Consider an orthonormal basis $(\varphi\_k)$ of $L^2(\mathbb R)$ with Lebesgue measure. 


I came along a nice number theoretic question in analysis:


Write $$f\_k(x):=\int\_{\left\lvert y \right\rvert \ge x } \left\lvert \varphi\_k(z) \right\rvert^2 \ dz.$$


Clearly, $f\_k$ are continuous monotonically decreasing functions such that $f\_k(0)=1$ and $\lim\_{x \rightarrow \infty} f\_k(x)=0.$


I ask: Is it possible that $f\_k(x)$ are for all $x$- $\mathbb Q$ linearly dependent numbers? 


We say $(f\_k(x))$ are linearly dependent for all $x$ if:
For every $x \in [0,\infty)$ there is $N\in \mathbb N$ and a non-zero vector $(q\_1,...,q\_N)$ of rationals such that $$\sum\_{k=1}^{N} q\_i f\_k(x)=0.$$


Typical examples of orthonormal bases( Think of an ONB that has fairly disjoint support) over $L^2(\mathbb R)$ seem to suggest that this is not possible and that we have rational linear dependence everywhere, but I do not know whether this is a general fact?


To strengthen the hypothesis that there must be a set of rationally independent numbers, please take into account that they are dense. [click me.](https://math.stackexchange.com/questions/657039/is-the-span-of-rationally-independent-real-numbers-dense-in-mathbbr)


If there are any further questions, please let me know.



Number theory on Banach space $L^2(\mathbb R)$ meets linear independence?
You can have an orthonormal basis of $L^2(\mathbb R)$ with all $f\_n$ equal.
Start with the Walsh functions, which are an orthonormal basis of $L^2([0,1])$ with absolute value $1$ everywhere. Then use an isometry of $L^2([0,1])$ to $L^2(\mathbb R)$ given by $Tf(t) = a(t) f(b(t))$ for suitable functions $a(t)$ and $b(t)$.


