Picard Theorems
March 24, 2016
Abstract
In this dissertation, I will provide a full proof of both Picard’s Little
Theorem and Picard’s Great Theorem. I will also describe various
alternative proofs which often require less preparation.
Picard’s Little Theorem. If f : C → C is an entire function whose image
omits more than one value, then it is constant.
Picard’s Great Theorem. If f is a non-constant analytic function with an
isolated essential singularity at z0 , then on every deleted neighbourhood of
z0 , f attains every finite value, with one possible exception, infinitely often.
1
Picard’s Little Theorem
Introduction
In this project I aim to provide a detailed account of the proofs of the Picard
Theorems, relying on numerous texts and books, but often proving theorems
myself. The idea behind approaching a proof of Picard’s Little Theorem is to
construct the modular function λ(τ ), on the half plane Im(τ ) > 0, and prove
the result from an application of the monodromy theorem to λ(τ ). This will
occupy the main section of my project. The latter stages of my project will
prove Picard’s Great Theorem, by establishing the idea of normal families of
functions, and also briefly discuss alternative proofs whilst comparing them
to the main proofs in the project. When citations appear at the end of a
proof, this indicates that the proof of the theorem is contained in that source,
and that source was used heavily in my account of the proof.
1
1.1
Monodromy Theorem
Analytic Continuation
Before stating and proving the Monodromy Theorem, we will introduce the
concept of analytic continuation. This will broaden our definition of an
analytic function and allow us to permit multi-valued relations without any
issue.
An analytic function f (z) defined on a region Ω will be written as a “function
element”, (f, Ω), and we will show what it means for a function to be globally
analytic by examining a relation between a collection of function elements.
Definition 1.1.1. A function element (f2 , Ω2 ) is said to be a direct analytic
continuation of (f1 , Ω1 ) in the region Ω2 , if Ω1 ∩ Ω2 6= ∅ and f1 (z) = f2 (z)
in Ω1 ∩ Ω2 .
It is easy to see that if (f2 , Ω2 ) and (g2 , Ω2 ) are direct analytic continuations of
(f1 , Ω1 ) then f2 = g2 in Ω1 ∩Ω2 and thus throughtout Ω2 . Hence if there exists
a direct analytic continuation of (f1 , Ω1 ) to Ω2 then it is uniquely determined
(Ahlfors, 1966). Thus our definition of direct analytic continuation establishes
a link between two function elements.
More generally we consider chains of function elements (f1 , Ω1 ), (f2 , Ω2 ),...,
(fn , Ωn ) such that (fk , Ωk ) is a direct analytic continuation of (fk−1 , Ωk−1 ).
Elements belonging to this chain are said to be analytic continuations of
2
each other.
We can now adequately define a global analytic function:
Definition 1.1.2. A global analytic function is a non-empty collection F
of function elements (f, Ω) such that any two elements in F are analytic
continuations of each other and the chain between them consists of function
elements from F (Ahlfors, 1966).
The issue with many interesting functions is that they are multi-valued. One
way that you can work around this is to establish the concept of a branch of
a function. This leads to the following definition:
Definition 1.1.3. For a given function element (f, Ω) and a point z0 ∈ Ω
we define a branch of f at z0 to be the collection of function elements (g, Ω0 )
such that z0 ∈ Ω0 and f = g in the neighbourhood of z0 (Conway, 1973).
Proposition 1.1.4. The branch relation between function elements described above is an equivalence relation.
Proof. Define f ∼ g if f is related to g as in the above definition. We need
to prove the following:
1. f ∼ f .
2. f ∼ g ⇒ g ∼ f .
3. f ∼ g and g ∼ h ⇒ f ∼ h.
Definition 1.1.3 is equivalent to saying that two function elements are in the
same branch collection if and only if they share the same Taylor expansion
about z0 . And thus the proof is trivial with this result (Ahlfors, 1966).
We call the equivalence classes in question the analytic branches at z0 . We
denote the branch at z0 determined by (f, Ω) as [f, z0 ].
1.2
Analytic Continuation Along Curves
Definition 1.2.1 Let γ : [0, 1] → C be a path and suppose that for each t
in [0, 1] there is a function element (ft , Ωt ) such that:
1. γ(t) ∈ Ωt .
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2. For each t in [0, 1] there is a δ > 0 such that |s − t| < δ implies that
γ(s) ∈ Ωt and [fs , γ(s)] = [ft , γ(s)].
Then (f1 , Ω1 ) is the analytic continuation of (f0 , Ω0 ) along the path γ
(Conway, 1973).
This definition requires a little explanation. The first condition simply forces
the path in question to run through successive regions Ωt as t goes from 0 to
1. For the second condition we can see that as γ is continuous and included in
the open set Ωt then we can find a δ such that γ(s) ∈ Ωt whenever |s − t| < δ.
When this condition is satisfied we have that ft = fs on Ωt ∩ Ωs .
Definition 1.2.2. If γ : [0, 1] → C is a path from a to b and {(ft , Ωt ) :
0 ≤ t ≤ 1} is an analytic continuation along γ then the branch [f1 , b] is the
analytic continuation of [f0 , a] along γ (Conway, 1973).
Now by looking at the branches of two continuations along the same arc, we
can prove they are identical or different for all t. To do this we need the
following lemma first:
Lemma 1.2.3. If A is a clopen (closed and open) subset of a connected
topological space X, then A is equal to the space X or the empty set.
Proof. By the axioms of topology X and ∅ are clopen. Now assume that
there exists a different clopen subset A. Then A ∩ Ac = ∅ and A ∪ Ac = X
and therefore X is disconnected. This is a contradiction and therefore A
must be the space X or the empty set.
Proposition 1.2.4. Two continuations (represented by their respective
branches) [f1 , z(t)] and [f2 , z(t)] of a globally analytic function f along the
same arc γ are either identical, or differ for all t.
Proof. We consider the subset A of the arc interval (α, β) in which:
[f1 , z(t)] = [f2 , z(t)].
This subset describes all values of t for the scenairo in which the continuations
are identical. We aim to prove that this set is clopen and use the above
lemma to show that this means that this subset is either the empty set
(the continuations differ for all t) or the full space (the continuations are
identical).
First we choose t0 ∈ A. We note therefore that the branches in question are
determined by the elements (f10 , Ω1 ), (f20 , Ω2 ) respectively. Now there exists
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a neighbourhood around z(t0 ) where f1 = f2 . This is due to our assumption
that we are inside the set A. If we now take t sufficiently close to t0 , then
we acheive that z(t) is inside the neighbourhood of z(t0 ). Now f1 = f10 and
f2 = f20 ⇒ [f1 , z(t)] = [f2 , z(t)]. This shows that the condition holds in every
neighbourhood of our arbitrary point t0 ∈ A and hence A is open.
Now we prove that the complement of A is open also. Let t0 6∈ A. We use the
exact same argument as above but on Ac . Now we can see that there exists
a neighbourhood of z(t0 ) such that f10 (z) 6= f20 (z). Hence for t sufficiently
close to t0 we have z(t) inside the neighbourhood of z(t0 ) and f1 = f10 and
f2 = f20 . Therefore the branches in question are different due to f1 and f2
being different in the neighbourhood of z(t0 ). This gives that Ac is open
implying that A is clopen. Thus from the connectedness of intervals and the
lemma above we complete the proof (Ahlfors, 1966).
This means that a continuation is uniquely defined from its starting branch.
If we have the same starting branch for two continuations then the set above
is not empty indicating that the continuations are identical.
Now we only require one last important idea and few extra specific lemmas
in order to tackle the Monodromy Theorem.
1.3
Homotopy of Curves
We define an arc to be homotopic to another (with respect to the region Ω)
if they have the same end points and one can be continuously deformed into
the other.
Definition 1.3.1. Two arcs γ1 (t) and γ2 (t), defined over the same interval
α ≤ t ≤ β are homotopic with respect to Ω if there exists a continuous
deformation function z(t, s), defined for α ≤ t ≤ β and 0 ≤ s ≤ 1 with the
following properties:
1. z(t, s) ∈ Ω for all (t, s).
2. z(t, 0) = γ1 (t) and z(t, 1) = γ2 (t) for all t.
3. z(α, s) = γ1 (α) = γ2 (α) and z(β, s) = γ1 (β) = γ2 (β) for all s (Ahlfors,
1966).
For the deformation to take place within Ω, the values of the function z(t, s)
must be inside Ω for all (t, s). Condition 2 stipulates that the effect of the
deformation is just that; the function deforms γ1 (t) into γ2 (t) as s goes from
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0 to 1. Condition 3 enforces that the two arcs, and the deformation function
itself, need to have the fixed endpoints.
Proposition 1.3.2. Homotopy of paths with fixed endpoints is an equivalence relation.
Proof. 1. Reflexivity: Trivial.
2. Symmetry: if γ1 ∼ γ2 then we have z(t, s) as the deformation function
from γ1 (t) to γ2 (t). We let our new deformation function to be:
w(t, s) = z(t, 1 − s).
Immediately we can see that this deforms γ2 (t) into γ1 (t) and satisfys the
remaining two conditions.
3. Transitivity: if γ1 ∼ γ2 and γ2 ∼ γ3 then we have z1 (t, s) and z2 (t, s) as
the two deformation functions. We now let our new function (deforming
γ1 (t) into γ3 (t)) as:
z1 (t, 2s) s ≤ 12
w(t, s) =
z2 (t, 2s − 1) s > 12
This new function deforms γ1 (t) into γ3 (t), is continous as z1 (t, 1) = z2 (t, 0)
and satisfys the other two conditions.
This results in homotopic classes; arcs with common endpoints which deform
into each other. It is also useful to prove that different parametrisations of
the same arc are homotopic.
For γ1 = γ1 (t) to be a reparametrisation of γ2 = γ2 (t) there must exist a
nondecreasing function φ(t) such that γ1 (t) = γ2 (φ(t)). Now the function
z(t, s) = γ1 ((1 − s)t + sφ(t)).
takes values on the arc itself, and therefore in the space Ω. Also for s = 0 and
s = 1 we achieve z(t, 0) = γ1 (t) and z(t, 1) = γ1 (φ(t)) = γ2 (t) respectively as
required. As they describe the same curve the endpoints are clearly fixed
(Ahlfors, 1966).
We now describe a few more results:
Definition 1.3.3. Let γ : [0, 1] → C be a path from a to b. We say that the
function element (f, Ω) is analytically continuable along γ if for each t there
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is a convergent power series
ft (z) =
∞
X
an (t)(z − γ(t))n , |z − γ(t)| < r(t).
n=0
such that f0 (z) is the power series representing f (z) at z0 , and such that
when s is sufficiently close to t, then fs (z) = ft (z) for z in the intersection
of the disks of convergence.
We can say that the behaviour of this power series near any fixed point
z0 is akin to the branch of f at z0 . By Propostion 1.2.4. the series f1 (z)
is uniquely determined by f0 (z) and thus we refer to f1 (z) as the analytic
continuation of f (z) along γ.
We can immediately see that the coefficients an (t) depend continuously on
t. The following lemma tells us that the same is true for the radius of
convergence.
Lemma 1.3.4. Let D be a disk and (f, D) a function element. If R(z1 ) is
the radius of convergence of the power series expansion of f (z) about a point
z1 ∈ D then
|R(z1 ) − R(z2 )| ≤ |z1 − z2 |.
Proof. We describe R(z1 ) as the radius of the largest disk centered at z1 for
which f can be analytically continued. By the triangle inequality we have
that
DR(z1 ) (z1 ) ⊂ DR(z2 ) (z2 ) ⇔ R(z2 ) ≥ R(z1 ) + |z1 − z2 |.
Therefore if we have R(z2 ) > R(z1 ) + |z1 − z2 | then DR(z2 ) (z2 ) would contain
DR(z1 )+ (z1 ) for suffciently small > 0. This would imply that the Taylor expansion of f about z1 would coverge on DR(z1 )+ (z1 ) which is a contradiction.
Hence we have
R(z2 ) ≤ R(z1 ) + |z1 − z2 |.
Interchanging z1 and z2 gives
R(z1 ) ≤ R(z2 ) + |z1 − z2 |.
giving the result we need:
|R(z1 ) − R(z2 )| ≤ |z1 − z2 |.
(Gamelin, 2001)
We now describe a “local” version of the Monodromy Theorem which will be
used to prove the global classic result.
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Lemma 1.3.5. Let γ : [0, 1] → C be a path from a to b and let {(ft , Dt ) :
0 ≤ t ≤ 1} be an analytic continuation of f along γ. If σ : [0, 1] → C is
another path from a to b for which there exists a sufficiently small δ > 0 such
that |σ(t) − γ(t)| < δ for all t then there is an analytic continuation gt (z) of
ft (z) along σ and the terminal series g1 (z), centered at σ(1) = b, coincides
with f1 (z).
Proof. We can immediately define gt (z) to be the power series expansion
of ft (z) about σ(t). This is possible (i.e. the series converges) as σ(t) is
within the disk of convergence of ft (z) for δ suffciently small. Also σ(s)
lies inside the disk of convergence for s sufficiently near to t which implies
that fs (z) = ft (z) for s near to t. From this we see that gs (z) = gt (z) for s
suffciently close to t and therefore gt (z) is indeed an analytic continuation.
