Weak convergence in some space.

Question

I have the sequence $\{u_{k}\}_{k}$ weak convergent in the space $L^{2}(0,T; W^{1,2}(\Omega))$. What exactly does it mean? Does it imply weak convergence $\{u_{k}\}_{k}$ in $L^{2}(0,T;\Omega)$ or $\{ \bigtriangledown u_{k}\}_{k}$ in $[L^{2}(0,T;\Omega)]^{n}$? I would be grateful if somebody explained this to me.

Answer 1

Weak convergence in $L^2(0,T;W^{1,2}(\Omega))$ means by definition that we have $u^*(u_k) \to u^*(u)$ for any functional $u^*\in L^2(0,T; W^{1,2}(\Omega))^* = L^2(0,T;W^{1,2}(\Omega)^*)$. To see that this implies weak convergence in $L^2(0,T;L^2(\Omega))$, note that $W^{1,2}(\Omega) \to L^2(\Omega)$ is a continuous injection. Hence, its dual, the restriction map $L^2(\Omega)^* \to W^{1,2}(\Omega)^*$ is continuous. Given $u^* \in L^2(0,T;L^2(\Omega)^*)$, composition with the restriction map gives a functional $u^* \in L^2(0,T;W^{1,2}(\Omega)^*)$, weak convergence of $(u_k)$ gives what was to prove.

To see the weak convergence of the gradient, note that the map $\nabla\colon L^2(0,T;W^{1,2}) \to L^2(0,T;L^2(\Omega)^n)$ is linear and continuous, composing its dual with any given linear functional $u^* \in L^2(0,T;L^2(\Omega)^n)^*$ gives the statement about the gradients.