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Convergence of the Probabilistic Interpretation ofModulus
Nathan Albin, Joan Lind and Pietro Poggi-Corradini
June 23, 2021
Abstract
Given a Jordan domain Ω ⊂ C and two arcs A,B on ∂Ω, the modulus of the curve family connectingA and B in Ω is famously related, via the conformal map φ mapping Ω to a rectangle R = [0, L]× [0, 1]so that A and B are sent to the vertical sides, to the corresponding modulus in R. Moreover, in thecase of the rectangle the family of horizontal segments connecting the two sides has the same modulusas the entire connecting family. Pulling these segments back to Ω via φ yields a family of extremalcurves (also known as horizontal trajectories) connecting A to B in Ω. In this paper, we show thatthese extremal curves can be approximated by some discrete curves arising from an orthodiagonalapproximation of Ω. Moreover, we show that these curves carry a natural probability mass function(pmf) deriving from the theory of discrete modulus and that these pmf’s converge to the uniformdistribution on the set of extremal curves. The key ingredient is an algorithm that, for an embeddedplanar graph, takes the current flow between two sets of nodes A and B, and produces a unique pathdecomposition with non-crossing paths. Moreover, some care was taken to adapt recent results forharmonic convergence on orthodiagonal maps, due to Gurel-Gurevich, Jerison, and Nachmias, to ourcontext.
Contents
1 Introduction and results 2
2 Background and examples 32.1 Discrete modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Discrete harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.4 Orthodiagonal maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.5 Linearly connected and John domains . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Non-crossing minimal subfamilies 113.1 Non-crossing minimal subfamily algorithm . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Scaling limit of non-crossing minimal subfamilies . . . . . . . . . . . . . . . . . . . . . 153.3 Further consequences of the minimal non-crossing subfamily . . . . . . . . . . . . . . . 183.4 A numerical example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.4.1 Approximating the modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.4.2 Non-crossing curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
4 Convergence of discrete harmonic functions 224.1 Modified toolbox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.2 Proof of Theorem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
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1 Introduction and results
Many objects in complex analysis are approximated by discrete analogues. In this paper, we establishthe convergence of three discrete objects, discrete modulus, discrete paths that are extremal for themodulus, and discrete harmonic functions, to their continuous counterparts, the continuous modulus,the extremal curves (or horizontal trajectories) for the modulus, and harmonic functions.
Consider a Jordan domain Ω ∈ C and two arcs on ∂Ω. The modulus of the curve family connectingthese arcs in Ω is well understood. In particular, there is a subfamily of extremal curves (or horizontaltrajectories) that carry all the information for this modulus. We wish to approximate Ω with a planegraph G and approximate the boundary arcs with sets of vertices A and B. As previously studiedin [2, 3, 4], the discrete modulus of the family of paths between A and B in G arises from putting adensity on the edges of G. See Section 2.1 for the definitions. Our first task is to identify a subfamilyof paths between A and B in G that is the candidate approximating family for the extremal curvesin the classical setting. This is accomplished in the result below.
For a finite plane graph G, the sets of boundary vertices A and B are defined as follows: A = S1∩Vand B = S2 ∩ V , where S1, S2 are two disjoint arcs from a simple curve in the closed outer face of G.We always assume that A and B are nonempty.
Proposition 1. Let G be a finite plane graph, and let A and B be defined as above. Then the familyof all paths in G from A to B has a unique minimal subfamily with non-crossing paths. Moreover, thisfamily supports a pmf µ that is optimal for the dual discrete modulus problem, and both are outputfrom the Non-crossing Minimal Subfamily Algorithm described in Section 3.
Our next goal is to show that the candidate subfamily identified in Proposition 1 does in factapproximate the family of classical extremal curves. For this, we specialize to plane graphs thatarise from finite orthodiagonal maps, a class of graphs that generalize the square grid. Roughly, anothodiagonal map is a graph where every interior face is a quadrilateral with orthogonal diagonals.Associated with the othodiagonal map is the so-called primal graph G• that we utilize. See Section2.4 for definitions and notation.
The setting for the following result is as follows: Let Ω be a Jordan domain whose boundarycan be decomposed into four analytic arcs that do not meet in cusps. Let A and B be two of thesenon-adjacent arcs. Let Gε be an orthodiagonal map with edge length at most ε that approximates Ωwithin distance ε and has boundary vertices Aε and Bε approximating A and B, respectively. (SeeSection 2.4 for more precise definitions.)
Theorem 2. In the setting above, consider the family of all paths in G•ε from Aε to Bε, and let Γε bethe unique minimal subfamily with non-crossing paths obtained from Proposition 1, with correspondingpmf µε. Then, as ε → 0, µε converges to the uniform measure on the classical extremal curves fromA to B in Ω.
The convergence in the above theorem is weak convergence and the underlying metric is theHausdorff metric on curves. This theorem relies on the following result about discrete harmonicfunctions approximating continuous harmonic functions, which is an adaptation of the main theoremof [8].
Theorem 3. Let Ω be a Jordan domain whose boundary is decomposed into four analytic arcs that donot meet in cusps, and let A and B be two of these non-adjacent arcs. Let ψ be a conformal map ofΩ onto the rectangle R = [0, 1]× [0,m] taking the arcs A and B to the vertical sides of the rectangle,and let hc = Re ψ. Let ε, δ > 0 be small enough. Let G = (V • t V , E) be a finite orthodiagonalmap with boundary arcs S1, T1, S2, T2 with maximal edge length at most ε, so that the quadrilateralG approximates Ω within distance δ (in the sense of Definition 14). Let hd : V • → R be discreteharmonic on Int(V •), with the following boundary values: hd(z) = 0 for z ∈ V • ∩ S1 and hd(z) = 1for z ∈ V • ∩ S2. Then, for z ∈ V • ∩ Ω,
|hd(z)− hc(z)| ≤C
log1/2 (diam(Ω)/(ε ∨ δ)),
2
where C depends only on Ω.
An immediate consequence to this theorem is the following corollary, which generalizes a result byNorah Alrayes in [5] from the square grid setting to the orthodiagonal setting.
Corollary 4. In the setting of Theorem 2, as ε → 0 the discrete modulus of the path family betweenAε and Bε in G•ε converges to continuous modulus of the curve family between A and B in Ω.
We end with a note about the organization of this paper. In Section 2, we introduce the neededbackground, including discrete modulus, discrete harmonic functions, and orthodiagonal maps, andwe consider two illuminating examples. We begin Section 3 by discussing the Non-crossing MinimalSubfamily Algorithm and proving Proposition 1. Then we prove Theorem 2, using Theorem 3 as atool. We end by discussing an interpretation of Proposition 1 in terms of current flow and a rectangle-packing result. Section 4 contains the proof of Theorem 3 and is independent of Section 3.
2 Background and examples
2.1 Discrete modulus
Let G = (V,E, σ) be a finite weighted graph, which is also referred to as a network, with vertex set V ,edge set E, and edge weights σ : E → (0,∞). Let Γ be a family of objects in G, such as a collectionof paths in G. The usage matrix N is a |Γ| × |E| matrix with N (γ, e) giving the usage of edge e inobject γ. In the case that γ is a path, then N (γ, e) = 1e∈γ.
A density ρ : E → [0,∞) on G is called admissible for Γ if
`ρ(γ) :=∑e∈EN (γ, e)ρ(e) ≥ 1 for all γ ∈ Γ.
When Γ is family of paths, we consider `ρ(γ) to be the length of γ under ρ. In this setting ρ isadmissible for Γ if every path has length at least 1. For 1 ≤ p < ∞, the (discrete) p-modulus of Γ isdefined as
Modp(Γ) := inf∑e∈E
σ(e)ρ(e)p,
where the infimum is taken over all admissible densities ρ. See [2, 3, 4] for further background onthe discrete modulus. In the rest of this paper, we will specialize to the case when p = 2 and Γ is acollection of paths.
The extremal density ρ∗ is an admissible density that satisfies
Mod2(Γ) =∑e∈E
σ(e)ρ∗(e)2.
The extremal density is unique in this case and is characterized in the following theorem, which is thep = 2 case of a well-known result (see for instance, [4, Theorem 2.1], [1, Theorem 4-4], and also [6]).
Theorem 5. (Beurling’s Criterion) Let G = (V,E, σ) be a network and Γ a family of paths on G. Anadmissible density ρ is extremal for Mod2(Γ) if there is a subfamily Γ ⊂ Γ that satisfies the followingtwo conditions:
(i) `ρ(γ) = 1 for all γ ∈ Γ.
(ii) If h : E → R satisfies `h(γ) =∑
e∈γ h(e) ≥ 0 for all γ ∈ Γ, then∑e∈E
h(e)ρ(e)σ(e) ≥ 0.
Furthermore, in this case Mod2(Γ) = Mod2(Γ).
3
A subfamily Γ that satisfies the two conditions of Beurling’s Criterion for ρ∗ is called a Beurlingsubfamily of Γ. A minimal subfamily of Γ is a subfamily Γ ⊂ Γ so that Mod2(Γ) = Mod2(Γ) and suchthat removing any path from Γ decreases the modulus, i.e. for any γ ∈ Γ we have Mod2(Γ \ γ) <Mod2(Γ). As shown in [4], Γ will always contain a minimal subfamily, although it may not be unique,and any minimal subfamily is also a Beurling subfamily for Γ.
The following is the main property of a minimal subfamily that we will be using repeatedly.Although it can be deduced from [4, Theorem 3.5 and Section 5], it is not explicitly formulated there,and so we summarize the result here, for clarity.
Proposition 6 ([4]). Let G = (V,E, σ) be a finite weighted graph, and let Γ be a finite (non-empty)family of paths in G. If Γ ⊂ Γ is a minimal subfamily for Γ, then there is a probability mass function(or pmf) µ∗ supported on Γ that is associated with the extremal density ρ∗(e) for Mod2(Γ) as follows
σ(e)ρ∗(e)
Mod2(Γ)= Eµ∗
[N (γ, e)
]=∑γ: e∈γ
µ∗(γ). (1)
We call µ∗ an optimal pmf because by convex duality it can be shown to satisfy an optimizationproblem. For instance, when σ ≡ 1, this problem is especially nice and µ∗ is a pmf that minimizesthe expected overlap between two independent randomly-chosen paths:
Eµ∗ |γ ∩ γ′| = minµ
Eµ|γ ∩ γ′| =1
Mod2(Γ),
where the minimum is taken over all possible pmf’s µ on Γ.
Proof of Proposition 6. Theorem 3.5 (ii) of [4], shows that if Γ is a minimal subfamily of Γ, then thereis a unique set of Lagrange multipliers λ∗ that are optimal for the dual problem to modulus, so thatΓ = γ ∈ Γ : λ∗(γ) > 0. Then, in Section 5 of [4], the probabilistic interpretation of modulus is givenin [4, Theorem 5.1], showing that
Mod2(Γ) =
(minµ∈P(Γ)
µTNN Tµ
)−1
(2)
where P(Γ) is the family of all pmf’s supported on Γ. As seen in the proof of [4, Theorem 5.1],the minimization in (2) is derived from the dual problem to modulus by normalizing the Lagrangianmultipliers λ so as to have total sum 1Tλ equal to 1. Therefore, the optimal pmf µ∗ for Γ is obtainedby dividing λ∗ by 1Tλ∗.
2.2 Discrete harmonic functions
For a network G = (V,E, σ), when v, w ∈ V are adjacent, we will use the notation vw to refer to theedge incident to v and w. A function f : V → R is discrete harmonic at v ∈ V if
f(v)∑
w: vw∈Eσ(vw) =
∑w: vw∈E
f(w)σ(vw),
or in other words, the value of f at v is a weighted average of the values of f at the neighbors of v.The existence and uniqueness of discrete harmonic functions with specified boundary values is well
known; see for instance Section 2.1 of [10].
Proposition 7. (Existence and uniqueness of discrete harmonic functions) Let G = (V,E, σ) be afinite connected network. For U ⊂ V and g : V \ U → R, there is a unique function h : V → R suchthat h = g on V \ U and h is discrete harmonic on U .
4
The function h given by Proposition 7 is called the solution to the discrete Dirichlet problem on Uwith boundary data g. Viewed from the electric network perspective, h is called the voltage function.From its discrete gradient, we obtain the current flow f = σdh given by
f(vw) = σ(vw) [h(w)− h(v)] for all vw ∈ #»
E.
In the above, we use#»
E for the set of directed edges, which contains both possible orientations for allthe edges of E.
More generally, a flow on U ⊂ V is a function θ on#»
E so that each v ∈ U satisfies Kirchhoff’s nodelaw, i.e. ∑
w: vw∈ #»E
θ(vw) = 0.
The energy of a function θ :#»
E → R is given (with a slight abuse of notation) by
E(θ) =∑e∈E
1
σ(e)θ2(e).
The energy of h : V → R is defined to be the energy of its discrete gradient, i.e.
