Let $X$ be an infinite-dimensional Banach space and let $p\in(1,+\infty)$. We may define $L^p(\mathbb R;X)$. Is it always true that the topological dual of $L^p(\mathbb R;X)$ is $L^{p'}(\mathbb R;X^*)$? Maybe some reflexivity or separability is needed for the Radon-Nikodym argument to work. Is there a simple realization of the dual of $L^1(\mathbb R;X)$?
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$\begingroup$ Typo: you mean $L^{p'}({\mathbb R}, X^*)$, right? $\endgroup$– Yemon ChoiCommented Jan 28, 2015 at 17:58
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1 Answer
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Diestel-Uhl, Vector measures, Section IV.1., Theorem 1:
Let $(\Omega,\mu)$ be a $\sigma$-finite measure space, $1\leq p<\infty$ and $\frac{1}{p}+\frac{1}{p'}=1$.
The dual of $L^p(\mu;X)$ is $L^{p'}(\mu,X^\ast)$ if and only if $X^\ast$ has the Radon-Nikodym property.
This is especially the case if $X$ is reflexive.
answered Jan 28, 2015 at 18:01
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1$\begingroup$ Just to clarify: presumably there is a missing quantifier, and the full result is: the dual has the desired form for all $\sigma$-finite $(\Omega,\mu)$ if and only if $X^*$ has RNP? $\endgroup$ Commented Jan 28, 2015 at 18:03
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$\begingroup$ Yes, I add the details to the answer. Thank you. $\endgroup$ Commented Jan 28, 2015 at 18:04
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$\begingroup$ Another important special case (where $X^*$ has the RNP) is when $X^*$ is separable. For example, if $X$ is separable and $X^*$ is not separable, we would expect the dual of $L^{p}(\mu, X)$ to include some non-Bochner-measurable functions. $\endgroup$ Commented Jan 28, 2015 at 18:25
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$\begingroup$ What is the Radon-Nikodym property? Is it easy to say? $\endgroup$ Commented Jan 29, 2015 at 11:14
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2$\begingroup$ That Diestel--Uhl is a fine book. Have a look. $\endgroup$ Commented Jan 29, 2015 at 13:09