(Liouville’s Theorem) A bounded entire function is constant

1. Liouville Theorem

Theorem 1.1. (Liouville’s Theorem) A bounded entire function is constant.

To prove this theorem, we need the following Lemma:

Lemma 1.1. Let f be a holomorphic function on a domain (open connected) Ω of C.

Suppose f⁰ = 0. Then f is a constant function.

Proof. To show that f is a constant function, we need to show that f (z₀) = f (z₁) for all z0, z1∈ Ω. Since Ω is a domain, Ω is path connected. Then we can choose a curve γ : I → Ω so that γ(0) = z0 and γ(1) = z1. Then

Z

γ

f⁰(w)dw = f (z1) − f (z0).

Since f⁰= 0 on Ω, f (z1) = f (z0).

 Let us go back to the proof of Liouville’s theorem.

Let f be an entire function. Suppose |f (z)| ≤ M for all z ∈ C. To prove f is a constant, we only need to prove f⁰(z) = 0 on C and use the fact that C is a connected space.

Let z₀ ∈ C. Using Cauchy-integral formula, f⁰(z₀) = 1

2πi I

CR(z0)

f (z) (z − z₀)²dz for R > 0. Then

f⁰(z0) = 1 2π

Z 2π 0

f (z0+ Re^it) = − 1 2πR

Z 2π 0

f (z0+ Re^it)e^−itdt.

Using the boundedness of f, we see

|f⁰(z0)| ≤ 1 2πR

Z 2π 0

|f (z₀+ Re^it)|dt ≤ M R.

Letting R → ∞, we find f⁰(z0) = 0. Since z0 is arbitrary, f⁰ = 0 on C. Using Lemma 1.1, we complete the proof of our assertion.

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