Let $\Omega$ be an open bounded domain in $\mathbb{R}^n$ and let $T>0$.
Assume $\lVert \nabla u \rVert_{L^\infty ( 0,T; L^1 (\Omega))} \leq C_1$. Does this imply $\lVert u \rVert_{L^\infty (0,T;BV(\Omega))}\leq C_2$ ? Here $C_1,C_2$ are positive constants.
Does it hold if I make the additional assumption on $u \in L^\infty (0,T; W^{1,1})$?
Because then by definition of this space:
$u \in L^\infty(0,T;L^2(\Omega)) \subset L^\infty(0,T;L^1(\Omega))$,
$\nabla u \in L^\infty(0,T;L^1(\Omega))$.
Then it holds: $\lVert u \rVert_{L^\infty (0,T; BV (\Omega))} = \lVert u \rVert_{L^\infty (0,T; W^{1,1})} = \lVert u \rVert_{L^\infty(0,T;L^1(\Omega))} + \lVert \nabla u \rVert_{L^\infty(0,T;L^1(\Omega))} \leq \tilde{C} \nabla u \rVert_{L^\infty(0,T;L^1(\Omega))} + C_1$ and since $u \in L^\infty(0,T;L^1(\Omega))$, this implies $\lVert u \rVert_{L^\infty (0,T; BV (\Omega))} \leq C_2$.
Where I used the Poincaré inequality and the assumption $\lVert \nabla u \rVert_{L^\infty ( 0,T; L^1 (\Omega))} \leq C_1$. Is this proof correct?