Convolution of two sums (Fourier transform) This question is from the book "Advanced Engineering Mathematics" by Stroud. 


I can't seem to get the required answer for this. I've derived the two Fourier transform equations for them.

. 
U and V are dummy variables.
I've been told I need to use the convolution theorem by multiplying the functions in the frequency domain and finding the inverse transform of the solution, but I don't know how to proceed with the equations I've got.
 A: The Fourier transform is a linear operator, so
$$
F_1(\omega) = \sum_{n=-\infty}^{\infty} c_n \mathcal{F}(e^{jn\omega_0 t}) = \sum_{n=-\infty}^{\infty} c_n \delta( \omega-n\omega_0)
$$
Use the above to multiply the two Fourier transforms. Then see if you can take the take the inverse transform to finish the proof.
EDIT:
To see how to find the product of the two sums note that
$$
\delta( \omega - n \omega_0) \cdot \delta( \omega - m \omega_0) \neq 0 \Leftrightarrow n = m.
$$
Furthermore,
$$
m=n \Rightarrow \delta( \omega - n \omega_0) \cdot \delta( \omega - m \omega_0) = \delta( \omega - n \omega_0) = \delta( \omega - m \omega_0).
$$
Thus,
$$
d_m \delta( \omega - m\omega_0) \cdot \sum_{n=-\infty}^{\infty} c_n \delta( \omega-n\omega_0) = d_m \delta( \omega - m\omega_0) \cdot c_m \delta( \omega - m\omega_0) = d_m c_m \delta( \omega - m\omega_0)
$$
EDIT 2:
We know that
$$
F_1(\omega) = \sum_{n=-\infty}^{\infty} c_n \delta( \omega-n\omega_0)
$$
$$
F_2(\omega) = \sum_{m=-\infty}^{\infty} d_m \delta( \omega-m\omega_0)
$$
Their product is then
$$
F_1 (\omega) F_2 (\omega) = \left( \sum_{n=-\infty}^{\infty} c_n \delta( \omega-n\omega_0) \right) \left( \sum_{m=-\infty}^{\infty} d_m \delta( \omega-m\omega_0) \right)
$$
By the definition of the Dirac delta function, we know that
$$
\delta(\omega - a) \cdot \delta( \omega - b) =
\left\{
\begin{array}{lr}
\delta(x-a), & a = b \\
0, & a \neq b
\end{array}
\right.
$$
Thus, if we consider an individual term in the series $F_1 (\omega)$ by fixing $n$ and take its product with $F_2(\omega)$, we get
$$
c_n \delta(\omega - n\omega_0) \left( \sum_{m=-\infty}^{\infty} d_m \delta( \omega-m \omega_0) \right) = c_n d_n \delta(\omega - n\omega_0).
$$
Since this is true for all $n$, we have that
$$
F_1 (\omega) F_2 (\omega) = \sum_{n=-\infty}^{\infty} c_n d_n \delta( \omega-n \omega_0).
$$
