A conjugation C defined on a complex Hilbert space \(\mathcal {H}\) is an antilinear operator that is involutive \((C^{2} = I_{\mathcal{H}})\) and isometric, meaning that the following equality holds;
$$\begin{aligned} \langle C\xi, C\eta \rangle =\langle \eta, \xi \rangle\quad \text{for all $\xi, \eta \in \mathcal{H}$}. \end{aligned}$$
(3)
Thus it follows from (3) that \(\langle C\xi, C\eta \rangle =\overline{\langle \xi, \eta \rangle}\). Chō and Tanahashi [2] introduced a conjugation C on a complex Banach space \(\mathcal{B}\) as the operator satisfying the following relations;
$$\begin{aligned} C^{2}=I_{\mathcal{B}},\qquad \Vert C \Vert \leq 1,\qquad C(x+y)=Cx+Cy\quad \text{and}\quad C(\lambda x)=\overline{\lambda}Cx \end{aligned}$$
(4)
for all \(x, y \in \mathcal{B}\) and \(\lambda \in \mathbb{C}\).
Like in a Hilbert space, we will define a conjugation on a semiinnerproduct space using a semiinnerproduct. Throughout this section, \(\mathcal{X}\) denotes a semiinnerproduct space with a semiinnerproduct \([\cdot,\cdot ]\), unless specified otherwise.
Definition 3.1
An operator \(C: \mathcal{X} \rightarrow \mathcal{X}\) is a conjugation if it is involutive \((C^{2} = I_{\mathcal{X}})\) and
$$\begin{aligned}{} [ Cx, Cy ]=\overline{[ x, y ]}\quad \text{for all $x, y \in \mathcal{X}$}. \end{aligned}$$
(5)
Proposition 3.2
If C is a conjugation on \(\mathcal{X}\), then relation (4) holds for all \(x, y \in \mathcal{X}\) and \(\lambda \in \mathbb{C}\).
Proof
By the Cauchy–Schwarz inequality for a semiinnerproduct, we have that \(\Cx\^{2}=[Cx, Cx]=\overline{[x, x]}\leq \x\^{2}\) for every \(x\in \mathcal{X}\), which implies that \(\C\\le 1\). Since a semiinnerproduct is linear in the first variable, we have
$$\begin{aligned} \bigl[ C(x+y), Cz \bigr]&= \overline{[ x+y, z ]} = \overline{[ x, z ]}+ \overline{[ y, z ]} \\ &= [ Cx, Cz ]+[ Cy, Cz ] = [ Cx+Cy, Cz ] \end{aligned}$$
for all \(x, y, z \in \mathcal{X}\). Since the operator C is surjective, we can take \(z \in \mathcal{X}\) such that
$$\begin{aligned} Cz:= C(x+y)CxCy. \end{aligned}$$
Then we get that \(0=[ C(x+y)CxCy, Cz ]=[ Cz, Cz]\), so that \(Cz=0\), that is, \(C(x+y)=Cx+Cy\). To show that \(C(\lambda x)=\overline{\lambda}Cx\) for any \(x \in \mathcal{X}\) and \(\lambda \in \mathbb{C}\), take any element \(y \in \mathcal{X}\). Then we have
$$\begin{aligned} \bigl[ C(\lambda x), y \bigr]&= \overline{[ \lambda x, Cy]} = \overline{\lambda} \overline{[ x, Cy ]} \\ &= \overline{\lambda}[ Cx, y ] = [ \overline{\lambda}Cx, y ], \end{aligned}$$
which means that \(C(\lambda x)=\overline{\lambda}Cx\). Therefore C satisfies relation (4). □
Let C be a conjugation on a complex Hilbert space \(\mathcal{H}\). A bounded linear operator T on \(\mathcal{H}\) is Csymmetric if \(T= CT^{\ast}C\), where \(T^{\ast}\) is a Hilbert space adjoint of T, which is equivalent to \(\langle Tx, y \rangle =\langle x, CTCy \rangle \) for all \(x,y \in \mathcal{H}\). Chō et al. [1] have extend the notion of Csymmetric operators to Banach space operators via linear functionals in its dual space. However, we would like to extend the notion of the complex symmetry to semiinnerproduct space operators without using linear functionals. Even though semiinnerproducts in general are not additive in the second variables, we will use a semiinnerproduct to define the Csymmetric operator on a semiinnerproduct space.
