# Explicit element of $(\ell^{\infty})^* - \ell^1$? [duplicate]

Possible Duplicate:
What’s an example of a space that needs the Hahn-Banach Theorem?

It is well known that the dual of $\ell^{\infty}$ properly contains $\ell^1$ (over $\mathbb{N}$, say). For instance, any Banach limit is an example. However, the construction of a Banach limit that I've seen is nonconstructive: one defines the functional on a closed subspace and extends via the Hahn-Banach theorem (or some variation thereof, to bring about translation-invariance).

Nevertheless, one can see without the axiom of choice that $(\ell^{\infty})^* \neq \ell^1$, because $\ell^{\infty}$ is not separable and therefore (I think!) its dual cannot be, while $\ell^1$ is. However, I was recently thinking about the dual of $\ell^{\infty}$ together with the action of translation operators, and I realized I didn't have a good picture of what this object was like (though I had heard something vague about finitely additive measures).

Question: What's an explicit example of an element of $(\ell^{\infty})^* - \ell^1$?

Since my wondering about the nonuniqueness in the definition of Banach limits prompted this question, I also have a follow-up:

Question$^\prime$: Can one get a Banach limit explicitly (or at least, can one prove its existence without the axiom of choice)?

Closed as sort-of-duplicate: For the answer, see point #2 of Greg's answer to another question.

• Akhil: have you looked at mathoverflow.net/questions/5351/… ? I think the answers indicate (as does my hazy memory) that no such "explicit" functional exists. Commented Apr 27, 2010 at 2:12
• In fact: although your question isn't a duplicate per se of the one I linked to, I'm not sure it's worth having both of them open. Would anyone object if I voted to close this one, perhaps after a link is added to the older question? Or have I misunderstood Akhil's question? Commented Apr 27, 2010 at 2:14
• @Yemon: Yes, I think point #2 in Greg Kuperberg's answer constitutes an answer to this question. Commented Apr 27, 2010 at 2:51
• Akhil, generally one identifies $\ell_\infty$ with the continuous functions on $\beta(N)$, the Stone-Cech compactification of the natural numbers, so that $\ell_\infty^*$ is identified with the finite signed measures on $\beta(N)$. However, it is not hard to check that $\ell_\infty^*$ can also be identified with the finitely additive finite set functions on the natural numbers. See, e.g., Example 8 and Problem 36 in Section 6.4 of A. Wilansky "Functional Analysis". Commented Apr 27, 2010 at 17:50
• @Yemon: Thank you for the pointer to Greg Kuperberg's answer (which is relevant). I have also voted to close now. Commented Apr 27, 2010 at 19:28