# Fourier inverse transform of absolutely integrable function

Given that $f$ and its fourier transform $\mathcal{F}f$ are in $L^1(\mathbb{R^n})$, how to show that $\mathcal{F^{-1}} \mathcal{F} f=f$?.

I tried using Schwartz class functions to aproximate $f$ in $L^{1}$ as fourier transform is isometric isomorphism on Schwartz class functions.

• This is the ($L^1$) Inversion Theorem. It's one of the two basic results about the Fourier transform; see Rudin Real and Complex Analysis or Folland Real Analysis or the the start of any other treatment of the Fourier transform in the universe. – David C. Ullrich Mar 4 '18 at 18:39
You can show that, if $f \in L^1$, then the following holds for all $t > 0,\; x\in\mathbb{R}$: $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(s)e^{-ts^2}e^{isx}ds = \frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}f(y)e^{-(x-y)^2/4t}dy.$$ (This comes from solving Fourier's Heat Equation.) As $t\downarrow 0$, the right side tends to $f$ in the $L^1$ sense, which implies that there is a sequence $\{ t_n \}$ tending monotonically down to $0$ from above such that the right side evaluated at $t_n$ converges pointwise a.e. to $f$. Now if $\hat{f}$ is also in $L^1$, then the left side evaluated at $t_n$ tends pointwise everywhere to $\hat{f}^{\vee}(x)$, which proves $\hat{f}^{\vee}=f$ a.e.. In particular, $f$ can be changed on a set of measure zero so that it is continuous. Then, for the continuous $f$, $\hat{f}^{\vee}=f$ holds everywhere.
The argument in $\mathbb{R}^n$ works basically the same way.