Orthogonality and norm attainment of operators in semi-Hilbertian spaces

Abstract

We study the semi-Hilbertian structure induced by a positive operator A on a Hilbert space ${\mathbb {H}}.$ Restricting our attention to $A-$bounded positive operators, we characterize the norm attainment set and also investigate the corresponding compactness property. We obtain a complete characterization of the $A-$Birkhoff–James orthogonality of $A-$bounded operators under an additional boundedness condition. This extends the finite-dimensional Bhatia-$\breve{S}$emrl Theorem verbatim to the infinite-dimensional setting.

1 Introduction

The purpose of the paper was to explore the orthogonality and the norm attainment of bounded linear operators in the context of semi-Hilbertian structure induced by positive operators on a Hilbert space. Such a study was initiated by Krein in [10] and it remains an active and productive area of research till date. We refer the readers to [2, 3, 8, 18] and the references therein for more information on this. Let us now mention the relevant notations and the terminologies to be used in the article.

We use the symbol ${\mathbb {H}}$ to denote a Hilbert space. Finite-dimensional Hilbert spaces are also known as Euclidean spaces. Unless mentioned specifically, we work with both real and complex Hilbert spaces. The scalar field is denoted by ${\mathbb {K}},$ which can be either ${\mathbb {R}}$ or ${\mathbb {C}}.$ The underlying inner product and the corresponding norm on ${\mathbb {H}}$ are denoted by $\langle ~,~ \rangle$ and $\Vert \cdot \Vert ,$ respectively. In general, inner products on ${\mathbb {H}}$ are defined as positive definite, conjugate symmetric forms which are linear in the first argument. It should be noted that apart from the underlying inner product $\langle ~,~\rangle$ on ${\mathbb {H}},$ there may be many other inner products defined on ${\mathbb {H}},$ generating different norms. In order to avoid any confusion, whenever we talk of a topological concept on ${\mathbb {H}},$ we explicitly mention the norm that generates the corresponding topology. Let $B_{{\mathbb {H}}} = \left\{ x \in {\mathbb {H}} : \Vert x\Vert \le 1\right\}$ and $S_{{\mathbb {H}}} = \left\{ x \in {\mathbb {H}} : \Vert x\Vert = 1\right\}$ be the unit ball and the unit sphere of ${\mathbb {H}},$ respectively. We use the symbol $\theta$ to denote the zero vector of any Hilbert space other than the scalar fields ${\mathbb {R}}$ and ${\mathbb {C}}.$ For any complex number z,  Re(z) and Im(z) denote the real part and the complex part of z,  respectively. For any set $G \subset {\mathbb {H}},$ ${\overline{G}}$ denotes the norm closure of G. Let ${\mathbb {L}}({\mathbb {H}})({\mathbb {K}}({\mathbb {H}}))$ denote the Banach space of all bounded (compact) linear operators on ${\mathbb {H}}$, endowed with the usual operator norm. Given any $A \in {\mathbb {L}}({\mathbb {H}}),$ we denote the null space of A by N(A) and the range space of A by R(A). The symbol I is used to denote the identity operator on ${\mathbb {H}}.$ For $A \in {\mathbb {L}}({\mathbb {H}}),$ $A^*$ denotes the Hilbert adjoint of A. An operator $A \in {\mathbb {L}}({\mathbb {H}})$ can be represented as $A = Re A + i Im A,$ where $Re A = \frac{1}{2}(A + A^*)$ and $Im A =\frac{1}{2i}(A - A^*).$ Recall that $A \in {\mathbb {L}}({\mathbb {H}})$ is said to be a positive operator if $A = A^*$ and $\langle Ax,x\rangle \ge 0$ for all $x \in {\mathbb {H}}$. A positive operator A is said to be positive definite if $\langle Ax,x\rangle > 0$ for all $x \in {\mathbb {H}} \setminus \{\theta \}$. It is well known [2] that any positive operator $A \in {\mathbb {L}}({\mathbb {H}})$ induces a positive semi-definite sesquilinear form $\langle ~,~\rangle _A$ on ${\mathbb {H}},$ given by $\langle x,y\rangle _A = \langle Ax,y\rangle ,$ where $x , y \in {\mathbb {H}}.$ It is easy to see that $\langle ~,~\rangle _A$ induces a semi-norm $\Vert \cdot \Vert _A$ on ${\mathbb {H}},$ given by $\Vert x \Vert _A = \sqrt{\langle Ax,x\rangle }.$ Moreover, when A is positive definite, it can be verified that $\langle ~,~\rangle _A$ is an inner product on ${\mathbb {H}}$ and $\Vert \cdot \Vert _A$ is a norm on ${\mathbb {H}}.$ In fact, given any $A \in {\mathbb {L}}({\mathbb {H}}),$ it is natural to ask when the functions $\langle ~,~\rangle _A$ and $\Vert \cdot \Vert _A,$ defined as above, are an inner product and a norm on ${\mathbb {H}},$ respectively. We explore this question and some related topics in the first part of our main results. We refer the readers to [1, 4, 7, 11] for some more interesting results in this direction.

Given a Hilbert space $({\mathbb {H}}, \Vert \cdot \Vert )$ and a positive $A \in {\mathbb {L}}({\mathbb {H}}),$ it is clear that $ker \Vert \cdot \Vert _A = \{x \in {\mathbb {H}} : \Vert x\Vert _A = 0\}$ is a closed linear subspace of ${\mathbb {H}}$. Then there is a closed linear subspace $W \subseteq {\mathbb {H}}$ such that $W\bot ker \Vert \cdot \Vert _A$ and ${\mathbb {H}} = W + ker \Vert \cdot \Vert _A.$ Let P be the linear projection on W such that $ker P = ker \Vert \cdot \Vert _A.$ Then it follows from [17] that $\Vert x\Vert _A = \Vert Px\Vert _A.$ In other words, the restriction of $\Vert \cdot \Vert _A$ to the subspace W is indeed a norm which satisfies the parallelogram property and so $(W, \Vert \cdot \Vert _A)$ is an inner product space. The investigations for the space ${\mathbb {H}}$ equipped with the seminorm $\Vert \cdot \Vert _A$ are very closely connected to the investigations for the inner product space $(W, \Vert \cdot \Vert _A).$ Furthermore, we consider $A-$bounded linear operator $T : {\mathbb {H}} \longrightarrow {\mathbb {H}}.$ Next, we define linear operator ${\hat{T}} : W \longrightarrow W$ by ${\hat{T}}(w) := T(w).$ Now, it is very easy to see that we can think of the $A-$norm on ${\mathbb {L}}({\mathbb {H}})$ as the classical operator norm in the operator space ${\mathbb {L}}(W)$. Of course, in this case, W is equipped with the norm $\Vert \cdot \Vert _A : W \longrightarrow [0,\infty ).$ Recently, Zamani [18] investigated the orthogonality relation induced by a positive linear operator on a Hilbert space and obtained some interesting results. In particular, he generalized Theorem 1.1 of [5], also known as the Bhatia-$\breve{S}$emrl Theorem, that characterizes the Birkhoff-James orthogonality of matrices on Euclidean spaces. Let us now recall some relevant definitions from [2] and [18].

