Identification of $L^2(0,T;H^2(\Omega))$ with $L^\infty(Q_T)$

Question

Let $T>0$ and $\Omega \subset \mathbb{R}^n$ be a bounded set with smooth boundary. Define $Q_T:=\Omega \times (0,T).$ In which case can the Hilbert space $L^2(0,T;H^2(\Omega))$ be identified with $L^\infty(Q_T)$?

Answer 1

Never.

If $f\in L^2((0,T))$ is unbounded then $f\cdot\chi_{\Omega}$ is in $L^2((0,T);H^2(\Omega))$ but not in $L^{\infty}(\Omega_T)$.

If $u\in L^{\infty}(\Omega)$ is not in $H^2(\Omega_T)$, then $u\cdot\chi_{(0,T)}$ is in $L^{\infty}(\Omega_T)$ but not in $L^2((0,T);H^2(\Omega))$.