Birkhoff--James orthogonality of operators in semi-Hilbertian spaces and its applications

TL;DR

This paper generalizes Birkhoff--James orthogonality for operators in semi-Hilbertian spaces, providing characterizations, formulas, and extending classical theorems to this broader context.

Contribution

It introduces a new orthogonality relation in semi-Hilbertian spaces, extending existing theorems and deriving new formulas for operator distances.

Findings

Characterization of Birkhoff--James orthogonality via sequences of $A$-unit vectors.

Extension of a theorem by Bhatia and Semrl to semi-Hilbertian spaces.

Derivation of $A$-distance formulas involving supremum over orthogonal vectors.

Abstract

In this paper, the concept of Birkhoff--James orthogonality of operators on a Hilbert space is generalized when a semi-inner product is considered. More precisely, for linear operators $T$ and $S$ on a complex Hilbert space $\mathcal{H}$, a new relation $T\perp^B_A S$ is defined if $T$ and $S$ are bounded with respect to the seminorm induced by a positive operator $A$ satisfying ${\|T + \gamma S\|}_A\geq {\|T\|}_A$ for all $\gamma \in \mathbb{C}$. We extend a theorem due to R. Bhatia and P. \v{S}emrl, by proving that $T\perp^B_A S$ if and only if there exists a sequence of $A$-unit vectors $\{x_n\}$ in $\mathcal{H}$ such that $\displaystyle{\lim_{n\rightarrow +\infty}}{\|Tx_n\|}_A = {\|T\|}_A$ and $\displaystyle{\lim_{n\rightarrow +\infty}}{\langle Tx_n, Sx_n\rangle}_A = 0$. In addition, we give some $A$-distance formulas. Particularly, we prove \begin{align*} \displaystyle{\inf_{\gamma \in \mathbb{C}}}{\|T + \gamma S\|}_{A} = \sup\Big\{|{\langle Tx, y\rangle}_A|; \, {\|x\|}_{A} = {\|y\|}_{A} = 1, \, {\langle Sx, y\rangle}_A = 0\Big\}. \end{align*} Some other related results are also discussed.

Equations

$$\displaystyle\displaystyle{\inf_{\gamma\in\mathbb{C}}}{\|T+\gamma S\|}_{A}=\sup\Big{\{}|{\langle Tx,y\rangle}_{A}|;\,{\|x\|}_{A}={\|y\|}_{A}=1,\,{\langle Sx,y\rangle}_{A}=0\Big{\}}.$$

$${⟨x,y⟩}_A=⟨Ax,y⟩,x,y\in \mathcal{H}.$$

$$\displaystyle\mathbb{B}_{A^{1/2}}(\mathcal{H})=\Big{\{}T\in\mathbb{B}(\mathcal{H}):\,\exists\,c>0\,\forall x\in\mathcal{H};{\|Tx\|}_{A}\leq c{\|x\|}_{A}\Big{\}}.$$

$$\displaystyle{\|T\|}_{A}=\displaystyle{\sup_{x\in\overline{\mathcal{R}(A)},x\neq 0}}\frac{{\|Tx\|}_{A}}{{\|x\|}_{A}}=\inf\Big{\{}c>0;\,{\|Tx\|}_{A}\leq c{\|x\|}_{A},x\in\mathcal{H}\Big{\}}<\infty.$$

$$\displaystyle{\|T\|}_{A}=\displaystyle{\sup_{x\in\mathcal{H},{\|x\|}_{A}=1}}{\|Tx\|}_{A}=\sup\Big{\{}|{\langle Tx,y\rangle}_{A}|;\,x,y\in\mathcal{H},{\|x\|}_{A}={\|y\|}_{A}=1\Big{\}}.$$

$$\Vert T+\gamma S\Vert \ge \Vert T\Vert \text{mbox}forall\gamma \in \mathbb{C}.$$

$$\underset{n\to \infty }{\lim }\Vert T{x}_n\Vert =\Vert T\Vert \text{mbox}and\underset{n\to \infty }{\lim }⟨T{x}_n,S{x}_n⟩=0.$$

$${\Vert T+\gamma S\Vert }_A\ge {\Vert T\Vert }_A\text{mbox}forall\gamma \in \mathbb{C}.$$

$$\underset{n\to +\infty }{\lim }{\Vert T{x}_n\Vert }_A={\Vert T\Vert }_A\text{mbox}and\underset{n\to +\infty }{\lim }{⟨T{x}_n,S{x}_n⟩}_A=0.$$

$$\displaystyle\displaystyle{\inf_{\gamma\in\mathbb{C}}}{\|T+\gamma S\|}_{A}=\sup\Big{\{}|{\langle Tx,y\rangle}_{A}|;\,{\|x\|}_{A}={\|y\|}_{A}=1,\,Sx\perp_{A}y\Big{\}}.$$

$${\Phi }_A^{(T,S)}(x)=\{\begin{array}{ll}{\Vert Tx\Vert }_A^2-\frac{|{⟨Tx,Sx⟩}_A{|}^2}{{\Vert Sx\Vert }_A^2}& \text{if ∥Sx∥A=0,}\\ {\Vert Tx\Vert }_A^2& \text{if ∥Sx∥A=0}.\end{array}$$