Definition 1.1

Let ${\mathbb {H}}$ be a Hilbert space. Let $A \in {\mathbb {L}}({\mathbb {H}})$ be positive. An element $x \in {\mathbb {H}}$ is said to be $A-$orthogonal to an element $y \in {\mathbb {H}},$ denoted by $x \bot _A y,$ if $\langle x, y\rangle _A = 0.$

Note that if $A=I$, then the above definition coincides with the usual notion of orthogonality in Hilbert spaces.

Let $B_{A^{1/2}}({\mathbb {H}} )= \big \{ T \in {\mathbb {L}}({\mathbb {H}}) :~ \exists ~ c > 0 ~\text {such that}~\Vert Tx\Vert _A \le c\Vert x\Vert _A ~\forall x\in {\mathbb {H}} \big \}$. The $A-$norm of $T \in B_{A^{1/2}}({\mathbb {H}})$ is given as follows:

$$\begin{aligned} \Vert T\Vert _A = \sup _{x\in {\mathbb {H}}, \Vert x\Vert _A =1} \Vert Tx\Vert _A = \sup \left\{ |\langle Tx,y\rangle _A|: x,y \in {\mathbb {H}}, \Vert x\Vert _A= \Vert y\Vert _A =1\right\} . \end{aligned}$$

An operator $T \in {\mathbb {L}}({\mathbb {H}})$ is said to be $A-$bounded if $T \in B_{A^{1/2}}({\mathbb {H}} ).$

Definition 1.2

$T \in B_{A^{1/2}}({\mathbb {H}})$ is said to be $A-$Birkhoff–James orthogonal to $S \in B_{A^{1/2}}({\mathbb {H}}),$ denoted by $T\bot _A^B S,$ if $\Vert T+\gamma S\Vert _A \ge \Vert T\Vert _A~\text {for all }\gamma \in {\mathbb {C}}.$

Note that the above definition gives a generalization of the Birkhoff–James orthogonality of bounded linear operators on a Hilbert space. For more information on Birkhoff–James orthogonality in normed linear spaces, we refer the readers to the pioneering articles [6, 9]. Birkhoff–James orthogonality of bounded linear operators and some related applications have been explored in recent times in [5, 12, 13, 15, 16]. We also make use of the following notations:

Given a positive operator $A \in {\mathbb {L}}({\mathbb {H}}),$ let $B_{{\mathbb {H}}(A)}$ and $S_{{\mathbb {H}}(A)}$ denote the $A-$unit ball and the $A-$unit sphere of ${\mathbb {H}},$ respectively, i.e., $B_{{\mathbb {H}}(A)} = \left\{ x \in {\mathbb {H}} : \Vert x\Vert _A \le 1\right\}$ and $S_{{\mathbb {H}}(A)} = \left\{ x \in {\mathbb {H}} : \Vert x\Vert _A = 1\right\}$. For any $T \in B_{A^{1/2}}({\mathbb {H}}),$ the $A-$norm attainment set $M^T_A$ of T was considered in [18]:

$$\begin{aligned} M^T_A = \left\{ x \in {\mathbb {H}} : \Vert x\Vert _A = 1,~\Vert Tx\Vert _A = \Vert T\Vert _A \right\} . \end{aligned}$$

We study the structure of the $A-$norm attainment set of an $A-$bounded operator $T \in {\mathbb {L}}({\mathbb {H}})$ and also explore the corresponding compactness property. As the most important result of the present article, we obtain a complete characterization of the $A-$Birkhoff–James orthogonality of compact and $A-$bounded operators on ${\mathbb {H}}$ under an additional condition. This extends the Bhatia–$\breve{S}$emrl Theorem to the setting of semi-Hilbertian spaces, induced by a positive operator.

2 Main Results

We begin this section with a characterization of the norm-generating operators on a Hilbert space.

Theorem 2.1

Let ${\mathbb {H}}$ be a Hilbert space and let $A \in {\mathbb {L}}({\mathbb {H}}).$ Then $\Vert \cdot \Vert _A$ is a norm on ${\mathbb {H}}$ if and only if $\langle Ax,x \rangle > 0$ for all $x \in {\mathbb {H}} \setminus \{ \theta \}.$

Proof

As the necessary part of the theorem follows trivially, we only prove the sufficient part.

Clearly, $\Vert x+y\Vert _A ^2 = \Vert x\Vert _A^2 + \Vert y\Vert _A^2 + \langle Ax,y \rangle +\langle Ay,x \rangle .$ This shows that $\langle Ax,y \rangle +\langle Ay,x \rangle$ is real. It is easy to see that $Re \langle Ax,y \rangle + Re \langle Ay,x \rangle =\langle (Re A)x,y \rangle + \langle (Re A)y,x \rangle ,$ where $Re A = \frac{1}{2}(A + A^*).$

