Consider $f$ holomorphic on $B'(0,1) := B(0,1)\backslash\{0\}$ and $|f(z)|\le \ln(1/|z|),\forall z\in B'(0,1)$. Show that $f=0$ on $B'(0,1)$.

Question

Consider $f$ holomorphic on $B'(0,1) := B(0,1)\backslash\{0\}$ and $|f(z)|\le \ln(1/|z|),\forall z\in B'(0,1)$. Show that $f=0$ on $B'(0,1)$.

My attempt:

The function $f$ is holomorphic on $B'(0,1)$, thus $f(z)=\sum_{n\in \mathbb{Z}} a_nz^n, \forall z\in B'(0,1)$. Here,

$$ a_n=\frac1{2\pi i}\int_{\partial B_+(0,r)} \frac{f(z)}{z^{n+1}}dz,\quad \forall n\in\mathbb{Z}, 0<r<1.$$ I want to find an upper bound for these coefficients:

$$ |a_n|\le \frac1{2\pi}\int_{\partial B_+(0,r)} \frac{|f(z)|}{|z|^{n+1}}|dz|\le \frac1{2\pi}\int_{\partial B_+(0,r)} \frac{\ln(1/|z|)}{|z|^{n+1}}|dz|=\frac{\ln(1/r)}{r^n}.$$

Then, $|a_n|\to \ln(1)=0$ as $r\to 1-$. This shows that $|a_n|=0,\forall n\in \mathbb{N}$ and therefore $f=0$ on the punctured disk $B'(0,1)$.

Is this a good approach? Also, is this a common method to show that a function is identically $0$ on some set? Are there other appropriate and interesting strategies to tackle such problems?

Thanks.

Answer 1

Your approach is fine; only the ${\mathbb N}$ should be replaced by ${\mathbb Z}$. – Here is a hint for another approach:

Consider the function $$g(z):=z\>f(z)\qquad(0<|z|<1)\ ,$$ and argue about $g$ when $z\to0$ and $|z|\to1$.