Characterization of a-Birkhoff–James orthogonality in $C^*$-algebras and its applications

Abstract

Let ${\mathcal {A}}$ be a unital $C^*$-algebra with unit $1_{{\mathcal {A}}}$ and let $a\in {\mathcal {A}}$ be a positive and invertible element. Suppose that ${\mathcal {S}}({\mathcal {A}})$ is the set of all states on $\mathcal {{\mathcal {A}}}$ and let

$$\begin{aligned} {\mathcal {S}}_a ({\mathcal {A}})=\left\{ \dfrac{f}{f(a)} \, : \, f \in {\mathcal {S}}({\mathcal {A}}), \, f(a)\ne 0\right\} . \end{aligned}$$

The norm $\Vert x\Vert _a$ for every $x \in {\mathcal {A}}$ is defined by

$$\begin{aligned} \Vert x\Vert _a = \sup _{\varphi \in {\mathcal {S}}_a ({\mathcal {A}}) } \sqrt{\varphi (x^* ax)}. \end{aligned}$$

In this paper, we aim to investigate the notion of Birkhoff–James orthogonality with respect to the norm $\Vert \cdot \Vert _a$ in ${\mathcal {A}},$ namely a-Birkhoff–James orthogonality. The characterization of a-Birkhoff–James orthogonality in ${\mathcal {A}}$ by means of the elements of generalized state space ${\mathcal {S}}_a({\mathcal {A}})$ is provided. As an application, a characterization for the best approximation to elements of ${\mathcal {A}}$ in a subspace ${\mathcal {B}}$ with respect to $\Vert \cdot \Vert _a$ is obtained. Moreover, a formula for the distance of an element of ${\mathcal {A}}$ to the subspace ${\mathcal {B}}={\mathbb {C}}1_{{\mathcal {A}}}$ is given. We also study the strong version of a-Birkhoff–James orthogonality in ${\mathcal {A}} .$ The classes of $C^*$-algebras in which these two types orthogonality relationships coincide are described. In particular, we prove that the condition of the equivalence between the strong a-Birkhoff–James orthogonality and ${\mathcal {A}}$-valued inner product orthogonality in ${\mathcal {A}}$ implies that the center of ${\mathcal {A}}$ is trivial. Finally, we show that if the (strong) a-Birkhoff–James orthogonality is right-additive (left-additive) in ${\mathcal {A}},$ then the center of ${\mathcal {A}}$ is trivial.

1 Introduction and preliminaries

Let ${\mathcal {A}}$ be a unital $C^{*}$-algebra with unit $1_{{\mathcal {A}}}.$ We denote by ${\mathcal {A}}^{\prime }$ and ${\mathcal {Z}}({\mathcal {A}})$ the topological dual space and the center of ${\mathcal {A}},$ respectively. The adjoint of any element $x \in {\mathcal {A}}$ is denoted, as usual, by $x^* .$ Also, ${\textrm{Re}}(x)=\dfrac{1}{2}(x+x^*)$ is reserved to indicate the real part of x. An element a of ${\mathcal {A}}$ is called positive (written by $a\ge 0),$ if a is selfadjoint whose spectrum $\sigma (a)$ is contained in $[0,\infty ).$ It is known that if $a\in {\mathcal {A}}$ is positive, then there exists a unique positive element $b\in {\mathcal {A}}$ such that $a=b^2.$ Such an element b is called the positive square root of a and is denoted by $a^{\frac{1}{2}}.$ The symbol ${\mathcal {A}}^+$ stands for the cone of positive elements in ${\mathcal {A}} .$ If in addition a is invertible, then $a^{\frac{1}{2}}$ is invertible too and its inverse is denoted by $a^{-\frac{1}{2}}.$ A linear functional f on ${\mathcal {A}}$ is called positive if $f(a)\ge 0$ for every positive element $a\in {\mathcal {A}}.$ Given a positive functional f on ${\mathcal {A}},$ the following well-known version of the Cauchy–Schwarz inequality holds for every $x,y\in {\mathcal {A}}$: $|f(x^*y)|^2\le f(x^*x)f(y^*y).$

A state on ${\mathcal {A}}$ is a positive linear functional whose norm is equal to one. It is well-known that a linear functional on ${\mathcal {A}}$ is positive if and only if $f(1_{{\mathcal {A}}})=\Vert f\Vert ;$ see [17, Corollary 3.3.4]. Let ${\mathcal {S}}({\mathcal {A}})$ be the set of all states on ${\mathcal {A}} .$ Then

$$\begin{aligned} {\mathcal {S}}({\mathcal {A}}) = \{ f\in {\mathcal {A}}^{'} \, : \, f(1_{{\mathcal {A}}})= \Vert f \Vert =1 \}. \end{aligned}$$

Birkhoff–James orthogonality of elements in a normed linear space was introduced by Birkhoff in [11] and developed by James [14] to generalize the concept of orthogonality in inner product spaces. If x and y are vectors of a normed linear space $(X,\Vert \cdot \Vert ),$ then x is said to be Birkhoff–James orthogonal to y,  in short $x\perp _{BJ}y,$ if

$$\begin{aligned} \Vert x+\lambda y\Vert \ge \Vert x\Vert \quad (\forall \lambda \in {\mathbb {C}}). \end{aligned}$$

The concept of the strong Birkhoff–James orthogonality in $C^*$-algebras as a natural generalization of Birkhoff–James orthogonality was introduced and studied in [4, 6]. Let $x,y\in {\mathcal {A}}.$ Then x is said to be strong Birkhoff–James orthogonal to y,  denoted by $x\perp _{S-BJ}y,$ if

$$\begin{aligned} \Vert x+yb\Vert \ge \Vert x\Vert \quad (\forall b\in {\mathcal {A}}). \end{aligned}$$

We also recall that two elements x and y of ${\mathcal {A}}$ are orthogonal with respect to the ${\mathcal {A}}$-valued inner product $\langle x,y \rangle := x^* y$ if $\langle x,y \rangle =0 .$ It was shown in [4] the following relation between the strong and the classical Birkhoff–James orthogonality:

$$\begin{aligned} \langle x,y\rangle =0\Rightarrow x\perp _{S-BJ}y\Rightarrow x\perp _{BJ}y\quad (\forall x,y\in {\mathcal {A}}). \end{aligned}$$

It is well-known that the Birkhoff–James orthogonality of vectors in normed linear spaces can be characterized in terms of linear functionals [14]. Over the years, the problem of finding characterizations of Birkhoff–James orthogonality of matrices and generally of the elements of $C^*$-algebras has been considered by many mathematicians. A complete characterization of Birkhoff–James orthogonality of bounded linear operators defined on Hilbert spaces obtained by Bhatia and S̆emrl [10] (see also, [3, 9, 18]). Some famous and useful Characterizations of the (strong) Birkhoff–James orthogonality in $C^*$-algebra ${\mathcal {A}}$ and in a more general setting Hilbert $C^*$-modules over ${\mathcal {A}}$ in terms of the elements of state space ${\mathcal {S}}({\mathcal {A}})$ have been obtained in [5, 9, 16]. The characterization of the (strong) Birkhoff–James orthogonality for elements of a $C^*$-algebra by means of its state space were obtained as follows:

Theorem 1.1

[5, Theorem 2.7] An element $x\in {\mathcal {A}}$ is Birkhoff–James orthogonal to another element $y\in {\mathcal {A}},$ if and only if there is $f\in {\mathcal {S}}({\mathcal {A}})$ such that $f(x^*x)=\Vert x\Vert ^2$ and $f(x^*y)=0.$

Theorem 1.2

[4, Theorem 2.5] An element $x\in {\mathcal {A}}$ is strong Birkhoff–James orthogonal to another element $y\in {\mathcal {A}}$ if and only if there is $f\in {\mathcal {S}}({\mathcal {A}})$ such that $f(x^*x)=\Vert x\Vert ^2$ and $f(\langle x,y\rangle \,\langle y,x\rangle )=0$ if and only if there is $f\in {\mathcal {S}}({\mathcal {A}})$ such that $f(x^*x)=\Vert x\Vert ^2$ and $f(\langle x,y\rangle b)=0$ for all $b\in {\mathcal {A}}.$

The classes of $C^*$-algebras in which any two of these orthogonality relationships coincide have been described in [4, 6]. More precisely,

Theorem 1.3

[6, Corollary 4.10] Let ${\mathcal {A}}$ be a nonzero $C^*$-algebra. Then the following statements are equivalent : 

• (1) For all $x,y \in {\mathcal {A}} ,$ $x \perp _{S-BJ} y$ if and only if $\langle x, y\rangle =0 .$

• (2) For all $x,y \in {\mathcal {A}} ,$ $x \perp _{BJ} y$ if and only if $x \perp _{S-BJ} y .$

• (3) ${\mathcal {A}}$ is isomorphic to ${\mathbb {C}} .$

Let a be a nonzero positive element of ${\mathcal {A}} .$ A generalization of state space of ${\mathcal {A}}$ was introduced in [1] as follows:

$$\begin{aligned} {\mathcal {S}}_{a}({\mathcal {A}}):= \{ \varphi \in {\mathcal {A}}^{'} \, : \, \varphi \ge 0, \, \varphi ( a ) = 1\}=\left\{ \dfrac{f}{f(a)} \, : \, f\in {\mathcal {S}}({\mathcal {A}}), \, f(a) \ne 0 \right\} . \end{aligned}$$

Observe that if $a=1_{{\mathcal {A}}},$ then ${\mathcal {S}}_a({\mathcal {A}})={\mathcal {S}}({\mathcal {A}}).$ It has been proved in [1] that ${\mathcal {S}}_a({\mathcal {A}})$ is a nonempty convex and $w^*$-closed subset of ${\mathcal {A}}^{\prime }.$ But, unlike ${\mathcal {S}}({\mathcal {A}}),$ the set ${\mathcal {S}}_a({\mathcal {A}})$ may not be $w^*$-compact. In fact, according to the following result, ${\mathcal {S}}_a({\mathcal {A}})$ is $w^*$-compact if and only if a is invertible.

Proposition 1.4

[1, Proposition 2.3] Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let $a\in {\mathcal {A}}$ be a positive element. Then the following statement are equivalent : 

• (1) ${\mathcal {S}}_a({\mathcal {A}})$ is $w^*$-compact.

• (2) a is invertible.

