Extremal length

Remarks

The value does not change if $\Omega$ is replaced by a larger open set and $\rho$ is extended by $0$. If $f \colon \Omega \to \Omega'$ is a conformal map, then $\lambda(f(\Gamma)) = \lambda(\Gamma)$: the metric $\rho$ on $\Omega'$ pulls back to $(\rho \circ f)\abs{f'}$ with the same lengths and area. Example: let $R = (0, a) \times (0, b)$ and let $\Gamma$ be the curves in $R$ that join the left side to the right side. For $\rho = 1$, $L_\rho = a$ and $A(\rho) = ab$, so $\lambda(\Gamma) \ge a/b$. For any metric, each horizontal segment gives $\int_0^a \rho(x + iy)\, dx \ge L_\rho$, so by integrating in $y$ and the Cauchy–Schwarz inequality, $b L_\rho \le \int_R \rho \le \sqrt{ab\, A(\rho)}$, and $\lambda(\Gamma) = a/b$. Non-example: $\rho = 0$ is not a metric, since its area is $0$; it would give the meaningless ratio $0/0$. If every curve of $\Gamma_1$ contains a curve of $\Gamma_2$, then $\lambda(\Gamma_1) \ge \lambda(\Gamma_2)$.