Riemann Removable Singularities Theorem

Theorem
Let $U \subset \C$ be a domain, let $z_0 \in U$, and let $f: U \setminus \{z_0\} \to \C$ 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 $z_0$.
 * $(4): \quad f(z) = o(1/|z-z_0|)$ as $z\to z_0$.

Proof
Without loss of generality, we may assume that $U = \Bbb D := \{z \in \C: |z| < 1\}$ 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 Cauchy Integral Theorem. (For completeness, we sketch a self-contained argument below.)

Now assume that $(4)$ holds; i.e. $z\cdot f(z) \to 0$ as $z\to 0$. We need to show that (b) holds.

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

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

By applying the direction (b) $\implies$ (a) to this function, we see that $g$ is holomorphic in $\Bbb D$. We have :
 * $\displaystyle g'(0) = \lim_{z\to 0} \frac{g(z)} z = \lim_{z\to 0} f(z)$

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

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

By Morera's theorem, we need to show that $\displaystyle \int_C f(z)dz = 0$ for every closed curve $C$ in $\Bbb D$.

It follows from the Cauchy integral theorem that we only need to check that $\displaystyle \int_C f(z) = 0$ for a simple closed loop surrounding $0$, and that this integral is independent of the loop $C$.

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

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

This completes the proof.