Thus by the definition of gt (z) we have that g1 (z) must coincide with f1 (z)
at b = σ(1) = γ(1) (Gamelin, 2001).
Theorem 1.3.6.(The Monodromy Theorem). Let (f, D), D ⊂ Ω, be a
function element which can be analytically continued along all paths contained
in Ω. Let γ0 : [0, 1] → C and γ1 : [0, 1] → C be two paths from a to b which
are homotopic with respect to Ω. Then the analytic continuation of (f, D)
along γ0 (t) and γ1 (t) lead to the same terminal element at b.
Proof. As γ0 and γ1 are fixed endpoint homotopic they can be deformed by
the continuous path function (s, t) → γs (t), 0 ≤ s ≤ 1, 0 ≤ t ≤ 1. As this
path is within our region Ω we note that (f, D) can be continued analytically
along γs (t).
If we denote fs,t (z) as the analytic continuation of (f, D) along γs then we
see that the power series fs,t (z) and fs0 ,t0 (z) converge to the same function
for s0 sufficiently close to s and t0 sufficiently close to t. Therefore we have
that the radius of convergence Rs,t of fs,t varies continuously with s and t
(by Lemma 1.3.4, page 7) and hence there exists a δ > 0 such that Rs,t > δ
for all (s, t).
Now we can split the interval [0, 1] into n pieces and choose parameters;
0 = s0 < s1 < ... < sn such that |γsj (t) − γsj−1 (t)| < δ for 0 ≤ t ≤ 1. Thus
by the previous lemma we have that the analytic continuation along γsj−1
gives the same power series (and therefore branch) at b as the continuation
along γsj .
By repeating the above argument successively for γ0 = γs0 , γs1 ,..., γsn = γ1
we see that after n repetitions we obtain that the analytic continuation along
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γ0 leads to the same power series as the continuation along γ1 . Hence the
theorem is proved (Gamelin, 2001).
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2
2.1
The Modular Function λ(τ )
Periodic Functions
Definition 2.1.1. A function f (z) is said to be periodic if there exists a
non zero constant ω such that
f (z + ω) = f (z).
holds for all z.
Here we call the number ω a period. If no other integer divisor is a period,
then ω is known as the fundamental period (Copson, 1935).
Seeing this, one would wonder whether there exists functions that could
be described as being multi-periodic (having more than one fundamental
period).
Lemma 2.1.2. A non-constant analytic function f (z) cannot have arbitrarily
small periods.
Proof. Let f (z) be an analytic function with a set of periods ω, where |ω| > 0.
Let z0 be a point for which f (z) is regular (analytic and single-valued) then
the function f (z) − f (z0 ) has zeros at every z0 + ω. This is easy to see:
f (z0 + ω) − f (z0 ) = f (z0 ) − f (z0 ) = 0.
Since the lower bound of |ω| is zero there exists a point of the set z0 + ω in
every neighbourhood of z0 . Hence we have a sequence of zeros of f (z) − f (z0 )
which has a limit point inside the neighbourhood of z0 . This is impossible
unless f (z) − f (z0 ) is identically zero (Copson, 1935).
Lemma 2.1.3. If f (z) is a non-constant analytic function that has periods
ω and λω, then each of these periods is a multiple of a single fundamental
period if and only if λ ∈ R.
Proof. First we prove ⇐.
Taking λ ∈ R, if λ ∈ Z then there is nothing to prove. Each of the periods ω
and λω are multiples of a single fundamental period, say ω1 . If λ ∈
/ Z then
we can express the period λω in the form mω + αω, m ∈ Z, 0 < α < 1.
This gives that αω is itself a period as it is the difference between two periods.
Therefore for any period λω, not a multiple of ω, we have a corresponding
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period αω. However there must be only finitely many periods of the form
αω as otherwise one could describe a sequence of periods α1 ω, α2 ω, ... with
α1 > α2 ... by finding smaller and smaller differences between two periods of
the form αω. This sequence has a limit point by the Bolzano-Weierstrauss
Theorem which would imply that there exist periods of arbitrarily small
modulus and by Lemma 1.2 this is impossible.
Therefore now let us take ω1 =α1 ω where α1 is defined to be the smallest of
all the αi . Then we have ω and λω are multiples of ω1 . If not then by the
previous argument we have
λω = kω1 + α2 ω1
for k ∈ Z and 0 < α2 < 1. Similarly for ω. Thus α2 ω1 is a period with
modulus smaller than ω1 . This contradicts our definition of ω1 and so the
result follows.
Now we need to prove ⇒.
We have only one fundamental period, say ω 0 . Thus if the function has
periods ω and λω this implies that ω = `ω 0 and λω = kω 0 with `, k ∈ Z.
This implies that λ` = k ⇒ λ = k` and as `, k ∈ Z ⇒ λ ∈ Q ⊂ R(Copson,
1935).
Theorem 2.1.4. Let f (z) be a non constant analytic function which has two
fundamental periods ω1 and ω2 , whose ratio is not real. Then this function
is doubly periodic, i.e. every period is a sum of multiples of a pair of periods,
(known as primitive periods) not necessarily ω1 and ω2 .
Proof. From Lemma 2.1.3 we can see that f (z) is not simply periodic. Now,
since we have that ωω12 is not real we can express every period ω uniquely in the
form λω1 + µω2 with λ and µ real. This is done by solving the simultaneous
equations
Re(ω) = λRe(ω1 ) + µRe(ω2 ).
Im(ω) = λIm(ω1 ) + µIm(ω2 ).
We can clearly see that this gives us back the desired result as
ω = Re(ω) + iIm(ω) = λ(Re(ω1 ) + iIm(ω1 )) + µ(Re(ω2 ) + iIm(ω2 ))
= λω1 + µω2 .
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The determinant of this system is

 

Re(ω2 ) Re(ω1 ) Im(ω1 ) Re(ω1 )

 

Re(ω2 ) Im(ω2 ) = Im(ω1 ) × Im(ω2 ) = |ω1 ||ω2 | sin(arg(ω2 /ω1 ))
0
0
Where we take ω1 and ω2 to be the vectors with their real and imaginary
values as components in the above cross product. This is non zero, as
arg(ω2 /ω1 ) is strictly greater than zero, and therefore the system gives
unique solutions for λ and µ. However for a solution set (λ, µ), we see that
(λ, µ) is also a solution and it follows that λ and µ are real.
Now note if λ and µ are integers or zero then there is nothing to be proved.
However if λ and µ are not both integers then we can express, as we did
before, λω1 + µω2 in the form:
`ω1 + mω2 + αω1 + βω2
where `, m ∈ Z and 0 ≤ α, β < 1.
We can see that αω1 +βω2 is itself a period. Also we now find that α and β are
not equal to zero as if say β = 0, then α is not and thus f (z) has periods ω1
and αω1 , contradicting that ω1 is a fundamental period. Therefore for every
period ω, not of the form `ω1 +mω2 , we have a corresponding point αω1 +βω2
representing a period. This point lies inside the parallelogram described with
vertices 0, ω1 , ω2 , ω1 + ω2 (this fact will become more important later on).
There can only be finitely many of these points as otherwise this leads to the
existence of periods with arbtrarily small modulus by the same reasoning as
in the previous theorem which is impossible by Lemma 1.2. Also we cannot
have two periods for which the value of β (or α) is the same. Otherwise this
gives that αω1 + βω2 − (α0 ω1 + βω2 ) = |α − α0 |ω1 is a period which again
contradicts that ω1 is a fundamental period.
Hence there is a unique point of the set of periods αω1 + βω2 for which β is
the smallest. Let us call this ω2 0 . Now we claim that every period of f (z)
can be expressed as a sum of multiples of ω1 and ω2 0 . If not, then by the
previous argument we can find a period of the form α0 ω1 + β 0 ω2 0 for 0 < α0 ,
β 0 < 1.
Now
α0 ω1 + β 0 ω2 0 = α0 ω1 + β 0 (αω1 + βω2 )
= (α0 + αβ 0 )ω1 + ββ 0 ω2 .
As ββ 0 < β this contradicts the definition of ω2 0 . Hence we have proved
that if f (z) is a non-constant analytic function that possesses periods with
non-real ratio, it is necessarily doubly-periodic (Copson, 1935).
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2.2
The Period Module
We now describe an alternative way to look at periodicity by examining the
module of the periods of a function.
Definition 2.2.1. A module M is a set of elements which obey the following
rules:
(1) ω ∈ M ⇒ nω ∈ M with n ∈ Z.
(2) ω1 , ω2 ∈ M ⇒ ω1 + ω2 ∈ M
Hence we can look at the set of periods of a non-constant meromorphic
function as a module. This period module is discrete, i.e. all the points of
M are isolated. This is clear as since f (ω) = f (0) for all ω ∈ M , if there
exists a finite accumulation point this would imply that f is constant.
Notice that we proved earlier that there are three distinct discrete modules.
The module either consists either of zero alone (i.e. f not periodic), of integral
multiples nω of a single complex number ω 6= 0, or all linear combinations
of n1 ω1 + n2 ω2 with integer coefficients of two numbers ω1 and ω2 with non
real ratio (Ahlfors, 1966).
For the remainder of this project we will only be interested in the module
which is spanned by the two periods ω1 and ω2 . Since every ω = nω1 + mω2
we have that (ω1 , ω2 ) is a basis for our module M .
Imagine we have another basis for the module (ω10 , ω20 ). We wish to see the
relation between these two bases. We see that the following is true (proof
found in (Ahlfors, 1966), omitted here).
Proposition 2.2.2. Any two bases of the same module are connected by a
unimodular transformation (Ahlfors, 1966).
We note that a unimodular transformation is a matrix with determinant ±1:
a b
a, b, c, d ∈ Z.
c d
Hence would like to determine if a canonical basis of the period module
exists. By canonical we mean a basis such that the ratio τ = ωω21 is unique
under a unimodular transformation
ω20 = aω2 + bω1
, a, b, c, d ∈ Z
ω10 = cω2 + dω1
for ad − bc = ±1.
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Proposition 2.2.3 There exists a basis (ω1 , ω2 ) such that the ratio τ =
ω2 /ω1 satisfies the following conditions:
(i) Im(τ ) > 0;
(ii) − 12 < Re(τ ) ≤ 12 ;
(iii) |τ | ≥ 1;
(iv) Re(τ ) ≥ 0 if |τ | = 1;
And the ratio τ is uniquely defined by these conditions.
Proof. If our period module is non-empty it contains an element ω 6= 0.
Therefore it contains another element, say ω1 , with minimum absolute value.
Our discrete module also contains a number ω 0 which is not an integer
multiple of ω1 . We take the element of minimum absolute value which is not
an integer multiple of ω1 and call it ω2 .
Using this selection of ω1 and ω2 we have that |ω1 | ≤ |ω2 | and as ω1 + ω2 and
ω1 − ω2 are both not integer multiples of ω1 we also have that |ω2 | ≤ |ω1 + ω2 |
and |ω2 | ≤ |ω1 − ω2 |.
By this choice of basis we have immediately that |τ | ≥ 1. Using |w2 | ≤
|ω1 + ω2 | we have
|ω2 |2 ≤ |ω1 + ω2 |2
⇒ ω2 ω2 ≤ ω1 ω2 + ω2 ω1 + ω2 ω2
ω1 ω2 + ω1 ω2
⇒
≥ −1
ω1 ω1
and using |ω2 | ≤ |ω1 − ω2 | we obtain similarly
ω1 ω2 + ω1 ω2
≤1
ω1 ω1
ω2 ω2 ⇒ + ≤ 1
ω1 ω1
⇒ |Re(τ )| ≤
1
2
(1)
Moreover if Re(τ ) = − 21 we replace the basis (ω1 , ω2 ) by (ω1 , ω1 + ω2 ) giving
2
Re( ω1ω+ω
) = 12 . Therefore condition (ii) is satisfied. If Im(τ ) < 0 then we
1
replace the basis by (−ω1 , ω2 ); which ensures Im(τ ) > 0 without changing
the condition (1) on Re(τ ). Finally if |τ | = 1 and Re(τ ) ≤ 0 we replace the
14
basis by (−ω2 , ω1 ). By introducing all these changes when necessary the
basis satisfies the conditions on τ .
Since the most general change of basis is by a unimodular transformation,
we denote a new ratio by τ 0 and obtain
τ0 =
ω20
aω2 + bω1
aτ + b
=
=
0
ω1
cω2 + ω1
cτ + d
for ad − bc = ±1.
Separating into real and imaginary parts we obtain
Re(τ 0 ) + iIm(τ 0 ) =
=
=
aRe(τ ) + b + iaIm(τ )
cτ + d
(aRe(τ ) + b + iaIm(τ ))(cRe(τ ) + d − icIm(τ ))
|cτ + d|2
ac(Re(τ )2 + Im(τ )2 ) + ad(Re(τ ) + iIm(τ )) + bc(Re(τ ) − iIm(τ ))
|cτ + d|2
Equating imaginary parts we arrive at
Im(τ 0 ) =
±Im(τ )
|cτ + d|2
Where we used that ad − bc = ±1 and hence the sign in front of Im(τ ) is the
same as the sign in ad − bc = ±1.
