How to construct an “explicit” element of $(\ell^\infty(\mathbb N))^* \setminus \ell^1(\mathbb N)$? [duplicate]

Everything is in the title:

How to construct an "explicit" element of $(\ell^\infty(\mathbb N))^* \setminus \ell^1(\mathbb N)$?

To be more precise:

I define the Banach spaces $(\ell^1(\mathbb N),\|\cdot \|_1)$ and $(\ell^\infty(\mathbb N),\|\cdot \|_\infty)$ as usual by \begin{align} \ell^1(\mathbb N) & :=\big\{ (x_n)_{n\in\mathbb N}\in \mathbb R^\mathbb N \,\big|\, \sum_{n\in\mathbb N} |x_n|<\infty \big\}\,, & \|x\|_1 & := \sum_{n\in\mathbb N} |x_n|\,, \\ \ell^\infty(\mathbb N) & :=\big\{ (x_n)_{n\in\mathbb N}\in \mathbb R^\mathbb N \,\big|\, \sup_{n\in\mathbb N} |x_n|<\infty \big\}\,, & \|x\|_\infty & := \sup_{n\in\mathbb N}|x_n|\,. \end{align} The application \begin{align} f:\ell^1(\mathbb N) & \to (\ell^\infty(\mathbb N))^* \\ x & \mapsto \big(y\mapsto\sum_{n\in\mathbb N}x_ny_n\big) \end{align}is not surjective: I know how prove the existence of an element of $(\ell^\infty(\mathbb N))^* \setminus f(\ell^1(\mathbb N))$ using the Hahn-Banach theorem.

But I would like to construct an "explicit" example of such a functional.

Does someone know how to do that?

marked as duplicate by David Mitra, Chris Eagle, Ross Millikan, Henry T. Horton, rschwiebJan 24 '13 at 17:11

It is consistent without the axiom of choice that Hahn-Banach fails, and $(\ell_\infty)^\ast=\ell_1$. (It should be pointed out that the Hahn-Banach theorem is vastly weaker than the full axiom of choice.)