A conclusion from Aubin-Lions lemma

Question

Suppose that $u\in L^{\infty}(0,T;H_{0}^{1}(\Omega)\cap H^{2}(\Omega))$, and $u_{t}\in L^{\infty}(0,T;H_{0}^{1}(\Omega))$. Can we say from Aubin-Lions lemma (https://en.wikipedia.org/wiki/Aubin%E2%80%93Lions_lemma) that $$ u\in C(0,T;H_{0}^{1}(\Omega))\quad\&\quad u_{t}\in C(0,T;L^{2}(\Omega))? $$

Answer 1

The regularity $u_t \in L^\infty(0,T; H_0^1(\Omega))$ directly yields $u \in C(0,T;H_0^1(\Omega))$ (just integrate over $t$). You do not even need Aubin-Lions.

However, you cannot achieve $u_t \in C(0,T;L^2(\Omega))$ under your requirements: Take $\varphi \in C^{0,1}([0,T])$, such that $\varphi$ is weakly differentiable, but $\varphi' \not\in C([0,T])$. For arbitrary $\psi \in H_0^1(\Omega)$ consider $$u(t,x) = \varphi(t) \, \psi(x).$$