Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thank you for reading!

share|improve this question
    
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

1 Answer 1

up vote 2 down vote accepted

When proving boundedness, I prefer to avoid arguments by contradiction so that the bound can be made explicit if needed later. In this case, I would proceed as follows.

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.