Riemann Removable Singularities Theorem

Riemann Removable Singularities 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:


 * (a) $$f$$ extends to a holomorphic function $$f:U\to\C$$.
 * (b) $$f$$ extends to a continuous function $$f:U\to\C$$.
 * (c) $$f$$ is bounded in a neighborhood of $$z_0$$.
 * (d) $$f(z) = o(1/|z-z_0|)$$ as $$z\to z_0$$.

Proof
Without loss of generality, we may assume that $$U = \mathbb{D} := \{z \in \C: |z| < 1\}$$ and that $$z_0=0$$.

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

Clearly (a) $$\Rightarrow$$ (b) $$\Rightarrow$$ (c) $$\Rightarrow$$ (d).

That (b) implies (a) follows from the proof of the Cauchy Integral Theorem. (For completeness, we sketch a self-contained argument below.)

Now assume that (d) 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 $$\mathbb{D}$$ and holomorphic in $$\mathbb{D}\setminus\{0\}$$.

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

so $$f$$ extends continuously to $$\mathbb{D}$$, as claimed.

To prove (b) $$\Rightarrow$$ (a), we use Morera's Theorem and the Cauchy Integral Theorem.

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

It follows from the Cauchy integral theorem that we only need to check that $$\int_C f(z) = 0$$ 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
 * $$\left|\int_{C_{\varepsilon}} f(z)dz\right| \leq 2\pi \varepsilon \max_{z\in C_{\varepsilon}} |f(z)| \to 0$$

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

This completes the proof.