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.
Contents |
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
where (by Cauchy's integral formula)
and Cr is the circle about 0 of radius r > 0. We can estimate directly
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.
There is a short proof of the fundamental theorem of algebra based upon Liouville's theorem.
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.
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.
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]
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.
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,
The argument used during the proof shows that
So, if k > n,
Therefore, ak = 0.