is a curve in $\Omega$ joining $[A,B]_{\partial{\Omega^{\ast}}}$ and
$[C,D]_{\partial{\Omega^{\ast}}}$}\\}$ $\displaystyle d^{\prime}$
$\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$ is a curve in
$\Omega$ joining $[B,C]_{\partial{\Omega^{\ast}}}$ and
$[D,A]_{\partial{\Omega^{\ast}}}$}\\}$

Then:

$-\frac{8\pi}{\log\big{(}\frac{d^{\prime}}{2\delta}\big{)}}\leq{L^{\prime}-L}\leq{\frac{16\pi{L}^{2}}{\log\big{(}\frac{d}{2\delta}\big{)}}}$

###### Proof.

Let $\phi:\Omega\rightarrow{\mathcal{R}_{L}}$ be the conformal map from
$\Omega$ to the rectangle $\mathcal{R}_{L}$ so that the prime ends $A,B,C,D$
of $\Omega$ are mapped to the four corners of $\mathcal{R}_{L}$ and in
particular, $\phi(A)=i$. We write $\phi=h+i\widetilde{h}$ where $h$ is the
real part and $\widetilde{h}$ is the imaginary part of $\phi$. Then
$\|\nabla{h}\|=\|\nabla{\widetilde{h}}\|$ is the extremal metric giving us the
extremal length between $[A,B]_{\partial{\Omega^{\ast}}}$ and
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$. By Theorem 3.1:

$\displaystyle h(z)$
$\displaystyle\leq{\frac{2\pi}{\log\big{(}\frac{d^{\prime}}{2\delta}\big{)}}}\hskip
5.0pt\text{for $z\in{[A^{\prime},B^{\prime}]_{\partial{\Omega^{\prime}}}}$}$
$\displaystyle h(z)$
$\displaystyle\geq{L-\frac{2\pi}{\log\big{(}\frac{d^{\prime}}{2\delta}\big{)}}}\hskip
5.0pt\text{for $z\in{[C^{\prime},D^{\prime}]_{\partial{\Omega^{\prime}}}}$}$
$\displaystyle\widetilde{h}(z)$
$\displaystyle\leq{\frac{2\pi{L}}{\log\big{(}\frac{d}{2\delta}\big{)}}}\hskip
5.0pt\text{for $z\in{[B^{\prime},C^{\prime}]_{\partial{\Omega^{\prime}}}}$}$
$\displaystyle\widetilde{h}(z)$
$\displaystyle\geq{1-\frac{2\pi{L}}{\log\big{(}\frac{d}{2\delta}\big{)}}}\hskip
5.0pt\text{for $z\in{[D^{\prime},A^{\prime}]_{\partial{\Omega^{\prime}}}}$}$

The condition that $\delta\leq{\frac{d^{\prime}}{2}e^{-\frac{4\pi}{L}}}$
ensures that the values of $h$ on
$[C^{\prime},D^{\prime}]_{\partial{\Omega^{\prime}}}$ are larger than the
values of $h$ on $[A^{\prime},B^{\prime}]_{\partial{\Omega^{\prime}}}$.
Analogously, the condition that
$\delta\leq{\frac{d^{\prime}}{2}e^{-16\pi{L}}}$ ensures that the values of
$\widetilde{h}$ on $[D^{\prime},A^{\prime}]_{\partial{\Omega^{\prime}}}$ are
larger than the values of $\widetilde{h}$ on
$[B^{\prime},C^{\prime}]_{\partial{\Omega^{\prime}}}$. Plugging the metric
$\|\nabla{h}\|$ into the variational formula for $L^{\prime}$, we have that:

$L^{\prime}\geq{\frac{\left(L-\frac{4\pi}{\log\big{(}\frac{d^{\prime}}{2\delta}\big{)}}\right)^{2}}{L}}\geq{L-\frac{8\pi}{\log\big{(}\frac{d^{\prime}}{2\delta}\big{)}}}$
(3.7)

Similarly, plugging $\|\nabla{\widetilde{h}}\|$ into the variational formula
for the dual problem, we have that:

$\frac{1}{L^{\prime}}\geq{\frac{\left(1-\frac{4\pi{L}}{\log\big{(}\frac{d}{2\delta}\big{)}}\right)^{2}}{L}}\hskip
20.0pt\implies\hskip
20.0ptL^{\prime}\leq{\frac{L}{\left(1-\frac{8\pi{L}}{\log\big{(}\frac{d}{2\delta}\big{)}}\right)}}\leq{L+\frac{16\pi{L}^{2}}{\log\big{(}\frac{d}{2\delta}\big{)}}}$
(3.8)

Combining Equations 3.7 and 3.8, we arrive at the desired result. ∎

######  Proposition 3.7.

Suppose $(H,W^{\bullet},X^{\bullet},Y^{\bullet},Z^{\bullet})$ is a
suborthodiagonal rectangle of
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ so that
$(\widehat{H},W^{\bullet},X^{\bullet},Y^{\bullet},Z^{\bullet})$ is a
$\delta$\- good interior approximation of
$(\widehat{G},A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ and
$G=(V^{\bullet}\sqcup{V^{\circ}},E)$ is an orthodiagonal map with edges of
length at most $\varepsilon$. Let $L$ denote the extremal length between
$[A^{\bullet},B^{\bullet}]$ and $[C^{\bullet},D^{\bullet}]$ in $G^{\bullet}$
and let $L^{\prime}$ denote the extremal length between
$[W^{\bullet},X^{\bullet}]$ and $[Y^{\bullet},Z^{\bullet}]$ in $H^{\bullet}$.
Suppose also that $\delta$ and $\varepsilon$ satisfy
$(\delta\vee{\varepsilon})\leq{d^{\prime}e^{-\frac{2K}{L}}}$ and
$(\delta\vee{\varepsilon})\leq{de^{-8KL}}$, where:

$\displaystyle d$ $\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$
is a curve in $\Omega$ joining
$[A^{\bullet},B^{\bullet}]_{\partial{\widehat{G}}}$ and
$[C^{\bullet},D^{\bullet}]_{\partial{\widehat{G}}}$}\\}$ $\displaystyle
d^{\prime}$ $\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$ is a
curve in $\Omega$ joining $[B^{\bullet},C^{\bullet}]_{\partial{\widehat{G}}}$
and $[D^{\bullet},A^{\bullet}]_{\partial{\widehat{G}}}$}\\}$

and $K>0$ is the absolute constant from Theorem 3.3. Then:

$-\frac{4K}{\log\big{(}\frac{d^{\prime}}{\delta\vee{\varepsilon}}\big{)}}\leq{L^{\prime}-L\leq{\frac{8KL^{2}}{\log\big{(}\frac{d}{\delta\vee{\varepsilon}}\big{)}}}}$

###### Proof.

Let $h:V^{\bullet}\rightarrow{\mathbb{R}}$ and
$\widetilde{h}:V^{\circ}\rightarrow{\mathbb{R}}$ be the real and imaginary
parts of the tiling map associated to the orthodiagonal rectangle
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$. Then the metric
$\rho^{\bullet}$ on $G^{\bullet}$ given by the formula
$\rho^{\bullet}(u,v)=|h(u)-h(v)|$ for any edge $\\{u,v\\}\in{E^{\bullet}}$ is
the extremal metric giving us the extremal length between
$[A^{\bullet},B^{\bullet}]$ and $[C^{\bullet},D^{\bullet}]$ in $G^{\bullet}$.
Similarly, the metric $\rho^{\circ}$ on $G^{\circ}$ given by the formula
$\rho^{\circ}(u,v)=|\widetilde{h}(u)-\widetilde{h}(v)|$ for any edge
$\\{u,v\\}\in{E^{\circ}}$ is the extremal metric giving us the extremal length
between $[B^{\circ},C^{\circ}]$ and $[D^{\circ},A^{\circ}]$ in $G^{\circ}$. By
Theorem 3.3,

$\displaystyle h(z)$
$\displaystyle\leq{\frac{K}{\log\big{(}\frac{d^{\prime}}{\delta\vee{\varepsilon}}\big{)}}}\hskip
5.0pt\text{for $z\in{[W^{\bullet},X^{\bullet}]}$}$ $\displaystyle h(z)$
$\displaystyle\geq{L-\frac{K}{\log\big{(}\frac{d^{\prime}}{\delta\vee{\varepsilon}}\big{)}}}\hskip
5.0pt\text{for $z\in{[Y^{\bullet},Z^{\bullet}]}$}$
$\displaystyle\widetilde{h}(z)$ $\displaystyle\leq{\frac{K\hskip
1.0ptL}{\log\big{(}\frac{d}{\delta\vee{\varepsilon}}\big{)}}}\hskip
5.0pt\text{for $z\in{[X^{\circ},Y^{\circ}]}$}$ $\displaystyle\widetilde{h}(z)$
$\displaystyle\geq{1-\frac{K\hskip
1.0ptL}{\log\big{(}\frac{d}{\delta\vee{\varepsilon}}\big{)}}}\hskip
5.0pt\text{for $z\in{[Z^{\circ},W^{\circ}]}$}$

The condition that
$(\delta\vee{\varepsilon})\leq{d^{\prime}e^{-\frac{2K}{L}}}$ ensures that the
values of $h$ on $[Y^{\bullet},Z^{\bullet}]$ are larger than the values of $h$
on $[W^{\bullet},X^{\bullet}]$. Similarly, the condition that
$(\delta\vee{\varepsilon})\leq{de^{-8KL}}$ ensures that the values of
$\widetilde{h}$ on $[Z^{\circ},W^{\circ}]$ are larger than the values of
$\widetilde{h}$ on $[X^{\circ},Y^{\circ}]$. Plugging the metric
$\rho^{\bullet}$ into the variational formula for $L^{\prime}$, we have that:

$L^{\prime}\geq{\frac{\left(L-\frac{2K}{\log\big{(}\frac{d^{\prime}}{\delta\vee{\varepsilon}}\big{)}}\right)^{2}}{L}}\geq{L-\frac{4K}{\log\big{(}\frac{d^{\prime}}{\delta\vee{\varepsilon}}\big{)}}}$
(3.9)

Similarly, plugging $\rho^{\circ}$ into the variational formula for the dual
problem, we have that:

$\frac{1}{L^{\prime}}\geq{\frac{\left(1-\frac{2KL}{\log\big{(}\frac{d}{\delta\wedge{\varepsilon}}\big{)}}\right)^{2}}{L}}\hskip
20.0pt\implies\hskip
20.0ptL^{\prime}\leq{\frac{L}{\left(1-\frac{4KL}{\log\big{(}\frac{d}{\delta\wedge{\varepsilon}}\big{)}}\right)}}\leq{L+\frac{8K{L}^{2}}{\log\big{(}\frac{d}{\delta\vee{\varepsilon}}\big{)}}}$
(3.10)

Combining Equations 3.9 and 3.10, we arrive at the desired result. ∎

##  4 Limits of Discrete Holomorphic Functions are Holomorphic

Given an orthodiagonal map $G=(V^{\bullet}\sqcup{V^{\circ}},E)$ recall that a
function $F:V^{\bullet}\sqcup{V^{\circ}}\rightarrow{\mathbb{C}}$ is discrete
holomorphic on $G$ if for any interior face $Q$ of $G$ with primal diagonal
$e^{\bullet}=\\{u_{1},u_{2}\\}$ and dual diagonal
$e^{\circ}=\\{v_{1},v_{2}\\}$ we have that:

$\frac{F(u_{2})-F(u_{1})}{u_{2}-u_{1}}=\frac{F(v_{2})-F(v_{1})}{v_{2}-v_{1}}$

In particular, notice that if $F|_{V^{\bullet}}$ is strictly real then
$F|_{V^{\circ}}$ is strictly imaginary up to an additive constant. This
situation is typical of applications of discrete complex analysis. That is,
the real part of our discrete holomorphic function typically lives on the
primal graph $G^{\bullet}=(V^{\bullet},E^{\bullet})$ and the imaginary part
lives on the dual graph $G^{\circ}=(V^{\circ},E^{\circ})$.

