Suppose $\Omega\subset R^N$ is open bounded with Lipschitz boundary. Then for $u\in W^{1,p}(\Omega)$, there exists a linear operator $T$: $W^{1,p}(\Omega)\to L^{p}(\partial\Omega)$ such that the integration by parts formula will hold. And in addition, we have $$ \|T[u]\|_{L^p}\leq C\|u\|_{W^{1,p}}$$ This is the common result and we can find the prove in Evans & Gapriep book for example.
Now let us move to $BV(\Omega)$. We can also define the same trace operator and hence the integration by parts formula will work. However, I am really interested in the question that whether we have $$ \|T[u]\|_{L^1}\leq C\|u\|_{BV(\Omega)}\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ Is inequality $(1)$ hold? Yes, from the comment below by @John we just need to notice that $T$ is actually an linear BOUNDED operator.
Moreover, for same $\Omega$, we can approach $u\in W^{1,p}(\Omega)$ by a sequence $(u_n)\subset C^\infty(\bar{\Omega})$ in the norm of $W^{1,p}$ for $p<\infty$. Now, for $BV(\Omega)$, can we have $(u_n)\subset C^\infty(\bar{\Omega})$ such that $u_n\to u$ in $L^1$ and $\|Du_n\|(\Omega)\to \|Du\|(\Omega)$? Again, I don't think so, but could you help me to come up a counterexample? (Certainly $u\in BV(\Omega)$ can be approximated by $(u_n)\subset C^\infty({\Omega})$)
Thx!
Update: Now I have a prove for my second question. It is true actually. The way to prove this is to using extension operator $E$ and noticing the fact that $\|\partial E[u]\|(\partial \Omega)=0$