Theorem: Let $f:D^*(z_0,r)=D(z_0,r)-\{z_0\}\to \Bbb C$ be holomorphic and bounded. Then $\lim_{z\to z_0}f(z)$ exists and the function $\hat{f}:D(z_0,r)\to \Bbb C$ defined by $$\hat{f}(z) = \begin{cases} f(z) & \text{if }z\in D^*(z_0,r) \\ \lim_{z\to z_0}f(z) & \text{if }z=z_0 \end{cases}$$, is holomorphic on $D(z_0,r)$.
Proof:
Define $g:D(z_0,r)\to \Bbb C$ by $$g(z) = \begin{cases} (z-z_0)^2f(z) & \text{if }z\in D^*(z_0,r) \\ 0 & \text{if } z=z_0 \end{cases}$$
After proving that $g\in H(D(z_0,r))$, the author says that since $f$ is bounded by some positive number $M$, $\lvert g\rvert\le M\lvert z-z_0\rvert^2$ on $D(z_0,r)$, it follows that the power series expansion of $g$ at $z_0$ has the form $$g(z)= \sum_{n=2}^\infty \text{$a_n$}(z-z_0)^n \;\;\forall z\in D(z_0,r)$$
But I have no idea about the relation between $g$ is bounded and $g$ has a power series expansion. (The power series starts from 2 because $g(z_0)=0$ and $\frac{\partial g}{\partial z}(z_0)=0 $)
I know that $g$ is holomorphic on $D(z_0,r)$ and the radius of convergence of the power series is at least $r$. I think $g$ has power series expansion at $z_0$ because $g$ is holomorphic on $D(z_0,r)$. Do I miss something?
Thanks for your helping!
Yes, since $g$ is holomorphic in a neighbourhood of $z_0$ it has a Taylor series about $z_0$, and that series converges to $g(z)$ in a neighbourhood of $z_0$.
Let that series be $a_0 + a_1 (z-z_0) + a_2 (z-z_0)^2 + \ldots$. We then have $a_0 = \lim_{z \to 0} g(z)$. But we know that limit is $0$ because $|g(z)| \le M |z - z_0|^2$, so $a_0 = 0$. Next, $$g'(z_0) = \lim_{z \to z_0} \frac{g(z) - g(z_0)}{z - z_0} = \lim_{z \to z_0} \frac{g(z)}{z - z_0} = 0$$ because $$\left| \frac{g(z) }{z - z_0} \right| \le M |z - z_0|$$ and since the series of $g'(z)$ is the term-by-term derivative of the series of $g$, this says $a_1 = g'(z_0) = 0$.