# Continuity of the Restriction Maps on Smirnov Classes

Authors: Yüksel Soykan
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We prove the restriction maps define continuous linear operators on the Smirnov classes for some certain domain with analytic boundary.


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## Body

## 1. Introduction

As usual, we define the Hardy spaceH
2
=
H
2
(
Δ
) as the space of all functions f
:
z
→
∑
n
=
0
∞
‍
a
n
z
n for which the norm (
∥
f
∥
=
∑
n
=
0
∞
‍
|
a
n
|
2
)
1
/
2 is finite. Here, Δ is the open unit disc. For a more general simply connected domain D in the sphere or extended plane C
¯
=
C
∪
(
∞
) with at least two boundary points, and a conformal mapping φ from D onto Δ (i.e., a Riemann mapping function, abbreviation is RMF), a function g analytic in D is said to belong to the Smirnov class E
2
(
D
) if and only if g
=
(
f
∘
φ
)
φ
′
1
/
2 for some f
∈
H
2
(
Δ
) where φ
′
1
/
2 is an analytic branch of the square root of φ
′. The reader is referred to [1–7] and references therein for the basic properties of these spaces.LetC
=
(
C
1
,
C
2
,
C
3
,
…
,
C
N
) be an N-tuple of closed distinct curves on the sphere C
¯ and suppose that, for each i,  1
≤
i
≤
N,  C
i is a circle, a line ∪
{
∞
}, an ellipse, a parabola ∪
{
∞
}, or a branch of a hyperbola ∪
{
∞
}. Let D
i be the complementary domain of C
i. Recall that a complementary domain of a closed F
⊆
C
¯ is a maximal connected subset of C
¯
-
F, which must be a domain. For 1
≤
i
≤
N, suppose that φ
i
:
D
i
→
Δ is a conformal equivalence (i.e., RMF) and let ψ
i
:
Δ
→
D
i be its inverse. For 1
≤
i
≤
N, let us keep the notations of C
i,  D
i,  φ
i,  ψ
i fixed until the end of the paper.In this paper we prove the following.Theorem 1.
Let1
≤
i,  j
≤
N. Suppose that Γ is an open subarc of  C
j and suppose also that Γ
⊆
D
i  if  i
≠
j. Then the restriction f
→
f
|
Γ defines a continuous linear operator mapping E
2
(
D
i
) into L
2
(
Γ
).For similar work regarding restriction maps, see [8, 9]. Our conjecture is that Theorem 1 is valid if, for each j,  1
≤
j
≤
N,  C
j is a σ-rectifiable analytic Jordan curve.There are some similar results for rectifiable curves in Havin’s paper [10]. Also the Cauchy projection operator from L
p to E
p is bounded on all Carleson regular curves; compare the papers of David, starting with [11].We need the following Theorem to simplify the proof of Theorem1.Theorem 2 (Theorem 1 in [12]).
LetD be a complementary domain of  ∪
i
=
1
N
C
i and suppose that D is simply connected so that D
i is the complementary domain of C
i which contains D. Then(i)
∂
D is a σ-rectifiable closed curve and every f
∈
E
2
(
D
) has a nontangential limit function f
~
∈
L
2
(
∂
D
);
(ii)
(Parseval’s identity) the mapf
→
f
~  (
E
2
(
D
)
→
L
2
(
∂
D
)
) is an isometric isomorphism onto a closed subspace E
2
(
∂
D
) of L
2
(
∂
D
), so(1)
∥
f
∥
E
2
(
D
)
2
=
∥
f
~
∥
L
2
(
∂
D
)
2
=
1
2
π
∫
∂
D
‍
|
f
~
(
z
)
|
2
|
d
z
|
,
(
f
∈
E
2
(
D
)
)
.IfΓ
⊆
C
i is an open subarc, then
(2)
∥
f
~
|
Γ
∥
L
2
(
Γ
)
2
≤
∥
f
~
|
C
i
∥
L
2
(
C
i
)
2
=
∥
f
∥
E
2
(
D
i
)
2
,
because Parseval’s identity is true for the trivial chain (C
i) of curves. Hence Theorem 1 will be proved if the following theorem can be proved.Theorem 3.
Let1
≤
i
≠
j
≤
N. Suppose that Γ is an open subarc of C
j and that Γ
⊆
D
i. Then the restriction f
→
f
|
Γ defines a continuous linear operator mapping E
2
(
D
i
) into L
2
(
Γ
).