Definition 3.3
Let C be a conjugation on \(\mathcal{X}\). We say that \(T\in \mathcal{L}(\mathcal{X})\) is Csymmetric if
$$\begin{aligned}{} [ Tx, y ]=[ x, CTCy]\quad \text{for all $x,y \in \mathcal{X}$}. \end{aligned}$$
(6)
Remark 3.4
In Definition 3.3, equation (6) is equivalent to
$$\begin{aligned}{} [ x, Ty ]=[ CTCx, y] \quad\text{for all $x,y \in \mathcal{X}$}. \end{aligned}$$
(7)
Indeed, by putting \(Cx, Cy\) into (6) instead of \(x, y\) we obtain that \([ TCx, Cy ]=[ Cx, CTy]\). It follows from the definition of a conjugation C that \([ CTCx, y ]=\overline{[ Cx, CTy]} = [ x, Ty]\).
Proposition 3.5
Let C be a conjugation on \(\mathcal{X}\), and let \(T \in \mathcal{L}(\mathcal{X})\) be a Csymmetric operator.

(i)
λT is Csymmetric for any complex number λ.

(ii)
If T is invertible, then \(T^{1}\) is also Csymmetric.

(iii)
If \(S\in \mathcal{L}(\mathcal{X})\) is Csymmetric and commutes with T, then so is TS.
Proof
(i) For any complex number λ, we have
$$\begin{aligned} \bigl[(\lambda T)x, y\bigr]&=\lambda [Tx, y]=\lambda [x, CTCy] \\ &=[x, \overline{\lambda}CTCy]=\bigl[x, C(\lambda T)Cy\bigr], \end{aligned}$$
so that λT is Csymmetric.
(ii) For any \(y \in \mathcal{X}\), there exists \(z \in \mathcal{X}\) such that \(y=CTCz\). Indeed, since T is invertible and C is a conjugation, \(CT^{1}C\) is also invertible. Putting \(z:=CT^{1}Cy\), we get \(y=CTCz\). For any \(x,y \in \mathcal{X}\), we have
$$\begin{aligned} \bigl[T^{1}x, y\bigr] =\bigl[T^{1}x, CTCz\bigr]= \bigl[TT^{1}x, z\bigr]=[x, z]=\bigl[x, CT^{1}Cy\bigr], \end{aligned}$$
where the second equality follows from the Csymmetry of T. Thus \(T^{1}\) is Csymmetric, which completes the proof.
(iii) If \(S\in \mathcal{L}(\mathcal{X})\) commutes with T and is Csymmetric, then it follows that
$$\begin{aligned} \bigl[(TS)x, y\bigr]=[Sx, CTCy]=\bigl[x,CSC (CTCy)\bigr]=\bigl[x,C(ST)Cy\bigr]= \bigl[x, C(TS)Cy\bigr]. \end{aligned}$$
Hence TS is Csymmetric. □
Let \(T \in \mathcal{L}(\mathcal{X})\) and \(y \in \mathcal{X}\). By the Riesz representation theorem in a semiinnerproduct space [7], there is a unique vector \(T^{\dagger }y\) such that \([Tx, y] = [x, T^{\dagger }y]\) for all \(x \in \mathcal{X,}\) where \(T^{\dagger}\) is a generalized adjoint, which is not usually linear [11]. On the other hand, if C is a conjugation on \(\mathcal{X}\) and if \(T\in \mathcal{L}(\mathcal{X})\) is Csymmetric, then we obtain that
$$\begin{aligned} \bigl[x, T^{\dagger }y\bigr]=[Tx, y]=[x, CTCy] \quad\text{for all $x,y\in \mathcal{X}$}, \end{aligned}$$
so that \(T^{\dagger}=CTC\). Since \(CTC\) is linear, \(T^{\dagger}\) becomes a linear operator on \(\mathcal{X}\). It follows from (7) that \([x, Ty]=[CTCx, y]=[T^{\dagger }x, y]\) for all \(x,y\in \mathcal{X}\). Furthermore, \(T^{\dagger}\) is also Csymmetric. Indeed, for all \(x,y \in \mathcal{X}\),
$$\begin{aligned} \bigl[T^{\dagger }x, y\bigr]=[CTCx, y]=\overline{[TCx, Cy]}= \overline{ \bigl[Cx, T^{\dagger }Cy\bigr]}=\bigl[x, CT^{\dagger }Cy\bigr]. \end{aligned}$$
