Condition for the $1$-form $f\omega$, being $f \in C(\Omega)$ and $\omega = \omega_1\text{d}x_1 + \omega_2\text{d}x_2$, to be closed.

I am tryingto solve the following problem:

Let $\Omega$ be an open set of $\mathbb{R}^2$ and $\omega = \omega_1\text{d}x_1 + \omega_2\text{d}x_2 \in \text{F}_1(\Omega)$. Define $$L = \omega_2 \frac{\partial}{\partial x_1}-\omega_1\frac{\partial}{\partial x_2}.$$ Show that
$\exists \ f \in C^{\infty}(\Omega), f > 0$ satisfying $\text{d}(f\omega) = 0$ if and only if there exists exists $u \in C^{\infty}(\Omega)$ satifying $Lu = \frac{\partial\omega_2}{\partial x_1}-\frac{\partial\omega_1}{\partial x_2}$.

Now, I did the following:

Let $f$ have the desired properties. Then, since $f$ is a $0$-form: $$\text{d}(f\omega) = \text{d}f \wedge \omega + (-1)^0f\wedge \text{d}\omega = \text{d}f \wedge \omega + f \text{d}\omega$$ $$\therefore \text{d}(f\omega) = 0 \Longleftrightarrow \text{d}f\wedge\omega = -f\text{d}\omega$$ But: $$\text{d}f = \frac{\partial f}{\partial x_1}\text{d}x_1+\frac{\partial f}{\partial x_2}\text{d}x_2$$ $$\therefore \text{d}f\wedge\omega = \left(\omega_2\frac{\partial f}{\partial x_1} - \omega_1\frac{\partial f}{\partial x_2}\right)\text{d}x_1\wedge\text{d}x_2 = Lf\text{d}x_1\wedge\text{d}x_2$$ On the other side: $$-f\text{d}\omega = -f\left(\frac{\partial \omega_2}{\partial x_1} -\frac{\partial \omega_1}{\partial x_2} \right)\text{d}x_1\wedge\text{d}x_2$$ We then end up with $$\text{d}(f\omega) = 0 \Longleftrightarrow Lf = -f \left(\frac{\partial \omega_2}{\partial x_1} -\frac{\partial \omega_1}{\partial x_2} \right)$$ which is almost what we wanted.

Any hints on what I may have missed or on where to go from here will be the most appreciated.

With $f > 0$ and

$L = \omega_2 \dfrac{\partial}{\partial x_1} - \omega_1 \dfrac{\partial}{\partial x_2}, \tag 1$

it is easy to see that

$Lf = -f( \dfrac{\partial \omega_2 }{\partial x_1} - \dfrac{\partial \omega_1 }{\partial x_2}) \tag 2$

may be written

$-f^{-1}Lf = \dfrac{\partial \omega_2 }{\partial x_1} - \dfrac{\partial \omega_1 }{\partial x_2}; \tag 3$

also, with $f > 0$ we have

$-f^{-1}Lf = -L(\ln f), \tag 4$

since $L$ only involves first derivatives. Set

$u = -\ln f; \tag 5$

then

$Lu = L(-\ln f) = -f^{-1}Lf = \dfrac{\partial \omega_2 }{\partial x_1} - \dfrac{\partial \omega_1 }{\partial x_2} \tag 6$

as desired.

• Note that the proof would work equally well if we had $f < 0$ everywhere; just define $u = - \ln (-f)$. The condition that $f > 0$ is perhaps best interpreted as "$f$ exists and can be chosen to be positive", not "$f$ exists and must be positive." – Michael Seifert Oct 31 '17 at 20:50
• @MichaelSeifert: duly noted, with a tip of the hat to what was literally stated. – Robert Lewis Oct 31 '17 at 20:51

HINT: The fact that $f>0$ should be a hint. Try using $f=e^{-u}$ when $u$ satisfies the given equation.