54.3.2 Regions With Square Root Property
Theorem 54.3.6 Let Ω≠ ℂ for Ω a region and suppose Ω has the square root
property. Then for z 0 ∈ Ω there exists h : Ω → B
such that h is one to one,
onto, analytic, and h = 0
.
Proof: Define ℱ to be the set of functions, f such that f : Ω → B
is one to one and
analytic. The first task is to show
ℱ is nonempty. Then, using Montel’s theorem it will be
shown there is a function in
ℱ ,
h, such that
≥ for all
ψ ∈ℱ . When
this has been done it will be shown that
h is actually onto. This will prove the
theorem.
Claim 1: ℱ is nonempty.
Proof of Claim 1: Since Ω≠ ℂ it follows there exists ξ
Ω
. Then it follows
z − ξ and
are both analytic on Ω. Since Ω has the square root property,
there exists an analytic function,
ϕ : Ω
→ ℂ such that
ϕ 2 =
z − ξ for all
z ∈ Ω
, ϕ =
. Since
z − ξ is not constant, neither is
ϕ and it follows from the
open mapping theorem that
ϕ is a region. Note also that
ϕ is one to one because if
ϕ =
ϕ , then you can square both sides and conclude
z 1 − ξ =
z 2 − ξ implying
z 1 =
z 2 .
Now pick a ∈ ϕ
. Thus
=
a. I claim there exists a positive lower
bound to
for
z ∈ Ω. If not, there exists a sequence,
⊆ Ω such
that
∘ ------ ∘ ------ ∘ ------
zn − ξ +a = zn − ξ + za − ξ ≡ εn → 0.
Then
∘ ------ ( ∘ -----)
zn − ξ = εn − za − ξ (54.3.6)
(54.3.6)
and squaring both sides,
∘------
zn − ξ = ε2n + za − ξ − 2εn za − ξ.
Consequently,
=
ε n 2 − 2
ε n which converges to 0. Taking the limit in
54.3.6 , it follows 2
= 0 and so
ξ =
z a , a contradiction to
ξ Ω. Choose
r > 0 such
that for all
z ∈ Ω
, > r > 0
. Then consider
ψ(z) ≡ √---r----. (54.3.7)
z − ξ + a
(54.3.7)
This is one to one, analytic, and maps Ω into B
(
> r ). Thus
ℱ is not
empty and this proves the claim.
Claim 2: Let z 0 ∈ Ω. There exists a finite positive real number, η, defined
by
η ≡ sup {|ψ ′(z )| : ψ ∈ ℱ } (54.3.8)
0
(54.3.8)
and an analytic function, h ∈ℱ such that
=
η . Furthermore,
h = 0
.
Proof of Claim 2: First you show η < ∞ . Let γ
=
z 0 +
re it for
t ∈ and
r
is small enough that
B ⊆ Ω. Then for
ψ ∈ℱ , the Cauchy integral formula for the
derivative implies
′ 1 ∫ ψ(w)
ψ (z0) = 2πi -------2dw
γ (w − z0)
and so
≤ 2
πr = 1
∕r . Therefore,
η < ∞ as desired. For
ψ defined
above in
54.3.7
′ − rϕ′(z0) − r(1∕2)(√z0-−-ξ)− 1
ψ (z0) =----------2 = -------------2----⁄= 0.
(ϕ (z0)+ a) (ϕ (z0)+ a)
Therefore, η > 0. It remains to verify the existence of the function, h .
By Theorem 54.3.1 , there exists a sequence,
, of functions in
ℱ and an analytic
function,
h, such that
and
ψn → h,ψ′n → h′, (54.3.10)
(54.3.10)
uniformly on all compact subsets of Ω. It follows
|h′(z0)| = lim |ψ′n (z0)| = η (54.3.11)
n→ ∞
(54.3.11)
and for all z ∈ Ω,
|h(z)| = lim |ψ (z)| ≤ 1. (54.3.12)
n→∞ n
(54.3.12)
By 54.3.11 , h is not a constant. Therefore, in fact,
< 1 for all
z ∈ Ω in
54.3.12
by the open mapping theorem.
Next it must be shown that h is one to one in order to conclude h ∈ℱ . Pick z 1 ∈ Ω
and suppose z 2 is another point of Ω. Since the zeros of h − h
have no limit point,
there exists a circular contour bounding a circle which contains
z 2 but not
z 1 such that
γ ∗ contains no zeros of
h − h .
Using the theorem on counting zeros, Theorem 51.6.1 , and the fact that ψ n is one to
one,
1 ∫ ψ ′(w)
0 = nli→m∞ --- ------n-------dw
2∫πi γψ′n (w)− ψn (z1)
= -1- ---h-(w)----dw,
2πi γh (w )− h(z1)
which shows that
h − h has no zeros in
B . In particular
z 2 is not a zero of
h − h . This shows that
h is one to one since
z 2 ≠ z 1 was arbitrary. Therefore,
h ∈ℱ . It
only remains to verify that
h = 0.