Now suppose that both τ and τ 0 satisfy the conditions in the proposition.
We will show that they must be equal.
By condition (i) we have that the positive sign is valid in the above equation
and hence ad − bc = 1. Moreover since τ and τ 0 have identical roles we can
assume without loss of generality that Im(τ 0 ) ≤ Im(τ ). We then have that
|cτ + d|2 ≤ 1. Since c and d are integers there are a finite number of possible
values for c and d.
Case 1. c = 0, d = ±1.
Here ad − bc = 1 becomes ad = 1 and as a and d are integers we have that
either a = d = 1 or a = d = −1. The equation for τ 0 then becomes τ 0 = τ ± b
and by condition (ii) we have that
|b| = |Re(τ 0 ) − Re(τ )| < 1
As b is an integer, b = 0 and τ 0 = τ .
15
Case 2. d = 0
The equation ad − bc = 1 becomes −bc = 1. Hence we have that either b = 1,
c = −1 or b = −1, c = 1. Under these restrictions the inequality |cτ + d| ≤ 1
becomes |τ | ≤ 1 and by condition (iii) we have that |τ | = 1. The equation
for τ 0 becomes τ 0 = ±a − τ1 and since |τ | = 1 we have that τ 0 = ±a − τ By
condition (ii)
|a| = |Re(τ 0 ) + Re(τ )| < 1
unless τ 0 = τ = eiπ/3 as then |Re(τ 0 ) + Re(τ )| = 1. Aside from this value we
obtain a = 0 and thus τ 0 = −τ . Therefore if τ = α + iβ then τ 0 = −α + iβ
and by condition (iv) we have that α = 0 and hence β = 1. Therefore the
equation can only hold if τ 0 = τ = i.
Case 3. c 6= 0 and d 6= 0.
As c and d are integers this gives that |cd| ≥ 1. Therefore by use of condition
(ii) and (iii)
|cτ + d|2 = c2 |τ |2 + d2 + 2cdRe(τ ) ≥ c2 + d2 − |cd| = (|c| − |d|)2 + |cd| ≥ 1.
(2)
And since |cτ + d| ≤ 1 we have that equality must hold for all c 6= 0, d 6= 0
and τ under conditions (i) to (iv).
From (2) and |cτ + d| = 1,
c2 + d2 − |cd| ≤ 1.
This is only valid for c = ±1 and d = ±1 at which we obtain equality. Hence
|cτ + d| = c2 |τ |2 + d2 + 2cdRe(τ ) = |τ |2 + 1 ± 2Re(τ ) = 1
⇒ |τ |2 ± 2Re(τ ) = 0
The equation with the positive sign has no solutions by conditions (ii) and (iii).
The negative sign equation has just one solution at |τ | = 1 and Re(τ ) = 1/2
by the same conditions. Therefore we must have that τ = eiπ/3 and by
conditions (i) to (iv) we have a strict minimum of Im(τ ). Looking at Figure 1
we see that this minimum is achieved at the intersection of the line Re(τ ) = 12
and the unit circle. As Im(τ 0 ) ≤ Im(τ ), τ 0 is also confined to the point of
intersection and we have that τ 0 = τ = eiπ/3
Thus we have shown that under the conditions (i) to (iv) τ is unique under
any change of basis. The region for τ can be seen in Figure 1 on the next
page (Ahlfors, 1966).
16
Figure 1: The τ plane
2.3
Elliptic Functions
Definition 2.3.1. An elliptic function is a doubly periodic meromorphic
function over C.
Before explicitly setting up an elliptic function, we can prove some nice
results about them. Here we state Liouville’s Theorem without proof:
Lemma 2.3.2. Liouville’s Theorem. A analytic function that is bounded in
the whole plane must reduce to a constant (Ahlfors, 1966).
Theorem 2.3.3. An elliptic function that has no poles is constant Ahlfors
(1966)
Proof. Let f (z) be an elliptic function with periods ω1 and ω2 . Let Pα be a
parallelogram with vertices α, α + ω1 , α + ω2 , α + ω1 + ω2 . Without poles
the elliptic function is bounded on this parallelogram. We can tile the entire
17
plane by constructing a lattice from parallelograms with differing values of
α because we can always add and subtract periods to get back to Pα . Hence
the function is bounded in the whole plane. By Liouville’s Theorem the
function is constant.
Theorem 2.3.4 The sum of the residues of the poles inside a complete set
of representatives of an elliptic function is zero (Price, 2001).
By “complete set of representatives” we mean all the points in Pα . As for
these points, for every z ∈ C we have f (z) = f (z 0 ) for some z 0 ∈ Pα .
Proof. The number of poles inside Pα is finite as otherwise the set of poles
would have an accumulation point, say z0 in this compact region. This z0
would then be a pole, however would not be isolated, since every neighbourhood of z0 contains another pole. This contradicts the definition of a pole
and therefore we have a finite number of poles.
Since the number of poles is finite inside Pα we can choose α in Pα such that
there are no poles on the boundary ∂Pα . By the Cauchy Residue Theorem,
with poles at every zj 6∈ ∂Pα we have:
2πi
X
Z
res(f, zj ) =
f (z)dz
∂Pα
j
Z
α+ω1
=
Z
α+ω1 +ω2
f (z)dz +
α
Z
α+ω2
f (z)dz +
α+ω1
Z
α
f (z)dz +
α+ω1 +ω2
f (z)dz
α+ω2
Now using f (z) = f (z − ω1 ) and f (z) = f (z − ω2 ) as f is periodic gives:
Z α+ω1
Z α+ω1 +ω2
Z α+ω2
Z α
f (z)dz+
f (z−ω1 )dz+
f (z−ω2 )dz+
f (z)dz
α
α+ω1
α+ω1 +ω2
α+ω2
And then changing variables in the second and third integrals gives:
Z α+ω1
Z α+ω2
Z α
Z α
=
f (z)dz +
f (z)dz +
f (z)dz +
f (z)dz = 0
α
α
α+ω1
(Brubaker, 2008)
18
α+ω2
Theorem 2.3.5. (The argument principle) Let f (z) be a meromorphic in
Ω, a simply connected open region, with zeros at aj and poles at bk . Let C
be a closed curve in Ω which does not pass through any of the zeros or poles.
Let N and P denote respectively the number of zeros and poles of f , inside
the contour, counted with multiplicity. Then
Z
1
f 0 (z)
dz = N − P
2πi C f (z)
We will not prove this. A proof is gived in (Ahlfors, 1966). However this
then gives the theorem:
Theorem 2.3.6. The number of zeros and poles of a nonconstant elliptic
function are equivalent counting multiplicity.
Proof. We use the above argument principle with C = Pα . A full set of
mutally incongruent poles and zeros is contained within Pα as the elliptic
function is completely determined by its values on Pα (see“complete set of
representatives” comment above).
If f (z) is elliptic then:
d
d
d
f (z)|z=z0 =
f (z + ω1,2 )|z=z0 =
f (z)|z=z0 +ω1,2
dz
dz
dz
And so f 0 (z) is elliptic. Hence f 0 /f is elliptic. By the same method as the
above theorem we can choose α in Pα such that none of the poles or zeros
are on the boundary ∂Pα . Then by the same method again we show that
the integral is zero on the boundary and thus
N −P =0
and we have our result (Ahlfors, 1966).
Now as f (z) and f (z) − c have the same poles however a different set of zeros
we have that the number of solutions to the equation f (z) = c is a constant
independent of c. We will define the number of solutions to this equation as
the order of the elliptic function.
By the above results we have that an elliptic function cannot have a single
simple pole. Therefore the simplest, non trivial elliptic function is of order
two and thus has either a single irreducible double pole, or two poles of equal
but opposite residue. We will be concerned with the former.
19
2.4
Weierstrass ℘ function
An elliptic function with a single irreducible double pole was first constructed
by Weierstrass. Note that we may take the double pole to be at the origin.
Definition 2.4.1. The Weierstrass ℘ function is defined as
X
1
1
1
℘(z) = 2 +
− 2.
2
z
(z − ω)
ω
ω∈Ω\{0}
Ω := {ω ∈ C | ω = n1 ω1 + n2 ω2 , n1,2 ∈ N ,
ω1
ω2
6∈ R}.
Theorem 2.4.2. The Weierstrass ℘ function is doubly periodic with distinct
periods ω1 and ω2 .
Proof. To prove this we need to show that the series above converges uniformly for z restricted to any compact set and that the function is indeed
doubly periodic.
First we prove convergence:
1 ω 2 − (z − ω)2 1
−
=
(z − ω)2 ω 2 ω 2 (z − ω)2 2
ω − z 2 + 2zω − ω 2 =
ω 2 (z − ω)2
z(2ω − z) .
= 2
ω (z − ω)2 We would like to examine the behaviour of the above expression as ω becomes
arbitrarily large. Hence we stipulate that |ω| > 2|z|. Using this we can bound
the summand for ω suffciently large, leading to convergence by the comparison
test.
Using the reverse triangle inequality and |ω| > 2|z|:
1
1
|z − ω| ≥ |ω| − |z| > |ω| − |ω| > |ω|.
2
2
Now going back to the summand we want to bound we use the triangle
inequality:
z(2ω − z) zω + z(ω − z) z(ω − z) zω
ω 2 (z − ω)2 = ω 2 (z − ω)2 ≤ ω 2 (z − ω)2 + ω 2 (z − ω)2 20
Now we bring the modulus signs inside the fraction:
=
|z||ω − z|
|z||ω|
+
2
− ω|
|ω|2 |z − ω|2
|ω|2 |z
Now using our inequality |z − ω| > 12 |ω|:
≤
|z||ω|
|z|
6|z|
.
+
=
|ω|3
|ω|2 · 14 |ω|2 |ω|2 · 12 |ω|
The entire summand is bounded:
1 6|z|
1
(z − ω)2 − ω 2 ≤ |ω|3 .
Therefore for z contained on any compact set, ℘ converges uniformly if
X
ω∈Ω\{0}
1
< ∞.
|ω|3
We can relate this summand to a special case of the Riemann-zeta function.
Define f : R2 \ {0, 0} → R by
f (n1 , n2 ) =
|n1 ω1 + n2 ω2 |
|n1 | + |n2 |
As f (tn1 , tn2 ) = f (n1 , n2 ) for all t =
6 0, we can assume that n1 and n2 lie on
the unit circle and thus restrict the domain of f .
Now f is continuous on an compact set and so attains its infimum, say k.
Hence for all real pairs (n1 , n2 ),
|n1 ω1 + n2 ω2 |
≥k
|n1 | + |n2 |
If we let n = |n1 | + |n2 | and restrict n, n1 and n2 to the integers we see that
there is 4n pairs (n1 , n2 ) that give the value n.
If n1 and n2 are positive then there are n + 1 ways to sum to n; i.e. pairs
(n, 0), (n − 1, 1),...,(0, n). However since we allow negatives the pairs from
(n − 1, 1), (n − 2, 2), ..., (1, n − 1) give 4(n − 1) pairs and the two end pairs,
(n, 0) and (0, n), only admit two pairs each since they contain 0. Hence we
have 4(n − 1) + 2 + 2 = 4n pairs.
21
Now we have
|ω| ≥ kn
1
1
⇒
≤
3
|ω|
(kn)3
Taking the sum:
X
ω∈Ω\{0}
≤
∞
X
1
=
|ω|3

X

n=1
|n1 +|n2 |=n
X
(n1 ,n2 )∈Z2 \{0}
1
=
|ω|3
∞
X
X

n=1


∞
1  X 1 
=
(kn)3
(kn)3
n=1


|n1 |+|n2 |=n

X
|n1 +|n2 |=n
1 =
1 
|ω|3
∞
X
n=1
1
· 4n
(kn)3
Where in the last step we used that there are 4n pairs of (n1 , n2 ) that sum
to n.
Therefore we attain
X
ω∈Ω\{0}
∞
∞
n=1
n=1
X 4n
X 1
1
≤
= 4k −3
< 8k −3 < ∞.
3
3
|ω|
(kn)
n2
Where we used a special case of the Riemann zeta function:
∞
X
1
π2
=
< 2.
n2
6
n=1
We now need to prove that ℘ is doubly periodic with periods ω1 and ω2 .
For this we differentiate ℘(z). As we have proved uniform convergence we
can differentiate the series term by term (see Corollary 4.1.5 on page 41) to
give:
X
−2
−2
℘0 (z) = 3 +
z
(z − ω)3
ω∈Ω\{0}
=
X
ω∈Ω
Where we have brought the
−2z −3
−2
(z − ω)3
inside the sum as the term where ω = 0.
We can see that this function is doubly periodic:
X
X
X
−2
−2
−2
℘0 (z+ω1 ) =
=
=
= ℘0 (z).
3
3
(z + ω1 − ω)
(z − (ω − ω1 ))
(z + ω)3
ω∈Ω
ω−ω1 ∈Ω
22
ω∈Ω
And similarly for ω2 gives ℘0 (z + ω2 ) = ℘0 (z).
