Smoothness in the space of bounded linear operators on semi-Hilbert space

TL;DR

This paper introduces and characterizes the concept of A-smoothness for bounded linear operators on semi-Hilbert spaces induced by positive operators, linking it to differentiability and analyzing specific operator classes.

Contribution

It defines A-smoothness in semi-Hilbert spaces, provides characterizations for A-bounded and A-compact operators, and relates A-smoothness to Gâteaux differentiability.

Findings

A-smoothness is characterized for A-bounded operators.

A-smoothness of A-compact operators is analyzed via A-norm attainment.

Gâteaux differentiability of the semi-norm is equivalent to A-smoothness.

Abstract

Given a nonzero positive operator $A$ on a Hilbert space $\mathbb{H}$, a semi-inner product is naturally induced on $\mathbb{H}$. In this work, we introduce the notion of \emph{$A$-smoothness} for bounded linear operators on the resulting semi-Hilbert space and investigate its various properties. We provide a comprehensive characterization of the $A$-smoothness for $A$-bounded operators and further analyze the $A$-smoothness of $A$-compact operators in terms of their $A$-norm attainment sets. Utilizing these characterizations, we establish that G\^{a}teaux differentiability of the semi-norm $\|\cdot\|_A$ at an $A$-bounded operator is equivalent to its $A$-smoothness. Furthermore, we characterize the $A$-smoothness of $2\times 2$ block diagonal matrices.

Equations

$$⟨A^{1/2}x,A^{1/2}y{⟩}_{\text{R}(A^{1/2})}=⟨P_{\overline{R(A)}}x,P_{\overline{R(A)}}y⟩$$

$$\Vert T{\Vert }_A=\underset{\Vert x{\Vert }_A=1}{\sup }\Vert Tx{\Vert }_A=\sup \{|⟨Tx,y{⟩}_A|:x,y\in \mathbb{H},\Vert x{\Vert }_A=\Vert y{\Vert }_A=1\}.$$

$${\mathbb{B}}_A(\mathbb{H})=\{T\in \mathbb{B}(\mathbb{H}):R(T^{\ast }A)\subset R(A)\}.$$

$$\Vert \tilde{T}{\Vert }_{\mathbb{B}(\text{R}(A^{1/2}))}=\Vert T{\Vert }_A=\Vert Tx{\Vert }_A=\Vert ATx{\Vert }_{\text{R}(A^{1/2})}=\Vert \tilde{T}Ax{\Vert }_{\text{R}(A^{1/2})}.$$

$$\Vert Ax{\Vert }_{\text{R}(A^{1/2})}^2$$

$$W_{A}(T,S)=\big{\{}\sigma\in\mathbb{C}:\|x_{n}\|_{A}=1,\|Tx_{n}\|_{A}\to\|T\|_{A}\,\,\mbox{and}\,\langle Tx_{n},Sx_{n}\rangle_{A}\to\sigma\big{\}}.$$

$$W(T,S)=\big{\{}\sigma\in\mathbb{C}:\|x_{n}\|=1,\|Tx_{n}\|\to\|T\|\,\,\mbox{and}\,\langle Tx_{n},Sx_{n}\rangle\to\sigma\big{\}}.$$

$$\displaystyle\|Tx\|_{A}^{2}=\big{\|}\sum_{n=1}^{\infty}x_{n}Te_{n}\big{\|}_{A}^{2}$$

$$\Vert T{\Vert }_A=\underset{\Vert x{\Vert }_A=1}{\sup }\Vert Tx{\Vert }_A=\underset{\Vert x{\Vert }_A=1}{\sup }\left|\sum _{n=1}^{\infty }\frac{x_n}{n^{3/2}}\right|.$$

$$\Vert T{\Vert }_A=\underset{\Vert y\Vert =1}{\sup }|⟨y,z⟩|=\Vert z\Vert =\frac{\pi }{\sqrt{6}}.$$

$$\lim \Vert T{z}_n{\Vert }_A^2$$

$$\sup \{\Vert Ty{\Vert }_A:⟨x_0,y{⟩}_A=0\text{mbox}with\Vert y{\Vert }_A=1\}<\Vert T{\Vert }_A.$$

$$\displaystyle\big{\{}\|\widetilde{T}z\|_{\textbf{R}(A^{1/2})}:\|z\|_{\textbf{R}(A^{1/2})}=1,\,\langle y_{0},z\rangle_{\textbf{R}(A^{1/2})}=0\big{\}}<\|\widetilde{T}\|_{\mathbb{B}(\textbf{R}(A^{1/2}))}.$$

$$\displaystyle\sup\big{\{}\|\widetilde{T}Au\|_{\textbf{R}(A^{1/2})}:\|Au\|_{\textbf{R}(A^{1/2})}=1,\,\langle Ax_{0},Au\rangle_{\textbf{R}(A^{1/2})}=0\big{\}}<\|T\|_{A}$$

$${\rho }_{\pm }^A(T,S)=\underset{t\to 0^{\pm }}{\lim }\frac{\Vert T+tS{\Vert }_A-\Vert T{\Vert }_A}{t}$$

$$\lambda$$

$$\frac{\Vert T-tS{\Vert }_A-\Vert T{\Vert }_A}{-t}=-\frac{\Vert T+t(-S){\Vert }_A-\Vert T{\Vert }_A}{t}$$

$$\mu$$

$$⟨\tilde{T}{x}_n,\tilde{S}{x}_n{⟩}_{\text{R}(A^{1/2})}\to {\lambda }_0,⟨\tilde{T}{y}_n,\tilde{S}{y}_n{⟩}_{\text{R}(A^{1/2})}\to {\sigma }_0,$$

$$\underset{n\to \infty ,k\to \infty }{\lim }\Vert \tilde{T}A{u}_{n,k}{\Vert }_{\text{R}(A^{1/2})}=\Vert \tilde{T}{\Vert }_{\text{R}(A^{1/2})}=\Vert T{\Vert }_A,$$

$$\underset{n\to \infty ,k\to \infty }{\lim }⟨\tilde{T}A{u}_{n,k},\tilde{S}A{u}_{n,k}{⟩}_{\text{R}(A^{1/2})}={\lambda }_0.$$

$$\underset{n\to \infty ,k\to \infty }{\lim }\Vert \tilde{T}A{v}_{n,k}{\Vert }_{\text{R}(A^{1/2})}=\Vert \tilde{T}{\Vert }_{\text{R}(A^{1/2})}=\Vert T{\Vert }_A,$$

$$\underset{n\to \infty ,k\to \infty }{\lim }⟨\tilde{T}A{v}_{n,k},\tilde{S}A{v}_{n,k}⟩={\sigma }_0.$$

$$z_r=\frac{u_{n_r,k_r}}{\Vert A{u}_{n_r,k_r}{\Vert }_{\text{R}(A^{1/2})}},w_s=\frac{v_{n_s,k_s}}{\Vert A{v}_{n_s,k_s}{\Vert }_{\text{R}(A^{1/2})}}.$$

$$\Vert \tilde{T}A{z}_r{\Vert }_{\text{R}(A^{1/2})}\to \Vert \tilde{T}{\Vert }_{\text{R}(A^{1/2})}$$

$$⟨\tilde{T}A{z}_r,\tilde{S}A{z}_r{⟩}_{\text{R}(A^{1/2})}\to {\lambda }_0$$

$$\underset{n\to \infty }{\lim }{\left(\Vert M{x}_n{\Vert }_A^2+\Vert N{y}_n{\Vert }_A^2\right)}^{1/2}=\Vert M{\Vert }_A$$

