I'm attempting to show that the Fourier transform of Dyson's equation for a constant potential V,
\begin{equation}
G(\mathbf{r},t,\mathbf{r}_0,t_0)
=
G^0(\mathbf{r},t,\mathbf{r}_0,t_0) + \dfrac{1}{\hbar} \int G^0(\mathbf{r},t,\mathbf{r}',t') VG(\mathbf{r}',t',\mathbf{r}_0,t_0) d^3 \mathbf{r}'d t'
\end{equation}
is
\begin{equation}
G(\mathbf{k},\omega)
=
G^0 (\mathbf{k},\omega) + \hbar^{-1} G^0 (\mathbf{k},\omega) V G(\mathbf{k},\omega)
\end{equation}
I'm having a bit of trouble showing the convolution portion of this, and could use a hand with the integration. My work is the following:
Since we are dealing with a constant potential, $ V(\mathbf{r}',t') = V $ and $G(\mathbf{r},t,\mathbf{r}_0,t_0) \rightarrow G(\mathbf{r}-\mathbf{r}_0,t-t_0) $. Applying the Fourier transform of both sides, we have \begin{align} \mathcal{F} \big \{ G(\mathbf{r}-\mathbf{r}_0,t-t_0) \big \} &= \int G(\mathbf{r}-\mathbf{r}_0,t-t_0) e^{i\omega t - i\mathbf{k}\cdot\mathbf{r}} d^3 \mathbf{r} d t \nonumber = G(\mathbf{k},\omega) \nonumber \\[.5em] &= \int \bigg [ G^0(\mathbf{r}-\mathbf{r}_0,t-t_0) + \dfrac{1}{\hbar} \int G^0(\mathbf{r}-\mathbf{r}',t-t') VG(\mathbf{r}'-\mathbf{r}_0,t'-t_0) d^3 \mathbf{r}'d t' \bigg ] e^{i\omega t - i\mathbf{k}\cdot\mathbf{r}} d^3 \mathbf{r} d t \nonumber \\[.5em] &= G^0(\mathbf{k},\omega) + \dfrac{V}{\hbar} \int \int G^0(\mathbf{r}-\mathbf{r}',t-t') G(\mathbf{r}'-\mathbf{r}_0,t'-t_0) d^3 \mathbf{r}'d t' e^{i\omega t - i\mathbf{k}\cdot\mathbf{r}} d^3 \mathbf{r} d t \label{FT_ConstantV} \end{align}
From here, I need to show that \begin{equation} \dfrac{V}{\hbar} \int \int G^0(\mathbf{r}-\mathbf{r}',t-t') G(\mathbf{r}'-\mathbf{r}_0,t'-t_0) e^{i\omega t - i\mathbf{k}\cdot\mathbf{r}} d^3 \mathbf{r}'d t' d^3 \mathbf{r} d t = \dfrac{V}{\hbar} G^0 (\mathbf{k},\omega) G(\mathbf{k},\omega) \end{equation}
I know this involves a convolution, but I am unsure how to show this. Any help would be appreciated.
Edit: Because this is a constant potential, $G(\mathbf{r},t,\mathbf{r}_0,t_0) \rightarrow G(\mathbf{r}-\mathbf{r}_0,t-t_0) $. I have edited my original work to reflect this.
Edit: $G^0 (\mathbf{k},\omega)$ and $G(\mathbf{k},\omega)$ represent the Fourier transform of $G(\mathbf{r}-\mathbf{r}_0,t-t_0)$ and $G^0(\mathbf{r}-\mathbf{r}_0,t-t_0)$, respectively.