I am trying to proof the following statements:
Let $f \colon \{ z \in \mathbb{C} : 0 < \vert z \vert < r_0 \}$ be a holomorphic function. For $0<r<r_0$ we define $M_r(f):= \operatorname{max} \{ \vert f(z) \vert : \vert z \vert = r\}$. The following statements hold true:
i) $0$ is removable iff $M_r(f)$ is bounded for $r \rightarrow 0$.
ii) $0$ is a Pole iff $M_r(f) \rightarrow \infty$ and $\exists l \in \mathbb{N}: M_r(z^lf)$ is bounded for $r \rightarrow 0$.
iii) $0$ is an essential singularity iff $\forall l \in \mathbb{N}: M_r(z^lf)\rightarrow \infty $
I have managed to prove i) and ii). Also, because of that the direction "<=" is clear to me for iii), but I am struggling with the other one.
I would be very thankful for help.
Best Answer
The $\Leftarrow$ implication relies on the following observation. Denote $M_r = M_r(z^l f)$, and take any sequence $\{ r_i \}$ of decreasing radii $r_i \to 0$. Then $\limsup_{i \to \infty} M_{r_i} = \limsup_{r \to 0} M_r$.
First let's see how the observation implies the $\Leftarrow$ direction. By contrapostion we need to prove that if for some $l$ we've got $M_r(z^l f) \nrightarrow \infty$, then the singularity is either a pole or removable. Since $M_r(z^l f) \nrightarrow \infty$, we can choose a decreasing sequence of radii $r_i \to 0$ such that $M_{r_i}(z^l f)$ converges to a finite limit. The observation then implies, that $\limsup_{r \to 0} M_r(z^l f)$ is equal to the same limit, hence $M_r$ is bounded, hence we find ourselves in one of the conditions i) or ii). $\square$
The proof of the observation looks as follows.
Take $0 < r_1 < r_2 < r_3$ and consider open annulus $D = \{z : r_1 < |z| < r_3 \}$. The maximum modulus principle implies that for every closed annuli $\overline{D}$ the maximum modulus lies on the set of boundary circles. Hence the $(\star)$ estimate $M_{r_2} \leq \max \left( M_{r_1}, M_{r_3} \right)$.
To prove the observation we need to show that for any decreasing sequence $\{ r_i \}$ of real numbers converging to $0$ the equality $\limsup_{i \to \infty} M_{r_i} = \limsup_{r \to 0} M_r$ holds. Inequality $$\limsup_{i \to \infty} M_{r_i} \leq \limsup_{r \to 0} M_r$$ is trivial (and does not require any properties of $M_r$), hence to prove the equality we need to prove the $\geq$ inequality. Take a decreasing sequence $\{ k_i \}$ of numbers converging to $0$. We need to show that $$\limsup_{i \to \infty} M_{r_i} \geq \limsup_{i \to \infty} M_{k_i}. $$
For every $i \in \mathbb{N}$ there exists $N(i) \in \mathbb{N}$ such that $r_{N(i)+1} \leq k_i \leq r_{N(i)}$. Hence the $(\star)$ estimate, gives $$M_{k_i} \leq \max \left( M_{r_{N(i)+1}}, M_{r_{N(i)}} \right). $$ Let $P(i)$ be either of numbers $\{ N(i), N(i)+1 \}$ such that $$ M_{r_{P(i)}} = \max \left( M_{r_{N(i)}}, M_{r_{N(i)+1}} \right). $$ Then for every $i$ the following inequality holds $$ M_{r_{P(i)}} \geq M_{k_i}. $$ Hence $$ \limsup_{i \to \infty} M_{k_i} \leq \limsup_{i \to \infty} M_{r_{P(i)}} \leq \limsup_{i \to \infty} M_{r_i}.$$
$\square$