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

Question

Possible Duplicate:
Nonnegative linear functionals over $l^\infty$
An explicit functional in $(l^\infty)^*$ not induced by an element of $l^1$?

Everything is in the title:

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

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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?

Answer 1

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.)

You can't do it without appealing to the intangible which is the axiom of choice (or some of its weakened versions, such as the Hahn-Banach theorem).