MHB Can you find all entire, doubly periodic functions on the complex plane?

[Ackbach]

Here is my first offering for the University POTW! I hope you will enjoy it.

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A complex-valued function $f(z)$ on the complex plane is doubly periodic if there are two periods $\omega_0$ and $\omega_1$ of $f(z)$ that do not lie on the same line through the origin (that is, $\omega_0$ and $\omega_1$ are linearly independent over the reals, and $f(z+\omega_0) =f(z+\omega_1)=f(z)$ for all complex numbers $z$.) Find all the entire (analytic on the whole complex plane) doubly periodic functions.

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[Ackbach]

This is problem IV.5.3 in Complex Analysis, by Theodore Gamelin.

Congratulations to mathbalarka for his correct solution, which follows:

Spoiler

Claim : Bounded entire functions on $\Bbb C$ are constant.

Proof : Consider an entire function $f$ on $\Bbb C$ such that $|f(z)|<M$ for all $z\in \Bbb C$. By analyticity, $$f(z) = \sum_{n = 0}^\infty a_n z^n$$ where $\displaystyle 2\pi ia_n = \int_{|z|=R} \frac{f(z)}{z^{n+1}} dz$ for any positive real $R$ by definition. Thus, $$|a_n| \leq \frac1{2\pi} \int_{|z|=R} \frac{|f(z)|}{|z|^{n+1}} |dz| \leq \frac{M}{2\pi} \int_{|z|=R} \frac{|dz|}{|z|^{n+1}} = \frac{M}{2\pi R^{n+1}} \int_{|z|=R} |dz| = \frac{M}{2\pi R^{n+1}} \cdot 2 \pi R = \frac{M}{R^n}$$ Letting $R \to \infty$ results $a_n = 0$ for all $n > 0$, and using this in the series expansion above leaves us with $f(z) = a_0$, our desired conclusion. $\blacksquare$

Let $f$ be a complex entire function satisfying $f(z) = f(z + w_0) = f(z + w_1)$ for all $z \in \Bbb C$ for $\Bbb R$-linearly independent elements $w_0$, $w_1$. This implies that $f(z)$ is defined everywhere on the period parallelogram $\Lambda$ with vertices on $0, w_0, w_1$ and $w_0 + w_1$ in the complex plane. But $f$ is continuous (as it's analytic) and $\Lambda$ is a compact set in $\Bbb C$, so $f(\Lambda)$ is also compact, i.e., bounded. Extending by periodicity, $f(\Bbb C)$ is also bounded, which implies $f$ is identically constant all over $\Bbb C$ by the lemma above. Thus every doubly periodic function on $\Bbb C$ is identically constant.

Addendum (contributes nothing whatsoever to the solution of the original problem) : It should be noted that such function do exist when one lifts the condition "entire on $\Bbb C$", or rather, replaces it by "entire on $\Bbb P^1$". In that case, the function

$$\wp(z) = \frac1{z^2} + \sum_{(m, n) \in \Bbb Z^2 \setminus (0, 0)} \left [ \frac1{(z + mw_1 + n w_2)^2} - \frac1{(mw_1 + nw_2)^2} \right ]$$

is doubly periodic with periods $w_1$ and $w_2$ all over it's domain, and has two poles sitting at the vertices of the period parallelogram. In general, these functions are called elliptic functions and they arise in a major branch of mathematics focused on studying elliptic curves and modular forms.

An alternative approach would be the following:

Spoiler

If the function is analytic, it is continuous. Functions continuous on compact domains such as the parallelogram in question are bounded. But you can tile the entire complex plane with those parallelograms. Hence, the function is bounded everywhere. By Liouville's Theorem, such functions must be constant.