A uniform semiinnerproduct space means a uniformly continuous semiinnerproduct space where the induced normed vector space is complete and uniformly convex. Here the (uniform) continuity implies that
$$\begin{aligned} \operatorname{Re}\bigl\{ [y,x+ty]\bigr\} \to \operatorname{Re}\bigl\{ [y,x]\bigr\} \quad\text{(uniformly) as $t \in \mathbb{R} \to 0$}. \end{aligned}$$
Giles [7, Theorem 7] proved that for a uniform semiinnerproduct space \(\mathcal{X}\), the dual space \(\mathcal{X}^{\star}\) is also a uniform complex semiinnerproduct space with respect to the semiinnerproduct defined by \([x^{\star}, y^{\star}]_{\star}=[y, x]\). Moreover, he proved that for every continuous linear functional \(x^{\star}\) in a dual space \(\mathcal{X}^{\star}\), there exists a unique vector \(x\in \mathcal{X}\) such that
$$\begin{aligned} x^{\star}(z)=[z, x] \quad\text{for all } z\in \mathcal{X}, \end{aligned}$$
so that the map \(x \mapsto x^{\star}=[\cdot, x]\) is a onetoone mapping from \(\mathcal{X}\) onto \(\mathcal{X}^{\star}\). For any \(T\in \mathcal{L}(\mathcal{X})\), the dual operator \(T^{\star}\in \mathcal{L}(\mathcal{X}^{\star})\) is given by \(T^{\star }y^{\star}(z)=y^{\star}(Tz)\) for all \(y^{\star }\in \mathcal{X}^{\star}\) and \(z \in \mathcal{X}\).
If C is a conjugation on a uniform semiinnerproduct space \(\mathcal{X}\), then we define the dual operator \(C^{\star}:\mathcal{X^{\star}} \to \mathcal{X^{\star}}\) by
$$\begin{aligned} \bigl(C^{\star}\bigl(x^{\star}\bigr)\bigr) (z):= \overline{x^{\star}(Cz)} \quad\text{for all $z\in \mathcal{X}$}. \end{aligned}$$
(8)
We have that \((C^{\star}(x^{\star}))(z)=\overline{x^{\star}(Cz)}= \overline{[Cz, x]}=[z, Cx]=(Cx)^{\star}(z)\), so that \(C^{\star}(x^{\star})=(Cx)^{\star}\). Thus we have the following commutative diagram:
Moreover, the dual operator \(C^{\star}\) is a conjugation on \(\mathcal{X^{\star}}\). Indeed, for any \(x^{\star}, y^{\star} \in \mathcal{X}^{\star}\), there exist unique vectors x and y in \(\mathcal{X}\) such that
$$\begin{aligned} \bigl[C^{\star}x^{\star}, C^{\star}y^{\star} \bigr]_{\star}=\bigl[(Cx)^{\star},(Cy)^{ \star} \bigr]_{\star} =[Cy, Cx]=\overline{[y, x]}= \overline{\bigl[x^{\star}, y^{\star}\bigr]_{\star}}. \end{aligned}$$
Since \((Cx)^{\star}=C^{\star}(x^{\star})\) for all \(x\in \mathcal{X}\), we observe that relation (4) implies equation (5).
Proposition 3.6
Let C be a conjugation on a uniform semiinnerproduct space \(\mathcal{X}\).

(i)
If \(T\in \mathcal{L}(\mathcal{X})\) is Csymmetric, then \(T^{\star}\in \mathcal{L}(\mathcal{X}^{\star})\) is also \(C^{\star}\)symmetric.

(ii)
If \(T\in \mathcal{L}(\mathcal{X})\) is Csymmetric, then \((T^{\dagger})^{\star}=(T^{\star})^{\dagger}\).

(iii)
If \(\{T_{n} \}\) is a sequence of Csymmetric operators such that \(T_{n} \to S\) in the strong topology, then S is Csymmetric.
