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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.

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Do you have $\|h\|_H\leq C\|h\|_V$ for some fixed positive constant $C$ and $h\in V$? –  Tomás Jun 27 '13 at 13:39
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Yes, embedding is continuous –  aere Jun 27 '13 at 13:42

2 Answers 2

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 $$\|P_nv\|^2_V\leq c,\ \forall\ v\in V,\ \|v\|_V\leq 1$$

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Thanks. Is your last inequality right? It doesnt imply boundedness on its own I think –  aere Jun 27 '13 at 14:00
    
Suppose that $T$ is a linear operator satisfying $\|Tx\|\leq c$ for all $x$ with $\|x\|=1$. Take any $y$ and note that $\|T(y/\|y\|)\|\leq c$, so $\|T(y)\|\leq c\|y\|$. –  Tomás Jun 27 '13 at 14:03
    
Thanks. As I wrote below, do you think the result will still hold if $b_i$ is just basis, not orthogonal? The projection would be defined by the $H$ inner product in that case $(P_n u-u,v_n) =0$ for all $v_n \in \text{span}(b_1,...,b_n)$ –  aere Jun 27 '13 at 15:51
    
In this case, you can write $P_nu=\sum _{i=1}^n \alpha_i b_i$. By solving a system you have that $\alpha_i$ can be written in terms of $(b_i,b_j)_H$ and $(P_nu,b_i)_H$. Then, if $P_n:H\to H$ is bounded and $V$ is continuously embedded in $H$, by the same argument as in the above proof, you can conclude the boundedness of $P_n:V\to V$. –  Tomás Jun 27 '13 at 16:04
    
Clear answer.+1 (oops, daily voting limit reached, will +1 later) –  Shuhao Cao Jun 28 '13 at 16:17

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.

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Thanks. Do you think the result will still hold if $b_i$ is just basis, not orthogonal? The projection would be defined by the $H$ inner product in that case $(P_n u-u,v_n) =0$ for all $v_n \in \text{span}(b_1,...,b_n)$ –  aere Jun 27 '13 at 15:51
    
Yes, with the given form of $P_n$, it factors through the orthogonal projection onto $\operatorname{span}(b_1,\,\ldots,\,b_n)$ regardless of whether the $b_i$ are orthogonal. So you have $P_n = \varphi \circ \pi_n$, $\pi_n$ is continuous as an othogonal projection, $\varphi$ is continuous because its domain is finite-dimensional. –  Daniel Fischer Jun 27 '13 at 16:42

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