## 2. Preliminaries for the Proof of Theorem3

Let us keep the notation of Theorem3 fixed for the rest of the paper and let us also agree to use l for arc-length measure.An arc or closed curveγ is called σ-rectifiable if and only if it is a countable union of rectifiable arcs in C, together with (
∞
) in the case when ∞
∈
γ. For instance, a parabola without ∞ is σ-rectifiable arc, and a parabola with ∞ is σ-rectifiable Jordan curve. The following definition will simplify the language.Definition 4.
Letγ
⊆
C be a simple σ-rectifiable arc contained in a simply connected domain G
⊆
C
¯. We say that γ has the restriction property in G  if and only if the map g
→
g
|
γ defines a continuous linear operator mapping E
2
(
G
) into L
2
(
γ
).
Thus, the last sentence of Theorem3 reads “Γ has the restriction property in D
i.”Lemma 5 (Invariance Lemma (Lemma 4 in [9])).
LetG
1
,
G
2
⊆
C
¯ be simply connected domains and suppose that γ
1
⊆
G
1
∩
C,  γ
2
⊆
G
2
∩
C are simple σ-rectifiable arcs. If χ
:
G
1
→
G
2 is a conformal equivalence onto G
2 and χ
(
γ
1
)
=
γ
2, then γ
1 has the restriction property in G
1 if and only if γ
2 has the restriction property in G
2.Corollary 6.
Theorem3 is true; that is, Γ has the restriction property in D
i, if and only if φ
i
(
Γ
) has the restriction property in Δ, for some RMF φ
i
:
D
i
→
Δ.Asubarc
γ of Γ has the restriction property in D
i if and only if φ
i
(
γ
) has the restriction property in Δ. Corollary 6 will be used in the following way. Γ will be written as the union of finitely many subarcs and we will show that each of these subarcs has the restriction property in D
i; it will then follow that Γ itself has the required restriction property. Three different kinds of subarc will be considered.Definition 7.
A subarcγ
⊆
Γ is said to be of type I if and only if γ
¯
⊆
D
i (i.e., both of its end-points a
,
b belong to D
i).Lemma 8 (Lemma 6 in [9]).
Letγ be a subarc of Γ and suppose that φ
i,  θ
i are Riemann mapping functions for D
i.(i)
φ
i
(
γ
) has the restriction property in Δ if and only if  θ
i
(
γ
) has the restriction property in Δ;
(ii)
φ
i
(
γ
) is rectifiable if and only if θ
i
(
γ
) is rectifiable;
(iii)
ifγ is of type I, then φ
i
(
γ
)
¯
⊆
Δ and φ
i
(
γ
) is rectifiable;
(iv)
ifγ is of type I, it has the restriction property in D
i.We can now “ignore” subarcs ofΓ whose closure (in C
¯) is contained in D
i. We will now restrict our attention to subarcs of Γ with a single end-point a
∈
∂
D
i, the other being in D
i. There are two types, depending on whether a
∈
C or a
=
∞.Definition 9.
(i) An open subarcγ of Γ is of type II if and only if it has an end-point a
∈
∂
D
i
∩
C and γ
¯
-
(
a
)
⊆
D
i
∩
C.
(ii) In the case whereC
i is unbounded (so that ∞
∈
∂
D
i) an open subarc γ
⊆
Γ is of type III if and only if ∞ is an end-point of γ and γ
¯
-
(
∞
)
⊆
D
i.Modulo a finite subset ofD
i
,
Γ is the union of at most three open subarcs, each of which is of type I, II, or III; see Figure 1.Figure 1
Type I, II, and III arcs.Ifγ is a type II or type III subarc of Γ then φ
i
(
γ
) is a simple open analytic arc in Δ with one end-point on the circle T and the other in Δ. We will show that φ
i
(
γ
) has the restriction property in Δ using the powerful Carleson theorem (Theorem 11 below).Definition 10 (see [1, p.157]).
For0
<
h
<
1 and 0
≤
θ
<
2
π, let C
θ
h
=
{
z
∈
C
:
1
-
h
≤
|
z
|
≤
1
,
θ
≤
arg
z
≤
θ
+
h
}. A positive regular Borel measure μ on Δ is called a Carleson measure if there exists a positive constant M such that μ
(
C
θ
h
)
≤
M
h, for every h and every θ.Theorem 11 (see [1, p. 157, Theorem 9.3] or see [13, p. 37]).
Letμ be a finite positive regular Borel measure on Δ. In order that there exists a constant C
>
0 such that
(3)
∫
Δ
‍
|
f
(
z
)
|
2
d
μ
(
z
)
≤
C
∥
f
∥
2
,
∀
f
∈
H
2
(
Δ
)
,
it is necessary and sufficient that μ be a Carleson measure.
To complete the proof of Theorem3 it is sufficient to show that arc-length measure on φ
i
(
γ
) is a Carleson measure whenever γ is of type II or III.It will be useful to use arc-length to parametrizeγ and φ
i
(
γ
). Recall that a compact arc σ is calledsmooth if there exists some parametrization g
:
[
a
,
b
]
→
σ such that g
∈
C
1
[
a
,
b
] and g
′
(
t
)
≠
0, ∀
t
∈
[
a
,
b
]. Note that if σ is smooth, then it is rectifiable; that is,
(4)
l
(
σ
)
=
∫
a
b
‍
|
g
′
(
t
)
|
d
t
<
∞
.To define the arc-length parametrization ofσ put s
=
s
(
t
)
=
∫
a
t
‍
|
g
′
(
u
)
|
d
u for a
≤
t
≤
b so that 0
≤
s
≤
l
(
σ
). Then s
′
(
t
)
=
|
g
′
(
t
)
| and t
→
s
(
t
)  (
[
a
,
b
]
→
[
0
,
l
]
) is C
1 with strictly positive derivative. Hence also its inverse s
→
t
(
s
)  (
[
0
,
l
]
→
[
a
,
b
]
) is C
1 with strictly positive derivative. Recall that the arc-length parametrization of the smooth arc σ is the map h
:
[
0
,
l
]
→
σ satisfying h
(
s
)
=
{the point on σ length s from the initial point (
g
(
a
)
)
}; that is, h
(
s
)
=
g
(
t
(
s
)
)  0
≤
s
≤
l.Sinceh
′
(
s
)
=
g
′
(
t
(
s
)
)
t
′
(
s
), h
∈
C
1
[
0
,
l
], with nonzero derivative, necessarily |
h
′
(
s
)
|
=
1 since
(5)
h
′
(
s
(
t
)
)
=
g
′
(
t
)
t
′
(
s
)
=
g
′
(
t
)
s
′
(
t
)
=
g
′
(
t
)
|
g
′
(
t
)
|
.
We need the following lemma.Lemma 12 (Theorem 1 in [14]).
Letσ
⊆
Δ
¯ be a smooth simple arc with arc-length parametrization g
∈
C
1
[
0
,
l
]. Suppose that |
g
(
0
)
|
=
1,  |
g
(
s
)
|
<
1  for 0
<
s
≤
l. Then arc-length measure on σ
∩
Δ is a Carleson measure; hence σ
∩
Δ has the restriction property in Δ.


## 3. Type II Subarcs

The following lemma gives the continuity of the restriction map for finite end-points.Lemma 13.
A type II arcγ
⊆
Γ
⊆
D
i has the restriction property in D
i.Proof.
By Lemmas12 and 5 it is sufficient to show that φ
i
(
γ
)
¯ is a smooth arc in Δ
¯. Suppose that γ has end-points a
∈
∂
D
i
∩
C and b
∈
D
i
∩
C, so that γ
¯
=
γ
∪
(
a
)
∪
(
b
). Clearly γ
¯ is a smooth arc. Because C
i is an open analytic arc, φ
i can be continued analytically into a neighbourhood U of a so as to be conformal in D
i
∪
U. This means that φ
i is conformal in a neighbourhood of γ
¯ and so φ
i
(
γ
)
¯
=
φ
i
(
γ
¯
) is a smooth arc in Δ
¯ with |
φ
i
(
a
)
|
=
1 and φ
i
(
γ
¯
-
(
a
)
)
⊆
Δ. The result now follows from Lemmas 12 and 5.We have now made a good deal of progress because of the following.Lemma 14.
Theorem3 is true if C
i is a circle or an ellipse.Proof.
In this caseΓ is a finite union of type I and type II arcs only, so the result follows by Lemma 8(iv) and Lemma 13.


