Prove $H^k(\mathbb{R}^n) = W^{k,2}(\mathbb{R}^n)$ I want to show $H^k(\mathbb{R}^n) = \{ u \in S'(\mathbb{R}^n): (1+|\xi|^2)^\frac{k}{2}\widehat u \in L^2\}$ is equivalent to the Sobolev space $W^{k,2}(\mathbb{R}^n)$.
If $u \in H^k(\mathbb{R}^n)$ then $\widehat u = \frac{f}{(1+|\xi|^2)^\frac{k}{2}} \in L^2$ so by Plancherel's theorem $u \in L^2$. Moreover $\forall |\alpha| \leq k: |\widehat{\partial^{\alpha} u}| = |\xi^\alpha|\widehat u \leq (1+|\xi|^2)^\frac{k}{2} |\widehat u| \in L^2$. So $H^k \subseteq W^{k,2}$. The other direction seems harder to me. Any idea on how to proceed? Many thanks.
 A: Take $d=1$, $k=1$ for simplicity; the idea generalizes easily. If $u\in W^{k,2}$ then $u\in L^2$ and has a weak derivative $g\in L^2$, i.e.
$$
\int_{\mathbb{R}} g\phi~dx = -\int_{\mathbb{R}} u\phi_x~dx
$$
for any Schwartz function $\phi$.
We now claim that $i\xi \widehat{u}\in L^2$, and show this by proving $\widehat{g} = -i\xi\widehat{u}$ almost everywhere, whereby Plancherel implies the claim since $g\in L^2$. Intuitively this claim is natural since $g = u_x$ for classically differentiable functions, but since $u$ is only weakly differentiable we need to proceed without ever directly referring to $u_x$. To prove the claim, we take the definition of weak differentiability and apply Plancherel, using the fact that $g,u,\phi,\phi_x$ all belong to $L^2$. This yields
$$
\int_{\mathbb{R}} \widehat{g}\widehat{\phi}~d\xi = -\int_{\mathbb{R}} \widehat{u}\widehat{\phi_x}~d\xi = -\int_{\mathbb{R}} i\xi\widehat{u}\widehat{\phi}~d\xi.
$$
Since $\phi$ is an arbitrary Schwartz function, so is $\widehat{\phi}$. Therefore by the fundamental lemma of calculus of variations, we conclude that $\widehat{g} = -i\xi\widehat{u}$ almost everywhere.
Since $\widehat{u}\in L^2$ and $\xi \widehat{u}\in L^2$, it now follows that $(1+|\xi|^2)^{\frac{1}{2}}\widehat{u}\in L^2$ from unpacking the latter definition.
For higher dimensions the proof is essentially identical. For higher derivatives you will need to show that $|\xi|^j\widehat{u}\in L^2$ for $j=1,\ldots,k$, and use this to conclude that $(1+|\xi|^2)^{\frac{k}{2}}\widehat{u}\in L^2$; the reason becomes clear when you expand out $(1+|\xi|^2)^k$.
