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I was wondering what the dual space of $L^2(\mathbb{R},L^1(0,1))$ is? (equipped with Lebesgue measures) Formally, one would suspect that it is just $L^2(\mathbb{R},L^{\infty}(0,1))$. But this may be a bit too quick, as there many references that suggest that this formally replacing each space by its dual is only possible in the reflexive case.

Here, is already a good answer at Math.stackexchange, but it only works for finite measure spaces: click me.

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  • $\begingroup$ A probably stupid question from an algebraist with little knowledge of Bochner integrals: can $L^2(\mathbb{R},L^1(0,1))$ be identified with the set of Borel functions (modulo null sets) $f$ on $\mathbb{R}\times(0,1)$ such that $\int_{\mathbb{R}}\left(\int_0^1|f|\right)^2<+\infty$? Can it be identified with $L^1((0,1),L^2(\mathbb{R}))$? $\endgroup$ – Gro-Tsen Mar 3 '17 at 22:26
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"Finite measure" is no restriction. Either change the measure on $\mathbb R$ to an equivalent finite one, or write $L^2(\mathbb R)$ as the $l^2$-direct sum of spaces $L^2([n,n+1])$.

As in the link you give, since $L^\infty(0,1)$ fails the Radon-Nikidym property, the dual of $L^2(\mathbb R,L^1(0,1))$ is strictly larger than $L^2(\mathbb R,L^\infty(0,1))$.

When you start with a bounded linear functional on $L^2(\mathbb R, L^1(0,1))$, you come up with a vector measure with values in $L^\infty$. If that vector measure has a Radon-Nikodym derivative, it is the member of $L^2(\mathbb R,L^\infty(0,1))$ that you want. But if not, then your functional does not correspond to an element of $L^2(\mathbb R,L^\infty(0,1))$.

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  • $\begingroup$ This is a wonderful answer! $\endgroup$ – Bombyx mori Mar 20 '17 at 9:43
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$\def\rmd{{\kern.4mm\rm d\kern.4mm}}\def\sp{\kern.4mm}$For $1\le p<+\infty$ and $I=[\sp 0\sp,1\sp]$ and $F=L^p(\mathbb R,L^1(I))\sp$, a simple example of a continuous linear functional on $F$ that is not given by any element in $L^{p^*}(\mathbb R,L^{+\infty}(I))$ is $\vec x\mapsto\int_0^1\int_0^t x(t,s)\rmd s\rmd t$ when $x:\mathbb R\times I\to\mathbb K$ represents the element $\vec x$ of $F\sp$. The proof is left an exercise to the reader.

This is mainly an additional comment to Gerald Edgar's answer.

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Your question has been partially answered in that it has been shown that the naive version is false. However, there is a concrete representation of the dual which might be of interest. The result becomes more transparent in a more general setting which might also be of interest to you. Consider the space $L^p(\mu,E)$ where $\mu$ is a finite measure, $1\leq p<\infty$ and $E$ is a Banach space. Then its dual is naturally identifiable with $L^q(\mu,E')$ if and only if $E'$ has RNP ($q$ the conjugate of $p$). This is treated in detail in the beginning of chapter IV of "Vector measures" by Diestel and Uhl. However, there is a representation which works for any Banach space. I have not found this result explicitly in the literature but the basic idea is due to L. Schwartz (Sem. d'anal. fonctionelle (1974-75), Exp. 4---readily available online). I will state the result for the case of separable $E$: The dual is the space $L^q_{w^\ast}(\mu,E')$, where the subscript $w\ast$ refers to the fact that we are using functions which are weak star rather than norm measurable. Thus in your case, the dual is $L^2_{w\ast}(\mu,L^\infty)$. (As pointed out above there are standard methods to extend such results to the case of $\sigma$-finite measures, in particular the real line with Lebesgue measure).

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