Proof
(i) Suppose that T is a Csymmetric operator on \(\mathcal{X}\). Let f and g be arbitrary elements in the dual space \(\mathcal{X}^{\star}\). Since \(\mathcal{X}\) is a uniform semiinnerproduct space, there exist unique vectors x and y in \(\mathcal{X}\) such that \(x^{\star}=f\) and \(y^{\star}=g\). First, we observe that \(T^{\star}x^{\star}=(T^{\dagger}x)^{\star}\). Indeed, for any \(z\in \mathcal{X}\), we have
$$\begin{aligned} \bigl(T^{\star}x^{\star}\bigr) (z)=x^{\star}(Tz)=[Tz, x]= \bigl[z, T^{\dagger}x\bigr]=\bigl(T^{\dagger}x\bigr)^{ \star}(z). \end{aligned}$$
Moreover, for any \(z \in \mathcal{X}\) and \(y^{\star} \in \mathcal{X}^{\star}\), we see that
$$\begin{aligned} \bigl(C^{\star}T^{\star}C^{\star}\bigr)y^{\star}(z)=y^{\star}(CTCz)=(CTC)^{ \star}y^{\star}(z), \end{aligned}$$
so \(C^{\star}T^{\star}C^{\star} = (CTC)^{\star}\). Thus we have
$$\begin{aligned} \bigl[T^{\star}x^{\star}, y^{\star}\bigr]_{\star}&= \bigl[y, T^{\dagger}x\bigr]=\bigl[CT^{\dagger}Cy, x\bigr] \\ &=\bigl[x^{\star}, \bigl(CT^{\dagger}Cy\bigr)^{\star} \bigr]_{\star}=\bigl[x^{\star}, (CTC)^{\star}y^{ \star} \bigr]_{\star} \\ &=\bigl[x^{\star}, C^{\star}T^{\star}C^{\star}y^{\star} \bigr]_{\star}, \end{aligned}$$
which means that \(T^{\star}\) is \(C^{\star}\)symmetric.
(ii) For any \(z \in \mathcal{X}\) and \(y^{\star }\in \mathcal{X}^{\star}\), we obtain that
$$\begin{aligned} \bigl(T^{\dagger}\bigr)^{\star }y^{\star}(z)=(CTC)^{\star }y^{\star}(z)= \bigl(C^{\star }T^{ \star }C^{\star}\bigr)y^{\star }(z). \end{aligned}$$
On the other hand, it follows from (i) that \(T^{\star}\) is \(C^{\star}\)symmetric. Hence, for all \(x^{\star }\in \mathcal{X}^{\star}\),
$$\begin{aligned} \bigl[x^{\star}, \bigl(T^{\star}\bigr)^{\dagger}y^{\star} \bigr]_{\star}=\bigl[T^{\star}x^{\star}, y^{ \star} \bigr]_{\star}=\bigl[x^{\star}, \bigl(C^{\star }T^{\star }C^{\star} \bigr)y^{\star}\bigr]_{ \star}. \end{aligned}$$
This means that \((T^{\dagger})^{\star}=(T^{\star})^{\dagger}\).
(iii) Since \(\(S  T_{n})x\ \to 0\) for all \(x\in \mathcal{X}\), for all \(x,y\in \mathcal{X}\), we have
$$\begin{aligned}{} [CSCx, y] = \lim_{n\to \infty}[CT_{n}Cx, y] = \lim _{n\to \infty}[x, T_{n}y] = [x, Sy], \end{aligned}$$
where the third equality follows from uniform continuity. Thus S is a Csymmetric operator. □
Now we compute the numerical range of a conjugation on \(\ell _{n}^{p}(\mathbb{C})\).
Example 3.7
Let C be a complex conjugation on \(\ell _{n}^{p}(\mathbb{C})\) \((1\leq p<\infty )\) given by \(Cx=\overline{x}=(\overline{x}_{1},\ldots,\overline{x}_{n})\) for \(x\in \ell _{n}^{p}(\mathbb{C})\). Then we have:

(1)
\(W(C)=\{\lambda \in \mathbb{C}: \lambda =1 \}\) for \(n=1\),

(2)
\(W(C)=\{\lambda \in \mathbb{C}: \lambda \leq 1 \}\) for \(n\geq 2\).
