Do functions exist and are they dense? Or does it depend on the basis?

Question

Consider an orthonormal basis $(\varphi_n)_{n \in \mathbb N}$ of $L^2(\mathbb R).$

We consider the functionals $\Phi_n$ given by $$ C^b(\mathbb R) \ni f \mapsto \left\langle \varphi_n, f \varphi_{n+1} \right\rangle$$

for any $n \in \mathbb N$ where $C^b(\mathbb R)$ are the continuous and bounded functions on $\mathbb R.

I would like to know:

Consider the union $X:=\bigcup_{n \in \mathbb N} \operatorname{ker}(\Phi_n).$ Is the complement of this set $X$ non-empty. Is the complement perhaps even dense in $C^b(\mathbb R)$ or does it depend on the basis?

Answer 1

The complement of $X$ may be empty.

Consider a basis such that $\mathrm{supp}\,\varphi_k \cap \mathrm{supp}\,\varphi_{k+1}=\emptyset$ for some $k$ (for example, consisting of Walsh functions on each interval $[n, n+1]$). Then obviously $\mathrm{Ker}\,\Phi_k = C^b(\mathbb{R})$.