Unbounded operator on $\ell^2$

Question

If we take $T\in B(\ell^2(\mathbb N))$ given by $Te_n=n e_n$ on the canonical orthonormal basis, that is

$$ T(x_1,x_2,\ldots)=(x_1,2 x_2,3 x_3,\ldots), $$ with domain $$D(T)=\{x\in \ell^2;Tx \in \ell^2 \}.$$

I want to prove that $(T, D(T))$ is unbounded.

A linear operator $T:D(T)\subset E\to F$ is said to be bounded if there exists a constant $c>0$ such that $$\|Tx\|_{F}\leq c\|x\|_{E},$$ for all $x\in D(T)$. Note that $D(T)$ is called the domain of $T$.

Answer 1

Let $e_n$ be the element of $\ell^2$ defined as follows: the $n$-th coordinate of $e_n$ is one and all the others are zero. Then $Te_n=n\cdot e_n$ and $e_n$ has norm one. For all $M\gt 0$, pick an integer $n$ such that $n\geqslant M$ in order to see that the definition of bounded is not satisfied.

Answer 2

Take $\{e_n\} \subset D(T)$ , then $||T(e_n)|| = n \to \infty$