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Zinkin
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Orthonormal basis and decay

Take a family of intervals $(I_i)$ of constant finite length, i.e. $\left\lvert I_i \right\rvert=\text{const}$ with the same constant independent of $i$. Those intervals are assumed to be almost disjoint in the sense that $I_i \cap I_j=\emptyset$ if and only if $\left\lvert i-j \right\rvert >1,$ so we only have a nearest neighbor overlap.

So you can think of the intervals as ordered on the real line in the sense that $I_1$ is the outer left interval that intersects $I_2$ but no longer $I_3$. Yet, $I_2$ intersects $I_3$ and so on.

There is a $L^2$ function $f:I_1 \rightarrow \mathbb{R}$ we would like to study.

To do so, we consider a family of functions $h_i:I_i \rightarrow \mathbb{R}$ that are $L^2$ normalized (but not orthogonal).

Those function are assumed to communicate with its nearest neighbour in the sense that $c < \left\lvert \langle h_i,h_{i+1} \rangle \right\rvert < C$ with $c>0$ and $C<1$ independent of $i$.

This avoids some trivialities later on.

The way we constructed the $I_i$, there are only two functions $h_1$ and $h_2$ that can possibly satisfy $\langle h_1,f\rangle \neq 0$ and $\langle h_2,f\rangle \neq 0.$ Otherwise $\langle h_i ,f \rangle =0.$

We would like to study $f$ by projecting it onto the span $V:=\overline{span}(h_i).$ The projection of $f$ onto this space will be denoted as $\text{proj}_V(f).$

Now intuitively, the $h_i$ with $i>2$ will not be able to tell us much about $f$ because their inner product with $f$ is zero.

However, we now consider the Gram-Schmidt orthonormalized sequence $(g_i)$ of the $h_i$, i.e. $g_1:=h_1$ and $g_2:=\frac{h_2-\langle h_2,g_1\rangle g_1}{\left\lVert h_2-\langle h_2,g_1\rangle g_1 \right\rVert}$ and so on. Note that also $g_3,g_4,...$ and so on may now have non-vanishing overlap with $f$, although this one should intuitively still be small.

Clearly, $\text{proj}_V(f)=\sum_{n=1}^{\infty} a_n g_n$ for some $a_n$.

I would like to know: How rapidly do the coefficients $a_n$ decay? In particular, is it true that (the following notation means up to a constant) $\left\lvert a_n \right\rvert \lesssim L^n$ for some $L\in (0,1)$?

If anything is unclear, please let me know.

Zinkin
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