In this section, we prove the following result of independent interest:

######  Theorem 4.1.

Let $\Omega$ be a subdomain of $\mathbb{C}$ and let
$\Omega_{n}=(V_{n}^{\bullet}\sqcup{V_{n}^{\circ}},E_{n})$ be a sequence of
orthodiagonal maps so that the edges of $\Omega_{n}$ are of length at most
$\varepsilon_{n}$ and $\varepsilon_{n}\rightarrow{0}$ as
$n\rightarrow{\infty}$. Suppose that for any compact set $K\subseteq{\Omega}$
there exists $N\in{\mathbb{N}}$ such that for all $n\geq{N}$,
$K\subseteq{\widehat{\Omega}_{n}}$. For each $n\in{\mathbb{N}}$, let
$F_{n}:V_{n}^{\bullet}\sqcup{V_{n}^{\circ}}\rightarrow{\mathbb{C}}$ be a
discrete holomorphic function on $\Omega_{n}$ so that:

$\text{Re}(F_{n}(z))=0\hskip 5.0pt\text{for all $z\in{V^{\circ}}$},\hskip
20.0pt\text{Im}(F_{n}(z))=0\hskip 5.0pt\text{ for all $z\in{V^{\bullet}}$}$

That is, the real part of $F_{n}$ lives on the primal graph
$\Omega_{n}^{\bullet}$ and the imaginary part of $F_{n}$ lives on the dual
graph $\Omega_{n}^{\circ}$. Suppose also that the Dirichlet energies of the
$F_{n}$’s are uniformly bounded on compacts. That is, for any compact set
$K\subseteq{\Omega}$ and $N\in{\mathbb{N}}$ such that $K\subseteq{\Omega_{n}}$
for all $n\geq{N}$ we have that:

$\sup_{n\geq{N}}\mathcal{E}^{\bullet}_{K}(\text{Re}(F_{n}))=\frac{1}{2}\sup_{n\geq{N}}\Big{(}\sum_{\begin{subarray}{c}Q\in{F_{in}(\Omega_{n})}\\\
Q\subseteq{K}\end{subarray}}\frac{|e^{\circ}_{Q}|}{|e^{\bullet}_{Q}|}\Big{(}\text{Re}(F_{n}((e_{Q}^{\bullet})^{+}))-\text{Re}(F_{n}((e_{Q}^{\bullet})^{-}))\Big{)}^{2}\Big{)}<\infty$

Or equivalently, since $\text{Re}(F_{n})$ and $\text{Im}(F_{n})$ are conjugate
harmonic functions:

$\sup_{n\geq{N}}\mathcal{E}^{\circ}_{K}(\text{Im}(F_{n}))=\frac{1}{2}\sup_{n\geq{N}}\Big{(}\sum_{\begin{subarray}{c}Q\in{F_{in}(\Omega_{n})}\\\
Q\subseteq{K}\end{subarray}}\frac{|e^{\bullet}_{Q}|}{|e^{\circ}_{Q}|}\Big{(}\text{Im}(F_{n}((e_{Q}^{\circ})^{+}))-\text{Im}(F_{n}((e_{Q}^{\circ})^{-}))\Big{)}^{2}\Big{)}<\infty$

Let $\widehat{F}_{n}:\widehat{\Omega}_{n}\rightarrow{\mathbb{C}}$ be a
sequence of continuous functions on $\widehat{\Omega}_{n}$ so that:

$\text{Re}(\widehat{F}_{n}(z))=F_{n}(z)\hskip 5.0pt\text{for all
$z\in{V^{\bullet}_{n}}$},\hskip
20.0pt\text{Im}(\widehat{F}_{n}(z))=F_{n}(z)\hskip 5.0pt\text{for all
$z\in{V^{\circ}_{n}}$}$

That is, for each $n\in{\mathbb{N}}$, $\widehat{F}_{n}$ is some sort of
sensible extension of $F_{n}$ to a continuous function on
$\widehat{\Omega}_{n}$. If:

$\widehat{F}_{n}\rightarrow{F}\hskip 4.0pt\text{uniformly on compacts in
$\Omega$}$

Then $F:\Omega\rightarrow{\mathbb{C}}$ is holomorphic.

In Section 5 we will use this result to show that any subsequential limit of
our tiling maps is holomorphic. This is crucial to showing that our tiling
maps converge to the relevant conformal map.

######  Remark 4.2.

Note that while Theorem 4.1 is sufficient for our purposes, stronger results
already exist in the literature. Namely, in [9], Chelkak, Laslier, and
Russkikh prove that local uniform limits of discrete holomorphic functions on
t- embeddings are holomorphic (see Proposition 6.15 of [9]). Since every
orthodiagonal map is a t-embedding (see Section 8.1 of [9]), it follows that
the same is true of discrete holomorphic functions on orthodiagonal maps.

Having been initially unfamiliar with their work, we found an independent
proof of this result in the more restrictive orthodiagonal setting, with the
additional condition that the Dirichlet energies of your discrete holomorphic
functions must be uniformly bounded on compacts. While this is a strictly
weaker result in a less general setting, we still think it is worthwhile to
present the proof. The reasons for this are twofold:

  1. 1.

The proof is elementary and takes place in a simpler setting.

  2. 2.

It gives us an excuse to introduce Lemma 4.3 which tells us that we can
approximate continuous contours by discrete contours in our orthodiagonal map,
that are close to the corresponding continuous contour in Hausdorff distance
and have comparable length. We use this in [23] to show that the discrete
Dirichlet problem on orthodiagonal maps converges with a polynomial rate to
the corresponding continuous Dirichlet problem, for Hölder boundary data.

As per the discussion above, to prove Theorem 4.1, we first need the following
lemma:

######  Lemma 4.3.

Suppose $G=(V^{\bullet}\sqcup{V^{\circ}},E)$ is an orthodiagonal map with mesh
at most $\varepsilon$, and $\delta$ is a positive real number so that
$\delta\geq{4\varepsilon}$. Suppose $\ell$ is a line segment in $\mathbb{C}$
so that $L=\text{length}(\ell)\geq{8\varepsilon}$ and $\widehat{G}$ contains a
$\delta$\- neighborhood of $\ell$. Then there exist nearest- neighbor paths
$\gamma^{\bullet}$ in $G^{\bullet}$ and $\gamma^{\circ}$ in $G^{\circ}$ that
are $\delta$\- close to $\ell$ in Hausdorff distance, and:

$\text{length}(\gamma^{\bullet}),\text{length}(\gamma^{\circ})\leq{2L\Big{(}1+\frac{4\varepsilon}{\delta}\Big{)}}$

Furthermore, the endpoints of $\gamma^{\bullet}$ and $\gamma^{\circ}$ both lie
within $\delta$ of the endpoints of $\ell$.

###### Proof.

Let $G=(V^{\bullet}\sqcup{V^{\circ}},E)$ be an orthodiagonal map with edges of
length at most $\varepsilon$. It follows that the edges of $G^{\bullet}$ and
$G^{\circ}$, which correspond to diagonals of inner faces of $G$, have length
at most $2\varepsilon$. Let $\ell$ be a line segment in $\mathbb{C}$ so that
$\widehat{G}$ contains a $\delta$\- neighborhood of $\ell$. Without loss of
generality, suppose $\ell=[0,L]$, the line segment between $0$ and $L$ in
$\mathbb{C}$. Let $R=[0,L]\times{[-\delta,\delta]}$. Since $\widehat{G}$
contains a $\delta$\- neighborhood of $\ell$, $R\subseteq{\widehat{G}}$.

Let $H=(V_{H}^{\bullet}\sqcup{V_{H}^{\circ}},E_{H})$ be the suborthodiagonal
map of $G$ formed by taking the union of all the inner faces of $G$ contained
in $R$. Since $R$ is convex, $H$ is simply- connected with a unique,
distinguished boundary face. Moreover, observe that any pair of neighboring
vertices along the boundary of $H$, must be part of an inner face of $G$ that
intersects the boundary of the rectangle $R$. Thus, the vertices and edges on
the boundary of $H$ all lie within $\varepsilon$ of $\partial{R}$. Let
$A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet}$ be the points of
$\partial{V}_{H}^{\bullet}$ closest to the four corners
$(0,-\delta),(0,\delta),(L,\delta),(L,-\delta)$ of $R$. Then
$(H,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a
$(2\varepsilon,\varepsilon)$\- good interior approximation of $R$.

Figure 12: The $(\varepsilon,2\varepsilon)$\- good approximation of the
conformal rectangle $R$ (with its four corners as the distinguished boundary
points) that we construct in the proof of Lemma 4.3. For simplicity, our
orthodiagonal map $G$ here is just a chunk of the square grid but, in general,
this need not be the case.

By Proposition 3.5:

$\frac{(L-4\varepsilon)^{2}}{4\delta{L}}\leq{\lambda([A^{\bullet},B^{\bullet}]\leftrightarrow{[C^{\bullet},D^{\bullet}]};H^{\bullet})}\leq{\frac{4\delta{L}}{(2\delta-4\varepsilon)^{2}}}$

Plugging the metric $\rho(e^{\bullet})=|e^{\bullet}|$ into the variational
problem for
$\lambda([A^{\bullet},B^{\bullet}]\leftrightarrow{[C^{\bullet},D^{\bullet}]};H^{\bullet})$
we have that:

$\displaystyle\frac{4\delta{L}}{(2\delta-4\varepsilon)^{2}}$
$\displaystyle\geq{\lambda([A^{\bullet},B^{\bullet}]\leftrightarrow{[C^{\bullet},D^{\bullet}]};G^{\bullet})}$
(4.1)
$\displaystyle\geq{\frac{\big{(}\inf\\{\text{length}(\gamma^{\bullet}):\text{$\gamma^{\bullet}$
is a curve in $H^{\bullet}$ joining $[A^{\bullet},B^{\bullet}]$ and
$[C^{\bullet},D^{\bullet}]$}\\}\big{)}^{2}}{A(\rho)}}$ (4.2)

where:

$\displaystyle A(\rho)$
$\displaystyle=\sum_{e^{\bullet}\in{E^{\bullet}}}\frac{|e^{\circ}|}{|e^{\bullet}|}|e^{\bullet}|^{2}\stackrel{{\scriptstyle(\ast)}}{{=}}2\sum_{Q\in{F_{in}(H)}}\text{Area}(Q)=2\hskip
1.0pt\text{Area}(\widehat{H})\leq{4\delta{L}}$

Equality “$(\ast)$" follows from the fact that if $Q$ is an inner face of $G$
with primal diagonal $e^{\bullet}$ and dual diagonal $e^{\circ}$, then
$\text{Area}(Q)=\frac{1}{2}|e^{\bullet}||e^{\circ}|$. Plugging our estimate
for $A(\rho)$ into Equation 4.2, we have that:

$\displaystyle\inf\\{\text{length}(\gamma^{\bullet}):\text{$\gamma^{\bullet}$
is a curve in $H^{\bullet}$ joining $[A^{\bullet},B^{\bullet}]$ and
$[C^{\bullet},D^{\bullet}]$}\\}\leq{\frac{2L}{\big{(}1-\frac{2\varepsilon}{\delta}\big{)}}}\leq{2L\Big{(}1+\frac{4\varepsilon}{\delta}\Big{)}}$

Since any curve in $H^{\bullet}$ joining $[A^{\bullet},B^{\bullet}]$ and
$[C^{\bullet},D^{\bullet}]$ is $\delta$\- close to $\ell$ in Hausdorff
distance, this completes our proof. The proof that we can find a curve
$\gamma^{\circ}$ in $G^{\circ}$ with the desired properties follows by the
same argument, verbatim, with $G^{\circ}$ in place of $G^{\bullet}$. ∎

Armed with this lemma, we are now ready to prove Theorem 4.1:

###### Proof.

Since $F$ is the local uniform limit of continuous functions, $F$ must be
continuous. Hence, to prove that $F$ is holomorphic, by Problem 3 in Chapter 2
of [29], it suffices to check that:

$\oint_{\gamma}F(z)dz=0$

for every simple closed curve $\gamma$ in $\Omega$ that traces out the
boundary of a rectangle.

With this in mind, let $\gamma$ be a simple closed curve in $\Omega$, oriented
counterclockwise, that traces out the boundary of a rectangle $R$ with length
$l$ and width $w$. Let $d$ be the distance between the rectangle $R$ and the
boundary of $\Omega$. Let $K$ be a compact subset of $\Omega$ that contains a
$d/2$\- neighborhood of $R$ and let $N$ be a natural number so that
$K\subseteq{\widehat{\Omega}_{n}}$ for all $n\geq{N}$.