$$\displaystyle W_{A}(T,S)=\Big{\{}\xi\in\mathbb{C};\,\exists\,\{x_{n}\}\subset\mathcal{H},\,{\|x_{n}\|}_{A}=1,\,$$

$$\displaystyle\qquad\qquad\mbox{and}\,\displaystyle{\lim_{n\rightarrow+\infty}}{\langle Tx_{n},Sx_{n}\rangle}_{A}=\xi\Big{\}}$$

$${\Vert T\Vert }_A=\sup \{{\Vert Tx\Vert }_A;x\in \overline{\mathcal{R}(A)},{\Vert x\Vert }_A=1\},$$

$$\underset{n\to +\infty }{\lim }{\Vert T{x}_n\Vert }_A={\Vert T\Vert }_A.$$

$$|{⟨T{x}_n,S{x}_n⟩}_A|\le {\Vert T{x}_n\Vert }_A{\Vert S{x}_n\Vert }_A\le {\Vert T\Vert }_A{\Vert S\Vert }_A.$$

$$\displaystyle W_{A}(T,S)\subset\big{\{}\xi\in\mathbb{C};\,|\xi|\leq{\|T\|}_{A}{\|S\|}_{A}\big{\}}.$$

$$\displaystyle\big{|}{\|Tx^{n}_{m}\|}_{A}-{\|T\|}_{A}\big{|}<\varepsilon$$

$$\displaystyle\big{|}{\langle Tx^{n}_{m},Sx^{n}_{m}\rangle}_{A}-\xi_{n}\big{|}<\frac{\varepsilon}{2}$$

$$\displaystyle\big{|}{\|Tx^{n}_{m}\|}_{A}-{\|T\|}_{A}\big{|}<\varepsilon$$

$$\displaystyle\big{|}{\langle Tx^{n}_{m},Sx^{n}_{m}\rangle}_{A}-\xi\big{|}\leq\big{|}{\langle Tx^{n}_{m},Sx^{n}_{m}\rangle}_{A}-\xi_{n}\big{|}+|\xi_{n}-\xi|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$

$$W_A(T,S)$$

$$\displaystyle\displaystyle{\lim_{n\rightarrow+\infty}}{\|T(y_{n}+z_{n})\|}_{A}={\|T\|}_{A},\,\mbox{and}\,\displaystyle{\lim_{n\rightarrow+\infty}}{\langle Ty_{n},Sz_{n}\rangle}_{A}+{\langle Tz_{n},Sz_{n}\rangle}_{A}=\xi\Big{\}}.$$

$$W_A(T,S)$$

$$\displaystyle\qquad\qquad\displaystyle{\lim_{n\rightarrow+\infty}}{\|Tz_{n}\|}_{A}={\|T\|}_{A},\,\mbox{and}\,\displaystyle{\lim_{n\rightarrow+\infty}}{\langle Tz_{n},Sz_{n}\rangle}_{A}=\xi\Big{\}}$$

$$\displaystyle=\Big{\{}\xi\in\mathbb{C};\,\exists\,\{z_{n}\}\subset\overline{\mathcal{R}(A)},\,{\|z_{n}\|}_{A}=1,\,$$

$$\displaystyle\qquad\qquad\displaystyle{\lim_{n\rightarrow+\infty}}{\|PTz_{n}\|}_{A}={\|PT\mid_{\overline{\mathcal{R}(A)}}\|}_{A},\,\mbox{and}\,\displaystyle{\lim_{n\rightarrow+\infty}}{\langle PTz_{n},PSz_{n}\rangle}_{A}=\xi\Big{\}}$$

$$=W_{A_0}(\tilde{T},\tilde{S}),$$

$$\displaystyle m(S)=\inf\Big{\{}\|Sx\|:\,x\in\mathcal{H},\|x\|=1\Big{\}}.$$

$$\displaystyle m_{A}(S)=\inf\Big{\{}{\|Sx\|}_{A}:\,x\in\mathcal{H},{\|x\|}_{A}=1\Big{\}}.$$

$$\underset{n\to +\infty }{\lim }{\Vert T{x}_n\Vert }_A={\Vert T\Vert }_A\text{mbox}and\underset{n\to +\infty }{\lim }{⟨T{x}_n,S{x}_n⟩}_A=0.$$

$${\Vert T+\gamma S\Vert }_A^2$$

$$={\Vert T{x}_n\Vert }_A^2+\overline{\gamma }{⟨T{x}_n,S{x}_n⟩}_A+\gamma {⟨S{x}_n,T{x}_n⟩}_A+|\gamma {|}^2{\Vert S{x}_n\Vert }_A^2$$

$${\Vert T+\gamma S\Vert }_A^2\ge {\Vert T\Vert }_A^2+|\gamma {|}^2\underset{n\to \infty }{\lim }\sup {\Vert S{x}_n\Vert }_A^2\ge {\Vert T\Vert }_A^2+|\gamma {|}^2{m}_A^2(S)$$

Full text

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Some other related results are also discussed.