E(h) = E(σdh) =∑vw∈E
σ(vw) [h(w)− h(v)]2 .
We are now able to explain the close connection between the discrete 2-modulus and discreteharmonic functions. Suppose G is network with disjoint nonempty sets A,B ⊂ V . Let Γ be the familyof paths in G from A to B, and let h be the solution to the discrete Dirichlet problem on V \ (A∪B)with h|A = 0 and h|B = 1. Then,
Mod2(Γ) = E(h),
and the extremal density is related to the solution of the Dirichlet problem as follows
ρ∗(vw) = |h(w)− h(v)|. (3)
See Theorem 4.2 in [2].
2.3 Examples
We wish to illustrate the concepts from the previous sections and our main results by taking a lookat two simple examples.
Example 8. For the first example, let R be the rectangle [0, L] × [0, 1], where L ∈ N. Let G be thesquare grid in R with lattice size 1/n and weights σ ≡ 1. Let Γ be the family of all paths in G fromA = v ∈ V : Re(v) = 0 to B = v ∈ V : Re(v) = L.
For this example, we claim the following:
1. The extremal density is ρ∗(e) =
1Ln if e is a horizontal edge
0 if e is a vertical edge.
2. The subfamily of horizontal paths Γ is a Beurling subfamily.
3. Mod2(Γ) = Mod2(Γ) = 1L
(1 + 1
n
).
4. Γ is the unique minimal subfamily.
5. The optimal pmf µ∗ is unique and is uniform on Γ.
5
We will use Beurling’s criterion to establish the first two claims. Note that each horizontal pathγ ∈ Γ has Ln horizontal edges, and so `ρ∗(γ) = 1. To establish the second condition of Beurling’scriterion, assume that h : E → R satisfies
∑e∈γ h(e) ≥ 0 for all γ ∈ Γ. Then, since ρ∗(e) = 0 when e
is vertical, ∑e∈E
h(e)ρ∗(e) =1
Ln
∑e horizontal
h(e) =1
Ln
∑γ∈Γ
∑e∈γ
h(e) ≥ 0.
Thus Beurling’s criterion (Theorem 5) shows that ρ∗ is the extremal density and Mod2(Γ) = Mod2(Γ).We compute
Mod2(Γ) =∑e∈E
ρ∗(e)2 =
(1
Ln
)2
(# of horizonal edges) =1
L
(1 +
1
n
),
since there are Ln(n+ 1) horizontal edges. Notice that as n→∞, the discrete modulus converges to1L , which is the corresponding continuous modulus.
An alternate approach to finding ρ∗ would be to check that h(v) = 1LRe(v) is discrete harmonic on
V \ (A ∪ B). Thus h is the solution to the discrete Dirichlet problem with boundary values h|A = 0and h|B = 1, and we could obtain ρ∗ from (3).
We observe that Γ is a minimal subfamily, since removing any curve γ ∈ Γ will decrease themodulus. To see this, change ρ∗ to be zero on the removed path γ and use Beurling’s criterion toshow that the new density is extremal for Γ \ γ. To show that Γ is the unique minimal subfamily,assume to the contrary that Γ is another minimal subfamily. Then the optimal pmf µ on Γ satisfiesthat the expected edge usage of e under µ is given by ρ∗(e)/Mod2(Γ) by (1). This implies that theexpected edge usage for any vertical edge must be zero, and the paths in Γ cannot use any verticaledges. Therefore Γ ⊂ Γ. However, since Mod2(Γ) = Mod2(Γ), we must have Γ = Γ.
Let µ∗ be the optimal pmf on Γ. From (1), we see that if e is a horizontal edge and γ is thecorresponding horizontal path, then
µ∗(γ) = Eµ∗ [N (γ, e)] =ρ∗(e)
Mod2(Γ)=
1
n+ 1.
This implies that µ∗ is uniform on Γ.We wish to vary n, and so we update our notation to show the dependence on n, i.e. let µ∗n = µ∗
which is supported on Γn = Γ. For this example, the result of Theorem 2 is fairly intuitive: as n→∞,µ∗n converges to the uniform measure on all horizontal curves in R, which are the extremal curves inthe continuous setting.
Example 9. In our second example, we will look at a rotated rectangle. Let R be the rectangle[0,√
2L]× [0,√
2] rotated by π/4 (i.e. the corners of R are 0, L+ iL, L−1+ i(L+1),−1+ i.) In R, takethe (non-rotated) square grid with lattice size 1/n and weights σ ≡ 1. Let Γ be the family of all pathsin R from the left side A = v : Im(v) = −Re(v) to the right side B = v : Im(v) = −Re(v) + 2L.See Figure 1.
In contrast to Example 8, there is not a unique minimal subfamily in this case, and we will describetwo such subfamilies. Consider first the subfamily Γz of “zig-zag” paths, which alternate betweenincreasing the real component and increasing the imaginary component (or vice-versa). These pathseach have 2Ln steps, and each edge is used in exactly 1 path. By arguing as before, Beurling’s criterioncan be used to show that the extremal density is 1
2Ln on all edges and Γz is a Beurling subfamily.Further we see that Γz is also a minimal subfamily, since removing any path will decrease the modulus.From the extremal density and the fact that there are 4n2L total edges, we see that Mod2(Γ) = 1
L ,which is equal to the corresponding continuous modulus.
Another minimal subfamily Γr consists of the “reflected” paths which keep increasing the samecomponent (whether real or imaginary) until they hit the boundary of R and then they switch toincreasing the other component. The paths in this family will also have 2Ln steps each, and each edge
6
Figure 1: The graph for Example 9 with L = 2, n = 3, the vertices of A shown in red, and thevertices of B shown in blue.
is used in exactly 1 path. One can check that this will also be a Beurling subfamily and a minimalsubfamily.
Each subfamily will have an associated optimal pmf: let µzn be the pmf on the zig-zag subfamilyΓz, and let µrn be the pmf on the reflected subfamily Γr. Both measures will be uniform, giving the2n paths in the relevant subfamily weight 1
2n . Further both µzn and µrn will have a limit as n → ∞,but the limiting objects will be different. The limit for µzn can be thought of as the uniform measureon the straight line paths of slope 1 in R (which are the extremal curves in the continuous setting).However, the limit for µrn will be supported on paths that move in a straight line either horizontallyor vertically until reflecting off the boundary of the rectangle.
To illustrate the point that not all Beurling subfamilies are minimal, we wish to examine onefurther subfamily. Let
Γ = γ ∈ Γ : γ has 2Ln steps.
In other words, this is the collection of all curves that have length 1 under the extremal density. Thisfamily contains both Γz and Γr. In a moment, we will show that this is a Beurling subfamily, andhence it will be the largest Beurling subfamily by construction.
Let µ be the pmf that arises from viewing the paths in Γ as 1-dim simple random walk (SRW)trajectories started on the left side of R and reflected to stay in the rectangle. In other words, we canthink of creating the paths dynamically as follows: Each of the 2n initial edges has equal probability1
2n . Once the first edge is chosen, we chose the next edge with probability 1/2 of going up or going tothe right. Then we repeat. If we happen to be on the boundary, then we take the only available edgewith probability 1. One can show for any fixed edge e that Pµ[e ∈ γ] = 1
2n , see Claim 10 below.
With this understanding of µ, we will be able to show that Γ is a Beurling subfamily by verifyingBeurling’s Criterion. Let h ∈ RE with
∑e∈γ h(e) ≥ 0 for all γ ∈ Γ. Then
∑e∈E
h(e) · 1
2nL=∑e∈E
h(e) · 1
L
∑γ∈Γ
µ(γ)1e∈γ =1
L
∑γ∈Γ
µ(γ)∑e∈γ
h(e) ≥ 0.
A short computation will show that µ is a pmf on Γ that minimizes the expected overlap of two
7
independent randomly chosen paths in Γ:
Eµ|γ ∩ γ′| = Eµ
[∑e∈E
1e∈γ∩γ′
]=∑e∈E
Pµ[e ∈ γ ∩ γ′
]=
(1
2n
)2
· 4n2L
= L.
SinceL = Mod2(Γ)−1 = min
µ∈P(Γ)Eµ|γ ∩ γ′|,
µ is a pmf on Γ that minimizes expected overlap. Hence, µ is an optimal pmf on Γ. Note that it maynot be unique, since we are only guaranteed a unique optimal pmf for a minimal subfamily, a factthat follows from [4, Theorem 3.5(ii)].
To highlight the fact that the limit in Theorem 2 depends on the correct choice of subfamily, wemention that the limit of the pmf µ as n → ∞ yields 1-dim reflected Brownian motion trajectoriesstarted at a uniform point on the left side of R.
Claim 10. Pµ[e ∈ γ] = 12n for any edge e.
Proof. This claim can be shown by induction. In particular, we can subdivide the edges into 2nLdistinct sets of 2n edges
Ek = all the edges that could be step k in a SRW trajectory ,
and we do induction on k to show that Pµ[e ∈ γ] is the same for all edges e in Ek. The base casefollows by definition, since each of the initial edges has equal probability. We will show the basicidea behind the induction proof by explaining the reasoning for E2 and E3. Consider the vertices thatedges from E1 lead into (we consider the edges oriented so that the SRW trajectories flow from the leftside to the right side of R). Each of these vertices has exactly two edges leading into it, meaning thatour SRW induces a uniform distribution on these vertices. Since each of these vertices has two edgesleading out, and we will choose between them with equal probability, we have uniform probability forall edges in E2. Now consider the vertices that the edges from E2 lead into. There are two specialvertices, one on the top boundary and one on the bottom boundary, that only have one edge leadinginto them and one edge leading out. All the rest have two edges leading in and two edges leading out.While we no longer have a uniform distribution induced on these vertices, we still obtain the uniformdistribution on E3. (Reason: We have the same chance of arriving at any internal vertex, and wehave exactly half this probability of arriving at a boundary vertex. From the internal vertex, we havetwo edges to choose from with equal probability, while at the boundary vertex we must take the onlyavailable edge.)
Note that the paths in Γz are non-crossing, while in Γr and Γ they do cross. In Section 3 we willgive an algorithm that, in the case of plane graphs, always produces the unique non-crossing minimalsubfamily.
2.4 Orthodiagonal maps
A finite orthodiagonal map is a finite connected plane graph that satisfies the following:
• Each edge is a straight line segment.
• Each inner face is a quadrilateral with orthogonal diagonals.
• The boundary of the outer face is a simple closed curve.
8
See the top left image in Figure 2 for an illustration.We will briefly describe some of the structure associated with an orthodiagonal map G = (V,E).
Since G is a bipartite graph, we obtain a bipartition of the vertices V = V • t V . Each inner faceQ is determined by the four vertices on its boundary, which must alternate between V • and V . Wewill write Q = [v1, w1, v2, w2] with the convention that v1, v2 ∈ V •, w1, w2 ∈ V , and the verticesare ordered counterclockwise around Q. We can create two additional graphs associated with G, theprimal graph G• = (V •, E•) and the dual graph G = (V , E). There is one primal edge and one dualedge for each inner face of G. In particular, e•Q ∈ E• is an edge between v1 and v2 in Q; it is definedto be the line segment v1v2, if this segment is contained in Q, and otherwise it is the union of the linesegments v1p and pv2, where p = (w1 + w2)/2 is the midpoint of w1w2. The edge eQ ∈ E betweenw1 and w2 in Q is defined similarly. For an edge e ∈ E•, the dual edge to e is the unique edge e ∈ Eso that there is an inner face Q with e = e•Q and e = eQ. The primal network (G•, σ) and the dualnetwork (G, σ) have the following edge weights:
σ(e•Q) =|w1w2||v1v2|
and σ(eQ) =|v1v2||w1w2|
,
where the notation |ab| is the Euclidian distance between a and b. Note that this distance is notnecessarily the length of the edge.
One way to interpret these weights is through a finite-volume approximation of the Laplace equa-tion. Let v1 be an interior vertex in the primal graph, and consider the collection of quadrilateralscontaining this vertex. The bottom right picture in Figure 2 shows an example of such a collectionof quadrilaterals (light edges in the figure). The edges of the dual graph surround a polygononalregion, R (thick dashed lines in the figure). If h is the potential for the continuum problem, then thedivergence theorem implies that
0 =
∫R
∆h dx =
∫∂R
∂h
∂ndS.
The integral on the right is a sum of integrals along the edges of R. Let w1w2 be some such edge,as indicated in the figure. There is a corresponding edge v1v2 in the primal graph. Since these edgesare orthogonal, (h(v2)−h(v1))/|v1v2| provides an approximation of the normal derivative along w1w2.Thus, the contribution of w1w2 to the boundary integral can be approximated as∫
w1w2
∂h
∂ndS ≈ |w1w2|
|v1v2|(h(v2)− h(v1)).
Performing a similar approximation on the remaining edges of R yields the approximation
0 ≈∑
u:v1u∈E
|w1w2||v1u|
(h(u)− h(v1)).