For any element $x\in {\mathcal {A}},$ the a-operator semi-norm $\Vert \cdot \Vert _a :{\mathcal {A}}\rightarrow [0,\infty )$ is defined by

$$\begin{aligned} \Vert x \Vert _{a} := \sup \left\{ \sqrt{\varphi (x^* a x )} \, : \varphi \in {\mathcal {S}}_a ({\mathcal {A}}) \right\} . \end{aligned}$$

Due to the Proposition 1.4, if a is not invertible, then ${\mathcal {S}}_a({\mathcal {A}})$ is not $w^*$-compact, and so it may happen that $\Vert x\Vert _a=\infty$ for some $x\in {\mathcal {A}};$ see [1, Example 3.2]. Denote by ${\mathcal {A}}^a:=\{x\in {\mathcal {A}} : \Vert x\Vert _a<\infty \}.$ It was shown in [1] that $\Vert \cdot \Vert _a$ is a submultiplicative semi-norm on ${\mathcal {A}}^a;$ i.e., $\Vert xy\Vert _a\le \Vert x\Vert _a\Vert y\Vert _a$ for all $x,y\in {\mathcal {A}}^a.$ Also, $\Vert x\Vert _a=0$ if and only if $ax=0.$ In addition, if a is invertible, then $\Vert \cdot \Vert _a$ is a norm on ${\mathcal {A}} .$ Consequently, $\Vert \cdot \Vert _{1_{{\mathcal {A}}}}$ is equal to the $C^*$-norm $\Vert \cdot \Vert$ of ${\mathcal {A}}.$

An element $x^\sharp \in {\mathcal {A}}$ is called an a-adjoint of $x\in {\mathcal {A}}$ if $ax^\sharp =x^*a.$ The set of all a-adjointable elements of ${\mathcal {A}}$ is denoted by ${\mathcal {A}}_a.$ Note that ${\mathcal {A}}_a={\mathcal {A}}$ if ${\mathcal {A}}$ is commutative.

In [1, Corollary 4.9] it was proved that if $x \in {\mathcal {A}}_a$ and $x^\sharp$ is an a-adjoint of it, then

$$\begin{aligned} \Vert x\Vert _{a}^2 = \Vert xx^\sharp \Vert _{a} = \Vert x^\sharp x\Vert _{a} =\Vert x^\sharp \Vert _{a}^2 . \end{aligned}$$

(1.1)

An element $x\in {\mathcal {A}}$ is said to be a-selfadjoint if ax is selfadjoint; i.e., $ax=x^*a.$ Moreover, any element $x \in {\mathcal {A}}_a$ can be written as $x=x_1 + i x_2 ,$ where $x_1$ and $x_2$ are a-selfadjoint. In fact, if $x^{\sharp }$ is an a-adjoint of x,  then

$$\begin{aligned} x = \dfrac{x+x^\sharp }{2} + i \dfrac{x-x^\sharp }{2i}. \end{aligned}$$

(1.2)

This decomposition is not unique, since there might be many (or none) a-adjoints $x^\sharp$ of x;  see e.g., [1, 8]. Note that if we assume that a is invertible, then x has the unique a-adjoint $x^\sharp =a^{-1}x^*a,$ and, therefore, the decomposition (1.2) is unique.

The notions, ${\mathcal {S}}_a({\mathcal {A}})$ and $\Vert \cdot \Vert _a$ were introduced in [1] to generalize algebraic numerical range and algebraic numerical radius of elements of $C^*$-algebra ${\mathcal {A}}.$ To study abundant results related to these concepts the reader is referred to [1, 2, 5].

In this paper, we investigate the notions of Birkhoff–James orthogonality and its strong version in an unital $C^*$-algebra ${\mathcal {A}}$ with respect to the norm $\Vert \cdot \Vert _a,$ whenever $a\in {\mathcal {A}}$ is a positive and invertible element.

In Sect. 2 first, the main properties of a-Birkhoff–James orthogonality are studied and a variety of examples in simple $C^*$-algebra ${\mathbb {M}}_n({\mathbb {C}})$ are presented to illustrate the relationship between a-Birkhoff–James orthogonality and Birkhoff–James orthogonality. Next, a complete characterization of a-Birkhoff–James orthogonality in terms of elements of the generalized state space ${\mathcal {S}}_a({\mathcal {A}})$ is presented. As an application, a characterization for the best approximation to elements of ${\mathcal {A}}$ in a subspace ${\mathcal {B}}$ is obtained with respect to $\Vert \cdot \Vert _a.$ Moreover, a generalization of the well-known distance formula which obtained by Williams in [19] is given.

Section 3 is devoted to the study of strong a-Birkhoff–James orthogonality in unital $C^*$-algebras. The classes of unital $C^*$-algebras in which the a-Birkhoff–James orthogonality coincides with the strong a-Birkhoff–James orthogonality are described.

In particular, we prove that if $x\perp ^a_{S-BJ}y$ implies $\langle x,y\rangle _a:=x^*ay=0,$ for all $x,y\in {\mathcal {A}},$ then the center of ${\mathcal {A}}$ is trivial, i.e., the only central elements of ${\mathcal {A}}$ are multiplies of the identity. Moreover, we prove that the right additivity (left additivity) of (strong) a-Birkhoff–James orthogonality in ${\mathcal {A}}$ concludes that ${\mathcal {Z}}({\mathcal {A}})\cong {\mathbb {C}}1_{{\mathcal {A}}}.$

2 a-Birkhoff–James orthogonality in $C^*$-algebras

Let ${\mathcal {B}}({\mathcal {H}})$ be the $C^*$-algebra of all bounded linear operators on a complex Hilbert space ${\mathcal {H}}$ with inner product $\langle \cdot , \cdot \rangle .$ If $\mathrm{dim {\mathcal {H}}}=n,$ then we identify ${\mathcal {B}}({\mathcal {H}})$ with the simple $C^*$-algebra ${\mathbb {M}}_n({\mathbb {C}})$ of all $n\times n$ complex matrices and denote the identity matrix by $I_n.$ Assume that $A\in {\mathcal {B}}({\mathcal {H}})$ is a positive operator, which induces a positive semi-definite sesquilinear form $\langle \cdot ,\cdot \rangle _A:{\mathcal {H}}\times {\mathcal {H}}\rightarrow {\mathbb {C}}$ defined by $\langle x,y\rangle _A=\langle Ax,y\rangle .$ The semi-norm $\Vert \cdot \Vert _A$ induced by $\langle \cdot ,\cdot \rangle _A$ is defined by $\Vert x\Vert _A=\sqrt{\langle Ax,x\rangle }$ for every $x\in {\mathcal {H}}.$ Furthermore, the set of all A-bounded operators on ${\mathcal {H}}$ is defined by

$$\begin{aligned} {\mathcal {B}}_{A^{\frac{1}{2}}} ( {\mathcal {H}}) := \{ T \in {\mathcal {B}}({\mathcal {H}}) \, : \, \exists \, c > 0, \, \Vert T x \Vert _{A} \le c \Vert x \Vert _{A}, \quad \forall x \in {\mathcal {H}} \}. \end{aligned}$$

In fact, ${\mathcal {B}}_{A^{\frac{1}{2}}} ( {\mathcal {H}})$ is a unital subalgebra of ${\mathcal {B}}({\mathcal {H}})$ which is equipped with the semi-norm

$$\begin{aligned} \gamma _A (T) := \sup _{\Vert x\Vert _A=1} \sqrt{\langle ATx , Tx \rangle }\quad (T \in {\mathcal {B}}_{A^{\frac{1}{2}}} ( {\mathcal {H}})). \end{aligned}$$

The Birkhoff–James orthogonality with respect to the semi-norm $\gamma _A(\cdot )$ (called A-Birkhoff–James orthogonality) was studied by Zamani in [20]. An operator $T \in {\mathcal {B}}_{A^{\frac{1}{2}}} ( {\mathcal {H}})$ is called A-Birkhoff–James orthogonal to the operator $S \in {\mathcal {B}}_{A^{\frac{1}{2}}} ( {\mathcal {H}}) ,$ denoted by $T \perp _{BJ}^{A} S ,$ if $\gamma _{A} (T + \lambda S) \ge \gamma _{A} (T)$ for all $\lambda \in {\mathbb {C}}.$ The following characterization of A-Birkhoff–James orthogonality which extends the Bhatia and S̆emrl Theorem for A-bounded operators has been obtained as follow:

Theorem 2.1

[20, Theorem 2.2] Let $T,S \in {\mathcal {B}}_{A^{\frac{1}{2}}} ( {\mathcal {H}}) .$ The following conditions are equivalent : 

• (1) $T \perp _{BJ}^{A} S .$

• (2) There exists a sequence $\{ h_n \}$ of A-unit vectors $(\Vert h_n\Vert _A=1)$ in ${\mathcal {H}}$ such that

  $$\begin{aligned} \lim _{n \rightarrow \infty } \Vert T h_n \Vert _{A} = \gamma _{A} (T) \quad \text {and} \quad \lim _{n \rightarrow \infty } \langle T h_n , S h_n \rangle _A = 0 . \end{aligned}$$

  (2.1)

We recall that by the Gelfand–Naimark Theorem, any unital $C^*$-algebra ${\mathcal {A}}$ can be considered as a norm closed $*$-subalgebra of ${\mathcal {B}}({\mathcal {H}})$ for some Hilbert space ${\mathcal {H}}$ . In fact, there exists an unital faithful $*$-representation $\pi :{\mathcal {A}}\rightarrow {\mathcal {B}}({\mathcal {H}})$ such that $\Vert x\Vert =\Vert \pi (x)\Vert$ for all $x\in {\mathcal {A}};$ see e.g., [12, 17]. It was proved that in [1], if $\pi :{\mathcal {A}}\rightarrow {\mathcal {B}}({\mathcal {H}})$ is a unital faithful $*$-representation of ${\mathcal {A}},$ then

$$\begin{aligned} \Vert x \Vert _a = \gamma _{\pi (a)} ( \pi (x)) \end{aligned}$$

(2.2)

for any $x\in {\mathcal {A}}.$ As a direct consequence of this fact, we have ${\mathcal {A}}^A={\mathcal {B}}_{A^{\frac{1}{2}}} ( {\mathcal {H}})$ for ${\mathcal {A}}={\mathcal {B}}({\mathcal {H}})$ and all positive operator $A\in {\mathcal {B}}({\mathcal {H}}).$

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let $a\in {\mathcal {A}}$ be a nonzero positive element. It was proved in [1, Theorem 3.9] that ${\mathcal {A}}_a\subset {\mathcal {A}}^a.$ Now, if we assume that a is a positive and invertible element of ${\mathcal {A}},$ then for every $x\in {\mathcal {A}}$ the equation $ax^\sharp =x^*a$ has the unique solution $x^\sharp =a^{-1}x^*a,$ and so every $x\in {\mathcal {A}}$ is a-adjointable. Therefore ${\mathcal {A}}^a={\mathcal {A}}.$

From now on we assume that ${\mathcal {A}}$ is a unital $C^*$-algebra and $a\in {\mathcal {A}}$ is a positive and invertible element. Let us introduce the concept of a-Birkhoff–James orthogonality with respect to the norm $\Vert \cdot \Vert _a$ in $C^*$-algebras.

