The problem:
we have two functions $f(x), g(x)\in C^{1}[-\pi, \pi]$, and we know that
\begin{align} \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \cos\left(ky\right) \sin\left(k\left\lvert y-z\right\rvert\right)\:f\left(y\right)g\left(z\right) \,dy\, dz &= 0, \\ \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \sin\left(ky\right) \sin\left(k\left\lvert y-z\right\rvert\right)\:f\left(y\right)g\left(z\right) \,dy\, dz &= 0, \end{align}
for all $k\in\mathbb{R}$.
What I want to prove is:
- if $g(z)$ is not zero function, then $f(y)$ must be $0$.
Some results:
the problem's difficult part is the absolute value, otherwise it is simply Fourier transform.
if we take the above first equation as $H(k)$, then differentiating at $k=0$ will give $\forall p\in \mathbb{N}$, the first equation gives
$$\iint \sum_{m+n=p} \frac{y^{2m} \left\lvert y-z\right\rvert^{2n+1}}{\left(2m\right)!\left(2n+1\right)!} \:f\left(y\right)g\left(z\right) \,dy\, dz = 0,$$
- the second gives a similar one.
Thanks in advance.
