Numerical ranges and complex symmetric operators in semi-inner-product spaces

Abstract

We introduce the numerical range of a bounded linear operator on a semi-inner-product space. We compute the numerical ranges of some operators on $\ell _{2}^{p}(\mathbb{C})$ $(1\le p < \infty )$ and show that the numerical range of the backward shift on an infinite-dimensional space $\ell ^{p}$ is the open unit disc. We define a conjugation and a complex symmetric operator on a semi-inner-product space and discuss complex symmetry in the dual space. We prove some properties of a generalized adjoint of a complex symmetric operator. We also show that the numerical range of the complex conjugation on $\ell _{n}^{p}$ $(n \ge 2)$ is the closed unit disc. Finally, we discuss the sequentially essential numerical ranges of operators on a semi-inner-product space.

1 Introduction

For the study of operator theory in Banach spaces, Lumer [12] introduced a semi-inner-product, which is different from an inner product in that it is in general not conjugate symmetric. Thus a semi-inner-product is generally nonlinear with respect to its second variable. Giles [7] showed that in a fairly large class of Banach spaces, it is possible to construct a semi-inner-product with some desirable properties of the inner product. He proved that every normed space is a semi-inner-product space on which the semi-inner-product satisfies an extra homogeneity condition and gave fundamental properties extending Hilbert space type arguments to Banach spaces. Recently, semi-inner-products have been used as a useful tool in establishing the concept of reproducing kernel Banach spaces for machine learning [14].

On a separable complex Hilbert space $\mathcal{H}$, a conjugation is an isometric antilinear involution C from $\mathcal{H}$ to $\mathcal{H}$. A simple example of a conjugation on a Hilbert space is the pointwise complex conjugation on $L^{2}(\Omega,\mu )$, where $(\Omega,\mu )$ is a measure space with a positive measure μ. Garcia et al. [6, Lemma 2.11] proved that there exists an orthonormal basis $\{e_{n}\}_{n\ge 1}$ in $\mathcal{H}$ such that $Ce_{n}=e_{n}$ for any positive integer n, which asserts that every conjugation is unitarily equivalent to the canonical conjugation on an $\ell ^{2}$-space of the appropriate dimension. Takagi [13] studied the antilinear eigenvalue problem $Tx=\lambda \overline{x}$ where T is an $n\times n$ symmetric complex matrix and x̅ denotes the complex conjugation of the vector x in $\mathbb{C}^{n}$. The result of Godič and Lucenko [8] states that any unitary operator U on $\mathcal{H}$ can be factored as $U=CJ$ and is both C-symmetric and J-symmetric, where C and J are conjugations on $\mathcal{H}$. This generalizes the well-known fact that any planar rotation can be factored as the product of two reflections. Chō and Tanahashi [2] defined a conjugation on a complex Banach space and studied some spectral properties of complex symmetric operators.

The numerical range of $T \in \mathcal{L(H)}$ is the collection of complex numbers of the form $\langle T\xi, \xi \rangle$ with ξ ranging through the unit vectors in $\mathcal{H}$. The numerical range is very useful in studying operators and has many applications (see [9] for details), for example, numerical ranges are regarded as a rough estimate of eigenvalues, and generalizations of the numerical range are used to study quantum computing [3]. Recently, Hur and Lee [10] also studied the numerical ranges of conjugations and antilinear operators acting on a Hilbert space.

We now give a brief outline of the paper. In Sect. 2, we study the numerical range of a bounded linear operator on a semi-inner-product space. Using the standard semi-inner-product on $\ell _{n}^{p}(\mathbb{C})$, we compute numerical ranges of several operators, where $\ell _{n}^{p}(\mathbb{C})$ is the complex n-dimensional space with the $\ell ^{p}$-norm $(1\le p<\infty )$. Particularly, we compute numerical ranges of some operators acting on $\ell _{2}^{p}(\mathbb{C})$ and show that the numerical range of the backward shift on $\ell ^{p}(\mathbb{C})$ is the open unit disc, where $\ell ^{p}(\mathbb{C})$ is an infinite-dimensional space. In Sect. 3, we introduce a conjugation and a complex symmetric operator on a semi-inner-product space and investigate their basic properties. We prove some properties of a generalized adjoint of a complex symmetric operator on a semi-inner-product space. Moreover, we show that the numerical range of the complex conjugation on $\ell _{n}^{p}(\mathbb{C})$ $(n\ge 2)$ is the closed unit disc. Finally, we discuss the sequentially essential numerical ranges of operators multiplied by a conjugation in a semi-inner-product space.

