# Prove that $2\text{Area}(\omega) \leq \text{length}(\gamma)\,\text{distance}(\gamma)$

So basically, I am given the following to prove:

Let $$+\gamma$$ be a positively oriented smooth Jordan arc, and let $$\omega$$ denote the interior of $$+\gamma$$. Recall that if $$F = (F_1, F_2):D \to \mathbb{R}^2$$ is a continuously differentiable vector field in an open set $$D$$ containing $$\omega \cup(+\gamma)$$, then

$$\iint_\omega \left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) dxdy = \oint_{+\gamma}F \cdot \overrightarrow{ds}$$ where the right hand-side is the line-integral of $$F$$ along the path $$+\gamma$$.

By suitably choosing $$F$$, prove that $$\DeclareMathOperator{\Area}{Area} \DeclareMathOperator{\diameter}{diameter} \DeclareMathOperator{\length}{length} 2\Area(\omega) \leq \diameter(+\gamma) \length(+\gamma)$$ where

$$\diameter(+\gamma) = \sup\{|z(s)-z(t)| : s,t \in [a,b]\}$$ and $$\length(+\gamma) = \int_{a}^{b} |z'(t)| dt$$.

The only thing I know so far is that I need to find an $$F$$ such that $$\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} =1$$ because then $$\iint_{\omega} dxdy = \Area(\omega).$$ However, I do not know where to proceed from there!

I assume that $$z$$ is $$\gamma$$ in the definition of $$\operatorname{diameter}(+\gamma)$$. Without loss of generality, we can assume that $$\gamma(a)=(0,0)$$ (otherwise, we can translate $$\gamma$$ until the point $$\gamma(a)$$ hits the origin).

I would take $$\vec{F}=(F_1,F_2)$$ with $$F_1=-y$$ and $$F_2=x$$, so that $$\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=2.$$ By Green's theorem, we have $$2\operatorname{Area}(\omega)=\iint_\omega \left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)\ dx\ dy=\oint_{+\gamma}\vec{F}\cdot \vec{dr}.$$ Hence, $$2\operatorname{Area}(\omega)\leq \oint_{+\gamma}\left\vert\vec{F}\cdot \vec{dr}\right\vert\leq \oint_{+\gamma}\left\Vert\vec{F}\right\Vert\ dr.$$ Because $$\left\Vert\vec{F}\right\Vert\leq \operatorname{diameter}(+\gamma)$$, we conclude that $$2\operatorname{Area}(\omega)\leq \oint_{+\gamma}\operatorname{diameter}(+\gamma)\ dr=\operatorname{diameter}(+\gamma)\oint_{+\gamma}dr=\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma).$$

This inequality is a quite weak. If we define $$\operatorname{radius}(+\gamma)$$ to be $$\inf\Big\{r>0:\exists p\in\mathbb{R}^2,\ \forall u\in[a,b],\ \big\vert\gamma(u)-p\big\vert< r\Big\}\,,$$ then we have $$2\operatorname{Area}(\omega)\leq \operatorname{radius}(+\gamma)\operatorname{length}(+\gamma).$$ The equality holds iff $$\gamma$$ traces a circle (once). You can use Jung's theorem to show that $$2\operatorname{Area}(\omega)\leq \frac{1}{\sqrt{3}}\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma),$$ which is stronger than the required result, but is still weak. So, it is an interesting question to find the infimum $$\lambda_{\min}$$ of all $$\lambda>0$$ such that $$\operatorname{Area}(\omega)\leq \lambda\operatorname{diameter}(+\gamma)\operatorname{length}(+\gamma).$$ We already know that $$\lambda_{\min} \leq \frac1{2\sqrt{3}}$$.

• In the case where $\gamma$ is a loop based at the origin $\lvert F \rvert \leq \mathrm{diam}(\gamma)$. How can you be sure that this is true for a general loop? Commented Oct 18, 2018 at 14:20
• @Gibbs I don't understand the question. Can you elaborate?
– user593746
Commented Oct 18, 2018 at 14:24
• Did you miss the assumption that I made in the first paragraph, namely that I assumed the origin lies in the image of $\gamma$?
– user593746
Commented Oct 18, 2018 at 14:25
• We have practically the same solution. I don't see why translating $\gamma$ is worse than translating $F$.
– user593746
Commented Oct 18, 2018 at 14:34
• To each his own, but I think assuming that $\gamma$ passing through the origin makes it quite easy to see that $\left\Vert\vec{F}\right\Vert\leq \operatorname{diameter}(+\gamma)$.
– user593746
Commented Oct 18, 2018 at 14:38

Choose $$F(x,y) = (-y+y(a),x-x(a))$$ and define a $$1$$-form $$\alpha := (x-x(a))\mathrm{d}y+(y(a)-y)\mathrm{d}x = F_2(x,y)\mathrm{d}y+F_1(x,y)\mathrm{d}x$$. Then $$\mathrm{d}\alpha = 2\mathrm{d}x \wedge \mathrm{d}y$$ and so $$\int_{\omega} \mathrm{d}\alpha= \int_{\omega}\left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) \mathrm{d}x\mathrm{d}y= 2\int_{\omega} \mathrm{d}x \wedge \mathrm{d}y = 2\mathrm{Area}(\omega).$$ By Stokes theorem (which reduces exactly to the curl theorem you are recalling in this case), you also get \begin{align} \int_{\omega} \mathrm{d}\alpha &= \int_{\partial \omega} \alpha = \int_{\gamma+} \alpha\\ &= \int_a^b (x(t)-x(a))y'(t)+(y(a)-y(t))x'(t)\,\mathrm{d}t = \int_a^b \langle F(x,y),(x'(t),y'(t))\rangle\,\mathrm{d}t, \end{align} where $$\langle \cdot, \cdot \rangle$$ denotes the dot product. By Cauchy-Schwarz inequality you get \begin{align} \int_a^b \langle F(x,y),(x'(t),y'(t))\rangle\,\mathrm{d}t & \leq \int_a^b \sqrt{(x(t)-x(a))^2+(y(t)-y(a))^2}\sqrt{x'(t)^2+y'(t)^2}\, \mathrm{d}t \\ & \leq \mathrm{diameter}(\omega)\int_a^b \lvert z'(t)\rvert \, \mathrm{d}t\\ &\leq \mathrm{diameter}(\omega)\mathrm{length}(\gamma+). \end{align}