Problem: Describe the class of all holomorphic functions on $\mathbb{C}-\{0\}$ such that

$$\sup_{(x,y)\neq (0,0)}\frac{|f(x+iy)|}{|\log(x^2+y^2)|}<\infty.$$

Attempt at a solution:

Let $z=x+iy$, then we have:

$$\frac{|f(z)|}{|\log|z|^2|}\leq c$$. So,

$|f(z)|\leq c|\log|z|^2|.$

For large enough $z$, we have $|\log|z|^2|\leq |z|^2$ so we get:

$$|f(z)|\leq c|z|^2.$$ Now by extented Liouville's Theorem, $f(z)$ must reduce to a polynomial of degree at most two.

Is this correct?


  • $\begingroup$ Why stop at $|z|^2$? And even if you stop at $|z|$, you still might not get the right answer ;) $\endgroup$ – Evan Aug 14 '13 at 23:43
  • 2
    $\begingroup$ Do polynomials of degree at most two satisfy your constraint? $\endgroup$ – Patrick Da Silva Aug 14 '13 at 23:43
  • $\begingroup$ And Liouville's theorem applies on functions that are holomorphic over the whole complex plane, not over functions who are missing a point of holomorphicness. (The proof relies on a Taylor expansion around $0$ and then using Cauchy's theorem, so you expect the pole at $0$ to play a role here...) :P $\endgroup$ – Patrick Da Silva Aug 14 '13 at 23:45

We also need to consider the fact that $f$ might have a singularity at $z=0$. However, we can handle that because $$ \begin{align} \lim_{z\to0}|zf(z)| &\le\lim_{z\to0}|z|2c\log(|z|)\\ &=0 \end{align} $$ Riemann's Theorem says that the singularity at $z=0$ is removable. Now, using Cauchy's Integral Formula we get that $$ f^{(n)}(z)=\frac{n!}{2\pi i}\oint\frac{f(w)\,\mathrm{d}w}{(w-z)^{n+1}} $$ where the integral is over a counter-clockwise circle of radius $R$. This means that $$ \left|\,f^{(n)}(z)\,\right|\le\frac{n!}{R^n}2c\log(R) $$ Just as with the proof of Liouville's Theorem, we show that $f^{(n)}(z)=0$ and $f$ must be constant, and since $f(z)=0$ when $|z|=1$, we know that $f(z)=0$.

Note added

I mentioned above that your conditions imply that $f(z)=0$ when $|z|=1$. This in itself implies that $f(z)=0$ everywhere since $\{z:f(z)=0\}$ has an accumulation point (see this section).

  • $\begingroup$ I already showed that $f$ had a removable singularity, but I should have included it in my attempt I guess. Thanks for the solution, i like it! $\endgroup$ – V-B Aug 14 '13 at 23:59
  • $\begingroup$ In this case as well, it suffices to take $n=1$. $\endgroup$ – V-B Aug 15 '13 at 0:01
  • $\begingroup$ @V-B: Yes, it does, but I got carried away :-) $\endgroup$ – robjohn Aug 15 '13 at 0:02
  • $\begingroup$ @V-B: I apologize for the accusatory tone of my first sentence. It has been toned down. $\endgroup$ – robjohn Aug 15 '13 at 0:06
  • $\begingroup$ Don't worry about the accusatory tone :) To your note: That is actually fantastic! I can't believe I missed that. That follows directly from the Identity Theorem (or the fact that holomorphic functions, which do not vanish identically, have isolated zeros). $\endgroup$ – V-B Aug 15 '13 at 0:21

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