Characterizations of operator Birkhoff-James orthogonality

TL;DR

This paper characterizes Birkhoff-James orthogonality in Hilbert C*-modules and operators, introduces a Pythagorean relation for bounded operators, and explores new approximate orthogonality concepts.

Contribution

It provides novel characterizations of operator orthogonality, a Pythagorean relation, and a new approximate orthogonality in the context of Hilbert C*-modules and bounded operators.

Findings

Characterizations of strong Birkhoff-James orthogonality for Hilbert C*-modules.

A Pythagorean relation for bounded linear operators.

Conditions for orthogonality involving norm attaining sets and subspace structures.

Abstract

In this paper, we obtain some characterizations of the (strong) Birkhoff--James orthogonality for elements of Hilbert $C^*$-modules and certain elements of $\mathbb{B}(\mathscr{H})$. Moreover, we obtain a kind of Pythagorean relation for bounded linear operators. In addition, for $T\in \mathbb{B}(\mathscr{H})$ we prove that if the norm attaining set $\mathbb{M}_T$ is a unit sphere of some finite dimensional subspace $\mathscr{H}_0$ of $\mathscr{H}$ and $\|T\|_{{{\mathscr{H}}_0}^\perp} < \|T\|$, then for every $S\in\mathbb{B}(\mathscr{H})$, $T$ is the strong Birkhoff--James orthogonal to $S$ if and only if there exists a unit vector $\xi\in {\mathscr{H}}_0$ such that $\|T\|\xi = |T|\xi$ and $S^*T\xi = 0$. Finally, we introduce a new type of approximate orthogonality and investigate this notion in the setting of inner product $C^*$-modules.

Equations

$${⟨x,y⟩}^{\ast }⟨x,y⟩\le \Vert ⟨x,x⟩\Vert ⟨y,y⟩(x,y\in V).$$

$$|\phi (⟨x,y⟩){|}^2\le \phi (⟨x,x⟩)\phi (⟨y,y⟩)(x,y\in V).$$

$$\Vert x+\lambda y\Vert \ge \Vert x\Vert (\lambda \in \mathbb{C}).$$

$$\Big{\|}(T+\gamma S)+\lambda S\Big{\|}^{2}\geq\Big{\|}T+\gamma S\Big{\|}^{2}+|\lambda|^{2}m^{2}(S)\qquad(\lambda\in\mathbb{C}).$$

$$\Vert x+ya\Vert \ge \Vert x\Vert (a\in \mathcal{A}),$$

$$x\perp y⟹x{\perp }_B^{s}y⟹x{\perp }_{B}y,(x,y\in V),$$

$$x{\perp }_B^{s}y\iff x{\perp }_{B}y⟨y,x⟩(x,y\in V).$$

$$x{\perp }_B^{s}y\iff x{\perp }_{B}y(x,y\in V).$$

$$\displaystyle x\perp_{B}\big{(}\|x\|^{2}\,y-y\langle x,x\rangle\big{)}\qquad(x,y\in V),$$

$$\displaystyle x\perp^{s}_{B}\big{(}\|x\|^{2}x-x\langle x,x\rangle\big{)}\qquad(x\in V).$$

$$\Vert x+\lambda y{\Vert }^2\ge \Vert x{\Vert }^2-2\epsilon |\lambda |\Vert x\Vert \Vert y\Vert (\lambda \in \mathbb{C}).$$

$$\Vert x+ya{\Vert }^2\ge \Vert x{\Vert }^2-2\epsilon \Vert a\Vert \Vert x\Vert \Vert y\Vert (a\in \mathcal{A}).$$

$$x{\perp }^{\epsilon }y⟹x{\perp }_{B^{\epsilon }}^{s}y⟹x{\perp }_B^{\epsilon }y,(x,y\in V),$$

$$(1-16\epsilon )\Vert T\Vert \Vert x\Vert \le \Vert Tx\Vert \le \Vert T\Vert \Vert x\Vert (x\in V).$$

$$\displaystyle x\perp_{B}y\Leftrightarrow\big{(}\exists\,\varphi\in\mathcal{S}(\mathscr{A}):\varphi(\langle x,x\rangle)=\|x\|^{2}\,\,\mbox{and}\,\,\varphi(\langle x,y\rangle)=0\big{)}$$

$$\displaystyle x\perp^{s}_{B}y\Leftrightarrow\big{(}\exists\,\varphi\in\mathcal{S}(\mathscr{A}):\varphi(\langle x,x\rangle)=\|x\|^{2}\,\,\mbox{and}\,\,\varphi(\langle x,y\rangle a)=0\,\,\forall\,a\in\mathscr{A}\big{)}.$$

$$\Vert x+\lambda y{\Vert }^2\ge \Vert x{\Vert }^2+|\lambda {|}^2\mathrm{tr}(P|y{|}^2)(\lambda \in \mathbb{C}).$$

$$\Vert x+\lambda y{\Vert }^2$$

$$=\phi (⟨x,x⟩)+\lambda \phi (⟨x,y⟩)+\overline{\lambda \phi (⟨x,y⟩)}+|\lambda {|}^2\phi (⟨y,y⟩)$$

$$=\Vert x{\Vert }^2+|\lambda {|}^2\phi (|y{|}^2).$$

$$\Vert x+\lambda y{\Vert }^2\ge \Vert x{\Vert }^2+|\lambda {|}^2\phi (|y{|}^2)(\lambda \in \mathbb{C}).$$

$$\Vert x+\lambda y{\Vert }^2\ge \Vert x{\Vert }^2+|\lambda {|}^2\phi (|y{|}^2)=\Vert x{\Vert }^2+|\lambda {|}^2\mathrm{tr}(P|y{|}^2)(\lambda \in \mathbb{C}).$$

$$\Vert x+\lambda y\Vert \ge \sqrt{\Vert x{\Vert }^2+|\lambda {|}^2\mathrm{tr}(P|y{|}^2)}\ge \Vert x\Vert (\lambda \in \mathbb{C}).$$

