What is the dual to $L^\infty (I,H^1(M))$?, where $I$ is an interval in the real line; $H^1(M)$ is Sobolev space of degree 1, and $M$ is a compact manifold like the torus.
Any references that show that is appreciated.
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$\begingroup$ My guess would be the space $\mathcal{M}(I, H^{-1}(M))$, where $\mathcal{M}$ denotes the space of bounded Radon measures and $H^{-1}$ is the dual of $H^1$. In short, "compose" the dual spaces in the same order. For a proof, I guess you could try to mimic the usual one which shows that $(L^{\infty})^* = \mathcal{M}$. (At least, it is rather clear that the suggested space acts canonically on your space.) $\endgroup$– HachinoCommented Mar 18, 2015 at 11:29
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$\begingroup$ Do you have a reference for such a proof as you described? $\endgroup$– AlanCommented Mar 18, 2015 at 11:48
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1$\begingroup$ Various questions on M.SE and MO asked for such results, here is one of them. (And now I wonder whether I messed up the dual of $L^{\infty}$ with that of $\mathcal{C}_0$. That's quite possible.) $\endgroup$– HachinoCommented Mar 18, 2015 at 11:50
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2$\begingroup$ @Hachino The dual of $L^\infty$ is much bigger than ${\mathcal M}$. ${\mathcal M}$ would be the dual of $C(I)$ (assuming $I$ is compact). $\endgroup$– Yemon ChoiCommented Mar 18, 2015 at 13:41
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1$\begingroup$ @Alan: That is a very different question - you aren't looking for the dual but the predual. The answer is almost certainly going to be $L^1(I, H^1(M))$ and the proof will probably look very similar to the proof that $L^1(I, \mathbb{R})^* = L^\infty(I, \mathbb{R})$. The canonical reference for any question like "what is the dual of Banach space $X$" is Dunford and Schwartz, Linear Operators. Another possible place to look is Dinculeanu's Vector Measures. $\endgroup$– Nate EldredgeCommented Mar 18, 2015 at 15:11
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1 Answer
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It seems clear from the above comments, in particular your mention of Alaoglu, that what you are looking for is a Banach space whose dual is naturally identifiable with the one in your query. Such a space is the corresponding $L^1$-space with values in $H^1$. This follows immediately from Theorem 1 on p. 96 of the classic "Vecor measures" by Diestel and Uhl. (The specific form of your image space is a bit of a red herring---the important fact is that it is a Hilbert space and so has the Radon Nikodym property).
Two remarks: If you really want a dual, rather than a predual, then there is an explicit description as a space of vector-valued finitely additive measures. And there is a slight ambivalence when talking about the dual of $H^1$. As a Hilbert space, it is self-dual but in the context of Sobolev spaces it is more usual to regard $H^{-1}$ as the dual. This does not materially affect the above response.