## 4. Type III Subarcs

The proof of Theorem3 will be completed by showing that every type III arc in D
i has the restriction property in D
i. We have an open subarc γ of an open subarc Γ of C
j and Γ
⊆
D
i. In this case ∞ is an end-point of γ and ∞
∈
∂
D
i, so both C
i and C
j are unbounded. We will use the same strategy we used for type II arcs in Lemma 13; we show that σ
=
φ
i
(
γ
)
¯ is a smooth arc in Δ as in Lemma 12, so that φ
i
(
γ
) has the restriction property in Δ and so γ has the restriction property in D
i. The proof is more complicated because conformality of φ
i at ∞ cannot necessarily be used. Instead we make use of the fact thatas
z
→
∞
along
γ,the unit tangent vector of
γ
at
z
tends to a limit. The following two Lemmas help us exploit this fact.Lemma 15.
Letg
∈
C
1
[
0
,
∞
) with g
′
(
t
)
≠
0  (
t
≥
0
). Suppose that c
∈
C and
(6)
lim
⁡
t
→
∞
g
(
t
)
=
c
,
lim
⁡
t
→
∞
g
′
(
t
)
|
g
′
(
t
)
|
=
ω
,
(
|
ω
|
=
1
)
exist. Define σ
=
g
(
[
0
,
∞
)
)
∪
(
c
). Then(i)
σ is a compact arc,
(ii)
σ is rectifiable,
(iii)
σ is smooth.Proof.
(i) Definef on [
0,1
] by
(7)
f
(
t
)
=
{
g
(
tanh
-
1
t
)
0
≤
t
<
1
c
t
=
1
.
Then f
∈
C
[
0,1
] is a continuous parametrization of σ.
(ii) To prove thatσ is rectifiable, it suffices to show that, for some T
>
0,  ∫
T
∞
‍
|
g
′
(
u
)
|
d
u
<
∞. Let ε
(
t
)
=
ω
-
(
g
′
(
t
)
/
|
g
′
(
t
)
|
). So ε
(
t
)
→
0 as t
→
∞. Choose T
≥
0 such that |
ε
(
t
)
|
≤
1
/
2 for t
≥
T. Then, for t
≥
T,
(8)
|
g
′
(
t
)
|
(
1
-
ω
¯
ε
(
t
)
)
=
ω
¯
g
′
(
t
)
.
Hence(9)
∫
T
t
‍
|
g
′
(
u
)
|
(
1
-
ω
¯
ε
(
u
)
)
d
u
=
ω
¯
(
g
(
t
)
-
g
(
T
)
)
,
(
t
>
T
)
,
|
ε
|
≤
1
2
⟹
Re
(
1
-
ω
¯
ε
)
≥
1
2
⟹
2
Re
(
1
-
ω
¯
ε
)
≥
1
.
So(10)
∫
T
t
|
g
′
(
u
)
|
d
u
≤
2
∫
T
t
|
g
′
(
u
)
|
Re
(
1
-
ω
¯
ε
(
u
)
)
d
u
=
2
Re
(
ω
¯
(
g
(
t
)
-
g
(
T
)
)
)
⟶
2
Re
(
ω
¯
(
c
-
g
(
T
)
)
)
as
t
⟶
∞
,
and hence
(11)
∫
T
∞
‍
|
g
′
(
u
)
|
d
u
<
∞
,
which establishes the rectifiability of σ.
(iii) Leth
:
[
0
,
l
]
→
σ be the arc-length parametrization of σ. Then h
∈
C
[
0
,
l
], h
(
s
)
=
g
(
t
) where ∫
0
t
‍
|
g
′
(
u
)
|
d
u
=
s and s
′
(
t
)
=
|
g
′
(
t
)
|. Therefore the map t
→
s  (
[
0
,
∞
)
→
[
0
,
l
)
) is C
1 with strictly positive derivative. So the inverse map s
→
t  (
[
0
,
l
)
→
[
0
,
∞
)
) is C
1. Since t
(
s
(
t
)
)
≡
t and t
′
(
s
)
=
1
/
s
′
(
t
) where 0
≤
t
≤
∞ and 0
≤
s
≤
l, it follows that
(12)
lim
⁡
s
→
l
h
′
(
s
)
=
lim
⁡
t
→
∞
g
′
(
t
)
t
′
(
s
)
=
lim
⁡
t
→
∞
g
′
(
t
)
s
′
(
t
)
=
lim
⁡
t
→
∞
g
′
(
t
)
|
g
′
(
t
)
|
=
ω
.
Hence h
′ is continuous and so h
∈
C
1
[
0
,
l
].Lemma 16.
Letk
∈
C
1
[
0
,
∞
) with k
′
(
t
)
≠
0  (
t
≥
0
) and suppose that k
(
t
)
→
∞ as t
→
+
∞. Then, if |
ω
|
=
1,
(13)
k
′
(
t
)
|
k
′
(
t
)
|
⟶
ω
⟹
k
(
t
)
|
k
(
t
)
|
⟶
ω
.Proof.
Writeω
=
e
i
α. Choose T
′ such that t
≥
T
′
⇒
Re
e
-
i
α
(
k
′
(
t
)
/
|
k
′
(
t
)
|
)
>
0. Then using arg
^ to denote the principal value of arg we see that
(14)
θ
(
t
)
=
α
+
arg
^
e
-
i
α
k
′
(
t
)
|
k
′
(
t
)
|
is a branch of arg
⁡
(
k
′
/
|
k
′
|
) and hence also of arg
⁡
k
′ on [
T
′
,
∞
) which tends to α as t
→
∞. We will find a branch ϑ of arg
k which also tends to α as t
→
∞.
Letε
>
0. Choose T such that t
≥
T
≥
T
′
⇒
α
-
ε
/
2
≤
θ
≤
α
+
ε
/
2. Now k
(
t
)
-
k
(
T
)
=
∫
T
t
‍
k
′
(
u
)
d
u is a limit of Riemann sums ∑
‍
(
t
i
+
1
-
t
i
)
k
′
(
ξ
i
).
The sectorS (see Figure 2) is closed under addition and multiplication by positive scalars; therefore
(15)
k
(
t
)
-
k
(
T
)
∈
S
for
t
≥
T
.
So there is an argument μ
(
t
) of k
(
t
)
-
k
(
T
) satisfying
(16)
α
-
ε
2
≤
μ
(
t
)
≤
α
+
ε
2
(
t
≥
T
)
.
Now k
(
t
)
/
(
k
(
t
)
-
k
(
T
)
)
→
1 as t
→
∞. So
(17)
∃
T
1
≥
T
such
that
t
≥
T
1
⟹
-
ε
2
<
arg
^
k
(
t
)
k
(
t
)
-
k
(
T
)
<
ε
2
.
If we define
(18)
ϑ
(
t
)
=
μ
(
t
)
+
arg
^
k
(
t
)
k
(
t
)
-
k
(
T
)
(
t
≥
T
1
)
,
then ϑ
(
t
) is an argument of k
(
t
) and
(19)
t
≥
T
1
⟹
|
ϑ
(
t
)
-
α
|
<
ε
2
+
ε
2
=
ε
.
Hence also
(20)
|
k
(
t
)
|
k
(
t
)
|
-
ω
|
=
|
e
i
ϑ
(
t
)
-
e
i
α
|
<
ε
.
Consequently,
(21)
k
(
t
)
|
k
(
t
)
|
⟶
ω
=
e
i
α
,
and our Lemma is proved.Figure 2
The sectorS.There are now four cases to prove depending on the geometry ofC
i and D
i.


### 4.1. Case  1:D
i Is a Half-Plane

The following lemma will be needed here and in Case2.Lemma 17.
LetG be the open right half-plane Re
z
>
0 and let θ
(
z
)
=
(
z
-
1
)
/
(
z
+
1
) so that θ is a Riemann mapping function for G. Let k
:
[
0
,
∞
)
→
G be an injective C
1 function such that k
′
(
t
)
≠
0, for all t
≥
0, and lim
⁡
t
→
∞
k
(
t
)
=
∞. Let ρ be the (simple) arc parametrized by k. If  lim
⁡
t
→
∞
(
k
′
(
t
)
/
|
k
′
(
t
)
|
)
=
ω  (with |
ω
|
=
1), then σ
=
θ
(
ρ
)
¯ satisfies the hypothesis of Lemma 12 and, hence, ρ has the restriction property in G.Proof.
Putg
=
θ
∘
k, so that g
∈
C
1
[
0
,
∞
) parametrizes θ
(
ρ
). Clearly g
(
t
)
→
1 as t
→
∞. Now g satisfies the hypothesis of Lemma 15, for we can show that g
′
(
t
)
/
|
g
′
(
t
)
|
→
ω
-
1 as t
→
∞. Since θ
′
(
z
)
=
2
/
(
z
+
1
)
2 it follows that
(22)
g
′
(
t
)
|
g
′
(
t
)
|
=
|
1
+
k
(
t
)
|
2
(
1
+
k
(
t
)
)
2
k
′
(
t
)
|
k
′
(
t
)
|
=
k
′
(
t
)
|
k
′
(
t
)
|
|
k
(
t
)
|
2
(
k
(
t
)
)
2
|
1
+
1
/
k
(
t
)
|
2
(
1
+
1
/
k
(
t
)
)
2
⟶
ω
-
1
,
using Lemma 16.
Soσ
=
g
[
0
,
∞
)
∪
(
ω
-
1
) satisfies Lemma 12; hence g
[
0
,
∞
) has the restriction property in Δ. But g
[
0
,
∞
)
=
θ
(
ρ
) and, therefore, by Lemma 5, ρ has the restriction property in G.Now suppose thatC
i is a line and D
i is a half-plane. By Invariance Lemma 5 with a linear equivalence χ
(
z
)
=
α
z
+
β  (
α
≠
0
) we can assume that C
i is the imaginary axis and that D
i
=
G, the open right half-plane, as above. If γ
⊆
D
i is a type III arc, it is a subarc of a line, parabola, or hyperbola component. Obviously γ has a parametrization k as in Lemma 17. Hence γ has the restriction property in D
i.


