Let $f(z)$ be an entire function such that $$|f(z)|<\frac{1}{|\text{Im}(z)|},\qquad z\in\Bbb C-\Bbb R.$$ The question asked me to prove that $f(z)=0$. At least looking at it, it really seems to have an application of Liouville's theorem lurking around somewhere, but I haven't found it.
My thoughts first led me to think about doing this by contradiction and using Picard's little theorem. So, I've considered a strip containing the real axis (say of width $2$ for simplicity). This will imply the complement maps into the unit disk, which implies that this strip has to map almost everywhere else on the complex plane. Now, on the imaginary axis, I know that $f(z)$ will vanish as $z\to\infty$, and this seems like it might be useful. I'm sort of under the impression $f(z)$ might have an essential singularity at $\infty$, which would mean $f(1/z)$ has an essential singularity at $0$. Either way, it of course can't have a pole at $\infty$ because of $f(z)$ vanishing on the imaginary axis, and if it is a removable singularity, it must be $0$, which still gives the solution. Therefore, I think it must have something to do with $f(z)$ having an essential singularity at $\infty$.
I was hoping to combine this with the above inequality and deduce a contradiction, but I haven't thought of one yet.
Any hints or suggestions? I'm studying for a prelim. exam, so this isn't homework.
Update: So, in the spirit of searching methods using the essential singularity at $z=0$, we have $$\left|f\left(\frac{1}{z}\right)\right|\leq\frac{1}{\left|\text{Im}\left(\frac{1}{z}\right)\right|}=\frac{|z|^2}{|y|}.$$ Thus, for $|y|>1$, we have $$\left|f\left(\frac{1}{z}\right)\right|<|y|\left|f\left(\frac{1}{z}\right)\right|<|z|^2.$$ Thus, if I can show that $|f(1/z)|$ is a polynomial, we'll know it can't have an essential singularity at the origin, concluding my proof.
Best Answer
Let $\sum_{n=0}^\infty a_n z^n$ be the Taylor expansion of $f(z)$. Also denote $a_{-1}=a_{-2}=0$. By definition, $$a_n=\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)}{z^{n+1}}dz,\quad\forall n\ge -2,\, \forall R>0.\tag{1}$$
Note that $\mathrm{Im} (z)=\frac{z-\bar{z}}{2i}$, so when $|z|=R$, $$\mathrm{Im} (z)=\frac{z-\frac{R^2}{z}}{2i}.\tag{2}$$ Substituting $(2)$ into $(1)$, we have $$a_{n-1}-R^2 a_{n+1}=\frac{1}{\pi}\int_{|z|=R}\frac{\mathrm{Im} (z)\cdot f(z)}{z^{n+1}}dz,\quad\forall n\ge -1,\, \forall R>0.\tag{3}$$ By the assumption of $f$ and continuity, $|\mathrm{Im} (z)\cdot f(z)|\le 1$ on $\mathbb{C}$. Then from $(3)$ we know $$|a_{n-1}-R^2a_{n+1}|\le \frac{2}{R^n},\quad\forall n\ge -1,\, \forall R>0.\tag{4}$$
Given $n\ge -1$, dividing both sides of $(4)$ by $R^2$ and letting $R\to \infty$, we have $a_{n+1}=0$. Therefore, all the coefficients of the Taylor expansion of $f$ are $0$, i.e. $f\equiv 0$.