2 Numerical ranges of semi-inner-product space operators

After introducing a semi-inner product space by Lumer [12], semi-inner-products have widely been applied to study bounded linear operators on Banach spaces [4]. Many properties of semi-inner-products were discovered by many authors, in particular, Giles [7]. We first recall the definitions of the semi-inner-product and the numerical range of a bounded operator on a semi-inner-product space and point out elementary properties of the numerical range.

Definition 2.1

Let $\mathcal{X}$ be a complex vector space. A semi-inner-product on $\mathcal{X}$ is a function $[\cdot,\cdot ]: \mathcal{X}\times \mathcal{X} \to \mathbb{C}$ satisfying the following properties: for any $x,y,z \in \mathcal{X}$,

• (1) $[x+y, z]=[x, z]+[y, z]$,

• (2) $[\lambda x, y]=\lambda [x, y]$ for all $\lambda \in \mathbb{C}$,

• (3) $[x, x]>0$ for $x\neq 0$,

• (4) $|[x, y]|^{2}\leq [x, x][y, y]$.

We say that $\mathcal{X}$ equipped with a semi-inner-product is a semi-inner-product space.

Lumer [12] proved that a semi-inner-product space is a normed vector space with norm $\|x\| = [x, x]^{1/2}$ and every normed linear space can be made into a semi-inner-product space. We see that a semi-inner-product is an inner product if and only if the induced norm satisfies the parallelogram law. Giles [7] showed that every normed vector space can be represented as a semi-inner-product space with the homogeneity property, that is,

$$\begin{aligned}{} [x, \lambda y]=\overline{\lambda}[x, y]\quad \text{for all $x,y \in \mathcal{X}$ and $\lambda \in \mathbb{C}$}. \end{aligned}$$

(1)

He proved the Riesz representation theorem for Hilbert space in the context of uniform semi-inner-product spaces, which says that if f is a continuous linear functional on $\mathcal{X}$, then there is a unique vector y in $\mathcal{X}$ such that $f(x) = [x, y]$ for all x in $\mathcal{X}$.

It is well known that for $1 < p<\infty$, the space $\ell _{n}^{p}(\mathbb{C})$ has the semi-inner-product defined by

$$\begin{aligned}{} [x,y]_{p}= \frac {1}{ \Vert y \Vert _{p}^{p-2}}\sum_{j=1}^{n} x_{j} \overline{y_{j}} \vert y_{j} \vert ^{p-2} \quad\text{for $x,y \ne 0$}, \end{aligned}$$

which is consistent with the $\ell ^{p}$-norm $\|\cdot \|_{p}$. For $p=1$, the semi-inner-product is given by

$$\begin{aligned}{} [x,y]_{1}= \Vert y \Vert _{1} \sum _{j=1}^{n} x_{j} \operatorname{sgn}( \overline{y_{j}}), \end{aligned}$$

where $\operatorname{sgn}(z)$ is ${z}/{|z|}$ if $z\in \mathbb{C} \setminus \{0\}$, and 0 if $z =0$.

In a semi-inner-product space $\mathcal{X}$, the numerical range $W(T)$ of $T \in \mathcal{L}(\mathcal{X})$ was defined in [12] as the set of numbers

$$\begin{aligned} W(T):=\bigl\{ [Tx, x]: [x, x]=1, x\in \mathcal{X} \bigr\} . \end{aligned}$$

This definition extends the classical one in a Hilbert space. It is well known that the numerical range of an operator in a Hilbert space is always convex; the proof can be done by reducing the problem to considering the numerical range of $2 \times 2$ matrices. However, the numerical range in a semi-inner-product space is not convex in general [12, Theorem 15].

