Existence and uniqueness

3.2 Definition

3.2.2 Existence and uniqueness

has the law µthen L restricted to any chart Σ is a local HRLCκ(D;w0 →p0) in Σ.

This µ is called the (global) HRLC_κ(D;w₀ →p₀).

There are some other possible definitions of P_t,W^ξ in (3.2.5) such that Theorem 3.2.2 and Corollary 3.2.1 hold with no changes or minor changes. For example, Let I be a side arc of Dsuch thatw0 6∈I. We may defineP_t,W^ξ to be the harmonic measure function ofI inD\W(K_t^ξ) as long as there is a neighborhood ofIin∂Dthat does not intersect W(K_t^ξ). Then the law of L defined before Corollary 3.2.1 is called the local HRLC_κ(D;w₀ → I) in Σ. Letw₁ be a prime end of D other than w₀, and J maps a neighborhood of w1 conformally onto a neighborhood of 0 in H such thatJ(w1) = 0.

If there is a neighborhood of w₁ in Db that does not intersect W(K_t^ξ), then We may define P_t,W^ξ to be the minimal function in D\W(K_t^ξ) with the pole w1, normalized by J. Then the law of L is called the local HRLC_κ(D;w₀ → w₁) in Σ, normalized by J. In fact, two local HRLCκ(D;w0 → w1) in the same chart differ only by a constant speed time-change. We may also define the (global) HRLC_κ(D;w₀ →I) or HRLC_κ(D;w₀ →w₁) normalized byJ, similarly as the case that the target is a prime end.

Proposition 3.2.1 Suppose(K_t^r) has the law of the radial SLE_κ(D;w₀ →p₀). Then t 7→K_t/(4π^r 2) has the law of the HRLCκ(D;w0 →p0). Suppose (K_t^c) has the law of a chordal SLE_κ(D;w₁ → w₂). Then t 7→ K_t^c has the law of an HRLC_κ(D;w₁ → w₂).

Suppose(K_t^s) has the law of the strip SLEκ(D;w0 →I). Thent 7→K_t/π^s 2 has the law of the HRLC_κ(D;w₀ → I). Suppose (K_t^a) has the law of the annulus SLE_κ(D;w₀ → S). Then t7→K_p−√3

p³−3t has the law of the HRLCκ(D;w0 →S).

Proof. The proof is similar as that of Theorem 3.2.1. 2

sup_0≤s≤t|ζ(s)−η(s)|< δ, then for any z ∈F, ϕ^ζ_t ◦(ϕ^η_t)⁻¹(z) is well defined, and

|z−ϕ^ζ_t ◦(ϕ^η_t)⁻¹(z)| ≤Ctkζ−ηk_t, ∀z ∈F,

where ϕ^η_t and ϕ^ζ_t are standard radial LE maps driven by η and ζ, respectively.

Proof. Choose ε∈(0,1) such that F ⊂(1−ε)D. Letd >0 be the distance between F and (1−ε)∂D. There is C_ε >0 such that for χ₁, χ₂ ∈∂D and x₁, x₂ ∈(1−ε)D,

¯¯

¯¯x₁χ₁+x₁ χ1 −x1

−x₂χ₂+x₂ χ2−x2

¯¯

¯¯≤C_ε(|χ₁−χ₂|+|x₁−x₂|). (3.2.7)

Fix z ∈F and t ∈[0, a]. Let s0 be the first s >0 that is equal to t or

f(s) :=|ϕ^η_s◦(ϕ^η_t)⁻¹(z)−ϕ^ζ_s◦(ϕ^η_t)⁻¹(z)|

is equal tod. Then for 0≤s≤s0,

|ϕ^η_s◦(ϕ^η_t)⁻¹(z)| ≤ |ϕ^η_t ◦(ϕ^η_t)⁻¹(z)|=|z|.

From the definition of d, we see that for 0≤ s≤s₀,ϕ^η_s ◦(ϕ^η_t)⁻¹(z), ϕ^ζ_s ◦(ϕ^η_t)⁻¹(z)∈ (1−ε)D. So ϕ^ζ_s ◦(ϕ^η_t)⁻¹(z) is well defined. Since ϕ^η₀ = ϕ^ζ₀ = id, f(0) = 0. From (2.2.1) and (3.2.7), we conclude that

f(s)≤C_ε(kζ−ηk_ts+ Z _s

0

f(r)dr), 0≤s≤s₀.

