The fourier transform of $f \in L^1(\mathbb R)$ is given by $\displaystyle\hat f ( s) = \int_{\mathbb R} f(x) \exp(-2\pi ix s) \mathrm{d}x.$ Suppose $\hat f \in L^2,$ is it true that $f \in L^2?$
As I did not get any counter examples to disprove, I have tried prove this unsuccessfully :
We have $\displaystyle\lVert \hat f \rVert_2 = \int |\hat f(s) |^2\mathrm{d}s = \int \left|\int f(x) \exp(-2\pi ix s) \mathrm{d}x\right|^2 \mathrm{d}s $ is finte.
We have to show that the integral $\displaystyle\int |f(x)|^2 \mathrm{d}x$ is finite.
If Fourier inversion holds we may write $\displaystyle\int |f(x)|^2\mathrm{d}x = \int \left| \int \hat f (s) \exp(2i \pi s x)\mathrm{d}s\right|^2\mathrm{d}x$ but I am not able to proceed here. Also, I am not sure, if fourier inversion holds for $L^1$ functions.
In brief, yes, for $f\in L^1$, if $\widehat{f}\in L^2$, then $f$ itself is/was in $L^2$.
As you rightfully worry, this does not follow just by computing the Fourier inversion integral... since $\widehat{f}$ may fail to be in $L^1$, so the integral has convergence problems.
As @JohnDoe commented, in fact the general definition of the Plancherel extension to $L^2$ of the Fourier transform is not by an integral, but by taking $L^2$ limits (of Fourier transforms computed by literal integrals) from $L^2\cap L^1$.
More abstractly, if one wants, $L^1$ functions give tempered distributions, so lots of Fourier transform stuff works out fine, though not by direct, literal integrals.