AMS Subject Classification (2020). 47B02, 47 B15.
Key words: conjugation, antilinear operator, antilinear normal operator

1 Introduction↩︎

The theory of bounded or unbounded linear operators on Hilbert space is well developed and there exists a huge quantity of research papers and books on linear operators. In constrast, conjugate-linear operators, or equivalently, antilinear operators (that is, operators \(T\) satisfying \[T(\alpha x+\beta y)=\overline{\alpha}\, T(x)+\overline{\beta}\, T(y) \quad \textrm{for} \quad\alpha, \beta\in \mathbb{C}, \, x, y\in {\mathcal{D}}(T))\] are rarely investigated in the literature.

Antilinear operators occur in the study of unbounded \(C\)-symmetric operators and \(C\)-self-adjoint operators. A densely defined linear operator \(T\) on a Hilbert space \({\mathcal{H}}\) is called \(C\)-symmetric (resp. \(C\)-self-adjoint) with respect to a conjugation \(C\) on \({\mathcal{H}}\) if \(CTC\subseteq T^*\) (resp. \(CTC= T^*).\) For such operators, the antilinear operators \(TC\) and \(T^*C\) play an important role (see [1], [2], [3], [4]). In the finite dimensional case, \(C\)-self-adjoint operators can be described by complex symmetric matrices, a well-studied class of matrices (see e.g. [5], [6]).

Antilinear unitary operators appear in quantum physics since the work of E. Wigner. By Wigner’s theorem [7], each symmetry of the set of pure states of a quantum mechanical system is given by a unitary or by an antiunitary operator. Wigner [8] also classified the antiunitary operators \(A\) for which the unitary operator \(A^2\) has a complete set of eigenvectors. Further, the \(\mathcal{P}\mathcal{T}\)- symmetry operator in quantum mechanics is an antilinear operator (see e.g. [9]).

The subject of the present paper are densely defined closed antilinear unbounded operators on general complex Hilbert spaces. Our particular emphasize is on the structure of unbounded antilinear normal operators.

Let us describe the contents of this paper. In Section 2, we develop basic definitions on antilinear operators. For a densely defined antilinear operator \(T\) its adjoint \(T^\dagger\) is defined by \[\langle Tx,y\rangle= \langle T^\dagger y, x\rangle \quad \textrm{for}\quad x\in {\mathcal{D}}(T), y\in {\mathcal{D}}(T^\dagger ).\] This is Wigner’s classical definition [8]. We show that for each densely defined closed antilinear operator \(T\) the linear operator \(T^\dagger T\) is self-adjoint and positive, so the definition  \(|T|:=\sqrt{T^\dagger T}\)  of the absolute value of \(T\) makes sense.

In Section 3, the polar decompositon of densely defined closed antilinear operators is obtained (Theorem 3) and we introduce some natural classes of antilinear operators. As in the linear case, we define that \(T\) is self-adjoint if \(T=T^\dagger\) and \(T\) is normal if \(T^\dagger T=TT^\dagger\). Antilinear normal operators are closely related to \(C\)-normal operators as defined in [10], see Remark 2. A densely defined closed antilinear operator \(T\) if self-adjoint if and only if \(T=C|T|=|T| C\) for some conjugation \(C\) (Theorem 4).

In Section 4, we give some basic characterizations of antilinear normal operators. It is shown that an antilinear operator \(T\) is normal if and only if there exists a conjugation \(C\) such the product \(TC\) is a linear normal operator (Proposition 8 and Theorem 9). Further, a densely defined closed antilinear operator \(T\) is normal if and only if there exists an antilinear unitary operator \(V\) such that \(T=V |T|=|T|V\) and \(T^\dagger =V^\dagger |T|=|T|V^\dagger\) (Theorem 11).

Section 5 is devoted to antilinear normal operators on the two-dimensional Hilbert space \(\mathbb{C}^2\). This simple case illustrates the general case and it delimits possible results.

In Section 6, we study an important model of antilinear normals. Let \(N_1\) and \(N_2\) be normal linear operators on Hilbert spaces \({\mathcal{H}}_1\) and \({\mathcal{H}}_2\), respectively, and define a normal linear operator \({\mathcal{N}}\) on \({\mathcal{H}}:={\mathcal{H}}_1\oplus {\mathcal{H}}_2\) by the operator matrix \[\begin{gather} {\mathcal{N}}=\left( \begin{array}{ll} N_1 & ~ 0 \\ ~0 & N_2 \end{array}\right). \end{gather}\] Further, let \({\mathcal{C}}\) be a conjugation of the Hilbert space \({\mathcal{H}}_1\oplus {\mathcal{H}}_2\) of the form \[\begin{gather} {\mathcal{C}}=\left( \begin{array}{ll} ~0 & C_2\\ C_1 & ~0 \end{array}\right) \end{gather}\] such that \({\mathcal{N}}{\mathcal{C}}={\mathcal{C}}{\mathcal{N}}\). Then the operator matrix \[\begin{gather} T:={\mathcal{N}}\, {\mathcal{C}}=\left( \begin{array}{ll} ~~~ 0 & N_1C_2 \\ N_2C_1 & ~~~ 0 \end{array}\right). \end{gather}\] defines an antilinear normal operator \(T\) on \({\mathcal{H}}={\mathcal{H}}_1\oplus {\mathcal{H}}_2\). Such operators \(T\) are investigated in Section 6 in detail.

The main result of this paper is Theorem 22 which describes the structure of antilinear normal operators. It is stated and discussed in Section 7 and proved in Section 8. According to this result each antilinear normal operator \(T\) decomposes as an orthogonal direct sum of an antilinear self-adjoint operator \(T_{\mathrm{sa}}\) and an antilinear completely non-self-adjoint operator \(T_{\mathrm{nsa}}\). The non-self-adjoint part \(T_{\mathrm{nsa}}\) is of the form of operators treated in Section 6. In particular, the results obtained in Section 6 provide criteria for the unitary equivalence and the irreducibility of such operators \(T_{\mathrm{nsa}}\).

In the finite-dimensional case, antilinear normal operators were studied and their structure was classified by F. Herbut and M. Vujicic [11] adapting Wigner’s description of antiunitaries. As note above, the present paper is about antilinear operators on general Hilbert spaces including in particular unbounded operators.

Let us fix some notational conventions that will be used throughout this paper. All Hilbert spaces are complex and their scalar products are linear in the first variables and conjugate-linear in the second. The symbols \({\mathcal{H}}\), \({\mathcal{H}}_1\), \({\mathcal{H}}_2\) always stand for Hilbert spaces. For an operator \(T\), its domain, its kernel and its range are denoted by \({\mathcal{D}}(T)\), \({\mathcal{N}}(T)\) and \({\mathcal{R}}(T)\), respectively. The spectral measure of a normal or self-adjoint linear operator \(N\) is denoted by \(E_N(\cdot)\).

For an antilinear operator \(A\) on \({\mathcal{H}}\) we define the resolvent set \(\rho(A)\) to be the set of numbers \(\lambda\) for which the operator \(A-\lambda I\) has a bounded inverse \((A-\lambda I)^{-1}\) defined on \({\mathcal{H}}\). Note that \((A-\lambda I)^{-1}\) is then a real-linear operator ([12]), that is, \[(A-\lambda I)^{-1}(\alpha x+\beta y)=\alpha\, (A-\lambda I)^{-1}x +\beta\, (A-\lambda I)^{-1}y \quad \textrm{for} ~\alpha, \beta\in \mathbb{R}, \,x, y\in {\mathcal{H}}.\] As in the linear case, the spectrum \(\sigma(A)\) is the complement set \(\mathbb{C}\backslash \rho(A).\)

For a subset \(M\) of \(\mathbb{C}\) let \(\overline{M}\) denote the closure of \(M\) and \(M^c:=\{\overline{z}: z\in M\}\) the complex conjugate of \(M\).

