I am having trouble understanding this:
I have a function
$$ \delta (t_1-t_2) $$
but I want to prove that in the frequency domain, it is:
$$\delta(\omega_1+\omega_2) $$
So, we have:
$$F(t0,w_{1})=\int _{-\infty }^{\infty }\!\delta \left( {\it t_1}-{\it t_0} \right) { {\rm e}^{-iw_{{1}}t_{{1}}}}{dt_{{1}}}$$
$$ F(w_1,w_2)=\int _{-\infty }^{\infty }\!{{\rm e}^{-i \left( w_{{2}}+w_{{1}} \right) t_{{0}}}}{dt_{{0}}}$$
$$F(w_1,w_2)=2\pi \delta \left( w_{{2}}+w_{{1}} \right)$$
This is not a formal answer, just a hint of how to proceed.
The two Fourier transform of $f:\mathbb{R}^2 \to \mathbb{R}$ is given by $\hat{f}(\omega_1,\omega_2) = \int_{\mathbb{R}^2} f(t_1,t_2) e^{-i(\omega_1 t_1 + \omega_2 t_2)} d t_1 d t_2$.
With the distribution given by $f(t_1,t_2) = \delta(t_1-t_2)$, we have $f:\mathbb{R}^2 \to \mathbb{R}$ is given by \begin{eqnarray} \hat{f}(\omega_1,\omega_2) &=& \int_{\mathbb{R}} \left( \int_{\mathbb{R}} f(t_1,t_2) e^{-i(\omega_1 t_1 + \omega_2 t_2)} d t_1 \right) d t_2 \\ &=& \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \delta(t_1-t_2) e^{-i(\omega_1 t_1 + \omega_2 t_2)} d t_1 \right) d t_2 \\ &=& \int_{\mathbb{R}} e^{-i(\omega_1 t_2 + \omega_2 t_2)} d t_2 \\ &=& \int_{\mathbb{R}} e^{-i((\omega_1 + \omega_2) t)} d t \\ &=& 2 \pi \delta(\omega_1 + \omega_2) \end{eqnarray} The last equation follows from the fact the transform of the distribution $t \mapsto \delta(t)$ is $\omega \to 1$. Taking the inverse transform gives the desired result.