Positivity of $\int_0^1 \sin^2(kx)f(x) dx$

Question

I'm interested in proving restrictions on $f$ necessary for the integral \begin{equation*} \int_0^1 \sin^2 (kx) f(x) dx \end{equation*} to be positive for all $k > 0$. Obviously a nonnegative $f$ would suffice (provided it was positive on a set of nonzero measure). I'm wondering if there are less restrictive conditions that may be placed on $f$. Thanks in advance.

Answer 1

A partial answer: the question is less trivial than I initially suspected.

It is practical to consider that $\sin^2(kx)=\frac{1}{2}\left(1-\cos(2kx)\right)$.
Assuming $f\in L^2(0,1)$ we are allowed to write $$ f(x) \stackrel{L^2(0,1)}{=} c_0 + \sum_{n\geq 1} c_n \cos(2\pi n x)+\sum_{n\geq 1} s_n \sin(2\pi n x) $$ where $c_0$ has to be non-negative as a consequence of the Riemann-Lebesgue lemma.

By considering $k=\pi n$ with $n\in\mathbb{N}^+$ we get that each $c_n$ has to be $\leq 2c_0$.
This leaves enough room for plenty of somewhere-negative functions fulfilling the constraints for each $k\in \pi\mathbb{N}$, like

$$ \pi^2 x(1-x)+1-\frac{\pi^2}{6} = 1-\sum_{n\geq 1}\frac{\cos(2\pi n x)}{n^2} $$ which can be shown to fulfill the constraints for any $k>0$, ouch!

So my actual conjecture is that the claim holds iff the coefficients of the Fourier cosine series of $f(x)+f(-x)$ over $(0,1)$ fulfill some set of inequalities to be discovered. $f\geq 0$ is definitely not necessary.