2 Antilinear Operators↩︎

In this paper, we prefer to speak about antilinear operators rather than conjugate-linear operators.

Let \({\mathcal{H}}\) be a Hilbert space with scalar product \(\langle\cdot,\cdot\rangle_{\mathcal{H}}\). We define another Hilbert space \({\mathcal{H}}^{\mathrm{anti}}\) as follows. The sets of \({\mathcal{H}}\) and \({\mathcal{H}}^{\mathrm{anti}}\) are the same and also the additions of vectors. The complex multiple \(\lambda \circ x\) of \(x\in{\mathcal{H}}\) in \({\mathcal{H}}^{\mathrm{anti}}\) is equal to the multiple \(\overline{\lambda} \, x\) in \({\mathcal{H}}\) and the scalar product of \({\mathcal{H}}^{\mathrm{anti}}\) is given by \[\begin{align} \langle x,y\rangle_{{\mathcal{H}}^{\mathrm{anti}}}=\langle y,x\rangle_{\mathcal{H}},\quad x,y \in {\mathcal{H}}. \end{align}\] To avoid confusion, for \(\lambda \in\mathbb{C}\) and \(x\in {\mathcal{H}}\), the symbol \(\lambda \circ x\) means the complex multiple of \(x\) in \({\mathcal{H}}^{\mathrm{anti}}\) and \(\lambda x\) denotes the complex multiple of \(x\) in \({\mathcal{H}}\).

We verify (for instance) linearity and conjugate-linearity of the new scalar product \(\langle \cdot,\cdot \rangle_{{\mathcal{H}}^{\mathrm{anti}}}\): For \(\lambda, \mu\in \mathbb{C}\) and \(x,y\in {\mathcal{H}}\), we derive \[\begin{align} \langle \lambda \circ x,\mu\circ y\rangle_{{\mathcal{H}}^{\mathrm{anti}}}=\langle \overline{\lambda}\, x,\overline{\mu}\, y\rangle_{{\mathcal{H}}^{\mathrm{anti}}}=\langle \overline{\mu}\, y,\overline{\lambda}\, x\rangle_{\mathcal{H}}=\overline{\mu}\, \lambda\, \langle y, x\rangle_{\mathcal{H}}=\lambda\, \overline{\mu} \langle x,y\rangle_{{\mathcal{H}}^{\mathrm{anti}}}. \end{align}\] Note that \(({\mathcal{H}}^{\mathrm{anti}})^{\mathrm{anti}}\) is just the Hilbert space \({\mathcal{H}}\). The embedding of \(x\in {\mathcal{H}}\) into \({\mathcal{H}}^{\mathrm{anti}}\) is denoted by \(\iota(x)\). In particular, for \(y\in {\mathcal{H}}^{\mathrm{anti}}\), \(\iota(y)\) denotes the embedding of \(y\) into \({\mathcal{H}}=({\mathcal{H}}^{\mathrm{anti}})^{\mathrm{anti}}\).

Let \(T\) be a mapping of a Hilbert space \({\mathcal{H}}_1\) into a Hilbert space \({\mathcal{H}}_2\). Clearly, \(T:{\mathcal{H}}_1\to {\mathcal{H}}_2\) is an antilinear operator if and only if \(T:{\mathcal{H}}_1^{\mathrm{anti}}\to {\mathcal{H}}_2\) is a linear operator, or equivalently, \(T:{\mathcal{H}}_1\to {\mathcal{H}}_2^{\mathrm{anti}}\) is a linear operator. This enables us to study antilinear operators by means of linear operators.

For mappings \(T:{\mathcal{H}}_1\to {\mathcal{H}}_2\) and \(S:{\mathcal{H}}^{\mathrm{anti}}_1\to {\mathcal{H}}_2\) we define mappings \(j(T):{\mathcal{H}}_1^{\mathrm{anti}}\to {\mathcal{H}}_2\) and \(j(S):{\mathcal{H}}_1\to {\mathcal{H}}_2\) by \[\begin{align} j(T)(y):&= T(\iota(y)), \quad y\in {\mathcal{H}}_1^{\mathrm{anti}},\\ j(S)(x):&= S(\iota(x)), \quad x\in{\mathcal{H}}_1. \end{align}\] Notions such as closedness, boundedness or density of domains are the same for \(T\) and \(j(T)\) and for \(S\) and \(j(S)\), because \({\mathcal{H}}_1\) and \({\mathcal{H}}_1^{\mathrm{anti}}\) have the same norms.

Let us recall the definition of the adjoint of a densely defined linear operator \(S:{\mathcal{H}}_1\to {\mathcal{H}}_2\). The domain of \(S^*\) is \[{\mathcal{D}}(S^\ast){=}\{ y \in {\mathcal{H}}_2:\text{There exists u \in {\mathcal{H}}_1 such that \langle Sx, y \rangle_2 {=} \langle x,u \rangle_1 forx \in {\mathcal{D}}(S)} \}.\] Since \({\mathcal{D}}(S)\) is dense in \({\mathcal{H}}_1\), then the vector \(u \in {\mathcal{H}}_1\) is uniquely determined by \(y\). Setting \(S^\ast y= u\), \(S^*\) becomes a linear operator \(S^\ast:{\mathcal{H}}_2\to {\mathcal{H}}_1\) such that \[\begin{align} \label{tstar} \langle Sx,y \rangle_2 = \langle x, S^\ast y \rangle_1 \quad\text{for}~~ x \in {\mathcal{D}}(S),~~ y \in {\mathcal{D}}(S^\ast). \end{align}\tag{1}\]

Now suppose that \(T\) is a densely defined antilinear operator of \({\mathcal{H}}_1\) into \({\mathcal{H}}_2\). Then \(j(T):{\mathcal{H}}_1^{\mathrm{anti}}\to{\mathcal{H}}_2\) is a linear operator with dense domain in \({\mathcal{H}}_1^{\mathrm{anti}}\), so its adjoint \(j(T)^*:{\mathcal{H}}_2\to {\mathcal{H}}_1^{\mathrm{anti}}\) is a linear operator which is defined by the preceding. We define the adjoint \(T^\dagger\) of \(T\) by \[T^\dagger y:=\iota(j(T)^*y), \quad y\in {\mathcal{H}}_2.\] This means that \[\begin{align} {\mathcal{D}}(T^\dagger){=}\{ y \in {\mathcal{H}}_2:&\text{There exists u \in {\mathcal{H}}_1 such that \langle Tx, y \rangle_2 {=} \langle u,x \rangle_1 forx \in {\mathcal{D}}(T)} \}, \\ & \quad\quad\quad\quad \quad T^\dagger y=u, \quad y\in {\mathcal{D}}(T^\dagger), \end{align}\] Since \(j(T)^*: {\mathcal{H}}_2\to {\mathcal{H}}_1^{\mathrm{anti}}\) is linear, \(T^\dagger=\iota\circ j(T)^*:{\mathcal{H}}_2\to {\mathcal{H}}_1\) is an antilinear operator and equation (1 ) reads as \[\begin{align} \label{tstar1} \langle Tx,y \rangle_2 = \langle T^\dagger y,x \rangle_1 \quad\text{for}~~ x \in {\mathcal{D}}(T),~~ y \in {\mathcal{D}}(T^\dagger). \end{align}\tag{2}\]

Note that (2 ) coincides with the definition of the adjoint of an antilinear operator according to Wigner [8]. The preceding approach to the definition of \(T^\dagger\) can be considered as a mathematical justification of Wigner’s definition.