Since $(\widehat{F}_{n})_{n=1}^{\infty}$ is a sequence of continuous functions
that converges to $F$ uniformly on compacts, by Arzela-Ascoli, it follows that
the functions $\widehat{F}_{n}$ are equicontinuous and uniformly bounded on
compacts. With this in mind, for any $\delta>0$, let $\omega(\delta)$ denote
the modulus of continuity of the family of functions
$(\widehat{F}_{n})_{n=N}^{\infty}$ on $K$. That is, for any $n\geq{N}$ and any
points $x,y\in{K}$, we have that:

$|F_{n}(y)-F_{n}(x)|\leq{\omega(|x-y|)}$

where:

$\lim\limits_{\delta\rightarrow{0^{+}}}\omega(\delta)=0$

Recall that by discrete Morera’s theorem, for each $n\in{\mathbb{N}}$, the
fact that $F_{n}$ is discrete holomorphic on $\Omega_{n}$ tells us that for
any simple closed directed curve $\eta_{n}$ in $\Omega_{n}$:

$\sum_{\begin{subarray}{c}\vec{e}\in{\eta_{n}}\\\
\vec{e}=(e^{-},e^{+})\end{subarray}}\big{(}F_{n}(e^{-})+F_{n}(e^{+})\big{)}(e^{+}-e^{-})=0$

By Lemma 4.3, for $n\geq{N}$ sufficiently large, we can pick simple closed
contours $\gamma_{n}$ and $\eta_{n}$ in $\Omega_{n}$ so that:

  1. 1.

$\gamma_{n}$ and $\eta_{n}$ are both oriented counterclockwise.

  2. 2.

$\gamma_{n}$ lies outside of the rectangle $R$ and
$d_{Haus}(\gamma_{n},\partial{R})=O(\varepsilon_{n})$.

  3. 3.

$\eta_{n}$ lies inside the rectangle $R$ and
$d_{Haus}(\eta_{n},\partial{R})=O(\varepsilon_{n})$.

  4. 4.

If $\gamma_{n}=(w_{1},x_{1},w_{2},x_{2},...,w_{k_{n}},x_{k_{n}},w_{1})$ where
$w_{1},w_{2},...,w_{k_{n}}\in{V_{n}^{\bullet}}$ and
$x_{1},x_{2},...,x_{k_{n}}\in{V_{n}^{\circ}}$, then $w_{i}\sim{w_{i+1}}$ for
all $i$, where our indices $i$ are viewed modulo $k_{n}$, so that
$\gamma_{n}^{\bullet}=(w_{1},w_{2},...,w_{k_{n}},w_{1})$ is a simple closed
contour in $\Omega_{n}^{\bullet}$. Furthermore:

$\text{length}(\gamma_{n}^{\bullet})=\sum_{i=1}^{k_{n}}|w_{i+1}-w_{i}|=O(l+w)$

  5. 5.

If $\eta_{n}=(y_{1},z_{1},y_{2},z_{2},..,y_{m_{n}},z_{m_{n}},y_{1})$ where
$y_{1},y_{2},...,y_{m_{n}}\in{V_{n}^{\bullet}}$,
$z_{1},z_{2},...,z_{m_{n}}\in{V_{n}^{\circ}}$, then $z_{i}\sim{z_{i+1}}$ for
all $i$, where our indices are viewed modulo $m_{n}$, so that
$\eta_{n}^{\circ}=(z_{1},z_{2},...,z_{m_{n}},z_{1})$ is a simple closed
contour in $\Omega_{n}^{\circ}$. Furthermore:

$\text{length}(\eta_{n}^{\circ})=\sum_{i=1}^{m_{n}}|z_{i+1}-z_{i}|=O(l+w)$

Figure 13: The contours $\gamma$, $\gamma_{n}$, $\gamma_{n}^{\bullet}$,
$\eta_{n}$, and $\eta_{n}^{\circ}$ for some orthodiagonal map.

Consider the discrete contour integral of $F_{n}$ over $\gamma_{n}$:

$\sum_{\begin{subarray}{c}\vec{e}\hskip 1.0pt\in{\gamma_{n}}\\\
\vec{e}=(e^{-},e^{+})\end{subarray}}\big{(}F_{n}(e^{-})+F_{n}(e^{+})\big{)}(e^{+}-e^{-})=\sum_{i=1}^{k_{n}}F_{n}(x_{i})(w_{i+1}-w_{i})+\sum_{i=1}^{k_{n}}F_{n}(w_{i})(x_{i}-x_{i-1})=0$
(4.3)

Th first equality follows by rewriting the original sum over directed edges as
a sum over vertices. The second equality is just discrete Morera. Since
$\text{Im}(\widehat{F}_{n})$ agrees with $F_{n}$ on $V^{\circ}$, for each
directed edge $(w_{i},w_{i+1})$, we have that:

$\big{|}\int_{w_{i}}^{w_{i+1}}\text{Im}(\widehat{F}_{n}(z))dz-
F_{n}(x_{i})(w_{i+1}-w_{i})\big{|}\leq{\omega(\varepsilon_{n})\cdot|w_{i+1}-w_{i}|}$

provided that $n$ is large enough so that the contour $\gamma_{n}$ is
contained in $K$. Summing over directed edges we have that:

$\Big{|}\oint_{\gamma_{n}^{\bullet}}\text{Im}(\widehat{F}_{n}(z))dz-\sum_{i=1}^{k_{n}}F_{n}(x_{i})(w_{i+1}-w_{i})\Big{|}=O(\omega(\varepsilon_{n})\cdot{(l+w)})$
(4.4)

In other words, we see that the discrete contour integral of
$\text{Im}(F_{n})$ over $\gamma_{n}$ is close to the continuous contour
integral of $\text{Im}(\widehat{F}_{n})$ over $\gamma_{n}^{\bullet}$. In
contrast, since the $x_{i}$’s don’t form a simple closed contour in
$\Omega_{n}^{\circ}$ and we don’t have any control over the quantity:

$\sum_{i=1}^{k_{n}}|x_{i+1}-x_{i}|$

it is not clear that we have a similar result comparing the discrete contour
integral of $\text{Re}(F_{n})$ over $\gamma_{n}$ to some continuous contour
integral of $\text{Re}(\widehat{F}_{n})$. This is where our second contour
$\eta_{n}$ comes in.

By the same argument we used to handle the behavior of $\text{Im}(F_{n})$ on
$\gamma_{n}$, since $\text{Re}(\widehat{F}_{n})$ agrees with $F_{n}$ on
$V^{\bullet}$, for any directed edge $(z_{i-1},z_{i})$ we have that:

$\big{|}\int_{z_{i-1}}^{z_{i}}\text{Re}(\widehat{F}_{n}(z))dz-
F_{n}(y_{i})(z_{i}-z_{i-1})\big{|}\leq{\omega(\varepsilon_{n})\cdot|z_{i}-z_{i-1}|}$

Summing over directed edges we have that:

$\Big{|}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz-\sum_{i=1}^{m_{n}}F_{n}(y_{i})(z_{i}-z_{i-1})\Big{|}=O(\omega(\varepsilon_{n})\cdot{(l+w)})$
(4.5)

Let $H$ be the suborthodiagonal map of $G$ bounded by the curves $\eta_{n}$
and $\gamma_{n}$. By discrete Green’s theorem applied to the function
$\text{Re}(F_{n})$ on $H$, we have that:

$\sum_{i=1}^{k_{n}}F_{n}(w_{i})(x_{i}-x_{i-1})-\sum_{i=1}^{m_{n}}F_{n}(y_{i})(z_{i}-z_{i-1})=\sum_{\begin{subarray}{c}Q\in{F_{in}(H)}\\\
Q=[u_{1},v_{1},u_{2},v_{2}]\end{subarray}}\big{(}F_{n}(u_{2})-F_{n}(u_{1})\big{)}(v_{2}-v_{1})$

Applying the Cauchy-Schwarz inequality, we have that:

$\displaystyle\hskip 2.0pt\Big{|}\sum_{\begin{subarray}{c}Q\in{F_{in}(H)}\\\
Q=[u_{1},v_{1},u_{2},v_{2}]\end{subarray}}\big{(}F_{n}(u_{2})-F_{n}(u_{1})\big{)}(v_{2}-v_{1})\Big{|}$
$\displaystyle\leq$ $\displaystyle\hskip
2.0pt\Big{(}\sum_{\begin{subarray}{c}Q\in{F_{in}(H)}\\\
Q=[u_{1},v_{1},u_{2},v_{2}]\end{subarray}}\frac{|v_{2}-v_{1}|}{|u_{2}-u_{1}|}\big{(}F_{n}(u_{2})-F_{n}(u_{1})\big{)}^{2}\Big{)}^{1/2}\Big{(}\sum_{\begin{subarray}{c}Q\in{F_{in}(H)}\\\
Q=[u_{1},v_{1},u_{2},v_{2}]\end{subarray}}\frac{|u_{2}-u_{1}|}{|v_{2}-v_{1}|}|v_{2}-v_{1}|^{2}\Big{)}^{1/2}$
$\displaystyle\leq$ $\displaystyle\hskip
2.0pt\mathcal{E}_{K}^{\bullet}(\text{Re}(F_{n}))^{1/2}\Big{(}\sum_{\begin{subarray}{c}Q\in{F_{in}(H)}\\\
Q=[u_{1},v_{1},u_{2},v_{2}]\end{subarray}}|u_{2}-u_{1}||v_{2}-v_{1}|\Big{)}^{1/2}=\sqrt{2}\hskip
2.0pt\mathcal{E}_{K}^{\bullet}(\text{Re}(F_{n}))^{1/2}(\text{Area}(\widehat{H}))^{1/2}$
$\displaystyle=$ $\displaystyle\hskip
2.0pt\mathcal{E}_{K}^{\bullet}(\text{Re}(F_{n}))^{1/2}(l+w)^{1/2}\cdot{O(\varepsilon_{n}^{1/2})}\leq{\big{(}\sup_{n\geq{N}}\mathcal{E}_{K}^{\bullet}(\text{Re}(F_{n}))\big{)}^{1/2}}(l+w)^{1/2}\cdot{O(\varepsilon_{n}^{1/2})}$

Thus:

$\Big{|}\sum_{i=1}^{k_{n}}F_{n}(w_{i})(x_{i}-x_{i-1})-\sum_{i=1}^{m_{n}}F_{n}(y_{i})(z_{i}-z_{i-1})\big{|}\leq{\big{(}\sup_{n\geq{N}}\mathcal{E}_{K}^{\bullet}(\text{Re}(F_{n}))\Big{)}^{1/2}}(l+w)^{1/2}\cdot{O(\varepsilon_{n}^{1/2})}$
(4.6)

By 4.4, we have that:

$\lim_{n\rightarrow{\infty}}\Big{(}\oint_{\gamma_{n}^{\bullet}}\text{Im}(\widehat{F}_{n}(z))dz-\sum_{i=1}^{k_{n}}F_{n}(x_{i})(w_{i+1}-w_{i})\Big{)}=0$
(4.7)

Similarly, since the Dirichlet energies of the $F_{n}$’s are uniformly bounded
on compacts, combining 4.5 and 4.6, we have that:

$\lim_{n\rightarrow{\infty}}\Big{(}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz-\sum_{i=1}^{k_{n}}F_{n}(w_{i})(x_{i}-x_{i-1})\Big{)}=0$
(4.8)

Combining 4.7 and 4.8, we have that:

$\lim_{n\rightarrow{\infty}}\Big{(}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz+\oint_{\gamma_{n}^{\bullet}}\text{Im}(\widehat{F}_{n}(z))dz\Big{)}=0$

Thus, to prove that:

$\oint_{\gamma}F(z)dz=\oint_{\gamma}\text{Re}(F(z))dz+\oint_{\gamma}\text{Im}(F(z))dz=0$

It suffices to show that:

$\lim_{n\rightarrow{\infty}}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz=\oint_{\gamma}\text{Re}(F(z))dz$

and:

$\lim_{n\rightarrow{\infty}}\oint_{\gamma_{n}^{\bullet}}\text{Im}(\widehat{F}_{n}(z))dz=\oint_{\gamma}\text{Im}(F(z))dz$

To this effect, let $\psi$ be any smooth, real- valued function on $\Omega$.
By the triangle inequality:

$\displaystyle\Big{|}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz-\oint_{\gamma}\text{Re}(F(z))dz\Big{|}\leq$
$\displaystyle\hskip
2.0pt{\Big{|}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz-\oint_{\eta_{n}^{\circ}}\text{Re}(F(z))dz\Big{|}}+\Big{|}\oint_{\eta_{n}^{\circ}}\text{Re}(F(z))dz-\oint_{\eta_{n}^{\circ}}\psi(z)dz\Big{|}$
$\displaystyle+\Big{|}\oint_{\eta_{n}^{\circ}}\psi(z)dz-\oint_{\gamma}\psi(z)dz\Big{|}+\Big{|}\oint_{\gamma}\psi(z)dz-\oint_{\gamma}\text{Re}(F(z))dz\Big{|}$

We bound the first term as follows:

