex Hilbert spaces. It will be shown that the result does not hold for all linear relations; an example will reveal this fact. However, some necessary and sufficient conditions will be provided for the validity of the result in the context of selfadjoint linear relations.

The mechanism of proving the results in the case of linear operators mainly uses the spectral theory of selfadjoint operators. The main ingredient for the new results within this study is a new notion, namely the notion of the $\mathfrak{L}$-stability of a linear operator in linear subspaces. This new notion plays a fundamental role in the extension of the results from linear operators to linear relations.

This study is organized as follows. In Section 2, some preparatory material concerning linear relations in complex Hilbert spaces are stated. Section 3 presents an example which can be seen as a motivation for this study. The stability of linear operators in certain linear subspaces is introduced in Section 4. Finally, Section 5 is devoted to the main results. An example within this section reveals the potential of the obtained results as tools in the study of Dirac structures in infinite-dimensional spaces; some recent developments in the theory of Dirac structures can be found in [12,13].

2. Linear Relations in Linear Spaces

Assume that $\mathfrak{H}$ is a complex Hilbert space whose inner product is denoted by $\langle ·,·\rangle$. A linear relation, or relation for short, A, in $\mathfrak{H}$ is a linear subspace of $\mathfrak{H}\times \mathfrak{H}$, the Cartesian product of $\mathfrak{H}$ and itself. The domain and the range of A are linear subspaces of $\mathfrak{H}$; they are defined by the following:

$$\mathrm{dom}A=\{x:\{x,y\}\in A\},\mathrm{ran}A=\{y:\{x,y\}\in A\}.$$

Furthermore, $\ker A$ and $\mathrm{mul}A$ stand for the kernel and the multi-valued part of A, which are the linear spaces defined by

$$\ker A=\{x:\{x,0\}\in A\},\mathrm{mul}A=\{y:\{0,y\}\in A\}.$$

The linear relation A is the graph of a linear operator if and only if $\mathrm{mul}A=\left\{0\right\}$. The inverse relation, $A^{-1}$, of A is given by

$$A^{-1}=\{\{y,x\}:\{x,y\}\in A\}.$$

The following identities express the duality of A and its inverse, respectively:

$$\mathrm{dom}{A}^{-1}=\mathrm{ran}A,\mathrm{ran}{A}^{-1}=\mathrm{dom}A,\ker A^{-1}=\mathrm{mul}A,\mathrm{mul}{A}^{-1}=\ker A.$$

For the relations A and B in $\mathfrak{H}$, the operator-like sum $A+B$ is the relation in $\mathfrak{H}$, defined by

$$A+B=\left\{\right\{x,y+z\}:\{x,y\}\in A,\{x,z\}\in B\},$$

while the product $BA$ is a linear relation, also in $\mathfrak{H}$, given by

$$BA=\left\{\right\{x,z\}:\{x,y\}\in A,\{y,z\}\in B\}.$$

For $\lambda \in \mathbb{C}$, the relation $\lambda A$ in $\mathfrak{H}$ is defined by

$$\lambda A=\left\{\right\{x,\lambda y\}:\{x,y\}\in A\},$$

while $A-\lambda I$, where I is the identity operator in $\mathfrak{H}$, is given by

$$A-\lambda I=\left\{\right\{x,y-\lambda x\}:\{x,y\}\in A\}.$$

The notation $\rho \left(A\right)$ stands for the resolvent set of the linear relation A in a Hilbert space, $\mathfrak{H}$; it consists of all complex numbers, $\lambda$, for which ${(A-\lambda I)}^{-1}$ is a bounded, everywhere-defined linear operator. The adjoint of the linear relation A in $\mathfrak{H}$ is the closed linear relation $A^{\ast }$ defined by

$$A^{\ast }=\{\{f,f^{\prime }\}\in \mathfrak{H}\times \mathfrak{H}:\langle f^{\prime },h\rangle =\langle f,h^{\prime }\rangle \mathrm{for}\mathrm{all}\{h,h^{\prime }\}\in A\}.$$

It is clear that the double adjoint $A^{\ast \ast }$ is the closure of the relation A. Furthermore, for any linear relation, A, in a Hilbert complex space, $\mathfrak{H}$, it is known that

$$\ker A^{\ast }\oplus \overline{\mathrm{ran}}A=\mathfrak{H}.$$

The linear relation A in the complex Hilbert space $\mathfrak{H}$ is said to be nonnegative if $\langle f^{\prime },f\rangle \ge 0$ for all $\{f,f^{\prime }\}\in A$. The linear relation A in the Hilbert space $\mathfrak{X}$ is said to be selfadjoint if $A^{\ast }=A$, so that it is automatically closed. The linear operator A in $\mathfrak{H}$ is said to be positive if $\langle Af,f\rangle >0$ for all nonzero $f\in \mathfrak{H}$ and nonnegative if $\langle Af,f\rangle \ge 0$ for all $f\in \mathfrak{H}$. Any nonnegative linear operator, A, admits a unique square root; it is a nonnegative linear operator denoted by $A^{{\displaystyle \frac{1}{2}}}$ which satisfies the condition ${\left(A^{{\displaystyle \frac{1}{2}}}\right)}^2=A$. Moreover, it is well known that $\ker A^{{\displaystyle \frac{1}{2}}}=\ker A$. More information concerning the square root of nonnegative linear operators and relations can be found, for instance, in [14], while for the general theory of linear relations, the reader may consult the monograph [8].