As usual, the adjoint of a linear operator \(S\) is denoted by \(S^*\). Note that for an antilinear operator \(T\) and a linear operator \(S\) we have \((cT)^\dagger= c\, T^\dagger\) and \((cS)^*=\overline{c}\, S^*\) for \(c\in \mathbb{C}\), \(c\neq 0.\)

If \(A\) is a linear operator and \(S\) and \(T\) antilinear operators, then \(AT\), \(TA\) are antilinear, \(ST\) is linear and \[\begin{align} T^\dagger A^*\subseteq (AT)^\dagger,~~~ A^*T^\dagger \subseteq (TA)^\dagger,~~~ S^\dagger T^\dagger \subseteq (TS)^*. \end{align}\]

Let \(A:{\mathcal{H}}_1\to {\mathcal{H}}_2\) be a densely defined antilinear operator. Then \[\begin{align} j(A)^*j(A):{\mathcal{H}}_1^{\mathrm{anti}}\to {\mathcal{H}}_1^{\mathrm{anti}}, \quad j(A)j(A)^*:{\mathcal{H}}_2\to {\mathcal{H}}_2. \end{align}\] From the corresponding definitions we obtain \[\begin{align} \label{adaggera} A^\dagger A=\iota\circ (j(A)^* j(A))\circ\iota \end{align}\tag{3}\]

One could try to study antilinear operators by passing to linear operators on the Hilbert space \({\mathcal{H}}^{\mathrm{anti}}\) as done in the proof of Proposition 1. This has the disadvantage that an antilinear operator \(T:{\mathcal{H}}\to {\mathcal{H}}\) becomes a linear operator of \({\mathcal{H}}\) into another Hilbert space \({\mathcal{H}}^{\mathrm{anti}}\). We will use another way and try to express an antilinear operator as a product of some appropriate linear operator and some conjugation.

3 Polar Decomposition and Classes of Operators↩︎

For a densely defined closed linear operator \(T\) its absolute value \(|T|\) is the unique positive square root \(\sqrt{T^*T}\) of the positive self-adjoint operator \(T^*T\).

As in the linear case we have the following fact.

That is, \(U\) is an antilinear isometric map of the closed subspace \({\mathcal{K}}_1:={\mathcal{N}}(U)^\bot\), the initial space of \(U\), on the closed subspace \({\mathcal{R}}(U)\), the final space of \(U\), and \(U\) is zero on the complement of \({\mathcal{K}}_1\).

The following formula (5 ) is the polar decomposition for antilinear operators.

Many standard notions for linear operators carry over almost verbatim to antilinear operators. It is very easy to “guess" the following definitions.

Note that an antilinear densely defined operator \(T\) is formally normal if and only if \({\mathcal{D}}(T)\subseteq {\mathcal{D}}(T^\dagger)\) and \(\|Tx\|=\| T^\dagger x\|\) for all \(x\in {\mathcal{D}}(T)\); likewise \(T\) is normal if and only if \(T\) is formally normal and \({\mathcal{D}}(T)={\mathcal{D}}( T^\dagger)\).

The following theorem clarifies the structure of antilinear self-adjoint operators. For bounded operators, this result was stated as [14]. The unbounded case requires additional arguments.

The next proposition implies that the spectrum of antilinear self-adjoint operators is not empty and invariant under multiplication by complex numbers of modulus one.

4 Antilinear Normal Operators↩︎

As mentioned in the introduction, our main emphasize in this paper is on antilinear normal operators.

A simple characterization of normality is given in the following lemma.

A general method for the study of antilinear normals is to write them as products of linear normal operators with conjugations.

The following auxiliary lemma is used in the proof of Theorem 11.

5 Antilinear Normal Operators on \(2\)-dimensional Hilbert Space↩︎

For \(z\in \mathbb{C}\) we define an antilinear mapping \(T_z\) of \(\mathbb{C}^2\) by \[\begin{align} \label{defTz} T_z(x_1,x_2)=(\overline{z}\, \overline{x_2}, z\, \overline{x_1}), \quad (x_1,x_2)\in \mathbb{C}^2. \end{align}\tag{8}\] Let \(C\) denote the complex conjugation on \(\mathbb{C}\). Then \(T_z\) can be written as \[\begin{gather} T_z=\left( \begin{array}{ll} \overline{z} & 0\\ ~0 & z \end{array}\right) \left( \begin{array}{ll} 0 & C\\ C & 0 \end{array}\right)=\left( \begin{array}{ll} 0 & C\\ C & 0 \end{array}\right) \left( \begin{array}{ll} \overline{z} & 0\\ ~ 0& z \end{array}\right)=\left( \begin{array}{ll} 0 &\overline{z}\\ z & 0 \end{array}\right) \left( \begin{array}{ll} C & 0\\ 0 & C \end{array}\right). \end{gather}\] One easily verifies that \[\begin{align} \label{Tz2} T_z^\dagger=T_{\overline{z}}~~\textrm{and}~~~ T_vT_z=\left( \begin{array}{ll} \overline{vz}& ~0 \\ ~~ 0 & vz \end{array}\right) ~~~ \textrm{for}~~ v,z\in \mathbb{C}. \end{align}\tag{9}\] In particular, \[\begin{align} \label{TdaggerT} T_z^\dagger T_z=T_zT_z^\dagger =\overline{z}\, z\left( \begin{array}{ll} 1& 0 \\ 0 & 1 \end{array}\right), ~~~ z\in \mathbb{C}. \end{align}\tag{10}\] Thus, for any \(z\in\mathbb{C}\), \(T_z\) is an antilinear normal operator on \(\mathbb{C}^2\).

Obviously, \(T_z\) is an antilinear self-adjoint operator if and only if \(z\in \mathbb{R}\) and \(T_z\) is an antilinear unitary operator if and only if \(|z|=1.\)

Let \(\lambda\in \mathbb{C}\) and \((y_1,y_2)\in \mathbb{C}^2\). We consider the equation \[\begin{align} T_z(x_1,x_2)-\lambda (x_1,x_2)=(y_1,y_2), \end{align}\] that is, \[\begin{align} \overline{z}\, \overline{x_2}-\lambda x_1=y_1, \quad z\overline{x_1} -\lambda x_2=y_2. \end{align}\] These equations imply that \[\begin{align} (\overline{z}^2 -|\lambda|^2) x_1=\overline{z}\, \overline{ y_2} +\overline{\lambda}\, y_1,\quad (z^2 -|\lambda|^2) x_2=z\, \overline{y_1} +\overline{\lambda} \, y_2. \end{align}\] Therefore, if \[\begin{align} \label{lambdacond} |\lambda|\neq z, - z, \overline{z}, -\overline{z}, \end{align}\tag{11}\] then \(\lambda\) belongs to the resolvent set of \(T_z\) and \[\begin{align} \label{resolventz} x_1=(\overline{z}^2- |\lambda|^2)^{-1}(\overline{z}\, \overline{y_2}+\overline{\lambda}\, y_1),\quad x_2=(z^2-|\lambda|^2)^{-1}(z\, \overline{y_1}+\overline{\lambda}\, y_2). \end{align}\tag{12}\] In particular, we have the following result.

Formula (12 ) shows that for \(\lambda \neq 0\) the operator \((T_z-\lambda I)^{-1}\) is neither linear nor antilinear, but it is real-linear.

