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# 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)| ≤ 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. Since f is entire, 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. We can estimate directly

$$| a_k | \le \frac{1}{2 \pi} \oint_{C_r} \frac{ | f ( \zeta ) | }{ | \zeta |^{k+1} } \, |d\zeta| \le \frac{1}{2 \pi} \oint_{C_r} \frac{ M }{ r^{k+1} } \, |d\zeta| = \frac{M}{2 \pi r^{k+1}} \oint_{C_r} |d\zeta| = \frac{M}{2 \pi r^{k+1}} 2 \pi r = \frac{M}{r^k},$$

where in the second inequality we have invoked the assumption that |f(z)| ≤ M for all z and 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

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

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 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 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

$$(\forall z\in\mathbb{C}):|g(z)|=\frac1{|f(z)-w|}<\frac1r\cdot$$

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 C ⊂ C ∪ {∞}, 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,

$$f(z) = \sum_{k=0}^\infty a_k z^k.$$

The argument used during the proof shows that

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

So, if k > n,

$$|a_k|\leqslant\lim_{r\rightarrow+\infty}Mr^{n-k}=0.$$

Therefore, ak = 0.
References

^ Liouville, Joseph (1847), "Leçons sur les fonctions doublement périodiques", Journal für die Reine und Angewandte Mathematik 88: 277–310, 1879, ISSN 0075-4102
^ Cauchy, Augustin-Louis (1844), "Mémoires sur les fonctions complémentaires", Œuvres complètes d'Augustin Cauchy, 1, 8, Paris: Gauthiers-Villars (published 1882)
^ Lützen, Jesper (1990), Joseph Liouville 1809–1882: Master of Pure and Applied Mathematics, Studies in the History of Mathematics and Physical Sciences, 15, Springer-Verlag, ISBN 3-540-97180-7