Clearly, $\Vert \cdot \Vert _A$ trivially satisfies all the properties for being a norm, except possibly the triangle inequality. The triangle inequality is satisfied if for all $x,y \in {\mathbb {H}},$

$$\begin{aligned} \Vert x+y\Vert _A\le & {} \Vert x\Vert _A + \Vert y\Vert _A \\ i.e., \ if, \langle A(x+y), x+y \rangle\le & {} \langle Ax,x \rangle + \langle Ay,y \rangle + 2 \langle Ax,x \rangle ^{1/2} \langle Ay,y \rangle ^{1/2}\\ i.e., \ if, \langle Ax,y \rangle + \langle Ay,x \rangle\le & {} 2 \langle Ax,x \rangle ^{1/2} \langle Ay,y \rangle ^{1/2}\\ i.e., \ if, Re \langle Ax,y \rangle + Re \langle Ay,x \rangle\le & {} 2 \langle Ax,x \rangle ^{1/2} \langle Ay,y \rangle ^{1/2}\\ i.e., \ if, \langle (Re A)x,y \rangle + \langle (Re A)y,x \rangle\le & {} 2 \langle Ax,x \rangle ^{1/2} \langle Ay,y \rangle ^{1/2}. \end{aligned}$$

Note that for all $x \in {\mathbb {H}}, \ \langle Re Ax, x \rangle = \frac{1}{2} ( \langle Ax, x \rangle + \langle A^*x,x \rangle \rangle ) = \frac{1}{2} ( \langle Ax, x \rangle +\overline{\langle Ax, x \rangle } ) = \langle Ax, x \rangle$. This proves that ReA is positive definite and so there exists a unique positive operator B on ${\mathbb {H}}$ such that $Re A = B^2.$ Now, we have

$$\begin{aligned} |\langle (Re A)x,y \rangle |= & {} |\langle B^2x,y \rangle | = |\langle Bx,By \rangle | = \Vert Bx\Vert \Vert By\Vert \\= & {} \langle B^2x,x \rangle ^{1/2} \langle B^2y,y \rangle ^{1/2}= \langle (Re A)x,x \rangle ^{1/2} \langle (Re A)y,y \rangle ^{1/2}\\= & {} \langle Ax,x \rangle ^{1/2} \langle Ay,y \rangle ^{1/2}. \end{aligned}$$

Similarly, we can show that $|\langle (Re A)y,x \rangle | \le \langle Ax,x \rangle ^{1/2} \langle Ay,y \rangle ^{1/2}.$ Therefore,

$$\begin{aligned} \langle (Re A)x,y \rangle + \langle (Re A)y,x \rangle \le |\langle (Re A)x,y \rangle + \langle (Re A)y,x \rangle | \le 2 \langle Ax,x \rangle ^{1/2} \langle Ay,y \rangle ^{1/2}. \end{aligned}$$

This completes the proof of the fact that $\Vert \cdot \Vert _A$ is a norm on ${\mathbb {H}}.$ $\square$

As mentioned in the introduction, if A is a positive definite operator on a Hilbert space ${\mathbb {H}},$ then A generates an inner product $\langle ~ , ~\rangle _A$ on ${\mathbb {H}}$ defined as $\langle x, y \rangle _A = \langle Ax, y \rangle$ for all $x, y \in {\mathbb {H}}.$ On the other hand, suppose that $A \in {\mathbb {L}}({\mathbb {H}})$ is such that $\langle x, y \rangle _A$ is an inner product on ${\mathbb {H}}.$ From the conjugate-symmetry of inner product, it follows that A must be self adjoint and from the positive definiteness of inner product, it follows that A must be positive definite. This is mentioned in the following proposition:

Proposition 2.1

Let ${\mathbb {H}}$ be a Hilbert space and let $A \in {\mathbb {L}}({\mathbb {H}}).$ Then $\langle ~,~ \rangle _A$ is an inner product on ${\mathbb {H}}$ if and only if A is positive definite.

Remark 2.1

In view of the above theorem, there is a subtle difference in the description of the norm generating operators, depending on whether the underlying Hilbert space is complex or real. This is illustrated in the following two points:

• 1. If ${\mathbb {H}}$ is a complex Hilbert space then $\langle ~,~\rangle _A$ and $\Vert \cdot \Vert _A$ are inner product and norm on ${\mathbb {H}}$, respectively, if and only if A is a positive definite operator on ${\mathbb {H}}$. This is because of the well-known fact that in case of a complex Hilbert space ${\mathbb {H}},$ if $A \in {\mathbb {L}}({\mathbb {H}})$ is such that $\langle Ax,x\rangle \ge 0$ for all $x \in {\mathbb {H}},$ then $A=A^*.$

• 2. If ${\mathbb {H}}$ is real, then there may exist $A \in {\mathbb {L}}({\mathbb {H}})$ such that $A \ne A^{*}$ (and consequently, A is not positive definite) but $\Vert \cdot \Vert _A$ is a norm on ${\mathbb {H}}.$ As for example, consider the operator A on the Hilbert space $\ell _2^2({\mathbb {R}})$ defined as $A(x,y)=(x-y,x+y)$ for all $(x,y) \in {\mathbb {R}}^2.$ Then it is easy to see that $\langle Ax, x \rangle > 0$ for all $x \ne \theta$ but $A \ne A^{*}$. A generates a norm given by $\Vert x \Vert _A = \langle Ax, x \rangle ^{1/2}$ on $\ell _2^2({\mathbb {R}})$ but $\langle x, y \rangle _A = \langle Ax, y \rangle$ is not an inner product on $\ell _2^2({\mathbb {R}})$. The inner product that induces the norm $\Vert \cdot \Vert _A$ is given by $\langle (Re A) x, y \rangle$. In fact, given any $A \in {\mathbb {L}}({\mathbb {H}})$ with $\langle Ax, x \rangle > 0$ for all $x \ne \theta ,$ the positive definite operator ReA always generates an inner product $\langle x, y \rangle _{Re A} = \langle (Re A) x, y \rangle$ which induces the norm $\Vert \cdot \Vert _A.$

Our next theorem guarantees that under a suitable condition, given any inner product on an infinite-dimensional separable Hilbert space ${\mathbb {H}}$, there exists a unique positive definite operator that generates the given inner product.

