Is this operator bounded? Hilbert space projection Let $V \subset H$ be Hilbert spaces (different inner products) with $V$ dense in $H$. Let $b_n$ be an orthonormal basis for $H$ and an orthogonal basis for $V$.
Define $$P_n:H \to \text{span}(b_1,...,b_n)$$ by $$P_n h = \sum_{i=1}^n (h,b_j)_Hb_j$$
by truncation.
Is it true that
$$P_n:V \to V$$ is bounded? How do I show that? If not, what assumptions does one need? Thanks.
 A: Let $v\in V$ and $P_nv=\sum_{i=1}^n(v,b_i)_Hb_i$. Note 
\begin{eqnarray}
 \|P_nv\|_V^2 &=& (\sum_{i=1}^n(v,b_i)_Hb_i,\sum_{i=1}^n(v,b_i)_Hb_i)_V      \nonumber \\
   &=& \sum_{i,j=1}^n(v,b_i)_H(v,b_j)_H(b_i,b_j)_V \nonumber \\
   &=& \sum_{i=1}^n (v,b_i)^2_H(b_i,b_i)_V \\
  &\leq & \sum_{i=1}^n\|v\|_H^2\|b_i\|_H^4\|b_i\|_V^2
\end{eqnarray}
If $\|v\|_V\leq 1$, then because the embedding is continuous, we conclude that $\|v\|_H\leq C$ for some constant $C$. This combined with the last inequality implies that $$\|P_nv\|^2_V\leq c,\ \forall\ v\in V,\ \|v\|_V\leq 1$$ 
A: 
Yes, embedding is continuous

With a continuous embedding, we need no computation to see that $P_n \colon V \to V$ is continuous, since then we can write
$$ P_n = F_n \circ \pi_n \circ j$$
where $j \colon V \to H$ is the continuous embedding, $\pi_n \colon H \to \operatorname{span}(b_1,\,\ldots,\,b_n)$ is the orthogonal projection [hence continuous], and
$$F_n \colon \operatorname{span}(b_1,\,\ldots,\,b_n) \to V;\quad F_n (x) = \sum_{\nu = 1}^n (x,b_\nu)_H\cdot b_\nu $$
is a linear map with finite dimensional domain, hence continuous.