### 4.2. Case  2:D
i Is the Concave Complementary Domain of a Parabola

Any two parabolas are conformally equivalent via a linear equivalence:μ
(
z
)
=
a
z
+
b  (
a
,
b
∈
C
,
a
≠
0
). So assume that C
i is the parabola
(23)
y
2
=
4
(
1
-
x
)
and that D
i is the complementary domain to the “right” of C
i.The function(24)
w
⟶
(
1
+
w
)
2
maps the open right half-plane G conformally onto D
i and the imaginary axis onto C
i. Its inverse is the function
(25)
ϑ
(
z
)
=
z
1
/
2
-
1
,
(
z
∈
D
i
)
,
where z
1
/
2 is the principal square-root of z (here and throughout all standard multivalued functions will take their principal values).Now letγ
⊆
D
i be a type III arc. Because G is conformally equivalent to D
i via ϑ it will be sufficient to show that the arc ϑ
(
γ
)
⊆
G has a parametric function k as in Lemma 17. Letting h be the arc-length parametrization of γ, then h
∈
C
1
[
0
,
∞
),  |
h
′
(
t
)
|
≡
1 and h
(
t
)
→
∞ as t
→
∞, and h is injective.Nowγ is a subarc of a line, parabola, or hyperbola component. Hence as z
→
∞ along γ the unit tangent vector at z tends to a limit ω (|
ω
|
=
1). Thus
(26)
lim
⁡
t
→
∞
h
′
(
t
)
|
h
′
(
t
)
|
=
lim
⁡
t
→
∞
h
′
(
t
)
=
ω
,
and therefore
(27)
lim
⁡
t
→
∞
h
(
t
)
|
h
(
t
)
|
=
ω
,
by Lemma 16.Putk
=
ϑ
∘
h. Then k is an injective parametric function for ϑ
(
γ
). Clearly k
∈
C
1
[
0
,
∞
),  k
(
t
)
→
∞ as t
→
∞, and
(28)
k
′
(
t
)
=
ϑ
′
(
h
(
t
)
)
h
′
(
t
)
≠
0
,
∀
t
≥
0
.Moreover,(29)
k
′
(
t
)
|
k
′
(
t
)
|
=
|
h
(
t
)
|
1
/
2
h
(
t
)
1
/
2
h
′
(
t
)
|
h
′
(
t
)
|
⟶
ω
1
/
2
.Sok is as in Lemma 17, which shows that γ has the restriction property in D
i.Remark 18.
The notationω
1
/
2 is ambiguous when ω
=
-
1 (γ could be part of another parabola). But, because type I arcs can be ignored, we can assume that either γ is contained entirely in the upper half-plane, in which case (
-
1
)
1
/
2
=
i, or else γ is in the lower half-plane and (
-
1
)
1
/
2
=
-
i.


### 4.3. Case  3:D
i Is the Convex Complementary Domain of a Parabola

In this case the parabola(30)
y
2
=
4
(
π
4
)
2
(
(
π
4
)
2
-
x
)
will be chosen for C
i, and D
i will be the complementary domain to the “left” of C
i. This choice is made because then we have the relatively simple Riemann mapping function
(31)
φ
i
(
z
)
=
tan
2
(
z
1
/
2
)
,
(
z
∈
D
i
)
.
This function maps the real interval (
-
∞
,
(
π
/
4
)
2
) in an increasing fashion onto (
-
1,1
), and so it maps the upper/lower half of D
i onto the upper/lower half of Δ. The formula for φ
i is indeterminate on (
-
∞
,
0
], but these singularities are removable and the formula
(32)
φ
i
(
x
)
=
-
tanh
2
(
-
x
)
1
/
2
can be used to define φ
i
(
x
), for negative x. This mapping will be examined in detail in a moment, but first we dispose of a trivial case and make some simple observations.Letγ
⊆
D
i be a type III arc. If γ is a real interval (
-
∞
,
a
), with a
<
(
π
/
4
)
2, then φ
i
(
γ
) is a subinterval of (
-
1,1
) which obviously has the restriction property in Δ. So this case is trivial and needs no more attention.The following observations are elementary.(i)
Ifγ is part of another line, then it must be parallel to R and certainly disjoint from (
-
∞
,
0
].
(ii)
Ifγ is part of another parabola C
j, then C
j must be symmetric about R and have an equation of the form
(33)
y
2
=
4
a
(
b
-
x
)
,
where0
<
a
≤
(
π
/
4
)
2,  b
≤
(
π
/
4
)
2.
(iii)
Ifγ is part of a hyperbola, then its asymptote must be parallel to R.
(iv)
In all (nontrivial) casesγ intersects (
-
∞
,
0
] in at most two points. So, because type I arcs can be ignored there is no loss of generality in assuming that Im
⁡
z has constant sign on γ and that Re
z
<
0 on γ.
(v)
Hence, for definiteness, we can assume thatγ is contained in the open second quadrant.
(vi)
In all casesy
2
/
x tends to a limit as z
→
∞ along γ. If γ is part of a line or hyperbola, the limit is 0, and if γ is part of the parabola in (ii) above the limit is -
4
a. For future reference let us note that
(34)
0
≤
lim
⁡
⁡
y
2
4
|
x
|
≤
(
π
4
)
2
.
(vii)
Because thelim
⁡ in (34) exists and because type I arcs can be ignored, we can assume that
(35)
y
2
x
2
<
1
,
on
⁡
γ
.Now letγ be type III arc in D
i as in (v) and (vi). We will show that φ
i
(
γ
) has the restriction property in Δ. To elucidate φ
i
(
γ
) it is convenient to work backwards, examining the mapping properties of the square map (
z
→
z
2
), then tan, and then the principal square root.Lemma 19.
LetΔ
+ be the open semidisc
(36)
Δ
+
=
{
z
∈
C
:
|
z
|
<
1
,
x
>
0
}
.
If σ
′ is a smooth simple arc in Δ
+
¯, if i is an end-point of σ
′, and if σ
′
-
{
i
}
⊆
Δ
+, then the arc
(37)
σ
=
{
z
2
:
z
∈
σ
′
}
is a smooth simple arc in Δ
¯ satisfying the hypothesis of Lemma 12, so that σ
-
{
-
1
} has the restriction property in Δ.Proof.
This is clear: the square mapz
→
z
2 is conformal in a neighbourhood of σ
′.Now letS be the open strip
(38)
S
=
{
z
∈
C
:
0
<
x
<
π
4
}
.
It is well known that tan maps S conformally onto Δ
+. The imaginary axis is mapped to the vertical part of ∂
Δ
+, and the line π
/
4
+
i
R is mapped to the semicircular part of ∂
Δ
+. Moreover, if z tends to infinity in S in such a way that y
→
+
∞, then tan
⁡
z
→
i.Lemma 20.
Letk
∈
C
1
[
0
,
∞
) be injective and satisfy k
′
(
t
)
≠
0, for t
≥
0. Suppose also that(i)
k
(
t
)
∈
S for all t
≥
0,
(ii)
Im
⁡
k
(
t
)
→
+
∞ as t
→
+
∞,
(iii)
lim
⁡
t
→
∞
Re
k
(
t
)
=
x
0 exists (
0
≤
x
0
≤
π
/
4
),
(iv)
lim
⁡
t
→
∞
(
k
′
(
t
)
/
|
k
′
(
t
)
|
)
=
i.
Ifγ
′ is the arc parametrized by k, then σ
′
=
(
tan
⁡
γ
′
)
∪
{
i
} satisfies the hypothesis of Lemma 19, so that tan
2
γ
′ has the restriction property in Δ.Proof.
Letg
=
tan
∘
k, so that g parametrizes γ
′ and tan
⁡
γ
′
=
g
[
0
,
∞
). Now g
∈
C
1
[
0
,
∞
),  g
′
(
t
)
≠
0, for all t
≥
0, and g
(
t
)
→
i as t
→
+
∞. Lemma 15 will be used to show that σ
′
=
g
[
0
,
∞
)
∪
(
i
) satisfies the hypothesis of Lemma 19. For all t
≥
0,
(39)
g
′
(
t
)
|
g
′
(
t
)
|
=
|
cos
⁡
⁡
k
(
t
)
|
2
(
cos
⁡
⁡
k
(
t
)
)
2
k
′
(
t
)
|
k
′
(
t
)
|
.
Letk
(
t
)
=
x
(
t
)
+
i
y
(
t
). Since x
(
t
)
→
x
0 and y
(
t
)
→
+
∞, as t
→
+
∞, and because cos
⁡
x
,
cosh
⁡
y
>
0 on γ,
(40)
|
cos
⁡
⁡
k
(
t
)
|
2
cos
⁡
2
k
(
t
)
=
(
|
cos
⁡
⁡
x
(
t
)
cosh
⁡
y
(
t
)
-
i
sin
⁡
x
(
t
)
sinh
⁡
y
(
t
)
|
cos
⁡
⁡
x
(
t
)
cosh
⁡
y
(
t
)
-
i
sin
⁡
x
(
t
)
sinh
⁡
y
(
t
)
)
2
=
|
1
-
i
tan
⁡
x
(
t
)
tanh
y
(
t
)
|
2
(
1
-
i
tan
⁡
x
(
t
)
tanh
y
(
t
)
)
2
⟶
|
1
-
i
tan
⁡
x
0
|
2
(
1
-
i
tan
⁡
x
0
)
2
.
So lim
⁡
t
→
∞
(
g
′
(
t
)
/
|
g
′
(
t
)
|
) exists.The function(41)
ϑ
(
z
)
=
z
1
/
2
maps D
i
-
(
-
∞
,
0
] conformally onto the vertical strip S as above. The limiting values of ϑ from above and below a point x on (
-
∞
,
0
] are at ±
i
(
-
x
)
1
/
2, respectively. Now tan maps S conformally onto Δ
+ and tan
±
i
(
-
x
)
1
/
2
=
±
i
tanh
(
-
x
)
1
/
2. Finally the square function maps Δ
+ conformally onto Δ
-
(
(
-
1,0
]
), and it maps both of ±
i
tanh
(
-
x
)
1
/
2 and  -
tanh
2
(
-
x
)
1
/
2. Thus the cut made by ϑ is repaired by the square function (by Schwarz’s Reflection Principle): φ
i is continuous at all points of (
-
∞
,
0
] and therefore analytic on D
i. Because φ
i
(
z
)
∈
(
-
1,0
] if and only if z
∈
(
-
∞
,
0
] the injectivity of φ
i on D
i is clear.Letγ
⊆
D
i be a type III arc. Assume that y
>
0 and x
<
0 when z
=
x
+
i
y
∈
γ. Let γ
′
=
ϑ
(
γ
) so that γ
′
⊆
S. We show that γ
′ is as in Lemma 20 so that tan
2
γ
′ has the restriction property in Δ and, hence, γ has the restriction property in D
i.Letz
=
x
+
i
y be an arbitrary point of γ and write
(42)
z
1
/
2
=
u
+
i
v
,
for the corresponding point ϑ
(
z
)
∈
γ
′; then
(43)
x
+
i
y
=
u
2
-
v
2
+
2
i
u
v
.
Eliminating v, and remembering that x
<
0, we see that
(44)
u
2
=
1
2
(
x
+
(
x
2
+
y
2
)
1
/
2
)
=
|
x
|
2
(
(
1
+
y
2
x
2
)
1
/
2
-
1
)
.Sincey
2
/
x
2
<
1 (observation (vii)), the binomial series implies that
(45)
u
2
=
y
2
4
|
x
|
-
1
16
y
4
|
x
|
3
+
⋯
~
y
2
4
|
x
|
,
as z tends to ∞ along γ. It follows from (34) that
(46)
lim
⁡
t
→
∞
u
2
=
a
exists
,
0
≤
a
≤
(
π
4
)
2
.Now leth be the arc-length parametrization of γ and write h
(
t
)
=
x
(
t
)
+
i
y
(
t
). Let k
=
ϑ
∘
h
=
h
1
/
2 so that k parametrizes γ
′. Write k
(
t
)
=
u
(
t
)
+
i
v
(
t
). (i), (ii), (iii), and (iv) of Lemma 20 can now be verified.Obviouslyk
(
t
)
∈
S, for all t
≥
0, so (i) is true. As t
→
∞,  |
k
(
t
)
|
=
|
h
(
t
)
|
1
/
2
→
∞, but since 0
≤
u
(
t
)
≤
π
/
4 we must have v
(
t
)
→
+
∞, so that (ii) is true. Item (iii) follows from (46). Now h
(
t
)
→
∞ as t
→
∞,  |
h
′
(
t
)
|
≡
1, and h
′
(
t
)
→
-
1 as t
→
∞. So, by Lemma 16,
(47)
lim
⁡
t
→
∞
k
′
(
t
)
|
k
′
(
t
)
|
=
|
h
(
t
)
|
1
/
2
h
(
t
)
1
/
2
h
′
(
t
)
|
h
′
(
t
)
|
⟶
-
i
(
-
1
)
=
i
.
So (iv) is true and we have now completed the proof.