Definition 2.2

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and $a\in {\mathcal {A}}$ be a positive and invertible element. We say that an element $x \in {\mathcal {A}}$ is Birkhoff–James orthogonal with respect to the norm $\Vert \cdot \Vert _{a}$ (a-Birkhoff–James orthogonal) to an element $y \in {\mathcal {A}} ,$ in short $x \perp _{BJ}^{a} y ,$ if

$$\begin{aligned} \Vert x + \lambda y \Vert _{a} \ge \Vert x \Vert _{a}\quad (\forall \lambda \in {\mathbb {C}}). \end{aligned}$$

First, note that a-Birkhoff–James orthogonality reduces to the Birkhoff–James orthogonality when $a=1_{{\mathcal {A}}}.$ Also, it is easy to see that a-Birkhoff–James orthogonality is homogenous; i.e., if $x \perp _{BJ}^{a} y ,$ then $\alpha x \perp _{BJ}^{a} \beta y$ for all $\alpha , \beta \in {\mathbb {C}} .$ It is trivial for $\alpha =0$ or $\beta =0 .$ So, suppose that $\alpha$ and $\beta$ are nonzero complex numbers. For each $\lambda \in {\mathbb {C}} ,$ we have

$$\begin{aligned} \Vert \alpha x + \lambda \beta y \Vert _a = \left\| \alpha \left( x + \dfrac{\beta }{\alpha } \lambda y \right) \right\| _a = |\alpha | \left\| \left( x +\dfrac{\beta }{\alpha } \lambda y\right) \right\| _a \ge | \alpha | \, \Vert x \Vert _a = \Vert \alpha x \Vert _a. \end{aligned}$$

It follows that $\alpha x \perp _{BJ}^{a} \beta y .$

Also, a-Birkhoff–James orthogonality is non-degenerate. Indeed, let $0\ne x \in {\mathcal {A}}$ and $x \perp _{BJ}^{a} x .$ Then $\Vert x+ \lambda x\Vert _a \ge \Vert x\Vert _a$ for all $\lambda \in {\mathbb {C}} .$ For $\lambda = -1 ,$ we get $\Vert x\Vert _a =0 ,$ and so $ax =0 .$ Therefore $x=0 ,$ since a is invertible.

Moreover, for any two nonzero elements $x,y\in {\mathcal {A}},$ if x is orthogonal to y in the a-Birkhoff–James sense, then x and y are linearly independent. In fact, if we assume to the contrary that there exists $k \in {\mathbb {C}}$ such that $y=k x ,$ then $x \perp _{BJ}^{a} kx .$ It follows that $x \perp _{BJ}^{a} x ,$ since a-Birkhoff–James orthogonality is homogenous. Hence $ax =0 ,$ and so $x=0 ,$ which is a contradiction.

Let $f\in {\mathcal {S}}({\mathcal {A}}).$ According to [1], the linear functional defined by

$$\begin{aligned} \varphi (z)=f\left( a^{-\frac{1}{2}}za^{-\frac{1}{2}}\right) \quad (z\in {\mathcal {A}}) \end{aligned}$$

(2.3)

belongs to ${\mathcal {S}}_a({\mathcal {A}}).$ Now, let $x \in {\mathcal {A}}$ and let $a \in {\mathcal {A}}$ be positive and invertible such that $ax=xa .$ Then

$$\begin{aligned} \Vert x\Vert _a^2=\sup _{\varphi \in {\mathcal {S}}_a({\mathcal {A}})}\varphi (x^*ax) =\sup _{f\in {\mathcal {S}}({\mathcal {A}})} f \left( a^{-\frac{1}{2}} x^* a x a^{-\frac{1}{2}} \right) =\sup _{f\in {\mathcal {S}}({\mathcal {A}})} f(x^*x)=\Vert x\Vert ^2. \end{aligned}$$

(2.4)

Also, note that $x^\sharp = a^{-1} x^* a$ is a-adjoint of x,  and so it follows from (1.1) that

$$\begin{aligned} \Vert x\Vert _{a}^2 = \Vert x a^{-1} x^* a \Vert _{a} = \Vert a^{-1} x^* a x\Vert _a = \Vert a^{-1} x^* a \Vert _{a}^2. \end{aligned}$$

Hence

$$\begin{aligned} \Vert x\Vert _{a}^2 = \Vert x x^*\Vert _{a} = \Vert x^* x \Vert _{a} = \Vert x^*\Vert _{a}^2. \end{aligned}$$

Since there is at most one norm on a $*$-algebra making it a $C^*$-algebra, the following result is obtained.

Corollary 2.3

If ${\mathcal {A}}$ is a commutative and unital $C^*$-algebra and $a\in {\mathcal {A}}$ is positive and invertible,  then $\Vert \cdot \Vert _a$ agrees with the $C^*$-norm of $C^*$-algebra ${\mathcal {A}}.$ In this case, the a-Birkhoff–James orthogonality and the Birkhoff–James orthogonality are equivalent on ${\mathcal {A}}.$

It should be noted that $\Vert \cdot \Vert _a$ does not satisfy to the $C^*$-condition in noncommutative $C^*$-algebra, even when a is invertible. To make this clear, we present the following example.

Example 2.4

Let ${\mathbb {M}}_2 ({\mathbb {C}})$ be the $C^*$-algebra of all $2 \times 2$ complex matrices, and let ${\textrm{Tr}}$ be the usual trace functional on ${\mathbb {M}}_2 ({\mathbb {C}}) .$ According to the Example 2.2 of [1], for any positive matrix $h \in {\mathbb {M}}_2 ({\mathbb {C}}) ,$ let $\varphi _h$ be the positive linear functional given by

$$\begin{aligned} \varphi _h (x) := {\textrm{Tr}}(hx), \quad (x \in {\mathbb {M}}_2 ({\mathbb {C}})). \end{aligned}$$

It is known that any state on ${\mathbb {M}}_2 ({\mathbb {C}})$ is of the form $\varphi _h$ with ${\textrm{Tr}}(h) =1 .$ For a positive matrix $a \in {\mathbb {M}}_2 ({\mathbb {C}}) ,$ we have

$$\begin{aligned} {\mathcal {S}}_{a} ({\mathbb {M}}_{2}({\mathbb {C}})) = \{ \varphi _h \, : \, h \in {\mathbb {M}}_2 ({\mathbb {C}})^{+} \quad \text {and} \quad {\textrm{Tr}}(ha) =1\}. \end{aligned}$$

Now, let $a=\left[ \begin{array}{cccc} 2&{}0 \\ 0&{}1 \end{array}\right] .$ Then with some simple matrix computations, we conclude that

$$\begin{aligned} {\mathcal {S}}_{a} ({\mathbb {M}}_{2}({\mathbb {C}})) =\{ \varphi _h \, : \, h\in {\mathcal {L}}_a\}, \end{aligned}$$

where

$$\begin{aligned} {\mathcal {L}}_a:=\left\{ h =\left[ \begin{array}{cccc} h_{11}&{}h_{12} \\ \overline{h}_{12}&{} h_{22} \end{array}\right] \in {\mathbb {M}}_2 ({\mathbb {C}})^{+}\, : \, h_{12} \in {\mathbb {C}}, \, h_{11}, h_{22} \ge 0 \ \text {and}\ 2h_{11} +h_{22} =1\right\} . \end{aligned}$$

Hence for $x=\left[ \begin{array}{cccc} 0&{}2 \\ 1&{}0 \end{array}\right] ,$ we get

$$\begin{aligned} \Vert x \Vert _{a}^{2} = \sup _{\varphi _{h} \in {\mathcal {S}}_a ({\mathbb {M}}_2 ({\mathbb {C}}))} \varphi _{h} (x^* a x)&= \sup _{h\in {\mathcal {L}}_a} {\textrm{Tr}}\left( \left[ \begin{array}{cccc} h_{11}&{} 8h_{12} \\ \overline{h}_{12}&{} 8h_{22} \end{array}\right] \right) \\&=\sup _{2h_{11}+h_{22}=1, h_{11}, h_{22} \ge 0} h_{11} +8h_{22} = 8. \end{aligned}$$

But similarly, we have

$$\begin{aligned} \Vert x^* x \Vert _{a}^{2}= \sup _{\varphi _{h} \in {\mathcal {S}}_a( {\mathbb {M}}_2 ({\mathbb {C}}))} \varphi _{h}( (x^* x) a (x^* x)) =\sup _{2h_{11}+h_{22}=1, h_{11}, h_{22} \ge 0} 2h_{11} +16h_{22} = 16. \end{aligned}$$

The following provide us with examples reveal that the a-Birkhoff–James orthogonality is independent from the Birkhoff–James orthogonality in unital and noncommutative $C^*$-algebras, even when a is positive and invertible.