Theorem 2.2

Let $({\mathbb {H}}, \langle ~,~\rangle )$ be a separable Hilbert space. Let $\langle ~,~ \rangle _1$ be another inner product on ${\mathbb {H}}.$ Then the following two conditions are equivalent:

• (i) there exists a positive definite operator A on ${\mathbb {H}}$ such that $\langle ~,~ \rangle _1 = \langle ~,~ \rangle _A.$

• (ii) there exists $M >0$ such that $\Vert x\Vert _1 \le M \Vert x\Vert$ for all $x \in {\mathbb {H}},$ where $\Vert \cdot \Vert _1$ is the norm induced by the inner product $\langle ~,~ \rangle _1$ on ${\mathbb {H}}.$

Proof

(i$) \Rightarrow ($ii) :  Clearly, $\Vert x\Vert _1^2 = \langle x,x \rangle _1 = \langle x,x \rangle _A = \langle Ax,x \rangle \le \Vert A\Vert \Vert x\Vert ^2.$

(ii$) \Rightarrow ($i) :  Since $\Vert x\Vert _1 \le M\Vert x\Vert$ for all $x \in {\mathbb {H}}$, it follows that ${\mathbb {H}}$ is a separable inner product space with respect to $\langle ~,~ \rangle _1.$ Let $({\mathcal {H}}, \langle ~,~\rangle _{{\mathcal {H}}})$ be the completion of $({\mathbb {H}}, \langle ~,~ \rangle _1).$ Clearly, $\langle x,y\rangle _{{\mathcal {H}}}= \langle x,y\rangle _1$ for all $x,y \in {\mathbb {H}}.$ Since ${\mathbb {H}}$ is separable with respect to $\langle ~,~\rangle _1,$ it is easy to deduce that ${\mathcal {H}}$ is separable with respect to $\langle ~,~\rangle _{{\mathcal {H}}}.$ Let $B = \{e_1,e_2,e_3,\ldots \}$ be an orthonormal basis of ${\mathbb {H}}$ with respect to $\langle ~,~ \rangle$ and let $B_1 = \{f_1,f_2,f_3,\ldots \}$ be an orthonormal basis of ${\mathcal {H}}$ with respect to $\langle ~,~\rangle _{{\mathcal {H}}}$. Consider the map ${\widetilde{T}} : ({\mathcal {H}},\langle ~,~\rangle _{{\mathcal {H}}}) \rightarrow ({\mathbb {H}}, \langle ~,~ \rangle )$ defined by ${\widetilde{T}}(\sum _{i=1}^{\infty }a_i f_i) =\sum _{i=1}^{\infty }a_i e_i$ where $a_i \in {\mathbb {K}}(={\mathbb {R}},{\mathbb {C}})$ for all $i \in {\mathbb {N}}.$ It can be verified easily that ${\widetilde{T}}$ is well-defined and linear. Let $T ={\widetilde{T}}\mid _{({\mathbb {H}}, \langle ~,~\rangle _1)}.$ It is easy to see that $\langle x, y\rangle _1 = \langle Tx, Ty\rangle$ for all $x,y \in {\mathbb {H}}$. Thus $\Vert Tx\Vert ^2 = \langle x, x\rangle _1 \le M^2 \Vert x\Vert ^2.$ In particular, T is bounded and, therefore, the adjoint operator $T^* : ({\mathbb {H}},\langle ~,~\rangle ) \longrightarrow ({\mathbb {H}},\langle ~,~\rangle _1)$ exists. Let $A = T^*T$. Then it is easy to see that A is a positive definite operator on $({\mathbb {H}}, \langle ~,~\rangle )$ such that $\langle x, y \rangle _1 = \langle Ax,y \rangle$ for all $x,y \in {\mathbb {H}}.$

The uniqueness of A follows from the fact that if B is any positive definite operator that generates the inner product $\langle ~,~ \rangle _1$ then $\langle Ax, y \rangle = \langle Bx, y \rangle$ for all $x, y \in {\mathbb {H}}$ and so $A=B.$ $\square$

In light of the above theorem, let us make the following two remarks:

Remark 2.2

In case ${\mathbb {H}}$ is finite-dimensional, Condition (ii) of the above theorem holds true automatically. Therefore, we obtain a complete description of the set of all inner products defined on an Euclidean space, in terms of positive definite operators on ${\mathbb {H}}.$ Following the usual matricial representation of linear operators on Euclidean spaces, it seems convenient to say that every positive definite matrix defines an inner product on ${\mathbb {K}}^{n}$ and conversely.

Remark 2.3

We note that if $\langle ~, ~\rangle _1$ is an inner product on ${\mathbb {H}}$ such that Condition (ii) of the above theorem is satisfied, it is not necessarily true that $({\mathbb {H}}, \langle ~,~\rangle _1)$ is complete. Such an example will be constructed explicitly in the proof of Theorem 2.3 (iv).

The unit ball $B_{{\mathbb {H}}}$ is convex and bounded with respect to $\Vert \cdot \Vert$. Also, it is compact (in the topology induced by $\Vert \cdot \Vert$) if and only ${\mathbb {H}}$ is finite-dimensional. We next study some analogous geometric and topological properties of the $A-$unit ball $B_{{\mathbb {H}}(A)}$ with respect to the norm $\Vert \cdot \Vert$. We begin with the following proposition, the proof of which is omitted as it follows rather trivially from the convexity of the $A-$norm and the continuity of the inner product.

Proposition 2.2

Let ${\mathbb {H}}$ be a Hilbert space and let $A \in {\mathbb {L}}({\mathbb {H}})$ be positive. Then $B_{{\mathbb {H}}(A)}$ is convex and closed with respect to $\Vert \cdot \Vert$.

We would like to describe the boundedness properties of the $A-$unit ball and the $A-$unit sphere with respect to the norm $\Vert \cdot \Vert .$ We require the following proposition which is particularly useful in our study. The proof is omitted, as it can be obtained quite easily.

Proposition 2.3

Let ${\mathbb {H}}$ be a Hilbert space. Let $A \in {\mathbb {L}}({\mathbb {H}})$ be positive. Then ${\mathbb {H}} = N(A)\oplus \overline{R(A)}.$

We describe the boundedness properties of the $A-$unit ball and the $A-$unit sphere in the next theorem.

Theorem 2.3

Let ${\mathbb {H}}$ be a Hilbert space and let $A \in {\mathbb {L}}({\mathbb {H}})$ be positive. Then the following hold true:

• (i) If $N(A) \ne \{\theta \}$, then both $S_{{\mathbb {H}}(A)}$ and $B_{{\mathbb {H}}(A)}$ are unbounded with respect to $\Vert \cdot \Vert$.