$$\Vert M{x}_n{\Vert }_A^2+\Vert N{y}_n{\Vert }_A^2$$

$$\underset{n\to \infty }{\lim }\left(\Vert M{x}_n{\Vert }_A^2+\Vert N{y}_n{\Vert }_A^2\right)\le \Vert M{\Vert }_A^2-(\Vert M{\Vert }_A^2-\Vert N{\Vert }_A^2)\underset{n\to \infty }{\mathrm{limsup}}\Vert y_n{\Vert }_A^2$$

$$⟨M{x}_n,R{x}_n{⟩}_A\to \lambda \text{and}⟨M{y}_n,R{y}_n{⟩}_A\to \mu ,$$

$$\lim\Bigg{\langle}T\begin{pmatrix}x_{n}\\ 0\end{pmatrix},S\begin{pmatrix}x_{n}\\ 0\end{pmatrix}\Bigg{\rangle}_{\mathbb{A}}=\lim\langle Mx_{n},Rx_{n}\rangle_{A}=\lambda.$$

$$\lim\Bigg{\langle}T\begin{pmatrix}y_{n}\\ 0\end{pmatrix},S\begin{pmatrix}y_{n}\\ 0\end{pmatrix}\Bigg{\rangle}_{\mathbb{A}}=\lim\langle My_{n},Ry_{n}\rangle_{A}=\mu.$$

Full text

1. introduction

The concept of Gâteaux differentiability of the norm function plays a central role in understanding the geometry of Banach spaces and the space of bounded linear operators. It corresponds to the notion of smoothness of a point, meaning that the point admits a unique supporting functional. Several works have contributed to this area of study (see [6, 13, 15, 18, 19]). In this note, we aim to explore the Gâteaux differentiability of a semi-norm within the setting of semi-Hilbertian operators. This investigation is carried out through a newly introduced concept known as A-smoothness. Before presenting the main results, we begin by establishing some notations and terminology that will be used throughout this work.

Let $(\mathbb{H},\langle\cdot,\cdot\rangle)$ be a Hilbert space over the scalar field $\mathbb{R}$ or $\mathbb{C}.$ We denote $\Re(z)$ as the real part of $z\in\mathbb{C}.$ Let $\mathbb{B}(\mathbb{H})$ be the collection of all bounded linear operators on $\mathbb{H}.$ An operator $A\in\mathbb{B}(\mathbb{H})$ is said to be positive if $\langle Ax,x\rangle\geq 0$ for all $x\in\mathbb{H}$. For any $T\in\mathbb{B}(\mathbb{H}),$ the range and the null space of $T$ are denoted by $R(T)$ and $N(T)$, respectively. Suppose that $A$ is a positive nonzero operator which generates a semi-inner product $\langle\cdot,\cdot\rangle_{A}$ on $\mathbb{H}$ , defined as $\langle x,y\rangle_{A}=\langle Ax,y\rangle.$ The semi-inner product induces a semi-norm $\|\cdot\|_{A}$, defined as $\|x\|_{A}=\sqrt{\langle x,x\rangle_{A}}=\|A^{1/2}x\|$ for all $x\in\mathbb{H}.$ Note that, $\|x\|_{A}=0$ if and only if $x\in N(A)$. The vector space $\mathbb{H}$, equipped with the semi-inner product $\langle\cdot,\cdot\rangle_{A}$, is referred to as a semi-Hilbert space. This study was initiated by Krein in [10]. The semi-Hilbert space is complete if and only if $R(A)$ is closed in $\mathbb{H}$. The semi-inner product $\langle\cdot,\cdot\rangle_{A}$ induces an inner product on the quotient space $\mathbb{H}/N(A)$ defined as $[\bar{x},\bar{y}]=\langle x,y\rangle_{A}$ for all $\bar{x},\bar{y}\in\mathbb{H}/N(A)$. The completion of $(\mathbb{H}/N(A),[\cdot,\cdot])$ is isometrically isomorphic to the Hilbert space $\left(R(A^{1/2}),\langle\cdot,\cdot\rangle_{\textbf{R}(A^{1/2})}\right)$, where the inner product is defined as

[TABLE]

for all $x,y\in\mathbb{H}.$ The Hilbert space $\left(R(A^{1/2}),\langle\cdot,\cdot\rangle_{\textbf{R}(A^{1/2})}\right)$ is denoted by $\textbf{R}(A^{1/2})$ and the norm associated with the inner product $\langle\cdot,\cdot\rangle_{\textbf{R}(A^{1/2})}$ is written as $\|\cdot\|_{\textbf{R}(A^{1/2})}$. Note that, $R(A)$ is dense in $\textbf{R}(A^{1/2})$ (see [1]). We direct the reader to [1, 2, 3] for more insights into the space $\textbf{R}(A^{1/2})$. Let $B_{\mathbb{H}(A)}=\{x\in\mathbb{H}:\|x\|_{A}\leq 1\}$ and $S_{\mathbb{H}(A)}=\{x\in\mathbb{H}:\|x\|_{A}=1\}$ be the $A$-unit ball and the $A$-unit sphere of the semi-Hilbert space $\left(\mathbb{H},\langle\cdot,\cdot\rangle_{A}\right),$ respectively. An operator $T\in\mathbb{B}(\mathbb{H})$ is said to be $A$-bounded if there exists a positive constant $c$ such that $\|Tx\|_{A}\leq c\|x\|_{A}$ for all $x\in\mathbb{H}.$ The set of all such operators is denoted by $\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ For any $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$, the $A$-norm is given by

[TABLE]

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ The $A$-norm attainment set of $T$ is defined as $M_{A}(T)=\{x\in\mathbb{H}:\|x\|_{A}=1,\ \|Tx\|_{A}=\|T\|_{A}\}.$ Whenever $A=I,$ we denote $M(T)$ as the usual norm attainment set of $T.$ For $T\in\mathbb{B}(\mathbb{H}),$ an operator $W\in\mathbb{B}(\mathbb{H})$ is called an $A$-adjoint of $T$ if $\langle Tx,y\rangle_{A}=\langle x,Wy\rangle_{A}$ for all $x,y\in\mathbb{H},$ equivalently, the equation $AX=T^{*}A$ has a solution. Not every $T\in\mathbb{B}(\mathbb{H})$ admits an $A$-adjoint. By Douglas theorem [7], $T$ admits a unique $A$-adjoint if and only if $R(T^{*}A)\subset R(A).$ From now on, we denote by $\mathbb{B}_{A}(\mathbb{H})$ the set of all $T\in\mathbb{B}(\mathbb{H})$ admitting a unique $A$-adjoint:

[TABLE]

If $T\in\mathbb{B}_{A}(\mathbb{H}),$ its $A$-adjoint is denoted by $T^{\sharp}$ and satisfies $R(T^{\sharp})\subset\overline{R(A)}$. Note that $T^{\sharp}=A^{\dagger}T^{*}A$, where $A^{\dagger}$ is the Moore-Penrose inverse of $A,$ see [17]. Moreover, $\mathbb{B}_{A}(\mathbb{H})\subset\mathbb{B}_{A^{1/2}}(\mathbb{H})\subset\mathbb{B}(\mathbb{H})$. We direct readers to [1, 2, 3, 20, 21] for further insights in this direction. $T\in\mathbb{B}(\mathbb{H})$ is said to be $A$-compact [5] if every bounded sequence in $(\mathbb{H},\|\cdot\|_{A})$ has a convergent subsequence in $(\mathbb{H},\|\cdot\|_{A})$.

