$14.3.$ Lemma. Suppose $D'\Subset D$ are open subsets of $\mathbb{C}$. Then given any $\epsilon >0$, there exists a closed vector subspace $A\subset L^2(D,\mathcal{O})$ of finite codimension such that $||f||_{L^2(D')}\leq \epsilon ||f||_{L^2(D)}$ for every $f\in A$
After this Forster defines Square Integrable Cochains. I am not being able to understand the claim he made in the next paragraph:
$14.5.$ If $V_i\Subset U_i, i = 1,\dots, n$ are relatively compact open subsets and $\mathcal{B}=(V_i)_{1\leq i\leq n}$, then to simplify the notation we will write $\mathcal{B}\ll \mathcal{U}$. For any cochain $\xi\in C^q(\mathcal{U},\mathcal{O})$ one has $||\xi||_{L^2(\mathcal{B})}<\infty$. It then follows directly from Lemma $(14.3)$ that given any $\epsilon>0$, there exists a closed vector subspace $A \subset Z^1_{L^2}(\mathcal{U},\mathcal{O})$ of finite codimension such that $||\xi||_{L^2(\mathcal{B})}\leq\epsilon ||\xi||_{L^2(\mathcal{U})}$ for every $\xi\in A$
Using $14.3$ I can see that for each $i$ we get the inequality, from there how do I prove this for a cochain? Also how are we having the closed subspace $A$ to be inside $Z^1_{L^2}(\mathcal{U},\mathcal{O})$ ? A detailed answer is most welcome. Thanks in advance.
This was actually not so hard. @AlanMuniz's suggestion solves it. Given an open cover we apply the lemma on each cover to get a bunch of subspaces like in the lemma. Their direct sum is a subspace with finite codimension in $1$st cochain level, taking the intersection of that set with $Z^1_{L^2}(\mathcal{U},\mathcal{O})$ will be our desired set. Notice that this set has finite codimension in $Z^1_{L^2}(\mathcal{U},\mathcal{O})$.