$$F(\omega)=\frac{\sin \omega}{\omega}$$ $$G(\omega)=\frac{\sin \omega}{\omega}e^{-j2\omega}$$
\begin{align} F(\omega )*G(\omega)&=\int^{+ \infty}_{-\infty} \frac{\sin \tau}{\tau}\frac{\sin (\omega-\tau)}{(\omega-\tau)}e^{-j2(\omega-\tau)}d\tau \\ &=0? \end{align}
$F(\omega)$ and $G(w)$ is the Fourier transform of $f(t)$ and $g(t)$.
