## Qual-type problem

Problem: Let $f$ be entire and suppose that for all $z\in{\mathbb C}$:

$\displaystyle |f(z)|\leq \frac{1}{|{\mathcal Re} z|}.$

Show that $f(z)\equiv 0$.

Solution: For $k=0,1,2,...,$ let

$\displaystyle I(r,k)=\int_{|z|=r}\left(1+\frac{z^2}{|z|^2}\right)\frac{f(z)}{z^{k+1}}\frac{dz}{2\pi i}.$

On one hand, using the fact that

$\displaystyle \left|1+\frac{z^2}{|z|^2}\right|=\frac{2|{\mathcal Re} z|}{|z|},$

we get that $|I(r,k)|\rightarrow 0$ as $r\rightarrow\infty$, for each $k$.

And on the other hand, direct integration shows that $I(r,k)\rightarrow a_k$ where $f(z)=\sum_{n=0}^\infty a_n z^n$. $\Box$.

Questions: Is there a different proof, say, using the maximum principle? What if $f$ is simply harmonic?

Update: The same power series approach seems to give the harmonic case.

As an aside, I didn’t know before that $1/|{\mathcal Re} z|$ is actually subharmonic. In fact it seems that $\Delta |{\mathcal Re} z|^\alpha=\alpha(\alpha-1)|{\mathcal Re} z|^{\alpha-2}$, which is positive unless $0<\alpha<1$.
3. Oops I guess there are some integrability issues with $1/|{\mathcal Re} z|$….