I am studying analysis by using Elon Lages Lima's book "Curso de Análise (vol. 2)" (the book is written in portuguese). I am trying to solve the following problem about path integrals but I got stuck in some items.
Let $\omega$ be a closed $1$-form, different than zero in all points of the open set $U \subset \mathbb{R^m}$, and $f: U \rightarrow \mathbb{R}$ a function of class $C^1$ with no critical points. Prove that the following affirmations are equivalent:
(a) - the form $f \cdot \omega$ is closed;
(b) - $df$ is a multiple of $\omega$;
(c) - for every $x \in U$, and every $v \in \mathbb{R^m}$, $\omega(x) \cdot v = 0 \iff \langle \text{grad } f(x), v \rangle = 0$.
Here's my attempt to solve the problem:
$(a) \implies (b)$:
Let $\omega = \sum{a_i dx_i}$. Then, $f \cdot w = \sum f a_i dx_i$. We have:
$$ \text{$f \cdot w$ is closed } \implies \frac{\partial (f a_i)}{\partial x_j} = \frac{\partial (f a_j)}{\partial x_i} \implies \frac{\partial f}{\partial x_j} a_i + \frac{\partial a_i}{\partial x_j} f = \frac{\partial f}{\partial x_i} a_j + \frac{\partial a_j}{\partial x_i} f $$
Since $\omega$ is a closed form we have $\frac{\partial a_i}{\partial x_j} = \frac{\partial a_j}{\partial x_i}$ and the expression above is simplified to:
$$ \frac{\partial f}{\partial x_j} a_i = \frac{\partial f}{\partial x_i} a_j$$
and therefore:
$$ a_i(x) [\frac{\partial f}{\partial x_i}(x)]^{-1} = a_j(x) [\frac{\partial f}{\partial x_j}(x)]^{-1} = k(x) $$
where $k(x)$ does not depend on the choice of the index $i$. I think I should have obtained a $k$ that also does not depend on the point $x$ to conclude that $df$ is a multiple of $\omega$, right? Or the dependency means that at every point $x$ we have $df(x) \cdot v = k(x) (\omega(x) \cdot v)$ and in this case the calculations above are enough?
$(b) \implies (c)$ - I think I was able to solve this one:
Indeed for every $x \in U$ and every $v \in \mathbb{R^m}$ we have $\omega(x) \cdot v = 0 \iff df(x) \cdot v = 0$ (because $df$ is a multiple of $\omega$) $\iff \langle \text{grad} f(x), v \rangle = 0$.
$ (c) \implies (a)$:
I couldn't go very far here. I notice that by the calculations done in item $(a) \implies (b)$ it suffices to prove:
$$ \frac{\partial f}{\partial x_j} a_i = \frac{\partial f}{\partial x_i} a_j$$
and I also noticed that by hypothesis, picking $v = e_i$ we have that $a_i(x) = 0 \iff \frac{\partial f}{\partial x_i}(x) = 0$. I was not able to proceed any further.
Can anyone help me in cases $(a) \implies (b)$ and $(c) \implies (a)$? Thanks in advance.
HINTS: Don't write coordinates. Just write the product rule for $d(f\omega)$ and use the fact that $\omega$ is closed.
I honestly think that (c) is an inconvenient detour. I would just check (b)$\implies$ (a) (immediately) and (c)$\implies$ (b) (immediately).