Liouville’s Theorem

Theorem

If (f:\mathbb C\to\mathbb C) is entire (holomorphic on all of (\mathbb C)) and bounded, then (f) is constant:

$$(\exists M>0\text{such that }|f(z)|\le M\text{ for every }z\in \mathbb{C})⟹f\equiv c.$$

This is a global result in complex analysis: a nonconstant entire function must become arbitrarily large somewhere. It is not true for differentiable functions of a real variable; for example, (\sin x) is bounded and nonconstant on (\mathbb R).

Terminology

• Holomorphic at (z_0): complex differentiable in a neighbourhood of (z_0).
• Entire: holomorphic at every point of (\mathbb C).
• Bounded: there is one constant (M) satisfying (|f(z)|\le M) on the whole plane.

The hypotheses are essential. For instance, (\sin z) is entire but not bounded on (\mathbb C), while (f(z)=1/z) is bounded on ({z:|z|\ge1}) but is not entire.

Proof using Cauchy’s estimates

Fix an arbitrary point (a\in\mathbb C). Since (f) is entire, it is holomorphic on and inside every circle

$$C_R=\{z:|z-a|=R\},R>0.$$

Cauchy’s integral formula for derivatives gives

$$f^{\prime }(a)=\frac{1}{2\pi i}{\int }_{C_R}\frac{f(z)}{(z-a{)}^2}dz.$$

Assume $(|f(z)|\le M)$ for all (z\in\mathbb C). On (C_R), (|z-a|=R), and the length of the circle is (2\pi R). Therefore,

$$\begin{array}{rl}|f^{\prime }(a)|& \le \frac{1}{2\pi }(2\pi R)\frac{M}{R^2}\\ & =\frac{M}{R}.\end{array}$$

This inequality holds for every (R>0). Sending (R\to\infty) yields (f’(a)=0).

Because (a) was arbitrary, (f’\equiv0) on (\mathbb C). Hence (f) is constant.

Higher-derivative form

Cauchy’s estimate more generally states

$$|f^{(n)}(a)|\le \frac{n!M}{R^n}.$$

Letting (R\to\infty) shows (f^{(n)}(a)=0) for every (n\ge1). Thus the Taylor series of (f) has only its constant term.

Consequence: the Fundamental Theorem of Algebra

Liouville’s theorem quickly proves the Fundamental Theorem of Algebra.

Let (p) be a nonconstant complex polynomial and suppose, to obtain a contradiction, that it has no zero in (\mathbb C). Then

$$g(z)=\frac{1}{p(z)}$$

is entire. If (p(z)=a_nz^n+\cdots+a_0) with (n\ge1), then

$$|p(z)|\to \infty (|z|\to \infty ).$$

Consequently (g(z)\to0) as (|z|\to\infty), so (g) is bounded outside a sufficiently large disk. It is continuous, hence bounded, on the closed disk as well. Therefore (g) is bounded on (\mathbb C).

By Liouville’s theorem, (g) is constant, which would make (p) constant—a contradiction. Thus every nonconstant polynomial in (\mathbb C[z]) has a complex root.

Polynomial-growth version

A useful extension replaces boundedness by controlled growth.

Generalized Liouville theorem

Let (f) be entire. If there are constants (C>0) and an integer (n\ge0) such that

$$|f(z)|\le C(1+|z|{)}^n\text{for all }z\in \mathbb{C},$$

then (f) is a polynomial of degree at most (n).

Proof idea

For (k>n), Cauchy’s estimate on (|z|=R) gives

$$|f^{(k)}(0)|\le \frac{k!}{R^k}\underset{|z|=R}{\max }|f(z)|\le k!C\frac{(1+R{)}^n}{R^k}.$$

The right-hand side tends to (0) as (R\to\infty), so (f^{(k)}(0)=0). All Taylor coefficients above degree (n) vanish, and therefore (f) is a polynomial of degree at most (n).

The original theorem is the case (n=0).

Related results

• Maximum modulus principle: a nonconstant holomorphic function cannot attain a maximum of (|f|) at an interior point. Liouville’s theorem can be viewed as its global counterpart.
• Cauchy’s integral formula: supplies the derivative estimates that drive the proof.
• Removable singularity at infinity: an entire function bounded near (\infty) extends holomorphically to the Riemann sphere; holomorphic functions on the sphere are constant.
• Picard’s theorem: a much stronger result: a nonconstant entire function omits at most one complex value.

Common pitfalls

1. “Bounded” must hold on all of (\mathbb C), not merely on a disk, a half-plane, or the real axis.
2. An entire function need not be bounded: (e^z), (\sin z), and every nonconstant polynomial are counterexamples.
3. The conclusion is “constant,” not necessarily zero. Any constant function satisfies the hypotheses.
4. The theorem concerns complex differentiability. Real differentiability alone is far too weak.