1. Introduction and preliminaries

Let $\mathbb{B}(\mathcal{H})$ denote the $C^{\ast}$-algebra of all bounded linear operators on a complex Hilbert space $\mathcal{H}$ with an inner product $\langle\cdot,\cdot\rangle$ and the corresponding norm $\|\cdot\|$. The symbol $I$ stands for the identity operator on $\mathcal{H}$. If $T\in\mathbb{B}(\mathcal{H})$, then we denote by $\mathcal{R}(T)$ and $\mathcal{N}(T)$ the range and the kernel of $T$, respectively, and by $\overline{\mathcal{R}(T)}$ the norm closure of $\mathcal{R}(T)$. Throughout this article, we assume that $A\in\mathbb{B}(\mathcal{H})$ is a positive operator and that $P$ is the orthogonal projection onto $\overline{\mathcal{R}(A)}$. Recall that $A$ is called positive if $\langle Ax,x\rangle\geq 0$ for all $x\in\mathcal{H}$. Such an $A$ induces a positive semidefinite sesquilinear form ${\langle\cdot,\cdot\rangle}_{A}:\,\mathcal{H}\times\mathcal{H}\rightarrow\mathbb{C}$ defined by

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Denote by ${\|\cdot\|}_{A}$ the seminorm induced by ${\langle\cdot,\cdot\rangle}_{A}$, that is, ${\|x\|}_{A}=\sqrt{{\langle x,x\rangle}_{A}}$ for every $x\in\mathcal{H}$. It can be easily seen that ${\|\cdot\|}_{A}$ is a norm if and only if $A$ is an injective operator, and that $(\mathcal{H},{\|\cdot\|}_{A})$ is a complete space if and only if $\mathcal{R}(A)$ is closed in $\mathcal{H}$. For $x,y\in\mathcal{H}$, we say that $x$ and $y$ are $A$-orthogonal, denoted by $x\perp_{A}y$, if ${\langle x,y\rangle}_{A}=0$. Note that this definition is a natural extension of the usual notion of orthogonality, which represents the $I$-orthogonality case. Furthermore, we put

[TABLE]

We say that an operator $T\in\mathbb{B}(\mathcal{H})$ is $A$-bounded if $T$ belongs to $\mathbb{B}_{A^{1/2}}(\mathcal{H})$. It can be shown that $\mathbb{B}_{A^{1/2}}(\mathcal{H})$ is a unital subalgebra of $\mathbb{B}(\mathcal{H})$ which, in general, is neither closed nor dense in $\mathbb{B}(\mathcal{H})$ (see [2]). We equip $\mathbb{B}_{A^{1/2}}(\mathcal{H})$ with the seminorm ${\|\cdot\|}_{A}$ defined as follows:

[TABLE]

In addition, for $T\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$, we have

[TABLE]

Of course, many difficulties arise. For instance, it may happen that ${\|T\|}_{A}=\infty$ for some $T\in\mathbb{B}(\mathcal{H})$. In addition, not any operator admits an adjoint operator for the semi-inner product ${\langle\cdot,\cdot\rangle}_{A}$. For more details about this class of operators, we refer the reader to [2]. In recent years, several results covering some classes of operators on a complex Hilbert space \big{(}\mathcal{H},\langle\cdot,\cdot\rangle\big{)} are extended to \big{(}\mathcal{H},{\langle\cdot,\cdot\rangle}_{A}\big{)}; see [2, 3] and their references.

The notion of orthogonality in $\mathbb{B}(\mathcal{H})$ can be introduced in many ways ( see, e.g., [13]). When $T,S\in\mathbb{B}(\mathcal{H})$, we say that $T$ is Birkhoff–James orthogonal to $S$, denoted $T\perp^{B}S$, if

[TABLE]

In Hilbert spaces, this orthogonality is equivalent to the usual notion of orthogonality. This notion of orthogonality plays a very important role in the geometry of Hilbert space operators. For $T,S\in\mathbb{B}(\mathcal{H})$, Bhatia and Šemrl in [4, Remark 3.1] and Paul in [14, Lemma 2] independently proved that $T\perp^{B}S$ if and only if there exists a sequence of unit vectors $\{x_{n}\}$ in $\mathcal{H}$ such that

[TABLE]

It follows then that if the Hilbert space $\mathcal{H}$ is finite-dimensional, $T\perp^{B}S$ if and only if there is a unit vector $x\in\mathcal{H}$ such that $\|Tx\|=\|T\|$ and $\langle Tx,Sx\rangle=0$.

Recently, some authors extended the well known result of Bhatia–Šemrl (see [6, 18, 19]). Moreover, the papers [18] and [19] show another ways to obtain the Bhatia–Šemrl theorem. Some other authors studied different aspects of orthogonality of operators on various Banach spaces and elements of an arbitrary Hilbert $C^{*}$-module; see, for instance,[1, 5, 7, 10, 11, 15, 17, 20].