Throughout this paper, $\mathcal{X}$ and $\mathcal{L}(\mathcal{X})$ denote a semi-inner-product space with semi-inner-product $[\cdot, \cdot ]$ and the set of bounded linear operators on $\mathcal{X}$, respectively, unless specified otherwise. We always assume that every semi-inner-product space has this homogeneity property.

The following elementary properties were observed by Lumer [12]. Let $T, S \in \mathcal{L}(\mathcal{X})$ and $\alpha, \beta \in \mathbb{C}$. We denote by $\sigma _{a}(T)$ the approximate point spectrum and by $\partial \sigma (T)$ the boundary of the spectrum.

• (i) $\frac{1}{4}\|T\|\leq w(T) \leq \|T\|$ for the numerical radius $w(T)=\sup \{|[Tx, x]|: x\in \mathcal{X} \}$,

• (ii) $W(T)=\{ \lambda \}$ if and only if $T=\lambda I$,

• (iii) $W(T)$ contains all of the eigenvalues of T,

• (iv) $W(\alpha T+\beta I)=\alpha W(T)+\beta$,

• (v) $W(T+S)\subseteq W(T)+W(S)$,

• (vi) $\sigma _{a}(T)\subseteq \operatorname{cl}[W(T)]$, i.e., $\partial \sigma (T)\subseteq \operatorname{cl}[W(T)]$ where $\operatorname{cl}[W(T)]$ denotes the closure of $W(T)$.

In this section, we explicitly compute the numerical range in a finite-dimensional semi-inner-product space. We denote by $\ell _{n}^{p}(\mathbb{C})$ $(p \ge 1)$ the complex n-dimensional space $\mathbb{C}^{n}$ equipped with the $\ell ^{p}$-norm. We first consider the numerical ranges of $2 \times 2$ matrices acting on the space $\ell _{2}^{p}(\mathbb{C})$.

Example 2.2

• (1) If $T=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ acts on $\ell _{2}^{p}(\mathbb{C})$ $(1\le p<\infty )$, then the numerical range $W(T)$ is the closed interval $[0, 1]$ since $[Tu, u]_{p}=|x|^{p}$ for any unit vector $u=\left(\begin{array}{c}x\\ y\end{array}\right)$.

• (2) For $T=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ acting on $\ell _{2}^{1}(\mathbb{C})$, the numerical range $W(T)$ is the open unit disc. Indeed, let $u=\left(\begin{array}{c}x\\ y\end{array}\right)$ in $\ell _{2}^{1}(\mathbb{C})$ be any unit vector, so $|x|+|y|=1$. Then we have that

  $$\begin{aligned}{} [Tu,u]_{1}=y \operatorname{sgn}(\overline{x}) = \textstyle\begin{cases} 0 & \text{if } x=0, \\ y e^{i\theta} & \text{if } x \ne 0, \end{cases}\displaystyle \end{aligned}$$

  where θ is a real number such that $e^{i\theta} = \frac {\overline{x}}{|x|}$. If $|[Tu,u]_{1}|=1$ for nonzero x, then we have $1=|[Tu,u]_{1}|= |y||\operatorname{sgn}(\overline{x})|=|y|$. However, we should have $|x|=0$ because of $|x|+|y|=1$, which contradicts to assumption. Thus the numerical range $W(T)$ is the open unit disc.

When $1< p<\infty$, we now investigate the numerical range of $T=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ acting on $\ell _{2}^{p}(\mathbb{C})$.

Lemma 2.3

For $1 < p<\infty$, the range of the function $f(t):=(\tan t)^{\frac{2}{p}}(\cos t)^{2}$ defined on $[0, {\pi}/2)$ is the interval $[0,\frac{1}{p}(p-1)^{\frac{p-1}{p}} ]$.