This means that the derivatives of the functions f1 (z) = ℘(z + ω1 ) − ℘(z)
and f2 (z) = ℘(z + ω2 ) − ℘(z) are zero implying that f1 (z) and f2 (z) are
constant. As ℘(z) is clearly an even function we take z = −ω1 /2 to give:
f1 (−
ω1
ω1
ω1
ω1
ω1
) = ℘(ω1 − ) − ℘(− ) = ℘( ) − ℘( ) = 0.
2
2
2
2
2
Similarly for ω2 .
Hence we have shown that ℘(z) converges and is doubly periodic with distinct
periods ω1 and ω2 (Ahlfors, 1966; Brubaker, 2008).
As ℘(z) has a pole of multiplicity 2 we have from Theorem 2.3.6 (page 17)
that the order of ℘(z) is 2. Similarly as ℘0 (z) has a triple pole, the order of
℘0 (z) is 3.
Next we would like to introduce ζ(z) the antiderivative of ℘(z). We let
ζ 0 (z) = −℘(z) and define:
X 1
1
z
1
+ +
.
ζ(z) = +
z
z − ω ω ω2
ω∈Ω\{0}
We must again prove convergence:
2
1
ω + ω(z − ω) + z(z − ω) 1
z
|z|2
=
=
+
+
z − ω ω ω2 |ω|2 |z − ω|
ω 2 (z − ω)
Now again we use the condition |ω| > 2|z| ⇒ |z − ω| > 12 |ω|.
|z|2
|z|2
2|z|2
≤
=
|ω|2 |z − ω|
|ω|3
|ω|2 · 12 |ω|
Therefore the summand is bounded:
1
1
z 2|z|2
z − ω + ω + ω 2 ≤ |ω|3 .
And we have seen previously that
X
ω∈Ω\{0}
1
.
|ω|3
converges for any z contained in a compact set. Hence ζ(z) converges (Price,
2001).
23
Now we wish to construct the Laurent expansion of ζ(z) about zero. This
will then give us the Laurent expansion of ℘(z) about zero by differentiation
(Ahlfors, 1966).
∞
1
−1
−1 X z n
=
=
z−ω
ω(1 − ωz )
ω
ωn
n=0
Therefore
1
1
z
z2
z3
+ + 2 = − 3 − 4 − ...
z−ω ω ω
ω
ω
We have that if k is odd then ω k = −(−ω)k . Thus once we sum over all the
periods, for each ω, there is a corresponding −ω term which cancels with
with it if the power is odd. Hence all terms with periods to an odd power
vanish and we obtain:
∞
ζ(z) =
1 X
−
Fk z 2k−1
z
k=2
where we take:
Fk =
X
ω∈Ω\{0}
1
ω 2k
Since ℘(z) = −ζ 0 (z), we obtain
∞
℘(z) =
X
1
+
(2k − 1)Fk z 2k−2 .
z2
k=2
(Ahlfors, 1966)
Proposition 2.4.3. The Weierstrass ℘ function satisfies the following
differential equation:
℘0 (z)2 = 4℘(z)3 − 60F2 ℘(z) − 140F3
The below proof is found in (Ahlfors, 1966).
Proof. Using our Laurent expansion for ℘(z) which we have just found we
calculate the required terms:
℘(z) =
1
+ 3F2 z 2 + 5F3 z 4 + ...
z2
2
+ 6F2 z + 20F3 z 3 + ...
z3
2
2
4
24F2
℘0 (z)2 = − 3 + 6F2 z + 20F3 z 3 + ... = 6 − 2 − 80F3 + ...
z
z
z
℘0 (z) = −
24
3
4℘(z) = 4
3
1
4
36F2
2
4
+ 3F2 z + 5F3 z + ... = 6 + 2 + 60F3 + ...
2
z
z
z
Now we note:
℘0 (z)2 − 4℘(z)3 = −
60F2
− 140F3 + O(z)
z2
⇒ ℘0 (z)2 − 4℘(z)3 + 60F2 ℘(z) = −140F3 + O(z)
The left hand side is an elliptic function since it is a polynomial in two elliptic
functions ℘0 (z) and ℘(z). The right hand side has no poles. Hence we have
that it should be equal to a constant and we conclude:
℘0 (z)2 − 4℘(z)3 + 60F2 ℘(z) = −140F3
We can write this differential equation as
℘0 (z)2 = 4(℘(z) − e1 )(℘(z) − e2 )(℘(z) − e3 )
where e1 , e2 , e3 are the roots to the equation 4w3 − 60F2 w − 140F2 . We
would like to determine the zeros of ℘0 (z). From the periodicity and eveness
of ℘(z) we have ℘(ω1 − z) = ℘(z) and hence we achieve ℘0 (ω1 − z) = −℘0 (z).
Letting z = ω1 /2 we obtain ℘0 (ω1 /2) = 0. Similarly ℘0 (ω2 /2) = 0 and
℘0 ((ω1 + ω2 )/2) = 0. As these results are mutally incongruent modulo the
periods, from Theorem 2.3.6 (page 17), and the fact that ℘0 (z) has a pole
of order 3 (page 21) it follows that these are the three zeros of ℘0 (Ahlfors,
1966).
We note that we can set:
e1 = ℘(ω1 /2),
(3)
e2 = ℘(ω2 /2),
(4)
e3 = ℘((ω1 + ω2 )/2).
(5)
It is easy to see that the ek are distinct. As ℘(z) is an even function, the
equation
℘(z) − ek = 0
is solved by
z = ±ω1 /2 , for k = 1,
z = ±ω2 /2 , for k = 2,
z = ±(ω1 + ω2 )/2 , for k = 3.
25
Hence each ek has multiplicity 2. If two values were the same then this value
would be assumed four times, contradicting the fact that the order of ℘(z)
is two and so the number of solutions to ℘(z) − c = 0 should be 2 for any
constant c (Theorem 2.3.6) (Ahlfors, 1966).
Furthermore if we substitute the values z = ω1 /2, ω2 /2 and (ω1 + ω2 )/2 into
the equation for ℘(z) then we see that multiplication of the periods by a
constant m, multiplies the ek by m−2 . Therefore the modular function:
λ(τ ) =
e3 − e2
e1 − e2
depends only on the ration τ = ω2 /ω1 (Ahlfors, 1966).
It is then immediately clear that λ(τ ) is defined upon the upper half plane
Im(τ ) > 0, and is analytic as e1 6= e2 . It is also never equal to 0 as e3 6= e2
and never equal to 1 as e3 6= e1 .
Next we note that under a general unimodular transformation:
ω20 = aω2 + bω1
ω10 = cω2 + dω1
the roots ek can at most be permuted since the ℘ function is invariant (due
to being periodic with periods ω1 and ω2 ) under the transformation. Now
from (3), (4) and (5) we see that under the transformation
e1 = ℘((cω2 +dω1 )/2) , e2 = ℘((aω2 +bω1 )/2) , e3 = (((a+c)ω2 +(b+d)ω1 )/2)
if a, d ≡ 1 mod 2 and b, c ≡ 0 mod 2 we have
e1 = ℘(k1 ω2 + k2 ω1 + ω1 /2) = ℘(ω1 /2),
e2 = ℘(k3 ω2 + k4 ω1 + ω2 /2) = ℘(ω2 /2),
e3 = (k5 ω2 + k6 ω1 + (ω1 + ω2 )/2) = ℘((ω1 + ω2 )/2).
where kj ∈ Z and ℘ is periodic in ω1 and ω2 . Hence the ek are not permuted
for this condition on a, b, c, d. In other words we have:
aτ + b
a b
1 0
λ
= λ(τ ) for
≡
mod 2
c d
0 1
cτ + d
(Ahlfors, 1966)
We may ask what happens when this is not the case. For this we can just
consider matrices which
are
equivalent
(when the entries of the matrix are
1 1
0 1
taken modulo 2) to
and
, as all others can be composed of
0 1
1 0
26
these. Clearly in the first case we obtain that ω20 ≡ ω1 + ω2 and ω10 ≡ ω1 and
therefore we have that e2 and e3 are permuted giving
τ0 =
ω20
=τ +1
ω10
e2 − e3
λ(τ + 1) =
=
e1 − e3
e3 −e2
e1 −e2
e3 −e2
e1 −e2 −
1
=
λ(τ )
λ(τ ) − 1
(6)
In the second case, we have that ω20 ≡ ω1 and ω10 ≡ ω2 so that e1 and e2 are
interchanged. The determinant of this matrix is −1 and so the transformation
τ → τ1 would send the upper half plane to the lower half plane, where λ is
not defined. As such we introduce a negative giving
τ0 = −
1
τ
1
e3 − e1
e1 − e2 − (e3 − e2 )
λ −
=
=1−
= 1 − λ(τ ).
τ
e2 − e1
e1 − e2
(7)
These functional equations will be useful in describing the behaviour of λ(τ )
at degenerate points.
2.5
Conformal Mapping and Convergence of λ(τ )
The aim of this section is to prove that λ(τ ) maps conformally to the upper
half plane from some specific region Ω (see Figure 2, page 32). First however
we need to examine the behaviour of λ(τ ) as Im(τ ) → ∞.
We use the periods ω1 = 1 and ω2 = τ for simplicity, and from the definition
of ek we obtain:
!
∞
X
1
1
−
,
e3 − e2 =
(n1 − 12 + (n2 − 12 )τ )2 (n1 + (n2 − 21 )τ )2
n ,n =−∞
1
e1 − e2 =
2
∞
X
n1 ,n2 =−∞
1
1
−
1
2
(n1 − 2 + n2 τ )
(n1 + (n2 − 12 )τ )2
!
.
To evaluate these sums we note (without proof, which can be found in
(Ahlfors, 1966)) that
∞
X
π2
1
=
.
2
sin (πz) n =−∞ (z − n1 )2
1
27
Now by summing over n1 we see that for e3 we take z = 1/2 − (n2 − 1/2)τ ,
for e2 we take z = −(n2 − 1/2)τ and for e1 we take z = 1/2 − n2 τ . Using
the angle formulae for cosine and sine we have
1
1
2 π
2
sin
τ = cos π n2 −
τ .
− π n2 −
2
2
2
Plugging all these values in gives:
e3 − e2 = π 2
∞
X
cos2
n2 =−∞
e1 − e2 = π
2
∞
X
1
1
−
1
2
sin π n2 − 12 τ
π n2 − 2 τ
1
1
−
cos2 (πn2 τ ) sin2 π n2 − 21 τ
n2 =−∞
!
,
!
.
We will evaluate these sums and describe the behaviour as Im(τ ) → ∞. To
start with we will consider:
∞
X
n=1
cos2 (π
1
n − 12 τ )
Noting that
1
1
|eiπ(n− 2 )τ | = e−π(n− 2 )Im(τ )
1
1
1
⇒ 2 cos π n −
τ = eiπ(n− 2 )τ + e−iπ(n− 2 )τ 2
1
1
≥ eπ(n− 2 )Im(τ ) − e−π(n− 2 )Im(τ )
Now if we take n ≥ n0 where n0 is an integer such that e−(2n0 −1)πIm(τ ) ≤ 1/2
we have
1
1
2 cos π n −
τ ≥ eπ(n− 2 )Im(τ ) 1 − e−π(2n−1)Im(τ )
2
1
1
1
≥ eπ(n− 2 )Im(τ ) ≥ eπ(n−1)Im(τ )
2
2
Hence we can show absolute convergence:
∞ X
1
2
1
cos π n − 2 τ n=1
28
≤
n0
X
n=1
4
e
π (n− 12 )Im(τ )
−e
−π (n− 12 )Im(τ )
2 +
∞
X
16e−2(n−1)πIm(τ ) < ∞
n=n0 +1
The last thing we need to show is uniform convergence in the limit Im(τ ) →
∞.
For any > 0 if Im(τ ) ≥ δ > 0 we can take n0 (only depending on ) such
that
∞
∞
X
X
16e−2(n−1)πIm(τ ) ≤
16e−2(n−1)πδ ≤ n=n0 +1
n=n0 +1
Therefore
n0
∞ X
X
4
1
≤
2
2 + 1
1
cos π n − 21 τ π n− 2 )Im(τ )
n=1 e (
n=1
− e−π(n− 2 )Im(τ )
And taking limits as Im(τ ) → ∞
∞ X
1
lim
2
cos π n − 12 τ Im(τ )→∞
n=1
≤
n0
X
lim
Im(τ )→∞
n=1
4
e
π(
n− 12
)Im(τ ) − e−π(n− 12 )Im(τ )
2 + ≤ As is arbitrary we have that
∞ X
1
lim
2
=0
1
Im(τ )→∞
cos π n − 2 τ n=1
which implies
∞
X
lim
Im(τ )→∞
Now for
n=1
cos2
0
X
n=−∞
cos2
1
=0
π n − 12 τ
1
π n − 12 τ
we can take n → −n and then we have the sum is equal to
∞
X
n=0
cos2
1
π n + 12 τ
29
where we used that cos(−z) = cos(z).
We can now use the same method from above (we omit the working) to show
that
∞
X
1
=0
lim
2
Im(τ )→∞
cos π n + 12 τ
n=0
and thus
∞
X
lim
Im(τ )→∞
n=−∞
cos2
1
=0
π n − 12 τ
We can employ the same method to the remaining summands and obtain
lim
Im(τ )→∞
∞
X
2
n=−∞
sin
1
=0,
π n − 21 τ
lim
Im(τ )→∞
∞
X
n=−∞,n6=0
1
=0
cos2 (πnτ )
Therefore we have that as Im(τ ) → ∞, e3 − e2 → 0, e1 − e2 → π 2 and
thus λ(τ ) → 0 uniformly with respect to the real part of τ . Moreover, from
equation (7) on page 24 we see that λ(τ ) → 1 as Im(τ ) → 0 (Ahlfors, 1966).
