- \frac{1}{n} \|Ax_n\|$ $>\|T\|- \frac{1}{n^3} - \frac{1}{n} \|A\|$ $\rightarrow \|T\|$ as $n \rightarrow \infty. $ Since $ x_n \in S_{\mathbb{X}},$ $ \| Tx_n \| \leq \| T \|  $. This proves our claim.\
Since $(a)$ does not hold, we assume that, $\inf \limits_{n \in \mathbb{N}}\|Ax_n\|= c > 0$.\
Choose $n_1 \in \mathbb{N}$ such that $n_1 > \frac{2 \|T\|}{c}$. Since $\|Tx_n\| \rightarrow \|T\|>0,$ there exists $n_2\in \mathbb{N}$ such that $\|Tx_n\| > \frac{\|T\|}{2} > 0$ for all $n \geq n_2$. Choose $n_3 \in \mathbb{N}$ such that $n_3 > \frac{2}{\|T\|}$. Let $n_0 =$ max$\{ n_1,n_2, n_3\}$. Then for all $n\geq n_0$, $0< \frac{1}{n \|Tx_n\|} < \frac{2}{n \|T\|}< 1$, which implies that for all $n\geq n_0$, $0< 1 - \frac{1}{n \|Tx_n\|}< 1$.\
Choose $\epsilon_n=\sqrt[]{1- (1- \frac{1}{n \|Tx_n\|})^2}$. 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 $0 \leq \lambda < \frac{1}{n}$,\
$\|Tx_n + \lambda Ax_n\| \geq \|Tx_n\|- \lambda \|Ax_n\| \geq \|Tx_n\|-\frac{1}{n}$.\
For $\frac{1}{n} \leq \lambda \leq n$, we claim that $\|Tx_n +\lambda Ax_n\| \geq \|Tx_n\|- \frac{1}{n}$.\
Suppose on the contrary, we have, $\|Tx_n +\lambda Ax_n\| < \|Tx_n\|- \frac{1}{n}$ for some $\frac{1}{n} \leq \lambda \leq n$. Now, $Tx_n + 