The notion of Birkhoff–James orthogonality has been explored in recent years for the study of geometric and analytic properties in the space of bounded linear operators (see [12]). An operator $T\in\mathbb{B}(\mathbb{H})$ is said to be Birkhoff-James orthogonal to $S\in\mathbb{B}(\mathbb{H})$ if $\|T+\lambda S\|\geq\|T\|$ for all scalars $\lambda.$ Analogously, in the semi-Hilbertian setting, $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ is $A$-Birkhoff–James orthogonal to $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}),$ denoted by $T\perp_{A}^{B}S,$ (see [19, 20]) if $\|T+\lambda S\|_{A}\geq\|T\|_{A}$ for all scalars $\lambda.$ It is well known that the right additivity of Birkhoff-James orthogonality at a point of a normed linear space characterizes smoothness of the norm at that point. Motivated by this we introduce the notion of smoothness of operators in $\mathbb{B}_{A^{1/2}}(\mathbb{H})$ as follows:

Definition 1.1**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ be such that $\|T\|_{A}\neq 0.$ Then $T$ is said to be $A$-smooth if for any $S_{1},S_{2}\in B_{A^{1/2}}(\mathbb{H})$ such that $T\perp_{A}^{B}S_{1}$ and $T\perp_{A}^{B}S_{2}$ imply $T\perp_{A}^{B}S_{1}+S_{2}.$

In this work, we begin by characterizing the notion of $A$-smoothness for $A$-bounded operators in a semi-Hilbert space. We then focus on the case where the set $M_{A}(T)$ is non-empty, providing a refined characterization of $A$-smoothness under this condition. We establish that an $A$-compact operator is $A$-smooth if and only if the intersection $M_{A}(T)\cap\overline{R(A)}$ is a singleton up to scalar multiples. Furthermore, we demonstrate the equivalence between $A$-smoothness and the existence of the Gâteaux derivative of the associated semi-norm function in semi-Hilbertian operator spaces. We also explore the relation between the $A$-smoothness of operators in $\mathbb{B}_{A^{1/2}}(\mathbb{H})$ and the smoothness of operators in $\mathbb{B}(\textbf{R}(A^{1/2}))$. Finally, we examine $A$-smoothness in the context of block diagonal matrices.

2. Main Results.

We first aim to characterize the newly introduced notion of $A$-smoothness. Before that we note the following known results, which will be useful throughout this article.

Lemma 2.1**.**

[1]** For any $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H}),$ there exists a unique $\widetilde{T}\in\mathbb{B}(\textbf{R}(A^{1/2}))$ such that $\widetilde{T}W_{A}=W_{A}T$, where $W_{A}:\mathbb{H}\to\textbf{R}(A^{1/2})$ satisfying $W_{A}x=Ax$ for all $x\in\mathbb{H}.$

Lemma 2.2**.**

• (i)

[9]** Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ Then $\|\widetilde{T}\|_{\mathbb{B}(\textbf{R}(A^{1/2}))}=\|T\|_{A}$.

• (ii)

[11]** Let $T\in\mathbb{B}_{A}(\mathbb{H}).$ Then $\widetilde{T^{\sharp}}=(\widetilde{T})^{*}$ .

• (iii)

[5]** Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ and $R(A)$ be closed. Then $T$ is $A$-compact if and only if $\widetilde{T}$ is compact.

In this connection we observe a relation between the $A$-norm attainment set of $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ and the norm attainment set of $\widetilde{T}\in\mathbb{B}(\textbf{R}(A^{1/2})).$

Proposition 2.3**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ and $\widetilde{T}\in\mathbb{B}(\textbf{R}(A^{1/2}))$ be as given in Lemma 2.1. Then $x\in M_{A}(T)\cap\overline{R(A)}$ if and only if $Ax\in M(\widetilde{T}).$

Proof.

Suppose that $x\in M_{A}(T)\cap\overline{R(A)}.$ This implies $\|Tx\|_{A}=\|T\|_{A}.$ By using Lemma 2.1 and Lemma 2.2 (i), we have

[TABLE]

Also, note that

[TABLE]

Therefore, $Ax\in M(\widetilde{T}).$ Using similar argument we can obtain the converse part. ∎

Next, we mention the characterization of $A$-Birkhoff-James orthogonality, which plays a significant role in our whole scheme of things.

Theorem 2.4**.**

[21, Th. 2.2]** Let $T,S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ Then the following conditions are equivalent:

• (i)

$T\perp_{A}^{B}S$.

• (ii)

There exists a sequence $\{x_{n}\}\in B_{\mathbb{H}(A)}$ such that $\lim\limits_{n\to\infty}\|Tx_{n}\|_{A}=\|T\|_{A}$ and $\lim\limits_{n\to\infty}\langle Tx_{n},Sx_{n}\rangle=0.$

In the following theorem we completely characterize $A$-smoothness of an $A$-bounded operator.

Theorem 2.5**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ be such that $\|T\|_{A}\neq 0$. Then $T$ is $A$-smooth if and only if for any $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}),$ $W_{A}(T,S)$ is singleton, where

[TABLE]

Proof.

Since $A$-orthogonality is homogeneous, without loss of generality we may assume that $\|T\|_{A}=\|S\|_{A}=1.$ We first prove the necessary part. Suppose on the contrary that $W_{A}(T,S)$ is not singleton for some $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$. Let $\alpha,\beta\in W_{A}(T,S)$ be such that $\alpha\neq\beta.$ Therefore, there exist $A$-norming sequnces $\{x_{n}\}$ and $\{y_{n}\}$ of $T$ such that $\langle Tx_{n},Sx_{n}\rangle_{A}\to\alpha$ and $\langle Ty_{n},Sy_{n}\rangle_{A}\to\beta.$ Suppose that $\alpha,\beta$ both are nonzero. Now consider $S_{1}=S-\alpha T$ and $S_{2}=S-\beta T.$ Then it is easy to see that $\langle Tx_{n},S_{1}x_{n}\rangle_{A}\to 0.$ This implies $T\perp_{A}^{B}S_{1}.$ Similarly, we can show that $T\perp_{A}^{B}S_{2}.$ As $T$ is $A$-smooth, it follows that $T\perp_{A}^{B}S_{1}-S_{2},$ i.e., $T\perp_{A}^{B}(\beta-\alpha)T.$ Since $\alpha\neq\beta,$ we get $\|T\|_{A}=0,$ which is a contradiction. Now suppose that at least one of $\alpha$ or $\beta$ zero. Let $\alpha=0.$ This implies $T\perp_{A}^{B}S.$ From previous argument we can show that $T\perp_{A}^{B}S-\beta T.$ Therefore, by $A$-smoothness of $T$ we get $T\perp_{A}^{B}S-\beta T-S.$ This implies $\|T\|_{A}=0,$ which is again a contradiction.

To prove the sufficient part, let $T\perp_{A}^{B}S_{1}$ and $T\perp_{A}^{B}S_{2}$ for some $S_{1},S_{2}\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ From Theorem 2.4 we see that there exists $\{x_{n}\}$ with $\|x_{n}\|_{A}=1$ such that $\|Tx_{n}\|_{A}\to\|T\|_{A}$ and $\langle Tx_{n},S_{1}x_{n}\rangle_{A}\to 0.$ Moreover, we have $W_{A}(T,S_{1})$ is singleton. So we obtain that $W_{A}(T,S_{1})=\{0\}$. Similarly, $W_{A}(T,S_{2})=\{0\}.$ Therefore, we can take a common sequence $\{x_{n_{k}}\}\subset\{x_{n}\}$ with $\|Tx_{n_{k}}\|_{A}\to\|T\|_{A}$ such that $\langle Tx_{n_{k}},S_{1}x_{n_{k}}\rangle_{A}\to 0$ and $\langle Tx_{n_{k}},S_{2}x_{n_{k}}\rangle_{A}\to 0.$ This implies that $\langle Tx_{n_{k}},(S_{1}+S_{2})x_{n_{k}}\rangle_{A}\to 0$ and thus $T\perp_{A}^{B}S_{1}+S_{2}.$ This shows that $T$ is $A$-smooth. ∎

If we consider $A=I\in\mathbb{B}(\mathbb{H})$ then we have the following corollary regarding classical smoothness of operators defined on a Hilbert space, which is also proved in [19].

