Riemann's removable singularity theorem

hbghlyj

complex.pdf page78
Suppose that $U$ is an open subset of $\Bbb C$ and $z_0 ∈ U$. If $f : U \setminus\{z_0\} →\Bbb C$ is holomorphic and bounded near $z_0$, then $f$ extends to a holomorphic function on all of $U$.

Proof. Define $h(z)$ by
\[
h(z)=\left\{\begin{array}{cc}
\left(z-z_0\right)^2 f(z), & z \neq 0 ; \\
0, & z=z_0
\end{array}\right.
\]
Then clearly $h(z)$ is holomorphic on $U \setminus\left\{z_0\right\}$, using the fact that $f$ is and standard rules for complex differentiablility. On the other hand, at $z=z_0$ we see directly that
\[
\frac{h(z)-h\left(z_0\right)}{z-z_0}=\left(z-z_0\right) f(z) \rightarrow 0
\]
as $z \rightarrow z_0$ since $f$ is bounded near $z_0$ by assumption. It follows that $h$ is in fact holomorphic everywhere in $U$. But then if we chose $r>0$ is such that $\bar{B}\left(z_0, r\right) \subset U$, then by Corollary 7.11 $h(z)$ is equal to its Taylor series centred at $z_0$, thus
\[
h(z)=\sum_{k=0}^{\infty} a_k\left(z-z_0\right)^k .
\]
But since we have $h\left(z_0\right)=h^{\prime}\left(z_0\right)=0$ we see $a_0=a_1=0$, and hence $\sum_{k=0}^{\infty} a_{k+2}\left(z-z_0\right)^k$ defines a holomorphic function in $B\left(z_0, r\right)$. Since this clearly agrees with $f(z)$ on $B\left(z_0, r\right) \setminus\{0\}$, we see that by redefining $f\left(z_0\right)=a_2$, we can extend $f$ to a holomorphic function on all of $U$ as required.

hbghlyj

This doesn't hold for real functions: $\sin\frac1x$ is differentiable on $\Bbb R\setminus\{0\}$ and bounded near 0 but cannot extend to a differentiable function over $\Bbb R$
Following the proof, define $h(z)=\begin{cases}x^2\sin\frac1x&x≠0\\0&x=0\end{cases}$
$h(0)=h'(0)=0$ but $h''(0)$ doesn't exist.