Suppose $X$ is a topological space, and $\mu$ is a Borel measure on $X$. Also suppose we have an $n$-dimensional vector bundle $E \to X$, with an inner product $\langle \cdot,\cdot \rangle\_x$ on the fibre $E\_x$ for all $x \in X$, in such a way that each $E\_x$ is complete and such that there exists a vector bundle trivialisation which is compatible with the fibrewise inner products. The inner product $\langle \cdot, \cdot \rangle\_x$ determines a norm $||\cdot||\_x$ on $E\_x$.


Say that a (not necessarily continuous) section $\sigma \colon X \to E$ is *measurable* if its restriction to each trivialising open set $U \subset X$ is given by a measurable function $U \to \mathbb{F}^n$ (here $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$, depending on how you feel). Denote the set of measurable sections (understood as being defined up to measure zero) by $\Gamma(E)$.


Given a section $\sigma \in \Gamma(E)$ and a number $p \in (0,\infty)$, we can define
$$||\sigma||\_p := \left( \int\_X ||\sigma(x)||\_x^p \; d\mu(x) \right)^{1/p},$$
and we can then define the corresponding $E$-valued Lebesgue space $L^p(X;E)$ in the obvious way.



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> **Question:** do we have the usual duality relations for Lebesgue spaces, i.e. $(L^p(X;E))^\* \cong L^{p^\prime}(X;E)$ where $1 = \frac{1}{p} + \frac{1}{p^\prime}$?
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As one would expect, there is a kind of Hölder inequality: if $\sigma \in L^p(X;E)$ and $\tau \in L^{p^\prime}(X;E)$, then the function $\langle \sigma,\tau \rangle$ on $X$, given by
$$\langle \sigma,\tau \rangle(x) := \int\_X \langle \sigma(x), \tau(x) \rangle\_x \; d\mu(x),$$
satisfies $|| \langle \sigma,\tau \rangle||\_{L^1(X)} \leq ||\sigma||\_p ||\tau||\_{p^\prime}$. It follows that the pairing $\langle \cdot,\cdot \rangle$ can be used to isometrically embed $L^{p^\prime}(X;E)$ into $(L^p(X;E))^\*$.


However, I haven't been able to prove the reverse containment - that each functional $\phi \in (L^p(X;E))^\*$ is given by pairing with an element of $L^{p^\prime}(X;E)$ - without additional assumptions, such as the existence of a finite trivialising cover for $E$ with uniform norm control (for example, when $X$ is compact). In this case, one can recover the result from the corresponding result for trivial bundles - which is essentially the case of vector-valued Lebesgue spaces - but constants appear which depend on the cardinality of a trivialising cover, which is somewhat unexpected.


Has this been explicitly proven anywhere? Is it even true in general? (I'll be very surprised if it isn't)



Duality relations for Lebesgue spaces of sections of vector bundles
Maybe the following helps:
Theorem 3.12 (page 20) in the following source has such a related result, albeit for higher Sobolev spaces.
There are quite subtle requirements for the trivialising atlas and the partition of unity which are used in the proof.


* [MR2343536](http://www.ams.org/mathscinet-getitem?mr=2343536); Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.


You can access the beginning of the book via scholar.google.com.


Second Edit:
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It seems to me now it is much simpler.


Proof: Measure theoretically, there are no nontrivial bundles. So you can find a global orthonormal frame by measurable sections $s\_1,\dots,s\_n$ of your bundle so that any section $f$ is of the form $f=\sum \_i f^i.s\_i$ where $(f^i)\_{i=1}^n \in L^p(X,\mathbb R^n)$. This gives an isometry between $L^p$-sections of the bundle and a usual $\mathbb R^n$-valued $L^p$-space.


