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(I copy and paste and edit from Is this operator bounded? Hilbert space projection, my question is almost the same)

Let $V \subset H$ be Hilbert spaces (different inner products) with $V$ dense and continuously embedded in $H$. Let $\{b_j\}$ be a basis for $H$ and for $V$. Define $$P_n:H \to \text{span}(b_1,...,b_n)$$ by $$(P_n h-h,v_n)_H = 0\quad\text{for all $v_n \in \text{span}(b_1,...,b_n)$}$$ by truncation.

Is it true that $$P_n:V \to V$$ is a bounded operator where the constant that bounds it does not depend on $n$?

It is true as an operator from $H$ to $H$ (take $v_n = P_nh$ in the definition) but not sure for $V$.

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  • $\begingroup$ Why do you say that $P_n$ maps $V$ to $V$? If $b_1\notin V$, then $P_1$ takes values outside of $V$. $\endgroup$ – ˈjuː.zɚ79365 Jun 28 '13 at 14:24
  • $\begingroup$ @ˈjuː.zɚ79365 sorry $b_i$ are in $V$ too. They're basis for both spaces. $\endgroup$ – michael_faber Jun 28 '13 at 18:57
  • $\begingroup$ In what sense is $\{b_j\}$ a basis? It can't be orthonormal in both inner products. $\endgroup$ – Nate Eldredge Oct 11 '13 at 16:02

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