Liouville by Maximum Modulus Principle

cn2_71828182846

Prove Liouville's theorem (i.e. a bounded entire function is constant) using the maximum modulus principle. That is, DON'T use the Cauchy estimate.

**Posted:** Sun Dec 12, 2004 3:23 pm

Kent Merryfield

Note that the maximum modulus principle alone is insufficient. You must use at least some other property of holomorphic functions.

Let $M$ be the set of smooth functions from $\mathbb{C}$ to $\mathbb{C}$ that satisfy the maximum modulus property: that is, if $f\in M,$ then $|f|$ has no local maxima. The (entire) holomorphic functions form a subset of $M$ as do also the harmonic functions. Both the holomorphic and harmonic functions do satisfy Liouville's Theorem. But $f(x+iy)=\tan^{-1}x+i\tan^{-1}y$ is also in $M$ and it's bounded and not constant.

As for the proof itself, let $f$ be a bounded holomorphic function, and consider $g(z)=f\left(\frac1z\right)$ . This a bounded function with an isolated singularity at $0$ . We make use of the theorem that says such a singularity must be removable. Thus, $\lim_{|z|\to\infty}f(z)$ exists, and we can uniquely extend $f$ to a continuous function on the one-point compactification of $\mathbb{C}$ (the "Reimann sphere.")

That makes this newly extended version of $f$ a continuous function on a compact set, so its modulus must assume a maximum. But if $f$ is not constant, this would be a local maximum of $|f|$ (or $|g|$ ), which is not permitted by the maximum modulus principle.

We do need that theorem on removable singularities.

**Posted:** Mon Dec 13, 2004 9:17 am

cn2_71828182846

How about this proof:

Let f be entire, and define g(z) = (f(z) - f(0))/z, which is also entire. Since f is bounded, as |z| goes to infinity, |g(z)| goes to 0. Thus we can choose R > 0 s.t. there exists :varepsilon: s.t. |g(z)| < :varepsilon: for all |z| > R and the maximum value of |g(z)| for |z| <= R is > :varepsilon: . Such a maximum exists because |g(z)| is continuous and the closed disc is compact. Let w = z be the maximum of |g(z)| in this closed disc. Then |g(w)| >= |g(z)| for all z in C. Thus by the global maximum modulus principle, since C is open, g must be constant. Since |g(z)| goes to 0 as |z| goes to infinity, g must be identically 0. Thus f(z) = f(0) for all z in C, whereby f is constant.

**Posted:** Mon Dec 13, 2004 5:57 pm

Kent Merryfield

That works, too. I prefer cn2_71828182846's proof.

Note that that proof also makes use of a removable singularity, but the fact that it is removable is more elementary than the version I used.

**Posted:** Mon Dec 13, 2004 6:21 pm

Myth

Very nice proof, cn2_71828182846!

**Posted:** Mon Dec 13, 2004 10:16 pm

_________________
Myth is out of here

cn2_71828182846

Thanks!

**Posted:** Tue Dec 14, 2004 8:39 pm