Corollary 2.6**.**

Let $T\in\mathbb{B}(\mathbb{H})$ be nonzero. Then $T$ is smooth if and only if $W(T,S)$ is singleton for every $S\in\mathbb{B}(\mathbb{H}),$ where

[TABLE]

Now we establish that $A$-compact operators possess a nonempty $A$-norm attainment set whenever $R(A)$ is closed.

Proposition 2.7**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ be $A$-compact and let $R(A)$ be closed. Then $M_{A}(T)\neq\emptyset.$

Proof.

Since $T$ is $A$-compact and $R(A)$ is closed, from Lemma 2.1 and Lemma 2.2 (iii), there exists a unique $\widetilde{T}\in\mathbb{B}(\textbf{R}(A^{1/2}))$ such that $\widetilde{T}$ is compact. Note that $\|T\|_{A}=\|\widetilde{T}\|_{\textbf{R}(A^{1/2})}.$ As $\textbf{R}(A^{1/2})$ is a Hilbert space and $\widetilde{T}$ is compact, one can easily observe that there exists $y_{0}\in\textbf{R}(A^{1/2})$ with $\|y_{0}\|_{\textbf{R}(A^{1/2})}=1$ such that $\|\widetilde{T}y_{0}\|_{\textbf{R}(A^{1/2})}=\|\widetilde{T}\|_{\textbf{R}(A^{1/2})}.$ Since $R(A)$ is closed, there exists $x_{0}\in\mathbb{H}$ such that $Ax_{0}=y_{0}.$ Clearly, $\|x_{0}\|_{A}=\|y_{0}\|_{\textbf{R}(A^{1/2})}=1.$ So we have $\|\widetilde{T}Ax_{0}\|_{\textbf{R}(A^{1/2})}=\|\widetilde{T}\|_{\textbf{R}(A^{1/2})}.$ In other words, $\|Tx_{0}\|_{A}=\|T\|_{A}.$ This shows that $M_{A}(T)\neq\emptyset.$ ∎

We show by an example that the closedness of $R(A)$ cannot be omitted from the above proposition.

Example 2.8**.**

Let $A:\ell_{2}\to\ell_{2}$ be defined as $Ae_{n}=\frac{1}{n}e_{n},$ where $\{e_{n}\}$ denotes the standard orthonormal basis of $\ell_{2}$. Clearly, $A$ is nonzero positive operator. Also, $R(A)$ is not closed. Consider $T\in\mathbb{B}(\ell_{2})$ defined as $Te_{n}=\frac{1}{n^{3/2}}e_{1}.$ We first show that $T$ is $A$-bounded. Given any $x=(x_{n})\in\ell_{2},$ $\|x\|_{A}=\|A^{1/2}x\|=\left(\sum\limits_{n=1}^{\infty}\frac{|x_{n}|^{2}}{n}\right)^{1/2}.$ Now

[TABLE]

Thus, we get $\|Tx\|_{A}\leq\frac{\pi}{\sqrt{6}}\|x\|_{A}.$ This shows that $T$ is $A$-bounded, i.e., $T\in\mathbb{B}_{A^{1/2}}(\ell_{2}).$ Next, we claim that $\|T\|_{A}=\frac{\pi}{\sqrt{6}}.$ Note that

[TABLE]

Let us consider that $y=(y_{n})=(\frac{x_{n}}{\sqrt{n}})\in\ell_{2}$ and $z=(\frac{1}{n})\in\ell_{2}.$ Observe that $y=A^{1/2}x$ and therefore, $\|x\|_{A}=1\iff\|y\|=1.$ Then following the above relation we get

[TABLE]

If possible let there exists $x=(x_{n})\in\ell_{2}$ with $\|x\|_{A}=1$ such that $\|Tx\|_{A}=\frac{\pi}{\sqrt{6}}.$ Now, $\left|\sum\limits_{n=1}^{\infty}\frac{x_{n}}{n^{3/2}}\right|=|\langle y,z\rangle|=\|z\|=\|y\|\|z\|.$ Now by the equality condition of Cauchy-Schwarz inequality, we have $y=\lambda z,$ for some positive scalar $\lambda.$ Thus we obtain $x=(x_{n})=(\frac{\lambda}{\sqrt{n}}).$ But note that $x\notin\ell_{2}.$ This shows that $M_{A}(T)=\emptyset.$ On the other hand, since $R(T)$ is finite-dimensional, one can easily check that $T$ is $A$-compact. Thus, being $A$-compact, $T$ does not attain its $A$-norm whenever $R(A)$ is not closed.

In the following theorem, we completely characterize the $A$-smoothness of $A$-bounded operators provided that $M_{A}(T)$ is nonempty. This will lead to the characterization of the $A$-smoothness of $A$-compact operators.

Theorem 2.9**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ be such that $\|T\|_{A}\neq 0$ and let $M_{A}(T)\neq\emptyset.$ Then the following are equivalent:

• (i)

$T$* is $A$-smooth.*

• (ii)

$M_{A}(T)\cap\overline{R(A)}=\{\mu x_{0}:|\mu|=1\},$* for some $x_{0}\in S_{\mathbb{H}(A)}$ and $\sup\{\|Ty\|_{A}:\langle x_{0},y\rangle_{A}=0\,\,\mbox{with}\,\,\|y\|_{A}=1\}<\|T\|_{A}.$*

Proof.

(i) $\implies$ (ii): Suppose on the contrary that $z_{1},z_{2}\in M_{A}(T)\cap\overline{R(A)}$ are such that $z_{1}\neq\lambda z_{2}$ for any $\lambda\in\mathbb{C}.$ From [20, Th. 2.4] we note that $M_{A}(T)\cap\overline{R(A)}$ is $A$-unit sphere of some subspace of $\mathbb{H}.$ Thus, without loss of generality we may assume that $\langle z_{1},z_{2}\rangle_{A}=0.$ Let $H_{0}=span\{z_{1},z_{2}\}$ and $H_{0}^{\perp_{A}}=\{y\in\mathbb{H}:\langle x,y\rangle_{A}=0,x\in H_{0}\}.$ Observe that $H_{0}\cap H_{0}^{\perp_{A}}=\{0\}.$ So we can write $\mathbb{H}=H_{0}\oplus H_{0}^{\perp_{A}}.$ Given any $z\in\mathbb{H}$ we write $z=\alpha z_{1}+\beta z_{2}+h$ for some $\alpha,\beta\in\mathbb{C}$ and $h\in H_{0}^{\perp_{A}}.$ Now we define $S_{1},S_{2}\in\mathbb{B}(\mathbb{H})$ as $S_{1}z=\alpha_{1}Tz_{1}$ and $S_{2}z=\beta Tz_{2}+Th.$ As $T$ is $A$-bounded, then clearly $S_{1},S_{2}\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$. Note that $S_{1}z_{2}=0$ and $S_{2}z_{1}=0.$ This implies that $T\perp_{A}^{B}S_{1}$ and $T\perp_{A}^{B}S_{2}.$ Since $S_{1}+S_{2}=T,$ we arrive at a contradiction that $T$ is $A$-smooth. Thus, $M_{A}(T)\cap\overline{R(A)}=\{\mu x_{0}:|\mu|=1\}$ for some $x_{0}\in S_{\mathbb{H}(A)}.$

Now again suppose on the contrary that $\sup\{\|Ty\|_{A}:\langle x_{0},y\rangle_{A}=0\,\,\mbox{with}\,\,\|y\|_{A}=1\}=\|T\|_{A}.$ For any $z\in\mathbb{H}$ we write $z=\alpha x_{0}+h$ where $\alpha\in\mathbb{C}$ and $h\in H_{0}:=\{y\in\mathbb{H}:\langle x_{0},y\rangle_{A}=0\}.$ Define $S_{1},S_{2}$ as $S_{1}z=\alpha Tx_{0}$ and $S_{2}z=Th$ for all $z\in\mathbb{H}.$ From our assumption there exists $\{h_{n}\}\subset H_{0}$ with $\|h_{n}\|_{A}=1$ satisfying $\lim\|Th_{n}\|_{A}\to\|T\|_{A}$, we have $\langle Th_{n},S_{1}h_{n}\rangle_{A}=0.$ This implies $T\perp_{A}^{B}S_{1}$. Whereas, $\langle Tx_{0},S_{2}x_{0}\rangle_{A}=0$ implies $T\perp_{A}^{B}S_{2}.$ But observe that $T\not\perp_{A}^{B}S_{1}+S_{2}=T,$ which contradicts the fact that $T$ is $A$-smooth. This completes the proof of the theorem.

