I have two problems, the first one I think I've proved, but I have problems on the second one. Let $\omega$ a closed $1$-form defined on a open $U\subset \mathbb{R^2}$ and let $\gamma:[0,1]\rightarrow U$ a differentiable curve such that $\left | \omega_{\gamma(t)} \right |<c$ for all $t\in[0,1]$. Prove that $\left | \int _{\gamma }\omega \right |<ck$ for some $k=\text{constant}$.
My proof is, let $w=a(x)dx_{1}+b(x)dx_{2}$ so
$$\begin{align} \left | \int _{\gamma }\omega \right | &= \left | \int _{\gamma } a(x)dx_{1}+b(x)dx_{2} \right |\\ &=\left | \int_{0}^{1}a(\gamma (t))\gamma'(t) +b(\gamma (t))\gamma'(t) \right |\\ &\leq \left | \int_{0}^{1}a(\gamma (t))\gamma'(t)\right | + \left | \int_{0}^{1}b(\gamma (t))\gamma'(t) \right |\\ &=c_{1}\gamma'(1)-c_{1}\gamma'(0)+c_{2}\gamma'(1)-c_{2}\gamma'(0)\\ &\leq c(\gamma'(1)-\gamma'(0)) \end{align}$$ so let $k=(\gamma'(1)-\gamma'(0))$ Is that right?
Now, I´m stuck on the second part of the exercise; let $\omega$ a closed $1$-form in $\mathbb{R^{2}}\setminus\{(0,0)\}$, if $\omega$ is bounded on a disk centered on the origin then prove that $\omega$ is exact on $\mathbb{R^{2}}\setminus\{(0,0)\}$.
Thanks for any help!
If you look to $\omega$ as a vector rather than a 1-form, that is, $\omega(x,y) = (a(x,y), b(x,y)),$ then the following relation holds:
$$ \int_\gamma \omega = \int_{t_0}^{t_1} \langle \omega(\gamma(t)),\gamma'(t)\rangle dt,$$
where $t_0$ and $t_1$ are the frontiers of the interval where $t$ lies. You can do this because there is an isomorphism between $\mathbb{R}^2$ and its dual space. Now, you can use that $|\int f(x)dx| \leq \int|f(x)|dx$ together with the Cauchy-Schwarz inequality, and you will conclude that $k$ must be the length of $\gamma$.
For the second part, you can choose $\gamma$ to be a circumference centered in the origin and contained in the disk where $\omega$ is bounded, and calculate $\int_\gamma \omega$. Then you show, using the previous result and taking the radius of $\gamma$ arbitrarily small, that this integral must be zero. Then you use this to conclude the same for every closed curve in $\mathbb{R}^2-\{0\}$.