Showing $\widehat{f \star g}(\xi) = \widehat{f}(\xi) \widehat{g}(\xi)$

Question

Goal. Show that $\widehat{f \star g}(\xi) = \widehat{f}(\xi) \widehat{g}(\xi)$ in the context of Fourier Analysis.

Assumptions.

\begin{equation} \widehat{f}(\xi) = \int_{\mathbb{R}} e^{-2 \pi i x \xi} f(x) dx \tag{∧} \end{equation}

\begin{equation} (f ⋆ g)( \xi ) = ∫f(\xi -y)g(y) dy \tag{⋆} \end{equation}

Attempt.

\begin{align*} \widehat{f ⋆ g}(\xi) &= ∧ ∘ (f ⋆ g)(\xi)\\ &= ∫_\mathbb{R} \left( e^{-2 \pi i x \left( ∫f (\xi - y) g(y) dy\right)}∫f(x - y) g(y)dy\right)dx \end{align*}

and

\begin{align*} \widehat{f}(\xi) \widehat{g}(\xi) &= \left( \int_{\mathbb{R}} e^{-2 \pi i x \xi} f(x) dx \right) \left( \int_{\mathbb{R}} e^{-2 \pi i x \xi} g(x) dx \right) \end{align*}

From here it is not so clear how these converge to the same identity.

Answer 1

\begin{align*} \displaystyle \widehat{f \ast g} &:= \int_{\mathbb{R}^n} e^{-2\pi i(x,\xi)} (f \ast g)(x) dx \\ &= \int_{\mathbb{R}^n} e^{-2\pi i(x,\xi)} \left(\int_{\mathbb{R}^n} g(x-y)f(y) dy \right)dx \\ &= \int_{\mathbb{R}^n} e^{-2\pi i(y,\xi)} f(y)dy \int_{\mathbb{R}^n} e^{-2\pi i(x-y,\xi)} g(x-y) dx \\ &= \widehat{f} \cdot \widehat{g} \end{align*} by Fubini theorem, properties of the scalar product and with the change of variable $z=x-y$.

Answer 2

You seem to have some weird notion as to what is the argument of what. The following is an identity of numbers

$$\widehat H(\xi):=\int_{\Bbb R}e^{-2\pi ix\xi}H(x)\,dx $$

Therefore, we may see $\widehat H$ as a function $\widehat H:\Bbb R\to\Bbb C$ and, consequently, the Fourier transform as a function $$\mathcal F:L^1_{\Bbb C}(\Bbb R)\to \Bbb C^{\Bbb R}\\ \mathcal F(H)= \widehat H$$

Or, if it's more clear to you, you may see the entire construct $\widehat{(\bullet_1)}(\bullet_2)$ as a function of two variables - the first one being a function $H:\Bbb R\to \Bbb C$ and the second one being a real number $\xi$ - which assigns to $(H,\xi)$ the number $\mathcal F(H)(\xi)$.

In particular, however, $\mathcal F\circ H(\xi)$ is not a thing, because $\mathcal F\circ H$ simply does not exist. It's $\mathcal (\mathcal F(H))(\xi)$.

In your case, you are asked to evaluate the instance where $H=f\star g$. Specifically $$\left(\mathcal F\left(\boxed{f\star g}\right)\right)(\xi)=\int_{\Bbb R} e^{-2\pi i x\xi}\left(\boxed{f\star g}\right)(x)\,dx=\\=\int_{\Bbb R}e^{-2\pi i x\xi}\left(\int_{\Bbb R} f(y)g(y-x)\,dy\right)\,dx$$ And here it's really just a job for Fubini-Tonelli.