Proof

We see that $f(0)=0$ and $\lim_{t\rightarrow \frac{\pi}{2}^{-}}f(t)=0$. Since

$$\begin{aligned} f'(t)=2(\tan t)^{\frac{2}{p}}\cos t \biggl(\frac {1}{p\sin t}-\sin t \biggr), \end{aligned}$$

f has the absolute maximum when $\sin t=1 / \sqrt{p}$. In this case, we have

$$\begin{aligned} \tan t=\frac {1}{\sqrt{p-1}} \quad\text{and}\quad \cos t=\sqrt{ \frac {p-1}{p}}, \end{aligned}$$

so that the maximum value of f is equal to $\frac{1}{p}(p-1)^{\frac{p-1}{p}}$. By the intermediate value theorem the range of f is $[0,\frac{1}{p}(p-1)^{\frac{p-1}{p}} ]$. □

Example 2.4

For $1 < p<\infty$, let $D_{p}$ be the closed disc of radius $\frac{1}{p}(p-1)^{\frac{p-1}{p}}$ centered at the origin. For $T=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ acting on $\ell _{2}^{p}(\mathbb{C})$, we see that $W(T)$ is the closed disc $D_{p}$.

To show that $W(T)=D_{p}$, we take any unit vector $u=\left(\begin{array}{c}x\\ y\end{array}\right)$ in $\ell _{2}^{p}(\mathbb{C})$ with $|x|^{p}+|y|^{p}=1$. Since $[Tu, u]_{p}=y\overline{x}|x|^{p-2}$ and $|y|^{p} =1-|x|^{p}$, we have

$$\begin{aligned} \bigl\vert [Tu, u]_{p} \bigr\vert ^{p}=\bigl(1- \vert x \vert ^{p}\bigr) \vert x \vert ^{p(p-1)}. \end{aligned}$$

Since the function $g(t)=(1-t^{p})t^{p(p-1)}$ has the maximum value $\frac{1}{p^{p}} (p-1)^{p-1}$ at $t= (\frac{p-1}{p} )^{1/p}$, $|[Tu, u]_{p} |$ has the maximum value $\frac{1}{p}(p-1)^{\frac{p-1}{p}}$ when $|x|^{p}=\frac{p-1}{p}$. Thus we see that $W(T)$ is contained in the closed disc $D_{p}$.

To show the reverse inclusion, let λ be any complex number in $D_{p}$. We can write $\lambda =re^{i\theta}$ for some $0\leq r \leq \frac{1}{p}(p-1)^{\frac{p-1}{p}}$ and $\theta \in \mathbb{R}$. We take a unit vector

$$\begin{aligned} u= \bigl( \vert \cos \alpha \vert ^{\frac{2}{p}}, e^{i\theta} \vert \sin \alpha \vert ^{ \frac{2}{p}} \bigr)^{t} \in \ell _{2}^{p}(\mathbb{C}) \end{aligned}$$

for any real number α. We note that the arctangent function $\arctan: \mathbb{R} \rightarrow (-\frac {\pi}{2}, \frac {\pi}{2})$ is 1–1 and onto. By Lemma 2.3 there exists a unique real number $\alpha \in [0, \arctan ((p-1)^{\frac{p-1}{2}}p^{-\frac{p}{2}}) ]$ such that $r=|\tan \alpha |^{\frac{2}{p}}|\cos \alpha |^{2}$. Consequently, for such a unit vector u, we have

$$\begin{aligned}{} [Tu, u]_{p}=e^{i\theta} \vert \sin \alpha \vert ^{\frac{2}{p}} \vert \cos \alpha \vert ^{ \frac{2p-2}{p}} =e^{i\theta} \vert \tan \alpha \vert ^{\frac{2}{p}} \vert \cos \alpha \vert ^{2}=re^{i \theta}= \lambda, \end{aligned}$$

which completes the proof.

