The Little Picard Theorem and Its Proof
November 22, 2011
Little Picard Theorem. If f is entire and its image omits 2 values, then it
is constant.
Lemma 1. Let U be a holomorphically simply connected domain. Let f be
holomorphic on U and its image omits the value of 0 and 1. Then there exists
a holomorphic g on U such that
f (z) = − exp(iπ cosh[2g(z)])
and g(C) contains no disc of radius 1.
it Proof. The function g indeed can be constructed by taking (a branch) of
“log” and “square root”. Here is the construction: Since f never vanishes and
U is holomorphically simply connected, there is a holomorphic function h on
1
U such that eh = f on U . Let F := 2πi
h. Since f omits the value 1, F
does not assume any integer values, inparticular F omits the value 0 and 1.
So there are holomorphic functions H1 , H2 on U (again, use the fact that U
is holomorphically simply connected) such that H12 = F, H22 = F − 1. Let
H = H1 − H2 , then H is never zero on U , so there is a holomorphic function g
such that eg = H. From the definition
cosh(2g) + 1 =
1 2g
1
1
(e + e−2g ) + 1 = (eg + e−g )2 = (H + H −1 )2 = 2F.
2
2
2
Thus
f (z) = − exp(iπ cosh[2g(z)]).
√
√
Next, we claim that g does not assume, on U , the values of π log( n+ n − 1)+
1
2 mπi for all positive integers n and all integers m due to the fact that the image
√
of
√ f omits 1the value 1. Indeed, if there is z0 ∈ U with g(z0 ) = π log( n +
n − 1) + 2 mπi. Then it is easy to verify that
2 cosh[2g(z0 )] = (−1)m [2(2n − 1)]
1
so f (z0 ) = 1 which contradicts the assumption that f omits the value 1.
√
√
Finally we notice that points π log( n + n − 1) + 21 mπi, n is any positive
integer and m is any integer, from a vertices of a grid of rectangles. The height
of an arbitrary rectangle is
√
1
mπ − 1 (m + 1)π = 1 π < 3.
2
2
2
√
√
√
√
The√wideth is
√ log( n +
√1 + √ n) − log( n + n − 1) > 0. Now φ(x) =
log( x + 1 + x) − log( x + x −√1) is a decreasing function so that its width
of any rectangle ≤ φ(1) = log(1 + 2) < log e = 1. Hence the diagonal of the
rectangle < 2. we thus see that g(C) contains no disc of radius 1. This proves
the Lemma 1.
In order to use the above lemma, we need to study the range of (any) holomorphic functions. We prove the following important and interesting theorem.
Bloch’s theorem. Let f be analytic on a region containing D(0, 1) and f (0) =
0, f 0 (0) = 1. Then there f (D(0, 1)) constains a disk of radius 1/72 (which is
independent of f !).
remark: From the open mapping theorem, we know that f (D(0, 1)) is open and
from f (0) = 0 we know that 0 ∈ f (D(0, 1)). This theorem tells us how large a
disk (neighborhood of 0) can f (D(0, 1)) contain.
Assume that the Bloch’s theorem holds, and assume that g is not constant (otherwise the theorem holds), we now derive a contradiction. Since g is not constant, there some z0 with g 0 (z0 ) 6= 0. By considering g(z+z0 )−g(z0 ) if necessary,
we can assume g(0) = 0, g 0 (0) 6= 0. Consider h(z) := g(Rz)/(Rg 0 (0)). Then h
satisfy the condition in Bloch’s theorem. It then gives that h(D(0, 1)) contains
a disk of radius 1/72, or g(D(z0 , R)) contains a disc of redius (1/72)R|g 0 (0)|,
which contains a disc of radius 1 when R is bigger enough. This contradicts
with that fact that g(U ) contains no disc of radius 1 derived in Lemma 1. This
finishes the proof of the Little Picard Theorem.
It remains to prove Bloch’s thoerem. To do so, we first prove
Lemma 2. Let f be holomorphic on D(0, 1) with f (0) = 0, f 0 (0) = 1. Assume that |f (z)| ≤ M for all z ∈ D(0, 1). Then M ≥ 1 and f (D(0, 1)) ⊃
D(0, 1/(6M )).
2
Proof. Let 0 < r < 1 and f (z) = z + a2 z 2 + · · · , according to Cauchy’s
estimate
M
|an | ≤ n , n ≥ 1.
r
By lettign r → 1, we get |an | ≤ M . In particular, we have 1 = |a1 | ≤ M . This
proves that M ≥ 1. Now for |z| = (4M )−1 ,
|f (z)| ≥ |z|−
∞
X
|an Z n | ≥ (4M )−1 −
n=2
∞
X
M (4M )−n = (4M )−1 −(16M −4)−1 ≥ (6M )−1 .
n=2
Hence for any w with |w| < (6M )−1 , on |z| = (4M )−1 ,
|f (z)| > |w|.
By Rouche’s theorem, the functions f − w and f have the same number of
zeros in |z| < (4M )−1 . However, since f (0) = 0, f has at least one zero in
|z| < (4M )−1 . Thus f (z) = w has at least one solution in |z| < (4M )−1 . This
proves Lemma 2.
Corollary. Let g be holomorphic on D(0, R) with g(0) = 0, g 0 (0) = µ. Assume
that |g(z)| ≤ M for all z ∈ D(0, R). Then g(D(0, R)) ⊃ D(0, R2 µ2 /(6M )).
Proof: Consider
f (z) :=
g(Rz)
.
Rg 0 (0)
Then f is holomorphic on D(0, 1), f (0) = 0, f 0 (0) = 1 and
|f (z)| ≤
M
M
=
.
0
R|g (0)|
µR
By the above lemma, f (D(0, 1)) contains the disc D(0, σ) with
σ=
1
M
6 µR
=
µR
,
6M
which translates in term of the orignal g that g(D(0, R)) ⊃ D(0, R2 µ2 /(6M )).
This proved the Colloary.
Proof of Bloch’s theorem: To apply Corollary to proof Bloch’s theorem, the
main task is to drop the condition that |f (z)| ≤ M . Here is the trick: Let
K(r) = max{|f 0 (z)| : |z| = r} and consider h(r) = (1 − r)K(r). Then h is
continuous and h(0) = 1, h(1) = 0. Let r0 = sup{r : h(r) = 1}. Then h(r0 ) = 1
and h(r) ≤ 1 for r < r0 (so we get a bound for K(r) thus the derivative f (thus
the value of |f )). Here is how to do it: Take a with |a| = r0 and |f 0 (a)| = K(r0 ).
Then
|f 0 (a)| = (1 − r0 )−1 .
3
Now for |z − a| ≤
r0 ,
1
2 (1
− r0 ) := ρ0 (so |z| ≤
1
2 (1
+ r0 ), from the definition of
1
1
1
1
|f 0 (z)| ≤ K( (1 + r0 ) = h( (1 + r0 )[1 − (1 + r0 )]−1 < [1 − (1 + r0 )]−1 = 1/ρ0 .
2
2
2
2
Consider g(z) = f (z + a) − f (a). Notice that for z ∈ D(0, 12 ρ0 ), the line segment
γ[a, z + a] lies in D(a, ρ0 ). Hence
Z
1
|g(z)| ≤ | f 0 (w)dw| ≤ |z| < 1/2.
ρ0
γ
Corollary implies that g(D(0, 21 ρ0 )) ⊃ D(0, σ) where
σ=
2
1
3 ρ0
6
1
2
1
3
4
ρ0
2
=
1
.
72