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Let $\Omega \subset \mathbb R^d$ an open set. Let $(u_n)\subset W^{1,p}(\Omega )$.

Q1) We suppose that $u_n\to u$ strongly in $L^p(\Omega )$ and $\nabla u_n\to v$ strongly in $L^p(\Omega )$. Do we have that $\nabla u=v$ a.e. ?

Q2) We suppose that $u_n\to u$ weakly in $L^p(\Omega )$ and $\nabla u_n\to v$ strongly in $L^p(\Omega )$. Do we have that $\nabla u=v$ a.e. ?

Q3) We suppose that $u_n\to u$ strongly in $L^p(\Omega )$ and $\nabla u_n\to v$ weakly in $L^p(\Omega )$. Do we have that $\nabla u=v$ a.e. ?

Q3) We suppose that $u_n\to u$ weakly in $L^p(\Omega )$ and $\nabla u_n\to v$ weakly in $L^p(\Omega )$. Do we have that $\nabla u=v$ a.e. ?


Attempts

1) I would say

$$\|v-\nabla u\|_{L^p(\Omega )}\leq \|v-\nabla u_n\|_{L^p(\Omega )}+\|\nabla u_n-\nabla u\|_{L^p(\Omega )}=\|v-\nabla u_n\|_{L^p(\Omega )}+\|\nabla (u_n- u)\|_{L^p(\Omega )},$$ how can I conclude that $\|\nabla (u_n-u)\|_{L^p(\Omega )}\to 0$ ? I was thinking about something as Poincaré, but hypothesis are not strong enough.

2),3),4) Since for the 1) I have problem, I can't concude for the other as well.

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    $\begingroup$ The answer to Q3 (and the others) is "yes"; just pass to the limit in the definition of weak derivatives. $\endgroup$ – Michał Miśkiewicz Dec 25 '17 at 22:05

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