Deriving integral formula for Fourier transform of $L^1$ functions

Question

How should I derive the formula of the Fourier transform

$$\hat{f}(\kappa)=\int_{\mathbb{R}^n}f(x)e^{-2\pi i\kappa\cdot x}dx$$

Given that $f$ is in $L^1(\mathbb{R}^n)$ regarded as a tempered distribution?

Answer 1

It comes from the fact that $$\int_{\mathbb R^n} \hat f(\xi) \phi(\xi) d\xi = \int_{\mathbb R^n} f(x) \hat \phi(x) dx $$ for all $\phi \in \mathcal S(\mathbb R^n)$. It can be shown using Fubini's theorem.

Answer 2

First, the definition of FT for tempered distributions:

$$\langle \hat f, \varphi \rangle = \langle f , \hat \varphi\rangle$$

Second, the representation of $L^1$ distributions as integrals: $$\langle f , \hat \varphi\rangle = \int_\Bbb R f(\xi) \hat \varphi(\xi)d\xi$$

Afterwards, nothing but real analysis (Foubini theorem helps): $$\int_\Bbb R f(\xi) \hat \varphi(\xi)d\xi = \int_\Bbb R f(\xi) \left(\int_\Bbb R e^{-2\pi i\xi x}dx \varphi(x)\right)d\xi $$ $$=\int_\Bbb R \left(\int_\Bbb R f(\xi) e^{-2\pi i\xi x} d\xi \right)\varphi(x) dx = \left\langle\int_\Bbb R f(\xi) e^{-2\pi i\xi x} d\xi,\varphi(x)\right\rangle$$