ON THE REGULARITY OF THE RIEMANN
MAPPING FUNCTION IN THE PLANE
S. Cho† and S.R. Pai
Abstract. We prove the regularity properties of the Riemann mapping function on
simply bounded domains in the plane
1. Introduction.
We say two regions Ω1 , Ω2 in the plane are conformally equivalent, if there exists
a conformal 1-to-1 mapping ψ of Ω1 onto Ω2 . It follows that two conformally
equivalent regions are homeomorphic. But there is a much more important relation
between conformally equivalent regions: if ψ is as above, then f −→ f ◦ ψ is a
1-to-1 mapping of H(Ω2 ) onto H(Ω1 ) which preserves sums and products. Hence
problems about H(Ω2 ) can be transferred to problems in H(Ω1 ) with the aid of
the mapping ψ. In [2], Bell solved Dirichlet problem in the plane by means of the
Cauchy integral. Here he used these invariant properties.
The most important case of this is based on the Riemann mapping theorem:
Theorem R. Given any simple connected region Ω which is not the whole plane,
there exists a 1-to-1 holomorphic function f of Ω onto the unit disc ∆. Furthermore
the function f is determined uniquely by conditions f (a) = 0, f 0 (a) > 0 a ∈ Ω.
We call the function f in Theorem R as Riemann mapping function. In [3,4],
Chung studied some of the properties for the domains in the plane using these
biholomorphic mappings.
Also it is important to investigate the boundary behavior of the Riemann mapping function to study the mapping properties of the domain on the boundary. In
the present work, we consider the regularity properties of the Riemann mapping
functions on the boundary. That is, we will prove the following three theorems.
Theorem 1.1. Suppose that Ω is a bounded simply connected region whose boundary points are simple. Then the Riemann mapping function f of Ω onto ∆ extends
to a 1-to-1 continuous function F of Ω onto ∆. Futhermore, the function F maps
bΩ 1-to-1 onto b∆.
Theorem 1.2. Suppose that Ω is a bounded simply connected region which contains
a real analytic arc ζ = ζ(t) in its boundary. Then the Riemann mapping function
f : Ω −→ ∆ extends holomorphically across ζ.
2000 Mathematics Subject Classification. 30C75.
Key words and phrases. Riemann mapping function, Conformal mappings.
†This work was supported by Pusan National University Research Grant and by BSRI-96-1411
Typeset by AMS-TEX
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2
S. CHO† AND S.R. PAI
Theorem 1.3. Let Ω be a simply connected region bounded by a C ∞ smooth simple
closed curve. Then the Riemann mapping function f from Ω to ∆ extends C ∞
smoothly upto Ω.
Even though some of these regularity problems have been known, we present
here a new and nicer method to prove the regularity theories.
2. Continuous extension.
In this section, we prove the continuous extension of the Riemann mapping
function. First, we need the following lemma.
Lemma 2.1. Let D ⊂ C be a domain in the plane. If f : D −→ C is a 1-to-1
holomorphic function, then
Z
|f 0 |2 dA = Area of f (D)
D
Proof. Let f = u + iv, where u and v are real valued functions on D. Then
f 0 = fx = ux + ivx and by the Cauchy Riemann equation, we have
Ã
!
u
u
x
y
|f 0 |2 = ux ux + vx vx = ux vy − uy vx = det
.
vx vy
Hence
R
D
|f 0 |2 dA =
R
f (D)
dA = Area of f (D).
Definition 2.2. Let Ω be a simply connected region in the plane. A boundary
point b ∈ bΩ is called simple if b has the following property: To every sequence
{an } in Ω converging to b, there correspond a curve γ = γ(t), 0 ≤ t ≤ 1, and a
sequence {tn }, 0 < t1 < t2 · · · < tn → 1, such that γ(tn ) = an (n = 1, 2, 3, · · · )
and γ(t) ∈ Ω for 0 ≤ t < 1. In other words, there exists a curve in Ω which passes
through all an ’s and which ends at b.
Lemma 2.3. Let Ω be a bounded simply connected region. Suppose that f is a
1-to-1 holomorphic function of Ω onto ∆. Define a function ψ on Ω by
(
dist(f (z), b∆) if z ∈ Ω
ψ(z) =
0
if z ∈ bΩ
Then ψ is a continuous function on Ω.
Proof. For z, w in Ω, we have
¯
¯
|ψ(z) − ψ(w)| = ¯(1 − |f (z)|) − (1 − |f (w)|)¯
¯
¯
= ¯|f (z)| − |f (w)|¯ < |f (z) − f (w)|
So the continuity of ψ on Ω is clear.
