Weak Convergence in $H^{1}_{0,\sigma}$

Question

This is a simple question. If a sequence $\{ u_{m} \}_{m \in \mathbb{N}}$ converges weakly to some function $u$ in $H^{1}_{0,\sigma}$, (i.e. the set of functions in $W^{1,2}_{0}$ with divergence $0$) then does the sequence $\{ \nabla u_{m} \}_{m \in \mathbb{N}}$ converge weakly to $\nabla u$ in $L^{2}$?

Answer 1

I have two ways (choose your favorite):

• The operator $T:u \mapsto \nabla u$ is linear and continuous from $H_0^1$ to $[L^2]^d$, hence it is continuous wrt to the weak topology. Hence you get $\nabla u_m \rightharpoonup \nabla u$.

• $H_0^1$ and the product space $L^2\times[L^2]^d$ are isometric under the operator $Tu=(u,\nabla u)$, and hence weak convergence in $H_0^1$ is equivalent to $(u_m,\nabla u_m) \rightharpoonup (u,\nabla u)$ in $L^2\times[L^2]^d$.