TITLE: $f$ entire, $f$ satisfies $|f(x+iy)|\leq\frac{1}{|y|}$ for all $x,y\in\mathbb{R}$. Prove that $f\equiv 0$.
QUESTION [18 upvotes]: Let $f$ be an entire function. Suppose that $f$ satisfies
  $$
|f(x+iy)|\leq\frac{1}{|y|}.
$$
  for all $x,y\in\mathbb{R}$. Prove that $f$ is identically zero.

I'm having some trouble with this, but I'm probably just overthinking it. The first instinct as usual is to try and make something happen with Liouville's theorem (and then since $f$ tends to zero on the imaginary axis we can conclude that $f\equiv 0$), but since $f$ is potentially unbounded on the real axis it's inapplicable. I'm pretty sure it's still impossible for $|f|\to\infty$ as $x\to\pm\infty$ while the function is bounded everywhere else, as it seems there would be some topological obstruction, but I could be wrong. I've also thought of getting an estimate on the $|f'|$ as in the proof of Liouville's theorem, but it fails for the same reasons. Any hints or tips are greatly appreciated. Thank you

REPLY [4 votes]: Here's another approach: Note that $z-\bar z=2iy.$ Thus
$$\tag 1 |(z-\bar z)f(z)| \le 2$$
for all $z.$ Let's write $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Setting $z=re^{it},$ we get
$$\tag 2  (z-\bar z)f(z)  = (re^{it} - re^{-it})(\sum_{n=0}^{\infty}a_nr^ne^{int}).$$
Play with $(2)$ a little to see it equals
$$\tag 3  \sum_{n=0}^{\infty}(a_nr^{n+1} - a_{n+2}r^{n+3})e^{i(n+1)t} -a_0re^{-it} - a_1r^2.$$
Let's now integrate the modulus squared of $(3)$ over $[0,2\pi]$using the orthogonality of the exponentials. We get, recalling $(1),$
$$\sum_{n=0}^{\infty}|a_nr^{n+1} - a_{n+2}r^{n+3}|^2 +|a_0r|^2 +|a_1r^2|^2 \le 4.$$
This inequality holds for all $r>0.$ Verify that this holds iff $a_n=0$ for all $n,$ i.e., iff $f\equiv 0.$