To prove the continuity of ψ on bΩ, suppose ψ is not continuous at z0 ∈
bΩ. Then there exists a sequence {zn } in Ω converging to z0 such that ψ(zn ) =
dist(f (zn ), b∆) > λ > 0 for all n, for some constant λ. Since {f (zn )} is bounded,
ON THE REGULARITY OF RIEMANN FUNCTION
3
there exists a subsequence {f (znk )} of {f (zn )} which converges to w0 . Here w0
must satisfy that dist(w0 , b∆) ≥ λ, and hence that w0 is an interior point of Ω. Let
F = f −1 . Then znk = F (f (znk )) and hence
z0 = lim znk = lim F (f (znk )) = F (w0 ).
k→∞
k→∞
Since F is a 1-to-1 function from ∆ onto Ω and since w0 ∈ int(∆), there exists a
point in Ω which is the image of w0 under F . Thus
0 = ψ(z0 ) = ψ(F (w0 )) = dist(w0 , b∆) ≥ λ > 0
This is a contraction. Therefore ψ is also continuous on bΩ. ¤
Using lemma 2.3, we can prove the following theorem.
Theorem 2.4. Let Ω be a bounded simple connected region and let f be the Riemann mapping function of Ω onto ∆.
(1) If b is a simple boundary point of Ω, then f can be extended continuously
to Ω ∪ {b} and moreover |f (b)| = 1.
(2) If b1 and b2 are distinct simple boundary points of Ω, then f can be extended
continuously to Ω ∪ {b1 } ∪ {b2 } and f (b1 ) 6= f (b2 ).
Proof. (1) Suppose that (1) is false. Then there exists a sequence {an } in Ω converging to b such that f (a2n ) → w1 , f (a2n+1 ) → w2 , w1 6= w2
Choose a curve γ = γ(t) passing through all an ’s and tending to b and set
Γ(t) = f (γ(t)). From Lemma 2.3, |Γ(t)| can be extended continuously to [0, 1] and
|Γ(t)| −→ 1 as t −→ 1. So |w1 | = |w2 | = 1. Therefore it follows that one of two
open arcs A whose union is S − ({w1 } ∪ {w2 }) (where S is the unit circle in C) has
the property that every radius of ∆ which ends at a point of A intersects the range
of Γ in the infinite set which has a limit point on S.
Set F = f −1 . Note that F is a bounded holomorphic function on ∆ and that
F (Γ(t)) = γ(t), 0 ≤ t < 1. Thus F has radial limits a.e, and moreover
lim F (reiθ ) = b
(2.1)
r→1
a.e on A. From (2.1), we have limr→1 (F − b)(reiθ ) = 0 a.e on A, and hence
F (z) ≡ b on ∆ because F is a holomorphic function. Note that F is 1-to-1 because
f is 1-to-1. This is a contradiction and hence w1 = w2 and |f (b)| = |w1 | = 1.
(2) Suppose that the extension f satisfies f (b1 ) = f (b2 ) = w0 . Let γ1 and γ2 be
curves in Ω tending to b1 and b2 , respectively. We may assume that dist(γ1 , γ2 ) >
1
2 |b1 − b2 |. Let δ > 0 be so small that Γ1 (0) = f (γ1 (0)) and Γ2 (0) = f (γ2 (0))
are not in D(w0 , δ) ∩ ∆. Let 0 < r < δ. Since Γi (t) = f (γi (t)) → w0 as t → 1,
i = 1, 2, it follows that w1 , w2 ∈ D(w0 , r). Set w1 = w0 + reiθ1 , w2 = w0 + reiθ2 ,
−π ≤ θ1 , θ2 < π, and let η = max{|θ1 |, |θ2 |}. Then
¯Z
¯
¯ θ2
¯
¯
¯
0
iθ
iθ
|F (w1 ) − F (w2 )| = ¯
F (w0 + re )ire dθ¯
¯ θ1
¯
Z η
¯ 0
¯
¯F (w0 + reiθ )¯ dθ
(2.2)
≤r
−η
µZ
η
≤r
−η
¯ 0
¯
¯F (w0 + reiθ )¯2 dθ
¶ 12 µZ
¶ 12
η
dθ
−η
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S. CHO† AND S.R. PAI
by Hölder’s inequality. But by the choice of γ 0 s, we have:
(2.3)
dist(F (w1 ), F (w2 )) = |F (w1 ) − F (w2 )| >
1
|b1 − b2 |.
