\mathbb{N}$. Therefore letting $n\rightarrow \infty$, we have, $\|T+ \lambda A\| \geq \|T\|$, since $\| Tx_n\| \rightarrow \|T\|$ as $n\rightarrow \infty$.\
Similarly, $Ay_n\in (Ty_n)^{-}$ implies that, for any $\lambda \leq 0$, $\|T+ \lambda A\| \geq \|Ty_n+ \lambda Ay_n\| \geq \|Ty_n\|$ for all $n \in \mathbb{N}$. Therefore letting $n\rightarrow \infty$, we have, $\|T+ \lambda A\| \geq \|T\|$, since $\| Ty_n\| \rightarrow \|T\|$ as $n\rightarrow \infty$. This completes the proof of the fact that $T\bot_B A$.

In the next example we illustrate the fact that the conditions stated in Proposition $ 2.2 $ are only sufficient but not necessary for $ T \perp_B A. $

Define $T, A : l_1 \rightarrow l_1$ by $Te_n = (1- \frac{1}{n+1})e_n$, $n\geq 1$, and $Ae_n = \frac{1}{n+1} e_n$, $n\geq 1$, where $e_n =(0,0, \ldots ,0,1,0, \ldots ),$ with $1$ in the $n$-th position and $0$ elsewhere.\
Let $x= \sum \limits_{n=1}\limits ^\infty a_n e_n \in l_1$ where $a_n \in \mathbb{R}$ for all $n$. Then $\|Tx\| =\sum \limits_{n=1}\limits ^\infty |a_n| |1- \frac{1}{n+1}| \leq \sum \limits_{n=1}\limits ^\infty |a_n| = \|x\|$. So $\|T\| \leq 1$. Also $\|Te_n\|= (1- \frac{1}{n+1}) \rightarrow 1$ as $n \rightarrow \infty$. Hence $\|T\|= 1$.\
First we show that $ T \perp_B A.$ For any scalar $ \lambda  