Complex Analysis – Proving a Doubly-Periodic Entire Function is Constant

analyticitycomplex-analysisentire-functionsholomorphic-functions

Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.$$Prove that $f$ is constant.

The intuition seems clear to me, we have the three vertices of a triangle given by $0$, $\omega_1$ and $\omega_2$. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be $f(0)$, fine. Throwing values for $z$ there, I have found that $$f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.$$

I don't know how to improve the above for, say, rationals (at least). Some another ideas would be:

  • Checking that $f' \equiv 0$. I don't have a clue of how to do that.

  • Write $w = a\omega_1+b\omega_2$, with $a,b \in \Bbb R$, do stuff and conclude that $f(w) = f(0)$. This approach doesn't seem good, because I only have a weak result with integers above.

  • Finding that $f$ coincides with $f(0)$ on a set with an accumulation point. This seems also bad: the set on with $f$ coincides with $f(0)$ by which I found above is discrete.

Nothing works and this is getting annoying… And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea.

(On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.)

Best Answer

The values the function take on the plane are the values it takes in the compact parallelogram with vertices on $0,\omega_1,\omega_2,\omega_1+\omega_2$. Therefore the entire function is bounded, and hence constant by Liouville's theorem.


Every point $z$ of the plane can be written as $z=x\omega_1+y\omega_2$ with $x,y$ reals, since $\omega_1,\omega_2$ are independent over the reals. Then $$f(z)=f(\{x\}\omega_1+\{y\}\omega_2),$$ where $0\leq\{x\}<1$ is such that $x-\{x\}$ is integer. The point $\{x\}\omega_1+\{y\}\omega_2$ is inside the compact parallelogram with vertices $0,\omega_1,\omega_2,\omega_1+\omega_2$.

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