Is there a way, without using much extra theory (other than the basic ideas used in textbooks deriving the Fourier transform for the first time, and ideally just using general theorems about integrals), of showing that
$$1.\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}g(t)h(x-t)dt\right)e^{-i2 \pi k x}dx=\int_{-\infty}^{\infty}g(x)e^{-i2 \pi k x}dx\int_{-\infty}^{\infty}h(x)e^{-i2 \pi k x}dx$$ $$2.\int_{-\infty}^{\infty}g(x)h(x)e^{-i2 \pi k x}dx=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{-i2 \pi t x}dx \int_{-\infty}^{\infty}f(x)e^{-i2 \pi (k-t) x}dx dt?$$
The two above threorems are the long-widned versions of the below: $$1.f(x)=g(x)\otimes h(x)\Leftrightarrow F(k)=G(k)H(k)$$ $$2.f(x)=g(x) h(x)\Leftrightarrow F(k)=G(k) \otimes H(k).$$
Where $$g(x)\otimes h(x)=\int_{-\infty}^{\infty}g(t)h(x-t)dt$$ $$f(x)=\int_{-\infty}^{\infty}F(k)e^{i2\pi k x}dk \Leftrightarrow F(k)=\int_{-\infty}^{\infty}f(x)e^{-i2 \pi k x}dx.$$
If $h(t),g(t)\in L_1( \mathbb{R} )$, then we have
$$ \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}g(t)h(x-t)dt\right)e^{-i2 \pi k x}dx = \int_{-\infty}^{\infty}g(t)\int_{-\infty}^{\infty}h(x-t)e^{-i2 \pi k x}dx dt .$$
Using the change of variables $x-t=y$ gives
$$ = \int_{-\infty}^{\infty}g(t)\int_{-\infty}^{\infty}h(y)e^{-i2 \pi k (y+t) }dy dt = \int_{-\infty}^{\infty}g(t)e^{-i2 \pi k t }dt\int_{-\infty}^{\infty}h(y)e^{-i2 \pi k y } dy \,. $$