Given $f \in L^2(\mathbb{R})$, exists dual function where $f \in W^{1, 2}(\mathbb{R})$ iff $\int_\mathbb{R} |\xi|^2 |\hat{f}(\xi)|^2\,d\xi < +\infty$?

Question

Let$$W^{1, 2}(\mathbb{R}) := \left\{ (f, f') \in L^2(\mathbb{R}) \times L^2(\mathbb{R}): f \text{ has a continuous representative and for all }x,\,y \in \mathbb{R},\text{ }f(x) - f(y) = \int_y^x f'(z)\,dz\right\}.$$If $(f, f') \in W^{1, 2}$, then for all $x$, $y \in \mathbb{R}$, we have$$|f(x) - f(y)| \le \|f'\|_{L^2} |x - y|^{1/2}.$$(This follows directly from Cauchy-Schwarz applied to the product of $f′$ and the function that is constant and equal to $1$.)

Question. Given $f \in L^2(\mathbb{R})$, is there an $f' \in L^2(\mathbb{R})$ so that $(f, f') \in W^{1, 2}(\mathbb{R})$ if and only if$$\int_\mathbb{R} |\xi|^2 |\hat{f}(\xi)|^2\,d\xi < +\infty?$$

Answer 1

I don't understand why is this isn't a nontrivial consequence of the characterizations of Sobolev spaces. If $f\in W^{1,2}$, then $|\widehat{\partial_{x}f}(\xi)|=|\xi||\widehat{f}(\xi)|\in L^{2}(\mathbb{R})$. If the Fourier condition holds, then by distributional calculus, $c(\xi\widehat{f})^{\vee}(x)=f'(x)\in L^{2}(\mathbb{R})$, where $c$ is some absolute constant.