It is easy to prove (1). Indeed, for any \(x\in \ell _{1}^{p}(\mathbb{C})\) with \(x=1\), we write \(x=e^{i\theta}\) for some real number θ. Obviously, we have \([Cx, x]_{p}=[\overline{x}, x]_{p}=e^{2i\theta}\), and so \(W(C)=\{\lambda \in \mathbb{C}: \lambda =1 \}\).
To show the second statement, let \(x\in \ell _{n}^{p}(\mathbb{C})\) be any unit vector, i.e., \(\x\_{p}^{2} = [x, x]_{p}=1\). By the Cauchy–Schwarz inequality we have
$$\begin{aligned} \bigl\vert [Cx, x]_{p} \bigr\vert ^{2}\leq [Cx, Cx]_{p} [x, x]_{p}= \overline{[x, x]_{p}} [x, x]_{p}=1, \end{aligned}$$
which implies that \(W(C)\subseteq \{\lambda \in \mathbb{C}: \lambda \leq 1 \}\).
For the reverse inclusion, let λ be any complex number with \(\lambda \leq 1\). We write a polar form \(\lambda =\lambda e^{i\theta}\) for some real number θ. Now we take a unit vector \(x\in \ell _{n}^{p}(\mathbb{C})\) given by
$$\begin{aligned} x= \biggl( {\biggl(}\frac{1+ \vert \lambda \vert }{2} {\biggr)}^{\frac{1}{p}}e^{ \frac{i\theta}{2}}, {\biggl(}\frac{1 \vert \lambda \vert }{2} {\biggr)}^{ \frac{1}{p}}ie^{\frac{i\theta}{2}}, 0, \ldots, 0 \biggr). \end{aligned}$$
Then we have
$$\begin{aligned}{} [Cx, x]_{p}=[\overline{x}, x]_{p}= \biggl( \frac{1+ \vert \lambda \vert }{2} \frac{1 \vert \lambda \vert }{2} \biggr) e^{i\theta} = \vert \lambda \vert e^{i\theta}= \lambda, \end{aligned}$$
which implies that \(W(C)\) contains the closed unit disc. Therefore the numerical range \(W(C)\) is the closed unit disc.
Let C be the usual complex conjugation given in Example 3.7. Then we see that
$$\begin{aligned} w(C)=\sup \bigl\{ \bigl\vert [Cx, x]_{p} \bigr\vert : [x, x] =1, x \in \ell _{n}^{p} \bigr\} = 1 \quad\text{for all $n \ge 1$}, \end{aligned}$$
where \(w(C)\) is the numerical radius of C. Moreover, we can find infinitely many unit vectors x that attain the numerical radius of the complex conjugation C on \(\ell _{n}^{1}(\mathbb{C})\) \((n \geq 1)\), that is, vectors x with \([Cx,x]_{1}=1\). We explicitly construct vectors attaining the numerical radius \(w(C)\) in the following example.
Example 3.8
Let \(n \geq 2\). For any \(\lambda \in \mathbb{C}\) with \(0<\lambda \leq \frac{1}{n1}\), we take the vector \(x=(x_{1}, x_{2}, \ldots, x_{n})^{t} \in \ell _{n}^{1}(\mathbb{C})\) given by
$$\begin{aligned} x_{j}= \textstyle\begin{cases} \overline{\lambda} & \text{if } 1\leq j \leq n1, \\ ( \frac {1}{ \vert \lambda \vert }n+1 ) \overline{\lambda} & \text{if } j=n. \end{cases}\displaystyle \end{aligned}$$
(9)
Then we have that
$$\begin{aligned}{} [x, x]_{1}=1\quad \text{and}\quad \bigl\vert [Cx, x]_{1} \bigr\vert = \biggl\vert \biggl(\frac{\lambda}{ \vert \lambda \vert } \biggr) \biggr\vert ^{2}=1. \end{aligned}$$
For any complex number λ with \(\lambda \geq n1\), we put \(x=(x_{1}, x_{2}, \ldots, x_{n})\in \ell _{n}^{1}(\mathbb{C})\), where
$$\begin{aligned} x_{j}= \textstyle\begin{cases} \overline{\lambda}^{1} & \text{if } 1\leq j \leq n1, \\ \overline{\lambda}^{1}( \vert \lambda \vert n+1) & \text{if } j=n. \end{cases}\displaystyle \end{aligned}$$
(10)
Then it follows that
$$\begin{aligned}{} [x, x]_{1}=1 \quad\text{and}\quad \bigl\vert [Cx, x]_{1} \bigr\vert = \biggl\vert \biggl( \frac{ \vert \lambda \vert }{\lambda} \biggr) \biggr\vert ^{2}=1. \end{aligned}$$
Similarly, we also have infinitely many numerical radius attaining vectors in the infinitedimensional space \(\ell ^{1}(\mathbb{N})\) in the same way as (9) and (10) except for jth terms with 0 \((j>n)\).