### 4.4. Case  4:C
i Is a Hyperbola Component

We can deal simultaneously with the convex and concave complementary domains of a hyperbola component as follows. Let-
π
/
2
<
α
<
π
/
2 and let C
i
=
sin
(
α
+
i
R
). If α
<
0,  C
i is the arc
(48)
C
i
=
{
z
=
x
+
i
y
∈
C
:
x
<
0
,
x
2
sin
2
α
-
y
2
cos
⁡
2
α
=
1
}
,
and if α
>
0,  C
i is the arc
(49)
C
i
=
{
z
=
x
+
i
y
∈
C
:
x
>
0
,
x
2
sin
2
α
-
y
2
cos
⁡
2
α
=
1
}
.LetD
i be the complementary domain to the “left” of C
i; then D
i is convex when α
<
0 and concave when α
>
0. Linear equivalence will be used as before to reduce the general case to this one.The functionsin
-
1 maps the double cut plane C
-
{
(
-
∞
,
-
1
]
∪
[
1
,
∞
)
} conformally onto the vertical strip |
x
|
<
π
/
2, mapping the upper/lower parts of the first domain onto the upper/lower parts of the second. The upper and lower limits of sin
-
1 at a point -
x
∈
(
-
∞
,
-
1
] are -
π
/
2
±
i
cosh
⁡
-
1
x. The arc C
i
=
sin
(
α
+
i
R
) is mapped to the line Re
z
=
α. Therefore sin
-
1 maps D
i
-
(
-
∞
,
-
1
] conformally onto the strip
(50)
D
α
=
{
z
=
x
+
i
y
∈
C
:
-
π
2
<
x
<
α
}
.If(51)
λ
(
z
)
=
π
4
z
+
(
π
/
2
)
α
+
(
π
/
2
)
,
then λ maps D
α conformally onto the strip
(52)
S
=
{
z
=
x
+
i
y
∈
C
:
0
<
x
<
π
4
}
.Therefore(53)
φ
i
(
z
)
=
tan
2
λ
(
sin
-
1
z
)
is a Riemann mapping function for D
i. Now let γ be a type III arc in D
i. As in Case 3 the case γ
⊆
R is trivial, so we can assume that γ lies entirely in the upper half-plane. It will be sufficient for us to show that λ
(
sin
-
1
γ
) has a parametric function k as in Lemma 20.Letz
=
x
+
i
y be arbitrary point of γ and write sin
-
1
z
=
u
+
i
v for the corresponding point of sin
-
1
γ. Clearly, by (50),
(54)
u
+
i
v
∈
D
α
.Now(55)
z
=
x
+
i
y
=
sin
⁡
(
u
+
i
v
)
=
sin
⁡
u
cosh
⁡
v
+
i
cos
⁡
⁡
u
sinh
⁡
v
,
so that
(56)
|
z
|
2
=
sin
2
u
cosh
⁡
2
v
+
cos
⁡
2
u
sinh
⁡
2
v
=
sin
2
u
+
sinh
⁡
2
v
.Asz
→
∞ along γ, |
z
|
2
→
+
∞ and sin
2
u remains bounded; therefore
(57)
v
⟶
+
∞
as
z
⟶
∞
along
γ
.It now follows from (56) and (57) that
(58)
sin
⁡
u
=
x
|
z
|
(
tanh
2
v
+
sin
2
u
cosh
⁡
2
v
)
1
/
2
~
x
|
z
|
as
z
⟶
∞
.Leth be the arc-length parametrization of γ. As z
→
∞ along γ its unit tangent vector has a limit e
i
θ, say. The asymptotes of C
i are the rays arg
⁡
z
=
±
(
π
/
2
-
α
). Therefore
(59)
lim
⁡
t
→
∞
h
′
(
t
)
|
h
′
(
t
)
|
=
lim
⁡
t
→
∞
h
′
(
t
)
=
e
i
θ
,
where
π
2
-
α
≤
θ
≤
π
.So, by (57) and Lemma 16,
(60)
lim
⁡
t
→
∞
h
(
t
)
|
h
(
t
)
|
=
e
i
θ
.Nowg
=
sin
-
1
∘
h is a parametric function for sin
-
1
γ. By (54) it follows that(i)
g
(
t
)
∈
D
α  (
t
≥
0
),  and (57) shows that
(ii)
Im
⁡
g
(
t
)
→
+
∞ as t
→
∞.Equation (60) shows that(iii)
lim
⁡
t
→
∞
Re
g
(
t
)
=
sin
-
1
cos
⁡
θ
=
(
π
/
2
)
-
θ  and we notice that -
π
/
2
≤
(
π
/
2
)
-
θ
≤
α, by (59).Finally observe that(61)
g
′
(
t
)
|
g
′
(
t
)
|
=
|
1
-
h
(
t
)
2
|
1
/
2
(
1
-
h
(
t
)
2
)
1
/
2
h
′
(
t
)
|
h
′
(
t
)
|
.Now in the upper half-plane(
1
-
w
2
)
1
/
2
~
-
i
w, as w
→
∞. So, as t
→
∞,
(62)
g
′
(
t
)
|
g
′
(
t
)
|
~
|
h
(
t
)
|
-
i
h
(
t
)
h
′
(
t
)
|
h
′
(
t
)
|
,
and therefore(iv)
lim
⁡
t
→
∞
(
g
′
(
t
)
/
|
g
′
(
t
)
|
)
=
i.It follows easily thatk
=
λ
∘
g satisfies the hypothesis of Lemma 20, and therefore φ
i
(
γ
) has the restriction property in Δ.


