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)


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.

  • $\begingroup$ thanks for the reference - a quick look hasn't made things any clearer to me, so I'll give it a longer look and see how it goes! $\endgroup$ – Alex Amenta Aug 6 '13 at 9:20
  • $\begingroup$ i'm not sure if this is applicable, but maybe i'm just a bit lost. doesn't Theorem 3.12 just show (when restricted to $L^p$ spaces) that $L^{p^\prime}$ is contained in $(L^p)^*$, but not the reverse containment? $\endgroup$ – Alex Amenta Aug 6 '13 at 9:53
  • $\begingroup$ regarding your edit: i can fill some of the gaps, but i still can't make it to the end of the proof. writing $\lambda_\alpha$ for the function in $L^{p^\prime}(U_\alpha;E)$ induced by the local action of $\lambda$, we need to prove that the sums $\sum^N \lambda_\alpha$ are bounded (uniformly in $N$) in $L^{p^\prime}$. i don't see how this follows from your argument - in particular i don't see how the partition of unity comes into play. any further tips are much appreciated! $\endgroup$ – Alex Amenta Aug 6 '13 at 23:35
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    $\begingroup$ @ Peter Michor: Obviously some additional assumptions on the base space are required to get such a global trivialization by orthonormal measurable sections. For example those given in my answer are sufficient. In view of this, your argument is simpler than mine provided that the scalar and hence the finite-dimensional vector valued case is assumed to be known. @ Alex Amenta: In view of the preceding, I will not waste my time to add the details for the proof of $\|y\|_{p'}\le\|u\|$ in my answer. $\endgroup$ – TaQ Aug 7 '13 at 17:26
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    $\begingroup$ @ Alex Amenta: The base $X$ being Lindelöf indeed is sufficient to guarantee the bundle being globally measurably trivial but this does not relate in any way to the given Borel measure on $X$ since no regularity of the measure is inherent in it just being defined on the $\sigma$-algebra of Borel sets. So some additional assumptions are also needed in order for the Riesz representation theorem to be applicable to $L^p(X,\mathbb R^{\,n})$. In Dudley's statement of the theorem, $\sigma$-finiteness of the measure is assumed, but if I recall correctly, this is not necessary for $1<p<+\infty$. $\endgroup$ – TaQ Aug 8 '13 at 11:50

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.

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    $\begingroup$ indeed, the argument in the linked question doesn't give uniform norm control - and in fact there are trivial bundles which don't admit finite trivialising covers with uniform norm control. (consider the trivial n-dimensional bundle over the interval $(0,\infty)$, with the inner product over the point $t$ being given by $t^{-1}$ times the standard inner product) $\endgroup$ – Alex Amenta Aug 6 '13 at 9:06
  • $\begingroup$ nevertheless, your comment is interesting and worth making! thanks $\endgroup$ – Alex Amenta Aug 6 '13 at 9:07

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}}\ $.

  • $\begingroup$ i don't understand the last part of your proof - how do you choose $x$ such that $|u(x)| \leq ||u|| ||x||_p$ turns into the inequality for $||y||_{p^\prime}$? $\endgroup$ – Alex Amenta Aug 6 '13 at 23:45
  • $\begingroup$ @Alex Amenta: The inequality $\|y\|_{p'}^{\,p'}\le\|u\|\,\|y\|_{p'}^{\,\frac{p'}p}$ only gives the basic idea behind the actual argument for which I refer you to see e.g. page 163 in the proof of the Riesz Representation Theorem 6.4.1 in Richard M. Dudley's book "Real Analysis and Probability" (Wadsworth, 1989; there is also a later edition with possibly different numberings) where the exact measure theoretic details are given in the case of scalar valued functions. $\endgroup$ – TaQ Aug 7 '13 at 0:46
  • $\begingroup$ This argument readily generalizes to the case of sections of a finite-rank (or even Hilbert space valued) vector bundle under the additional assumptions on the base topological space given at the beginning of my answer. $\endgroup$ – TaQ Aug 7 '13 at 0:47
  • $\begingroup$ here is where the issue lies with this approach: in the final approximation step, one must take an increasing sequence of sets $E(n)$ of finite measure such that $\cup_n E(n) = X$. in the usual setting, we know that the function $y$ induced by $u \in (L^p)^*$ satisfies $||y||_{L^p(U)} \leq ||u||$ for all sets $U$ of finite measure - in particular, we can take $U = E(n)$. (continued below) $\endgroup$ – Alex Amenta Aug 7 '13 at 1:22
  • $\begingroup$ in my setting, we don't have the above $L^{p^\prime}$ bound on all sets $U$ - we only know it on trivialising open sets. even though the space is $\sigma$-finite, we generally can't write the space as an increasing union of trivialising open sets. $\endgroup$ – Alex Amenta Aug 7 '13 at 1:24

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