Extremal Length of Union

Theorem

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

Then the extremal length of their union satisfies

$\dfrac{1}{\lambda(\Gamma_1\cup \Gamma_2)} \leq \dfrac{1}{\lambda(\Gamma_1)} + \dfrac{1}{\lambda(\Gamma_2)}$

Suppose that additionally $\Gamma_1$ and $\Gamma_2$ are disjoint in the following sense: there exist disjoint Borel subsets $A_1,A_2\subseteq X$ such that $\bigcup\Gamma_1\subset A_1$ and $ \bigcup \Gamma_2\subset A_2$.

Then

$\dfrac{1}{\lambda(\Gamma_1\cup \Gamma_2)} = \dfrac{1}{\lambda(\Gamma_1)} + \dfrac{1}{\lambda(\Gamma_2)}$

Proof

Set $\Gamma := \Gamma_1\cup \Gamma_2$.

Let $\rho_1$ and $\rho_2$ be conformal metrics as in the definition of extremal length, normalized such that

$ L( \Gamma_1, \rho_1)=L(\Gamma_2,\rho_2)=1$

We define a new metric by $\rho := \max(\rho_1,\rho_2)$. Then $L(\Gamma,\rho)\geq 1$ and $A(\rho)\leq A(\rho_1)+A(\rho_2)$.

Hence

$\dfrac{1}{\lambda(\Gamma)} \leq \dfrac{A(\rho)}{L(\Gamma,\rho)} \leq A(\rho) = A(\rho_1) + A(\rho_2) = \dfrac{1}{L(\Gamma_1,\rho_1)} + \dfrac{1}{L(\Gamma_2,\rho_2)}$

Taking the infimum over all metrics $\rho_1$ and $\rho_2$, the claim follows.

Now suppose that the disjointness assumption holds, and let $\rho$ again be a Borel-measurable conformal metric, normalized such that $L(\Gamma,\rho)=1$.

We can define $\rho_1$ to be the restriction of $\rho$ to $A_1$, and likewise $\rho_2$ to be the restriction of $\rho$ to $A_2$.

By this we mean that, in local coordinates, $\rho_j$ is given by

$ \rho_j(z)|dz| = \begin{cases} \rho(z)|dz| & z\in A_j \\ 0 |dz| & \text{otherwise}\end{cases}$

Then $A(\rho)=A(\rho_1)+A(\rho_2)$ and $L(\Gamma_1,\rho_1),L(\Gamma_2,\rho_2)\geq 1$.

Hence

$ A(\rho)=A(\rho_1)+A(\rho_2) \geq \dfrac{A(\rho_1)}{L(\Gamma_1,\rho)} + \dfrac{A(\rho_2)}{L(\Gamma_2,\rho)} \geq \dfrac{1}{\lambda(\Gamma_1)} + \dfrac{1}{\lambda(\Gamma_2)}$

Taking the infimum over all metrics $\rho$, we see that

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

Together with the first part of the Proposition, this proves the claim.

$\blacksquare$