Sobolev space and partial derivative. Showing that $D_j:H^{s}\to H^{s-1}$.

Question

To show that $D_j u\in H^{s-1}$ I need to see that $\langle \xi \rangle^{s-1}D_j u\in L^2$ when $\langle \xi \rangle^{s}u\in L^2$

Why $D_ju\in H^{s-1}$?

Answer 1

Everything's done in terms of the Fourier transform, so it's no surprise that this should be too. Just remember that the derivative $D_j$ on the Fourier transform side is multiplication by $\xi_j$ so that $$ |\xi_j|\langle \xi\rangle^{s-1}|\hat{u}| \leq \langle \xi \rangle^s|\hat{u}|, $$ where we used that $|\xi_j|\leq \langle \xi\rangle$. The $L^2$ norm of the left hand side is $\| D_ju\|_{H^{s-1}}$, while the right hand side's is $\| u\|_{H^s}$.