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.
