Liouville’s Theorem - Complex Analysis, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

This article is about Liouville's theorem in 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    for all z in C is constant. Equivalently, non-constant holomorphic functions on C have dense 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:

where (by Cauchy's integral formula)

and C_r 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   

where in the second inequality we have used the fact that |z|=r on the circle C_r. 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 a_k = 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{\displaystyle \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

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. 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 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

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