The last thing we need before we can show the conformal mapping of λ(τ )
is to find the order to which λ(τ ) vanishes with e−πiτ (Ahlfors, 1966). This
result will be used later to establish the conformal mapping.
We look at the leading terms of the summation of e3 − e2 which are the
terms corresponding to n = 0 and n = 1. These will be sufficient to describe
the behaviour as Im(τ ) → ∞ due to the periodicity of sin and cos. The sum
of these leading terms is
"
#
"
#
1
1
1
1
−
+ π2
−
π2
cos2 − 21 τ π
sin2 − 12 τ π
cos2 12 τ π
sin2 21 τ π
= 2π
2
4eiτ π
4eiτ π
+
(1 + eiτ π )2 (1 − eiτ π )2
where we used that 2 cos(x) = eix + e−ix and 2i sin(x) = eix − e−ix . Therefore
we obtain that for Im(τ ) arbitrarily large:
2π 2 e−iτ π
4eiτ π
4eiτ π
−iτ π
λ(τ )e
≈
+
π2
(1 + eiτ π )2 (1 − eiτ π )2
4
4
=2
+
(1 + eiτ π )2 (1 − eiτ π )2
30
Now taking the limit as Im(τ ) → ∞ and noting that
lim
eiτ π =
Im(τ )→∞
lim
eiπRe(τ ) e−πIm(τ ) = 0
Im(τ )→∞
we have that
λ(τ )e−iτ π → 16
as Im(τ ) → ∞ (Ahlfors, 1966).
We can now look to showing that, for a specfic domain which will be defined,
the modular lambda function maps conformally to the upper half plane.
Definition 2.4.3. A subset Rλ of the upper half plane is called the fundamental domain of λ(τ ) if
1. There are no two distinct points, τ 6= τ 0 of Rλ such that λ(τ ) = λ(τ 0 ).
2. If τ is in the upper half plane, then there is a point τ 0 in the closure of
Rλ such that λ(τ ) = λ(τ 0 ).
It can be shown that the fundamental domain for λ in question is the region
Im(τ ) > 0 , −1 < Re(τ ) ≤ 1 , |τ − 1/2| ≥ 1/2 , |τ + 1/2| > 1/2.
However we are interested in the open subset of the fundamental domain for
which Re(τ ) > 0, |τ − 1/2| > 1/2.
This is the open region Ω (Figure 2 on next page) which is bounded by
(but does not include) the imaginary axis, the line Re(τ ) = 1 and the circle
centered at τ = 1/2 with radius 1/2.
Theorem 2.4.4. The modular function λ effects a one to one conformal
mapping of the open region Ω onto the upper half plane.
Proof. We first look at what happens on the boundary of the closure of Ω.
Looking back at the sums for ek it can be shown that they are real when τ
is purely imaginary. Hence we have that λ(τ ) is real on the imaginary axis.
Also we can transform the imaginary axis to the line Re(τ ) = 1 by τ → τ + 1.
From the functional equations ((6) and (7) on page 24) we have that λ(τ + 1)
is real due to λ(τ ) being real. Next we can transform the line Re(τ ) = 1 into
the circle |τ − 1/2| = 1/2 by τ → 1 − τ1 . Again λ 1 − τ1 is real. Hence we
have that λ takes real values on the boundary of the closure of Ω.
31
Figure 2: The region Ω
We would like to see how many times λ(τ ) takes a non real value, say ω0 , in
Ω. Now we imagine the horizontal line segment Im(τ ) = t0 for any arbitrarily
large t0 . As the horizontal line segment Im(τ ) = t0 does not contain τ = 0,
its image under the linear transformations − τ1 and 1 − τ1 are circles (the first
tangent to the real axis at τ = 0 and the second at τ = 1, see Figure 2 page
32). However as the real axis does contain zero, then we do not obtain a
finite denominator for all values of τ and thus under these transformations it
is again the real line. And since these two horizontal lines never meet, they
cannot intersect under the transformation. We then truncate the region Ω
by removing these regions.
For any non-real value ω0 , we take t0 sufficently large, so that we can be
certain that the point ω0 is not in the portions we cut off. i.e. for any
non-real value we can take t0 sufficiently large such that ω0 is not inside the
circle at τ = 0 or τ = 1 and that Im(ω0 ) < t0 .
1
Now we map the circle near τ = 1 on a curve λ = λ(1 − τ1 ) = 1 − λ(τ
) with
−πiτ
τ = s + it0 for 0 ≤ s ≤ 1. Using that λ(τ )e
→ 16 for Im(τ ) arbitrarily
large we have
1−
1
eπt0 −πis
1
e−πiτ
≈1−
=
1
−
=
1
−
λ(τ )
16eπiτ
16
16
32
Here eπt0 is the radius of the circle which is arbitrarily large. The value of s
determines the angle and as this varies from 0 to 1 the angle varies from 0
to −π. However the negative sign reflects everything in the real axis and we
obtain a semicircle with infinite radius lying in the upper half plane only.
Now if we look at the image of the contour of the truncated region Ω we see
that under the mapping λ(τ ) the imaginary axis, Re(τ ) = 1 and the circle
|τ − 1/2| = 1/2 map to the real line. And as the horizontal line segment maps
to the semicircle described above we have a contour which approximately
encloses the upper half plane.
Before continuing we briefly discuss the concept of winding numbers. Intuitively a winding number measures the number of times a given curve “winds”
around a given point. The curve cannot pass through the point. We define
the winding number of a closed curve γ : [0, 1] → Ω about the origin as the
complex number
Z
1
1
wγ =
dz.
2πi γ z
We can then talk about the winding number of a point where γ is the image
of a curve under a holomorphic function f : Ω :→ C \ {0} (Conway, 1973).
Z 0
1
f (z)
wf ◦γ =
dz
2πi γ f (z)
We prove this as follows, from the definition given above for the winding
number we have:
Z
Z 1 0
Z 0
1
1
1
f (γ(t))γ 0 (t)
1
f (z)
wf ◦γ =
dz =
dt =
dz
2πi f ◦γ z
2πi 0
f (γ(t))
2πi γ f (z)
Therefore if we took f (z) = z − a, this would give a general winding number
about any point a. We notice that this contour integral is the exact same as
the one used in the Argument Principle (Theorem 2.3.5, page 17). As f was
defined as holomorphic, it has no poles and so we can infer:
wf ◦γ = N
where N is the number of zeros of f (counted with multiplicity) inside the
contour f ◦ γ.
Now moving back to our modular function and the mapping. We take the
contour of Ω to be our γ and we take the function f = λ − ω0 , where λ is our
modular function. Now from our understanding of the intuitive nature of
33
the winding number and since our image of the contour of Ω approximately
encloses the upper half plane we have that the winding number of ω0 is 1 if
Im(ω0 ) > 0 and is 0 if Im(ω0 ) < 0.
Hence the number of zeros of λ(τ ) − ω0 is 1 if Im(ω0 ) > 0 and is 0 if
Im(ω0 ) < 0. Since ω0 was arbitrary we have that λ(τ ) assumes every value
in the upper half plane exactly once and no value in the lower half plane.
Since λ is injective and analytic, it has non vanishing derivative λ0 (τ ) for all
τ in Ω. Hence the map is conformal (Ahlfors, 1966).
We now ask what happens if we include the boundary of Ω and extend the
domain by reflection along the imaginary axis. This new domain is the region
Ω ∪ Ω0 where we take the closure of Ω in the upper half plane and hence
{0, 1} 6∈ Ω. By reflection the region Ω0 that is symmetric to Ω with respect
to the imaginary axis maps to the lower half plane and since τ = 0, 1, ∞
correspond to λ = 1, ∞, 0; we have that Ω ∪ Ω0 maps under λ to the entire
plane minus the points 0 and 1. Therefore we obtain the universal covering
of the twice punctured plane; λ : Ω ∪ Ω0 → C − {0, 1} (Ahlfors, 1966).
Definition 2.4.5. The congruence subgroup mod 2 (Γ(2)) is defined as the
group
a b
Γ(2) =
: a ≡ d ≡ ±1, b ≡ c ≡ 0 mod 2
c d
Theorem 2.4.6. Every point τ in the upper half plane is equivalent under
the congruence subgroup mod 2 to exactly one point in Ω ∪ Ω.
Proof. In Figure 3 (page 36) we note that the region ∆ ∪ ∆0 is the region
described in Proposition 2.2.2.
Looking at Figure 3 we can verify that the transformations τ , − τ1 , τ − 1,
1
τ −1
τ
1−τ , τ , 1−τ map the region ∆ to the shaded regions. We can do this
√
by examining how the the points i, 12 + i 23 and ∞ are changed under
the transformations. The same can be shown for ∆0 mapping to the nonshaded regions. We denote the matrices of the transformations by Sk for
k = {1, 2, ...6}. Hence the union of these six sets Sk (∆), k = 1, ..., 6, is all
the shaded parts of Ω ∪ Ω0 . Then the matrices of the inverse transformations
Sk−1 are, in order,
1 0
0 −1
1 1
1 −1
0 1
1 0
,
,
,
,
,
0 1
1 0
0 1
1 0
−1 1
1 1
Notice that every matrix with determinant ±1 is equivalent to one of these
matrices when you take the entries mod 2. Hence any unimodular matrix
34
is congruent to exactly one of them. Again the same can be shown for the
0
matrices Sk0 which map the the region ∆ to the non shaded regions of Ω ∪ Ω0 .
And so collectively the images of ∆ and ∆0 cover the set Ω ∪ Ω0 .
Now let τ be any point in the upper half plane. By Proposition 2.2.2 there
exists a unimodular transformation which perfoms a change of basis and
ensures that the resulting τ 0 is in ∆ ∪ ∆0 . We call that transformation S and
we therefore have Sτ is in ∆ ∪ ∆0 and thus in ∆ ∪ ∆0 . Suppose that Sτ is in
∆. The matrix S is congruent to one of the Sk−1 . Hence the matrix T = Sk S
is congruent to the identity matrix and thus in the congruence subgroup.
We can then see that T τ = Sk (Sτ ) is in Ω ∪ Ω0 . This is done equivalently if
Sτ is in ∆0 . Therefore there is always a T τ in Ω ∪ Ω0 .
However we wish to show that τ is equivalent under the congruence subgroup
mod 2 to a unique point in Ω ∪ Ω0 . Hence we note that the boundary of Ω0
contains Re(τ ) = −1, Re(τ ) = 0 and |τ + 1/2| = 1/2.
Now the line Re(τ ) = 0 lies in Ω. The transformation τ → τ + 2 maps the
line Re(τ ) = −1 to the line Re(τ ) = 1 which is in Ω. The transformation
τ → 2ττ+1 maps the circle |τ + 1/2| = 1/2 to the circle |τ − 1/2| = 1/2 which
is the circle in Ω. As these transformations are congruent to the identity
matrix under the congruence subgroup mod 2 it follows that the points
on the boundary are equivalent under the subgroup to a point in the set
Ω ∪ Ω0 . And since the Sk and the Sk0 are all mutually incongruent under the
congruence subgroup mod 2 the points T τ define a unique point in Ω ∪ Ω0
(Ahlfors, 1966).
35
Figure 3: Mapping of ∆ ∪ ∆0 to Ω ∪ Ω0
36
Theorem 2.4.7. (Picard’s Little Theorem). If a entire function f : C → C
omits two values, then it is constant.
Proof. Let the two values omitted by f be α and β. Then we can define
f1 (z) = f (z)−α
β−α which is entire and omits 0 and 1. Hence there is no restriction
in assuming f omits 0 and 1.
We define a global analytic function h, whose function elements (h, Ω) have
the following two properties Im(h(z)) > 0 and λ(h(z)) = f (z) for all z ∈ Ω.
First we wish to show that h can be continued along all paths. From Theorem
2.4.4 (see the first paragraph on pg. 29) and the following paragraph there
exists τ0 in Ω ∪ Ω0 such that λ(τ0 ) = f (0). Since λ0 (τ0 ) 6= 0, due to the same
theorem, there exists a local inverse of λ defined in a neighbourhood ∆0 of
−1
f (0) and denoted λ−1
0 , such that λ(λ0 (w)) = w for w ∈ ∆0 and
λ−1
0 (f (0)) = τ0
By continuity of f we have a neighbourhood Ω0 of the origin in which
f (z) ∈ ∆0 . Hence we can define
h(z) = λ−1
0 (f (z)) for z ∈ Ω0 .
We have to show that (h, Ω0 ) can be continued along any path and that
Im(h(z)) remains positive. If this were not true then there would exist a
path γ = γ(t), 0 ≤ t ≤ t1 , such that h could be continued and Im(h) remains
positive up to any t < t1 , however either h cannot be continued to t1 or that
Im(h) → 0 as t → t1 .