$$\Vert x+ya{\Vert }^2\ge \Vert x{\Vert }^2+\mathrm{tr}(P|ya{|}^2)(a\in \mathcal{A}).$$

$$\Vert a+bc{\Vert }^2\ge \Vert a{\Vert }^2+\Vert \pi (bc)\xi {\Vert }^2(c\in \mathcal{A}).$$

$$\Vert a+bc{\Vert }^2$$

$$\displaystyle=\|a\|^{2}+\big{[}\pi(bc)\xi,\pi(bc)\xi\big{]}=\|a\|^{2}+\|\pi(bc)\xi\|^{2},$$

$$\Vert e+ab{\Vert }^2\ge 1+\Vert \pi (ab)\xi {\Vert }^2(b\in \mathcal{A}).$$

$$|\phi (eab)|\le \sqrt{\phi (ea{a}^{\ast }{e}^{\ast })\phi (b^{\ast }b)}=\sqrt{\phi (a^2)\phi (b^{\ast }b)}=0(b\in \mathcal{A}).$$

$$\underset{n\to \infty }{\lim }\Vert T{\xi }_n\Vert =\Vert T\Vert \text{mbox}and\underset{n\to \infty }{\lim }[T{\xi }_n,S{\xi }_n]=0(\text{mbox}resp.\underset{n\to \infty }{\lim }{S}^{\ast }T{\xi }_n=0).$$

$$\Vert T\xi \Vert =\Vert T\Vert \text{mbox}and[T\xi ,S\xi ]=0(\text{mbox}resp.S^{\ast }T\xi =0).$$

$$0\in \left\{\lambda \in \mathbb{C}:\exists ({\xi }_n)\subset \mathcal{H},\Vert {\xi }_n\Vert =1,\underset{n\to \infty }{\lim }[T^{\ast }S{\xi }_n,{\xi }_n]=\lambda \right\},$$

Full text

1. Introduction and preliminaries

Let $\mathbb{B}(\mathscr{H},\mathscr{K})$ denote the linear space of all bounded linear operators between Hilbert spaces $(\mathscr{H},[.,.])$ and $(\mathscr{K},[.,.])$. By $I$ we denote the identity operator. When $\mathscr{H}=\mathscr{K}$, we write $\mathbb{B}(\mathscr{H})$ for $\mathbb{B}(\mathscr{H},\mathscr{K})$. By $\mathbb{K}(\mathscr{H})$ we denote the algebra of all compact operators on $\mathscr{H}$, and by $\mathcal{C}_{1}(\mathscr{H})$ the algebra of all trace–class operators on $\mathscr{H}$. Let $\mathbb{S}_{\mathscr{H}}=\{\xi\in\mathscr{H}:\,\|\xi\|=1\}$ be the unit sphere of $\mathscr{H}$. For $T\in\mathbb{B}(\mathscr{H})$, let $\mathbb{M}_{T}$ denote the set of all vectors in $\mathbb{S}_{\mathscr{H}}$ at which $T$ attains norm, i.e., $\mathbb{M}_{T}=\{\xi\in\mathbb{S}_{\mathscr{H}}:\,\|T\xi\|=\|T\|\}.$ For $T\in\mathbb{B}(\mathscr{H},\mathscr{K})$ the symbol $m(T):=\inf\{\|T\xi\|:\,\xi\in\mathbb{S}_{\mathscr{H}}\}$ denotes the minimum modulus of $T$.

Inner product $C^{*}$-modules generalize inner product spaces by allowing inner products to take values in an arbitrary $C^{*}$-algebra instead of the $C^{*}$-algebra of complex numbers.

In an inner product $C^{*}$-module $(V,\langle\cdot,\cdot\rangle)$ over a $C^{*}$-algebra $\mathscr{A}$ the following Cauchy–Schwarz inequality holds (see also [1]):

[TABLE]

Consequently, $\|x\|=\|\langle x,x\rangle\|^{\frac{1}{2}}$ defines a norm on $V$. If $V$ is complete with respect to this norm, then it is called a Hilbert $\mathscr{A}$-module, or a Hilbert $C^{*}$-module over $\mathscr{A}$. Any $C^{*}$-algebra $\mathscr{A}$ can be regarded as a Hilbert $C^{*}$-module over itself via $\langle a,b\rangle:=a^{*}b$. For every $x\in V$ the positive square root of $\langle x,x\rangle$ is denoted by $|x|$. In the case of a $C^{*}$-algebra, we get the usual notation $|a|=(a^{*}a)^{\frac{1}{2}}.$ By $\mathcal{S}(\mathscr{A})$ we denote the set of all states of $\mathscr{A}$, that is, the set of all positive linear functionals of $\mathscr{A}$ whose norm is equal to one.

Furthermore, if $\varphi\in\mathcal{S}(\mathscr{A})$, then $(x,y)\mapsto\varphi(\langle x,y\rangle)$ gives rise to a usual semi-inner product on $V$, so we have the following useful Cauchy -Schwarz inequality:

[TABLE]

We refer the reader to [11, 17, 20] for more information on the basic theory of $C^{*}$-algebras and Hilbert $C^{*}$-modules.

A concept of orthogonality in a Hilbert $C^{*}$-module can be defined with respect to the $C^{*}$- valued inner product in a natural way, that is, two elements $x$ and $y$ of a Hilbert $C^{*}$-module $V$ over a $C^{*}$-algebra $\mathscr{A}$ are called orthogonal, in short $x\perp y$, if $\langle x,y\rangle=0$.

