I am working on a problem stating as below:
Consider a holomorphic function $f$ defined on the puncture disc $D(0,1)\setminus\{0\}$. Show that $0$ is a removable singularity of $f$ if $f$ is square integrable.
This question is similar to the post here: Singularities in the punctured unit disc and square integrability
In fact, I've solved it following the idea in above post.
Below is my proof:
We can write $f(z)$ as Laurent Expansion around $z_{0}=0$, such that $$f(z)=\sum_{n=-\infty}^{\infty}a_{n}z^{n}.$$
Then, we have $$f(re^{i\theta})=\sum_{n=-\infty}^{\infty}a_{n}r^{n}e^{in\theta},\ \overline{f(re^{i\theta})}=\sum_{n=-\infty}^{\infty}\overline{a_{n}}r^{n}e^{-in\theta}.$$
Note that for the integral $$\int_{0}^{2\pi}e^{in\theta}e^{im\theta}d\theta,$$ if $n=-m$, then the above integral is $2\pi$, but if $n\neq -m$, then the above integral is a complex integral of a holomorphic function along a circle and thus by Cauchy's Theorem, the above integral is $0$.
Now, with this in mind, we have
\begin{align*}
\int_{0}^{2\pi}|f(re^{i\theta}|^{2}d\theta&=\int_{0}^{2\pi}\Big(\sum_{n=-\infty}^{\infty}a_{n}r^{n}e^{in\theta}\Big)\Big(\sum_{n=-\infty}^{\infty}\overline{a_{n}}r^{n}e^{-in\theta}\Big)d\theta \\
&=2\pi\sum_{n=-\infty}^{\infty}|a_{n}|^{2}r^{2n}.\\
\end{align*}
On the other hand, since $\|f\|_{L_{2}}<\infty$, for any disc $D_{z_{0}}(R)$ centered at $z_{0}=0$ with radius $R$, we have
\begin{align*}
\infty>\int_{D}|f(z)|^{2}dz&=\int_{0}^{R}\int_{0}^{2\pi}|f(re^{i\theta})|^{2}4d\theta dr\\
&=2\pi\int_{0}^{R}\sum_{n=-\infty}^{\infty}|a_{n}|^{2}r^{2n+1}dr\\
&=2\pi\sum_{n=-\infty}^{\infty}|a_{n}|^{2}\int_{0}^{R}r^{2n+1}dr\\
\end{align*}
Now, for all $2n+1\geq 0$, $\int_{0}^{R}r^{2n+1}dr<\infty$, but for all $2n+1<0$, $\int_{0}^{R}r^{2n+1}dr=\infty$.
Thus, the only way to make the above inequality hold is that $2n+1\geq 0$, which means that $n\geq 0$ since $n\in\mathbb{Z}$.
This implies that in the Laurent series, $a_{n}=0$ for all $n\leq -1$. This implies that $z_{0}=0$ is a removable singularity.
However, this question is the part (c) of a problem, and I am wondering if there is another way to prove it, by using part (a) and (b).
Here is the part (a) and part (b):
(a) Show that $0$ is a removable singularity if $|f(z)|\leq C|z|^{-\alpha}$, with $\alpha<1$.
(b) Show that, for any holomorphic function $g$ on the disc of center $b$, radius $\epsilon$, we have $$|g(b)|\leq\dfrac{C}{\epsilon}\Big(\int_{D(b,\epsilon)}|g(x+iy)|^{2}dxdy\Big)^{1/2}.$$
I have proven those two parts and they both of a generalization in Stein Chapter 3 Exercise 13 and 20, respectively.
However, I have no idea about how to apply those two to part (c). Perhaps they are really not connected to each other.
Best Answer
You can use (b) and the following version of Riemann's theorem on removable singularities:
For $0 < r < 1/2$ and $|z|= r$ apply (b) to $f$ on the disk $D(z, r)$: $$ |f(z)| \le \frac {C}{r}\left(\int_{D(z,r)}|f(x+iy)|^{2}dxdy\right)^{1/2} \\ \implies |zf(z)| \le C \left(\int_{\dot D(0, 2r)}|f(x+iy)|^{2}dxdy\right)^{1/2} $$ where $\dot D(0, 2r) = D(0, 2r) \setminus \{ 0 \}$ shall denote the punctured disk.
The right-hand side converges to zero for $r \to 0$ (this follows e.g. from the Lebesgue theorem on dominated convergence), and then (a') implies that the singularity is removable.