Let \(B\subset\mathbb C\) be an annulus bounded by Jordan curves \(C_1\) and \(C_2\). Prove that there are an arc \(\alpha\subset B\) joining \(C_1\) to \(C_2\) and a loop \(\beta\subset B\) separating \(C_1\) from \(C_2\) such that \[ L(\alpha)L(\beta)\le\operatorname{area}(B). \]
Hint: take a conformal map \(f:A(R)\to B\) and consider \[ \int_{A(R)} |f'(z)|\,|z|^{-1}\,|dz|^2. \]
Solution
Let \[ f:A(R)=\{1<|z|<R\}\longrightarrow B \] be a conformal map, and set \[ I=\int_{A(R)}\frac{|f'(z)|}{|z|}\,dA(z). \] We will select a radial segment and a concentric circle whose image lengths are controlled by two different averages of the same integral \(I\).
1. For \(\theta\in[0,2\pi)\), let\[ \ell(\theta) =\int_1^R|f'(re^{i\theta})|\,dr. \]Then\[ I=\int_0^{2\pi}\ell(\theta)\,d\theta. \]
In polar coordinates \(z=re^{i\theta}\), \[ dA=r\,dr\,d\theta. \] Therefore \[ \begin{aligned} I &=\int_0^{2\pi}\int_1^R \frac{|f'(re^{i\theta})|}{r}\,r\,dr\,d\theta\\ &=\int_0^{2\pi}\int_1^R |f'(re^{i\theta})|\,dr\,d\theta\\ &=\int_0^{2\pi}\ell(\theta)\,d\theta. \end{aligned} \]
2. There is an angle \(\theta_0\) for which the image of the radial segment\[ \{re^{i\theta_0}:1<r<R\} \]has length at most \(I/(2\pi)\).
By 1, the average value of \(\ell(\theta)\) over \([0,2\pi)\) is \[ \frac{I}{2\pi}. \] Hence for some \(\theta_0\), \[ \ell(\theta_0)\le\frac{I}{2\pi}. \]
The image radial segment approaches the two boundary components of \(B\); since these are Jordan curves, the conformal map extends to the boundary components, so its closure is an arc \(\alpha\) joining \(C_1\) to \(C_2\). Its length is \[ L(\alpha)=\ell(\theta_0)\le\frac{I}{2\pi}. \]
3. For \(1<r<R\), let\[ m(r)=\int_0^{2\pi}|f'(re^{i\theta})|\,r\,d\theta. \]Then\[ I=\int_1^R m(r)\,\frac{dr}{r}. \]
The quantity \(m(r)\) is exactly the Euclidean length of the loop \[ \theta\longmapsto f(re^{i\theta}). \] Also \[ \begin{aligned} \int_1^R m(r)\frac{dr}{r} &=\int_1^R\int_0^{2\pi} |f'(re^{i\theta})|\,r\,d\theta\,\frac{dr}{r}\\ &=\int_1^R\int_0^{2\pi}|f'(re^{i\theta})|\,d\theta\,dr\\ &=I \end{aligned} \] by 1.
4. There is a radius \(r_0\in(1,R)\) such that\[ m(r_0)\le\frac{I}{\log R}. \]
The measure \(dr/r\) of the interval \((1,R)\) is \[ \int_1^R\frac{dr}{r}=\log R. \] Thus 3 says that the average of \(m(r)\) with respect to this measure is \(I/\log R\). Hence some \(r_0\) satisfies \[ m(r_0)\le\frac{I}{\log R}. \]
The circle \(|z|=r_0\) separates the two boundary components of \(A(R)\), so its image \[ \beta=f(\{|z|=r_0\}) \] is a loop in \(B\) separating \(C_1\) from \(C_2\), and \[ L(\beta)=m(r_0). \]
5. Cauchy–Schwarz gives\[ I^2\le 2\pi\log R\,\operatorname{area}(B). \]
By Cauchy–Schwarz on \(A(R)\), \[ \begin{aligned} I^2 &=\left( \int_{A(R)}|f'(z)|\frac1{|z|}\,dA \right)^2\\ &\le \left(\int_{A(R)}|f'(z)|^2\,dA\right) \left(\int_{A(R)}\frac{1}{|z|^2}\,dA\right). \end{aligned} \]
Because \(f\) is conformal and bijective, \[ \int_{A(R)}|f'(z)|^2\,dA =\operatorname{area}(B). \] Also \[ \int_{A(R)}\frac1{|z|^2}\,dA =\int_0^{2\pi}\int_1^R\frac1{r^2}r\,dr\,d\theta =2\pi\log R. \] Combining these gives the claimed inequality.
6. The selected curves satisfy\[ L(\alpha)L(\beta)\le\operatorname{area}(B). \]
By 2 and 4, \[ L(\alpha)L(\beta) \le \frac{I}{2\pi}\cdot\frac{I}{\log R} =\frac{I^2}{2\pi\log R}. \] Applying 5 gives \[ L(\alpha)L(\beta) \le\operatorname{area}(B). \]