3. The Motivation

Firstly, a class of selfadjoint linear relations in complex Hilbert spaces is presented; it will be used in the next example.

Lemma 1.

Assume that $\mathfrak{H}$ is a complex Hilbert space and ${\mathfrak{L}}_1$ and ${\mathfrak{L}}_2$ are two linear subspaces of $\mathfrak{H}$. Define the linear relation A in $\mathfrak{H}$ by $A={\mathfrak{L}}_1\times {\mathfrak{L}}_2$. Then, A is selfadjoint in $\mathfrak{H}$ if and only if ${\mathfrak{L}}_1$ and ${\mathfrak{L}}_2$ are both closed subspaces and they are orthogonal to each other. In particular, if $\mathfrak{L}$ is a closed linear subspace of $\mathfrak{H}$, then the linear relation $A=\mathfrak{L}\times {\mathfrak{L}}^{\perp }$ is selfadjoint in $\mathfrak{H}$.

Proof. 

It is easily seen that $A^{\ast }={\mathfrak{L}}_2^{\perp }\times {\mathfrak{L}}_1^{\perp }$. Therefore, A is selfadjoint in $\mathfrak{H}$ if and only if ${\mathfrak{L}}_1={\mathfrak{L}}_2^{\perp }$ and ${\mathfrak{L}}_2={\mathfrak{L}}_1^{\perp }$ or, equivalently, ${\mathfrak{L}}_1$ and ${\mathfrak{L}}_2$ are both closed subspaces and they are orthogonal to each other. □

Based on Lemma 1, an example that motivates the results of this study is presented next. Actually, it is shown that the result of Devinatz, Nussbaum, and von Neumann presented at the beginning of this study is not valid for the general case of selfadjoint linear relations.

Example 1.

Assume that $\mathfrak{H}$ is a complex Hilbert space and let $\mathfrak{M}$, $\mathfrak{L}$, and $\mathfrak{N}$ three closed subspaces of $\mathfrak{H}$ such that

$$\mathfrak{M}⊊\mathfrak{L}⊊\mathfrak{N}.$$

(1)

We define the following linear relations: $A=\mathfrak{L}\times {\mathfrak{L}}^{\perp }$; $A_1=\mathfrak{M}\times {\mathfrak{M}}^{\perp }$; and $A_2=\mathfrak{N}\times {\mathfrak{N}}^{\perp }$. Lemma 1 shows that A, $A_1$, and $A_2$ are selfadjoint linear relations in $\mathfrak{H}$. Using (1), one has

$${\mathfrak{N}}^{\perp }⊊{\mathfrak{L}}^{\perp }⊊{\mathfrak{M}}^{\perp }.$$

(2)

Since $A_1{A}_2=\mathfrak{N}\times {\mathfrak{M}}^{\perp }$, it follows from (1) and (2) that

$$A⊊A_1{A}_2.$$

(3)

4. The Stability of Linear Operators in Linear Subspaces

A notion which was needed in the last section is now introduced; an elementary example follows.

Definition 1.

Let $\mathfrak{H}$ be a complex Hilbert space. Assume that A is a linear operator in $\mathfrak{H}$ and $\mathfrak{L}$ a linear subspace of $\mathfrak{H}$. One says that A is $\mathfrak{L}$-stable if

$$\{x\in \mathfrak{L}:\langle Ax,x\rangle =0\}=\left\{0\right\}.$$

Example 2.

Let $A:{\mathbb{R}}^2\to {\mathbb{R}}^2$, given by $A(x_1,x_2)=(x_1+2x_2,3x_1+4x_2)$, for all $\overline{x}=(x_1,x_2)\in {\mathbb{R}}^2$. We denote by $\langle ·,·\rangle$ the usual inner product of ${\mathbb{R}}^2$. Then,

$$\langle A\overline{x},\overline{x}\rangle =x_1^2+5x_1{x}_2+4x_2^2=(x_1+x_2)(x_1+4x_2).$$

It is easily seen that the condition $\langle A\overline{x},\overline{x}\rangle =0$ leads to $x_1=-x_2$ or $x_1=-4x_2$. Consequently, A is not stable if ${\mathfrak{L}}_1=\mathrm{span}\left\{(1,-1)\right\}$; ${\mathfrak{L}}_2=\mathrm{span}\left\{(4,-1)\right\}$; and ${\mathfrak{L}}_1+{\mathfrak{L}}_2={\mathbb{R}}^2$. But, it is stable in any one-dimensional subspace $\mathfrak{L}$ of ${\mathbb{R}}^2$ different from ${\mathfrak{L}}_1$ and ${\mathfrak{L}}_2$.