$\displaystyle\Big{|}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz-\oint_{\eta_{n}^{\circ}}\text{Re}(F(z))dz\Big{|}$
$\displaystyle\leq{\text{length}(\eta_{n}^{\circ})\cdot{\|\text{Re}(\widehat{F}_{n})-\text{Re}(F)\|}}=O\big{(}(l+w)\cdot\|\text{Re}(\widehat{F}_{n})-\text{Re}(F)\|\big{)}$

where $\|\text{Re}(\widehat{F}_{n})-\text{Re}(F)\|$ is the sup norm of
$\text{Re}(\widehat{F}_{n})-\text{Re}(F)$ on $K$. We can handle the second and
fourth terms analogously:

$\displaystyle\Big{|}\oint_{\eta_{n}^{\circ}}\text{Re}(F(z))dz-\oint_{\eta_{n}^{\circ}}\psi(z)dz\Big{|}$
$\displaystyle\leq{\text{length}(\eta_{n}^{\circ})\cdot{\|\text{Re}(F)-\psi\|}}=O\big{(}(l+w)\cdot{\|\text{Re}(F)-\psi\|}\big{)}$
$\Big{|}\oint_{\gamma}\psi(z)dz-\oint_{\gamma}\text{Re}(F(z))dz\Big{|}\leq{\text{length}(\gamma)\cdot{\|\psi-\text{Re}(F)\|}}=2(l+w)\cdot{\|\psi-\text{Re}(F)\|}$

To bound the third term, recall that if $\sigma$ is a simple closed Lipschitz
curve, oriented counterclockwise, and $h$ is a complex- valued function whose
real and imaginary part are both $C^{2}$, Green’s theorem tells us that:

$\oint_{\sigma}h(z)dz=\int_{D}\partial_{\overline{z}}h(x+iy)dxdy$

where $D$ is the region bounded by $\sigma$. We know that $\gamma$ bounds the
rectangle $R$. Let $R_{n}$ denote the region bounded by $\eta_{n}^{\circ}$. By
Green’s theorem,

$\displaystyle\Big{|}\oint_{\gamma}\psi(z)dz-\oint_{\eta_{n}^{\circ}}\psi(z)dz\Big{|}$
$\displaystyle=\Big{|}\int_{R}\partial_{\overline{z}}\psi(x+iy)dxdy-\int_{R_{n}}\partial_{\overline{z}}\psi(x+iy)dxdy\Big{|}$
$\displaystyle=O\big{(}(l+w)\cdot\|\partial_{\overline{z}}\psi\|\cdot\varepsilon_{n}\big{)}$

Letting $n\rightarrow{\infty}$, the first and third terms vanish, leaving us
with:

$\limsup_{n\rightarrow{\infty}}\Big{|}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz-\oint_{\gamma}\text{Re}(F(z))dz\Big{|}=O\big{(}(l+w)\cdot\|\text{Re}(F)-\psi\|\big{)}$

Since smooth functions are dense in the space of continuous functions on a
compact set with the sup norm, we can choose $\psi$ so that
$\|\text{Re}(F)-\psi\|$ is arbitrarily small. Thus:

$\limsup_{n\rightarrow{\infty}}\Big{|}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz-\oint_{\gamma}\text{Re}(F(z))dz\Big{|}=0$

From which we conclude that:

$\lim_{n\rightarrow{\infty}}\oint_{\eta_{n}^{\circ}}\text{Re}(\widehat{F}_{n}(z))dz=\oint_{\gamma}\text{Re}(F(z))dz$

Applying the same argument to
$\big{(}\oint_{\gamma_{n}^{\bullet}}\text{Im}(\widehat{F}_{n}(z))dz-\oint_{\gamma}\text{Im}(F(z))dz\big{)}$,
we get that:

$\lim_{n\rightarrow{\infty}}\oint_{\gamma_{n}^{\bullet}}\text{Im}(\widehat{F}_{n}(z))dz=\oint_{\gamma}\text{Im}(F(z))dz$

∎

##  5 Convergence of Tilings to Conformal Maps

###  5.1 Proof of Theorem 2.15

###### Proof.

Suppose $\Omega\subsetneq{\mathbb{C}}$ is a simply connected domain with
distinguished prime ends $A,B,C,D$ listed in counterclockwise order and
$\delta_{n},\varepsilon_{n}>0$ are sequences of positive reals so that:

$(\delta_{n},\varepsilon_{n})\rightarrow{(0,0)}\hskip 5.0pt\text{as
$n\rightarrow{\infty}$}$

For each $n\in{\mathbb{N}}$, let
$\Omega_{n}=(V_{n}^{\bullet}\sqcup{V_{n}^{\circ}},E)$ be an orthodiagonal map
so that
$(\Omega_{n},A_{n}^{\bullet},B_{n}^{\bullet},C_{n}^{\bullet},D_{n}^{\bullet})$
is a $(\delta_{n},\varepsilon_{n})$\- good interior approximation to
$(\Omega,A,B,C,D)$ for some choice of distinguished boundary points
$A_{n}^{\bullet},B_{n}^{\bullet},C_{n}^{\bullet},D_{n}^{\bullet}\in{\partial{V_{n}^{\bullet}}}$.
Let $\phi_{n}$ be the tiling map associated to the orthodiagonal rectangle
$(\Omega_{n},A_{n}^{\bullet},B_{n}^{\bullet},C_{n}^{\bullet},D_{n}^{\bullet})$
with real part $h_{n}$ and imaginary part $\widetilde{h}_{n}$. That is,
$\widetilde{h}_{n}:V_{n}^{\circ}\rightarrow{\mathbb{R}}$ solves the following
boundary value problem on $\Omega_{n}^{\circ}$:

$\displaystyle\widetilde{h}_{n}(x)$ $\displaystyle=1\hskip 5.0pt\text{for all
$x\in{[B_{n}^{\circ},C_{n}^{\circ}]}$}$ $\displaystyle\widetilde{h}_{n}(x)$
$\displaystyle=0\hskip 5.0pt\text{for all
$x\in{[D_{n}^{\circ},A_{n}^{\circ}]}$}$
$\displaystyle\Delta^{\circ}\widetilde{h}_{n}(x)$ $\displaystyle=0\hskip
5.0pt\text{for all
$x\in{V_{n}^{\circ}\setminus{\big{(}[B_{n}^{\circ},C_{n}^{\circ}]\cup{[D_{n}^{\circ},A_{n}^{\circ}]}\big{)}}}$}$

and $h_{n}:V^{\bullet}\rightarrow{\mathbb{R}}$ is the harmonic conjugate of
$\widetilde{h}_{n}$ which solves the following boundary value problem on
$\Omega_{n}^{\bullet}$:

$\displaystyle h_{n}(x)$ $\displaystyle=0\hskip 5.0pt\text{for all
$x\in{[A_{n}^{\bullet},B_{n}^{\bullet}]}$}$ $\displaystyle h_{n}(x)$
$\displaystyle=L_{n}\hskip 5.0pt\text{for all
$x\in{[C_{n}^{\bullet},D_{n}^{\bullet}]}$}$
$\displaystyle\Delta^{\bullet}h_{n}(x)$ $\displaystyle=0\hskip 5.0pt\text{for
all
$x\in{V_{n}^{\bullet}\setminus{\big{(}[A_{n}^{\bullet},B_{n}^{\bullet}]\cup{[C_{n}^{\bullet},D_{n}^{\bullet}]}\big{)}}}$}$

where $L_{n}$ is the extremal length from $[A_{n}^{\bullet},B_{n}^{\bullet}]$
to $[C_{n}^{\bullet},D_{n}^{\bullet}]$ in $\Omega_{n}^{\bullet}$. Let $F_{n}$
be the discrete holomorphic function on $\Omega_{n}$ that agrees with $h_{n}$
on $V_{n}^{\bullet}$ and agrees with $i\widetilde{h}_{n}$ on $V_{n}^{\circ}$.
Let $\widehat{F}_{n}$ be any sensible extension of $F_{n}$ to a continuous
function on $\widehat{\Omega}_{n}$ so that:

$\text{Re}(\widehat{F}_{n}(z))=h_{n}(z)\hskip 5.0pt\text{for all
$z\in{V^{\bullet}_{n}}$}\hskip
20.0pt\text{Im}(\widehat{F}_{n}(z))=\widetilde{h}_{n}(z)\hskip 5.0pt\text{for
all $z\in{V^{\circ}_{n}}$}$

One natural way to do this is to triangulate the faces of
$\Omega_{n}^{\bullet}$ and define the real part of $\widehat{F}_{n}$ on each
triangle by interpolating linearly between the values of $h_{n}$ at the corner
vertices. Analogously, triangulating each face of $\Omega_{n}^{\circ}$, we can
define the imaginary part of $\widehat{F}_{n}$ on each triangle by
interpolating linearly between the values of $\widetilde{h}_{n}$ at the corner
vertices. If $\phi_{n}$ is the tiling map associated with the orthodiagonal
rectangle
$(\Omega_{n}^{\bullet},A^{\bullet}_{n},B^{\bullet}_{n},C^{\bullet}_{n},D^{\bullet}_{n})$,
our estimates for the modulus of continuity of $h_{n}$ and $\widetilde{h}_{n}$
in Theorem 3.3 tell us that for any $n\in{\mathbb{N}}$ and any
$z\in{\widehat{\Omega}_{n}}$, we have that:

$|\phi_{n}(z)-\widehat{F}_{n}(z)|\leq{\frac{K(L_{n}\vee{1})}{\log\big{(}\frac{d_{n}\wedge{d_{n}^{\prime}}}{\varepsilon_{n}}\big{)}}}$
(5.1)

where:

$\displaystyle d_{n}$
$\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$ is a curve in
$\widehat{\Omega}_{n}$ joining
$[A_{n}^{\bullet},B_{n}^{\bullet}]_{\partial{\widehat{\Omega}}_{n}}$ and
$[C_{n}^{\bullet},D_{n}^{\bullet}]_{\partial{\widehat{\Omega}}_{n}}$}\\}$
$\displaystyle d_{n}^{\prime}$
$\displaystyle=\inf\\{\text{diameter}(\gamma):\text{$\gamma$ is a curve in
$\widehat{\Omega}_{n}$ joining
$[B_{n}^{\bullet},C_{n}^{\bullet}]_{\partial{\widehat{\Omega}_{n}}}$ and
$[D_{n}^{\bullet},A_{n}^{\bullet}]_{\partial{\widehat{\Omega}_{n}}}$}\\}$

and $K>0$ is an absolute constant. By Proposition 3.5, the sequence
$(L_{n})_{n=1}^{\infty}$ is uniformly bounded. Meanwhile, Lemma A.4 tells us
that $d_{n}\wedge{d_{n}^{\prime}}\rightarrow{d\wedge{d^{\prime}}}$ as
$n\rightarrow{\infty}$. Hence, the quantity on the RHS of this inequality
tends to $0$ uniformly in $z$ as $n\rightarrow{\infty}$. Thus, to show that
the tiling maps $\phi_{n}$ converge to the relevant conformal map $\phi$, it
suffices to show this is true of the functions $\widehat{F}_{n}$.

The uniform boundedness of the $L_{n}$’s also tells us that the functions
$\widehat{F}_{n}$ are uniformly bounded. By Theorem 3.3, these functions are
also equicontinuous on compacts. Hence, by Arzela-Ascoli, the functions
$\widehat{F}_{n}$ are precompact with respect to the topology of uniform
convergence on compacts in $\Omega$. Since $L_{n}$ is precisely the discrete
Dirichlet energy of $\widehat{F}_{n}$ and the sequence
$(L_{n})_{n=1}^{\infty}$ is uniformly bounded, Theorem 4.1 tells us that any
subsequential limit of our discrete holomorphic functions $\widehat{F}_{n}$ is
holomorphic.

