The domain $\Omega \subset \mathbf{R}^d (d=2,3)$ with smooth boundary $\Gamma $.
Is this implication is true:
$u\left(x,t\right)\in W^{1,1}\left(\left[0,T\right],L^2\left(\Omega^{d}\right)\right) \Rightarrow u\in L^{\infty}\left(\left[0,T\right],L^2\left(\Omega^{d}\right)\right) $
Let $u \in W^{1,1}(0,T;X)$ for some Banach space $X$, i.e., $u' \in L^1(0,T;X)$. By the fundamental theorem of calculus we have in $X$ $$u(t)=u(s)+\int_s^t u'(\tau) d\tau,$$ for almost all $s,t \in (0,T)$. Taking the norm of $X$ $$\|u(t)\|_X = \left\|u(s)+\int_s^t u'(\tau) d\tau\right\|_X \leq \|u(s)\|_X + \int_s^t \|u'(\tau)\|_Xd\tau.$$ Taking the essential supremum over $t \in (0,T)$ and integrating over $s \in (0,T)$ gives $$\| u\|_{L^\infty(0,T;X)} \leq \|u\|_{L^1(0,T;X)} + T \|u'\|_{L^1(0,T;X)} \leq \max(1,T) \|u\|_{W^{1,1}(0,T;X)}. $$