In a normed linear space there are several notions of orthogonality, all of which are generalizations of orthogonality in a Hilbert space. One of the most important is concept of the Birkhoff–James orthogonality: if $x,y$ are elements of a complex normed linear space $(X,\|\cdot\|)$, then $x$ is orthogonal to $y$ in the Birkhoff–James sense [6, 16], in short $x\perp_{B}y$, if

[TABLE]

The central role of the Birkhoff–James orthogonality in approximation theory, typified by the fact that $T\in\mathbb{B}(\mathscr{H})$ is a best approximation of $S\in\mathbb{B}(\mathscr{H})$ from a linear subspace $M$ of $\mathbb{B}(\mathscr{H})$ if and only if $T$ is a Birkhoff–James orthogonal projection of $S$ on to $M$. By the Hahn -Banach theorem, if $x,y$ are two elements of a normed linear space $X$, then $x\perp_{B}y$ if and only if there is a norm one linear functional $f$ of $X$ such that $f(x)=\|x\|$ and $f(y)=0$. If we have additional structures on a normed linear space $X$, then we obtain other characterizations of the Birkhoff -James orthogonality see [3, 5, 13, 22, 25] and the references therein.

In Section 2, we present some characterizations of the Birkhoff–James orthogonality for elements of a Hilbert $\mathbb{K}(\mathscr{H})$-module and elements of $\mathbb{B}(\mathscr{H})$. Next, we will give some applications. In particular, for $T,S\in\mathbb{B}(\mathscr{H})$ with $m(S)>0$, we prove that there exists a unique $\gamma\in\mathbb{C}$ such that

[TABLE]

As a natural generalization of the notion of Birkhoff–James orthogonality, the concept of strong Birkhoff–James orthogonality, which involves modular structure of a Hilbert $C^{*}$-module was introduced in [2]. When $x$ and $y$ are elements of a Hilbert $\mathscr{A}$-module $V$, $x$ is orthogonal to $y$ in the strong Birkhoff–James sense, in short $x\perp^{s}_{B}y$, if

[TABLE]

i.e. if the distance from $x$ to $\overline{y\mathscr{A}}$, the $\mathscr{A}$-submodule of $V$ generated by $y$, is exactly $\|x\|$. This orthogonality is “between” $\perp$ and $\perp_{B}$, i.e.,

[TABLE]

while the converses do not hold in general (see [2]). It was shown in [2] that the following relation between the strong and the classical Birkhoff -James orthogonality is valid:

[TABLE]

In particular, by [3, Proposition 3.1], if $\langle x,y\rangle\geq 0$, then

[TABLE]

If $V$ is a full Hilbert $\mathscr{A}$-module, then the only case where the orthogonalities $\perp^{s}_{B}$ and $\perp_{B}$ coincide is when $\mathscr{A}$ is isomorphic to $\mathbb{C}$ (see [3, Theorem 3.5]), while orthogonalities $\perp^{s}_{B}$ and $\perp$ coincide only when $\mathscr{A}$ or $\mathbb{K}(V)$ is isomorphic to $\mathbb{C}$ (see [3, Theorems 4.7, 4.8]). Further, by [3, Lemma 4.2], we have

[TABLE]

and

[TABLE]

In Section 2, we obtain a characterization of the strong Birkhoff–James orthogonality for elements of a $C^{*}$-algebra. We also present some characterizations of the strong Birkhoff–James orthogonality for certain elements of $\mathbb{B}(\mathscr{H}).$ In particular, for $T\in\mathbb{B}(\mathscr{H})$ we prove that if $\mathbb{S}_{{\mathscr{H}}_{0}}=\mathbb{M}_{T}$, where ${\mathscr{H}}_{0}$ is a finite dimensional subspace of $\mathscr{H}$ and $\|T\|_{{{\mathscr{H}}_{0}}^{\perp}}<\|T\|$, then for every $S\in\mathbb{B}(\mathscr{H})$, $T\perp^{s}_{B}S$ if and only if there exists a unit vector $\xi\in{\mathscr{H}}_{0}$ such that $\|T\|\xi=|T|\xi$ and $S^{*}T\xi=0$.

For given $\varepsilon\geq 0$, elements $x,y$ in an inner product $\mathscr{A}$-module $V$ are said to be approximately orthogonal or $\varepsilon$-orthogonal, in short $x\perp^{\varepsilon}y$, if $\|\langle x,y\rangle\|\leq\varepsilon\|x\|\|y\|$. For $\varepsilon\geq 1$, it is clear that every pair of vectors are $\varepsilon$-orthogonal, so the interesting case is when $\varepsilon\in[0,1)$.

In an arbitrary normed space $X$, Chmieliński [7, 8] introduced the approximate Birkhoff–James orthogonality $x\perp^{\varepsilon}_{B}y$ by

[TABLE]

Inspired by the above the approximate Birkhoff–James orthogonality, we propose a new type of approximate orthogonality in inner product $C^{*}$-modules.

Definition 1.1**.**

For given $\varepsilon\in[0,1)$ elements $x,y$ of an inner product $\mathscr{A}$-module $V$ are said to be approximate strongly Birkhoff -James orthogonal, denoted by $x\perp^{s}_{B^{\varepsilon}}y$, if

[TABLE]

In Section 3, we investigate this notion of approximate orthogonality in inner product $C^{*}$-modules. In particular, we show that

[TABLE]

while the converses do not hold in general.

As a result, we show that if $T\,:V\longrightarrow W$ is a linear mapping between inner product $\mathscr{A}$-modules such that $x\perp_{B}y\Longrightarrow Tx\perp^{s}_{B^{\varepsilon}}Ty$ for all $x,y\in V$, then

[TABLE]

Some other related topics can be found in [14, 15, 23, 24].

2. Operator (strong) Birkhoff -James orthogonality

The characterization of the (strong) Birkhoff -James orthogonality for elements of a Hilbert $C^{*}$-module by means of the states of the underlying $C^{*}$-algebra are known. For elements $x,y$ of a Hilbert $\mathscr{A}$-module $V$ the following results were obtained in [2, 5]:

[TABLE]

and

[TABLE]

In the following result we establish a characterization of the Birkhoff–James orthogonality for elements of a Hilbert $\mathbb{K}(\mathscr{H})$-module.

