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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 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. (1)

    \([x+y, z]=[x, z]+[y, z]\),

  2. (2)

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

  3. (3)

    \([x, x]>0\) for \(x\neq 0\),

  4. (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.

  1. (i)

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

  2. (ii)

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

  3. (iii)

    \(W(T)\) contains all of the eigenvalues of T,

  4. (iv)

    \(W(\alpha T+\beta I)=\alpha W(T)+\beta \),

  5. (v)

    \(W(T+S)\subseteq W(T)+W(S)\),

  6. (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. (1)

    If T= ( 1 0 0 0 ) 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= ( x y ) .

  2. (2)

    For T= ( 0 1 0 0 ) acting on \(\ell _{2}^{1}(\mathbb{C})\), the numerical range \(W(T)\) is the open unit disc. Indeed, let u= ( x y ) 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= ( 0 1 0 0 ) 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= ( 0 1 0 0 ) 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= ( x y ) 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= ( a b 0 a ) 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= ( 1 i 0 1 ) 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.

figure b

Proposition 2.7

Let T= ( a b 0 c ) 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.

  1. (i)

    λT is C-symmetric for any complex number λ.

  2. (ii)

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

  3. (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 \(\mathcal{X}\).

  1. (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.

  2. (ii)

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

  3. (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. (1)

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

  2. (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 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.

  1. (i)

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

  2. (ii)

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

  3. (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 \((U^{\star})^{\dagger}=(U^{\star})^{-1}\).

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Acknowledgements

The authors would like to thank Professor Woo Young Lee for his valuable comments.

Funding

I.J. An was supported by the National Research Foundation of Korea (NRF) funded by Ministry of Science and ICT (NRF-2020R1A2C1A01006036). J. Heo was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science and ICT (NRF-2020R1A4A3079066).

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An, I.J., Heo, J. Numerical ranges and complex symmetric operators in semi-inner-product spaces. J Inequal Appl 2022, 150 (2022). https://doi.org/10.1186/s13660-022-02886-x

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  • DOI: https://doi.org/10.1186/s13660-022-02886-x

MSC

  • 46C50
  • 47A05
  • 47A12

Keywords

  • Semi-inner-product space
  • Numerical range
  • Conjugations
  • Complex symmetric operators
  • Generalized adjoint