Thus, when the continuous potential is restricted to the graph, it approximately satisfies Kirchhoff’scurrent law on each interior vertex, with the edge conductances defined above. Moreover, this approx-imation improves as the lengths of the graph edges converge to zero.
We list some further notation relating to an orthodiagonal map G:
∂G = the topological boundary of the outer face of G
Int(G) = the open subset enclosed by ∂G
G = Int(G) ∪ ∂GG = the union of the convex hulls of the closures of the inner faces of G
In [8], Gurel-Gurevich, Jerison, and Nachmias show that one can approximate continuous harmonicfunctions by discrete harmonic functions on an orthodiagonal map, generalizing the work of [15] and[17].
9
S1
T1
S2
T2
Figure 2: Top left: An example of an orthodiagonal map with the bipartation of the vertices.Top right: The topological boundary of the map decomposed into four paths: S1, S2 (solid) andT1, T2 (dashed). Bottom left: The primal graph G• (black/red vertices and solid edges) with∂V • shown in red and the dual graph G (white vertices and dashed edges). Bottom right:Finite volume interpretation of the edge weights assigned to an orthodiagonal map.
Theorem 11 ([8]). Let Ω ⊂ C be a bounded simply connected domain, and let g : C → R be a C2
function. Given ε, δ ∈ (0,diam(Ω)), let G = (V • t V , E) be a finite orthodiagonal map with maximaledge length at most ε such that the Hausdorff distance between ∂G and ∂Ω is at most δ. Let hc : Ω→ Rbe the solution to the continuous Dirichlet problem on Ω with boundary data g|∂Ω, and let hd : V • → Rbe the solution to the discrete Dirichlet problem on V • \ ∂G with boundary data g|∂G∩V •. Set
C1 = supz∈Ω
|∇g(z)| and C2 = supz∈Ω
||Hg(z)||2,
where Ω = conv(Ω ∪ G). Then there is a universal constant C <∞ such that for all z ∈ V • ∩ Ω,
|hd(z)− hc(z)| ≤C diam(C1 + C2ε)
log1/2(diam(Ω)/(δ ∨ ε).
In the above result, the boundary condition involves the vertices of the primal graph that are in thetopological boundary of G. In our work, we need a similar result with different boundary conditions.This leads us to the following definition.
Definition 12. The graph G = (V • t V , E) is a finite orthodiagonal map with boundary arcsS1, T1, S2, T2, if G is a finite orthodiagonal map with the following boundary condition:
• (Boundary condition) The topological boundary of G can be subdivided into four nonemptypaths S1, T1, S2, T2, listed counterclockwise, with Sj ∩ Tk equal to a single vertex, for j, k = 1, 2.
This boundary condition allows us to think of G as a conformal rectangle with “sides” S1, S2 and“top/bottom” T1, T2. For the associated primal and dual graphs, we define the (graph) boundaries asfollows:
∂V • := V • ∩ (S1 ∪ S2) and ∂V := V ∩ (T1 ∪ T2) .
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Note that ∂V • and ∂V do not include all vertices in the topological boundary of G. Further, wedefine
Int(V •) := V • \ ∂V • and Int(V ) := V \ ∂V ,
and we point out that these are different from the collections of vertices in the topological interior ofG• and G. See Figure 2.
Definition 13. Let Ω be a simply connected Jordan domain in the plane such that its boundary canbe decomposed into four analytic arcs A, τ1, B, τ2, listed in counter-clockwise order, with the propertythat the arcs do not form cusps at the points where they connect. We will call Ω a smooth quadrilateral.
Definition 14. We wish to approximate a smooth quadrilateral Ω with a finite orthodiagonal mapG with boundary arcs S1, T1, S2, T2. We say that G approximates Ω within distance δ when Ω ⊂ Gand distG(S1, A), distG(S2, B), and distG(Tk, τk) are bounded by δ.
The notation distG(x, y) means the minimum length of all paths between x and y that remain in G.So the notation distG(S1, A) ≤ δ means that for all x ∈ S1, there is a point y ∈ A with distG(x, y) ≤ δand vice versa.
2.5 Linearly connected and John domains
A simply connected domain Ω is linearly connected if there exists K > 0 so that any two pointsz, w ∈ Ω can be connected by a curve γ ⊂ Ω with diam(γ) ≤ K|z−w|. On the other hand, a boundedsimply connected domain Ω is a John domain if there exists J > 0 so that every crosscut κ of Ωsatisfies diam(H) ≤ J diam(κ) for one of the two components H of Ω \ κ. The inner domain of apiecewise smooth Jordan curve is linearly connected if and only if it has no inward-pointing cusps,and it is John domain if and only if there are no outward pointing cusps. (See Chapter 5 of [11].)Thus a smooth quadrilateral Ω, as defined in the previous subsection, is a linearly connected Johndomain. This geometric control gives us the Holder continuity of some relevant functions.
Proposition 15. Let Ω be a smooth quadrilateral with boundary arcs A, τ1, B, τ2 Let ψ be a conformalmap of Ω onto the rectangle R = [0, 1] × [0,m] taking the arcs A and B to the vertical sides of therectangle, and let hc = Re ψ. Then there exists C > 0 so that
|hc(z)− hc(w)| ≤ C|z − w|1/4 for all z, w ∈ Ω.
Further, there exists p ∈ (0, 1/2] and C > 0 so that
|ψ−1(z)− ψ−1(w)| ≤ C|z − w|p for all z, w ∈ R.
Proof. Let FΩ : D→ Ω and FR : D→ R be conformal maps with ψ = FR F−1Ω . The conformal map
FR is a Holder continuous with exponent 1/2. Since Ω is a linearly connected domain, Theorem 5.7 of[11] implies that FΩ satisfies a lower Holder condition with exponent b < 2 on ∂D, which is equivalentto a Holder condition on F−1
Ω with exponent 1/b > 1/2 on ∂Ω. Hence,
|hc(x)− hc(y)| ≤ C|x− y|1/4
on ∂Ω. Then Theorem 1 in [9] proves the first result.For the second statement, we use the same argument applied to ψ−1 = FΩ F−1
R . Recall that everyJohn domain is a Holder domain, which implies that FΩ is Holder continuous.
3 Non-crossing minimal subfamilies
In this section we assume that the finite network G = (V,E, σ) is a plane graph (i.e. G is embeddedin the plane with no edges crossing.) Further, we assume that there are nonempty sets of boundary
11
vertices A and B defined as follows: A = S1 ∩ V and B = S2 ∩ V , where S1, S2 are two disjoint arcsfrom a simple curve C in the closed outer face of G.
Let F be the conformal map from the interior of C to a rectangle [0, L] × [0, 1] that takes S1 tothe left side and S2 to the right side. We use F to obtain an orientation for G. Given a path γ fromA to B, the slit vertical strip z : Re(z) ∈ [0, L] \ F (γ) has two connected components. Let Uγ bethe connected component containing −i (i.e. the component “under” F (γ)). Then for a subgraph Hof G and a path γ in H from A to B, we say that γ is the top path in H from A to B if F (H) ⊂ Uγ .We say that two paths from A to B are non-crossing when one path is the top path in the subgraphinduced by these two paths. A family of paths is non-crossing when every pair of paths in the familyis non-crossing. We note that two non-crossing paths are allowed to share vertices and edges. Alsowe say that an edge e is below a path γ if γ is the top path of the graph induced by e ∪ γ.
Remark 16. Two paths γ1 and γ2 are crossing if one can find e1 ∈ γ1 \ γ2 and e2 ∈ γ2 \ γ1, so thate1 is below γ2 and e2 is below γ1.
3.1 Non-crossing minimal subfamily algorithm
To establish the existence and uniqueness of the non-crossing minimal subfamily in Proposition 1,we will use the following algorithm that outputs the desired minimal subfamily given the extremaldensity ρ∗ for the family of paths from A to B.
Non-crossing Minimal Subfamily Algorithm.Input: planar network G = (V,E, σ), sets of boundary vertices A and B as above, and the extremaldensity ρ∗ for the family of curves from A to B in G.
Initialization: set r(e) = σ(e)ρ∗(e) for each edge e ∈ E.
Iteration:
1. Remove any edges e for which r(e) = 0 from G (call such edges zero edges).
2. If there is one, add the top path γ in G to a growing family Γ; if not, terminate.
3. For the top path γ identified in (2), set m(γ) = mine∈γ r(e).
4. For each edge e in the top path γ identified in (2), update r(e) to be r(e)−m(γ).
5. Repeat.
Output: curve family Γ and the weights m : Γ→ (0,∞).
In our proof of Proposition 1, we will show that this algorithm outputs a minimal subfamily Γof non-crossing paths, such that its corresponding optimal pmf µ∗, guaranteed by Proposition 6, isproportional to m. Before getting to the proof, we show that we can direct the non-zero edges of Gto obtain a directed acyclic graph (once we remove any edges with ρ∗(e) = 0).
Lemma 17. Let ρ∗ be the unique extremal density for the family of paths from A to B in G, and letE0 = e ∈ E : ρ∗(e) = 0. Then there exists an orientation of all edges in E \E0 so that the followinghold:
(i) At any vertex v not in A ∪B, the sum of σ(e)ρ∗(e) over the edges leading into v equals the sumof σ(e)ρ∗(e) over the edges leading out of v.
(ii) All edges in E \ E0 with exactly one vertex in A (resp. B) are oriented away from A (resp.towards B).
(iii) Let γ be a path in in G \ E0 from A to B. Then γ is a directed path from A to B if and only ifγ has length 1, i.e. `ρ∗(γ) = 1.
(iv) G \ E0 is a directed acyclic graph.
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Proof. Let h be the solution to the discrete Dirichlet problem on V \(A∪B) with h|A = 0 and h|B = 1,and recall that ρ∗(vw) = |h(w) − h(v)|. For each edge e incident to vertices v, w with ρ∗(e) 6= 0, weorient e from v to w when h(w)− h(v) > 0.
Let f be the corresponding current flow: f(vw) = σ(vw) [h(w)− h(v)] for vw ∈ #»
E. For v ∈V \ (A ∪B), let Ein
v be the edges directed into v and Eoutv be the edges directed out of v. Then since
f is a flow,
0 =∑
w: vw∈Ef(vw) =
∑e∈Eout
v
σ(e)ρ∗(e)−∑e∈Ein
v
σ(e)ρ∗(e),
which proves (i).Statement (ii) follows from the discrete maximum principle, which implies that 0 ≤ h ≤ 1. To
show (iii), let γ be a path from A to B in G \ E0 given by vertices v0, v1, · · · , vk. Then
`ρ∗(γ) =
k∑j=1
∣∣h(vj)− h(vj−1)∣∣ ≥ k∑
j=1
h(vj)− h(vj−1) = 1.
There is equality in the above equation if and only if γ is a directed path from A to B.To show that G \E0 has no cycles (respecting the edge directions), we will assume for the sake of
contradiction that there is a cycle of directed edges e1, e2, · · · , ek with each ej directed from vj−1 tovj . Then
0 <
k∑j=1
ρ∗(ej) =
k∑j=1
h(vj)− h(vj−1) = h(vk)− h(v0) = 0,
since vk = v0. This is a contradiction.
Proof of Proposition 1. We begin by analyzing the Non-crossing Minimal Subfamily Algorithm. Notethat this algorithm must terminate, because we remove at least one edge from the graph during everyiteration (except possibly the first one). We will show that when the algorithm terminates, then thereare no edges remaining in the graph.
We claim that after each time we complete step (1), the only possible connected components ofthe resulting graph are singleton vertices or connected subgraphs that contain vertices from both Aand B. The first time we complete step (1), we remove all edges with ρ∗(e) = 0. Let H be one of theresulting connected components with at least one non-zero edge. Since H is a directed acyclic graphby Lemma 17, H must contain at least one source vertex and at least one sink vertex. Properties (i)and (ii) of Lemma 17 imply that the source vertex must be in A and the sink vertex must be in B.Hence H must contain vertices from both A and B.
Now let’s suppose that after some later iteration of step (1) we obtain a connected component Hwith at least one edge but without vertices from both A and B. For the sake of simplicity, let’s assumethat H does not contain a vertex from A (as a similar argument applies in the other case.) Since H isa directed acyclic graph, it must contain at least one source vertex v and v 6∈ A. Let γ1, · · · , γn be allthe paths in the current Γ that pass through v. Then since all of the edges directed towards v havebeen removed at this step,
n∑i=1
m(γi) =∑e∈Ein
v
σ(e)ρ∗(e) =∑
e∈Eoutv
σ(e)ρ∗(e),
where the last equality is by Lemma 17 (i). However, this means that we have also removed all theedges directed out from v, contradicting the fact that H is is a connected component larger than asingleton vertex.
