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(a) Let $\omega$ a $1-$form defined on the open set $ U \subset \mathbb R ^n$ and $ c:[a,b] \to U$ a $ C^1 -$differentiable curve such that $ |\omega (c(t))| \leq M \quad \forall t \in [a,b]$

Prove that $$ \displaystyle{\Bigg| \int_c \omega \Bigg| \leq ML}$$

where $L$ is the length of the curve $c$.

(b) Let $\omega=a_1dx +a_2dy$ a closed $1-$form defined on $\mathbb R^2 -\{(0,0)\}$. If $\omega$ is bounded ( which means that $a_1 ,a_2$ are bounded) on a disk with center the origin $O(0,0)$ prove that:

$$ \omega \text{ is exact on } \mathbb R^2 -\{(0,0)\}$$

(c) If $\omega$ is a closed $1-$form defined on $\mathbb R^2 -\{(0,0)\}$ such that $\displaystyle{ \lim_{x^2 + y^2 \to 0} \left( \sqrt{x^2 + y^2} \omega \right) =0 }$ prove that :

$$ \omega \text{ is exact on } \mathbb R^2 -\{(0,0)\}$$

I have done (a) but I am stuck in (b) and (c). I think I have to use (a) and Poincare's lemma for $1-$ forms but I don't know how.

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1 Answer 1

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I don't know what your proof of ii) looks like, but I'd expect that you estimate $$| \int_{\gamma} \omega | \le C M r $$

for closed curves $\gamma$ surrounding the origin, e.g. $\gamma(t)= r(\cos t , \sin t)$ and then let $r\rightarrow 0$. That approach should do the trick, for iii), as well. (in ii) you know the integrals approach $0$ as fast as $r$, but you don't need that rate of decay, convergence to $0$ is sufficient).

Edit (added explanation): if you write down the integral with the curves I defined earlier, you get $$\left| \int_\gamma \omega\,\right| =\left| \int_0^{2\pi}r(\omega_1\circ\gamma\sin(t) + \omega_2\circ\gamma \cos(t)) dt \right|\le \int_0^{2\pi}|\cdots |\le C r|\sup_{x=r} \omega |$$ for some constant $C$ not depending on $r$ or $\omega$ which swallows the integral over $\sin, \cos$. Note $r=\sqrt{x^2+y^2}$, so the expression on the rhs tends to $0$ when $r$ does. Since $\omega$ is closed this implies that $\int_\gamma \omega$ vanishes for all these $\gamma$, since they are homotopic to each other in the domain of definition of $\omega$, so the integral is the same for all these curves. You should be in the posession of a theorem which allows you to conclude from this that $\omega$ is exact. If this is not true you should explain how you solved ii)

Another edit: In simply connected domains every closed form is exact, this you seem to know. Recall the proof of this. In general, this is not true. What remains true is that a closed differential form $\omega$ is exact if the integral along any closed curve vanishes. This applies here, as I outlined earlier.

(You should have seen a theorem claiming something like that if you are given this kind of exercise. If not, it is not too hard to see, though a bit technical if you want to make it rigorous. You pick and fix any point $p$ in the domain you are looking at, $U$ say, (assume it's open and connected, then it's path connected), and for $q$ in $U$ choose any smooth curve $\gamma_{pq}$ joining $p$ and $q$. Then define $$F(p) = 0; \,\, F(q) :=\int_{\gamma_{pq}} \omega $$ You need to verify that this is well defined, i.e. does not depend on the particular choice of $\gamma_{pq}$, but this is what follows from the fact that the integral over any closed curve vanishes. The reasoning above does show this only for curves winding once around the origin, but is easily generalizd accordingly. I don't go into the details of this particular claim since I don't know which tools you have available.

Once you know $F$ is well defined you show $dF=\omega$, this is similar to the case of simply connected domains).

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This is, btw, a nice little exercise which shows that one should become wary if $|\omega|$ grows faster than $1/r$ when $r\rightarrow 0$. –  user20266 Jun 18 '12 at 11:40
    
I can't understand how from this I will prove that $\omega$ is exact. Can you explain it with a little more details? Thank you! –  passenger Jun 18 '12 at 12:53
    
@passenger I added an explanation to the answer. –  user20266 Jun 18 '12 at 13:43
    
I didn't said that I solved (b) ! Anyway if we were in a simple connected set then this is true but since $ \mathbb R^2$ is not I can't see how I conclude this. –  passenger Jun 18 '12 at 15:00
1  
@passenger: you can use brute force here, since $\sin$ and $\cos$ are both bounded by $1$ in absolute value, so, for example, $|r\omega_1 \circ \gamma(t) \sin(t)| \le |r\omega_1\circ \gamma(t) | \le r \sup_{x=r}| \omega_1 |$. This last rhs is constant if $r$ is fixed, so integrating everything gives you this constant times $2\pi$. The same can be done for the $\omega_2$ term, so you arrive at something like $\cdots \le 2 \pi r (\sup_r |\omega_1 | + \sup_r |\omega_2 |) \le 4\pi r \sup_r |\omega|$ –  user20266 Jun 18 '12 at 17:46

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