From a standard argument of differential equations, we get

f(s0)≤3(e^s0Cε−1)kζ−ηkt ≤Cs0kζ−ηkt, (3.2.8)

whereC := 3(e^Cεa−1)/a. Let δ=d/(Ca). Ifkζ−ηk_t < δ, thenf(s₀)< d, sos₀ =t.

This ends the proof. 2

Lemma 3.2.2 Suppose a > 0 and β is a Jordan curve in Ω such that the doubly

connected domain bounded by β and ∂D is contained in Ω. There are δ, ε, C > 0 depending on D, p₀, W, Ω, β and a, such that if t ∈(0, a] and ζ, η ∈ C[0, t] satisfy kζ −ηkt < δ and K_a^ζ ∩ β = ∅, then for any z ∈ D with |z| ≥ 1 − ε, we have (ϕ^ζ_t)⁻¹(z),(ϕ^η_t)⁻¹(z)∈Ω, and

|P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹(z)−P_t,W^η ◦W ◦(ϕ^η_t)⁻¹(z)| ≤Ctkζ−ηkt.

Proof. We can find another Jordan curve γ in Ω disjoint from β such that the doubly connected domain bounded by γ and ∂D is contained in Ω and contains β. Let m be the modulus of the doubly connected domain bounded by β and γ.

By conformal invariance and the comparison principle of moduli, the modulus of the doubly connected domain bounded by ϕ^ζ_t(γ) and ∂D is at least m. So there is ε₀ =ε₀(m)>0 such that ϕ^ζ_t(γ)⊂(1−ε₀)D. Let J_k = (1−ε₀/k)∂D, k= 1,2,3, and J₄ =p

1−ε₀/3∂D.

Let F = p

1−ε₀/3D. From Lemma 3.2.1, we have C₁, δ₁ > 0 depending on a and F such that ifkζ −ηk_t < δ₁, then for all z ∈ F, ϕ^ζ_t ◦(ϕ^η_t)⁻¹(z) , ϕ^η_t ◦(ϕ^ζ_t)⁻¹(z) exist, and

|z−ϕ^ζ_t ◦(ϕ^η_t)⁻¹(z)|,|z−ϕ^η_t ◦(ϕ^ζ_t)⁻¹(z)| ≤C₁tkζ−ηk_t. (3.2.9) Since J₁ lies between ϕ^ζ_t(γ) and ∂D, so does J₃. Thus (ϕ^ζ_t)⁻¹(J₃) lies between γ and ∂D. So for any z ∈ D such that |z| ≥ 1−ε0/3, (ϕ^ζ_t)⁻¹(z) lies in the domain bounded byγ and ∂D, which is contained in Ω. Ifkζ−ηk_t< δ = min{δ₁, ε₀/(6C₁a)}, from ϕ^ζ_t(γ) ⊂ F and (3.2.9) we have |ϕ^η_t(z)−ϕ^ζ_t(z)| ≤ C1tkζ−ηkt < ε0/6 for all z ∈ γ. Note ϕ^ζ_t(γ)⊂(1−ε₀)D. Thus ϕ^η_t(γ) ⊂(1−ε₀/2)D. This means that J₂ lies between ϕ^η_t(γ) and ∂D. Similarly, we have (ϕ^η_t)⁻¹(z) lies in Ω, for any z ∈ D such that |z| ≥ 1−ε₀/3. Thus P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹ and P_t,W^η ◦W ◦(ϕ^η_t)⁻¹ are well defined on a domain that contains {z ∈ D : |z| ≥ 1−ε0/3} when kζ−ηkt < δ. It is clear that they are harmonic, and they have vanished continuation at ∂D. Now suppose

kζ−ηkt< δ, define

M = sup

z∈J3

|P_t,W^η ◦W ◦(ϕ^η_t)⁻¹(z)−P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹(z)|;

N = sup

z∈J4

|P_t,W^η ◦W ◦(ϕ^η_t)⁻¹(z)−P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹(z)|.