Now, let us introduce the notion of $A$-Birkhoff–James orthogonality of operators in semi-Hilbertian spaces.

Definition 1.1**.**

An element $T\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$ is called an $A$-Birkhoff–James orthogonal to another element $S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$, denoted by $T\perp^{B}_{A}S$, if

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It is a generalization of the notion of Birkhoff–James of Hilbert space operators. Notice that the $A$-Birkhoff–James orthogonality is homogenous, that is, $T\perp^{B}_{A}S\,\Leftrightarrow\,(\alpha T)\perp^{B}_{A}(\beta S)$ for all $\alpha,\beta\in\mathbb{C}$.

The paper is organized as follows: In the next section, we obtain some characterizations of $A$-Birkhoff–James orthogonality for bounded linear operators in semi-Hilbertian spaces. In particular, for $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$, we show that $T\perp^{B}_{A}S$ if and only if there exists a sequence of $A$-unit vectors $\{x_{n}\}$ in $\mathcal{H}$ such that

[TABLE]

Furthermore, for the finite-dimensional Hilbert space $\mathcal{H}$, we show that $T\perp^{B}_{A}S$ if and only if there exists an $A$-unit vector $x\in\mathcal{H}$ such that ${\|Tx\|}_{A}={\|T\|}_{A}$ and ${\langle Tx,Sx\rangle}_{A}=0$. The mentioned property extends the Bhatia–Šemrl theorem.

In the last section, some formulas for the $A$-distance of an operator to the class of multiple scalars of another one in semi-Hilbertian spaces are given. In particular, we show that

[TABLE]

We then apply it to prove that $\displaystyle{\inf_{\gamma\in\mathbb{C}}}{\|T+\gamma S\|}^{2}_{A}=\displaystyle{\sup_{{\|x\|}_{A}=1}}\Phi^{(T,S)}_{A}(x)$, where

[TABLE]

Our results cover and extend the works of Fujii and Nakamoto in [9] and Bhatia and Šemrl in [4].

2. $A$-Birkhoff–James orthogonality of operators

We first prove a technical lemma that we need in what follows. We use some techniques of [3, Theorem 3.2] to prove this result. In fact, the following lemma extends Magajna’s theorem [12].

Lemma 2.1**.**

Let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$. Then the set

[TABLE]

is nonempty, compact, and convex.

Proof.

Since the seminorm of $T\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$ is given by

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there exists a sequence of $A$-unit vectors $\{x_{n}\}$ in $\overline{\mathcal{R}(A)}$ such that

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Furthermore, using the Cauchy–Schwarz inequality, we have

[TABLE]

Hence, $\{{\langle Tx_{n},Sx_{n}\rangle}_{A}\}$ is a bounded sequence of complex numbers, so there exists a subsequence $\{{\langle Tx_{n_{k}},Sx_{n_{k}}\rangle}_{A}\}$ that converges to some $\xi_{0}\in\mathbb{C}$. Thus $\xi_{0}\in W_{A}(T,S)$ and hence $W_{A}(T,S)$ is nonempty.

On the other hand, considering the definition of $W_{A}(T,S)$ follows that

[TABLE]

Therefore, to prove that $W_{A}(T,S)$ is compact, it is enough to show that $W_{A}(T,S)$ is closed. Let $\xi_{n}\in W_{A}(T,S)$ and let $\displaystyle{\lim_{n\rightarrow+\infty}}\xi_{n}=\xi$. Since $\xi_{n}\in W_{A}(T,S)$, there exists a sequence of $A$-unit vectors $\{x^{n}_{m}\}$ in $\mathcal{H}$ such that $\displaystyle{\lim_{m\rightarrow+\infty}}{\|Tx^{n}_{m}\|}_{A}={\|T\|}_{A}$ and $\displaystyle{\lim_{m\rightarrow+\infty}}{\langle Tx^{n}_{m},Sx^{n}_{m}\rangle}_{A}=\xi_{n}$. Now, let $\varepsilon>0$. Hence

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and also

[TABLE]

for all sufficiently large $m$. From (2.1) and (2.2), we get

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and

[TABLE]

for all sufficiently large $m$. Therefore we deduce that $\displaystyle{\lim_{m\rightarrow+\infty}}{\|Tx^{n}_{m}\|}_{A}={\|T\|}_{A}$ and $\displaystyle{\lim_{m\rightarrow+\infty}}{\langle Tx^{n}_{m},Sx^{n}_{m}\rangle}_{A}=\xi$. Thus $\xi\in W_{A}(T,S)$ and so $W_{A}(T,S)$ is closed.

