Math 207 - Spring ’17 - François Monard
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Lecture 11 - Montel’s theorem, Riemann mapping theorem
Normal families and Montel’s theorem
Compact subsets of Rn are those that automatically generate limit points out of any sequence,
making them perfect generators of existence theorems. In this section, we generalize this concept
to subsets of infinite-dimensional spaces, namely,
• Spaces of analytic functions defined on a compact subset K of C, equipped with the topology
of uniform convergence.
• Spaces of analytic functions defined on an open subset Ω of C, equipped with the topology
of normal convergence (i.e., uniform convergence on every compact subset of Ω).
Without any complex analytic consideration, the famous Arzelá-Ascoli theorem states a sufficient condition for a family of functions to be “compact”. Here and below, we say that a family of
functions F = {f : K → C, f ∈ F} is uniformly equicontinuous on K if
∀ε > 0,
∃δ > 0,
|w − z| < δ
=⇒
|f (z) − f (w)| < ε,
∀f ∈ F.
A general theorem, which holds in a much more general compact space without requiring complex analytic tools, is the following:
Theorem 1 (Arzelá-Ascoli). If F is uniformly equicontinuous on K, then every sequence fn in F
has a subsequence which converges uniformly in K.
Let Ω ⊂ C open, fixed. A family F = {f : Ω → C analytic, f ∈ F} is called normal if for every
sequence fn ∈ F, fn has a subsequence converging normally on Ω. The theorem below says that
it takes very little for a sequence of analytic functions to be normal: the functions only have to be
uniformly bounded on any compact K ⊂⊂ Ω by the same constant.
Theorem 2 (Montel’s theorem). If F is a family of analytic functions defined on an open set
Ω ⊂ C, uniformly bounded on every compact subset of Ω, then F is normal.
Example 1. For any open set Ω ⊂ C, a sequence {fn }n satisfying fn : Ω → D for every n ∈ N is
uniformly bounded by 1, so it has a normally convergent subsequence.
For the proof, see also [SS, Ch. 8 §3.3].
Proof. We split the proof into three steps:
(i) For every compact K ⊂⊂ Ω, the family F is uniformly equicontinuous.
(ii) For every compact K ⊂⊂ Ω and every sequence fn ∈ F, fn has a uniformly convergent
subsequence on K.
(iii) For every sequence fn ∈ F, fn has a normally convergent subsequence in Ω.
Math 207 - Spring ’17 - François Monard
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S
Proof of (i). Let r > 0 such that for every z ∈ K, D3r (z) ⊂ Ω and let Kr = z∈K D2r (z). Fix
w ∈ K and z ∈ Dr (w). Setting γ = C2r (w), we have for every ζ ∈ γ, |ζ − w| = 2r and |ζ − z| ≥ r.
In particular,
1
1 |z − w|
|z − w|
ζ − z − ζ − w = |ζ − z||ζ − w| ≤ 2r2 .
Then using Cauchy’s integral formula, we deduce that for any f ∈ F
Z
2π(2r) |z − w|
1
1
1 ≤
f
(ζ)
−
dζ
sup |f (z)| ≤ C|z − w|,
|f (z) − f (w)| =
2π γ
ζ −z ζ −w
2π
2r2 f ∈F ,z∈Kr
where C is uniform in f ∈ F and z ∈ K. Therefore, F is uniformly Lipschitz on K thus uniformly
equicontinuous on K.
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Proof of (ii). Let {wj }∞
j=1 a dense sequence in K. Let fn a sequence in F. The sequence
fn (w1 ) is bounded in C and as such, has a convergent subsequence. In other words, there exists
a subsequence fn,1 of fn such that fn,1 (w1 ) converges. Next, there exists a subsequence fn,2 of
fn,1 such that fn,2 (w2 ) converges. Then fn,2 (w1 ) converges as well, as a subsequence of fn,1 (w1 ).
