Extremal distance

Question

I'm reading Conformally Invariant Processes in the Plane by Lawner and I have a doubt, he takes $R_L=(0,L)\times(0,i\pi) \subset \mathbb C$ and define $\partial_1=[0,i\pi]$, $\partial_2 = [L,L+i\pi]$, then define $f(z)=2 \min \{\mathbb P_z(B_\tau \in \partial_1),P_z(B_\tau \in \partial_2)\}$, where $B$ is a Brownian motion starting at $z$, $\tau=\inf\{t\ge 0\colon B(t)\in \partial (R_L)\}$, and $$ \Omega(R_L,\partial_1,\partial_2)= \sup\{f(z) \colon z \in R_L \}. $$ Then he takes a Jordan domain D (a.k.a. a domain bounded with boundary a jordan curve) and 4 points in the boundary ordered clockwise ($z_1,z_2,z_3,z_4$). If $A_1,A_2$ are the arcs between $z_1,z_2$ and $z_3,z_4$ then he define $\Omega(D,A_1,A_2)$ in the same way as above. At least he define the $\pi$-extremal distance $L(D; A_1,A_2)$ as the only $L$ such that $\Omega(R_L,\partial_1,\partial_2)=\Omega(D,A_1,A_2)$. I think this $L$ is the only number in $\mathbb R$ such that exists a conformal map $f\colon D \to R_L$ such that $f(z_1)=i\pi$, $f(z_2)=0$, $f(z_3)=L$, $f(z_4)=L+i\pi$.

My question is: How can i prove that this $L$ exists and how can i prove that such $f$ exist too?

Answer 1

By the Riemann Mapping Theorem, $D$ and $R_L$ are conformally equivalent for any $L$. Now, there is a theory of when such a map $\varphi:B(0,1)\to D$ extends to the boundary: For instance, it extends continuously to all of the boundary when $D$ is a Jordan Domain, making the map a homeomorphism from the closed ball onto the closure of $D$. For details on this, see, for instance, Rudin or Complex Analysis: An Invitation by Murali Rao.

Now, fix the unique conformal equivalence $\psi_L: B(0,1)\to R_L$ such that $0$ is mapped to the middle of $R_L$ and $\psi'(0)>0$. By the above, $\psi_L$ extends to the boundary, and you can simply check $\psi_L^{-1}$ at the four corners and $\varphi^{-1}$ at $z_1,z_2,z_3,z_4$.

The classification of the holomorphic automorphisms of the unit disk should now tell you that it is only possible to find a conformal equivalence $\xi_L:D\to R_L$ matching the correct boundary points if there exists $\theta\in [0,2\pi)$ such that $\varphi^{-1}(z_n)=e^{i\theta}\psi_L^{-1}(c_n)$ where $c_n$ is the corresponding corner of $R_L$ for all $n$.

This should establish both existence and uniqueness. I'm sorry it took someone six months to find you post.

Answer 2

Okay. It took some time, but I have now completed what I find to be a more full, fleshed out proof. I don't know if anyone cares anymore, but for completeness' sake: Here is a proof of the fact for points in the disk (the general case then follows from the details mentioned above):

Note: Throughout, $Cay(z)=\frac{z-i}{z+i}$ denotes the Cayley transform (which is a known conformal equivalence between the disk and the upper half-plane). Also, it is assumed that we know that the extremal length of the family of horizontal crossings of $[0,L]\times i[0,1]$ is $L$.

Proposition:

Let $e^{i\theta_1},e^{i\theta_2},e^{i\theta_3},e^{i\theta_4}$ be four boundary points of $S^1$ labelled in counter-clockwise order. Then, there exists a unique $L>0$ such that there is a conformal equivalence $\varphi:B(0,1)\to (0,L)\times i(0,1)$ which extends to $S^1$ and maps $e^{i\theta_1}$ to $c_1:=0,$ $e^{i\theta_2}$ to $c_2:=L,$ $e^{i\theta_3}$ to $c_3:=L+i$ and $e^{i\theta_4}$ to $c_4:=i$.

Proof: Let $L>0.$ Since $(0,L)\times i(0,1)$ is a Jordan domain, any conformal equivalence from $B(0,1)$ extends continuously to a homeomorphism of $B[0,1]$.