Theorem 2.1**.**

Let $V$ be a Hilbert $\mathbb{K}(\mathscr{H})$-module and $x,y\in V$. Then the following statements are equivalent:**

• (i)

$x\perp_{B}y$.

• (ii)

There exists a positive operator $P\in\mathcal{C}_{1}(\mathscr{H})$ of trace one such that

[TABLE]

Proof.

Let $x\perp_{B}y$. By (2.1), there exists a state $\varphi$ over $\mathbb{K}(\mathscr{H})$ such that $\varphi(\langle x,x\rangle)=\|x\|^{2}$ and $\varphi(\langle x,y\rangle)=0.$ For every $\lambda\in\mathbb{C}$, we therefore have

[TABLE]

Thus

[TABLE]

Now, by [20, Theorem 4.2.1], there exists a positive operator $P\in\mathcal{C}_{1}(\mathscr{H})$ of trace one such that $\varphi(T)={\rm tr}(PT)$, $T\in\mathbb{K}(\mathscr{H})$. Thus we have

[TABLE]

Conversely, if (ii) holds then, since $|\lambda|^{2}{\rm tr}(P|y|^{2})\geq 0$ for all $\lambda\in\mathbb{C}$, we get

[TABLE]

Hence $x\perp_{B}y$. ∎

Remark 2.2*.*

Let $V$ be a Hilbert $\mathbb{K}(\mathscr{H})$-module and $x,y\in V$. Using the same argument as in the proof of Theorem 2.1 and (2.2) we obtain $x\perp^{s}_{B}y$ if and only if there exists a positive operator $P\in\mathcal{C}_{1}(\mathscr{H})$ of trace one such that

[TABLE]

In the following result we establish a characterization of the strong Birkhoff–James orthogonality for elements of a $C^{*}$-algebra.

Theorem 2.3**.**

Let $\mathscr{A}$ be a $C^{*}$-algebra, and $a,b\in\mathscr{A}$. Then the following statements are equivalent:**

• (i)

$a\perp^{s}_{B}b$.

• (ii)

There exist a Hilbert space $\mathscr{H}$, a representation $\pi:\mathscr{A}\to\mathbb{B}(\mathscr{H})$ and a unit vector $\xi\in\mathscr{H}$ such that

[TABLE]

Proof.

Suppose that $a\perp^{s}_{B}b$. By (2.2) applied to $V=\mathscr{A}$ and using the same argument as in the proof of Theorem 2.1, there exists a state $\varphi$ of $\mathscr{A}$ such that $\|a+bd\|^{2}\geq\|a\|^{2}+\varphi(|bd|^{2})$ for all $d\in\mathscr{A}$. Now, by [11, Proposition 2.4.4] there exist a Hilbert space $\mathscr{H}$, a representation $\pi:\mathscr{A}\to\mathbb{B}(\mathscr{H})$ and a unit vector $\xi\in\mathscr{H}$ such that for any $c\in\mathscr{A}$ we have $\varphi(c)=[\pi(c)\xi,\xi].$ Hence

[TABLE]

for all $c\in\mathscr{A}.$

The converse is obvious. ∎

Corollary 2.4**.**

Let $\mathscr{A}$ be a unital $C^{*}$-algebra with the unit $e$. For every self-adjoint noninvertible $a\in\mathscr{A}$, there exist a Hilbert space $\mathscr{H}$, a representation $\pi:\mathscr{A}\to\mathbb{B}(\mathscr{H})$ and a unit vector $\xi\in\mathscr{H}$ such that

[TABLE]

Proof.

Since $a$ is noninvertible, $a^{2}$ is noninvertible as well. Therefore there is a state $\varphi$ of $\mathscr{A}$ such that $\varphi(a^{2})=0$. We have $\varphi(ee^{\ast})=\|e\|^{2}=1$ and

[TABLE]

Thus by (2.2) we get $e\perp^{s}_{B}a$. Hence, by Theorem 2.3, there exist a Hilbert space $\mathscr{H}$, a representation $\pi:\mathscr{A}\to\mathbb{B}(\mathscr{H})$ and a unit vector $\xi\in\mathscr{H}$ such that $\|e+ab\|^{2}\geq 1+\|\pi(ab)\xi\|^{2}$ for all $b\in\mathscr{A}$. ∎

Now, we are going to obtain some characterizations of the (strong) Birkhoff–James orthogonality in the Hilbert $C^{*}$-module $\mathbb{B}(\mathscr{H})$. Let $T,S\in\mathbb{B}(\mathscr{H})$. It was proved in [4, Theorem 1.1 and Remark 3.1] and [2, Proposition 2.8], that $T\perp_{B}S$ (resp. $T\perp^{s}_{B}S$) if and only if there is a sequence of unit vectors $({\xi}_{n})\subset\mathscr{H}$ such that

[TABLE]

When $\mathscr{H}$ is finite dimensional, it holds that $T\perp_{B}S$ (resp. $T\perp^{s}_{B}S$) if and only if there is a unit vector $\xi\in\mathscr{H}$ such that

[TABLE]

The following results are immediate consequences of the above characterizations.

Corollary 2.5**.**

Let $T\in\mathbb{B}(\mathscr{H})$ be an isometry and $S\in\mathbb{B}(\mathscr{H})$ be an invertible positive operator. Then $T\not\perp_{B}TS$.

Corollary 2.6**.**

Let $S\in\mathbb{B}(\mathscr{H})$. Then the following statements are equivalent:**

• (i)

$S$* is non-invertible.*

• (ii)

$T\perp_{B}S$* for every unitary operator $T\in\mathbb{B}(\mathscr{H})$.*

Proof.