Since a plane Brownian motion started from z ∈ J₄ has probability 1/2 to hit J₃ before ∂D, so N_t ≤M_t/2.

Note that P_t,W^η −P_t,W^ζ is harmonic inD\W(K_t^ζ∪K_t^η) including the polep₀, and has vanished continuation on the sides of D other than α which contains w₀. It is clear that (ϕ^η_t)⁻¹(J₃) and (ϕ^η_t)⁻¹(J₄) are disjoint from K_t^η. Since J₃, J₄ ⊂ F, and kζ−ηk_t< δ ≤δ₁, soϕ^ζ◦(ϕ^η_t)⁻¹(J₃) andϕ^ζ◦(ϕ^η_t)⁻¹(J₄) are defined. Thus (ϕ^η_t)⁻¹(J₃) and (ϕ^η_t)⁻¹(J₄) are disjoint from K_t^ζ. Since J₃, J₄ ⊂ {z ∈ D : |z| ≥ 1−ε₀/3}, so from the last paragraph, (ϕ^η_t)⁻¹(J₃) and (ϕ^η_t)⁻¹(J₄) lie in Ω. Thus W ◦(ϕ^η_t)⁻¹(J₃) andW◦(ϕ^η_t)⁻¹(J₄) are two Jordan curves inD, and the latter disconnects the former fromW(K_t^ζ))∪W(K_t^η) and the side that w₀ lies on. Now define

M⁰ = sup

z∈J3

|P_t,W^η ◦W ◦(ϕ^η_t)⁻¹(z)−P_t,W^ζ ◦W ◦(ϕ^η_t)⁻¹(z)|;

N⁰ = sup

z∈J4

|P_t,W^η ◦W ◦(ϕ^η_t)⁻¹(z)−P_t,W^ζ ◦W ◦(ϕ^η_t)⁻¹(z)|.

From the maximal principle, we haveM⁰ ≤N⁰.

LetA₁ be the closure of the domain bounded byγ and J₁. Since |(ϕ^ζ_t)⁻¹(z)|<|z|, andJ₁ lies betweenϕ^ζ_t(γ) and∂D, so (ϕ^ζ_t)⁻¹(J₁)⊂A₁. It is clear thatP_t,W^ζ ≤P₀, the Green function in D with the pole at p₀. So P_t,W^ζ ◦W ◦(ϕ^ξ_t)⁻¹ onJ₁ is bounded by the maximum ofP₀◦W on A₁. Let A₂ be the domain bounded by J₂ and ∂D. Since P_t,W^ζ ◦W ◦(ϕ^ξ_t)⁻¹ vanishes on∂D, by reflection principle and Harnack principle, the gradient of P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹ onA₂ is bounded by some C₂ >0 depending on D, p, W, A₁ and ε₀. Since J₃, J₄ ⊂F, so by (3.2.9), if kζ−ηk_t < δ, then

|z−ϕ^ζ_t ◦(ϕ^η_t)⁻¹(z)|< C1tδ < ε0/6, ∀z ∈J3∪J4.

It follows that the line segment [z, ϕ^ζ_t ◦(ϕ^η_t)⁻¹(z)] is contained in A2, forz ∈J3∪J4. We now have

|M −M⁰| ≤ sup

z∈J3

|P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹(z)−P_t,W^ζ ◦W ◦(ϕ^η_t)⁻¹(z)|.

≤ sup

w∈A2

|∇(P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹)(w)| ·sup

z∈J3

|z−ϕ^ζ_t ◦(ϕ^η_t)⁻¹(z)| ≤C₁C₂tkζ−ηk_t. Similarly, |N −N⁰| ≤C₁C₂tkζ−ηk_t. Thus

M ≤M⁰+C₁C₁tkζ−ηk_t≤N⁰+C₁C₂tkζ−ηk_t

≤N + 2C₁C₂tkζ−ηk_t≤M/2 + 2C₁C₂tkζ−ηk_t,

which implies that M ≤4C₁C₂tkζ−ηk_t. LetC = 4C₁C₂ and ε=ε₀/3. The proof of this lemma is completed by the maximal principle. 2

