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 {displaystyle |f(z)|leq M} for all z {displaystyle z} in C {displaystyle mathbb {C} } is constant. Equivalently, non-constant holomorphic functions on C {displaystyle mathbb {C} } have unbounded images. 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 {displaystyle |f(z)|leq M} for all z {displaystyle z} in C {displaystyle mathbb {C} } is constant. Equivalently, non-constant holomorphic functions on C {displaystyle 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. 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 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 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. 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. Consider that for g = 0 the theorem is trivial so we assume g ≠ 0. {displaystyle g eq 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. 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