As per our discussion in Section 1, in 2D statistical physics, if you have a
holomorphic observable that arises as the limit of some discrete holomorphic
observables, you can typically recover the identity of the limiting object
from the boundary conditions. This is what we will do here. Suppose
$(\widehat{F}_{n_{k}})_{k\geq{1}}$ is a subsequence of
$(\widehat{F}_{n})_{n\geq{1}}$ such that $\widehat{F}_{n_{k}}$ converges
uniformly on compacts to some holomorphic function $F$ and $L_{n_{k}}$
converges to some $L>0$ as $k\rightarrow{\infty}$. We can always pick such a
subsequence because the sequence $(\widehat{F}_{n})_{n\geq{1}}$ is precompact
and the sequence $(L_{n})_{n\geq{1}}$ is uniformly bounded away from $0$ and
$\infty$. Fix $x,y\in{\Omega}$. By Theorem 3.3:

$|\widehat{F}_{n_{k}}(y)-\widehat{F}_{n_{k}}(x)|\leq{\frac{K(L_{n_{k}}\vee{1})}{\log{\Big{(}\frac{d_{n_{k}}\wedge{d_{n_{k}}^{\prime}}}{(d_{\widehat{\Omega}_{n_{k}}}(x,y)\wedge{d_{cc}^{\widehat{\Omega}_{n_{k}}}(x,y)})\vee{\varepsilon_{n_{k}}}}\Big{)}}}}$

for all $k$ large enough so that $x,y\in{\widehat{\Omega}_{n_{k}}}$. Taking
the limit as $k\rightarrow{\infty}$ of both sides of this inequality, by
Corollary A.3 and Lemma A.4 of Appendix A.1, we have that:

$|F(y)-F(x)|\leq{\frac{K(L\vee{1})}{\log\Big{(}\frac{d\wedge{d^{\prime}}}{d_{\Omega}(x,y)\wedge{d_{cc}^{\Omega}(x,y)}}\Big{)}}}$
(5.2)

Thus, $F:\Omega\rightarrow{\mathbb{C}}$ extends continuously to
$\Omega^{\ast}$. What is more, as we alluded to earlier, our understanding of
the boundary behavior of the $\widehat{F}_{n_{k}}$’s will allow us to recover
enough of the boundary behavior of $F$ to uniquely identify $F$ as the
conformal map $\phi$ in the statement of Theorem 2.15.

Suppose $z\in{(A,B)_{\partial{\Omega^{\ast}}}}$. Since $z$ does not lie on the
boundary arcs $[B,C]_{\partial{\Omega^{\ast}}}$, $[C,D]$ or
$[D,A]_{\partial{\Omega^{\ast}}}$:

$c(z)=\min\\{d_{cc}^{\Omega}(z,[B,C]_{\partial{\Omega^{\ast}}};[A,D]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(z,[C,D]_{\partial{\Omega^{\ast}}};[A,B]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(z,[D,A]_{\partial{\Omega^{\ast}}};[B,C]_{\partial{\Omega^{\ast}}})\\}>0$

Let $(z_{m})_{m=1}^{\infty}$ be a sequence of points in $\Omega$ so that
$z_{m}\rightarrow{z}$ in Carathèodory distance. Equivalently, fixing some
point $z_{0}\in\Omega$, for each $z_{m}$ there exists a crosscut $\gamma_{m}$
of $\Omega$ so that $z$ and $z_{m}$ lie in the same connected component of
$\Omega^{\ast}\setminus{\gamma_{m}}$, $z_{0}$ lies in the other connected
component, and $\ell_{m}=\text{length}(\gamma_{m})\rightarrow{0}$. By the same
reasoning we used to arrive at Equation 5.2, restricting our attention to the
real part of $F$, we have that:

$|Re(F)(z)-Re(F)(z_{m})|\leq{\frac{K}{\log\big{(}\frac{d^{\prime}}{\ell_{m}}\big{)}}}$

where $K>0$ is an absolute constant. Since $\widehat{F}_{n_{k}}$ converges to
$F$ uniformly on compacts:

$|Re(F)(z_{m})-Re(\widehat{F}_{n_{k}})(z_{m})|\rightarrow{0}\hskip
10.0pt\text{as $k\rightarrow{\infty}$}$

Choose $m$ large enough so that $\ell_{m}<\frac{c(z)}{3}$. Having done this,
choose $k$ large enough so that $z_{m}\in{\widehat{\Omega}_{n_{k}}}$ and
$\delta_{n_{k}},\varepsilon_{n_{k}}<\frac{c(z)}{3}$. Since
$\partial{\widehat{\Omega}}_{n_{k}}$ is a closed simple curve separating
$z_{m}$ from $\partial{\Omega^{\ast}}$, $\gamma_{m}$ must intersect
$\partial{\widehat{\Omega}}_{n_{k}}$. The resulting points of intersections
must all lie on the arc
$[A_{n_{k}}^{\bullet},B_{n_{k}}^{\bullet}]_{\partial{\widehat{\Omega}_{n_{k}}}}$.
To understand why this is the case, suppose to the contrary that one of these
points of intersection lies on the arc
$[B^{\bullet}_{n_{k}},C^{\bullet}_{n_{k}}]_{\partial{\widehat{\Omega}_{n_{k}}}}$.
Let $x_{k}$ be a point of intersection between $\gamma_{m}$ and
$\partial{\widehat{\Omega}}_{n_{k}}$ that lies on
$[B^{\bullet}_{n_{k}},C^{\bullet}_{n_{k}}]_{\partial{\widehat{\Omega}_{n_{k}}}}$.
Then we can find a point
$x_{k}^{\prime}\in{[B^{\circ}_{n_{k}},C^{\circ}_{n_{k}}]}\subseteq{\partial{V^{\circ}}}$
that lies within $\varepsilon_{n_{k}}$ of $x_{k}$. Since
$(\Omega_{n_{k}},A_{n_{k}}^{\bullet},B_{n_{k}}^{\bullet},C_{n_{k}}^{\bullet},D_{n_{k}}^{\bullet})$
is a $(\delta_{n_{k}},\varepsilon_{n_{k}})$\- good interior approximation of
$(\Omega,A,B,C,D)$, $x_{k}^{\prime}$ is $\delta_{n_{k}}$\- close to
$[B,C]_{\partial{\Omega^{\ast}}}$ in crosscut distance. That is, we can find a
crosscut $\eta_{k}$ of $\Omega$ of length at most $\delta_{n_{k}}$, joining
$x_{k}^{\prime}$ to $[B,C]_{\partial{\Omega^{\ast}}}$, and separating
$x_{k}^{\prime}$ from $[D,A]_{\partial{\Omega^{\ast}}}$. Stitching together
$\gamma_{m}$, $\eta_{k}$, and the line segment between $x_{k}$ and
$x_{k}^{\prime}$, we conclude that $z$ is
$(\ell_{m}+\delta_{n_{k}}+\varepsilon_{n_{k}})$\- close to
$[B,C]_{\partial{\Omega^{\ast}}}$ in crosscut distance, where:

$\ell_{m}+\delta_{n_{k}}+\varepsilon_{n_{k}}<c(z)$

This is a contradiction, since:

$c(z)\leq{d_{cc}^{\Omega}}(z,[B,C]_{\partial{\Omega^{\ast}}};[D,A]_{\partial{\Omega^{\ast}}})$

where the quantity on the RHS of this inequality is precisely the crosscut
distance between $z$ and $[B,C]_{\partial{\Omega^{\ast}}}$ in $\Omega$. The
exact same argument allows us to rule out points of intersection lying on the
arcs
$[C^{\bullet}_{n_{k}},D^{\bullet}_{n_{k}}]_{\partial{\widehat{\Omega}_{n_{k}}}}$
and
$[D^{\bullet}_{n_{k}},A^{\bullet}_{n_{k}}]_{\partial{\widehat{\Omega}_{n_{k}}}}$.
Hence,
$(\gamma_{m}\cap{\partial{\widehat{\Omega}_{n_{k}}}})\subseteq{[A_{n_{k}}^{\bullet},B_{n_{k}}^{\bullet}]_{\partial{\widehat{\Omega}_{n_{k}}}}}$.

Since the points of intersection of $\gamma_{m}$ and
$\partial{\widehat{\Omega}_{n_{k}}}$ all lie along
$[A_{n_{k}}^{\bullet},B_{n_{k}}^{\bullet}]_{\partial{\widehat{\Omega}_{n_{k}}}}$,
$z_{m}$ is $\ell_{m}$\- close to the boundary arc
$[A_{n_{k}}^{\bullet},B_{n_{k}}^{\bullet}]_{\partial{\widehat{\Omega}_{n_{k}}}}$
of $\widehat{\Omega}_{n_{k}}$, along which $Re(\widehat{F}_{n_{k}})$ is
identically $0$. Hence, by our regularity estimates in Theorem 3.3:

$|Re(\widehat{F}_{n_{k}})(z_{m})|\leq{\frac{K}{\log\big{(}\frac{d^{\prime}}{\ell_{m}}\big{)}}}$

Putting all this together, we have that:

$|Re(F)(z)|\leq{\frac{2K}{\log\big{(}\frac{d^{\prime}}{\ell_{m}}\big{)}}}$

for any $m$ sufficiently large. Since $\ell_{m}\rightarrow{0}$ as
$m\rightarrow{\infty}$, we conclude that $Re(F)(z)=0$. Since the point
$z\in{(A,B)_{\partial{\Omega^{\ast}}}}$ was arbitrary, $Re(F)(z)=0$ for all
$z\in{[A,B]_{\partial{\Omega^{\ast}}}}$. The fact that
$Re(F)|_{[C,D]_{\partial{\Omega^{\ast}}}}=L$,
$Im(F)|_{[B,C]_{\partial{\Omega^{\ast}}}}=0$, and
$Im(F)|_{[D,A]_{\partial{\Omega^{\ast}}}}=1$, all follow by the same argument.
Via the argument principle, this information about the boundary behavior of
$F$ tells us that $F$ is a conformal map from $\Omega$ to the rectangle
$\mathcal{R}_{L}$. Since $F(A)=i$, $F(B)=0$, $F(C)=L$, and $F(D)=L+i$, we
conclude that $F=\phi$. Furthermore, $L_{n}$ converges to $L$, where $L$ is
the extremal length between $[A,B]_{\partial{\Omega^{\ast}}}$ and
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$. ∎

##  Appendix A Appendix

###  A.1 Elementary Results in Topology and Complex Analysis

In this section, we prove Proposition 2.14 which tells us that any conformal
rectangle admits a $\delta$-good interior approximation, along with a number
of elementary results in topology and complex analysis that are needed for the
proof of our main result, Theorem 2.15. While the proofs of these results are
all fairly straightforward, so as not to distract the reader from the main
thrust of our argument, we’ve relegated them to this Appendix.

######  Lemma A.1.

For each $n\in{\mathbb{N}}$, let $(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$ be a
$\delta_{n}$\- good approximation of the conformal rectangle
$(\Omega,A,B,C,D)$, where $\delta_{n}\rightarrow{0}$ as
$n\rightarrow{\infty}$. Then for any compact set $K\subseteq{\Omega}$, there
exists $N\in{\mathbb{N}}$ such that $K\subseteq{\Omega_{n}}$ for all
$n\geq{N}$.

###### Proof.

Let $\phi:\Omega\rightarrow{\mathcal{R}_{L}}$ be the conformal map from
$\Omega$ to the rectangle $\mathcal{R}_{L}$ so that the prime ends $A,B,C,D$
of $\Omega$ are mapped to the four corners of $\mathcal{R}_{L}$ and in
particular, $\phi(A)=i$. Here, $L$ is the extremal length between
$[A,B]_{\partial{\Omega^{\ast}}}$ and $[C,D]_{\partial{\Omega^{\ast}}}$ in
$\Omega$. We write $\phi=h+i\widetilde{h}$ where $h$ and $\widetilde{h}$ are
the real and imaginary parts of $\phi$. Since
$[A_{n},B_{n}]_{\partial{\Omega_{n}}}$ and
$[C_{n},D_{n}]_{\partial{\Omega_{n}}}$ are $\delta_{n}$-close to
$[A,B]_{\partial{\Omega^{\ast}}}$ and $[C,D]_{\partial{\Omega^{\ast}}}$ in
crosscut distance, our regularity estimates for $h$ in Theorem 3.1 tell us
that:

$\displaystyle h(z)$
$\displaystyle\leq{\frac{2\pi}{\log\big{(}\frac{d^{\prime}}{2\delta_{n}}\big{)}}}\hskip
5.0pt\text{for all $z\in{[A_{n},B_{n}]_{\partial{\Omega_{n}}}}$},$
$\displaystyle h(z)$
$\displaystyle\geq{L-\frac{2\pi}{\log\big{(}\frac{d^{\prime}}{2\delta_{n}}\big{)}}}\hskip
5.0pt\text{for all $z\in{[C_{n},D_{n}]_{\partial{\Omega_{n}}}}$}$

Similarly, since $[B_{n},C_{n}]_{\partial{\Omega_{n}}}$ and
$[D_{n},A_{n}]_{\partial{\Omega_{n}}}$ are $\delta_{n}$\- close to $[B,C]$ and
$[D,A]$ in crosscut distance, our regularity estimates for $\widetilde{h}$ in
Theorem 3.1 tell us that:

$\displaystyle\widetilde{h}(z)$
$\displaystyle\geq{1-\frac{2\pi{L}}{\log\big{(}\frac{d}{2\delta_{n}}\big{)}}}\hskip
5.0pt\text{for all $z\in{[B_{n},C_{n}]_{\partial{\Omega_{n}}}}$},$
$\displaystyle\widetilde{h}(z)$
$\displaystyle\leq{\frac{2\pi{L}}{\log\big{(}\frac{d}{2\delta_{n}}\big{)}}}\hskip
5.0pt\text{for all $z\in{[B_{n},C_{n}]_{\partial{\Omega_{n}}}}$}$

