Let $I=(-1,1)$ be an interval in one dimension. Let $u\in BV(I)$ be defined as $$ u(x)= \begin{cases} 0,&\text{ if }x\in(-1,0)\\ 1,&\text{ if }x\in(0,1) \end{cases} $$ Clearly, we have $u\in BV$ and $S_u$, the jump point of $u$, is at $0$. We shall use $Du$ to dentoe the weak derivative of $u$ in $BV$ space, and $u'$ the absolutely continuous part of $Du$.
Now let $u_n$ be a sequence of $BV$ function such that $u_n\to u$ strong in $L^1$.
My question: Does there exists a function $f_n$ defined on $\mathbb R^+$ such that
$f_n\to 1$ a.e.,
if $$ \limsup_{n\to\infty} \int_{I}f_n(|u_n'(x)|)dx<\infty $$ then we have $$ \liminf_{n\to\infty} \int_{I}f_n(|u_n'(x)|)dx\geq\int_I|u'|dx $$ and for sufficit large $n$, we have $S_{u_n}\Delta S_u=\varnothing$ and there exists $x_n$, $y_n\in S_{u_n}$ such that $$ x_n<0<y_n $$ and $\lim_{n\to\infty}x_n=\lim_{n\to\infty}y_n=0$? (Recall $S_u=\{0\}$)
Thank you very much!
