# Liouville's theorem (complex analysis)

In complex analysis, Liouville's theorem, named after Joseph Liouville, states that every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number M such that $|f(z)|\leq M$ for all $z$ in $\mathbb {C}$ is constant. Equivalently, non-constant holomorphic functions on $\mathbb {C}$ have unbounded images.

The theorem is considerably improved by Picard's little theorem, which says that every entire function whose image omits two or more complex numbers must be constant.

## Proof

The theorem follows from the fact that holomorphic functions are analytic. If f is an entire function, it can be represented by its Taylor series about 0:

$f(z)=\sum _{k=0}^{\infty }a_{k}z^{k}$

where (by Cauchy's integral formula)

$a_{k}={\frac {f^{(k)}(0)}{k!}}={1 \over 2\pi i}\oint _{C_{r}}{\frac {f(\zeta )}{\zeta ^{k+1}}}\,d\zeta$

and Cr is the circle about 0 of radius r > 0. Suppose f is bounded: i.e. there exists a constant M such that |f(z)| ≤ M for all z. We can estimate directly

$|a_{k}|\leq {\frac {1}{2\pi }}\oint _{C_{r}}{\frac {|f(\zeta )|}{|\zeta |^{k+1}}}\,|d\zeta |\leq {\frac {1}{2\pi }}\oint _{C_{r}}{\frac {M}{r^{k+1}}}\,|d\zeta |={\frac {M}{2\pi r^{k+1}}}\oint _{C_{r}}|d\zeta |={\frac {M}{2\pi r^{k+1}}}2\pi r={\frac {M}{r^{k}}},$

where in the second inequality we have used the fact that |z| = r on the circle Cr. But the choice of r in the above is an arbitrary positive number. Therefore, letting r tend to infinity (we let r tend to infinity since f is analytic on the entire plane) gives ak = 0 for all k ≥ 1. Thus f(z) = a0 and this proves the theorem.

## Corollaries

### Fundamental theorem of algebra

There is a short proof of the fundamental theorem of algebra based upon Liouville's theorem.

### No entire function dominates another entire function

A consequence of the theorem is that "genuinely different" entire functions cannot dominate each other, i.e. if f and g are entire, and |f| ≤ |g| everywhere, then f = α·g for some complex number α. Consider that for g = 0 the theorem is trivial so we assume $g\neq 0.$  Consider the function h = f/g. It is enough to prove that h can be extended to an entire function, in which case the result follows by Liouville's theorem. The holomorphy of h is clear except at points in g−1(0). But since h is bounded and all the zeroes of g are isolated, any singularities must be removable. Thus h can be extended to an entire bounded function which by Liouville's theorem implies it is constant.

### If f is less than or equal to a scalar times its input, then it is linear

Suppose that f is entire and |f(z)| is less than or equal to M|z|, for M a positive real number. We can apply Cauchy's integral formula; we have that

$|f'(z)|={\frac {1}{2\pi }}\left|\oint _{C_{r}}{\frac {f(\zeta )}{(\zeta -z)^{2}}}d\zeta \right|\leq {\frac {1}{2\pi }}\oint _{C_{r}}{\frac {|f(\zeta )|}{\left|(\zeta -z)^{2}\right|}}|d\zeta |\leq {\frac {1}{2\pi }}\oint _{C_{r}}{\frac {M|\zeta |}{\left|(\zeta -z)^{2}\right|}}\left|d\zeta \right|={\frac {MI}{2\pi }}$

where I is the value of the remaining integral. This shows that f′ is bounded and entire, so it must be constant, by Liouville's theorem. Integrating then shows that f is affine and then, by referring back to the original inequality, we have that the constant term is zero.

### Non-constant elliptic functions cannot be defined on ℂ

The theorem can also be used to deduce that the domain of a non-constant elliptic function f cannot be $\mathbb {C} .$  Suppose it was. Then, if a and b are two periods of f such that a/b is not real, consider the parallelogram P whose vertices are 0, a, b and a + b. Then the image of f is equal to f(P). Since f is continuous and P is compact, f(P) is also compact and, therefore, it is bounded. So, f is constant.

The fact that the domain of a non-constant elliptic function f can not be $\mathbb {C}$  is what Liouville actually proved, in 1847, using the theory of elliptic functions. In fact, it was Cauchy who proved Liouville's theorem.

### Entire functions have dense images

If f is a non-constant entire function, then its image is dense in $\mathbb {C} .$  This might seem to be a much stronger result than Liouville's theorem, but it is actually an easy corollary. If the image of f is not dense, then there is a complex number w and a real number r  > 0 such that the open disk centered at w with radius r has no element of the image of f. Define

$g(z)={\frac {1}{f(z)-w}}.$

Then g is a bounded entire function, since for all z,

$|g(z)|={\frac {1}{|f(z)-w|}}<{\frac {1}{r}}.$

So, g is constant, and therefore f is constant.

## On compact Riemann surfaces

Any holomorphic function on a compact Riemann surface is necessarily constant.

Let $f(z)$  be holomorphic on a compact Riemann surface $M$ . By compactness, there is a point $p_{0}\in M$  where $|f(p)|$  attains its maximum. Then we can find a chart from a neighborhood of $p_{0}$  to the unit disk $\mathbb {D}$  such that $f(\phi ^{-1}(z))$  is holomorphic on the unit disk and has a maximum at $\phi (p_{0})\in \mathbb {D}$ , so it is constant, by the maximum modulus principle.

## Remarks

Let $\mathbb {C} \cup \{\infty \}$  be the one point compactification of the complex plane $\mathbb {C} .$  In place of holomorphic functions defined on regions in $\mathbb {C}$ , one can consider regions in $\mathbb {C} \cup \{\infty \}.$  Viewed this way, the only possible singularity for entire functions, defined on $\mathbb {C} \subset \mathbb {C} \cup \{\infty \},$  is the point . If an entire function f is bounded in a neighborhood of , then is a removable singularity of f, i.e. f cannot blow up or behave erratically at . In light of the power series expansion, it is not surprising that Liouville's theorem holds.

Similarly, if an entire function has a pole of order n at —that is, it grows in magnitude comparably to zn in some neighborhood of —then f is a polynomial. This extended version of Liouville's theorem can be more precisely stated: if |f(z)| ≤ M|zn| for |z| sufficiently large, then f is a polynomial of degree at most n. This can be proved as follows. Again take the Taylor series representation of f,

$f(z)=\sum _{k=0}^{\infty }a_{k}z^{k}.$

The argument used during the proof using Cauchy estimates shows that for all k ≥ 0,

$|a_{k}|\leq Mr^{n-k}.$

So, if k > n, then

$|a_{k}|\leq \lim _{r\to \infty }Mr^{n-k}=0.$

Therefore, ak = 0.

Liouville's theorem does not extend to the generalizations of complex numbers known as double numbers and dual numbers.