Since $A_{n}$ lies on both $[A_{n},B_{n}]_{\partial{\widehat{\Omega}_{n}}}$
and $[C_{n},D_{n}]_{\partial{\widehat{\Omega}_{n}}}$, we have that:

$|\phi(A_{n})-\phi(A)|\leq{\frac{2\pi}{\log\big{(}\frac{d^{\prime}}{2\delta_{n}}\big{)}}+\frac{2\pi{L}}{\log\big{(}\frac{d}{2\delta_{n}}\big{)}}}\leq{\frac{4\pi(L\vee{1})}{\log\big{(}\frac{d\wedge{d^{\prime}}}{2\delta_{n}}\big{)}}}$

By the same reasoning:

$|\phi(B_{n})-\phi(B)|,|\phi(C_{n})-\phi(C)|,|\phi(D_{n})-\phi(D)|\leq{\frac{4\pi(L\vee{1})}{\log\big{(}\frac{d\wedge{d^{\prime}}}{2\delta_{n}}\big{)}}}$

Consider the simple closed curve $\phi(\partial{\Omega_{n}})$ whose interior
is precisely $\phi(\Omega_{n})$. We can split this curve into four pieces as
follows:

$\phi(\partial{\Omega}_{n})=\phi([A_{n},B_{n}]_{\partial{\Omega_{n}}})\cup{\phi([B_{n},C_{n}]_{\partial{\Omega_{n}}})}\cup{\phi([C_{n},D_{n}]_{\partial{\Omega_{n}}})}\cup{\phi([D_{n},A_{n}]_{\partial{\Omega_{n}}})}$

By our estimates from earlier:

  * •

The curve $\phi([A_{n},B_{n}]_{\partial{\Omega}_{n}})$ travels from
$\phi(A_{n})$ to $\phi(B_{n})$, where $\phi(A_{n})$ lies within
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}$ of $i$
and $\phi(B_{n})$ lies within
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}$ of $0$.
In between, the curve $\phi([A_{n},B_{n}]_{\partial{\Omega}})$ lies within
$\frac{2\pi}{\log(\frac{d^{\prime}}{2\delta_{n}})}$ of the boundary arc
$\\{0\\}\times{[0,1]}$ of $\mathcal{R}_{L}$.

  * •

The curve $\phi([B_{n},C_{n}]_{\partial{\Omega}_{n}})$ travels from
$\phi(B_{n})$ to $\phi(C_{n})$, where $\phi(B_{n})$ lies within
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}$ of $0$
and $\phi(B_{n})$ lies within
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}$ of $L$.
In between, the curve $\phi([B_{n},C_{n}]_{\partial{\Omega}})$ lies within
$\frac{2\pi{L}}{\log(\frac{d}{2\delta_{n}})}$ of the boundary arc
$[0,L]\times{\\{0\\}}$ of $\mathcal{R}_{L}$.

  * •

The curve $\phi([C_{n},D_{n}]_{\partial{\Omega}_{n}})$ travels from
$\phi(C_{n})$ to $\phi(D_{n})$, where $\phi(C_{n})$ lies within
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}$ of $L$
and $\phi(D_{n})$ lies within
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}$ of
$L+i$. In between, the curve $\phi([C_{n},D_{n}]_{\partial{\Omega}})$ lies
within $\frac{2\pi}{\log(\frac{d^{\prime}}{2\delta_{n}})}$ of the boundary arc
$\\{L\\}\times{[0,1]}$ of $\mathcal{R}_{L}$.

  * •

The curve $\phi([D_{n},A_{n}]_{\partial{\Omega}_{n}})$ travels from
$\phi(D_{n})$ to $\phi(A_{n})$, where $\phi(D_{n})$ lies within
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}$ of
$L+i$ and $\phi(A_{n})$ lies within
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}$ of $i$.
In between, the curve $\phi([D_{n},A_{n}]_{\partial{\Omega}})$ lies within
$\frac{2\pi{L}}{\log(\frac{d}{2\delta_{n}})}$ of the boundary arc
$[0,L]\times{\\{1\\}}$ of $\mathcal{R}_{L}$.

Putting all this together, it follows that:

$\left[\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})},L-\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}\right]\times{\left[\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})},1-\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}\right]}\subseteq{\phi(\Omega_{n})}$

Fix a compact set $K\subseteq{\Omega}$ and let
$D=\text{dist}(\phi(K),\mathcal{R}_{L})$. Then for any $n$ sufficiently large
so that
$\frac{4\pi(L\vee{1})}{\log(\frac{d\wedge{d^{\prime}}}{2\delta_{n}})}\leq{D}$,
we have that $K\subseteq{\Omega_{n}}$. ∎

######  Remark A.2.

Notice that if the orthodiagonal rectangle
$(G,A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a
$(\delta_{n},\varepsilon_{n})$\- good interior approximation of the conformal
rectangle $(\Omega,A,B,C,D)$ then
$(\widehat{G},A^{\bullet},B^{\bullet},C^{\bullet},D^{\bullet})$ is a
$(\delta_{n}+2\varepsilon_{n})$\- good interior approximation of
$(\Omega,A,B,C,D)$. Hence, the analogue of Lemma A.1 for discrete
approximations of a continuous domain also holds.

######  Lemma A.3.

Suppose $(\Omega,A,B,C,D)$ is a conformal rectangle and for each
$n\in{\mathbb{N}}$, $(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$ is a
$\delta_{n}$-good interior approximation of $(\Omega,A,B,C,D)$, where
$\delta_{n}\rightarrow{0}$ as $n\rightarrow{\infty}$. Then:

$\displaystyle\lim_{n\rightarrow{\infty}}d_{\Omega_{n}}(x,y)$
$\displaystyle=d_{\Omega}(x,y),$
$\displaystyle\lim_{n\rightarrow{\infty}}d_{cc}^{\Omega_{n}}(x,y)$
$\displaystyle=d_{cc}^{\Omega}(x,y)$

for any $x,y\in{\Omega}$.

###### Proof.

First we address the convergence of ambient distance. Fix $x,y\in{\Omega}$. By
Lemma A.1, we know that for any $n$ sufficiently large, $x,y\in{\Omega_{n}}$.
For such $n$, $d_{\Omega}(x,y)\leq{d_{\Omega_{n}}(x,y)}$, since
$\Omega_{n}\subseteq{\Omega}$. Hence:

$d_{\Omega}(x,y)\leq{\liminf_{n\rightarrow{\infty}}d_{\Omega_{n}}(x,y)}$

To get the reverse inequality, fix $\varepsilon>0$ and let $\gamma$ be a
smooth curve from $x$ to $y$ in $\Omega$ so that
$\text{length}(\gamma)\leq{d_{\Omega}(x,y)+\varepsilon}$. Since $\gamma$ is a
compact subset of $\Omega$, Lemma A.1 tells us that for all $n$ sufficiently
large, $\gamma\subseteq{\Omega_{n}}$, and so
$d_{\Omega_{n}}(x,y)\leq{d_{\Omega}(x,y)+\varepsilon}$. Hence:

$\limsup_{n\rightarrow{\infty}}d_{\Omega_{n}}(x,y)\leq{d_{\Omega}(x,y)+\varepsilon}$

Since $\varepsilon>0$ was arbitrary, letting $\varepsilon$ tend to $0$, we
have that:

$\limsup_{n\rightarrow{\infty}}d_{\Omega_{n}}(x,y)\leq{d_{\Omega}(x,y)}$

Regarding the convergence of crosscut distance, again, fix $x,y\in{\Omega}$
and suppose $n$ is sufficiently large so that $x,y\in{\Omega_{n}}$. We will
show that:

$\lim_{n\rightarrow{\infty}}d_{cc}^{\Omega_{n}}(\\{x,y\\},[A_{n},B_{n}]_{\partial{\Omega_{n}}};[C_{n},D_{n}]_{\partial{\Omega_{n}}})=d_{cc}^{\Omega}(\\{x,y\\},[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})$
(A.1)

The corresponding result for the other three pairs of boundary arcs follows by
the exact same argument. Hence, if we can verify Equation A.1, it follows
that:

$\lim_{n\rightarrow{\infty}}d_{cc}^{\Omega_{n}}(x,y)=d_{cc}^{\Omega}(x,y)$

Suppose $\gamma_{n}$ is a crosscut of $\Omega_{n}$ that joins $x$ and $y$ to
$[A_{n},B_{n}]_{\partial{\Omega_{n}}}$ and separates $x$ and $y$ from
$[C_{n},D_{n}]_{\partial{\Omega_{n}}}$. Let $u_{n}$ and $v_{n}$ be the
endpoints of $\gamma_{n}$. Then there are three possibilities:

  1. 1.

Both $u_{n}$ and $v_{n}$ lie on $[A_{n},B_{n}]_{\partial{\Omega_{n}}}$.

  2. 2.

One point lies on $[A_{n},B_{n}]_{\partial{\Omega_{n}}}$, while the other lies
on a neighboring arc: either $[B_{n},C_{n}]_{\partial{\Omega_{n}}}$ or
$[D_{n},A_{n}]_{\partial{\Omega_{n}}}$.

  3. 3.

One points lies on $[B_{n},C_{n}]$ and the other point lies on
$[D_{n},A_{n}]_{\partial{\Omega_{n}}}$.

In all three cases, the argument is analogous, so for simplicity, let’s assume
we are in the first situation. Since $(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$ is
a $\delta_{n}$-good interior approximation of $(\Omega,A,B,C,D)$, there exist
crosscuts $\eta_{n}^{(1)}$ and $\eta_{n}^{(2)}$ of $\Omega$ so that:

  * •

$\eta_{n}^{(1)}$ joins $u_{n}$ to $[A,B]_{\partial{\Omega^{\ast}}}$ and
separates $u_{n}$ from $[C,D]_{\partial{\Omega^{\ast}}}$

  * •

$\eta_{n}^{(2)}$ joins $v_{n}$ to $[A,B]_{\partial{\Omega^{\ast}}}$ and
separates $v_{n}$ from $[C,D]_{\partial{\Omega^{\ast}}}$

  * •

$\text{length}(\eta_{n}^{(1)}),\text{length}(\eta_{n}^{(2)})<{\delta_{n}}$

Stitching together $\eta_{n}^{(1)},\gamma_{n}$ and $\eta_{n}^{(2)}$, we can
produce a crosscut of $\Omega$ that joins $x$ and $y$ to
$[A,B]_{\partial{\Omega^{\ast}}}$ and separates $x$ and $y$ from
$[C,D]_{\partial{\Omega^{\ast}}}$ with length at most
$\text{length}(\gamma_{n})+2\delta_{n}$. Since this is true for any such
crosscut $\gamma_{n}$ of $\Omega_{n}$, it follows that:

$d_{cc}^{\Omega}(\\{x,y\\},[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})<d_{cc}^{\Omega_{n}}(\\{x,y\\},[A_{n},B_{n}]_{\partial{\Omega_{n}}};[C_{n},D_{n}]_{\partial{\Omega_{n}}})+2\delta_{n}$

for all $n$ sufficiently large so that $x,y\in{\Omega_{n}}$. Hence:

$d_{cc}^{\Omega}(\\{x,y\\},[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})\leq{\liminf_{n\rightarrow{\infty}}d_{cc}^{\Omega_{n}}(\\{x,y\\},[A_{n},B_{n}]_{\partial{\Omega_{n}}};[C_{n},D_{n}]_{\partial{\Omega_{n}}})}$

Figure 14: Stitching together $\eta_{n}^{(1)}$, $\eta_{n}^{(2)}$, and
$\gamma_{n}$ gives us a crosscut of $\Omega$ that joins $x$ and $y$ to
$[A,B]_{\partial{\Omega^{\ast}}}$ and separates $x$ and $y$ from
$[C,D]_{\partial{\Omega^{\ast}}}$.

To get the reverse inequality, suppose $\gamma$ is a crosscut of $\Omega$ that
joins $x$ and $y$ to $[A,B]_{\partial{\Omega^{\ast}}}$ and separates $x$ and
$y$ from $[C,D]_{\partial{\Omega^{\ast}}}$. Let
$\gamma:[0,1]\rightarrow{\Omega^{\ast}}$ be an explicit parametrization of
$\gamma$ and let $u,v\in{\partial{\Omega^{\ast}}}$ be the endpoints of
$\gamma$: $u=\gamma(0)$, $v=\gamma(1)$. Since any crosscut of $\Omega$ that
starts or ends at a corner $A,B,C,D$ of our conformal rectangle is close to a
crosscut of $\Omega$ with almost the same length that doesn’t use any of the
corners of $(\Omega,A,B,C,D)$, we can assume without loss of generality that
neither $u$ nor $v$ is a corner of $(\Omega,A,B,C,D)$. From here, there are
three possibilities:

  1. 1.

