Parallel Law for Extremal Length

Theorem

Let $X$ be a Riemann surface.

Let $\Gamma_1, \Gamma_2$ be families of rectifiable curves (or, more generally, families of disjoint unions of rectifiable curves) on $X$.

Suppose that $\Gamma_1$ and $\Gamma_2$ are disjoint, in the sense that there exist disjoint Borel subsets $A_1, A_2 \subseteq X$ such that, for any $\gamma_1 \in \Gamma_1$ and $\gamma_2 \in \Gamma_2$, we have $\gamma_1 \subseteq A_1$ and $\gamma_2 \subseteq A_2$.

Suppose that $\Gamma$ is a third curve family, with the property that every element $\Gamma_1$ and every element of $\Gamma_2$ contains some element of $\Gamma$.

Then the extremal length of $\Gamma$ satisfies

$\dfrac 1 {\lambda(\Gamma)} \geq \dfrac 1 {\lambda(\Gamma_1)} + \dfrac 1 {\lambda(\Gamma_2)}$

Proof

The assumption means that every element of $\Gamma_1 \cup \Gamma_2$ contains some element of $\Gamma$.

Hence:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac 1 {\lambda (\Gamma)}\)	\(\geq\)	\(\displaystyle \frac 1 {\lambda (\Gamma_1 \cup \Gamma_2)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by the comparison principle)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 {\lambda (\Gamma_1)} + \frac 1 {\lambda (\Gamma_2)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by the extremal length formula for unions).

$\blacksquare$

Sources

• L.V.Ahlfors: Conformal Invariants: Topics in Geometric Function Theory (1973)