If h
≠ 0
, consider
ϕ h ∘ h where
ϕ α is the fractional linear transformation defined
in Lemma
54.3.3 . By this lemma it follows
ϕ h ∘ h ∈ℱ . Now using the chain rule,
| | | |
||(ϕh(z0) ∘ h)′(z0)|| = ||ϕ′h(z0)(h(z0))|||h′(z0)|
| |
||----1----|| ′
= ||1 − |h (z0)|2|||h (z0)|
|| ||
= ||----1----||η > η
|1 − |h (z0)|2|
Contradicting the definition of
η . This proves Claim 2.
Claim 3: The function, h just obtained maps Ω onto B
.
Proof of Claim 3: To show h is onto, use the fractional linear transformation of
Lemma 54.3.3 . Suppose h is not onto. Then there exists α ∈ B
∖ h . Then
0
≠ ϕ α ∘ h for all
z ∈ Ω because
ϕ ∘ h(z) =-h(z)−-α-
α 1 − αh(z)
and it is assumed α
h . Therefore, since Ω has the square root property, you can
consider an analytic function
z → . This function is one to one because both
ϕ α and
h are. Also, the values of this function are in
B by Lemma
54.3.3 so it is in
ℱ .
Now let
------- ∘ ------
ψ ≡ ϕ√ ϕα∘h(z0) ∘ ϕα ∘ h. (54.3.13)
(54.3.13)
Thus
∘ ---------
ψ (z0) = ϕ √ϕα∘h(z0) ∘ ϕα ∘h (z0) = 0
and ψ is a one to one mapping of Ω into B
so
ψ is also in
ℱ . Therefore,
| |
|ψ′(z)| ≤ η,||(∘ ϕ-∘h)′(z )|| ≤ η. (54.3.14)
0 | α 0|
(54.3.14)
Define s
≡ w 2 . Then using Lemma
54.3.3 , in particular, the description of
ϕ α − 1 =
ϕ − α , you can solve
54.3.13 for
h to obtain
h(z) = ϕ ∘s∘ ϕ √ ------∘ ψ
−α − ϕα∘h(z0)
( ◜------≡F◞◟-------◝ )
= ( ϕ−α ∘ s∘ ϕ √------∘ ψ) (z)
− ϕα∘h(z0)
= (F ∘ψ)(z) (54.3.15)
Now
−1
F (0) = ϕ−α ∘s ∘ϕ−√ ϕα∘h(z0)(0) = ϕ α (ϕα ∘ h(z0)) = h(z0) = 0
and F maps B
into
B . Also,
F is not one to one because it maps
B to
B and has
s in its definition. Thus there exists
z 1 ∈ B such that
ϕ − =
− and another point
z 2 ∈ B such that
ϕ − =
.
However, thanks to
s,F =
F .
Since F
=
h = 0
, you can apply the Schwarz lemma to
F . Since
F is not one
to one, it can’t be true that
F =
λz for
= 1 and so by the Schwarz lemma it must
be the case that
< 1. But this implies from
54.3.15 and
54.3.14 that
η = |h′(z)| = |F′(ψ (z ))||ψ′(z )|
′ 0 ′ 0′ 0
= |F (0)||ψ (z0)| < |ψ (z0)| ≤ η,
a contradiction. This proves the theorem.
The following lemma yields the usual form of the Riemann mapping theorem.
Lemma 54.3.7 Let Ω be a simply connected region with Ω≠ ℂ . Then Ω has the
square root property.
Proof: Let f and
both be analytic on Ω
. Then
is analytic on Ω so by Corollary
50.7.23 , there exists
, analytic on Ω such that
′ =
on Ω
. Then
′ = 0 and so
f =
Ce
=
e a + ib e
. Now let
F =
+
a +
ib. Then
F is still a primitive of
f ′ ∕f and
f =
e F . Now let
ϕ ≡ e F . Then
ϕ is the desired square root and so Ω has the
square root property.
Corollary 54.3.8 (Riemann mapping theorem) Let Ω be a simply connected region
with Ω≠ ℂ and let z 0 ∈ Ω. Then there exists a function, f : Ω → B
such that f
is one to one, analytic, and onto with f = 0
. Furthermore, f − 1 is also analytic.
Proof: From Theorem 54.3.6 and Lemma 54.3.7 there exists a function,
f : Ω → B
which is one to one, onto, and analytic such that
f = 0
. The
assertion that
f − 1 is analytic follows from the open mapping theorem.