• (ii) If ${\mathbb {H}}$ is finite-dimensional, then $B_{{\mathbb {H}}(A)}\cap \overline{R(A)}(=B_{{\mathbb {H}}(A)}\cap {R(A)})$ is bounded with respect to $\Vert \cdot \Vert$.

• (iii) If H is finite-dimensional, then $B_{{\mathbb {H}}(A)}$ is bounded with respect to $\Vert \cdot \Vert$ if and only if $N(A)=\{\theta \}.$

• (iv) Both (ii) and (iii) fail to hold if ${\mathbb {H}}$ is infinite-dimensional.

Proof

We first observe that

$$\begin{aligned}\Vert Ax\Vert ^2 =\langle Ax,Ax\rangle =\langle A^2x,x\rangle =\langle Ax,x\rangle _A \end{aligned}$$

and so it follows that $\Vert x\Vert _A =0$ if and only if $x \in N(A)$:

• (i) Let $x \in N(A)$ be such that $x \ne \theta .$ Then $\Vert x\Vert _A = 0$ and so $\Vert \lambda x\Vert _A = 0$ for all $\lambda \in {\mathbb {K}}(={\mathbb {R}},{\mathbb {C}}).$ Next we claim that if $z \in S_{{\mathbb {H}}(A)},$ then $z+\lambda x \in S_{{\mathbb {H}}(A)}$ for all $\lambda \in {\mathbb {K}}(={\mathbb {R}},{\mathbb {C}}).$ Clearly, $\Vert z+\lambda x\Vert _A \le \Vert z\Vert _A + |\lambda | \Vert x\Vert _A =1.$ Again, $\Vert z+\lambda x\Vert _A \ge \Vert z\Vert _A - |\lambda | \Vert x\Vert _A =1.$ Thus $z+\lambda x \in S_{{\mathbb {H}}(A)}$ for all $\lambda \in {\mathbb {K}}(={\mathbb {R}},{\mathbb {C}}).$ Therefore, $S_{{\mathbb {H}}(A)}$ is unbounded with respect to $\Vert \cdot \Vert$ and so $B_{{\mathbb {H}}(A)}$ is also unbounded with respect to $\Vert \cdot \Vert$.

• (ii) Suppose on the contrary that $B_{{\mathbb {H}}(A)}\cap {R(A)}$ is unbounded with respect to $\Vert \cdot \Vert$. Then for each $n\in {\mathbb {N}}$, there exists $v_n\in B_{{\mathbb {H}}(A)}\cap {R(A)}$ such that $\Vert v_n\Vert \ge n$. Let $w_n=\frac{v_n}{\Vert v_n\Vert }$. Then $\Vert w_n\Vert =1$ and $\Vert w_n\Vert _A\le \frac{1}{n}$. Clearly, $\{w_n\}\subseteq S_{\mathbb {H}}$. Since ${\mathbb {H}}$ is finite-dimensional, $S_{{\mathbb {H}}}$ is compact. Without loss of generality we may assume that $w_n\longrightarrow w,$ where $w\in S_{{\mathbb {H}}}$. By Proposition 2.2, $B_{{\mathbb {H}}(A)}\cap {R(A)}$ is a closed set with respect to $\Vert \cdot \Vert$ and hence $w\in B_{{\mathbb {H}}(A)}\cap {R(A)}$. It is easy to check that $\Vert w_n\Vert _A\longrightarrow \Vert w\Vert _A$. Therefore, $\Vert w\Vert _A=0$ and so $w\in N(A).$ This shows that $w \in N(A) \cap R(A)$ and so $w=\theta$, a contradiction to our assumption that $w\in S_{{\mathbb {H}}}$. Therefore, $B_{{\mathbb {H}}(A)}\cap {R(A)}$ is bounded with respect to $\Vert \cdot \Vert$.

• (iii) As ${\mathbb {H}}$ is finite-dimensional, ${\mathbb {H}} = N(A)\oplus R(A).$ Therefore, any $x \in B_{{\mathbb {H}}(A)}$ can be uniquely written as $x = u+v,$ where $u \in B_{{\mathbb {H}}(A)}\cap {N(A)}$ and $v \in B_{{\mathbb {H}}(A)}\cap R(A).$ From (ii), it follows that $B_{{\mathbb {H}}(A)}$ is bounded with respect to $\Vert \cdot \Vert$ if and only if $B_{{\mathbb {H}}(A)}\cap {N(A)}$ is bounded with respect to $\Vert \cdot \Vert$. From (i), it follows that $N(A)=\{\theta \}$ if $B_{{\mathbb {H}}(A)}$ is bounded with respect to $\Vert \cdot \Vert$. On the other hand, if $N(A)=\{\theta \}$ then $B_{{\mathbb {H}}(A)} = B_{{\mathbb {H}}(A)}\cap R(A)$ is bounded with respect to $\Vert \cdot \Vert$ by applying (ii).

• (iv) Consider the Hilbert space $\ell _2.$ Let $A \in {\mathbb {L}}(\ell _2)$ be defined by $A(x_1,x_2,x_3,\ldots ) = (x_1, \frac{x_2}{2},\frac{x_3}{3},\ldots ),$ where $(x_1,x_2,x_3,\ldots ) \in \ell _2.$ It is easy to check that A is positive definite and $N(A)=\{\theta \}.$ Therefore, $\ell _2=\overline{R(A)}(\ne R(A))$. Consider the sequence $\{v_n\} \subseteq \ell _2$ where $\{v_n\} = \{\sqrt{n}e_n\},$ where $\{e_n\}$ is the usual orthonormal basis of $\ell _2.$ Clearly, $\Vert v_n\Vert _A^2 = \langle Av_n,v_n\rangle = 1$ for each $n \in {\mathbb {N}}$ but $\Vert v_n\Vert = \sqrt{n}$ for each $n \in {\mathbb {N}}$.

$\square$

In view of the above theorem, we make the following remark on the geometry of semi-Hilbertian spaces.

Remark 2.4

Let A be a positive operator on a Hilbert space ${\mathbb {H}}.$ If $\Vert x\Vert _A =0$ for some $x \ne \theta ,$ then by (i) of Theorem 2.3, the $A-$unit sphere of ${\mathbb {H}}$ contains a straight line. In other words, the semi-normed space $({\mathbb {H}}, \Vert \cdot \Vert _A)$ is not strictly convex whenever A is not positive definite.

