This question is about hyperbolic equations of divergence form. The result has been applied directly, without proof, in Evans' book. I do not think this question is trivial. Please help by giving a proof to the following statement.
Let $\Omega$ be a bounded region with smooth boundary. Suppose that $u\in L^2(0,T,H_0^1(\Omega))$, $u_t\in L^2(0,T,L^2(\Omega))$ and $u_{tt}\in L^2(0,T,H^{-1}(\Omega))$. Suppose further that $v\in H^1(0,T,H_0^1(\Omega))$, then $\langle u_t,v\rangle$ admits weak derivative $\langle u_{tt},v\rangle+\langle u_t,v_t\rangle$ on $[0,T]$, and is therefore absolutely continuous. Here $\langle\cdot,\cdot\rangle$ denotes the dual product on $H^{-1}(\Omega)\times H_0^1(\Omega)$ and is reduced to $L^2$ inner product if both arguments belong to $L^2$.
$\textbf{My attempt}$: If $v$ does not depend on the time variable, by definition we have$$\int_0^T\langle u_t,v\rangle\eta'(t)dt=\langle \int_0^Tu_t\eta'(t)d t,v\rangle=-\langle \int_0^Tu_{tt}\eta(t)d t,v\rangle=-\int_0^T\langle u_{tt},v\rangle\eta(t)dt$$
for arbitrary $\eta(\cdot)\in C_0^\infty((0,T))$ and the proof is complete. However, with $v\in H^1(0,T,H_0^1(\Omega))$, it seems the order of the integral and the dual product cannot be changed similarly. I cannot find a way to fix this argument.