Next we wish to apply Beurling’s Criterion (Theorem 5) to show that the algorithm’s output pathfamily Γ is a Beurling subfamily. Our first step is to show that the paths in Γ all have length 1. Thiswill follow from Lemma 17 (iii) once we show that each output path is a directed path from A to B.Let γ ∈ Γ be obtained during the kth iteration of the algorithm. Let H be the connected component(of the graph obtained after step (1) of the kth iteration) that contains γ. Then H is a directed acyclic
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graph and γ is the top path of H. Further, from the argument above, we know that all source verticesof H are in A and all sink vertices are in B. Suppose γ is not a directed path from A to B, and let ebe an edge in γ that is oriented “to the left”. Since all the sources (resp. sinks) are in A (resp. B),there must be some directed path in H from A to B that contains e. However, the planarity of H andthe fact that γ is the top path imply that H must contain a directed cycle. This contradicts Lemma17 (iv). Therefore, γ is a directed path from A to B and hence has length 1.
Now we will check the second condition for Γ to be a Beurling subfamily. Let h ∈ RE with∑e∈γ
h(e) ≥ 0
for all γ ∈ Γ. We apply this assumption and the fact that
σ(e)ρ∗(e) =∑γ∈Γ
m(γ)1e∈γ,
to obtain the needed condition:∑e∈E
h(e)ρ∗(e)σ(e) =∑e∈E
h(e)∑γ∈Γ
m(γ)1e∈γ
=∑γ∈Γ
m(γ)∑e∈γ
h(e)
≥ 0.
Thus Γ is a Beurling subfamily, which implies that Mod2(Γ) is the same as the modulus of the familyof all curves from A to B in G.
We now know the existence of a minimal subfamily of non-crossing curves, since Γ must containsuch a subfamily. However, we wish to show that Γ is in fact a minimal subfamily and the only suchone with non-crossing paths. Let Γ be any minimal subfamily with non-crossing paths, and let µ bethe optimal pmf supported on Γ guaranteed by Proposition 6. We will show that Γ ⊂ Γ, which willthen imply that Γ = Γ.
Let γ1, γ2, · · · , γn be the paths of Γ in the order that they are identified in the algorithm. We willbegin by proving that γ1 must be in Γ. To show this by contradiction, assume γ1 is not in Γ. All theedges of γ1 must be contained in paths of Γ by (1) since ρ∗ is nonzero on these edges, by constructionduring the algorithm. Let γ be a path from Γ that contains the longest initial segment in commonwith γ1. Let e0 be the edge in γ1 immediately after this initial segment (such an edges must existsince γ 6= γ1). Further γ must contain an edge below γ1 after the initial segment. Now there must besome other path in Γ that contains e0, and by the choice of γ, this path must have some edge belowthe initial segment of γ. By Remark 16, this implies that Γ contains two paths that cross, which is notpossible. Hence γ1 ∈ Γ. The non-crossing condition further implies that there must be some edge e inγ1 that is contained in no other path in Γ (by the argument above). Recall that σ(e)ρ∗(e)/Mod2(Γ)gives the expected edge usage of e with respect to the optimal pmf µ. Therefore,
σ(e)ρ∗(e) = mina∈γ1
σ(a)ρ∗(a) = m(γ1), (4)
because other edges a ∈ γ1 may be used in multiple paths from Γ. We can deduce that µ(γ1) =m(γ1)/Mod2(Γ).
For our inductive step, assume that γ1, · · · , γk−1 are in Γ with µ(γj) =m(γj)
Mod2(Γ) for j = 1, · · · , k−1.Assume we are at the point in the algorithm where we have just identified γk as the top path, and lete ∈ γk. Then since e was not removed yet,
r(e) = σ(e)ρ∗(e)−∑
j<k: e∈γj
m(γj) > 0.
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Therefore ∑j<k: e∈γj
µ(γj) =1
Mod2(Γ)
∑j<k: e∈γj
m(γj) <σ(e)ρ∗(e)
Mod2(Γ)=∑γ: e∈γ
µ(γ),
where the last equality is from (1). This means that there must be a path in Γ \ γ1, · · · , γk−1that contains e, since we have not accumulated enough mass from the paths γ1, · · · , γk−1. The samecontradiction proof as with γ1 above shows that γk ∈ Γ. Further, as with γ1, the non-crossing conditionimplies that there must be some edge in γk that is contained in no other path in Γ \ γ1, · · · , γk−1.Such edges e ∈ γk will satisfy
r(e) = mina∈γk
r(a),
and therefore
µ(γk) =m(γk)
Mod2(Γ). (5)
This completes the induction proof and establishes that Γ = Γ and (5) holds for all γk ∈ Γ.
3.2 Scaling limit of non-crossing minimal subfamilies
In this section we will prove Theorem 2 after detailing the set-up. Let Ω be a smooth quadrilateral,as defined in Definition 13, with boundary arcs A, τ1, B, τ2 listed counterclockwise. Let εn > 0 satisfyεn → 0 as n → 0. Let Gn be an othordiagonal map with boundary arcs An, αn, Bn, βn that approx-imates Ω within distance εn, in the sense of Definition 14, and has edge length bounded by εn. Letσn be the weights on the edges of G•n. Consider the family of all paths in G•n from An to Bn, and letΓn be the unique minimal subfamily with non-crossing paths. Recall that there is a unique extremaldensity ρn on the edges of G•n and a unique optimal pmf µn on Γn, which are related by
σn(e)ρn(e)
Mod2(Γn)= Eµn [N (γ, e)]. (6)
We will prove that µn converges to the uniform probability measure on the family of classical extremalcurves in Ω as εn → 0.
We begin with a lemma about a useful dual minimal subfamily. Since this lemma only relates tothe orthodiagonal map (and not to the underlying domain), for ease of reading, we omit the subscriptn.
Lemma 18. Consider the family of all paths in G from α to β, and let Γ be the unique minimalsubfamily with non-crossing paths provided by Proposition 1. We claim that
Mod2(Γ) ·Mod2(Γ) = 1. (7)
Further the unique extremal density η for Γ satisfies
η(e) =σ(e)ρ(e)
Mod2(Γ)= Eµ[N (γ, e)] =
∑γ∈Γ: e∈γ
µ(γ), (8)
where e ∈ E is the dual edge to e ∈ E•.
Proof. From Fulkerson duality, see [3, Section 4.2], we have that
Mod2(Γ) ·Mod2(Γ) = 1, (9)
where Γ is the family of all AB-cuts. Recall that S ⊂ V • is an AB-cut if A ⊂ S and B ∩ S = ∅.Moreover, the usage of S is defined by considering the boundary ∂S = e = xy ∈ E• : |S∩x, y| = 1.Then, the usage is set to N (S, e) = 1e∈∂S. Note that we can define a bijective map r : E• → E bysetting r(e) := e which is the dual edge to e. Therefore, applying the transformation r to each edgein ∂S, we obtain an object r(S) on the dual graph G. We want to show that r(S) contains a path
15
connecting α and β and conversely every such path comes from a cut S. First, every node in V • \ ∂Gis contained in the interior of a closed dual face for G. The only problem arises when the node v isin ∂G. In this case, let C be the closure of the component of v in the intersection of the unboundedface for G with G. Finally, consider a small ball Dv centered at v that is disjoint from any othervertex and define C ∪Dv to be the dual face containing v. By taking the union of all the dual facescontaining a node in S and then taking the interior of this set, we obtain an open set U containing S(and hence A) in its interior, such that B ⊂ Int(U c).
Now we can apply Lemma 2.4 in [7], where the quadrilateral is our set G, and obtain a simple curveγ connecting α and β with γ ⊂ ∂U ∩ G. By construction we see that γ is a union of dual edges, andhence is a simple path in G. Conversely, every such path can be continued to get a Jordan domainwith A in its interior and B in its exterior. Therefore, Mod2(Γ) = Mod2(Γ) and (7) follows from(9). Moreover, (8) follows from (1) and [3, Theorem 4 and Theorem 7], by letting η• be the extremaldensity for Γ and setting η(e) = η•(r(e)) (thinking of r as an idempotent operation).
Proof of Theorem 2. Let φ be the conformal map from Ω onto the rectangle R = (0, L) × (0, 1) sothat (after extending φ to the boundary) φ(A) is the left side of R and φ(B) is the right side of R.
Let hn be the solution to the discrete Dirichlet problem on Int(V •n ) with hn|An = 0 and hn|Bn = L.Similarly let hn be the solution to the discrete Dirichlet problem on Int(V n ) with hn|αn = 0 andhn|βn = 1. Note that
ρn(v1v2) =1
L|hn(v1)− hn(v2)| and ηn(w1w2) = |hn(w1)− hn(w2)|.
By Theorem 3, we know that hn approximates Re φ and hn approximates Im φ.The non-crossing condition for Γn allows us to order the paths in Γn: γ < γ′ means γ 6= γ′ and γ′
is the top path in the subgraph consisting of these two paths. Equivalently, γ < γ′ means that γ isidentified after γ′ in the Non-crossing Minimal Subfamily Algorithm.
Fix γ′ ∈ Γn and set
p =∑γ≤γ′
µn(γ).
Let e be an edge in γ′, with dual edge e. Now consider a curve γ ∈ Γn that contains e. First wewill show that every path γ ∈ Γn contains exactly one edge a with a ∈ γ. Note that each γ ∈ Γncontains at least one such edge, since we can obtain an AB-cut from γ. By (8), since γ has length1 under ηn,
1 =∑
a:a∈γηn(a) =
∑a:a∈γ
∑γ∈Γn:a∈γ
µn(γ)
=∑
γ∈Γn:a∈γµn(γ) [ # of edges a ∈ γ with a ∈ γ] ≥ 1.
Therefore, the number of edges a ∈ γ with a ∈ γ must be identically equal to one.Under the orientation of edges induced by hn in Lemma 17, γ is an oriented path from αn to βn
by Lemma 17 (iii). Let e = xy, oriented with respect to this orientation. Consider the portion of γ
before x and let this be given by vertices u0, · · · , uk = x. Then
hn(x) =
k∑j=1
hn(uj)− hn(uj−1) =
k∑j=1
ηn(ujuj−1) ≤∑γ≤γ′
µn(γ) = p.
Similarly,hn(y) ≥ p.
Further by Theorem 3 and Proposition 15,
|hn(y)− hn(x)| ≤ |hn(y)− Im φ(y)|+ |Im φ(y)− Im φ(x)|+ |Im φ(x)− hn(x)|
≤ C
log1/2 (diam(Ω)/εn)+ Cε1/4n (10)
≤ δn,
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where δn = Clog1/2(1/εn)
for some constant C depending on Ω. So
p ≤ hn(y) ≤ p+ δn and p− δn ≤ hn(x) ≤ p.
Now let u be a point on e. Then |u− y| ≤ 2εn and by arguing as in (10) we obtain
Im φ(u) ≤ hn(y) + |Im φ(y)− hn(y)|+ |Im φ(u)− Im φ(y)|≤ p+ 3δn.
Similarly,
Im φ(u) ≥ hn(x)− |Im φ(x)− hn(x)| − |Im φ(u)− Im φ(x)|≥ p− 3δn.
Since this hold for all edges of γ′, we have that for any point u on γ′
Im φ(u) ∈ [p− 3δn, p+ 3δn]. (11)
We are now ready to consider the deterministic limit points of Γn as n→∞, using the Hausdorffdistance as our metric on curves. We claim that these limiting curves are the classical extremal curvesin Ω:
Γ := φ−1(Im(z) = p) : p ∈ [0, 1].
This will be accomplished in two steps. We will first show that the limits points of Γn are containedin Γ and then we will show that every curve of Γ is a limit point of Γn.
Suppose that γ? is a limit point of Γn. Then there is a sequence γk ∈ Γnk with γk converging toγ?. By Proposition 15 this implies that φ(γk) must converge to φ(γ?). By (11), the oscillation of theimaginary part of φ(γk) must converge to zero, implying that Im[φ(γ?)] must be constant. This showsthat γ? ∈ Γ.
To show that every curve in Γ is a possible limit, we must choose q ∈ [0, 1] and find a γn ∈ Γn thatis close to φ−1(Im(z) = q). Choose the unique γn = γn(q) ∈ Γn so that∑
γ<γn
µn(γ) < q and q ≤∑γ≤γn
µn(γ) =: pn. (12)
Note that if γ′ ∈ Γn and e ∈ γ′, then by (10) we have
µn(γ′) ≤∑γ:e∈γ
µ(γ) = ηn(e) = |hn(y)− hn(x)| ≤ δn.
Therefore pn − q ≤ δn, and this together with (11) implies that
Im φ(γn) ∈ [q − 4δn, q + 4δn]. (13)
Thus φ(γn) converges to z ∈ R : Im(z) = q and so γn converges to φ−1(Im(z) = q).We are now ready to prove the convergence of the probability measures µn. The underlying
metric space is X = (⋃n Γn) ∪ Γ, endowed with the Hausdorff metric. We claim that the set of
probability measures µn is tight, i.e., for each ε > 0 there exists a compact subset Kε of X suchthat µn(Kε) > 1− ε for all n. Since X is compact by construction, µn satisfies this condition withKε = X. Therefore, Prohorov’s Theorem (see [12]) implies that µn is conditionally compact (i.e.its closure is compact) in the space of all probability measures on X equipped with the topology ofweak convergence. This implies the existence of subsequential limits of µn.
