An equality about the Sobolev space $W^{1,2}$

Question

From the Plancherel identity, we know that $$\int_{\mathbb{R}} |f(x)|^2\,dx=\int_{\mathbb{R}}|\widehat{f}(\xi)|^2\,d\xi$$ is valid for all $L^2(\mathbb{R})$ functions and in particular for Schwartz functions $f\in \mathcal{S}(\mathbb{R})$. Also we know that the Fourier transform of the derivative $f'$ of $f$ is $2\pi i\xi\widehat{f}(\xi)$. From this we see that \begin{equation} \int_{\mathbb{R}} |f'(x)|^2\,dx=\int_{\mathbb{R}} (2\pi \xi)^2|\widehat{f}(\xi)|^2\,d\xi. \tag{*} \end{equation} for all $f\in\mathcal{S}(\mathbb{R})$. Now I want to show that $(*)$ is valid for all $f\in W^{1,2}(\mathbb{R})$. The hint says that we can use a limiting argument and use the fact that $L^2$ is complete. I know that $\mathcal{S}(\mathbb{R})$ is dense in $W^{1,2}(\mathbb{R})$, but I don't know how to use the limiting argument.

Answer 1

Take a sequence of Schwartz functions $(f_n)_n$ which converges in $W^{1,2}$ to $f$. This implies $f_n ' \to f'$ in $L^2$ and hence $(2\pi i \xi) F[f_n] = F[f_n '] \to F[f']$ in $L^2$.

But we also know $F[f_n] \to F[f]$ in $L^2$. Now, use that convergence in $L^2$ yields a subsequence which converges almost everywhere.

From this, it is not hard to see that $F[f'] = (2 \pi i \xi) F[f]$. Now use that $\int |f'|^2 = \int | F[f']|^2$ to conclude the proof.

The same argument as given here is very often useful.