The next result characterizes the case of nonnegative linear operators in complex Hilbert spaces; it motivates one of the main results presented within the last section.

Proposition 1.

Assume that $\mathfrak{H}$ is a complex Hilbert space, A is a nonnegative selfadjoint linear operator in $\mathfrak{H}$, and $\mathfrak{L}$ is a linear subspace of $\mathfrak{H}$. Then, A is $\mathfrak{L}$-stable if and only if $\mathfrak{L}$ is a linear subspace of $\mathrm{dom}A\cap \mathrm{ran}A$.

Proof. 

Let $x\in \mathrm{dom}A$ such that $\langle Ax,x\rangle =0$. Then,

$$0=\langle Ax,x\rangle ={\displaystyle \langle }{\left(A^{{\displaystyle \frac{1}{2}}}\right)}^{2}x,x{\displaystyle \rangle }={\displaystyle \langle }{A}^{{\displaystyle \frac{1}{2}}}x,A^{{\displaystyle \frac{1}{2}}}x{\displaystyle \rangle }={\parallel A^{{\displaystyle \frac{1}{2}}}x\parallel }^2,$$

so that $A^{{\displaystyle \frac{1}{2}}}x=0$, which implies that

$$x\in \ker A^{{\displaystyle \frac{1}{2}}}=\ker A={\left(\mathrm{ran}{A}^{\ast }\right)}^{\perp }={\left(\mathrm{ran}A\right)}^{\perp }.$$

(4)

Assume now that A is $\mathfrak{L}$-stable and let $x\in \mathfrak{L}$. Then, $x\in \mathrm{dom}A$ such that $\langle Ax,x\rangle =0$. It follows from (4) that $x\in {\left(\mathrm{ran}A\right)}^{\perp }$. Since x must be 0, it follows that $x\in \mathrm{ran}A$, which leads to $x\in \mathrm{dom}A\cap \mathrm{ran}A$. Consequently, $\mathfrak{L}$ is a subspace of $\mathrm{dom}A\cap \mathrm{ran}A$.

Conversely, assume that $\mathfrak{L}$ is a linear subspace of $\mathrm{dom}A\cap \mathrm{ran}A$. Let $x\in \mathfrak{L}$ such that $\langle Ax,x\rangle =0$. Then, it follows from (4) that

$$x\in \mathrm{dom}A\cap \mathrm{ran}A\cap {\left(\mathrm{ran}A\right)}^{\perp }=\left\{0\right\},$$

which shows that, in fact, A is $\mathfrak{L}$-stable. This completes the proof. □

As one of the reviewers pointed out, it should be mentioned that the concept of L-stability is strongly connected to the concept of sesquilinear form in functional analysis. In the generous context of nonnegative or positive operators, such forms play an important role in the understanding of some important properties of the associated operator. For instance, if A is a selfadjoint operator, then the associated quadratic form is real-valued, and if the operator is positive, then the corresponding quadratic form is also positive.

5. Main Results

The following results are of some general interest.

Lemma 2.

Assume that $\mathfrak{H}$ is a complex Hilbert space and let A and B be two linear relations in $\mathfrak{H}$ such that $A\subset B$. If there exists $\lambda \in \rho \left(A\right)\cap \rho \left(B\right)$, then $A=B$.

Proof. 

It follows from $A\subset B$ that $A-\lambda I\subset B-\lambda I$ so that

$${(A-\lambda I)}^{-1}\subset {(B-\lambda I)}^{-1}.$$

(5)

Since both ${(A-\lambda I)}^{-1}$ and ${(B-\lambda I)}^{-1}$ are bounded, everywhere-defined linear operators, it follows from (5) that ${(A-\lambda I)}^{-1}={(B-\lambda I)}^{-1}$ or, equivalently, $A-\lambda I=B-\lambda I$, which further leads to $A=B$, as desired. □

A characterization of the equality of two selfadjoint linear relations is provided next.

Proposition 2.

Assume that $\mathfrak{H}$ is a complex Hilbert space and let A and B be two linear relations in $\mathfrak{H}$ such that $A\subset B$. Also, assume that A is a selfadjoint linear relation. Then, $A=B$ if and only if there exists $\lambda \in (\mathbb{C}\setminus \mathbb{R})\cap \rho \left(B\right)$.

Proof. 