Lemma 3.2.3 Leta andβ be as in Lemma 3.2.1. Then there areδ, C >0depending on D, p₀, W, Ω, β and a, such that if t∈(0, a]and ζ, η∈C[0, t]satisfy kζ−ηk_t< δ and K_a^ζ∩β =∅, then

|(∂_x∂_y/∂_y)(P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹◦eⁱ)(ζ(t))

−(∂_x∂_y/∂_y)(P_t,W^η ◦W ◦(ϕ^η_t)⁻¹◦eⁱ)(η(t))| ≤ C(1 +t)kζ−ηk_t. (3.2.10)

Proof. Letε₀,J_k,k = 1, . . . ,4, andA_l,l = 1,2, be defined as in the proof of Lemma 3.2.2. In that proof we see that for allξ∈C[0, t], 0< t≤a,P_t,W^ξ ◦W◦(ϕ^ξ_t)⁻¹ onJ₁ is uniformly bounded. By reflection principle and Harnack principle, this implies that

∂_x^k∂_y(P_t,W^ξ ◦W◦(ϕ^ξ_t)⁻¹◦eⁱ),k = 0,1,2, are uniformly bounded onR. LetP(z) denote the Green function with the pole at p₀, in the subdomain of D that is bounded by W(J₂) and the sides of D that does not contains w₀. Then P_t,W^ξ (z)≥ P(z)≥ C₀ on W(J₁), for someC₀ >0. This implies that|∂_y(P_t,W^ξ ◦W◦(ϕ^ξ_t)⁻¹◦eⁱ)|onRis uniformly bounded from below byC₀/(−ln(1−ε₀)). Since∂_x(∂_x∂_y/∂_y) =∂_x²∂_y/∂_y−(∂_x∂_y/∂_y)²,

we proved that for some C >0,

|∂_x(∂_x∂_y/∂_y)(P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹◦eⁱ)(z)| ≤C. (3.2.11)

By Lemma 3.2.2 there are δ, C1 >0 such that

|P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹(z)−P_t,W^η ◦W ◦(ϕ^η_t)⁻¹(z)| ≤C₁tkζ−ηk_a,

for t ∈ (0, a] and z ∈ J₃. By Harnack principle, there is C₂ > 0 such that for all z ∈R,

|∂_y(P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹◦eⁱ)(z)−∂_y(P_t,W^η ◦W ◦(ϕ^η_t)⁻¹◦eⁱ)(z)| ≤C₂tkζ−ηk_t,

|∂x∂y(P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹◦eⁱ)(z)−∂x∂y(P_t,W^η ◦W ◦(ϕ^η_t)⁻¹◦eⁱ)(z)| ≤C2tkζ−ηkt. From the first part of the proof, ∂y and ∂x∂y are uniformly bounded, and ∂y is uni- formly bounded from zero, so we have

|(∂_x∂_y/∂_y)(P_t,W^ζ ◦W ◦(ϕ^ζ_t)⁻¹◦eⁱ)(z)

−(∂_x∂_y/∂_y)(P_t,W^η ◦W ◦(ϕ^η_t)⁻¹◦eⁱ)(z)| ≤ Ctkζ−ηk_t. (3.2.12) The proof is completed by combining (3.2.11) and (3.2.12). 2

Lemma 3.2.4 There are a, C > 0 such that for z ∈R, t ∈(0, a], and ζ, η ∈C[0, t], the inequality (3.2.10) holds.

Proof. There is a Jordan curve β in Ω such that the doubly connected domain bounded by β and ∂D is contained in Ω. We can find a >0 such that any hull in D of capacity w.r.t. 0 less a is disjoint from β. Thus for any t ∈ (0, a] and ξ ∈ C[0, t], K_t^ξ ∩β = ∅. From Lemma 3.2.3, the inequality (3.2.10) holds when ζ, η ∈ C[0, t]

with kζ − ηk_t < δ for some δ > 0. The condition kζ −ηk_t < δ can be dropped because we can choose ξ₀ =ζ, ξ₁, . . . , ξ_n =η in C[0, t] such that kξ_j−1−ξ_jk_t< δ and P_n

1kξ_j−1−ξ_jk_t=kζ−ηk_t. 2

Proof of Theorem 3.2.2. Letξ0 =A. Define ξn inductively:

