Let $V \subset H$ be Hilbert spaces with a continuous, compact and dense imbedding. Let $\{w\_j\}\_j \subset V$ be a basis of $V$ and of $H$ (so finite linear combinitions are dense) which is **not** orthogonal. Define $V\_n = \text{span}(w\_1, ..., w\_n)$.


Is it possible to find an operator $Q\_n:V \to V\_n$ satisfying


1. $(Q\_n v-v, v\_n)\_H = 0 $ for all $v \in V$ and $v\_n \in V\_n$
2. $\lVert Q\_n v \rVert\_V \leq C\lVert v \rVert\_V$ where $C$ is independent of $n$


?


This kind of stuff comes up in Galerkin approximations but usually the basis used is orthogonal wrt. $H$ and orthonormal wrt. $V$. In my situation, I cannot use such a nice o.g/o.n basis because it is preferable to use a different special basis, and so the task is much harder. (When the basis is orthogonal in $V$ and o.n in $H$, the standard proof of this uses compactness of $V \subset H$ and Hilbert-Schmidt theory to get the result 2. after defining $Q\_n$ as satisfying 1.)



Existence of a projection operator onto subspace of Hilbert space
I think the answer, in general, is no. Here's a counter-example; perhaps someone else can find a simpler one. Take $V = \ell^2$, and let the inner product for $H$ restricted to $V$ correspond to the infinite tridiagonal matrix
$$
\begin{bmatrix} 2 & -1 & & \\ -1 & 2 & -1 & \\ & \ddots & \ddots & \ddots \end{bmatrix}.
$$
I'll call this matrix $H$ so that $(u,v)\_H = (Hu,v)$ for $u,v \in \ell^2$. Now let $V\_n$ just be the span of the first $n$ standard basis vectors. Let me abuse notation and also use $V\_n$ for the matrix with these vectors as columns. Then
$$
 Q\_n = V\_n (V\_n^\* H V\_n)^{-1} V\_n^\* H
$$
This infinite matrix is zero outside of the $n \times n+1$ upper-left block, which is given by $\begin{bmatrix} I\_n & u \end{bmatrix}$ with $u\_i = -\tfrac{i}{n+1}$. The square of the $\ell^2$ norm of $Q\_n$ works out to be $1+\Vert u \Vert^2$, which is
$$
 \Vert Q\_n \Vert^2 = \tfrac{n}{3} + \tfrac{5}{6} + \tfrac{1}{6} \tfrac{1}{n+1},
$$
and so $\Vert Q\_n \Vert \to \infty$ as $n \to \infty$.


The embedding above of $V$ into $H$ is not compact, but this can be fixed, as follows. Take $D$ to be the infinite diagonal matrix with $n$th diagonal entry equal to $1/n$. Set $A = D^{1/2} H D^{1/2}$. $A$ is compact as an operator on $V = \ell^2$, and can be taken to define the inner product for a Hilbert space in which $V$ is compactly embedded. The analysis for the corresponding projection $Q\_n = V\_n (V\_n^\* A V\_n)^{-1} V\_n^\* A$ proceeds as before, $u\_i$ now given by $-(\tfrac{i}{n+1})^{3/2}$. The squared norm of the projection in this case is then
$$
\Vert Q\_n \Vert^2 = \tfrac{n}{4}+\tfrac{3}{4}+\tfrac{1}{4}\tfrac{1}{n+1},
$$
which still diverges to infinity.


