Comparison Principle for Extremal Length

From ProofWiki
Jump to: navigation, search


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 $X$.

Let every element of $\Gamma_1$ contain some element of $\Gamma_2$.


Then the extremal lengths of $\Gamma_1$ and $\Gamma_2$ are related by:

$\lambda \left({\Gamma_1}\right) \ge \lambda \left({\Gamma_2}\right)$


More precisely, for every conformal metric $\rho$ as in the definition of extremal length, we have:

$L \left({\Gamma_1, \rho}\right) \ge L \left({\Gamma_2, \rho}\right)$


Proof

We have:

\(\displaystyle L \left({\Gamma_1, \rho}\right)\) \(=\) \(\displaystyle \inf_{\gamma \mathop \in \Gamma_1} L \left({\gamma, \rho}\right)\) $\quad$ by definition $\quad$
\(\displaystyle \) \(\ge\) \(\displaystyle \inf_{\gamma \mathop \in \Gamma_2} L \left({\gamma, \rho}\right)\) $\quad$ since every curve of $\Gamma_1$ contains a curve of $\Gamma_2$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle L \left({\Gamma_2, \rho}\right)\) $\quad$ by definition $\quad$

This proves the second claim.

The second claim implies the first by definition.

$\blacksquare$


Sources