Now we look when two operators \(T_z\) and \(T_v\) are unitarily equivalent. We compute \[\begin{gather} \label{tztz} \left( \begin{array}{ll} ~0 & 1 \\ -1 & 0 \end{array}\right) T_z\left( \begin{array}{ll} 0 & - 1 \\ 1 & ~ 0 \end{array}\right)=T_{-\overline{z}}, \end{gather}\tag{13}\] \[\begin{gather} \label{tzt-z} \left( \begin{array}{ll} 1 &~ 0 \\ 0 & -1 \end{array}\right) T_z\left( \begin{array}{ll} 1 & ~0 \\ 0 & -1 \end{array}\right)=T_{-z}. \end{gather}\tag{14}\] This shows that \(T_z\) is unitarily equivalent to \(T_{-\overline{z}}\) and also to \(T_{-z}\). Therefore, each operator \(T_z\) is unitarily equivalent to an operator \(T_v\) with \(v\) from the closed right upper quarter plane \[\begin{align} \overline{K}:=\{v\in \mathbb{C}: 0\leq \arg (v)\leq \pi/2\}, \end{align}\] where we have set \(\arg(0):=0\).

6 A Model for Antilinear Normal Operators↩︎

Throughout this section, \(N_1\) and \(N_2\) are normal linear operators on Hilbert spaces \({\mathcal{H}}_1\) and \({\mathcal{H}}_2\), respectively, and \({\mathcal{N}}\) denotes the normal linear operator on \({\mathcal{H}}:={\mathcal{H}}_1\oplus {\mathcal{H}}_2\) given by the operator matrix \[\begin{gather} {\mathcal{N}}=\left( \begin{array}{ll} N_1 & ~ 0 \\ ~0 & N_2 \end{array}\right). \end{gather}\] Further, we assume that \({\mathcal{C}}\) be a conjugation on \({\mathcal{H}}_1\oplus {\mathcal{H}}_2\) of the form \[\begin{gather} {\mathcal{C}}=\left( \begin{array}{ll} ~0 & C_2\\ C_1 & ~0 \end{array}\right) \end{gather}\] such that \[\begin{align} \label{nccommute} {\mathcal{N}}{\mathcal{C}}={\mathcal{C}}{\mathcal{N}}. \end{align}\tag{15}\] Then we define an antilinear operator \(T\) on \({\mathcal{H}}_1\oplus {\mathcal{H}}_2\) by \[\begin{gather} \label{defiT} T:={\mathcal{N}}\, {\mathcal{C}}=\left( \begin{array}{ll} ~~~ 0 & N_1C_2 \\ N_2C_1 & ~~~ 0 \end{array}\right). \end{gather}\tag{16}\] This is the setup which will be kept in this section. In particular, we assume throughout that (15 ) holds.

The operator \(T\) appears as operator \(T_{\mathrm{nsa}}\) in the structure Theorem 22 in Section 7. Also, note that the operator \(T_z\) defined by (8 ) is a special case.

Since \({\mathcal{C}}\) is a conjugation, \(C_1\) is the adjoint of \(C_2:{\mathcal{H}}_2\to {\mathcal{H}}_1\), \(C_2\) is the adjoint of \(C_1:{\mathcal{H}}_1\to {\mathcal{H}}_2\), and we have \[\begin{align} \label{csquared} C_1C_2=I_{{\mathcal{H}}_2},~~~C_2C_1=I_{{\mathcal{H}}_1}. \end{align}\tag{17}\] Equation (15 ) is equivalent to \[\begin{align} \label{nccommute1} N_1C_2=C_2N_2~~ \textrm{and}~~ N_2C_1=C_1N_1. \end{align}\tag{18}\] Combined with (17 ) the latter implies that \[\begin{align} \label{n2n1} C_1N_1C_2=N_2~~ \textrm{and}~~ C_2N_2C_1=N_1. \end{align}\tag{19}\] Clearly, \(T\) is closed and densely defined and we have \[\begin{gather} T^\dagger:={\mathcal{C}}{\mathcal{N}}^*=\left( \begin{array}{ll} ~~~ 0 & C_1N_2^* \\ C_2N_1^* & ~~~ 0 \end{array}\right). \end{gather}\] It follows from \({\mathcal{N}}{\mathcal{C}}={\mathcal{C}}{\mathcal{N}}\) that \({\mathcal{C}}{\mathcal{N}}{\mathcal{C}}={\mathcal{N}}\) and hence \({\mathcal{C}}{\mathcal{N}}^*{\mathcal{C}}={\mathcal{N}}^*,\) so that \({\mathcal{C}}{\mathcal{N}}^*={\mathcal{N}}^*{\mathcal{C}}.\) Therefore \[\begin{align} TT^\dagger= {\mathcal{N}}{\mathcal{C}}{\mathcal{C}}{\mathcal{N}}^*={\mathcal{N}}{\mathcal{N}}^*={\mathcal{N}}^*{\mathcal{N}},~~T^\dagger T={\mathcal{C}}{\mathcal{N}}^*{\mathcal{N}}{\mathcal{C}}={\mathcal{N}}^*{\mathcal{C}}{\mathcal{C}}{\mathcal{N}}={\mathcal{N}}^*{\mathcal{N}}. \end{align}\] Hence \(T\) is an antilinear normal operator on \({\mathcal{H}}_1\oplus {\mathcal{H}}_2.\) Further, (15 ) implies that \[\begin{align} \label{T2} T^2= {\mathcal{N}}{\mathcal{C}}{\mathcal{N}}{\mathcal{C}}={\mathcal{N}}{\mathcal{N}}{\mathcal{C}}{\mathcal{C}}={\mathcal{N}}^2. \end{align}\tag{20}\]

For the next considerations restrictions on the spectrum of the operator \(N_1\) are useful. For this the following definition is convenient.

We will use this notion for the set \[\begin{align} K:=\{z\in \mathbb{C}: 0<\arg z\leq \pi/2\} . \end{align}\] (We adopt the convention that the origin \(0\) is not in \(K\).)

Now we investigate the resolvent of \(T\). Let \(\lambda\in \mathbb{C}\). Given \((y_1,y_2)\in {\mathcal{H}}_1\oplus {\mathcal{H}}_2\), we consider the equation \[\begin{align} \label{Tlambda} T(x_1,x_2)-\lambda (x_1,x_2)=(y_1,y_2), \end{align}\tag{22}\] or equivalently, \[\begin{align} N_1\, C_2x_2-\lambda x_1=y_1, \quad N_2 C_1x_1 -\lambda x_2=y_2. \end{align}\] Applying \(N_1C_2\) to the second equation, multiplying the first by \(\overline{\lambda}\) and adding both equations we obtain \[\begin{align} \label{nc1} (N_1C_2N_2C_1 -\lambda \overline{\lambda}\, )x_1=N_1C_2y_2 + \overline{\lambda}\, y_1. \end{align}\tag{23}\] Similarly, we apply the operator \(N_2C_1\) to the first equation, multiply the second by \(\overline{\lambda}\) and add both equations. This yields \[\begin{align} \label{nc2} (N_2C_1 N_1C_2 -\lambda \overline{\lambda}\, ) x_2=N_2C_1y_1 +\overline{\lambda}\, y_2. \end{align}\tag{24}\] Using (18 ) and (17 ) it follows from (23 ) and (24 ) that \[\begin{align} \label{nc3} (N_1^2 -|\lambda|^2 )x_1=N_1C_2y_2 + \overline{\lambda}\, y_1,~~~ (N_2^2 -|\lambda|^2 )x_2=N_2C_1y_1 + \overline{\lambda}\, y_2. \end{align}\tag{25}\] By Lemma 16, \(\rho(N_2^2)\) is the complex conjugate of the set \(\rho(N_1^2)\), so \(|\lambda|^2\in \rho(N_1^2)\) implies \(|\lambda|^2\in \rho(N_2^2)\). Therefore, if \(|\lambda|^2\in \rho(N_1^2)\) it follows from (25 ) that \[\begin{align} \tag{26} x_1&=(N_1^2- |\lambda|^2)^{-1}(N_1C_2 y_2+\overline{\lambda}\, y_1),\\ x_2&=(N_2^2-|\lambda|^2)^{-1}(N_2C_1 y_1+\overline{\lambda}\, y_2),\tag{27} \end{align}\] and therefore \(\lambda \in \rho(T)\). Further, if \(|\lambda|^2\in \sigma(N_1^2)\), we conclude from (25 ) that \(\lambda \notin \rho(T)\), so that \(\lambda \in \sigma(T)\).

