Any $\mathbb{R}$-linear mapping $X: C^\infty(M, \mathbb{R}) \to \mathbb{R}$ with $X(fg) = X(f)g(x) + f(x)X(g)$ given by $X(f) = Df_x(v)$?

Question

Let $M$ be a smooth manifold, and let $C^\infty(M, \mathbb{R})$ denote the collection of smooth real valued functions on $M$. For $x \in M$, how do I see that any $\mathbb{R}$-linear mapping $X: C^\infty(M, \mathbb{R}) \to \mathbb{R}$ satisfying $X(fg) = X(f)g(x) + f(x)X(g)$ is given by $X(f) = Df_x(v)$ for some uniquely determined vector $v \in DM_x$?

Answer 1

Let's start by recalling one form of Taylor's theorem:

Let $f$ be a smooth function on $\Bbb R^n$. Then, at $0$, $f$ may be written as $$f(x) = f(0) + \sum x_i \frac{\partial f}{\partial x_i}(0) + \sum x_ix_j h(x),$$ where $h(x)$ is some smooth function.

For convenience I denote $\partial f/\partial x_i$ as $f_i$.

First, your derivations are local objects, so we may as well pick a chart around the point $x$ such that it's $0 \in \Bbb R^n$. Then the Leibniz rule says that $X(1) = X(1 \cdot 1) = 2X(1)$. The equation $X(1) = 2X(1)$ implies that $X(1)$, and hence by linearity $X(f(0))$, is zero. Now the Leibniz rule again says that $$X(x_i x_j h) = x_i(0) X(x_j h) + x_j(0)h(0) X(x_i) = 0,$$ because $x_i(0) = x_j(0) = 0$. What we've determined, then, is that $$X(f) = X\left(\sum x_i f_i(0)\right) = \sum f_i(0)X(x_i);$$ so $X$ is completely determined by where it sends $x_i$. This shows that the space of derivations is at most $n$-dimensional. Now remember that (given a chart) we have a map $\Bbb R^n \to DM_x$, sending a vector to the directional derivative in that direction. It's injective - different directions give different directional derivatives. But by the dimension constraint we just got above, it must then be surjective, too. So it's an isomorphism, as desired.