In $l_1$ space we may define the Fourier transform, but the inverse Fourier transform doesn't always exist. If $f$ and $g$ have the same Fourier transform function, do we necessarily have that $f=g\quad a.e.$?
(I think there exists this question in the forum but I failed to find it...)
Did you really mean $\ell_1$ ($=\ell_1(\mathbb Z)$) or $L^1$ ($=L^1(\mathbb R)$)? See below for why I wonder. It doesn't matter to your question, the answer is the same for any locally compact Abelian group:
The Inversion Theorem says that if $h$ and $\hat h$ are both integrable then $h$ is the inverse Fourier transform of $\hat h$. Let $h=f-g$; then $\hat h=0$ is integrable, so the Inversion Theorem says that $h=0$.
The reason I suspect perhaps you didn't mean $\ell_1$ is that if $a\in\ell_1(\mathbb Z)$ then the inverse transform does always exist, and equals $a$.
Say $a=(a_j)\in\ell_1(\mathbb Z)$. The Fourier transform is the periodic function $f=\hat a$ defined by $$f(t)=\sum_j a_je^{-ijt}.$$Uniform convergence shows that $$a_j=\frac1{2\pi}\int_0^{2\pi}f(t)e^{ijt}\,dt,$$which is to say that $a$ is the inverse transform of $f$.
