55.2.4 Great And Glorious Theorem, Simply Connected Regions
Here is given a laundry list of properties which are equivalent to an open set being simply
connected. Recall Definition 50.7.21 on Page 5605 which said that an open set, Ω is
simply connected means
^ℂ
∖ Ω is connected. Recall also that this is not the same thing at
all as saying ℂ ∖ Ω is connected. Consider the outside of a disk for example. I will
continue to use this definition for simply connected because it is the most convenient one
for complex analysis. However, there are many other equivalent conditions.
First here is an interesting lemma which is interesting for its own sake. Recall
n
(p,γ)
means the winding number of γ about p. Now recall Theorem 50.7.25
implies the following lemma in which B^{C} is playing the role of Ω in Theorem
50.7.25.
Lemma 55.2.4Let K be a compact subset of B^{C}, the complement of a closed set.Then there exist continuous, closed, bounded variation oriented curves
{Γ j}
_{j=1}^{m}forwhich Γ_{j}^{∗}∩ K = ∅ for each j, Γ_{j}^{∗}⊆ Ω, and for all p ∈ K,
∑m
n (Γ k,p) = 1.
k=1
while for all z ∈ B
∑m
n (Γ k,z) = 0.
k=1
Definition 55.2.5Let γ be a closed curve in an open set, Ω,γ :
[a,b]
→ Ω.Then γ is said to be homotopicto a point, p in Ω if there exists a continuousfunction, H :
[0,1]
×
[a,b]
→ Ω such that H
(0,t)
= p,H
(α,a)
= H
(α,b)
, andH
(1,t)
= γ
(t)
. This function, H is called a homotopy.
Lemma 55.2.6Suppose γ is a closed continuous bounded variation curve in anopen set, Ω which is homotopic to a point. Then if a
∕∈
Ω, it follows n
(a,γ)
= 0.
Proof: Let H be the homotopy described above. The problem with this is that it is
not known that H
(α,⋅)
is of bounded variation. There is no reason it should be.
Therefore, it might not make sense to take the integral which defines the winding
number. There are various ways around this. Extend H as follows. H
(α,t)
= H
(α,a)
for
t < a,H
(α,t)
= H
(α,b)
for t > b. Let ε > 0.
1-∫ t+(2b−εa)(t−a)
H ε(α, t) ≡ 2ε −2ε+t+-2ε-(t−a)H (α,s)ds,H ε(0,t) = p.
(b−a)
Thus H_{ε}
(α,⋅)
is a closed curve which has bounded variation and when α = 1, this
converges to γ uniformly on
[a,b]
. Therefore, for ε small enough, n
(a,Hε(1,⋅))
= n
(a,γ)
because they are both integers and as ε → 0,n
(a,Hε(1,⋅))
→ n
(a,γ)
. Also,
H_{ε}
(α,t)
→ H
(α,t)
uniformly on
[0,1]
×
[a,b]
because of uniform continuity of H.
Therefore, for small enough ε, you can also assume H_{ε}
(α,t)
∈ Ω for all α,t. Now
α → n
(a,H ε(α,⋅))
is continuous. Hence it must be constant because the winding number
is integer valued. But
∫
lim -1- --1--dz = 0
α→0 2πi H ε(α,⋅)z − a
because the length of H_{ε}
(α,⋅)
converges to 0 and the integrand is bounded because
a
∕∈
Ω. Therefore, the constant can only equal 0. This proves the lemma.
Now it is time for the great and glorious theorem on simply connected regions. The
following equivalence of properties is taken from Rudin [?]. There is a slightly different
list in Conway [?] and a shorter list in Ash [?].
Theorem 55.2.7The following are equivalent for an open set, Ω.
Ω is homeomorphic to the unit disk, B
(0,1)
.
Every closed curve contained in Ω is homotopic to a point in Ω.
If z
∕∈
Ω, and if γ is a closed bounded variation continuous curve in Ω, thenn
(γ,z)
= 0.
Ω is simply connected, (
^ℂ
∖ Ω is connected and Ω is connected. )
Every function analytic on Ω can be uniformly approximated by polynomialson compact subsets.