The essential numerical range for a bounded linear operator on a Hilbert space is defined as the closure of the numerical range of the image in the Calkin algebra, and many equivalent conditions are known [5]. We now introduce the sequentially essential numerical range of T on a semiinnerproduct space \(\mathcal{X}\) by
$$\begin{aligned} W_{e}(T)=\Bigl\{ z\in \mathbb{C}: \lim_{n} [Tx_{n}, x_{n}]=z \text{ for some } \{x_{n}\}\subset \mathcal{X} \text{ with } [x_{n},x_{n}]=1, x_{n} \stackrel{w}{\longrightarrow} 0 \Bigr\} . \end{aligned}$$
Theorem 3.9
Let \(T\in \mathcal{L}(\mathcal{X})\), and let C be a conjugation on \(\mathcal{X}\). Then we have
$$\begin{aligned} W(CTC)=\overline{W(T)}\quad \textit{and}\quad W_{e}(T)= \overline{W_{e}(CTC)}, \end{aligned}$$
where S̅ denotes the complex conjugation of S.
Proof
If \(z\in W(CTC)\), then there exists a vector \(x\in \mathcal{X}\) with \([x, x]=1\) such that
$$\begin{aligned} z=[CTCx, x]=\overline{[TCx, Cx]}\in \overline{W(T)}. \end{aligned}$$
This means that \(W(CTC)\subset \overline{W(T)}\). Since \(W(T)=W(C^{2}TC^{2})\subset \overline{W(CTC)}\), we get the reverse inclusion. Therefore we have \(W(CTC)=\overline{W(T)}\).
If \(z\in W_{e}(CTC)\), then there exists a sequence \(\{x_{n}\}\subset \mathcal{X}\) with \([x_{n},x_{n}]=1\) and \(x_{n} \stackrel{w}{\longrightarrow} 0\). Since \(\lim_{n} x_{n}=0\) in the weak sense, we obtain that \(\lim_{n} f(x_{n})=0\) for all \(f\in \mathcal{X}^{\star}\). Since \(C^{\star}f \in \mathcal{X}^{\star}\) for all \(f\in \mathcal{X}^{\star}\), we have \(\lim_{n} f(Cx_{n})=\lim_{n} C^{\star}f(x_{n})=0\), which implies that \(Cx_{n}\stackrel{w}{\longrightarrow}0\). Thus we have
$$\begin{aligned} z=\lim_{n} [CTCx_{n}, x_{n}]=\lim _{n} \overline{[TCx_{n}, Cx_{n}]}\in \overline{W_{e}(T)}. \end{aligned}$$
This implies that \(W_{e}(CTC)\subset \overline{W_{e}(T)}\). The reverse inclusion follows from
$$\begin{aligned} W_{e}(T)=W_{e}\bigl(C^{2}TC^{2}\bigr) \subset \overline{W_{e}(CTC)}, \end{aligned}$$
which completes the proof. □
Corollary 3.10
Let C be a conjugation on \(\mathcal{X}\), and let \(T\in \mathcal{L}(\mathcal{X})\) be Csymmetric.

(i)
\(W(T^{\dagger})=\overline{W(T)}\) and \(W_{e}(T^{\dagger})=\overline{W_{e}(T)}\).

(ii)
\(W(T)=\{\overline{[x, Tx]}: [x, x]=1, x\in \mathcal{X} \}\).