It remains to show that the only possible subsequential limit of µn is the uniform measure on Γ.To that end, let µ be a limit of µnk , let r ∈ (0, 1), and set
Hr = γ ∈ X : φ(γ) ⊂ [0, L]× [0, r].
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To characterize µ as the uniform measure on Γ, we must simply show that µ(Hr) = r for all r ∈ (0, 1).Since µ is the limit of µnk under weak convergence, we have that µnk(Hr)→ µ(Hr). Let q1 = r− 4δnand q2 = r + 4δn, and let γn(q1), γn(q2) be defined by (12). Then by (13), Hr contains the set
γ ∈ Γn : γ ≤ γn(q1),
and Hr ∩ Γn is contained in the set
γ ∈ Γn : γ ≤ γn(q2).
Thus for pn(q1) defined by (12),
µn(Hr) ≥ pn(q1) ≥ q1 = r − 4δn,
implying that µ(Hr) ≥ r. Similarly
µn(Hr) ≤ pn(q2) ≤ q2 + δn = r + 5δn,
implying that µ(Hr) ≤ r. Thus we have shown that µ(Hr) = r for all r ∈ (0, 1), which characterizesthe limit measure as the uniform measure on Γ.
3.3 Further consequences of the minimal non-crossing subfamily
As mentioned in the Introduction, Proposition 1 has an interpretation in terms of current flow. LetG = (V,E, σ) be a finite weighted plane graph. As before, assume that there are nonempty sets ofboundary vertices A and B defined as follows: A = S1 ∩ V and B = S2 ∩ V , where S1, S2 are twodisjoint arcs from a simple curve in the closed outer face of G. Solve the (weighted) Dirichlet problemand find a potential h : V → R, so that h|A = 0, h|B = 1, and h is discrete harmonic on V \ (A ∪B).The voltage potential h produces a current flow f via Ohm’s law:
f(xy) = σ(xy) [h(y)− h(x)] for all xy ∈ #»
E. (14)
Mathematically, f is the gradient flow of h. Under the given hypothesis, the current flow f from Ato B is unique and minimizes flow energy, which is known as Thomson’s principle. The current flowhas a couple of properties: it satisfies the node law at every node not in A or B, and it satisfies thecycle law. In general, flows can often be decomposed in the so-called path formulation. Given a pathγ from A to B let xγ be the unit flow along γ. Then the current flow f can be written as a sum∑aγxγ of path flows. What Proposition 1 shows is that in the planar case, there is a unique way of
decomposing the current flow f so that the paths γ in the path formulation do not cross (which makesintuitive sense physically). Moreover, the algorithm also produces the coefficients aγ for each path.
Proposition 19. With these notations, the current flow from A to B can be decomposed uniquelyas∑
γ∈Γmγxγ, where the family Γ and the weights mγ are output by the non-crossing minimalsubfamily algorithm.
Proof. It is well-known that the modulus Mod2(Γ(A,B)) of the family of all walks from A to B inG coincides with the effective conductance of the pair AB. In other words, if we solve the Dirichletproblem and find a potential h : V → R, so that h|A = 0, h|B = 1, and h is discrete harmonic onV \(A∪B), then Mod2(Γ(A,B)) equals the value of the current flow f = σdh generated by h, which isalso known as the effective conductance effC(A,B) for the pair AB. (A proof that Mod2(Γ(A,B)) =effC(A,B) in the case when A and B are points can be found in [2, Theorem 4.2]). Let Γ be thesubfamily given by the non-crossing minimal subfamily algorithm. For γ ∈ Γ, let xγ be the unit flowalong γ. Since the weights m(γ) for γ ∈ Γ add up to Mod2(Γ(A,B)), then the current flow f can bewritten as a sum
∑m(γ)xγ of path flows, without changing its value.
18
Another consequence of the non-crossing minimal subfamily algorithm is a rectangle-packing resultakin to the famous square uniformization of Oded Schramm [13].
Proposition 20. Let G = (V,E, σ) be a plane graph with sets of boundary vertices A,B, as in thesetting of the non-crossing minimal subfamily algorithm. Let h : V → R be the discrete harmonic onV \ (A ∪ B) with h|A = 0 and h|B = 1. Set M := E(h) = Mod2 Γ(A,B) and consider the rectangleR = [0, 1]×[0,M ]. Then there is a unique way to tile R with a collection of smaller rectangles Ree∈Eso that the boundaries of two rectangles share a vertical segment if and only if the corresponding edgesin G share a node, and they abut on the same horizontal line if and only if the corresponding edges inG share a face.
Moreover, each rectangle Re is of the form [h(e), h(e) + ρ(e)]× [h(e), h(e) + σ(e)ρ(e)], where ρ(e)is the extremal density for Mod2 Γ(A,B), h(e) := minh(x), h(y) : e = xy, and h(e) can be computedwith the non-crossing minimal subfamily algorithm.
Remark 21. Note that the area of each rectangle Re is σ(e)ρ2(e), which is the contribution of e tothe modulus: ∑
e∈Eσ(e)ρ2(e) = Mod2 Γ(A,B).
In other words, σ(e) is the aspect ratio of Re and ρ and M solve the problem of packing a collection ofrectangles with given aspect ratio into a larger rectangle (respecting the combinatorics of the primalgraph).
Note also that the rectangle R(e) degenerates when the corresponding ρ(e) is equal to zero.Finally, Proposition 20 is stated for general planar graphs G. When, G is the primal graphs of an
othodiagonal map, then the quantity h(e) coincides with the harmonic function defined on the dualgraph with h = 0 on the bottom side and h = M on the top side.
A B
H
J
N
K P
U
C
G
E O
I
L
R
T
A B B J J N
J P
H K K P
P UN P
Figure 3: This is an example of Proposition 20 in the case of orthodiagonal maps. For the mapabove the primal graph (in blue) is given by the nodes A,B,H, J,K,N, P, U . Here we computethe modulus M of the family of curves connecting the set A,H to the set U. In this casethe quantity h(e) coincides with the harmonic function defined on the dual graph with h = 0on the set C,L, T and h = M on the set I, R.
19
Proof. Here we perform the non-crossing minimal subfamily algorithm from the bottom up and writeγ1, . . . , γn for the output curves in the order that they are added to Γ. So γ1 is the bottom curve,etc...We also write mj := m(γj) for the corresponding weights computed by the algorithm.
To construct the rectangle packing, we begin by mapping γ1 to the strip (0, 1)× (0,m1). Writingγ1 = x1
0 · · · x1k1
we trace the horizontal segments [h(x1j ), h(x1
j+1)] × 0, for j = 0, . . . , k1 − 1, as
well as the vertical segments h(x1j ) × [0,m1], for j = 0, . . . , k1. Finally, for every edge e = x1
jx1j+1
that attains the minimum in step 3 of the algorithm, set h(e) = m1 and trace the horizontal segment[h(x1
j ), h(x1j+1)]× m1, thus closing up the rectangle corresponding to these edges.
Rather than a formal proof by induction, we now describe the second step. Consider the secondcurve γ2 and add the strip (0, 1)×(m1,m2). Observe that γ2 coincides with γ1 except for the edges thatrealized the minimum in the previous step. More specifically, γ2 will replace groups of consecutive edgesof the form e = x1
jx1j+1 with h(e) = m1, with new subpaths of the form x2
k1, . . . , x2
kl, . . . , x2
kL, where the
first and last node belong to A∪B ∪x1j. In particular, we will have h(x2
k1) < h(x2
kl) < · · · < h(x2
kL).
Therefore, we now subdivide the tops of this group of rectangles that we closed up in the previous step,and add the vertical segments of the form h(x2
kl) × [m1,m2]. Finally extend the vertical segments
from the previous step that were not bounding an edge that realized the minimum.When the algorithm ends we will have packed the rectangle [0, 1]× [0,M ]. Uniqueness follows by
observing that, if a rectangle-packing as above is given, then one could get a non-crossing minimalsubfamily simply by considering the horizontal segments connecting the left and right vertical sidesof [0, 1]× [0,M ].
3.4 A numerical example
We have implemented the non-crossing minimal subfamily algorithm, and some of its consequences,numerically. Consider the family of paths connecting the arcs ab and cd of the cross domain on theleft of Figure 4.
Figure 4: Left: a Delaunay triangulation of a cross domain; Center: the triangulation showntogether with its dual Voronoi graph; Right: the corresponding orthodiagonal map.
The figure shows a Delaunay triangulation of the domain, produced by the software packageTriangle [14]. The circumcenters of the triangles form a dual grid known as the Voronoi diagram.In the center of Figure 4, we show the same triangulation along with its Voronoi dual. the edges ofthe triangulation are blue, and the edges connecting two circumcenters are red. Note that the blueand red edges are orthogonal, hence can be thought as the orthogonal diagonals of a quadrilateral.In particular, an orthodiagonal map is obtained by connecting each blue node with a neighboring rednode. We have drawn the corresponding orthodiagonal map on the right in Figure 4. With someminor caveats (see below), the Delaunay triangulation can be thought as the primal graph of theorthodiagonal map and the Voronoi diagram as the dual graph. See also [16, Section 5.1].
20
3.4.1 Approximating the modulus
In order to approximate the modulus of the family of curves, we first choose a size parameter, ε > 0,which controls the size of the triangles in the triangulation. The triangulation is constructed to satisfythe Delaunay property as well as two additional constraints: no triangle may have area larger thanε2 and no angle in any triangle may be smaller than 20. These additional constraints ensure thatthe edges of the triangulation are of order O(ε), which is needed to ensure convergence of the discretemodulus to the continuous modulus.
Next, we need to assign appropriate conductances to the edges in the triangulation (Figure 4, left).As stated above, the triangulation is almost the graph that is obtained from the blue vertices of thequadrilateral mesh (Figure 4,right) by connecting vertices that belong to the same quadrilateral. Thisdoes not quite reproduce the triangulation on the left; notably absent are the edges on the boundaryof the domain. As a consequence, any vertex of the triangulation that is incident only upon boundaryedges is absent from the quadrilateral mesh. In the example shown, all convex “corners” are excludedfrom the quadrilateral mesh. Fortunately, there is a relatively simple way to address these issues atthe boundary.
Most edges of the triangulation will correspond to diagonals in the quadrilateral mesh. For these,we can use the formulas specified in Section 2.4. Namely, we assign the conductance, σ, on suchan edge to be the ratio of the length of the dual edge to the length of the edge in question. To allboundary edges, we assign σ = ε/|Eb|, where Eb is the set of boundary edges. Since the potential his bounded between 0 and 1, this means, by Ohm’s law, that the contribution of all boundary edgesto the current flow is O(ε), ensuring that the contribution of the extraneous boundary edges to theapproximation is negligible as ε→ 0. With this choice of conductances, we can then solve the discreteDirichlet problem to approximate the potential function, h.
3.4.2 Non-crossing curves
Once the discrete potential is computed, the non-crossing minimal family can be determined using thealgorithm described in Section 3.1. The left side of Figure 5 shows the result of this numerical studyfor the choice ε = 0.005. The algorithm produced a sequence of 34403 non-crossing paths from theboundary segment ab to the boundary segment cd. We have plotted 21 of these paths, plotting roughlyevery 1720-th path in the sequence. The first and last path trace the boundary segments between aand d and between b and c. In the middle of Figure 5, we show the corresponding curves as computedby Driscoll’s Schwarz–Christoffel Toolbox for Matlab. Using the toolbox code, we approximated aconformal mapping between the cross and a rectangle with ab and cd sent to the vertical sides of therectangle. Inverting that map on horizontal lines produced the image. The image on the right side ofthe figure shows the superposition of both sets of curves.
Using the non-crossing path decomposition, we are able to assign a two-dimensional coordinate toeach vertex in the graph. The x-coordinate of each vertex is simply the value of the discrete potentialh on that vertex. To obtain the y-coordinate of a vertex v, we locate the first path in the sequencepassing through v and sum the flows of all previous paths. The rectangle packing for this choice of ε istoo dense to draw, but the left side of Figure 6 shows a rectangle packing for ε = 0.075. This choice ofcoordinates provides a mapping on the vertices of the triangulation that approximates the conformalmap to the rectangle. The right side of the figure shows the non-crossing paths in these coordinates (forthe finer, ε = 0.005, triangulation). Except near a and d, where there is large distortion, these curvesare nearly horizontal, as we would expect. Incidentally, the rectangle packing gives an indication ofwhere one may refine the original triangulation, since the largest distortion corresponds to very largerectangles in the packing.
21
Figure 5: Left: non-crossing curves for a triangulation of the cross domain; Center: the corre-sponding curves as computed by the Schwarz-Christoffel Toolbox in Matlab; Right: the super-position of both sets of curves.
Figure 6: Left: the rectangle packing associated with a triangulation of the cross domain shownin Figure 4; Right: the images of the non-crossing curves from Figure 5 in coordinates given bythe rectangle packing.