For every f analytic on Ω and every closed continuous bounded variation curve,γ,
∫
γf (z)dz = 0.
Every function analytic on Ω has a primitive on Ω.
If f,1∕f are both analytic on Ω, then there exists an analytic, g on Ω such thatf = exp
(g)
.
If f,1∕f are both analytic on Ω, then there exists ϕ analytic on Ω such thatf = ϕ^{2}.
Proof: 1⇒2. Assume 1 and let γ be a closed curve in Ω. Let h be the homeomorphism,
h : B
∖ Ω is not connected, there exist disjoint
nonempty sets, A and B such that A∩B = A∩B = ∅. It follows each of these sets must
be closed because neither can have a limit point in Ω nor in the other. Also, one and only
one of them contains ∞. Let this set be B. Thus A is a closed set which must also be
bounded. Otherwise, there would exist a sequence of points in A,
{an}
such that
lim_{n→∞}a_{n} = ∞ which would contradict the requirement that no limit points of A
can be in B. Therefore, A is a compact set contained in the open set, B^{C}≡
{z ∈ ℂ : z ∕∈ B }
. Pick p ∈ A. By Lemma 55.2.4 there exist continuous bounded variation
closed curves
{Γ k}
_{k=1}^{m} which are contained in B^{C}, do not intersect A and such
that
∑m
1 = n(p,Γ k)
k=1
However, if these curves do not intersect A and they also do not intersect B then they
must be all contained in Ω. Since p
5⇒6 Every polynomial has a primitive and so the integral over any closed bounded
variation curve of a polynomial equals 0. Let f be analytic on Ω. Then let
{fn}
be a
sequence of polynomials converging uniformly to f on γ^{∗}. Then
be a bounded variation continuous curve joining z_{0}
to z in Ω, you define a primitive for f as follows.
∫
F (z) = γ(z0,z)f (w )dw.
This is well defined by 6 and is easily seen to be a primitive. You just write the difference
quotient and take a limit using 6.
F (z + w )− F (z) 1 (∫ ∫ )
liwm→0--------------- = lwim→0 -- f (u)du− f (u)du
w w ∫ γ(z0,z+w ) γ(z0,z)
= lim 1- f (u)du
w→0 w γ(z,z+w)
1 ∫ 1
= lim -- f (z + tw)wdt = f (z).
w→0 w 0
7⇒8 Suppose then that f,1∕f are both analytic. Then f^{′}∕f is analytic and so it has a
primitive by 7. Let this primitive be g_{1}. Then
(e−g1f)′ = e−g1 (− g′)f + e−g1f′
( 1′)
= − e−g1 f- f + e−g1f ′ = 0.
f
Therefore, since Ω is connected, it follows e^{−g1}f must equal a constant. (Why?) Let the
constant be e^{a+ibi}. Then f
(z)
= e^{g1(z)
}e^{a+ib}. Therefore, you let g
(z)
= g_{1}
(z)
+ a + ib.
8⇒9 Suppose then that f,1∕f are both analytic on Ω. Then by 8f
(z)
= e^{g(z)
}. Let
ϕ
(z)
≡ e^{g(z)
∕2}.
9⇒1 There are two cases. First suppose Ω = ℂ. This satisfies condition 9 because if
f,1∕f are both analytic, then the same argument involved in 8⇒9 gives the
existence of a square root. A homeomorphism is h
(z)
≡
√-z--2
1+|z|
. It obviously maps
onto B
(0,1)
and is continuous. To see it is 1 - 1 consider the case of z_{1} and
z_{2} having different arguments. Then h
(z1)
≠h
(z2)
. If z_{2} = tz_{1} for a positive
t≠1, then it is also clear h
(z1)
≠h
(z2)
. To show h^{−1} is continuous, note that if
you have an open set in ℂ and a point in this open set, you can get a small
open set containing this point by allowing the modulus and the argument to lie
in some open interval. Reasoning this way, you can verify h maps open sets
to open sets. In the case where Ω≠ℂ, there exists a one to one analytic map
which maps Ω onto B
(0,1)
by the Riemann mapping theorem. This proves the
theorem.