Riemann Removable Singularities Theorem

Theorem
Let $U \subset \C$ be a domain, let $z_0 \in U$, and let $f: U \setminus \left\{ {z_0}\right\} \to \C$ be holomorphic.

Then the following are equivalent:


 * $(1): \quad f$ extends to a holomorphic function $f: U \to \C$.
 * $(2): \quad f$ extends to a continuous function $f: U \to \C$.
 * $(3): \quad f$ is bounded in a neighborhood of $z_0$.
 * $(4): \quad f \left({z}\right) = o \left({\dfrac 1 {\left|{z - z_0}\right|} }\right)$ as $z \to z_0$.

Proof
, we may assume that:
 * $U = \Bbb D := \left\{{z \in \C: \left|{z}\right| < 1}\right\}$

and that $z_0 = 0$.

Otherwise, restrict $f$ to a suitable disk centered at $z_0$, and precompose with a suitable affine map.

A holomorphic function is continuous.

A continuous function is locally bounded.

So $(1) \implies (2) \implies (3)$.

Also, $(3) \implies (4)$ by definition.

That $(2) \implies (1)$ follows from the proof of the Residue Theorem.

(For completeness, we sketch a self-contained argument below.)

Now assume that $(4)$ holds.

That is:
 * $z \cdot f \left({z}\right) \to 0$ as $z \to 0$.

We need to show that (b) holds.

By assumption, the function:
 * $g \left({z}\right) := \begin{cases} z \cdot f \left({z}\right) & : z \ne 0 \\ 0 & : \text{otherwise} \end{cases}$

is continuous in $\Bbb D$ and holomorphic in $\Bbb D \setminus \left\{ {0}\right\}$.

By applying the direction (b) $\implies$ (a) to this function, we see that $g$ is holomorphic in $\Bbb D$.

We have:
 * $\displaystyle g' \left({0}\right) = \lim_{z \to 0} \frac {g \left({z}\right)} z = \lim_{z \to 0} f \left({z}\right)$

so $f$ extends continuously to $\Bbb D$, as claimed.

To prove $(2) \implies (1)$, we use Morera's Theorem and the Residue Theorem.

By Morera's Theorem, we need to show that:
 * $\displaystyle \int_C f \left({z}\right) \, \mathrm d z = 0$

for every closed curve $C$ in $\Bbb D$.

It follows from the Cauchy integral theorem that we only need to check that:
 * $\displaystyle \int_C f \left({z}\right) \, \mathrm d z = 0$

for a simple closed loop surrounding $0$, and that this integral is independent of the loop $C$.

Letting $C = C_\epsilon \left({0}\right)$ be the circle of radius $\epsilon$ around $0$, we see that:
 * $\displaystyle \left|{\int_{C_\epsilon} f \left({z}\right) \, \mathrm d z}\right| \le 2 \pi \epsilon \max_{z \in C_\epsilon} \left|{f \left({z}\right)}\right| \to 0$

as $\epsilon\to 0$ (because $f$ is continuous in $0$ by assumption).

This completes the proof.