Hence we need only prove that h in fact can be continued to t1 and that
Im(h) remains positive as we approach t1 . Since γ and t1 are arbitrary this
will show that h can be continued along all paths.
We can determine τ1 such that λ(τ1 ) = f (γ(t1 )) and therefore we can again
define a local inverse λ−1
in a neighbourhood ∆1 of f (γ(t1 )) such that
1
λ−1
(f
(γ(t
)))
=
τ
.
Let
Ω
be
a neighbourhood of γ(t1 ) in which f (z) ∈ ∆1
1
1
1
1
and choose t2 < t1 such that γ(t) ∈ Ω1 for t2 ≤ t ≤ t1 . Now we assumed
that (h, Ω0 ) can be continued for t < t1 and so we have that h(γ(t2 )) is
well defined; i.e. γ(t2 ) is inside the domain of the analytical continuation of
(h, Ω0 ) for t < t1 .
Therefore we also know that λ(τ ) has the same value f (γ(t2 )) at τ = h(γ(t2 ))
and at τ = λ−1
1 (f (γ(t2 ))). Hence by Theorem 2.4.5 there exists a modular
transformation S in the congruence subgroup mod 2 such that
S λ−1
1 (f (γ(t2 ))) = h(γ(t2 ))
37
We define h1 (z) = S λ−1
1 (f (z)) for z ∈ Ω1 . We then obtain that h1 (γ(t2 )) =
h(γ(t2 )) where γ(t2 ) is an arbitrary element inside the intersection of the
domain of the analytical continuation of (h, Ω0 ) and Ω1 . Thus we have
that (h1 , Ω1 ) is an analytic continuation of (h, Ω0 ) along γ(t) up to t1 which
satisifes λ(h1 (z)) = f (z) and Im(h1 ) > 0. We conclude that h can be
continued along all paths.
Because of this, and that the plane is simply connected, by the monodromy
theorem h defines an entire function h(z). However |eih(z) | = e−Im(h) < 1 as
h has all its values in the upper half plane. Hence by Liouville’s theorem
eih(z) must reduce to a constant, and thus so must h(z) and consequently so
must f (z) = λ(h(z)). (Ahlfors, 1966)
38
Picard’s Great Theorem
Unfortunately when establishing a proof of Picard’s Great Theorem many
more non trivial proofs are required and as such a concrete starting point
from ”first principles” is out of reach for this short dissertation. Therefore
this chapter will assume some rather non trivial results from the outset which
will be stated when needed.
3
Montel’s Theorem
We now introduce the linear space C(Ω); the space of all continous functions
on the open set Ω. Let Ω ⊂ C. We will be interested in subsets of the space
C(Ω).
Definition 3.1.1. A set F ⊂ C(Ω) is normal if every sequence of functions
in F contains a subsequence which converges uniformly on compact sets of
Ω to a function f ∈ C(Ω) or ∞ (Conway, 1973).
This limit f of the subsequence is not required to belong to F.
Definition 3.1.2 A set F ⊂ C(Ω) is equicontinuous at a point z0 in Ω if
and only if for every > 0 there is a δ > 0 such that for |z − z0 | < δ
|f (z) − f (z0 )| < for every f in F (Conway, 1973).
In the following definition we denote H(Ω) as the collection of analytic
functions on Ω.
Definition 3.1.3. A set F ⊂ H(Ω) is locally bounded if for each point a in
Ω there are constants M and r > 0 such that for all f in F.
|f (z)| ≤ M
for |z − a| < r. Or equivalently
sup{|f (z)| : |z − a| < r, f ∈ F} < ∞.
(Conway, 1973)
This leads us nicely into the following lemma:
Lemma 3.1.4 A set F in H(Ω) is locally bounded if and only if for each
compact set K ⊂ Ω there is a constant M such that
|f (z)| ≤ M
39
for all f in F and z in K.
Proof. First we prove ⇒. Denote the open ball {z ∈ C : |z − a| < r} by
Br (a). Let K ⊂ Ω be a compact subset. Now the distance d = ρ(K, ∂Ω) is
non zero so we can take the open cover of K in Ω giving:
[
K⊂
Bd (a) ⊂ Ω
a∈K
As K is compact then we can pick a finite subcover of open disks {Bd (aj )}j .
K⊂
n
[
Bd (aj )
j=1
For each Bd (aj ) there is a constant Mj such that |f (z)| ≤ Mj for all z ∈
Bd (aj ). Taking M = maxj {Mj } we have that |f (z)| ≤ M for all z in K.
Now we prove ⇐. For all a ∈ Ω take r > 0 such that B2r (a) ⊂ Ω. This
ensures that the closed ball Br [a] is a compact subset of Ω. Thus we have
that |f (z)| ≤ M for z in Br [a]. If we hadn’t stipulated the first condition
the open ball could be the entire space Ω and thus the closure would not be
contained inside Ω. Now
Br (a) ⊂ Br [a]
and as |f (z)| ≤ M for all z ∈ Br [a] we have that |f (z)| ≤ M for all z ∈ Br (a).
Hence the function is locally bounded in every open disk in Ω. As Ω is open,
it is can be expressed as a union of open balls with centres in Ω. Hence
|f (z)| ≤ M for all z ∈ Ω.
We now state an important lemma without proof (proof in Conway (1973)):
Lemma 3.1.5. (Arzela-Ascoli Theorem.) A set F ⊂ C(Ω) is normal if and
only if the following conditions are satisfied:
(a) F is locally bounded;
(b) F is equicontinuous at each point of Ω.
(Conway, 1973)
40
We now prove the theorem named in the title of this section:
Montel’s Theorem 3.1.6. A family F in H(Ω) is normal if F is locally
bounded.
Proof. As F is locally bounded we need only prove equicontinuity at every
point in Ω. The result then follows from Lemma 3.1.5.
Let a ∈ Ω and take > 0. Since F is locally bounded we can take the
compact closed ball Br [a] ⊂ Ω and by hypothesis there exists M > 0 and
r > 0 such that |f (z)| ≤ M for all z in Br [a] and f in F. Now using the
Cauchy Integral Formula with γ(t) = a + reit :
Z
1
f (w)
f (w)
f (a) − f (z) =
−
dw
2πi γ (w − a) (w − z)
Z
f (w)(a − z)
1 dw
|f (a) − f (z)| =
2π γ (w − a)(w − z)
Since |f (w)| ≤ M for all w ∈ Br [a] we have:
Z
M |z − a| 1
|f (a) − f (z)| =
dw
2π
γ (w − a)(w − z)
Now if we limit z to the disk with radius r/2 centered at a then:
|z − a| <
r2
r
⇒ (w − a)(w − z) ≥
2
2
for all w in Br [a].
Z
M |z − a| dw
|f (a) − f (z)| ≤
r2 π
γ
Evaluating the integral gives
|f (a) − f (z)| ≤
Hence letting δ < min
1
2M |z − a|
r
r
2 r, 2M then |z − a| < δ ⇒ |f (a) − f (z)| < for all f in F.
This proves equicontinuity on the 12 r radius disk. (Conway, 1973)
41
4
Hurwitz’s Theorem
We now look at proving another important result towards attempting Picard’s
Great Theorem. First we must prove a few intermediary results. From here
on when I write fn → f , I am describing that fn converges uniformly to f
on all compact sets K ⊂ Ω.
Proposition 4.1.1. (Morera’s Theorem) Let Ω be a simply
R connected region
and let f : Ω → C be a continuous function such that C f = 0 for every
simple closed path C in Ω, then f is analytic in Ω.
Proof. Note that if we prove that f is analytic on every open disk contained
within Ω then it follows that f is analytic on all of Ω. Hence we can assume
Ω is an open disk.
As the integral over any simple closed path is equal to zero this implies that
we can achieve a well defined function F : Ω → C by setting
Z z
F (z) =
f (w) dw
a
Otherwise we would receive a different answer (and thus a different function)
depending on the path taken.
We want to prove that f has a primitive, namely F . Then we can conclude
that F is analytic and thus so is f .
We then fix z0 in Ω and see that
Z z0
Z
F (z) =
f (w) dw +
a
z
f (w) dw
z0
Z
z0
F (z0 ) =
f (w) dw
a
⇒
F (z) − F (z0 )
1
=
z − z0
z − z0
Z
z
f (w) dw
z0
Now notice that
f (z0 )
f (z0 ) =
z − z0
Z
z
z0
1
dw =
z − z0
Z
z
f (z0 ) dw
z0
Therefore using this and taking modulus:
Z z
F (z) − F (z0 )
1
−
f
(z
)
=
f
(w)
−
f
(z
)
dw
0
0
(z − z0 )
z − z0
z0
42
Now as z → z0 we have that w → z0 as w follows a continous path from z0
to z. Hence as f is a continuous function ⇒ f (w) → f (z0 ) as w → z0 . Due
to this we have:
∀ > 0 ∃δ > 0 such that |f (w) − f (z0 )| < whenever |z − z0 | < δ
Z z
F (z) − F (z0 )
= .
dw
⇒ − f (z0 ) <
(z − z0 )
|z − z0 | z0
Therefore we have shown
lim
z→z0
F (z) − F (z0 )
= f (z0 )
z − z0
Hence F 0 = f , we know that F is analytic. Also as f is the derivative of an
analytic function it is itself analytic (Conway, 1973).
Lemma 4.1.2. Let γ be a rectifiable (finite in length) curve in C and
suppose that fn and f are continous functions on {γ}. If fn → f as n → ∞
then
Z
Z
f = lim
fn
n→∞ γ
γ
Proof. As the length of γ is finite, denote V (γ) as the length. Now as fn → f
as n → ∞ (see top of page 37) there exists an N such that for all n > N we
have
|fn (z) − f (z)| ≤
, ∀z ∈ {γ}
V (γ)
Now
Z Z
f ≤ |f | |dz| ≤ V (γ) sup[|f (z)| : z ∈ {γ}]
γ
γ
R
R
We need to show that γ f − γ fn ≤ .
Z
Z
Z
Z
f − fn = f − fn dz ≤ |f − fn | |dz|
γ
γ
γ
γ
≤ V (γ) sup[|f − fn | : z ∈ {γ}] = V (γ) ·
Z
⇒
Z
f = lim
γ
n→∞ γ
(Conway, 1973)
43
fn .
= .
V (γ)
Lemma 4.1.3. Let f be analytic in Br [a] and suppose that |f (z)| ≤ M for
all z in Br [a]. Then
M n!
|f (n) (a)| ≤ n
r
Proof. By Cauchy Integral Formula for C the circle of radius r centered at
a:
Z
f (w)
n!
f (n) (a) =
dw
2πi C (w − a)n+1
Parametrise C with w = a + reiθ with 0 ≤ θ < 2π
Z 2π
Z 2π iθ )rieiθ
f (a + reiθ )rieiθ n!
f
(a
+
re
n!
(n)
dθ
|f (a)| = dθ ≤
2πi 0
2π 0 (reiθ )n+1
(reiθ )n+1
Z
Z
n! 2π M
M n!
n! 2π |f (a + reiθ )|
dθ
≤
dθ = n
=
2π 0
rn
2π 0 rn
r
Where we used that |f | ≤ M on C (Conway, 1973).
We can now prove a result needed for the proof of Hurwitz’s Theorem.
Theorem 4.1.4. If {fn } is a sequence in H(Ω) and f belongs to C(Ω) such
(k)
that fn → f then f is analytic and fn → f (k) for each integer k ≥ 1.
Proof. We will use Morera’s Theorem to show that f is analytic. Let C be a
compact closed curve contained within a disk D ⊂ Ω. Since C is compact fn
converges to f uniformly. Therefore using Lemma 4.1.2 we have
Z
Z
f = lim
fn = 0
T
n→∞ C
where the integral is equal to zero by Cauchy’s Theorem as each fn analytic.
Therefore by Morera’s Theorem f must be analytic in every disk D ⊂ Ω.
This shows that f is analytic in Ω.
(k)
Now we show that fn → f (k) .
First let D = Br [a] ⊂ Ω and then we let R > r so that BR [a] ⊂ Ω. The
reason we construct two closed balls is we want to restrict |z − a| ≤ r however
we want |z − a| to be strictly less than R as the Cauchy Integral Formula
only works for z inside C and not on the boundary. Hence if γ is the circle
|z − a| = R then Cauchy’s Integral Formula gives:
Z
k!
fn (w) − f (w)
(k)
(k)
dw
fn (z) − f (z) =
2πi γ (w − z)k+1
44
for z in D.
Now we use Lemma 4.1.3 with the extra restriction of |z − a| ≤ r. Hence we
can parametrise w = a + Reiθ , z = a + reiθ for 0 ≤ θ < 2π. We obtain
|fn(k) (z) − f (k) (z)| ≤
Mn k!R
(R − r)k+1
for |z − a| ≤ r.
Here Mn = sup[|fn (w) − f (w)| : |w − a| = R] and since fn → f ⇒
(k)
limn→∞ Mn = 0. Therefore it follows that fn converges uniformly to
f (k) on Br [a].
Now if K is any compact subset of Ω and 0 < r < ρ(K, ∂Ω) then we can
construct an open subcover of K. This means that there are a1 , ..., an in K
such that
n
[
K⊂
Br (aj )
j=1
(k)
As clearly fn converges uniformly to f on every open ball, the convergence
is uniform on an arbitrary compact subset K (Conway, 1973).