There is another nice way to obtain Remark 2.4. Namely, now suppose that A is positive, but not positive definite. Since $\Vert \cdot \Vert _A$ is a seminorm, it follows from Theorem 3.2 from [17] that any $x \in B_{{\mathbb {H}}(A)}$ can be uniquely written as $x = u+v$, where $u \in B_{W}$ and $v \in ker \Vert \cdot \Vert _A.$ Note that $B_{W}$ is the closed unit ball in the inner product space $(W, \Vert \cdot \Vert _A)$, and $ker \Vert \cdot \Vert _A$ is a linear subspace. Therefore, it is easy to see that $A-$unit sphere of ${\mathbb {H}}$ contains a straight line and the seminormed space $({\mathbb {H}}, \Vert \cdot \Vert _A)$ is not strictly convex whenever A is not positive definite.

In Theorem 2.2 of [14], the authors studied the norm attainment sets of bounded linear operators on a Hilbert space. In particular, it was proved that in an inner product space ${\mathbb {H}}$, for any operator $T \in {\mathbb {L}}({\mathbb {H}}),$ the norm attainment set $M_T$ is either the empty set $\phi ,$ or, $M_T$ is the unit sphere of some subspace of ${\mathbb {H}}.$ Our next result generalizes this, in case of $A-$bounded operators.

Theorem 2.4

Let ${\mathbb {H}}$ be a Hilbert space. Let $A \in {\mathbb {L}}({\mathbb {H}})$ be positive and let $T \in B_{A^{1/2}}({\mathbb {H}}).$ Then either $M_A^T = \phi$ or $M_A^T \cap \overline{R(A)}$ is the $A-$unit sphere of some subspace of ${\mathbb {H}}.$

Proof

If $M_A^T = \phi ,$ then we have nothing to prove. Let us assume that $M_A^T \ne \phi .$ Let $x \in M_A^T.$ As ${\mathbb {H}} = N(A)\oplus \overline{R(A)},$ x can be uniquely written as $x=u+v,$ where $u \in N(A)$ and $v \in \overline{R(A)}.$ Hence $\Vert u \Vert _A = 0$ and $\Vert x \Vert _A = \Vert v \Vert _A.$ As $T \in B_{A^{1/2}}({\mathbb {H}}),$ it follows that $\Vert Tu\Vert _A = 0$ and, therefore, $\Vert Tx\Vert _A = \Vert Tv\Vert _A = \Vert T\Vert _A.$ This proves that $M_A^T \cap \overline{R(A)} \ne \phi .$

To prove that $M_A^T \cap \overline{R(A)}$ is the $A-$ unit sphere of some subspace of ${\mathbb {H}},$ it is enough to show that $\frac{\lambda _1 e_1 \pm \lambda _2 e_2}{\Vert \lambda _1 e_1 \pm \lambda _2 e_2\Vert _A} \in M_A^T \cap \overline{R(A)}$, whenever $e_1,~e_2 \in M_A^T \cap \overline{R(A)}$ and $\lambda _1,~\lambda _2 \in {\mathbb {K}} (={\mathbb {R}},~{\mathbb {C}}).$ Let $e_1,~ e_2 \in M_A^T \cap \overline{R(A)},$ then $\Vert Te_1\Vert _A = \Vert Te_2\Vert _A = \Vert T\Vert _A$ and $\Vert e_1\Vert _A = \Vert e_2\Vert _A = 1.$ First we claim that $\Vert \cdot \Vert _A$ satisfies the parallelogram law for all $x,~y \in {\mathbb {H}}.$ Let $x,y \in {\mathbb {H}}.$ Then we have

$$\begin{aligned} \Vert x+y\Vert _A^2 +\Vert x-y\Vert _A^2= & {} \langle x+y,x+y\rangle _A + \langle x-y,x-y\rangle _A \\= & {} \langle A(x+y),x+y\rangle + \langle A(x-y),x-y\rangle \\= & {} 2(\langle Ax,x\rangle + \langle Ay,y\rangle ) \\= & {} 2(\Vert x\Vert _A^2 +\Vert y\Vert _A^2). \end{aligned}$$

This proves our claim. Therefore,

$$\begin{aligned} 2(|\lambda _1|^2 +|\lambda _2|^2)\Vert T\Vert _A^2= & {} 2(\Vert \lambda _1 Te_1\Vert _A^2 + \Vert \lambda _2 Te_2\Vert _A^2) \\= & {} \Vert \lambda _1Te_1 + \lambda _2 Te_2\Vert _A^2 + \Vert \lambda _1 Te_1 - \lambda _2 Te_2\Vert _A^2 \\= & {} \Vert T(\lambda _1 e_1 + \lambda _2 e_2)\Vert _A^2 + \Vert T(\lambda _1 e_1 - \lambda _2 e_2)\Vert _A^2\\\le & {} \Vert T\Vert _A^2(\Vert \lambda _1 e_1 + \lambda _2 e_2\Vert _A^2 + \Vert \lambda _1 e_1 - \lambda _2 e_2\Vert _A^2)\\= & {} 2(|\lambda _1|^2 +|\lambda _2|^2)\Vert T\Vert _A^2.\\ \end{aligned}$$

Hence the above inequality is actually an equality. Since $\Vert T(\lambda _1 e_1\pm \lambda _2 e_2)\Vert _A \le \Vert T\Vert _A\Vert \lambda _1 e_1 \pm \lambda _2 e_2\Vert _A,$ it follows that

$$\begin{aligned} \Vert T(\lambda _1 e_1\pm \lambda _2 e_2)\Vert _A = \Vert T\Vert _A\Vert \lambda _1 e_1 \pm \lambda _2 e_2\Vert _A. \end{aligned}$$

This establishes the theorem. $\square$

In the next theorem we study the compactness property of $M_A^T \cap \overline{R(A)}.$

Theorem 2.5

Let ${\mathbb {H}}$ be a Hilbert space and let $A \in {\mathbb {L}}({\mathbb {H}})$ be a positive operator such that $B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ is bounded with respect to $\Vert \cdot \Vert$. Let $T \in {\mathbb {K}}({\mathbb {H}}) \cap B_{A^{1/2}}({\mathbb {H}}).$ Then $M_A^T \cap \overline{R(A)}$ is compact with respect to $\Vert \cdot \Vert$.

