Let $V$ be a Banach space with dual $V'$ and suppose that $V$ is included into the Hilbert space $H$ so that the inclusion is continuous and dense. Then after identification $H = H'%$ we have $V \subset H \subset V'$. Let $p \in (1,+\infty)$, $\frac 1 p + \frac{1}{p'} = 1$, $T>0$, and define the Banach space $$ W^p(0,T) = \bigl\{ u \in L^p(0,T,V) \colon \dot u \in L^{p'}(0,T,V') \bigr\}, $$ where the derivative is supposed to exist a.e. and the norm is given by $\|u\|_{W^p} =\|u\|_{L^p}+\|\dot u\|_{L^{p'}}$. Then it is known that $W^p(0,T) \subset C(0,T,H)$ and the inclusion is continuous. From this we can derive two following statements:
- If $u_n \in W^p(0,T)$, $u_n \to u$ in $L^p(0,T,V)$ and $\dot u_n \to v$ in $L^{p'}(0,T,V')$ then $u$ is differentiable a.e. and $\dot u = v$. This follows from completeness of $W^p(0,T)$ as a Banach space and from the form of its norm.
- If $u_n \to u$ in $W^p(0,T)$ then $u_n \to u$ in $C(0,T,H)$. This follows from the continuity of the inclusion of $W^p(0,T)$ in $C(0,T,H)$.
But suppose now that $u_n \in W^p(0,T)$, $u_n \to u$ in $L^p(0,T,V)$ weakly, $\dot u_n \to v$ in $L^{p'}(0,T,V')$ weakly. Is it possible to say that $u$ is differentiable a.e. and $\dot u = v$?
Further, suppose that $u_n \to u$ in $W^p(0,T)$ weakly. Is it possible to say that $u_n \to u$ pointwise?
The graph $\{(u,\dot u):u\in W^p(0,T)\}$ of the time derivative is a linear and closed subset of $L^p(0,T,V)\times L^{p'}(0,T,V^*)$, hence convex and weakly closed. Therefore, if $u_n\rightarrow u$ weakly in $L^p(0,T,V)$ and $\dot u_n\rightarrow v$ weakly in $L^{p'}(0,T,V^*)$, then $u\in W^p(0,T)$ and $v=\dot u$.
I thinks the question concerning differentiability a.e. can also be answered affirmative - see the appendix of Brezis' "Opérateurs maximaux monotones".
If $V$ is even compactly embedded into $H$, then $W^p(0,T)$ embedds compactly into $L^2(0,T;H)$ and therefore $u_n\rightarrow u$ in $W^p(0,T)$ implies $u_n\rightarrow u$ a.e. in $H$. This is version of Aubin's lemma.