Liouville's theorem from complex analysis states that a holomorphic function $f(z)$ on the plane that is bounded in magnitude is constant. The usual proof uses the Cauchy integral formula. But this has always struck me as indirect and unilluminating. There is a proof via harmonic function theory, but this also seems to involve an unnecessarily large amount of prior buildup. So one might seek a more direct proof as below.


Assume that $f(z)$ is nonconstant. The fact that $f(z)$ is holomorphic at every point implies that at any given point, there is a direction such that moving in that direction makes $|f(z)|$ larger. But this doesn't prove that $|f(z)|$ is unbounded, because *a priori* its magnitude could behave like $5 - \frac{1}{|z|}$ or some such thing. 


In the case of $f(z) = \frac{1}{P(z)}$ where $P(z)$ is a polynomial, one knows that $|f(z)|$ tends toward $0$ as $|z| \to \infty$ so that there's some closed disk such that if $|f(z)|$ is bounded, then it has a maximum in the interior of the disk, which contradicts the fact that one can always make $f(z)$ larger by moving in a suitable direction. But for general $f(z)$, one doesn't have this argument.


One can try to reason based on the power series expansion of a holomorphic function $f(z)$ that is not a polynomial. Because polynomials are unbounded as $|z| \to \infty$ and grow in magnitude in a way that's proportional to their degree, one might think that a power series, which can be regarded as an infinite degree polynomial, would also be unbounded as $|z| \to \infty$. This is of course false: take $f(z) = \sin(z)$, then as $|z| \to \infty$ along the real axis, $f(z)$ remains bounded. The point is that the dominant term in the partial sums of the power series varies with $|z|$, and that the relevant coefficients change, alternating in sign and tending toward zero rapidly, so that the gain in size corresponding to moving to the next power of $z$ is counterbalanced by the change in coefficient. But there's *some* direction that one can move in for which $f(z)$ is unbounded: in particular, for $f(z) = \sin(z)$, $f(z)$ is unbounded along the imaginary axis.


This suggests that we write $a\_n = s\_{n}e^{i \theta\_n}$ for the coefficient of $z^n$ in the power series expansion of $f(z)$ and write $z = re^{i \theta}$ (where $s, r > 0$) so that


$$f(z) = \sum\_{n = 0}^{\infty} {a\_n}z^n = \sum\_{n = 0}^{\infty} sr^n e^{n\theta + \theta\_n} $$


and try to find a function $\theta = g(r)$ such that $f(z)$ is unbounded as $r \to \infty$ if one takes $\theta = g(r)$. 


But I don't know what to do next. Any ideas? Any ideas for *other* strategies of proving Liouville's theorem that are more direct than the ones using Cauchy's theorem?



Liouville's theorem with your bare hands
I think the most illuminating proof of Liouville's theorem uses Riemann surfaces. Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be a bounded holomorphic function, and set $g(z) = f(1/z)$. Then $g : \mathbb{C} \setminus 0 \rightarrow \mathbb{C}$ is a bounded holomorphic function, so Riemann's removable singularities theorem says that $g$ can be extended over $0$. Translated into the language of Riemann surfaces, this says that $f$ extends to a holomorphic function $F : \mathbb{P}^1 \rightarrow \mathbb{C}$. Since $\mathbb{P}^1$ is compact, $F$ must have a global maximum. However, the maximum modulus principle says that a nonconstant holomorphic function cannot have a local maximum, so $F$ must be constant.


