Let $f\in L^{1}(\mathbb R)$ and we define its Fourier transform as follows: $\hat{f}(\xi)=\int_{\mathbb R} f(x)e^{-2\pi i \xi\cdot x} dx, (\xi \in \mathbb R);$ and we define $f^{\vee}(x):=\hat{f}(-x)= \int_{\mathbb R} f(\xi) e^{2\pi i \xi\cdot x} d\xi, (x\in \mathbb R).$
By Fourier inversion Theorem we get: If $f\in L^{1}(\mathbb R)$ and $\hat{f}\in L^{1}(\mathbb R)$, then $(\hat{f})^{\vee}=f.$
Let $f\in L^{1}(\mathbb R).$ Clearly, $\hat {f}\in L^{\infty}(\mathbb R).$
My Questions:
(1) Can we expect $(\hat{f})^{\vee}= f$; without assuming $\hat{f}\in L^{1}(\mathbb R)$ ?
(2)(Motivation behind first question) Put, $\mathcal{F}L^{1}:= \{f\in L^{\infty}(\mathbb R):\hat{f}\in L^{1}(\mathbb R)\}$ and define the map $\mathcal{F}:\mathcal{F}L^{1}\to L^{1}(\mathbb R)$ such that $ \mathcal{F}(f)= \hat{f}, (f\in \mathcal{F}L^{1}).$ Can we expect the map $\mathcal{F}$ is onto ? (This may be possible to answer without answering first question )
Thanks,