Liouville's Theorem (complex Analysis) - Proof

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)

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

| 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 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.

Read more about this topic:  Liouville's Theorem (complex Analysis)

Famous quotes containing the word proof:

    “There are some persons in this world, who, unable to give better proof of being wise, take a strange delight in showing what they think they have sagaciously read in mankind by uncharitable suspicions of them.”
    —Herman Melville (1819–1891)

    “The source of Pyrrhonism comes from failing to distinguish between a demonstration, a proof and a probability. A demonstration supposes that the contradictory idea is impossible; a proof of fact is where all the reasons lead to belief, without there being any pretext for doubt; a probability is where the reasons for belief are stronger than those for doubting.”
    —Andrew Michael Ramsay (1686–1743)

    “To cease to admire is a proof of deterioration.”
    —Charles Horton Cooley (1864–1929)