Given $f$ is holomorphic on the domain $ U := \mathbb{C} \backslash \{0\}$ and that $|f(z)| \leq |z|^{1/2}$ for all $z \in U$.
How does one prove $f$ is zero everywhere?
Given $f$ is holomorphic on the domain $ U := \mathbb{C} \backslash \{0\}$ and that $|f(z)| \leq |z|^{1/2}$ for all $z \in U$.
How does one prove $f$ is zero everywhere?
We have
$|z| \cdot |f(z)| \leq |z|^{\frac{3}{2}} \to 0 \qquad (z \to 0)$
By applying Riemann's theorem we conclude that $f$ is holomorphically extandable over 0, hence $f$ is an entire function, and $f(0)=0$. There exists a sequence $(a_n)_{n \geq 0}$ such that
$f(z) = \sum_{n \geq 0} a_n \cdot z^n \tag{1}$
where $a_0=0$ since $f(0)=0$. Since
$|a_n| \leq \frac{1}{r^n} \cdot \max \{|f(\xi)|; |\xi|=r\}$
for all $r>0$ (a consequence of Cauchy's differentiation formula) we obtain
$|a_n| \leq r^{\frac{1}{2}-n} \to 0 \qquad (r \to \infty)$
for all $n \geq 1$, thus $a_n=0$ for all $n \geq 1$. From (1) we conclude $f(z)=0$.