Extension of Fourier transform to $L^1$

Question

If $f: \mathbb{R}^n \to \mathbb{R} $ is a function on the Schwartz space $\mathcal{S} (\mathbb{R}^n) $ we define the Fourier transform of $f$ as the function $$ \hat f(\xi)= \int_{\mathbb{R}^n} f(x)e^{2 \pi ix\xi} dx $$ How can I prove that if $f \in L^1(\mathbb{R}^n)$, then the same formula for $\hat f$ holds? I'd like to prove that that formula extends the Fourier transform on $L^1$.

Answer 1

If $f \in L^1(\mathbb{R}^n)$, then $\|f(x) e^{2\pi i x \zeta}\|_{L^1} \leq \|f(x)\|_{L^1} \|e^{2 \pi i x \zeta}\|_{L^\infty} = \|f(x)\|_{L^1} < \infty$. So the equation is well defined on $L^1$. We used Holder's inequality.

Or as was pointed out by Surb, more elementarily: $$ \int |f(x) e^{2\pi i \zeta x}| \leq \int |f(x)| < \infty$$ where the last inequality is by assumption.

It's worth noting that defining the Fourier transform on $L^1$ is the most natural starting place. No reason to restrict to the Schwarz space, though that can be useful when trying to prove nice properties such as interaction with differentiation.

Answer 2

It is known that $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^1(\mathbb{R}^n)$. Hence every function of $L^1(\mathbb{R})$ can be approximated by a function of $\mathcal{S}(\mathbb{R}^n).$ I wanted to prove that the formula can be seen as limit of a sequence of Fourier transform in $\mathcal{S}(\mathbb{R}^n)$.