The following sentence appears in the functional analysis book by Daya Reddy: "Note that if $u$ belongs to $H^s(\Omega),$ then $Au\in H^{s-2m}(\Omega)$ since $A$ is a differential operator of order $2m$".
Here, $A$ refers to a continuous differential operator, and $s\geq 2m.$ I was trying to reason this using weak derivatives but am not able to come up with a convincing explanation. I'd appreciate any help to this end.
On
I just realized that the reason is simple: as $u\in H^s(\Omega),$ we have $D^\alpha u\in L^2(\Omega)\,\forall \lvert\alpha\rvert\leq s,$ which implies $D^{\alpha_1}\left(D^{\alpha_2} u\right)\in L^2(\Omega)\,\forall \lvert\alpha_1\rvert\leq s-2m,\lvert\alpha_2\rvert\leq 2m,$ and therefore, $D^{\alpha_1}\left(Au\right)\in L^2(\Omega)\,\forall \lvert\alpha_1\rvert\leq s-2m.$ Thus, $Au\in H^{s-2m}(\Omega).$
For $s \in \mathbb R$ the Sobolev space $H^s$ is defined to be $$H^s = \{ f \in \mathscr S' : (1+|\xi|^2)^{s/2} \hat f \in L^2 \}$$
Take $f \in H^s$. Then $f \in \mathscr S'$ and $(1+|\xi|^2)^{s/2} \hat f \in L^2$. Now, $\mathscr S'$ is closed under differentiation, so $f' \in \mathscr S'$, and since $\widehat{f'}(\xi) = i \xi \, \hat f\!(\xi)$ we have to reduce $s$ with $1$ to get $(1+|\xi|^2)^{s/2} \widehat{f'} \in L^2$: $$\begin{align} \int \left| (1+|\xi|^2)^{(s-1)/2} \widehat{f'} \right|^2 & = \int (1+|\xi|^2)^{s-1} \, |\xi|^2 \, |\hat f|^2 \\ & \leq \int (1+|\xi|^2)^{s-1} \, (1+|\xi|^2) \, |\hat f|^2 \\ & = \int (1+|\xi|^2)^{s} \, |\hat f|^2 \\ & = \int \left| (1+|\xi|^2)^{s/2} \, \hat f \right|^2 \\ & < \infty \end{align}$$