# Liouville's theorem (complex analysis)

{{#invoke:Hatnote|hatnote}} 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)| ≤ M for all z in C is constant.

The theorem is considerably improved by Picard's little theorem, which says that every entire function whose image omits at least two 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:

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

where (by Cauchy's integral formula)

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

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

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

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. To show this, 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, 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

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

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 C. Suppose it was. Then, if a and b are two periods of f such that ab 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 C is what Liouville actually proved, in 1847, using the theory of elliptic functions.[1] In fact, it was Cauchy who proved Liouville's theorem.[2][3]

### Entire functions have dense images

If f is a non-constant entire function, then its image is dense in 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) = 1/(f(z) − w). Then g is a bounded entire function, since

${\displaystyle (\forall z\in {\mathbb {C} }):|g(z)|={\frac {1}{|f(z)-w|}}<{\frac {1}{r}}\cdot }$

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

## Remarks

Let C ∪ {∞} be the one point compactification of the complex plane C. In place of holomorphic functions defined on regions in C, one can consider regions in C ∪ {∞}. Viewed this way, the only possible singularity for entire functions, defined on CC ∪ {∞}, 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 at ∞, i.e. blows up like 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,

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

The argument used during the proof using Cauchy estimates shows that

${\displaystyle (\forall k\in \mathbb {N} ):|a_{k}|\leqslant Mr^{n-k}.}$

So, if k > n,

${\displaystyle |a_{k}|\leqslant \lim _{r\rightarrow +\infty }Mr^{n-k}=0.}$

Therefore, ak = 0.