Clearly, if $A=B$, then B is selfadjoint, so that one can choose $\lambda \in (\mathbb{C}\setminus \mathbb{R})\subset \rho \left(A\right)=\rho \left(B\right)$. Conversely, let $\lambda \in (\mathbb{C}\setminus \mathbb{R})\subset \rho \left(B\right)$. Since A is selfadjoint, it follows that $\lambda \in \rho \left(A\right)$. Applying Lemma 2, one now has $A=B$. □

A direct consequence of Proposition 2 reads as follows. It offers a first result of Devinatz, Nussbaum, and von Neumann’s type in the case of linear relations.

Corollary 1.

Assume that $\mathfrak{H}$ is a complex Hilbert space and let A, $A_1$, and $A_2$ be three selfadjoint linear relations in $\mathfrak{H}$ such that $A\subset A_1{A}_2$. Then, $A=A_1{A}_2$ if and only if there exists $\lambda \in (\mathbb{C}\setminus \mathbb{R})\cap \rho \left(A_1{A}_2\right)$.

Proof. 

The result follows immediately from Proposition 2 with $B=A_1{A}_2$. □

Under some natural assumptions, a new result concerning the case of the equality in certain multi-valued operators’ inclusion is proven next.

Theorem 1.

Let $\mathfrak{H}$ be a complex Hilbert space; let A be a linear relation in $\mathfrak{H}$ such that there exists a non-real complex number, $\lambda \in \rho \left(A\right)$; let $A_1$ be a selfadjoint linear relation in $\mathfrak{H}$; and let $A_2$ be a selfadjoint linear operator in $\mathfrak{H}$. If $A\subset A_1{A}_2$ and $A_2$ is $\mathrm{ran}{A}_1$-stable, then $A=A_1{A}_2$.

Proof. 

It follows from $A\subset A_1{A}_2$ that $A-\lambda I\subset A_1{A}_2-\lambda I$, so that

$${(A-\lambda I)}^{-1}\subset {(A_1{A}_2-\lambda I)}^{-1}.$$

(6)

Since $\lambda \in \rho \left(A\right)$, it follows that ${(A-\lambda I)}^{-1}$ is a bounded, everywhere-defined linear operator in $\mathfrak{H}$. It will be shown that ${(A_1{A}_2-\lambda I)}^{-1}$ is also an operator.

Let $m\in \mathrm{mul}{(A_1{A}_2-\lambda I)}^{-1}$, so that $m\in \ker (A_1{A}_2-\lambda I)$, which implies that $\{m,\lambda m\}\in A_1{A}_2$. Then, $\{m,x\}\in A_2$ and $\{x,\lambda m\}\in A_1$, where $x=A_{2}m$. Since both $A_1$ and $A_2$ are selfadjoint, it follows that $\langle x,m\rangle \in \mathbb{R}$ and also that

$$\langle \lambda m,x\rangle =\langle x,\lambda m\rangle =\overline{\lambda }\langle x,m\rangle \in \mathbb{R}.$$

Consequently, $\langle x,m\rangle =0$, which can be rewritten as

$$\langle A_{2}m,m\rangle =0.$$

(7)

Since $\lambda m\in \mathrm{ran}{A}_1$, it also follows that

$$m\in \mathrm{ran}{A}_1.$$

(8)

Using the fact that $A_2$ is $\mathrm{ran}{A}_1$-stable, one obtains from (7) and (8) that $m=0$, so that ${(A_1{A}_2-\lambda I)}^{-1}$ is an operator. Furthermore,

$$\mathfrak{H}=\mathrm{dom}{(A-\lambda I)}^{-1}\subset \mathrm{dom}{(A_1{A}_2-\lambda I)}^{-1},$$

so that

$$\mathrm{dom}{(A-\lambda I)}^{-1}=\mathrm{dom}{(A_1{A}_2-\lambda I)}^{-1}=\mathfrak{H}.$$

(9)

Since both ${(A-\lambda I)}^{-1}$ and ${(A_1{A}_2-\lambda I)}^{-1}$ are operators, the equality in (9) shows that, in fact, ${(A-\lambda I)}^{-1}={(A_1{A}_2-\lambda I)}^{-1}$. Then, $A-\lambda I=A_1{A}_2-\lambda I$, which leads to the desired conclusion, $A=A_1{A}_2$. □

The next characterization follows from Theorem 1.

Corollary 2.

Let $\mathfrak{H}$ be a complex Hilbert space; let A be a linear relation in $\mathfrak{H}$ such that there exists a non-real complex number, $\lambda \in \rho \left(A\right)$; and let $A_1$ and $A_2$ be two selfadjoint linear relations in $\mathfrak{H}$. If $\ker A_1=\left\{0\right\}$, $A\subset A_1{A}_2$ and $A_1^{-1}$ are $\mathrm{dom}{A}_2$-stable, then $A=A_1{A}_2$.

