Below is a word by word copy from my textbook of a certain derivation for the Fourier transform of a Convolution:
Let $g_1(\alpha)$ and $g_2(\alpha)$ be the Fourier transforms of $f_1(x)$ and $f_2(x)$ respectively. Assuming that $$\int_{x=-\infty}^{\infty}\mid f_1(x)f_2(x)\mid\mathrm{d}x$$ is finite, then by the definition of the Fourier transform:$$g(\alpha)=\frac{1}{2\pi}\int_{x=-\infty}^{\infty}f(x)e^{-i\alpha x}\,\mathrm{d}x$$ Then $$\begin{align}g_1(\alpha)\cdot g_2(\alpha) & =\frac{1}{2\pi}\int_{v=-\infty}^{\infty}f_1(v)e^{-i \alpha v}\,\mathrm{d}v\cdot\frac{1}{2\pi}\int_{u=-\infty}^{\infty}e^{-i\alpha u}\,\mathrm{d}u \\ & = \left(\frac{1}{2\pi}\right)^2\int_{v=-\infty}^{\infty}\int_{u=-\infty}^{\infty}e^{-i\alpha(v+u)}f_1(v)f_2(u)\mathrm{d}v\,\mathrm{d}u\end{align}$$ Where we have used different dummy integration variables. Next we make the change of variables $x=v+u$, $\mathrm{d}x=\mathrm{d}v$ in the $v$ integral, to get $$\begin{align}g_1(\alpha)\cdot g_2(\alpha)& =\left(\frac{1}{2\pi}\right)^2\int_{x=-\infty}^{\infty}\int_{u=-\infty}^{\infty}e^{-i\alpha x}f_1(x-u)f_2(u)\mathrm{d}x\,\mathrm{d}u \\ & = \left(\frac{1}{2\pi}\right)^2\int_{x=-\infty}^{\infty}e^{-i\alpha x}\color{blue}{\left(\int_{u=-\infty}^{\infty}f_1(x-u)f_2(u)\,\mathrm{d}u\right)}\tag{1}\end{align}$$ If we define the convolution of $f_1(x)$ and $f_2(x)$ by $$f_1\ast f_2=\color{blue}{\int_{u=-\infty}^{\infty}f_1(x-u)f_2(u)\,\mathrm{d}u}$$ then $(1)$ becomes $$g_1\cdot g_2=\frac{1}{2\pi}\left(\frac{1}{2\pi}\int_{x=-\infty}^{\infty}f_1\ast f_2 \,e^{-i\alpha x}\,\mathrm{d}x\right)=\frac{1}{2\pi}\cdot(\text{Fourier transform of}\,f_1\ast f_2)$$ In other words $\color{#180}{\fbox{${g_1\cdot g_2}$}}$ and $\color{#180}{\fbox{${\dfrac{1}{2\pi}f_1\ast f_2}$}}$ are a pair of Fourier transforms.
Because of the symmetry of the $f(x)$ and $g(\alpha)$ integrals, there is a similar relating $f_1\cdot f_2$ and the convolution of $g_1$and $g_2$.
Namely:
$\color{red}{\fbox{${g_1\ast g_2}$}}$ and $\color{red}{\fbox{${f_1\cdot f_2}$}}$ are a pair of Fourier transforms.
It is left as an exercise in my book for me to show that $\color{red}{\fbox{${g_1\ast g_2}$}}$ and $\color{red}{\fbox{${f_1\cdot f_2}$}}$ are a pair of Fourier transforms by following the similar steps used to show that $\color{#180}{\fbox{${g_1\cdot g_2}$}}$ and $\color{#180}{\fbox{${\dfrac{1}{2\pi}f_1\ast f_2}$}}$ are a pair of Fourier transforms.
My Attempt:
Letting $f_1(x)$ and $f_2(x)$ be the Fourier transforms of $g_1(\alpha)$ and $g_2(\alpha)$ respectively. Using the definition of the Inverse Fourier transform: $$f(x)=\int_{x=-\infty}^{\infty}g(\alpha)e^{i\alpha x}\,\mathrm{d}x$$
Therefore $$\begin{align}f_1(x)\cdot f_2(x)&=\int_{r=-\infty}^{\infty}g_1(r)e^{i\alpha r}\,\mathrm{d}r\cdot\int_{s=-\infty}^{\infty}g_2(s)e^{i\alpha s}\,\mathrm{d}s\\ & = \int_{r=-\infty}^{\infty}\int_{s=-\infty}^{\infty}e^{i\alpha (r + s)}g_1(r)g_2(s)\,\mathrm{d}r\,\mathrm{d}s\end{align}$$
Where $r$ and $s$ are the dummy variables of integration. Now making the change of variables $x = r + s$, so $\mathrm{d}x=\mathrm{d}r$ in the $r$ integral, to get $$\begin{align}f_1(x)\cdot f_2(x)&=\int_{x=-\infty}^{\infty}\int_{s=-\infty}^{\infty}e^{i\alpha x}g_1(x-s)g_2(s)\,\mathrm{d}x\,\mathrm{d}s\\ & =\int_{x=-\infty}^{\infty}e^{i\alpha x}\color{#F80}{\left(\int_{s=-\infty}^{\infty}g_1(x-s)g_2(s)\,\mathrm{d}s\right)}\tag{2} \end{align}$$ Defining the convolution of $g_1$ and $g_2$ by $$g_1\ast g_2=\color{#F80}{\int_{s=-\infty}^{\infty}g_1(x-s)g_2(s)\,\mathrm{d}s}$$ Then $(2)$ becomes $$f_1\cdot f_2=\int_{x=-\infty}^{\infty}g_1\ast g_2\,e^{i\alpha x}\,\mathrm{d}x=\text{Inverse Fourier transform of}\,g_1\ast g_2$$
My question is: Can I now conclude that $\color{red}{\fbox{${g_1\ast g_2}$}}$ and $\color{red}{\fbox{${f_1\cdot f_2}$}}$ are Fourier pairs? Also, could someone please provide verification that my proof is correct?
Thanks.