Example 2.5

In the context of the same $a=\left[ \begin{array}{cccc} 2 &{}0 \\ 0&{}1 \end{array}\right]$ as, and similarly to the method we applied in the previous example, let $x=\left[ \begin{array}{cccc} 0 &{}-1 \\ 0&{}1 \end{array}\right]$ and $y=\left[ \begin{array}{cccc} 0 &{}1 \\ 0&{}1 \end{array}\right]$ be matrices in ${\mathbb {M}}_2({\mathbb {C}}).$ Then

$$\begin{aligned} \Vert x\Vert _{a}^2&= \sup _{\varphi _{h} \in {\mathcal {S}}_a ({\mathbb {M}}_2 ({\mathbb {C}}))} \varphi _{h} (x^* a x)= \sup _{h\in {\mathcal {L}}_a}{\textrm{Tr}} (h (x ^* a x)) \\&= \sup _{h\in {\mathcal {L}}_a} {\textrm{Tr}} \left( \left[ \begin{array}{cccc} h_{11}&{} h_{12} \\ \overline{h}_{12} &{} h_{22} \end{array}\right] \left[ \begin{array}{cccc} 0&{} 0 \\ 0&{} 3 \end{array}\right] \right) =\sup _{2h_{11}+h_{22}=1, h_{11}, h_{22} \ge 0} 3h_{22} = 3. \end{aligned}$$

Also, for every $\lambda \in {\mathbb {C}} ,$ we have

$$\begin{aligned} \Vert x + \lambda y \Vert _{a}^{2}&=\sup _{h\in {\mathcal {L}}_a}\varphi _h((x+\lambda y)^*a(x+\lambda y))\\&=\sup _{2h_{11}+h_{22}=1, h_{11}, h_{22} \ge 0} \big (3(1+|\lambda |^2) -2{\textrm{Re}}(\lambda )\big ) h_{22}\\&=3(1+|\lambda |^2) -2{\textrm{Re}}(\lambda ). \end{aligned}$$

However, for $\lambda = \dfrac{1}{3} ,$ we see that $\Vert x+ \lambda y \Vert _{a}^2 = \dfrac{8}{3} < 3 = \Vert x\Vert _{a}^2,$ which yields that $x \not \perp _{BJ}^a y .$ On the other hand, it can easily be seen that $\Vert x\Vert ^2 = 2$ and $\Vert x+ \lambda y \Vert ^2 = 2+2|\lambda |^2$ for all $\lambda \in {\mathbb {C}} .$ Hence $\Vert x+ \lambda y \Vert ^2= 2(1+|\lambda |^2) \ge 2=\Vert x\Vert ,$ for all $\lambda \in {\mathbb {C}} .$ Thus $x \perp _{BJ} y .$

Now, let $x=\left[ \begin{array}{cccc} 0&{}\frac{1}{2} \\ 0&{}-1 \end{array}\right]$ and $y=\left[ \begin{array}{cccc} 0 &{}1 \\ 0&{}1 \end{array}\right] .$ Then $x \perp _{BJ}^a y ,$ since $\langle x , y\rangle _a =x^* ay =0 .$ But, for every $\lambda \in {\mathbb {C}} ,$ we have

$$\begin{aligned} \Vert x+\lambda y\Vert ^2=\left| \dfrac{5}{4}+2|\lambda |^2-{\textrm{Re}}(\lambda ) \right| . \end{aligned}$$

So for $\lambda = \dfrac{1}{4},$ we have $\Vert x+\lambda y\Vert ^2 = \dfrac{9}{8} \,<\dfrac{5}{4}=\Vert x\Vert ^2 ,$ and therefore $x \not \perp _{BJ} y .$

Assume that ${\mathcal {A}}$ is a unital and commutative $C^*$-algebra, $x,y \in {\mathcal {A}}$ and $a\in {\mathcal {A}}$ is positive and invertible. If $x \perp _{BJ} y ,$ then by Theorem 1.1, there must exist $f\in {\mathcal {S}}({\mathcal {A}})$ such that $f(x^*x)=\Vert x\Vert ^2$ and $f(x^*y)=0.$ Since ${\mathcal {A}}$ is commutative, by Corollary 2.3 we conclude that

$$\begin{aligned} \varphi (x^* a x) = f \left( a^{-\frac{1}{2}} x^* a x a^{-\frac{1}{2}} \right) = f (x^* x) = \Vert x\Vert ^2= \Vert x\Vert _{a}^2 , \end{aligned}$$

and

$$\begin{aligned} \varphi (x^* a y) = f \left( a^{-\frac{1}{2}} x^* a y a^{-\frac{1}{2}} \right) = f (x^* y) =0, \end{aligned}$$

where $\varphi \in {\mathcal {S}}_a({\mathcal {A}})$ is defined in (2.3). This fact motivates us to obtain a similar characterization for a-Birkhoff–James orthogonality in unital $C^*$-algebras. More precisely, we shall present a characterization of a-Birkhoff–James orthogonality in a unital $C^*$- algebra ${\mathcal {A}}$ based on the elements of its generalized state space ${\mathcal {S}}_a({\mathcal {A}}).$ In fact, we use a simple way to obtain the next fundamental result through the standard Gelfand–Naimark representation of ${\mathcal {A}}$ as a concrete $C^*$-subalgebra of ${\mathcal {B}}({\mathcal {H}})$ and displayed formula (2.2). However, for completion of the subject and the convenience of the reader, we present a short proof for it. Note that this characterization is a generalization of the well-known Theorem 1.1 when we take $a=1_{{\mathcal {A}}},$ and plays a fundamental role to achieve our forthcoming main results.

Theorem 2.6

Let ${\mathcal {A}}$ be a unital $C^*$-algebra,  $x,y\in {\mathcal {A}}$ and let a be positive and invertible element of ${\mathcal {A}}.$ Then the following statements are equivalent : 

• (1) $x \perp _{BJ}^{a} y .$

• (2) There is $\varphi \in {\mathcal {S}}_{a}({\mathcal {A}} )$ such that $\varphi (x^* a x) = \Vert x \Vert _{a}^{2}$ and $\varphi (y^{*} a x )= 0$ $(\varphi (x^*ay)=0).$

Proof

$(1) \Rightarrow (2)$ Let $x \perp _{BJ}^{a} y$ and let $\pi : {\mathcal {A}} \rightarrow {\mathcal {B}}({\mathcal {H}})$ be a unital faithful $*$-representation of ${\mathcal {A}} .$ Since a is invertible, it follows from (2.2) that $\pi (x) , \pi (y) \in {\mathcal {B}}_{\pi (a)^{\frac{1}{2}}} ( {\mathcal {H}}) ,$ and so $\pi (x)\perp ^{\pi (a)}_{BJ} \pi (y).$ Hence Theorem 2.1, concludes that there exists a sequence of $\pi (a)$-unit vectors $\{ h_n\} \in {\mathcal {H}}$ such that

$$\begin{aligned}&\lim _{n \rightarrow \infty } \Vert \pi (x ) h_n \Vert _{\pi (a)} = \gamma _{\pi (a)} (\pi (x)), \end{aligned}$$

(2.5)

$$\begin{aligned}&\lim _{n \rightarrow \infty } \langle \pi (x) h_n , \pi (y) h_n \rangle _{\pi (a)} =0. \end{aligned}$$

(2.6)

The linear functionals $\varphi _{n} : {\mathcal {A}} \rightarrow {\mathbb {C}}$ defined by $\varphi _n (z) = \langle \pi (z ) h_n , h_n \rangle$ belong to ${\mathcal {S}}_a ({\mathcal {A}})$ for all $n\in {\mathbb {N}}$ (see [1, Theorem 3.5]). Now, (2.2) and (2.5) imply that

$$\begin{aligned} \lim _{n \rightarrow \infty } \varphi _n ( x^* a x )&= \lim _{n \rightarrow \infty } \langle \pi (x^* a x) h_n , h_n \rangle = \lim _{n \rightarrow \infty } \langle \pi (a) \pi (x) h_n , \pi (x) h_n \rangle \\&=\lim _{n \rightarrow \infty } \Vert \pi (x ) (h_n) \Vert _{\pi (a)}^{2} = \gamma ^{2 }_{\pi (a)}(\pi (x)) = \Vert x \Vert _{a}^2 . \end{aligned}$$

In addition, from (2.6), we infer that

$$\begin{aligned} \lim _{n \rightarrow \infty }\varphi _n (y^* ax)&= \lim _{n \rightarrow \infty }\langle \pi (y^* a x) h_n , h_n \rangle =\lim _{n \rightarrow \infty }\langle \pi (a) \pi (x) h_n , \pi (y) h_n \rangle \\&=\lim _{n \rightarrow \infty }\langle \pi (x) h_n , \pi (y) h_n \rangle _{\pi (a)} =0 . \end{aligned}$$

Thus

$$\begin{aligned} \lim _{n \rightarrow \infty } \varphi _n (x^* a x) = \Vert x\Vert _{a}^{2},\quad \lim _{n \rightarrow \infty } \varphi _n (y^* a x ) =0. \end{aligned}$$

(2.7)

In addition, by Proposition 1.4, ${\mathcal {S}}_a ({\mathcal {A}})$ is $w^*$-compact. So there is $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi _n \overset{w^*}{\longrightarrow }\ \varphi .$ Therefore, (2.7) implies that

$$\begin{aligned} \varphi ( x^* ax) = \Vert x \Vert _{a}^{2}\quad \text{ and } \quad \varphi ( y^* ax ) =0. \end{aligned}$$

$(2) \Rightarrow (1 )$ Assume that there is $\varphi \in {\mathcal {S}}_{a}({\mathcal {A}} )$ such that $\varphi (x^* a x) = \Vert x \Vert _{a}^{2}$ and $\varphi (y^{*} a x )= 0 .$ Then for each $\lambda \in {\mathbb {C}} ,$ we get

$$\begin{aligned} \Vert x + \lambda y \Vert _{a}^{2}&\ge \varphi ((x + \lambda y)^* a (x + \lambda y)) \\&= \varphi ( x^* a x) + 2 {\textrm{Re}} ( {\overline{\lambda }} \varphi (y^* a x)) + | \lambda |^2 \varphi ( y^* a y) \\&=\varphi ( x^* a x) + |\lambda |^2 \varphi ( y^* a y) \ge \varphi ( x^* ax) = \Vert x \Vert _{a}^{2}. \end{aligned}$$

Therefore $x\perp _{BJ}^ay.$ $\square$

As the first direct consequence of Theorem 2.6, it is easy to see that for given linearly independent vectors $x,y\in {\mathcal {A}},$ there exists a unique $\alpha \in {\mathbb {C}}$ such that $x\perp ^{a}_{BJ}(\alpha x+ y).$ Indeed, we take $\alpha =0$ if $x\perp ^{a}_{BJ}y.$ Now, suppose that $x\not \perp ^a_{BJ}y.$ Since $a \in {\mathcal {A}}$ is invertible, there exists $\varphi \in {\mathcal {S}}_a({\mathcal {A}})$ such that $\varphi (x^*ax)=\Vert x\Vert ^2_a,$ by Proposition 1.4. Furthermore, $\varphi (x^*ay)\ne 0,$ by Theorem 2.6. Let $\alpha =-\dfrac{\varphi (x^*ay)}{\varphi (x^*ax)}.$ Then

$$\begin{aligned} \varphi (x^*a(\alpha x+y))=-\dfrac{\varphi (x^*ay)}{\varphi (x^*ax)}\varphi (x^*ax)+\varphi (x^*ay)=0. \end{aligned}$$

Therefore $x\perp ^a_{BJ}(\alpha x+y).$

Further, the next result gives us some more examples of a-Birkhoff–James orthogonality for elements of ${\mathcal {A}}$ to some appropriate elements.

