# Boundary behavior of a Periodic Holomorphic Function on the Upper Half Plane

This might be a really silly question, but here it is anyway.

Let $f$ be a holomorphic function with $f(z+1)=f(z)$ on the upper half plane satisfying the following:

1. $(z-\bar{z})^2 f(z)$ is bounded.
2. $(z-\bar{z})^2 f'(z)$ is bounded.

Are these two conditions enough to conclude that $(z-\bar{z}) f(z)$ is bounded?

Here's an argument for why I think this should be true.

Let $g(z)=(z-\bar{z})f(z)$.

$g$ is bounded on every $H_r$ for $r>0$, where $H_r = \{ (x,y) : y>r\}$, so the only place $g$ could blow up is near the boundary.

(EDIT: This is where I am using the periodicity. For $H_r$, where $r\ge 1$, I have that $|z-\bar{z}|^2|f(z)| \ge |g(z)|$. If $r<1$, then $H_r = H_1 \cup (-\infty,\infty)\times [r,1]$, but periodicity of $g$ allows us to conclude that $g$ is bounded on the bottom rectangle.)

Suppose that $g$ goes to infinity as $y\to 0$. Then, it must be true that $f\to \infty$ as $y\to 0$, using L'Hopital's Rule, we have

$\lim_{y\to 0} y f(x,y) = -\lim_{y\to 0} y^2 f_y(x,y) = (-i) \lim_{y\to 0} y^2 f'$

But since $(z-\bar{z})^2 f(z)$ is bounded, we have a contradiction.

Please help me see what I need to do to make this argument correct, or if you could see a different approach, or even a counterexample, or even a reference.

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Seems right to me. Why are you unsure? –  Sam Dec 7 '11 at 1:32
My use of L'Hopital's rule seems very naive. I was actually expecting a lot of comments explaining to me that I'm not using L'Hopital correctly :p –  Braindead Dec 7 '11 at 1:38
Where do you use periodicity, and what did you mean by that? Some version of $f(z+1) = f(z)?$ –  Will Jagy Dec 7 '11 at 2:46
Yes, I mean exactly $f(z+1)=f(z)$. Now that you mention it, it seems like I don't use it at all! (Actually, condition 1 is a result of periodicity of $f$ and some other thing. So maybe I can drop periodicity altogether in lieu of condition 1.) –  Braindead Dec 7 '11 at 2:58

For any $z=x+iy$ with $y>0$, $|x|\le 1/2$, write $f(z)=f(x+i)+\int_{x+i}^z f'(\zeta)\,d\zeta$. Property 2 says that $|f'(\zeta)|\le A(\mathrm{Im}\,\zeta)^{-2}$ for some constant $A$. Therefore, $$|f(z)|\le |f(x+i)|+A\int_{x+i}^z (\mathrm{Im}\,\zeta)^{-2}\,|d\zeta| \\ \le \sup_{|x|\le 1/2}|f(x+i)| + A+\frac{A}{y}$$ By periodicity, the estimate applies for all $z$. When $y\le 1$, it yields a uniform bound on $y|f(z)|$. For large $y$ apply Property 1 as you did.