of [@SPJ].

\[lemma:rotund\] Let $\mathbb{X}$ be a normed linear space.Then $x\in S_\mathbb{X}$ is a rotund point if and only if $x\bot_B y \Rightarrow x\bot_{SB} y$ for any $y\in \mathbb{X} \setminus \{\theta\}$.

In the following theorem, we obtain the characterization of Birkhoff-James orthogonality of rotund points in the space of bounded linear operators defined between infinite dimensional normed linear spaces.

\[theorem:rotund\] Let $\mathbb{X}$ and $\mathbb{Y}$ be two normed linear spaces. Let $T$ be a rotund point of $\mathbb{B} (\mathbb{X}, \mathbb{Y})$. Then for any $A \in \mathbb{B} (\mathbb{X}, \mathbb{Y}) , ~T \bot_B A$ if and only if there exists two sequences $\{x_n\}$, $\{y_n\}$ in $S_\mathbb{X}$ and two sequences of positive real numbers $\{\epsilon_n\}$ , $\{\delta_n\}$ such that\
$(i)  \epsilon_n \rightarrow 0$ , $\delta_n\rightarrow 0$ as $n\rightarrow \infty$ .\
$(ii) \| Tx_n\| \rightarrow \|T\|$ and $\|Ty_n\|\rightarrow\|T\|$ as $n\rightarrow \infty$ .\
$(iii)Ax_n\in(Tx_n)^{+(\epsilon_n)}$ and $Ay_n\in(Ty_n)^{-(\delta_n)}$ for all $n\in \mathbb{N}$.

Let us first prove the easier sufficient part.\
Since $ Ax_n\in(Tx_n)^{+(\epsilon_n)} $ for all $n\in \mathbb{N}, $ $\|Tx_n+\lambda Ax_n\|\geq \sqrt[]{1-\epsilon_n^2}~\|Tx_n\|$ for all $\lambda\geq0.$\
This implies, given any $ \lambda \geq 0, $ $ \|T+\lambda A\|\geq\|(T+\lambda A)x_n\| \geq\sqrt[]{1-\epsilon_n^2} \|Tx_n\|$. Since $ \epsilon_n \rightarrow 0 $ and $ \|Tx_n\| \rightarrow \| T \| $ as $ n \rightarrow 