Proof

Clearly, $\overline{R(A)}$ is a Hilbert space with respect to $\Vert \cdot \Vert .$ It is easy to see that A is positive definite on $\overline{R(A)}.$ Therefore, $\Vert \cdot \Vert _A$ is a norm on $\overline{R(A)}.$ We claim that $\Vert \cdot \Vert _A$ and $\Vert \cdot \Vert$ are equivalent norms on $\overline{R(A)}.$ Clearly, $\frac{1}{\sqrt{\Vert A\Vert }}\Vert x\Vert _A \le \Vert x\Vert$ for all $x\in {\mathbb {H}}$. Let $x\in \overline{R(A)}$. Then $\frac{x}{\Vert x\Vert _A}\in B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$. Since $B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ is bounded, there exists $M >0$ such that $\Vert z\Vert \le M$ for all $z \in B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$. Therefore, $\frac{\Vert x\Vert }{\Vert x\Vert _A} \le M$. Thus $\frac{1}{\sqrt{\Vert A\Vert }}\Vert x\Vert _A \le \Vert x\Vert \le M\Vert x\Vert _A$ for all $x\in \overline{R(A)}$. Thus our claim is established. Therefore, $\overline{R(A)}$ is a Hilbert space with respect to $\Vert \cdot \Vert _A.$ Next, let $\{v_n\}$ be a sequence in $M_A^T\cap \overline{R(A)}$. We show that $\{v_n\}$ has a convergent subsequence in $M_A^T\cap \overline{R(A)}$ with respect to $\Vert \cdot \Vert$. Since ${\mathbb {H}}$ is reflexive and $B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ is closed, convex and bounded with respect to $\Vert \cdot \Vert$, it follows that $B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ is weakly compact with respect to $\Vert \cdot \Vert$. Thus the sequence $\{v_n\}$ has a weakly convergent subsequence $\{ v_{n_k}\}$ with respect to $\Vert \cdot \Vert .$ Suppose $v_{n_k} \rightharpoonup v$ for some $v \in B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ with respect to $\Vert \cdot \Vert$. Since $T \in {\mathbb {K}}({\mathbb {H}}),$ it follows that $T v_{n_k} \longrightarrow Tv$ with respect to $\Vert \cdot \Vert$. It is easy to see that

$$\begin{aligned} \Vert T\Vert _A^2 = \lim _{k \rightarrow \infty }\Vert T v_{n_k}\Vert _A^2 =\lim _{k \rightarrow \infty }\langle AT v_{n_k},T v_{n_k}\rangle = \langle AT v,T v\rangle =\Vert Tv\Vert _A^2. \end{aligned}$$

As $\Vert v\Vert _A \le 1,$ we conclude that $v \in M_A^T \cap \overline{R(A)}$ and $1 = \Vert v_{n_k}\Vert _A \longrightarrow \Vert v\Vert _A = 1.$ As $v_{n_k} \rightharpoonup v$ with respect to $\Vert \cdot \Vert ,$ clearly, $v_{n_k} \rightharpoonup v$ with respect to $\Vert \cdot \Vert _A.$ Since $(\overline{R(A)}, \langle ~,~ \rangle _A)$ is a Hilbert space, it follows that $v_{n_k} \longrightarrow v$ with respect to $\Vert \cdot \Vert _A.$ As $\Vert \cdot \Vert _A$ and $\Vert \cdot \Vert$ are equivalent norms on $\overline{R(A)}$, therefore, $v_{n_k} \longrightarrow v$ with respect to $\Vert \cdot \Vert .$ This establishes the theorem. $\square$

Remark 2.5

Note that, $M_A^T \cap \overline{R(A)}$ is also compact with respect to $\Vert \cdot \Vert _A$ in $\overline{R(A)}$, due to the fact that $\Vert \cdot \Vert _A$ and $\Vert \cdot \Vert$ are equivalent norms on $\overline{R(A)}.$

In [18], the author has characterized the $A-$Birkhoff–James orthogonality of $A-$bounded operators on a Hilbert space with the help of $A-$norming sequences. In the finite-dimensional case, the Bhatia-$\breve{S}$emrl Theorem follows from the said characterization, as shown in Theorem 2.4 of [18]. The main difference between the characterizations of $A-$Birkhoff–James orthogonality of operators in the infinite-dimensional case and the finite-dimensional case is that the approximate orthogonality of the images of norming sequences in the former case can be strengthened to the exact orthogonality of the images of a norming vector in the later case. For the convenience of the readers, let us mention the relevant results from [18] and [5].

Theorem 2.6

(Zamani, Theorem 2.2 of [18]). Let $T,S \in B_{A^{1/2}}({\mathbb {H}}).$ Then the following conditions are equivalent:

• (i) there exists a sequence of $A-$unit vectors $\{x_n\}$ in ${\mathbb {H}}$ such that

  $\lim _{n \rightarrow \infty }\Vert Tx_n\Vert _A =\Vert T\Vert _A$ and $\lim _{n \rightarrow \infty }\langle Tx_n,Sx_n \rangle _A = 0.$

• (ii) $T \bot _A^B S$.

Theorem 2.7

(Bhatia and $\breve{S}$emrl, Theorem 1.1 of [5]) A matrix A is orthogonal to a matrix B if and only if there exists a unit vector $x \in {\mathbb {H}}$ such that $\Vert Ax\Vert =\Vert A\Vert$ and $\langle Ax,Bx\rangle = 0.$

In our next theorem, we show that under certain additional conditions, the said strengthening of the $A-$Birkhoff–James orthogonality of $A-$bounded operators can be preserved even in the infinite-dimensional case.