(ii)$\implies$ (i): Let $S_{1},S_{2}\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ be such that $T\perp_{A}^{B}S_{i}$ for each $i\in\{1,2\}.$ Now we claim that $\langle Tx_{0},S_{i}x_{0}\rangle_{A}=0$ for each $i\in\{1,2\}.$ We only show it for $S_{1}$ as the case for $S_{2}$ follows similarly. From Theorem 2.4 we note that there exists $\{z_{n}\}\subset S_{\mathbb{H}(A)}$ such that $\|Tz_{n}\|_{A}\to\|T\|_{A}$ and $\langle Tz_{n},S_{1}z_{n}\rangle_{A}\to 0.$ Suppose that $H_{0}=\{y\in\mathbb{H}:\langle x_{0},y\rangle_{A}=0\}.$ Note that $x_{0}\notin H_{0}$ otherwise $\|x_{0}\|_{A}=0.$ Therefore, we may write $z_{n}=\alpha_{n}x_{0}+h_{n},$ where $\alpha_{n}\in\mathbb{C}$ and $h_{n}\in H_{0}.$ As $\|z_{n}\|_{A}=1,$ it is easy to see that $|\alpha_{n}|\leq 1$ and $\|h_{n}\|_{A}^{2}=1-|\alpha_{n}|^{2}.$ Since $\{\alpha_{n}\}\subset\mathbb{C}$ is bounded, without loss of generality we can consider $\alpha_{n}\to\alpha_{0}$ with $|\alpha_{0}|\leq 1.$ Next, we show that $\lim\|h_{n}\|_{A}=0$. On the contrary suppose that $\lim\|h_{n}\|_{A}\neq 0$. Without loss of generality, let $\|h_{n}\|_{A}\neq 0$ for all $n\in\mathbb{N}.$ From [14, Th. 2.6], whenever $x_{0}\in M_{A}(T),$ we have $\langle x_{0},y\rangle_{A}=0\implies\langle Tx_{0},Ty\rangle_{A}=0.$ Now

[TABLE]

From the above equality we get $\lim\|Th_{n}\|_{A}^{2}=(1-|\alpha_{0}|^{2})\|T\|_{A}^{2}.$ Previously we already have $\|h_{n}\|_{A}^{2}=1-|\alpha_{n}|^{2}$ which implies $1-|\alpha_{0}|^{2}=\lim\|h_{n}\|_{A}^{2}.$ So, we get $\lim\|Th_{n}\|_{A}^{2}=\lim\|h_{n}\|_{A}^{2}\|T\|_{A}^{2}.$ This implies that $\lim\left\|T\left(\frac{h_{n}}{\|h_{n}\|_{A}}\right)\right\|_{A}=\|T\|_{A},$ which contradicts our hypothesis. Thus, we have $\lim\|h_{n}\|_{A}=0$ and so $|\alpha_{0}|=1.$ Therefore, $\|Tz_{n}\|_{A}\to\|Tx_{0}\|_{A}=\|T\|_{A}$ will obtain $\langle Tz_{n},S_{1}z_{n}\rangle_{A}\to 0\implies\langle Tx_{0},S_{1}x_{0}\rangle_{A}=0.$ This completes the proof of our claim. Now one can easily observe that $T\perp_{A}^{B}S_{1}$ and $T\perp_{A}^{B}S_{2}$ will imply $T\perp_{A}^{B}S_{1}+S_{2}.$ This means $T$ is $A$-smooth.

∎

The characterization of Birkhoff-James orthogonality of operators in $\mathbb{B}(\mathbb{H}),$ popularly known as Bhatia-Šemrl Theorem was proved in [4] and [16] independently. Zamani generalizes this Bhatia-Šemrl Theorem in the framework of semi-Hilbertian structure. An operator $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ is said to satisfy the Bhatia-Šemrl Property (in short, BŠ Property) if for any $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}),$ there exists $x\in M_{A}(T)$ such that $T\perp_{A}^{B}S\iff\langle Tx,Sx\rangle_{A}=0.$ In this thread, we observe a connection between $A$-smooth operator and Bhatia-Šemrl Property.

Theorem 2.10**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ be such that $\|T\|_{A}\neq 0$ and $M_{A}(T)\neq\emptyset.$ Then $T$ is $A$-smooth if and only if the following conditions hold:

• (i)

$M_{A}(T)\cap\overline{R(A)}=\{\mu x_{0}:|\mu|=1\}.$**

• (ii)

For any $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$, $T\perp_{A}^{B}S\iff\langle Tx_{0},Sx_{0}\rangle_{A}=0$, i.e., $T$ satisfies BŠ Property.

Proof.

To prove the necessary part, we only prove (ii) as (i) follows directly from Theorem 2.9. To prove “$\implies$”, suppose on the contrary there exists $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ such that $T\perp_{A}^{B}S$ but $\langle Tx_{0},Sx_{0}\rangle_{A}\neq 0.$ Consider $S^{\prime}=T-\frac{\|T\|_{A}^{2}}{\langle Tx_{0},Sx_{0}\rangle_{A}}S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ One can clearly observe that $T\perp_{A}^{B}S^{\prime}.$ As $T$ is $A$-smooth, we obtain that $T\perp_{A}^{B}(S^{\prime}+\frac{\|T\|_{A}^{2}}{\langle Tx_{0},Sx_{0}\rangle_{A}}S),$ i.e., $T\perp_{A}^{B}T,$ which is a contradiction. The reverse implication trivially follows.

Next we show the sufficient part. Suppose that $S_{1},S_{2}\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ are such that $T\perp_{A}^{B}S_{1}$ and $T\perp_{A}^{B}S_{2}.$ Then we have $\langle Tx_{0},S_{i}x_{0}\rangle_{A}=0$ for each $i\in\{1,2\}.$ From this we obtain $\langle Tx_{0},(S_{1}+S_{2})x_{0}\rangle_{A}=0.$ This proves that $T\perp_{A}^{B}S_{1}+S_{2},$ i.e., $T$ is $A$-smooth.This completes the proof. ∎

Next, we characterize the $A$-smoothness of $A$-compact operators by applying Theorem 2.9.

Theorem 2.11**.**

Let $T$ be $A$-compact and let $R(A)$ be closed. Then the following are equivalent:

• (i)

$T$* is $A$-smooth.*

• (ii)

$M_{A}(T)\cap{R(A)}=\{\mu x_{0}:|\mu|=1\},$* for some $x_{0}\in S_{\mathbb{H}(A)}.$*

Proof.

