Suppose I have a function $f,$ and after computing its Fourier transform using $\hat{f}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-i\omega x}dx$ (which is the definition for $L^1$ functions), I show that $\hat{f}\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$. Does it automatically mean $f\in L^1(\mathbb{R}) \cap L^2(\mathbb{R})?$
I was thinking it should, since the Fourier transform definition for $L^1$ and $L^2$ coincides for functions in the intersection, and the Fourier transform defines a bijection on these two spaces.