Proposition 2.5

Let $T=\left(\begin{array}{cc}a& b\\ 0& -a\end{array}\right)$ act on $\ell _{2}^{1}(\mathbb{C})$ where $a\in \mathbb{R}\setminus \{0\}$ and $b\in \mathbb{C}$. Then

$$\begin{aligned} W(T) = \textstyle\begin{cases} \{x+iy\in \mathbb{C}: (x+a)^{2}+y^{2}\leq 4a^{2}\} & \textit{if } \vert b \vert =2 \vert a \vert , \\ \{x+iy\in \mathbb{C}: y^{2}\leq \frac { \vert b \vert ^{2}}{ \vert 4a^{2}- \vert b \vert ^{2} \vert }(x-a)^{2} \} & \textit{if } \vert b \vert \neq 2 \vert a \vert . \end{cases}\displaystyle \end{aligned}$$

Proof

For any unit vector $u=(e^{i\alpha}\cos ^{2}\theta, e^{i\beta}\sin ^{2}\theta )^{t}\in \ell _{2}^{1}(\mathbb{C})$ with $0\leq \alpha, \beta < 2\pi$ and $0\leq \theta <\pi$, we have

$$\begin{aligned}{} [Tu, u]_{1}&=a\cos ^{2}\theta +be^{i(\beta -\alpha )}\sin ^{2}\theta -a \sin ^{2}\theta \\ &=a\cos 2\theta +\frac{b}{2}(1-\cos 2\theta )e^{i(\beta -\alpha )}. \end{aligned}$$

Letting $[Tu, u]_{1}=: x+iy$ with $x,y \in \mathbb{R}$, we have the equation

$$\begin{aligned} (x-a\cos 2\theta )^{2}+y^{2}=\frac { \vert b \vert ^{2}}{4}(1-\cos 2 \theta )^{2}, \end{aligned}$$

so that the following quadratic equation for $\cos 2\theta$ holds;

$$\begin{aligned} \bigl(4a^{2}- \vert b \vert ^{2}\bigr)\cos ^{2}2\theta -2\bigl(4ax- \vert b \vert ^{2}\bigr)\cos 2 \theta +4x^{2}+4y^{2}- \vert b \vert ^{2}=0. \end{aligned}$$

(2)

In the case of $|b|=2|a|$, we have the equation $2a(x-a)\cos 2\theta =x^{2}+y^{2}-a^{2}$. If $x=a$, then we have $y=0$. Assumet that $x\neq a$. Then it follows that

$$\begin{aligned} \vert \cos 2\theta \vert = \biggl\vert \frac {x^{2}+y^{2}-a^{2}}{2a(x-a)} \biggr\vert \leq 1. \end{aligned}$$

In the case of $a(x-a)>0$, we have $W(T)=\emptyset$. On the other hand, in the case of $a(x-a)<0$, we get the inequality $(x+a)^{2}+y^{2}\leq 4a^{2}$, so that

$$\begin{aligned} W(T)=\bigl\{ x+iy\in \mathbb{C}: (x+a)^{2}+y^{2}\leq 4a^{2}\bigr\} \setminus \bigl\{ (a,0) \bigr\} . \end{aligned}$$

By combining these cases we have $W(T)=\{x+iy\in \mathbb{C}: (x+a)^{2}+y^{2}\leq 4a^{2}\}$.

In the case of $|b|< 2|a|$, since $\cos 2\theta$ must be real, the discriminant of equation (2) gives the inequality

$$\begin{aligned} \bigl(4ax- \vert b \vert ^{2}\bigr)^{2}- \bigl(4a^{2}- \vert b \vert ^{2}\bigr) \bigl(4x^{2}+4y^{2}- \vert b \vert ^{2}\bigr) \geq 0, \end{aligned}$$

so that

$$\begin{aligned} y^{2}\leq \frac { \vert b \vert ^{2}}{4a^{2}- \vert b \vert ^{2}}(x-a)^{2}. \end{aligned}$$

