Question Suppose $u$ and $v$ are in $L^1(0,T; X)$ where $X$ is Banach. Suppose v = u' in the distributional sense. I want to show that, for $w \in X^*$, that $\frac{d}{dt}\langle w, u \rangle = \langle w, v \rangle$ implies $\langle w, \int_0^T \phi' u + \int_0^T \phi v\rangle = 0$ for all $\phi \in C_c^\infty(0,T)$.
Attempt I can get no further than this (please check if I am right): By definition, we have $\int_0^T \langle w, u \rangle \phi' = -\int_0^T \langle w, v \rangle \phi$ hence $\int_0^T \langle w, u \rangle \phi' + \langle w, v \rangle \phi \;dt= 0$ which is $\int_0^T w(u) \phi' + w(v)\phi \;dt= 0$ and by linearity $\int_0^T w(\phi' u) + w(\phi v) \;dt= 0$ so $\int_0^T w(\phi' u) + \int_0^T w(\phi v) = \int_0^T \langle w, \phi' u\rangle + \int_0^T \langle w, \phi v\rangle $ Where to go from here? How can I put the integral inside the functional? Also I am not sure if I am using the linearity property of functionals correctly, since I am assuming that $\phi$ doesn't depend on $X$. Thanks