We next show that $W_{A}(T,S)$ is convex. Since $\mathcal{H}$ can be decomposed as $\mathcal{H}=\mathcal{N}(A)\oplus\overline{\mathcal{R}(A)}$, so every $x\in\mathcal{H}$ can be written in a unique way into $x=y+z$ with $y\in\mathcal{N}(A)$ and $z\in\overline{\mathcal{R}(A)}$. Furthermore, since $A\geq 0$, it follows that $\mathcal{N}(A)=\mathcal{N}(A^{1/2})$ which implies that ${\|x\|}_{A}={\|z\|}_{A}$. Thus

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Since $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$, then $T(\mathcal{N}(A))\subset\mathcal{N}(A)$ and $S(\mathcal{N}(A))\subset\mathcal{N}(A)$. Hence, we get

[TABLE]

where $A_{0}=A\mid_{\overline{\mathcal{R}(A)}}$, $\widetilde{T}=PT\mid_{\overline{\mathcal{R}(A)}}$ and $\widetilde{S}=PS\mid_{\overline{\mathcal{R}(A)}}$. By [12, Lemma 2.1], we conclude that $W_{A}(T,S)$ is convex. ∎

Recall that the minimum modulus of $S\in\mathbb{B}(\mathcal{H})$ is defined by

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This concept is useful in studying linear operators (see [13], and further references therein). The $A$-minimum modulus of $S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$ can be defined by

[TABLE]

We are now in a position to establish the main result of this section. To establish the following theorem, we use some ideas of [16, Theorem 2].

Theorem 2.2**.**

Let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$. Then the following conditions are equivalent:

• (i)

There exists a sequence of $A$-unit vectors $\{x_{n}\}$ in $\mathcal{H}$ such that

[TABLE]

• (ii)

${\|T+\gamma S\|}^{2}_{A}\geq{\|T\|}^{2}_{A}+|\gamma|^{2}m^{2}_{A}(S)$* for all $\gamma\in\mathbb{C}$.*

• (iii)

$T\perp^{B}_{A}S$.

Proof.

(i)$\Rightarrow$(ii) Suppose that (i) holds. We have

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for all $\gamma\in\mathbb{C}$ and $n\in\mathbb{N}$. Thus

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for all $\gamma\in\mathbb{C}$.

(ii)$\Rightarrow$(iii) This implication is trivial.

(iii)$\Rightarrow$(i) If ${\|S\|}_{A}=0$, then since $T$ is a seminorm, there exists a sequence of $A$-unit vectors $\{x_{n}\}$ in $\mathcal{H}$ such that $\displaystyle{\lim_{n\rightarrow+\infty}}{\|Tx_{n}\|}_{A}={\|T\|}_{A}$. So, the Cauchy–Schwarz inequality implies

[TABLE]

Hence, $\displaystyle{\lim_{n\rightarrow+\infty}}{\langle Tx_{n},Sx_{n}\rangle}_{A}=0$. Now, let ${\|S\|}_{A}\neq 0$. It is enough to show that $0\in W_{A}(T,S)$, where $W_{A}(T,S)$ is defined in Lemma 2.1. let $0\notin W_{A}(T,S)$. Lemma 2.1 implies that $W_{A}(T,S)$ is a nonempty compact and convex subset of the complex plane $\mathbb{C}$; hence because of the rotation, we may suppose that $W_{A}(T,S)$ is contained in the right half-plane. Therefore there is a line that separates [math] from $W_{A}(T,S)$. In other words, there exists $\tau>0$ such that $\mbox{Re}W_{A}(T,S)>\tau$. Let

[TABLE]

and

[TABLE]

We first claim that $\delta<{\|T\|}_{A}$. Suppose $\delta\geq{\|T\|}_{A}$. Hence $\delta={\|T\|}_{A}$. Thus there exists a sequence of vectors $\{x_{n}\}$ in $\mathcal{H}_{\tau}$ such that $\displaystyle{\lim_{n\rightarrow+\infty}}{\|Tx_{n}\|}_{A}={\|T\|}_{A}$. As $x_{n}\in\mathcal{H}_{\tau}$ so ${\|x_{n}\|}_{A}=1$ and $\mbox{Re}W_{A}(T,S)\leq\frac{\tau}{2}$. Now the sequence $\{{\langle Tx_{n},Sx_{n}\rangle}_{A}\}$ is bounded, and hence it has a convergent subsequence, without loss of generality, we can assume that $\{{\langle Tx_{n},Sx_{n}\rangle}_{A}\}$ is convergent. If we set $\xi=\displaystyle{\lim_{n\rightarrow+\infty}}{\langle Tx_{n},Sx_{n}\rangle}_{A}$, then $\mbox{Re}(\xi)\leq\frac{\tau}{2}$ and this contradicts the fact that $\mbox{Re}W_{A}(T,S)>\frac{\tau}{2}$. Thus $\delta<{\|T\|}_{A}$. Let $\gamma_{0}=\max\{\frac{-\tau}{2{\|S\|}^{2}_{A}},\frac{\delta-{\|T\|}_{A}}{2{\|S\|}_{A}}\}$. Then $\gamma_{0}<0$. We claim that ${\|T+\gamma_{0}S\|}_{A}<{\|T\|}_{A}$. Let $x$ be an $A$-unit vector in $\mathcal{H}$. If $x\in\mathcal{H}_{\tau}$, then

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and so ${\|(T+\gamma_{0}S)x\|}_{A}\leq\frac{\delta}{2}+\frac{{\|T\|}_{A}}{2}$.