In the case of $|b|>2|a|$, we similarly get the inequality

$$\begin{aligned} y^{2}\leq \frac { \vert b \vert ^{2}}{ \vert b \vert ^{2}-4a^{2}}(x-a)^{2}. \end{aligned}$$

These complete the proof. □

Example 2.6

If $T=\left(\begin{array}{cc}1& i\\ 0& -1\end{array}\right)$ acts on $\ell _{2}^{1}(\mathbb{C})$, then it follows form Proposition 2.5 that the numerical range $W(T)$ is the region satisfying $-\frac {1}{\sqrt{3}}(x-1)\leq y\leq \frac {1}{\sqrt{3}}(x-1)$ as follows.

Proposition 2.7

Let $T=\left(\begin{array}{cc}a& b\\ 0& c\end{array}\right)$ act on $\ell _{2}^{1}$, where $a, c\in \mathbb{R}\setminus \{0\}$, $a+c\neq 0$, and $b\in \mathbb{C}$. For any unit vector $u=(x, y)^{t}\in \ell _{2}^{1}$, we have

$$\begin{aligned}{} [Tu, u]_{1} = \textstyle\begin{cases} a & \textit{if } \vert x \vert =1 \textit{ and } \vert y \vert =0, \\ c & \textit{if } \vert x \vert =0 \textit{ and } \vert y \vert =1, \\ a \vert x \vert +c(1- \vert x \vert )+\frac {b\overline{x}y}{ \vert x \vert } & \textit{if } \vert x \vert \neq 0 \textit{ and } \vert y \vert =1- \vert x \vert . \end{cases}\displaystyle \end{aligned}$$

Proof

The proof can be obtained by simple computations, so we omit it. □

Remark 2.8

In Proposition 2.7, we suppose that $b, x, y$ are pure imaginary numbers. Let $b=\alpha i$ for a nonzero real number α. If $|x|=r$ for $0< r < 1$, then

$$\begin{aligned}{} [Tu, u]_{1}=\bigl\{ r a+(1-r)c \bigr\} +(1-r)\alpha i. \end{aligned}$$

So, $[Tu, u]_{1}$ converges to a as $r \rightarrow 1$. On the other hand, $[Tu, u]_{1}$ goes to $c\pm \alpha i$ as $r \rightarrow 0$.

Corollary 2.9

Let T be as in Proposition 2.7, and let $u=(x, y)^{t}\in \ell _{2}^{1}$ be a unit vector. If, in addition, b is a nonzero real number, then for $x \neq 0$ and y with $|y|=1-|x|$,

$$\begin{aligned} W(T)\subset \biggl\{ \lambda \in \mathbb{C}: \bigl\vert \operatorname{{Re}}( \lambda ) \bigr\vert \leq \frac{ \vert a+b+c \vert }{2} \textit{ and } \bigl\vert \operatorname{{Im}}(\lambda ) \bigr\vert \leq \frac{ \vert b \vert }{2} \biggr\} . \end{aligned}$$

Proof

If $x=x_{1}+i x_{2}$ and $y=y_{1}+i y_{2}$, where $x_{j}$ and $y_{j}$ are real numbers ($j=1, 2$), then we obtain from Proposition 2.7 that

$$\begin{aligned}{} [Tu, u]_{1}= \biggl\{ (a-c) \vert x \vert +c+\frac{b}{ \vert x \vert }(x_{1} y_{1}+x_{2} y_{2}) \biggr\} +i \frac{b}{ \vert x \vert }(x_{1} y_{2}-x_{2} y_{1}). \end{aligned}$$

By the Cauchy–Schwarz inequality, we get that for $t:=\sqrt{x_{1}^{2}+x_{2}^{2}}$,

$$\begin{aligned} (x_{1} y_{1}+x_{2} y_{2})^{2}& \leq \bigl(x_{1}^{2}+x_{2}^{2}\bigr) \bigl(y_{1}^{2}+y_{2}^{2}\bigr) \\ &=\bigl(x_{1}^{2}+x_{2}^{2}\bigr) \Bigl(1-\sqrt{x_{1}^{2}+x_{2}^{2}} \Bigr)^{2} \\ &=t^{2} (1-t)^{2}=:f(t). \end{aligned}$$

