BLOCH’S THEOREM
0
Lemma 0.1. Let f be analytic in ∆ = {|z| < 1} with f (0) = 0 and
√ f (0) =
1.
(z)| ≤ M for all z ∈ ∆, then f (∆) contains the disk |w| ≤ ( M + 1−
√ If |f
2
M) .
Proof.
P∞ BynSchwartz’s lemma, we implicitly have M ≥ 1. Let f (z) = z +
n=2 an z . Using CIF, we have |an | ≤ M for all n. Therefore,
|f (z)| ≥ |z| −
∞
X
|an ||z|n
n=2
(0.1)
M r2
=r−
1−r
for |z| = r < 1. Obviously, we can maximize the RHS of (0.1) by taking
r
M
(0.2)
r =ρ=1−
M +1
√
√
and correspondingly, |f (z)| ≥ ( M + 1 − M )2 for |z| = ρ.
For all |w| < ρ, |f (z)−(f (z)−w)| = |w| ≤ |f (z)| for all |z| = ρ. Therefore,
f (z) − w and f (z) have the same
of zeros in |z| < ρ. It follows that
√ number √
f (∆) contains the disk |w| ≤ ( M + 1 − M )2 .
Obviously, by “scaling”, we have the following:
Lemma 0.2. Let f be an analytic function on D = {|z − a| < R}. If
|fp
(z) − f (a)| ≤ M √
for all z ∈ D, then f (D) contains the disk |w − f (a)| ≤
( M + |f 0 (a)R| − M )2 .
Lemma 0.3. An analytic function f (z) on ∆ is 1-1 if |f 0 (z) − M | < |M |
for all z ∈ ∆ and a constant M ∈ C.
Proof. Let z1 and z2 be two distinct points in ∆ and let γ be the line joining
z1 and z2 . Then
Z
0
|f (z1 ) − f (z2 )| = f (z)dz γ
Z
Z
0
≥ M dz − (f (z) − M )dz (0.3)
γ
γ
Z
Z
≥ |M | |dz| − |f 0 (z) − M ||dz| > 0
γ
γ
and hence f (z) is 1-1.
1
2
BLOCH’S THEOREM
Theorem 0.4 (Bloch’s Theorem). Let f (z) be an analytic function on ∆
satisfying f 0 (0) = 1. Then there is a positive constant B (called Bloch’s
constant), independent of f , such that there exists a disk S ⊂ ∆ where f
is 1-1 and whose image f (S) contains a disk of radius B. In particular,
B > 1/72.
Proof. Obviously, it is enough to show this for f 0 (z) bounded on ∆.
Let
m(r, g) = max |g(z)|.
(0.4)
|z|=r
We let 0 ≤ r0 < 1 be the largest number such that (1 − r0 )m(r0 , f 0 ) = 1.
Such r0 exists since f 0 (z) is bounded on ∆.
Then (1 − r)m(r, f 0 ) < 1 for r > r0 and hence
|f 0 (z)| ≤
(0.5)
1
1 − |z|
for |z| ≥ r0 . And by principle of maximum modulus, we have
|f 0 (z)| ≤ m(r0 , f 0 ) =
(0.6)
1
1 − r0
for |z| ≤ r0 . In conclusion,
|f 0 (z)| ≤
(0.7)
1
1 − max(r0 , |z|)
for all z ∈ ∆.
Let a ∈ ∆ be a number such that |a| = r0 and |f 0 (a)| = 1/(1 − r0 ).
For 0 < ρ < 1 − r0 and |z − a| ≤ ρ, we have
|f 0 (z) − f 0 (a)| ≤
(0.8)
1
1
+
1 − r0 1 − r0 − ρ
and hence
(0.9)
|z − a|
|f (z) − f (a)| ≤
ρ
0
0
1
1
+
1 − r0 1 − r0 − ρ
by Schwartz’s lemma. Therefore, |f 0 (z) − f 0 (a)| < |f 0 (a)| for z in the disk
ρ(1 − r0 − ρ)
(0.10)
S = |z − a| <
.
2(1 − r0 ) − ρ
By Lemma 0.3, f√is 1-1 on S. Obviously, the radius of S is maximized when
we set ρ = (2 − 2)(1 − r0 ) and correspondingly,
n
o
√
(0.11)
S = |z − a| < (3 − 2 2)(1 − r0 ) .
Moreover, since
√
(0.12)
|f (z) − f (a)| ≤ ln
2+1
2
!
BLOCH’S THEOREM
3
for z ∈ S by (0.7), we conclude that f (S) contains a disk of radius
v
v
!
!2
u
u
√
√
u
u
√
1
2+1
2+1 
tln
> .
(0.13)
+ (3 − 2 2) − tln
2
2
72
Remark 0.5. The key to the proof of Bloch’s theorem is the existence of
a ∈ ∆ and positive constants C1 and C2 such that |f 0 (z)| ≤ C2 |f 0 (a)| for all
|z − a| ≤ C1 /|f 0 (a)|.