Riemann Removable Singularities Theorem

Theorem
Let $U \subset \C$ be a domain.

Let $z_0 \in U$.

Let $f: U \setminus \set {z_0} \to \C$ 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 deleted neighborhood of $z_0$.
 * $(4): \quad \map f z = \map o {\dfrac 1 {\cmod {z - z_0} } }$ as $z \to z_0$.

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

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.

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 Cauchy's Residue Theorem.

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

Now assume that $(4)$ holds.

That is:
 * $z \cdot \map f z \to 0$ as $z \to 0$.

We need to show that $(2)$ holds.

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

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

By applying the direction $(2) \implies (1)$ to this function, we see that $g$ is holomorphic in $\Bbb D$.

We have:
 * $\ds \map {g'} 0 = \lim_{z \mathop \to 0} \frac {\map g z} z = \lim_{z \mathop \to 0} \map f z$

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

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

By Morera's Theorem, we need to show that:
 * $\ds \int_C \map f z \rd z = 0$

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

It follows from the Cauchy Integral Theorem that we only need to check that:
 * $\ds \int_C \map f z \rd z = 0$

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

Letting $C = \map {C_\epsilon} 0$ be the circle of radius $\epsilon$ around $0$, we see that:
 * $\ds \size {\int_{C_\epsilon} \map f z \rd z} \le 2 \pi \epsilon \max_{z \mathop \in C_\epsilon} \size {\map f z} \to 0$

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

This completes the proof.