Proof. 

It follows from the hypothesis that

$$A^{-1}\subset A_2^{-1}{A}_1^{-1}.$$

(10)

Also, $A_1^{-1}$ and $A_2^{-1}$ are two selfadjoint linear relations. Furthermore, ${\lambda }^{-1}$ is a non-real complex number which is in the resolvent set of $A^{-1}$. It follows from $\ker A_1=\left\{0\right\}$ that $A_1^{-1}$ is a selfadjoint linear operator in $\mathfrak{H}$ which is $\mathrm{ran}{A}_2^{-1}=\mathrm{dom}{A}_2$-stable. The application of Theorem 1 with $A^{-1}$, $A_2^{-1}$, and $A_1^{-1}$ instead of A, $A_1$, and $A_2$, respectively, leads to

$$A^{-1}=A_2^{-1}{A}_1^{-1},$$

so that the conclusion simply follows. □

Proposition 3.

Let $\mathfrak{H}$ be a complex Hilbert space, let A and $A_1$ be two selfadjoint linear relations in $\mathfrak{H}$, and let $A_2$ be a selfadjoint linear operator in $\mathfrak{H}$. If $A\subset A_1{A}_2$ and $A_2$ is $\mathrm{ran}{A}_1$-stable, then $A=A_1{A}_2$.

Proof. 

Since A is a selfadjoint linear relation, it follows that there exists a non-real complex number, $\lambda \in \rho \left(A\right)$. The result now follows immediately from Theorem 1. □

The next result is a direct consequence of Corollary 2 and Proposition 3.

Corollary 3.

Let $\mathfrak{H}$ be a complex Hilbert space, and let A, $A_1$, and $A_2$ be three selfadjoint linear relations in $\mathfrak{H}$. If $\ker A_1=\left\{0\right\}$, $A\subset A_1{A}_2$, and $A_1^{-1}$ are $\mathrm{dom}{A}_2$-stable, then $A=A_1{A}_2$.

Proof. 

It follows from the hypothesis that

$$A^{-1}\subset A_2^{-1}{A}_1^{-1}.$$

(11)

Also, $A^{-1}$, $A_1^{-1}$ and $A_2^{-1}$ are three selfadjoint linear relations. It follows from $\ker A_1=\left\{0\right\}$ that $A_1^{-1}$ is a selfadjoint linear operator in $\mathfrak{H}$ which is $\mathrm{ran}{A}_2^{-1}=\mathrm{dom}{A}_2$-stable. The application of Proposition 3 with $A^{-1}$, $A_2^{-1}$, and $A_1^{-1}$ instead of A, $A_1$, and $A_2$, respectively, leads to

$$A^{-1}=A_2^{-1}{A}_1^{-1}.$$

so that the conclusion $A=A_1{A}_2$ simply follows. □

Corollary 4.

Let $\mathfrak{H}$ be a complex Hilbert space, let A and $A_1$ be two selfadjoint linear relations in $\mathfrak{H}$, and let $A_2$ be a positive selfadjoint linear operator in $\mathfrak{H}$. If $A\subset A_1{A}_2$, then $A=A_1{A}_2$.

Proof. 

Since any positive operator in $\mathfrak{H}$ is $\mathfrak{H}$-stable, the statement is now a direct consequence of Proposition 3. □

Under a condition concerning the kernels of the involved linear relations, another characterization is presented next. Its proof is quite similar to the one of Theorem 1. However, for the sake of completeness, a full proof is included.

Theorem 2.

Let $\mathfrak{H}$ be a complex Hilbert space, let A and $A_1$ be two selfadjoint linear relations in $\mathfrak{H}$, and let $A_2$ be a nonnegative selfadjoint linear operator in $\mathfrak{H}$. If $A\subset A_1{A}_2$ and $\ker A_2\subset \ker A_1$, then $A=A_1{A}_2$.

Proof. 

It follows from $A\subset A_1{A}_2$ that $A-iI\subset A_1{A}_2-iI$, so that

$${(A-iI)}^{-1}\subset {(A_1{A}_2-iI)}^{-1}.$$

(12)

Since $i\in \rho \left(A\right)$, it follows that ${(A-iI)}^{-1}$ is a bounded, everywhere-defined linear operator in $\mathfrak{H}$. It will be shown that ${(A_1{A}_2-iI)}^{-1}$ is also an operator.