## 4.1. Case  1:D
i Is a Half-Plane

The following lemma will be needed here and in Case2.Lemma 17.
LetG be the open right half-plane Re
z
>
0 and let θ
(
z
)
=
(
z
-
1
)
/
(
z
+
1
) so that θ is a Riemann mapping function for G. Let k
:
[
0
,
∞
)
→
G be an injective C
1 function such that k
′
(
t
)
≠
0, for all t
≥
0, and lim
⁡
t
→
∞
k
(
t
)
=
∞. Let ρ be the (simple) arc parametrized by k. If  lim
⁡
t
→
∞
(
k
′
(
t
)
/
|
k
′
(
t
)
|
)
=
ω  (with |
ω
|
=
1), then σ
=
θ
(
ρ
)
¯ satisfies the hypothesis of Lemma 12 and, hence, ρ has the restriction property in G.Proof.
Putg
=
θ
∘
k, so that g
∈
C
1
[
0
,
∞
) parametrizes θ
(
ρ
). Clearly g
(
t
)
→
1 as t
→
∞. Now g satisfies the hypothesis of Lemma 15, for we can show that g
′
(
t
)
/
|
g
′
(
t
)
|
→
ω
-
1 as t
→
∞. Since θ
′
(
z
)
=
2
/
(
z
+
1
)
2 it follows that
(22)
g
′
(
t
)
|
g
′
(
t
)
|
=
|
1
+
k
(
t
)
|
2
(
1
+
k
(
t
)
)
2
k
′
(
t
)
|
k
′
(
t
)
|
=
k
′
(
t
)
|
k
′
(
t
)
|
|
k
(
t
)
|
2
(
k
(
t
)
)
2
|
1
+
1
/
k
(
t
)
|
2
(
1
+
1
/
k
(
t
)
)
2
⟶
ω
-
1
,
using Lemma 16.
Soσ
=
g
[
0
,
∞
)
∪
(
ω
-
1
) satisfies Lemma 12; hence g
[
0
,
∞
) has the restriction property in Δ. But g
[
0
,
∞
)
=
θ
(
ρ
) and, therefore, by Lemma 5, ρ has the restriction property in G.Now suppose thatC
i is a line and D
i is a half-plane. By Invariance Lemma 5 with a linear equivalence χ
(
z
)
=
α
z
+
β  (
α
≠
0
) we can assume that C
i is the imaginary axis and that D
i
=
G, the open right half-plane, as above. If γ
⊆
D
i is a type III arc, it is a subarc of a line, parabola, or hyperbola component. Obviously γ has a parametrization k as in Lemma 17. Hence γ has the restriction property in D
i.


## 4.2. Case  2:D
i Is the Concave Complementary Domain of a Parabola

Any two parabolas are conformally equivalent via a linear equivalence:μ
(
z
)
=
a
z
+
b  (
a
,
b
∈
C
,
a
≠
0
). So assume that C
i is the parabola
(23)
y
2
=
4
(
1
-
x
)
and that D
i is the complementary domain to the “right” of C
i.The function(24)
w
⟶
(
1
+
w
)
2
maps the open right half-plane G conformally onto D
i and the imaginary axis onto C
i. Its inverse is the function
(25)
ϑ
(
z
)
=
z
1
/
2
-
1
,
(
z
∈
D
i
)
,
where z
1
/
2 is the principal square-root of z (here and throughout all standard multivalued functions will take their principal values).Now letγ
⊆
D
i be a type III arc. Because G is conformally equivalent to D
i via ϑ it will be sufficient to show that the arc ϑ
(
γ
)
⊆
G has a parametric function k as in Lemma 17. Letting h be the arc-length parametrization of γ, then h
∈
C
1
[
0
,
∞
),  |
h
′
(
t
)
|
≡
1 and h
(
t
)
→
∞ as t
→
∞, and h is injective.Nowγ is a subarc of a line, parabola, or hyperbola component. Hence as z
→
∞ along γ the unit tangent vector at z tends to a limit ω (|
ω
|
=
1). Thus
(26)
lim
⁡
t
→
∞
h
′
(
t
)
|
h
′
(
t
)
|
=
lim
⁡
t
→
∞
h
′
(
t
)
=
ω
,
and therefore
(27)
lim
⁡
t
→
∞
h
(
t
)
|
h
(
t
)
|
=
ω
,
by Lemma 16.Putk
=
ϑ
∘
h. Then k is an injective parametric function for ϑ
(
γ
). Clearly k
∈
C
1
[
0
,
∞
),  k
(
t
)
→
∞ as t
→
∞, and
(28)
k
′
(
t
)
=
ϑ
′
(
h
(
t
)
)
h
′
(
t
)
≠
0
,
∀
t
≥
0
.Moreover,(29)
k
′
(
t
)
|
k
′
(
t
)
|
=
|
h
(
t
)
|
1
/
2
h
(
t
)
1
/
2
h
′
(
t
)
|
h
′
(
t
)
|
⟶
ω
1
/
2
.Sok is as in Lemma 17, which shows that γ has the restriction property in D
i.Remark 18.
The notationω
1
/
2 is ambiguous when ω
=
-
1 (γ could be part of another parabola). But, because type I arcs can be ignored, we can assume that either γ is contained entirely in the upper half-plane, in which case (
-
1
)
1
/
2
=
i, or else γ is in the lower half-plane and (
-
1
)
1
/
2
=
-
i.


