Duality relations for Lebesgue spaces of sections of vector bundles

Question

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.

Question: do we have the usual duality relations for Lebesgue spaces, i.e. $(L^p(X;E))^* \cong L^{p^\prime}(X;E)$ where $1 = \frac{1}{p} + \frac{1}{p^\prime}$?

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)

Answer 1

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; 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:

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.

Answer 2

The existence of a finite trivialising cover is a less stringent condition than one would expect: see Does every vector bundle allow a finite trivialization cover?

(Sorry for the commentlike answer but it seems that by migrating to SE I lost reputation points, so I have not enough of them to properly comment)

Edit: however, the uniform norm control over the cover might be an issue when $X$ is not compact, so my comment is really not that helpful I guess.

Answer 3

At least if $X$ is $\sigma$-compact and if compact sets have finite measure and if the local vector bundle trivializations are homeomorphisms, I would expect the representation of $(L^p(E))^*$ as $L^{p'}(E)$ to hold. The proof would proceed by first dividing $X$ into countably many disjoint relatively compact sets $B_i$ over which the bundle is trivial. Given $u$ in $(L^p(E))^*$, first consider the functionals $u_i:z\mapsto u(\bar z_i)$ on $L^p(E_{\,|B_i})$ where $\bar z_i$ denotes the zero extension. Then you get the representation of $u_i$ by some $y_i$ in $L^{p'}(E_{\,|B_i})$ . Patch these together to get a measurable section $y$ of $E$ such that $u(x)=\int_X\langle y(t),x(t)\rangle_t\,d\mu(t)$ holds for $x$ in $L^p(E)$. Then you can show $\|y\|_{p'}\le\|u\|$ in the same way as this is done in the case of real or complex valued functions just by considering the inequality $|u(x)|\le\|u\|\,\|x\|_p$ for $x$ chosen so that it gets the form $\|y\|_{p'}^{\,p'}\le\|u\|\,\|y\|_{p'}^{\,\frac{p'}p}$ which directly gives the required inequality. Note here that you can get $\sum_{j=1}^n(\eta_j\xi_j)=(\sum_{j=1}^n\eta_j^2)^{\frac{p'}2}$ by choosing $\xi_j=\dfrac{\eta_j}{(\sum_{j=1}^n\eta_j^2)^{1-\frac{p'}2}}\ $.