i got the following question in my manifold analysis course and Im kinda stuck.
let $F : \mathbb{R}^n \setminus \{0\} \rightarrow \mathbb{R}^n$ be a smooth radial vector field, that is $F = f(||x||)x$ for some continuously differentiable $f : \mathbb{R}\setminus \{0\} \rightarrow \mathbb{R}$
show that there exists a continuously differentiable function $g : \mathbb{R}^n \setminus\{0\} \rightarrow \mathbb{R}$ such that $\nabla g = F$
I've seen proofs that use Poincaré lemma alas we have not seen it, nor have we seen (or even defined) the rotor/curl of vector field so such proofs which I've found were of no use to me either.
I do know that since f is continuous it has an antiderivative alas I didnt manage to reach a solution from that direction.
any help would be appreciated!
Don't worry if you're not familiar with the curl operator or Poincare's lemma. There's a much more elegant argument that doesn't use any of these concepts. The idea is to reduce our $n$-dimensional problem to a $1$-dimensional problem along the radial direction, by exploiting the rotational symmetry of the setup.
$\mathbf F(\mathbf x)$ is a radial vector field, so intuitively, you would expect that if $g(\mathbf x)$ is a scalar field whose gradient is $\mathbf F(\mathbf x)$, then $g(\mathbf x)$ ought to be symmetric under rotations about the origin, i.e. $g(\mathbf x) = \tilde g(\|\mathbf x\|)$ for some continuously differentiable function $\tilde g : \mathbb R^{> 0 } \to \mathbb R$.
Let's pursue this line of thought - let's see if we can find a $g(\mathbf x)$ of the form $g(\mathbf x) = \tilde g(\|\mathbf x\|)$ that satisfies $\nabla g(\mathbf x) = \mathbf F(\mathbf x)$.
Assuming that $g(\mathbf x) = \tilde g(\|\mathbf x\|)$ for some $\tilde g : \mathbb R^{> 0 } \to \mathbb R$, you should be able to use the chain rule to obtain the following expression for the gradient of $g$. $$ \nabla g(\mathbf x) = \left( \frac{\tilde g'(\|\mathbf x\|)}{\|\mathbf x\|} \right) \mathbf x$$ Please have a go at doing this calculation using the chain rule!
Anyway, in order for $\nabla g (\mathbf x)$ to be equal to $\mathbf F(\mathbf x)$, we need $$ \frac{\tilde g'(\|\mathbf x\|)}{\|\mathbf x\|} = f(\| \mathbf x \| )$$ to hold for all $\mathbf x \in \mathbb R^n \setminus \{ \mathbf 0 \}$.
Thus $\nabla g (\mathbf x) = \mathbf F(\mathbf x)$ for all $\mathbf x \in \mathbb R^n \setminus \{ \mathbf 0 \}$ if and only if $\tilde g : \mathbb R^{> 0 } \to \mathbb R$ satisfies the ordinary differential equation $$ g'(r) = rf(r)$$ for all $r \in \mathbb R^{> 0}$.
All that remains is for you to argue that, whatever $f(r)$ is, there exists a $g(r)$ that satisfies this ordinary differential equation.