Are weak-star limits in $L^\infty (0,T;L^2(\Omega))$ and in $L^\infty (\Omega \times \left[ 0,T \right])$ equal?

Question

Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$ and let $T>0$.

Let $u_n$ be a sequence such that there exists a subsequence (still denoted by $u_n$) \begin{equation} u_n \rightarrow u \text{ weakly-star in } L^\infty (0,T;L^2(\Omega)) \end{equation} and that there exists a subsequence (again denoted by $u_n$) such that

\begin{equation} u_n \rightarrow u^*\text{ weakly-star in } L^\infty (\Omega \times \left[ 0,T \right]) \end{equation}

Is it possible to imply from this that the limits are equal, i.e. $u = u^*$?

I need this property for a proof, but I have no approach how to show it, the only thing I know is that $L^\infty( 0,T; L^\infty (\Omega)) \subset L^\infty (\Omega \times \left[ 0,T \right])$, see question $L^\infty((0,T)\times\Omega)$ is not equal to $L^\infty(0,T;L^\infty(\Omega))$..

Answer 1

Let $I=(0,T)$. The space $L^\infty(I,L^2(\Omega))$ is the dual of $L^1(I,L^2(\Omega))$, while $L^\infty(\Omega \times I)$ is the dual of $L^1(\Omega\times I)$. Hence, the weak-star convergence is tested against functions in $L^1(I,L^2(\Omega))$ and $L^1(\Omega\times I)=L^1(I,L^1(\Omega))$. Now, $L^1(I,L^1(\Omega)) \supset L^1(I,L^2(\Omega))$, so weak-star convergence in $L^\infty(\Omega \times I)$ implies weak-star convergence in $L^\infty(I,L^2(\Omega))$, and the limits are equal.