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Possible Duplicate:
Nonnegative linear functionals over $l^\infty$

Setup: Let $l^\infty$ be the set of bounded sequences (with terms in $\mathbb{R}$), and let $l^1$ be the set of sequences of terms of absolutely convergent series. Any element $y=(\eta_i)$ of $l^1$ becomes a bounded linear functional $f_y:l^\infty\rightarrow \mathbb{R}$ by defining $f_y(x)=\sum \eta_i\xi_i$ for $x=(\xi_i)\in l^\infty$. However, it is well-known that not every bounded linear functional on $l^\infty$ can be realized this way. ("$l^1$ is not the dual of $l^\infty$.")

Proving this was an exercise in a class I am taking recently. Classmates and I obtained the result via the Hahn-Banach theorem. Other classmates argued that $l^1$ is separable and $l^\infty$ is not, so $l^1$ cannot be homeomorphic, let alone isometrically isomorphic, to $l^\infty$'s dual. However, all of us were unsatisfied because what we really wanted was an explicitly computable bounded functional on $l^\infty$ not induced as above by an element of $l^1$. I and another classmate attempted to generalize the Cesaro mean but he convinced me this cannot be made to work on all bounded sequences. None of us came up with an actual example. So:

Question 1: Are there explicitly computable elements of $(l^\infty)^*$ not induced as above by elements of $l^1$? If so, what is an example?

Question 2: If the answer to question 1 is "no," is this because the claim "$l^1$ is not the dual of $l^\infty$" hangs on some nonconstructive axiom (e.g. the axiom of choice)?

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marked as duplicate by t.b., Rudy the Reindeer, Davide Giraudo, Asaf Karagila, Grigory M Dec 13 '11 at 18:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ In my answer here I argue that the answer to question 1 is no and give references for the precise assumptions needed. It is consistent with ZF that $(l^\infty)^\ast = l^1$. The standard proof of the fact that $X^\ast$ separable implies $X$ separable needs some choice, so the argument involving this statement doesn't hold water. $\endgroup$ – t.b. Dec 13 '11 at 16:53
  • $\begingroup$ @t.b. - Thanks! That seems to fully answer the question! (Moderators: perhaps the question should be closed as an exact duplicate?) $\endgroup$ – Ben Blum-Smith Dec 13 '11 at 17:22
  • $\begingroup$ You're welcome. Following your proposal I'll vote to close as a duplicate. (By the way: I appreciate the way you ask your questions here very much. If only more questions of this quality were asked here!) $\endgroup$ – t.b. Dec 13 '11 at 17:45
  • $\begingroup$ Jonas Meyer just posted two relevant links as a comment in the duplicate thread: mathoverflow.net/questions/22661/… and mathoverflow.net/questions/5351/… $\endgroup$ – t.b. Dec 13 '11 at 21:59
  • $\begingroup$ @t.b.-thanks, this has been incredibly helpful; and thanks for the positive feedback. $\endgroup$ – Ben Blum-Smith Dec 22 '11 at 14:25