By [10, Proposition 3.3], $S\in\mathbb{B}(\mathscr{H})$ is not invertible if and only if

[TABLE]

for every unitary operator $T$. Hence, by using (2.3), the statements are equivalent. ∎

Corollary 2.7**.**

Let $T,S\in\mathbb{B}(\mathscr{H})$. Then the following statements hold:**

• (i)

If $\dim\mathscr{H}<\infty,$ then $T\perp_{B}S$ if and only if there is a unit vector $\xi\in\mathscr{H}$ such that $\|T\|\xi=|T|\xi$ and $[T\xi,S\xi]=0.$

• (ii)

If $\dim\mathscr{H}=\infty,$ then $T\perp_{B}S$ if and only if there is a sequence of unit vectors $({\xi}_{n})\subset\mathscr{H}$ such that \lim_{n\rightarrow\infty}\big{(}\|T\|{\xi}_{n}-|T|{\xi}_{n}\big{)}=0 and $\lim_{n\rightarrow\infty}[T{\xi}_{n},S{\xi}_{n}]=0.$

• (iii)

If $\dim\mathscr{H}<\infty,$ then $T\perp^{s}_{B}S$ if and only if there is a unit vector $\xi\in\mathscr{H}$ such that $\|T\|\xi=|T|\xi$ and $S^{*}T\xi=0.$

• (iv)

If $\dim\mathscr{H}=\infty,$ then $T\perp^{s}_{B}S$ if and only if there is a sequence of unit vectors $({\xi}_{n})\subset\mathscr{H}$ such that \lim_{n\rightarrow\infty}\big{(}\|T\|{\xi}_{n}-|T|{\xi}_{n}\big{)}=0 and $\lim_{n\rightarrow\infty}S^{*}T{\xi}_{n}=0.$

Proof.

(i) Let $T\perp_{B}S$. Take the same vector $\xi$ as in (2.4). So, we have

[TABLE]

This forces $|T|^{2}\xi=\|T\|^{2}\xi$ and thus $|T|\xi=\|T\|\xi$, as asserted.

The converse is trivial.

Using (2.3) and (2.4), we can similarly prove the statements (ii)-(iv). ∎

Theorem 2.8**.**

Let $S\in\mathbb{B}(\mathscr{H})$. Let $\mathscr{H}_{0}\neq\{0\}$ be a closed subspace of $\mathscr{H}$ and $P$ be the orthogonal projection onto $\mathscr{H}_{0}$. Then the following statements hold:**

• (i)

If $\dim\mathscr{H}<\infty,$ then $P\perp_{B}S$ if and only if there is a unit vector $\xi\in\mathscr{H}_{0}$ such that $[S\xi,\xi]=0.$

• (ii)

If $\dim\mathscr{H}=\infty,$ then $P\perp_{B}S$ if and only if there is a sequence of unit vectors $({\xi}_{n})\subset\mathscr{H}_{0}$ such that $\lim_{n\rightarrow\infty}[S{\xi}_{n},{\xi}_{n}]=0.$

Proof.

(i) Let $P\perp_{B}S$. By (2.4), there is a unit vector $\zeta\in\mathscr{H}$ such that $\|P\zeta\|=\|P\|=1$ and $[P\zeta,S\zeta]=0.$ We have $\zeta=\xi+\eta$ where $\xi\in\mathscr{H}_{0}$ and $\eta\in{{\mathscr{H}}_{0}}^{\perp}$. Since $\|\xi\|=\|P(\xi+\eta)\|=\|P\zeta\|=1$ and $\|\xi\|^{2}+\|\eta\|^{2}=1$, so we get $\eta=0$. Hence $[S\xi,\xi]=[S(\xi+\eta),\xi]=[S(\xi+\eta),P(\xi+\eta)]=\overline{[P\zeta,S\zeta]}=0.$

The converse is trivial.

(ii) Let $P\perp_{B}S$. Take the vector sequence $({\zeta}_{n})$ of $\mathscr{H}$ as in (2.3). We have ${\zeta}_{n}={\mu}_{n}+{\eta}_{n}$ where ${\mu}_{n}\in\mathscr{H}_{0}$ and ${\eta}_{n}\in{{\mathscr{H}}_{0}}^{\perp}$. Since $\lim_{n\rightarrow\infty}\|{\mu}_{n}\|=\lim_{n\rightarrow\infty}\|P({\mu}_{n}+{\eta}_{n})\|=\lim_{n\rightarrow\infty}\|P{\zeta}_{n}\|=1$ and $\|{\mu}_{n}\|^{2}+\|{\eta}_{n}\|^{2}=1$, so we get $\lim_{n\rightarrow\infty}\|{\eta}_{n}\|=0$. We may assume that $\|{\mu}_{n}\|\geq\frac{1}{2}$ for every $n\in\mathbb{N}$. Let us put ${\xi}_{n}=\frac{{\mu}_{n}}{\|{\mu}_{n}\|}$. We have

[TABLE]

whence

[TABLE]

Since $\lim_{n\rightarrow\infty}[P{\zeta}_{n},S{\zeta}_{n}]=0$ and $\lim_{n\rightarrow\infty}\|{\eta}_{n}\|=0$, from the above equality we get $\lim_{n\rightarrow\infty}[S{\xi}_{n},{\xi}_{n}]=0$.

The converse is trivial. ∎

Theorem 2.9**.**

Let $T,S\in\mathbb{B}(\mathscr{H})$. Then the following statements are equivalent:**

• (i)

$T\perp_{B}S$.

• (ii)

$\|T+\lambda S\|^{2}\geq\|T\|^{2}+|\lambda|^{2}\,m^{2}(S)\qquad(\lambda\in\mathbb{C}),$**

where $m(S)$ is the minimum modulus of $S$.

Proof.

