TITLE: Discontinuous linear functional
QUESTION [63 upvotes]: I'm trying to find a discontinuous linear functional into $\mathbb{R}$ as a prep question for a test. I know that I need an infinite-dimensional Vector Space. Since $\ell_2$ is infinite-dimensional, there must exist a linear functional from $\ell_2$ into $\mathbb{R}$. However, I'm having trouble actually coming up with it. 
I believe I'm supposed to find an unbounded function (although I'm not sure why an unbounded function is necessarily not continuous; some light in that regard would be appreciated too), so I thought of using the vectors $e^i$, which have all entries equal to zero, except for the $i$-th one. Then, you can define $f(e^i)=i$.  That'd be unbounded, but I'm not sure if it'd be linear, and even if it is, I'm not sure how to define it for all the other vectors in $\ell_2$. 
A friend mentioned that at some point the question of whether the set $E=\{e^i:i\in\mathbb{Z}^+\}$ is a basis would come up, but I'm not sure what a basis has to do with continuity of $f$. 
I'm just learning this topic for the first time, so bear with me please. 
The space of sequences that are eventually zero (suggested by a few people) turned out to be exactly what I needed.  It also helped to cement the notions of Hamel basis, not continuous, etc.

REPLY [16 votes]: As Robert Israel already mentioned, you cannot write down an explicit (free of the axiom of choice) unbounded linear functional on a Banach space.  But it's generally not hard for incomplete normed spaces.  Nobody has mentioned my favorite example: the functional $\ell: C^1[-1,1] \to \mathbb{R}$ given by $\ell(f) = f'(0)$.