2
From (2.2) and (2.3), we obtain an inequality
1
2
|b1 − b2 | < πr2
4
(2.4)
³R
Z
η
2
|F 0 (w0 + reiθ )| dθ,
−η
´ 12
√
since
dθ
≤ π. Now we integrate both sides of (2.4) with respect to r
variable. Then by Lemma 2.1, it follows that
η
−η
Z
∞=
0
Z
δ
1
2
|b1 − b2 | dr ≤
4πr
Z
0
δ
Z
η
2
|F 0 (w0 + reiθ )| rdθdr
−η
|F 0 (s)|2 dA = Area of F (D(w0 , δ) ∩ ∆)
≤
D(w0 ,δ)∩∆
≤ Area of Ω < ∞,
This is a contradiction and hence f (b1 ) 6= f (b2 ). ¤
Now we are ready to prove Theorem 1.1. Here we state Theorem 1.1 again.
Theorem 2.5. Suppose that Ω is a bounded simply connected region whose boundary points are simple. Then the Riemann mapping function f of Ω onto ∆ extends
to a 1-to-1 continuous function of Ω onto ∆. Futhermore, the extension f maps
bΩ 1-to-1 onto b∆.
Proof. Let z0 be a point on bΩ and assume that {zn } ⊂ Ω is a sequence converging
to z0 . Without loss of generality, we may assume that {zn } ⊂ bΩ. Since every
boundary point of Ω is simple, it follows that for each zn there is zn0 ∈ Ω so that
|zn − zn0 | < 1/n and that |f (zn ) − f (zn0 )| < 1/n. Note that {zn0 } also converges to
z0 . Therefore f (zn0 ) converges to f (b) by Theorem 2.4. Hence f (zn ) also converges
to f (b) and this proves the continuity of f on Ω.
Note that ∆ ⊂ f (Ω) ⊂ ∆. Since f is continuous, it follows that f (Ω) is compact
and hence f (Ω) = ∆. ¤
3. Smooth extension.
In this section, we consider the behavior of the Riemann mapping function on
the region, whose boundary contains a C ∞ smooth arc or a real analytic arc. First
we prove the following Generalized Schwarz Reflection Principle.
Lemma 3.1. Let Ω be a bounded region whose boundary contains a real analytic
curve γ = γ(t), t ∈ (a, b) with γ 0 (t) 6= 0. Suppose that u is harmonic in Ω,
continuous on Ω ∪ γ, and zero on γ. Then u extends harmonically across γ.
Proof. Let t0 ∈ (a, b) and assume that γ is real analytic. So we can write γ(t) =
P∞ γ (n) (tb )
(t − t0 )n , t ∈ (t0 − δ, t0 + δ), for some δ > 0. We replace t in the
n=0
n!
ON THE REGULARITY OF RIEMANN FUNCTION
5
power series by a complx value z, for |z − t0 | < δ. Then γ = γ(z) is a holomorphic
function in the disc. Consequently, we have a holomorphic function γ = γ(z) in a
region D, which consists of discs symmetric to the real axis and which contains the
segment (a, b).
We note that γ = γ(z) is locally 1-to-1 since γ 0 (z) 6= 0. Shink D again in
which γ is 1-to-1 (again denoted by D). So we can consider u ◦ γ on D. From
the properties of u and γ, it follows that u ◦ γ is defined on D and satisfies all the
properties in Classical Schwarz Reflection Principle. Let D+ be the portion that
satisfies γ(D+ ) ⊂ Ω. Then u ◦ γ has a harmonic extension to D− and therefore u
has a harmonic extension across γ. ¤
Lemma 3.2. Let Ω be a bounded simply connected region whose boundary is a real
analytic arc γ = γ(t), in the sense that only one half of the disc, whose center is
on γ, is inside Ω. Then every boundary point of Ω is simple.
Proof. As in Lemma 3.1, we can regard γ(t) as a function of complex variable t.
Let b ∈ bΩ with γ(tb ) = b. Then there exist D(tb , r0 ) and D(b, r) r, r0 > 0 such
(n)
P∞
that each z ∈ D(b, r) can be written as a power series z = n=0 γ n!(tb ) (t − tb )n
for some t ∈ D(tb , r0 ) because γ is holomorphic on D(tb , r0 ).
Let {an } be a sequence in Ω such that an −→ b. Then there exists an integer
N > 0 such that n > N =⇒ an ∈ D(b, r). Put bi = aN +i (i = 1, 2, · · · ). Then
(n)
P∞
we can represent bi as : bi = n=0 γ n!(tb ) (ti − tb ) for some ti ∈ D(tb , r0 ), and it
is clear that bi → b as i → ∞. Since γ is continuous on D(tb , r0 ), it follows that
bi → b as ti → tb .
Note that we can connect finite points by a curve. Therefore let us show that
there exists a curve in Ω passing all bi ’s and tending to b. That is, we assume
that there is a function ψ(s) 0 ≤ s ≤ 1 and a sequence {si } s1 < s2 · · · → 1 such
that ψ(si ) = bi and that ψ(s) ∈ Ω for 0 ≤ t < 1. In D(tb , r0 ), two points are path
connected. Therefore there exists a curve β that passes all ti . So there is a sequence
(n)
P∞
{si } so that s1 < s2 · · · → 1 and that β(si ) = ti . Let α(t) = n=0 γ n!(tb ) (t − tb ).