(i)$\Rightarrow$(ii) Let $T\perp_{B}S$ and $\dim\mathscr{H}=\infty$. By (2.3), there exists a sequence of unit vectors $({\xi}_{n})\subset\mathscr{H}$ such that $\lim_{n\rightarrow\infty}\|T{\xi}_{n}\|=\|T\|$ and $\lim_{n\rightarrow\infty}[T{\xi}_{n},S{\xi}_{n}]=0.$ We have

[TABLE]

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

[TABLE]

When $\dim\mathscr{H}<\infty$, by using (2.4), we can similarly prove the statement (ii).

(ii)$\Rightarrow$(i) This implication is trivial. ∎

Remark 2.10*.*

Notice that for $S\in\mathbb{B}(\mathscr{H})$ it is straightforward to show that $m(S)>0$ if and only if $S$ is bounded below, or equivalently, $S$ is left invertible. So in the implication (i)$\Rightarrow$(ii) of Theorem 2.9, if $S$ is left invertible then $m(S)>0$.

It is well known that Pythagoras’ equality does not hold in $\mathbb{B}(\mathscr{H})$. The following result is a kind of Pythagorean inequality for bounded linear operators.

Corollary 2.11**.**

Let $T,S\in\mathbb{B}(\mathscr{H})$ with $m(S)>0$. Then there exists a unique $\gamma\in\mathbb{C}$, such that

[TABLE]

Proof.

The function $\lambda\longmapsto\|T+\lambda S\|$ attains its minimum at, say, $\gamma$ (there may be of course many such points) and hence $T+\gamma S\perp_{B}S$. So, by Theorem 2.9, we have

[TABLE]

Now, suppose that $\xi$ is another point satisfying the inequality

[TABLE]

Choose $\lambda=\gamma-\xi$ to get

[TABLE]

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

Let $T\in\mathbb{B}(\mathscr{H})$. For every $S\in\mathbb{B}(\mathscr{H})$, it is easy to see that if there exists a unit vector $\xi\in\mathscr{H}$ such that $\|T\|\xi=|T|\xi$ and $S^{*}T\xi=0$ then $T\perp^{s}_{B}S$. The question is under which conditions the converse is true. When the Hilbert space is finite dimensional, it follows from Corollary 2.7 (iii) that there exists a unit vector $\xi\in\mathscr{H}$ such that $\|T\|\xi=|T|\xi$ and $S^{*}T\xi=0$.

The following example shows that the finite dimensionality in the statement (iii) of Corollary 2.7 is essential.

Example 2.12**.**

Consider operators $T,S\,:\ell^{2}\longrightarrow\ell^{2}$ defined by

[TABLE]

and

[TABLE]

One can easily observe that $T\perp_{B}S$ and $T^{*}S(\xi_{1},\xi_{2},\xi_{3},\cdots)=\frac{1}{2}{\xi_{1}}^{2}\geq 0$. So, by (1.1), we get $T\perp^{s}_{B}S$. But there does not exist $\xi\in\ell^{2}$ such that $\|T\|\xi=|T|\xi$.

We now settle the problem for any infinite dimensional Hilbert space. The proof of Theorem 2.13 is a modification of one given by Paul et al. [21, Theorem 3.1].

Theorem 2.13**.**

Let $\dim\mathscr{H}=\infty$ and $T\in\mathbb{B}(\mathscr{H})$. If $\mathbb{S}_{{\mathscr{H}}_{0}}=\mathbb{M}_{T}$, where ${\mathscr{H}}_{0}$ is a finite dimensional subspace of $\mathscr{H}$ and $\|T\|_{{{\mathscr{H}}_{0}}^{\perp}}=\sup\{\|T\xi\|:\,\xi\in{{\mathscr{H}}_{0}}^{\perp},\,\|\xi\|=1\}<\|T\|$, then for every $S\in\mathbb{B}(\mathscr{H})$ the following statements are equivalent:**

• (i)

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

• (ii)

There exists a unit vector $\xi\in{\mathscr{H}}_{0}$ such that $\|T\xi\|=\|T\|$ and $S^{*}T\xi=0$.

• (iii)

There exists a unit vector $\xi\in{\mathscr{H}}_{0}$ such that $\|T\|\xi=|T|\xi$ and $S^{*}T\xi=0$.

Proof.

(i)$\Rightarrow$(ii) Suppose (i) holds. By (2.3), there exists a sequence of unit vectors $\{\zeta_{n}\}$ in $\mathscr{H}$ such that

[TABLE]

For each $n\in\mathbb{N}$ we have

[TABLE]

where $\xi_{n}\in\mathscr{H}_{0}$ and $\eta_{n}\in{\mathscr{H}_{0}}^{\perp}$.

Since $\mathscr{H}_{0}$ is a finite dimensional subspace and $\|\xi_{n}\|\leq 1$, so $\{\xi_{n}\}$ has a convergent subsequence converging to some element of $\mathscr{H}_{0}$. Without loss of generality we assume that $\lim_{n\rightarrow\infty}\xi_{n}=\xi$. Since $\mathbb{S}_{{\mathscr{H}}_{0}}=\mathbb{M}_{T}$, so

[TABLE]

and

[TABLE]

Now for each non-zero element $\xi_{n}\in\mathscr{H}_{0}$, by hypothesis $\frac{\xi_{n}}{\|\xi_{n}\|}\in\mathbb{S}_{{\mathscr{H}}_{0}}=\mathbb{M}_{T}$ and so $\|T\xi_{n}\|=\|T\|\,\|\xi_{n}\|$. Thus

[TABLE]

Hence $[T^{*}T\xi_{n},\xi_{n}]=\|T^{*}T\xi_{n}\|\,\|\xi_{n}\|$. By the equality case of Cauchy–Schwarz inequality $T^{*}T\xi_{n}=\lambda_{n}\xi_{n}$ for some $\lambda_{n}\in\mathbb{C}$ and therefore

[TABLE]

By (2.5), (2.6) and (2.8) we have

[TABLE]

whence by (2.7) we reach

[TABLE]

By the hypothesis $\|T\|_{{\mathscr{H}_{0}}^{\perp}}<\|T\|$ and so by (2.9) there does not exist any non-zero subsequence of $\{\|\eta_{n}\|\}$. So we conclude that $\eta_{n}=0$ for all $n\in\mathbb{N}$. Hence (2.5), (2.7) imply

[TABLE]

(ii)$\Rightarrow$(iii) This implication follows from the proof of Corollary 2.7.

