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Let $V \subset H$ be separable Hilbert spaces with dense and continuous embedding. For each $n$, let $V_n$ and $H_n$ be finite-dimensional subspaces of $V$ and $H$ respectively with dimension $n$.

Let $h_j$ be a o.n basis of $H$.

Define a projection operator $P_n:H \to H_n$ by $$P_n (\sum_{j=1}^\infty (u,h_j)_Hh_j) = \sum_{j=1}^n (u,h_j)_Hh_j.$$

Suppose I know that for every $v \in H$, $P_nv \in V_N$ (because it turns out that $H_n \subset V_n$ by construction of these spaces). Does it then follow that $P_n: V \to V$ is bounded by a constant independent of $n$? I am not sure about this actually.

But does this situation appear stupid or contradictory in any way?

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  • $\begingroup$ So you are also assuming that $\{h_1,\ldots,h_n\}$ spans $H_n$? $\endgroup$ – Julien Jul 14 '13 at 15:58
  • $\begingroup$ And I guess (otherwise it is of course yes, these are uniformly bounded by $1$) you are asking about the operator norm of $P_n:V\longrightarrow V$ with respect to the original norm on $V$? Then the answer is no, with the same "canonical" example as in Etienne's answer here. $\endgroup$ – Julien Jul 14 '13 at 16:08
  • $\begingroup$ @julien I had hoped that the extra property I have here (that $P_n v \in V_n$ and not just in $H_n$ might give me a different answer.. $\endgroup$ – matt.w Jul 14 '13 at 17:19
  • $\begingroup$ also note his projection operator is defined on $V$ and not $H$. $\endgroup$ – matt.w Jul 14 '13 at 17:20
  • $\begingroup$ These projections are just the canonical projections of $L^2$ functions onto trigonometric polynomials of degree $\leq n$ by truncation of the Fourier series. That's perfectly defined on $H$. And uniformy bounded by $1$ as operators from $H$ to $H$ (Bessel, or Parseval). But not uniformly bounded when restricted to $V$ equipped with the sup norm into $V$ with the sup norm. That's exactly the situation you mention. With $H_n=V_n\subseteq V$. $\endgroup$ – Julien Jul 14 '13 at 17:29
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It is not quite clear, how $H$ and $V$ are related.

Note that $H_n \subset V_n$ implies $H_n = V_n$, since both spaces are of dimension $n$.

The following example shows that $H$ and $V$ could not be arbitrary for your conclusion: Take $H = H_0^1(-,1)$ and $V = L^2(-,1)$, $n = 1$, $h_n = 1-|x|$. As scalar product in $H$, we choose $(u,v) = \int_{-1}^1 u' \, v' \, dx$. Then, projection onto $H_1$ is not even defined on $V$, albeit $H_1 \subset V_1$.

Edit: In the case that $V$ is continuously embedded in $H$, i.e., $\|v\|_V \le C \, \|v\|_H$, your estimate simply follows: Since $H_n = V_n$ is finite dimensional, the $V$-norm and the $H$-norm are equivalent on this subspace. Hence, $P_n : H \to V_n$ is continuous. Since the $V$-norm is stronger than the $H$-norm, you get the continuity of $P_n : V \to V_n$.

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  • $\begingroup$ Thanks for replying. Sorry I should have said $V \subset H$. But, now I am not sure that I can get that bound on $V$.. Not sure if it follows. $\endgroup$ – matt.w Jul 13 '13 at 19:32

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