We illustrate this result by a simple example.

Let \(S\) be a linear or antilinear operator on \({\mathcal{H}}\). If \(S_0\) is a self-adjoint linear resp. antilinear operator acting a closed subspace \({\mathcal{H}}_0\) of \({\mathcal{H}}\) such that \(S_0\subseteq S\), we say that \(S_0\) is a self-adjoint part of \(A\). Note that in this case \({\mathcal{D}}(S_0)\subseteq {\mathcal{D}}(S)\), \({\mathcal{D}}(S_0)\) is dense in \({\mathcal{H}}_0\) and \(S_0\) maps \({\mathcal{D}}(S_0)\) into \({\mathcal{H}}_0\). The self-adjoint operator \(S_0=0\) on \({\mathcal{D}}(S_0)={\mathcal{H}}_0=\{0\}\) is called the trivial self-adjoint part of \(S\).

The following fact will be used later several times.

7 A Structure Theorem for Antilinear Normal Operators↩︎

Let us recall two definitions which have been used already earlier in this paper. Throughout we abbreviate \[\begin{align} K:=\{z\in \mathbb{C}: 0<\arg z\leq \pi/2\}~~ \textrm{and}~~ K_\mathbb{T}:=K\cap \mathbb{T}. \end{align}\] According to Definition 6 a normal operator \(N\) is called \(K\)-supported if \[E_N(\mathbb{C}\backslash K)=0.\]

The following theorem is the main result of this paper. By an antinormal operator we mean an antilinear normal operator.

Theorem 22 will be proved in Section 8. Here we discuss this result, derive some consequences, and collect a number of useful facts and formulas for an antilinear normal operator \(T\). For this we retain the notation of Theorem 22.

First we note that the operator \(T_{\mathrm{nsa}}\) is just the operator model treated in Section 6. Therefore, since \(N_1\) is supported by the set \(K\), all results and formulas from this section apply to \(T_{\mathrm{nsa}}\). Let us briefly recall some of them.

First, \(T_{\mathrm{nsa}}\) is completely non-self-adjoint according to Proposition 21 which explaines our notation \(T_{\mathrm{nsa}}\). Formula (19 ) yields \[\begin{align} N_2=C_1N_1C_2\quad \textrm{and}\quad N_1= C_2N_2C_1. \end{align}\] Next, by Proposition 17 we have \({\mathcal{N}}(T_{\mathrm{nsa}}-\lambda I)=\{0\}\) for all \(\lambda \in \mathbb{C}.\) Even more, under the stronger condition \(\sigma (N_1)\subseteq K\) the resolvent set \(\rho(T_{\mathrm{nsa}})\) is whole complex plane \(\mathbb{C}\) and the resolvent of \(T_{\mathrm{nsa}}\) is given by the formulas (26 ) and (27 ). Finally, Proposition 18 is about the unitary equivalence of operators \(T_{\mathrm{nsa}}\) and Proposition 19 characterizes when \(T_{\mathrm{nsa}}\) is irreducible.

Next we state some useful formulas. On the Hilbert space \({\mathcal{H}}= {\mathcal{H}}_{\mathrm{sa}}\oplus {\mathcal{H}}_{\mathrm{nsa}}\equiv {\mathcal{H}}_{\mathrm{nsa}}\oplus({\mathcal{H}}_1\oplus {\mathcal{H}}_2)\) we define a conjugation \[C =C_0\oplus {\mathcal{C}}= C_0\oplus \left( \begin{array}{ll} ~ 0 & C_2\\ C_1 & ~ 0 \end{array}\right)\] and a normal linear operator \[N=S\oplus {\mathcal{N}}= S\oplus \left( \begin{array}{ll} N_1 & ~ 0\\ ~ 0 & N_2 \end{array}\right).\] From \(T_{\mathrm{sa}}=SC_0=C_0S\) and \(T_{\mathrm{nsa}}={\mathcal{N}}{\mathcal{C}}={\mathcal{C}}{\mathcal{N}}\) we obtain \[\begin{align} \label{formNC}T =NC=CN=T_{\mathrm{sa}}\oplus T_{\mathrm{nsa}}= SC_0\oplus \left( \begin{array}{ll} ~ 0 & N_1C_2\\ N_2C_1 & ~0 \end{array}\right). \end{align}\tag{33}\] Formula (33 ) describes the structure of the antilinear normal operator \(T\).

Since \(N_1\) is supported by \(K\), the spectral measure \(E_N\) of the normal operator \(N\) is supported by the closed right half plane \(H:=\{z\in \mathbb{C}: {\rm Re\, z}\, \geq 0\}\). Then \[{\mathcal{H}}={\mathcal{H}}_{\mathrm{sa}}\oplus {\mathcal{H}}_1\oplus {\mathcal{H}}_2=E_N([0,+\infty)){\mathcal{H}}\oplus E_N(K){\mathcal{H}}\oplus E_N(K^c){\mathcal{H}}\] and \(N\) is given by \[\begin{align} N=S \oplus N_1\oplus N_2\equiv \int_0^{+\infty}t\, dE_N(t) \oplus \int_K z\, dE_N(z) \oplus \int_{K^c} z\, dE_N(z). \end{align}\] The conjugation \(C\) leaves \({\mathcal{H}}_{\mathrm{sa}}\) invariant and gives a bijection of \({\mathcal{H}}_1\) and \({\mathcal{H}}_2\). Moreover, \[\begin{gather} \tag{34} (T_{\mathrm{nsa}})^\dagger =\left( \begin{array}{ll} ~~~ 0 & N_1^*C_2\\ N_2^*C_1 & ~~~0 \end{array}\right), \quad (T_{\mathrm{nsa}})^2=\left( \begin{array}{ll} N_1^2 & ~~~0\\ 0 & N_2^2 \end{array}\right), \\ (T_{\mathrm{nsa}})^\dagger T_{\mathrm{nsa}}=\left( \begin{array}{ll} N_1^*N_1 & ~~~0\\ ~~~ 0 & N_2^*N_2 \end{array}\right),\quad | T_{\mathrm{nsa}}|=\left( \begin{array}{ll} |N_1| & ~~0\\ ~~0 & |N_2| \end{array}\right).\tag{35} \end{gather}\] Since \(T_{\mathrm{nsa}}={\mathcal{N}}{\mathcal{C}}={\mathcal{C}}{\mathcal{N}}\) implies \(T_{\mathrm{nsa}}^\dagger={\mathcal{N}}^*{\mathcal{C}}={\mathcal{C}}{\mathcal{N}}^*\), we compute for \(k,n\in \mathbb{N}_0\), \[\begin{gather} ((T_{\mathrm{nsa}})^\dagger)^k(T_{\mathrm{nsa}})^n =({\mathcal{N}}^*)^k{\mathcal{N}}^n{\mathcal{C}}^{k+n}=\left( \begin{array}{ll} (N_1^*)^kN_1^n & ~~~0\\ 0 & (N_2)^*)^kN_2^n \end{array}\right) {\mathcal{C}}^{k+n} . \end{gather}\] Note that \({\mathcal{C}}^{k+n}=I_{{\mathcal{H}}_{\mathrm{nsa}}}\) if \(k+n\) is even and \({\mathcal{C}}^{k+n}={\mathcal{C}}\) if \(k+n\) is odd.

