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)?
