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Which is the dual of the space $\ell^\infty([0,1]^d)$? Does the dual of such space consists of a set of functionals $\phi$ admitting the representation

$$ \phi(f)=\int_0^1 f\circ g(u) \psi_\phi(u) du, \quad f \in \ell^\infty([0,1]^d), $$

with $g:(0,1)\mapsto[0,1]^d$, $\psi_\phi \in L^1((0,1), \mathcal{B}((0,1)), \lambda)$, and $\mathcal{B}((0,1))$ and $\lambda$ denoting the Borel $\sigma$-algebra and the Lebesgue measure on the unit interval, respectively? If the above representation is too specific, does the dual consists of a set of functionals that can be expressed as integrals of elements $f$ of $\ell^\infty([0,1]^d)$ on appropriate spaces with repsect to the Lebesgue measure of such spaces?

Jack London
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  • No, that representation fails even for the case $d=1$. – GEdgar Apr 26 '18 at 11:43
  • Ok, thanks. Do you know any condition under which $\phi(f)$, with $\phi$ a continuous functional, can be given an integral representation? – Jack London Apr 26 '18 at 12:08
  • Elements of the duals of spaces like $l^{\infty}, L^{\infty} (\mu)$ etc have a known representation only when they are finite dimensional. – Kavi Rama Murthy Apr 26 '18 at 12:14
  • A subspace of the dual of $\ell^\infty([0,1]^d)$ is $\ell^1([0,1]^d)$. You can think of those functionals as integrals (with respect to a discrete measure). – GEdgar Apr 26 '18 at 12:15

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