55.5. THE PICARD THEOREMS 1763

55.5.6 The Great Big Picard TheoremThe next theorem is the main result which the above lemmas lead to. It is the Big Picardtheorem, also called the Great Picard theorem.Recall B′ (a,r) is the deleted ball consistingof all the points of the ball except the center.

Theorem 55.5.17 Suppose f has an isolated essential singularity at 0. Then for everyR > 0, and β ∈ C, f−1 (β )∩B′ (0,R) is an infinite set except for one possible exceptionalβ .

Proof: Suppose this is not true. Then there exists R1 > 0 and two points, α and β suchthat f−1 (β )∩B′ (0,R1) and f−1 (α)∩B′ (0,R1) are both finite sets. Then shrinking R1 andcalling the result R, there exists B(0,R) such that

f−1 (β )∩B′ (0,R) = /0, f−1 (α)∩B′ (0,R) = /0.

Now let A0 denote the annulus{

z ∈ C : R22 < |z|< 3R

22

}and let An denote the annulus{

z ∈ C : R22+n < |z|< 3R

22+n

}. The reason for the 3 is to insure that An ∩An+1 ̸= /0. This

follows from the observation that 3R/22+1+n > R/22+n. Now define a set of functions onA0 as follows:

fn (z)≡ f( z

2n

).

By the choice of R, this set of functions missed the two points α and β . Therefore, byLemma 55.5.16 there exists a subsequence such that one of the two options presented thereholds.

First suppose limk→∞

∣∣∣∣ fnk − f∣∣∣∣

K,∞= 0 for all K a compact subset of A0 and f is ana-

lytic on A0. In particular, this happens for γ0 the circular contour having radius R/2. Thusfnk must be bounded on this contour. But this says the same thing as f (z/2nk) is boundedfor |z|=R/2, this holding for each k = 1,2, · · · . Thus there exists a constant, M such that oneach of a shrinking sequence of concentric circles whose radii converge to 0, | f (z)| ≤M.By the maximum modulus theorem, | f (z)| ≤M at every point between successive circlesin this sequence. Therefore, | f (z)| ≤M in B′ (0,R) contradicting the Weierstrass Casoratitheorem.

The other option which might hold from Lemma 55.5.16 is that limk→∞ ρK(

fnk ,∞)= 0

for all K compact subset of A0. Since f has an essential singularity at 0 the zeros of f inB(0,R) are isolated. Therefore, for all k large enough, fnk has no zeros for |z| < 3R/22.This is because the values of fnk are the values of f on Ank , a small anulus which avoids allthe zeros of f whenever k is large enough. Only consider k this large. Then use the aboveargument on the analytic functions 1/ fnk . By the assumption that limk→∞ ρK

(fnk ,∞

)= 0,

it follows limk→∞

∣∣∣∣1/ fnk −0∣∣∣∣

K,∞= 0 and so as above, there exists a shrinking sequence

of concentric circles whose radii converge to 0 and a constant, M such that for z on anyof these circles, |1/ f (z)| ≤ M. This implies that on some deleted ball, B′ (0,r) wherer ≤ R, | f (z)| ≥ 1/M which again violates the Weierstrass Casorati theorem. This provesthe theorem.

As a simple corollary, here is what this remarkable theorem says about entire functions.