I read this:
For $v$, $w$ in $L^2(0,T;H^1(S))$ (with weak derivatives in $H^{-1}(S)$ for each time), the product $(v(t), w(t))_{L^2(S)}$ is absolutely continuous wrt. $t \in [0,T]$ and $\frac{d}{dt}\int_S v(t)w(t) = \langle v', w \rangle_{H^{-1}(S), H^1(S)} + \langle v, w' \rangle_{H^{1}(S), H^{-1}(S)}$ holds a.e. in $(0,T)$. As a consequence, the formula of partial integration holds $\int_0^T \langle v', w \rangle_{H^{-1}, H^1} = (v(T), w(T))_{L^2(S)} - (v(0), w(0))_{L^2(S)} - \int_0^T \langle v, w' \rangle_{H^{1}, H^{-1}}$
I am wondering what role absolute continuity plays here. I know if a real-valued (normal in the undergrad sense) function is abs. cont. then it satisfies an equation similar to the one above but this one involves dual space pairing so I can't see how it is analgous. I would appreciate someone explaining me this. Thanks.