ξ_n+1(t) = A(t) +λ Z _t

0

(∂_x∂_y/∂_y)(P_s,W^ξn ◦W ◦(ϕ^ξ_sⁿ)⁻¹◦eⁱ)(ξ_n(s))ds. (3.2.13)

CompareP_t+ε,W^ξ ◦W◦(ϕ^ξ_t+ε)⁻¹ withP_t,W^ξ ◦W◦(ϕ^ξ_t)⁻¹in the similar way as we compare P_t,W^ζ ◦W◦(ϕ^ζ_t)⁻¹ withP_t,W^η ◦W◦(ϕ^η_t)⁻¹, and we find that (∂_x∂_y/∂_y)(P_s,W^ξ ◦W◦(ϕ^ξ_s)⁻¹◦ eⁱ)(ξ_n(t)) is continuous in t. Thus ξ_n are well defined and continuous.

Choose ε₀ >0 such that (1−ε)∂D⊂Ω for ε ∈(0, ε₀]. Let a and C be in Lemma 3.2.4 depending on ε₀. We may chose a small enough such that Lemma 3.2.4 implies kξ_n+2 −ξ_n+1k_a ≤ kξ_n+1−ξ_nk_a/2. Thus {ξ_n} is a Cauchy sequence in C[0, a]. Let ξ ∈C[0, a] be the limit of {ξn}. Then ξsolves (3.2.5) for t∈[0, a]. If there is another solutionξ⁰ on [0, a] of (3.2.5), then a similar argument gives kξ−ξ⁰k_a≤ kξ−ξ⁰k_a/2, which forces that ξ⁰ =ξ. Thus the solution of (3.2.5) exists uniquely on [0, a].

Now suppose (3.2.5) has two different solutions ξ₁ andξ₂. Letb >0 be the biggest tsuch thatξ1(t) = ξ2(t). Letwbbe the prime end ofDb :=D\W(K_b^ξ1) determined by the Loewner chain t7→W(K_t^ξ1) at time b. Then Σ_b := Σ\W(K_b^ξ1) is a neighborhood of wb in Db. Moreover, Ωb := ϕ^ξ_b¹(Ω\K_b^ξ1) is a neighborhood of e^iξ1(b) in D, and W_b :=W ◦(ϕ^ξ_b)⁻¹ maps Ω_b conformally onto Σ_b and can be extended conformally to the boundary such that Wb(e^iξ1(b)) = wb and Wb(Σb ∩∂D) ⊂ ∂Dˆ b. Let Pe_t,W^ζ _b be the Green function in D_b\W_b(K_t^ζ) with the pole atp₀. For j = 1,2, letζ_j(t) :=ξ_j(b+t).

Then ϕ^ξ_b+t^j =ϕ^ζ_t^j ◦ϕ^ξ_b^j, and K_b+t^ξj =K_b^ξj ∪(ϕ^ξ_b)⁻¹(K_t^ζj). Thus Pe_t,W^ζj _b =P_b+t,W^ξj . So we have

Pe_t,W^ζj _b ◦Wb◦(ϕ^ζ_t^j)⁻¹ =P_b+t,W^ξj ◦W ◦(ϕ^ξ_b+t^j )⁻¹.

LetA(t) =e A(b+t)−A(b) +ξ1(b). Then for j = 1,2, ζj solves the equation

ζ_j(t) =A(t) +e λ Z _t

0

(∂_x∂_y/∂_y)(Pe_s^ζj◦W_b◦(ϕ^ζ_s^j)⁻¹◦eⁱ)(ζ_j(s))ds. (3.2.14)

From the first part of this proof, we see that ζ₁(t) = ζ₂(t) for t ∈ (0, c), for some c > 0. Thus ξ₁(t) = ξ₂(t) for 0 ≤ t ≤b+c, which contradicts the choice of b. Thus the solution of (3.2.5) is unique.

Now let [0, T) be the maximal definition interval of the solution ξ of (3.2.5).

