For any function $f ∈ F(M )$, $f X ∈ V(M)$

Question

Let $X ∈ V(M)$ be a vector field. To show: for any function $f ∈ F(M )$, $f X ∈ V(M)$.

I thought:

Given $f ∈ F(M)$ and $X ∈ V(M)$ we define $(fX)(x) := f(x)X(x)$ for all $x ∈ M$. And $F(M)$ is the set of the differentiable fonction.

Answer 1

In $C^\infty$ case, a vector field $X\in \mathcal{X}(M)$ is an $\mathbb R$-linear map

$$X\colon C^\infty(M)\to C^{\infty}(M)$$

such $X(h\cdot g) = h\cdot X(g) + g\cdot X(h)$ for every $g, h\in C^{\infty}(M)$.

Now, if $f\in C^\infty(M)$ and $X\in \mathcal X(M)$, $fX$ is still $\mathbb R$-linear and obeys $$(fX)(hg) = f\cdot X(hg) = f \cdot \big( h\cdot X(g) + g\cdot X(h) \big) = h\cdot (fX)(g) + g\cdot (fX)(h)$$