Show that there exists a holomorphic function $g:D\to \Bbb C$ such that $f=g$ on $D\setminus K$. Let $D=\{z\in \Bbb C: |z|<1\}$ and $K\subset D$ be a compact subset. Suppose that $f:D\setminus K \to \Bbb C$ is holomorphic and that there is a sequence of polynomials $\{p_n\}_{n=1}^\infty$ such that $p_n \to f$ as $n\to\infty$ on compact subsets of $D\setminus K$.
Show that there exists a holomorphic function $g:D\to \Bbb C$ such that $f=g$ on $D\setminus K$.
Can I use the Laurent series expansion here? Let the limit of the coefficient of $p_n$ be the coefficient of our Laurent series expansion. Then is this the $g$ we need?
However, I do not know how the uniformly convergence is used here. Could anyone kindly help? Thanks!
 A: Consider $0<r<R<1$ such that $\{ z: \; r<|z|<R\}=: A(r,R)$ is contained in $D \backslash K$. Let $\sum_{-\infty}^{\infty} a_k z^k$ denote the Laurent series expansion of $f$ on $A(r,R)$.
Let $\gamma$ be the boundary of the disk of radius $R$ around $0$. Then for $k \in \Bbb N$, we have (by construction of the Laurent series) that $$a_{-k} = \frac{1}{2\pi i}\int_{\gamma} f(z)z^{k-1}dz$$
Now fix some $\epsilon>0$ and fix some $N \in \Bbb N$ such that $\sup_{z \in \gamma} |f(z)-p_N(z)|<\epsilon$. We know that since $p_N$ is analytic on $D$ it follows that $\int_{\gamma} p_N(z)z^{k-1}dz = 0$. Therefore for $k \in \Bbb N$ $$|a_{-k}| = \frac{1}{2\pi}\bigg| \int_{\gamma} f(z)z^{k-1}dz - \int_{\gamma} p_N(z)z^{k-1}dz \bigg| \leq \frac{1}{2\pi}\int_{\gamma} \big| f(z)-p_N(z) \big| \cdot |z|^{k-1} d|z|< \epsilon R^{k-1}$$ Letting $\epsilon \to 0$, we see that $a_{-k}=0$ for all $k \in \Bbb N$, and therefore $f(z) = \sum_{k=0}^{\infty} a_kz^k$ for $z \in A(r,R)$. Hence $f$ can be analytically extended to the entire disk of radius $R$. Letting $R \to 1$, we see $f$ can be extended to the entire disk.
