How to simplify this expression including integral, modulus and convolution?

Question

I encountered this expression during simplification: $$ \int \lvert exp(ixy)g(y)\star f(x)\lvert^2dy $$ $i$ is the imaginary unit, $g(y)$ is a real function while $f(x)$ is a complex function. $\star$ stands for convolution. Could anyone show me how to simplify this expression? For example, can the order of integration and convolution be exchanged? Or under what condition can I move the integration inside the $\lvert ... \lvert ^2$?

Answer 1

$$\int_y \lvert e^{ixy}g(y)\star f(x)\lvert^2dy$$

Assuming it is a convolution with respect to variable $x$ (the single common variable), one can transform this expression into:

$$\int_y \lvert \int_t e^{i(x-t)y}g(y)f(t)dt \lvert^2dy$$

$$=\int_y \lvert \int_t g(y) e^{-ity}e^{ixy}f(t)dt \lvert^2dy$$

$$=\int_y \lvert g(y) e^{ixy}\lvert^2 \times \lvert \int_t e^{-ity} f(t)dt \lvert^2dy$$

$$=\int_y \lvert g(y) \lvert^2 \times \underbrace{\lvert e^{ixy}\lvert^2 }_1 \times \lvert \underbrace{\int_t e^{-ity} f(t)dt}_{\mathscr{F}(f)(y)} \lvert^2dy$$

where $\mathscr{F}$ is the Fourier transform operator.

I don't see further simplification.

Remark: In the very particular case where $g$ is a constant, one can apply Parseval relationship (the norm of $\mathscr{F}(f)$ is equal to the norm of $f$).