We state an important corollary without proof, (proof may be found in
(Conway, 1973):
P
Corollary 4.1.5. If the series
an (x) converges uniformly to f (x) on [a, b],
0
and the derivatives an (z) are continuous and converge uniformly on the same
interval then
∞
X
0
f (x) =
a0n (z)
n=0
i.e. the series may be differentiated term by term.
We can now prove Hurwitz’s Theorem, whose corollary is important in the
proof of Picard’s Great Theorem.
Theorem 4.1.6. (Hurwitz’s Theorem) Let Ω be an open connected subset
Ω ⊂ C and suppose {fn } in H(Ω) converges uniformly on compact sets to f .
If f 6≡ 0, BR [a] ⊂ Ω and f (z) 6= 0 for |z − a| = R then there is an integer N
such that for n ≥ N , f and fn have the same number of zeros in BR (a).
Proof. As f (z) 6= 0 for |z − a| < R there exists
δ = inf[|f (z)| : |z − a| = R] > 0
45
and as fn → f uniformly on {z : |z − a| = R} there exists an integer n1 such
that |fn (z)| > 12 δ for n ≥ n1 and |z − a| = R.
Now we have
fn (z) − f (z) 1
1
2
f (z) − fn (z) = fn (z)f (z) ≤ δ 2 |fn (z) − f (z)|.
Now as fn → f we have that f1n → f1 uniformly on {z : |z − a| = R}. Also
by Theorem 4.1.4 we have fn0 → f 0 which implies that
f0
fn0
→
fn
f
uniformly on the circle γ = {z : |z − a| = R} and by Lemma 4.1.2 we have
Z 0
Z 0
1
f (z)
1
fn (z)
dz = lim
dz
n→∞
2πi γ f (z)
2πi γ fn (z)
These integrals describe the number of zeros of f and fn respectively. Since
they are integers there must exist an N such that for all n ≥ N we have
Z 0
Z 0
f (z)
1
fn (z)
1
dz =
dz.
2πi γ f (z)
2πi γ fn (z)
(Conway, 1973)
Corollary 4.1.7. If {fn } ⊂ H(Ω) converges to f in H(Ω) and each fn never
vanishes on Ω then either f ≡ 0 or f never vanishes.
Proof. As each fn never vanishes on Ω then each of the fn have no zeros in
Ω. Hence if f 6≡ 0 then the number of zeros of f must be zero. Hence f
never vanishes (Conway, 1973).
In order to prove the next theorem, we make a small detour back to the
modular function. We proved that the modular function maps the upper
half plane to the twice punctured plane. We will now describe a conformal
mapping from the unit disk to the upper half plane, so upon composition we
can map from the unit disk to the twice punctured plane.
We consider the map
z→
z+1
i(z − 1)
46
we can check explicitly that this maps the unit disk to the upper half plane:
z+1
1
z+1
1 − |z|2
z+1
Im
=
=
−
>0
i(z − 1)
2i i(z − 1) −i(z − 1)
|z − 1|2
The last inequality follows from |z| < 1, as z is in the unit disc, and so the
numerator is greater than zero. If we call this mapping φ and denote the
unit disk by D then we see that λ ◦ φ : D → C \ {0, 1}.
In the next theorem, when we refer to the modular function being defined
(and analytic) on the unit disc, we are referring to the composition λ ◦ φ
above.
5
Montel-Caratheodory Theorem
Theorem 5.1.1. (Montel-Caratheodory Theorem) If F is the family of all
analytic functions on a region Ω that omit the values 0 and 1, then F is
normal in Ω.
Proof. Since normality is a local property we may assume that Ω is the unit
disk |z| < 1. Let us consider the modular function once again. We have
that the modular function is analytic on the unit disc (see the comment
before this theorem). For z = λ(τ ), τ is a multi-valued function of z. Fixing
z0 ∈ Ω we define ν(z) = λ−1 (z). We want to consider ν(f (z)) and thus for
any f ∈ F we choose the single valued analytic branch for which ν(f (z0 )) is
a principal value (i.e. ν(f (z0 )) is a value of our chosen branch).
From this starting point we can uniquely define ν(f (z)) in a suffciently small
neighbourhood of z0 . As each f (z) is analytic in Ω and only takes values
in the domain of ν(z), we can analytically continue ν(f (z)) into the disk Ω.
Therefore we obtain a single valued analytic function by the Monodromy
Theorem (1.3.6) which is bounded inside Ω; f˜(z) = ν(f (z)) with |f˜(z)| < 1
(see comment before this theorem) for z ∈ Ω, f ∈ F.
We want to be able to prove locally boundedness for all f ∈ F which
would imply normality by Montel’s Theorem (3.1.6). To do this let {fn }
be an arbitrary sequence of F, and let α be an accumulation point of the
sequence {fn (z0 )}. Due to the nature of the modular function, there are
three degenerate cases α ∈ {0, 1, ∞} and so we consider four cases which
cover all possible values of α.
Case 1: α 6∈ {0, 1, ∞}
47
Consider a subsequence {fn0 } ⊆ {fn } such that fn0 (z0 ) → α as n0 → ∞.
Then the sequence {f˜n0 } has itself a subsequence {f˜nk } which converges
uniformly on compact subsets of Ω to a function f˜∞ . Clearly |f˜∞ | ≤ 1, and
if equality is assumed for any z ∈ Ω then f˜∞ ≡ eiθ for some real θ. This
implies that |f˜nk (z)| → 1 as k → ∞. This is equal to saying |ν(fnk (z))| → 1.
Since the modular function maps the open unit disk to the plane minus
{0, 1}, as |τ | → 1, (i.e. as we approach the boundary of the unit disk) the
modular function approaches one of its degenerate values {0, 1, ∞}. Now
ν(z) = λ−1 (z) = τ and so |ν(fnk (z))| → 1 if and only if fnk (z) → 0, 1 or
∞. This is a contradiction as we stipulated α 6∈ {0, 1, ∞} and thus we have
|f˜∞ | < 1 in Ω.
Now consider an arbitrary compact subset K ⊆ Ω. Then for some constant
m, |f˜∞ | ≤ m < 1 on K and due to the uniform convergence of f˜n to f˜∞ on
k
K there exists m0 such that
|f˜nk | ≤ m0 < 1 , z ∈ K for k suffciently large.
Now as the modular function is analytic on the unit disc, it is bounded for
values |τ | ≤ m0 and we have |λ(τ )| ≤ M for some constant M . Therefore as
|ν(fnk (z))| = |f˜nk (z)| ≤ m0 we have
|fnk (z)| = |λ(ν(fnk (z)))| ≤ M
Therefore we have shown that {fnk } is locally bounded and by Montel’s
Theorem (3.1.6) F is a normal family.
Case 2: α = 1
Let {fn0 } ⊆ {f
such that fn0 (z0 ) → 1 as n0 → ∞. We depn } be a subsequence
fine gn0 (z) = fn0 (z), n0 = 1, 2, ..., to be a branch such that limn0 →∞ gn0 (z0 ) =
−1. Then all gn0 are analytic in Ω, omit the values 0 and 1, with limit point
at z0 equal to −1.
Hence Case 1 applies and there is a subsequence {gnk } ⊆ {gn0 } which
converges uniformly on compact sets of Ω to an analytic function. Therefore
so does {gn2 k } = {fnk }.
Case 3: α = 0
As in Case 1, let {fn0 } ⊆ {fn } be a subsequence such that fn0 (z0 ) → 0
as n0 → ∞. We define gn0 (z) = 1 − fn0 (z), n0 = 1, 2, ..., and as such the
functions gn0 are analytic on Ω, omit 0 and 1 and limn0 →∞ gn0 (z0 ) = 1
Therefore Case 2 applies and there is a subsequence {gnk } ⊆ {gn0 } which
48
converges uniformly on compact sets of Ω to an analytic function. And again
so does 1 − {gnk } = {fnk }.
Case 4: α = ∞
Let {fn0 } ⊆ {fn } be a subsequence such that fn0 (z0 ) → ∞ as n0 → ∞. Then
we define gn0 (z) = 1/fn0 (z), n0 = 1, 2, ..., which is analytic on Ω, omits 0 and
1 and limn0 →∞ gn0 (z0 ) = 0.
Therefore by Case 3 there is a subsequence {gnk } ⊆ {gn0 } which converges
uniformly on compact sets of Ω to the analytic function g, with g(z0 ) = 0.
Now as all gn0 never vanish, it follows from Corollary 4.1.6 that g ≡ 0.
Therefore we have fnk → ∞ uniformly on compact sets of Ω.
Since we have covered all values of α it follows that F is normal in Ω (Schiff,
1993).
6
Picard’s Great Theorem
Lemma 6.1.1. (Maximum Modulus Principle: Local) Let R > 0. If f is
holomorphic on all neighbourhoods of a point a, and
|f (z)| ≤ |f (a)| ∀z ∈ BR (a)
then f is constant.
Proof. Fix r, 0 < r < R and use the Cauchy Integral Formula for γ the circle
|z − a| < r.
Z
1
f (w)
f (a) =
dw
2πi γ w − a
Z 2π
f (a + reiθ )rieiθ
1
=
dθ
2πi 0
reiθ
Z 2π
1
=
f (a + reiθ ) dθ
2π 0
Now
1
|f (a)| ≤
2π
Z
2π
0
Z 2π
1
iθ |f (a)| dθ = |f (a)|
f (a + re ) dθ ≤
2π 0
Therefore we obtain that
Z 2π
|f (a)| − |f (a + reiθ )| dθ = 0.
0
49
Since f is holomorphic the integrand is continuous and it is ≥ 0 by hypothesis.
Therefore
|f (a)| − |f (a + reiθ )| = 0 , ∀θ ∈ [0, 2π] , r ∈ (0, R)
And |f | is constant.
Now we have f = u + iv and |f |2 = u2 + v 2 = c2 where c is a constant.
Differentiating |f |2 with respect to x and y gives 2uux + 2vvx = 0 and
2uuy + 2vvy = 0 respectively.
Now using the Cauchy-Riemann equations we can verify that
(u2 + v 2 )ux = 0 ⇒ c2 ux = 0
If c = 0, then f = 0 which is constant. If ux = 0 then by the same method
uy = 0 and u is constant. Similarly for v. Hence f = u + iv is constant
(Wang, 2013).
Theorem 6.1.2 (Maximum Modulus Theorem) If D is a bounded domain
and f is holomorphic on D and continuous on its closure D. Then |f | attains
its maximum on the boundary of D.
Proof. D is bounded and so its closure D is both closed, bounded and hence
compact. As |f | is continuous on the compact set D we have that |f | attains
its suprenum on D, at a point say a.
Assume a 6∈ ∂D. Then a is inside the open set D and hence is inside the
neighbourhood N (a, R) for some R > 0 such that N (a, R) ⊂ D. Therefore
|f | attains its maximum on N (a, R) at a. Hence by the preceding Lemma, f
is constant on N (a, R). And by the Identity Theorem, see (Wang, 2013), we
have that f ≡constant.
Now if f is not constant then we have the required contradiction and |f |
attains its maximum on the boundary of D. Also if f is constant then all
points, on the boundary or otherwise, are maxima (Wang, 2013).
We have now proved all that is needed to prove Picard’s Great Theorem.
50
Theorem 6.1.3 (Picard’s Great Theorem) Let f be an analytic function
with an essential singularity at z = z0 . Then in every deleted neighbourhood
of z0 , f attains every complex number, with one possible exception, an
infinite number of times.
Proof. We may assume that z0 = 0 and so the deleted neighbourhood is the
disk D := {z : 0 < |z| < R}. We suppose that the function omits two values
a and b and aim to show that f cannot have an essential singularity.
We define fn : D → C by
fn (z) = f
z
n
These functions are analytic and omit values a and b. Therefore according to
Theorem 5.1.1 the family {fn } is normal on D. Let {fnk } be a subsequence
that converges uniformly on the compact set |z| = R2 to a function F (z),
where F (z) is either analytic or ≡ ∞.
If F (z) is analytic let M = max{|F (z)| : |z| = 21 R}. Then
f z = |fn (z)| ≤ |fn (z) − F (z)| + |F (z)| ≤ 2M
k
k
nk for suffciently large nk and |z| = 21 R. This implies that
|f (z)| ≤ 2M
for sufficiently large nk and |z| =
1
2nk R.
Since we have that |f (z)| ≤ 2M for all z on concentric circles about the origin
it follows from the Maximum Modulus Theorem (6.1.2) that |f (z)| ≤ 2M
on concentric annuli about zero. This is due to the maximum of |f | being
attained on the boundary of the annuli, i.e. the circles themselves. Hence
we can choose the any two circles and consider the area between them and
the condition holds:
1
|f (z)| ≤ 2M , 0 < |z| ≤
R.
2n1
Therefore f (z) is bounded by 2M on a deleted neighbourhood of zero and
therefore z = 0 cannot be an essential singularity.
n
o
If F (z) ≡ ∞ then fn 1−a converges uniformly on compact subsets of D to
k
zero. Setting
1
h(z) =
f (z) − a
we conclude that, as above, h(z) is bounded in a deleted neighbourhood of
zero and thus f (z) again does not have an essential singularity.
