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. In
the disc the solution u
is then expressed in
terms o

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 that U is an
open disc. We let G be
a holomorphic funct

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 all
points in at distance
1/ from the boundary
of . I

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 u, defined by
u(z)=u(G(z), is
harmonic in the strip.
M

(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 : H
D found in Section 1.1
allow us to determine
the

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 a polygonal
region P, namely a
bounded, simply
connect

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 fixed 0 <
x0 < 1 we have |fn(1)
fn(x0)| 1 as n tends

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 gave
various explicit
formulas in the case of
regions th

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
mind, we define f1()
= f1(F(ei) i)
whenever < 0,
and f0

) 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 is mapped
to A and the segment B
is mapped to B , with

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
upon us: since D is
simply connected, the
same must be

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 uniformly on
K. Given > 0, choose
as in the definition of

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 (the limit need
not be in F). The proof
that a family

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
continuity and
Cauchys theorem, we
may take the path of
i

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
uniformly bounded as f
ranges in F. This follows
from t

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 F in the compact
set consisting of all
points in at a

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 answer to the
second question
requires some
regularity of .

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
MAPPINGS uniformly
bounded. However,
this family is not
equ

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 must have zn w.
But this implies f(zn)
f(w), which is a

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 221 Corollary 2.3
The only
automorphisms of the
unit d

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, Area(f(U) =
f(U) dx dy, and the fact
that the determinant

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
automorphisms of the
disc. Theorem 2.2 If f is
an automorph

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 of variables
shows that u(x, y) = sin
y 2 f0(x t)
cosh

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 the steady-
state heat equation. In
these specific
exampl

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 introduced in
Exercise 7 of Chapter 1,
appear in a number of
pro

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 branch (defined in
the complex plane slit
along the infin

, 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 be
seen by making the
change of variables x =
1/ku in

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 does not
vanish on or at any
point of its interior
beside

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 this is
true whenever |z| r.
Letting r 1 gives the
first

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 two
matrices in G, then fM
fM = fMM . This
follows fr