If $x\notin\mathcal{H}_{\tau}$, then we can write $Tx=(r+it)Sx+y$ with $r,t\in\mathbb{R}$ and $Sx\perp_{A}y$. Thus

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and hence $2r+\gamma_{0}>0$. Now, let us put

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Since $\gamma_{0}^{2}+2r\gamma_{0}<0$, we obtain

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Hence ${\|(T+\gamma_{0}S)x\|}^{2}_{A}\leq{\|T\|}^{2}_{A}+(\gamma_{0}^{2}+2r\gamma_{0})\theta$. Thus in all cases

[TABLE]

whence

[TABLE]

Since \max\Big{\{}\big{(}\frac{\delta}{2}+\frac{{\|T\|}_{A}}{2}\big{)}^{2},{\|T\|}^{2}_{A}+(\gamma_{0}^{2}+2r\gamma_{0})\theta\Big{\}}<{\|T\|}^{2}_{A}, we obtain ${\|T+\gamma_{0}S\|}_{A}<{\|T\|}_{A}$. Therefore we deduce that $T\not\perp^{B}_{A}S$ which contradicts our hypothesis and the proof is completed. ∎

The following corollary gives a direct application of Theorem 2.2 for the case $A=I$.

Corollary 2.3**.**

$($see [4, Remark 3.1] and [14, Lemma 2]$)$ Let $\mathcal{H}$ be a complex Hilbert space and let $T,S\in\mathbb{B}(\mathcal{H})$. Then the following statements are equivalent:

• (i)

$T\perp^{B}S$.

• (ii)

There exists a sequence of unit vectors $\{x_{n}\}$ in $\mathcal{H}$ such that

[TABLE]

In what follows, for $T\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$, we denote $\mathbb{M}^{T}_{A}$ the set of all $A$-unit vectors at which $T$ attains the seminorm ${\|\cdot\|}_{A}$, that is,

[TABLE]

For more information on norm-attaining sets, see [8]. In the next theorem, we consider a finite-dimensional Hilbert space and characterize the $A$-Birkhoff–James orthogonality of operators in semi-Hilbertian spaces.

Theorem 2.4**.**

Let $\mathcal{H}$ be a finite-dimensional Hilbert space and let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$. Then the following conditions are equivalent:

• (i)

There exists $x\in\mathbb{M}^{T}_{A}$ such that $Tx\perp_{A}Sx$.

• (ii)

$T\perp^{B}_{A}S$.

Proof.

(i)$\Rightarrow$(ii) Suppose that (i) holds. Then there exists an $A$-unit vectors $x\in\mathcal{H}$ such that ${\|Tx\|}_{A}={\|T\|}_{A}$ and $Tx\perp_{A}Sx$. Put $x_{n}=x$ for all $n\in\mathbb{N}$. So, by the equivalence (i)$\Leftrightarrow$(iii) in Theorem 2.2, we deduce that $T\perp^{B}_{A}S$.

(ii)$\Rightarrow$(i) First note that, by using the decomposition $\mathcal{H}=\mathcal{N}(A)\oplus\overline{\mathcal{R}(A)}$ and letting $A_{0}=A\mid_{\overline{\mathcal{R}(A)}}$, it can be seen that the set $\{x\in\overline{\mathcal{R}(A)};\,{\|x\|}_{A_{0}}=1\}$ is homeomorphic to the set $\{x\in\overline{\mathcal{R}(A)};\,\|x\|=1\}$, which is compact since $\overline{\mathcal{R}(A)}$ is finite-dimensional. Thus we get the set $\{x\in\overline{\mathcal{R}(A)};\,{\|x\|}_{A_{0}}=1\}$ is compact.

Now, suppose that (ii) holds. Put $\widetilde{T}=PT\mid_{\overline{\mathcal{R}(A)}}$ and $\widetilde{S}=PS\mid_{\overline{\mathcal{R}(A)}}$. Therefore, by the equivalence (i)$\Leftrightarrow$(iii) in Theorem 2.2, there exists a sequence of $A_{0}$-unit vectors $\{x_{n}\}$ in $\overline{\mathcal{R}(A)}$ such that

[TABLE]

Since the set $\{x\in\overline{\mathcal{R}(A)};\,{\|x\|}_{A_{0}}=1\}$ is compact, hence $\{x_{n}\}$ has a subsequence $\{x_{n_{k}}\}$ that converges to some $x\in\overline{\mathcal{R}(A)}$ with ${\|x\|}_{A_{0}}=1$. This yields ${\|\widetilde{T}x\|}_{A_{0}}=\displaystyle{\lim_{k\rightarrow+\infty}}{\|\widetilde{T}x_{n_{k}}\|}_{A_{0}}={\|\widetilde{T}\|}_{A_{0}}$ and ${\langle\widetilde{T}x,\widetilde{S}x\rangle}_{A_{0}}=\displaystyle{\lim_{k\rightarrow+\infty}}{\langle\widetilde{T}x_{n_{k}},\widetilde{S}x_{n_{k}}\rangle}_{A_{0}}=0$. From this it follows that $x\in\mathbb{M}^{T}_{A}$ and $Tx\perp_{A}Sx$. ∎

As an immediate consequence of Theorem 2.4, we have the following result.