On the interval $(0,1)$, f has the maximum $\frac{1}{16}$ at $t=\frac{1}{2}$, so that we get $|x_{1} y_{1}+x_{2} y_{2}|\leq \frac{1}{4}$, and this gives the inequality

$$\begin{aligned} \bigl\vert \operatorname{Re}\bigl([Tu, u]_{1}\bigr) \bigr\vert \leq \frac{ \vert a+b+c \vert }{2}. \end{aligned}$$

In this case, we also get the inequality $|\operatorname{Im}([Tu, u]_{1}) |=|2b(x_{1} y_{2}-x_{2} y_{1})|\leq \frac {|b|}{2}$ by a similar method. This completes the proof. □

Theorem 2.10

Let T be the backward shift on an infinite dimensional Banach space $\ell ^{p}(\mathbb{C})$ for $1\le p< \infty$. Then the numerical range $W(T)$ is the open unit disc.

Proof

Let $x=(x_{1}, x_{2}, x_{3}, \ldots )$ be any unit vector in $\ell ^{p}(\mathbb{C})$, and let $k = \min \{ i \ge 1: x_{i} \ne 0\}$. Then we have

$$\begin{aligned} \bigl\vert [Tx, x]_{p} \bigr\vert & \le \sum _{j=k}^{\infty} \vert x_{j+1} \vert \vert x_{j} \vert ^{p-1} \\ &\leq \frac{1}{p}\sum_{j=k}^{\infty} \bigl\{ \vert x_{j+1} \vert ^{p}+(p-1) \vert x_{j} \vert ^{p} \bigr\} \\ &=\frac{1}{p} \Biggl\{ (p-1) \vert x_{k} \vert ^{p}+p\sum_{j=k+1}^{\infty} \vert x_{j} \vert ^{p} \Biggr\} \\ &= \frac{1}{p} \Biggl\{ p\sum_{j=k}^{\infty} \vert x_{j} \vert ^{p}- \vert x_{k} \vert ^{p} \Biggr\} =1-\frac{ \vert x_{k} \vert ^{p}}{p}< 1, \end{aligned}$$

where the second inequality follows from the inequality of arithmetic and geometric means. Hence we obtain that $|[Tx, x]_{p}|<1$ for any unit vector $x\in \ell ^{p}(\mathbb{C})$, which implies that $W(T)$ is contained in the open unit disc.

To show the reverse inclusion, let $\lambda =re^{i\theta}$ be any vector in the open unit disc with $0\leq r<1$. We take the vector $x \in \ell ^{p}(\mathbb{C})$ given by

$$\begin{aligned} x= {\bigl(}\bigl(1-r^{p}\bigr)^{\frac{1}{p}}, r\bigl(1-r^{p} \bigr)^{\frac{1}{p}}e^{i\theta}, r^{2}\bigl(1-r^{p} \bigr)^{\frac{1}{p}}e^{2i\theta}, r^{3}\bigl(1-r^{p} \bigr)^{\frac{1}{p}}e^{3i \theta}, \ldots {\bigr)}. \end{aligned}$$

Then we see that $\|x\|_{p} =1$, so that $[x, x]_{p}= \|x\|_{p}^{2} = 1$. Moreover, we get that

$$\begin{aligned}{} [Tx, x]_{p}=re^{i\theta}\bigl(1-r^{p}\bigr)\sum _{k=1}^{\infty}r^{(k-1)p}=re^{i \theta}=\lambda, \end{aligned}$$

which implies that $W(T)$ contains the open unit disc. This completes the proof. □

3 Conjugations and complex symmetric operators

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 semi-inner-product space using a semi-inner-product. Throughout this section, $\mathcal{X}$ denotes a semi-inner-product space with a semi-inner-product $[\cdot,\cdot ]$, 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 semi-inner-product, 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 semi-inner-product 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

$$\begin{aligned} Cz:= C(x+y)-Cx-Cy. \end{aligned}$$

Then we get that $0=[ C(x+y)-Cx-Cy, Cz ]=[ Cz, Cz]$, 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 C-symmetric 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 C-symmetric 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 semi-inner-product space operators without using linear functionals. Even though semi-inner-products in general are not additive in the second variables, we will use a semi-inner-product to define the C-symmetric operator on a semi-inner-product space.

