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:
- $(z-\bar{z})^2 f(z)$ is bounded.
- (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!