Trace theorem for Bochner space.

Question

Let $ \Omega $ be a smooth bounded domain in $ \mathbb{R}^n $, it is well-known that the trace theroem implies that there is a compact embeding $ H^1(\Omega)\hookrightarrow L^2(\partial\Omega) $. Precisely speaking, the trace operator $ T:H^1(\Omega)\to L^2(\partial\Omega) $ is compact, i.e. all bounded sequence in $ H^1(\Omega) $ is mapped to a precompact sequence in $ L^2(\partial\Omega) $ by $ T $. I wonder if there are similar results for the Bochner space $ L^{p}(0,T;H^1(\Omega)) $, where $ T>0 $ is a positive number and $ 1\leq p\leq \infty $. For $ L^{p}(0,T;H^1(\Omega)) $, the trace operator also makes sense, i.e. $$ T:L^{p}(0,T;H^1(\Omega))\to L^{p}(0,T;L^2(\partial\Omega)). $$ I want to ask if $ T $ is compact. Can you give me some hints or references?

Answer 1

No, $T$ is not compact. Take a sequence $(\phi_n)$ in $L^p(0,T)$ that has no strongly convergent subsequence, take $v\in H^1 \setminus \{0\}$. Then $\phi_n\cdot v$ is bounded in $L^p(H^1)$, and $T(\phi_nv)$ is has no converging subsequence.