Uniqueness from orthogonality relation?

Question

This question was posted yesterday on MathOverflow by Michael Smith and received a number of upvotes. I too think the question was interesting. However, for some unknown to me reasons, it has been deleted by the author. Perhaps, the answer to the question was found to be "no". Even then, it would be interesting to know more about the matter. So, I am re-posting the question (slightly re-phrased), along with the suggested approach to it that I posted earlier.

Suppose that functions $f$ and $g$ in $C^1[-\pi,\pi]$ are such that \begin{equation} \int_{-\pi}^\pi\int_{-\pi}^\pi dy\,dz\, \cos(ky)\sin(k|y-z|)f(y)g(z)=0 \end{equation} and \begin{equation} \int_{-\pi}^\pi\int_{-\pi}^\pi dy\,dz\, \sin(ky)\sin(k|y-z|)f(y)g(z)=0 \end{equation} for all real $k$. Does it follow then that either $f$ or $g$ is the zero function?

Answer 1

This is not a complete answer. Rather, it is a suggestion on how to reduce the problem to a possibly simpler one.

For each $c\in(-\pi/2,\pi/2)$, multiply each of your two identities by $e^{i c k - \epsilon k^2/2}$ with $\epsilon>0$, integrate in $k\in\mathbb R$, let $\epsilon\to0$, and then subtract one of the resulting identities from the other. Unless my calculations are mistaken (which is not unlikely), you thus get the following convolution-type identity (for all $c\in(-\pi/2,\pi/2)$): \begin{equation} \int_{-\pi}^\pi dy\,f(y)[g(c)+g(2y-c)I\{|2y-c|<\pi\}]\,\text{sgn}(y-c)=0, \end{equation} where $I$ is the indicator function; this identity can of course be rewritten as \begin{equation} \int_{-\pi}^\pi dy\,f(y)\,\text{sgn}(y-c)\,g(c)+\int_{(c-\pi)/2}^{(c+\pi)/2} dy\,f(y)g(2y-c)\,\text{sgn}(y-c)=0. \end{equation} I don't yet know if it implies that either $f$ or $g$ is the zero function. (Also, I think the assumption that $f$ and $g$ are in $C^1[-\pi,\pi]$ can be relaxed at least to that of $f$ and $g$ being in $C[-\pi,\pi]$.)