Continuing by induction for any j, we extract fn,j a subsequence of fn,j−1 such that fn,j (wk )
converges for every 1 ≤ k ≤ j. We then define gn = fn,n the diagonal sequence. gn is a subsequence
of fn,j for every j and thus gn (wk ) converges for every integer k. We now show that gn is uniformly
Cauchy on K, which will prove that it converges uniformly on K.
Let ε > 0. By (i), since gn is equicontinuous, there exists δ > 0 such that
|z − w| < δ
=⇒
|gn (z) − gn (w)| < ε,
∀n ∈ N.
Now since K is compact, it is covered by finitely many Dδ (wj ) for 1 ≤ j ≤ J. And the J
Cauchy sequences {gn (wj )}n clearly satisfy jointly that there exists M such that n, m ≥ M implies
|gn (wj ) − gm (wj )| < ε for every 1 ≤ j ≤ J. Now for any z ∈ K, there exists 1 ≤ j ≤ J such that
|z − wj | < δ. Then we have, for n, m ≥ M
|gn (z) − gm (z)| ≤ |gn (z) − gn (wj )| + |gn (wj ) − gm (wj )| + |gm (wj ) − gm (z)| ≤ 3ε,
where the first and third term are less than ε due to equicontinuity, and the middle one, by Cauchyness of {gn (wj )}n . Claim (ii) is proved.
Proof of (iii). By means of a compact exhaustion of Ω, this is another diagonalization
argument. For ` ∈ N, define K` = {z ∈ C, dist(z, ∂Ω) ≥ 1/`} ∩S
D` (0). The sequence K` is such
that K` is included in the interior of K`+1 for every `, and Ω = `∈N K` . In particular, for every
compact K ⊂⊂ Ω, there exists some j ∈ N such that K ⊂ Kj .
Now let fn a sequence in F. By (ii), there exists a subsequence fn,1 of fn converging uniformly
on K1 . There exists a subsequence fn,2 of fn,1 converging uniformly on K2 . Repeating for every
j ∈ N, there exists fn,j a subsequence of fn,j−1 converging uniformly on Kj . Extracting the diagonal
sequence gn := fn,n , such a sequence converges uniformly on every Kj , and since every compact
subset of Ω is included in some Kj , the sequence gn converges normally on Ω.
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one make take for instance an enumeration of the countable set K ∩ (Q + iQ)
Math 207 - Spring ’17 - François Monard
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The Riemann mapping theorem
We now have all the tools required to prove the Riemann mapping theorem, stating that if Ω ( C
is simply connected, then there exists a conformal equivalence between Ω and the unit disc D.
The proof requires a few steps, which we now break down. Let Ω ( C an open, simply connected
domain, and let z0 ∈ Ω fixed. Denote by F the family of functions
F = {f : Ω → D analytic, one-to-one,
f (z0 ) = 0}.
We will show the following:
Lemma 3. The set F is non-empty.
Proof. Since Ω 6= C, let λ ∈
/ Ω fixed. Then the function f (z) = z − λ is analytic, one-to-one and
non-vanishing on Ω, therefore it admits an analytic squareroot g on Ω. g is clearly one-to-one (if it
wasn’t, then neither would be f ).
We claim g(Ω) is open. Indeed, since g is one-to-one, then it is bijective onto g(Ω), and the
relation g(z)2 = z − λ implies that the inverse of g admits the expression g −1 (w) = w2 + λ for
w ∈ g(Ω), which is clearly analytic, hence continuous. Therefore, since Ω is open and g −1 is
continuous, g(Ω) = (g −1 )−1 (Ω) is open.
Since g(Ω) is open, there exists Dr (w0 ) ⊂ g(Ω) for some r > 0, but then since g 2 = f and f
is one-to-one, we necessarily have Dr (−w0 ) ∩ g(Ω) = ∅, otherwise if two opposite points are in the
range of g, then their images by f are the same, contradicting injectivity. This implies the relation
|g(z) + w0 | > r,
therefore the function p(z) =
r
g(z)+z0
∀z ∈ Ω,
is analytic, one-to-one on Ω such that |p(z)| < 1, so p : Ω → D.
