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.)

  • 2
    $\begingroup$ Riesz's projection theorem does not depend on the choice of a basis, so I don't see your point ? Take $Q_n$ to be the orthogonal projection on $V_n$ and your two conditions are satisfied (with $C = 1$ btw). $\endgroup$
    – Hachino
    Apr 1, 2015 at 12:06
  • 2
    $\begingroup$ @Hachino Note that 2. is asking for boundedness in $V$ to $V$, not in $H$ and 1. uses the inner product in $H$. $\endgroup$
    – Charpe
    Apr 1, 2015 at 12:31
  • $\begingroup$ Is $\{w_j\}$ supposed to be countable? $\endgroup$ Apr 1, 2015 at 19:21
  • $\begingroup$ Unless I am missing something, doesn't 1 force $Q_n$ to be precisely the $H$-orthogonal projection onto $V_n$, restricted to $V$? $\endgroup$ Apr 1, 2015 at 19:32
  • 1
    $\begingroup$ @NateEldredge Ok, I agree. $\endgroup$
    – Charpe
    Apr 1, 2015 at 20:01

1 Answer 1


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.

  • $\begingroup$ While I'm confident the embedding of $V$ into $H$ here is continuous, I'm doubting now that it's compact. This may disqualify this answer, as the question is now worded. $\endgroup$
    – James
    Apr 1, 2015 at 21:52

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