Note: $<\xi>=(1+|\xi|^2)^{1/2}$
Why $u\in H^k(\mathbb{R}^n)\Leftrightarrow <\xi>^k\hat{u}\in L^2(\mathbb{R}^n)$?
To think that to use the density of the space of Schwartz in the space of Sobolev but I do not arrive at anything ... My idea $S:=S(\mathbb{R}^n)$ is dense in $H^s(\mathbb{R}^n)$ then for $u$ in $H^s(\mathbb{R}^n)$ exists $(u_j)\in S$ such that $u_j\to u$. Then $D^{\alpha}u_j\to D^{\alpha}u$ and $\left\|(1+|x|^2)^{k}D^{\alpha}(u_j)\right\|_{L^{\infty}}\leq C$... I also know that $u\in L^2$ if and only if $\hat{u}\in L^2$
For $\Rightarrow$, let $\alpha$ be such that $|\alpha| \leq k$, let $f=D^{\alpha}u$, let $\hat{u}$ be the $L^2$ (and distributional) Fourier transform of $u$. $f$ is both a $\mathcal{S}’$ distribution and a $L^2$ function, hence its (distributional) Fourier transform is (up to a scalar function) $\xi^{\alpha}\hat{u}$, and it must also be a $L^2$ function.
Now, $\langle \xi\rangle^k |\hat{u}| \leq C\sum_{|\alpha| \leq k}{|\xi^{\alpha}\hat{u}|}$, hence $\langle \xi\rangle^k |\hat{u}| \in L^2$.
The proof is similar in the other sense, if you recall $|\xi^{\alpha}| \leq \langle \xi \rangle^k$.