(iii)$\Rightarrow$(i) This implication is trivial. ∎

Corollary 2.14**.**

Let $\dim\mathscr{H}=\infty$ and $T\in\mathbb{B}(\mathscr{H})$. If $\mathbb{S}_{{\mathscr{H}}_{0}}=\mathbb{M}_{T}$, where ${\mathscr{H}}_{0}$ is a finite dimensional subspace of $\mathscr{H}$ and $\|T\|_{{{\mathscr{H}}_{0}}^{\perp}}<\|T\|$, then there exists a unit vector $\xi\in{\mathscr{H}}_{0}$ such that $\|T\|\xi=|T|\xi$ and $\|T\|^{2}\,T^{*}T\xi=(T^{*}T)^{2}\xi$.

Proof.

By (1.2), T\perp^{s}_{B}\big{(}\|T\|^{2}T-TT^{*}T\big{)}. So, by Theorem 2.13, there exists a unit vector $\xi\in{\mathscr{H}}_{0}$ such that $\|T\|\xi=|T|\xi$ and \big{(}\|T\|^{2}T-TT^{*}T\big{)}^{*}T\xi=0. Thus $\|T\|^{2}\,T^{*}T\xi=(T^{*}T)^{2}\xi$. ∎

Corollary 2.15**.**

Let $\dim\mathscr{H}=\infty$ and let $T\in\mathbb{B}(\mathscr{H})$ be a nonzero positive operator. If $\mathbb{S}_{{\mathscr{H}}_{0}}=\mathbb{M}_{T}$, where ${\mathscr{H}}_{0}$ is a finite dimensional subspace of $\mathscr{H}$ and $\|T\|_{{{\mathscr{H}}_{0}}^{\perp}}<\|T\|$, then for every $S\in\mathbb{B}(\mathscr{H})$ the following statements are equivalent:**

• (i)

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

• (ii)

There exists a unit vector $\xi\in{\mathscr{H}}_{0}$ such that $T\xi=\|T\|\xi$ and $S^{*}\xi=0$.

Proof.

Obviously, (ii)$\Rightarrow$(i).

Suppose (i) holds. By Theorem 2.13, there exists a unit vector $\xi\in{\mathscr{H}}_{0}$ such that $\|T\xi\|=\|T\|$ and $S^{*}T\xi=0$. Since $T\geq 0$, $\|T\xi\|=\|T\|\Leftrightarrow T\xi=\|T\|\xi$. Therefore, $S^{*}T\xi=0\Leftrightarrow S^{*}\xi=0$, as $T\neq 0$. ∎

3. An approximate strong Birkhoff -James orthogonality

Recall that in an inner product $\mathscr{A}$-module $V$ and for $\varepsilon\in[0,1)$, we say $x,y$ are approximate strongly Birkhoff -James orthogonal, in short $x\perp^{s}_{B^{\varepsilon}}y$, if

[TABLE]

The following proposition states some basic properties of the relation $\perp^{s}_{B^{\varepsilon}}$.

Proposition 3.1**.**

Let $\varepsilon\in[0,\frac{1}{2})$ and $V$ be an inner product $\mathscr{A}$-module. Then the following statements hold for every $x,y\in V$:**

• (i)

$x\perp^{s}_{B^{\varepsilon}}x\Leftrightarrow x=0$.

• (ii)

$x\perp^{s}_{B^{\varepsilon}}y\Rightarrow\alpha x\perp^{s}_{B^{\varepsilon}}\beta y$* for all $\alpha,\beta\in\mathbb{C}$.*

• (iii)

$x\perp^{\varepsilon}y\Rightarrow x\perp^{s}_{B^{\varepsilon}}y$.

• (iv)

$x\perp^{s}_{B^{\varepsilon}}y\Rightarrow x\perp^{\varepsilon}_{B}y$.

• (v)

$x\perp^{s}_{B^{\varepsilon}}y\Leftrightarrow x\perp^{\varepsilon}_{B}ya$* for all $a\in\mathscr{A}$.*

Proof.

(i) Let $x\perp^{s}_{B^{\varepsilon}}x$. Also, suppose that $(e_{i})_{i\in I}$ is an approximate unit for $\mathscr{A}$. We have

[TABLE]

Since $\lim\limits_{i}\|x-xe_{i}\|=0$ and $\|e_{i}\|=1$, we get $(1-2\varepsilon)\|x\|^{2}\leq 0$. Thus $x=0$.

The converse is obvious.

(ii) Let $x\perp^{s}_{B^{\varepsilon}}y$ and let $\alpha,\beta\in\mathbb{C}$. Excluding the obvious case $\alpha=0$ we have

[TABLE]

Hence $\alpha x\perp^{s}_{B^{\varepsilon}}\beta y$.

(iii) Let $x\perp^{\varepsilon}y$. For any $a\in\mathscr{A}$ we have

[TABLE]

Thus $\|x+ya\|^{2}\geq\|x\|^{2}-2\varepsilon\,\|a\|\,\|x\|\,\|y\|$, or equivalently, $x\perp^{s}_{B^{\varepsilon}}y$.

(iv) Let $x\perp^{s}_{B^{\varepsilon}}y$. Hence for any $\lambda\in\mathbb{C}$ and an approximate unit $(e_{i})_{i\in I}$ for $\mathscr{A}$ we have

[TABLE]

Since $\lim\limits_{i}\|ye_{i}-y\|=0$, whence we get $\|x+\lambda y\|^{2}\geq\|x\|^{2}-2\varepsilon\,|\lambda|\,\|x\|\,\|y\|$, or equivalently, $x\perp^{\varepsilon}y$.