Corollary 2.7

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let $a \in {\mathcal {A}}$ be positive and invertible. For each $x, y\in {\mathcal {A}},$ we have

$$\begin{aligned} x\perp _{BJ}^a \left( \Vert x\Vert _{a}^2 y a^{\frac{1}{2}} - y a^{-\frac{1}{2}} \langle x,x\rangle _a \right) . \end{aligned}$$

Proof

For the convenience, $x^*ax$ and $x^*ay$ are shown with the symbols $\langle x,x \rangle _a$ and $\langle x,y \rangle _a,$ respectively for all $x,y\in {\mathcal {A}}.$ For $x=0$ or $y=0 ,$ the statement is trivial. Now, assume that x, y are nonzero elements of ${\mathcal {A}}.$ Since a is invertible, there exists $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi (\langle x,x \rangle _a ) =\Vert x\Vert _{a}^2$ by Proposition 1.4. The Cauchy–Schwarz inequality and (1.1) tell us

$$\begin{aligned}&\big | \varphi \big ( \big \langle x , \Vert x\Vert _{a}^2\, y\, a ^{\frac{1}{2}} - y \, a ^{-\frac{1}{2}} \langle x,x \rangle _a \big \rangle _a \big )\big |^2\\&\quad =\big | \varphi \big ( \Vert x\Vert _{a}^2\, \langle x,y\rangle _a \, a ^{\frac{1}{2}} -\langle x,y\rangle _a \, a ^{-\frac{1}{2}} \langle x,x \rangle _a \big )\big |^2 \\&\quad = | \varphi \big ( \langle x,y\rangle _a ( \Vert x\Vert _{a}^2 \, a ^{\frac{1}{2}} - a ^{-\frac{1}{2}}\, \langle x,x \rangle _a \big ) |^2 \\&\quad \le \, \varphi ( \langle x,y\rangle _a \langle y,x\rangle _a ) \, \varphi (\Vert x\Vert _{a}^4 \, a -2 \Vert x\Vert _{a}^2 \, \langle x,x\rangle _a + \langle x,x\rangle _a \, a^{-1} \langle x,x \rangle _a ) \\&\quad = \varphi ( \langle x,y\rangle _a \langle y,x\rangle _a ) \, \big ( \Vert x\Vert _{a}^4 \, \varphi (a) - 2 \Vert x\Vert _{a}^2 \, \varphi ( \langle x,x\rangle _a ) + \varphi ((x^\sharp x)^* a(x^\sharp x) )\big ) \\&\quad \le \varphi ( \langle x,y\rangle _a \langle y,x\rangle _a ) \, (\Vert x\Vert _{a}^4 - 2 \Vert x\Vert _{a}^4 + \Vert x^\sharp x \Vert _{a}^2) \\&\quad = \varphi ( \langle x,y\rangle _a \langle y,x\rangle _a ) (\Vert x\Vert _{a}^4 - 2 \Vert x\Vert _{a}^4 + \Vert x \Vert _{a}^4 ) =0 . \end{aligned}$$

It follows that $\varphi (\langle x , \Vert x\Vert _{a}^2\, y\, a ^{\frac{1}{2}} - y \, a ^{-\frac{1}{2}} \langle x,x \rangle _a \rangle _a)=0.$ Consequently, Theorem 2.6 implies that

$$\begin{aligned} x\perp _{BJ}^a \left( \Vert x\Vert _{a}^2 y a^{\frac{1}{2}} - y a^{-\frac{1}{2}} \langle x,x\rangle _a \right) . \end{aligned}$$

$\square$

The a-algebraic numerical range of any element $x\in {\mathcal {A}}$ is defined by

$$\begin{aligned} V_a(x)=\{\varphi (ax) : \varphi \in {\mathcal {S}}_a({\mathcal {A}})\}. \end{aligned}$$

It has been proved in [1, Theorem 4.7] that $V_a(x)$ is a nonempty convex and compact subset of complex numbers for all $x\in {\mathcal {A}}^a={\mathcal {A}},$ since a is invertible. An extension of the William’s Theorem [19, Theorem 1] is obtained in [2, Theorem 2.14].

The following direct result of Theorem 2.6 gives us an alternative proof for this fact.

Corollary 2.8

Let $x\in {\mathcal {A}}.$ Then $0\in V_a(x)$ if and only if $\Vert x-\lambda 1_{{\mathcal {A}}}\Vert _a\ge |\lambda |$ for all $\lambda \in {\mathbb {C}}.$

Proof

Since $0\in V_a(x),$ there is $\varphi \in {\mathcal {S}}_a({\mathcal {A}})$ such that $\varphi (1_{{\mathcal {A}}}ax)=\varphi (ax)=0.$ Also, we have $\varphi (1^*_{{\mathcal {A}}}a1_{{\mathcal {A}}})=\varphi (a)=1=\Vert 1_{{\mathcal {A}}}\Vert ^2_a.$ It follows from Theorem 2.6 that $1_{{\mathcal {A}}}\perp ^a_{BJ}x,$ which implies that $\Vert x-\lambda 1_{{\mathcal {A}}}\Vert _a\ge |\lambda |$ for all $\lambda \in {\mathbb {C}}.$

Now, if $\Vert \lambda 1_{{\mathcal {A}}}-x\Vert _a=\Vert x-\lambda 1_{{\mathcal {A}}}\Vert _a\ge |\lambda |$ for all $\lambda \in {\mathbb {C}},$ then $1_{{\mathcal {A}}}\perp ^a_{BJ}x,$ by the homogeneity of the Birkhoff–James orthogonality. So there is $\varphi \in {\mathcal {S}}_a({\mathcal {A}})$ such that $\varphi (ax)=0.$ Therefore $0\in V_a(x).$ $\square$

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let $x\in {\mathcal {A}}.$ Suppose that ${\mathcal {B}}$ is a subspace of ${\mathcal {A}}.$ An element $y_0\in {\mathcal {B}}$ is said to be a best approximation to x in ${\mathcal {B}}$ if

$$\begin{aligned} \Vert x-y_0\Vert = {\textrm{dist}}(x,{\mathcal {B}}):=\inf \{\Vert x-y\Vert : y\in {\mathcal {B}}\} . \end{aligned}$$

The problem of finding characterizations of orthogonality of an element to subspace ${\mathcal {B}}$ is closely related to the best approximation problems. A specific question is when is the zero vector a best approximation to x from ${\mathcal {B}}?$ This is the same as asking when is x orthogonal to ${\mathcal {B}}?$ Due to the Theorem 1.1, it has been proved in [5, 9] that for any elements x and y of $C^*$-algebra ${\mathcal {A}},$ 0 is a best approximation to x in ${\mathcal {B}}={\mathbb {C}}y$ if and only if there exists $f\in {\mathcal {S}}({\mathcal {A}})$ such that $f(x^*x)=\Vert x\Vert ^2$ and $f(x^*y)=0.$ Moreover, a generalized version of this fact has been proved in [13] for any element x and for any subspace ${\mathcal {B}}$ of ${\mathcal {A}}.$ As an application of Theorem 2.6, we present the following characterization of the best approximation for an element of ${\mathcal {A}}$ with respect to the norm $\Vert \cdot \Vert _a.$ To achieve this goal, we need the following nice result from [2].

Theorem 2.9

[2, Theorem 2.13] Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let a be a positive element of ${\mathcal {A}}.$ Let $f:a{\mathcal {A}}^a\rightarrow {\mathbb {C}}$ be a linear functional such that $f(a)=1$ and $| f(az) | \,\le \Vert z\Vert _a$ for all $z\in {\mathcal {A}}^a.$ Then there exists $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi (az)=f(az)$ for all $z\in {\mathcal {A}}^a.$

Theorem 2.10

Let ${\mathcal {A}}$ be a unital $C^*$-algebra,  $a \in {\mathcal {A}}$ be a positive and invertible element and let ${\mathcal {B}}$ be a subspace of ${\mathcal {A}}.$ Then $y_0 \in {\mathcal {B}}$ is a best approximation to an element $x \in {\mathcal {A}}$ with respect to $\Vert \cdot \Vert _a$ if and only if there exists $\varphi \in {\mathcal {S}}_{a} ({{\mathcal {A}}})$ such that

$$\begin{aligned} \varphi ( ( x - y_0)^{*} a (x - y_0 ) )= \Vert x - y_0 \Vert _{a}^{2} \end{aligned}$$

and

$$\begin{aligned} \varphi (x^* a y) = \varphi ( y_{0}^* a y )\quad (\forall y\in {\mathcal {B}}). \end{aligned}$$

Proof

If ${\mathcal {A}}$ is commutative, then the desired result immediately follows from [13, Theorem 1.1] and Corollary 2.3. Now, suppose that ${\mathcal {A}}$ is a noncommutative $C^*$-algebra and $y_0 \in {\mathcal {B}}$ is a best approximation to x with respect to $\Vert \cdot \Vert _a.$ Since

$$\begin{aligned} {\textrm{dis}}(x - y_0 , {\mathcal {B}}) = \inf _{y^{\prime }\in {\mathcal {B}}} \Vert x - y_0 - y^{\prime } \Vert _a = \inf _{y^{\prime }\in {\mathcal {B}}} \Vert x - (y_0 + y^{\prime }) \Vert _a = {\textrm{dis}} (x , {\mathcal {B}}), \end{aligned}$$

without loss of generality, we may assume that $y_0 = 0 .$

Now, suppose that $x\in {\mathcal {A}}$ and there exists $\varphi \in {\mathcal {S}}_{a} ({\mathcal {A}})$ such that $\varphi (x^* a x) = \Vert x\Vert _{a}^{2}$ and $\varphi (x^* a y) = 0$ for all $y \in {\mathcal {B}} .$ By Theorem 2.6 and homogeneity of a-Birkhoff–James orthogonality, we conclude that $\Vert x - \lambda y \Vert _a \, \ge \Vert x \Vert _a$ for all $y \in {\mathcal {B}}$ and all $\lambda \in {\mathbb {C}} .$ Hence

$$\begin{aligned} \Vert x\Vert _a = \inf _{ y \in {\mathcal {B}}} \Vert x - y \Vert _a = {\textrm{dis}} ( x , {\mathcal {B}} ). \end{aligned}$$

Therefore $y_0 =0$ is a best approximation to x in ${\mathcal {B}} .$

Conversely, suppose that $y_0=0$ is a best approximation to x in ${\mathcal {B}} .$ Then we have $\Vert x\Vert _a \, \le \Vert x+ \lambda y\Vert _a$ for all $y \in {\mathcal {B}}$ and all $\lambda \in {\mathbb {C}} .$ Theorem 2.6 tells us for each $y \in {\mathcal {B}}$ there exists $\varphi _{y} \in {\mathcal {S}}_{a} ({\mathcal {A}})$ such that $\varphi _{y} (x^* a x) = \Vert x\Vert _{a}^{2}$ and $\varphi _{y} ( x^* a y ) =0 .$