Since \({\mathcal{N}}={\mathcal{C}}{\mathcal{N}}{\mathcal{C}}\), it follows from Lemma 20 that for the spectral measure \(E_{\mathcal{N}}\) of \({\mathcal{N}}\) and each Borel subset \(M\) of \(\mathbb{C}\) we have \[\begin{align} \label{enc}{\mathcal{C}}E_{\mathcal{N}}(M){\mathcal{C}}=E_{\mathcal{N}}(M^c). \end{align}\tag{36}\]

Now we formulate some consequences of Theorem 22. First we restate the special case of Theorem 22 for antiunitaries.

The next proposition deals with irreducible antilinear normal operators.

Finally, we characterize the unitary equivalence of antinormal operators.

8 Proof of Theorem 22↩︎

The following auxiliary lemma is used in the proof of Theorem 22. It is a slight generalization of a lemma in [16]:

The implication (ii)\(\to\)(i) of Theorem 22 is easily verified. Now we begin with the proof of the main implication (i)\(\to\)(ii).

By Theorem 11, \(T^2\) is a normal linear operator and there exist an antiunitary operator \(V_T\) and a positive self-adjoint linear operator \(|T|\) on \({\mathcal{H}}\) such that \[\begin{align} \label{tsquared} T^2=V_T^2 |T|^2=|T|^2V_T^2 . \end{align}\tag{38}\] Since \(V_T\) is an antilinear unitary operator, \(V_T^2\) is a unitary linear operator on \({\mathcal{H}}\).

Our first main aim is to desribe the structure of the antiunitary operator \(V_T\), that is, we treat the case \(T=V_T\). The crucial ingredient is the spectral theory of the linear unitary operator \(V_T^2\). By the spectral theorem there is a spectral measure \(E_{V_T^2}\) on \(\mathbb{T}\) such that \[\begin{align} V_T^2=\int_{\mathbb{T}} z\, dE_{V_T^2}(z).\label{spectralvt2} \end{align}\tag{39}\]

Let \(\lambda \in \rho(V_T^2)\). Using that \(V_T\) is antilinear we derive \[\begin{align} V_T(V_T^2-\lambda I)x=(V_T^2-\overline{\lambda} I)V_T x \end{align}\] for all \(x\in {\mathcal{H}}\). Setting \(y=(V_T^2-\lambda I)x\) we obtain \[\begin{align} \label{tresolvent} (V_T^2-\overline{\lambda} I)^{-1} V_Ty=V_T(V_T^2-\lambda I)^{-1} y \quad \textrm{for all}~~y\in {\mathcal{H}}, \,\lambda \in \rho(V_T^2). \end{align}\tag{40}\] Recall that \(M^c:= \{ \overline{z}: z\in M\}\) denotes the complex onjugate of a set \(M\). From (40 ) it follows that for each Borel subset \(M\) of \(\mathbb{T}\) we have \[\begin{align} \label{EVA2}E_{V_T^2}(M^c )\, V_T=V_T E_{V_T^2}(M). \end{align}\tag{41}\]

First we treat the restriction of \(V_T\) to the subspace \[{\mathcal{G}}:={\mathcal{N}}(V_T^2+I)= {E_{V_T^2}}( \{-1 \}){\mathcal{H}}.\] The map \(J:= V_T\lceil {\mathcal{G}}\) of the Hilbert space \({\mathcal{G}}\) into itself is antilinear and isometric. Since \(J^2x=V_T^2x=-x\) for \(x\in {\mathcal{G}}\), we have \(J^2=-I_{\mathcal{G}}\), that is, \(J\) is a skew-conjugation on \({\mathcal{G}}\).

For later use we suppose that \(S\) is a bounded self-adjoint linear operator on \({\mathcal{G}}\) which commutes with \(J\). When we treat the case of a general antilinear normal operator in the later stages of the proof this operator \(S\) will be specified. Using that \(V_T\) is antiunitary we derive \[\begin{align} \langle & JS^m x,x\rangle=\langle Jx,J^2S^m x\rangle=\langle Jx,S^mJ^2x\rangle\\ &=\langle Jx,-S^m x \rangle =- \langle S^mJx,x\rangle =- \langle JS^mx,x\rangle, \end{align}\] so that \(\langle JS^mx,x\rangle=0\) for \(x\in {\mathcal{G}}\) and \(m\in \mathbb{N}_0\). This shows that the assumptions of Lemma 26 are satisfied. By this result there exists a closed linear subspace \({\mathcal{R}}\) of \({\mathcal{G}}\) such that \(S{\mathcal{R}}\subseteq {\mathcal{R}}\), \({\mathcal{R}}\, \bot\, J{\mathcal{R}}\) and \({\mathcal{G}}={\mathcal{R}}\oplus J {\mathcal{R}}\). We define closed linear subspaces \({\mathcal{K}}_1\) and \({\mathcal{K}}_2\) of \({\mathcal{G}}\) by \[\begin{align} \label{defk1k2} {\mathcal{K}}_1:=J{\mathcal{R}}\equiv V_T{\mathcal{R}}, \quad {\mathcal{K}}_2:={\mathcal{R}}. \end{align}\tag{42}\] Then \({\mathcal{G}}\equiv {\mathcal{N}}(V_T^2+I)={\mathcal{K}}_1\oplus {\mathcal{K}}_2\). Clearly, \(V_T:{\mathcal{K}}_1\to {\mathcal{K}}_2\) and \(V_T:{\mathcal{K}}_2\to {\mathcal{K}}_1\).

Now we define \[\begin{align} \label{defhsahnsa} {\mathcal{H}}_{\mathrm{sa}}:=E_{V_T^2}(\{ 1\}){\mathcal{H}}~~ \textrm{and}~~{\mathcal{H}}_{\mathrm{nsa}}:=E_{V_T^2}(\mathbb{T}\backslash \{1\}){\mathcal{H}}. \end{align}\tag{43}\] Since we consider the case \(T=V_T\), we then have \({\mathcal{H}}={\mathcal{H}}_{\mathrm{sa}}\oplus {\mathcal{H}}_{\mathrm{nsa}}.\)

Let us abbreviate \[C_+:=\{z\in \mathbb{T}:0<\arg z<\pi\},~~ C_-:=\{z\in \mathbb{T}:\pi<\arg z<2\pi\}.\] We define two unitary linear operators \(U_1\) and \(U_2\) on the Hilbert subspaces \[\begin{align} {\mathcal{H}}_1 :={E_{V_T^2}}(C_+)& {\mathcal{H}}\oplus {\mathcal{K}}_1,~ {\mathcal{H}}_2:={E_{V_T^2}}(C_-){\mathcal{H}}\oplus {\mathcal{K}}_2, \end{align}\] respectively, of \({\mathcal{H}}_{\mathrm{nsa}}\) by \[\begin{align} U_1:=&\int_{C_+} z_+^{1/2} \, d{E_{V_T^2}}(z)\oplus {\rm i}\,I_{{\mathcal{K}}_1},\tag{44} \\ U_2:=&\int_{C_-} z_+^{1/2} \, d{E_{V_T^2}}(z)\oplus (- {\rm i})I_{{\mathcal{K}}_2},\tag{45} , \end{align}\] where \(z_+^{1/2}\) denotes the square root of \(z\in \mathbb{C}\backslash \mathbb{R}\) which is uniquely determined by the property that \({\mathrm{sign}}~ {\mathrm{Im}}\, z ={\mathrm{sign}}~ {\mathrm{Im}}\, z_+^{1/2}\). Clearly, \({\mathcal{H}}_1\bot {\mathcal{H}}_2\) and since \({\mathcal{K}}_1\oplus {\mathcal{K}}_2={\mathcal{N}}(V_T^2+I)={E_{V_T^2}}(\{-1\}){\mathcal{H}}\), we have \[{\mathcal{H}}_{\mathrm{nsa}}={\mathcal{H}}_1\oplus {\mathcal{H}}_2.\]

Let \(U_1'\) and \(U_2'\) denote the operator integrals in (44 ) and (45 ) acting on the corresponding Hilbert subspaces \({\mathcal{H}}_1'\) and \({\mathcal{H}}_2'\), respectively.

