Almost orthogonal collection of $L^{2}$ functions Suppose I had an infinite sequence of smooth compactly supported functions $\{f_{n}\}_{n \geq 1}$ such that for each $n$, $f_{n}$ is mutually orthogonal to all but a fixed absolute constant $C$ of the other $f_{m}$. That is, $\int_{\mathbb{R}}f_{m}f_{n}\, dx = 0$ for all but $C$ of the other functions in $\{f_{n}\}_{n \geq 1}$ (where $C$ is independent of $n$).
Is it true that $$\left\|\sum_{n \geq 1}f_{n}\right\|_{L^{2}(\mathbb{R}^n)} \leq C'\left(\sum_{n \geq 1}\|f_{n}\|_{L^{2}(\mathbb{R}^n)}^{2}\right)^{1/2}$$
for some absolute constant $C'$?
 A: Your claim is indeed true. For the proof, we first need the following
auxilliary result (which might also be of independent interest):
Lemma: Let $X\neq\emptyset$ be a set and let $\sim\subset X\times X$
be a relation which is reflexive and symmetric (but not necessarily
transitive). For $x\in X$, let
$$
\left[x\right]:=\left\{ y\in X\,\middle|\, y\sim x\right\} 
$$
and assume that $N:=\sup_{x\in X}\left|\left[x\right]\right|$ (the
largest cardinality of all $\left[x\right]$) is finite.
Then there is a partition $X=\biguplus_{\ell=1}^{N}X_{\ell}$ with
$x\not\sim y$ for $x,y\in X_{\ell}$ with $x\neq y$ and arbitrary
$\ell\in\left\{ 1,\dots,N\right\} $.
Proof: Let $X_{1}\subset X$ be maximal with the following property:
For $x,y\in X_{1}$ with $x\neq y$, we have $x\not\sim y$. Existence
of such a set is an easy consequence of Zorn's Lemma.
Now, if $X_{1},\dots,X_{m}$ are already constructed for some $1\leq m\leq N-1$,
let $X_{m+1}\subset X\setminus\left(X_{1}\cup\dots\cup X_{m}\right)$
be maximal with the same property as above. Again, existence follows
from Zorn's Lemma.
Clearly, $\left(X_{\ell}\right)_{\ell=1,\dots,N}$ are pairwise disjoint
and satisfy the property that $x\not\sim y$ for $x,y\in X_{\ell}$
with $x\neq y$ and arbitrary $\ell\in\left\{ 1,\dots,N\right\} $.
It remains to show that $X=\bigcup_{\ell=1}^{N}X_{\ell}$. Suppose
that this is not the case. Then there is some $x\in X\setminus\bigcup_{\ell=1}^{N}X_{\ell}$.
Thus, $X_{\ell}\cup\left\{ x\right\} \subset X\setminus\left(X_{1}\cup\dots\cup X_{\ell-1}\right)$
is a strict superset of $X_{\ell}$. By maximality, we see that there
must be some $y_{\ell}\in X_{\ell}\cup\left\{ x\right\} $ with $x\neq y_{\ell}$
and $x\sim y_{\ell}$. Note $y_{\ell}\in X_{\ell}$ since $x\neq y_{\ell}$.
By disjointness of the $\left(X_{\ell}\right)_{\ell=1,\dots,N}$,
we see that $x,y_{1},\dots,y_{N}$ are pairwise distinct. Hence,
$$
N\geq\left|\left[x\right]\right|\geq\left|\left\{ x,y_{1},\dots,y_{N}\right\} \right|=N+1,
$$
a contradiction. This proves our Lemma. $\square$
Remark: For a countable index set, we can avoid direct usage of Zorn's
Lemma. This is left as an exercise.
Now, we can assume without loss of generality that $f_{n}\neq0$ for
all $n\in\mathbb{N}$ (otherwise, simply drop those $f$). Define
$$
n\sim m\qquad:\Leftrightarrow\qquad f_{n}\not\perp f_{m},\text{ i.e. if }f_{n}\text{ and }f_{m}\text{ are not orthogonal}.
$$
Because of $f_{n}\neq0$ for all $n$, this is easily seen to be a
reflexive, symmetric relation on $\mathbb{N}$. By your assumption,
$\left|\left[n\right]\right|\leq C$ for all $n\in\mathbb{N}$, so
that the Lemma above yields a partition $\mathbb{N}=\bigcup_{\ell=1}^{C}I_{\ell}$
with $n\not\sim m$ (i.e. $f_{n}\perp g_{m}$) for all $n,m\in I_{\ell}$
with $n\neq m$ for arbitrary $\ell\in\left\{ 1,\dots,C\right\} $.
Now, the triangle inequality, Pythagoras theorem and the Cauchy Schwarz
inequality yield
\begin{align*}
\left\Vert \sum_{n\in\mathbb{N}}f_{n}\right\Vert _{L^{2}}^{2} & =\left\Vert \sum_{\ell=1}^{C}\sum_{n\in I_{\ell}}f_{n}\right\Vert _{L^{2}}^{2}\\
 & \leq\left(\sum_{\ell=1}^{C}\left\Vert \sum_{n\in I_{\ell}}f_{n}\right\Vert _{L^{2}}\right)^{2}\\
 & =\left(\sum_{\ell=1}^{C}\sqrt{\sum_{n\in I_{\ell}}\left\Vert f_{n}\right\Vert _{L^{2}}^{2}}\right)^{2}\\
 & =\left(\sqrt{\sum_{\ell=1}^{C}1^{2}}\cdot\sqrt{\sum_{\ell=1}^{C}\sum_{n\in I_{\ell}}\left\Vert f_{n}\right\Vert _{L^{2}}^{2}}\right)^{2}\\
 & =C\cdot\sqrt{\sum_{n\in\mathbb{N}}\left\Vert f_{n}\right\Vert _{L^{2}}^{2}}.
\end{align*}
Note that this proof also yields convergence of the series $\sum_{n\in\mathbb{N}}f_{n}$
in the first place (given that $\sum_{n\in\mathbb{N}}\left\Vert f_{n}\right\Vert _{L^{2}}^{2}<\infty$).
Altogether, we see
$$
\left\Vert \sum_{n\in\mathbb{N}}f_{n}\right\Vert _{L^{2}}\leq\sqrt{C}\cdot\sqrt{\sum_{n\in\mathbb{N}}\left\Vert f_{n}\right\Vert _{L^{2}}^{2}}.
$$
As a final remark, note that exactly the same arguments work in any
Hilbert space, not only in $L^{2}$.
A: I don't know, although I tend to doubt it. There's a related fact that who knows may be good enough for your purposes: If $f_n$ is orthogonal to $f_m$ except when $|n-m|\le C$ then the inequality holds:
$$\left|\left|\sum f_n\right|\right|_2^2=\sum_n\sum_{j=-C}^C\langle f_n,f_{n+j}\rangle
\le\sum_{j=-C}^C\sum_n||f_n||_2||f_{n+j}||_2\le(2C+1)\sum||f_n||_2^2.$$
