Let $T$ be a linear operator between two normed spaces $X$ and $Y$. Show that
- a) If $T$ is not bounded then for $\varepsilon>0$ , $\sup \Vert Tx\Vert = \infty$ where $\Vert x \Vert < \varepsilon$.
- b) If $\{T(x_{n})\}$ is bounded for any sequence $\{x_{n}\}$ such that $x_n\to 0$, then $T$ is bounded.
Please help me if you have a good idea for my question. I do not understand this concept. Thank you.
(a) If $\sup_{\|x\|<\epsilon} \|Tx\|=M<\infty $ then $\sup_{x\in X,\|x\|>0}\frac{\|Tx\|}{\|x\|}=\sup_{x\in X,\|x\|>0}\|T(\epsilon x/\|x\|)\|/\epsilon\leq \sup_{\|x\|<\epsilon} \|Tx\|/\epsilon=M/\epsilon<\infty. $ Hence $\sup_{\|x\|<\epsilon} \|Tx\|<\infty$ implies $T$ is bounded.
(b) Let $T$ is unbounded. So, by (a) for any $n\in\mathbb{N}$, $\sup_{\|x\|<1/n}=\infty$. So there exists an $x_n$ with $\|x_n\|<1/n$ such that $\|T(x_n)\|\geq n$. Hence $T$ unbounded implies there is a sequence $(x_n)$ converging to zero such that $\|T(x_n)\|\rightarrow \infty.$