Convolution Theorem and Fourier Transform

Question

How do I prove that $F(f(x)g(x)) = \frac{1}{\sqrt{2\pi}}\tilde{f}(k)*\tilde{g}(k)$?

Answer 1

Let $F, G$ the Fourier Transform of $f$ and $g$. The choice of the coefficient follows by how one defines the Fourier Transform. I will avoid id, according to my definition, but the results are the same.

Now:

\begin{align*} F(u)G(u) & = \left(\int_{-\infty}^{+\infty} e^{-2\pi i u x}\ \text{d}x\right) G(u)\\ & = \int_{-\infty}^{+\infty} e^{-2\pi i u x} G(u)\ \text{d}x \\ & = \int_{-\infty}^{+\infty} f(x) \left(\int_{-\infty}^{+\infty} g(y-x) e^{-i 2\pi u y}\ \text{d}y\right)\text{d}x \\ & = \int_{-\infty}^{+\infty} \left(\int_{-\infty}^{+\infty} f(x) g(y-x)\ \text{d}x \right)e^{-2\pi i u y}\text{d}y \\ & = \int_{-\infty}^{+\infty} (f(x)*g(x)) e^{-2\pi i u y}\ \text{d}y \\ & = \mathcal{F}\left({f(x)*g(x)}\right) \end{align*}

Now you can read the proof backwards, operating with the inverse Fourier Transform to obtain what you asked for:

$$\underbrace{\mathcal{F}^{-1}\left(\mathcal{F}\left(f(x)*g(x)\right)\right)}_{\equiv f(x)*g(x)} = \mathcal{F}^{-1}\left(F(u)G(u)\right)$$

That is

$$ f(x)*g(x) = \mathcal{F}^{-1}\left(F(u)G(u)\right)$$