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 $(\ell^\infty(\mathbb N))^* \setminus \ell^1(\mathbb N)$?

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, rschwieb Jan 24 '13 at 17:11

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  • 1
    $\begingroup$ You might be interested in Banach Limits. $\endgroup$ – JSchlather Jan 24 '13 at 16:42
  • $\begingroup$ @ChrisEagle Indeed it seems to be a duplicate... But I didn't find it when I did my search... $\endgroup$ – Sebastien B Jan 24 '13 at 16:45

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).

  • $\begingroup$ And so do you have an example using the axiom of choice? (If it is the most constructive thing we can hope...) $\endgroup$ – Sebastien B Jan 24 '13 at 16:49
  • $\begingroup$ You already remarked that you know how to use the Hahn-Banach theorem to work that out, so to use the axiom of choice... first prove the Hahn-Banach theorem, then apply it. :-) $\endgroup$ – Asaf Karagila Jan 24 '13 at 16:50

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