## 4.3. Case  3:D
i Is the Convex Complementary Domain of a Parabola

In this case the parabola(30)
y
2
=
4
(
π
4
)
2
(
(
π
4
)
2
-
x
)
will be chosen for C
i, and D
i will be the complementary domain to the “left” of C
i. This choice is made because then we have the relatively simple Riemann mapping function
(31)
φ
i
(
z
)
=
tan
2
(
z
1
/
2
)
,
(
z
∈
D
i
)
.
This function maps the real interval (
-
∞
,
(
π
/
4
)
2
) in an increasing fashion onto (
-
1,1
), and so it maps the upper/lower half of D
i onto the upper/lower half of Δ. The formula for φ
i is indeterminate on (
-
∞
,
0
], but these singularities are removable and the formula
(32)
φ
i
(
x
)
=
-
tanh
2
(
-
x
)
1
/
2
can be used to define φ
i
(
x
), for negative x. This mapping will be examined in detail in a moment, but first we dispose of a trivial case and make some simple observations.Letγ
⊆
D
i be a type III arc. If γ is a real interval (
-
∞
,
a
), with a
<
(
π
/
4
)
2, then φ
i
(
γ
) is a subinterval of (
-
1,1
) which obviously has the restriction property in Δ. So this case is trivial and needs no more attention.The following observations are elementary.(i)
Ifγ is part of another line, then it must be parallel to R and certainly disjoint from (
-
∞
,
0
].
(ii)
Ifγ is part of another parabola C
j, then C
j must be symmetric about R and have an equation of the form
(33)
y
2
=
4
a
(
b
-
x
)
,
where0
<
a
≤
(
π
/
4
)
2,  b
≤
(
π
/
4
)
2.
(iii)
Ifγ is part of a hyperbola, then its asymptote must be parallel to R.
(iv)
In all (nontrivial) casesγ intersects (
-
∞
,
0
] in at most two points. So, because type I arcs can be ignored there is no loss of generality in assuming that Im
⁡
z has constant sign on γ and that Re
z
<
0 on γ.
(v)
Hence, for definiteness, we can assume thatγ is contained in the open second quadrant.
(vi)
In all casesy
2
/
x tends to a limit as z
→
∞ along γ. If γ is part of a line or hyperbola, the limit is 0, and if γ is part of the parabola in (ii) above the limit is -
4
a. For future reference let us note that
(34)
0
≤
lim
⁡
⁡
y
2
4
|
x
|
≤
(
π
4
)
2
.
(vii)
Because thelim
⁡ in (34) exists and because type I arcs can be ignored, we can assume that
(35)
y
2
x
2
<
1
,
on
⁡
γ
.Now letγ be type III arc in D
i as in (v) and (vi). We will show that φ
i
(
γ
) has the restriction property in Δ. To elucidate φ
i
(
γ
) it is convenient to work backwards, examining the mapping properties of the square map (
z
→
z
2
), then tan, and then the principal square root.Lemma 19.
LetΔ
+ be the open semidisc
(36)
Δ
+
=
{
z
∈
C
:
|
z
|
<
1
,
x
>
0
}
.
If σ
′ is a smooth simple arc in Δ
+
¯, if i is an end-point of σ
′, and if σ
′
-
{
i
}
⊆
Δ
+, then the arc
(37)
σ
=
{
z
2
:
z
∈
σ
′
}
is a smooth simple arc in Δ
¯ satisfying the hypothesis of Lemma 12, so that σ
-
{
-
1
} has the restriction property in Δ.Proof.
This is clear: the square mapz
→
z
2 is conformal in a neighbourhood of σ
′.Now letS be the open strip
(38)
S
=
{
z
∈
C
:
0
<
x
<
π
4
}
.
It is well known that tan maps S conformally onto Δ
+. The imaginary axis is mapped to the vertical part of ∂
Δ
+, and the line π
/
4
+
i
R is mapped to the semicircular part of ∂
Δ
+. Moreover, if z tends to infinity in S in such a way that y
→
+
∞, then tan
⁡
z
→
i.Lemma 20.
Letk
∈
C
1
[
0
,
∞
) be injective and satisfy k
′
(
t
)
≠
0, for t
≥
0. Suppose also that(i)
k
(
t
)
∈
S for all t
≥
0,
(ii)
Im
⁡
k
(
t
)
→
+
∞ as t
→
+
∞,
(iii)
lim
⁡
t
→
∞
Re
k
(
t
)
=
x
0 exists (
0
≤
x
0
≤
π
/
4
),
(iv)
lim
⁡
t
→
∞
(
k
′
(
t
)
/
|
k
′
(
t
)
|
)
=
i.
Ifγ
′ is the arc parametrized by k, then σ
′
=
(
tan
⁡
γ
′
)
∪
{
i
} satisfies the hypothesis of Lemma 19, so that tan
2
γ
′ has the restriction property in Δ.Proof.
Letg
=
tan
∘
k, so that g parametrizes γ
′ and tan
⁡
γ
′
=
g
[
0
,
∞
). Now g
∈
C
1
[
0
,
∞
),  g
′
(
t
)
≠
0, for all t
≥
0, and g
(
t
)
→
i as t
→
+
∞. Lemma 15 will be used to show that σ
′
=
g
[
0
,
∞
)
∪
(
i
) satisfies the hypothesis of Lemma 19. For all t
≥
0,
(39)
g
′
(
t
)
|
g
′
(
t
)
|
=
|
cos
⁡
⁡
k
(
t
)
|
2
(
cos
⁡
⁡
k
(
t
)
)
2
k
′
(
t
)
|
k
′
(
t
)
|
.
Letk
(
t
)
=
x
(
t
)
+
i
y
(
t
). Since x
(
t
)
→
x
0 and y
(
t
)
→
+
∞, as t
→
+
∞, and because cos
⁡
x
,
cosh
⁡
y
>
0 on γ,
(40)
|
cos
⁡
⁡
k
(
t
)
|
2
cos
⁡
2
k
(
t
)
=
(
|
cos
⁡
⁡
x
(
t
)
cosh
⁡
y
(
t
)
-
i
sin
⁡
x
(
t
)
sinh
⁡
y
(
t
)
|
cos
⁡
⁡
x
(
t
)
cosh
⁡
y
(
t
)
-
i
sin
⁡
x
(
t
)
sinh
⁡
y
(
t
)
)
2
=
|
1
-
i
tan
⁡
x
(
t
)
tanh
y
(
t
)
|
2
(
1
-
i
tan
⁡
x
(
t
)
tanh
y
(
t
)
)
2
⟶
|
1
-
i
tan
⁡
x
0
|
2
(
1
-
i
tan
⁡
x
0
)
2
.
So lim
⁡
t
→
∞
(
g
′
(
t
)
/
|
g
′
(
t
)
|
) exists.The function(41)
ϑ
(
z
)
=
z
1
/
2
maps D
i
-
(
-
∞
,
0
] conformally onto the vertical strip S as above. The limiting values of ϑ from above and below a point x on (
-
∞
,
0
] are at ±
i
(
-
x
)
1
/
2, respectively. Now tan maps S conformally onto Δ
+ and tan
±
i
(
-
x
)
1
/
2
=
±
i
tanh
(
-
x
)
1
/
2. Finally the square function maps Δ
+ conformally onto Δ
-
(
(
-
1,0
]
), and it maps both of ±
i
tanh
(
-
x
)
1
/
2 and  -
tanh
2
(
-
x
)
1
/
2. Thus the cut made by ϑ is repaired by the square function (by Schwarz’s Reflection Principle): φ
i is continuous at all points of (
-
∞
,
0
] and therefore analytic on D
i. Because φ
i
(
z
)
∈
(
-
1,0
] if and only if z
∈
(
-
∞
,
0
] the injectivity of φ
i on D
i is clear.Letγ
⊆
D
i be a type III arc. Assume that y
>
0 and x
<
0 when z
=
x
+
i
y
∈
γ. Let γ
′
=
ϑ
(
γ
) so that γ
′
⊆
S. We show that γ
′ is as in Lemma 20 so that tan
2
γ
′ has the restriction property in Δ and, hence, γ has the restriction property in D
i.Letz
=
x
+
i
y be an arbitrary point of γ and write
(42)
z
1
/
2
=
u
+
i
v
,
for the corresponding point ϑ
(
z
)
∈
γ
′; then
(43)
x
+
i
y
=
u
2
-
v
2
+
2
i
u
v
.
Eliminating v, and remembering that x
<
0, we see that
(44)
u
2
=
1
2
(
x
+
(
x
2
+
y
2
)
1
/
2
)
=
|
x
|
2
(
(
1
+
y
2
x
2
)
1
/
2
-
1
)
.Sincey
2
/
x
2
<
1 (observation (vii)), the binomial series implies that
(45)
u
2
=
y
2
4
|
x
|
-
1
16
y
4
|
x
|
3
+
⋯
~
y
2
4
|
x
|
,
as z tends to ∞ along γ. It follows from (34) that
(46)
lim
⁡
t
→
∞
u
2
=
a
exists
,
0
≤
a
≤
(
π
4
)
2
.Now leth be the arc-length parametrization of γ and write h
(
t
)
=
x
(
t
)
+
i
y
(
t
). Let k
=
ϑ
∘
h
=
h
1
/
2 so that k parametrizes γ
′. Write k
(
t
)
=
u
(
t
)
+
i
v
(
t
). (i), (ii), (iii), and (iv) of Lemma 20 can now be verified.Obviouslyk
(
t
)
∈
S, for all t
≥
0, so (i) is true. As t
→
∞,  |
k
(
t
)
|
=
|
h
(
t
)
|
1
/
2
→
∞, but since 0
≤
u
(
t
)
≤
π
/
4 we must have v
(
t
)
→
+
∞, so that (ii) is true. Item (iii) follows from (46). Now h
(
t
)
→
∞ as t
→
∞,  |
h
′
(
t
)
|
≡
1, and h
′
(
t
)
→
-
1 as t
→
∞. So, by Lemma 16,
(47)
lim
⁡
t
→
∞
k
′
(
t
)
|
k
′
(
t
)
|
=
|
h
(
t
)
|
1
/
2
h
(
t
)
1
/
2
h
′
(
t
)
|
h
′
(
t
)
|
⟶
-
i
(
-
1
)
=
i
.
So (iv) is true and we have now completed the proof.