Corollary 2.5**.**

Let $\mathcal{H}$ be finite dimensional and let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$. Then the following statements are equivalent:

• (i)

$T\perp^{B}_{A}S$.

• (ii)

There exists $x\in\mathbb{M}^{T}_{A}$ such that for every $\gamma\in\mathbb{C}$

[TABLE]

3. Some $A$-distance formulas

In this section we give some formulas for the $A$-distance of an operator to the class of multiple scalars of another one in semi-Hilbertian spaces. For $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$ we have, by definition, $d_{A}(T,\mathbb{C}S):=\displaystyle{\inf_{\gamma\in\mathbb{C}}}{\|T+\gamma S\|}_{A}$. The following auxiliary lemma is needed for next results.

Lemma 3.1**.**

Let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$. Then there exists $\zeta_{0}\in\mathbb{C}$ such that

[TABLE]

Proof.

If ${\|S\|}_{A}=0$, then

[TABLE]

for all $\gamma\in\mathbb{C}$. It is therefore enough to put $\zeta_{0}=0$. If ${\|S\|}_{A}\neq 0$, then put $\mathbb{D}:=\left\{\gamma\in\mathbb{C};\,|\gamma|\leq\frac{2{\|T\|}_{A}}{{\|S\|}_{A}}\right\}$ and define $f:\,\mathbb{D}\rightarrow\mathbb{R}$ by the formula $f(\gamma)={\|T+\gamma S\|}_{A}$. Clearly, $f$ is continuous and attains its minimum at, say, $\zeta_{0}\in\mathbb{D}$ (of course, there may be many such points). Then ${\|T+\gamma S\|}_{A}\geq{\|T+\zeta_{0}S\|}_{A}$ for all $\gamma\in\mathbb{D}$. If $\gamma\notin\mathbb{D}$, then $|\gamma|>\frac{2{\|T\|}_{A}}{{\|S\|}_{A}}$. Since $0\in\mathbb{D}$, we obtain

[TABLE]

Thus ${\|T+\gamma S\|}_{A}\geq{\|T+\zeta_{0}S\|}_{A}$ for all $\gamma\notin\mathbb{D}$. Therefore, ${\|T+\gamma S\|}_{A}\geq{\|T+\zeta_{0}S\|}_{A}$ for all $\gamma\in\mathbb{C}$. So, we conclude that $\displaystyle{\inf_{\gamma\in\mathbb{C}}}{\|T+\gamma S\|}_{A}={\|T+\zeta_{0}S\|}_{A}$ and hence $d_{A}(T,\mathbb{C}S)={\|T+\zeta_{0}S\|}_{A}$. ∎

The following result is a kind of the Pythagorean relation for bounded operators in semi-Hilbertian spaces.

Theorem 3.2**.**

Let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$ with $m_{A}(S)>0$. Then there exists a unique $\zeta_{0}\in\mathbb{C}$, such that

[TABLE]

for every $\gamma\in\mathbb{C}$.

Proof.

By Lemma 3.1, there exists $\zeta_{0}\in\mathbb{C}$ such that

[TABLE]

or equivalently,

[TABLE]

Thus $(T+\zeta_{0}S)\perp^{B}_{A}S$. So, by the equivalence (i)$\Leftrightarrow$(ii) in Theorem 2.2, for every $\gamma\in\mathbb{C}$, we have

[TABLE]

Now, suppose that $\zeta_{1}$ is another point satisfying the inequality

[TABLE]

Choose $\gamma=\zeta_{0}-\zeta_{1}$ to get

[TABLE]

Hence $0\geq|\zeta_{0}-\zeta_{1}|^{2}\,m^{2}_{A}(S)$. Since $m^{2}_{A}(S)>0$, we get $|\zeta_{0}-\zeta_{1}|^{2}=0$, or equivalently, $\zeta_{0}=\zeta_{1}$. This shows that $\zeta_{0}$ is unique. ∎

Here, we establish one of our main results. In fact, in what follows, we provide a version of the Bhatia–Šemrl theorem (see [4, p. 84]) in the setting of operators in semi-Hilbertian spaces.

Theorem 3.3**.**

Let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$. Then

[TABLE]

Proof.

Let $x,y\in\mathcal{H}$, ${\|x\|}_{A}={\|y\|}_{A}=1$ and let $Sx\perp_{A}y$. The Cauchy–Schwarz inequality implies

[TABLE]

for all $\gamma\in\mathbb{C}$. Thus

[TABLE]

for all $\gamma\in\mathbb{C}$ and so

[TABLE]

Hence

[TABLE]