Theorem 2.8

Let ${\mathbb {H}}$ be a Hilbert space and let $A \in {\mathbb {L}}({\mathbb {H}})$ be positive such that $B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ is bounded with respect to $\Vert \cdot \Vert .$ Let $T,S \in {\mathbb {K}}({\mathbb {H}}) \cap B_{A^{1/2}}({\mathbb {H}}).$ Then $T \bot _A^BS$ if and only if there exists $v \in M^T_A$ such that $Tv\bot _ASv.$

Proof

The sufficient part of the theorem follows easily. Indeed, suppose that there exists $v \in M^T_A$ such that $Tv\bot _ASv.$ Then

$$\begin{aligned} \Vert T+\lambda S\Vert _A\ge & {} \Vert Tv+\lambda Sv\Vert _A \\\ge & {} \Vert Tv\Vert _A \\= & {} \Vert T\Vert _A~\text {for all }\lambda \in {\mathbb {K}} (={\mathbb {R}},~{\mathbb {C}}). \end{aligned}$$

Let us prove the necessary part of the theorem. By Theorem 2.2 of [18], there exists a sequence $\{x_n\} \subseteq S_{{\mathbb {H}}(A)}$ such that

$$\begin{aligned} \lim _{n \rightarrow \infty }\Vert Tx_n\Vert _A =\Vert T\Vert _A~\text {and} \lim _{n \rightarrow \infty }\langle Tx_n,Sx_n \rangle _A = 0.\end{aligned}$$

Since ${\mathbb {H}} = N(A)\oplus \overline{R(A)}$, it follows that $x_n = u_n + v_n$ for each $n \in {\mathbb {N}}$, where $u_n \in N(A)$ and $v_n \in \overline{R(A)}.$ Clearly, $\Vert u_n\Vert _A =0$ for all $n \in {\mathbb {N}}$. Thus $\Vert x_n\Vert _A = \Vert v_n + u_n\Vert _A \le \Vert v_n\Vert _A + \Vert u_n\Vert _A=\Vert v_n\Vert _A$. Again, $\Vert x_n\Vert _A = \Vert v_n + u_n\Vert _A \ge \Vert v_n\Vert _A - \Vert u_n\Vert _A=\Vert v_n\Vert _A$. Therefore, $\Vert x_n\Vert _A = \Vert v_n\Vert _A$ for each $n \in {\mathbb {N}}.$ As $\{x_n\} \subseteq S_{{\mathbb {H}}(A)}$, we conclude that $\{v_n\} \subseteq S_{{\mathbb {H}}(A)} \cap \overline{R(A)}.$ Since $T,S \in B_{A^{1/2}}({\mathbb {H}})$, $\Vert Tu_n\Vert _A = \Vert Su_n\Vert _A = 0$ for all $n \in {\mathbb {N}}$. Hence $\Vert Tx_n\Vert _A = \Vert Tv_n\Vert _A$ and $\Vert Sx_n\Vert _A = \Vert Sv_n\Vert _A$ for each $n \in {\mathbb {N}}.$ Since ${\mathbb {H}}$ is reflexive and $B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ is closed, convex and bounded with respect to $\Vert \cdot \Vert ,$ therefore, $B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ is weakly compact with respect to $\Vert \cdot \Vert .$ Thus the sequence $\{v_n\}$ has a weakly convergent subsequence. Without loss of generality we may assume that $v_n \rightharpoonup v$ with respect to $\Vert \cdot \Vert$ on ${\mathbb {H}}$, for some $v \in B_{{\mathbb {H}}(A)} \cap \overline{R(A)}.$ Since $T,S \in {\mathbb {K}}({\mathbb {H}}),$ it follows that $Tv_n \longrightarrow Tv$ and $Sv_n \longrightarrow Sv$ with respect to $\Vert \cdot \Vert$ in ${\mathbb {H}}$. Therefore,

$$\begin{aligned} \Vert T\Vert ^2_A= & {} \lim _{n \rightarrow \infty }\Vert Tx_n\Vert ^2_A = \lim _{n \rightarrow \infty }\Vert Tv_n\Vert ^2_A\\= & {} \lim _{n \rightarrow \infty } \langle ATv_n, Tv_n \rangle = \Vert Tv\Vert ^2_A. \end{aligned}$$

As $\Vert v\Vert _A \le 1,$ we conclude that $v \in M_A^T \cap \overline{R(A)}.$

Next we show that $Tv\bot _A Sv.$ As $\Vert Tu_n\Vert _A = \Vert Su_n\Vert _A = 0,$ it is immediate that $Tu_n,Su_n \in N(A)$ for all $n \in {\mathbb {N}}.$ Since A is positive, it follows that $N(A)=N(A^{1/2}).$ Hence $A^{1/2}(Tu_n)=A^{1/2}(Su_n)= \theta$ for all $n \in {\mathbb {N}}.$ Therefore, we have

$$\begin{aligned} 0= & {} \lim _{n \rightarrow \infty }\langle Tx_n,Sx_n \rangle _A = \lim _{n \rightarrow \infty }\langle Tu_n + Tv_n,Su_n + Sv_n \rangle _A \\= & {} \lim _{n \rightarrow \infty }\langle A(Tu_n + Tv_n),Su_n + Sv_n \rangle \\= & {} \lim _{n \rightarrow \infty }\langle A^{1/2}(Tu_n + Tv_n), A^{1/2}(Su_n + Sv_n) \rangle \\= & {} \lim _{n \rightarrow \infty }\langle A^{1/2}Tv_n, A^{1/2}Sv_n \rangle = \lim _{n \rightarrow \infty }\langle ATv_n,Sv_n \rangle = \langle ATv,Sv \rangle = \langle Tv,Sv \rangle _A. \end{aligned}$$

Thus $Tv\bot _ASv.$ This completes the proof of the theorem. $\square$

We end this article with the following closing remark:

Remark 2.6

Note that in Theorem 2.8, if ${\mathbb {H}}$ is finite-dimensional and $A = I,$ then the Bhatia-$\breve{S}$emrl Theorem (Theorem 1.1 of [5]) follows immediately. In particular, the finite-dimensional Bhatia-$\breve{S}$emrl Theorem can be extended verbatim to the infinite-dimensional setting of semi-Hilbertian spaces, provided certain additional conditions are satisfied. We further observe that Theorem 2.4 of [18] follows as a corollary to Theorem 2.8, since in a finite-dimensional Hilbert space, $B_{{\mathbb {H}}(A)} \cap \overline{R(A)}$ is bounded with respect to $\Vert \cdot \Vert$ and every linear operator is compact.