Then α is continuous on D(tb , r0 ). Since only one half of the disc (whose center
is on γ) is inside Ω, the image of the curve β under the funtion α is a curve in Ω
tending to b such that α(ti ) = bi . If we consider α ◦ β, then this is the desired curve
passing thru bi ’s and converging to b. ¤
Now we prove that the Riemann mapping function can be extended holomorphically to the boundary which contains a real analytic arc.
Theorem 3.3. Suppose that Ω is bounded by a simply connected curve which contains a real analytic arc ζ = ζ(t). Then the Riemann mapping function f : Ω −→ ∆
extends holomorphically across ζ.
Proof. By Lemma 3.2, f can be extended continuously to the curve ζ. Let z0 ∈ ζ.
By Theorem 2.4, it follows that f (z0 ) 6= 0. We choose a disc D centered at z0 and
that f (z) 6= 0 on D. Let us consider the function h(z) = log |f (z)| on D. Then h(z)
is harmonic on D+ where D+ is the subset of D in Ω. Since f is continuous on D+ ,
h(z) = log |f (z)| is also continuous on D+ . Note that D+ = D+ ∪ ζD , where ζD
is the part of ζ in D. Moreover h ≡ 0 on ζD , since |f (z)| = 1 on ζ. Hence by the
Generalized Schwarz Reflection Principle, h extends harmonically accross ζ. Shink
disc again so that h is harmonic on D(z0 , ε). Let v be the harmonic conjugate of h
6
S. CHO† AND S.R. PAI
on D(z0 , ε) and set H(z) = h(z) + iv(z). Then H is holomorphic on D(z0 , ε) and
satisfies
|eH(z) | = eh(z) = |f (z)|.
Hence
f (z)
= eiθ ,
H(z)
e
or f (z) = eiθ eH(z) ,
for some θ ∈ [0, 2π). We note that eiθ eH(z) is holomorphic and hence that eiθ eH(z)
extends f holomorphically in D(z0 , ε) ∩ Ω. ¤
Theorem 3.4. Let Ω be a simply connected region bounded by a C ∞ smooth simple
closed curve. Then the Riemann mapping function f from Ω to ∆ extends C ∞
smoothly to Ω.
Proof. Note that all of the boundary points of the region that is bounded by a
C ∞ smooth simple closed curve are simple. Hence f extends continuously upto the
boundary by Theorem 2.5. Let a ∈ Ω and f (a) = 0. Define a function ψ : Ω −→ R
as follows: ψ(z) = log |f (z)| for all z except a and ψ(a) = 0. Since ψ(z) = log |f (z)|
is the real part of log f (z) which is analytic in Ω − {a}, ψ is harmonic in Ω − {a}.
From Lemma 2.3, it follows that dist(f (z), b∆) = 1 − |f (z)| is continuous on Ω,
and equal to zero on bΩ. Thus log |f (z)| is also continuous on Ω − {a} and equal
to zero on bΩ.
Choose a small ε > 0 so that D(a, ε) ⊂ Ω. Choose a C ∞ cut-off function ξ
defined on Ω so that ξ ≡ 0 on D(a, 2ε ) and ξ ≡ 1 in Dc (a, ε). Set u(z) = ξ(z)ψ(z) =
ξ(z) log |f (z)|, z ∈ Ω. Then it is clear that u is continuous on Ω, equal to zero on
bΩ and that 4u ∈ C ∞ (Ω). This last property shows that u itself is smooth on Ω
by the regularity theory of the Laplace operator [2]. Note that u(z) = log |f (z)|
near bΩ. Thus log |f (z)| extends C ∞ smoothly upto bΩ and so does
³
elog |f (z)|
´2
= |f (z)|2 = f (z)f¯(z).
Thus f is a smooth function on Ω and this proves our theorem.
¤
References
1. L. Ahlfors,, Complex analysis, McGraw-Hill, Singapore 3rd. ed. (1979).
2. S.R. Bell, Solving the Dirichlet problem in the plane by means of the Cauchy integral, Indiana
Univ. Math. J. 39 (1990), 1–16.
3. Y. B. Chung, The Bergman kernel function and the Ahlfors Mapping in the plane, Indiana
Univ. Math. J. 42 (1993), 1-10.
4. Y. B. Chung, The Ahlfors Mapping function and an extremal problem in the plane, Houston
J. of Math. 19 (1993), 263-273.
(†)Department of Mathematics Education, Pusan National University, Pusan 609–
735, Korea
E-mail address: [email protected]