show/hide this revision's text 4 added 195 characters in body; added 23 characters in body; added 4 characters in body

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.

show/hide this revision's text 3 insert duplicate link

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)?

    Post Closed as "exact duplicate" by Andrew Stacey, Yemon Choi, Akhil Mathew, Pete L. Clark, Scott Morrison

show/hide this revision's text 2 added banach-spaces tag
show/hide this revision's text 1