Let $m\in \mathrm{mul}{(A_1{A}_2-iI)}^{-1}$, so that $m\in \ker (A_1{A}_2-iI)$, which implies that $\{m,im\}\in A_1{A}_2$. Then, $\{m,x\}\in A_2$ and $\{x,im\}\in A_1$, where $x=A_{2}m$. Since both $A_1$ and $A_2$ are selfadjoint, it follows that $\langle x,m\rangle \in \mathbb{R}$ and also that

$$\langle im,x\rangle =\langle x,im\rangle =-i\langle x,m\rangle \in \mathbb{R}.$$

Consequently, $\langle x,m\rangle =0$, which can be rewritten as

$$\langle A_{2}m,m\rangle =0.$$

(13)

As in the proof of Proposition 1, it follows from (13) that

$$m\in \ker A_2.$$

(14)

Since $im\in \mathrm{ran}{A}_1$, it also follows that

$$m\in \mathrm{ran}{A}_1.$$

(15)

Using the hypothesis, (14), and (15), one obtains that

$$m\in \ker A_2\cap \mathrm{ran}{A}_1\subset \ker A_1\cap \mathrm{ran}{A}_1=\left\{0\right\},$$

so that $m=0$, which shows that ${(A_1{A}_2-iI)}^{-1}$ is an operator. Furthermore,

$$\mathfrak{H}=\mathrm{dom}{(A-iI)}^{-1}\subset \mathrm{dom}{(A_1{A}_2-iI)}^{-1},$$

so that

$$\mathrm{dom}{(A-iI)}^{-1}=\mathrm{dom}{(A_1{A}_2-iI)}^{-1}=\mathfrak{H}.$$

(16)

Since both ${(A-iI)}^{-1}$ and ${(A_1{A}_2-iI)}^{-1}$ are operators, the equality in (9) shows that, in fact, ${(A-iI)}^{-1}={(A_1{A}_2-iI)}^{-1}$. Then, $A-iI=A_1{A}_2-iI$, which leads to the desired conclusion, $A=A_1{A}_2$ □

A direct consequence of Theorem 2 is now stated; its proof is similar to the one of Corollary 2.

Corollary 5.

Let $\mathfrak{H}$ be a complex Hilbert space, let A and $A_2$ be two selfadjoint linear relations in $\mathfrak{H}$, and let $A_1$ be a nonnegative selfadjoint linear relation in $\mathfrak{H}$. If $A\subset A_1{A}_2$, $\ker A_1=\left\{0\right\}$, and $\mathrm{mul}{A}_1\subset \mathrm{mul}{A}_2$, then $A=A_1{A}_2$.

Proof. 

It follows from the hypothesis that

$$A^{-1}\subset A_2^{-1}{A}_1^{-1}.$$

(17)

Also, $A^{-1}$, $A_1^{-1}$, and $A_2^{-1}$ are three selfadjoint linear relations. It follows from $\ker A_1=\left\{0\right\}$ that $A_1^{-1}$ is, in fact, a nonnegative linear operator in $\mathfrak{H}$. Furthermore,

$$\ker A_1^{-1}=\mathrm{mul}{A}_1\subset \mathrm{mul}{A}_2=\ker A_2^{-1}.$$

The application of Theorem 2 with $A^{-1}$, $A_2^{-1}$, and $A_1^{-1}$ instead of A, $A_1$, and $A_2$, respectively, leads to

$$A^{-1}=A_2^{-1}{A}_1^{-1},$$

so that the conclusion $A=A_1{A}_2$ simply follows. □

The next result can be seen as a direct generalization of the classical Devinatz–Nussbaum–von Neumann theorem to the case of linear relations.

Theorem 3.

Assume that $\mathfrak{H}$ is a complex Hilbert space; A is a linear relation in $\mathfrak{H}$ such that $\rho \left(A\right)$ contains some $\lambda \in \mathbb{C}\setminus \mathbb{R}$ and its conjugate, $\overline{\lambda }$; $A_1$ is a symmetric linear relation in $\mathfrak{H}$ and $A_2$ is a symmetric linear operator in $\mathfrak{H}$ such that $\mathrm{mul}{A}_1\cap \ker A_2=\left\{0\right\}$. Then, $A\subset A_1{A}_2$ implies that $A=A_1{A}_2$.

Proof. 

Clearly, $A-\lambda I\subset A_1{A}_2-\lambda I$, so that

$${(A-\lambda I)}^{-1}\subset {(A_1{A}_2-\lambda I)}^{-1}.$$

(18)

Since $\lambda \in \rho \left(A\right)$, it follows that ${(A-\lambda I)}^{-1}$ is a bounded, everywhere-defined linear operator in $\mathfrak{H}$, which shows that $\mathrm{dom}{(A-\lambda I)}^{-1}=\mathfrak{H}$. Consequently, the inclusion of (18) leads to

$$\mathrm{dom}{(A-\lambda I)}^{-1}=\mathrm{dom}{(A_1{A}_2-\lambda I)}^{-1}=\mathfrak{H}.$$

(19)

