An Elliptic function can not be holomorphic/analytic?

Question

I was reading about elliptic functions on the wiki and it said that a doubly periodic meromorphic function in contention of being an elliptic function can not be analytic/holomorphic as it would then be bounded. Can anyone explain this to me and thankfully with a rigorous proof?

Answer 1

Let $X$ denote the fundamental parallelogram for your elliptic function $f$; that is, if $f$ has periods $1$ and $\tau$, let $X$ denote the parallelogram with vertices $0, 1, \tau$, and $1 +\tau$. The supremum of $|f|$ on $\mathbb{C}$ is clearly the supremum of $|f|$ on $X$, which is finite if $f$ is analytic (and thus continuous everywhere) because $X$ is compact. But any bounded, entire function must be constant.

Answer 2

It is the Liouville theorem that leads to that conclusion. Assume we have a doubly periodic function that is holomorphic on a cell of the lattice. The cell being compact the function is bounded on the cell. Because it is doubly periodic it is holomorphic and bounded on all the complex plane. Liouville theorem says that any entire function that is bounded is constant. This is a contradiction.

Answer 3

I think you're trying to say that a nonconstant doubly periodic function $f$ cannot be entire (not analytic as in your question), is that right?

Firstly, a doubly periodic meromorphic function (that is, a function that is doubly periodic with two periods (say $\omega_1$ and $\omega_2$, with $\frac{\omega_1}{\omega_2}$ to be a non-real number), and that is analytic everywhere except for the poles(which have to be in a finite number)).

So on the primary parallelogram with vertices 0, $\omega_1$, $\omega_2$ and $\omega_1+\omega_2$, the function is repeated elsewhere, on the other parallelograms, defined by the integer combinations of the 4 numbers above.

So if this function were entire, then note that on the primary parallelogram, the function $|f|$ is continuous (because $f$ is analytic) and its domain is compact. So it must have a maximum. Hence, $f$ is bounded on that parallelogram, as well as the other parallelograms, since $f$ is doubly periodic, so $f$ is bounded on the entire complex plane. So by Liouville's Theorem, an entire, bounded function must be constant.