From Proposition 2.7 we note that $M_{A}(T)\neq\emptyset.$ (i)$\implies$(ii) follows from Theorem 2.9. To prove (ii)$\implies$(i), we only show that for any $A$-compact operator $T$,

[TABLE]

Since $T$ is $A$-compact and $R(A)$ is closed, then from Lemma 2.1 and Lemma 2.2 (iii) we get a unique $\widetilde{T}\in\mathbb{B}(\textbf{R}(A^{1/2}))$ is also compact. As $M_{A}(T)\cap R(A)=\{\mu x_{0}:|\mu|=1,\|x_{0}\|_{A}=1\}$, using Proposition 2.3 we get $M(\widetilde{T})=\{\mu Ax_{0}:|\mu|=1\}.$ Let $Ax_{0}=y_{0}.$ As $\widetilde{T}$ is compact, one can observe that (c.f. [12, Remark 6.2.4])

[TABLE]

Since $R(A)$ is closed, we can take $z=Au$ for some $u\in\mathbb{H}.$ Then we have

[TABLE]

Now applying Theorem 2.9 again, we prove the desired result. ∎

The following corollary is immediate.

Corollary 2.12**.**

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

• (i)

$T$* is $A$-smooth.*

• (ii)

$M_{A}(T)\cap R(A)=\{\mu x_{0}:|\mu|=1,\|x_{0}\|_{A}=1\}.$**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$. The $A$-operator semi-norm $\|\cdot\|_{A}$ is a continuous convex function on the space $\mathbb{B}_{A^{1/2}}(\mathbb{H})$. Then for $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}),$

[TABLE]

are said to be right-hand and left-hand Gâteaux derivative of $\|\cdot\|_{A}$ at $T$ in the direction $S.$ $\|\cdot\|_{A}$ is Gâteaux differentiable at $T$ if $\rho_{+}^{A}(T,S)=\rho_{-}^{A}(T,S)$ for all $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ We refer the readers to [8] for more detailed study on this topic. Using Theorem 2.5 we show that the Gâteaux derivative of the semi-norm in $\mathbb{B}_{A^{1/2}}(\mathbb{H})$ at a point is equivalent to the $A$-smoothness of that point. Before this we need the following lemma.

Lemma 2.13**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ with $\|T\|_{A}\neq 0.$ Then for every $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ and $\lambda\in\mathbb{R},$ the following are equivalent:

• (i)

$\lambda\in\Re(W_{A}(T,S)).$**

• (ii)

$\rho_{-}^{A}(T,S)\leq\lambda\leq\rho_{+}^{A}(T,S).$**

Proof.

Suppose that (i) holds true. Then there exists a sequence $\{x_{n}\}$ such that $\|x_{n}\|_{A}=1$ and $\lim\|Tx_{n}\|_{A}=\|T\|_{A}$ satisfying $\Re(\lim\langle Tx_{n},Sx_{n}\rangle_{A})=\lambda.$ For any $t>0,$ we have

[TABLE]

Therefore, $\lambda\leq\rho_{+}^{A}(T,S).$ On the other hand, we get

[TABLE]

This shows that $\lambda\geq\rho_{-}^{A}(T,S),$ which proves (ii).

Now let (ii) hold true. For each $n,$ consider $x_{n}^{\lambda}\in\mathbb{H}$ with $\|x_{n}^{\lambda}\|_{A}=1$ such that $\langle Tx_{n}^{\lambda},rSx_{n}^{\lambda}\rangle_{A}=r(1-\frac{1}{n})\lambda$ for all $r\in\mathbb{R}.$ In other words, $x_{n}^{\lambda}\otimes Tx_{n}^{\lambda}$ is a linear functional on the subspace $\{rS:r\in\mathbb{R}\}$ dominated by the sublinear function $(1-\frac{1}{n})\rho_{+}^{A}(T,S).$ By Hahn-Banach extension, for any $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}),$ we observe that $(1-\frac{1}{n})\rho_{-}^{A}(T,S)\leq\Re(\langle Tx_{n}^{\lambda},Sx_{n}^{\lambda}\rangle_{A})\leq(1-\frac{1}{n})\rho_{+}^{A}(T,S).$ Since $\rho_{+}^{A}(T,T)=\rho_{-}^{A}(T,T)=\|T\|_{A}^{2},$ it follows that $\lim\|Tx_{n}^{\lambda}\|_{A}^{2}=\|T\|_{A}^{2}.$ Together with the fact that $\lambda=\Re(\lim\langle Tx_{n}^{\lambda},Sx_{n}^{\lambda}\rangle_{A})$ we obtain $\lambda\in\Re(W_{A}(T,S)).$ This completes the proof of the lemma. ∎

Theorem 2.14**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ be such that $\|T\|_{A}\neq 0.$ Then $\|\cdot\|_{A}$ is Gâteaux differentiable at $T$ if and only if $T$ is $A$-smooth.

Proof.

Suppose that $\|\cdot\|_{A}$ is Gâteaux differentiable at $T$. Then for any $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ we have $\rho_{-}^{A}(T,S)=\rho_{+}^{A}(T,S).$ Then from Lemma 2.13 we have $\Re(W_{A}(T,S))$ is singleton for each $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ If possible let $\mu,\sigma\in W_{A}(T,S_{0})$ for some $S_{0}\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ Let $\mu=\lim\langle Tx_{n},S_{0}x_{n}\rangle_{A}$ and $\sigma=\lim\langle Ty_{n},S_{0}y_{n}\rangle_{A}$ for some $\{x_{n}\},\{y_{n}\}\subset S_{\mathbb{H}(A)}$ satisfying $\lim\|Tx_{n}\|_{A}=\lim\|Ty_{n}\|_{A}=\|T\|_{A}.$ Now

[TABLE]

This proves that $W_{A}(T,S)$ is singleton for each $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ and therefore, from Theorem 2.5 we get $T$ is $A$-smooth.

Conversely, if $T$ is $A$-smooth then from Theorem 2.5 we have $W_{A}(T,S)$ is singleton for every $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ From [21, Lemma 2.1] we get $\Re(W_{A}(T,S))$ is compact. Thus applying Lemma 2.13 we get $\rho_{-}^{A}(T,S)=\rho_{+}^{A}(T,S)$ for every $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}).$ This proves that $\|\cdot\|_{A}$ is Gâteaux differentiable at $T.$ Hence the theorem. ∎

Theorem 2.14 reveals that the problem of whether the $A$-operator semi-norm at a point $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ is Gâteaux differentiable or not can be tackled by the concept of $A$-smoothness of $T.$ In this connection we note the following examples:

Example 2.15**.**

(i) Consider that $\mathbb{H}=\mathbb{R}^{2}$ endowed with its usual inner product norm. Let $T\in\mathbb{B}(\mathbb{H})$ be the identity matrix. As $M(T)=S_{\mathbb{H}},$ $T$ is not smooth. So the operator norm on $\mathbb{B}(\mathbb{H})$ is not Gâteaux differentiable at $T.$ Let $A=\begin{pmatrix}1&0\\ 0&0\end{pmatrix}$. Clearly, $A$ is positive. Also, note that $R(A)=\{(x,0)^{t}:x\in\mathbb{R}\}.$ For any $x\in\mathbb{H}$ satisfying $\|x\|_{A}=1,$ it follows that $x=\pm(1,0)^{t}.$ So $M_{A}(T)\cap R(A)=\{\pm(1,0)^{t}\}.$ From Corollary 2.12, we get $T$ is $A$-smooth. Therefore, from Theorem 2.14, we observe that the $A$-operator semi-norm is Gâteaux differentiable at $T.$ In fact, it is easy to observe that whenever $rank(A)=1$, every $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$ with $\|T\|_{A}\neq 0$ is $A$-smooth.

