$\{\epsilon_n\}$ , $\{\delta_n\}$ 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 $n\in \mathbb{N}$.

We first prove the easier sufficient part.\
Suppose $(a)$ holds. Now for any scalar $\lambda$, $\|T + \lambda A\| \geq \|Tx_n + \lambda Ax_n\|\geq \|Tx_n\|- |\lambda | \|Ax_n\| \rightarrow \|T\|$ as $n \rightarrow \infty$. Therefore $T \bot_{B} A$.\
Now suppose $(b)$ holds. Then following the same line of arguments as in the proof of the sufficient part of Theorem \[theorem:rotund\], we obtain $T\perp_B A$. This completes the proof of the sufficient part.\
Let us now prove the comparatively trickier necessary part.\
Suppose $(a)$ does not hold.\
Without loss of generality let us assume that $\|A\| \leq 1$.\
Since $T \bot_{B} A$, for any nonzero scalar $\lambda$, $\|T + \lambda A\| \geq \|T\|$. In particular, for each $n \in \mathbb{N}$, $$\|T + \frac{1}{n} A\| > \|T\| - \frac{1}{n^3}.$$ Therefore, for each $n \in \mathbb{N}$, there exists a sequence $\{x_n\} $ in $ S_{\mathbb{X}}$ such that $\|(T+ \frac{1}{n} A) x_n \| > \|T\|- \frac{1}{n^3} \geq \|Tx_n\|- \frac{1}{n^3}$.\
We claim that $\|Tx_n\|\rightarrow \|T\|$. Indeed, $\|Tx_n\|=\|(T+\frac{1}{n}A)x_n - \frac{1}{n}Ax_n\|$ $\geq \|(T+\frac{1}{n}A)x_n\| 