Let $u_n \in H^1(\Omega)$
Then, $u_n$ is said to converge weakly in $H^1$ to $u$ if $u_n$ converges weakly to $u$ in $L^2$ and $\nabla u_n$ converges weakly to $\nabla u$ in $L^2$.
What does it mean by $\nabla u_n$ converging weakly to $\nabla u$ in $L^2$?
Does this mean that every "component" of the gradient of $u_n$ converges weakly to the gradient of $u$?
Best Answer
The "real" definition for weak convergence is:
$u_n\rightharpoonup u$ in $H^1(\Omega)$ if for each bounded linear functional $f$ on $H^1(\Omega): \langle f,u_n\rangle\rightarrow \langle f,u\rangle$ as $n\to\infty$. We want to show that both deffinitons are equivalent.
1) Note that for arbitrary functions $g_0,g_1,...,g_n\in L^2(\Omega)$ the functional defined by $\langle g,u\rangle :=\int\limits_{\Omega}{g_0u+\sum\limits_{i=1}^{n}{g_iu_{x_i}}}$ is a bounded linear functional on $H^1(\Omega)$. Now, choosing $g_0\in L^2(\Omega)$ arbitrarily and $g_1,g_2,...,g_n\equiv 0$ you get that $u_n\rightharpoonup u$ in $L^2(\Omega)$ (here we used the Riesz representation theorem for the functionals in $L^2$). Analogously, choosing $g_k\in L^2(\Omega)$ arbitrarily and $g_i\equiv 0,i\in\{0,1,2,...,n\},i\neq k$ you get $u_{n_{x_k}}\rightharpoonup u_{x_k}$ in $L^2(\Omega)$, so indeed each component of the gradient also tends weakly in $L^2$ to the corresponding component from $\nabla u$.
2) Conversly, if your definition is satisfied, we want to show that the "real" one is also true. Here. we use the Riesz representation theorem for the Hilbert space $H^1(\Omega)$: if $f\in(H^1(\Omega))^*$ then there exists a function $w\in H^1(\Omega): \langle f,v\rangle=(w,v)=\int\limits_{\Omega}{(wv+\sum\limits_{i=1}^{n}{w_{x_i}}v_{x_i})dx},\,\forall v\in H^1(\Omega)$ where, obviously, $w,w_{x_i}\in L^2(\Omega),\,i=1,2,...,n$. By your definition $u_n\rightharpoonup u$ and $u_{n_{x_i}}\rightharpoonup u_{x_i}$ in $L^2(\Omega)$ means (using Riesz representation for $L^2$) $\int\limits_{\Omega}{gu_ndx}\to\int\limits_{\Omega}{gudx}$ and $\int\limits_{\Omega}{gu_{n_{x_i}}dx}\to\int\limits_{\Omega}{gu_{x_i}dx}$ for each $g\in L^2(\Omega)$. Therefore for each $f\in(H^1(\Omega))^*\,\,$ $\int\limits_{\Omega}{wu_ndx}\to\int\limits_{\Omega}{wudx}$ and $\int\limits_{\Omega}{w_{x_i}u_{n_{x_i}}dx}\to\int\limits_{\Omega}{w_{x_i}u_{x_i}dx}\,$ implying $\langle f,u_n\rangle =\int\limits_{\Omega}{(wu_n+\sum\limits_{i=1}^{n}{w_{x_i}}u_{n_{x_i}})dx}\to \int\limits_{\Omega}{(wu+\sum\limits_{i=1}^{n}{w_{x_i}}u_{x_i})dx}=\langle f,u\rangle$