Both $u$ and $v$ lie on $(A,B)_{\partial{\Omega^{\ast}}}$.

  2. 2.

One point lies on $(A,B)_{\partial{\Omega^{\ast}}}$ while the other lies on a
neighboring arc: either $(B,C)_{\partial{\Omega^{\ast}}}$ or
$(D,A)_{\partial{\Omega^{\ast}}}$.

  3. 3.

One point lies on $(B,C)_{\partial{\Omega^{\ast}}}$ and the other point lies
on $(D,A)_{\partial{\Omega^{\ast}}}$.

The argument in all three cases is analogous, so for simplicity, suppose we
are in the first situation. Then:

$\displaystyle c(u)$
$\displaystyle=\min\\{d_{cc}^{\Omega}(u,[B,C]_{\partial{\Omega^{\ast}}};[A,D]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(u,[C,D]_{\partial{\Omega^{\ast}}};[A,B]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(u,[D,A]_{\partial{\Omega^{\ast}}};[B,C]_{\partial{\Omega^{\ast}}})\\}>0$
$\displaystyle c(v)$
$\displaystyle=\min\\{d_{cc}^{\Omega}(v,[B,C]_{\partial{\Omega^{\ast}}};[A,D]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(v,[C,D]_{\partial{\Omega^{\ast}}};[A,B]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(v,[D,A]_{\partial{\Omega^{\ast}}};[B,C]_{\partial{\Omega^{\ast}}})\\}>0$

Choose $s,t\in(0,1)$ so that:

$\displaystyle\text{length}(\gamma[0,s])$ $\displaystyle<\frac{c(u)}{4\pi},$
$\displaystyle\text{length}(\gamma[t,1])$ $\displaystyle<\frac{c(v)}{4\pi}$

Since $\gamma[s,t]$ is a compact subset of $\Omega$, by Lemma A.1,
$\gamma[s,t]\subseteq{\Omega_{n}}$ for all $n$ sufficiently large. In this
case, since $\gamma$ starts and ends outside $\Omega_{n}$, $\gamma$ must
intersect $\Omega_{n}$ at least twice: once when it enters $\Omega_{n}$ and
once when it exits. Since $\gamma[s,t]\subseteq{\Omega_{n}}$, the points of
intersection of $\gamma$ with $\partial{\Omega_{n}}$ must all lie in
$\gamma[0,s]$ or $\gamma[t,1]$. Since
$u=\gamma(0)\in{(A,B)_{\partial{\Omega^{\ast}}}}$ and
$\text{diam}(\gamma[0,s])<\frac{c(u)}{4\pi}$, there is a subarc $\chi$ of
$\\{z\in{\Omega^{\ast}}:|z-u|=\frac{c(u)}{4\pi}\\}$ so that
$[C,D]_{\partial{\Omega^{\ast}}}$ and $\gamma[0,s]$ lie in distinct connected
components of $\Omega^{\ast}\setminus{\chi}$. In other words:

$d_{cc}^{\Omega}(\gamma[0,s],[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})<\frac{c(u)}{2}$

Figure 15: On the left: The situation we are trying to avoid: the crosscut
$\gamma$ which joins $x$ and $y$ to $[A,B]_{\partial{\Omega^{\ast}}}$ and
separates $x$ and $y$ from $[C,D]_{\partial{\Omega^{\ast}}}$ intersects
$\partial{\Omega_{n}}$ along the arc $[D_{n},A_{n}]_{\partial{\Omega_{n}}}$
rather than $[A_{n},B_{n}]_{\partial{\Omega_{n}}}$. On the right: For a fixed
crosscut $\gamma$, we can avoid the picture on the left by taking $m>n$
sufficiently large.

Hence, the points of intersection of $\gamma[0,s]$ and $\partial{\Omega_{n}}$
must all lie on $[A_{n},B_{n}]_{\partial{\Omega_{n}}}$. To see why, suppose
$w_{n}$ is a point of intersection of $\partial{\Omega_{n}}$ and
$\gamma[0,s]$, and $w_{n}$ lies on one of the other boundary arcs of
$(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$. Without loss of generality, suppose
$w_{n}\in{[B_{n},C_{n}]_{\partial{\Omega_{n}}}}$. Since
$(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$ is a $\delta_{n}$-good interior
approximation of $(\Omega,A,B,C,D)$, there exists a crosscut $\zeta_{n}$ of
$\Omega$ joining $w_{n}$ to $[B,C]_{\partial{\Omega^{\ast}}}$, and separating
$w_{n}$ from $[D,A]_{\partial{\Omega^{\ast}}}$ so that
$\text{length}(\zeta_{n})<\delta_{n}$. Stitching together $\chi$ and
$\zeta_{n}$ we can produce a crosscut of $\Omega$ with length at most
$\delta_{n}+\frac{c(u)}{2}$ that joins $u$ to
$[B,C]_{\partial{\Omega^{\ast}}}$ and separates $u$ from
$[D,A]_{\partial{\Omega^{\ast}}}$. Hence, for $n$ sufficiently large so that
$\delta_{n}<\frac{c(u)}{2}$, we have that:

$d_{cc}^{\Omega}(u,[B,C]_{\partial{\Omega^{\ast}}},[D,A]_{\partial{\Omega^{\ast}}})<\delta_{n}+\frac{c(u)}{2}<c(u)$

This contradicts the definition of $c(u)$. Thus, the points of intersection of
$\gamma[0,s]$ and $\partial{\Omega_{n}}$ all lie on
$[A_{n},B_{n}]_{\partial{\Omega^{\ast}}}$. By the exact same argument, the
points of intersection of $\gamma[t,1]$ and $\partial{\Omega_{n}}$ also lie on
$[A_{n},B_{n}]_{\partial{\Omega^{\ast}}}$, for $n$ sufficiently large. Hence,
there exists a subarc $\gamma_{n}$ of $\gamma$ so that $\gamma_{n}$ is a
crosscut of $\Omega_{n}$ that joins $x$ and $y$ to
$[A_{n},B_{n}]_{\partial{\Omega_{n}}}$ and separates $x$ and $y$ from
$[C_{n},D_{n}]_{\partial{\Omega_{n}}}$. Since
$\text{length}(\gamma_{n})\leq{\text{length}(\gamma)}$, and we can build such
a crosscut $\gamma_{n}$ of $\Omega_{n}$ for any crosscut $\gamma$ of $\Omega$
that joins $x$ and $y$ to $[A,B]_{\partial{\Omega^{\ast}}}$ in $\Omega$ and
separates $x$ and $y$ from $[C,D]_{\partial{\Omega^{\ast}}}$, provided that
$n$ is sufficiently large, it follows that:

$\limsup_{n\rightarrow{\infty}}d^{\Omega_{n}}_{cc}(\\{x,y\\},[A_{n},B_{n}]_{\partial{\Omega_{n}}};[C_{n},D_{n}]_{\partial{\Omega_{n}}})\leq{d^{\Omega}_{cc}(\\{x,y\\},[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})}$

∎

######  Lemma A.4.

Suppose $(\Omega,A,B,C,D)$ is a conformal rectangle and for each
$n\in{\mathbb{N}}$, $(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$ is a
$\delta_{n}$-good interior approximation of $(\Omega,A,B,C,D)$, where
$\delta_{n}\rightarrow{0}$ as $n\rightarrow{\infty}$. Suppose:

$d=\inf\\{\text{diam}(\gamma):\text{$\gamma$ is a curve in $\Omega$ joining
$[A,B]_{\partial{\Omega^{\ast}}}$ and $[C,D]_{\partial{\Omega^{\ast}}}$}\\}$

and for each $n\in{\mathbb{N}}$:

$d_{n}=\inf\\{\text{diam}(\gamma):\text{$\gamma$ is a curve in $\Omega_{n}$
joining $[A_{n},B_{n}]_{\partial{\Omega_{n}}}$ and
$[C_{n},D_{n}]_{\partial{\Omega_{n}}}$}\\}$

Then:

$\lim_{n\rightarrow{\infty}}d_{n}=d$

###### Proof.

Suppose $\gamma_{n}$ is a curve in $\Omega_{n}$ from
$[A_{n},B_{n}]_{\partial{\Omega_{n}}}$ to
$[C_{n},D_{n}]_{\partial{\Omega_{n}}}$. Let $u_{n}$ be the endpoint of
$\gamma_{n}$ on $[A_{n},B_{n}]_{\partial{\Omega_{n}}}$ and let $v_{n}$ be the
endpoint of $\gamma_{n}$ on $[C_{n},D_{n}]_{\partial{\Omega_{n}}}$. Since
$(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$ is a $\delta_{n}$-good interior
approximation of $(\Omega,A,B,C,D)$, we can find crosscuts $\eta_{n}^{(1)}$
and $\eta_{n}^{(2)}$ of $\Omega$ so that $\eta_{n}^{(1)}$ joins $u_{n}$ to
$[A,B]_{\partial{\Omega^{\ast}}}$ and separates $u_{n}$ from
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$, $\eta_{n}^{(2)}$ joins $v_{n}$
to $[C,D]_{\partial{\Omega^{\ast}}}$ and separates $v_{n}$ from
$[A,B]_{\partial{\Omega^{\ast}}}$ in $\Omega$, and
$\text{length}(\eta_{n}^{(1)}),\text{length}(\eta_{n}^{(2)})<\delta_{n}$.
Stitching together $\eta_{n}^{(1)},\gamma_{n}$ and $\eta_{n}^{(2)}$ we can
produce a simple curve from $[A,B]_{\partial{\Omega^{\ast}}}$ to
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$ with diameter at most
$\text{diam}(\gamma_{n})+2\delta_{n}$. Hence, $d\leq{d_{n}+2\delta_{n}}$ for
all $n\in{\mathbb{N}}$, which tells us that:

$d\leq{\liminf_{n\rightarrow{\infty}}d_{n}}$

Figure 16: Given an curve $\gamma_{n}$ in $\Omega_{n}$ from
$[A_{n},B_{n}]_{\partial{\Omega_{n}}}$ to
$[C_{n},D_{n}]_{\partial{\Omega_{n}}}$ with endpoints
$u_{n}\in{[A_{n},B_{n}]_{\partial{\Omega_{n}}}}$ and
$v_{n}\in{[C_{n},D_{n}]_{\partial{\Omega_{n}}}}$, we can stitch together
$\gamma_{n}$ with the crosscuts $\eta_{n}^{(1)}$ and $\eta_{n}^{(2)}$ to
produce a curve from $[A,B]_{\partial{\Omega^{\ast}}}$ to
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$ whose diameter is close to that
of $\gamma_{n}$.

