I have come across a problem from mathematical physics which asks me to find all real (smooth, say) functions $g(\cdot)$ such that
$$ \frac{1}{2\pi}\int_\mathbb{R}e^{ig(xy)}dx=\delta (y)\,, $$
where $\delta (\cdot )$ is Dirac's distribution.
It is obvious that we can pick $$ g(xy)=\pm xy \, , $$
but I need to know whether these are the only solutions. Unfortunately, I am unable to conclude that the relation given above forces $g (\cdot )$ to be a linear function - maybe it does not!
The difficulty stems from the fact that one cannot use the (generalized) orthogonality of the exponentials $e^{ixy}$ to "undo" the integral on the left.
Update 1
Assuming $g(u)=p+qu+ru^2$ with non-zero $p,q,r \in \mathbb R$, the integral on the left can be evaluated using the Fourier transform of a complex Gaussian. The result is not a Dirac-$\delta$, so the function $g(\cdot )$ cannot be a quadratic polynomial.
Clinging at straws... A conceptual insight seems necessary rather than a computation.
Appendix
The original post started with a little detour which I reproduce here: I was looking for all (smooth, say) functions $f(x,y)$ such that
$$ \frac{1}{2\pi}\int_\mathbb{R}e^{if(x,y)}dx=\delta (y)\,. $$
Rescaling $y\to ay$ with $a\neq 0$, one can derive $$ \frac{1}{2\pi}\int_\mathbb{R}e^{if(\frac{x}{a},\,ay)}dx=\delta (y)\,, $$
which suggests that we should have $f(x,y)=g(xy)$ - otherwise the left-hand-side depends on $a$ while the right-hand-side would not (one can probably make this argument more rigorous).