Is it true in general that if $h \in L^{1}(\mathbb{R})$ such that $\hat{h} \in L^{2}(\mathbb{R})$, where $\hat{h}$ is the Fourier transform of $f$, then in fact $h \in L^{2}(\mathbb{R})$? If so how one can go by proving it? Otherwise is there a counter-example?
Thanks.
Yes, it is, due to Plancherel theorem. The Fourier transform is defined for functions in $L^1$ initially; however, Plancherel theorem states that $\|h\|_2 = \|\hat{h}\|_2$ (or, more generally, that $\langle f,g \rangle = \langle \hat{f},\hat{g} \rangle$). Since $\hat{h} \in L^2$, one has thus $\|h\|_2 = \|\hat{h}\|_2 < \infty$, hence $h \in L^2$. This theorem is actually responsible for the Fourier transform $\mathscr{F} : L^2(\Bbb{R}) \to L^2(\Bbb{R})$ to be an isometry inside $L^2$ (i.e. a unitary automorphism of $L^2$).