55.5. THE PICARD THEOREMS 1759

Similarly, ∣∣∣∣ log( f (0))πi

−2∣∣∣∣ ≤

(∣∣∣∣ lnβ

π

∣∣∣∣2 +(2+1)2

)1/2

=

(∣∣∣∣ lnβ

π

∣∣∣∣2 +9

)1/2

It follows from 55.5.28 that

2 |H (0)| ≤ ln

2

(∣∣∣∣ lnβ

π

∣∣∣∣2 +9

)1/2+2π.

Hence from 55.5.27| f (z)| ≤

exp

π exp

ln

2

(∣∣∣∣ lnβ

π

∣∣∣∣2 +9

)1/2+2π

exp(

48C ln(

11−θ

))and so, letting M (β ,θ) be given by the above expression on the right, the lemma is proved.

The following theorem will be referred to as Schottky’s theorem. It looks just like theabove lemma except it is only assumed that f is analytic on B(0,R) rather than on an openset containing B(0,R). Also, the case of an arbitrary center is included along with arbitrarypoints which are not attained as values of the function.

Theorem 55.5.8 Let f be analytic on B(z0,R) and suppose that f does not take on eitherof the two distinct values a or b. Also suppose | f (z0)| ≤ β . Then letting θ ∈ (0,1) , itfollows

| f (z)| ≤M (a,b,β ,θ)

for all z ∈ B(z0,θR) , where M (a,b,β ,θ) is a function of only the variables β ,θ ,a,b. (Inparticular, there is no dependence on R.)

Proof: First you can reduce to the case where the two values are 0 and 1 by considering

h(z)≡ f (z)−ab−a

.

If there exists an estimate of the desired sort for h, then there exists such an estimate forf . Of course here the function, M would depend on a and b. Therefore, there is no loss ofgenerality in assuming the points which are missed are 0 and 1.

Apply Lemma 55.5.7 to B(0,R1) for the function, g(z) ≡ f (z0 + z) and R1 < R. Thenif β ≥ | f (z0)|= |g(0)| , it follows |g(z)|= | f (z0 + z)| ≤M (β ,θ) for every z ∈ B(0,θR1) .Now let θ ∈ (0,1) and choose R1 < R large enough that θR = θ 1R1 where θ 1 ∈ (0,1) .Then if |z− z0|< θR, it follows

| f (z)| ≤M (β ,θ 1) .

Now let R1→ R so θ 1→ θ .