4 Convergence of discrete harmonic functions
In this section, we prove Theorem 3 by adapting the arguments in [8] to our setting.We begin by recalling our assumptions:
• The Jordan domain Ω is a smooth quadrilateral with boundary arcs A, τ1, B, τ2.
• The orthodiagonal map G = (V • t V , E) with boundary arcs S1, T1, S2, T2 has maximal edgelength at most ε and approximates Ω within distance δ. Recall that by Definition 14, we askthat Ω ⊂ G.
• We assume λ := ε ∨ δ is small enough.
• The continuous harmonic function hc is defined by hc = Re ψ, where ψ is a conformal map of Ωonto a rectangle R = [0, 1]× [0,m], taking the arcs A and B to the vertical sides of the rectangle.
• The function hd : V • → R is discrete harmonic on Int(V •) with hd(z) = 0 for z ∈ S1 ∩ V • andhd(z) = 1 for z ∈ S2 ∩ V •.
• By scale invariance, we may assume diam(Ω) = 1 (and so λ < 1).
22
AB
τ1
τ2
∂G1
Figure 7: A sketch of the setting encountered in the proof of Theorem 3. A smooth quadrilateralΩ with boundary arcs A,B (orange) and τ1, τ2 (black) is approximated by an orthodiagonal mapG. The boundary of the sub-orthodiagonal map G1 is dashed black and ∂G \ ∂G1 is dashedgray.
Under these assumptions, our goal is to show for all z ∈ V • ∩ Ω
|hd(z)− hc(z)| ≤C
log1/2 (1/λ), (15)
where λ = ε∨ δ and C depends only on Ω. Note that in our proofs we will use C to refer to constantsthat are independent of λ but may depend on Ω. The exact value of these constants may change fromline to line.
4.1 Modified toolbox
We will be able to use some of the tools of [8] as they are stated in that paper. However, we need tomodify a handful of the results to apply to our setting, and that is the goal of this section.
There is one last piece of terminology that we introduce before we begin. Let G = (V •tV , E) be afinite orthodiagonal map with boundary arcs S1, T1, S2, T2. We say that G1 is a sub-orthodiagonal mapof G when G1 is an othordiagonal map contained in G and the topological boundary ∂G1 intersectsT1 ∪ T2. The boundary ∂V •1 are the points from V •1 that are in ∂G1 \ (T1 ∪ T2) and the boundary∂V 1 are the points in V 1 ∩ (T1 ∪ T2). When we prove Theorem 3 in Section 4.2, we will work witha sub-orthodiagonal map G1 with dist(G1, A ∪ B) ≥ λ1/4. We require that ∂G1 intersects T1 ∪ T2,since otherwise we will be in the setting of [8] and there will be nothing for us to prove. Lastly, wenote that we will want G1 to be compactly contained in the domain of ψ, which will be accomplishedby reflecting ψ over the arcs τk (and taking λ small enough). Since it may help to have a visual, seeFigure 7 for a sketch of this setting.
We begin with a modified version of Proposition 5.2 of [8], which tells us that the energy of hc−hdis small.
Proposition 22. Assume we are in the setting of Theorem 3, and G1 = (V • t V , E) is a sub-orthodiagonal map of G. Assume that ψ can be extended to an open set O so that dist(G1, ∂O) ≥ 2ε.
Let hc = Reψ, and let hd : V • → R be discrete harmonic on Int(V •) and satisfy hd = hc on ∂V •.Set
L = ||∇hc||∞,G1and M = ||Hhc||∞,G1
.
ThenE•(hc − hd) ≤ C
[M2ε2 + L2(ε ∨ δ)
],
where C depends only on area(G) and length(τ1 ∪ τ2).
23
Proof. We begin by following the initial steps in the proof of Proposition 5.2 in [8]. In particular, wewish to define the bijective map e → e† from
#»
E• to#»
E. For e ∈ #»
E•, let Q = [v1, w1, v2, w2] be theinner face of G containing e. We set e† = eQ, and we choose its orientation based on the orientation of
e: if e− = v1 then choose the orientation with (e†)− = w1, and in the other case choose the oppositeorientation.
Defineφ(e) = σ(e)[hc(e
+)− hc(e−)].
We wish to define a flow θ on G• using the harmonic conjugate hc of hc (i.e. hc = Imψ). However,we must take a little more care near the boundary arcs τk. To this end, let Uk be the component ofG1 \ Ω that intersects τk, for k = 1, 2. Next, we define a projection p : V → Ω as follows: for w ∈ Ω,set p(w) = w; for w ∈ Uk ∩ V , set p(w) to be a point in τk that is closest to w. For e ∈ #»
E•, define
θ(e) = hc(p((e†)+))− hc(p((e†)−)).
Note that if e† is not incident to a vertex in U1 ∪ U2, then our definition gives
θ(e) = hc((e†)+)− hc((e†)−)
matching the definition in [8].Now we want to show that θ satisfies the node law on Int(V •). Let v ∈ Int(V •). Suppose first v is
contained within an interior face fv of the dual graph G, i.e. there is a cycle of dual edges surroundingv. Let w1, w2, · · · , wk be the vertices incident to fv in counterclockwise order. Then (taking indicesmod k) ∑
e∈ #»E• : e−=v
θ(e) =
k∑j=1
[hc(p(wj+1))− hc(p(wj))
]= 0.
If v is not in an interior face f , then the edges e† such that e ∈ #»
E• with e− = v form a path withboth endpoints in ∂V (and in fact both endpoints will be in the same path Tk). Let w1, w2, · · · , wkbe the vertices of this path, indexed so that connecting wk to w1 in the outer face of G would give acounterclockwise cycle. Then
∑e∈ #»E• : e−=v
θ(e) =k−1∑j=1
[hc(p(wj+1))− hc(p(wj))
]= hc(p(wk))− hc(p(w1)) = 0,
where we used that hc = Imψ is constant on τk.Next, we apply Proposition 4.9 in [8] with f = hc and h = hd to obtain that
E•(hc − hd) ≤ E•(hc − hd) + E•(θ − σ dhd) = E•(φ− θ),
and we note that it suffices to bound E•(φ− θ). As noted in [8], the contribution to E•(φ− θ) from aface Q = [v1, w1, v2, w2] is
|v1v2||w1w2|
(|w1w2||v1v2|
[hc(v2)− hc(v1)]− [hc(p(w2))− hc(p(w1))]
)2
= |v1v2| · |w1w2| ·
(hc(v2)− hc(v1)
|v1v2|− hc(p(w2))− hc(p(w1))
|w1w2|
)2
. (16)
Therefore, assuming that neither w1 nor w2 are in U1 ∪U2, the quantity in the parentheses (using theCauchy-Riemann equations) can be rewritten as
hc(v2)− hc(v1)− 〈∇hc(q),v〉|v1v2||v1v2|
− hc(z(w2))− hc(z(w1))− 〈∇hc(q),w〉|w1w2||w1w2|
, (17)
24
where v and w are the unit vectors in the directions of # »v1v2 and # »w1w2, respectively, and q ∈ Q is theintersection point of e•Q and eQ. Both terms in (17) can be bounded as in [8] by∣∣∣∣hc(v2)− hc(v1)− 〈∇hc(q),v〉|v1v2|
|v1v2|
∣∣∣∣ ≤ 2Mε,∣∣∣∣∣ hc(w2)− hc(w1)− 〈∇hc(q),w〉|w1w2||w1w2|
∣∣∣∣∣ ≤ 2Mε.
Therefore if w1, w2 are not in U1 ∪ U2, the contribution to E•(φ− θ) is at most
|v1v2| · |w1w2| · (4Mε)2 = 32 area(Q)M2ε2.
Here, however, we need to deal with some extra cases. Assume that at least one of w1 and w2 arein U1 ∪ U2. We will bound the terms in the parentheses of (16) directly. Note first that∣∣∣∣hc(v2)− hc(v1)
|v1v2|
∣∣∣∣ ≤ L.If w1 and w2 are both in Uk, then hc(p(w2))− hc(p(w1)) = 0. If w1 is in Uk, but w2 is not, then theline segment connecting w1 to w2 must cross τk and |p(w1)− w1| ≤ |w1 − w2|, by the definition of p.(And a similar statement is true if w2 is in Uk+1, taking the indices mod 2). Then
|p(w1)− p(w2)| = |p(w1)− w1|+ |w1 − w2|+ |w2 − p(w2)| ≤ 3|w1 − w2|,
which implies that ∣∣∣∣∣ hc(p(w2))− hc(p(w1))
|w1w2|
∣∣∣∣∣ ≤ L |p(w2)− p(w1)||w2 − w1|
≤ 3L.
Thus if Q is incident to a vertex in U1 ∪ U2, then the contribution to E•(φ− θ) is at most
|v1v2| · |w1w2| · (4L)2 ≤ 32 area(Q)L2.
The faces Q that are incident to a vertex in U1∪U2 must be in a narrow band containing τk for eitherk = 1 or k = 2, and the area of these bands is bounded by 2 length(τ1 ∪ τ2)(ε ∨ δ). Thus summingover all inner faces Q of G gives
E•(φ− θ) ≤ 32 area(G)M2ε2 + 64 length(τ1 ∪ τ2)L2(ε ∨ δ).
The next proposition generalizes Proposition 6.1 from [8] to our setting.
Proposition 23. Assume we are in the setting of Theorem 3, and G1 = (V • t V , E) is a sub-orthodiagonal map of G. Assume that ψ can be extended to an open set O so that dist(G1, ∂O) ≥ 2ε.
Fix z0 ∈ Int(V •) ∩ Ω, and let Dr be the closed disk of radius r ≥ 4ε centered at z0. Assume thatDr ∩ ∂V • = ∅. Then there is a unit flow θ in G•1 from V • ∩Dr to ∂V • so that
E•(θ) ≤ C log
(diam(Ω)
ε
).
Proof. Recall that ψ is a conformal map of Ω onto the rectangle R = [0, 1]× [0,m] taking the arcs τ1
and τ2 to the horizontal sides of the rectangle. Let ψ(z0) = x0 + iy0. We may assume that y0 ≥ m/2,by rotating the rectangle 180 around its midpoint if needed. Let
g(z) = exp
(π
y0z
)+ exp
(πx0
y0
).
25
Note that g maps R conformally strictly into the annulus A with inner radius 1 and outer radius eπ/y0
centered at eπx0/y0 , and g maps R′ = z ∈ R : Im(z) < y0 onto A∩H. Further, g takes the horizontalsides of R′ to intervals in R, and
g(ψ(z0)) = g(x0 + iy0) = eπx0y0 eiπ + e
πx0y0 = 0.
Let F = gψ. We will define a flow φ on#»
E• from z0 to ∂V • as follows. Given a face Q = [v1, w1, v2, w2],let e be the orientation of e•Q with e− = v1. If F (eQ) ⊂ H, set
φ(e) = arg(F (w2))− arg(F (w1)) and φ(−e) = arg(F (w1))− arg(F (w2)),
where we use a branch of the argument that is well-defined on H. If F (eQ) does not intersect H, thenwe define φ(e) = 0 = φ(−e). If F (eQ) intersects H, we define φ(e) to be the change of argument along
F (eQ) ∩H, respecting the orientation above.
Let v ∈ Int(V •). Consider the set N of edges to e ∈ #»
E• with e− = v. The edges dual to the edgesin N either form a cycle around v (in the case v is in the topological interior of G) or a path startingand ending on a boundary arc Tk. The image of these dual edges under F either form a cycle or apath in A, which we will call γv. We will first consider the case that γv is a cycle contained in H. Letw1, w2, · · · , wk be the corresponding vertices in V , ordered counter-clockwise with wk = w1. Then∑
e∈ #»E• : e−=v
φ(e) =
k−1∑j=1
[arg(F (wj+1))− arg(F (wj))] .
This is equal to the change of argument as we move around γv counter-clockwise, which is 0. Next, inthe case that γv does not intersect H, we also obtain
∑e∈ #»E• : e−=v φ(e) = 0, since φ(e) is zero for each
e in the sum. Lastly consider the case that γv intersects H but is not a cycle completely contained inH. Then γv ∩H consists of one or more paths which all start and end in R. If v = z0, then one pathwill start in R+ and end in R−. All other paths will start and end with the same argument. Since∑
e∈ #»E• : e−=v φ(e) gives the sum of the change in argument over the path(s) of γv ∩H, this is equal to
π when v = z0 and it is equal to 0 otherwise. Thus φ is a flow of strength π from z0 to ∂V •. Define
θ(e) =
0 if e+ and e− ∈ Dr
1πφ(e) else
.
Then θ is a unit flow from from V • ∩Dr to ∂V •.It remains to bound the energy of θ. Let Q = [v1, w1, v2, w2] be a face of G1 so that v1, v2 are not
both in Dr. We first consider the case that F (eQ) is contained in H. Then the contribution of e•Q toE•(θ) is
|v1v2||w1w2|
|θ(e•Q)|2 =1
π2|v1v2| · |w1w2|
(arg(F (w2))− arg(F (w1))
|w1w2|
)2
.