Now, take one such conformal equivalence, $\psi$ such that $\psi(0)=\frac{L+i}{2}$ and $\psi'(0)>0$ and let $z_j=\psi^{-1}(c_j)$.

Note that the map $r(z)=-z+L+i$ is an automorphism of $[0,L]\times i[0,1]$ which maps $c_j$ to $c_{j+2}$, where addition is taken mod $4$. By the uniqueness part of the Riemann Mapping Theorem, this implies that $r(\psi(z))=\psi(-z)$. Accordingly, $z_j=-z_{j+2}$.

For $a>0$ let $$ \Psi_a(z)=Cay \left(a \frac{Cay^{-1}(z)-Cay^{-1}(e^{i\theta_4})}{Cay^{-1}(z)-Cay^{-1}(e^{i\theta_2})}\right), $$ which is an automorphism of $B(0,1)$ mapping $e^{i\theta_4}$ to $-1$ and $e^{i\theta_2}$ to $1$. Furthermore, the expression is continuous in $a$. In particular, let $x_a$ denote the intersection of the line $t\Psi_a (e^{i\theta_1})+(1-t\Psi_a (e^{i\theta_3}))$ with $\mathbb{R}$, which is again continuous. Now, as $a\to 0,$ we see that $x_a\to -1$ and as $a\to \infty,$ we have $x_a\to 1,$ implying, by the Intermediate Value theorem, that there is some $a$ such that $x_a=0.$ This happens if and only if $\Psi_a(e^{i\theta_1})=-\Psi_a(e^{i\theta_3})$.

Then, however, since the only automorphisms of $B(0,1)$ that fix $0$ are the rotations, the desired conformal equivalence exists if and only if the rotation taking $e^{i\theta_1}$ to $z_1$ also happens to take $e^{i\theta_2}$ to $z_2$. Now it only remains to establish that any such set of four points can be realised as pre-images of corners.

Fixing $z_1=1$ and $z_3=-1,$ we wish to argue that $L_0\mapsto z_2(L_0)$ is continuous on $[0,L]$. Indeed, applying $\psi^{-1}$ to $[0,L_0]\times i[0,1]$ yields two points $z_2'(L_0):=\psi^{-1}(L)$ and $z_3'(L_0):=\psi^{-1}(L_0+\frac{i}{2})$. Then, there is an $a(L_0)>0$ such that $$ \zeta_{L_0}(z)=Cay(a(L_0) (Cay^{-1}(z)-Cay^{-1}(z_3'(L_0))) $$ maps $z_2'(L_0)$ to a point diametrically opposite the image of $z_4$. By the previous uniqueness statement, $\zeta_{L_0}(\psi^{-1}(z))$ must be a rotation of the map from $[0,L_0]\times i[0,1]$ to $B(0,1)$ mapping $\frac{L_0+i}{2}$ to $0$.

However, $z_3'(L_0)$ and $z_2'(L_0)$ are continuous in $L_0,$ since $\psi^{-1}$ is continuous, and, fixing $z_3'(L_0),$ then $a(L_0)$ is clearly a continuous function of $z_2'(L_0)$. In conclusion, $L_0\mapsto z_2(L_0)$ is continuous on $[0,L_0]$. Since $L$ was arbitrary, we get that the position of $z_2$ depends continuously on $L$, given that we fix $z_1$ and $z_3$.

Now, if $\Gamma$ is the family of curves crossing from the arc $I_1=(z_1,z_2)$ to the arc $I_2(z_3,z_4),$ we have $\rho(z)=1_{B(0,1)}\frac{1}{d(I_1,I_2)}\in Adm(\Gamma)$ by definition, and $$ \int_{\mathbb{C}} \rho^2(z)\textrm{d}z=\frac{\pi^2}{d(I_1,I_2)}, $$ implying $m(\Gamma)\geq \frac{d(I_1,I_2)}{\pi^2}.$ Hence, as $\Gamma$ is conformally equivalent to the curve family associated to $Q_{1/L}$ (it simply crosses $[0,L]\times i[0,1]$ from the other side), $L=m(Q_L)\leq \frac{\pi^2}{d(I_1,I_2)},$ implying that $d(I_1,I_2)\to 0$ as $L\to \infty,$ which, by symmetry, happens if and only if $z_2\to z_3$. This establishes the desired.