Let ${\mathcal {M}}=\{ \alpha x^* a x + \beta a + x^* a y \, : \, \alpha , \beta \in {\mathbb {C}}, \, y \in {\mathcal {B}} \}$ be a subspace of ${\mathcal {A}}$ generated by $x^* a x,$ a and $x^* a {\mathcal {B}} .$ Since a is invertible, it is known that x has a unique decomposition $x = x_1 + i x_2$ such that $x_1$ and $x_2$ are a-selfadjoint. In fact, $x_1 = \dfrac{x+x^\sharp }{2}$ and $x_2= \dfrac{x-x^\sharp }{2i}.$ Hence

$$\begin{aligned} {\mathcal {M}}&=\{ \alpha (x_1+i x_2)^* a (x_1+i x_2) + \beta a + (x_1+i x_2)^* a y \, : \, \alpha , \beta \in {\mathbb {C}}, \, y \in {\mathcal {B}} \} \\&= \{ \alpha (x_1 ^* a - i x_2^* a) (x_1+i x_2) + \beta a + (x_1 ^* a - i x_2^* a) y \, : \, \alpha , \beta \in {\mathbb {C}}, \, y \in {\mathcal {B}} \} \\&=\{ \alpha (a x_1 - i a x_2) (x_1+i x_2) + \beta a + (a x_1 - i a x_2) y \, : \, \alpha , \beta \in {\mathbb {C}}, \, y \in {\mathcal {B}} \} \\&=\{ a\left( \alpha ( x_1 - i x_2) (x_1+i x_2) + \beta 1_{{\mathcal {A}}} +(x_1 - i x_2) y\right) \, : \, \alpha , \beta \in {\mathbb {C}}, \, y \in {\mathcal {B}} \}. \end{aligned}$$

Define the mapping $\psi : {\mathcal {M}} \rightarrow {\mathbb {C}}$ by

$$\begin{aligned} \psi (\alpha x^* a x + \beta a + x^* a y) = \alpha \Vert x\Vert _{a}^{2} + \beta . \end{aligned}$$

Clearly $\psi$ is a linear mapping. To show that $\psi$ is well defined, it is enough to prove that if $\alpha x^* a x + \beta a + x^* a y = 0 ,$ then $\psi (\alpha x^* a x + \beta a + x^* a y)=0 .$ Note that for each $\alpha ,\beta \in {\mathbb {C}}$ and any $y \in {\mathcal {B}} ,$ we have $\varphi _y ( \alpha x^* a x + \beta a + x^* a y) = \alpha \Vert x\Vert _{a}^{2} +\beta ,$ since $\varphi _y (a)=1 ,$ $\varphi _y (x^* ax)=\Vert x\Vert _{a}^2$ and $\varphi _y ( x^* a y)=0 .$ Now, let $u(\alpha ,\beta ,y)=\alpha (x_1 - i x_2)(x_1 + i x_2 ) + \beta 1_{{\mathcal {A}}} + (x_1 - i x_2) y$ for all $\alpha ,\beta \in {\mathbb {C}}$ and all $y\in {\mathcal {B}}.$ Then by the Cauchy-Schwartz inequality, we get

$$\begin{aligned}&| \psi (\alpha x^* a x + \beta a + x^* a y) | = |\, \alpha \, \Vert x\Vert _{a}^{2} + \beta \, | =| \varphi _{y} (\alpha x^*ax + \beta a + x^*a y) \nonumber |\\&\quad =| \varphi _{y} (a u(\alpha ,\beta ,y)) |=| \varphi _{y} (a^{\frac{1}{2}}a^{\frac{1}{2}} u(\alpha ,\beta ,y)) | \nonumber \\&\quad \le \sqrt{\varphi _{y}(a)} \sqrt{\varphi _{y} (u(\alpha ,\beta ,y)^* a \, u(\alpha ,\beta ,y)) } \nonumber \\&\quad =\sqrt{\varphi _{y} (u(\alpha ,\beta ,y)^* a \, u(\alpha ,\beta ,y)) } \nonumber \\&\quad \le \Vert u(\alpha ,\beta ,y)\Vert = \Vert \alpha (x_1 - i x_2)(x_1 + i x_2) + \beta 1_{{\mathcal {A}}} + (x_1 - i x_2)y \Vert _{a} . \end{aligned}$$

(2.8)

If $\alpha x^* a x + \beta a + x^* a y=0 ,$ then $a(\alpha (x_1 - i x_2)(x_1 + i x_2) + \beta 1_{{\mathcal {A}}} + (x_1 - i x_2)y) =0,$ and so $\Vert \alpha (x_1 - i x_2)(x_1 + i x_2) + \beta 1_{{\mathcal {A}}} + (x_1 - i x_2)y\Vert _a =0 .$ Thus $\psi (\alpha x^* a x + \beta a + x^* a y) = \psi (0) = 0 ,$ by (2.8).

Define ${\mathcal {N}}:a{\mathcal {A}}\rightarrow [0,\infty )$ by ${\mathcal {N}} (az)= \Vert z \Vert _a$ for all $z\in {\mathcal {A}}$ and note that ${\mathcal {N}}$ is a norm on $a{\mathcal {A}} .$ Moreover, (2.8) follows that

$$\begin{aligned} | \psi (\alpha x^* a x + \beta a + x^* a y ) |&\le {\mathcal {N}} (a (\alpha (x_1 - i x_2)(x_1 + i x_2) + \beta 1_{{\mathcal {A}}} + (x_1 - i x_2)y) )\\&= {\mathcal {N}} ( \alpha x^* a x + \beta a + x^* a y ). \end{aligned}$$

Hence $\Vert \psi \Vert \, \le \, 1$ with respect to the norm ${\mathcal {N}} ,$ and therefore $\psi :({\mathcal {M}},{\mathcal {N}}(\cdot ) ) \subseteq a {\mathcal {A}} \rightarrow {\mathbb {C}}$ is a bounded linear functional. The Hahn–Banach Theorem tells us $\psi$ can be extend to a linear functional $f : a {\mathcal {A}} \rightarrow {\mathbb {C}}$ such that $\Vert f \Vert = \Vert \psi \Vert \, \le \, 1 ,$ $f |_{({\mathcal {M}} , {\mathcal {N}}(\cdot ) )} = \psi$ and $f (a) =1 .$ In addition,

$$\begin{aligned} |f(az)| \, \le \, \Vert f\Vert \,{\mathcal {N}}(a z ) \, \le \, {\mathcal {N}}(a z ) = \Vert z \Vert _a \quad ( \forall z \in {\mathcal {A}}). \end{aligned}$$

Taking the above considerations into account, by Theorem 2.9 one can find $\varphi \in {\mathcal {S}}_{a} ({\mathcal {A}})$ such that $\varphi ( az) =f(az)$ for all $z\in {\mathcal {A}}.$ Therefore, there exists $\varphi \in {\mathcal {S}}_{a} ({\mathcal {A}})$ such that

$$\begin{aligned} \varphi (x^* ax)&= \varphi ( (x_1 + i x_2 )^* a (x_1 + i x_2) ) = \varphi (a (x_{1} - i x_{2} ) (x_1 + i x_2) ) \\&= f (a (x_{1} - i x_{2} ) (x_1 + i x_2)) = f ((x_1 + i x_2 )^* a (x_1 + i x_2)) \\&= f(x^* ax) = \psi (x^*ax)=\Vert x\Vert ^2_a , \end{aligned}$$

and

$$\begin{aligned} \varphi (x^* ay)&=\varphi ((x_1 + i x_2 )^* ay)=\varphi (a(x_1 - i x_2 ) y)=f (a(x_1 - i x_2 ) y)\\&=f ((x_1 + i x_2 )^* ay ) =f(x^*ay)=\psi (x^*ay)=0\quad (\forall y\in {\mathcal {B}}). \end{aligned}$$

$\square$

As a direct consequence of Theorem 2.10, we get the following characterization of a-Birkhoff–James orthogonality to a subspace in a unital $C^*$-algebra.

Corollary 2.11

Let ${\mathcal {B}}$ be a subspace of a unital $C^*$-algebra ${\mathcal {A}}$ and let x be an element of ${\mathcal {A}}.$ Then x is a-Birkhoff–James orthogonal to ${\mathcal {B}}$ if and only if there is $\varphi \in {\mathcal {S}}_{a} ({\mathcal {A}})$ such that $\varphi (x^*ax)=\Vert x\Vert ^2_a$ and $\varphi (x^*ay)=0$ for all $y\in {\mathcal {B}}.$

The next result present a generalization of the well-known distance formula which obtained by Williams in [19].

Corollary 2.12

Let ${\mathcal {A}}$ be a unital $C^*$-algebra,  $a\in {\mathcal {A}}$ be a positive and invertible element and let $x \in {\mathcal {A}} .$ Then

$$\begin{aligned} {\textrm{dist}}^{2} (x, {\mathbb {C}}1_{{\mathcal {A}}})=\min _{\lambda \in {\mathbb {C}}} \Vert x - \lambda 1_{{\mathcal {A}}} \Vert _{a}^2 = \max \{ \varphi ( x^* ax ) - | \varphi (ax) |^2 \, : \, \varphi \in {\mathcal {S}}_a ({\mathcal {A}}) \}. \end{aligned}$$

Proof

Let $\alpha \in {\mathbb {C}}$ be such that $\Vert x - \alpha 1_{{\mathcal {A}}} \Vert _{a} = {\textrm{dist}} ( x , {\mathbb {C}} 1_{{\mathcal {A}}} ) .$ For any $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi (a x)=\alpha ,$ we have

$$\begin{aligned} \varphi ( x^* ax) - | \varphi ( ax)|^2 = \varphi ((x - \alpha 1_{{\mathcal {A}}})^* a (x - \alpha 1_{{\mathcal {A}}})) \le \Vert x - \alpha 1_{{\mathcal {A}}} \Vert _{a}^2 = {\textrm{dist}}^{2} (x, {\mathbb {C}}1_{{\mathcal {A}}}). \end{aligned}$$

Hence

$$\begin{aligned} \max \{ \varphi ( x^* ax) - | \varphi ( ax)|^2 \, : \, \varphi \in {\mathcal {S}}_a ({\mathcal {A}}) \} \le {\textrm{dist}}^{2} (x, {\mathbb {C}}1_{{\mathcal {A}}}) = \min _{\lambda \in {\mathbb {C}}} \Vert x - \lambda 1_{{\mathcal {A}}} \Vert _{a}^2 . \end{aligned}$$

On the other hand, by Theorem 2.10, there is $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that

$$\begin{aligned} \varphi ((x- \alpha 1_{{\mathcal {A}}} )^* a (x- \alpha 1_{{\mathcal {A}}})) = \Vert x- \alpha 1_{{\mathcal {A}}} \Vert _{a}^2 \quad \text {and} \quad \varphi (ax) = \alpha . \end{aligned}$$

Therefore $\varphi ( x^* ax) - | \varphi ( ax)|^2 = {\textrm{dist}}^{2} (x, {\mathbb {C}}1_{{\mathcal {A}}}) ,$ and so

$$\begin{aligned} \max \{\varphi ( x^* ax) - | \varphi ( ax)|^2 \, : \, \varphi \in {\mathcal {S}}_a ({\mathcal {A}}) \} \ge {\textrm{dist}}^{2} (x, {\mathbb {C}}1_{{\mathcal {A}}}). \end{aligned}$$

$\square$

3 Strong a-Birkhoff–James orthogonality in $C^*$-algebras

Our main goal in this section is to introduce and study the notion of strong Birkhoff–James orthogonality with respect to the norm $\Vert \cdot \Vert _a$ in unital $C^*$-algebras. It should be noted that what is obtained in this section is an extension and modification of some results of [4, 6]. We start this section with introducing the concept of strong a-Birkhoff–James orthogonality.

