The convolution theorem states that the Fourier transform of the product of two functions is the convolution of their Fourier transforms. In the [-1,-1] notation, the Fourier transform of the product of the two functions $f\left(\vec{r}\right)$ and $g\left(\vec{r}\right)$ is,

$$\mathcal{F}_{-1,-1}\{f\left(\vec{r}\right)g\left(\vec{r}\right)\}=\frac{1}{\left( 2\pi \right)^d}\int\limits_{-\infty}^{\infty} f\left(\vec{r}\right)g\left(\vec{r}\right)e^{-i\vec{k}\cdot\vec{r}}d\vec{r},$$

where $d$ is the number of dimensions. Expressing $f\left(\vec{r}\right)$ and $g\left(\vec{r}\right)$ in terms of their Fourier transforms yields,

$$\mathcal{F}_{-1,-1}\{f\left(\vec{r}\right)g\left(\vec{r}\right)\}=\frac{1}{\left( 2\pi \right)^d}\int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} F_{-1,-1}\left(\vec{k'}\right)e^{i\vec{k'}\cdot\vec{r}}d\vec{k'}\int\limits_{-\infty}^{\infty} G_{-1,-1}\left(\vec{k''}\right)e^{i\vec{k''}\cdot\vec{r}}d\vec{k''}e^{-i\vec{k}\cdot\vec{r}}d\vec{r}.$$

The exponential factors can be combined so that this can be expressed as some function of $\vec{k'}$ and $\vec{k''}$ times the Fourier transform of the plane wave $e^{i(\vec{k'}+ \vec{k''})\cdot\vec{r}}$. The Fourier transform of a plane wave is a $\delta$-function.

$$\delta(\vec{k}-\vec{k}')=\frac{1}{(2 \pi)^d}\int\limits_{\infty}^{\infty}e^{i(\vec{k}-\vec{k}')\cdot\vec{r}}d\vec{r}.$$ $$\mathcal{F}_{-1,-1}\{f\left(\vec{r}\right)g\left(\vec{r}\right)\}=\int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \delta \left(\vec{k}-\vec{k'}-\vec{k''}\right) F_{-1,-1}\left(\vec{k'}\right) G_{-1,-1}\left(\vec{k''}\right)d\vec{k'}d\vec{k''}.$$

Integrating over $\vec{k''}$ yields,

$$\mathcal{F}_{-1,-1}\{f\left(\vec{r}\right)g\left(\vec{r}\right)\}= \int\limits_{-\infty}^{\infty} F_{-1,-1}\left(\vec{k'}\right) G_{-1,-1}\left(\vec{k'}-\vec{k}\right)d\vec{k'}.$$

This last line states that the Fourier transform of the product of two functions is the convolution of their Fourier transforms.

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• Convolution theorem on Wikipedia