Find the Minimum of $ F(u)= \int\limits_{-2}^{+2}|u(x) - \chi_{[0,2]}(x)|dx + |Du|(\mathbb R)$. Let $F: BV(\mathbb R) \to \mathbb R$ be a functional defined as:
\[
F(u)=  \int\limits_{-2}^{+2}|u(x) - \chi_{[0,2]}(x)|dx + |Du|(\mathbb R).
\]
Show that there is no minimum on $W^{1,1}$, but the infimum is exactly the minimum on $BV$.
Attempt: $F(\chi_{[0,2]})=2$. I think that $F(u) \geq 2$ for every $u \in BV$ but I don't know why. Given that we have a Theorem which says that it exists $\{u_n\} \in C^{\infty}$ such that $u_n \to \chi_{[0,2]}$ in $L^1$ and $\|u'_n\|_1 \to |Du|(\mathbb R)$ thus we get $F(u_n)\to 2$ and so this is the infimum in $W^{1,1}$. I don't know why there is no $u \in W^{1,1}$ such that $F(u)=2$, I can "understand" why in some sense but can't prove it.
Thanks!

First of all, $\chi_{(0,2)}$ isn't optimal; the minimum (on $BV$) is assumed for $u=\chi_{(0,\infty)}$, and $F(u)=1$.
Now let me show that $F(u)\ge 2$ for all $u\in W^{1,1}$ (so the assertion about the infimum is incorrect). Consider $F_-(u)=\int_{-2}^0 |u(x)|\, dx + |Du|(-\infty,0)$. Since the derivative exists and is in $L^1$, the total variation equals $\|u'\|_{L^1(-\infty,0)}$, which I'll denote by $N_-$ for easier reference.
Let's say $u(0)>0$. Now there are two cases: in the first case, $u(x)=0$ for some $x\in (-2,0)$. This implies that $N_-\ge u(0)$. On the other hand, if $N_-<u(0)$, then $\int_{-2}^0 u\, dx > 2(u(0)-N_-)$ (by comparing with the extreme case where $u$ drops to its minimal value on $(-2,0)$ instantaneously).
Notice that $F_-(u)> u(0)$ in either case. A similar of discussion of $F_+(u)=\int_0^2|u(x)-1|\, dx+|Du|(0,\infty)$ will show that $F_+(u)\ge 2-u(0)$ (if $u(0)\le 2$), and our claim follows.