51
Lastly, if there are two values say α and β which are assumed only finitely
often by f (z), then there exists a suffciently small deleted neighbourhood of
zero in which f (z) omits α and β. Hence the result follows (Conway, 1973;
Schiff, 1993).
52
Comparison of Alternative Proofs
There are many proofs of both Picard’s Little and Great Theorem. Our proof
of Picard’s Little Theorem we explained earlier was the original construction
of the proof by Picard himself. Since then many more proofs have been
found. There are two main ideas for proving Picard’s Little Theorem, the
first uses the monodromy theorem and the modular function as we have
found, and the second uses results of the behaviour of holomorphic functions
evaluated on the unit disk.
Considering the former, the mapping properties of the modular function
are the only important ingredient, along with the monodromy theorem, to
prove Picard’s Little Theorem. Hence some authors, such as Markushevich
(Markushevich, 1967), aim to develop the modular function by using the
symmetry principle to explicitly construct a conformal map of the unit disk to
whole plane. Choosing 3 points and connecting them with arcs perpendicular
to the circle he creates a triangular region which maps to the upper half
plane. By using the reflection principle, he creates a new set of smaller
triangles, (outside of the orginal triangle but inside the disk) which then
map to the lower half plane. By performing this reflection operation over
and over, and swapping between mapping to the upper or lower half plane,
he proves that you can then exhaust the unit disk (Markushevich, 1967).
Hence there exists a mapping from the unit disk to the whole plane. This
mapping omits 3 points at which you may take to be at 0, 1 and ∞. He then
defines the multi-valued inverse function, take a respective branch and prove
Picard’s Little Theorem (Markushevich, 1967).
Whilst this proof is shorter, as much of the theory of elliptic functions is not
needed to get to the modular function, I believe Picard’s original proof is
more, as Lars Ahlfors (Ahlfors, 1966) states, “penetrating”. Picard’s proof
allows one to develop a concrete understanding of why the theorem holds, as
the result does not appear to come from some magical construction.
The other proof using the behaviour of holomorphic functions, is much more
interesting. Since it would be relatively hard for me to explain in words this
proof, I have given a short review of the results below.
7
Bloch’s Little Picard Proof
We now state (and prove in parts) an alternative proof for Picard’s Little
Theorem. This proof rests upon the amazing theorem of Bloch:
53
Theorem 7.1.1. (Bloch). If f is holomorphic in a neighbourhood of the
1
closed unit disk D and f 0 (0) = 1 then f (D) contains a disk of radius 72
(Conway, 1973).
We will not prove this here, (proof found in (Conway, 1973)) however we
will prove an important corollary:
Corollary 7.1.2. Let f be a holomorphic function on a open simply connected domain containing the closed ball of radius R, BR [0], then f (BR (0))
1
contains a disk of radius 72
R|f 0 (0)|.
Proof. If f 0 (0) = 0 then the result holds and so assume f 0 (0) 6= 0. Then
considering the function
f (Rz)
g(z) =
Rf 0 (0)
g(z) is holomorphic in the neighbourhood of the closed unit disk as f is, and
g 0 (0) =
Rf 0 (Rz)
=1
Rf 0 (0)
We can now apply Bloch’s theorem and so g(D) contains a disk of radius
1
1
0
72 . Hence f (RD) = f (BR (0)) contains a disk of radius 72 R|f (0)| (Conway,
1973).
We now state (without proof, proof found in (Conway, 1973)) the second
important ingredient in this proof:
Theorem 7.1.3. (Landau). Let G ⊂ C be a simply connected domain and
let f : G → C be holomorphic such that f does not assume the values 0 or 1
on G. Then there is a holomorphic function g : G → C such that
f (z) = − exp(iπ cosh[2g(z)])
Moreover g(G) does not contain a disk of radius 1, see (Conway, 1973).
We may now prove Picard’s Little Theorem, see (Conway, 1973).
Theorem 7.1.4. (Picard’s Little Theorem). If an entire function f : C → C
omits two values, then it is constant.
Proof. We may again assume f omits the values 0 and 1. From the previous
theorem this gives an entire function g(z) such that g(C) does not contain a
disk of radius 1. If f is not constant then neither is g and as such there is
54
a point z0 such that g 0 (z0 ) 6= 0. Without loss of generality, by considering
g(z + z0 ), we may assume that g 0 (0) 6= 0. By the corollary to Bloch’s theorem
1
we see that g(BR (0)) contains a disk of radius 72
R|f 0 (0)|. By choosing R
arbitrarily large we have that g(C) does contain a disk of radius 1. This is a
contradiction and thus f must be constant (Conway, 1973).
This is the proof many authors refer to when they describe an elementary
proof of Picard’s Little Theorem. The omitted proofs above are not difficult
to formulate and Bloch’s Theorem only depends on Schwarz’s Lemma and
a few intermediary lemmas. Hence this proof is both short and elegant.
Therefore it is a signifcant improvement on the length of the method of
Picard, however again I feel that it does not afford the reader as much
intuition.
With regards to Picard’s Great Theorem, the most well known proof involves
the use of Montel’s theorem, along with Schottky’s theorem (see Theorem
8.1.3. below), to prove the Montel-Caratheodory theorem and then onto
Picard’s Great Theorem. In this dissertation since I had already defined
the modular function, and more notably the inverse, I could prove MontelCaratheodory theorem without the need for Schottky’s theorem (see Theorem
5.1.1).
There is another standard proof of Picard’s Great Theorem which can be
seen in Narasimhan’s complex analysis text (Narasimhan, 1985). This uses
Schottky’s theorem along with the Casorati-Weierstrass theorem, a result
very closely linked to Picard’s Great Theorem. Again we give a short review
of the results.
8
Alternative Great Picard Proof
Theorem 8.1.2. (Riemann Extension Theorem). Let a ∈ C, r > 0,
D = {z ∈ C : |z − a| < r}. Let f be a function holomorphic on the set
D \ {a}. Then the following are equivalent:
(1) f is holomorphically extendable over a;
(2) f is continuously extendable over a;
(3) There exists a neighbourhood of a on which f is bounded;
(4) limz→a (z − a)f (z) = 0.
We will not give the proof of this theorem. The proof of this is found in
55
(Narasimhan, 1985). Note that the last condition (4) is sufficient that the
point at a is a removable singularity.
We now prove the Casorati-Weierstrass Theorem- a close neighbour to
Picard’s Great Theorem.
Theorem 8.1.3. (Casorati-Weierstrass Theorem). Let a ∈ C, r > 0,
D∗ = {z ∈ C : 0 < |z − a| < r}, and D be as above. Let f be holomorphic
on any open subset U ⊃ D of the complex plane, and suppose that a is an
essential singularity of f . Then f (D∗ ) is dense in C.
Proof. We assume that f (D∗ ) is not dense, i.e. that there exists some value
b ∈ C that the function can never get close to. Therefore ∃ > 0 such that
|f (z) − b| ≥ for all z ∈ D∗ . We now define the function g(z):
g(z) =
1
f (z) − b
This is holomorphic on D∗ and bounded by −1 . Therefore by the Riemann
extension theorem we can holomorphically extend g to the region D, where
this is the region D∗ including the point a. As such we can rearrange for
f (z):
1
f (z) =
+b
g(z)
Now since g(z) was holomorphic on D we can examine limz→z0 g(z). If the
limit is equal to zero then f has a pole at z0 and if the limit is not equal to
zero then f has a removable singularity at z0 . In both these cases we obtain
a contradiction and so f (D∗ ) must be dense in C (Narasimhan, 1985).
We now state Schottky’s Theorem without proof, proof found in (Conway,
1973):
Theorem 8.1.4. Let R > 0 and C > 0. Let F be the family of functions such
that f ∈ H(DR ) and omit the values 0 and 1, and FC = {f ∈ F : |f (0)| ≤ C}.
Then, for 0 < r < R, there exists a constant M which depends only on r, R
and C such that
|f (z)| ≤ M for |z| < r and all f ∈ FC
And thus we can prove Picard’s Great Theorem:
Theorem 8.1.5. (Picard’s Great Theorem). Let f be an analytic function
with an essential singularity at z = a. Then in every deleted neighbourhood
56
of a, f attains every complex number, with one possible exception, an infinite
number of times.
Proof. Let D∗ be the same as in Theorem 8.1.2. We assume, by way of
contradiction, that f (D∗ ) omits two values α and β. Then 0, 1 6∈ g(D∗ )
∗
where g(z) = f (z)−α
β−α . Then g ∈ H(D ) and has an essential singularity at a
if f does.
Thus by the Casorati-Weierstrass theorem there exists a sequence {zn } of
complex numbers such that zn → 0, |zn+1 | < |zn | < Re−2π for n ≥ 1 such
that |g(a + zn )| < 1. This is because Casorati-Weierstrass says that g can
get arbitrarily close to any complex number in any neighbourhood of an
essential singularity. Here we stipulate that the complex number in question
is 0 and the distance we want to achieve is 1.
We then define the family Fn ∈ H(D), D = {w ∈ C : |w| < 1} by Fn (w) =
g(a+zn e2πiw ). Hence Fn maps D into C\{0, 1} and |Fn (0)| = |g(a+zn )| < 1.
Applying Schottky’s theorem, there is a constant M > 0 such that |Fn (w)| ≤
M for |w| ≤ 1/2. This implies |Fn (w)| ≤ M for w ∈ R, −1/2 ≤ w ≤ 1/2.
Hence we have:
|g(a + zn eiθ )| ≤ M for − π ≤ θ ≤ π
Now, as zn eiθ applies a direction within the unit circle to zn , we have that
|g(a + z)| ≤ M for all |z| = |zn |. By the maximum principle we proved earlier
(in Theorem 6.1.2) the condition holds between any two circles z = |zk | and
therefore |g(a + z)| ≤ M for |zn+1 | ≤ |z| ≤ |zn |. As this holds for all n ≥ 1
and |zn | → 0 as n → ∞, we obtain |g(a + z)| ≤ M for 0 ≤ |z| ≤ z1 . This in
turn gives:
|g(z)| ≤ M for 0 ≤ |z − a| ≤ |z1 |
As such this implies that a is a removable singularity which is a contradiction.
Hence f must assume at least one of α or β. As before if they are assumed
only finitely many we can choose a neighbourhood D∗ such that they are
both omitted and the result follows (Narasimhan, 1985).
This different proof circumvents the problem of having to examine normal
families and only focuses on the strengthening of Casorati-Weierstrass. Since
that significantly shortens the proof it is often described as more elementary
as it also allows someone not well versed in complex analysis to easily
understand it. However to fully explain this method, we would need more
definitions and a proof of Schottky’s Theorem (8.1.4) (Conway, 1973) which
is not especially trivial.
57
Conclusion.
In this report, I have provided a proof of Picard’s Little and Great Theorem.
Working closely with several texts, I developed the concept of analytic
continuation and used that to prove the Monodromy Theorem. Starting
with periodic functions, I constructed the Weierstrass ℘ function in order
to explicitly define the modular function. Then, after proving that the
modular function maps a specific subset of the upper half plane conformally
to the twice punctured plane, I proved, using the monodromy theorem,
Picard’s Little Theorem. In the second section, I introduced the concept of
normality in families of functions and proved various results linking normality
to whether the function is locally bounded and equicontinuous. Then using
these results I proved the Montel-Caratheodory Theorem, which was then
used to prove Picard’s Great Theorem. In the last section, I aimed to show
different ways one can approach the proof of the Picard Theorems, with these
methods often being vastly shorter and more elementary. The majority of
the results in this project are self-contained, with clear and coherent proofs.
There are however some theorems detailed in this report which I do not
prove, however citations explain which source one can find these results in.
On conclusion, I have thoroughly enjoyed creating this piece of work, and I
have increased my comprehension of many areas of analysis greatly.
58
References
Ahlfors, L.V. (1966). Complex Analysis: An Introduction to the Theory
of Analytic Functions of One Complex Variable. 2rd Edition. New York:
McGraw-Hill.
Brubaker, B. (2008). Handout in Elliptic Functions 18.784: The Mathematical
Legacy of Ramanujan. Massachusetts Institute of Technology. Available
at:
http://math.mit.edu/ brubaker/Math784/elliptic.pdf
Conway, J.B. (1973). Functions of One Complex Variable. New York:
Springer-Verlag.
Copson, E.T. (1935). Theory of Functions of a Complex Variable. Oxford:
Oxford Univeristy Press.
Gamelin, T.W. (2001). Complex Analysis. New York: Springer-Verlag.
Markushevich, A.I. (1967). Theory of Functions of a Complex Variable:
Volume 3. New Jersey: Prentice-Hall.
Narasimhan, R. (1985). Complex Analysis in One Variable. Stuttgart:
Birkhäuser.
Price, M. (2001). Elliptic Functions. Ph. D. Warwick University. Available
at: http://www.cs.bath.ac.uk/MPrice/downloads/Elliptic.pdf
Schiff, J.L. (1993). Normal Families. New York: Springer-Verlag.
Wang, Y. (2013). Complex Analysis: Review. Ph. D. Purdue University.
Available at:
https://www.math.purdue.edu/ wang838/notes/complex.pdf
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