From the transformation formula of spectral measures (see e.g. [13]) it follows that the spectral measure of the unitary operator \(U_1'\) on \({\mathcal{H}}_1'\) is supported by the set \(\{z\in \mathbb{T}:0<\arg z<\pi/2\}\). Since \(U_1\lceil {\mathcal{K}}_1= {\rm i}\, I_{{\mathcal{K}}_1}\), \(U_1\) is supported by \(\{z\in \mathbb{T}:0<\arg z\leq \pi/2\}=K_\mathbb{T}\).

Let \(x\in {\mathcal{H}}_1'\). We consider the integral \(U_1'x\) as a limit of sums of terms \((z_j)_+^{1/2} E_{V_T^2}(M_j)x\), with \(z_j\in C_+\) and \(M_j\subseteq C_+\). Using (41 ) and the fact that \(V_T\) is antilinear we obtain \[\begin{align} V_T \big((z_j)_+^{1/2} E_{V_T^2}(M_j)\big) x=\big( \overline{z_j}_+^{1/2} E_{V_T^2}(M_j^c )\big) V_T x. \end{align}\] Since \(\overline{z_j}_+^{1/2} E_{V_T^2}(M_j^c )V_T x\) are the corresponding terms of a sum for the integral \(U_2'(V_T x)\), we conclude that \(V_Tx\in {\mathcal{H}}_2'\) and \(V_TU_1'x=U_2' V_T x\) for \(x\in {\mathcal{H}}_1'\). Further, \(V_T\) maps \({\rm i}x \in {\mathcal{K}}_1\) to \((-{\rm i}) V_1x\in {\mathcal{K}}_2\) for \(x\in {\mathcal{K}}_1\). Therefore \(V_T:{\mathcal{H}}_1\to {\mathcal{H}}_2\) and \[\begin{align} \label{uvrelation1} V_TU_1x=U_2V_Tx, \quad x\in {\mathcal{H}}_1. \end{align}\tag{46}\] A similar reasoning yields that \(V_T:{\mathcal{H}}_2\to {\mathcal{H}}_1\) and \[\begin{align} \label{uvrelation2} V_TU_2y=U_1V_Ty, \quad y\in {\mathcal{H}}_2. \end{align}\tag{47}\]

Let \(V_{\mathrm{nsa}}\) denote the restriction of \(V_T\) to \({\mathcal{H}}_{\mathrm{nsa}}\). Then \(V_{\mathrm{nsa}}:{\mathcal{H}}_{\mathrm{nsa}}\to {\mathcal{H}}_{\mathrm{nsa}}\) and the antlinear operator \(V_{\mathrm{nsa}}\) on \({\mathcal{H}}_{\mathrm{nsa}}\) is given by a operator block matrix \[\begin{gather} V_{\mathrm{nsa}}=\left( \begin{array}{ll} ~0 & V_2\\ V_1 & ~0 \end{array}\right). \end{gather}\] From their definitions it is clear that \(U_1\) and \(U_2\) are linear unitary operators on the Hilbert spaces \({\mathcal{H}}_1\) and \({\mathcal{H}}_2\), respectively. Let \(U\) denote the unitary linear operator on \({\mathcal{H}}_{\mathrm{nsa}}={\mathcal{H}}_1\oplus {\mathcal{H}}_2\) defined by the block matrix \[\begin{gather} \label{defiU} U=\left( \begin{array}{ll} U_1 & ~ 0\\ ~ 0 & U_2 \end{array}\right). \end{gather}\tag{48}\] From (46 ) and (47 ) we obtain \[\begin{align} \label{uvrelation3} V_{\mathrm{nsa}}U=UV_{\mathrm{nsa}}. \end{align}\tag{49}\]

Next we verify that \[\begin{align} \label{vnsu2} V_{\mathrm{nsa}}^2 =U^2. \end{align}\tag{50}\] Indeed, since \(V_T^2x=-x\) for \(x\in {\mathcal{K}}_1\), it follows from the definitions of \(U_1\) and \({\mathcal{H}}_1\) and the spectral decomposition (39 ) of \(V_T^2\) that \[\begin{align} U_1^2=\int_{C_+} z \, d{E_{V_T^2}}(z)\oplus ( -I_{{\mathcal{K}}_1}) =V_T^2\lceil {\mathcal{H}}_1=V_{\mathrm{nsa}}^2\lceil {\mathcal{H}}_1. \end{align}\] Similarly, \(U_2^2=V_{\mathrm{nsa}}^2\lceil {\mathcal{H}}_2\). Since \({\mathcal{H}}_{\mathrm{nsa}}={\mathcal{H}}_1\oplus {\mathcal{H}}_2\), these equalities imply (50 ).

Define \[\begin{gather} {\mathcal{C}}:= V_{\mathrm{nsa}}U^{-1}=\left( \begin{array}{ll} ~~ 0 & V_2U_2^{-1}\\ V_1U_1^{-1} & ~~ 0 \end{array}\right). \end{gather}\] Using that \(V_{\mathrm{nsa}}U^{-1}=U^{-1}V_{\mathrm{nsa}}\) by (49 ) and (50 ) we derive \[{\mathcal{C}}^2=V_{\mathrm{nsa}}U^{-1}V_{\mathrm{nsa}}U^{-1}=V_{\mathrm{nsa}}^2 U^{-2}= I_{{\mathcal{H}}_{\mathrm{nsa}}}.\] Further, since \(V_{\mathrm{nsa}}\) is antiunitary and \(U^{-1}\) is unitary, \({\mathcal{C}}\) is antilinear and we have \(\|{\mathcal{C}}x\|=\|V_{\mathrm{nsa}}U^{-1}x\|=\|U^{-1}x\|=\|x\|\) for \(x\in {\mathcal{H}}_{\mathrm{nsa}}\), so \({\mathcal{C}}\) is isometric. Hence \({\mathcal{C}}\) is a conjugation on the Hilbert space \({\mathcal{H}}_{\mathrm{nsa}}\) such that the operators \(V_{\mathrm{nsa}}\), \(U\), \({\mathcal{C}}\) pairwise commute and \[\begin{align} \label{vnsnsuc} V_{\mathrm{nsa}}= U{\mathcal{C}}={\mathcal{C}}U. \end{align}\tag{51}\] This proves the assertion for \(V_{\mathrm{nsa}}\) in the case \(T=V_T\), that is, for the special case of an antiunitary operator.

Now we show the assertion for \(T_{\mathrm{nsa}}\) in the case of a general antinormal operator \(T\). Recall from Theorem 11 that the positive self-adjoint operator \(|T|\) commutes with \(V_T\) and so with \(V_T^2\), because \(T\) is antinormal. Hence \(|T|\) commutes with the spectral projections of \(V_T^2\), that is, \[E_{V_T^2}(M)\, |T|\subseteq|T| E_{V_T^2}(M)\] for each Borel subset \(M\) of \(\mathbb{T}\). Therefore, the closed subspaces \({\mathcal{H}}_{\mathrm{sa}}=E_{V_T^2}(\{ 1\}){\mathcal{H}}\), \({\mathcal{H}}_{\mathrm{nsa}}=E_{V_T^2}(\mathbb{T}\backslash \{1\}){\mathcal{H}}\), \(E_{V_T^2}(C_+){\mathcal{H}}\),  and \(E_{V_T^2}(C_-){\mathcal{H}}\) reduce the self-adjoint operator \(|T|\).

