Let $V \subset H$ be separable Hilbert spaces with continuous embedding and suppose $\{v_n\}$ be a (non-orthogonal) basis for $V$. If we let $V_n = \text{span}(v_1, ..., v_n)$ and given $h \in H$ we define an operator $P$ by $$(P_n(h)-h, v)_H = 0$$ for all $v \in V_n$, then what space does $P_n(h)$ lie in?
I hoped it was in $V_n$. But I don't know how to prove that.
Fix $n$. Let $\{ e_{1},e_{2},\cdots,e_{n} \}$ be an equivalent orthonormal basis obtained from $\{ v_{1},v_{2},\cdots,v_{n}\}$ by Graham-Schmidt using the inner product on $H$. Each $e_{j} \in V_{n}$ for $1 \le j \le n$ because $e_{j}$ is a linear combination of $\{ v_{1},v_{2},\cdots,v_{j}\}$; therefore, $$ P_{n}h = \sum_{j=1}^{n}(h,e_{j})_{H}e_{j} \in V_{n},\;\;\; h \in H. $$ This is true for any fixed $n =1,2,3,\cdots$.