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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.

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.

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)?

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.

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)?

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)?