Definition 3.1

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and $a \in {\mathcal {A}}$ be a positive and invertible element. An element $x \in {\mathcal {A}}$ is said to be strongly a-Birkhoff–James orthogonal to an element $y \in {\mathcal {A}} ,$ in short $x \perp _{S-BJ}^{a} y ,$ if

$$\begin{aligned} \Vert x + y b \Vert _{a} \ge \Vert x \Vert _{a}\quad (\forall b \in {\mathcal {A}}). \end{aligned}$$

Obviously, $x \perp _{S-BJ}^{a} y$ implies $x \perp _{BJ}^{a} y$ for all $x,y\in {\mathcal {A}}.$ So for every $x,y \in {\mathcal {A}} ,$ we obtain:

$$\begin{aligned} \langle x , y \rangle _a :=x^*ay=0 \, \Rightarrow \, x \perp _{S-BJ}^{a} y \, \Rightarrow \, x \perp _{BJ}^{a} y. \end{aligned}$$

(3.1)

Indeed, if $\langle x, y \rangle _a =0 ,$ then for each $b \in {\mathcal {A}} ,$ we have

$$\begin{aligned} \Vert x + yb \Vert _{a}^{2}&= \sup _{\varphi \in {\mathcal {S}}_a ({\mathcal {A}})} \varphi ((x+yb)^* a (x+yb)) \\&=\sup _{\varphi \in {\mathcal {S}}_a ({\mathcal {A}})} \big ( \varphi (\langle x,x\rangle _a ) + 2{\textrm{Re}}\varphi (\langle x,y\rangle _a b ) + \varphi (\langle yb,yb \rangle _a ) \big ) \\&=\sup _{\varphi \in {\mathcal {S}}_a ({\mathcal {A}})} \big ( \varphi (\langle x,x\rangle _a ) + \varphi (\langle yb , yb \rangle _a )\big ) \, \ge \, \sup _{\varphi \in {\mathcal {S}}_a ({\mathcal {A}})} \varphi ( \langle x,x\rangle _a )= \Vert x\Vert _{a}^{2} . \end{aligned}$$

Also, note that

$$\begin{aligned} x \perp _{S-BJ}^{a} y \, \Leftrightarrow \, x \perp _{BJ}^{a} yb \quad (\forall \, b \in {\mathcal {A}}). \end{aligned}$$

(3.2)

The converses in (3.1) do not hold in general. The following example explains this fact.

Example 3.2

Let $a=\left[ \begin{array}{cccc} 2&{}0 \\ 0&{}1 \end{array}\right] .$ If $x=I_2$ and $y=\left[ \begin{array}{cccc} 0&{}1 \\ 1&{}0 \end{array}\right] ,$ then for every $\lambda \in {\mathbb {C}} ,$ we have

$$\begin{aligned} \Vert x + \lambda y \Vert _{a}^{2}&=\sup _{h\in {\mathcal {L}}_a}\varphi _h((x+\lambda y)^*a(x+\lambda y))\\&=\sup _{h\in {\mathcal {L}}_a} \big ( (2+|\lambda |^2 ) h_{11} + 2 \textrm{Re }((2{\overline{\lambda }}+\lambda ) h_{12}) +(2|\lambda |^2 +1) h_{22} \big ) \\&\ge 1+\dfrac{|\lambda |^2}{2}\ge 1=\Vert x\Vert ^2_a, \end{aligned}$$

since $h_0=\left[ \begin{array}{cccc} \frac{1}{2} &{}0 \\ 0&{}0 \end{array}\right] \in {\mathcal {L}}_a .$ Hence $x\perp ^a_{BJ}y.$ But, we may easily check that $x \not \perp _{S-BJ}^a y .$ To this end, note that for $b=\left[ \begin{array}{cccc} \frac{1}{2}&{}-\frac{1}{2} \\ -\frac{1}{2} &{}-\frac{1}{4} \end{array}\right] ,$ we get

$$\begin{aligned} \Vert x + yb \Vert _{a}^{2}&= \sup _{h \in {\mathcal {L}}_a } {\textrm{Tr}} (h (x +yb)^* a (x+yb)) \\&=\sup _{h \in {\mathcal {L}}_a } {\textrm{Tr}} \left( \left[ \begin{array}{cccc} h_{11} &{}h_{12} \\ \overline{h}_{12}&{}h_{22} \end{array}\right] \left[ \begin{array}{cccc} \frac{3}{4} &{} 0 \\ 0 &{} \frac{3}{8} \end{array}\right] \right) \\&=\sup _{2h_{11}+h_{22}=1, h_{11},h_{22} \ge 0}\left( \dfrac{3}{4}h_{11} +\dfrac{3}{8}h_{22} \right) = \dfrac{3}{8} <1= \Vert x\Vert _{a}^2. \end{aligned}$$

Now, let $x=I_2$ and $y=\left[ \begin{array}{cccc} 0&{}0 \\ 1&{}0 \end{array}\right] .$ If $b=\left[ \begin{array}{cccc} b_{11}&{}b_{12} \\ b_{21}&{}b_{22} \end{array}\right] \in {\mathbb {M}}_2({\mathbb {C}})$ is arbitrary, then

$$\begin{aligned} \Vert x + yb \Vert _{a}^{2}&=\sup _{h\in {\mathcal {L}}_a}\varphi _h( (x +yb)^* a (x+yb))\\&=\sup _{h \in {\mathcal {L}}_a} \big ((2+|b_{11}|^2) h_{11} +2{\textrm{Re}}( {b}_{11}(1+\overline{b}_{12}) h_{12}) +|1+b_{12} |^2 h_{22} \big ) \\&\ge 1+\dfrac{|b_{11}|^2}{2}\ge 1=\Vert x\Vert ^2_a. \end{aligned}$$

So $x \perp _{S-BJ}^{a} y ,$ while clearly, $\langle x,y \rangle _a = \left[ \begin{array}{cccc} 0 &{} 0 \\ 1&{} 0 \end{array}\right] \ne 0 .$

Our next result gives us a characterization of strong a-Birkhoff–James orthogonality based on elements of generalized state space ${\mathcal {S}}_a({\mathcal {A}})$ of unital $C^*$-algebra ${\mathcal {A}}.$ Actually, this result extend Theorem 2.5 of [4] for the norm $\Vert \cdot \Vert _a$ on ${\mathcal {A}}.$

Theorem 3.3

Let ${\mathcal {A}}$ be a unital $C^*$-algebra,  $x , y \in {\mathcal {A}}$ and let $a \in {\mathcal {A}}$ be a positive and invertible element. Then the following statements are equivalent : 

• (1) $x \perp _{S-BJ}^{a} y ;$

• (2) $x \perp _{BJ}^{a} y\langle y , x \rangle _a ;$

• (3) There is $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi ( x^* ax) = \Vert x\Vert _{a}^{2}$ and $\varphi ( \langle x, y \rangle _a \langle y , x\rangle _a ) =0 ;$

• (4) There is $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi ( x^* ax) = \Vert x\Vert _{a}^{2}$ and $\varphi ( \langle x, y \rangle _a b ) =0 ,$ for all $b \in {\mathcal {A}} .$

Proof

$(1) \Rightarrow (2 )$ If $x \perp _{S-BJ}^{a} y,$ then $x\perp _{BJ}^ayb$ for all $b \in {\mathcal {A}} ,$ by (3.2). Now, let $b=\langle y , x \rangle _a .$ Then $x \perp _{BJ}^{a} y\langle y , x \rangle _a .$

$(2) \Rightarrow (3)$ If $x \perp _{BJ}^{a} y\langle y , x \rangle _a ,$ then it follows from Theorem 2.6 that there is $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi ( x^* ax) = \Vert x\Vert _{a}^{2}$ and $\varphi ( \langle x , y \rangle _a \langle y , x\rangle _a )= \varphi (\langle x , y \langle y , x\rangle _a \rangle _a ) = 0 .$

$(3) \Rightarrow (4)$ If there exists $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi ( x^* ax) = \Vert x\Vert _{a}^{2}$ and $\varphi ( \langle x, y \rangle _a \langle y , x\rangle _a ) =0 ,$ then by the Cauchy–Schwarz inequality, we have

$$\begin{aligned} | \varphi ( (x^* a y ) b ) |^2 \, \le \, \varphi ( (x^* a y )(y^* a x )) \, \varphi ( b^* b) = 0, \quad (\forall b \in {\mathcal {A}}), \end{aligned}$$

which follows that $\varphi ( \langle x, y \rangle _a b ) =0$ for all $b \in {\mathcal {A}} .$

$(4) \Rightarrow (1 )$ It follows directly from Theorem 2.6 and the definition of strong a-Birkhoff–James orthogonality. $\square$

Proposition 3.4

Let ${\mathcal {A}}$ be a unital $C^*$-algebra, $x , y \in {\mathcal {A}}$ and let $a \in {\mathcal {A}}$ be a positive and invertible element. If $\langle x , y \rangle _a \, \ge \, 0 ,$ then

$$\begin{aligned} x \perp _{S-BJ}^{a} y\Leftrightarrow x \perp _{BJ}^{a} y. \end{aligned}$$

