If $c_{00}$ be the space of sequences with a finite number of non-zero elements with $\|.\|_\infty$ norm and $\Gamma : c_{00} \to c_{00}$ a linear operator. Is there any example of unbounded linear operator?
I was thinking to give the following if $x = (x_1,x_2,...)$, then $\Gamma x_i = a_i x_i$, where $\{a_i\}_{i\ge 1}$ be I be an unbounded sequence of real numbers.
But then such a transformation is not $c_{00} \to c_{00}$, right? Since $\Gamma x \notin c_{00}$.
Also , I have a confusion in general. We have that the operator is a self map $\Gamma : c_{00} \to c_{00}$ , it means that $\Gamma x \in c_{00}$ , since $c_{00} \subset l^\infty \implies \sup\{\Gamma x_i, \forall i\}<\infty$. Then how could the operator to be unbounded?
Define $T:C_{00}\rightarrow C_{00}$ by $(Tx)(n)=nx_{n}$. Clearly $T$ is a well-defined map, and is linear. For each $n\in\mathbb{N}$, let $\delta^{(n)}\in C_{00}$ be defined as $\delta_{k}^{(n)}=\begin{cases} 1, & \mbox{if }k=n\\ 0, & \mbox{if }k\neq n \end{cases}.$ Note that $||\delta^{(n)}||=1$ but $||T\delta^{(n)}||=n$. Hence, $T$ is unbounded.