(iii)
If, in addition, \(\mathcal{X}\) is a uniform semiinnerproduct space, then
$$\begin{aligned} W\bigl(C^{\star}T^{\star}C^{\star}\bigr)=\overline{W \bigl(T^{\star}\bigr)} =\bigl\{ \bigl[x^{\star}, T^{\star}x^{\star}\bigr]_{\star}: \bigl[x^{\star}, x^{\star}\bigr]_{\star}=1, x^{ \star}\in \mathcal{X^{\star}} \bigr\} . \end{aligned}$$
Proof
It immediately follows from Proposition 3.6 and Theorem 3.9. □
We say that \(T \in \mathcal{L}(\mathcal{X})\) is an isometry if \([Tx, Ty]=[x, y]\) for \(x, y\in \mathcal{X}\), a unitary if it is isometric and surjective, and a Hermitian operator if \(W(T)\subset \mathbb{R}\). For a conjugation C on \(\mathcal{X}\), we have that T is an isometry (a unitary or a Hermitian operator, respectively) if and only if \(CTC\) is an isometry (a unitary or a Hermitian operator, respectively). Indeed, if T is an isometry, then for \(x, y\in \mathcal{X}\),
$$\begin{aligned}{} [CTCx, CTCy]=\overline{[TCx, TCy]}=\overline{[Cx, Cy]}=[x, y], \end{aligned}$$
which implies that \(CTC\) is an isometry. Conversely, if \(CTC\) is an isometry, then for \(x, y\in \mathcal{X}\),
$$\begin{aligned}{} [Tx,Ty]=[TCz,TCw]=\overline{[CTCz, CTCw]} = \overline{[z,w]}=[Cz, Cw]=[x,y], \end{aligned}$$
where \(z=Cx\) and \(w=Cy\). Similarly, we can see that T is a unitary if and only if \(CTC\) is a unitary. It follows from Theorem 3.9 that T is Hermitian if and only if \(CTC\) is also Hermitian.
In [8, Lemma 3.1] and [6, Theorem 3.1], it has been proved that any unitary operator on a Hilbert space can be constructed by gluing together two copies of essentially the same antilinear operator. The following proposition provides a perspective on the structure of unitary operators in a semiinnerproduct space.
Proposition 3.11
If C and G are conjugations on \(\mathcal{X}\), then \(U=CG\) is a unitary and is both Csymmetric and Gsymmetric.
Proof
For any \(x, y \in \mathcal{X}\), we have
$$\begin{aligned}{} [Ux, Uy]=[CGx, CGy]=\overline{[Gx, Gy]}=[x, y], \end{aligned}$$
which means that U is isometric. Since C and G are conjugations on \(\mathcal{X}\), it is obvious that U is surjective, so that it is a unitary. Moreover, we have
$$\begin{aligned}{} [CUCx, y]=[GCx, y]=\overline{[Cx, Gy]}=[x, CGy]=[x, Uy] \end{aligned}$$
and
$$\begin{aligned}{} [GUGx, y]=[GCx, y]=\overline{[Cx, Gy]}=[x, CGy]=[x, Uy]. \end{aligned}$$
Thus U is both Csymmetric and Gsymmetric. □
Remark 3.12
(i) In Proposition 3.11, if \(U^{\dagger}\) is a generalized adjoint of a unitary \(U=CG\), then we get from Csymmetry of U that \(U^{\dagger}=CUC=GC\). Hence we have \(UU^{\dagger}=U^{\dagger }U=I_{\mathcal{X}}\). This means that \(U^{\dagger}=U^{1}\).
(ii) Suppose that \(\mathcal{X}\) in Proposition 3.11 is a uniform semiinnerproduct space. Let \(C^{\star}\) and \(G^{\star}\) be conjugations on \(\mathcal{X}^{\star}\) corresponding to C and G, which are given by (8). By Propositions 3.6 and 3.11, \(U^{\star}=C^{\star }G^{\star}\) is a unitary on \(\mathcal{X}^{\star}\) and is both \(C^{\star}\)symmetric and \(G^{\star}\)symmetric. It also follows from \(C^{\star}\)symmetry of \(U^{\star}\) that \((U^{\star})^{\dagger }=G^{\star }C^{\star}\), so that \((U^{\star})^{\dagger}=(U^{\star})^{1}\).