On the other hand, by Lemma 3.1, there exists $\zeta_{0}\in\mathbb{C}$ such that $d_{A}(T,\mathbb{C}S)={\|T+\zeta_{0}S\|}_{A}$. We assume that $\zeta_{0}=0$ (otherwise we just replace $T$ by $T+\zeta_{0}S$). Thus $d_{A}(T,\mathbb{C}S)={\|T\|}_{A}$, or equivalently, $T\perp^{B}_{A}S$. Then, by the equivalence (i)$\Leftrightarrow$(iii) in Theorem 2.2, there exists a sequence of $A$-unit vectors $\{x_{n}\}$ in $\mathcal{H}$ such that $\displaystyle{\lim_{n\rightarrow+\infty}}{\|Tx_{n}\|}_{A}={\|T\|}_{A}$ and $\displaystyle{\lim_{n\rightarrow+\infty}}{\langle Tx_{n},Sx_{n}\rangle}_{A}=0$. Now, let $Tx_{n}=\alpha_{n}Sx_{n}+\beta_{n}y_{n}$ with $Sx_{n}\perp_{A}y_{n}$, ${\|y_{n}\|}_{A}=1$, and $\alpha_{n},\beta_{n}\in\mathbb{C}$. Then we have

[TABLE]

Consequently, we obtain

[TABLE]

whence

[TABLE]

From (3.1) and (3.2), we conclude that

[TABLE]

∎

For $T\in\mathbb{B}(\mathcal{H})$, Fujii and Nakamoto in [9] proved that $d_{A}(T,\mathbb{C}I)$ can be written in the following form:

[TABLE]

which shows that $d_{A}(T,\mathbb{C}I)$ is the supremum over the lengths of all perpendiculars from $Tx$ to $x$, where $x$ passes over the set of unit vectors. In the following theorem, for $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$, we show that $d_{A}(T,\mathbb{C}S)$ can also be expressed in the form generalizing of (3.3).

Theorem 3.4**.**

Let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$. Then

[TABLE]

where

[TABLE]

Proof.

For every $\gamma\in\mathbb{C}$ and every $A$-unit vector $x\in\mathcal{H}$ such that ${\|Sx\|}_{A}\neq 0$, we have

[TABLE]

Thus

[TABLE]

Also, in the case ${\|Sx\|}_{A}=0$ we have

[TABLE]

Hence we obtain $\Phi^{(T,S)}_{A}(x)\leq{\|T+\gamma S\|}^{2}_{A}$ for every $A$-unit vector $x\in\mathcal{H}$ and every $\gamma\in\mathbb{C}$. Therefore, $\displaystyle{\sup_{{\|x\|}_{A}=1}}\Phi^{(T,S)}_{A}(x)\leq{\|T+\gamma S\|}^{2}_{A}$ for every $\gamma\in\mathbb{C}$ and consequently,

[TABLE]

Thus

[TABLE]

Now, take $A$-unit vectors $x,y\in\mathcal{H}$ such that $Sx\perp_{A}y$. If ${\|Sx\|}_{A}=0$, then

[TABLE]

If ${\|Sx\|}_{A}\neq 0$, then

[TABLE]

So, we conclude that $|{\langle Tx,y\rangle}_{A}|^{2}\leq\displaystyle{\sup_{{\|x\|}_{A}=1}}\Phi^{(T,S)}_{A}(x)$ for all $A$-unit vectors $x,y\in\mathcal{H}$ such that $Sx\perp_{A}y$. Therefore, Theorem 3.3 implies that

[TABLE]

Now, the result follows from (3.4) and (3.5). ∎

We close this paper with the following Inf-sup equality in semi-Hilbertian spaces.

Theorem 3.5**.**

Let $T,S\in\mathbb{B}_{A^{1/2}}(\mathcal{H})$. Then

[TABLE]

Proof.

Let $x\in\mathcal{H}$ with ${\|x\|}_{A}=1$. If ${\|Sx\|}_{A}=0$, then

[TABLE]

for all $\gamma\in\mathbb{C}$. Thus

[TABLE]

whence $\displaystyle{\inf_{\gamma\in\mathbb{C}}}{\|(T+\gamma S)x\|}^{2}_{A}={\|Tx\|}^{2}_{A}$. Hence $\displaystyle{\inf_{\gamma\in\mathbb{C}}}{\|(T+\gamma S)x\|}^{2}_{A}=\Phi^{(T,S)}_{A}(x)$.

If ${\|Sx\|}_{A}\neq 0$, then simple computations show that

[TABLE]

Thus ${\|(T+\gamma S)x\|}^{2}_{A}$ achieves its minimum at $-\frac{{\langle Tx,Sx\rangle}_{A}}{{\|Sx\|}^{2}_{A}}$ and the minimum value is ${\|Tx\|}^{2}_{A}-\frac{|{\langle Tx,Sx\rangle}_{A}|^{2}}{{\|Sx\|}^{2}_{A}}$. Hence $\displaystyle{\inf_{\gamma\in\mathbb{C}}}{\|(T+\gamma S)x\|}^{2}_{A}=\Phi^{(T,S)}_{A}(x)$ for every $A$-unit vector $x\in\mathcal{H}$. From this, by Theorem 3.4, we conclude that

[TABLE]

∎

Acknowledgments. The author would like to thank the referees for their valuable comments which helped to improve the paper. This work was supported by a grant from Shanghai Municipal Science and Technology Commission (18590745200).