Does the Fourier transform of a smooth $L^2$ function decay rapidly at infinity For $L^1$ functions, there is a correspondence through the Fourier Transform between functions which are smooth and functions which decay rapidly at infinity. More precisely, if $f$ is a smooth $L^1$ function, $\hat{f}$ is an $L^\infty$ function which decays faster than any polynomial at infinity, and conversely, if $g$ is an $L^\infty$ function which decays faster than any polynomial at infinity, then the inverse fourier transform of $g$ is $L^1$ and smooth.
For $L^2$ functions, there is a difficulty. The Fourier transform formula is no longer valid on individual functions. It only is defined on the $L^2$ classes of functions.
I have two questions:


*

*Is there a standard way to extend "decays rapidly at infinity" to $L^2$ classes of functions?

*Assuming (1), is there a 1-1 correspondence between classes of $L^2$ functions with a smooth representative and classes of $L^2$ functions which decay rapidly at infinity?
 A: 
if $f$ is a smooth $L^1$ function, $\hat{f}$ is an $L^\infty$ function which decays faster than any polynomial at infinity

This is not true as stated. Unquantified smoothness does not lead to quantitative conclusions. Indeed, the properties of $\hat f$ that you claimed imply $\hat f\in L^2$, hence $f\in L^2$. But a smooth $L^1$ function need not be in $L^2$. Another line of counterexamples comes from functions like $f(x) = \sin(e^x)/(1+x^4)$. This is a smooth function in $L^1\cap L^2$, but since $f'\notin L^2$, it follows that $\xi \hat f(\xi)\notin L^2$, hence $\hat f$ does not decay all that quickly. 
What is true is that a smooth $L^1$ function such that all of its derivatives are also in $L^1$ has rapidly decaying Fourier transform.
Now consider a smooth $L^2$ function $f$ whose derivatives are in $L^2$.  Then $$(1+|\xi|)^k \hat f(\xi)\in L^2\quad \text{ for all }  k\ge 0\tag{1} $$ and the converse holds too. Indeed, by the Cauchy-Schwarz inequality, 
$$
\int |g(\xi)|\le \sqrt{\int (1+|\xi|)^{-2}\,d \xi}\sqrt{\int (1+|\xi|)^2 |g(\xi)|^2\,d\xi}
$$
Hence $(1)$ implies $(1+|\xi|)^k \hat f(\xi)\in L^1 $ for all $k\in L^1$, and the latter property yields the smoothness of $f$. 
But one can build $\hat f$ that satisfies $(1)$ and is not even bounded: e.g., 
$$
\hat f(\xi) = \sum_{n=1}^\infty n \chi_{[n,n+e^{-n}]}
$$
(and get $f$ from it by the inverse transform).  So, the answer to the question posed is negative.  