(v) Let $x\perp^{s}_{B^{\varepsilon}}y$ and let $(e_{i})_{i\in I}$ be an approximate unit for $\mathscr{A}$. We have

[TABLE]

for all $a\in\mathscr{A}$ and all $\lambda\in\mathbb{C}$. Since $\lim\limits_{i}\|yae_{i}-ya\|=0$, we obtain from the above inequality

[TABLE]

for all $a\in\mathscr{A}$ and all $\lambda\in\mathbb{C}$. Thus $x\perp^{\varepsilon}_{B}ya$ for all $a\in\mathscr{A}$.

The converse is trivial. ∎

Proposition 3.1 shows that in an arbitrary inner product $C^{*}$-module the relation $\perp^{\varepsilon}$ is weaker than the relation $\perp^{s}_{B^{\varepsilon}}$ and this relation is weaker than the relation $\perp^{\varepsilon}_{B}$, but the converses are not true in general (see the example below).

Example 3.2**.**

Suppose that $\varepsilon\in[0,\frac{1}{2})$. Consider $\mathbb{M}_{2}(\mathbb{C})$, regarded as an inner product $\mathbb{M}_{2}(\mathbb{C})$-module. Let $I=\begin{bmatrix}1&0\\ 0&1\end{bmatrix},A=\begin{bmatrix}-1&0\\ 0&1\end{bmatrix}$ and $B=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}$. Then

[TABLE]

for all $\lambda\in\mathbb{C}$. Hence $I\perp^{\varepsilon}_{B}A$. But not $I\perp^{s}_{B^{\varepsilon}}A$ since

[TABLE]

On the other hand, for any $C=\begin{bmatrix}c_{1}&c_{2}\\ c_{3}&c_{4}\end{bmatrix}$ we have

[TABLE]

Therefore $I\perp^{s}_{B^{\varepsilon}}B$. But not $I\perp^{\varepsilon}B$ since

[TABLE]

By combining Proposition 3.1 (iv) and [19, Theorem 3.5] we obtain the following result (see also [9, 12, 18]).

Corollary 3.3**.**

Let $V,W$ be inner product $\mathscr{A}$-modules, $\varepsilon\in[0,\frac{1}{2})$ and $T\,:V\longrightarrow W$ a linear mapping satisfying $x\perp_{B}y\Longrightarrow Tx\perp^{s}_{B^{\varepsilon}}Ty$. Then

[TABLE]

Proposition 3.4**.**

Let $\varepsilon\in[0,1)$. Let $x,y$ be elements in an inner product $\mathscr{A}$-module $V$ such that $\langle x,x\rangle\perp^{s}_{B^{\varepsilon}}\langle x,y\rangle$, then $x\perp^{s}_{B^{\varepsilon}}y$.

Proof.

We assume that $x\neq 0$. Since $\langle x,x\rangle\perp^{s}_{B^{\varepsilon}}\langle x,y\rangle$ therefore for every $a\in\mathscr{A}$ we have

[TABLE]

or equivalently,

[TABLE]

Hence we get

[TABLE]

Since $\|x\|^{2}\neq 0$ we obtain from the above inequality

[TABLE]

Thus $x\perp^{s}_{B^{\varepsilon}}y$. ∎

Proposition 3.5**.**

Let $x,y$ be two elements in an inner product $\mathscr{A}$-module $V$ and let $\varepsilon\in[0,1)$. If there exists a state $\varphi$ on $\mathscr{A}$ such that $\varphi(\langle x,x\rangle)=\|x\|^{2}$ and $|\varphi(\langle x,y\rangle a)|\leq\varepsilon\,\|a\|\,\|x\|\,\|y\|$ for all $a\in\mathscr{A}$, then $x\perp^{s}_{B^{\varepsilon}}y$.

Proof.

We assume that $x\neq 0$. Let $a\in\mathscr{A}$. By the Cauchy–Schwarz inequality, we have

[TABLE]

Thus $\|x\|^{2}\leq\|x\|\,\|x+ya\|+\varepsilon\,\|a\|\,\|x\|\,\|y\|$, i.e, $\|x+ya\|\geq\|x\|-\varepsilon\,\|a\|\,\|y\|$. We consider two cases:

If $\|x\|-\varepsilon\,\|a\|\,\|y\|\geq 0$, then we get

[TABLE]

If $\|x\|-\varepsilon\,\|a\|\,\|y\|<0$, then we reach

[TABLE]

Hence $x\perp^{s}_{B^{\varepsilon}}y$. ∎

Proposition 3.6**.**

Let $x,y$ be two elements in an inner product $\mathscr{A}$-module $V$ and let $\varepsilon\in[0,\frac{1}{2})$. If $x\perp^{s}_{B^{\varepsilon}}y$ then there exists a state $\varphi$ on $\mathscr{A}$ such that

[TABLE]

Proof.

Suppose that $x\perp^{s}_{B^{\varepsilon}}y$. Because of the homogeneity of relation $\perp^{s}_{B^{\varepsilon}}$ we may assume, without loss of generality, that $\|x\|=\|y\|=1$. Then, for arbitrary $a\in\mathscr{A}$ we have

[TABLE]

Since $\|-\langle y,x\rangle\|\leq\|y\|\,\|x\|=1$, hence for $a=-\langle y,x\rangle\in\mathscr{A}$ we get

[TABLE]

On the other side, by Theorem 3.3.6 of [20], there is $\varphi\in\mathcal{S}(\mathscr{A})$ such that

[TABLE]

Also, we have

[TABLE]

so, we get

[TABLE]

Therefore $\varphi(\langle x,y\rangle\langle y,x\rangle)\leq 2\varepsilon$. Now, by the Cauchy–Schwarz inequality, we reach

[TABLE]

∎