Riemann Removable Singularities Theorem

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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 neighborhood of $z_0$.
$(4): \quad \map f z = \map o {\dfrac 1 {\cmod {z - z_0} } }$ as $z \to z_0$.


Proof

Without loss of generality, 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.

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:

$z \cdot \map f z \to 0$ as $z \to 0$.

We need to show that (b) 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 (b) $\implies$ (a) 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 the 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.

$\blacksquare$


Source of Name

This entry was named for Georg Friedrich Bernhard Riemann.