Background
Suppose we have a set of orthonormal basis functions $\{\psi_j\}_{j=1}^{\infty}$ of $L^2([0,1])$ (for example trigonometric basis), we know that $$ \int_0^1 \psi_i(x)\psi_j(x)\,\mathrm dx = \delta_{ij} $$ where $\delta_{ij} = 1$ when $i=j$, else $\delta_{ij}=0$.
Question
I am wondering if it is possible to find a pretty well-behaved function $f(x)$ (for example it is uniformly bounded and continuous over $[0,1]$) such that there exists an $\epsilon >0$ that $$ \int_0^1\psi_i(x)\psi_j(x)f(x)\,\mathrm dx > \epsilon $$ for all $i\neq j$. That is, a fixed function significantly change the orthogonality of $\{\psi_j\}_{j=1}^{\infty}$
It does not feel quite possible to me, but I do not know any road towards analyzing this, thank you for your help.
This is impossible in general. For example, consider a basis formed as follows: Start with $\psi_1(x) = 2\chi_{[0,1/2]}$ and $\psi_2(x) = 2\chi_{(1/2,1]}$, where $\chi_A$ is the indicator function for $A$ (i.e. it is 1 on A and 0 elsewhere). These are orthogonal, so they can be extended to a Hilbert basis for $L^2([0,1])$. Yet for any $f\in L^2([0,1])$ we have $\int_{[0,1]} \psi_1 \psi_2 f dx = \int_{[0,1]} 0 \cdot f dx = 0$