$n\geq n_0$, $0< 1 - \frac{ \|A\|}{n \|Tx_n\|}< 1$.\
Choose $\epsilon_n=\sqrt[]{1- (1- \frac{\|A\|}{n \|Tx_n\|})^2}$. Then clearly $\epsilon_n \rightarrow 0$ as $n \rightarrow \infty$ .\
We claim that $Ax_n\in(Tx_n)^{+(\epsilon_n)}$ for all $n\geq n_0$ .\
Let $n\geq n_0$. Then for $0< \lambda \leq \frac{1}{n}$,\
$\|Tx_n+\lambda Ax_n\|\geq \|Tx_n\|-|\lambda | \|Ax_n\|\geq \|Tx_n\|-\frac{1}{n} \|A\|=\sqrt{1-\epsilon_n^2} \|Tx_n\|$.\
For $\lambda > \frac{1}{n} > 0$, we claim that $\|Tx_n+ \lambda Ax_n\| \geq \|Tx_n\|. $\
Suppose on the contrary, we have, $\|Tx_n+ \lambda Ax_n\| < \|Tx_n\| $ for some $ \lambda > \frac{1}{n}. $ Now, there exists $ t \in (0, 1) $ such that $Tx_n+  \frac{1}{n} Ax_n = t(Tx_n)+ (1-t)(Tx_n+\lambda Ax_n)$. This implies that, $\|Tx_n+ \frac{1}{n} Ax_n \| \leq t \|Tx_n\|+ (1-t)\|Tx_n+ \lambda Ax_n\|< t \|Tx_n\| +(1-t) \|Tx_n\|= \|Tx_n\| \leq \|T\|$, a contradiction.\
Therefore, for all $\lambda \geq 0$, $\|Tx_n+ \lambda Ax_n\| \geq \|Tx_n\|-\frac{1}{n} \|A\|=\sqrt[]{1-\epsilon_n^2} ~\|Tx_n\|$. This completes the proof of our claim.\
Similarly, considering $\|T-\frac{1}{n} A\|> \|T\|$ 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 $ (i), (ii) $ and $ (iii) $ are satisfied.

Next we illustrate with an example to show that the conditions mentioned in Theorem 