\infty, $ we obtain, $$\|T+\lambda A \|\geq \|T\|.$$ Similarly, $ Ay_n\in(Ty_n)^{-(\delta_n)} $ for all $n\in \mathbb{N}$ implies that, for any $ \lambda < 0, $ $$\|T+\lambda A \|\geq \|T\|.$$ This completes the proof of the sufficient part.\
Let us now prove the more involved necessary part.\
Since $T\in \mathbb{B} (\mathbb{X}, \mathbb{Y})$ is a rotund point, clearly $ T $ is nonzero. Also by Lemma \[lemma:rotund\], $T\bot _B A\Rightarrow T\bot_{SB} A$ for any nonzero $ A \in \mathbb{B} (\mathbb{X}, \mathbb{Y}). $\
Therefore for each $n\in \mathbb{N}$, $$\|T+ \frac{1}{n}A\| >\|T\|.$$ This implies that for each $ n \in \mathbb{N}, $ there exists $x_n\in S_\mathbb{X}$ such that $\|(T+\frac{1}{n}A)x_n\|>\|T\|.$ We claim that $\|Tx_n\|\rightarrow \|T\|$. Indeed, $\|Tx_n\|=\|(T+\frac{1}{n}A)x_n - \frac{1}{n}Ax_n\|$ $\geq \|(T+\frac{1}{n}A)x_n\| - \frac{1}{n} \|Ax_n\|$ $>\|T\|- \frac{1}{n} \|A\|$ $\rightarrow \|T\|$ as $n \rightarrow \infty. $ Since $ x_n \in S_{\mathbb{X}},$ $ \| Tx_n \| \leq \| T \|  $. This proves our claim.\
Since $\|Tx_n\| \rightarrow \|T\|>0,$ there exists $n_1\in \mathbb{N}$ such that $\|Tx_n\| > \frac{\|T\|}{2} > 0$ for all $n \geq n_1$. Choose $n_2 \in \mathbb{N}$ such that $n_2 > \frac{2 \|A\|}{\|T\|}$. Let $n_0 =$ max$\{ n_1,n_2\}$. Then for all $n\geq n_0$, $0< \frac{\|A\|}{n \|Tx_n\|} < \frac{2 \|A\|}{n \|T\|}< 1$, which implies that for all 