Suppose that ∪_0≤t<TK_t^ξ does not intersect some Jordan curve J₁ in Ω such that the doubly connected domain bounded byJ1 and∂Dis contained in Ω. Then we can find another Jordan curveJ₂that has the above properties thatJ₁has, and is disconnected byJ1 from∂D. Letm >0 be the modulus of the doubly connected domain bounded by J₁ and J₂. Then for anyt ∈[0, T), the modulus of the doubly connected domain bounded by ∂D∪K_t^ξ and J2 is greater than m. There is ε0 = ε0(m)> 0 such that ϕ^ξ_t(J₂) ⊂ (1−ε₀)D for all 0 ≤ t < T. Let a depending on ε₀ be as in the first part of this proof. Choose b ∈ [0, T) so that T < b+a. Let Ωb := ϕ^ξ_t(Ω\K_b^ξ). Then (1−ε)∂D⊂Ω_bfor 0< ε≤ε₀. Thus (3.2.14) has a solutionζon [0, a]. Letξ⁰(t) =ξ(t) for 0 ≤ t ≤ b, and ξ⁰(t) = ζ(t−b) for b ≤ t ≤ b+a. Then ξ⁰ solves (3.2.5). Note T < b+a. By the uniqueness of the solution, ξ⁰(t) = ξ(t) for 0 ≤ t < T. So the solution ξ can be extended to [0, b+a]. This contradicts the assumption that [0, T) is the maximal definition interval of the solution.

Suppose ξ₀ is the solution forA₀ and is defined on [0, a]. To prove thatS_ais open in k · k_a and S_a 3 A 7→ ξ|_[0,a] is (k · k_a,k · k_a) continuous, it suffices to show that if A_n →A₀ in k · k_a, and ξ_n are solutions for A_n, then ξ_n is defined on [0, a] for n big enough andkξ_n−ξ₀k_a →0. Letβ be a Jordan curve in Ω disjoint fromK_a^ξ0 such that the doubly connected domain bounded by β and ∂D is contained in Ω. Letδ >0 be as in Lemma 3.2.3. If kξ_n−ξ₀k_t< δ for t∈(0, a], then from Lemma 3.2.3,

|ξn(s)−ξ0(s)| ≤ kAn−A0ka+C Z _s

0

|ξn(r)−ξ0(r)|dr,

for some C > 0 and s∈ [0, t]. This implies that |ξ_n(s)−ξ₀(s)| ≤ e^CskA_n−A₀k_a. If n is big enough, then kAn−A0ka < e^−Caδ. Suppose ξn is not defined on [0, a], then there is some t₀ ∈[0, a) such that K_t^ξ₀ⁿ∩β 6=∅. However, since

kξ_n−ξ₀k_t0 ≤e^Ct0kA_n−A₀k_a < δ,

and K_t^ξ₀⁰ ∩β = ∅, from Lemma 3.2.1, K_t^ξ₀ⁿ does not intersect β. The contradiction shows thatξ_n is defined on [0, a]. From a similar argument, we conclude that there is not∈[0, a] such that|ξn(t)−ξ0(t)| ≥δ. It follows thatkξn−ξ0ka≤e^CakAn−A0ka. Thusξ_n→ξ₀ ink · k_a. From the proofs of the Lemmas it is clear that ξ7→∂_y(P_t,W^ξ ◦ W◦(ϕ^ξ_t)⁻¹◦eⁱ)(ξ(t))|_t∈[0,a]is (k · ka,k · ka) continuous on the set ofξ∈[0, a] such that K_a^ξ ⊂ Ω. So it is also true if the first ξ in ghe above sentence is replaced by A ∈S_a. 2

Suppose P_t,W^ξ is defined to be the harmonic function of some side arc I of D in D\W(L^ξ_t) such that w0 6∈ I, or the minimal function in D\W(L^ξ_t) with the pole at some prime end w₁ 6= w₀, normalized by (J, q,1). If I or w₁ does not lie on α, then Theorem 3.2.2 holds without any change, and the proof is also the same. If I orw₁ lies on α, then Theorem 3.2.2 holds with the change of the long-range behavior of L^ξ_t, which is ∪_0≤t<TL^ξ_t either intersects every Jordan curve β in Ω such that the doubly connected domain bounded byβ and∂Dis contained in Ω,or intersects every neighborhood of I or w₁. The proof is a bit more complicated, but the basic idea is the same. And we choose to omit the proof.