Hi Math StackExchange!
My question is somewhat related to the one I asked yesterday here.
Consider the following setting:
$\Omega$ a bounded domain with smooth boundary.
$u_n$ is a bounded sequence of functions in $L^\infty(0,T;H^1(\Omega))$
$T$ is the trace operator;
By Banach-Alaoglu we can extract a subsequence of $u_n$ (not relabeled) such that $u_n$ converges weak-* in $L^\infty(0,T;H^1(\Omega))$. I want to know whether there exists a subsequence (not relabeled) such that $Tu_n \overset{\ast}{\rightharpoonup} Tu$ in $L^2(0,T;L^2(\partial \Omega)$, where $u$ is the weak* limit of $u_n$ in $L^\infty(0,T;H^1(\Omega))$ given by an application of Banach-Alaoglu.
Although the general statement of my previous question is not true, it seems to me that the statement above is true.
Indeed, on the one hand $Tu_n$ is bounded in $L^\infty(0,T;L^2(\Omega))$, so there exists a subsequence (not relabeled) such that $Tu_n \overset{\ast}{\rightharpoonup} v$ in $L^\infty(0,T;L^2(\partial\Omega))$.
On the other hand, the embedding $L^\infty(0,T;H^1(\Omega)) \subset L^2(0,T;H^1(\Omega))$ is continuous and $T$ is continuous from the Hilbert space $L^2(0,T;H^1(\Omega))$ to the Hilbert space $L^2(0,T;L^2(\partial \Omega))$, so
$u_n \rightharpoonup u$ in $L^2(0,T;H^1(\Omega))$ and
$Tu_n\rightharpoonup Tu$ in $L^2(0,T;L^2(\partial \Omega))$.
So $v = Tu$.
Is this reasoning correct?
Thank you very much.
