Is the embedding $L^2(0,T;H^1) \subset L^2(0,T;L^2)$ compact?

Question

Is the embedding $L^2(0,T;H^1(\Omega)) \subset L^2(0,T;L^2(\Omega))$ compact?

Answer 1

No. Here is a counterexample, inspired by the answer to this question:

If $V \subset H$ compact, is $L^2(0,T;V) \subset L^2(0,T;H)$ compact too?

Take $v\in H^1$, $T=\pi$, $$ \phi_n(t) = \sin(n t)v. $$ Since $\sin(n \circ)$ converges weakly but not strongly to zero in $L^2(0,T)$, it follows that $\phi_n$ converges weakly to zero in $L^2(0,T;H^1)$.

If there would be a subsequence of $(\phi_n)$ converging strongly to zero in $L^2(0,T;L^2)$, then this would imply $\sin(n \circ) \to 0$ in $L^2(0,T)$, which is a contradiction.

Hence, the embedding is not compact. You need some control of the time derivative (see Aubin-Lions theorem).