Show that linear Operator on $\ell^2$ is unbounded Currently, I am preparing for a next semester course and trying to figure out some basic concepts in functional analysis.
Let $T:\mathcal{D}(T)\to \ell^2$ be defined by
$$T((x_n)_{n\in\mathbb{N}})=(nx_n)_{n\in\mathbb{N}}$$
where $\mathcal{D}(T)\subset \ell^2$ consist of all sequences $(x_n)$ with finitely many nonzero elements. (cf. Erwin Kreyszig: Introductory Functional Analysis, problem 10.1 8)
(a) Show that $T$ is unbounded. (b) Does $T$ have proper linear extensions? (c) Can $T$ be linearly extended to the whole space $\ell^2$?  
(a) First I am interested in proving the unboundedness: Basically I need to show $\lVert T\rVert=\infty$. My first attempt, where I am stuck at, was to prove it directly.
$$\lVert Tx \rVert=\left(\sum_n|nx_n|^2\right)^{1/2}=\left(\sum_nn^2|x_n|^2\right)^{1/2}$$
But I am unable to work it out. Any help is appreciated.
 A: For the unboundedness, consider the elements $e_k = (0,...,0,1,0,...)$ where the $1$ is in the $k$-th position. Then $e_k \in D(T)$ and $\|e_k\| = 1$ for all $k$. Notice that we have that $Te_k = (0,...,0,k,0,...)$ the $k$ being in the $k$-th position and thus $\|Te_k\| \to \infty$ as $k\to \infty$. But this means that
$$\|T\| = \sup_{\|x\|=1}\|Tx\| \geq \sup_{k}\|Te_k\| = \infty.$$
For part (b), think about whether there exist sequences in $x \in \ell^2$ which do not belong to $D(T)$ such that $Tx \in \ell^2$. For example, you could take $x_n = \frac{1}{n^2}$. Then $(Tx)_n = \frac{1}{n}$ so that $Tx \in \ell^2$. You can use this information to figure out some subspaces to which $T$ extends which properly include $D(T)$. Taking into account the reasoning above, one candidate is the subspace
$$M = \{x \in \ell^2: \mbox{there exists }C > 0\mbox{ such that } |x_n| \le \frac{C}{n^2}\}.$$
$T$ cannot be linearly extended to the whole space, however. Notice that for each $x \in D(T)$, we have $\|Tx\| \geq \|x\|$. If $\tilde{T}$ is an extension of $T$ to $\ell^2$, then the same estimate is true of $\tilde{T}$. Hence $\tilde{T}$ would have closed range (see here). Since $\tilde{T}$ is one-to-one, it is invertible
when considered as an operator from $\ell^2 \to T(\ell^2)$ and the estimate $\|\tilde{T}x\| \geq \|x\|$ shows that $\tilde{T}^{-1}$ would be bounded. But then the Open Mapping Theorem would imply that $\tilde{T}$ is bounded, a contradiction.
A: Take the sequence of sequences:
$u_0: 0, \dots$
$u_1:1,0,\dots$
$u_2: 1,1,0,\dots$
$u_3: 1,1,1,0,\dots$
They all are in $D(T)$
The image of these sequences by $T$ are
$Tu_0:0,\dots$
$Tu_1:0,1,0,\dots$
$Tu_2:0,1,2,0,\dots$
$Tu_3:0,1,2,3,0,\dots$
Take $x=u_i$ for some $i$,
$\left\|x\right\|=\left(\sum\limits_n|x_n|^2\right)^{1/2}=\left(\sum\limits_{n=0}^i1\right)^{1/2}=\sqrt{i}$
$\left\|Tx\right\|=\left(\sum\limits_nn^2|x_n|^2\right)^{1/2}=\left(\sum\limits_{n=0}^in^2\right)^{1/2}=\sqrt{\cfrac{i(i+1)(2i+1)}{6}}$
$\cfrac{\left\|Tu_i\right\|}{\left\|u_i\right\|}=\cfrac{\sqrt{\cfrac{i(i+1)(2i+1)}{6}}}{\sqrt{i}}=\sqrt{\cfrac{(i+1)(2i+1)}{6}}\underset{i\to+\infty}{\longrightarrow}+\infty$
