TITLE: Uniqueness of Fourier transform in $L^1$
QUESTION [8 upvotes]: The Fourier transform of an $L^1$ function is defined by $$\hat{f}(y)=\int_\mathbb{R}f(x)e^{-ixy}dx$$
Is it true that for functions $f,g\in L^1$, if $\hat{f}=\hat{g}$, then $f=g$?

REPLY [6 votes]: $$ \int_{\mathbb R} e^{-a|x|^2/2 + i x x_0} \hat f(x) \, dx = \int_{\mathbb R}  f(y) \int_{\mathbb R} \frac1{2\pi} e^{-a|x|^2/2} e^{-ix(y-x_0)} \, dx \, dy = \int_{\mathbb R} \frac1{\sqrt{2 \pi a}} e^{-|y-x_0|^2/2a} f(y) \, dy .$$
The first equality is by substitution and Fubini.  The second uses standard tables of Fourier transforms.  The last quantity converges to $f(x_0)$ in $L_1$ as $a\to 0^+$ (see first that it converges if $f$ is a continuous and compactly supported, then use the usual tricks of approximating $f$ by a such a function).  I might be off by a factor of $2\pi$.