Consider the following expression $$H(\vec{x},t) = f_{tt}(\vec{x},t)g(\vec{x},t) - f(\vec{x},t)g_{tt}(\vec{x},t)$$ where f and g are functions of time and space and the subscript " $_{tt}$ " denotes a double time derivative. If I take (maybe naively) the temporal Fourier transform of $H$ and I use
- the derivative property of the Fourier transform
- the fact that the transform of a product is a convolution of transforms
I can write $$\hat{H}(\vec{x},\omega) = -\omega^2(\hat{f}*\hat{g})(\vec{x},\omega) + \omega^2(\hat{g}*\hat{f})(\vec{x},\omega)$$ where $\omega$ is the pulsation variable, $\hat{f}$ and $\hat{g}$ are the Fourier transform of $f$ and $g$, and $*$ is the convolution product relative to the variable $\omega$. With the commutative property of the convolution product I found $\hat{H}(\vec{x},\omega) = 0$, and so $$H(\vec{x},t) = 0 \text{ .}$$ I believe that this is not true, what have I done wrong ? I am using this expression in a physics context so $f$ and $g$ are supposed to be causal functions and belong to $\mathcal{L}^2$.
The problem is with the equation after "I can write". If you calculate the Fourier transform more carefully, you will notice that the $\omega^2$ does not come out of the convolution, but you need to convolve the functions $\omega^2\hat f(x,\omega)$ and $\hat g(x,\omega)$ (and the other way around for the other term). Write down the convolution and you will see that you can't pull the $\omega^2$ outside the integral.