Proving discontinuity of linear operator using the def.

Question

Let $T:C_{00}\rightarrow C_{00}$ by $(Tx)(n)=nx_{n}$. Then, $T$ is a well-defined map, and is linear. For each $n\in\mathbb{N}$, let $\delta^{(n)}\in C_{00}$ be defined as $\delta_{k}^{(n)}=\begin{cases} 1, & \mbox{if }k=n\\ 0, & \mbox{if }k\neq n \end{cases}.$

So, $||\delta^{(n)}||=1$ but $||T\delta^{(n)}||=n$. Hence, $T$ is unbounded.

I want to show that it is discontinuous using the definition , wondered to know if it is correct?

let $x_n,y_n∈C_{00}$ , s.t. $\|x_n−y_n\|_\infty \le \eta, n \in \mathbb{N}$ and for some $\eta>0$.

Then, $(x_n−y_n)\in C_{00}$ and ; $$ \|Tx_n−Ty_n\|_\infty = \|T(x_n−y_n)\|_\infty = n \to \infty$$ , so it is not continuous.

Answer 1

Your argument doesn't make sense. How did you get $\|T(x_n-y_n)\|=n$?

A general argument: Suppose $T$ is a linear map on a normed linear space such that there exists a sequence $(x_n)$ with $\|x_n\|=1$ for all $n$ and $\|Tx_n\| \to \infty$. Let $y_n=\frac {x_n} {\sqrt {\|Tx_n\|}}$. Then $y_n \to 0$ and $\|Ty_n\| \to \infty$. Hence $T$ is not continuous.