Note that |v1v2| · |w1w2| = 2 area(Q). Since Q has diameter at most 2ε and has at least one vertexnot in Dr, the distance of Q from z0 is at least r − 2ε ≥ 2ε. Along with the assumption thatdist(G1, ∂O) ≥ 2ε, this gives that Q has distance at least 2ε from the boundary of O \ z0. Hence thederivative of log F is comparable at every point in Q. Thus
|v1v2||w1w2|
|θ(e•Q)|2 =2
π2area(Q)
(arg(F (w2))− arg(F (w1))
|w2 − w1|
)2
≤ 2
π2area(Q) max
z∈eQ
∣∣∣∣ ddz log F (z)
∣∣∣∣2≤ C
∫Q
1
|F (z)|2|F ′(z)|2 dz
= C
∫F (Q)
1
|z|2dz.
26
If F (eQ) intersects H but is not fully contained in it, the above estimate still holds, since there arepoints w′1, w
′2 on the edge eQ so that the contribution of e•Q to E•(θ) is
2
π2area(Q)
(arg(F (w′2))− arg(F (w′1))
|w1w2|
)2
≤ 2
π2area(Q)
(arg(F (w′2))− arg(F (w′1))
|w′2 − w′1|
)2
.
The distortion theorem implies that for |z − z0| = 2ε,
|F (z)| ≥ C ε|F ′(z0)| ≥ Cε|ψ′(z0)| ≥ C εdist(ψ(z0), ∂ψ(O))
diam(Ω).
Let w0 ∈ ∂O so that dist(ψ(z0), ∂ψ(O)) = |ψ(z0)−ψ(w0)|. Then by Proposition 15, there is p ∈ (0, 1/2]so that
dist(ψ(z0), ∂ψ(O)) = |ψ(z0)− ψ(w0)| ≥ C|z0 − w0|1/p ≥ C (dist(z0, ∂O))1/p ≥ Cε1/p.
Therefore we have that for |z−z0| = 2ε, then |F (z)| ≥ cε1+1/p/ diam(Ω) =: d. This implies F (Q) doesintersect a disc of radius d centered at 0. To bound F (Ω), we note that |F (z)| ≤ 2e2π/m for z ∈ Ω.Putting this all together, we find that
E•(θ) ≤ C∑
Q:dist(Q,z0)≥2ε
∫F (Q)
1
|z|2dA(z)
≤ C∫ 2π
0
∫ 2e2π/m
d
1
s2s ds dt
≤ C log
(diam(Ω)
ε
).
Lastly we need modified versions of the three results in Section 7 of [8]. The logical relationshipbetween these results is the following:
Lemma 7.3 =⇒ Lemma 7.2 =⇒ Proposition 7.1.
Since these results make use of the fact that the boundary value problem, in their case, is defined onthe whole topological boundary, we will need to modify them. This will have a ripple effect, causingus to update all three results, although the proofs connecting the results will follow the logic of [8]and we will omit some of the details.
We begin with stating the modification of Proposition 7.1:
Proposition 24. Assume we are in the setting of Theorem 3, and G1 = (V • t V , E) is a sub-orthodiagonal map of G (which may be equal to G). Let x, y ∈ V •. Assume we may choose r,R sothat r ≥ 1
2 |x − y|, R ≥ 2r + 3ε, and there is a path from x to y in G• ∩ Dr, where the notation Dr
means the closed disk of radius r centered at 12(x + y). Further, we assume that either DR does not
intersect ∂V • or, if G1 = G, then R ≤ 14dist(T1, T2).
Let h : V • → R be discrete harmonic on Int(V •). Set
κ = maxv,v′∈∂V •∩DR
|h(v)− h(v′)|,
with κ = 0 if there are no such vertices. Then there exists a universal constant C <∞ such that
|h(x)− h(y)| ≤ C E•(h)1/2
log1/2 [R/(r + ε)]+ κ.
27
S1
S2
T1
T2
DR
xy
Figure 8: A sketch of the setting for Lemma 25: the topological boundary of H subdivided intopaths S1, T1, S2, T2 (solid black), the path γx from x to ∪Sk with vertices in X (dashed red),the path γy from y to ∪Sk with vertices in Y (dashed blue), and the path γxy from x to y inDr (dashed black).
The proof of this result is the same as the proof of Proposition 7.1 in [8], simply substitutingLemma 25 in the place of Lemma 7.2 in [8].
For the next lemma, we introduce one more definition. For ρ > 0, a ρ-edge of G• is an edge e ∈ E•whose dual edge e ∈ E connects two vertices w,w′ ∈ V with |w| < ρ ≤ |w′|. See Figure 8 for anillustration of the setting of Lemma 25.
Lemma 25. Under the assumptions of Proposition 24, suppose that the origin is located at 12(x+ y)
and that h(x) < h(y). Define
X = v ∈ V • : h(v) ≤ h(x)Y = v ∈ V • : h(v) ≥ h(y)
and set Y ′ = Y ∪ (∂V • ∩D). Assume that X ∩ Y ′ = ∅. Then for every ρ ∈ (r + ε, R − ε), there is apath in G• from X to Y ′ consisting entirely of ρ-edges.
This lemma will follow from the following result about plane graphs. We say a few words aboutthis after the proof.
Lemma 26. Let H be a finite connected plane graph with planar dual graph H†. Let W,W † be thevertex sets of H,H†. For each w ∈ W †, let f(w) denote the face of H that contains w. Assume thateach inner face f(w) is contained in the open disk of radius ε centered at w. Fix 0 < r < R withr+ ε < R− ε, and let DR be the closed disk of radius R centered at the origin. Let z ∈W † be the dualvertex contained in the outer face of H, and assume that z /∈ DR.
Assume that ∂H, the topological boundary of the outer face of H, can be subdivided into 2 or 4nonempty paths S1, T1 or S1, T1, S2, T2. Let x, y ∈ W satisfy |x|, |y| ≤ r. Let X,Y be subsets of Wsuch that there is a path in H from x to ∪Sk whose vertices are all in X, and there is a path in Hfrom y to ∪Sk whose vertices are all in Y . Assume there is a path in H from x to y that is containedin Dr, the closed disk of radius r centered at the origin. Further assume that Sk ∩Dc
R 6= ∅.Set Y ′ = Y ∪ [DR ∩ (∪Sk)]. If X ∩ Y ′ = ∅, then for each ρ ∈ (r + ε, R − ε), there is a path in
H \ (∂H) from X to Y ′ consisting entirely of ρ-edges.
Proof. We begin by following the proof of Lemma 7.3 in [8] to obtain a cycle K in H such that all ofits edges are ρ-edges. For convenience, we briefly repeat that argument here. Let u ∈ W † be a dualvertex contained in an inner face of H with x incident to this face. Then u 6= z and |u| ≤ r + ε < ρ.Let σρ be the set of vertices w ∈ W † so that there is a path from u to w in H† with all vertices ofthis path in the open disk of radius ρ centered at 0. Let H[σρ] be the subgraph of H whose verticesand edges make up the boundaries of the faces containing w for each w ∈ σρ. The graph H[σρ] is
28
connected and contained in DR. Let foutρ be the outer face of H[σρ], which contains the complement
of DR. Let K be the boundary of foutρ , which is a connected subgraph. One can check that all of the
edges in K are ρ-edges. Further, we note that the edges in K are either in ∂H, or within distance εof Cρ, the circle of radius ρ centered at 0. The former situation can happen when Cρ extends into theouter face of H. However, when an arc of Cρ is contained entirely within the interior of H, then therewill be edges of K within distance ε of this arc.
Let γx denote the path in H from x to ∪Sk whose vertices are all in X, let γy denote the pathin H from y to ∪Sk whose vertices are all in Y , and let γxy denote the path in H from x to y thatis contained in Dr. Since γx ⊂ X is disjoint from (∪Sk) ∩DR ⊂ Y ′, we must have that γx connectsx to a point in ∂H that is outside DR. Hence γx must cross the annulus DR \ Dr. For the sake ofsimplicity, assume that γy connects to a vertex in S2 (otherwise renumber the boundary paths.) Thenγy ∪ S2 must also cross the annulus DR \ Dr, since Sk ∩ Dc
R 6= ∅. Further γx and γy ∪ (∪Sk ∩ DR)must be disjoint within DR, since the vertices of the former are in X and the vertices of the later arein Y ′. Hence we can find a cycle J composed of three disjoint arcs: (1) an arc contained in γxy withinDr, (2) an arc is contained in γx that crosses DR \Dr, and (3) an arc that crosses DR \Dr and whoseintersection with DR is contained in γy ∪ (∪Sk).
Now the interior of J is contained in the interior of H by construction. Further it contains an arcof Cρ. Hence there must be a path contained in K that crosses the interior of J from a vertex in γxto a vertex in γy ∪ (∪Sk). This path could potentially contain some edges in ∂H, but the only wayfor this to happen is for the path to contain edges from γx or γy ∪ (∪Sk). Hence, we can shorten thepath by starting at the last vertex it encounters in γx and ending at the first vertex it encounters inγy∪(∪Sk). Then all the edges in the path are strictly within the interior of J and hence in the interiorof H. This gives us the desired path.
Since Lemma 25 follows from Lemma 26 by the same logic as in [8], we will only give a brief sketch.
Proof outline of Lemma 25.
• Modify G• and G to create graphs that are planar dual graphs.
• Use the discrete maximum principle to show the existence of the needed paths, i.e. a path fromx to ∪Sk whose vertices are all in X and a path from y to ∪Sk whose vertices are all in Y .
• Apply Lemma 26 to obtain a path in the modified graph from X to Y ′ that consists of ρ-edges.
• Verify that the path is in fact in G• and the edges are ρ-edges in G•. This will follow since thepath we obtain does not contain any edges from the topological boundary which was the onlypart affected by the modification.
4.2 Proof of Theorem 3
The proof of Theorem 3 follows the same structure as in [8]. We give an outline of the overall logicbelow and then conclude by establishing Claims 27-31.
The first step is to show that the desired bound holds near the boundary arcs A and B.
Claim 27. The bound (15) holds for all z ∈ V • ∩ Ω such that dist(z,A ∪B) ≤ 2λ1/4.
For the rest of the proof, we will assume that there are z ∈ V • ∩ Ω with dist(z,A ∪ B) > 2λ1/4.Since diam(Ω) = 1, this will imply that 2λ1/4 ≤ 1. This is equivalent to λ ≤ 1
16 or λ ≤ 18λ
1/4, whichwe will use at times.
The following construction and claim will provide us with a sub-orthodiagonal map, as in [8]. LetP be the set of all inner faces Q of G such that dist(Q,A ∪B) ≥ λ1/4, and let G[P ] be the subgraphof G that is the union of the boundaries of the faces in P .
29
Claim 28. The set P is nonempty. The inner faces of G[P ] are precisely the elements of P , andevery point p on the boundary of the outer face of G[P ] is either in T1 ∪ T2 or satisfies
dist(p,A ∪B) ≤ 5
4λ1/4. (18)
Each z ∈ V • ∩ Ω with dist(z,A ∪B) > 2λ1/4 is a vertex of a unique block of G[P ]. If H is a block ofG[P ], then the inner faces of H are all elements of P and every point p on the boundary of the outerface of H is either in T1 ∪ T2 or satisfies (18).
Claim 28 implies that each block of H of G[S] is a sub-orthodiagonal map. Fix z0 ∈ V • ∩ Ω withdist(z0, A ∪ B) > 2λ1/4, and let G1 be the block of G[P ] containing z0. Note that if the topologicalboundary ∂G1 does not intersect T1∪T2, we are in the situation of [8] and the rest of the proof followsas in their proof. Hence we will assume that ∂G1 does intersect T1∪T2. We take ∂V •1 to be the pointsfrom V •1 that are in ∂G1 \ (T1 ∪ T2) and Int(V •1 ) = V •1 \ ∂V •1 .
Claim 29. Let G1 denote the union of the convex hulls of the closures of the inner faces of G1. Then
||∇hc||∞,G1≤ Cλ−1/4 and ||Hhc||∞,G1
≤ Cλ−1/2.
Define h(1)d : V •1 → R to be the solution to the discrete Dirichlet problem on Int(V •1 ) with boundary
data hc on ∂V •1 . Then
|hd(z0)− hc(z0)| ≤ |hd(z0)− h(1)d (z0)|+ |h(1)
d (z0)− hc(z0)|. (19)
Claim 30. The first term of (19) is bounded as follows:
|hd(z0)− h(1)d (z0)| ≤ C
log1/2(1/λ).
It remains to bound the second term of (19). Set f = h(1)d − hc, and note that f = 0 on ∂V •1 . Let
D4ε be the disk of radius 4ε centered at z0.
