Liouville's theorem with your bare hands 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?
 A: There is a truly elementary proof. Nothing but high school mathematics + the notion of limit is used.
First one proves Cauchy's inequality for polynomials:
$$|f(0)|\leq M(r),$$
where $M(r)$ is the maximum of $|f|$ on the circle $|z|=r$.
This is proved as follows: let $\epsilon_k=\exp(2\pi ik/n),$
where $n>$ the degree of the polynomial.
Then
$f(0)=\frac{1}{n}\sum_{k=1}^nf(\epsilon_kz),$
because
$\sum_k^n(\epsilon_k)^j=0$ for all $j\in[1,n-1]$, by the Geometric Progression formula,
so all other terms except the constant term, in the right hand side cancel. 
Taking absolute values, we obtain the above inequality.
Then by passing to a limit this inequality is true for all entire functions.
Applying this to $(f-f(0))/z$, we obtain
$$|f'(0)|\leq (M(r)+|f(0)|)/r,$$
for all entire functions.
If the function is bounded, we conclude that $f'(0)=0$.
Applying this to $f(z-a)$ we obtain $f'(a)=0$ for all $a$, that is $f$
is constant.
A: 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.
A: I beg to differ with Andy and propose Nelson's proof as THE book proof that a bounded analytic function is constant. In fact Nelson's little gem is for harmonic functions, but the proof is incredibly beautiful and, of course, applies to analytic functions by considering their real and imaginary parts.
There are just two ingredients: 
1. The mean value theorem: the value of a harmonic function in the plane (or n-space) at some point is the average of the function over any disc centered at that point. We can even take that as definition of a harmonic function if we have a family of "discs" and a measure with which to average.
2. A geometric property of metric discs on the plane: Given any two points $x$ and $y$, for sufficiently large radii $R$, the symmetric difference of the disc of radius $R$ centered at $x$ and the disc of the same radius centered at $y$ has negligible area compared with the area of the discs. 
Now the proof goes as follows: take any two points $x$ and $y$ on the plane and choose discs of a very large radius $R$ centered at each of these points. If the harmonic function is bounded, its average over the two discs, that by (2) basically coincide, has to be almost the same. Let $R$ go to infinity and you're done.
Remark/Question Clearly the proof makes sense for a class of metric measure spaces. Has this sort of spaces (symmetric difference between large metric balls having small relative measure) been studied on its own? Normed spaces are in this category.
A: Here's a proof via character theory; in some ways it's a recasting of the Cauchy proof, but it has the advantage of being purely algebraic.  In particular, this proof will work over any algebraically closed complete valued field, for example (some mild care must be taken over a field of positive characteristic when one takes the limit $k\to\infty$).
Let $$f(z)=\sum_{n=0}^\infty a_nz^n$$ be a bounded entire function; we wish to show that $a_n=0$ for $n>0$.  We use the fact that 
$$\sum_{\zeta \text{ a $k$-th rooth of unity}} \zeta^n$$ equals $k$ if $k$ divides $n$ and equals $0$ otherwise.  So in particular setting $g_n(z)=f(z)/z^n$ we have for $k>n$ that
$$\frac{1}{k}\sum_{\zeta \text{ a $k$-th rooth of unity}} g_n(r\zeta)=a_n+\sum_{m=1}^\infty a_{km+n}r^{km}.$$
The sum on the right tends to zero with $k$, as the $a_{km+n}$ decrease very rapidly by the entireness of $f$.  So we have $$a_n=\lim_{k\to\infty}\frac{1}{k}\sum_{\zeta \text{ a $k$-th rooth of unity}} g_n(r\zeta).$$  But taking the limit as $|r|\to \infty$ and using the boundedness of $f$, this is zero for $n>0$.
I say that this is a recasting of the Cauchy integral proof because the character sum we use is essentially a Riemann sum for the integral of $g_n$ around a circular contour.  Indeed, one may prove the Cauchy integral formula for circular contours this way.
A: If your definition of holomorphic is that there is an absolutely convergent power series $\sum_{n=0}^{\infty} a_n z^n$ then Parseval's theorem implies that the mean square of the function on the circle of radius $R$ is $\sum_{n=0}^{\infty} |a_n^2| R^{2n}$ and it follows that $a_n=0$ for all $n>0$.
Edit (new approach)
Without loss of generality let's assume that $f(0)=0$ and $f'(0)=0$.  Consider the function $f(x)/x$, continuous on all of $\mathbb{C}$.  It is analytic on the punctured plane and converges to 0 on its boundary, hence by the maximum principle must be 0.
You can drop the $f'(0)=0$ condition above if you assume instead that $f$ is twice differentiable at 0, so then $f(x)/x$ is holomorphic on all of $\mathbb{C}$ and decreases to 0, so must be identically 0.
This does presuppose the maximum principle.
