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Let's have a look to the unit interval $[0,1]$ and a Banach space $X$ and then to the space $$ E:=L^{\infty}([0,1],X), $$ i.e. all essentially bounded Banach-valued functions $f:[0,1]\rightarrow X$. My question is, what is the dual $E'$ of $E$? Is there an identification as in the case of $L^{\infty}(\Omega,\mu)$ and $\ell^{\infty}$ by the mean of bounded additively functions of bounded variations? Is there such an duality result? If so can also give references for that?

Thank you very much :)

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    $\begingroup$ See also mathoverflow.net/q/250677/454 $\endgroup$
    – Gerald Edgar
    Commented Apr 13, 2018 at 10:41
  • $\begingroup$ @GeraldEdgar: you are right with this reference but I think there is no explicit answer given since the reference there use not the definition of $L^{\infty}$ as space of essentially bounded measurable functions from $[0,1]$ to $X$ (which I well use). $\endgroup$
    – Miguel Chapman
    Commented Apr 13, 2018 at 12:15
  • $\begingroup$ @MiguelChapman: true, I didn't explicitly define $L^\infty$ in my question, but my defintion (in my head) is the same as yours. This problem has been bothering me for some time, feel free to contact me if you want to discuss it! $\endgroup$
    – user14166
    Commented Apr 16, 2018 at 14:56

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I take it from your question and comments that it is important that one considers the Banach space of bounded meaurable functions, not equivalence classes thereof. In the scalar case, the dual is the space of finitely additive bounded measures. In the vector valued case, the dual is, at least for the case of a separable Banach space, naturally identifiable with the space of bounded, finitely additive measures with values in the dual of $E$. The case of measurable functions with values in a non-separable Banach space is scary, at least for me.The classic "Vector measures" by Diestel and Uhl is probably still the best reference for such topics---an earlier classic is that of Dunford and Schwatrz.

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    $\begingroup$ I don't think this is true (see my question linked above, which still has no answer). The problem is that the $X$-valued simple functions aren't dense in $L^\infty([0,1];X)$ when $X$ is infinite dimensional. Thus there is no natural way to define the action of a measure $\mu$ on $f \in L^\infty([0,1];X)$ (i.e. no integral $\int f d\mu$). Some authors get around this problem by defining $L^\infty([0,1];X)$ to be the closure of the simple functions (i.e. the 'totally measurable' functions), but this isn't such a useful definition, at least to me... $\endgroup$
    – user14166
    Commented Apr 16, 2018 at 14:54

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