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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.

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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 $n =1,2,3,\cdots$.

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  • $\begingroup$ Why does $P_nh$ as defined in the OP equal the right hand side in your answer? Your formula defines the the projection onto the first $n$ basis functions of $h$ of course but I don't see why it's the same as the OP's projection. $\endgroup$ – riem Sep 29 '14 at 18:01
  • $\begingroup$ @riem : Verify that $(P_{n}h-h,e_{j})_{H}=0$ for $1 \le j \le n$. It then follows that $(P_{n}h-h,v)_{H}=0$ for all $v \in V_{n}$. $\endgroup$ – DisintegratingByParts Sep 29 '14 at 18:29

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