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I have the following question: given $m$, $p$, and $\Omega \subset \mathbb R^n$ bounded or unbounded, if the sequence $$u_i \rightharpoonup u$$ weakly in $W^{k,p}(\Omega)$, then is it true that $$\nabla^{k-m}u_i \rightharpoonup \nabla^{k-m}u$$ weakly in $W^{m,p}(\Omega)$ with the convention $$ \nabla^q = \left\{ \begin{aligned} & \Delta^{q/2} & & \text{ if } \; q \; \text{ is even},\\ & \nabla\Delta^{(q-1)/2} & & \text{ if } \; q \; \text{ is odd}. \end{aligned} \right. $$

A particular case of this question is the following: if $u_i \rightharpoonup u$ weakly in $H^1(\Omega)$, then $\nabla u_i \rightharpoonup \nabla u$ weakly in $L^2(\Omega)$. I believe that the answer is positive.

Please advise.

QA Ngô
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1 Answers1

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Yes it is true. Taking the weak derivative is a bounded and linear operation, hence it is weakly continuous (weak-weak continuous). E.g. see here.

Take $\nabla^{k-m} : W^{k,p} \to W^{m,p}$, linearity is clear and boundedness follows from $\|\nabla^{k-m}u\|_{W^{m,p}} \leq \|u\|_{W^{k,p}}$ (on the right side are just more terms).

Hence, from $u_m \rightharpoonup u$ in $W^{k,p}$ it follows $\nabla^{k-m}u \rightharpoonup \nabla^{k-m} u$ in $W^{m,p}$.

Cahn
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