In Evans PDE book, I have the next Theorem:
If $u \in W^{1,p}([0,T],X)$ then:
i) $u(t) = u(s) + \int_{s}^tu'(\tau) d\tau $ for $0\leq s\leq t \leq T$
ii) $\max_{0\leq t \leq T} \| u(t)\|_X \leq C \| u \|_{W^{1,p}([0,T],X)}$
With $C$ a constant depending of $T$, now my question is how can I prove ii) using i)?
First, from (i) we get $$ \|u(t)-u(s)\|\le\int_s^t \|u'(\tau)\|d\tau, $$ which proves that $u\in C([0,T],X)$. It remains to show the norm bound.
Second, integrating $$ u(t)-u(s) = \int_s^t u'(\tau)d\tau. $$ with respect to $s$ from $0$ to $T$ yields $$ \int_0^T u(t)-u(s)ds = \int_0^T\int_s^t u'(\tau)d\tau\, ds\\ = \int_0^t\int_s^t u'(\tau)d\tau\, ds - \int_t^T\int_t^s u'(\tau)d\tau\, ds\\ =\int_0^t\int_0^\tau u'(\tau)ds\,d\tau - \int_t^T\int_\tau^T u'(\tau)ds\,d\tau,\\ =\int_0^t\tau u'(\tau)d\tau - \int_t^T(T-\tau) u'(\tau)ds\,d\tau,\\ $$ hence $$ \|\int_0^T u(t)-u(s)ds\|\le 2 T\int_0^T \|u'(\tau)\|d\tau. $$ This shows $$ \|u(t)\| \le \frac1T\|\int_0^T u(t)-u(s)ds\| + \frac1T \|\int_0^T u(s)ds\| \le 2 \int_0^T \|u'(\tau)\|d\tau + \frac1T \int_0^T \|u(s)\|ds\\ \le 2 \|u'\|_{L^1(0,T;X)} + T^{-1} \|u\|_{L^1(0,T;X)}. $$ The bound follows from the embedding $L^p(0,T)\hookrightarrow L^1(0,T)$.