If $\hat{f}\in L^2(\mathbb{R})$ then $\hat{f}$ is rapidly decreasing. This is problem 2.3.6 from the book of Mckean, "Fourier Series and Integrals".

Problem:
Check that if $\hat{f}\in L^2(\mathbb{R}^1)$ then $\hat{f}$ is rapidly decreasing in the sense of $\gamma^n\hat{f}(\gamma)\in L^2(\mathbb{R}^1)$ for every $n\geq 0$ iff $f\in C^{\infty}$ and all its derivatives $D^nf$ belong to $L^2(\mathbb{R^1})$. Also check the formula $\widehat{D^nf}=(2\pi i\gamma)^n \hat{f}$.

There are a lot of things I do not understand here, and I would be really happy if you could help me sort things out :).
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Firstly: I thought the space of rapidly decreasing functions, also known as Schwartz space, was the space
$$S(\mathbb{R})=\{f\in C^{\infty}(\mathbb{R}):\|f\|_{\alpha,\beta}<\infty,\qquad \alpha,\beta\in\mathbb{Z}_+\},$$
where $C^{\infty}(\mathbb{R})$ denotes the set of smooth functions from $\mathbb{R}$ to $\mathbb{C}$ and
$$\|f\|_{\alpha,\beta}=\sup_{\alpha,\beta\in\mathbb{R}}\Big | x^{\alpha}D^{\beta}f(x) \Big|.$$
That is, if I hadn't read the 'definition' they gave in the exercise, I would have tried to show that $\hat{f}\in S(\mathbb{R})$. Do you know if this equivalent to being in $S(\mathbb{R})$? Since I cannot see how both of the definitions can be equivalent, I guess I have to stick with the first one (the one they gave in the exercise).
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Secondly: Perhaps a silly question, if I want to check that $\hat{f}$ is rapidly decreasing according to the above definition, should I both check that 
"$\gamma^n\hat{f}(\gamma)\in L^2(\mathbb{R}^1)$ for every $n\geq 0$ implies $f\in C^\infty$ and all its derivatives $D^nf$ belong to $L^2(\mathbb{R^1})$"
and that
"$f\in C^{\infty}$ and all its derivatives $D^nf$ belong to $L^2(\mathbb{R^1})$ implies $\gamma^n\hat{f}(\gamma)\in L^2(\mathbb{R}^1)$ for every $n\geq 0$"?
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Thirdly: A "hint" is given in the book. They say that "One half of the proof is a little tricky if $f$ is not compact: You need to show, e.g., that $f(x)\to 0$ as $x\to\infty$ if $f$ and $f'$ belong to $L^2(\mathbb{R})$. To do this, write
$$f^2(x)=f^2(0)+\int_0^x (f^2)' \, dt=f^2(0)+2\int_0^x ff' \, dt$$
in order to convince yourself that the limit exists, and then argue separately that it had better be zero."
I don't understand anything of the above hint. Why should i consider $f$, when not compact, as a separate case? Why do I want to show that $f(x)\to 0$ as $x\to\infty$? How does the above expression for $f^2(x)$ help me?

I would be really happy if you could help me lay out a solution to this problem (and explain to me what I am supposed to prove). Perhaps you could give me the steps, and some hints, for me to fill out. I really don't know where to begin and how to tackle this problem. :)
 A: Essentially, the result follows from the identity $\widehat{D^nf}=(2\pi i\gamma)^n\widehat{f}$ (whenever either of the terms it's well defined).
If $\gamma^n\widehat{f}(\gamma)\in L^2(\mathbb{R})$ for all $n$ then $\widehat{f}\in L^1(\mathbb{R})$ since in particular $|\widehat{f}(\gamma)|\leqslant C(1+|\gamma|)^{-2}$ a.e. for some constant $C>0$. Then the inverse Fourier transform formula gives $f(\gamma)=\int_\mathbb{R}e^{2\pi i x\gamma}\widehat{f}(x)\mathrm{d}x$ for all $\gamma\in\mathbb{R}$. Then, using again that $\gamma^n\widehat{f}(\gamma)\in L^2(\mathbb{R})$ we can differentiate inside the integral to get $D^nf(\gamma)=\int_{\mathbb{R}}e^{2\pi i x\gamma}(2\pi\gamma)^n\widehat{f}(x)\mathrm{d}x$.
For $f\in C^\infty$ such that $D^nf\in L^2$ for all $n$, consider first the case when $f\in\mathcal{S}$. Then $\lim_{\gamma\to\pm\infty}D^mf(\gamma)=0$ for all $m$ and integration by parts gives $\widehat{D^nf}(\gamma)=(2\pi i\gamma)^n\widehat{f}(\gamma)$ so that $\gamma^n\widehat{f}(\gamma)\in L^2$ for all $n$. In the general case, for a given $n$ you can always approximate $f$ and $D^nf$ in $L^2$ by $\{f_k\}_k$ and $\{D^nf_k\}_k$ respectively with $f_k\in\mathcal{S}$.
