Number theory on Banach space $L^2(\mathbb R)$ meets linear independence?

Question

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.

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

Answer 1

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 $b(t)$.

Answer 2

not a solution (see comment)
So $f_1$ and $f_2$ are both decreasing continuous functions on $[0,+\infty)$. I assume it likely that $f_1$ is not equal to $f_2$. But then $f_1/f_2$, defined at least in a neighborhood of zero, is continuous and not constant, so for some $x$, we must have $f_1(x)/f_2(x)$ is rational. Thus for that $x$, we conclude that $f_1(x),f_2(x)$ is not linearly independent.