\frac{1}{n} Ax_n = (1- \frac{1}{n \lambda})Tx_n + \frac{1}{n \lambda}(Tx_n +\lambda Ax_n)$. This implies that, $\|Tx_n\|- \frac{1}{n^3} < \|Tx_n + \frac{1}{n} Ax_n\| \leq  (1- \frac{1}{n \lambda})\|Tx_n\| + \frac{1}{n \lambda}\|(Tx_n +\lambda Ax_n)\| < (1- \frac{1}{n \lambda})\|Tx_n\| + \frac{1}{n \lambda}(\|Tx_n\| - \frac{1}{n})= \|Tx_n\|- \frac{1}{n^2 \lambda}$. This implies that $\lambda > n$, a contradiction.\
Thus for $0 \leq \lambda \leq n$, $\|Tx_n +\lambda Ax_n\| \geq \|Tx_n\|- \frac{1}{n}$. Hence, $0 \leq \lambda \leq \frac{2 \|T\|}{c}$ gives that $\|Tx_n +\lambda Ax_n\| \geq \|Tx_n\|- \frac{1}{n}$.\
Now, for $\lambda > \frac{2 \|T\|}{c}$, $\|Tx_n +\lambda Ax_n\| \geq \lambda \|Ax_n\|- \|Tx_n\| \geq \lambda c- \|Tx_n\| > 2\|T\|-\|Tx_n\| \geq \|Tx_n\|- \frac{1}{n}$.\
Therefore for all $\lambda \geq 0$, $\|Tx_n +\lambda Ax_n\| \geq \|Tx_n\|- \frac{1}{n}= \sqrt[]{1-\epsilon_n^2} ~\|Tx_n\|$. This completes the proof of our claim.\
Similarly, considering $\|T-\frac{1}{n} A\|> \|T\|- \frac{1}{n^3}$ for each $n \in \mathbb{N}$, we can find the desired sequences $\{y_n\} $ in $ S_{\mathbb{X}}$ and $ \{\delta _n\} $ in $ \mathbb{R}^+$ such that all the conditions of $(b)$ are satisfied.

As an application of Theorem  \[theorem:rotund\] we give a complete characterization of Birkhoff-James orthogonality of bounded linear functionals on a normed linear space, whose dual is strictly convex. First, we need the following easy proposition.

\[prop:real\] For any two real 