Definition: Let $p$ be a seminorm on some space $X$. A sequence $(x_n)$ in $X$ is said to be Cauchy with respect to $p$ if for any $\epsilon > 0$, there is an $N > 0$ such that $p(x_m - x_n) < \epsilon$ for all $m,n>N$.
Problem: Let $V$ be an open subset of $\mathbb{C}^n$ and let $\mathcal{A}(V)$ be the set of all analytic functions on $V$. For any compact subset $K$ of $V$, define the seminorm $p_K$ omn $\mathcal{A}(V)$ by $$p_k(f) = \max_{x \in K} {|f(x)|}.$$
Suppose $(f_n)$ is a sequence in $\mathcal{A}(V)$ that is Cauchy with respect to all $p_K$. Show that there is an $f \in \mathcal{A}(V)$ such that for all $K$, $p_K(f_n -f) \to 0$ as $n \to \infty$.
My Attempt: For each $K$, denote by $E(K)$ the space of all functions continuous on $K$ and holomorphic in its interior. Then $p_k$ is in fact a norm on $E(K)$ and $E(K)$ is a Banach space under this norm. Clearly, $\mathcal{A}(V)$ is a subspace of $E(K)$ for any $K$.
If $(f_n)$ is Cauchy with respect to each $p_K$, then for each $K$, there is a $g_K \in E(K)$ such that $f_n \to g_K$ in $E(K)$.
... I'm stuck at this step. I think I should be able to use the $g_K$'s to explicitly define the limit $f$ but I can't seem to think of a way to do that. Any help would be appreciated. Thanks!
You must know the following result: Uniform Limit of Analytic Functions is Analytic.
Now, let $(f_n)$ be Cauchy with respect to each $p_K$. Since $C(K)=\left\{f:K\rightarrow\mathbb{C}:f\text{ is continuous}\right\}$ is complete with the supremum norm (which is simply $p_K$) and $f_n|_K\in C(K)$, there exists $g_K\in C(K)$ such that $f_n|_K\rightarrow g_K$ uniformly (in $K$). Since each $f_n$ is analytic in $K$, so is $g_K$.
It's easy to show that, given compacts $K$ and $L$, we have $g_K|_{K\cap L}=g_L|_{K\cap L}$.
Define $g:V\rightarrow\mathbb{C}$ by $g(x)=g_{\left\{x\right\}}(x)$. Then $g|_K=g_K$ for every compact $K$, so $g$ is analytic when restricted to compact sets. Since analyticity is a local property and $V$ is locally compact, then $g$ is analytic in $V$.