(ii) Let $\mathbb{H}=\mathbb{R}^{3}.$ Consider the diagonal matrix $T=diag\{2,1,1\}$ and let $A=diag\{0,1,1\}.$ Clearly, $A$ is positive and $R(A)=\{(0,y,z)^{t}:y,z\in\mathbb{R}\}.$ Also note that $M(T)=\{\pm(1,0,0)^{t}\}.$ Therefore, $T$ is smooth. On the other hand, $\|(0,1,0)^{t}\|_{A}=\|(0,0,1)^{t}\|_{A}=1$ and $\|T(0,1,0)^{t}\|_{A}=\|A^{1/2}(0,1,0)^{t}\|=1=\|T\|_{A}.$ So, $\pm(0,1,0)^{t}\in M_{A}(T).$ Similarly, we get $(0,0,1)^{t}\in M_{A}(T).$ It follows from Corollary 2.12 that $T$ is not $A$-smooth. Therefore, from Theorem 2.14 we get that the function $\|\cdot\|_{A}$ on $\mathbb{B}_{A^{1/2}}(\mathbb{H})$ is not Gâteaux differentiable at $T.$

In the following result we compare the $A$-smoothness in $\mathbb{B}_{A^{1/2}}(\mathbb{H})$ with that in $\mathbb{B}(\textbf{R}(A^{1/2}).$

Theorem 2.16**.**

Let $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H})$. Then $T$ is $A$-smooth if and only if $\widetilde{T}$ is smooth.

Proof.

We first prove the necessary part. Since $T\in\mathbb{B}_{A^{1/2}}(\mathbb{H}),$ it follows from Lemma 2.1 that there exists a unique $\widetilde{T}\in\mathbb{B}(\textbf{R}(A^{1/2}))$ such that $\widetilde{T}A=AT$. If possible suppose that $\widetilde{T}$ is not smooth. Then from Corollary 2.6 there exist $\widetilde{S}\in\mathbb{B}(\textbf{R}(A^{1/2}))$ and two sequences $\{x_{n}\},\{y_{n}\}$ with $\|x_{n}\|_{\textbf{R}(A^{1/2})}=\|y_{n}\|_{\textbf{R}(A^{1/2})}=1$ and $\|\widetilde{T}x_{n}\|_{\textbf{R}(A^{1/2})}\to\|\widetilde{T}\|_{\textbf{R}(A^{1/2})},\|\widetilde{T}y_{n}\|_{\textbf{R}(A^{1/2})}\to\|\widetilde{T}\|_{\textbf{R}(A^{1/2})}$ such that

[TABLE]

where $\lambda_{0}\neq\sigma_{0}.$ As for each $n,$ $x_{n},y_{n}\in R(A^{1/2}),$ we write $A^{1/2}u_{n}=x_{n}$ and $A^{1/2}v_{n}=y_{n}$ for some $u_{n},v_{n}\in\mathbb{H}.$ Also, note that $R(A)$ is dense in $\textbf{R}(A^{1/2}).$ Therefore, for each $n,$ there exist $\{u_{n,k}\},\{v_{n,k}\}\subset\mathbb{H}$ such that $\lim\limits_{k\to\infty}Au_{n,k}=A^{1/2}u_{n}$ and $\lim\limits_{k\to\infty}Av_{n,k}=A^{1/2}v_{n}$ in $\|\cdot\|_{\textbf{R}(A^{1/2})}.$ This implies $\lim\limits_{k\to\infty}\|Au_{n,k}\|_{\textbf{R}(A^{1/2})}=\lim\limits_{k\to\infty}\|Av_{n,k}\|_{\textbf{R}(A^{1/2})}=1$ for each $n.$ Also, we have

[TABLE]

Now we consider $A$-normalized subsequences of $\{u_{n,k}\}$ and $\{v_{n,k}\}$, respectively as the following:

[TABLE]

Note that $\|z_{r}\|_{A}=\|w_{s}\|_{A}=1$ for all $r$ and $s.$ From the above equations (2.3) and (2.4) together with the fact that $\lim\limits_{k\to\infty}\|Au_{n,k}\|_{\textbf{R}(A^{1/2})}=1$ we have

[TABLE]

and

[TABLE]

Similarly, from equations (2.5) and (2.6), we can show that $\|Tw_{s}\|_{A}\to\|T\|_{A}$ and $\langle Tw_{s},Sw_{s}\rangle_{A}\to\sigma_{0}.$ Hence $\lambda_{0},\sigma_{0}\in W_{A}(T,S).$ From Theorem 2.5 we get that $T$ is not $A$-smooth, which is a contradiction. Therefore, we obtain that $\widetilde{T}$ is smooth.

To prove the sufficient part, suppose on the contrary that $T$ is not $A$-smooth. Following Theorem 2.5, for some $S\in\mathbb{B}_{A^{1/2}}(\mathbb{H}),$ there exist $A$-norming sequences $\{x_{n}\},\{y_{n}\}$ of $T$ such that $\langle Tx_{n},Sx_{n}\rangle_{A}\to\lambda$ and $\langle Ty_{n},Sy_{n}\rangle_{A}\to\mu$ where $\lambda\neq\mu.$ As $\|Ax_{n}\|_{\textbf{R}(A^{1/2})}=\|x_{n}\|_{A},$ it is easy to see $\|\widetilde{T}Ax_{n}\|_{\textbf{R}(A^{1/2})}\to\|\widetilde{T}\|_{\textbf{R}(A^{1/2})}.$ Moreover, $\langle\widetilde{T}Ax_{n},\widetilde{S}Ax_{n}\rangle_{\textbf{R}(A^{1/2})}\to\lambda.$ Similarly, we obtain that $\|\widetilde{T}Ay_{n}\|_{\textbf{R}(A^{1/2})}=\|\widetilde{T}\|_{\textbf{R}(A^{1/2})}$ and $\langle\widetilde{T}Ay_{n},\widetilde{S}Ay_{n}\rangle_{\textbf{R}(A^{1/2})}\to\mu.$ This contradicts that $\widetilde{T}$ is smooth (see Corollary 2.6). Hence the theorem. ∎

Applying the above theorem we note the following corollary.

Corollary 2.17**.**

Let $T\in\mathbb{B}_{A}(\mathbb{H})$ be $A$-compact and let $R(A)$ be closed. Then $T$ is $A$-smooth if and only if $T^{\sharp}$ is $A$-smooth.

Proof.

Let $T$ be $A$-smooth. It follows from Theorem 2.16 that $\widetilde{T}$ is smooth. Also, from Lemma 2.2 (iii) we get $\widetilde{T}$ is compact. Thus $(\widetilde{T})^{*}$ is smooth (see [18, Th. 1]). Note that, $(\widetilde{T})^{*}=\widetilde{T^{\sharp}}.$ This implies that $\widetilde{T^{\sharp}}$ is smooth. Therefore, again following Theorem 2.16 we have $T^{\sharp}$ is $A$-smooth. The converse part follows using the same argument. ∎

We conclude this article by characterizing the $A$-smoothness of block diagonal matrices using Theorem 2.5. To this end, we first observe the following lemma.

