Prove there exists an unbounded linear operator $T:\ell^1\rightarrow \mathbb R^2$.. How to prove there exists an unbounded linear operator $T:\ell^1\rightarrow \mathbb R^2$?
 A: This must require the use of the axiom of choice. It is consistent that the axiom of choice fails and every linear operator between Banach spaces is continuous, i.e. bounded. So you can't just write something like this down in a formula and you need to appeal to some intangible objects, in this answer we appeal to the existence of a Hamel basis.
Fix a Hamel basis $\cal B$ and without loss of generality assume that every $b\in\cal B$ is such that $\|b\|_1=1$. Let $\{b_i\mid i\in\Bbb N\}$ a sequence from $\cal B$, now we define the following function:
$$F(b)=\begin{cases} (i,i)& \exists i\in\Bbb N\text{ such that }b=b_i\\0 &\text{otherwise}\end{cases}$$
Extend this $F$ to a linear operator and it is easy to see that it is unbounded.
A: $$(a_1,a_2,a_3,...,a_n,...) \mapsto \left(\sum_{n\ge 1} n\,a_n\,,\ 0\right)$$
is an example, if you mean unbounded operator as this wikipedia article.
A: This is not a direct aswer, but may be helpful.
Berci's answer shows that one can explicitly construct an unbounded (everywhere defined) operator on some infinite-dimensional normed space $\,X$ (so without using AC). On the other side, Asaf told us that this is impossible if $\,X$ is Banach (i.e. complete with respect to the distance induced by the norm). Why is that so?
The reason is that every infinite-dimensional Banach space has no countable Hamel basis (this is a useful exercise) and so one cannot exhibit one without AC.
