Derivative of a function and Sobolev space

Question

Let function $f$ belongs to the Sobolev space or order $\beta$ defined by $$ \mathcal{S}^{\beta}(\mathbb{R}) = \left\{u \in L^2(\mathbb{R}): \int_{\mathbb{R}}(1+|\xi|^2)^{\beta}|\hat u(\xi)|^2d\xi < \infty \right\}, $$ where $\hat u(\xi) = \int_{\mathbb{R}} e^{-ix\xi}u(x)dx$ is the Fourier transform of $u$. Can we deduce directly that the derivative $f'$ of $f$ belongs to $\mathcal{S}^{\beta - 1}(\mathbb{R})$? Thank in advance for any help.

Answer 1

By the answers from this question:

Fourier Transform of Derivative

It follows that $|\hat{u'}(\xi)|^2=\frac{1}{({2\pi|\xi|})^2}|\hat{u}(\xi)|^2$. Inserting this with $f\in\mathcal{S}^\beta(\mathbb{R})$ then brings

\begin{align*} \int_\mathbb{R} \frac{1}{4\pi^2}(1+|\xi|^2)^{\beta-1}|\hat{u'}(\xi)|^2d\xi\leq \int_\mathbb{R} \frac{(1+|\xi|^2)^{\beta}}{4\pi^2|\xi|^2}|\hat{u'}(\xi)|^2d\xi=\int_\mathbb{R}(1+|\xi|^2)^\beta|\hat{u}(\xi)|^2d\xi <\infty. \end{align*}

Answer 2

Using that $\hat{u'}(\xi) = i\xi\hat u(\xi)$ and $\lvert\xi\rvert \leq 1+\lvert\xi\rvert^2$: $$ \int_\mathbb R (1+\lvert\xi\rvert^2)^{\beta-1}\lvert\hat{u'}(\xi)\rvert^2d\xi = \int_\mathbb R \lvert\xi\rvert(1+\lvert\xi\rvert^2)^{\beta-1}\lvert\hat{u}(\xi)\rvert^2d\xi \leq \int_\mathbb R (1+\lvert\xi\rvert^2)^\beta\lvert\hat u(\xi)\rvert^2 d\xi. $$