# A nonconstant doubly periodic function cannot be analytic

I have been stuck on this problem for a long time:

A non constant $F(z)$ is such that $F(z+a)=F(z)$ and $F(z+bi)=F(z)$ where $a>0$ and $b>0$ are given constants. Prove that $F(z)$ cannot be analytic in the rectangle $\{(x,y) : 0<x<a , 0<y<b \}$.

The problem is under the section of Liouville's theorem: a bounded entire functions is constant. But how to apply this to a rectangle?

• I mean in the rectangle 0<x<a , 0<y<b . the problem oocurs under the section of Liouville theorem ... But how can that theorem be applied?? – herashefat Sep 18 '15 at 6:54

The problem as stated is false. As a counterexample, consider the function $$F(z)=\begin{cases} 1, & z=an + bmi\ \text{for}\ n,m\in \mathbb Z\\ 0, & else. \end{cases}$$ It is analytic in the rectangle $(0,a)\times (0,b)$, because it is identically $0$ there. Also it is periodic as required.

The function can be extended on the whole plane as follows: $$G(z+na+imb) = F(z) \qquad n,m\in \mathbb Z$$ to get an analytic bounded function.

added. The resulting function is bounded because it is periodic, hence its range $G(\mathbb C) = G(R) = F(R)$ where $R=[0,b]\times [0,b]$ is the given rectangle, and being $R$ compact and $F$ continuos, $F(G)$ is compact hence bounded. The resulting function is analytic because an analytic function in a point is uniquely determined by the values on a sequence of approximating points. Hence if $F$ is analytic and $F(z)=F(z+a)$ for some point $z$ in the left side of the rectangle, the extension in $F(z+a)$ coincides with the extension inn $F(z)$.

By Liouville Theorem the extended function is bounded and so must be the original function...

note. I assume that the function is analytic on the closed rectangle, which means that there is an open set containing the rectangle where the function has an analytic extension.

• In the problem statement, it is not given that $F$ is analytic on the boundary of the rectangle. Also we don't know it is continuous up to the boundary, so there isn't a way to extend. See my counterexample. – pre-kidney Sep 18 '15 at 7:07
• Hi there were also equal signs with x and y so x greater or equal to zero , less than or equal to a , same with y ....are you saying the problem in the text is wrong?? – herashefat Sep 18 '15 at 7:16
• @Emanuele Paolini how does this prove the function is NOT analytic on the CLOSED rectangle?? – herashefat Sep 18 '15 at 7:19
• @herashefat maybe the text is correct but you should put all the details in your question. My answer proves that the function is constant which is a contradiction with the hypothesis. – Emanuele Paolini Sep 18 '15 at 7:22
• @EmanuelePaolini thanks for your response Some questions 1.Why can we write F(z+na+imb )=F(z) ?? 2. why must this be bounded and analytic?? – herashefat Sep 18 '15 at 7:36