How to decide if a function is a Fourier transform of a function in $L^2(\mathbb{R},\mathbb{C})$

Question

The function in question is $$ g(x) = \begin{cases} e^x \mbox{ if } x<0,\\ 0 \mbox { if } x\geq0. \end{cases} $$ Is there an easy way to decide if there exists a function $f\in L^2(\mathbb R,\mathbb C)$ such that $\mathcal F(f)=g$? The inverse Fourier transform of $g(x)$ is $$ \frac 1{2\pi} \left( \frac 1{t^2+1}-i\frac t{t^2+1}\right). $$

Answer 1

To wrap it up we have $$ \int|g|^{2}dx=\int^{0}_{-\infty}e^{2x}dx=\frac{1}{2}e^{2x}|^{0}_{-\infty}=\frac{1}{2}<\infty $$ Therefore by Plancherels' theorem $\mathcal{G}g$ would work. In fact $g$ is also in $L^{1}$, so in fact lies in $L^{1}\cap L^{2}$, and Fourier transform is well defined, it needs no extension there. As PhoemueX pointed out the function is not in Schwartz class.