Liouville’s Theorem

Liouville’s theorem in complex analysis is given by a French mathematician and Engineer Joseph Liouville. This result was presented by Liouville in his lectures in 1847, although it was believed to be first proven by his famous contemporary mathematician A.L. Cauchy in 1844.

Liouville’s theorem is concerned with the entire function being bounded over a given domain in a complex plane. An entire or integral function is a complex analytic function that is analytic throughout the whole complex plane. For example, exponential function, sin z, cos z and polynomial functions.

Statement of Liouville’s Theorem

The statement of Liouville’s Theorem has several versions. Thus according to Liouville’s Theorem, if f is an integral function (entire function) satisfying the inequality |f(z)| ≤ M, where M is a positive constant, for all values of z in complex plane C, then f is a constant function.

In other words, if f(z) is an analytic function for all finite values of z and is bounded for all values of z in C, then f is a constant function.

Thus, Liouville’s Theorem states that:

Proof of Liouville’s Theorem

By the theorem hypothesis, f is bounded entire function such that for M be a positive constant |f(z)| ≤ M.

Let z₁ and z₂ be arbitrary points in z-plane. C be a circle in z-plane with z₁ as centre and radius R such that z₂ be any point inside the circle C. Then by Cauchy’s Integral Formula, we have

We can choose R large enough as f is analytic throughout z-plane, so that |z₂ – z₁| < R/2.

Since the circle C:|z – z₁| < R, we have

|z – z₂| = |(z – z₁) – (z₂ – z₁)| ≥ |z – z₁| – |z₂ – z₁| ≥ R – R/2 = R/2

⇒ |z – z₂| ≥ R/2

Now, from (1), we have

⇒ |f(z₂) – f(z₁)| ≤ 2M|z₂ – z₁|/R

The right-hand side of the above inequality tends to zero as R → ∞ . Hence, for the entire function f, R→ ∞ , therefore

f(z₂) – f(z₁) = 0 ⇒ f(z₁) = f(z₂).

Since z₁ and z_2, are arbitrarily chosen, this holds for every points in the complex z-plane.

Thus, f is a constant function.

Corollaries of Liouville’s Theorem

• A non constant entire function is not bounded.
• The fundamental theorem of Algebra: Every non constant complex polynomial has a root.
• If f is a non constant entire function, then w-image if f is dense in complex plane C.

Related Articles

• Complex Numbers
• Analytic Functions
• Limits and Continuity
• Differentiabilty
• Integration
• Complex Conjugate

Solved Examples on Liouville’s Theorem

Example 1:

Let f = u(z) + iv(z) be an entire function in complex plane C. If |u(z)| < M for every z in C, where M is a positive constant, then prove that f is a constant function.

Solution:

Given, f = u(z) + iv(z) is an entire function in complex plane C such that |u(z)| < M for every z in C.

Let g(z) = e^f(z)

since f is entire ⇒ g is also an entire function

Now, takin modulus on both side

|g(z)| = |e^f(z)| = |e^{u(z) + iv(z)}| = |e^u(z) . e^iv(z)| = |e^u(z)|

Since, |e^i𝜃| = 1

Therefore, |g(z)| = e^M, (as |u(z)| < M, hence e^M is constant)

⇒ g(z) is a bounded function

As g(z) is bounded entire function

By Liouville’s theorem, g is a constant function

⇒ e^f(z) is a constant function

⇒ f = u(z) + iv(z) is constant

⇒f is a constant function.

Example 2:

Let f be an entire function such that |f(z)| ≥ 1 for every z in C. Prove that f is a constant function.

Solution:

Given f is an entire function such that |f(z)| ≥ 1 for every z in C

Let g(z) = 1/f(z)

Since f is an entire function ⇒ g is an entire function

Now, |g(z)| = |1/f(z)| = 1/|f(z)|

As |f(z)| ≥ 1 ⇒ 1/|f(z)| ≤ 1

Therefore, |g(z)| ≤ 1

⇒ g is bounded

Thus, g is an bounded entire function

Then, by Liouville’s Theorem g is a constant function

Consequently, f is a constant function.