Definition 3.3

Let C be a conjugation on $\mathcal{X}$. We say that $T\in \mathcal{L}(\mathcal{X})$ is C-symmetric 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 C-symmetric operator.

• (i) λT is C-symmetric for any complex number λ.

• (ii) If T is invertible, then $T^{-1}$ is also C-symmetric.

• (iii) If $S\in \mathcal{L}(\mathcal{X})$ is C-symmetric 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 C-symmetric.

(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 C-symmetry of T. Thus $T^{-1}$ is C-symmetric, which completes the proof.

(iii) If $S\in \mathcal{L}(\mathcal{X})$ commutes with T and is C-symmetric, 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 C-symmetric. □

Let $T \in \mathcal{L}(\mathcal{X})$ and $y \in \mathcal{X}$. By the Riesz representation theorem in a semi-inner-product 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 C-symmetric, 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 C-symmetric. 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 semi-inner-product space means a uniformly continuous semi-inner-product 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 semi-inner-product space $\mathcal{X}$, the dual space $\mathcal{X}^{\star}$ is also a uniform complex semi-inner-product space with respect to the semi-inner-product 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 one-to-one 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 semi-inner-product 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 semi-inner-product space $\mathcal{X}$.

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

• (ii) If $T\in \mathcal{L}(\mathcal{X})$ is C-symmetric, then $(T^{\dagger})^{\star}=(T^{\star})^{\dagger}$.

• (iii) If $\{T_{n} \}$ is a sequence of C-symmetric operators such that $T_{n} \to S$ in the strong topology, then S is C-symmetric.

Proof

(i) Suppose that T is a C-symmetric operator on $\mathcal{X}$. Let f and g be arbitrary elements in the dual space $\mathcal{X}^{\star}$. Since $\mathcal{X}$ is a uniform semi-inner-product 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 C-symmetric 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 $0<|\lambda |\leq \frac{1}{n-1}$, we take the vector $x=(x_{1}, x_{2}, \ldots, x_{n})^{t} \in \ell _{n}^{1}(\mathbb{C})$ given by

$$\begin{aligned} x_{j}= \textstyle\begin{cases} \overline{\lambda} & \text{if } 1\leq j \leq n-1, \\ ( \frac {1}{ \vert \lambda \vert }-n+1 ) \overline{\lambda} & \text{if } j=n. \end{cases}\displaystyle \end{aligned}$$

(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 $|\lambda |\geq n-1$, we put $x=(x_{1}, x_{2}, \ldots, x_{n})\in \ell _{n}^{1}(\mathbb{C})$, where

$$\begin{aligned} x_{j}= \textstyle\begin{cases} \overline{\lambda}^{-1} & \text{if } 1\leq j \leq n-1, \\ \overline{\lambda}^{-1}( \vert \lambda \vert -n+1) & \text{if } j=n. \end{cases}\displaystyle \end{aligned}$$

(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 infinite-dimensional 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 semi-inner-product 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 C-symmetric.

• (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 semi-inner-product 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 semi-inner-product space.

Proposition 3.11

If C and G are conjugations on $\mathcal{X}$, then $U=CG$ is a unitary and is both C-symmetric and G-symmetric.

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 C-symmetric and G-symmetric. □

Remark 3.12

(i) In Proposition 3.11, if $U^{\dagger}$ is a generalized adjoint of a unitary $U=CG$, then we get from C-symmetry 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 semi-inner-product 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 $(U^{\star})^{\dagger}=(U^{\star})^{-1}$.