If p(z0 ) = 0, then p ∈ F, and if p(z0 ) 6= 0, then hp(z0 ) ◦ p ∈ F, where hp(z0 ) =
z−p(z0 )
1−p(z0 )z
is a conformal
automorphism of the unit disc taking p(z0 ) to 0.
In the next lemma, we will use the hyperbolic translations hw ∈ Aut D defined by hw (z) =
in particular the following
Lemma 4. h0w (0) = 1 − |w2 | and h0w (w) =
z−w
1−wz ,
1
.
1−|w|2
Proof. Straightforward computation.
Lemma 5. If f ∈ F and f is not onto, then there exists g ∈ F satisfying |g 0 (z0 )| > |f 0 (z0 )|.
Proof. Let f ∈ F not onto and let w ∈
/ f (Ω), then hw ◦ f is a non-vanishing analytic function on
Ω, therefore it admits an analytic squareroot q such that q 2 = hw ◦ f . q does not vanish either so
q(z0 ) = λ 6= 0 and λ2 = (q(z0 ))2 = hw (f (z0 )) = hw (0) = −w. Let us set g = hλ ◦ q and show that
g fulfills the lemma. We compute
g 0 (z0 ) = h0λ (q(z0 ))q 0 (z0 ) = h0λ (λ)q 0 (z0 ) =
1
q 0 (z0 ).
1 − |λ|2
Differentiating the relation q 2 = hw ◦ f at z0 , we arrive at
2q(z0 )q 0 (z0 ) = h0w (f (z0 ))f 0 (z0 ) = h0w (0)f 0 (z0 ) = (1 − |w|2 )f 0 (z0 ) = (1 − |λ|4 )f 0 (z0 ).
Math 207 - Spring ’17 - François Monard
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Concluding,
g 0 (z0 ) =
1 − |λ|4 0
1
1 + |λ|2 0
f
(z
)
=
f (z0 ).
0
1 − |λ|2 2λ
2λ
The proof is complete by checking that whenever |λ| =
6 1,
1+|λ|2
2|λ|
> 1.
Theorem 6 (Riemann mapping theorem). There exists an element h ∈ F that is onto. In particular, h is a conformal equivalence between Ω and D.
Proof. The family F is uniformly bounded, so by Montel’s theorem, it is normal. Let us denote
m = sup |f 0 (z0 )|.
f ∈F
Since conformal mappings have non-vanishing derivatives so m > 0, and m is finite by using
Cauchy estimate (there exists r > 0 such that Dr (z0 ) ⊂ Ω, then for every f ∈ F, |f 0 (z0 )| ≤
(z)|
max|z−z0 |=r |f R
≤ R1 ). From the characterization of the supremum, there exists a sequence fn ∈ F
such that limn→∞ |fn0 (z0 )| = m. Using Montel’s theorem, since the sequence fn is uniformly
bounded by 1 on Ω, fn admits a subsequence {fnk }k which converges normally to a function
h : Ω → C.
We claim that h ∈ F. Clearly, h(z0 ) = limk→∞ fnk (z0 ) = limk→∞ 0 = 0. h is analytic as a
uniform limit of analytic functions. |h0 (z0 )| = limk→∞ |fn0 k (z0 )| = m. Since m 6= 0, h is not a
constant. Then by Hurwitz’s theorem, since all fnk ’s are one-to-one, then h must be one-to-one.
We claim that h is onto. By contrapositive of the previous lemma, we have |h0 (z0 )| ≥ |g 0 (z0 )|
for every g ∈ F, so h is onto.
References
[SS] Complex Analysis, Elias M. Stein and Rami Shakarchi, Princeton Lectures in Analysis II. 1
[T] Complex Variables, Joseph L. Taylor. Pure and Applied Undergraduate Texts Vol. 16, 2011.