## 4.4. Case  4:C
i Is a Hyperbola Component

We can deal simultaneously with the convex and concave complementary domains of a hyperbola component as follows. Let-
π
/
2
<
α
<
π
/
2 and let C
i
=
sin
(
α
+
i
R
). If α
<
0,  C
i is the arc
(48)
C
i
=
{
z
=
x
+
i
y
∈
C
:
x
<
0
,
x
2
sin
2
α
-
y
2
cos
⁡
2
α
=
1
}
,
and if α
>
0,  C
i is the arc
(49)
C
i
=
{
z
=
x
+
i
y
∈
C
:
x
>
0
,
x
2
sin
2
α
-
y
2
cos
⁡
2
α
=
1
}
.LetD
i be the complementary domain to the “left” of C
i; then D
i is convex when α
<
0 and concave when α
>
0. Linear equivalence will be used as before to reduce the general case to this one.The functionsin
-
1 maps the double cut plane C
-
{
(
-
∞
,
-
1
]
∪
[
1
,
∞
)
} conformally onto the vertical strip |
x
|
<
π
/
2, mapping the upper/lower parts of the first domain onto the upper/lower parts of the second. The upper and lower limits of sin
-
1 at a point -
x
∈
(
-
∞
,
-
1
] are -
π
/
2
±
i
cosh
⁡
-
1
x. The arc C
i
=
sin
(
α
+
i
R
) is mapped to the line Re
z
=
α. Therefore sin
-
1 maps D
i
-
(
-
∞
,
-
1
] conformally onto the strip
(50)
D
α
=
{
z
=
x
+
i
y
∈
C
:
-
π
2
<
x
<
α
}
.If(51)
λ
(
z
)
=
π
4
z
+
(
π
/
2
)
α
+
(
π
/
2
)
,
then λ maps D
α conformally onto the strip
(52)
S
=
{
z
=
x
+
i
y
∈
C
:
0
<
x
<
π
4
}
.Therefore(53)
φ
i
(
z
)
=
tan
2
λ
(
sin
-
1
z
)
is a Riemann mapping function for D
i. Now let γ be a type III arc in D
i. As in Case 3 the case γ
⊆
R is trivial, so we can assume that γ lies entirely in the upper half-plane. It will be sufficient for us to show that λ
(
sin
-
1
γ
) has a parametric function k as in Lemma 20.Letz
=
x
+
i
y be arbitrary point of γ and write sin
-
1
z
=
u
+
i
v for the corresponding point of sin
-
1
γ. Clearly, by (50),
(54)
u
+
i
v
∈
D
α
.Now(55)
z
=
x
+
i
y
=
sin
⁡
(
u
+
i
v
)
=
sin
⁡
u
cosh
⁡
v
+
i
cos
⁡
⁡
u
sinh
⁡
v
,
so that
(56)
|
z
|
2
=
sin
2
u
cosh
⁡
2
v
+
cos
⁡
2
u
sinh
⁡
2
v
=
sin
2
u
+
sinh
⁡
2
v
.Asz
→
∞ along γ, |
z
|
2
→
+
∞ and sin
2
u remains bounded; therefore
(57)
v
⟶
+
∞
as
z
⟶
∞
along
γ
.It now follows from (56) and (57) that
(58)
sin
⁡
u
=
x
|
z
|
(
tanh
2
v
+
sin
2
u
cosh
⁡
2
v
)
1
/
2
~
x
|
z
|
as
z
⟶
∞
.Leth be the arc-length parametrization of γ. As z
→
∞ along γ its unit tangent vector has a limit e
i
θ, say. The asymptotes of C
i are the rays arg
⁡
z
=
±
(
π
/
2
-
α
). Therefore
(59)
lim
⁡
t
→
∞
h
′
(
t
)
|
h
′
(
t
)
|
=
lim
⁡
t
→
∞
h
′
(
t
)
=
e
i
θ
,
where
π
2
-
α
≤
θ
≤
π
.So, by (57) and Lemma 16,
(60)
lim
⁡
t
→
∞
h
(
t
)
|
h
(
t
)
|
=
e
i
θ
.Nowg
=
sin
-
1
∘
h is a parametric function for sin
-
1
γ. By (54) it follows that(i)
g
(
t
)
∈
D
α  (
t
≥
0
),  and (57) shows that
(ii)
Im
⁡
g
(
t
)
→
+
∞ as t
→
∞.Equation (60) shows that(iii)
lim
⁡
t
→
∞
Re
g
(
t
)
=
sin
-
1
cos
⁡
θ
=
(
π
/
2
)
-
θ  and we notice that -
π
/
2
≤
(
π
/
2
)
-
θ
≤
α, by (59).Finally observe that(61)
g
′
(
t
)
|
g
′
(
t
)
|
=
|
1
-
h
(
t
)
2
|
1
/
2
(
1
-
h
(
t
)
2
)
1
/
2
h
′
(
t
)
|
h
′
(
t
)
|
.Now in the upper half-plane(
1
-
w
2
)
1
/
2
~
-
i
w, as w
→
∞. So, as t
→
∞,
(62)
g
′
(
t
)
|
g
′
(
t
)
|
~
|
h
(
t
)
|
-
i
h
(
t
)
h
′
(
t
)
|
h
′
(
t
)
|
,
and therefore(iv)
lim
⁡
t
→
∞
(
g
′
(
t
)
/
|
g
′
(
t
)
|
)
=
i.It follows easily thatk
=
λ
∘
g satisfies the hypothesis of Lemma 20, and therefore φ
i
(
γ
) has the restriction property in Δ.



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*Source: 102169-2014-09-11.xml*