# Compactly supported functions and projections

Let $\Omega$ be an open subset of $\mathbb{R}^n$ and take a family of continuous compactly supported functions $f_n$ on $\Omega$ normalized to one (in the $L^2$ sense).

Then, these functions span a subspace $V$ of $L^2$. Now, consider two $L^2$ functions $g$ and $h$ with non-intersecting support. Assume that both $g$ and $h$ only overlap with $k$ of those functions $f_n$. It may happen now that although $\langle g,h \rangle =0$ due to the non-intersecting support of the two, $\langle P_V g,P_V h \rangle \neq 0$ where $P_V$ is the projection onto $V$. However, we expect that "morally" $\langle P_V g,P_V h \rangle$ should still be small if they are far apart and the $(f_n)$ are sufficiently localized.

Of course one could trivially estimate $\langle P_V g,P_V h \rangle \le \left\lVert g\right\rVert \left\lVert h \right\rVert,$ but this estimate is very bad in general as it does not capture the fact that most of the time, this should be something close to zero, since $\langle g,h \rangle =0.$

I ask: Does anybody see a better estimate that captures this $\langle P_V g,P_V h \rangle \neq 0$ being small? Especially, one would expect that the larger $V$ is, the closer should we get to zero.

Let me rephrase the questions like this: Assuming you are not allowed to evaluate $P_Vg$ or $P_Vh$ but only $g,h$ and all $(f_n)$ and inner-products among those objects. Can we get from this a better bound?

• It seems that in case $\|g\|=\|h\|$ we can choose $V$ to be 1-dimenional and close to the vector $g+h$. In this case $P_Vg\approx P_V h$ and $\langle P_Vg,P_Vh\rangle\approx\|g\|/\sqrt{2}$, so is not small. – Taras Banakh Aug 7 '17 at 9:19
• @TarasBanakh I said: The larger $V$ is, the closer should we get to zero. Now, you come up with a one-dimensional "counterexample". Do you notice something? – Zinkin Aug 7 '17 at 14:02
• I don't quite understand what you mean in the last paragraph. If you can take inner products of $g$ with the $f_n$ then you can compute $P_V g$, using Gram-Schmidt for instance. So what is really the restriction? – Nate Eldredge Aug 7 '17 at 14:35
• @Zinkin What do you mean by larger $V$? Of larger dimension? Then you can take this 1-dimensional $V_1$ and consider the infinite-dimensional space $V=V_1\oplus L^\perp$, where $L^\perp$ is the orthogonal complement of the linear span $L$ of the set $\{g,h\}\cup V$. Then this larger space $V$ is infinite-dimensional but still the projections of $g$ and $h$ have large inner product. But if "larger" means tends to the whole space in a suitable sense then the inner product tens to zero, but then this is trivial. – Taras Banakh Aug 7 '17 at 15:35

I think V is large means that V tends to $L^2(\Omega)$, i.e. you can select enough elements $f_n$ so that $V_n = span\{f_1,\cdots,f_n\}$ and the space span{$f_1,\cdots,f_{\infty}$} forms a complete basis of $L^2(\Omega)$. and (it should means that , otherwise you could add infinite k with support has no intersection with g ,h, then $\langle P_V g, P_V h \rangle$ is invariant).
Under this assumption, $\langle P_V g, P_V h \rangle$ should tend to 0, since
$\langle P_V g, P_V h \rangle - \langle g, h \rangle = \langle P_V g, P_V h \rangle - \langle P_V g, h \rangle + \langle P_V g, h \rangle - \langle g, h \rangle \leq \| P_{V^{\bot}} h\| + \| P_{V^{\bot}} g\|$
tends to zero if V is large in the sense V tends to $L^2(\Omega)$.