Here we consider $\mathbb{A}=\begin{pmatrix}A&0\\ 0&A\end{pmatrix}\in\mathbb{B}(\mathbb{H}\oplus\mathbb{H}),$ where $A\in\mathbb{B}(\mathbb{H})$ is a positive operator. Clearly, $\mathbb{A}$ is a positive operator on $\mathbb{H}\oplus\mathbb{H}$ and it generates the semi-inner product $\langle x,y\rangle_{\mathbb{A}}=\langle x_{1},y_{1}\rangle_{A}+\langle x_{2},y_{2}\rangle_{A}$ for all $x=\begin{pmatrix}x_{1}\\ x_{2}\end{pmatrix}\in\mathbb{H}\oplus\mathbb{H}$ and $y=\begin{pmatrix}y_{1}\\ y_{2}\end{pmatrix}\in\mathbb{H}\oplus\mathbb{H}.$

Lemma 2.18**.**

Let $T=\begin{pmatrix}M&0\\ 0&N\end{pmatrix}\in B_{\mathbb{A}^{1/2}}(\mathbb{H}\oplus\mathbb{H})$ be such that $\|M\|_{A}>\|N\|_{A}$ and let $\begin{pmatrix}x_{n}\\ y_{n}\end{pmatrix}$ be an $\mathbb{A}$-norming sequence of $T$. Then

• (i)

$\lim_{n\to\infty}\|x_{n}\|_{A}=1.$**

• (ii)

$\lim_{n\to\infty}\|y_{n}\|_{A}=0.$**

• (iii)

$\left\{\frac{x_{n}}{\|x_{n}\|_{A}}\right\}$* is an $A$-norming sequence of $M$.*

Proof.

Since $\begin{pmatrix}x_{n}\\ y_{n}\end{pmatrix}$ is an $\mathbb{A}$-norming sequence of $T$ and $\|M\|_{A}>\|N\|_{A},$ we have

[TABLE]

and $\|x_{n}\|_{A}^{2}+\|y_{n}\|_{A}^{2}=1.$ Now,

[TABLE]

Thus,

[TABLE]

So, $\lim_{n\to\infty}\|x_{n}\|_{A}=1.$ Now, $\lim_{n\to\infty}\|Mx_{n}\|_{A}^{2}=\|M\|_{A}^{2}-\lim_{n\to\infty}\|Ny_{n}\|_{A}^{2}=\|M\|_{A}^{2}.$ If we consider the sequence $\left\{\frac{x_{n}}{\|x_{n}\|_{A}}\right\}$ then it becomes the $A$-norming sequence of $M$. ∎

Remark 2.19**.**

It is easy to observe that if $\{x_{n}\}$ is an $A$-norming sequence of $M$ then $\begin{pmatrix}x_{n}\\ 0\end{pmatrix}$ is an $\mathbb{A}$-norming sequence of $T,$ where $\|M\|_{A}\geq\|N\|_{A}$.

Theorem 2.20**.**

Let $T=\begin{pmatrix}M&0\\ 0&N\end{pmatrix}\in\mathbb{B}_{\mathbb{A}}(\mathbb{H}\oplus\mathbb{H}).$

• (i)

If $\|M\|_{A}>\|N\|_{A}$ then $T$ is $\mathbb{A}$-smooth if and only if $M$ is $A$-smooth.

• (ii)

If $\|M\|_{A}<\|N\|_{A}$ then $T$ is $\mathbb{A}$-smooth if and only if $N$ is $A$-smooth.

• (iii)

If $\|M\|_{A}=\|N\|_{A}$ then $T$ is not $\mathbb{A}$-smooth.

Proof.

(i) Let $\|M\|_{A}>\|N\|_{A}.$ If possible suppose that $M$ is not $A$-smooth. Then there exist $R\in\mathbb{B}_{A}(\mathbb{H})$ and two sequences $\{x_{n}\},\{y_{n}\}$ with $\|x_{n}\|_{A}=\|y_{n}\|_{A}=1$ and $\|Mx_{n}\|_{A}\to\|M\|_{A},\|My_{n}\|_{A}\to\|M\|_{A}$ such that

[TABLE]

where $\lambda\neq\mu$. It follows from Remark 2.19 that $\begin{pmatrix}x_{n}\\ 0\end{pmatrix}$ and $\begin{pmatrix}y_{n}\\ 0\end{pmatrix}$ are two $\mathbb{A}$-norming sequences of $T.$ Now consider $S=\begin{pmatrix}R&Q\\ U&V\end{pmatrix}\in\mathbb{B}_{\mathbb{A}}(\mathbb{H}\oplus\mathbb{H}).$ One can clearly observe that

[TABLE]

Similarly we get

[TABLE]

As $\lambda\neq\mu,$ it follows from Theorem 2.5 that $T$ is not $\mathbb{A}$-smooth, which is a contradiction.

Conversely, let $M$ be $A$-smooth. Then $W_{A}(M,P)$ is singleton for all $P\in\mathbb{B}_{A}(\mathbb{H}).$ If possible suppose that $T$ is not $\mathbb{A}$-smooth. Then there exists $S=\begin{pmatrix}P&Q\\ U&V\end{pmatrix}\in\mathbb{B}_{\mathbb{A}}(\mathbb{H}\oplus\mathbb{H})$ and two $\mathbb{A}$-norming sequences $\begin{pmatrix}x_{n}\\ y_{n}\end{pmatrix},$ $\begin{pmatrix}u_{n}\\ v_{n}\end{pmatrix}$ of $T$ such that

[TABLE]

From Lemma 2.18, we have $\lim\limits_{n\to\infty}\|y_{n}\|_{A}=0$ and so

[TABLE]

Similarly, from $\left\langle T\begin{pmatrix}u_{n}\\ v_{n}\end{pmatrix},S\begin{pmatrix}u_{n}\\ v_{n}\end{pmatrix}\right\rangle_{\mathbb{A}}\to\mu,$ we have

[TABLE]

where $\lambda\neq\mu$. It follows from Lemma 2.18 that $\left\{\frac{x_{n}}{\|x_{n}\|_{A}}\right\}$ and $\left\{\frac{u_{n}}{\|u_{n}\|_{A}}\right\}$ are two $A$-norming sequences of $M.$ Since $M$ is $A$-smooth, we have $\langle Mx_{n},Px_{n}\rangle_{A}\to\mu_{0}$ and $\langle Mu_{n},Pu_{n}\rangle_{A}\to\mu_{0}$. From (2) and (2.9), we have $\lambda=\mu=\mu_{0}$. This contradicts the fact that $\lambda\neq\mu$. Therefore, $T$ is $\mathbb{A}$-smooth.

(ii) The proof of (ii) follows by using the same argument as in (i).

(iii) Let $\|M\|_{A}=\|N\|_{A}.$ Suppose $\{x_{n}\}$ and $\{y_{n}\}$ are two $A$-norming sequences of $M$ and $N,$ respectively. Let $P\in\mathbb{B}_{A}(\mathbb{H})$ be such that $\langle Mx_{n},Px_{n}\rangle_{A}\to\lambda(\neq 0).$ Since $\{x_{n}\}$ and $\{y_{n}\}$ are two $A$-norming sequences of $M$ and $N,$ then $\begin{pmatrix}x_{n}\\ 0\end{pmatrix}$ and $\begin{pmatrix}0\\ y_{n}\end{pmatrix}$ are two $\mathbb{A}$-norming sequences of $T.$ Suppose $S=\begin{pmatrix}P&0\\ 0&0\end{pmatrix}\in\mathbb{B}_{\mathbb{A}}(\mathbb{H}\oplus\mathbb{H}).$ Then $\langle Mx_{n},Px_{n}\rangle_{A}\to\lambda(\neq 0)$ and $\langle Ny_{n},Py_{n}\rangle_{A}\to 0.$ Therefore, from Theorem 2.5, we get $T$ is not $\mathbb{A}$-smooth. ∎

Acknowledgements

Somdatta Barik and Souvik Ghosh would like to thank UGC, Govt. of India and CSIR, Govt. of India, respectively, for the support in the form of Senior Research Fellowship under the mentorship of Professor Kallol Paul. The research of Professor Kallol Paul is supported by CRG Project bearing File no. CRG/2023/00716 of DST-SERB, Govt. of India.

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The authors have no relevant financial or non-financial interests to disclose.

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All authors contributed to the study. All authors read and approved the final version of the manuscript.