Also, the closed subspace \({\mathcal{G}}:= E_{V_T^2}(\{-1\}){\mathcal{H}}\) reduces the operator \(|T|\). Let \(S\) denote the inverse of the operator \((I+|T|)\lceil {\mathcal{G}}\) acting on the Hilbert space \({\mathcal{G}}\). Since \(V_T\) commutes with \(|T|\), \(J:=V_T\lceil {\mathcal{G}}\) commutes with the bounded self-adjoint operator \(S\) on \({\mathcal{G}}\). Therefore, Lemma 26 applies and the correponding subspace \({\mathcal{R}}\) is invariant under \(S\). Hence the closed subspaces \({\mathcal{K}}_1\) and \({\mathcal{K}}_2\) of \({\mathcal{G}}\) defined by (42 ) are invariant under \(S\). This implies that \({\mathcal{K}}_1\) and \({\mathcal{K}}_2\) reduce the operator \(|T|\). Hence \({\mathcal{H}}_1=E_{V_T^2}(C_+){\mathcal{H}}\oplus {\mathcal{K}}_1\) and \({\mathcal{H}}_2=E_{V_T^2}(C_-){\mathcal{H}}\oplus {\mathcal{K}}_2\) reduce \(|T|\). From the preceding we conclude that \[|T|\lceil {\mathcal{H}}_{\mathrm{nsa}}= |T|\lceil {\mathcal{H}}_1\oplus | T|\lceil {\mathcal{H}}_2~~~\textrm{on}~~ {\mathcal{H}}_{\mathrm{nsa}}={\mathcal{H}}_1\oplus {\mathcal{H}}_2.\] Let us abbreviate \(|T|_{\mathrm{nsa}}:=|T|\lceil {\mathcal{H}}_{\mathrm{nsa}}, |T|_1:=|T|\lceil {\mathcal{H}}_1, |T|_2:= | T|\lceil {\mathcal{H}}_2.\) Then \[\begin{gather} |T|_{\mathrm{nsa}}= \left( \begin{array}{ll} |T|_1 & ~0\\ ~0 & |T|_2 \end{array}\right). \end{gather}\] Since \(|T|\) commutes with the spectral projections \(E_{V_T^2}(\cdot)\), it also commutes with the spectral integrals \(U_1'\) and \(U_2'\). Obviously, \(|T|\) commutes with \({\rm i}\, I_{{\mathcal{K}}_1}\) and \(-{\rm i}\, I_{{\mathcal{K}}_2}\), so it commutes with the operators \(U_1\) and \(U_2\). Therefore, the block matrix \(|T|_{\mathrm{nsa}}\) commutes with the unitary operator block matrix \(U\) defined by (48 ), that is, \(U|T|_{\mathrm{nsa}}\subseteq |T|_{\mathrm{nsa}}U.\) Hence \(U|T|_{\mathrm{nsa}}U^*\subseteq |T|_{\mathrm{nsa}}.\) Since \(U|T|_{\mathrm{nsa}}U^*\) is self-adjoint, we get \(U|T|_{\mathrm{nsa}}U^*=|T|_{\mathrm{nsa}}\). Then \(|T|_{\mathrm{nsa}}=U^*|T|_{\mathrm{nsa}}U\). Thus, \[\begin{align} \label{comutns} U|T|_{\mathrm{nsa}}= |T|_{\mathrm{nsa}}U~~ \textrm{and}~~ U^*|T|_{\mathrm{nsa}}= |T|_{\mathrm{nsa}}U^*. \end{align}\tag{52}\] Define \(N_1:=U_1|T|_1\),  \(N_2:=U_2|T|_2\), and \[\begin{gather} {\mathcal{N}}:= \left( \begin{array}{ll} N_1 & ~0\\ ~0 & N_2 \end{array}\right) =U|T|_{\mathrm{nsa}}=|T|_{\mathrm{nsa}}U, \end{gather}\] where the last equality follows from (52 ). Since \(T=V_T|T|=|T|V_T\) and \(V_{\mathrm{nsa}}=U{\mathcal{C}}={\mathcal{C}}U\) by (51 ), we conclude that \(T_{\mathrm{nsa}}= |T|_{\mathrm{nsa}}V_{\mathrm{nsa}}=|T|_{\mathrm{nsa}}U {\mathcal{C}}={\mathcal{N}}{\mathcal{C}}\) and \(T_{\mathrm{nsa}}=V_{\mathrm{nsa}}|T|_{\mathrm{nsa}}={\mathcal{C}}U |T|_{\mathrm{nsa}}={\mathcal{C}}{\mathcal{N}}.\)

Now we show that the linear operators \(N_1\) and \(N_2\) are normal. Using (52 ) we derive \[\begin{align} \label{n1n2normal}{\mathcal{N}}^*{\mathcal{N}}=|T|_{\mathrm{nsa}}U^*U|T|_{\mathrm{nsa}}=|T|_{\mathrm{nsa}}^2,~~ {\mathcal{N}}{\mathcal{N}}^*=|T|_{\mathrm{nsa}}U U^*|T|_{\mathrm{nsa}}=|T|_{\mathrm{nsa}}^2. \end{align}\tag{53}\] Hence the diagonal operator block matrix \({\mathcal{N}}\) is normal and so are their diagonal entries \(N_1\) and \(N_2\).

Next we prove that \(N_1\) is supported by the set \(K\). Since \(N_1=U_1|T|_1\) and the unitary operator \(U_1\) and the positive self-adjoint operator \(|T|_1\) commute, using the product property of the spectral measure \(E_{N_1}\) of the normal operator \(N_1\) we obtain \[E_{N_1}(\mathbb{C}\backslash K)=E_{N_1}((\mathbb{T}\backslash K_{\mathbb{T}})\times [0,+\infty))=E_{U_1}(\mathbb{T}\backslash K_\mathbb{T})E_{|T|_1}([0,+\infty)).\] Since \(U_1\) is supported by \(K_\mathbb{T}\) as shown above, \(E_{U_1}(\mathbb{T}\backslash K_\mathbb{T})=0\) and hence \(E_{N_1}(\mathbb{C}\backslash K)=0\), that is, \(N_1\) is supported by \(K\).

The formulas (34 ) and (35 ) follow by straightforward computations. Note that (53 ) is just the first formula of (35 ). All assertions about \(T_{\mathrm{nsa}}\) are proved.

Now we turn to the case of \(T_{\mathrm{sa}}\) which is much simpler. Recall that \({\mathcal{H}}_{\mathrm{sa}}=E_{V_T^2}(\{ 1\}){\mathcal{H}}\) by (43 ). From (41 ) it follows that \(V_T\) leaves \({\mathcal{H}}_{\mathrm{sa}}\) invariant. Since \(V_T^2 x=x\) for \(x\in {\mathcal{H}}_{\mathrm{sa}}\) and \(V_T\) is antiunitary, \(C_0:=V_T\lceil {\mathcal{H}}_{\mathrm{sa}}\) is a conjugation and \(S:=|T|\lceil {\mathcal{H}}_{\mathrm{sa}}\) is a positive self-adjoint operator on \({\mathcal{H}}_{\mathrm{sa}}\). Since \(T=V_T |T|=|T|V_T\), we have \(T_{\mathrm{sa}}=SC_0=C_0S,\) see also Theorem 4.

This completes the proof of Theorem 22.

References↩︎