Proof

Assume that $x \perp _{BJ}^{a} y .$ By Theorem 2.6, there exists $\varphi \in {\mathcal {S}}_{a} ({\mathcal {A}})$ such that $\varphi ( \langle x ,x \rangle _a ) = \Vert x\Vert _{a}^{2}$ and $\varphi ( \langle x , y \rangle _a )=0 .$ Since $\langle x , y \rangle _a \, \ge \, 0 ,$ by the Cauchy–Schwarz inequality, for every $b \in {\mathcal {A}} ,$ we get

$$\begin{aligned} | \varphi ( \langle x ,y \rangle _a b ) |^2&= | \varphi ( \langle x ,y \rangle _{a}^{\frac{1}{2}} \langle x ,y \rangle _{a}^{\frac{1}{2} } b ) |^2 \\&\le \varphi ( \langle x ,y \rangle _{a}^{\frac{1}{2}} \langle x ,y \rangle _{a}^{\frac{1}{2}} ) \, \varphi ( b^* \langle x ,y \rangle _{a}^{\frac{1}{2}} \langle x ,y \rangle _{a}^{\frac{1}{2}} b) \\&\le \varphi ( \langle x ,y \rangle _{a} )\, \varphi ( b ^* \langle x ,y \rangle _{a} b ) =0 . \end{aligned}$$

Thus $\varphi ( \langle x ,y \rangle _ab ) =0$ for all $b\in {\mathcal {A}}.$ Therefore, Theorem 3.3 shows that $x \perp _{S-BJ}^{a} y .$ $\square$

Theorem 3.5

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let $a \in {\mathcal {A}}$ be positive and invertible. If

$$\begin{aligned} x \perp _{S-BJ}^{a} y \Leftrightarrow x \perp _{BJ}^{a} y\quad (\forall x,y\in {\mathcal {A}}), \end{aligned}$$

then the $C^*$-algebra ${\mathcal {A}}$ is commutative.

Proof

First, note that ${\mathcal {A}}^a={\mathcal {A}},$ since a is invertible. We shall show that for every $x, b \in {\mathcal {A}}$ there is a scalar $0\ne \alpha \in {\mathbb {C}}$ such that

$$\begin{aligned} x b \perp _{S-BJ}^{a} (xb^2 + \alpha x b). \end{aligned}$$

(3.3)

If $xb =0 ,$ obviously (3.3) holds. Now, let x be an element of ${\mathcal {A}}$ such that $xb \ne 0 .$ Then $xb \not \perp _{BJ}^{a} x .$ Indeed, if $xb \perp _{BJ}^{a} x ,$ then $xb \perp _{S-BJ}^{a} x ,$ by the assumption and thus $xb \perp _{BJ}^{a} xb ,$ by (3.2). It follows that $xb =0 ,$ which is not possible.

Moreover, by the definition of $\Vert \cdot \Vert _a$ and invertibility of a,  there is $\varphi \in {\mathcal {S}}_a({\mathcal {A}})$ such that $\varphi ( \langle xb , xb\rangle _a)=\Vert xb\Vert ^2_a.$ Hence by Theorem 2.6, we conclude that $\varphi (\langle xb , x \rangle _a) \ne 0 .$ Now, take $\alpha = \dfrac{- \Vert xb \Vert _a}{\varphi ( \langle xb , x\rangle _a)} .$ Thus

$$\begin{aligned} \varphi ( \langle xb , xb+ \alpha x \rangle _a ) = \Vert xb\Vert _{a}^{2} - \dfrac{\Vert xb\Vert _{a}^{2}}{\varphi (\langle xb , x \rangle _a)}\varphi (\langle xb , x \rangle _a ) =0 . \end{aligned}$$

The assumption and the Theorem 2.6 yields that $xb \perp _{S-BJ}^{a} (xb + \alpha x) .$ Hence $xb \perp _{BJ}^{a} (xb^2 + \alpha xb) ,$ by (3.2), and so

$$\begin{aligned} x b \perp _{S-BJ}^{a} (xb^2 + \alpha x b), \end{aligned}$$

by the hypothesis.

If ${\mathcal {A}}$ is not commutative, there will a nonzero $b\in {\mathcal {A}}$ with $b^2=0$ (see [12], p. 68). By (3.3), for $x=b^*$ there is a scalar $\alpha \ne 0$ such that $xb \perp _{S-BJ}^{a} \alpha xb .$ Hence $b^*b=xb=0,$ and so $b=0.$ This contradiction shows that ${\mathcal {A}}$ is commutative. $\square$

The next two results are direct consequences of Theorem 1.3, Corollary 2.3 and Theorem 3.5.

Corollary 3.6

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let $a \in {\mathcal {A}}$ be positive and invertible. The following statements are equivalent : 

• (1) For all $x,y \in {\mathcal {A}} ,$ $x \perp ^a_{BJ} y$ if and only if $\langle x,y \rangle _a =0 ;$

• (2) For all $x,y \in {\mathcal {A}} ,$ $x\perp ^a_{BJ} y$ if and only if $x \perp ^a_{S-BJ} y ;$

• (3) ${\mathcal {A}}$ is isomorphic to ${\mathbb {C}} .$

Corollary 3.7

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let $a \in {\mathcal {A}}$ be positive and invertible. If

$$\begin{aligned} x \perp _{S-BJ}^{a} y \Leftrightarrow \langle x,y\rangle _a =0\quad (\forall x,y \in {\mathcal {A}}), \end{aligned}$$

then ${\mathcal {Z}} ({\mathcal {A}}) \cong {\mathbb {C}} 1_{{\mathcal {A}}} .$

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and $a\in {\mathcal {A}}$ be positive and invertible. If $z \in {\mathcal {A}}$ is a noninvertible element of ${\mathcal {A}},$ then $zz^*a$ is not invertible, and so $0\in \sigma (zz^*a)=\sigma _a(zz^*a)\subseteq V_a(zz^*a),$ by [15, Remark 2.13 and Corollary 3.9]. Hence there exists $\varphi \in {\mathcal {S}}_a ({\mathcal {A}})$ such that $\varphi (\langle 1_{{\mathcal {A}}} , z \rangle _a \langle z , 1_{{\mathcal {A}}} \rangle _a )=\varphi (azz^*a)=0 .$ Also, we have $\varphi (1_{{\mathcal {A}}}^* a 1_{{\mathcal {A}}}) = \varphi (a)=1= \Vert 1_{\mathcal {A}}\Vert _{a}^2 .$ Consequently, Theorem 3.3 implies that

$$\begin{aligned} 1_{{\mathcal {A}}} \perp _{S-BJ}^a z,\quad \text{ and } \text{ so }\quad 1_{{\mathcal {A}}} \perp _{BJ}^a z. \end{aligned}$$

(3.4)

It has been shown in [7] that the left-additivity (right-additivity) of the (strong) Birkhoff–James orthogonality on a unital $C^*$-algebra implies that ${\mathcal {A}}$ is isomorphic to ${\mathbb {C}}1_{{\mathcal {A}}}.$ As a final result of this section, we will prove that if the (strong) a-Birkhoff–James orthogonality is right-additive on a unital $C^*$-algebra ${\mathcal {A}},$ then the center of ${\mathcal {A}}$ is trivial; i.e., ${\mathcal {Z}}({\mathcal {A}}) \cong {\mathbb {C}}1_{{\mathcal {A}}} .$

Theorem 3.8

Let ${\mathcal {A}}$ be a unital $C^*$-algebra and let a be a positive and invertible element of ${\mathcal {A}}.$ If (strong) a-Birkhoff–James orthogonality is right-additive on ${\mathcal {A}},$ then ${\mathcal {Z}}({\mathcal {A}}) \cong {\mathbb {C}}1_{{\mathcal {A}}} .$

Proof

First, assume that ${\mathcal {A}}$ is commutative. Then ${\mathcal {Z}}({\mathcal {A}})={\mathcal {A}}\cong {\mathbb {C}}1_{{\mathcal {A}}},$ by Corollary 2.3 and [7, Remark 2.8]. Now, suppose that ${\mathcal {A}}$ is noncommutative and $x \in {{\mathcal {A}}}$ is a noninvertible element of ${\mathcal {A}}.$ Then by (3.4), $1_{{\mathcal {A}}} \perp ^a_{BJ} x^* x .$ If we assume that $ax=xa ,$ then (2.4) follows that $\Vert x^* x \Vert _{a}=\Vert x^* x \Vert .$ Hence $\Vert x^* x \Vert _{a} 1_{{\mathcal {A}}} - x^* x$ is not invertible, since

$$\begin{aligned} \Vert x^* x \Vert _{a} = \Vert x^* x \Vert \in \sigma (x^* x) . \end{aligned}$$

Thus $1_{{\mathcal {A}}} \perp _{BJ}^a (\Vert x^* x \Vert _{a} 1_{{\mathcal {A}}} - x^* x) .$ The right-additivity of a-Birkhoff–James orthogonality follows that $1_{{\mathcal {A}}} \perp _{BJ}^a \Vert x^* x \Vert _{a} 1_{{\mathcal {A}}} .$ So $\Vert x^* x \Vert _{a} = \Vert x \Vert _{a}^{2} =0 ,$ because of the non-degeneracy of a-Birkhoff–James orthogonality. Hence $x=0 .$ Therefore we have proved that every nonzero element of $C^*$-subalgebra ${\mathcal {Z}}({\mathcal {A}})$ is invertible, and so ${\mathcal {Z}}({\mathcal {A}}) \cong {\mathbb {C}}1_{{\mathcal {A}}}$ by the Gelfand–Mazur Theorem. A similar argument works for strong a-Birkhoff–James orthogonality. $\square$

Remark 3.9

Suppose that a-Birkhoff–James orthogonality is left-additive in unital $C^*$-algebra ${\mathcal {A}}$ and let $x\in {\mathcal {A}}$ be positive and noninvertible such that $xa=ax.$ Then the $C^*$-subalgebra, ${\mathcal {B}}:=C^*(1_{{\mathcal {A}}},a,x),$ generated by $1_{{\mathcal {A}}},$ a and x is commutative. According to the Corollary 2.3, Birkhoff–James orthogonality is left-additive on ${\mathcal {B}}.$ Hence $x=0,$ by [7, Remark 2.8]. It follows that every nonzero element of ${\mathcal {Z}}({\mathcal {A}})$ is invertible, and so ${\mathcal {Z}}({\mathcal {A}})$ is trivial. It should be noted that the same proof works for right-additivity of a-Birkhoff–James orthogonality. However, a different approach is presented to study right-additivity in the previous Theorem.