Fourier transform of a Schwartz space function and norm The Schwartz space (of rapidly decreasing functions) is the set of all $C^{\infty}$ functions $f\colon\mathbb{R}\to\mathbb{C}$ such that 
$$
x^jf^{(k)}(x) \to 0
$$
for all integers $j,k\geq0$ as $x \to {\pm\infty}$.
Let $f$ be a rapidly decreasing function in Schwartz space and let $\mathcal{F}(f)$ its Fourier transform. If 
$$
\| f \|_2 = \| \mathcal{F}(f) \|_2 \quad,
$$ 
can we deduce that $\mathcal{F}(f)$ is a Schwartz function?

You don't need their norms to be equal - this is a consequence of the normalization factor chosen for the Fourier transform, and is called the Plancherel Theorem. 
To show that the Fourier transform of a Schwartz function is Schwartz, it suffices to show the following: 


*

*$k^\alpha\hat{f}(k)=i^{-\alpha}\mathcal{F}\left[f^{(\alpha)}\right]$

*$\frac{d^\alpha}{dk^\alpha}\hat{f}(k)=(-ik)^\alpha\hat{f}(k)$


Combine these two facts with the Riemann-Lebesgue lemma and you've got it.
Note: I'm using the following convention for the Fourier transform: 
$$
\hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ikx}dx
$$
