Let $\Omega$ be an open bounded subset in $\mathbb{R}^n$
Does weak convergence in $L^2(0,T;L^2(\Omega))$ imply weak convergence in $L^2(0,T;L^1(\Omega))$?
I tried to prove it the following way:
Let $u_n$ converge weakly to $u$ in $L^2(0,T;L^2)$, then it holds by definition $\int_0^T (u_n - u, \phi) \longrightarrow 0$ for alle $\phi \in L^2(0,T;L^2) = (L^2(0,T;L^2))'$. And the definition of weak convergence is $L^2(0,T;L^1)$ would be: $\int_0^T (u_n - u, \phi) \longrightarrow 0$ for all $\phi \in L^2(0,T;L^\infty) = (L^2(0,T;L^1))'$.
And since the space $L^2(0,T;L^\infty)$ is a subset of $L^2(0,T;L^2)$ we can say that weak convergence of a sequence in in $L^2(0,T;L^2(Ω))$ implies weak convergence in $L^2(0,T;L^1(Ω))$. Is this correct?