Claim 31. The set D4ε is disjoint from ∂V •1 , and for y ∈ V •1 ∩D4ε
|f(z0)− f(y)| ≤ C
log1/2(1/λ).
Proposition 22 and Claim 29 imply that
E•1 (f) ≤ C[M2ε2 + L2λ
]≤ Cλ1/2,
where L = ||∇hc||∞,G1and M = ||Hhc||∞,G1
. Applying Proposition 23 yields a unit flow θ in G•1 fromV •1 ∩D4ε to ∂V •1 such that
E•1 (θ) ≤ C log (1/λ) .
Then −θ is a unit flow from ∂V •1 to V •1 ∩D4ε. Applying Proposition 4.11 in [8] to f and −θ showsthat
gap∂V •1 ,V •1 ∩D4ε(f) ≤ E•1 (−θ)1/2E•1 (f)1/2 ≤ Cλ1/4 log1/2 (1/λ) ,
where gapX,Y (f) := miny∈Y
f(y) − maxx∈X
f(x). Note that Proposition 4.11 only applies in the case that
the gap is nonnegative, but the inequality is immediate in the other case. Since f = 0 on ∂V •1 , thereis y ∈ V •1 ∩D4ε so that
f(y) ≤ Cλ1/4 log1/2(1/λ).
Then combining this with Claim 31, we obtain
f(z0) = f(y) + [f(z0)− f(y)] ≤ Cλ1/4 log1/2(1/λ) +C
log1/2(1/λ)≤ C
log1/2(1/λ)
30
Applying the same argument to −f yields that
|f(z0)| ≤ C
log1/2(1/λ).
All that remains is to establish Claims 27-31. Two of these, Claims 28 and 30, follow the argumentsof the corresponding results, Claims 8.2 and 8.4, in [8] with only minor changes of constants. As aresult, we omit those proofs here.
Proof of Claim 27. Let z ∈ V • ∩ Ω such that dist(z,A ∪ B) ≤ 2λ1/4. Choose u ∈ A ∪ B such that|z−u| ≤ 2λ1/4. Then there is v ∈ S1∪S2 such that there is a path γuv from u to v in G with length atmost δ. Further by following the edge containing v to its endpoint in V •, there is w ∈ ∂V • such that|v−w| ≤ ε (and let γvw be the path from v to w along the edge). Then |z−w| ≤ 2λ1/4 +δ+ ε ≤ 4λ1/4.
Since hd(w) = hc(u), we have
|hd(z)− hc(z)| ≤ |hd(z)− hd(w)|+ |hc(u)− hc(z)|.
From Proposition 15,
|hc(u)− hc(z)| ≤ C|u− z|1/4 ≤ Cλ1/16 ≤ C
log1/2 (1/λ).
Thus it remains to bound |hd(z)− hd(w)|.Our goal will be to use Proposition 24, and so we must find a path from z to w in G•. Since Ω is
a linearly connected domain (with constant K), there is a path γzu in Ω ⊂ G from z to u so that
diam(γzu) ≤ K|z − u| ≤ 2Kλ1/4.
Therefore γzw = γzu ∪ γuv ∪ γvw is a path from z to w in G with
diam(γzw) ≤ 2Kλ1/4 + δ + ε ≤ (2K + 2)λ1/4.
Now the path γzw can be converted into a path γ in G• as in the proof of Lemma 3.2 in [8] (see theirFigure 4), so that dist(γzw, γ) ≤ ε. Thus
diam(γ) ≤ diam(γzu) + 2ε ≤ (2K + 4)λ1/4.
Now a disc centered at 12(z + w) will contain γ if its radius is at least
1
2|z − w|+ diam(γ) ≤ 2λ1/4 + (2K + 4)λ1/4 = (2K + 6)λ1/4.
Set r = (2K + 6)λ1/4 and R = (4K + 15)λ1/8. Note that, by taking λ small enough,
2r + 3ε ≤ R ≤ 1
4d,
where d := mindist(T1, T2),dist(S1, S2) and
R
r + ε=
(4K + 15)λ1/8
(2K + 6)λ1/4 + ε≥ λ−1/8.
The fact that R ≤ 14 d means that DR, the disc of radius R centered at 1
2 |z +w| can intersect at mostone of S1 or S2. Because hd is constant equal to 0 on S1 ∩V • and constant equal to 1 on S2 ∩V •, theterm κ from Proposition 24 will be zero. Therefore Proposition 24 gives
|hd(z)− hd(w)| ≤ C E•(hd)1/2
log1/2 (1/λ).
31
To complete the proof, it remains to bound E•(hd). We reflect ψ over τ1 and τ2, so that ψ isconformal in an open set O containing Ω. Define Omid = z ∈ O : 1/3 ≤ ψ(z) ≤ 2/3. Considerthe distance between τk ∩ Omid and ∂O. By taking λ small enough, we may assume that 3λ <dist (τk ∩ Omid, ∂O). Then the boundary of any connected component of G \ Omid will intersectexactly one of ψ−1(Re(z) = 1/3) or ψ−1(Re(z) = 2/3). Let Gleft be the union of all componentswhose boundary intersects ψ−1(Re(z) = 1/3), and similarly define Gright as the union of components
whose boundary intersects ψ−1(Re(z) = 2/3). Define the continuous function f on G as follows:
f(z) =
0 if z ∈ Gleft
1 if z ∈ Gright
3hc(z)− 1 else
.
Then by Proposition 4.8 in [8],E•(hd) ≤ E•(f).
Form Gmid from all the quadrilaterals that intersect Omid. Note that the only quadrilaterals in G thatcontribute to E•(f) are those in Gmid. Further, for these Q = [v1, w1, v2, w2] in Gmid, we have that
|f(v1)− f(v2)| ≤ 3 |hc(v1)− hc(v2)| .
ThereforeE•(f) ≤ 9E•mid(hc),
where the energy on the right is with respect to the graph Gmid. Since Gmid is contained within O,Proposition 5.2 in [8] implies that
E•mid(hc) ≤ 2
∫Gmid
|∇hc|2 + 2 area(Gmid)(10LMε+ 8M2ε2),
where L = ||∇hc||∞,Gmidand M = ||Hhc||∞,Gmid
. The constants L and M depend on Ω but areindependent of ε and δ. Since O and the extended ψ are defined by reflection, ψ(O) is contained inthe rectangle [0, 1] × [−m, 2m] and
∫Gmid|∇hc|2 ≤ 3
∫Ω |∇hc|
2. However,∫
Ω |∇hc|2 = m the modulus
of curves in R from Re(z) = 0 to Re(z) = 1. Thus we have shown that E•(hd) is bounded by aconstant (which depends only on Ω).
Proof of Claim 29. Note that supy∈Ω |hc(y)| ≤ 1, since hc is the real part of the conformal map ψonto [0, 1]× [0,m].
We reflect ψ over τ1 and τ2, so that ψ is conformal in an open set O containing Ω. Let z1, z2, z3, z4
be the four “corners” of Ω, i.e. where the arcs A,B and τ1, τ2 meet. Since the curves A,B, τ1, τ2
are smooth, there is an rk > 0 so that for z ∈ Drk(zk) ∩ Ω, dist(z, ∂O) ≥ 12dist(z,A ∪ B). Set
D = ∪4k=1Drk(zk). Then (τ1 ∪ τ2) \D is contained compactly within O, and we may assume that
dist ((τ1 ∪ τ2) \D, ∂O) ≥ λ1/4 (20)
by taking λ small enough.We wish to show that for all p ∈ G1
dist(p, ∂O) ≥ 5
16λ1/4. (21)
Let p ∈ G1. If the closest point to p in ∂O is in A ∪ B, then (21) holds since dist(p,A ∪ B) ≥ λ1/4.So we will assume that the closest point to p in ∂O is not contained in A∪B. First consider the casethat p ∈ Ω. The line segment connecting p to a closest point in ∂O must cross out of Ω at some pointz0 ∈ τ1 ∪ τ2. We have two possible cases: (1) z0 ∈ (τ1 ∪ τ2) \D or (2) z0 ∈ (τ1 ∪ τ2) ∩D. In case (1),we use (20) to obtain
dist(p, ∂O) ≥ dist(z0, ∂O) ≥ λ1/4.
32
In case (2),
dist(p, ∂O) ≥ dist(p, z0) + dist(z0, ∂O)
≥ 1
2[dist(p, z0) + dist(z0, A ∪B)]
≥ 1
2dist(p,A ∪B)
≥ 1
2λ1/4.
Next we assume that p /∈ Ω. Then there is some w0 ∈ τ1 ∪ τ2 so that dist(p, w0) ≤ δ. If w0 /∈ D,then
dist(p, ∂O) ≥ dist(w0, ∂O)− δ ≥ λ1/4 − δ ≥ 7
8λ1/4,
where we used that λ ≤ 18λ
1/4. If w0 ∈ D, then
dist(w0, ∂O) ≥ 1
2dist(w0, A ∪B) ≥ 1
2[dist(p,A ∪B)− dist(w0, p)] ≥
1
2
(λ1/4 − δ
)≥ 7
16λ1/4,
and so
dist(p, ∂O) ≥ dist(w0, ∂O)− dist(w0, p) ≥7
16λ1/4 − δ ≥ 5
16λ1/4.
This establishes (21).For all q ∈ G1,
dist(q, ∂O) ≥ 5
16λ1/4 − ε ≥ 3
16λ1/4. (22)
Proposition 3.7 of [8] and (22) imply that
||∇hc||∞,G1≤ Cλ−1/4.
We obtain the Hessian bound ||Hhc||∞,G1≤ Cλ−1/2 from Proposition 3.7 of [8] and (22) exactly as
in the proof of Claim 8.3 in [8].
Proof of Claim 31. Recall that we have z0 ∈ V •1 with dist(z0, A∪B) > 2λ1/4. From Claim 2, we knowthat every point p ∈ ∂G1 satisfies dist(p,A∪B) ≤ 5
4λ1/4. Thus the distance from z0 to ∂V •1 is at least
2λ1/4 − 54λ
1/4 = 34λ
1/4 ≥ 6λ, since λ1/4 ≥ 8λ, and this implies that D4ε cannot intersect ∂V •1 .Let y ∈ V •1 ∩D4ε. Then
|f(z0)− f(y)| ≤ |h(1)d (z0)− h(1)
d (y)|+ |hc(z0)− hc(y)|.
Our next step is to get a path from z0 to y in G1 with length bounded by Cλ. Then by Claim 3, wewill have that
|hc(z0)− hc(y)| ≤ ||∇hc||∞,G1· Cλ ≤ Cλ3/4.
Since y may not be in Ω, we will choose w in Ω so that there is a path γwy from w to y in G1 withthe length of γwy bounded by δ. Since Ω is a linearly connected domain (with constant K), there is apath γz0w in Ω ⊂ G1 from z0 to w so that
diam(γz0w) ≤ K|z0 − w| ≤ K (|z0 − y|+ δ) ≤ 4Kλ.
Therefore γz0y = γz0w ∪ γwy is a path from z0 to y in G1 with
diam(γz0y) ≤ (4K + 1)λ.
It remains to bound |h(1)d (z0)− h(1)
d (y)|. Propositions 4.8 and 5.1 of [8] give
E•1 (h(1)d ) ≤ E•1 (hc) ≤ 2
∫G1
|∇hc|2 + 2area(G1) (10LMε+ 8M2ε2)
33
where (using Claim 29)
L = ||∇hc||∞,G1≤ Cλ−1/4,
M = ||Hhc||∞,G1≤ Cλ−1/2.
As in the proof of Claim 27, ∫G1
|∇hc|2 ≤ 3
∫Ω|∇hc|2 = 3m,
where m is the continuous modulus of the family of curves from A to B in Ω. Thus,
E•1 (h(1)d ) ≤ C.
We will utilize Proposition 24 to finish the proof, and so we must verify that its hypotheses aresatisfied. The path γz0y from above can be converted into a path γ in G•1 as in the proof of Lemma3.2 in [8] (see their Figure 4), so that dist(γz0y, γ) ≤ ε. Thus
diam(γ) ≤ diam(γz0y) + 2ε ≤ (4K + 3)λ.
Thus a disc centered at 12(z0 + y) will contain γ if its radius is larger than
1
2|z0 − y|+ diam(γ) ≤ 3
2ε+ (4K + 3)λ ≤ (4K + 5)λ.
Set r = (4K + 5)λ and R = 14λ
1/4. We have chosen λ to be small enough so that 2r + 3ε ≤ R. Note
further that since the distance from z0 to ∂V •1 is at least 34λ
1/4, the distance from 12(z0 + y) to ∂V •1 is
at least 34λ
1/4 − 32ε ≥
12λ
1/4 (using that λ ≤ 18λ
1/4). Thus DR does not intersect ∂V •1 and so the termκ in Proposition 24 is equal to zero. Therefore Proposition 24 implies that
|h(1)d (z0)− h(1)
d (y)| ≤C E•1 (h
(1)d )1/2
log1/2 [R/(r + ε)]≤ C
log1/2 (1/λ).
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