If $ T : X \rightarrow \mathbb{R}$ is a non bounded linear operator then $\forall \alpha \in \mathbb{R^{+}} \ \exists\{x_n\}\subseteq X$ such that $x_n \rightarrow 0$ implies $Tx_n \rightarrow \alpha$
My teacher asked for the veracity of the previous assertion. I don't know yet if it's true. When attempt the problem (showing that it's true) I have difficulty to construct a sequence that converge to $\alpha$.
Any suggestions. Thanks.
The question is not properly worded. 'There exists' and 'implies' don't go together. The way I understand this is: show there exists $\{x_n\}$ such that $\|x_n\| \to 0$ and $Tx_n\to \alpha$ for all $n$. Since $T$ is not bounded there exists $\{x_n\}$ such that $|Tx_n| >n \|x_n\|$ . Let $t_n=\frac {n\|x_n\|} {Tx_n}$. Let $y_n=\frac {t_nx_n} {n\|x_n\|}$. Now verify that $Ty_n=1$ for all $n$ and $\|y_n\|\to 0$. This is the answer when $\alpha =1$. Just replace $y_n$ by $\alpha y_n$ for other values of $\alpha$.