Does a linear operator obeying Leibniz rule imply it is a differential operator?

At the moment I'm studying manifolds and in the definition of tangent space, the notion of derivation was used in the material I'm reading. I asked here about the matter and it was told me that it's not necessary to have a complete background in abstract algebra to proceed but instead just understand the definition of a derivation.

If I understood well a derivation is a way to extract the main properties of a differential operator: linearity and obeying the product rule. This line of thought leads me to a doubt: every differential operator should obey the product rule, but if a operator is linear and obey the product rule is possible to conclude it's a differential operator (it can calculate rates of change, be used to construct the taylor approximation to the function and every other properties we expect from a derivative)?

Solution 1:

It is really a problem in analysis. To simplify the matter I'll assume that the manifold is a convex (say) open subset $U\subset\mathbb{R}^n$ and that we consider tangent vectors at $0\in U$. We want to show: if $v:C^\infty(U)\to\mathbb{R}$ is a linear map satisfying $v(fg)=v(f)g(0)+v(g)f(0)$ then $v(f)=\sum_i v^i \partial f/\partial x^i\,(0)$ for some numbers $v^i$ (which can be determined by $v^i=v(x^i)$). We need an analytical lemma: there are functions $g_i\in C^\infty(U)$ such that $f=f(0)+\sum_i x^i g_i$. From there we get $\partial f/\partial x^i\,(0)=g_i(0)$ and (using $v(constant)=0$) $v(f)=\sum_i v(x^i) g_i(0)$. Together we get $v(f)=\sum_i v(x^i)\partial f/\partial x^i\,(0)$ as we wanted to show.

To prove the lemma we can just use $$f(x)=f(0)+\int_0^1 \frac{d}{dt}f(tx) dt=f(0)+\sum_i x^i\int_0^1\partial f/\partial x^i\,(tx)\,dt.$$

For a general manifold one needs a bit more work ($U$ is then a coordinate chart), but it's still just analysis.