The algebraic dual of a normed vector space is the space of all linear functionals to the ground field (either $\mathbb{R}$ or $\mathbb{C}$ for this question). The continuous dual is the subspace of all continuous linear functionals. Under certain conditions, it's quite easy to find an element of the algebraic dual of a space that is not in the continuous dual:

We can start with a particular vector space, say $\ell^0$, and put on it two inequivalent norms, say $\|-\|_1$ and $\|-\|_2$. As they are inequivalent, there will be some functional that is continuous with respect to the one but not the other, in this case the functional $(x_n) \to \sum x_n$ will do.

We can start with a normed vector space which has a topological basis which we then extend to a Hamel basis. Assuming that this extension is non-trivial, we can define a functional which is zero on the original topological basis but non-zero on some element of the extension. Unfortunately, the existence of a Hamel basis depends on some version of choice (not sure if it's equivalent to AC or not, but that's not important).

We can modify the previous construction to avoid Hamel bases by using the Hahn-Banach theorem.

What I would like is an example of a non-continuous functional on a **Banach** space that can be explicitly written down. So I don't want to use choice or HBT. If there were an example of a normed vector space with two inequivalent norms (may as well assume that $\|-\|_a$ is strictly weaker than $\|-\|_b$) so that the space is a Banach space with respect to the *weaker* norm then method (1) would work, but that would contradict the open mapping theorem (unless there's some subtlety that I can't see right now).

This builds on What’s an example of a space that needs the Hahn-Banach Theorem?. I'm not interested in throwing choice or HBT out of the window, but it's useful to know (from a pedagogical angle if nothing else) exactly where they are needed and how far one can go without them. One thing that surprised me a little in the answers to that question was that you need choice/HBT to see that $\ell^1$ is not the continuous dual to $\ell^\infty$. That made me wonder about the algebraic dual and whether the same is true for that, or not.