Question about linear functionals over C-infinity modules A remark in Nelson's "Tensor Analysis" implies there are no non-trivial linear functionals on the module of continuous vector fields on a manifold, when considered as a module over the ring of smooth real-valued functions. This is believable, but I can't see how to get a handle on this. It seems as though this could either be obvious, or deep!
 A: Let us answer the question in $\mathbb R^n$, which will be enough. A linear functional on the space of continuous tangent fields is a section $\omega = \sum_i \omega_i \, \mathrm d x^i$ of $T^* \mathbb R^n$. Since, by hypothesis, $\omega (\partial_i) \in C^\infty (\mathbb R^n)$, it follows that $\omega_i \in C^\infty (\mathbb R^n)$ for all $i$, so $\omega$ is smooth. We won't need this smoothness, though - only the underlying continuity.
For every $i$, consider now tangent fields of the form $f \partial_i$, with $f$ continuous on $\mathbb R^n$. Since $\omega (f \partial_i) \in C^\infty (\mathbb R^n)$, it follows that $f \omega_i \in C^\infty (\mathbb R^n)$ for every continuous $f$. If $\omega_i \ne 0$, there exists some $x \in \mathbb R^n$ such that $\omega_i (x) \ne 0 \in \mathbb R$. By continuity, there exists some small neighbourhood $U$ of $x$ such that $\omega_i \ne 0$ on $U$. Let $g$ be a continuous but not smooth function on $\mathbb R^n$ having the support in $U$. Define $h : \mathbb R^n \to \mathbb R$ by
$$ h(y) = \begin{cases} \frac {g(y)} {\omega_i (y)}, & y \in U \\ 0, & y \notin U \end{cases} $$
and notice that $h$ is continuous (that's why the support of $g$ was taken inside $U$). It follows that
$$ \omega (h \partial_i) = \begin{cases} g(y), & y \in U \\ 0, & y \notin U \end{cases}$$
is smooth (by hypothesis), which is not possible since $g$ has been chosen not smooth. Hence, $\omega_i = 0$.
Performing this analysis for every $i$ shows that $\omega = 0$.
To summarize, the argument consisted in two steps:

*

*show that $\omega$ is continuous

*show that if $f \omega_i$ is smooth for every continuous $f$ then $\omega_i = 0$.

