are continuous and
vanish at infinity, that
is, that lim|x| fj (x)
= 0 for j = 0, 1. The
method we shall follow
consists of relocating
the problem from the
strip to the unit disc via
a conformal map.
Lemma 1.3 Let V and U
be open sets in C and
F:VUa
holomorphic function.
If u : U C is a
harmonic function,
then u F is harmonic
on V . Proof. The thrust
of the lemma is purely
local, so we may
assume
all = 1, 2,. (b) Any
compact set K is
contained in K for
some . In particular =
=1 K . Lemma 3.4
Any open set in the
complex plane has an
exhaustion. Proof. If
is bounded, we let K
denote the set of
Poisson integral
formula is deduced in
Exercise 12 of Chapter
2 and Problem 2 in
Chapter 3 of this book.
216 Chapter 8.
CONFORMAL
MAPPINGS is the
Poisson kernel. Lemma
1.3 guarantees that the
function
(w1, w2). For more
information about this
subject, we refer the
reader to Problem 4.
2.2 Automorphisms of
the upper half-plane
Our knowledge of the
automorphisms of D
together with the
conformal map F
Similarly for the halfopen segments [z,w)
and (z,w] obtained by
restricting 0 t < 1 and
0 < t 1, respectively.
238 Chapter 8.
CONFORMAL
MAPPINGS 4.3
Boundary behavior In
what follows we shall
consider
uniformly bounded is
equicontinuous. This
follows directly from
the mean value
theorem. On the other
hand, note that the
family cfw_fn on [0, 1]
given by fn(x) = xn is
not equicontinuous
since for any
conformal map from
any proper, simply
connected open set to
the disc, or
equivalently to the
upper half-plane, but
this theorem gives
little insight as to the
exact form of this map.
In Section 1 we g
half-circle to R. More
precisely, as travels
from to 0, then
F(ei) goes from i +
to i , and as
travels from 0 to ,
then F(ei) goes from
to on the real
line. With the behavior
of F on the circle in
) to [0, ). On the
other hand, when z = x
is negative, then f (z) =
|x| 1ei(1) =
|x| 1ei , so f
maps the segment
(, 0] to (ei, 0].
The situation is
illustrated in Figure 3
where the infinite
segment A
to be a constant.
Therefore, a necessary
condition is to assume
that = C. Since D is
connected, we must
also impose the
requirement that be
connected. There is
still one more
condition that is forced
fn,j(wj ) converges.
Finally, let gn = fn,n
and consider the
diagonal subsequence
cfw_gn. By construction,
gn(wj ) converges for
each j, and we claim
that equicontinuity
implies that gn
converges unif
3.2 Montels theorem
Let be an open
subset of C. A family F
of holomorphic
functions on is said
to be normal if every
sequence in F has a
subsequence that
converges uniformly on
every compact subset
of
f(z) = z. Anticipating
the Schwarz-Christoffel
formula below, we
write z = f(z) = z 0 f ()
d = z 0 d with
+ = 1, and where the
integral is taken along
any path in the upper
half-plane. In fact, by
co
construction, since all
functions are required
to map into the unit
disc. Now, we turn to
the question of finding
a function f F that
maximizes |f (0)|. First,
observe that the
quantities |f (0)| are
d. Observe that 1
z 1 w = |z w| |
z| | w| |z w|
r2 since and |z
w| < r. Therefore |f(z)
f(w)| 1 2 2r r2 B|
z w|, 3. The Riemann
mapping theorem 227
where B denotes the
uniform bound for the
family
which we prove in the
next section. It is
completely general
(assuming only that
is a proper subset of C
that is simply
connected), and
necessitates no
regularity of the
boundary of . A
positive answ
holomorphic functions.
This conclusion is in
sharp contrast with the
real situation as
illustrated by the
family of functions
given by fn(x) = sin(nx)
on (0, 1), which is 226
Chapter 8.
CONFORMAL
MAPP
that contains no points
of the image f().
Otherwise, there exists
a sequence cfw_zn in
such that f(zn) f(w)
+2i. We exponentiate
this relation, and, since
the exponential
function is continuous,
we m
and using the fact that
( )(z) = z, we
deduce that f(z) =
ei(z), as claimed.
Setting = 0 in the
theorem yields the
following result. 2. The
Schwarz lemma;
automorphisms of the
disc and upper halfplane
limzz0 F(z) exists. To
prove this, we need a
preliminary result,
which uses the fact
that if f : U f(U) is
conformal, then
Area(f(U) = U |f (z)| 2
dx dy. This assertion
follows from the
definition, Ar
Another important
property of is that
it vanishes at z = ;
moreover it
interchanges 0 and ,
namely (0) = and
()=0. The next
theorem says that the
rotations combined
with the maps
exhaust all the
auto
translation of the
boundary condition by
x results in a
translation of the
solution by x as well.
We may therefore
apply the same
argument to f0(x + t)
and f1(x + t) (with x
fixed), and a final
change
log z f(z) = sin z 2 0
0 i 0 0 1 0 0 1 0 0 i
2 1 i 0 0 f(z) = log z f1(z)
= eiz f2(z) = iz f3(z) = 1
2 z + 1 z Figure 1.
Explicit conformal
maps upper half-plane,
where it arose in the
solution of th
unit disc whose inverse
is the rotation by the
angle , that is, r : z
eiz. More
interesting, are the
automorphisms of the
form (z) = z 1
z , where C with
| < 1. These
mappings, which
where introduce
the real axis arranged
in increasing order. The
exponents k will be
assumed to satisfy the
conditions k < 1 for
each k and 1 < n k=1
k. 6 The integrand in
(5) is defined as
follows: (z Ak)k is
that br
, K]. Next, when x >
1/k we have f (x) = 1
[(x2 1)(k2x2 1)]1/2 ,
and therefore, f(x) = K
+ iK x 1/k dx [(x2 1)
(k2x2 1)]1/2 .
However, 1/k dx
[(x2 1)(k2x2 1)]1/2
= 1 0 dx [(1 x2)(1
k2x2)]1/2 , as can
to g(z) = f(z) f(z1). If
g is not identically zero,
then z2 is an isolated
zero for g (because is
connected); therefore
1 = 1 2i g () g() d,
where is a small circle
centered at z2 chosen
so that g do
therefore f(z)/z is
holomorphic in D
(since it has a
removable singularity
at 0). If |z| = r < 1, then
since |f(z)| 1 we have
f(z) z 1 r , and by
the maximum modulus
principle, we can
conclude that th
group SL2(R) by G. Step
1. If M G, then fM
maps H to itself. This is
clear from the
observation that (4)
Im(fM(z) = (ad
bc)Im(z) |cz + d| 2 =
Im(z) |cz + d| 2 > 0
whenever z H. Step
2. If M and M are