Suppose $\gamma$ is a crosscut of $\Omega$ joining
$[A,B]_{\partial{\Omega^{\ast}}}$ and $[C,D]_{\partial{\Omega^{\ast}}}$ with
endpoints $u$ in $[A,B]_{\partial{\Omega^{\ast}}}$ and $v$ in
$[C,D]_{\partial{\Omega^{\ast}}}$. Since any path from
$[A,B]_{\partial{\Omega^{\ast}}}$ to $[C,D]_{\partial{\Omega^{\ast}}}$ in
$\Omega$ that starts or ends at a corner $A,B,C,D$ of our conformal rectangle
is close to a path from $[A,B]_{\partial{\Omega^{\ast}}}$ to
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$ that has almost the same
diameter and doesn’t use any of the corners of $(\Omega,A,B,C,D)$, we can
assume without loss of generality that $u\in{(A,B)}_{\partial{\Omega^{\ast}}}$
and $v\in{(C,D)_{\partial{\Omega^{\ast}}}}$. Hence:

$\displaystyle c(u)$
$\displaystyle=\min\\{d_{cc}^{\Omega}(u,[B,C]_{\partial{\Omega^{\ast}}};[A,D]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(u,[C,D]_{\partial{\Omega^{\ast}}};[A,B]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(u,[D,A]_{\partial{\Omega^{\ast}}};[B,C]_{\partial{\Omega^{\ast}}})\\}>0$
$\displaystyle c(v)$
$\displaystyle=\min\\{d_{cc}^{\Omega}(v,[B,C]_{\partial{\Omega^{\ast}}};[A,D]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(v,[C,D]_{\partial{\Omega^{\ast}}};[A,B]_{\partial{\Omega^{\ast}}}),d_{cc}^{\Omega}(v,[D,A]_{\partial{\Omega^{\ast}}};[B,C]_{\partial{\Omega^{\ast}}})\\}>0$

Let $\gamma:[0,1]\rightarrow{\Omega^{\ast}}$ be an explicit parametrization of
$\gamma$. Choose $s,t\in{(0,1)}$ so that:

$\displaystyle\text{diam}(\gamma[0,s])$ $\displaystyle<\frac{c(u)}{4\pi},$
$\displaystyle\text{diam}(\gamma[t,1])$ $\displaystyle<\frac{c(v)}{4\pi}$

Since $\gamma[s,t]$ is a compact subset of $\Omega$, by Lemma A.1,
$\gamma[s,t]\subseteq{\Omega_{n}}$ for all $n$ sufficiently large. In this
case, since $\gamma$ starts and ends outside $\Omega_{n}$, $\gamma$ must
intersect $\Omega_{n}$ at least twice: once when it enters $\Omega_{n}$ and
once when it exits. Since $\gamma[s,t]\subseteq{\Omega_{n}}$, the points of
intersection of $\gamma$ with $\partial{\Omega_{n}}$ must all lie in
$\gamma[0,s]$ or $\gamma[t,1]$. Since
$u=\gamma(0)\in{(A,B)_{\partial{\Omega^{\ast}}}}$ and
$\text{diam}(\gamma[0,s])<\frac{c(u)}{4\pi}$, there is a subarc $\chi$ of
$\\{z\in{\Omega^{\ast}}:|z-u|=\frac{c(u)}{4\pi}\\}$ so that
$[C,D]_{\partial{\Omega^{\ast}}}$ and $\gamma[0,s]$ lie in distinct connected
components of $\Omega^{\ast}\setminus{\chi}$. In other words:

$d_{cc}^{\Omega}(\gamma[0,s],[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})<\frac{c(u)}{2}$

Figure 17: On the left: The situation we are trying to avoid: the curve
$\gamma$ between $[A,B]_{\partial{\Omega^{\ast}}}$ and
$[C,D]_{\partial{\Omega^{\ast}}}$ in $\Omega$ enters $\Omega_{n}$ via the
boundary arc $[D_{n},A_{n}]_{\partial{\Omega_{n}}}$ rather than
$[A_{n},B_{n}]_{\partial{\Omega_{n}}}$. On the right: For a fixed curve
$\gamma$, we can avoid the picture on the left by taking $m>n$ sufficiently
large.

Hence, the points of intersection of $\gamma[0,s]$ and $\partial{\Omega_{n}}$
must all lie on $[A_{n},B_{n}]_{\partial{\Omega_{n}}}$. To see why, suppose
$w_{n}$ is a point of intersection of $\partial{\Omega_{n}}$ and
$\gamma[0,s]$, and $w_{n}$ lies on one of the other boundary arcs of
$(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$. Without loss of generality, suppose
$w_{n}\in{[B_{n},C_{n}]_{\partial{\Omega_{n}}}}$. Since
$(\Omega_{n},A_{n},B_{n},C_{n},D_{n})$ is a $\delta_{n}$-good interior
approximation of $(\Omega,A,B,C,D)$, there exists a crosscut $\zeta_{n}$ of
$\Omega$ joining $w_{n}$ to $[B,C]_{\partial{\Omega^{\ast}}}$, and separating
$w_{n}$ from $[D,A]_{\partial{\Omega^{\ast}}}$ so that
$\text{length}(\zeta_{n})<\delta_{n}$. Stitching together $\chi$ and
$\zeta_{n}$ we can produce a crosscut of $\Omega$ with length at most
$\delta_{n}+\frac{c(u)}{2}$ that joins $u$ to
$[B,C]_{\partial{\Omega^{\ast}}}$ and separates $u$ from
$[D,A]_{\partial{\Omega^{\ast}}}$. Hence, for $n$ sufficiently large so that
$\delta_{n}<\frac{c(u)}{2}$, we have that:

$d_{cc}^{\Omega}(u,[B,C]_{\partial{\Omega^{\ast}}},[D,A]_{\partial{\Omega^{\ast}}})<\delta_{n}+\frac{c(u)}{2}<c(u)$

This contradicts the definition of $c(u)$. Thus, the points of intersection of
$\gamma[0,s]$ and $\partial{\Omega_{n}}$ all lie on
$[A_{n},B_{n}]_{\partial{\Omega^{\ast}}}$. By the exact same argument, the
points of intersection of $\gamma[t,1]$ and $\partial{\Omega_{n}}$ all lie on
$[C_{n},D_{n}]_{\partial{\Omega^{\ast}}}$, for $n$ sufficiently large. Hence,
there exists a subarc $\gamma_{n}$ of $\gamma$ so that $\gamma_{n}$ is a
crosscut of $\Omega_{n}$ joining $[A_{n},B_{n}]_{\partial{\Omega_{n}}}$ and
$[C_{n},D_{n}]_{\partial{\Omega_{n}}}$ such that
$\gamma[s,t]\subseteq{\gamma_{n}}$, which tells us that
$\text{diam}(\gamma_{n})\leq{\text{diam}(\gamma)}$. Since we can do this for
any curve $\gamma$ from $(A,B)_{\partial{\Omega^{\ast}}}$ to
$(C,D)_{\partial{\Omega^{\ast}}}$ in $\Omega$, we conclude that:

$\limsup_{n\rightarrow{\infty}}d_{n}\leq{d}$

∎

In order to prove Proposition 2.14, we need the following analogue of Wolff’s
Lemma (see Proposition 2.2 of [24]) for conformal mappings of rectangles:

######  Lemma A.5.

Consider the conformal rectangle $(\mathcal{R}_{L},i,0,L,L+i)$, which is just
$\mathcal{R}_{L}$ with its four corners as the four distinguished boundary
points. Let $\Omega$ be a bounded simply connected domain and let
$\psi:\mathcal{R}_{L}\rightarrow{\Omega}$ be a conformal map from
$\mathcal{R}_{L}$ to $\Omega$. If $z_{0}$ is a point of $\mathcal{R}_{L}$ so
that $d=\text{dist}(z_{0},\partial{\mathcal{R}_{L}})<\frac{1\wedge{L}}{2}$ and
$\mathcal{I}$ and $\mathcal{J}$ are boundary arcs of
$(\mathcal{R}_{L},i,0,L,L+i)$ so that $\mathcal{I}$ is the boundary arc
closest to $z_{0}$, and $\mathcal{J}$ is opposite to $\mathcal{I}$, then there
exists a crosscut of $\Omega$ of length less than
$\sqrt{\frac{\pi\text{Area}(\Omega)}{2\log\big{(}\frac{1\wedge{L}}{2d}\big{)}}}$
joining $\psi(z_{0})$ to $\psi(\mathcal{I})$ and separating $\psi(z_{0})$ from
$\psi(\mathcal{J})$ in $\Omega$. Equivalently:

$d_{cc}^{\Omega}(\psi(z_{0}),\psi(\mathcal{I});\psi(\mathcal{J}))<\sqrt{\frac{\pi\text{Area}(\Omega)}{2\log\big{(}\frac{1\wedge{L}}{2d}\big{)}}}$

The proof of this result is identical to the proof of Wolff’s Lemma. For the
convenience of the reader, we reproduce the argument below:

###### Proof.

If $z_{0}$ is a point of $\mathcal{R}_{L}$ so that
$d=\text{dist}(z_{0},\partial{\Omega})=\text{dist}(z_{0},\mathcal{I})<\frac{1\wedge{L}}{2}$,
then for each $s\in{(d,\frac{1\wedge{L}}{2})}$,

$\gamma_{s}=\mathcal{R}_{L}\cap{\\{z:|z-z_{0}|=s\\}}$

is a crosscut of $\mathcal{R}_{L}$ that joins $z_{0}$ to $\mathcal{I}$ and
separates $z_{0}$ from $\mathcal{J}$. It follows that $\psi(\gamma_{s})$ is a
crosscut of $\Omega$ that joins $\psi(z_{0})$ to $\psi(\mathcal{I})$ and
separates $\psi(z_{0})$ from $\psi(\mathcal{J})$. We will now show that one of
these crosscuts $\psi(\gamma_{s})$ is sufficiently short, using the length-
area trick. Namely:

$\displaystyle\text{Area}(\Omega)$
$\displaystyle>\text{Area}(\psi\Big{(}\mathcal{R}_{L}\cap{\\{z:d<|z-z_{0}|<\frac{1\wedge{L}}{2}\\}}\Big{)})=\int_{\mathcal{R}_{L}\cap{\\{z:d<|z-z_{0}|<\frac{1\wedge{L}}{2}\\}}}|\psi^{\prime}(x+iy)|^{2}dxdy$
$\displaystyle=\int_{d}^{\frac{1\wedge{L}}{2}}\int_{\theta_{1}(s)}^{\theta_{2}(s)}s|\psi^{\prime}(z_{0}+se^{i\theta})|^{2}d\theta{ds}\overset{(\ast)}{\geq}{\int_{d}^{\frac{1\wedge{L}}{2}}\frac{2}{\pi{s}}\Big{(}\int_{\theta_{1}(s)}^{\theta_{2}(s)}s|\psi^{\prime}(z_{0}+se^{i\theta})|d\theta\Big{)}^{2}ds}$
$\displaystyle=\frac{2}{\pi}\int_{d}^{\frac{1\wedge{L}}{2}}\frac{\big{(}\text{length}(\psi(\gamma_{s}))\big{)}^{2}}{s}ds\geq{\frac{2}{\pi}\inf_{d<s<\frac{1\wedge{L}}{2}}\big{(}\text{length}(\psi(\gamma_{s}))\big{)}^{2}\hskip
1.0pt\log\Big{(}\frac{1\wedge{L}}{2d}\Big{)}}$

The inequality “$(\ast)$" follows by Cauchy-Schwarz and the observation that
since we’re working in $\mathcal{R}_{L}$,
$(\theta_{2}(s)-\theta_{1}(s))\geq{\frac{\pi}{2}}$ for all
$s\in{(d,\frac{1\wedge{L}}{2})}$. Rearranging, we get that:

$\inf_{d<s<\frac{1\wedge{L}}{2}}\text{length}(\psi(\gamma_{s}))<\sqrt{\frac{\pi\text{Area}(\Omega)}{2\log\big{(}\frac{1\wedge{L}}{2d}\big{)}}}$

∎

Having established Lemma A.5, Proposition 2.14 follows as an immediate
corollary:

###### Proof.

(of Proposition 2.14) Let $\phi:\Omega\rightarrow{\mathcal{R}_{L}}$ be the
conformal map from $\Omega$ to $\mathcal{R}_{L}$ so that the four
distinguished prime ends of $\Omega$ are mapped to the four corners of
$\mathcal{R}_{L}$ and in particular, $\phi(A)=i$. Fix
$r\in{(0,\frac{1\wedge{L}}{2})}$. Consider the subrectangle
$(\Omega_{r},A_{r},B_{r},C_{r},D_{r})$ of $(\Omega,A,B,C,D)$, where:

  * •

$\Omega_{r}=\phi^{-1}\big{(}(r,L-r)\times{(r,1-r)}\big{)}$

  * •

$A_{r}=\phi^{-1}(r+(1-r)i)$

  * •

$B_{r}=\phi^{-1}(r+ri)$

  * •

$C_{r}=\phi^{-1}((L-r)+ri)$

  * •

$D_{r}=\phi^{-1}((L-r)+(1-r)i)$

  *

In other words, $\Omega_{r}$ is the preimage of the rectangle
$(r,L-r)\times{(r,1-r)}$ under the conformal map $\phi$, and the prime ends
$A_{r},B_{r},C_{r},D_{r}$ are the preimages under $\phi$ of the four corners
of this rectangle. By Lemma A.5,

$d_{cc}^{\Omega}([A_{r},B_{r}]_{\partial{\Omega_{r}}},[A,B]_{\partial{\Omega^{\ast}}};[C,D]_{\partial{\Omega^{\ast}}})\leq{\sqrt{\frac{\pi\text{Area}(\Omega)}{2\log\big{(}\frac{1\wedge{L}}{2r}\big{)}}}}$

By the exact same reasoning, the analogous result is true for the other three
boundary arcs of $(\Omega_{r},A_{r},B_{r},C_{r},D_{r})$. Thus, choosing
$r<\big{(}\frac{1\wedge{L}}{2}\big{)}e^{-\frac{\pi\text{Area}(\Omega)}{2\delta^{2}}}$
for some fixed $\delta>0$, the desired result follows. ∎

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