It will be shown next that ${(A_1{A}_2-\lambda I)}^{-1}$ is a linear operator, i.e., its multi-valued part is trivial. Let $u\in \mathrm{mul}{(A_1{A}_2-\lambda I)}^{-1}=\ker (A_1{A}_2-\lambda I)$, so that $\{u,0\}\in A_1{A}_2-\lambda I$, which shows that $\{u,\lambda u\}\in A_1{A}_2$, so that

$$\{A_{2}u,\lambda u\}\in A_1.$$

(20)

Consider h, an arbitrary element of $\mathfrak{H}$. Since $\overline{\lambda }\in \rho \left(A\right)$, it follows that $\mathrm{ran}(A-\overline{\lambda }I)=\mathfrak{H}$. Consequently, there exists $x\in \mathrm{dom}A\subset \mathrm{dom}{A}_1{A}_2$ such that $\{x,h\}\in A-\overline{\lambda }I$. This further implies that $\{x,h+\overline{\lambda }x\}\in A\subset A_1{A}_2$, which leads to

$$\{A_{2}x,h+\overline{\lambda }x\}\in A_{1.}$$

(21)

Using the fact that $A_1$ is symmetric, it follows from (20) and (21) that

$$\langle \lambda u,A_{2}x\rangle =\langle A_{2}u,h+\overline{\lambda }x\rangle .$$

(22)

Furthermore, the fact that $A_2$ is symmetric leads to

$$\langle A_{2}u,\overline{\lambda }x\rangle =\langle \lambda u,A_{2}x\rangle .$$

(23)

A combination of (22) and (23) leads to

$$\langle A_{2}u,h\rangle =0,$$

(24)

for all $h\in \mathfrak{H}$, which implies that $A_{2}u=0$, so that $u\in \ker A_2$. Using (20), one now has $\{0,\lambda u\}\in A_1$, so that $\lambda u\in \mathrm{mul}{A}_1$, which shows that $u\in \mathrm{mul}{A}_1$. Therefore,

$$u\in \mathrm{mul}{A}_1\cap \ker A_2=\left\{0\right\},$$

so that $u=0$. Hence, ${(A_1{A}_2-\lambda I)}^{-1}$ is a linear operator. Using (18), (19), and the fact that both ${(A-\lambda I)}^{-1}$ and ${(A_1{A}_2-\lambda I)}^{-1}$ are linear operators, one now has

$${(A-\lambda I)}^{-1}={(A_1{A}_2-\lambda I)}^{-1}.$$

Thus, $A-\lambda I=A_1{A}_2-\lambda I$, so that $A=A_1{A}_2$, as desired. □

Theorem 3 is, in fact, an adaptation to the case of the linear relations of a result obtained and offered by the Academic Editor. This result can now be seen as a consequence of Theorem 3; it is stated below.

Corollary 6.

If A, $A_1$, and $A_2$ are linear operators in a complex Hilbert space, $\mathfrak{H}$, $\rho \left(A\right)$ contains some $\lambda \in \mathbb{C}$ and its conjugate, $\overline{\lambda }$, and $A_1$ and $A_2$ are symmetric operators, then $A\subset A_1{A}_2$ implies that $A=A_1{A}_2$.

Proof. 

Since $A_1$ is an operator, it follows that $\mathrm{mul}{A}_1=\left\{0\right\}$, so that the condition $\mathrm{mul}{A}_1\cap \ker A_2=\left\{0\right\}$ in Theorem 3 is satisfied. Now, the present result simply follows from Theorem 3. □

Finally, since any complex non-real number belongs to the resolvent set of any selfadjoint linear operator in a complex Hilbert space, the classical Devinatz–Nussbaum–von Neumann theorem, which can be found in ([1], Corollary 1), is now a direct consequence of Corollary 6; it is stated below.

Corollary 7.

If A, $A_1$, and $A_2$ are linear operators in a complex Hilbert space, $\mathfrak{H}$, A is a selfadjoint operator, and $A_1$ and $A_2$ are symmetric operators, then $A\subset A_1{A}_2$ implies that $A=A_1{A}_2$.

Example 3.

Consider two Hilbert spaces, $\mathcal{E}$ (the space of efforts) and $\mathcal{F}$ (the space of flows), and assume that there exists a unitary operator, $r_{\mathcal{E},\mathcal{F}}$, from $\mathcal{E}$ to $\mathcal{F}$; its inverse is denoted by $r_{\mathcal{F},\mathcal{E}}$. Let $\mathcal{F}\oplus \mathcal{E}$ be the Hilbert space determined by the Cartesian product space $\mathcal{F}\times \mathcal{E}$ equipped with the natural inner product:

$${\langle (f_1,e_1)\mid (f_2,e_2)\rangle }_{\mathcal{F}\oplus \mathcal{E}}={\langle f_1\mid f_2\rangle }_{\mathcal{F}}+{\langle e_1\mid e_2\rangle }_{\mathcal{E}},$$

