Picard's Theorem

Little Picard Theorem
A nonconstant entire function $$f:\C\to\C$$ omits at most one complex value $$a \in \C$$.

Big Picard Theorem
Suppose that:
 * $$U \subset \C$$ is a domain;
 * $$z_0\in U$$;
 * $$f: U \setminus \{z_0\} \to \C$$ is a holomorphic function with an essential singularity at $$z_0$$.

Then $$f$$ omits at most one complex value $$a\in\C$$.

Proof of the little Picard theorem using the universal cover of the twice-punctured plane
For this proof, we use the fact that there is a holomorphic covering map $$\phi:\mathbb{D}\to \C\setminus\{0,1\}$$, where $$\mathbb{D}=\{z\in\C:|z|<1\}$$.

This follows from the Riemann Uniformization Theorem, but is much easier to prove.

Indeed, such a covering is given by the elliptic modular function.

Now suppose that an entire function $$f:\C\to\C$$ omits two different complex values.

We may assume for simplicity that the omitted values are $$0$$ and $$1$$ (this can always be achieved by postcomposing $$f$$ with a suitable complex affine map.)

Because $$\C$$ is simply connected and $$\phi$$ is a covering, we can lift $$f$$ to a holomorphic function:
 * $$F: \C \to \mathbb{D}$$

with $$\phi \circ F = f$$. By Liouville's Theorem (Complex Analysis), $$F$$ is constant, and therefore $$f$$ is, too, proving the theorem.

Proof of the Big Picard Theorem using Montel's Theorem
Montel's Theorem is a very close relative of Picard's theorem.

It states that a family of holomorphic functions that all omit the same two values is normal.

(Recall that a family $$\mathcal{F}$$ of holomorphic functions $$f:U\to\C$$ is normal if every sequence in $$\mathcal{F}$$ has a subsequence that converges locally uniformly to a holomorphic function or to infinity.)

To prove the big Picard theorem, let us assume without loss of generality that $$U = \mathbb{D}= \{z \in \C: |z| < 1\}$$, and that $$z_0 = 0$$.

Suppose that $$f: \mathbb{D} \setminus \{0\} \to \C$$ is an analytic function that omits two different values $$a, b \in \C$$.

We need to show that $$0$$ is a pole or a removable singularity of $$f$$.

If $$f(z) \to \infty$$ as $$z \to 0$$, then $$0$$ is a pole.

So suppose that there is a sequence $$z_n \to 0$$ such that $$f (z_n)$$ converges to a value $$w \ne \infty$$.

We may assume that $$|z_n|$$ is strictly decreasing and that $$|z_n|<1/2$$ for all $$n$$.

Let us set $$\lambda_n := 2z_n$$, and consider the sequence of functions
 * $$f_n:\mathbb{D}\setminus\{0\}\to\C\setminus\{a,b\}; \quad f_{n}(z) := f(\lambda_n z).$$

By Montel's Theorem, this family is normal.

So we may assume, passing to a subsequence if necessary, that $$f_{n}$$ converges locally uniformly to a holomorphic function or to $$\infty$$.

Note that $$f_{n}(1/2) = f(z_n)\to w$$, so $$f_{n}$$ does not converge to $$\infty$$.

Let us set:
 * $$K_n := \max_{|z|=1/2} |f_n(z)| = \max_{|z|=|z_n|} |f(z)|$$.

It follows from the above that
 * $$K := \sup_{n\in\N} K_n < \infty$$.

But by the maximum principle, we have $$|f(z)| \le \max \left\{{K_{n-1},K_n}\right\} \le K$$ whenever $$|z_n| \le |z| \le |z_{n-1}|$$.

So $$|f(z)| \le K$$ for $$|z| < |z_0|$$, and hence $$0$$ is a removable singularity by the Riemann Removable Singularities Theorem.

Proof that big Picard implies little Picard
By the Fundamental Theorem of Algebra, a nonconstant polynomial does not omit any value.

Hence it suffices to prove the little Picard theorem when $$f$$ has an essential singularity at $$\infty$$, in which case it follows from the big Picard theorem.

Remarks

 * The above proof of the little Picard theorem using the universal cover of a twice-punctured plane can also be used to prove Montel's theorem. There are also more elementary proofs, using Zalcman's Rescaling Lemma.


 * The little Picard theorem also implies Montel's theorem, using the Rescaling Lemma. Hence all three theorems are equivalent.

This is an instance of the heuristic "Bloch principle", which states that properties that force entire functions to be constant also force families of functions to be normal.