(25)

where $f_1$, $f_2\in \mathcal{F}$ and $e_1$, $e_2\in \mathcal{E}$. Further, we define an indefinite inner product in $\mathcal{F}\times \mathcal{E}$ by

$${\left[(f_1,e_1)\mid (f_2,e_2)\right]}_{\mathcal{B}}={\langle f_1\mid r_{\mathcal{E},\mathcal{F}}{e}_2\rangle }_{\mathcal{F}}+{\langle e_1\mid r_{\mathcal{F},\mathcal{E}}{f}_2\rangle }_{\mathcal{E}}.$$

(26)

The Cartesian product $\mathcal{F}\times \mathcal{E}$ equipped with the inner product ${[·\mid ·]}_{\mathcal{B}}$ is called the bond space $\mathcal{B}$. We denote by $b=(f,e)$ an element of $\mathcal{B}$. For a linear space, $\mathcal{L}\subset \mathcal{B}$, the orthogonal complement ${\mathcal{L}}^{[\perp ]}$ of $\mathcal{L}$ is defined by

$${\mathcal{L}}^{[\perp ]}=\{b^{\prime }\in \mathcal{B},{[b\mid b^{\prime }]}_{\mathcal{B}}=0,\forall b\in \mathcal{L}\}.$$

(27)

The bond space $\mathcal{B}$ is non-degenerate (see [15]). Let $\mathcal{D}$ be a linear subspace of $\mathcal{B}$. Then, $\mathcal{D}$ is said to be a Dirac structure in $\mathcal{B}$ if $\mathcal{D}={\mathcal{D}}^{[\perp ]}$. Courant (see [16]) introduced Dirac structures for finite-dimensional systems and Dorfman (see [17]) generalized them for the first time to infinite-dimensional systems. A Dirac structure is a key concept for port-Hamiltonian systems, as it defines the internal geometric structure through which the physical system is interacting with the environment (see [18] and the references therein). Therefore, it is used to formalize the concept of a power-conserving interconnection in the modeling and control of physical systems such as mechanical systems and networks (see [19,20] and the references therein).

The theory of linear relations offers a natural framework to study the Dirac structures. In the language of linear relations, one has the following characterization:

$$\mathcal{D}isaDiracstructureon\mathcal{B}ifandonlyif\mathcal{D}=-r_{\mathcal{F},\mathcal{E}}{\mathcal{D}}^{\ast }{r}_{\mathcal{F},\mathcal{E}},$$

where ${\mathcal{D}}^{\ast }$ stands for the adjoint of $\mathcal{D}$. The computation of the adjoint for infinite-dimensional systems is, in many cases, very difficult. Therefore, one may interested in alternative criteria to verify that a physical structure is a Dirac structure.

The linear relation A in the complex Hilbert space $\mathfrak{H}$ is said to be skewadjoint if $A=-A^{\ast }$. Any skewadjoint linear relation can be seen as a Dirac structure in $\mathfrak{H}$. Furthermore, A is skewadjoint if and only if $iA$ is selfadjoint in the same Hilbert space, $\mathfrak{H}$.

Consequently, all the obtained results within the present study can provide similar results for Dirac structures by replacing the selfadjoint linear relation A with $\mathfrak{D}$, where $\mathfrak{D}$ is a Dirac structure.

6. Conclusions

Taking into account the important role of the theory of selfadjoint linear relations in Hilbert spaces in modeling phenomena in quantum physics, quantum chemistry, engineering, and control theory, the study of the equation $A=A_1{A}_2$ helps to decide when the product of two objects (selfadjoint operators or relations) involved in such a mathematical model is also a selfadjoint object.

The answer is obtained with the help of a third object, namely a selfadjoint relation which is assumed to be a subrelation of the product. More precisely, one starts with three linear relations, A, $A_1$, and $A_1$, acting on the same complex Hilbert space such that $A\subset A_1{A}_2$. The selfadjointness of the product $A_1{A}_2$ is assured by the equality in the aforementioned inclusion.

The present study offers two types of conditions for the equality $A=A_1{A}_2$ to hold. The first condition is given in terms of the resolvent sets of the objects involved. This one does not depend on the product structure of the right-hand side, $A_1{A}_2$. The results involving this condition are presented in Proposition 2 and Corollary 1.

The second condition is given in terms of the structure of the right-hand side, $A_1{A}_2$. This is based on a new notion, namely the notion of the $\mathfrak{L}$-stability of a linear operator in linear subspaces. This new notion plays a fundamental role in the extension of the results from linear operators to linear relations. The obtained statements involving this second condition are presented in Theorems 1–3 and their corollaries.

It should be pointed out that these results have potential as tools in the study of Dirac structures in infinite-dimensional spaces. Example 3 offers initial insight into such a kind of application.